Inter 1st Year

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 6 Biology in Human Welfare Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 6 Biology in Human Welfare

Very Short Answer Type Questions

Question 1.
Define parasitism and justify this term.
Answer:
An intimate association between two organisms of different species in which ‘one is benefited and the other one is often adversely affected’ is called parasitism. The word parasitism comes from the Greek word ‘parasitos’ (Para-at the side of Sitos – food or grain) which means one eating at another one’s table.

Question 2.
Distinguish between a vector and a reservoir host.
Answer:

Question 3.
Distinguish between mechanical vector and biological vector.
Answer:

Question 4.
What is a hyperparasite? Mention the name of one hyper-parasite.
Answer:
A parasite that parasitizes another parasite is called a Hyper parasite.
Ex: Nosema notabilis (a cnidosporan) is a parasite in Sphaerospora polymorpha (a cnidosporan parasite in the urinary bladder of the toadfish).

Question 5.
What do you mean by parasitic castration? Give one example.
Answer:
Some parasites cause the degeneration of gonads of the host making it sterile. This effect is called parasitic castration.
eg: Sacculina (root-headed barnacle, a crustacean) causes the degeneration of ovaries in the crab Carcinus maenas.

Question 6.
What are the endo-parasitic adaptations observed in Fasciola hepatica?
Answer:
The life cycle of Fasciola hepatica (sheep liver fluke) is very complex involving many developmental stages and two intermediate hosts to increase the chances of reaching a new definitive host.

Question 7.
Define Neoplasia. Give one example.
Answer:
Some cause abnormal growth of the host cells in a tissue to form new structures. This effect is called Neoplasia which leads to cancers.
Ex: Some Viruses.

Question 8.
Define the most accurate definition of the term ‘health’ and write any two factors that affect health.
Answer:
Health is a state of complete physical, mental and social well-being and not merely the absence of any disease or absence of physical fitness. Our health may be affected by crenetic disorders, infections, and lifestyle.

Question 9.
Distinguish between infectious and non-infectious diseases. Give two examples each.
Answer:

Question 10.
Entamoeba histolytica is an obligatory anaerobe justify.
Answer:
Mitochondria is absent in the endoplasm of Entamoeba histolytica. The absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica.

Question 11.
Distinguish between the precystic stage and the cystic stage of E.histolytica.
Answer:

Question 12.
What is the reserve food in the precystic and early cyst stages of Entamoeba histolytica?
Answer:
The cytoplasm of the precystic stage stores glycogen granules and chromatoid bars (made of ribonucleic protein) which act as reserve food.

Question 13.
A person is suffering from bowel irregularity, abdominal pain, blood and mucus in stool, etc. Based on these symptoms, name the disease and its causative organism.
Answer:

• The disease is Amoebiasis.
• The causative agent is the Trophozoite of “Entamoeba histolytica”.

Question 14.
On the advice of a doctor, a patient has gone to a clinical laboratory for the examination of a sample of faeces. The lab technician, on observing the stool of the patient diagnosed that the patient was suffering from amoebiasis. Write any two characteristic features based on which the technician came to that conclusion.
Answer:

1. Stool with blood and mucous.
2. Presence of a tetra nucleated cyst.

Question 15.
Define ‘asymptomatic cyst passers’ with reference to Entamoeba histolytica.
Answer:
Some people do not exhibit any symptoms, such people are called carriers of asymptomatic cyst passers as their stand contains the tetranuclear cysts. They help in spreading the parasites to their persons.

Question 16.
What are the stages of plasmodium vivax that infect the hepatocytes of man?
Answer:
Sporozoite, Cryptozoite, Macrometacryptozoite.

Question 17.
Define the prepatent period. What is its duration in the life cycle of plasmodium vivax?
Answer:
The interval between the first entry of plasmodium into the blood in the form of sporozoites and the second entry of plasmodium into the blood in the form of Cryptozoics is called a prepatent period. It lasts approximately 8 days.

Question 18.
Define incubation period. What is its duration in the life cycle of Plasmodium vivax?
Answer:
The period between the entry of Plasmodium into the blood in the form of sporozoite and the first appearance of symptoms of malaria in man is called the incubation period which is approximately 10 to 14 days.

Question 19.
What are Schuffner’s dots? What is their significance?
Answer:
Small red coloured dots appear in the cytoplasm of the RBC known as Schuffner’s dots. These are believed to be the antigens released by the plasmodium (Malaria) parasites.

Question 20.
What are hemozoin granules? What is their significance?
Answer:
The malaria parasite digests the globin part of the ingested hemoglobin and converts the soluble heam into insoluble crystalline hemozoin. It is called the ‘malaria pigment’ which is a disposable product.

Question 21.
What is exflagellation and what are the resultant products called?
Answer:
Male gapnetes show lashing movements like flagella and get separated from the cytoplasm of microgametocyte. This process is called exflagellation and resultant products are called male (or) microgametes.

Question 22.
Why is the syngamy found in plasmodium called anisogamy?
Answer:
Since two gametes are dissimilar in size, the syngamy found in plasmodium is called anisogamy.

Question 23.
What is Ookinete? Based on the sets of chromosomes how do you describe it?
Answer:
Ookinete is a long, splendor, motile, vermiform, two sets of chromosomes are present in it. So it is described as diploid form.

Question 24.
A person is suffering from chills and shivering and high temperature. These symptoms are cyclically followed by profuse sweating and a return to normal body temperature. Based on these symptoms name the disease and its causative organism.
Answer:
The disease is malaria and the causative organism is Plasmodium vivax.

Question 25.
Describe the methods of biological control of mosquitoes.
Answer:
Introduction of larvivorous fishes like Gambusia, and insectivorous plants like Utricularia into the places where mosquitoes breed.

Question 26.
The eggs of Ascaris are called “mammillated eggs”. Justify it.
Answer:
Each egg of Ascaris is surrounded by a protein coat with a rippled surface. Hence the eggs of Ascaris are called “mammilated eggs”.

Question 27.
What is meant by nocturnal periodicity with reference to the life history of a nematode parasite you have studied?
Answer:
Microfilaria larvae of W.brancrofti migrate to the peripheral blood circulation during nighttime between 10 P.M – 4 A.M. This tendency is called nocturnal periodicity.

Question 28.
Distinguish between lymphadenitis and lymphangitis.
Answer:

Question 29.
‘Elephantiasis is the terminal condition of filariasis’. Justify.
Answer:
Sweat glands of the skin in the affected region disintegrate and skin becomes rough so elephantiasis is the terminal condition of filariasis.

Question 30.
In which way does tobacco affect respiration? Name the alkaloid found in tobacco.
Answer:
Tobacco increases the carbon monoxide (CO) level and reduces the oxygen level in the blood. The alkaloid found in tobacco is “Nicotine”.

Question 31.
Define drug abuse.
Answer:
When drugs are used for a purpose other than medicinal use is called drug abuse.

Question 32.
From which substances ‘Smack’ and ‘coke’ are obtained?
Answer:
Smack is the common name for “Heroine”. It is obtained from the opium poppy plant. (Papaver somniferum).
Coke is the common name for “Cocaine”. It is obtained from the coca plant (Erythroxylum coca).

Question 33.
‘Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. justify the statement with an example.
Answer:
Many secondary metabolites of plants like opioids, cannabinoids, and coca alkaloids are abused nowadays. Even though they have medicinal properties they cause some effects.

Question 34.
Why are cannabinoids and anabolic steroids banned in sports and games?
Answer:
These days some sports persons take drugs such as cannabinoids and anabolic steroids to enhance their performance (Doping) and abuse of such drugs also causes side effects that’s why such drugs are banned in sports and games.

Question 35.
Mention the names of any four drugs which are used as medicines to treat patients with mental illnesses like depression, insomnia, etc., that are often abused.
Answer:
Barbiturates, Amphetamines, Benzodiazepines, Lysergic aciddiethyl amides (LSD).

Short Answer Type Questions

Question 1.
What is the need for parasites to develop special adarptations? Mention some special adaptations developed by the parasites.
Answer:
Parasites have to evolve mechanisms to counteract and neutralize the host’s defence in order to be successful within the host. For this purpose, the parasites have developed many special adaptations such as the loss of unnecessary sensory organs, formation of organs for adhesion, high reproductive capacity, etc.
Parasitic adaptations: Parasites have evolved special adaptations to meet the requirements and lead successful lives in the hosts.

• In order to live in the host, some parasites have developed structures like hooks, suckers, rostellum, etc., for anchoring, e.g: Taenia solium.
• Some intestinal parasites have developed protective cuticles to withstand the action of the digestive enzymes of the host, e.g: Ascaris lumbricoides.
• Some intestinal parasites produce anti enzymes to neutralize the effect of the host’s digestive enzymes, e.g: Taenia solium.
• Some parasites live as obligatory anaerobes as the availability of oxygen is very rare for them, e.g: Entamoeba histolytica, Taenia solium, etc.
• Some intestinal parasites live as facultative anaerobes, i.e., if oxygen is not available, they live anaerobically and if oxygen is available, they respire aerobically, e.g: Ascaris lumbricoides.

Question 2.
Distinguish between hypertrophy and hyperplasia with an example for each.
Answer:

Question 3.
Describe the structure of a trophozoite of Entamoeba histolytica.
Answer:
It is the most active, motile, feeding, and pathogenic stage that lives in the mucosa of the large intestine. It moves with the help of pseudopodium (lobopodium) which is produced anteriorly. The body of the trophozoite is surrounded by plasmalemma. Its cytoplasm is differentiated into an outer clear, viscous, non-granular ectoplasm and inner fluid-like granular, endoplasm.

Ribosomes, food vacuoles, and verticular cartwheel-shaped nucleus is present Absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica. It produces a proteolytic enzyme called histolytica which dissolves mucosas & sub-mucosa of the gut wall & releases blood, and tissue debris which are ingested by the trophozoite.

Hence food vacuoles are loaded with R.B.C & fragments of cells, and bacteria. The presence of R.B.C in the food vacuole and cart wheel-shaped nucleus are the characteristic features of the trophozoite.

Question 4.
Explain the life cycle of Entamoeba histolytica.
Answer:
The trophozoite undergoes binary fissions in the wall of the large intestine and produces a number of daughter entamoeba. They feed upon the bacteria and the hostly tissue elements, grow in size, and again multiply. After repeated binary fissions some of the young ones enter the lumen of the large intestine and transform into precystic stage.

Here, the precystic stage transforms into the cystic stage. Which in turn develops into tetranuclear cysts. The entire process is completed only in a few hours. These tetra nucleated cysts come out along with the faecal matter and can remain alive for about 10 days. These cysts reach new hosts through contaminated food and water. They pass into the small intestine of a new human host. Where the cyst wall gets, ruptured by the action of the enzyme trypsin releasing tetra nucleated amoeba. Such tetra nucleated exocyst amoeba is called metacyst. The four nuclei of the metacyst undergo mitotic divisions and produce eight nuclei. Each nucleus gets a bit of cytoplasm and thus eight daughter entamoeba are produced. The young ones develop into trophozoites and invade the large intestine.

Question 5.
Write a short note on the pathogenicity of Entamoeba histolytica.
Answer:
The trophozoites ‘dissolve’ the mucosal lining by histolysin going deep into the submucosa and causing ulcers. These ulcers contain cellular debris, lymphocytes, blood corpuscles, and bacteria. It leads to the formation of abscesses in the wall of the large intestine. Ultimately it results in stools with blood and mucous. This condition is called amoebic dysentery (or) Intestinal amoebiasis. Some people don’t exhibit any symptoms such people are called ‘carriers’ (or) asymptomatic cyst passers as their stools contain tetranucleotide cysts.

Question 6.
Describe the structure of the sporozoite of plasmodium vivax.
Answer:
The ultrastructure of the sporozoite of P. vivax was studied by barnham. It is sickle-shaped with a swollen middle part and pointed at both ends of it’s body. It measures about 15 microns in length and one micron in width. The body is covered by an elastic pellicle with microtubules which help in the curiggling movement of the sporozoite. The cytoplasm contains cell organelles such as the Golgi complex, E.R. mitochondria, and a nucleus.

The cytoplasm also shows many convoluted tubules of unknown function throughout the body. It contains a cup-like depression called an apical cup at the anterior end into which a pair of secretory organelles opens. They secrete a cytolytic enzyme which helps in the penetration of sporozoite into the liver cell.

Question 7.
Describe the cycle of Golgi in the life history of Plasmodium Vivax.
Answer:
It was first described by Camillo Golgi. Hence it is also called the Golgi cycle. This is initiated either by the trophozoites of the pre-erythrocytic cycle (or) the micro meta cryptozoites of the exo-erythrocytic cycle. In the fresh R.B.C, these stages assume the spherical shape and transform into trophozoite. It develops a small vacuole that gradually enlarges in size, and pushes the cytoplasm and nucleus to the periphery.

Now the plasmodium looks like a finger ring. Hence this stage is called the signet ring stage soon it loses the vacuole, develops pseudopodia, and becomes an amoeboid stage. With the help of pseudopodium, it actively feeds on the content of the R.B.C and increases in size. As a result, the R.B.C grows almost double the size. This process is called hypertrophy. The malaria parasite digests the globin part of the ingested hemoglobin and converts the soluble haem into insoluble Haemozine. It is called malaria pigment. During this stage, small red coloured dots appear in the cytoplasm of R.B.C known as “Schuffner’s dots’.

Now the parasite loses the pseudopodia and increases in size finally it occupies the entire R.B.C and becomes schizont. It undergoes schizogony and produces 12-24 erythrocytic merozoites. They are arranged in the form of a rose hence this stage is called the rosette stage. Finally, merozoites are released along with haemozoine into the blood.

Question 8.
Explain the pathogenicity of Wucheria bancrofti in Man.
Answer:
The infection causes filarial fever which is characterized by headache, mental depression, and an increase in the body temperature. In general, the infection of filarial worm causes inflammation effect in lymph vessels and lymph glands. Inflammation in the lymph vessels is called lymphangitis and that of lymph glands is called lymphadenitis. In the case of heavy infection, the accumulation of dead worms blocks the lymph vessels and lymph glands resulting in immense swelling of limbs, scrotum of males, and mammary glands in females. Fibroblasts accumulate in this tissue and form the fibrous tissue. In severe cases, the sweat glands of the skin in the affected region disintegrate and the skin becomes rough. This terminal condition is called elephantiasis.

Question 9.
Write short notes on typhoid fever and its prophylaxis.
Answer:
Typhoid fever: It is caused by salmonella typhi which is a gram-negative bacterium. It mainly lives in the small intestine of man and then migrates to other organs through blood. It can be confirmed by the Widal test.
Mode of infection: Contamination through food and water.
Symptoms: Sustained fever with high temperature upto 104°F. weakness, stomach pain, constipation, headache, and loss of appetite. Intestinal perforation and death may also occur in severe cases.
Prophylaxis: Advancements made in biological science have armed us to deal with many infections effectively. The immunization programme by the use of vaccines has enabled us to completely irradicate like typhoid. Biotechnology is making available never cheaper vaccines, and the discovery of antibiotics and various other drugs also enabled us to treat typhoid.

Question 10.
Write short notes on Pneumonia and its prophylaxis.
Answer:
Pneumonia: It is caused by gram-positive bacteria such as Streptococcus pneumonia and Haemophilus influenza. They infect the alveoli of the lungs in human beings.
Mode of infection: Contamination by inhaling the droplets/aerosols released by an infected person or even by sharing the utensils with an infected person.
Symptoms: The alveoli get filled with fluid leading to severe problems in respiration. In severe cases, the lips and fingernails may turn gray to bluish in colour.
Prophylaxis: Advancements made in biological science have armed us to deal with many infections effectively. The immunization programme by the use of vaccines has enabled us to completely irradicate pneumonia. Biotechnology is making available newer, cheaper vaccines, and the discovery of antibiotics and various other drugs also enabled us to treat pneumonia.

Question 11.
Write short notes on the common cold and its prophylaxis.
Answer:
Common cold: It is caused by a rhinovirus group of viruses. They infect the nose and respiratory passage but not the lungs.
Mode of infection: Contamination is by direct inhalation of the droplets resulting from cough or sneezes of an infected person or indirectly through contaminated objects such as pens, books, cups, door knobs, computer keyboards or mice, etc.
Symptoms: Nasal congestion, discharge from the nose, sore throat, hoar senses, cough, headache, tiredness, etc., which usually last for 3-7 days.
Prophylaxis: Advancements made in biological science have armed to deal with many infections effectively. The immunization programme by the use of vaccines has enabled us to completely irradicate like viral diseases common cold. Biotechnology is making available newer cheaper vaccines, the discovery of antibiotics and various other drugs also enabled use to treat viral diseases like the common cold.

Question 12.
Write short notes on ‘ringworm’ and its prophylaxis.
Answer:
Ringworm: It is one of the most common infectious diseases in man. It is caused by many fungi belonging to the genera, Microsporum, Trichophyton, and Epidermophyton. Heat and moisture help these fungi grow in the skin folds such as those in the groin or between the toes.
Mode of infection: Contamination is by using towels, clothes or combs of the infected persons or even from the soil.
Symptoms: Appearance of dry, scaly, usually round lesions accompanied by intense itching on various parts of the body such as skin, nails, and scalp.

Question 13.
What are the adverse effects of tobacco?
Answer:
Effect: Smoking increases the carbon monoxide (CO) level and reduces the oxygen level in the blood. Nicotine stimulates the adrenal gland to release adrenaline and nor-adrenaline into the blood. These hormones raise blood pressure and increase the heart rate. Smoking is associated with bronchitis, emphysema, coronary heart disease, and gastric ulcer and increases the incidence of cancers of the throat, lungs, urinary bladder, etc. Smoking also paves the way to hard drugs. Yet smoking is very prevalent in society, both among young and old-. Tobacco chewing is associated with an increased risk of cancer of the oral cavity.

Question 14.
Write short notes on opioids.
Answer:
Opioids: These are the drugs obtained from the opium poppy plant Papaver somniferous (vernacular name: Nallamandu mokka): They bind to specific opioid receptors present in our central nervous system and gastrointestinal tract. Some of them are morphine, heroin, etc.
Morphine: It is extracted from the dried latex of the unripe seed capsule (Pod) of the poppy plant. It occurs as colourless crystals or a white crystalline powder.
Mode of abuse: Generally it is taken orally or by injection.
Effect: It is effective as a sedative and painkiller. It is very useful in patients who have undergone surgery.
Heroin: It is a white, bitter, odourless, and crystalline compound, obtained by the acetylation of morphine. Chemically it is diacetylmorphine. It is commonly called a snack.
Mode of abuse: Generally it is taken by shorting and injection.
Effect: Heroin is a depressant and slows down body functions.

Question 15.
Write short notes on Cannabinoids.
Answer:
Cannabinoids: These are a group of chemicals obtained from the Indian temp, plant cannabis Sativa (vernacular name Ganjai mokka). They interact with cannabinoid receptors present in the brain. The flower tops, leaves, and the resin of this plant are used in various combinations to produce marijuana, hashish, charas, and ganja. These daufs, cannabinoids are being abused by even some sports – persons (doping).
Mode of abuse: These are generally taken by inhalation and oral ingestion.
Effect: Show their effects on the cardiovascular system of the body.

Question 16.
Write short notes on Cocaine.
Answer:
Coca alkaloid or cocaine: It is a white, crystalline alkaloid that is obtained from the leaves of the coca plant Erythroxylum coca, native to South America. It is commonly called coke or crack.
Mode of abuse: It is usually shorted.
Effect: It has a potent stimulating action on the central nervous system as it interferes with the transport of the neurotransmitter dopamine. Hence it produces a sense of euphoria and increased energy. Its excessive dosage causes hallucinations.

Question 17.
Why adolescence is considered a vulnerable phase?
Answer:
Adolescence: It is the time period between the beginning of puberty and the beginning of adulthood. In other words. It is the bridge linking childhood and adulthood. The age between 12-18 years is considered adolescence period. It is both a period and process during which a child becomes muture. It is accompanied by several biological and behavioural changes. Thus, adolescence is a very vulnerable phase of the mental and psychological development of an individual.

Question 18.
Distinguish between addiction and dependence.
Answer:
Addiction: It is a psychological attachment to certain effects such as euphoria. The most important thing one fails to realize it, the inherent addictive nature of tobacco, drugs, and alcohol, with the repeated use of TDA, the tolerance level of the receptors present in our body increases. Consequently, the receptors respond only to higher doses leading to greater intake and addiction. However, it should be clearly borne in mind that the use of TDA even once, can be a forerunner to addiction. Thus, the addictive potential of tobacco, drugs, and alcohol pull the users into a vicious circle leading to their regular use (abuse) from which they may not be able to get out. In the absence of any guidance or counseling, people get addicted and become dependent on them.

Dependence: It is the tendency of the body of manifest a characteristic unpleasant condition (withdrawal syndrome). The regular dose of drugs or alcohol is abruptly discontinued. The withdrawal syndrome is characterized by anxiety. Shakiness (tremors), nausea, and sweating may be relieved when regular use is resumed again. Dependence leads the patients to ignore all social norms.

Question 19.
‘Prevention is better than cure. justify with regard to TDA abuse.
Answer:
The age-old adage of prevention is better than cure holds true here also.
Some of the measures successful in the prevention and control of TDA abuse among adolescents are:

• Avoid undue parental pressure: Every child has his/her own choice. Capacity and personality. Parents should not force their children to perform beyond their capacity by comparing them with others in studies, games, etc.
• Responsibility of parents and teachers: They should look for the danger signs and counsel such students who are likely to get into the ‘trap’.
• Seeking help from peers: If peers find someone abusing drugs or alcohol immediately it should be brought to the notice of their parents or teachers so that they can guide them appropriately.
• Education and counseling: Educating and counseling the children to face problems, stress, and failures as a part of life.
• Seeking professional and medical help: A lot of help is available in the form of highly qualified psychologists, psychiatrists, and de-addiction and rehabilitation programmers.

Long Answer Type Questions

Question 1.
Explain the structure and life cycle of Entamoeba histolytica with the help of neat and labelled diagrams.
Answer:
Entamoeba histolytica (Gr. entos – within : amoiba – change histos – tissues ; lysis – dissolve) is a microscopic and monogenetic parasite that inhabits the large intestine and causes amoebic dysentery or amoebiasis in man.

It is cosmopolitan in distribution but more common in the tropical and subtropical regions of the world. It is common in the people of rural and densely populated urban areas wherever the hygienic conditions are poor.

Structure: Entamoeba histolytica passes through three distinct stages in its life cycle namely

• Trophozoite stage
• Precystic stage
• Cystic stage

(i) Trophozoit stage: It is the most active, motile, feeding, and pathogenic stage that lives in the mucosa and sub-mucosa membrane of the large intestine; It moves with the help of a lobopodium which is produced anteriorly. The body of the trophozoite is surrounded by plasma-lemma. Its cytoplasm is differentiated into outer clear, viscous non-granular ectoplasm and inner fluid like granular endoplasm.

Ribosomes, food vacuoles, and vesicular, cartwheel-shaped nuclei are present in the endoplasm. The absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica. It produces the proteolytic enzyme called histolysis due to which the species name histolytica was assigned to it. Due to the effect of this enzyme the mucosa and submucosa of the gut wall are dissolved releasing some amount of blood, and tissue debris that are ingested by the trophozoites. Hence the food vacuoles are with erythrocyte fragments of epithelial cells and bacteria. The presence of RBC in food vacuoles and cartwheel-shaped nuclei are the characteristic features of the trophozoites of Entamoeba histolytic.

(ii) Precystic stage: It is the non-feeding and non-pathogenic stage of Entamoeba histolytica that is found in the lumen of the large intestine. It is a small, spherical, or oval, non-motileform. The cytoplasm of the precystic stage stores glycogen granules and chromatid bars (made of ribonucleic protein) which act as reserve food.

(iii) Cystic stage: It is round in shape and is surrounded by a thin, delicate, and highly resistant cyst wall. It is found in the lumen of the large intestine. The process of development of the cyst wall is called encystation. Which is a means of tiding over the un¬favourable conditions that the parasite is going to encounter while passing to a new host. Soon after the encystation, the nucleus undergoes two successive mitotic divisions to form four daughter nuclei. This type of cystic stage is called tetranuclear cyst or mature cyst which is the stage infective to man.

Life cycle: The trophozoites undergo binary fissions in the wall of the large intestine and produce a number of daughter entamoeba. They feed upon the bacteria and the host’s tissue elements, grow in size, and again multiply. After repeated binary fission some of the young ones enter of the lumen of the large intestine and transform into precystic stages. Here, the precystic stages transform into cystic stages which in turn develop into tetranuclear cysts. The entire process is completed only in a few hours. These tetranuclear cysts come out along with the faecal matter and can remain alive for about 10 days. The cyst reaches a new host through contaminated food and water. In the small intestine of a new human host, the cyst wall gets ruptured releasing the tetranuclear amoebae. Such tetranuclear excystic amoebae are called metacysts.

The four nuclei of the metacyst undergo mitotic divisions and produce eight nuclei. Each nucleus gets a bit of the cytoplasm and thus eight daughter entamoeba or metacystic trophozoites are produced. These young ones develop into feeding stages called trophozoites. They invade the mucous membrane of the large intestine and grow into mature trophozoites.

Question 2.
Describe the life cycle of plasmodium vivax in man.
Answer:
The life cycle of plasmodium in man (The human phase): In man, the plasmodium reproduces by asexual reproduction called schizogony. It occurs in liver cells (hepatocytes) as well as in RBC. In liver cells, it is called hepatic schizogony and in RBC it is called erythrocytic schizogony.

Hepatic Schizogony: This was discovered by short and Cranham. Whenever a mosquito infected by plasmodium bites a man, nearly 2000 sporozoites are released into the blood of man through its saliva, within half an hour, they reach the hepatocytes where they undergo pre-erythrocytic and exo-erythrocytic cycles.

Pre-erythrocytic cycle: Whenever the sporozoites enter the liver cells they transform into trophozoites. They feed on the contents of the hepatic cells, assume a spherical shape, and attain the maximum size. This stage is called the schizont stage. Its nucleus divides several times Mitotically, followed by the cytoplasmic divisions resulting in approximately 12,000 daughter individuals called cryptozoites or the 1st generation merozoites. They enter the sinusoids of the liver by rupturing the cell membrane of the schizont and the liver cells. This entire process is completed approximately in 8 days. Now, these first-generation merozoites have two options, i.e., they can enter either fresh liver cells and continue the exo-erythrocytic cycle or they can enter RBC and continue the erythrocytic cycle.

Exo-erythrocytic cycle: If the trophozoites enter the fresh liver cells, they undergo changes similar to that of the pre-erythrocytic cycle and produce the second generation merozoites called meta cryptozoites. These are of two types the smaller micro-metacryptozoites and larger macro-metacry- photosites. This entire process is completed approximately in two days. The macro-metacryptozoites attack fresh liver cells and continue another exo-erythrocytic cycle, whereas the micro-metacryptozoites always enter the bloodstream and attack fresh RBC to continue the erythrocytic cycle.

Prepatent period: The interval between the first entry of plasmodium into the blood in the form of sporozoites and the second entry of plasmodium into the blood in the form of Cryptozoic is called a prepatent period. It lasts approximately 8 days. During this period, the host does not show any clinical symptoms of the disease. It is only a means of multiplication.

Erythrocytic cycle: It was first described by Camillo Golgi Hence it is also called Golgi cycle. This cycle is initiated either by the trophozoites of the pre-erythrocytic cycle or the micro metocryptozoites of the exo-erythrocytic cycle in the fresh RBC, these stages assume the spherical shape and transform into trophozoites. It develops a small vacuole that gradually enlarges in size pushing the cytoplasm and nucleus to the periphery.

Now the plasmodium looks like a fisher ring. Hence this stage is called the signet ring stage. Soon it loses the vacuole, develops pseudopodia, and becomes an amoeboid stage with the help of pseudopodia. It actively feeds on the contents of the RBC and increases in size. As a result, the RBC grows almost double its size. This process is called hypertrophy. The malaria parasite digests the globin part of the ingested hemoglobin and converts the soluble haem into insoluble crystalline haemozoin. It is called the ‘malaria pigment’ which is called a disposable product. During this stage, small red coloured dots appear in the cytoplasm of the RBC known as Schaffner’s dots. These are believed to be the antigens released by the parasite.

Now the plasmodium loses the pseudopodia, further increases in size, occupies the entire RBC, and becomes a schizont. It undergoes schizogony similar to that of the pre-erythrocytic cycle and produces 12 to 24 erythrocytic merozoites. They are arranged in the form of the petals of a rose in the RBC. Hence this stage is called the rosette stage] Finally the erythrocyte bursts and releases the merozoites along with haemozoin into the blood. This cycle is completed approximately in 48 hours.

Incubation period: The period between the entry of plasmodium into the blood in the form of sporozoite and the first appearance of symptoms of malaria in man is called the incubation period, which is approximately 10 to 14 days.

Formation of gametocytes: After repeated cycles of erythrocytic schizogony when the number of fresh RBC decreases, some merozoites enter the RBC and transform into gametocytes instead of continuing the erythrocytic cycle. This process generally takes place when the RBCs are present in the spleen and bone marrow.

The gametocytes are of two types namely smaller microgametocytes or male gametocytes and larger macrogametocytes or female gametocytes. The gametocytes cannot undergo further development in man as the temperature and PH of the blood man are not suitable for further development. These gametocytes reach the blood circulation and wait to reach the next host. They degenerate and die if they are not transferred to mosquitoes within a week.

Question 3.
Describe the life cycle of plasmodium vivax in mosquitoes.
Answer:
Life cycle of plasmodium in mosquito (The mosquito phase) Ross cycle: When a female Anopheles mosquito bite and sucks the blood of a malaria patient the gametocytes along with the other stages of the erythrocytic cycle reach the crop of mosquito. Here all the stages are digested except the gametocytes. Further part of the life cycle consists of

• Gametogony
• Fertilization
• Formation of ookinete & oocysts
• Sporogony

(i) Gametogony: The formation of male and female gametes from the gametocytes is called gametogony. It occurs in the lumen of the crop of mosquitoes.

Formation of male gametes: During this process, the nucleus of the microgametocyte divides into eight daughter nuclei called pronuclei which reach the periphery. The cytoplasm is pushed out in the form of eight flagella-like processes. Into each flagellum-like process, one pronucleus enters and forms a micro gamete or male gamete. These male gametes show lashing movements like flagella and get separated from the cytoplasm of microgametocyte. This process is called exflagellation.

Formation of female gamete: The female gametocyte undergoes a few changes and transforms into a female gamete. This process is called maturation. The nucleus of the female gamete moves towards the periphery and the cytoplasm at that point forms a projection. This projected region is called the fertilization cone.

Fertilization: The fusion of male and female gametes is called fertilization. It also occurs in the lumen of the crop of the mosquito. When an actively moving male gamete comes into contact with the fertilization cone of the female gamete, it enters it, and the pronuclei and cytoplasm of these two gametes fuse with each other, resulting in the formation of a synkaryon Since the two gametes are dissimilar in size this process is known as anisogamy. The female gamete that bears the synkaryon is called the zygote which is round and non-motile.

(iii) Formation of ookinete and oocysts: The zygote remains inactive for some time and then transforms into a long, slender, motile, vermiform ookinete or vermicule within 18 to 24 hours. It pierces the wall of the crop and settles beneath the basement membrane. It becomes round and secretes a cyst around its body. This encysted ookinete is now called an oocyst. About 50 to 500 oocysts are formed on the wall of the crop and appear in the form of small nodules.

(iv) Sporogony: The formation of sporozoites in the oocysts is called sporogony. According to Bano, the nucleus of the oocyst first undergoes reduction division followed by repeated mitotic divisions resulting in the formation of about 1,000 daughter nuclei. Each bit of the nucleus is surrounded by a little bit of the cytoplasm and transforms into a sickle-shaped sporozoite. Oocyst with such sporozoites is called sporocyst.

When this sporocyst raptures, the sporozoites are liberated into the haemocoel of the mosquito. From there, they travel into the salivary glands and are ready for infection. The life cycle of plasmodium in mosquitoes completes in about 10 to 24 days.

Question 4.
Describe the structure and life cycle of Ascaris lumbricoides with the help of a neat and labelled diagram.
Answer:
Ascaris lumbricoides is commonly called the common roundworm. It lives in the small intestine of man, more frequently in children. It is cosmopolitan in distribution. The mode of infection is through contaminated food and water. The infective stage is the embryonated: egg with the 2nd stage rhabditiform larva.

Structure: Sexes are separate and sexual dimorphism is distinct. In both males and females, the body is elongated and cylindrical. The mouth is present at the extreme anterior end and is surrounded by three chitinous lips close to the mouth. Mid ventrally there is a small aperture called an excretory pore.

Male: It has a curved posterior end which is considered the tail. The posterior end possesses a cloacal aperture and a pair of equal-sized chitinous pineal spicules or pineal setae which serve to transfer the sperms during copulation.

Female: It has a straight posterior end, the tail. The female genital pore or vulva is present mid-ventrally at about one-third the length from the mouth. The anus is present a little in front of the tail end.

Life history: Copulation takes place in the small intestine of a man. After copulation, the female releases approximately two lakh eggs per day. Each egg is surrounded by a protein coat with a rippled surface. Hence the eggs of Ascaris are described as mammilla eggs. The protein coat is followed by a chitinous shelf and a lipid layer internally. These eggs come out along with faecal matter. In the moist soil, development takes place inside the egg so that the 1st stage rhabditiform larva is produced. It undergoes the 1st moulting and becomes the 2nd stage rhabditiform larva which is considered the stage infective to man. They reach the alimentary canal of man through contaminated food and water.

In the small intestine, the shell gets dissolved so that the 2nd stage larva is released. Now it undergoes extra-intestinal migration. First, it reaches the liver through the hepatic portal vein. From there it reaches the heart through the post caval vein. It goes to the lungs through the pulmonary arteries.

In the alveoli of the lungs, it undergoes the 2nd moulting to produce the 3rd stage larva. It undergoes the 3rd moulting so that the 4th stage larva is produced in the alveoli only. It leaves the alveoli and reaches the small intestine again through the bronchi, trachea, larynx, glottis, pharynx, oesophagus, and stomach. In the small intestine. It undergoes the 4th and final moulting to become a young one which attains sexual maturity within 8 to 10 weeks.

Question 5.
Describe the life cycle of wucheria bancrofti.
Answer:
Wucheria bancrofti is commonly called the filarial worm as it causes filariasis in human beings. It is a digenetic parasite that lives in the lymph vessels of man. Sir Patrick Manson identified the female culex mosquito as its secondary host.

Life cycle: It completes its life cycle in two hosts namely man and female culex mosquito.

In man: Both male and female worms are found coiled together in the lymphatic vessels of man. After copulation, the female releases the sheathed microfilaria larvae into the lymph of the man. Each sheathed microfilaria larva measures 0.2 to 0.3 mm in length. It is surrounded by a loose cuticular sheath which is supposed to be the modified shell. They migrate to the blood circulation and reside in the deeper blood vessels during the daytime. They move to the peripheral blood circulation during the nighttime between 10.00 pm and 4.00 am. This tendency is referred to as nocturnal periodicity. When a female culex mosquito sucks the blood of an infected person. They enter the gut of mosquitoes. They die if they are not transferred to mosquitoes within 70 days.

In mosquito: In the midgut of a mosquito, the sheath of the larva is dissolved within 2 to 6 hours of the infection. The ex-sheathed microfilaria larva penetrates the gut wall and reaches the heamocoel of the mosquito. From there, it reaches the thoracic muscles and transforms into a sausage-shaped larva within two days. It is called the first stage larva or first stage microfilaria. This undergoes two moultings within 10 to 20 days and transforms into infective 3rd stage microfilaria. It reaches the labium of the mosquito.

In man after the infection: When an infected mosquito bites a man, the 3rd stage microfilaria larvae enter the blood circulation of the man and finally reach the lymphatic vessels. Here they undergo the 3rd and the 4th moultings to produce young filarial worms. They attain sexual maturity within 5 to 18 months.

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa

Very Short Answer Type Questions

Question 1.
Draw a labelled diagram of the T.S of the flagellum.
Answer:

Question 2.
List any two differences between a flagellum and a cilium.
Answer:

Question 3.
What are dynein arms? What is their significance?
Answer:
‘A’ tube of each peripheral doublet bears paired arms along its length called dynein arms made up of protein dynein.
The dynein arms of the ‘A’ tubule face the tubule ‘B’ of the adjacent doublet.

Question 4.
What is a kinety?
Answer:
In the ciliate protozoans, a longitudinal row of kinetosomes together with kinetodesmata constitute a unit called kinety.

Question 5.
Distinguish between synchronous and metachronous movements.
Answer:
Synchronous movement: Cilia in a transverse row beat simultaneously in one direction. It is called synchronous movement.
Metachronous movement: The sequential movement of cilia, in a longitudinal row, one after the other in one direction is called metachronous movement.

Question 6.
Why do we refer to the offspring, formed by the asexual method of reproduction, as a clone?
Answer:
As a result of the asexual method, the offsprings are, not only identical to one another but also exact copies of their parent. The term ‘clone’ is used to describe such morphologically and genetically similar individuals.

Question 7.
Distinguish between proter and opisthe.
Answer:
During transverse binary fission in Paramecium two daughters, individuals are formed. The anterior one is called proter and the posterior is called opisthe.

Question 8.
How is sexual reproduction advantageous in evolution?
Answer:
Sexual reproduction results the advantageous in evolution in genetic recombination occurs in sexual reproduction.

Question 9.
Distinguish between lobopodium and filopodium. Give an example to each of them.
Answer:
Lobopodium: The blunt and finger-like tubular pseudopodia containing both ectoplasm and endoplasm is called lobopodium.
Ex: Amoeba proteus
Filopodium: The slender filamentous pseudopodia with pointed tips, composed of only ectoplasm are called Filopodium.
Ex: Euglypha

Question 10.
Define conjugation with reference to ciliates. Give two examples.
Answer:
Conjugation is a temporary union between two senile ciliates that belong to two different ‘mating types’ for the exchange of nuclear material and its reorganization. – Wichterman.
Ex: Paramecium and Vorticella.

Short Answer Type Questions

Question 1.
Name the system that controls the fastest swimming movement of protozoans and write its components.
Answer:
Ciliary locomotion is the fastest swimming movement of protozoans, Hence, ciliates are the fastest protozoans.
Cilia are small hair-like structures found in ciliate protozoans like Paramecium.

Infraciliary system: It is located just below the pellicle in the ectoplasm of a ciliate. It includes kinetosomes, kinetodesmal fibrils, and kinetodesmata. The kinetosomes are present at the bases of cilia in transverse and longitudinal rows. The kinetodesmal fibrils are connected to the kinetosomes and run along the right side of each row of kinetosomes as a ‘cord of fibres’ called kinetodesmata. A longitudinal row of kinetosomes together with kinetodesmata constitute a unit called ‘kinety’.

All the kineties together form an infraciliary system, which is connected to a ‘motorium’, located near the cytopharynx. The infraciliary system and motorium form the ‘neuromotor system’ that controls and coordinates the movement of cilia.

Question 2.
Write the mechanism of bending the flagellum and explain effectively recovery strokes.
Answer:
The bending movement of a flagellum is brought about by the sliding of microtubules past each other due to the functioning of ‘dynein arms’ utilizing ATP. A flagellum pushes the fluid medium at right angles to the surface of its attachment, by its bending movement.

Bending movement of flagella and cilia: Dynein arms show a complex cycle of movements using energy provided by ATP (dynein arms are the sites of ATPase activity in the cilia and flagella). The dynein arms of each doublet attach to an adjacent doublet and pull the neighbouring doublet. So the doublets slide past each other in opposite directions. The arms release and reattach a little further on the adjacent doublet and again ‘puli’. As the doublets of a flagellum or cilium are physically held in place by the radial spokes, the doublets cannot slide past much. Instead, they curve and cause the bending of flagellum or cilium. Such bending movements of flagella and cilia play an important role in flagellar and ciliary locomotion.

The flagellar movement of many organisms is a sidewise-lash which consists of two strokes namely the effective or propulsive stroke and the recovery stroke.

• Effective stroke: Flagellum becomes rigid and starts bending to one side beating against the water. This beating against water is at right angles to the body axis and the organism moves forwards.
• Recovery stroke: Flagellum becomes comparatively soft so as to offer the least resistance to water and moves back to its original position. It is called ‘recovery stroke’.

Question 3.
What are lateral appendages? Based on their presence and absence, write the various types of flagella giving atleast one example for each type.
Answer:
Lateral appendages: Some flagella bear one or two or many rows of short, lateral hair-like fibrils called lateral appendages. They are of two types namely ‘mastigonemes’ and ‘flimmers’.
Types of Flagella: Based on the presence or absence and/or the number of rows of lateral appendages, five types of flagella are recognized.
(a) Stichonematic: This flagellum bears one row of lateral appendages on the axoneme.
E.g. Euglena and Astasia.
(b) Pantonematic: This flagellum has two or more rows of lateral appendages on the axoneme.
E.g. Peranema and Monas.
(c) Acronematic: This type of flagellum does not bear lateral appendages and the terminal part of the axoneme is naked without the outer sheath at its tip.
E.g. Chlamydomonas and Polytoma

(d) Pantacronematic: This type of flagellum is provided with two or more rows of lateral appendages and the axoneme ends in a terminal naked filament.
E.g. Urceolus.
(e) Anematic or simple: In this type of flagellum, lateral appendages and terminal filament are absent. Hence, it is called anematic (a-no; nematic-threads)
E.g. Chilomonas and Gryptomonas.

Question 4.
Describe the process of transverse binary fission in paramecium.
Answer:
Transverse binary fission is performed by Paramecium. Binary fission is the most common method of sexual reproduction in protozoans. During favourable conditions, Paramecium stops feeding after attaining its maximum growth.

At first, the micronucleus divides by mitosis and the macronucleus divides into two daughter nuclei by amitosis. The oral groove disappears. After karyokinesis, a transverse constriction appears in the middle of the body, which deepens and divides the parent cell into two daughter individuals, the anterior proper and the posterior opisthe. The proper receives the anterior contractile vacuole, cytopharynx, and cytosome from its parent individual. It develops a posterior contractile vacuole and a new oral groove.

The opisthe receives the posterior contractile vacuole of its parent. It develops a new anterior contractile vacuole, cytopharynx, cytostome and a new oral groove. Binary fission is completed in almost two hours, in favourable conditions and Paramecium can produce four generations of daughter individuals by binary fission in a day. The transverse binary fission is also called homothetogenic fission because the plane of fission is at right angles to the longitudinal axis of the body. As it occurs at right angles to the kineties, it is also called perkinetal fission.

Question 5.
Describe the process of longitudinal binary fission in Euglena.
Answer:
Binary fission is the most common method of asexual reproduction in protozoans. Longitudinal binary fission is performed by Euglena. In this type of binary fission, the body divides into two halves longitudinally, hence called longitudinal binary fission.

During the process of binary fission, the nucleus, basal granules, chromatophores, and cytoplasm undergo division. The nucleus divides by mitosis into two daughter nuclei. Then the kinetosomes and the chromatophores also divide. At first, a longitudinal groove develops in the middle of the anterior end. This groove extends gradually towards the posterior end until the two daughter individuals are separated. One daughter Euglena retains the parental flagella. The other daughter individual develops new flagella from the newly formed basal granules. The stigma, paraflagellar body, and contractile vacuole of the parent disappear. They develop afresh in both the daughter Euglenae. The longitudinal binary fission is known as symmetrogenic division because the two daughters Euglenae resemble each other like mirror images.

Question 6.
Write a short note on multiple fission.
Answer:
Multiple fission: It is the division Of the parent body into many smaller individuals (Multi-many; Fission- splitting). Normally multiple fission occurs during unfavourable conditions. During multiple fission, the nucleus first undergoes repeated mitotic divisions without cytokinesis. This causes the formation of many daughter nuclei. Then the cytoplasm also divides into as many bits as there are nuclei. Each cytoplasmic bit encircles one daughter nucleus. This results in the formation of many smaller individuals from a single-parent organism. There are different types of multiple fissions in protozoans such as Schizogony, malegametogony, sporogony in plasmodium, sporulation in Amoeba, etc.

Question 7.
Give an account of pseudopodia.
Answer:
Locomotion in protozoans is performed by cellular extensions such as pseudopodia found in rhizopods organisms. The pseudopodia are temporary extensions of cytoplasm that develop in the direction of the movement. These temporary structures are useful to move on the substratum as our legs do, hence the name ‘pseudopodia’. There are four kinds of pseudopodia in protozoans.

1. Lobopodia: blunt + finger like Pseudopodia. Ex: Amoeba, Entamoeba.
2. Filopodia: fiber like pseudopodia contain ectoplasm. Ex: Euglypha.
3. Reticulopodia: net like pseudopodia. Ex: Elphidium.
4. Axopodia or heliopodia: Sun ray-like pseudopodia. Ex: Actinophiys.

Question 8.
Give an account of the ultrastructure of an axoneme.
Answer:
Ultrastructure of flagellum: The axial filament or axoneme shows 9+2 organisation under the electron microscope. Two central singlets are enclosed by a fibrous inner sheath. Nine peripheral doublets form a cylinder between the inner sheath and the outer sheath. The “A” microtubule of each doublet is connected to the inner sheath by radial spokes. It also has pairs of arms all along the length and is directed towards the neighbouring doublet.

These arms are made of a protein called dynein. These arms create the sliding force. The peripheral doublets are surrounded by an outer membranous sheath called a protoplasmic sheath, which is the extension of the plasma membrane. Some flagella bear lateral appendages called flimmers or mastigonemes along the length of the axoneme above the level of the pellicle. Each flagellum arises from a basal granule that lies below the cell surface in the ectoplasm.

Question 9.
Draw a neat labelled diagram of Euglena.
Answer:

Question 10.
Draw a neat diagram of paramecium and label its important structures/components.
Answer:

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 5th Lesson Morphology of Flowering Plants Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 5th Lesson Morphology of Flowering Plants

Very Short Answer Questions

Question 1.
Differentiate fibrous roots from adventitious roots. [T.S. May, 18]
Answer:

Question 2.
Define modification. Mention how root is modified in Banyan tree and mangrove plants.
Answer:
Modification is defined as “A permanent morphological change in an organ in order to perform a special function”. In banyan tree, roots develop from the huge branches and grow into the soil become pillar like called prop roots or piller roots. In Mangroves – Many roots come out of the ground, grow vertically upwards called Pneumatophores, or Respiratory roots help in gaseous exchange.

Question 3.
What types of specialized roots are found in Epiphytic plants? What is their function?
Answer:
In Epiphytic plants, velamen roots are present. They help in absorption of moisutrefrom atmosphere.

Question 4.
How does the sucker of chrysanthemum differ from the stolon of Jasmine?
Answer:

Question 5.
What is meant by pulvinus leaf base? In members of which angiospermic family do you find them? [A.P. May, 18, 17. Mar, 14]
Answer:
The swollen leaf base is called pulvinous leaf base. It is seen is “Leguminasae” family.

Question 6.
Define venation. How do Dicots differ from Monocots with respect to venation. [A.P. Mar, 15]
Answer:
“The arrangement of veins and the veinlets in the lamina of the leaf” is called venation.

In Dicots, Midrib, lateral veins and veinlets form a network so called Reticulate venation.

In Monocots, the veins run parallel to each other within a lamina so called parallel venation.

Question 7.
How is Pinnately compound leaf is different from a palmately compound leaf? Explain with one example each.
Answer:
Lamina is divided into leaflets which are arranged on either side of the rachis is called pinnately compound leaf.
Ex : Neem

Lamina is divided into leaflets, which are arranged at the tip of the petiole, called Palmately compound leaf.
Ex : Bean, Citrus.

Question 8.
Which organ is modified to trap insects in Insectivorous plants? Give two examples. [Mar, 13]
Answer:
In Insectivorous plants, leaves (Lamina) are modified to trap insects.
Ex : Nepenthes, Drosera.

Question 9.
Differentiate between Racemose and Cymose inflorescences. [T.S. Mar, 15]
Answer:

Question 10.
What is the morphology of cup like structure in Cyathium? In which family it is found?
Answer:
In Cyathium, cup like structure is formed from Involucre of Bracts. It is seen in “Euphorbiaceae” family.

Question 11.
What type of Inflorescence is found in fig trees ? Why does the insect Balstophaga visits the Inflorescence of fig tree?
Answer:
In ‘Fig’ trees, ‘Hypanthodium’ Inflorescence is seen. An Insect called ‘Blastophaga’ visits this inflorescence, because it lay its eggs in the gall flowers.

Question 12.
Differentiate actinomorphic from zygomorphic flower.
Answer:

Question 13.
How do the petals in pea plant are arranged? What is such type of arrangement called?
Answer:
In Pea and Bean Flowers, there are five petals. The largest petal (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (Keel). This type of aestivation is called Descendingly Imbricate aestivation. This type of corolla is called vexillary or papilionaceous corolla.

Question 14.
What is meant by eipetalous condition? Give an example.
Answer:
When the stamens are attached to the petals that condition is called Epipetalous condition.
Ex : Brinjal, Datura.

Question 15.
Differentiate between apocarpous and syncarpores ovary.
Answer:

Question 16.
Define placentation. What type of placentation is found in Dianthus. [T.S. Mar, 15]
Answer:
“The arrangement of ovules with in the ovary is known as placentation”. In Dianthus – Free Central placentation is seen.

Question 17.
What is meant by parthenocarpic fruit? How is it useful?
Answer:
The fruit which is formed without fertilization of the ovary is called parthenocarpic fruit. They are useful commercial production of seedless fruits and also useful in Juice Indus tries.

Question 18.
What is the type of fruit found in mango ? How does it differ from that of coconut?
Answer:
In Mango, the type of fruit is Drupe. In Mango Epicarp is thin, Mesocarp is fleshy and edible and endocarp is stony. Where as in coconut, the mesocarp is fibrous.

Question 19.
Why certain fruits are called false fruits? Name two examples of plants having false fruits. [T.S. May, 18. Mar, 13]
Answer:
Fruits which are formed form other parts of the flower along with ovary are called false fruits.
Ex : Apple – Thalamus develops into false fruit.
Cashewnut – Pedicel develops into False fruit,

Question 20.
Name any two plants having single seeded dry fruits.
Answer:
(1) Rice (Caryopsis) (2) Cashew nut (nut) (3) Tridax (Cypsela)

Question 21.
Define Schizocarpic dry fruits. Give an example.
Answer:
The fruits which split into one-seeded bits, called mericarps and are called Schizocarpic fruits.
Ex : Acacia, Castor. ,

Question 22.
Define mericarp. In which plant you find it?
Answer:
The one seeded bits of Schizocarpic fruits are called mericarps.
Ex : Acacia.

Question 23.
What are aggregate fruits? Give two examples. [A.P. May, 18]
Answer:
In custard apple. Many carpels are present. Each carpel of apocarpous gynoecium develops into a fruitlet. Such bunch of fruitlets are called aggregate fruits.
Ex : Annona, Naravalia.

Question 24.
Name a plant that has single fruit developing from the entire inflorescence. What is such a fruit called? [Mar, 14]
Answer:
Pine apple. This fruit is called composite fruit.

Short Answer Type Questions

Question 1.
Explain different regions of root with neat labelled diagram.
Answer:

The tip of the root is covered by Thimble like structure called root cap which protect the root apex. A few m.m above the root cap, meristematic zone is present which consists of small, thin walled cells with dense protoplasm. These cells divide repeatedly. The cells proximal to this zone undergo rapid elongation and enlargement and are responsible for growth of the root in length.

This region is called elongation region. The cells of the elongation zone gradually differentiate and mature so called maturation zone. From this zone, some of the epidermal cells form very thin and delicate thread like structures, called root hairs which help in absorption of water and Minerals from the soil.

Question 2.
Justify the statement, “Underground parts of plants are not always roots”.
Answer:
In some plants, stems are underground and help in storage of food materials, organs of perennation to tide over unfavourable conditions, vegetative propagation and also protect themselves form grazing animals. More over, underground stems bear nodes, internodes, buds and scale leaves. Because of this, they are treated as stems but not roots, so underground parts of plants are not root.
Ex : Stem tuber of Potato, Rhizome of Zinger, Turmeric, corm of Amorphophallus, Colocasia and Bulb of onion.

Question 3.
Explain with examples different types of phyllotaxy.
Answer:
Arrangement of leaves on the stem or on the branchs is called Phyllotaxy. It is of three types.
A) Alternate Phyllotaxy : A single leaf arises at each node arranged in alternate manner.
Ex : Hibiscus rosa-sinensis, mustard.

B) Opposite Phyllotaxy : A pair of leaves arise at each node and lie opposite to each other.
Ex : Calotropis, Guava.

C) Whorled Phyllotaxy : More than two leaves arise at a node and form a whorl.
Ex : Nerium, Alstonia.

Question 4.
How do leaf modifications help plants?
Answer:
Leaf modifications will help to plants in Many ways. They are
a) In weak stemmed plants like peas, leaves are converted into long, coiled tendrils and help in climbing.
b) In Desert plants like Cacti, leaves are modified into spines, which gave protection from grazing animals.
c) In Onion and Garlic : The fleshy leaves help in storage of food materials.
d) In Australian acacia, the leaves are pinnately compound in which leaflets are small and short lived. The petioles of these plants expand, become green and synthesize food called phyllodes.
e) In Insectivores plants, like Nepenthes, Dionea, leaves are modified into traps which kill insects for their nitrogen requirement.
f) In Bryophyllum, epiphyllous buds which arise from the notches of the leaves develop into new plants, thus help in vegetative Propagation.

Question 5.
Describe any two special types of Inflorescences.
Answer:
1) Cyathium :

It is a single flower like special inflorescence found in Euphorbiaceae members. The inflorescence is covered by a deep cup like involucre of bracts. At the centre of the cup, there is a single female flower represented by a long stalked tricarpellary, Syncarpous pistil. Encircling this females flower, many male flowers are arranged in scorpioid cymes. Each male flower is represented by a single and stalked stamen. Male and female flowers are achlymadeous. Flowers are arranged in centrifugal manner.
Ex : Euphorbia and Poinsettia

2) Hypanthodium :

It is a fruit like inflorescence. The inflorescence axis is condensed and forms a fleshy, cup like structure with an apical opening. Small, sessile, unisexual flowers develop on the inner wall of the cup. Male flowers are near the apical opening and female flowers are at the base. In between these two types of flowers, some sterile female flowers called ‘gall flowers’ are present. The opening of flowers is not in a definite order.
Ex : Ficus species.

Question 6.
Describe the arrangement of floral membes in relation to their insertion on thalamus.
Answer:
Based on the position of Calyx, corolla and androecium on thalamus in respect to the ovary, the flowers are of three types.
They are :
A) Hypogynous,
B) Perigynous,
C) Epigynous

A) Hypogynous flower :
The gynoecium occupies the highest position, while the other parts are situated below it. The ovary is said to be superior.
Ex : Mustard, Brinjal.

B) Perigynous flower :
Gynoecium is present in the centre of the thalamus and other parts of the flower are located on the rim of the thalamus. The ovary is said to be half superior.
Ex : Rose, Pea.

C) Epigynous flower :
The ovary is completely embeded in the cup like thalamus. Other floral parts of the flower arise just above the ovary. In this, the ovary is said to be Inferior.
Ex : Guava, Cucumber.

Question 7.
“The flower of many angiospermic plants which show sepals and petals differe with respect to the arrangement of sepals and petals in respective whorls” – Explain.
Answer:
The mode of arrangement of sepals or petals in floral bud is knwon as Aestivation. It is of 4 types.
A) Valvate :
“When sepals or petals in a whorl just touch one another at the margins without overlapping”.
Ex : Calotropis.

B) Twisted :
“If the margin of the appendage overlaps that of the next one and so on”.
Ex : Cotton, Hibiscus. (Petals).

C) Imbricate :
If the margins of sepals or petals overlap one another but not in any particular direction.
Ex : Cassia, gulmohar.

D) Vexillary :
5 petals are present, of which the largest Petal (standard) overlaps the two lateral petals (Wings) which in turn overlap the two smallest anterior petals (keel). This is known as vexillary or papilionaceous aestivation.
Ex : Pea, Bean.

Question 8.
Describe any four types of placentations found in flowering plants.
Answer:
The arrangement of ovules with in the ovary is known as placentation. It is of 5 types.
A) Marginal :
“The placenta forms a ridge along the ventral suture of the ovary and the ovules are borne on this ridge forming two rows”.
Ex : Pea.

B) Axial :
The ovules are attached to central placenta in a multilocular ovary.
Ex : Tomato, lemon.

C) Parietal :
The ovules develop on the inner wall of the oary. Ovary is one chambered but it becomes two chambered due to the development of flase septum.
Ex : Mustard.

D) Free central :
The ovules are borne on central axis without septa.
Ex : Dianthus.

E) Basal :
The placenta develops at the base of the ovary and a single ovule is attached to it.
Ex : Sunflower.

Question 9.
Describe in brief fleshy fruits you studied.
Answer:
Fruits which become fleshy at maturity one are called fleshy fruits. They are 5 types.
1) Berry :
It develops from Bi or multicarpellary syncarpous Gynoecium. The mesocarp and endocarp are fused to form the pulp and the seeds are hard.
Ex : Tomato, Guava.

2) Pepo :
It develops from Tricarpellary syncarpous inferior ovary. The epicarp is rigid, the mesocarp is fleshy and the endocarp is smooth.
Ex : Cucumber.

3) Pome :
It develops from Bi of multicarpellary inferior ovary and is surrounded by fleshy thalamus. The endocarp is cartilegumous.
Ex : Apple.

4) Hesperidium :
It develops from Multicarpellary syncarpous, superior ovary. The epicarp is leathery with volatile giands, Mesocarp is papery and endocarp is several chambered with juicy hairs.
Ex : Citrus.

5) Drupe :
It develops from Monocarpellary superior ovary. In Mango, the epicarp is thin, a middle fleshy edible mesocarp and inner stony hard endocarp. In coconut, the mesocarp is fibrous.

Question 10.
Describe with examples, the various dry fruits you studied.
Answer:
The fruits which become dry at maturity are called Dry fruits. They are of 3 types.
1) Dry dehiscent fruits :
“The fruits which breaks and liberate the seeds”.
Ex : (a) In legumes the fruit dehisces dorsiventrally into two halves and liberate the seeds,
(b) In Cotton, Datura – The capsule dehisce in different ways to liberate the seeds.

2) Dry Indehiscent fruits :
The fruits do not break and liberate the seeds only after the disintegration of pericarp.
Ex : (a) In Rice, The fruit is caryopsis. In this the pericarp and seed coat fuse together.
(b) In Cashew, the fruit is nut, whiph develops from multicarpellary, Syncarpous ovary. The pericarp is stony.
(c) In Tridax, The fruit is cypsela, Identified by the presence of persistent pappus like calyx.

3) Schizocarpic fruits :
“The fruit at maturity split into one – seeded bits called mericarps”.

Long Answer Type Questions

Question 1.
Define root. Mention the types of root systems. Explain how root is modified to perform different functions. [A.P & T.S. Mar, 17, 15, 13]
Answer:
The Radicle of the embryo elongates leads to the formation of primary root which grows into the soil is called root. In Angiosperms, two types of root systems are present namely, (a) Tap root system (b) Fibrous root system.

a) Tap root system :
The primary root grows into the soil called tap root, which produce lateral roots and root lets constitutes tap root system. It is seen in Dicotyledonous plants.

b) Fibrous root system :
The primary root is short lived and is replaced by a large number of roots which orginate from the base of the stem constitutes fibrous root system. It is seen in Monocotyledonous plants.

Root Modifications :
In some plants, roots change their shape and structure to perform functions other than absorption and conduction of water and Minerals called root modifications. They are of different types.

1) Storage roots :
In Carrot, turnip (Tap roots), Sweat potato (Adventitious roots), Asparagus (Fibrous roots) become swollen due to storage of food called storage roots.

2) Prop roots :
In Banyan tree, Roots arise from the branches grow into the soil, become pillar like and give additional support called prop roots or pillar roots.

3) Stilt oots :
In Maize, sugarcane, roots arise from the lower nodes of the stem, give additional support called stilt roots.

4) Respiratorty roots :
In Mangroves like Rhizophora and Avicennia, Many roots come out of the ground and grow vertically upwards, help in respiration called Pneumatophores.

5) Epiphytic roots :
In Epiphytes like Vanda, special adventitious roots arise, help in absorption of moisture from almosphere called Velamen roots.

6) Parasitic roots :
In partial parasites like viscum and strga, some Haustorial roots enter into xylem of the host plant to get water and Minerals. In complete parasitic like cuscuta and Rafflesia, the haustorial roots enter into xylem and phloem of the host plant and obtain water and Minerals and food materials called Parasitic roots.

7) Nodular roots :
In the members of Fabaceae, the roots are inhabited by Rhizobium bacteria which helps in N2 fixation called Nodular roots.

8) Photosynthetic roots :
In some plants like Taeniophyllum, the roots are chlorophyllous and perform photosynthesis so called photosynthetic roots.

Question 2.
Explain how stem is modified variously to perform different functions. [Mar, 14] [A.P & T.S. Mar, 18]
Answer:
Stems are modified in several ways to perform different functions.

In some plants stem grows into soil, not only stores food materials but also acts as organs of pennations. Such stems are called underground stem modifications. Further, they also help in vegetative propagation.
Ex : The stem tuber of potato, Rhizome of zingiber, corm of colocasia and Bulb of onion.

In some plants, Aerial stems show many modifications.
1) Tendrils :
Slender, spirally coiled structures which may develop either from Auxiliary Bud [Cucumber, watermelon] or from Terminal Bud [Grapes] called stem Tendrils, help in climbing.

2) Thorns :
In some plants, buds are modified into woody, straight, pointed thorns, which protect plants from grazing animals.
Ex : Citrus, Bougainvillaea.

3) Phylloclade :
In some plants of arid regions, stems are modified into fleshy flattened (opuntia) or fleshy cylindrical (Euphorbia) or needle like structures (casuarina), called phylloclades. They carry out photosynthesis as their leaves are reduced to scales or spines to reduce the transpiration.

4) Bulbils :
In some plants, the vegetative buds (Diascorea) or floral buds (Agave) store food materials. At maturity they detach from the parent plants, develop adventitious roots and help in vegetative propagation. These are called Bulbils.

In some plants, some part of the stem is aerial and some part is underground. Such stems are called sub aerial stem modifications.
1. Runners :
Sub aerial stems of “oxalis” spread to new niches and form new plants when older parts die. Such plants are called runners.

2. Stolons :
In plants like Nerium, Jasmine, a slender lateral branch called ‘stolon’ arises from the base of the main axis, grow vertically, arches downwards and touch the ground, produce adventitious roots.

3. Off-sets :
In some aquatic plants like pistia and Eichhornia, a lateral branch of one internode length called offset which bears a rosette of leaves at each node and a tuft of adventitious roots arising from the base of the discoid stem.

4. Suckers :
In plants like banana, chrysanthemum, the lateral branches originate from the basal portion of the main stem, grows horizontally, beneath the soil and then come out obliquely upwards giving rise to leafy shoots. These branches are called suckers. All these sub-aerial stem modifications help in vegetative propagation.

Question 3.
Explain different types of Racemos inflorescences.
Answer:
Racemose Inflorescences are of several types. They are :
1) Simple Raceme :
The pedunde is simple, unbranched producing many pedicellate, bracteate flowers in Acropetal Manner.

2) Corymb :
The peduncle is long, and bears many flowers in Acropetal manner, but all the flowers are brought to the same level due to varied lengths of pedicels even through they are borne at different nodes.
Ex : Cassia, Cauliflower.

3) Umbel :
The flower appear to have arisen from the same point of the peduncle and is called umbel type. The Inflorescence is covered by a whorl of bracts called “Involucre”. Ex : Apiaceae, carrot.

4) Spike :
The peduncle is long and bears many sessile flowers, arranged in Aeropetal Manner.
Ex : Achy rant hus, poaceae.

5) Spadix :
In some plants, penduncle bears sessile unisexual and neutral flowers arranged in acropetal manner, covered and protected by modified Bract called spathe.
Ex : Cocos, colocasia.

6) Head :
In some plants, unisexual and bisexual sessile flowers develop centripetially on a condensed peduncle.
Ex : Tridax, sunflower.

Additional Questions & Answers

Question 1.
What is meant by Scutellum? In which type of seeds it is present?
Answer:
Large and shield shaped cotyledon is known as scutllum. It is seen in Monocot seeds.

Question 2.
Define with examples endospermic and non-endospermic seeds.
Answer:
Endosperm may be completely consumed by the developing.

embryo before seed maturation. Such seeds are called non-endospermic seeds.

The seeds with endosperm after maturation are called endospermic seeds.
Ex : Castor, Coconut.

Intext Questions

Question 1.
In which plant, the underground stem grows horizontally in soil and helps in perennation?
Answer:
Zingiber officinalis (ginger) – Rhizome.
Curcuma tonga (Turmeric) – Rhizome.

Question 2.
Needle like phylloclades are found in which plant?
Answer:
Casuarina.

Question 3.
Why do plants like Nepenthes trap Insects?
Answer:
Nepenthes generally grow in Nitrogen deficient soils. In this, leaves are modified into pitcher to trap insects for their nitrogen requirement.

Question 4.
What is the characteristic Inflorescence found in members of Asteraceae?
Answer:
Head or Capitulum.

Question 5.
Can you name a plant that has least number of flowers in its inflorescence?
Answer:
Hibiscus, Datura.

Question 6.
Which family shows naked flowers?
Answer:
Euphorbiaceae.

Question 7.
In which flowers of the fig trees does the Insect Blastophaga lay its eggs ?
Answer:
Gall flowers.

Question 8.
What type of symmetry is shown by the flowers of Canna?
Answer:
Asymmetric.

Question 9.
On which side of the flower do the flowers of pea have the keel petals?
Answer:
Anterior side.

Question 10.
What is the ratio of overlapping margins of petals to overlapped ones in imbricate aestivation?
Answer:
5 : 4

Question 11.
How many ovules are found attached in Basal placentation?
Answer:
One

Question 12.
Which part of the flower in Cashew plant forms the false fruit?
Answer:
Pedicel

Question 13.
Which plant has hard, stony endocarp and fleshy edible mesocarp?
Answer:
Mango

Question 14.
What is the morphology of spathe in spadix Inflorescence?
Answer:
Bract (enlarged)

Question 15.
What is the type of fruit known as if it develops from apocarpus ovary of a single flower?
Answer:
Aggregate fruits. (Anona squamosa)

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 3rd Lesson Science of Plants – Botany Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 3rd Lesson Science of Plants – Botany

Very Short Answer Questions

Question 1.
Explain how the term Botany has emerged.
Answer:
In ancient Greek language, the term Bouskein means Cattle feed. The term ‘Bouskein’ gave rise to “Botane” from which the term “Botany’’ is derived.

Question 2.
Name the books written Parasara and mention the important aspects discussed in those books. [A.P. Mar. 17]
Answer:
Parasara’s ‘Krishi Parasaram’ mentioned about agriculture and weeds. ‘Vrikshayurveda’ gives information about 14 types of forests; the external and internal characters of plants and about medicinal plants.

Question 3.
Who is popularly known as father of Botany ? What was the book written by him? [T.S. Mar, 18]
Answer:
“Theophrastus” is regarded as the Father of Botany. The book written by him was ‘De Historia plantarum”,

Question 4.
Who are Herbalists? What are the books written by them?
Answer:
The scientists who described the live Medicinal plants technically are called the Herbalists. The books written by them are called Herbals.

Question 5.
What was the contribution of Carolus Von Linnaeus for the development of plant taxonomy?
Answer:
“Carolus Von Linnaeus” popularised Binomial Nomenclature. He also proposed the sexual system of classification.

Question 6.
Why is Mendel considered as the father of Genetics?
Answer:
Mendel conducted Hybridization experiments on pea plants and also introduced the laws of inheritance in 1866. Hence, he is considered as the Father of Genetics.

Question 7.
Who discovered the cell and what was the book written by him? [A.P. May 17, Mar 14]
Answer:
Robert Hooke first discovered the cell. “Micrographia” was the book written by him.

Question 8.
What is Palaeobotany? What is its use? [T.S. Mar, 17, 15, 13]
Answer:
Palaeobotany deals with the study of fossil plants. It helps in understanding the course of evolution in plants.

Question 9.
Name the branches of Botany which deal with the chlorophylfous autotrophic thallophytes and non-chlorophyllous heterotrophic thallophytes.
Answer:
Study of chlorophyllous autotrophic thallophytes is called Phycology. Study of non- chlorophyllous heterotrophic thallophytes is called Mycology.

Question 10.
What are the group of plants that live as symbionts in lichens? Name the study of Lichens. [T.S. May, 18]
Answer:
Algae and Fungi are the plant groups that live as symbionts in lichens. The study of lichens is called lichenology.

Question 11.
Which group of plants are called Vascular cryptogams ? Name the branch of Botany which deals with them?
Answer:
“Pteridophytes” are called vascular cryptogams. The branch of Botany that deals with them is called Pteridology.

Question 12.
Which group of plants are called amphibians of plant kingdom? Name the branch of Botany which deals with them.
Answer:
Bryophytes are called amphibians of plant Kingdom. The branch that deals with Bryophytes is called Bryology.

Short Answer Type Questions

Question 1.
Explain in brief the scope of Botany in relation to agriculture, horticulture and medicine.
Answer:

1. The problems like decreasing resources and increasing population could be solved by enhancing crop yield through Green Revolution and.also by developing disease* pest resistance crops by utilizing the principles of Biotechnology.
2. Progress in Agriculture, Forestry horticulture is possible through hybridization experiments and Genetic Engineering technology.
3. New plant breeding techniques are useful to develop hybrid varieties in crop plants like Rice, Wheat, Maize, Sugarcane etc.
4. Phytopathology is helpful in the prevention and eradication of several plant diseases.
5. Knowledge on the role of plant hormones in plant growth and development is significant to improve Agriculture and Horticulture through herbicidal control of weed, breaking of seed dormancy etc., Experiments in tissue and organ culture is possible to produce large number of plants in the laboratory within a short duration.
6. Several industries could be developed based on Botany.
7. Studies in Medicinal plants like Arnica, Cinchona, Neem, Datura, Rauwolfia, Ocimum are also important to explore them for human health care.
8. Production of Antibiotics, Bioinsecticides, single cell proteins is also made possible by study of these product yielding plants.

Question 2.
Explain the scope of Botany taking plant physiology as example.
Answer:

1. The role of Minerals iri plant nutrition is useful in rational usage of chemical fertilizers and control of mineral deficiencies to improve Agricultural productivity.
2. Knowledge on the role of plant hormones in plant growth and development’s highly significant to improve Agriculture and horticulture through herbicidal control of weeds, breaking of seed dormancy, enhancement of shelf-life period of leafy vegetables like spinach, artificial ripening of fruits like apple, banana and rooting of stem cuttings for Vegetative propogation.

Question 3.
What are the different branches of botany that deal with morphology of plants ? Give their salient features.
Answer:
Morphology deals with the study of different organs of a plant. It is a fundamental requisite for classification of plants. It is of two types,
a) External Morphology
b) Internal Morphology.

a) External Morphology :
It is the study and description of external characters of plant organs like root, stem, leaf, flower, seed and fruit etc.

b) Internal Morphology :
It is the study of internal structure of different plant organs. It is of two types. They are i) Histology ii) Anatomy.
i) Histology :
Study of different tissues present in the plant body.

ii) Anatomy :
Study of gross internal details of plant organs like root, stem, leaf, flower, etc.

Long Answer Type Question

Question 1.
Give a comprehensive account of the scope of Botany in different fields giving an example for each.
Answer:
I) Agriculture :

1. Enhancement of crop yield through Green Revolution solves the problems like decreasing resources and increasing population.
2. New plant breeding techniques are useful to develop hybrid varieties in crop plants like Rice, Wheat, Maize, Sugarcane etc.
3. Phytopathology is helpful in prevention and eradication of several plant diseases.

II) Medicine :

1. Studies in Medicinal plants like Arnica, Cinchona, Neem, Datura, Rauwolfia, Ocimum are important to explore them for human health care.
2. Production of antibiotics, bioinsecticides, single cell proteins is made possible by study of these plants.

III) Environmental Issues :

1. Control of Green house effect by tree plantation and soil pollution by bio-remediation, recycling of nutrients by saprophytic organisms, usage of biofertilizers to avoid soil and water pollution caused by chemical fertilizers and prevention of soil erosion by sand binding plants.
2. Usage of algae (Chlorella) as food for astronauts in space research programmes and extraction of Iodine, agar-agar etc. from several sea weeds also indicate the scope of Botany for the Contemporary World.

IV) Commercial Products :
Knowledge of plants which yield commercially important and useful products like timber fibres, beverages such as coffe and tea, condiments, rubber, gums, resins, dyes, essential and aromatic oils is of great importance for their exploitation.

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals

Very Short Answer Type Questions

Question 1.
The body of sponges does not possess a tissue level of organisation, though it is made up of thousands of cells. Comment on it.
Answer:
Even though the sponge’s body is made up of thousands of cells, they exhibit a cellular grade of organization, due to the absence of sensory and nerve cells spongs do not possess a tissue level of organisation.

Question 2.
What is the tissue level of organisation among animals? Which metazoans exhibit this organisation?
Answer:
The tissue level of organisation is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like cnidarians.

Question 3.
Animals exhibiting which level of organisation lead the relatively more efficient way of life when compared to those of the other levels of organisation? Why?
Answer:
Animals exhibit an organ-system level of organisation of the animals and are exhibited by the triploblastic animals.
Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation.

Question 4.
What is monaxial heteropolar symmetry? Name the group of animals in which it is the principal symmetry.
Answer:
Monaxial heteropolar symmetry: When any plane passing through the central axis of the body divides an organism into two identical parts is called Monaxial heteropolar symmetry or radial symmetry. It is the principal symmetry of diploblastic animals such as Cnidarians and ctenophores.

Question 5.
Radial symmetry is an advantage to sessile or slow-moving organisms. Justify this statement.
Answer:
Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow-moving animals.

Question 6.
What is cephalization? How is it useful to its possessors?
Answer:
Cephalization: Concentration of nerve (Brain) and sensory cells at the anterior end of the body is called Cephalization. As a result of cephalization, these animals can sense the new environment and are more efficient than the other animals in seeking food, locating mates, and avoiding or escaping from predators.

Question 7.
Mention the animals that exhibited a ‘tube-within-a-tube’ organisation for the first time? Name their body cavity.
Answer:
Cnidarians and some flat warms are the first animals to exhibit a ‘tube-within-a-tube’ organisation. The body cavity is pseudocolor.

Question 8.
Why is the true coelom considered a secondary body cavity?
Answer:
During the embryonic development of the coelomates, the blastocoel is replaced by a true coelom derived from the mesoderm. So, the true coelom is also called ‘the secondary body cavity.

Question 9.
What are retroperitoneal organs?
Answer:
Certain organs such as the kidneys of the vertebrates are covered by the parietal peritoneum only on their ventral side. Such a peritoneum is called the ‘retroperitoneum’ and the organs lined by it are called ‘retroperitoneal organs.

Question 10.
If the mesentoblast cell is removed in the early embryonic development of protostomes what would be the fate of such animals?
Answer:
In the protostomes, the mesentoblast cell of the early embryo divides to form mesodermal blacks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal black leads to the formation of Schizocoelom. If the mesentoblast cell is removed in the early embryonic development of protostomes it will cause no ceolome in their animals.

Question 11.
What is enterocoelom? Name the enterocoelomate phyla in the animal kingdom.
Answer:
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called ‘enterocoelomates’.
Echinoderms, hemichordates, and chordates are enterocoelomates.

Question 12.
Stratified epithelial cells have a limited role in secretion. Justify their role in our skin.
Answer:
The main function of stratified epithelial cells is to provide protection against chemicals and mechanical stress. Hence the stratified epithelial cells have a limited role in secretion.

Question 13.
Distinguish between exocrine and endocrine glands with examples.
Answer:
Exocrine glands are provided with ducts. Secrete mucus, saliva, earwax, oil, milk, digestive enzymes, and other cell products.
Endocrine glands are ductless and their products are hormones that are not sent out via ducts but are carried to the target organs by blood.
Ex: Pituitary gland.

Question 14.
Distinguish between holocrine and apocrine glands.
Answer:
Apocrine glands in which the apical part of the gland cell in pinched off along with the secretory product.
Ex: Mammary glands
Holocrine glands, in which the entire cell disintegrates to discharge the contents.
Ex: Sebaceous glands

Question 15.
Mention any two substances secreted by mast cells and their functions.
Answer:
Mast cells secrete heparin – an anticoagulant, histamine, bradykinin – vasodilators, and serotonin – vasoconstrictor. Vasodilators cause inflammation in response to injury and infection.

Question 16.
Distinguish between a tendon and a ligament.
Answer:
Tendons are the collagen fibrous tissue of dense regular connective tissue which attaches the skeletal muscles to bones.
Ligaments are also the collagen fibers tissue of dense regular connective tissue which attach bones to other bones.

Question 17.
Distinguish between brown fat and white fat.
Answer:
White fat: It is the predominant type in adults, the adipocyte has a single large lipid droplet. White fat is metabolically not active.
Brown fat: It is found in fetuses and infants. Adipocyte of Brown fat has several small ‘lipid droplets’ and are metabolically active and generates heat to maintain body temperature required by infants.

Question 18.
What is the strongest cartilage? In which regions of the human body, do you find it?
Answer:
The fibrous cartilage is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

Question 19.
Distinguish between osteoblasts and osteoclasts.
Answer:
Osteoblasts are immature bone cells that secrete the organic components of the matrix and also play an important role in the mineralization of bone and become Osteocytes. Osteoclasts are phagocytic cells involved in the resorption of bone.

Question 20.
Define Osteon.
Answer:
In compact bone structure, a Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or Osteon.

Question 21.
What are Volkmann’s canals? What is their role?
Answer:
In compact bone structure, the Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals.

Question 22.
What is a Sesamoid bone? Give an example.
Answer:
Sesamoid bones are formed by ossification in tendons.
Eg: Patella (Knee cap) and Pisiform bone of the wrist of a mammal.

Question 23.
What is lymph? How does it differ from plasma?
Answer:
Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes.

Question 24.
What is the hematocrit value?
Answer:
The percentage of the total volume occupied by RBCs in the blood is called the hematocrit value.

Question 25.
What are intercalated discs? What is their significance?
Answer:
The dark lines across cardiac muscle are called intercalated discs (IDS). These discs are highly characteristic of the cardie muscle.

Question 26.
“Cardiac muscle is highly resistant to fatigue”. Justify.
Answer:
The cardiac muscle is highly resistant to fatigue because it has numerous acrosomes, many molecules of myoglobin (Oxygen storing pigment), and a copious supply of blood which facilitate continuous aerobic respiration. The muscles are immune to fatigue and work tirelessly from the embryonic state until death.

Question 27.
Distinguish between ‘nucleus’ and ‘ganglion’ with respect to the nervous system.
Answer:
A group of cell bodies in the Central Nervous System is called a ‘nucleus’, and in the Peripheral Nervous System, it is called a ‘ganglion’.

Question 28.
Distinguish between tracts and nerves with respect to the nervous system.
Answer:
Groups of axons (nerve fibers) in the central nervous system (CNS) are called tracts’ and in the peripheral nervous system (PNS) they are called ‘nerves’.

Question 29.
Name the glial cells that form the myelin sheath around the axons of the central nervous system and peripheral nervous system respectively.
Answer:
In the central nervous system, the glial cells are called ‘Oligodendrocytes’, in the peripheral nervous system, the glial cells are called Schwann cells.

Question 30.
Distinguish between white matter and Greymatter of ‘CNS’.
Answer:
Myelinated nerve fibers occur in the white matter of the CNS, and in most peripheral nerves, and Non-myelinated axons are commonly found in the grey matter of the CNS and autonomous nervous system.

Question 31.
What are microglia and what is their origin add a note on their function.
Answer:
Microglial cells are the Neurogila (supporting cells) of cells of CNS which are phagocytic cells, of mesodermal origin.

Question 32.
What are pseudounipolar neurons? Where do you find them?
Answer:
In a unipolar neuron, the soma or cyton is found in the dorsal root ganglion they are called pseudounipolar neurons. These are found in spinal nerves.

Short Answer Type Questions

Question 1.
Describe the four different levels of organization in metazoans.
Answer:
The levels of organisation in metazoans are as follows:
1. Cellular level of organisation: It is the lowest level of organisaiton among the metazoans and ‘ is exhibited by the sponges. Different types of cells are functionally isolated due to the absence of sensory and nerve cells. There is no uniformity of labor among the cells and they don’t form tissues.

2. Tissue level of organisation: This is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like the cnidarians. In these animals, the cells which perform the same function are arranged into tissues due to the presence of nerve cells and sensory cells.

3. Organ level of organisation: An aggregation of different kinds of tissues that are specialized for a particular function is called an organ. Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation. It is the first time appeared in the members of platyhelminths.

4. Organ-system level: It is the highest level of organisation among the animals and is exhibited by triploblastic animals such as the flatworms, nematodes, annelids, arthropods, molluscs, echinodermates, and chordates. High specilized sensory and nerve cells bring about a higher level of coordination and integration among the various organ systems to lead an efficient way of life.

Question 2.
In which group of bilaterians do you find solid bauplan? Why is it called so?
Answer:
The bilaterian’s body plan is the solid bauplan when only one plane that passes through the identical axis divides an organism into two identical parts, it is called bilateral symmetry. It is the principal type of symmetry in triploblastic animals.

Bilaterally symmetrical animals are more efficient than the other animals in seeking food locating mates and in avoiding or escaping from predators, because of cephalization. As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 3.
Mention the advantages of coelom over pseudocolor.
Answer:
Advantages of coelom over pseudocoelom:
1. Visceral organs of eucoelomates are muscular (because of their association with mesoderm) and so they can contract and relax freely independent of the muscular movements of the body wall in the coelomic space, e.g.: peristaltic movements of the alimentary canal.
2. Gametes are released into the coelom in some invertebrates (which do not have glnoducts) and in the female vertebrates.
3. Coelomic fluid receives excretory products and stores them temporarily before their elimination.
4. In the eucoelomates, the mesoderm comes into contact with the endoderm of the alimentary canal, and it causes ‘regional specialization of the gut, such as the development of gizzard, stomach, etc., This is referred to as ‘primary induction’. In the case of the pseudocoelomates, due to the absence of such a contract between the gut and the mesoderm, the wall of the gut does not show complex and highly specialized organs.

Question 4.
Describe the formation of schizocoelom and enterocoelom.
Answer:
Schizocoelom: During embryonic development, a specialized cell called 4d blastomere is formed. The cells formed from these cells divide and redivide and develop blocks of mesoderm in blastocoel. The blocks fuse and form the mesodermal band which later on splits to form the Schizocoelom. This type of coelom is present in Annelida, Arthropoda, and Mollusca.

Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are enterocoelomates. In these animals, mesodermal ouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.

Question 5.
Describe briefly the three types of intercellular junctions of epithelial tissues.
Answer:
The specialized ‘junctions’ provide both structural and functional links between the individual cells of an epithelium. They show different types of junctions so as to serve the specific needs of that tissue. They are
(A) Tight junctions: These junctions between epithelial cells prevent ‘leakages’ of body fluids. For example, they prevent leakages of water into the surrounding cells in our sweat glands. The plasma membranes of adjacent cells are tightly pressed against each other and are bound together by specific proteins.

(B) Desmosomes: Muscle cells are provided with “desmosomes” (anchoring junctions) which act as ‘rivets’ binding the cells together into strong sheets. Intermediate filaments made of protein ‘keratin’ anchor desmosomes in the cytoplasm.

(C) Gap junctions: They provide continuous cytoplasmic channels between adjacent cells. Various types of ions, sugar molecules amino acids, etc. can pass from a cell to an adjacent cell through ‘gap junctions. They occur in many types of tissues including the ‘cardiac muscles’ where they allow rapid conduction of impulses or depolarization.

Question 6.
Give an account of glandular epithelium.
Answer:
Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types.
Unicellular glands: Consisting of isolated glandular cells such as goblet cells of the gut.
Multicellular glands: Consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands.
Exocrine glands: These glands provided with ducts secrete mucus, saliva, earwax, oil, milk, digestive enzymes, etc.
Endocrine glands: These glands are ductless and their products are hormones, which are carried to the target organs by blood.
Based on the mode of secretion, Exocrine glands are further divided into three types.
1. Merocrine glands: Which release the secretory granules without the loss of other cellular material.
Ex: Pancreas
2. Apocrineglands: The apical part of the cell is pinched off along with the secretory product.
Ex: Mammary glands
3. Holocrine glands: The entire cell disintegrates to discharge the contents.
Ex: Sebaceous gland

Question 7.
Give a brief account of the cells of areolar tissue.
Answer:
Areolar tissue: It is one of the most widely distributed connective tissue in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells which secrete fibres. The inactive cells are called fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin vasoconstrictor). Vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, Phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. Tissue-fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.

Question 8.
Describe the three types of cartilage.
Answer:
Cartilage is a solid, but semi-rigid (flexible) connective tissue. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

Cartilage is a solid, but semi-rigid (flexible) connective tissue It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chocolates are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, larynx, trachea, and bronchi.

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the Intervertebral discs and pubic symphysis of the pelvis.

Question 9.
Explain the Haversian system.
Answer:
The compact bone consists of several structural units called Osteons or Haversian systems arranged around and parallel to the bone marrow cavities.

The Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein, and a lymphatic vessel. The Haversian canal is surrounded by concentric lamellae. Small fluid-filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with the Haversian canal.

Each lacuna encloses one osteocyte (an inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the perio¬steum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

Question 10.
Write short notes on lymph.
Answer:
Lymph: Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes. When compared to the tissue fluid, it contains very small amounts of nutrients and oxygen but has abundant CO2 and other metabolites.

The most important site of the formation of lymph is the interstitial space. As blood passes through the blood capillaries, some portion of blood that includes water, solutes, and proteins of low molecular weight passer: through the walls of capillaries, into the interstitial spaces due to hydrostatic pressure at the arteriolar ends. This fluid forms the interstitial fluid (tissue fluid). Most of the interstitial fluid is returned directly to the capillaries due to osmotic pressure at the venular ends.

A little amount of this tissue fluid passes through a system of lymphatic capillaries (lymph capillaries of the intestinal villi are called ‘lacteals’), vessels, and ducts and finally, reaches the blood through the subclavian veins. The extracellular ’tissue fluid’ that passes into the lymph capillaries and lymph vessels is called ‘lymph’. The lymphatic system represents an ‘accessory route’ by which interstitial fluid flows from tissue spaces into blood.

Question 11.
Describe the structure of a skeletal muscle.
Answer:
Skeletal (striped and voluntary) muscle:
It is usually attached to skeletal structures by ‘tendons’. In a typical muscle such as the ‘biceps’ muscle, skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle. It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ which is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet-like aponeurosis.

A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the ‘peripheral’ cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.

Question 12.
Describe the structure of a cardiac muscle.
Answer:
Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or ‘gap junctions. They permit ‘electrical impulses to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.

Question 13.
Give an account of the supporting cells of nervous tissue.
Answer:
Nervous tissue is of two types.
1. Nervous
2. Supporting cell or neuroglia

Neuroglia (supporting cells): These are the supporting and non-conducting cells that provide a microenvironment suitable for neuronal activity. Unlike neurons, they continue to divide throughout life. Neuroglial cells of the CNS include oligodendrocytes (that form myelin sheath as mentioned above); astrocytes (star-shaped cells) that form an interconnected network and bind neurons and capillaries (helping in providing the blood-brain barrier); ependymal cells, which are ciliated cells that line the cavities of the brain and spinal cord to bring movements in the cerebrospinal fluid; microglial cells, which are phagocytic cells, of mesodermal origin. Neuroglial cells of the peripheral nervous system include the satellite cells and Schwann cells. Satellite cells surround the cell bodies in ganglia, and Schwann cells form neurolemma around axons.

Question 14.
Describe the structure of a multipolar neuron.
Answer:
Multipolar neurons have one axon and two or more dendrites. Most neurons in our body are multipolar neurons.
A neuron usually consists of a cell body with one too many dendrites and a single axon.

Neurons: Neurons are the ‘functional units of nervous tissue. These are electrically excitable cells that receive, initiate, and conduct/transmit impulses. When a neuron is stimulated, an electric disturbance (action potential) is generated which swiftly travels along its plasma membrane. A neuron usually consists of a “cell body” with one to many dendrites and a single axon.

Cell body: It is also called perikaryon, cyton, or soma. It contains abundant granular cytoplasm and a large spherical nucleus. The cytoplasm has Nissl bodies (they represent RER, the sites of protein synthesis), neurofibrils, and lipofuscin granules.

Dendrites: Several short, branched processes that arise from the cyton are called dendrites. They also contain Nissl bodies and neurofibrils. They conduct nerve impulses towards the cell body (afferent processes).

Axon: An axon is a single, long, cylindrical process that originates from a region of the cyton called the axon hillock. The Plasmalemma of an axon is called axolemma, and the cytoplasm is called axoplasm, which contains neurofibrils. However, Nissl bodies are absent. An axon may give rise to collateral branches. Distally it branches into many fine filaments called telodendrites, (axon terminals), which end in bulb-like structures called synaptic knobs or terminal boutons. Synaptic knobs possess ‘synaptic vesicles’ containing chemicals called neurotransmitters.

Question 15.
Write short notes on (a) Platelets and (a) Synapse.
Answer:
(a) Blood platelets (Thrombocytes): These are colourless non- nucleated, round or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.

(b) Synapse: It is the place in between the two neurons or inter-neuronal or neuromuscular junctions. Nerve cells do not have direct contact with each other. There is a microscopic gap of about 200-300 A° present between the nerve cells called a synapse. The nerve cell present before synapse is called the presynaptic neuron and the behind is called the post-synaptic neuron. A neurotransmitter substance called acetylcholine is secreted in the synapse by presynaptic neurons’ telodendrites. Acetylcholine helps in the conduction of nerve impulses from one neuron to another neuron.

Long Answer Type Questions

Question 1.
What is coelom? Explain the different types of coelom with suitable examples and neat labelled diagrams.
Answer:
The body cavity, which is lined by mesoderm, is called coelom more elaborately, coelom is a fluid-filled space between the body wall and visceral organs and lined by mesodermal epithelium peritoneum. Based on the body cavity triploblastic animals can be classified into acoelomentes, pseudo-coelo mates, and coelomates.

Acoelomate bilaterians: The bilaterian animals in which the body cavity is absent are called acoelomates, e.g. Platy-helminthes (lowest bilaterians). In these animals, the mesenchyme derived from the thrid germinal layer, called mesoderm, occupies the entire blasto coel, between the ectoderm and the endoderm, so that the adults have neither the primary cavity (blasto-coelom) nor the secondary cavity (coelom). As there is no body cavity, the acoelomates exhibit a solid body plan.

Pseudocoelomate bilaterians: In some animals, the body cavity is not lined by mesodermal epithelia. Such animals are called Pseudocoelomates. They include the members of phylum Aschelminthes (Nematoda, Rotifera, and some minor phyla). During embryonic development mesoderm (mesenchyme) occupies only a part of the blastocoel adjoining the ectoderm. The unoccupied portion of the blastocoel persists as pseudocoelom.

Eucoelomate bilaterians: Coelom or ‘true coelom’ is a fluid-filled cavity, that lies between the body wall and the visceral organs and is lined by mesodermal epithelium, the peritoneum. The portion of the peritoneum that underlines the body wall is the parietal peritoneum or somatic peritoneum. The portion of the peritoneum that covers the visceral organs is the splanchnic peritoneum or visceral peritoneum. In coelomates, the visceral organs are suspended in the coelom by the peritoneum.

During the embryonic development of the eucoelomates, the blastocoel is replaced by the true coelom derived from the mesoderm. So, the true coelom is also called the ‘secondary body cavity. Based on the mode of formation of coelom, the eucoelomates are classified into two types:

I. Schizocoelomates: Animals in which the body cavity is formed by the ‘splitting of mesoderm’ are called schizocoelomates. Annelids, arthropods, and mollusks are schizocoelomates in the animal kingdom. All the schizocoelomates are protostomians and they show ‘holoblastic’, ‘spiral’, and ‘determinate’ cleavage. The 4d blastomere or mesentoblast cell of the early embryo divides to form mesodermal blocks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal block leads to the formation of Schizocoelom.

II. Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are the enterocoelomates. In these animals, mesodermal pouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.

Question 2.
What is symmetry? Describe the different types of symmetry in the animal kingdom with suitable examples.
Answer:
Symmetry: The concept of symmetry is fundamental in understanding the organisation of an animal. Symmetry in animals is the balanced distribution of paired body parts. The body plan of a vast majority of metazoans exhibits some kind of symmetry. However, most of the sponges and snails show asymmetry (lack of symmetry). The symmetry of an animal and its mode of life are correlated.

Asymmetry: The animals, which cannot be cut into two equal parts (antimeres) in any plane passing through the centre of the body is called asymmetrical, e.g.: most sponges and adult gastropods. In asymmetrical animals, the body lacks a definite form.

Symmetry: The regular arrangement of body parts in a geometrical design relative to the axis of the body is called symmetry. In a symmetrical animal, paired body parts are arranged on either side of the plane passing through the principal axis, such that they are equidistant from the plane. The unpaired body parts are located mostly on the plane, passing through the principal axis.
Basically, the symmetry in animals is of two kinds:
(i) Radial symmetry
(ii) Bilateral symmetry

(i) Radial Symmetry or Monaxial heteropolar Symmetry: When any plane passing through the central axis (oro-aboral axis/principal axis) of the body divides an organism into two identical parts, it is called radial symmetry. The animals with radial symmetry are either sessile or planktonic or sluggish forms. It is the principal symmetry of the diploblastic animals such as the cnidarians and ctenophores symmetrical (as it is five-angled, it is also called pentamerous radial symmetry). Radially symmetrical animals have many planes of symmetry, whereas pentamerous radially symmetrical animals have five planes of symmetry.

(ii) Bilateral symmetry: When only one plane (median sagittal plane) that passes through the central axis (anterior-posterior axis) divides an organism into two identical parts, it is called bilateral symmetry. It is the ‘principal type of symmetry’ in triploblastic animals. Among the triploblastic animals, some gastropods become secondarily asymmetrical though they have primarily bilaterally symmetrical larvae.

Bilaterally symmetrical animals are more efficient than other animals in seeking food, locating mates, and in avoiding or escaping from predators, because of cephalization (concentration of nerve and sensory cells at the anterior end). As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 3.
Classify and describe the epithelial tissues on the basis of structural modification of cells with examples.
Answer:
Epithelium (epi-upon; thelia – growing) forms the outer covering of the body and the living of internal organs and cavities. There are two types of epithelial tissues namely ‘simple epithelia1 and ‘compound epithelia’ based on the number of layers or strata.

Simple epithelium is composed of a single layer of cells and forms the lining of body cavities, ducts, and vessels. It helps in the diffusion, absorption, filtration, and secretion of substances. On the basis of the shape of the cells, it is further divided into three types:
(i) Simple squamous epithelium (Pavement epithelium): It is composed of a single layer of flat and tile-like cells, each with a centrally located ‘ovoid nucleus’. It is found in the endothelium of blood vessels, mesothelium of body cavities (pleura, peritoneum, and pericardium), wall of Bowman’s capsule of the nephron, lining of alveoli of lungs, etc.

(ii) Simple cuboidal epithelium: It is composed of a single layer of cube-like cells with centrally located spherical nuclei. It is found in germinal epi – thelium, proximal, and distal convoluted tubules of the nephron. Cuboidal epithelium of proximal convoluted tubule of nephron has ‘microvilli’.

(iii) Simple columnar epithelium: It is composed of a single layer of tall and slender cells with oval nuclei located near the base. It has mucus-secreting ‘goblet cells’ in some places. It is of two types:
(a) Ciliated columnar epithelium: Columnar epithelial cells have cilia on their free surface. It is mainly present in the inner surface of hollow organs like fallopian tubes, ventricles of the brain, the central canal of the spinal cord, bronchioles, etc.

(b) Non-ciliated columnar epithelium: Columnar cells are without cilia. It is found in the lining of the stomach and intestine. Microvilli are present in the columnar epithelium of the intestine to increase the surface area of absorption.

(II) Compound epithelium (stratified epithelium): It is made up of more than one layer of cells. Its main function is to provide protection against chemical and mechanical stress. It covers the dry surface of the skin as stratified, keratinized, squamous epithelium. It covers the moist surface of the buccal cavity, pharynx, esophagus, and vagina as stratified non-keratinized squamous epithelium. It forms the inner lining of the larger ducts of salivary glands, sweat glands, and pancreatic ducts as stratified cuboidal epithelium. It forms the wall of the urinary bladder as transitional epithelium.

(III) Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types – unicellular glands consisting of isolated glandular cells such as goblet cells of the gut, and multicellular glands, consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands. Exocrine glands are provided with ducts; that secrete mucus, saliva, earwax (cerumen), oil, milk, digestive enzymes, and other cell products. In contrast, endocrine glands are ductless and their products are ‘hormones’, which are not sent out via ducts, but are carried to the target organs by blood.

Based on the mode of secretion, exocrine glands are further divided into
(i) merocrine glands (e.g.: pancreas) which release the secretory granules without the loss of other cellular material.

(ii) Apocrine glands (e.g.: mammary glands) in which the apical part of the cell is pinched off along with the secretory product and
(iii) Holocrine glands (e.g.: sebaceous glands), in which the entire cell disintegrates to discharge the contents.

Question 4.
Describe the various types of connective tissue properly with suitable examples.
Answer:
Connective tissue proper is of two types as
(A) Loose connective tissue proper
(B) Dense connective tissue proper

(A) Loose connective tissue: Cells and fibres are loosely arranged in a semi-fluid ground substance there are three types of loose connective tissues-areolar tissue adipose tissue and reticular tissue.
(i) Areolar tissue: It is one of the most widely distributed connective tissues in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are the fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells with secreted fibres. The inactive cells are called
fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin (vasoconstrictor), vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. ‘Tissue fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.

(ii) Adipose tissue: It is specialized for fat storage. It consists of a large number of adipocytes and few fibres. The adipose tissue which is found beneath the skin provides. Adipose tissue is of two types: white adipose tissue, and brown adipose tissue. Excess nutrients which are not used immediately are converted into fats and stored in this tissue.
White adipose tissue (WAT): It is the predominant type in adults, and the adipocyte has a single large lipid droplet (monocular). White fat is metabolically not active.
Brown adipose tissue (BAT): It is found in fetuses and infants. Adipocyte of BAT has several small ‘lipid droplets’ (multilocular) and numerous mitochondria. Brown fat is metabolically active and generates ‘heat’ to maintain the body temperature required by infants.

(iii) Reticular tissue: It has specialized fibroblasts called reticular cells. They secrete ‘reticular fibers’ that form an interconnecting network. It forms the ‘supporting framework’ of lymphoid organs such as bone marrow, spleen, and lymph nodes and forms the reticular lamina of the ‘basement membrane’.

(B) Dense connective tissue: This tissue consists of more fibres, but fewer cells. It has very little ground substance. Based on the arrangement of fibres, the dense connective tissue is of three types.
(i) Dense regular connective tissue: In this tissue, collagen fibres are arranged parallel to one another in bundles. Tendons that attach the skeletal muscles to bones and ligaments which attach bones to other bones are examples of this type of connective tissue.

(ii) Dense irregular connective tissue: In this type of connective tissue, bundles of collagen fibres are irregularly arranged. Periosteum, endosteum, pericardium, heart valves, joint capsule, and deeper region of the dermis of skin contain/are made up of this type of connective tissue.

(iii) Elastic connective tissue: It is mainly made of yellow elastic fibres, capable of considerable extension and recoil. This tissue can recoil to its original shape, when the forces of stretch are released. It occurs in the wall of arteries, vocal cords, trachea, bronchi, and ‘elastic ligaments’ present between vertebrae.

In addition to the above-mentioned connective tissues, mucous connective tissue occurs as foetal or embryonic connective tissue. It is present in the umbilical cord as Wharton’s jelly.

Question 5.
What is skeletal tissue? Describe the various types of skeletal tissue.
Answer:
Skeletal tissue (supporting tissue): It forms the endoskeleton of the vertebrates. It supports the body, protects various organs, provides a surface for the attachment of muscles, and helps in locomotion. It is of two types:
(A) Cartilage (Gristle): Cartilage is a solid, but semi-rigid (flexible) connective tissue. It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chondrocytes are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium.

Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.
1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, Larynx, trachea, and bronchi.

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

(B) Bone (osseous) tissue: Bone is highly calcified (mineralized), solid, hard and rigid connective tissue. It is the major component of the endoskeleton of most adult vertebrates.

Bone has an outer fibrous connective tissue sheath called periosteum, the inner connective tissue sheath that lines the marrow cavity called the endosteum, a non-living extracellular matrix, living cells, and bone marrow. Bone cells include osteoblasts, osteocytes, and osteoclasts. Osteoblasts (immature bone cells) secrete the organic components (collagen fibres) of the matrix and also play an important role in the ‘mineralization of bone’ and become osteocytes (mature bone cells). Osteocytes are enclosed in fluid-filled lacunae. Osteoclasts are phagocytic cells involved in the resorption of bone.

Types of bones based on the method of formation:
(i) Cartilage bones (replacing bones or endochondral bones) are formed by ossification within the cartilage e.g. bones of limbs, girdles, and vertebrae.
(ii) Investing bones (membrane bones or dermal bones) are formed by the ossification in the embryonic mesenchyme e.g. most of the bones of the cranium.
(iii) Sesamoid bones are formed by ossification in tendons e.g.: patella (knee cap) and pisiform bone of the wrist of a mammal
(iv) are formed by ossification in the soft tissues, e.g.: Oscordis (inside the heart of ruminants), and Ospenis (inside the glans-penis of many mammals such as rodents, bats, and carnivores).

Types of bones based on the structure:
1. Spongy bone (Cancellous bone or trabecular bone): It occurs in the epiphyses and metaphyses
of long bones. It looks spongy and contains columns of bone called ‘trabeculae’ with irregular interspaces filled with red bone marrow.
2. Compact bone: The diaphysis of a long bone is made up of ‘compact bone. It has a dense continuous lamellar matrix between the periosteum and endosteum.

Structure of a compact bone: Diaphysis (shaft) is a part of a long bone that lies in between expanded ends. The diaphysis is covered by a dense connective fibrous tissue called the periosteum. The diaphysis of a long bone has a hollow cavity called marrow cavity which is lined or surrounded by the endosteum. In between periosteum and endosteum, the matrix of the bone is laid down in the form of ‘lamellae’. Outer circumferential lamellae are located immediately beneath the periosteum; inner circumferential lamellae are located around the endosteum.

Between the outer and inner circumferential lamellae, there are many Haversian systems (osteons – units of bone). The spaces between the Haversian systems are filled with interstitial lamellae. Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae. Small fluid filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae.

Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmanns canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

Question 6.
Give an account of the “formed elements” of Blood.
Answer:
Blood is a red-coloured, opaque, and slightly alkaline fluid. It is composed of blood plasma and formed elements or blood cells – the RBC, WBC, and platelets.
Formed elements of Blood cells: The blood corpuscles (RBC and WBC) constitute 45% of the total blood by volume.

(i) Red blood corpuscles (Erythrocytes): Erythrocytes of mammals are circular (elliptical in camels and Liamas), biconcave and enucleate. The biconcave shape provides a large surface area-to-volume ratio, thus providing more area for the exchange of gases. These are 7.8 pm in diameter. The number of RBCs per cubic millimeter of blood is about 5 million in a man and 4.5 million in a woman.

(ii) White blood corpuscles (Leucocytes): These are nucleate, colourless, complete cells. They are spherical or irregular in shape and are capable of exhibiting amoeboid movement into the extravascular areas by diapedesis. WBC are two main types: 1) Granulocytes and 2) Agranulocytes.

Granulocytes: They possess cytoplasmic granules that may take three different types of stains, neutral or acidic, or basic. The nucleus of the granulocytes is divided into lobes and assumes different shapes, hence, these are also called polymorphonuclear leucocytes. Based on the staining properties these are of three tvoes.

Basophils: They constitute about 0.4% of the total leucocytes. The nucleus is divided into irregular lobes. Cytoplasmic granules are ‘fewer’ and ‘irregular’ in shape. They take basic stains. They produce heparin, histamine, etc. They supplement the function of mast cells when needed.

Eosinophils (acidophils): They constitute about 2.3% of the total leucocytes. The nucleus is distinctly bilobed. The cytoplasm has large granules which stain with acidic dyes such as ‘eosin’.

Neutrophils: They constitute about 62% of the total leucocytes. The nucleus is many lobed (2-5). Specific cytoplasmic granules are small and abundant. They stain with ‘neutral dyes’.

Agranulocytes: Cytoplasmic granules are absent in agranulocytes. divided into lobes. These are of two types:
(a) Lymphocytes: They constitute about 30% of the total leucocytes. They are small, spherical cells with large spherical nuclei and scanty peripheral cytoplasm. There are functionally two types of lymphocytes – ‘B’ lymphocytes, which produce ‘antibodies’ and T lymphocytes which play a key role in the immunological reactions of the body. Some lymphocytes live only a few days while others survive for many years.

(b) Monocytes: They constitute about 5.3% of the leucocytes. The nucleus is kidney-shaped (reniform). These are the largest, motile phagocytes. They engulf bacteria and cellular debris. They differentiate into macrophages when they enter the connective tissues.

(iii) Blood platelets (Thrombocytes): These are colourless non-nucleated, round, or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.

Question 7.
Compare and contrast the three types of muscular tissues.
Answer:
Muscular tissue is mesodermal in origin. Muscles show three essential properties such as excitability, conductivity, and contractility. The study of muscular tissues is known as mycology. Muscular tissue has elongated cells called ‘muscle fibers’ (myocytes) which are surrounded by a connective tissue sheath. The extracellular matrix is absent. The plasma membrane of a muscle fiber is called sarcolemma.

The cytoplasm of a muscle fibre is called Sarco plasm, the endoplasmic reticulum, the sarcoplasmic reticulum, and the mitochondria, the Sarcosomes. The cytoplasm of a muscle fibre has several myofibrils. Each myofibril has thick (myosin) and thin (actin) myofilaments. The regular arrangement of myosin and actin filaments is responsible for the alternate dark and light bands of a ‘striated muscle’. Sarcoplasm also contains ATP, phosphocreatine, glycogen, and myoglobin. Muscles are of three types – skeletal, smooth, and cardiac.

Skeletal (striped and voluntary) muscle: It is usually attached to skeletal structures by ‘tendons’. A typical muscle such as the ‘biceps’ muscle/skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle.

It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ that is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet¬like aponeurosis.

A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the “peripheral” cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.

Smooth (unstriped and involuntary muscle): It is located in the walls of the visceral organs such as blood vessels, trachea, bronchi, stomach, intestine, excretory and genital ducts, and so this is also called ‘visceral muscle’. As cross striations are absent, it is called ‘smooth muscle. It is also found in the iris and ciliary body of the eye and in the skin as ‘arrector pili muscles that are attached to hair follicles.

Usually, smooth muscles are arranged in ‘layers’/’sheets’. A smooth muscle fibre is a spindle-shaped (fusiform), uninucleate cell. Myofibrils do not show alternate dark and light bands due to the irregular arrangement of actin and myosin fibres. They do not work under conscious control, and so they are called involuntary muscles. Smooth muscles exhibit ‘slow’ and ‘prolonged’ contractions. They may remain contracted for long periods without fatigue (show sustained involuntary contractions called ‘spasms’). The contraction of smooth muscles is under the control of the autonomous nervous system.

Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the ‘myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or gap junctions. They permit ‘electrical impulses’ to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 2nd Lesson Biological Classification Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 2nd Lesson Biological Classification

Very Short Answer Questions

Question 1.
What is the nature of cell-walls in diatoms?
Answer:
In Diatoms, the cell wall is made up of silica and thus the walls are Indestructible.

Question 2.
How are Viroids different from viruses? [Mar. 14]
Answer:

Question 3.
What do the terms phycobiont and mycobiont signify? [A.P. Mar, 17, Mar. 13]
Answer:
The Algal component in a lichen is called phycobiont. The fungal component in a lichen is called Mycobiont.

Question 4.
What do the terms ‘algal bloom’ and ‘Red Tides’ signify?
Answer:
In cyanobacteria :
The colonies and trichomes or filaments are surrounded by gelatinous sheath which often forms blooms in polluted water bodies called algal blooms.
Ex : Nostoc, Anabaena.

Red Dinoflagellates like “Gonyaulax” undergo rapid multiplication and make the sea “Red” so called Red Tides.

Question 5.
State two economically important uses of heterotrophic bacteria.
Answer:
Heterotrophic Bacteria help in making curd from milk, production of Antibiotics and Nitrogen fixation in legume roots.

Question 6.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement? [A.P. Mar. 15]
Answer:
Cyanobacteria are used in Agricultural fields for crop improvement because they help in Nitrogen fixation and make the soil fertile and also they show oxygenic photosynthesis.

Question 7.
Plants are autotrophic. Name some plants which are partially heterotrophic.
Answer:
Partial heterotrophic plants are viscum, loranthus, striga.

Question 8.
Who proposed five Kingdom classification? How many Kingdoms of this classification contain eukaryotes?
Answer:
R.H.Whittaker proposed five kingdom classification. In this classification, four kingdoms [Protista, Fungi, Plantae, Animalia are Eukaryotes.

Question 9.
Give the main criteria used for classification by Whittaker. [T.S. May. 18, Mar, 15]
Answer:
The main cricteria for five kingdom classification include cell structure, thallus organization. Mode of Nutrition, reproduction and phylogenetic relationships.

Question 10.
Name two diseases caused by Mycoplasmas. [A.P. May. 18]
Answer:
Mycoplasmas cause witches broom disease in plants, pleuropneumonia in cattle and v mycoplasmal urethritis in Humans.

Question 11.
What are slime moulds? Explain what is meant by plasmodium with reference to slime moulds.
Answer:
Slime moulds are saprophytic protists. The multinucleated mass of protoplasm is surrounded by a plasamamembrane under favourable conditions, they form an aggregation called plasmodium.

Short Answer Type Questions

Question 1.
What are the characteristic features of Euglenoids?
Answer:

1. Most of the euglenoids are Fresh Water organisms found in stagnant water.
2. They have a protein rich layer called pellicle which makes body flexible.
3. They have Two flagella, a short and a long one.
4. The anterior part of the cell bears an invagination consisting of cytostome, cytopharynx and reservoir.
5. Eye spot or photosensitive stigma is present in the reservoir.
6. They behave as heterotrophs when deprived of sunlight.
7. The reproduction is by lohgitudinal binary fission.

Question 2.
What are the advantages and disadvantages of Two kingdom classification?
Answer:
Advantages :

1. The two kingdom classification received considerable recognition from biologists and was in use for quite a long period of time.
2. Classification of organisms into plants, animals was easily done and was easy to understand.
3. The character that unified this whole kingdom was that all organisms included had a cell wall in their cells.

Disavantages :
1. There is no clear Cut distinction in the lower forms of life into plants and animals. Euglena for example is a unicellular organism having certain features of criminals and certain features of plants. If has a flagellum which is used for locomotion and food capturing. However it has chloroplasts like plant cells.

2. Chlomyclomonas is a unicellular algue. It is purely antotrophic but has 4 locomotor structures called flagella like protozoans.

3. The classification did not differentiate between the heterotrophic group, fungi and the autotrophic green plants through they also showed a characteristic difference in their wall composition. Hence two kingdom classification used for a lone time was found inadequate.

Question 3.
Give the salient features and importance of chrysophytes. [A.P. Mar, 15, 13]
Answer:
Chrysophytes includes diatoms and golden algae (desmids). They are found in freshwater as well as in marine water.’Most of them are photosynthetic. In diatoms the cell wall form two thin overlapping shells, epitheca over hypotheca which fit together as soap box. The walls are embedded with silica and thus the walls are indestructible. Diatoms leave larger amount of cell wall deposits in their habitat, this accumulation is referred as diatomaceous earth or kieselghur. They reproduce a sexually by Binary fission and sexually by gametes.

Importance :

1. The diatomaceous soil is used in polishing, filtration of oils 8 syrups.
2. Diatoms are the chief “producers” in the oceans.

Question 4.
Give a brief account of Dinoflagellates. [A.P. May, 18, Mar, 18] [T.S. Mar, 15]
Answer:

1. Dinoflagellates are mostly marine and photosynthetic.
2. The cell wall has stiff cellulose plates on the outer surface.
3. They have two flagella and produce spinning movements. So these protists are called “whirling whips”.
4. The nucleus has condensed chromosomes, which are without histones.
5. Some Dinoflagellates like Noctiluca show bioluminescence.
6. Red dinoflagellates like Gonyaulax undergo rapid multiplication and make the sea appear red
   (Redtides in Medeterranelian sea)
7. Toxins produced by them may kill fishes.

Question 5.
Write the role of Fungi in our daily life. [T.S. May, 18. Mar, 14]
Answer:
The role of Fungi in our dialy life show some uses and disuses. They are :
Uses :

1. Yeast are used to make bread and beer.
2. Some Fungi like penicillium are the source of antibiotics. (Penicilin)
3. Some Fungi like Agaricus are edible mushrooms.

Disuses :

1. Some Fungi cause rot of orange fruits
2. Spoilage of bread
3. White spots are seen in musterd leaves by Albugo.
4. Some Fungi (Puccinia) cause rust in wheat.

Long Answer Type Questions

Question 1.
Give the salient features and comparative account of different classes of fungi studied by you.
Answer:
A) Phycomycetes :

1. They are found in Aquatic habitats and on decaying wood or as obligate parasites on plants. They arealso called algal fungi.
2. The mycelium is aseptate and coenocytic.
3. A sexual reproduction takes place by zoospores or by aplanospores.
4. Zygospores are formed by the fusion of two gametes which may similar (Isogamous) or dissimilar (Anisogamous) or oogamous.
   Ex : Mucor, Rhizopus, Albugo.

B) Ascomycetes :

1. They are commonly called “sac fungi”
2. They are unicelluar (yeast) or multicellular (Penicillium).
3. They are saprophytic, decomposers, parasitic or coprophilous (growing on dung).
4. The mycelium is Branched and septate.
5. The reproduce asexually by conidia, produced on conidiophores.
6. Sexual spores are called ascospores Ex : Aspergillus, pencillium, ckiviceps.

C) Basidiomycetes :

1. They are commonly known as Mushrooms, bracket fungi or puffballs or club fungi.
2. They grow in soil, on logs and tree stumps and in living plant bodies as parasites.
   Ex : Rusts.
3. The mycelium is branched and septate.
4. The asexual spores are not found but common method is fragmentation.
5. Sex organs are absent.
6. Plasmogamy takes place by the fusion of two vegetative cells of different strains or genotypes.
7. Karyogamy and Meiosis take place in the basidium producing basidiospores.
   Ex : Agaricus, ustilago, polyporus Lycoperdon.

D) Deuteromycetes :

1. They are commonly known as imperfect fungi.
2. Some members are saprophytes or parasites, while a large members are decomposers.
3. The mycelium is branched and septate.
4. They reproduce asexually and vegetatively.
5. When the sexual stages were discovered, they were moved into another classes, asco or basidiomycetes.
   Ex : Alternaria, Colletotrichum, Trichoderma.

Question 3.
Describe briefly different groups of Monerans you have studied.
Answer:
They indued Alio prokaryotes like Archaebacteria, Eubacteria, Mycoplasma and Actinomycetes.

I. Archaebacteria :

1. They live in salty areas, hot springs or in Marshy places.
2. The cell wall contains pseudomurein.
3. The cell membrane contains branched chain lipids.
4. Methanogens are present in the gut of ruminant animals, and are responsible for the production of methane gas from the dung.

II. Eubacteria :

1. They occur almost everywhere, such as hotsprings, deserts, snow and deep oceans as parasites and as symbionts.
2. They are classified-into spherical coccus, rod shaped bacillus, the comma shaped vibrio and the spiral shaped spirillum, based on their shape.
3. The cell wall is made up of peptidoglycan.
4. The cell membrane shows infoldings called mesosomes.
5. They contain 70s type of ribosomes along with naked genetic material (nucleoid).
6. Some of eubacteria are autotrophic and the vast majority are heterotrophs.
7. Cyanobacteria like Nostoc and Anabaena are Unicellular, colonial or filamentous, aquatic or terrestrial algae.
8. They often forms blooms in polluted water bodies. Some of them can fix Atmospheric Nitrogen in specialised cells called Heterocysts.

III. Mycoplasma :

1. They completely lack a cell wall and are pleomorphic.
2. They are smallest living cells and can survive without oxygen.
3. They cause diseases in plants [witches broom]; in cattle (pleuropneumonia) and in humans (Mycoplasmal urethritis).

IV. Actinomycetes :

1. They are branched, filamentous bacteria.
2. The cell wall contains mycolic acid.
3. Most of them are saprophytic and decomposers.
4. Some are (Mycobacterium and corynebacterium) parasites.
5. A number of antibiotics are produced from the genus streptomyces.

Question 3.
Enumerate the salient features of different groups in protista.
Answer:
I. Chrysophytes :

1. Chrysophytes includes diatoms and golden algae (desmids).
2. They are found in freshwater as well as in marine water.
3. Most of them are photosynthetic.
4. In diatoms the cell walls form two thin overlapping shells, epitheca over hypotheca which fit together as soap box.
5. The walls are embedded with silica and thus the walls are indestructible.
6. Diatoms leave larger amount of cell wall deposits in their habitat, this accumulation is referred as diatomaceous earth or kieselghur.
7. They reproduce asexually by Binary fission and sexually by gametes. ‘

II. Dinoflagellates :

1. Dinoflagellates are mostly marine and photosynthetic.
2. The cell was has stiff cellulose plates on the outer surface.
3. They have two flagella and produce spinning movements. So these protists are called “whirling whips”.
4. The nucleus has condensed chromosomes, which are without histones.
5. Some Dinoflagellates like Noctiluca show bioluminescence.
6. Red dinoflagellates like Gonyaulax undergo rapid multiplication and make the sea appear red (Redtides in Medterranian sea)
7. Toxins produced by them may kill fishes.

III. Euglenoids :

1. Most of the euglenoids are Fresh Water organisms found in stagment water.
2. They have a protein rich layer called pellicle which makes body flexible.
3. They have Two flagella, a short and a long one.
4. The anterior part of the cell bears an invagination, consisting of cytostome, cytopharynx and reservoir.
5. Eye spot or photosensitive stigma is present in the reservoir.
6. They behave as heterotrophs when deprived of sunlight.
7. The reproduction is by longitudinal binary fission.

IV. Slime moulds :

1. They are saprophytic protists. The multinucleated mass of protoplasm is surrounded by a plasamamembrane.
2. Under suitable conditions, they form an aggregation called plasmodium.
3. During unfavourable conditions, the plasmodium differentiates and forms fruiting bodies bearing spores at their tips.
4. The spores are extremely resistant and are dispersed by air.

V. Protozoans :

1. They live as predators or parasites.
2. They do not contain cell wall.
3. The protoplason is surrounded by plasma membrane.
4. Among them. Amoeboid protozoans live in fersh water, sea water or moist soil. They move and capture their prey by pseudopodia.
   Ex : Entamoeba.
5. Flagellated protozoans are either free living or parasitic.
6. They have flagella.
7. The parasitic forms cause diaseases such as sleeping sickness.
   Ex : Trypanosoma.
8. Ciliated Protozoans : They are aquatic and actively moving organisms by having cilia.
9. They have a cavity that opens to outside of the cell surface.
   Ex: Paramoecium.

IV. Sporozoans :
They have an Infectious spore like stage in their life cycle.
Ex : Plasmodium (Malarial parasite) causes Malaria in humans.

Intext Questions

Question 1.
State two economically important uses of :
a) Heterotrophic bacteria b) Archaebacteria
Answer:
a) Heterotrophic bacteria :
They help in making curd from milk, production of Antibiotics and \ Nitrogen fixation in legume roots.

b) Archaebacteria :
1) Produce methane 2) Used by humans in biotechnoogy.

Question 2.
Give a comparative account of the classes of Kingdom Fungi on the basis of the following :
i) mode of nutrition
ii) mode of reproduction.
Answer:
i) mode of nutrition :
a) Phycomycetes : Obligate parasites.
b) Ascomycetes : Saprophytic, decomposers.
c) Basidiomycetes : Parasites.

d) Deuteromycetes :
Saprophytes, decomposers, ii) mode of reproduction :
a) Phycomycetes : Asexually by zoospores or aplanospores, sexually by gametes.
b) Ascomycetes : Asexually by conidia, sexually by ascospores.
c) Basidiomycetes : Asexually by fragmentation, sexually by fusion of two somatic cells.
d) Deuteromycetes : Asexually by conidia.

Question 3.
Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases.
Answer:
Viruses are made up of Nucleic acid and protein. The Nucleic acid could be either RNA or DNA. The protein part is called capsid which encloses the Nucleic acid. Tobacco Mosaic virus and Human Immuno Virus (HIV) contains RNA as genetic material. Bacteriophages, contain DNA as genetic material.

Common viral diseases are –

1. Tobacco Mosaic virus.
2. Potato spindle Tuber disease.
3. Human Immuno virus.
4. Scrapie disease of sheep.

Question 4.
Organise a discussion in your class on the topic – Are viruses living or non-living?
Answer:
Viruses are not truly ‘living’. They exist in crystal form outside the host. They are called obligate parasites.

Question 5.
Suppose you accidentally find an old preserved permanent slide without a label and in your effort to identify it, you place the slide under microscope and observe the following features :
a) unicellular body
b) well defined nucleus
c) biflagellate condition – one flagellum lying longitudinally and the other transversely.
What would you identify it as? Can you name the kingdom it belongs to?
Answer:
Protistian cell. They belongs to kingdom protista.

Question 6.
Polluted water bodies have usually high abundance of plants like Nostoc and Oscillatoria. Give reasons.
Answer:
Abundence of reserved food materials.
– Increase the level of dissolved O₂ in their environment.
– N₂ fixing capacity.

Question 7.
Cyanobacteria and heterotrophic bacteria have been clubbed together in Eubacteria of Kingdom Monera as per the five kingdom classification, even though the two are vastly different from each other. Is this grouping of the two types of taxa in the same kingdom justified? If so why?
Answer:
Grouping of the two types of tax a in the same kingdom is justified because both organisms contain ‘nif’ genes and are involved in Nitrogen fixation.

Question 8.
What observable features in Trypanosoma would make you classify it under kingdom protista?
Answer:

1. They are free living or parasitic.
2. They have flagella.

Question 9.
At a stage of their life cycle, ascomycetous fungi produce the fruiting bodies like cleistothecium perithecium or apothecium. How are these three types of fruiting bodies differ from each ohter?
Answer:
The globose ascocarp without opening is called cleistothecium.
The flask shaped ascocarp with an apical opening is called perithecium.
The cup or saucer shaped ascocarp is called apothecium.

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 1st Lesson The Living World Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 1st Lesson The Living World

Very Short Answer Questions

Question 1.
What does ICBN Stand for? [T.S. Mar, 18]
Answer:
ICBN stands for “International Code for Botanical Nomenclature”.

Question 2.
What is Flora? [T.S. Mar, 15]
Answer:
Actual account of habitat, distribution and systematic listing of plants of a given area is called Flora.

Question 3.
Define metabolism. What is the difference between anabolism and catabolism?
Answer:
Sum total of all the chemical reactions occuring in the body of an organism is called Metabolism.

Question 4.
Which is the largest botanical garden in the world? Name a few well known botanical gardens in India.
Answer:
Royal Botanical Garden (RBG) at kew (England) is the largest Botanical Garden in the world. Well Known Botanical gardens in India are

1. Indian Botanical gardens, Howrah.
2. National Botanical Research Institute, Lucknow.

Question 5.
Define the terms couplet and lead in taxonomic key. [A.P. Mar, 15]
Answer:
The contrasting characters generally in a pair is called “couplet”. Each statement in the key is called “Lead”.

Question 6.
What is meant by manuals and monographs?
Answer:
Manual is a small book specially designed for ready reference. Monographs contain Information on any one taxon.

Question 7.
What is systematics? [T.S. Mar, 18]
Answer:
“The study of different kinds of organisms, their diversities and also the relationship among them”.

Question 8.
Why are living organisms classified?
Answer:
Living organisms are classified to know their growth, reproduction, ability to sense environment and mount a suitable response, metabolism, ability to self replicate, interaction and emergence.

Question 9.
What is the basic unit of classification? Define it. [A.P. Mar, 17, 14, 13]
Answer:
“Species”.

“Species is defined as a group of Individual organisms with fundamental similarities”.

Question 10.
Give the scientific name of Mango. Identify the generic name and specific epithet. [T.S. Mar, 17]  [A.P. May, 18]
Answer:
Mangifera indica. Its Generic name is “Mangifera” and specific epithet is “indica”.

Question 11.
What is growth? What is the difference between the growth in living organisms and growth in non-living objects?
Answer:
Growth is a permanent and irreversible increase in the size of a living organism.

In Living organisms, growth is from Inside. Where as in Non-living objects like Mountains and sand mounds, growth occurs by accumulation on of material on the outer surface.

Short Answer Type Questions

Question 1.
What is meant by Identification and nomenclature? How is a key helpful in the identification & classification of an organism?
Answer:
Identification is defined as, “Whether a collected organism is entirely new or already known”. Nomenclature is defined as “providing a scientific name to an Identified organism”.

Key is a taxonomical aid used for Identification of plants and animals based on the similarities and dissimilarities. The keys are based on the contrasting characters generally in a pair called couplet. It represents the choice made between two opposite options. Each statement in the key is called lead, separate taxonomic keys are required for each taxonomic category such as family, genus and species for Identification purposes. Keys are generally analytical in nature.

Flora, Manuals, Monographs, and catelogues are also help in correct Identification of plants of a given area.

Question 2.
What are taxonomical aids? Give the importance of herbaria and museums.
Answer:
Taxonomical aids are the collection of actual specimens of plant and animal species which are useful in agriculture, forestry. Industry and in knowing bio-resources and their diversity.

Herbarium is a store house of collected plant specimens that are dried, pressed, and preserved on sheets. These sheets are used for future use and also carry a label providing information about data and place of collection, English, local and botanical names, family, collector’s name. Herbaria also serve as quick referal systems in taxonomical studies. Museums are generally set up in Educational Institutes. Museums have collections of preserved plant and animals specimens for study and reference.

Question 3.
Define a taxon. Give some examples of taxa at different hierarchial levels.
Answer:
Taxon is defined as, Any unit of category in a Taxonomic system. There are different taxonomic levels such as
a) Species – It is the lowest category in which a group of Individual organisms w|th fundamental similarities are placed.
Example : Mangifera indica (Mango), indica is the specific epithet.

b) Genus – A group of related species which has more characters in common in comparison to species of other genera.
Example : Potato and brinjal are two different species but belong to the genus solanum.

c) Family – It has a group of related genera with still less number of similarities as compared to genus and species.
Example : Solanum, Nicotiana and Datura are three different genera but placed in the family solanaceae.

d) Order – It is the assemblage of families which exhibit a few similar characters.
Example : Based on the floral characters.Solanaceae and Convolvulaceae are included in the order polemoniales.

Question 4.
How are botanical gardens useful in conserving biodiversity? Define the terms Flora, manuals, monographs and catalogues?
Answer:
Botanical gardens have collections of living plants for reference. Plant species in these gardens are grown for identification purposes and each plant is labelled indicating its botanical name and its family.

Flora :
It contains the actual account of habitat, distribution and systematic listing of plants of a given area.

Manual :
It is a small book containing the information for identification of names of species found in an area.

Monograph :
They contain information on any one taxon.

Catelogues :
Books which help in correct identification of plants.

Question 5.
Explain binomial nomenclature.
Answer:
Naming the plants with two words is called Binomial nomenclature. This system was given by Carolus Linnaeus. Naming is done by following some rules. They are

1. Biological names are generally in Latin and written in Italics.
2. The first word represents the Genus while the second word represents the species.
3. Both the words, when handwritten, are separately underlined.
4. The first word should starts with a capital letter while second word should starts with a small letter.
5. The name of the auther should be at the end of scientific name in abbreviated form.

Long Answer Type Questions

Question 1.
What is meant by living? Give a detailed account of any four defining features of life forms.
Answer:
The state of Material complex or individual characterized by the capacity to perform certain functional activities including metabolism, growth, and Reproduction.

1) Growth :
All living organisms increase in mass and increase in number. In plants, growth occurs by cell division, continuously throughout their life span. Unicellular organisms also grow in size until they divide by cell division.

2) Reproduction :
“Reproduction means progeny possessing feautures more or less similar to those of parents organisms, reproduce asexually and also by sexual methods. Fungi reproduce asexually by spores. Filamentous algae and Protonema of mosses reproduce by Fragmentation. In unicellular oganisms like Bacteria, unicellular algae or Amoeba reproduction is synonymous with growth. There are many organisms which do not reproduce. Hence reproduction also cannot be an all-inclusive defining characterestic of living organisms.

3) Metabolism :
The sum total of all the chemical reaction occuring in the body of a living organisms is called metabolism. All plants, Animals, Fungi and Microbes exhibit metabolism. No nonliving obejct exhibit metabolism.

4) Ability to sense their surroundings :
All living organisms, have the ability to sense their surroundings or environment and respond to Physical, Chemical or Biological stimuli. This response to environmental stimuli is called Irritability. Plants respond to external factors like light, water, temperature other organisms, pollutants etc. All organismsare aware of their surroundings and is called consciousness. Human beings has self consciousness except patients.

Question 2.
Define the following terms with examples.
i) Class ii) Family iii) Order iv) Genus v) Division
Answer:
i) Class :
A group of related orders constitutes class.
Ex : In plant kingdom, orders like Malvales, Rosales, Polemoniales are included in the class
Dicotyledonae.

ii) Family :
A group of related Genera’ is called family. They are characterised on the basis of both vegetative and reproductive features of plant species.
Ex : Solanum, Nicotiana and Datura are three different Genera but are placed in the family Solanaceae.

iii) Order :
Assemblage of families which exhibit a few similar characters is called order.
Ex : Families like convolvualeceae, Solanaceae are included in the order polemoniales based on the floral characters.

iv) Genus :
A groupo of related species which has more characters in common in comparison to species of other Genera is called Genus.
Ex : Potato (tuberosum) and Brinjal (melongena) are two different species but are placed in the same Genus, ‘Solanum’.

v) Divsion :
‘A group of related .classes’ is called Division.
Ex : Dicotyledonae and Monocotyl’edonae are two classes with a few similar characters are placed in Division spermatophyta.

Intext Questions

Question 1.
Some of the properties of tissues are not constituents of their cells. Give two examples to support the statement.
Answer:

1. Properties of tissues arise as a result of Interactions among the constituent cells.
2. Properties of cellular organelles arise as a result of Interactions among the molecular components comprising the organelle.

Question 2.
What do we learn from identification of individuals and populations?
Answer:
It is the prime source of taxonomic studies and also knows about or bioresources and their diversity.

Question 3.
Given below is the scientific name of Mango. Identify the correctly written name.
Mangifera Indica
Mangifera indica
Answer:
Mangifera indica.

Question 4.
Can you identify the correct sequence of taxonomical categories?
a) Species, Order, Division, Kingdom
b) Genus, Species, Order, Kingdom
c) Species, Genus, Order, Phylum
Answer:
C is correct.

Question 5.
Define the following terms :
i) Species ii) Class iii) Family iv) order v) Genus.
Answer:
i) Species :
A group of Individual organisms with fundamental similarities.

ii) Class :
An assemblage of related orders is called class.

iii) Family :
A group of related genera with less number of similarities as compared to genus and species.

iv) order :
An assemblage of families.

v) Genus :
A group of related species which has more characters in comparison to species of other Genera.

Question 6.
Illustrate the taxonomical hierarchy with suitable example of a plant.
Answer:
Kingdom = Plant kingdom
Division = Spermatophyta
Class = Dicotyledonae
Order = Sapindales
Family = Anacardiaceae
Genus = Mangifera
Species = Indica

Question 7.
What are the distinctive characteristics exhibited by living organisms? Describe them in brief.
Answer:
Growth, reproduction, ability to sense environment, Metabolism, ability to self duplicate.

Growth :
Increase in mass and Increase in number of Individuals.

Reproduction :
It is the production of progeny possessing features more or less similar to those of parents.

Metabolism :
It is the sum total of all the chemical reactions occurring in the body of an organism.

Question 8.
Life forms exhibit ‘unity in diversity’ – Discuss with your teacher.
Answer:
An important aspect of life is its vast diversity built on a base of underlying unity. For example, organisms as outwardly dissimilar as a bacterium, a human and an Oak tree are all composed of the same basic structural element, the cell, which in turn have many similar subcellular and molecular components.

Question 9.
List out the principles followed to provide scientific names for newly found organism?
Answer:

1. Identification of organisms whether it is entirely new or already known.
2. Particular organism should be named by same name all over the world. Thus nomenclature is providing a scientific name to an identified organism.

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata

Very Short Answer Type Questions

Question 1.
List out the characters shared by chordates and echinoderms.
Answer:
Chordates and echinoderms are enterococci, deuterostomes, and bilaterally symmetrical.

Question 2.
Write four salient features of cyclostomes.
Answer:

1. Cyclostomes are jawless aquatic forms.
2. The body is scaleless, long, slender, and eel-like in shape.
3. Endoskeleton is cartilaginous.
4. Vertebrae are represented by imperfect neural arches in some.
5. The mouth is circular and suctorial, Hence there are called Cyclostomes. Ex: Petromyzon.

Question 3.
What is the importance of endostyle in lancelets and ascidians?
Answer:
Endostyle is useful for accumulating and moving food particles to the oesophagus in lancelets and ascidians.

Question 4.
Name the type of caudal fin and scales that are present in a Shark and Catla respectively.
Answer:
The caudalfin in shark is heterocercal and scales are placoid. The caudal fin is catla is homocercal and scales are cycloid.

Question 5.
What is the importance of air bladder in fish?
Answer:
Fishes have an ‘air bladder’ acting as a ‘hydrostatic organ’ helping the fish float easily at the desired level without much expenditure of energy.

Question 6.
How do you justify the statement ‘heart in fishes is a branchial heart’?
Answer:
The heart of fish is two-chambered and is described as a branchial heart as it supplies blood only to the gills.

Question 7.
What are claspers? Which group of fishes possesses them?
Answer:
Claspers are formed from the posterior portion of pelvic fins in male cartilaginous fish. They serve as intermittent organs used to channel semen into the female’s cloaca during mating.
Ex: Chondrichthyes fishes possess Claspers.

Question 8.
How does the heart of an amphibian differ from that of a reptile?
Answer:
The heart of an amphibian is three-chambered. The heart of a reptile is incompletely four-chambered.

Question 9.
Name the structures that appeared for the first time in amphibians, in the course of evolution.
Answer:
The two pairs of pentadactyl limbs appeared for the first time in amphibians in the course of evolution.

Question 10.
How do you distinguish a male frog from a female frog?
Answer:
The male frog can be distinguished by the presence of sound amplifying vocal sacs and a copulatory pad on the first digit of each forelimb.

Question 11.
What is a ‘force pump’ in a frog? Why is named so?
Answer:
In frogs, during pulmonary respiration, the buccopharyngeal cavity acts like a ‘force pump’. Due to the elevation of the buccopharyngeal cavity the air forces the glottis to open and enter the lungs.

Question 12.
What are corporabigemina? Mention their chief function.
Answer:
Midbrain is represented by a pair of optic lobes called corpora bigemina. The optic lobes are associated with the sense of sight.

Question 13.
Distinguish between mesorchium and mesovarium.
Answer:
The testes are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesorchium.
The ovaries are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesovarium.

Question 14.
Distinguish between milt and spawn.
Answer:
During amplexus, the mass of eggs and the mass of sperm released by the female and male are called spawn and milt.

Question 15.
What are the ‘Golden ages’ of the first jawed vertebrates and the first amniotes?
Answer:
The Devonian period is considered the ‘golden age of first jawed vertebrates (Fishes).
The Mesozoic era is considered the golden age of amniotes (Reptiles).

Question 16.
Name two poisonous and non-poisonous snakes found in south India.
Answer:
Poisonous Snakes:

1. Naja naja (Cobra)
2. Bungarus (Krait)
3. Vipera russelli (Chain viper)

Non-poisonous Snakes:

1. Ptyas (rat snake)
2. Tropidonotus (Pond or grass snake).

Question 17.
In which features does the skin of a reptile differ from that of a frog?
Answer:

• The skin of reptiles is rough and dry, covered by horny epidermal scales, and shields.
• The skin of a frog is thin, scaleless, and moist.

Question 18.
Describe a cat and a lizard on the basis of their chief nitrogenous wastes excreted.
Answer:
Based on the nitrogenous wastes excreated by lizards are Uricotelic and cats are ‘Ureotelic’ animals.

Question 19.
Name the four extraembryonic membranes.
Answer:
The extraembryonic membranes namely amnion, allantois, chorion, and yolk sac.

Question 20.
What are Jacobson’s organs? What is their function?
Answer:
Jacobson’s organs are the specialized olfactory structures, that are highly developed in lizards and snakes.

Question 21.
What are pneumatic bones? How do they help birds?
Answer:
The main bones in birds are extensions of air sacs without bone marrow are called pneumatic bones. These are helpful in flying birds.

Question 22.
What is a ‘wishbone? What are the skeletal components that form it?
Answer:
In birds, both the clavicles are fused with the interclavicular to form a ‘V-shaped bone, called fiircula or ‘wish hone’ or ‘Merrythought bone’.

Question 23.
What is continuous oxygenation of the blood? How is it made possible in birds?
Answer:
The lungs of birds are compact, spongy, undistensible lungs associated with air sacs. Air sacs facilitate continuous air supply is called ‘continuous oxygenation of the blood.

Question 24.
Distinguish between the crop and the gizzard in birds.
Answer:

• The Oesophagus of birds is often dilated into a crop for the storage of food.
• The stomach is usually divided into glandular proventriculus and muscular gizzard a grinding mill.

Question 25.
Distinguish between altricial and precocial hatchlings.
Answer:

• Altricial: Young ones of flying birds’ hatchlings are altricial.
• Precocial: Young ones of flightless bird hatchlings are precocial.

Question 26.
In which group of animals do we find three ear ossicles on each side and what are their names from the innermost to the outermost?
Answer:
The middle ear possesses three ear ossicles in the Mammalia group of animals. They are malleus, incus and stapes.

Question 27.
How does a mature RBC of a mammal differ from that of other vertebrates?
Answer:

• In mammals, mature RBC is enucleated and biconcave.
• In other vertebrates RBC is nucleate.

Question 28.
Name the characteristic type of vertebrae found in reptiles, birds, and mammals.
Answer:

• Reptiles’ vertebrae are procoelons.
• Birds’ vertebrae are heterologous.
• Mammalian vertebrae are amphiplatyan.

Question 29.
Name the three meninges. In which group of animals do you find all of them?
Answer:
Mammals have three meninges. They are the outer dura mater, middle arachnoid mater, and inner diameter.

Question 30.
Name the vertebrate groups in which ‘renal portal system1 is absent.
Answer:
The renal portal system is absent in aves (birds) in vertebrate animals.

Short Answer Type Questions

Question 1.
Give three major differences between chordates and non-chordates and draw a sketch of a chordate’s body showing those features.
Answer:
The major differences between chordates and non-chordates.

Question 2.
Name the four ‘hallmarks’ of chordates and explain the principal function of each of them.
Answer:
The hallmarks of chordates: All the chordates exhibit four fundamental characteristics. They are Notochord, Dorsal tubular nerve cord, Pharyngeal slits or clefts, and Post-anal tail.

Notochord: It is a flexible rod-like structure situated along the mid-dorsal line between the gut and the nerve cord. It is derived from the embryonic chorda mesoderm. It is firm but flexible. It is present throughout life in the lancelets and cyclostomes. It is present in the tail of the tadpole larva of an ascidian, It is present in embryonic stages, but is replaced partly or wholly by the vertebral column in the adults of higher chordates. Remnants of notochord occur as nucleipulposi in the intervertebral discs of mammals.

Dorsal tubular nerve cord: A single, hollow tubular, and fluid-filled nerve cord is situated above the notochord and below the dorsal body wall. It is derived from the ectoderm of the embryo. In the higher chordates, it gets enlarged to form a distinct brain at the anterior end the rest of it becomes the spinal cord.

Pharyngeal slits of clefts: These are slit openings present on the Pharyngeal wall and meant for the exit of the water from the pharyngeal cavity. They are present throughout life in the protochordate, fishes, and some amphibians. These are present in larval stages in amphibians. They develop by in-pushing of ectoderm and corresponding out pursing of the endoderm. In land vertebrates, the gills become vestigial and nonfunctional and are restricted to embryonic stages only.

Post-anal tail: Chordates have a tail extending posteriorly to the anus. It is lost in many species during late embryonic development. It contains skeletal elements and muscles, coelom and visceral organs are absent in it.

Question 3.
Describe the features of a tunicate that reveals its chordate identity.
Answer:

1. The body of these animals is covered by cellulose-like covering tunicin, hence called tunicates.
2. These possess Notochord in the tail region during the larval stage, hence called Urochordata.
3. These are sedentary or pelagic marine forms.
4. The notochord is present only in larval tails and degenerated in adults.
5. Open type of blood vascular system with blood pigment vanadium.
6. Indirect development with tadpole larva.
7. Adults (mostly) show degenerate characters.
8. The nervous system is represented in the adult by a single dorsal ganglion.
9. They are bisexual or hermaphrodites.
10. Ex: Ascidia, Salpa, Doliolum, Pyrosoma and Oikopleura.

Question 4.
Compare and contrast sea squirts and lancelets.
Answer:
Sea squirts: These are included in class – Ascadiaceae of subphylum – Urochordata. These are sessile. These are solitary or colonial. The body is enclosed in a permanent test and un-segmented. All these are marine and occur from the surface water to greater depths. Coelom in the absent, pharynx is large and is perforated by numerous gill slits. Branchial aperture in anterior and atrial aperture is dorsal. The digestive tract is ‘complete’. The circulations system is of an open type, the heart is the tubular and the ventral heart. These are bisexual. Development generally includes a free-swimming tadpole larva. Notochordcontinued to the tail hence the name Urochordata.
Ex: Ascidia

Lancelets: Cephalochordates are also called Lancelets. These are marine animals and are small fish without paired fins. These are typical chordates because they possess the notochord, tubular nerve cord, and pharyngeal slits throughout this life. The coelom is enterocoelic respiration mostly across the external body surface. The circulatory system is of a closed type, the heart, blood corpuscles, and respiratory pigments are absent. Excretion by protonephridia fertilization is external and development is indirect.
Ex: Branchiostoma (amphioxus or Lancelet)

Question 5.
List out eight characteristics that help distinguish a fish from the other vertebrates.
Answer:
General characters:

1. Fishes are completely aquatic poikilothermic (cold-blooded) animals.
2. The body of a fish is usually streamlined and differentiated into the head, trunk, and tail.
3. The exoskeleton consists of mesodermal scales or bony plates. A few are scaleless.
4. The endoskeleton may be cartilaginous or bony. Skull is monocondylic. Vertebrae are amphicoelous. Centrum is concave at both anterior and posterior faces.
5. Locomotion is assisted by unpaired (median and caudal) fins along with paired (pectoral and pelvic) fins.
6. The mouth is ventral or terminal. Teeth are usually acrodont, homodont, and polyphyodont.
7. The exchange of respiratory gases is performed by the gills. The heart is ‘two-chambered’.
8. Kidneys are mesonephric. Fishes are mostly ammonotelic and some are ureotelic. (cartilaginous fishes).
9. Cranial nerves are 10 pairs, Meninx Primitiva is the only ‘meninx’ enveloping the central nervous system.
10. The internal ear consists of three semicircular canals. Lateral-line sensory system (to detect movement and vibration in the surrounding water) is well-developed.
11. Eyes are without eyelids and each eyeball is protected by a nictitating membrane.
12. Sexes are separate. Fertilization is internal or external. Development may be direct or indirect.

Question 6.
Compare and contrast cartilaginous and bony fishes.
Answer:

Question 7.
Describe the structure of the heart of the frog.
Answer:
The blood vascular system consists of the heart, blood vessels, and blood. The heart is a muscular organ situated in the upper part of the body cavity. It has two separate atria and a single undivided ventricle. It is covered by a double-layered membrane called the pericardium. A triangular chamber called sinus venosus joins the right atrium on the dorsal side. It receives blood through three vena cavae (caval veins). The ventricle opens into the conus arteriosus on the ventral side. The conus arteriosus bifurcates into two branches and each of them divides into three aortic arches namely carotid, systemic and pulmocutaneous. Blood from the heart is distributed to all parts of the body by the branches of the aortic arches. Three major veins collect blood from the different parts of the body and carry it to the sinus venosus.

Question 8.
Write eight salient features of the class – Amphibia.
Answer:
General characters of Amphibia:

1. They are the first tetrapods and lead a dual mode of life, i.e. on land and in freshwater.
2. The body is divided into distinct ‘head’ and ‘trunk’. The tail may or may not be present.
3. Skin is soft, scale-less (except for the members of Apoda), moist and glandular.
4. The body bears two pairs of equal or unequal pentadactyle limbs (caecilians are limbless).
5. Skull is dicondylic as in mammals. Vertebrae are mostly precocious (centrum is concave at its anterior face only) in the anurans, amphicoelous in the caecilians, and usually opisthocoelous (centrum is concave at its posterior face) in the urodeles. Sternum appeared for the first time in the amphibians.
6. The mouth is large; teeth are acrodont, homodont, and polyphyodont.
7. Respiratory gaseous exchange is mostly cutaneous; pulmonary and buccopharyngeal respiration also occurs. Branchial respiration is performed by larvae and some adult urodeles.
8. The heart is three-chambered with sinus venosus and conus arteriosus. Three pairs of aortic arches and well-developed portal systems are present; erythrocytes are nucleated.
9. Kidneys are mesonephric; ureotelic.
10. Meninges are the inner pia mater and outer dura mater; cranial nerves are 10 pairs.
11. The middle ear consists of a single ear ossicle, the columella Auris which is the modified ‘hyomandibula’ of the fishes.
12. Tympanum, lacrimal and harderian glands appeared for the first time in the amphibians.
13. Sexes are separate and fertilization is mostly external. Development is mostly indirect.
14. e.g. Bufo (toad), Rana (frog), Hyla (tree frog), Salamandra (salamander), Ichthyophis (limbless amphibian), Rhacophorus (flying frog).

Question 9.
Describe the male reproductive system of a frog with the help of a labelled diagram.
Answer:
Male Reproductive System of frog: The male reproductive system consists of a pair of yellowish and ovoid testes, which are attached to the kidneys and dorsal body wall by a double fold of peritoneum called mesorchium. Each testis is composed of innumerable seminiferous tubules which are connected to form 10 to 12 narrow tubules, the vasa efferentia. They enter the kidneys and open into the Bidders canal which is connected to the ureter through transverse canals of the kidney. The urinogenital ducts of both sides open into the cloaca.

Question 10.
Write short notes on organs of special senses in frogs.
Answer:
Special senses: Frog has sense organs such as the organs of touch, taste, smell, sight, and hearing. The well-organized structures among them are eyes, and internal ears, and the rest are ‘cellular aggregations’ around nerve endings. The receptors of touch occur in the skin. Organs of taste are called taste buds that lie on small papillae of the tongue. The organs of smell are a pair of nasal chambers.

The organs of sight are a pair of eyes located in the orbits of the skull. Eyes are protected by eyelids. The upper eyelid is immovable. The lower eyelid is folded into a transparent nictitating membrane, which can be drawn across the surface of the eye. The retina of the eye contains both rods and cones. Cones provide ‘colour vision’ and rods are helpful in ‘dim light vision’.

The ear is useful for hearing and balance. It consists of a middle ear closed externally by a large tympanic membrane (ear drum) and a columella that transmits vibrations to the inner ear. The inner ear consists of a utriculus with three semicircular canals and a small sacculus.

Question 11.
List out the salient features of Exo and endoskeleton in reptiles.
Answer:
The exoskeleton of reptiles occurs in the form of horny epidermal scales, shields, and claws.
Endoskeleton:

1. Skull is monocondylic and many have temporal fossae.
2. Each half of the lower jaws is formed by six bones.
3. Vertebrae are mostly procoelous.
4. The first two cervical vertebrae are specialized into the atlas and axis.
5. The vertebral column is distinguished into cervical, thoracic, lumbar, sacral, and caudal regions.
6. Most living reptiles possess two sacral vertebrae.
7. Interclavicular is associated with the pectoral girdle.
8. Ribs are single-headed except in crocodilians.

Question 12.
List out the extant orders of the Class – Reptilia. Give two examples for each Order.
Answer:

1. Chelonia – Chelone (marine green turtle), Testudo (terrestrial form), Trionyx (freshwater form)
2. Rhynchocephalia – Sphenodon (a ‘living fossil’, endemic to New Zealand)
3. Crocodilia – Crocodylus pulustris (Indian crocodile or mugger), Alligator (alligator), Gavialis gangeticus (Indian gavial or gharial)
4. Squamata
   • Lizards – Hemidactylus (wall lizard), Chameleon, Draco (flying lizard)
   • Snakes
     • Poisonous Snakes: Naja naja (cobra), Ophiophagus hannah (King cobra), Bungarus (krait), Daboia/Vipera russelli (chain viper)
     • Non-Poisonous Snakes: Ptyas (rat snake), Tropidonotus (grass snake or pond snake)

Question 13.
What are the modifications that are observed in birds that help them in flight?
Answer:
So many modifications are observed in birds that help them in flight.

1. Exo and endo skeletons and body structure features might have contributed to their successful aerial mode of life.
2. The exoskeleton consists of epidermal feathers. Feathers are unique to birds. They are useful for flight, particularly the Quill feathers help in flight.
3. The body is boat-shaped and streamlined.
4. Four limbs are modified into wings.
5. Many bones are neumatic with extensions of air sacs.
6. All modern flying birds are provided with powerful breast muscles (flight muscles) chiefly the pectoralis major and pectoralis minor.
7. Lungs are associated with air and seas.

Question 14.
What are the features peculiar to ratite birds? Give two examples of ratite birds.
Answer:
Ratite birds:

1. These are modern flightless running birds.
2. They are ‘discontinuous’ in their distribution like the lungfishes and marsupials.
3. They are characterized by the presence of reduced wings.
4. Feathers are without an interlocking mechanism.
5. Rectrices are absent or irregularly arranged.
6. Prren gland is absent.
7. Pygostyle is rudimentary or absent.
8. The sternum is like without a keel.
9. Clavicles are absent, and syrinx is absent.
10. The male animal has a penis.
11. Young ones are precocial.
12. Ex: Struthio camelus – (African ostrich); Dromaeus (Emu) Kiwi.

Question 15.
Mention the most important features of the nervous system and sense organs in mammals.
Answer:

• The nervous system and sense organs are well developed in mammals.
• Mammals have relatively large brains when compared to that other animals in relation to body size.
• The four optic lobes constitute corpora quadrigemina.
• The two halves of the cerebrum are connected by the corpus callosum.
• The central nervous system is enveloped by three meninges.
• Eyes have movable eyelids with eyelashes.
• The external ear has a large pinna middle ear and possesses three ear ossicles.
• They are malleus, incus, and stapes, Cochlea of the internal ear is spirally coiled and bears the organ of Corti which is the receptor of sound.
• Skin is one of the sense organs.

Question 16.
Write short notes on the following features of the eutherians.

1. Dentition
2. Endoskeleton

Answer:

1. Dentition: The Dental formula of eutherians is i 3/3; c 1/1j pm 4/4, m 3/3; dentition is the codon, heterodont, diphyodont.
2. Endoskeleton: Skull is dycondylic. Most mammals have seven cervical vertebrae; vertebrae are of the amphiplatyan type, sacral vertebrae are two or five, and ribs are double-headed.

Question 17.
Give an example for each of the following.

1. A viviparous fish
2. A fish possessing electric organs
3. A fish possessing poison sting
4. An organ that regulates buoyancy in the body of a fish
5. An oviparous animal with milk-producing glands.

Answer:

1. Scoliodon fish is viviparous fish.
2. Torpedo fish is possessing electric organs.
3. Dasyatis/Trygon fish possess poison sting.
4. The air bladder regulates buoyancy in the body of a fish.
5. Ornithorhynchus anatinus (Duck-billed platypus) is an oviparous animal with milk-producing glands.

Question 18.
Mention two similarities between
(a) Aves and mammals
(b) A frog and a crocodile
(c) A lizard and a snake
Answer:
(a) Aves and mammals:

• Aves and mammals are Triploblaste and bilaterally symmetrical.
• The heart is four-chambered in both.

(b) A frog and a crocodile:

• Erythrocytes are nucleated in both.
• Frogs and crocodiles are uriotelic animals.

(c) A lizard and a snake:

• Lizards and snakes are reptilian animals.
• The heart is incompletely four-chambered.
• Jacobson’s organs, the highly developed specialized olfactory structures are present.

Question 19.
Name the following animals.

1. A limbless amphibian
2. The largest of all living animals
3. An animal possessing dry and cornified skin
4. ‘National animal’ of India.

Answer:

1. Ichthyophis is a limbless amphibian.
2. Balaenoptera musculus (Blue whale) is the largest of all living animals.
3. Crocodylus is an animal possessing dry and cornified skin.
4. Panther Tigris (tiger) is the National animal of India.

Question 20.
Write the generic names of the following.

1. An oviparous mammal
2. Flying fox
3. Blue whale
4. Kangaroo

Answer:

1. An oviparous mammal’s generic name is Ornithorhynchus (Duckbilled platypus).
2. Flying fox’s generic name is Pteropus.
3. The blue whale’s generic name is Balaenoptera musculus
4. Kangaroo generic name is Macropus

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla

Very Short Answer Type Questions

Question 1.
What physical feature pertaining to the organism and its medium do you notice in a sponge body form in which sponges can be/were identified as animals and not plants? What do you call the region in the sponge body in which you noticed that feature?
Answer:
Sponges are primitive multicellular and sessile animals and have a cellular level of organisation. The body wall is composed of two layers separated by matrix mosohyl, and are heaving canal system for transport of water through Ostia, having a cavity in the body called a spongocoel hence the sponge are animals and are not plants.

Question 2.
What are the different structures that make up the internal skeleton of a sponge? What are the chemicals involved in the formation of these structures?
Answer:
The internal skeleton of a sponge is made up of different types of spicules.
Calcareous spicules made up of CaCO₃.
Ex: Sycon
Siliceous spicules – are made up of Silicon dioxide – glass.
Ex: Euplectella
Spongin fibres.
Ex: Spongilla

Question 3.
What are the functions of the canal system of sponges?
Answer:
The functions of the canal system of a sponge are gathering of food, respiratory exchange of gases, and removal of wastes.

Question 4.
What are the two chief morphological ‘body forms’ of cnidarians? What are their chief functions?
Answer:
The body form of Cnidarians is polyp and medusa. Polyp produces medusae by asexual reproduction. Medusae produce polyps by sexual reproduction.

Question 5.
What is metagenesis? Animals belonging to which phylum exhibit metagenesis?
Answer:
Cnidarians show two basic body forms called polyp and medusa. Cnidarians which exist in both forms exhibit alternation of generations called metagenesis.

Question 6.
What is the cnidarian group with quantitatively/relatively large mesoglea? What is the significance of such a well-developed mesoglea pertaining to the aquatic life of that group?
Answer:
The Scyphozoa of cnidarian animals have large mesoglea, it is the significance of these animals.

Question 7.
What is the chief difference between the hydrozoans and the rest, of the cnidarians regarding the germinal layer (s) in which its ‘defencive structures or cells of defence occur?
Answer:
The defencive structures Cnidocytes or Cnidoblasts occur only in the ectoderm, in the hydrozoans in the rest of the Cnidarians the cnidocytes occur in both ectoderm and endoderm.

Question 8.
What are the excretory cells of flatworms called? What is the other important function of these specialized cells?
Answer:
The excretory cells of flatworms are flame cells. Another important function of these specialized cells is osmoregulation.

Question 9.
Distinguish between amphids and phasmids.
Answer:
Amphids: These are the cuticular depressions present on the lips surrounding the mouth in the nematodes such as Aphasmidia animals and serve as Chemoreceptors.
Phasmids: These are the well-developed sensory organs and they occur in some nematodes such as phasmidia animals.

Question 10.
What is the essential difference between a ‘flat worm’ and a ’round worm’ with reference to the perivisceral area of the ‘bodies’.
Answer:
With the reference to the perivisceral area of the body, the flatworms have dorso-ventrally flattened bodies. The body is not segmented, but some of the animals exhibit pseudometamerism. In the Nematoda the body is circular in cross-section, hence the name roundworms, body is not segmented.

Question 11.
How do you account for the origin of the perivisceral space in the body of a nematode and an annelid?
Answer:
The perivisceral space in the body of a nematode is circular in cross-section. Hence the name ‘wound worms’ body unsegmented. In an annelid, the body is segmented by septa into segments or metameres (annulus – little rings) some of them (Nereis) possess lateral appendages parapodia.

Question 12.
What is metamerism? What is the essential difference between the mode of formation of individual morphological body units of a tapeworm and those of an earthworm?
Answer:
The body is divided into segments like units called metameres. Like the divination in known as metamerism.
Ex: Earthworm.
In tapeworm body segments are pseudometameres.
In earthworms, body segments are true segments or metameres.

Question 13.
How do you distinguish a ‘hirudineaun’ from the rest of the annelids, based on the morphological features pertaining to metamerism? How does the coelom of a leech differ from the coelom of an earthworm with reference to its contents?
Answer:
In hirudinean like Leach, the body is with a definite number of segments. The segments are externally sub-divided into annute, internal segmentation’ is absent.
In Leech coelom is filled with a characteristic tissue called botryoidal tissue. In earthworms, the coelom is filled with coelomic fluid.

Question 14.
What do you call the locomotor structures of Nereis? Why is Nereis called a polychaete?
Answer:
Locomotor structures of Nereies are parapodia. The parapodia bear many setae that help in locomotion hence the name Polychaeta.

Question 15.
What is botryoidal tissue?
Answer:
The coelom of Leech is filled with a characteristic tissue called botryoidal tissue, it is resembling a bunch of grapes. They range from excretion to storage of iron, calcium, and revascularization in areas of injury.

Question 16.
What is the difference between the epidermis of a Nematoda and that of an annelid? How does a nematode differ from an annelid with reference to the musculature of the body wall?
Answer:
The epidermis of Nematoda is syncytial and the epidermis of annelid animals is informed by one cell thick ectodermal epithelial cells.

Question 17.
What do you call the first and second pairs of cephalic appendages of a scorpion?
Answer:
The first and second pairs of cephalic appendages of a Scorpion are Chelicerae and Pedipalpi.

Question 18.
What is the uniqueness of the first two pairs of cephalic appendages of a crustacean compared to those of the other extant arthropods?
Answer:
In crustaceans, cephalic appendages are two pairs of antennae (antennules and antennae). It is the unique feature of Crustaceans compared to those of the other extent arthropods.

Question 19.
What is the sub-phylum to which ‘ticks’ and ‘mites’ belong? How do you distinguish them from insects with reference to their walking legs?
Answer:
Ticks and mites belongs to the sub-phylum Chelicerata and class Arachnida. These have four pairs of walking legs.

Question 20.
What are the respiratory structures of Limulus and Palamnaeus respectively?
Answer:
The respiratory structures of Limulus are book gills, and in palamnaeus are book-lungs.

Question 21.
What ara antennae? What is the arthropod group without antennae?
Answer:
Antennae are the sensory organs, of the animals of sub-phylum-Mandibulata of arthropod bear antennae.

Question 22.
What do you call the perivisceral cavity of an arthropod? Where from is it derived during development?
Answer:
The perivisceral cavity of an Arthropoda is a haemocoel, it is not true coelom, but derived from mostly the embryonic blastocoel.

Question 23.
Which arthropod, you have studied, is called a living fossil? Name its respiratory organs.
Answer:
The arthropod animal Limulus is called a living fossil, it is respiratory organs are book-gills.

Question 24.
How do you identify a Chiton from its external appearance? How many pairs of gills help in the respiration of Chiton?
Answer:
Chiton is bilaterally symmetrical and dorsoventrally flattened. Shell is dorsal and consists of eight transverse plates. Poat is ventral elongated and flat. Gills are 6 to 88 pairs helps in respiration.

Question 25.
What is the function of the radula? Give the name of the group of mollusks that do not possess a radula.
Answer:
The buccal cavity contains a file-like rasping organ called radula for feeding, except for the bivalves and tusk of Molluscs.

Question 26.
What is the other name for the gill of a mollusc? What is the function of osphradium?
Answer:
The other name for the gill of a mollusc is Ctenidia. The main function of Osphradium is to test the purity of water.

Question 27.
What is Aristotle’s lantern 7 Give one example of an animal possessing it?
Answer:
In the mouth of the sea Urchin a complex five Jawed masticatory apparatus called Aristotle’s Lantern.
Ex: Echinus.

Question 28.
What is the essential difference between the Juveniles and adults of echinoderms, symmetry-wise?
Answer:
The adult echinoderms are radially symmetrical (pentamerous radial symmetry), but Juveniles (Larvae) are bilaterally Symmetrical.

Question 29.
What are blood glands in pheretima?
Answer:
Blood glands are present in the 4th, 5th, and 6th segments of pheretima. They produce blood cells and haemoglobin which is dissolved in the plasma.

Question 30.
What are spermathecae on the body of pheretima?
Answer:
In Pheretima there are four pairs of spermothecae are located in the segments 6th to 9th as one pair in each segment. This receives and stores spermatozoa during copulation.

Short Answer Type Questions

Question 1.
Write short notes on the salient features of the anthozoans.
Answer:

1. Anthozoans are commonly referred to as sea anemones.
2. Anthozoa includes sea anemones, corals, and sea pens.
3. All are marine forms. These are solitary or colonial.
4. They are sedentary and only have polypoid information.
5. Coeienteron is divided into several compartments by vertical septa called mesenteries.
6. Mesoglea contains connective tissue.
7. Cnidocytes occur both in the ectoderm and endoderm and are cellular and contain amoebocytes.
8. Germ cells are derived from the endoderm. Ex: Adamsia (sea anemone), Gorgonia (sea fan), Pennatula (sea pen).

Question 2.
What is the class to which the flukes belong? Write short notes on the chief characters of the group.
Answer:
Flukes belong to the class Trematoda of Phylum-Platyhelminthes.

1. Trematoda organs are commonly called flukes.
2. These are parasitic on other animals.
3. The body is covered by a thick cuticle and bears two suckers, an oral and a ventral.
4. The mouth is anterior and the intestine is bifurcated.
5. These are bisexual (monoecious).
6. Life history is complex with many hosts and different types of stages – miracidium, sporocyst, redia, cercaria, etc. Ex: Fasciola (Liver fluke), Schistosoma (blood fluke).

Question 3.
What are the salient features exhibited by Polychaetes?
Answer:

1. These are commonly known as bristle worms.
2. All are marine. Many are burrowing, others are free swimming or crawling or tubicolous.
3. Head is distinct with sensory structures like eyes, antennae, palps, and cirri.
4. Clitellum is absent.
5. Each segment has a pair of lateral appendages called parapodia in which bundles of setae are arranged.
6. Animals are unisexual. Most segments bear glands. Gonoducts are absent.
7. Gametes are shed into the coelom.
8. Fertilization is external.
9. Development includes a trochophore larva. Ex: Nereis

Question 4.
How do the hirudineans differ from the polychaetes and oligochaetes?
Answer:

1. Definite number of body segments are present in hirudinean but many segments are present in polychaetes and earthworms.
2. Locomotion in leeches is by suckers but body setae in oligochaetes and parapodia in polychaetes. Parapodia also help in respiration.
3. Temporary clitellum during the breeding season is present in leeches but clitellum is absent in polychaetes and permanent clitellum is present in oligochaetes.
4. Hirudineans are bisexuals, oligochaetes are bisexual and polychaetes are unisexual animals.
5. Coelom is reduced on leeches, but coelom is spacious in oligochaetes and polychaetes.
6. Development is direct in leeches and earthworms but indirect in polychaetes.
7. Nutrient tissue called botryoidal tissue fills the coelom in hirudinean.
8. Anterior and posterior suckers are present in hirudineans. Such suckers are absent in polychaetes and oligochaetes. Ex: Pheretima, Tubifex

Question 5.
What are the chief characteristics of crustaceans?
Answer:

1. This includes prawns, crabs, lobsters, crayfishes, etc.
2. Mostly marine, a few are fresh water and some are adapted to terrestrial life.
3. In most species, the head and thorax fuse to form a cephalothorax.
4. Cephalic appendages are five pairs – first antennae (antennules) second antennae, mandibles, first maxillae and second maxillae.
5. Thoracic and abdominal appendages are typically biramous
6. Respiration is by gills.
7. Excretory organs are green glands or antennal glands.
8. Sense organs include statocysts, compound eyes, and antennae.
9. Gonopores are paired.
10. Development is direct or indirect involving several larval stages. The basic larva is nauplius. Ex: Palaemon (Prawn); Cancer (Crab).

Question 6.
Mention the general characters of Arachnida.
Answer:

1. This includes scorpions, spiders, ticks, and mites.
2. Primarily they are all terrestrial.
3. Prosoma bears one pair of pre-oral chelicerae, one pair of post-oral pedipalps, and four pairs of walking legs.
4. In spiders each chelicera bears a fang into which the poison gland opens.
5. Abdominal appendages are modified into book lungs, spinnerets, pectines, etc.
6. Telsun is usually absent. It is present as a sting in scorpions.
7. Respiration is by book lungs or tracheae.
8. Excretory organs are coaxial glands and malpighian tubules.
9. Scorpions are viviparous.
10. Development is direct. Ex: Palamnaeus (scorpion); Aranea (spider).

Question 7.
Compare briefly a centipede and a millipede.
Answer:

Question 8.
Cephalopods show several unique or advanced features when compared to the other molluscs. Discuss briefly.
Answer:

1. The class Cephalopoda includes cuttlefishes, squids, octopuses, nautilus, etc.
2. The Head is discrete and bears very conspicuous eyes.
3. Shell is either present (e.g.: Sepia) or absent (e.g.: Octopus). When present it may be multi-charactered and external (e.g.: Nautilus) or internal (e.g.: Loligo).
4. The foot is modified into eight to ten arms (tentacles) present around the mouth and siphons.
5. Some Cephalopods (e.g: Sepia) possess an ink gland as a defensive adaptation.
6. Ctenidia are two or four in number – dibranchiate. e.g.: Sepia and tetrabranchiates. (e.g.: nautilus)
7. The brain is complex and is protected by a cartilaginous cranium.
8. Eyes are superficially similar to those of vertebrates.
9. Development is direct. Ex: Architeuthis (giant squid).

Question 9.
Which class of Mollusca represents the primitive molluscs? What are their chief features?
Answer:
The primitive molluscs are represents the class Aplacophora of Phylum-Mollusca.
These are primitive forms with ‘worm-like’ bodies. These are marine forms without mantle, shell, foot, and nephridia. The Head is poorly developed. A rasping organ radula is present in the buccal cavity. Cuticle contains calcareous spicules. Eyes, statocysts, and tentacles are absent. The heart consists of a single auricle and a ventricle. A pair of gonads are present in some, there is a mid-ventral groove that is homologous to the foot of the other molluscs.
Ex: Neonmia, Chaetoderma.

Question 10.
What are the salient features of the echinoids?
Answer:

1. It includes sea urchins, heart urchins, sand dollars, etc. The body is ovoid or discoidal and covered by movable spines.
2. Arms are absent, tube feet are arranged in five bands, and bear suckers.
3. Ossicles of the body unite to form a rigid test or corona or case.
4. Pedicellaria is “three jawed”.
5. Anus and madreporite are aboral in position.
6. Ambulacral grooves are closed.
7. A complex five-jawed masticatory apparatus called Aristotle’s lantern is present just inside the mouth. It is absent in heart urchins.
8. Life history includes a larval form called echinopluteus.
9. Specialized gills called peristomial gills as present in sea urchins. Eg: Salmacis (Sea urchin), Echino Cardium (Heart urchin), Clypeastoer (Cake Urchin).

Question 11.
Mention the salient features of Holothuroidea.
Answer:
Holothuroidea: This class includes sea cucumbers. Body elongated in the oro-aboral axis. Arms, spines, and pedicellariae are absent skin are soft and leathery (Coriaceous). The dermis contains microscopic, isolated ossicles. The madreporite is internal, suspended in the perivisceral coelom. Tube feet are provided with suckers. The mouth is surrounded by retractile feeding tentacles, which are modified tube feet, chief gas exchange organs are a pair of respiratory trees that arise from the wall of the cloaca and form branched tubes in the perivisceral coelom. The development includes auricularia and doliolaria larvae.
Ex: Cueumaria, synaptic, Thyone.

Question 12.
What is the function of nephridia?
Answer:

• The nephridia of pheretima are ectodermal in origin and are metanephridia.
• Several types of nephridia occur in pheretima but are fundamentally similar in structure.
• Which opens outside through the nephridiopore – The nephridia tolled open nephridia. Ex: Septal nephridia.
• Those who do not have nephridiopore are called closed-type nephridia. Ex: Pharyngeal nephridia.
• Those open at the outer surface are called exonephridia.
• The nephridia play an important role in osmoregulation.
• Earthworms mostly excrete urea as the excretory product and are described as ureotelic animals.

Question 13.
How many types of nephridia occur in pheretima and how do you distinguish them?
Answer:
In Pheretima three types of nephridia are present.

1. Septal nephridia: The septal nephridia are present on the intersegmental septum from 15 & 16 segments onwards to last and are opened into the alimentary canal.
2. Integumentary nephridia: The integumentary nephridia attached to the inner body wall from the 3rd segment to the last. They open to the exterior on the body surface by nephridiopores.
3. Pharyngeal nephridia: The pharyngeal nephridia present three paired tufts in the segments 4tfl, 5th, and 6th. They open into the buccal cavity and pharynx.

Question 14.
Give an account of the hearts in the circulatory system of pheretima.
Answer:
Hearts in Pheretima: The dorsal blood vessel and the ventral blood vessel are connected by a pair of pulsatile hearts, in each of the seventh, ninth, twelfth, and thirteenth segments. Of these four pairs, the anterior two pairs connect only the dorsal blood vessel to the ventral blood vessel. Hence they are ’ called lateral hearts. The posterior two pairs connect both the dorsal blood vessel and the supra-oesophageal blood vessel with the ventral blood vessel. Hence, they are called lateral oesophageal hearts. These two types of hearts also differ in the number and arrangement of their valves. Four pairs of valves are present in each lateral heart, while three pairs of valves are present in each lateral oesophageal heart. Hearts allow the blood to flow into the ventral blood vessel only.

Long Answer Type Questions

Question 1.
Draw a labelled diagram of the reproductive organs of Pheretima.
Answer:
Reproductive organs of pheretima: Pheretima is a hermaphrodite (bisexual). There are two pairs of testes. One pair each present in the 10th and 11th segments. Their vasa deferentia run up to the 18th segment where they join the prostatic ducts. Two pairs of seminal vesicles present in the 11th and 12th segments are sacs in which spermatogonia mature into spermatozoa. The common prostatic and spermatic ducts open to the exterior by a pair of male genital pores on the ventrolateral sides of the 18th segment. Two pairs of accessory glands’ are present one pair each in the 17th and 19th segments. Four pairs of spermathecae are located in the segments 6th to 9th (one pair in each segment). They receive and store spermatozoa (spermatophores) during copulation.

One pair of ovaries is attached to the posterior face of the inter-segmental septum of the 12th and 13th segments. Oviducal funnels are present beneath the ovaries and they continue into oviducts (14th segment). They join together and open to the exterior on the ventral side of the 14th segment by a single median female genital pore.

Question 2.
Describe the digestive system and process of digestion in pheretima.
Answer:
The digestive system in pheretima: The alimentary canal is a straight tube and runs from the first to the last segment of the body. The mouth opens into the buccal cavity (1-3 segments) which leads into the muscular pharynx (4th segment). A small narrow tube, oesophagus (5-7 segments), continues into a muscular gizzard (8th segment). It helps in grinding the small particles of food in the decaying leaves (grinding mill). The stomach extends from segments 9 to 14. The food of earthworms is decaying leaves and other organic matter mixed with the soil. Calciferous glands, present in the stomach, neutralise the humic acid present in the humus of the soil. The intestine starts from the 15th segment and continues till the last segment.

A pair of short and conical intestinal caeca project from the intestine in the 26th segment. An internal median fold of the dorsal wall of the intestine called typhiosole, helping in increasing the area of absorption, is poorly developed in Pheretima (between the 26th and the rectum which occupies the last 23 to 28 segments). The alimentary canal opens to the exterior by a small rounded aperture called the anus. The ingested soil rich in organic matter passes through the digestive tract where digestive enzymes break down complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilized for various metabolic activities.

Process of digestion: Digestion in earthworms is extracellular. Earthworm obtains their nourishment from the organic debris (detritus) present in the soil. So it is called a detritivore. The pharynx is ejected due to the inside out of the buccal chamber. The pharynx, with the help of its radial-dilator muscles, works as a suction pump in feeding.

The organic food along with the swallowed soil particles is sucked into the pharynx, where it mixes with the salivary secretion. The mucin in the saliva lubricates the gut wall for the easy passage of food and also helps in the formation of the bolus. The proteolytic enzyme in the saliva partly digests the proteins. Then the food reaches the gizzard. Its circular muscle and the thick cuticle grind the food into fine particles. In this state, the food is easily acted upon by the digestive enzymes in the stomach and intestine.

The intestinal juice of an earthworm is comparable to the pancreatic juice of higher animals. All these enzymes like proteases, amylases, and lipases act upon the finely ground food and digest the organic matter in it. Proteases digest proteins into amino acids, amylases digest carbohydrates into glucose and lipases digest lipids into fatty acids and glycerol.

The digested food is absorbed by the intestinal epithelium in the typhlosolar region. The extensive capillary network of blood vessels of the intestine plays a vital role in absorption. The typholosole helps in increasing the area of absorption. The undigested food then passes to the rectum, where water is absorbed from the undigested food. Then the undigested matter is egested out through the anus in the form of worm castings.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 13th Lesson Organic Chemistry-Some Basic Principles and Techniques Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 13th Lesson Organic Chemistry-Some Basic Principles and Techniques

Very Short Answer Questions

Question 1.
Write the reagents required for the conversion of benzene to methyl benzene.
Answer:
Benzene reacts with methyl chloride in presence of anhydrous AlCl₃ to form methyl benzene.

This reaction is called as Friedal craft’s alkylation.

Question 2.
How is nitrobenzene prepared.
Answer:
Benzene undergoes nitration with nitration mixture (cone. HNO₃ + Cone H₂SO₄) at less than 60° C to form nitrobenzene.

Question 3.
Write the conformations of ethane.
Answer:

Question 4.
How do you prepare ethyl chloride from ethylene?
Answer:
Ethylene reacts with hydrogen chloride to form Ethyl chloride in presence of anhydrous AlCl₃ catalyst.

Question 5.
Write the IUPAC names of :
a) CH₃ – CH₂ – CH₂ – CH = CH₃
b)
c)
Answer:

Question 6.
Write the structures of Trichloroethanoic acid, Neopentane, P – nitro benzaldehyde. [Mar. ’13]
Answer:
Trichloro ethanoic acid – CCl₃ – COOH

Question 7.
Discuss Lassaigne’s test.
Answer:
Lassaigne’s Test: –

• A small dry sodium piece is taken in a fusion tube and heated gently till the sodium melts.
• To this molten sodium a small amount of organic compound is added and heated until it changes red hot.
• This red hot tube is plunged in a china dish containing water and this content is boiled, cooled and filtered.
• The obtained filtrate is known as Lassaigne’s extract.
• This test is used to detect the elements Nitrogen, Sulphur, halogens etc.,

Detection of Nitrogen : –
The Lassaigne’s extract is made alkaline by adding few drops of dil. NaOH. to this freshly prepared FeSO₄ solution is added and warmed then few drops of FeCl₃ are added followed by acidification with Cone. HQ (or) H₂SO₄. A bluish green colouration indicates the nitrogen.
Na + C + N → NaCN
2NaCN + FeSO₄ → Na₂SO₄ + Fe (CN)₂
Fe(CN)₂ + 4NaCN → Na₄ [Fe(CN)₆]
3Na₄ [Fe(CN)₆] 4- 4FeCl₃ → Fe₄ [Fe(CN)₆]₃ + 12 NaCl (Prussian blue colouration)

Question 8.
Explain the principle of chromatography.
Answer:
Chromatography has been developed as a method of separation components of a mixture generally between the stationary phase and a mobile phase.
Chromatography involves the three steps given below :
a) Adsorption and retention of a mixture of substances on the stationary phase and separation of adsorbed substances by the mobile phase to different distances on the stationary phase.
b) Recovery of the substances separated by a continuous flow of the mobile phase (known as elution) and
c) Qualitative and quantitative analysis of the eluted substances.

Question 9.
Explain why an organic liquid vaporizes at a temperature below it’s boiling point in it’s steam distillation.
Answer:
In steam distillation, the liquid boils when the sum of vapour pressures due to the organic liquid (P₁) and that due to water (P₂) becomes equal to the atmospheric pressure (P) i.e., P = P₁ + P₂.
∴ P₁ is lower than P the organic liquid vaporises at lower temperature than it’s boiling point.

Question 10.
Explain the following :
a) Crystallisation
b) Distillation
Answer:
a) Crystallisation:

• This technique is used for the purification of solid organic compounds.
• This method is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.
• The impure compound is dissolved in a solvent in which it is partially soluble at room temperature and  appreciably soluble at higher temperature.
• The solution is concentrated to get a nearly saturated solution on cooling the solution pure compound crystallises.
• On repeating this process finally gets the very pure compound.

b) Distillation :
This method is used to separate

1. 1. Volatile liquids from non volatile impurities and
   2. The liquids having enough difference in their boiling points.
2. Liquids having different boiling points vapourises at different temperatures.
3. These vapours are cooled and the liquids formed are collected separately.
   Eg : CHCl₃ (b.pt. 334 K) and C₆H₅NH₂ (b.pt. 457 K) are easily separated by distillation technique.

Short Answer Questions

Question 1.
Complete the following reaction and name the products A, B and C. [T.S. Mar. ’15] [A.P. Mar. 16]

Answer:
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
Acetylene (= A)

A = C₂H₂ (Acetylene)
B = C₆H₆ (Benzene)
C = C₆H₅CH₃ (Methyl benzene)

Question 2.
Name the products A, B and C formed in the following reactions. Give the equations for the reactions.

Answer:

A = 1, 2- dibromo ethane
B = Acetylene
C = 1, 1, 2, 2 – tetra bromo ethane

Question 3.
How does acetylene react with : a) Bromine b) Hydrogen ? Write the balanced equations for the above reactions. Name the products.
Answer:
a) Addition of Br₂: Acetylene reacts with bromine to form finally 1, 1, 2, 2 – tetra bromo ethane.

b) Addition of hydrogen : Acetylene undergoes addition reaction with hydrogen in the presence of Ni catalyst and gives ethylene and ethane.

Question 4.
What is substitution reaction ? Explain any two substitutuin reactions of benzene.
Answer:
Substitution : When an atom or a group in molecule is substituted by another atom or group then the reaction is called electrophillic substitution reaction.
e.g. : 1) Friedal craft’s alkylation : C₆H₆ reacts with chlorine in the presence of AlCl₃ and form chloro benzene.

e.g. : 2) Nitration : C₆H₆ when heated with a mixture of cone. HNO₃ and cone. H₂SO₄, below 60° C, to give nitrobenzene.

Question 5.
What is dehydrohalogenation ? Write the equation for the formation of alkene from alkyl halide.
Answer:
Removal of hydrogen and halogen from the adjacent carbon atoms is called dehydrohalogenation.
Alkene from alkyl halide : When ethyl chloride is heated with alcoholic KOH, ethylene is formed due to dehydrohalogenation

Question 6.
Which type of compounds react with Ozone ? Explain with one example.
Answer:
1) Unsaturated hydrocarbons react with ozone and form addition compound called ozonide. On hydrolysis the ozonide gives carbonyl compound. This is called ozonolysis.

2) Unsaturated hydrocarbons usually react with ozone. Ozonolysis is used for the location of the double bond in unsaturated compounds like alkene, alkyne and benzene.
Example : Ethylene undergoes addition reaction with ozone and form ozonide. If, on hydrolysis, in presence of Zn dust, gives formaldehyde and H₂O₂.

Question 7.
Give two examples each for position and functional isomerism. [A.P. Mar. ’16] [Mar. ’13]
Answer:
Position isomerism : This type of isomerism arises due to the difference in the position of substituent group or in the position of multiple bond.
e.g.:

e.g.: 2. CH₃ – CH₂ – C = CH 1 – butyne
CH₃ – C = C – CH₃ 2 – butyne

Functional group isomerism : This type of isomerism arises in carbon compounds having the same molecular formula but with different functional groups.
e.g.: 1. CH₃ – CH₂ – CH₂ – OH 1 – proparal (C₃H₃O)
CH₃ – CH₂ – O – CH₃ methoxy ethane (C₃H₈O)
e.g.: 2. CH₃ – CH₂ – OH Ethanol
CH₃ – O – CH₃ methoxy methane

Question 8.
Explain the mechanism of halogenations of methane.
Answer:
Methane undergoes halogenation (with Cl₂) to form finally tetrachloro methane. This is a substitution reaction and takes place in presence of light.

Mechanism:
Halogenation process involves the free radical chain mechanism
This mechanism involves 3 steps
i) Initiation
ii) Propagation
iii) Termination

i) Initiation : The reaction initiated by the cleavage of chlorine molecule in presence of light. Cl – Cl bond is weaker than the C – C and C – H bonds

ii) Propagation : Chlorine free radical attacks the CH₄ molecule and generates methyl free radical.

The methyl radical formed attacks the 2^nd molecule of chlorine to form another chlorine free radical.

Propagation takes place in several steps as follows.

iii) Termination : The reaction terminated after sometime due to consumption of reactants. The following are the possible chain terminating steps

Question 9.
How is ethylene prepared from ethyl alcohol ? Write the equation.
Answer:
Ethylalcohol reacts with Conc.H₂SO₄ at 170°C to form Ethylene.

The above reaction is dehydration reaction. (H₂O removal takes place)

Question 10.
Explain the reactions of acetylene with : a) Na in NH₃ b) chromic acid. Write the equations and name the products.
Answer:
a) With Na in NH₃ : Acetylene reacts with Na in liquid NH₃ to give monosodium acetylide and disodium acetylide.


b) Chromic acid : Acetylene is oxidised to acetic acid by chromic acid.

Question 11.
Explain crystallisation and sublimation phenomenon which are used in the purification of organic compounds.
Answer:
Crystallisation: The principle used here is that the impurities are insoluble or soluble in the given solvent at any temperature, but the substance to be purified is sparingly soluble at room temperature but highly soluble at highest temp. (i.e.,) almost near the boiling temp. of the solvent.
This method is useful to purify solid compounds.

Sublimation : Some solid substances when heated pass directly into vapour state without melting. Those vapour on cooling form directly solid with out condensing to liquid. This phenomenon is called sublimation.

Impure compound is taken in a beaker covered with a watch glass and heated on an electric plate. The compound sublimates and solidifies on the lower surface of the watch glass. Impurities settle in the beaker. Pure compounds in separated by scratching the watch glass.
This is also a purification method of solids.

Question 12.
Describe solvent extraction method to purify a compound.
Answer:
Solvent extractiân : Suppose an organic compound ‘A’ is more soluble in an organic solvent that in water but present in aq. solution. The aq. solution is shoken with the organic solvent. ‘A’ goes into the organic solvent which is immiscible with water. The organic layer is separated and distilled to remove the liquid solvent. The compound remains with distillation flask.

Question 13.
Explain the estimation of phosphorus and sulphur in the given organic compounds.
Answer:
Estimation of phosphorus : Known mass of organic compound (‘a’ gm) is heated with fuming HNO₃ in a carius tube. ‘P’ is oxidised to H₃PO₄ acid. This acid is precipitated as ammonium phosphomolybdate (‘b’ gm) by adding ammonia and ammonium molybdate solutions.
% g of phosphorus = \(\frac{100}{a}\) × \(\frac{b}{1877}\) × 31 gm
Molecular mass of (NH₄)₃ PO₄. 12 MO O₃ (ammonium phosphomolybdate) = 1877
Estimation of sulphur : Known mass of organic compound (‘a’ gm) is heated with sodium peroxide in a carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate (big) by adding excess of aq. BaCl₂ solution. The ppt. is filtered, washed, dried and weighed.
% g of sulphur = \(\frac{100}{a}\) × \(\frac{b \times 32}{233}\) g
Molecular mass of BaSO₄ = 233

Question 14.
Explain addition of HBr to propene with the ionic mechanism.
Answer:

Electrophilic addition mechanism : (Ionic Mechanism)
Step (i) ; Formation of carbonium ion

Step (ii) : Stable carbonium ion is attacked by Br^⊖

Question 15.
What is the product formed when sodium propionoate is heated with soda lime, (imp)
Answer:
Sodium propionate is heated with sodalime to form ethane. This reaction is decarboxylation reaction. NaOH + CaO is called as Decarboxylating agent.

Long Answer Questions

Question 1.
Explain the classification of hydrocarbons.
Answer:
Classification of hydrocarbons : Hydrocarbons are classified as aliphatic hydrocarbons and aromatic hydrocarbons. Aliphatic hydrocarbons are again classified as open chain hydrocarbons and closed chain hydrocarbons i.e., cyclohydrocarbons. Both open chain and closed chain hydrocarbons are again classified as hydrocarbons containing C – C single bonds. > C = C < compounds, – C = C – hydrocarbons. Aromatic compounds are called benzenoids as they are related to benzene.

Question 2.
Write IUPAC names of the following compounds :
a)
b) CH₂ = CH – C C – CH₃
c) CH₃ – CH = C (CH₃)₂
d) CH₂ – CH₂ – CH = CH₂
e)

Answer:
a) 1, 3 – Buta diene
b) CH₂ = CH – C ≡ C – CH₃
Pent, 1 -ene 3 – yne
c) CH₃ – CH = C (CH₃)₂
2 – Methyl 2 – Butene
d)
e)

4 – ethyl Dec 1, 5, 8 – Triene

Question 3.
Describe two methods of preparation of ethane. Give any three reactions of ethane.
Answer:
Methods of preparation of ethane:
1. Decarboxylation:
Ethane is prepared by heating sodium propionate with sodalime.

2. Kolbe’s electrolysis:
Ethane is obtained by the electrolysis of potassium acetate solution.

H₂ gas evolves at cathode.
Ethane and CO₂ are formed at anode.

Chemical properties of ethane :
1. Halogenation:
Ethane reacts with chlorine in the presence of sunlight. Hydrogen atoms are substituted by halogen atoms successively. The final product is Hexachloro ethane.

2. Nitration:
Ethane reacts with nitric acid at 400° C and gives nitro ethane.

3. Pyrolysis:
When ethane is heated in the absence of oxygen it decomposes giving ethylene and hydrogen.

Question 4.
Write the structural formulas and IUPAC names for all possible isomers having the number of double (or) triple bond as indicated :
a) C₄H₈ (one double bond)
b) C₅H₈ (one triple bond)
c) C₅H₁₂ (No multiple bonds).
Answer:
a) The possible Isomers of C₄H₈ (one double bond)

b) The possible Isomers of C₅H₈ (one triple bond)

c) The possible Isomers of C₅H₁₂ (No multiple bonds)

Question 5.
Write chemical equations for combustion reaction of the following hydrocarbons.
a) Butane
b) Pentene
c) Hexyne.
Answer:
a) Combustion of Butane :
C₄H₁₀ + \(\frac{13}{2}\) O₂ → 4CO₂ + 5H₂O + Energy

b) Combustion of Pentene:
C₅H₁₀ + \(\frac{15}{2}\) O₂ → 5CO₂ + 5H₂O + Energy

c) Combustion of Hexyne:
C₆H₁₀ + \(\frac{17}{2}\) O₂ → 6CO₂ + 5H₂O + Energy

Question 6.
Addition HBr to propene yields 2 – bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1 – bromo propane. Explain and give mechanism.
Answer:

• Addition of HBr to propene yields 2 – Bromopropene. Here the reaction mechanism is electrophillic. addition mechanism.
• Addition of HBr to propene in presence of benzoyl peroxide the reaction proceeds through free radical mechanism.
  

Electrophilic addition mechanism : (Ionic Mechanism)
Step (i) : Formation of carbonium ion
HBr → H⁺ + Br^⊖

Question 7.
Describe two methods of preparation of ethylene. Give equation for the reactions of ethylene with the following,
a) Ozone
b) Hypohalous acid
c) Cold and dil. alkKMno₄
d) Heated with O₂ at high pressure
Answer:
Preparation of ethylene:
1. Dehydrohalogenation of alkyl halide: When ethylbromide is heated with alcoholic KOH, ethylene is formed.

2. Dehydration of ethanol: When ethanol is heated with cone. H₂SO₄ at 170°C, ethylene is obtained.

Chemical properties :
a) With O₃: Ethylene undergoes addition reaction with ozone and gives a cyclic compound called ozonide. It undergoes hydrolysis in the presence of zinc dust to give formaldehyde.

b) With HOCl : Ethylene reacts with hypochlorous acid and gives ethylene chlorohydrin.

c) Cold and dil. alk. KMnO₄: Pink coloured cold and dil. alkaline KMn04 solution is Bayer’s reagent. Ethylene decolourises Bayers reagent to give ethylene glycol. This is test for unsaturation.

d) Polymerisation : When Ethylene is heated in the presence of 02 at a temperature of 200° C and high pressure of 1500 – 2000 atmospheres gives polyethene.

Question 8.
How does ethylene react with the following reagents ? Give the chemical equations and names of the products formed in the reactions,
a) Hydrogen halide
b) Hydrogen
c) Bromine
d) Water
e) Oxygen in presence of Ag at 200° C.
Answer:
(a) Reaction with hydrogen halide : Ethylene reacts with Hydrogen halides to give ethyl halides.
H₂C = CH₂ + HX → CH₃ – CH₂ X
e.g.: H₂C – CH₂ + HCl → C₂H₅Cl Ethyl chloride

(b) With Hydrogen : Ethylene reacts with H₂ in presence of Ni, Pt or Pd to give ethane.

(c) With Bromine : Ethylene decolourises red coloured Br₂ in CCl₄ to give 1,2- dibromo ethane.

(d) With water : Ethylene reacts with dil. H₂SO₄ to give ethyl alcohol.

(e)With Sulphur monochloride : Ethylene reacts with S₂Cl₂ to give musturd gas.

(f) With O₂ at presence of Ag: Air oxidises ethylene to ethylene oxide known as epoxide in presence of Ag catalyst at 200 – 400°C.

Question 9.
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan 3-one. Write the reaction, structure of the products and alkene – A. Give the IUPAC name of alkene – A.
Answer:

Question 10.
An alkene ‘A1 contains three C – C, eight C – H bonds and one C = C bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44o . Write IUPAC name of ‘A’.
Answer:

IUPAC name of ‘A’ is 2 – Butene

Question 11.
Give two methods of preparation of acetylene. How does it react with water and Ozone ?
Answer:
Preparation of acetylene :
1. Dehydrohalogenation : Acetylene is obtained when dibromo ethane is heated with alcoholic KOH.

2. Acetylene is obtained by heating iodoform with silver powder.

3. Addition of water : Acetylene undergoes addition reaction with water in the presence of mercuric sulphate and sulphuric acid. Vinyl alcohol formed in the reaction undergoes rearrangement and gives acetaldehyde . [Mar. 06, June 04]

4. Reaction with ozone : Acetylene reacts with ozone to form an ozonide which on hydrolysis in presence of Zn forms glyoxal. [June 04]

Question 12.
How does acetylene react with the following reagents ? Give the corresponding equations and name the products formed in the reactions
a) Acetic acid
b) Water
c) Hydrogen
d) Halogens
e) Hydrogen halide
f) Ammonical AgNO₃ and Cl₂Cl₂ [T.S. Mar. 16]
Answer:
(a) Reaction with Acetic acid : Acetylene on treatment with acetic acid gives vinyl acetate in the 1st step and then gives ethylidene acetate. Hg²⁺ ions act as catalyst.

(b) Addition of water: Acetylene undergoes addition reaction with water in the presence of mercuric sulphate and sulphuric acid. Vinyl alcohol formed in the reaction undergoes rearrangement gives acetaldehyde.

(c) Addition of hydrogen : Acetylene undergoes addition reaction with hydrogen in the presence of Ni catalyst and gives ethylene and ethane.

(d) Addition of halogens : Acetylene undergoes addition reaction with halogens to give 1, 1, 2, 2 tetra haloethane. Dihalo alkene is formed in the first step which then adds on another molecule of halogen to give tetra halo ethane.

(e) With Hydrogen halide : Acetylene on addition with HCl gives Vinyl chloride and finally 1, 1 – dichloroethane.

(f) With Ammonical AgNO₃ solution : Acetylene gas is passed through ammonical AgNO₃ solution to form a white precipitate of silver acetylide. .

With Ammonical Cu₂Cl₂ solution : Acetylene gas is passed through Ammonical cuprous chloride solution to give a red precipitate of cuprous acetylide.

Question 13.
Describe any two methods of preparation of benzene with corresponding equations. Benzene does not behave like an alkenes why ? How do we get methyl benzene from benzene ?
Answer:
Preparation of Benzene:
1. When sodium benzoate distilled with sodalime. Benzene is formed.

2. Polymerisation of Acetylene : When Acetylene gas is passed through red hot Cu or Fe tubes, it polymerises and gives Benzene.

Constitution of Benzene : Benzene molecule cannot be represented by a single structure. It has the following Resonance structures.

The resonance energy of Benzene is 150.48 KJ/mole. Thus, it has more stability. Hence does not undergo addition reactions like alkenes. Benzene is an aromatic molecule.
Benzene to toluene (Friedal Craft’s Alkylation) :
Benzene reacts with methyl chloride in presence of AlCl₃ and forms methyl benzene (Toluene).

Question 14.
How do we get benzene from acetylene ? Give the corresponding equation. Explain the halogenations, alkylation, acylation, nitration and Sulphonation of benzene.
Answer:
Preparation of Benzene from acetylene: On passing acetylene gas through red hot iron tubes, it trimerises to give benzene.

Chemical Properties:
(1) Halogenation : Benzene reacts with chlorine in the presence of FeCl₃ or AlCl₃ to give chioro-benzene.

(2) Friedel – Craft’s Alkylation : Benzene reacts with alkyl halides in the presence of AlCl₃ to give alkyl benzene.

(3) Friedel – Craft’s Acylation : Benzene reacts with acetyl chloride in the presence of AlCl₃ to give acetophenone.

(4) Nitration : Benzene when heated with a mixture of cone. H₂SO₄ and cone. HNO₃ below 60°C to give Nitrobenzene.

(5) Sulphonation : With fuming H₂SO₄, benzene, reacts to form benzene sulphonic acid.

Question 15.
Explain the differences between structural isomers and stereo isomers.
Answer:
Structural isomerism : When the isomerism is due to difference in the arrangement of atoms within the molecule, without any reference to space, the phenomenon is known as Structural isomerism. In this type of isomerism, the isomers possess the same molecular formula, but different structural formula.
Structural isomerism is further classified into different types :

Stereoisomerism : The stereoisomers have the same molecular and structural formula, but differ in the arrangement of atoms or groups in space. Thus the phenomenon exhibited by two or more compounds with the same molecular and structural formulae, but different spatial arrangement of atoms or groups. The spatial arrangement of atoms or groups is also referred to as configuration of the molecule. It is further classified into three types.

Question 16.
What is the differences between conformation and configuration in open chain molecules.
Answer:
Configurational Isomerism : (Optical and geometrical isomerism): These are the stereo – isomers which are discrete, stable and isolable substances. They cannot be inter-converted into one another without making and breaking of new bonds. These isomers cannot be superimposed on each other.
These isomers are further classified as enantiomers, diastereomers and geometrical isomers.

Conformational Isomers : These are the stereo – isomers that easily convert into one another by the rotation around “C – C” bonds. These are in dynamic equilibrium with one another and cannot be separated under ordinary conditions.
This type of isomerism is in alkanes such as n – butane.

Question 17.
What do you understand about geometrical isomerism ? Explain the geometrical isomers of 2 – butene.
Answer:
Geometrical isomerism (Cis – trans isomerism): The isomers which possess the same structural formula but differ in spatial arrangement of the groups around the double bond are called geometrical isomers and the phenomenon is known as geometrical isomerism.
When the same groups lie on the same side of double bond, it is Cis – isomer, while when the same groups lie on opposite side of double bond, the isomer is trans.

Question 18.
Explain the method of writing E – Z configurations for geometrical isomers taking CHCl = CFBr as your example.
Answer:
E – Z configurations for geometrical isomers : The following procedure is followed in specifying the configurations of compounds.
i) Arrange the atoms / groups attached to each doubly bonded carbon in the order of their atomic numbers.

ii) Choose the atom/group higher priority on each doubly bonded carbon. If the atoms/groups of higher priority on each carbon are on the same side of the molecule, the letter “Z” is used to denote the configuration of such isomer. When the atoms/groups of higher priority on each carbon are on the opposite sides of the molecule, the letter ‘E’ is prefixed before the name to indicate configuration.
Example : CHCl = CFBr
Among H and Cl, ₁₇Cl gets more priority than ₁H
Among H and Br, ₃₅Br gets more priority than ₉F

Question 19.
If an alkene contains on carbons at double bond Cl. Br and – CH₂ – CH₂ – OH – CH (CH₃)₂, Write the E and Z configurations of it.
Answer:
From the given data, the structural formulae of the molecule is

Question 20.
Write a note on :
a) Distillation
b) Fractional distillation
c) Distillation under reduced pressure
d) Steam distillation.
Answer:
a) Distillation : This process is useful for the purification of liquids contaminated with nonvolatile impurities. The impure liquid is boiled in a distillation flask and the vapours are condensed and collected in a receiver. This method can also be used to separate liquids if only their boiling points differ by above 40°C. However, in the case of liquids that have boiling point difference less than 40°C fractional distillation method is used.

b) Fractional distillation : Fractional distillation for liquids that have B.Pt difference less than 40°C. The technique here is that vapours of liquid mixture. When pass through long fractionating column vapours of the liquid with high B.Pt condense and those with low B.Pt pass over through the condenser get condensed and collected in the receiver.
Here, long tubes having different shapes and designs to fit for particular requirements are used.

These are called fractionating columns. The liquid mixture is taken in a distillation flask i.e., fitted with a fractionating column at its mouth. At the upper end of the column, there is a provision to connect it to the water condenser.

c) Distillation under reduced pressure : This method is useful to purify liquids that have very high boiling points and those which decompose at or below their boiling points. If the external pressure is reduced the liquid boils at lower temperature than its normal boiling point without decomposition.

d) Steam distillation : Here liquids which are immiscrible in water passes high boiling point and steam volatile are purified. In this method steam is passed into the hot liquid that is to be purified. The mixture of steam and vapour of volatile organic compound come out. This is because of the sum of the vapour pressures of steam and the liquid to be purified become equal to the atmospheric pressure. They are passed through the condenser and condensed and finally collected in the receiver. The water layer and the organic liquid layer are separated using a separating funnel.

Question 21.
Write a brief note on chromatography.
Explain the following :
a) Column chromatography
b) Thin layer chromatography
c) Partion chromatography.
Answer:
Chromatography has been developed as a method of separation components of a mixture generally between the stationary phase and a mobile phase.
Chromatography involves the three steps given below :
a) Adsorption and retention of a mixture of substances on the stationary phase and separation of adsorbed substances by the mobile phase to different distances on the stationary phase.

b) Recovery of the substances separated by a continuous flow of the mobile phase (known as elution) and

c) Qualitative and quantitative analysis of the eluted substances.

Classification :
Two general chromatography techniques are discussed below. They are :

1. adsorption chromatography
2. partition chromatography.

Adsorption chromatography is based on the adsorption of different compounds on an adsorbent to different degrees. Generally used adsorbents are silica gel and alumina. A mobile phase is allowed to move over a stationary phase, the adsorbent. The components of the mixture move to different distances over the stationary phase.

Differential adsorption principle is used in
a) column chromatography and
b) thin layer chromatography.

Column chromatography : In the column chromatography the components of a mixture are separated by a column of adsorbent packed in a glass tube. The column is fitted with a stopcock at its lower end. The mixture to be adsorbed on the adsorbent is placed at the top of the stationary phase. A suitable eluant, either a single solvent or a mixture of solvents is allowed to flow down the column slowly. Depending on the degree to which the compounds are adsorbed the components are separated. The most readily absorbed substances are retained near the top and other come down accordingly to various distances.

Thin layer chromatography (TLC): This also involves adsorption differences. Here the adsorbent, say silica gel or alumina is coated over a glass plate of suitable size in thin layer. The plate is called TLC plate or chromoplate. The solution of the mixture to be separated is applied as a small spot at about 2 cm from the bottom of the plate. The plate is then kept in a closed jar containing the eluant. As the eluant rises up the plate, the components of the mixture move up along with the eluant to various distances depending on their degree of adsorption.

The relative adsorption of a component of the mixture is expressed in terms of its RETARDATION FACTOR (R_f) value.
R_f = \(\frac{\text { Distance moved by the subs tance from base line }(\mathrm{X})}{\text { Dis tance moved by the solvent from base line }(\mathrm{Y})}\)
Partition chromatography : This is based on continuous differential partitioning of components of a mixture between the Stationary phase and the mobile phase. In paper chromatography, for example, a special paper called chromatography paper contains water trapped in it which acts as the stationary phase. The chromatography paper spotted with the solution of the mixture at the base is suspended in a suitable solvent or a mixture of solvents.

This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and moves over the spot. The paper selectively retains different components as per their differing partition in mobile and stationary phase. The paper strip so developed is known as chromatogram. The spots of the separated coloured compounds are detected and for colourless compounds other methods like spraying suitable reagent, are used.

Question 22.
Explain the estimation of nitrogen of an organic compound by
a) Duma’s method
b) Kjeldahl’s method.
(or)
Describe Duma’s and Kjeldahl’s method for the estimation of Nitrogen.
Answer:
Estimation of nitrogen : There are two methods to estimate nitrogen in the given organic compound known as
i) Duma’s method and
ii) Kjeldahed’s method.

i) Duma’s method : In this method a known weight of organic compound is heated strongly with coarse cupric oxide. Carbon and hydrogen get oxidised to carbon dioxide and water vapour respectively. Nitrogen if present is converted to nitrogen gas. Even if some nitrogen is converted to its oxides. They are reduced by hot copper gauze to nitrogen. The product gases are collected over a solution of potassium hydroxide. CO₂ is absorbed by KOH solution. Nitrogen is collected over potassium hydroxide solution and its volume is found out.

Suppose that ‘a’ g of organic compound gives V₁ ml of N₂ at room temperature TK. If atmosphere pressure is P and aqueous tension at TK is P, then the pressure of nitrogen gas at TK is (P – p). Let (P – p) = P₁ Reduce the volume of nitrogen to standard temperature 273 K and standard pressure 760 mm.
Volume of nitrogen at STP is v = \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{273}{760}\)
28 g of nitrogen at STP occupies 22400 ml
? g. of nitrogen at STP occupies Vml. of nitrogen.
28 × \(\frac{V}{22400}\)g
‘a’ g. of organic compound \(\frac{28 \times V}{22400}\) g of nitrogen.
100 g organic compound has ? of nitrogen = \(\frac{100}{a} \times \frac{28 \times V}{22400}\) g

ii) Kjeldah’s method : This is another method to estimate nitrogen. In this method, the compound is heated with concentrated sulphuric acid in the presence of small amount of CuSO₄.
Nitrogen is quantitatively converted into ammonium sulphate. The contents of the flask are transferred to another flask and heated with excess of sodium hydroxide solution to liberate ammonia gas. Ammonia gas so liberated is passed and absorbed in a known volume of known concentrated sulphuric acid that is relatively more in amount than that is required to neutralise NH₃ gas. Now, the excess of acid remained after the neutralisation by NH₃ is titrated against a standard solution of alkali. From the above, the amount of H₂SO₄ used to neutralise NH₃ is calculated. From this, the mass of ammonia formed is calculated and from that percentage of nitrogen is calculated.
Organic compound + H₂SO₄ →(NH₄)₂ SO₄
(NH₄)₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O + 2 NH₃
2 NH₃ + H₂SO₄ → (NH₄)₂SO₄
Calculation :
Let the mass of organic compound taken be ‘a’ g.
Let the volume of H₂SO₄ initially taken be ‘V_ml‘ and its molarity M.
After passing the NH₃ gas into the above acid, if the remaining acid is titrated with M molar NaOH and it consumes V₁ ml. of NaOH for complete neutralisation, then from the formula
\(\frac{\mathrm{MV}_1}{\mathrm{n}_1}\) (NaOH) = \(\frac{\mathrm{MV}_2}{\mathrm{n}_2}\) (H₂SO₄)
From the stoichiometric equation
2 NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
n₁ = number of moles of NaOH = 2; n₂ = number of moles of H₂SO₄ = 1
\(\frac{\mathrm{MV}_1}{2}\) = \(\frac{\mathrm{MV}_2}{1}\) or V₂ = \(\frac{\mathrm{V}_1}{2}\) ml
Therefore, the volume of H₂SO₄ neutralised by NH₃ is [V – \(\frac{\mathrm{V}_1}{2}\)] ml
(or) it is equal to 2 [V – \(\frac{\mathrm{V}_1}{2}\)] ml. of M molar NH₃ solution.
1000 ml. of 1M NH₃ solution contains 17g. of NH₃ or 14g. of N₂
2[V – \(\frac{\mathrm{V}_1}{2}\)] ml ml. of ‘M’ NH₃ solution contains
\(\frac{14 \times M \times 2\left[V-\frac{V_1}{2}\right]}{1000}\) g. of nitrogen.
Percentage of nitrogen = \(\frac{14 \times M \times 2\left[V-\frac{V_1}{2}\right]}{1000} \times \frac{100}{a}\)

Question 23.
Explain inductive effect with a suitable example.
Answer:
Inductive effect: The electron donating or electron with drawing effect of a groups an atom that is transmitted by the polarisation of electrons in σ bonds is called inductive effect.
Illustration : Consider the molecule CH₃ – CH₂ – CH₂ – Cl. There is a ‘σ’ covalent bond between carbon atom and chlorine atom. The electron pair between them is not equally shared. The more electronegative chlorine atom tends to attract the shared pair more towards itself. Due to this, the electron density tends to be greater nearer chlorine atom than carbon atom. It is generally represented
as But, carbon atom bonded to chlorine atom is itself attached to other carbon atoms. Therefore, the effect can be transitted further
“Inductive effect is defined as the polarisation of a bond caused by the polarization of adjacent σ bond”.
(-I) effect i.e., electron with drawing effect is in the order
⁺NH₃ > NO₂ > CN > SO₃H > CHO > CO > COOH > COCl > COOR > CONH₂ > F > G > Br > I > OH > OR > NH₂ > C₆H₅ > H.
+ I effect is – \(\bar{N} R\) > – \(\overline{O}\) ; – Se > S > -O ; – C(CH₃)₃ > – CH(CH₃)₂ > – CH₂CH₃ > – CH₃.

Question 24.
Write a note on mesomeric effect.
Answer:
Mesomeric effect:
“The electron pair displacement caused by an atom or group along a chain by a conjugative mechanism is called the mesomeric effect of that atom or group”.
Salient features of the mesomeric effect:
i) Permanent effect operating in the ground state of the molecule.
ii) Lone pairs and n electrons are involved and operate through conjugative mechanism of electron displacement.
iii) It influences the physical properties, reaction rates etc.

Groups which tend to increase the electron density of the rest of the molecule are said to have (+M) effect. Such groups tend to posses lone pairs of electrons.
e.g. :
Groups that decrease the electron density of the rest of the molecule are said to have (-M) effect. Unsaturated groups having polar character have – M effect.
. Here C = O group decreases the electron density of the remaining molecule. It has – M effect.
+ M effect is – F > – Cl > – Br > – I;
-NR₂ > OR > F;
– NH₂ > – OH > – F;
– OR > – SR > SeR
– O > – OR
– M effect is = O > = NR > = CR₂
= NR₂< =NR ≡ N > ≡ CR

Question 25.
Describe resonance effect with one example.
Answer:
Resonance effect: It is the polarity produced in a molecule by the interactions of two n bonds or between a π bond and a lone pair of electrons present on adjacent atoms. This effect is transmitted through the chain.
If the transfer of electron is away from the atoms or substituent group attached to the conjugated system, then the molecule gets some of its positions high electron density as in aniline and it is given (+ R). If the shift of electrons are towards the atom or substituent group it is (- R) as in nitrobenzene.

Groups showing (+ R) are X, – OH, – OR, – COOR, – NH₂, – NHR, – NR₂, – NHCOR etc.
(- R) effect are – COOH, CHO, > C = O, – CN, – NO₂.

Question 26.
Explain how many types of organic reactions are possible.
Answer:
Organic reactions are mainly classified into four types known as
i) Addition rections.
ii) Substitution reactions
iii) Elimination reactions
iv) Molecular rearrangements.

i) Addition reactions : In these reactions the reagent and the substract combine together to give a single product.
e.g.:
Depending on the reagent added in the slow rate determining step, addition reactions are again a) Electrophilic classified addition reactions, b) Nucleophilic addition reactions c) Free radical addition reactions.

ii) Substitution reactions : In these reactions on atom or a group of the substrate species is replaced by another atom or group. These are again classified as a) Electrophilic substitution b) Nucleophilic substitution and c) Free radical substitution reactions on the basis of the reagent involved in the rate determining step.
e.g.-: \(\overline{\mathrm{O}}\)H_(aq) + R – X → HO – R + X_(aq)

iii) Elimination reactions : In these reactions two or more atoms or groups of an organic substrate are removed to form multiple bonds.

iv) Molecular rearrangements : Here one organic species (generally less stable) rearranges to other species (generally more stable). For example. Fries rearrangement.

Question 27.
Write the possible conformations of ethane and explain which is more stable.
Answer:
Conformation of ethane : The conformational isomers of a given alkane are obtained by rotation about C – C bond and they are represented by Newman projections or line – Wedge or Sawhorse projections. Newman projections and energy diagrams for the various conformations of ethane are given below.

Staggered (S) occurs at Dihedral angle 60°, 180°, 300°
Rotation about the carbon – carbon bond in ethane is though very rapid, not completely free. Two conformations of ethane known as staggered conformation (S) and the eclipsed conformation (E) are very important though infinite number of confomation are possible.

The C – H bonds in the staggered conformation are arranged so that each one bisects the angle defined by two C – H bonds on adjacent carbon. In the eclipsed conformation each C – H bond is aligned with a C – H bond on adjacent carbon.

In the staggered one the distance between the hydrogen nuclei is 2.55 A° but in eclipsed 2.29 A°. The staggered and elipsed conformations are interconvertible by rotation of one carbon with respect to the other around the a bond that connects them. Different conformations of the same molecule are also called conformers or rotamers.
Stability :
Staggered (S) form is more stable than eclipsed. Because in case of eclipsed form electronic repulsions are high and in case of staggered form electronic repulsions are minimum.

Question 28.
Explain aromatic electrophillic substitution reactions of benzene.
Answer:
1) Halogenation : Benzene reacts with bromine or chlorine in the presence of Lewis acids like FeCl₃, AlCl₃ etc., to give corresponding halobenzene.
e.g. :

Similarly with bromine, bromobenzene is formed.

2) Nitration : Benzene undergoes nitration when heated with a mixture of 1 : 1 (by volume) concentrated nitric acid and concentrated sulphuric acid (nitration mixture) at a temperature below 60° C.

3) Benzene reacts with fuming sulphuric acid (oleum) and gives benzene sulphonic acid.

4) Friedel – Craft’s alkylation and acylation : Benzene reacts with alkyl halides and acylh in the presence of Lewis acids (AlCl₃, FeCl₃) and gives alkyl benzenes and acyl benzenes.

Benzene reacts with acetyl chloride in the presence of anhydrous aluminium, chloride and acetophenone.

General Mechanisms : Electrophilic substitution reaction (S_E) proceeds in two steps as
I. Generation of electrophile
II. a) Formation of carbocation intermediate
b) Removal of proton from carbocation intermediate.

I. Generation of electrophile E⁺: In the reactions halogenation, alkylation and acylatio benzene, anhydrous AlCl₃, the Lewis acid produces electrophile X⁺ say Cl⁺, R⁺ and RCO⁺ by reacting the reagent chlorine, alkyl halide and acylhalide respectively.

II. a) Formation of carbocation : Electrophile generated above attacks one of the benz carbons to change it to sp3 hybridised. The carbocation (Arrhenium ion) is stabilised through resonal.

Sigma complex loses aromatic character due to delocalisation of electrons stopping at sp³ carb

b) Losing of proton : To regain aromatic character the \(\stackrel{+}{\mathrm{C}}\) loses one proton to sp³ carbon on attack
of (AlCl₄)^– in case of halogenation, alkylation and acylation and HSO₄^– in case of nitration.

Question 29.
Explain electrophilic addition reactions of ethylene with mechanism.
Answer:
Electrophilic addition reactions of ethylene :
1. Addition of Hydrogen : Ethylene react with hydrozen in the presence of Pt, Pd or Ni catalyst to form ethane.

2. Addition of halogens : Halogens (Cl₂ or Br₂) react with ethylene in the presence of an inert solvent like CCl₄ to form dihalo derivative

Mechanism :

3. Addition of hydrogen halides : Ethylene reacts with HBr to form ethyl bromide

4. Addition of water : In presence of few drops of cone. H₂SO₄, water adds to ethylene to form alcohol.

Question 30.
With the help of mechanism explain free radical halogenations of alkanes.
Answer:
Methane undergoes halogenation (with Cl₂) to form finally tetrachloro methane. This is a substitution reaction and takes place in presence of light.

Mechanism :

• Halogenation process involves the free radical chain mechanism
• This mechanism involves 3 steps
  i) Initiation
  ii) Propagation
  iii) Termination

i) Initiation : The reaction initiated by the cleavage of chlorine molecule in presence of light. Cl – Cl bond is weaker than the C – C and C – H bonds

ii) Propagation : Chlorine free radical attacks the CH₄ molecule and generates methyl free radical.

The methyl radical formed attacks the 2^nd molecule of chlorine to form another chlorine free radical.

iii) Termination : The reaction terminated after sometime due to consumption of reactants.
The following are the possible chain terminating steps

Question 31.
Discuss Markownikov’s rule and Kharash effect.
Answer:
Statement of Markownikoff’s rule : The rule states that when an unsymmetrical reagent adds to a double bond, the positive part of the adding reagent attaches itself to a carbon of the double bond so as to yield the more stable carbocation as an intermediate.

Mechanism : A pair of electrons from the double bond attacks the electrophilic HX to produce an achiral trigonal planar carbocation intermediate. Thehalideion X then adds to either face of the positively charged carbon forming alkyl halide product.

Anti Mark Kownikoffs addition or peroxide effect or Kharasch effect : In the presence of peroxide (R – O – O – R) the addition of HBr to unsymmetrical alkene like propene takes place against Markowni- koff’s rule. As per Anti Markownkoff’s rule, the addition of HBr to an unsymmetrical alkene like propene takes place in such a way that the hydrogen atom becomes attached to the carbon atom with the fewer hydrogen atoms.


The 2° free radical is more stable than 1°. Therefore. 1 – bromopropane is major product.

Question 32.
How would you convert benzene in to following compounds ?
a) Chioro benzene
b) Toluene
c) p – nitrotoluene.
Answer:
a) Benzene reacts with chlorine inpresence of FeCl₃ to form chlorobenzene.

b) Benzene reacts with methyl chloride in presence of AlCl₃ to form toluene (Friedal Craft’s alkylation)

c) From benzene p – nitro Toluene is obtained as follows.

Question 33.
Why is wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms ? Illustrate with one example.
Answer:
Wurtz reaction Alkyl halides reacts with sodium metal in presence of dry ether to form higher alkanes.
R – X + 2Na + R – X R – R + 2NaX

• Wurtz reaction is preferred to prepare even no.of Carbons containing alkanes but not for odd no.of Carbons containing Alkanes.
• For getting odd no.of Carbons containing alkanes we have to consider two different alkyl halides.

The products obtained are in mixture form, So the percentage of the desired hydrocarbon is less, i.e., yield of the desired product is low.
Methane can not be prepared by this reaction.

Question 34.
Write the equations involved in the detection of Nitrogen, Halogens and sulphur in organic compounds.
Answer:
Lassaigne’s Test : –

• A small dry sodium piece is taken in a fusion tube and heated gently till the sodium melts.
• To this molten sodium a small amount of organic compound is added and heated until it changes red hot.
•  This red hot tube is plunged in a china dish containing water and this content is boiled, cooled and filtered.
• The obtained filtrate is known as Lassaigne’s extract.
• This test is used to detect the elements Nitrogen, Sulphur, halogens etc..

Detection of Nitrogen : –
The Lassaigne’s extract is made alkaline by adding few drops of dol. NaOH. to this freshly prepared FeSO₄ solution is added and warmed then few drops of FeCl₃ are added followed by acidification with Cone. HCl (or) H₂SO₄. A bluish green colouration indicates the nitrogen.
Na + C + N →NaCN
2NaCN + FeSO₄ → Na₂SO₄ + Fe (CN)₂
Fe(CN)₂ + 4NaCN → Na₄ [Fe(CN)₆]
3Na₄ [Fe(CN)₆] + 4FeCl₃ → Fe₄ [Fe(CN)₆]₃ + 12 NaCl (Prussian blue colouration)

b) Detection of Halogens : The sodium fusion extract is acidified with HNO₃ and is treated with AgNO₃ solution.
Ag + X → AgX

• White PPt indicates Cl^Θ ion
• Pale yellow PPt indicates Br^Θ ion
• Yellow PPt indicates I^Θ ion

c) Detection of Sulphur : To the sodium extract freshly prepared sodium nitroprusside solution. A deep violet colouration takes place.
S^-2 + [Fe(CN)₅NO]²⁺ → [Fe(CN)₅ NOS]^4- Violet

Question 35.
Explain how carbon and Hydrogen are quantitatively determined in an organic compound.
Answer:
Estimation of carbon and hydrogen : Both the elements are estimated in the same experiment simultaneously. A known weight of the organic substance is taken and completely burnt in excess of air and copper (II) oxide carbon changes to CO₂ and Hydrogen to H₂O.
C_xH_y + (x + \(\frac{y}{4}\))O₂ → x CO₂ + (\(\frac{y}{2}\)) H₂O
CO₂ and H₂O SO obtained are passed through already weighed ‘U‘ – tubes containing anhy, CaCl₂ and caustic potash respectively. The increased weigh is of these two tubes give the weigh of H₂O formed and weight of CO₂ formed.
Suppose ‘a’ g of organic compound on combustion gives ‘b’ grams of water vapour and. ‘C g of CO₂. Then
% g of Carbon = \(\frac{12}{44}\) × \(\frac{100}{a}\) × C g
% g of Hydrogen = \(\frac{2}{18}\) × \(\frac{100}{a}\) × b g

Question 36.
How do you determine sulphur, phosphorous and Oxygen are determined quantitatively in an organic compound ?
Answer:
Estimation of phosphorus : Known mass of organic compound (‘a’ gm) is heated with fuming HNO₃ in a carius tube. ‘P’ is oxidised to H₃PO₄ acid. This acid is precipitated as ammonium phosphomolybdate (‘b’ gm) by adding ammonia and ammonium molybdate solutions.
% g of phosphorus = \(\frac{100}{a}\) × \(\frac{b}{1877}\) × 31 gm
Molecular mass of (NH₄)₃ PO₄. 12 MO O₃ (ammonium phosphomoly bdate) = 1877
Estimation of sulphur : Known mass of organic compound (‘a’ gm) is heated with sodium peroxide in a carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate (big) by adding excess of aq. BaCl₂ solution. The ppt. is filtered, washed, dried and weighed.
%g of sulphur = \(\frac{100}{a}\) × \(\frac{b \times 32}{233}\) × g
Molecular mass of BaSO₄ = 233
Estimation of Oxygen : The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows :

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I₂O₅) when carbon monoxide is oxidised to carbon dioxide producing iodine.

On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.
Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.
Let the mass of organic compound taken be mg
Mass of carbon dioxide produced be m₁ g
∴ m₁ g carbon dioxide is obtained from \(\frac{32 \times m_1}{88}\) g O₂
∴ Percentage of oxygen = \(\frac{32 \times m_1 \times 100}{88 \times m}\)%

Question 37.
Explain Carius method for the determination of Halogens quantitatively in an organic compound.
Answer:
Estimation of halogens : (Carius method) Known mass of organic compound is heated with framing nitric acid in the presence of AgNO₃ in a hard glass tube called carius tube. Carbon and hydrogen of the compound are oxidised to CO₂ and H₂O. Halogens forms silver halide. So obtained silver halide (Agx) is filtered, washed, dried and weighed.
% g of halogen = \(\frac{100 \times b \times \text { atomic mass of ‘ } x \text { ‘ }}{a \times \text { molecular mass of } \mathrm{Agx}}\)
Where, a = mass of organic compound, b = mass of Agx formed
x = halogen atom.

Question 38.
What is carcinogenicity ? Explain with two examples.
Answer:
Benzene and several polynuclear hydrocarbons like 1, 3 – benzanthracene, 3 – methyl cholanthrene, 1, 2- benzpyrene etc., are toxic and said to carcinogenic (cancer producing).
Most of these are formed due to incomplete combustion of tobacco, coal, petroleum etc.
They undergo various chemical changes in human body and finally damage DNAto cause cancer.

Solved Problems

Question 1.
How many a and it bonds are present in each of the following molecules?
(a) HC ≡ CCH = CHCH₃
(b) CH₂ = C = CHCH₃
Solution:
(a) σ_C-C : 4; σ_C-H : 6; π_C=C: 1 ; π C ≡ C : 2
(b) σ_C-C : 3; σ_C-H : 6; π_C=C : 2.

Question 2.
What is the type of hybridisation of each carbon in the following compounds?
(a) CH₃Cl
(b) (CH₃)₂CO
(c) CH₃CN,
(d) HCONH₂
(e) CH₃CH = CHCN
Solution:
(a) sp³,
(b) sp³, sp²,
(c) sp³, sp,
(d) sp²,
(e) sp³, sp², sp², sp

Question 3.
Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.
(a)H₂C = O,
(b) CH₃F,
(c) HC ≡ N.
Solution:
(a) sp² hybridised carbon, trigonal planar;
(b) sp³ hybridised carbon, tetrahedral;
(c) sp hvbridised carbon, linear.

Question 4.
Expand each of the following condensed formulas into their complete structural formulas.
(a)CH³CH²COCH²CH³
(b) CH³CH = CH(CH²)³CH³
Solution:

Question 5.
For each of the following compounds, write a condensed formula and also their bond – line formula.
(a) HOCH₃CH₂CH₂CH(CH₃)CH(CH₃)CH₃
(b)

Solution:
Condensed formula:
(a) HO(CH₂)₃CH(CH₃)CH(CH₃)₂
(b) HOCH(CN)₂
Bond-line formula :

Question 6.
Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen.

Solution:

Question 7.
Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect.

Solution:
(a) Lowest locant number, 2, 5, 6 is lower than 3, 5, 7,
(b) substituents are in equivalent position; lower number is given to the one that comes first in the name according to alphabetical order.

Question 8.
Write the IUPAC names of the compounds i-iv from their given structures.

Solution:

• The functional group present is an alcohol (OH). Hence the suffix is ‘-01’.
• The longest chain containing – OH has eight carbon atoms. Hence the corresponding saturated hydrocarbon is octane.

The – OH is on carbon atom 3. In addition, a methyl group is attached at 6th carbon. Hence, the systematic name of this compound is 6-Methyloctan-3-ol.

i)

Solution:
The functional group present is ketone (>C = 0), hence suffix ‘-one’. Presence of two keto groups is indicated by ‘di’, hence suffix becomes ‘dione’. The two keto groups are at carbons 2 and 4. The longest chain contains 6 carbon atoms, hence, parent hydrocarbon is hexane. Thus, the systematic name is Hexane – 2, 4 – dione.

ii)

Solution:
Here, two functional groups namely ketone and carboxylic acid are present. The principal functional group is the cat ooxyuc acid group; hence the parent chain will be suffixed with ‘oic’ acid. Numbering of the chain starts from carbon of – COOH functional group. The keto group in the chain at carbon 5 is indicated by ‘oxo’. The longest chain including the principal functional group has 6 carbon- atoms; hence the parent hydrocarbon is hexane. The compound is, therefore, named as 5 Oxo- hexanoic acid.

iii)

Solution:
The two C=C functional groups are present at carbon atoms 1 and 3, while the C = C functional group is present at carbon 5. These groups are indicated by suffixes ‘diene’ and ‘yne’ respectively. The longest chain containing the functional groups has 6 carbon atoms; hence the parent hydrocarbon is hexane. The name of compound, therefore, is Hexa-1, 3-dien-5-yne.

Question 9.
Derive the structure of
(i) 2-Chlorohex- ane,
(ii) Pent-4-en-2-ol,
(iii) 3- Nitro-cydohexerte,
(iv) Cyclohex-2-en-l-ol,
(v) 6-Hydroxy-heptanal.
Solution:
(i) ‘hexane’ indicates the presence of 6 carbon atoms in the chain The functional group chloro is present at carbon 2. Hence, the structure of the compound is CH₃CH₂CH₂CH₂CH(Cl)CH₃.

(ii) ‘pent’ indicates that parent hydrocarbon contains 5 carbon atoms in the chain, ‘en’ and ‘ol’ correspond to the functional groups C=C and -OH at carbon atoms 4 and 2 respectively. Thus, the structure is CH₂ = CHCH₂CH (OH)CH₃.

(iii) Six membered ring containing a carbon- carbon double bond is implied by cyclo-hexene, which is numbered as shown in (I). The prefix 3-nitro means that a nitro group is present on C-3. Thus, complete structural formula of the compound is (II). Double bond is suffixed functional group whereas -NO₂ is prefixed functional group therefore double bond gets preference over -NO² group:

(iv) ‘1 -0l’ means that a -OH group is present at C-1. OH is suffixed functional, group and gets preference over C = C bond. Thus the structure is as shown in (II) :

(v) ‘heptanal’ indicates the compound to be an aldehyde containing 7 carbon atoms in the parent chain. The ‘6-hydroxy’ indicates that -OH group is present at carbon 6. Thus, the structural formula of the compound is : CH₃CH(OH)CH₂CH₂CH₂CH₂CHO. Carbon atom of – CHO group is included while numbering the carbon chain.

Question 10.
Write the structural formula of :
(a) o-Ethylanisole
(b)p-Nitroaniline,
(c) 2,3 – Dibromo -1 – phenylpentane,
(d) 4-Ethyl – 1 – fluoro-2-nitrobenzene.
Solution:

Question 11.
Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
(a) CH₃ – SCH₃,
(b) CH₃ – CN,
(c) CH₃ – Cu
Solution:

Question 12.
Giving justification, categorise the fol-lowing molecules/ions as nucleophile or electrophile :
HS^–, BF₃, C₂H₅O^–, (CH₃)₃ N :,

Solution:
Nucleophiles: HS^–, C₂H₅O^–, (CH₃)₃ N:, H²N^– These species have unshared pair of electrons, which can be donated and shared with an electrophile.
Electrophiles:
Reactive sites have only six valence electrons; can accept electron pair from a nucleophile.

Question 13.
Identify electrophilic centre in the fol-lowing: CH₃CH = O, CH₃CN, CH₃I.
Solution:
Among , the starred carbon atoms are electrophilic centers as they will have partial positive charge due to polarity of the bond.

Question 14.
Which bond is more polar in the following pairs of molecules:
(a) H₃C – H, H₃C – Br
(b) H₃C – NH₂, H₃C – OH
(c) H₃C – OH, H₃C – SH
Solution:
(a) C – Br, since Br is more electronegative than H, (b) C – O, (c) C – O

Question 15.
In which C – C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least ?
Solution:
Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence, the effect is least in the bond between . carbon-3 and hydrogen.

Question 16.
Write resonance structures of CH₃COO^– and show the movement of electrons by curved arrows.
Solution:
First, write the structure and put unshared pairs of valence electrons on appropriate atoms. Then draw the arrows one at a time moving the electrons to get the other structures.

Question 17.
Write resonance structures of CH₂ = CH – CHO. Indicate relative stability of the contributing structures.
Solution:
[I: Most stable, more number of covalent

bonds, each carbon and oxygen atom has an octet and no separation of opposite charge II: negative charge on more electronegative atom and positive charge on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable].

Question 18.
Explain why the following two structures, I and 11 cannot be the major contributors to the real structure of CH₃COOCH₃.

Solution:
The two structures are less important contributors as they involve charge separation. Additionally, structure I contains a carbon atom with an incomplete octet.

Question 19.
Explain why is more stable than and is the least stable cation.
Solution:
HyperconjugatiOn interaction in is greater than in as the has nine C – H bonds. In , vacant p orbital is perpendicular to the plane in which C – H bonds lie; hence cannot overlap with it. Thus, laoks hyper conjugative stability.

Question 20.
On complete combustion, 0.246 g of an organic compound gave 0.1989/ of carbon dioxide and 0.10149 of water. Determine the percentage composition of carbon and hydrogen in the compound.
Solution:
Percentage of carbon = \(\frac{12 \times 0.198 \times 100}{44 \times 0.246}\)
= 21 .95%
Percentage of hydrogen = \(\frac{2 \times 0.1014 \times 100}{18 \times 0.246}\)
= 4.58%

Question 21.
In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 5OmL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K = 15 mm)
Solution:
Volume of nitrogen collected at 300K and 715mm pressure is 50 mL
Actual pressure = 715 – 15 = 700 mm
Volume of nitrogen at STP = \(\frac{273 \times 700 \times 50}{300 \times 760}\)
= 41.9 ML
22,400 ml of N₂ at STP weighs = 28 g
41.9 mL of nitrogen weighs = \(\frac{28 \times 41.9}{22400}\) g
Percentage of nitrogen = \(\frac{28 \times 41.9 \times 100}{22400 \times 0.3}\)
= 17.46%

Question 22.
During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄. Find out the percentage of nitrogen in the compound?
Solution:
1 M of 10 mL H₂SO₄ = 1 M of 20 mL NH₃ 1000 mL of 1 M ammonia contains 14 g nitrogen
20 ml of 1 M ammonia contains \(\frac{14 \times 20}{1000}\) g nitrogen
Percentage of nitrogen = \(\frac{14 \times 20 \times 100}{1000 \times 0.5}\)
= 56.0%

Question 23.
In Canus method of estimation of halogen1 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.
Solution:
Molar mass of AgBr = 108 + 80
= 188 g mol^-1
188 g AgBr contains 80 g bromine
0.12 g AgBr contains \(\frac{80 \times 0.12}{188}\) g bromine
Percentage of bromine = \(\frac{80 \times 0.12 \times 100}{188 \times 0.15}\)
= 34.04%

Question 24.
In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound ?
Solution:
Molecular mass of BaSO₄ = 137 + 32 + 64
= 233g
233 g BaSO₄ contains 32 g sulphur
0.4813 g BaSO₄ contains \(\frac{32 \times 0.4813}{233}\) g sulphur
Percentage of sulphur = \(\frac{32 \times 0.4813 \times 100}{233 \times 0.157}\)
= 42.10%

Question 25.
Write structures of different chain isomers of alkanes corresponding to the molecular formula C₆H₁₄. Also write their IUPAC names.
Solution:

Question 26.
Write structures of different isomeric alkyl groups corresponding to the molecular formula C₅H₁₁. Write IUPAC names of alcohols obtained by attachment of -OH groups at different carbons of the chain.
Solution:

Question 27.
Write IUPAC names of the following compounds :

1. (CH₃)₃ C CH₂C(CH₃)₃
2. (CH₃)₂ C(C₂H₅)₂
3. tetra – tert-butylmethane

Solution:

1. 2, 2, 4, 4-Tetramethylpentane
2. 3 3-Dimethylpentane
3. 3, 3-Di-tert-butyl -2, 2, 4, 4 – tetramethylpentane

Question 28.
Write structural formulas of the following compounds:
(i) 3, 4, 4, 5-Tetramethylheptane
(ii) 2, 5-Dimethyhexane
Solution:

Question 29.
Write structures for each of the following compounds. Why are the given names incorrect? Write correct IUPAC names.
(i) 2-Ethylpentane
(ii) 5-Ethyl – 3-methylheptane
Solution:

Longest chain is of six carbon atoms and not that of five. Hence, correct name is 3-Methyl- hexane.

Numbering is to be started from the end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5-methyl-heptane.

Question 30.
Sodium salt of which acid will be needed for the preparation of propane? Write chemical equation for the reaction.
Solution:
Butanoic acid,

Question 31.
Write IUPAC names of the following compounds:

Solution:
(i) 2, 8 – Dimethyl – 3, 6 – decadiene;
(ii) 1, 3, 5, 7 Octatetraene;
(iii) 2 – n – Propylpent – 1 – ene;
(iv) 4 – Ethyl – 2, 6 – dimethyl – dec – 4 – ene;

Question 32.
Calculate number of sigma (σ) and pi (π) bonds in the above structures (i-iv).
Solution:
σ bonds : 33, π bonds : 2
σ bonds: 17, π bonds:4
σ bonds : 23, π bond : 1
σ bonds: 41, π bond : 1

Question 33.
Write structures and IUPAC names of different structural isomers of alkenes corresponding to C₂H₁₀.
Solution:

Question 34.
Draw cis and trans isomers of the following compounds. Also write their IUPAC names:
(i) CHCl = CHCl
(ii) C₂H₅CCH₃ = CCH₃C₂H₅
Solution:

Question 35.
Which of the following compounds will show cis-trans isomerism ?
(i) (CH₃)₂C = CH – C₂H₅
(ii) CH₂ = CBr₂
(iii) C₆H₅CH = CH – CH₃
(iv) CH₃CH = CCl CH₃
Solution:
(iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom.

Question 36.
Write IUPAC names of the products obtained by addition reactions of HBr to hex- 1-ene
(i) in the absence of peroxide and
(ii) in the presence of peroxide.
Solution:

Question 37.
Write structures of different isomers corresponding to the 5^th member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers ?
Solution:
5^th member of alkyne has the molecular formula C₆H₁₀. The possible isomers are :


Position and chain isomerism shown by different pairs.

Question 38.
How will you convert ethanoci acid into benzene?
Solution:

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 12th Lesson Environmental Chemistry Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 12th Lesson Environmental Chemistry

Very Short Answer Questions

Question 1.
Define the terms atmosphere, biosphere.
Answer:
Atmosphere : The blanket of gases present around the earth is called the atmosphere. It maintains the heat balance on earth. Atmosphere contains nitrogen and oxygen in large proportions.

Biosphere : The living organisms like plants, animals and human beings constitute the biosphere.
Biosphere is related to other environment segments.

Question 2.
Explain the terms Lithosphere, Hydrosphere.
Answer:
Lithosphere : The outer mantle of the solid earth consists of minerals present in earth gust and soil. The earth inner surface contains minerals and deeper inner layers contain natural gas and soil. Mountains and hills these are all constitutes lithosphere.

Hydrosphere : All the natural water resources together constitute the hydrosphere. Hydrosphere include oceans, seas, rivers, lakes, streams etc.

Question 3.
Define the term Soil Pollution.
Answer:
The concentration (or) accumulation of natural bodies is called as soil. Soil gets polluted due to industrial wastes, urban wastes, agricultural pollutants, chemical, radio active pollutants etc.

Question 4.
What is Chemical Oxygen demand (COD).     [A.P. Mar. 16] [Mar. 14]
Answer:
The amount of oxygen required to oxidise organic substances present in polluted water is called Chemical Oxygen Demand (COD).
It is an index for amount of organic substances present in water.

Question 5.
What is Bio Chemical Oxygen Demand (BOD) ? [A.P. Mar. 16] [Mar. 14]
Answer:
The amount of oxygen used by the suitable micro organisms present in water during five days at 20° C is called as Bio Chemical Oxygen Demand (BOD).

Question 6.
What are Troposphere and Stratosphere ?
Answer:
Troposphere : The major portion of the atmosphere which contains air is called troposphere.
It is present 0-11 Km from the earth.

Stratosphere : Stratosphere present 11 – 50 Km from the earth and it mainly contains ozone layer. It absorbs the harmful UV radiations coming from sun.

Question 7.
Name the major particulate pollutants present in Troposphere.
Answer:
The major particulate pollutants present in troposphere are dust, mist, fumes, smoke, smog etc.

Question 8.
List out four Gaseous Pollutants present in the polluted air.
Answer:
Oxides of Sulphur,Nitrogen and Carbon, Ozone, Hydrocarbons etc., are gaseous pollutants present in polluted air.

Question 9.
Green house effect is caused by and gases.
Answer:
Green house effect is caused by gases such as CO₂, CH₄, O₃, CFCs (Chloro Fluoro Carbons) and water vapour in the atmosphere.

Question 10.
Which oxides cause acid rain ? and What is its pH value ? [Mar. 13]
Answer

• Oxides of Nitrogen, Sulphur and Carbon dissolved in rain water forms acid rain.
• Acid rain has pH value lessthan 5.6.

Question 11.
Name two adverse effects caused by acid rains. [A.P. Mar. 16] [T.S. Mar. 15]
Answer:
Effects of acid rains

• Acid rains are harmful for agriculture, trees and plants because it dissolves and washes away nutrients needed for their growth.
• Acid rains affects the plants and animal life in aquatic ecosystem.
• Acid rains damages the old buildings and historical monuments like Taj mahal.
• Acid rains corrodes water pipes which lead to decrease the quality of drinking water.

Question 12.
What are smoke and mist ?
Answer:
Smoke : The solid particles (or) mixture of solid and liquid particles formed by the combustion of organic matter are called smoke particulates.
Eg : Cigarette smoke, oil smoke etc.

Mist: The particles produced by the spray liquids and by condensation of vapours in air is called mist.
Eg : H₂SO₄ – mist, herbicides, insecticides etc.
These miss their targets and travel through atmosphere to form mist.

Question 13.
What is classical smog ? and What is its Chemical Character (Oxidizing / reducing) ?
Answer:

• The mixture of smoke, fog and sulphur dioxide is called classical smog. It exists in cool humid climate.
• The chemical character of classical smog is reducing character. Hence it is also called as reducing smog.

Question 14.
Name the common components of Photo Chemical smog.
Answer:
The common components of Photo Chemical smog are O₃, NO, acrolein, formaldehyde and Peroxy Acetyl Nitrate (PAN).

Question 15.
What is PAN ? What effect is caused by it ?
Answer:

• Peroxy Acetyl Nitrate is called as PAN.
• Peroxy Acetyl Nitrate is (PAN) is a powerful eye irritant.

Question 16.
How is Ozone formed in the Stratosphere ?
Answer:

• UV radiations react with dioxygen (O₂) molecules and split into free oxygen (O) atoms. These free oxygen atoms (0) combined to form ozone molecule.
• The following are the reactions that takes place during the formation of ozone in stratosphere.
  

Question 17.
Give the Chemical equations involved in the Ozone depletion by CF₂Cl₂.
Answer:
CF₂Cl₂ which is released in the atmosphere mix with the normal atmospheric gases and reaches the stratosphere.

• In stratosphere CF₂Cl₂ react with powerful UV radiations and liberates chlorine free radicals.
  
• These chlorine radical reacts with ozone present in stratosphere to form chlorine monoxide radicals and molecular dioxygen.
  Cl* + O₃ → ClO* + O₂
• ClO radical react with atomic oxygen and produce more Cl – radicals.
  ClO* + O → Cl* + O₂

Question 18.
What is Ozone hole ? Where was it first observed ?
Answer:
The depletion of ozone layer is commonly known as ozone hole.

• It was first observed in Antarctica over the south pole.
• It was reported by atmospheric scientists working in Antarctica.

Question 19.
What is the value of dissolved Oxygen in pure cooled, water ?
Answer:
The value of dissolved oxygen in pure cooled water is around 10 ppm.

Question 20.
Give the possible BOD values of clean water and the polluted water.
Answer:

• The BOD value of clean water is less than 3 ppm.
• The BOD value of water is greaterthan 4 ppm then it is said to be polluted.
• Highly polluted water has BOD value more than 17 ppm.

Question 21.
Name three industrial Chemicals that pollute water.
Answer:
Detergents, paints, pesticides, dyes and pharmaceuticals etc.

Question 22.
What agrochemicals are responsible for water pollution ?
Answer:
Agrochemicals like chemical fertilisers, chemicals used for killing insects, fungi and weeds in crop etc., are responsible for water pollution.

Short Answer Questions

Question 1.
What are different segments of the earth’s environment ?
Answer:
Environment can be divided into four segments.

1. Atmosphere
2. Hydrosphere
3. Lithosphere
4. Biosphere

1) Atmosphere : The layer of air present around the Earth is called the atmosphere. The atmospheric air contains N₂ and O₂ in large proportions, while the rest of the gases like C02 are present only in smaller proportions. Atmosphere absorbs harmful radiations coming from the Sun. It plays an important role in maintaining the heat balance on Earth. If the proportions of the gases, especially O₂ and N₂ are disturbed by human activity, the equilibrium of the echo system is lost. It leads to disastrous consequences.

2) Hydrosphere : Hydrosphere includes all the surface and ground water resources i.e., oceans, rivers, lakes, polar ice caps etc., 97% of earth’s water is locked up in oceans. 3% is trapped in polar ice caps. Only small percentage of water is available for drinking, agricultural and industrial purpose. 80% of the earth’s surface is covered with water.

3) Lithosphere : One fifth of the total Earth surface is in the form of land. Inner layers of Earth contain minerals. Deeper inner layers of Earth contain Natural gas and oil. All these things, including hills and mountains come under Lithosphere. Plants, animals and human beings are occupied by it.

4) Biosphere : All living organisms like plants, animals and human beings constitute the Biosphere. Biosphere and other segments of the environment are interrelated. Biosphere is dependent on Atmosphere and Hydrosphere. Polluted atmosphere arrest the plant growth and bring health hazards among animals and human beings. Contaminated water causes many diseases and also death of aquatic animals.

Question 2.
Define the terms Sink, COD, BOD and TLV. [T.S. Mar. 16]
Answer:
Sink : The medium which retains and interacts with long lived pollutant is called the sink.
Eg : Oceans are important sinks for atmospheric CO₂.

COD : The amount of oxygen required to oxidise organic substances present in polluted water is called Chemical Oxygen Demand (COD).
It is an index for amount of organic substances present in water.

BOD : The amount of oxygen used by the suitable micro organisms present in water during five days at 20° C is called as Bio Chemical Oxygen Demand (BOD).

TLV : (Threshold Limit Value): The permissiable level of the toxic substances (or) pollutants in the atmosphere which affects a person adversly when he is exposed to this for 7 – 8 hrs. in a day is called TLV.

Question 3.
Name the gaseous pollutants present in the air and explain their formation.
Answer:
Gaseous pollutants present in air are
a) Oxides of sulphur
b) Oxides of nitrogen
c) Oxides of carbon
d) Hydro carbons

a) Oxides of sulphur :

• The oxides of sulphur formed by the fossil fuel containing sulphur are burnt.
  S + O₂ → SO₂
• Sulphur dioxide oxidises to form sulphur trioxide in presence of catalyst.
  2SO₂ + O₂ ⇌ 2SO₃
• Sulphur trioxide can also be formed by the reaction of SO₂ with O₃ (or) H₂O₂.
  SO₂ + O₃ ⇌ SO₃ + O₂
  SO₂ + H₂O₂ → H₂SO₄
• SO₂ is the most common oxide and causes the following adverse effects.
  a) It causes respiratory problems like asthma, bronchitis etc.
  b) It is poisonous to both animals and plants.
  c) It causes irritation to the eyes which results in tears and redness.

b) Oxides of nitrogen :
The major gases present in air are oxygen and nitrogen.

• These do not combine at normal temperature.
• Dinitrogen and dioxygen combined at high altitudes in presence of light to form oxides of nitrogen.
  
• No reacts with oxygen to form NO₂.
  2NO + O₂ → 2NO₂
• NO₂ is quickly formed by the reaction of NO with O₃.
  NO + O₃ → NO₂ + O₂

Adverse Effects

• NO₂ damages the leaves of plants and effects the efficiency of photosynthesis.
• NO₂ causes lung problems.
• NO₂ causes respiratory problems.
• NO₂ affects the textile fibres and metals.

c) Oxides of carbon : –
i) Carbon monoxide (CO)

• CO is one of the most harmful air pollutants.
• CO gas is produced by the incomplete combustion of carbon (coal, firewood, petrol etc.)
  C_(s) + \(\frac{1}{2}\)O_2(g) → C0(g)
• CO is mainly released into atmosphere by automobiles.

Adverse effects :

• It stops the transportation of oxygen to the organs and tissues.
• When it is inhaled by a human being it forms stable complex with haemoglobin of blood. This complex is named as carboxy haemoglobin.
  CO + Hb → CO – Hb
• This reduces the oxygen transportation in body and results into headache, eye problems, nerveous problems and heart problems.

ii) Carbondioxide :

• CO₂ is entered into atmosphere mainly by respiration process.
• CO₂ is also formed by the burning of fossil fuels.
  C + O₂ → CO₂ + heat energy
• CO₂ is formed by the decomposition of lime stone.
  CaCO₃ Cao + CO₂

Effects :

• Due to deforestation and burning of fossil fuels increases CO₂ release and the balance O₂ – CO₂ in atmosphere disturbed.
• The increased level of CO₂ causes global warming which causes several problems like dengue, malaria etc.
• Due to global warming efficiency of photosynthesis decreases.

d) Hydrocarbons:

• These mainly constitutes hydrogen and carbon.
• These are formed by the incomplete combustion of fuel used in auto mobiles.

Adverse Effects:

• These cause cancer.
• These are harmful to plants causing ageing, break down of tissues, flowers and leaves are shedded.

Question 4.
What is green house effect ? and how is it caused ?
Answer:
a) Global Warming : The Earth is heated by sunlight and some of the heat that is absorbed by the Earth is radiated back into space. But some gases like CO₂, CFCs, O₃, NO and water vapour present in the lower atmosphere do not allow the Earth to re-radiate the heat into space. A part of the heat so trapped in these atmospheric gases is re-emitted to the earth’s surface. This phenomenon is called the “Green house effect” (or) Global Warming.
The gases which are responsible for Green house effect are CO₂, CFC’s, O₃, NO, CH₄ water vapour etc. and they are termed as Green house gases. The effect of increase in CO₂ causes global warming.

b) Effects of global warming :

1. If there is a 1° C increase in the temperature, the ice caps of the polar region melt and level of the sea water increases by 90 cm. Due to this, so many coastal countries will be submerged.
2. Due to global warming, the rate of evaporation of water from the seas, rivers, ponds will increase. This leads to ultimately rains, cyclones and hurricanes.
3. Agriculture sector will be badly affected due to the fast evaporation of surface water. There will be a shortage in the supply of water for agricultural purpose.
4. Unseasonal rains.
5. Increase the infectious diseases like dengue, malaria, yellow fever, sleeping sickness etc.

Prevention : To reduce the level of CO₂ on the earth’s atmosphere, one must increase the number of sinks to absorb CO₂. Plants absorb a major portion of the CO₂. Therefore, more plants, trees, forests should be grown. The blue green algae present in the sea also gets extinct, due to water pollution. This should be prevented stopping the production of CFC etc.

Question 5.
Explain, with Chemical equations involved, the formation of acid rain.
Answer:
Acid rains are due to oxides of N, S and C (NO₂, SO₂ and CO₂).

• These oxides dissolve in rain water and formed as acids (HNO₃, H₂SO₄ and H₂CO₃).
• These come down to earth as rain and deposited on the earth’s surface.
• Acid – rain is more in industrial areas.

Chemical equations involved
NO₂ + NO₃ → N₂O₅
N₂O₅ + H₂O + 2HNO₃
CO₂ + H₂O → H₂CO₃
SO₃ + H₂O → H₂SO₄
The pH of acid rain is less than 5.6.

Question 6.
Explain in detail the adverse effects caused by the acid rain.
Answer:
Acid rains are harmful because

1. The life of old buildings will be considerable reduced.
2. The pH of the soil changes affecting its fertility.
3. Ammonium salts formed by acid rains can be seen as atmospheric haze.
4. Ammonium salts in rain drops result in wet decomposition.
5. Acid rains are harmful to agriculture, trees, plants etc.
6. Due to acid rains aquatic life disturbed.
7. Due to acid rains the historical monuments like Tajmahal are damaged.

Question 7.
How is Photochemical Smog formed ? What are its ill effects ?
Answer:

• When unsaturated hydrocarbons and nitrogen oxides produced from auto mobiles, factories reacts with sunlight and forms photo chemical smog.
• This occurs in warm, dry and hot climate.
• This has high concentration of oxidising agents. Hence it is called as photo chemical smog.

Formation :

• Fossil fuels are burnt, many pollutants are entered into troposphere.
• Out of these many pollutants two pollutants are main constituents of photo chemical smog. These are hydrocarbons and nitric oxide.
• These pollutants interacts with sun light and following reaction takes place.
  2NO_(g) + O_2(g) → 2NO_2(g)
  NO_2(g) → NO_(g) + O_(g)
• This oxygen atoms formed in the above reaction combine with O₂ to produce ozone.
  O_(g) + O_2(g) ⇌ O_3(g)
  NO_(g) + O_3(g) → NO_2(g) + O_2(g)
• O₃ is poisonous gas and both NO₂ and O₃ are strong oxidising agents.
• These react with unburnt hydro carbons in the polluted air to produce organic substances like HCHO, PAN etc.
• The common components of photo chemical smog are ozone, NO, acrolein, form aldehyde and PAN.

Question 8.
How is Ozone layer depleted in the atmosphere and what are the harmful effects caused by Ozone layer depletion ? [T.S. Mar. 16] [A.P. Mar. 15]
Answer:

• Ozone layer is present in stratosphere of the atmosphere.
• Ozone in the stratosphere is due to the following reactions.
  
• But due to industrialisation, certain chemical substances enter into stratosphere and destroy the ozone.
• The following are the major substances that causes depletion of ozone layer.
  1. CFCs (Chloro fluoro carbons (or) freons)
  2. NO
  3. Cl₂ (Chlorine)
• CFCs are colourless, odourless, lighter, non flammable, non toxic organic molecules which are used in refrigeratos, air conditioners, in the manufacturing of plastic foam and for cleaning computer parts.
• These CFCs enter in the stratosphere and deplete the ozone as follows.
  
• Chlorine radicals (Cl*) are continuously generated and decompose O₃.
• NO released by super sonic jet planes and that formed by burning fossil fuels enters into stratosphere and decompose ozone.
• Cl₂ decompose into Cl* radicals and these Cl* deplete O₃ by chain reactions.

Effects of depletion of the ozone layer (ozone hole): Due to depletion of ozone more U.V. enters into troposphere. This UV radiation leads to

1. ageing of skin
2. cataracts
3. skin burns
4. skin cancer
5. damage to fish production
6. killing many phytoplanktons
7. effect plant proteins leading to the harmful mutation of cells
8. evaporate surface water through the stomata of the leaves.
9. decrease moisture content of the soil.
10. damage paints, fibres, by fading them faster.
11. effect on photo synthesis etc.

Question 9.
List out the industrial wastes that cause water pollution and what are the international standards fixed for drinking water ?
Answer:
Industrial wastes from food processing plants and paper and pulp mills are oxygen – demanding wastes. So they cause depletion of D.0 from the water.

Salts, trace of elements like Copper, Zinc, Arsenic etc., metals coming out of chromium plating Industry pollute water. They effect the human health and aquatic animals.

In Japan, at Minimata, water was polluted with mercury released from industry. The mercury enter the fish and the people who consumed the fish are adversely effected.

Mining and Nuclear power plant pollute the water with radioactive substances.

International standards for drinking water

Fluoride concentration:

• The drinking water sample is tested for fluoride ion concentration. Its deficiency causes diseases like tooth decay etc.
• The permissible level of concentration of fluoride in water is up to 1 ppm.
• Fluoride concentration above 2 ppm causes brown mottling of teeth.
• Fluoride concentration greater than 10 ppm causes harmful effects to bones and teeth.

Lead :

• The prescribed upper limit of concentration limit of lead in drinking water is about 50 ppm.
• Lead can damage kidney, liver, reproductive system etc.

Sulphate :

• Sulphate concentration greater than 500 ppm in drinking water causes laxative effect.
  Nitrate :
• Nitrate concentration maximum limit in drinking water is about 50 ppm.
• Excess of nitrate in drinking causes disease such as methemoglobinemia.

Metals :

Maximum prescribed concentration of some common metals recommended in drinking water given in the following table.
Maximum prescribed concentration of Some Metals in Drinking Water.

Question 10.
Explain in detail the strategies adopted in Green Chemistry to avoid environment pollution.
Answer:
The ways of using the knowledge and the principles of chemistry and other sciences to develop methods to reduce as far as possible the pollution of the environment are known as green chemistry.
Ex : In the dry cleaning of clothes, earlier (CCl₂ = CCl₂) tetrachloroethane was used. This com-pound contaminates the ground water and is a suspected carcinogen. Therefore, using this compound is replaced by a process in which liquefied carbon dioxide with a suitable detergent is used. This would not pollute ground water much. Now-a-days hydrogen peroxide is used for bleaching clothes in laundries. This gives better results and decreases the consumption of water.
Ex : Synthesis of ethanol is now commercially prepared by one step by the oxidation of ethane in presence of catalyst.

Long Answer Questions

Question 1.
What is environmental pollution ? How many types of pollution are encountered ?
Answer:
Environmental pollution is defined as the addition of any external material like organic, Inorganic biological, radio logical (or) any change in nature which may harm (or) affect badly the living organism directly (or) indirectly, immediately (or) slowly.
The types of pollution encountered are the following :

1. Air pollution
2. Water pollution
3. Soil pollution
4. Oil pollution
5. Noise pollution etc.

Question 2.
Explain the following in detail.
a) Global warming
b) Ozone depletion
c) Acid rain
d) Eutrophication.
Answer:
a) Global Warming : The Earth is heated by sunlight and some of the heat that is absorbed by the Earth is radiated back into space. But some gases like CO₂, CFC’s, O₃, NO and water vapour present in the lower atmosphere do not allow the Earth to re-radiate the heat into space. A part of the heat so trapped in these atmospheric gases is re-emitted to the earth’s surface. This phenomenon is called the “Green house effect” (or) Global warming.
The gases which are responsible for Green house effect are CO₂, CFC’s, O₃, NO, CH₄ water vapour etc. and they are termed as Green house gases. The effect of increase in CO₂ causes global warming.

b) Effects of depletion of the ozone layer (ozone hole) : Due to depletion of ozone more U.V. enters into troposphere. This UV radiation leads to

1. ageing of skin
2. cataracts
3. skin burns
4. skin cancer
5. damage to fish production
6. killing many phytoplanktons
7. effect plant proteins leading to the harmful mutation of cells
8. evaporate surface water through the stomata of the leaves.
9. decrease moisture content of the soil.
10. damage paints, fibres, by fading them faster.
11. effect on photo synthesis etc.

c) Acid rain :

• Acid rains are due to oxides of N, S and C.(NO₂, SO₂ and CO₂)
• These oxides dissolve in rain water and formed as acids. (HNO₃, H₂SO₄ and H₂CO₃)
• These come down to earth as rain and deposited on the earth’s surface.
• Acid – rain is more in industrial areas.

Chemical equations involved:-
NO₂ + NO₃ → N₂O₅
N₂O₅ + H₂O + 2HNO₃
CO₂ + H₂O→ H₂CO₃
SO₃ + H₂O → H₂SO₄
The pH of acid rain is lessthan 5.6.

Acid rains are harmful because

1. The life of old buildings will be considerable reduced.
2. The pH of the soil changes affecting its fertility.
3. Ammonium salts formed by acid rains can be seen as atmospheric haze.
4. Ammonium salts in rain drops result in wet decomposition.
5. Acid rains are harmful to agriculture, trees, plants etc.
6. Due to acid rains aquatic life disturbed.
7. Due to acid rains the historical monuments like Tajmahal are damaged.

d) Eutrophication : Water present in ponds and lakes becomes over nutritious when organic sub-stances from agriculture and industry are thrown into it. It can support the luxuriant growth of algae and thus the lakes and ponds become marshy. This phenomenon is called Eutrophication.

Question 3.
Green Chemistry is to avoid environmental pollution. Explain.
Answer:
The ways of using the knowledge and the principles of chemistry and other sciences to develop methods to reduce as far as possible the pollution of the environment are known as green chemistry.
Ex : In the dry cleaning of clothes, earlier (CCl₂ = CCl₂) tetrachloroethane was used. This com-pound contaminates the ground water and is a suspected carcinogen. Therefore, using this com-pound is replaced by a process in which liquefied carbon dioxide with a suitable detergent is used. This would not pollute ground water much. Now-a-days hydrogen peroxide is used for bleaching clothes in laundries. This gives better results and decreases the consumption of water.
Ex : Synthesis of ethanol is now commercially prepared by one step by the oxidation of ethane in presence of catalyst.

Thus green chemistry is a cost effective approach and it involves reduction in material, energy consumption and waste greneration.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14 Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14

Very Short Answer Questions

Question 1.
Discuss the variation of oxidation states in the group -14 elements.
Answer:

• The common oxidation states exhibited by group – 14 elements are +4 and +2.
• Carbon exhibits negative oxidation states.
• Heavier elements exhibits +2 oxidation state.
• The tendency to show +2 oxidation state increases in the order Ge < Sn < pb.
• pb exhibits +2 oxidation state as stable state because of inert pair effect.

Question 2.
How the following compounds behave with water
a) BCl₃
b) CCl₄.
Answer:
a) BCl₃ reacts with water (hydrolysis) to form boric acid.

b) CCl₄ does not undergo hydrolysis due to lack of d-orbitals in the central atom ‘C1 and due to its highly non polar nature, CCl₄ does not acts as Lewis acid.

Question 3.
Are BCl₃ and SiCl₄ electron-deficient compounds ? Explain.
Answer:

• BCl₃ and SiCl₄ are electron-deficient compounds.
• These two compounds behave as Lewis acids.
• These compounds are electron pair acceptors.
• The following reacts support the electron deficiency of BCl₃ and SiCl₄.
  

Question 4.
Give the hybridization of carbon in
a) CO₃^-2
b) diamond
c) graphite
d) fullerene
Answer:
a) In C0₃^-2 carbon atom undergoes sp² hybridisation.
b) In Diamond carbon atom undergoes sp³ hybridisation.
c) In Graphite carbon atom undergoes sp² hybridisation.
d) In Fullerenes carbon atom undergoes sp² hybridisation.

Question 5.
Why is’CO’poisonous ? [T.S. Mar. 16]
Answer:
‘CO’ gas is highly poisonous because it has the ability to form a stable complex with haemoglobin.

Carboxy haemoglobin is 300 times more stable than oxyhaemoglobin.

Question 6.
What is allotropy ? Give the crystalline allotropes of carbon. [Mar. 13]
Answer:

• The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
• Crystalline allotropes of carbon are
  a) Diamond
  b) Graphite
  c) Fullerenes.

Question 7.
Classify the following oxides as neutral, acidic, basic or amphoteric,
a) CO
b) B₂O₃
c) SiO₂
d) CO₂
e) Al₂O₃
f) PbO₂
g) Tl₂O₃
Answer:
a) CO is neutral oxide.
b) B₂O₃ is acidic oxide.
c) SiO₂ is acidic oxide.
d) CO₂ is acidic oxide.
e) Al₂O₃ is amphoteric oxide.
f) PbO₂ is amphoteric oxide.
g) Tl₂O₃ is basic oxide.

Question 8.
Name any two manmade silicates.
Answer:
Glass and cement are man made silicates.

Question 9.
Write the outer electron configuration of group -14 elements.
Answer:
The general outer most electronic configuration of group – 14 elements is ns²np².
1) Carbon – [He] 2s²2p²
2) Silicon – [Ne] 3s²3p²
3) Germanium – [Ar] 3d¹⁰4s²4p²
4) Tin – [Kr] 4d¹⁰ 5s² 5p²
5) Lead – [Xe] 4f⁴ 5d¹⁰ 6s² 6p²

Question 10.
How does graphite function as a lubricant ?
Answer:
Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

Question 11.
Graphite is a good conductor – explain.
Answer:
In graphite carbon undergoes sp² hybridisation. Each carbon forms three a – bonds with three neighbouring carbon atoms. Fourth electron forms TC – bond and it is delocalised. Due to the presence of these moving (or) free electrons graphite acts as good conductor.

Question 12.
Explain the structure of silica.
Answer:

• Silica is a giant molecule with 3 – dimensional structure.
• In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
• Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
• Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
• In SiO₂ Silicon atom undergoes sp³ hybridisation.

Question 13.
What is ‘Synthesis gas’?
Answer:

• Water gas is also called as synthesis gas.
• It is a mixture of CO and H₂.
• It s prepared by passing steamover hot coke.
• It is used for the synthesis of methanol and a number of hydrocarbons. Hence it is called synthesis gas.

Question 14.
What is producer gas?
Answer:

• Producer gas is mixture of CO and N₂.
• It is prepared by passing air over hot coke.

Question 15.
Diamond has high melting point – Explain.
Answer:

• In Diamond each carbon undergoes sp³ hybridisation and it is surrounded by four other carbon atoms with strong a – bonds tetrahedrally.
• The C – C bond energy in diamond is very high and it has 3 – dimensional structure.
• Due to these reasons diamond has high melting point.
• It has melting point 4200 K.

Question 16.
Give the use of CO₂ in photosynthesis.
Answer:
The process of converting the atmospheric CO₂ into Carbohydrates by green plants is known as ‘photosynthesis’.
In Photosynthesis CO₂ changes to carbohydrates such as glucose.

Question 17.
How does CO₂ increase the green house effect ?
Answer:

• Green plants absorbs CO₂ gas for photosynthesis and releases O₂ gas.
• Due to deforestation, decomposition of lime stone and burning of fossil fuels CO₂ concentration is increased in atmosphere.
• The increase of CO₂ level disturbs the O₂ – CO₂ balance in the atmosphere and it is responsible for green house effect (or) global warming.

Question 18.
What are silicones ?
Answer:

• Silicones are the organo Silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – O – Si.
• Linkage preparation : These are formed by the hydrolysis of chlorosilanes.

Question 19.
Give the uses of silicones.
Answer:
Uses of silicones :

• These are used in surgical and cosmetic plants.
• These are used in preparation of silicone rubbers.
• These are used as sealoant, greases etc.
• These are used in preparing water proof clothes and papers.
• These are used as insulators.
• These are used in paints and enamels.

Question 20.
What is the effect of water on tin ?
Answer:

• Tin metal reacts with steam to form tin dioxide and dihydrogen gas.
• In this reaction steam is decomposed.
  Sn + 2H₂O SnO₂ + 2H₂

Question 21.
Write an account of SiCl₄.
Answer:

• Silicon tetrachloride (SiCl₄) is also called as tetra chloro silico methane.
• SiCl₄ can acts as Lewis acid due to availability of 3d orbital in ‘Si’.
• SiCl₄ undergoes hydrolysis due to presence of vacant 3d-orbital. Here water molecules forms dative bonds with empty 3d-orbitals of Siratom.

Uses :

• SiCl₄ and NH₃ mixture used to produce smoke screens.
• Ultra pure Silicon is used to make transistors.
• SiO₂ prepared from SiCl₄ used in epoxypaints, resis etc..

Question 22.
SiO₂ is a solid while CO₂ is a gas – explain.
Answer:

• Silica (SiO₂) has giant molecular structure.
• In SiO₂ ‘Si’ undergoes sp³ hybridisation.
•  It is a 3 – dimension structure in which each ‘Si’ atom is tetrahedrally surrounded by four oxygen atoms.
• Hence it exists as solid compound.
• CO₂ has linear structure.
• In CO₂ ‘C’ undergoes sp hybridisation.
• In between CO₂ molecule weak Vander Waal’s forces are present.
• In CO₂ molecule two double bonds are present.
•  Hence CO₂ exists as a gas.

Question 23.
Write the use of ZSM – 5.
Answer:

1. ZSM – 5 is a zeolite.
2. It is used to convert alcohols directly into gasoline.

Question 24.
What is the use of dry ice ?
Answer:

1. Solid CO₂ is called as dry ice.
2. It is used as refrigirent for frozen food and ice – creams.

Question 25.
How is water gas prepared ?
Answer:
Water gas is prepared by passing superheated steam over hot coke.

Question 26.
How is producer gas prepared ?
Answer:
Producer gas is prepared by passing air over white not coke.

Question 27.
C-C bond length in graphite is shorter than C-C bond length in diamond – explain.
Answer:

1. In graphite each carbon undergoes sp² hybridisation and hence bond length is 1.42 A° (or) 141.5 pm.
2. Graphite has hexagonal layer like lattice. It is a 2-dimensional structure.
3. In diamond each carbon undergoes sp3 – hybridisation and hence bond length is 1.54 A° (or) 154 pm.
4. Diamond has regular tetrahedral giant polymeric structure. It is a 3-dimensional structure.

Question 28.
Diamond is used as precious stone – explain.
Answer:

• Diamonds are used as precious stones.
• Diamonds are clear, colourless form of pure carbon.
• These are hardest substances occurring naturally.
• The weight of diamond expressed in carats.
  1 carat = 200 mg.

Question 29.
Carbon never shows co-ordination number greater than four while other members of carbon family show co-ordination number as high as six – explain.
Answer:
Carbon never shows co-ordination number greater than four because of absence of d-orbitals in carbon atom.

The other members of carbon family show co-ordination number as high as six because of availability of d – orbitals.

Question 30.
Producer gas is less efficient fuel than water gas – explain.
Answer:

1. Producer gas has calorific value 5439.2 KJ/m³
2. Water gas has calorific value 13000 KJ/m³.
3. Due to high calorific value of watergas, it is more efficient fuel than producer gas (or) producer gas is less efficient than watergas.

Question 31.
SiF₆^-2 is known while SiCl₆^-2 is not. Explain. [A.P. Mar. 16]
Answer:
SiF₆^-2 is known while SiCl₆^-2 is not because

• Si⁺⁴ has small size so it cannot be accomodate six large chloride ions.
• The interaction between lone pairs of Cl^– ion and Si⁺⁴ is not very strong.

Short Answer Questions

Question 1.
Explain the difference in properties of diamond and graphite on the basis of their structure.
Answer:
Diamond
a) Each carbon is sp³ hybridised.
b) Each carbon is bonded to 4 other carbons tetrahedrally.
c) It has a 3 dimensional structure.
d) C – C bond length is 1.54 Å and bond angle is 109° 28′.
e) Carbon atoms are firmly held with strong covalent bonds.
f) Diamond is very hard.
g) Density = 3.5 g/cc.
h) Graphite is a conductor due to the presence of free electrons.
i) It is transparent to light and X-rays. It has high refractive index (2.45).

Graphite
a) Each carbon is sp² hybridised.
b) Each carbon is bonded to 3 other carbon atoms to form hexagonal rings. It has sheet like structure.
c) It has a 2 dimensional structure.
d) C – C bond length in hexagonal rings is 1.42 A° and bond angle is 120°.
e) The distance between two adjacent layers is 3.35 A°. These layers are held by weak Vander Waal’s forces.
f) Graphite is soft.
g) Density – 2.2g/cc.
h) Diamond is an insulator due to the absence of free electrons.
i) It has layer, lattice. The layers are slippery. Hence it is greasy.

Question 2.
Explain the following.
a) PbCl₂ reacts with Cl₂ to give PbCl₄
b) PbCl₄ is unstable to heat,
c) Lead is not known to form PbI₄.
Answer:
a) PbCl₂ + Cl₂ → PbCl₂
But PbCl₄ is unstable than PbCl₂. Because compounds of lead in +2 oxidation state are stable than +4 oxidation state.

b) PbCl₄ is unstable to heat:

• In PbCl₄ lead exhibits +4 oxidation state.
• he compounds of lead in +2 oxidation state are stable than +4 oxidation state. Hence PbCl₄ is unstable to heat.

c) Lead is not known to form PbI₄:

• Pb – I bond formed initially during the reaction does not release enough energy to unpair the 6s electrons.
• Lead compounds in +2 state are stable than +4 state.
  Due to inert pair effect Pb exhibits stable +2 oxidation state.

Question 3.
Explain the following :
a) Silicon is heated with methyl chloride at high temperature in presence of copper.
b) SiO₂ is treated with HF.
c) Graphite is a Lubricant
d) Diamond is an abrasive.
Answer:
a)

• Methyl chloride reacts with silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si.
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (silicone) (Me = CH₃ – group)
  

b) SiO₂ is treated with HF to form SiF₄. Which on hydrolysis to form H₄SiO₄ and H₂SiF₆.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 4H₂O → H₄SiO₄ + 2H₂SiF₆

c) Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

d) The covalent bonds in diamond are very strong and difficult to break, Hence diamond is used as an abrasive for sharpening hard tpols, in making dyes and in the manufacturing of tungsten filaments etc.

Question 4.
What do you understand by
a) Allotropy
b) Inert pair effect
c) Catenation.
Answer:
a) Allotropy :
The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
Crystalline allotropes of carbon are
a) Diamond
b) Graphite.

b) Inert pair effect : The reluctance of ‘ns’ pair of electrons to take part in bond formation is known as inert pair effect.
(or)
The occurrence of oxidation states two units less than the group oxidation states is known as inert pair effect.
Eg : Lead exhibits +2 oxidation state as stable oxidation state due to inert pair effect. (Instead of +4 state).

c) Catenation : The phenomenon of self linkage of atoms among themselves to form long chains (or) rings is called as catenation.
Carbon has highest catenation tendency due to its high bond energy (348 KJ/mole).

Question 5.
If the starting material for the manufacturing of Silicons is RSiCi3. Write the structure of the product formed.
Answer:
When RSiCl₃ type of compound is used for the manufacturing of silicones a cross – linked silicon is formed.
Eg : When Methyl trichloro silane (CH₃SiCl₃) undergoes hydrolysis to give monomethyl silane triol. This undergoes polymerisation to form a very complex cross-linked polymer (Silicone).

Question 6.
Write a short note on Zeolites.
Answer:
Zeolites : The three dimensional structure contain no metal ions. If some of the Si⁺⁴ in them is replaced by Al⁺³ and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si₂O₈]^-2n form zeolites. Zeolites act as ion exchanges and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH₃, CO₂ and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca⁺² ions from hard water and replace them by Na⁺² ions.
Uses of Zeolites :
Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

Question 7.
Write a short note on Silicates.
Answer:
Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.
The silicates can be divided into six types. They are mentioned here.

1. Orthosilicate or Nesosilicates : Their general formula may be M₂¹¹ (SiO₄).
   Ex : Willemite Zn₂ (SiO₄).
2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO₇^-6 units. Pyrosilicates are rare.
   Ex: Thorveitite Ln₂ (Si₂O₇).
3. Chain silicates : They have the units (SiO₃)_n^2n-
   Ex : Spodumene LiA/ (SiO₃)₂.
   Amphiboles are one type chain silicates. Generally double chains are formed in them.
4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
   Ex : Beryl Be₃Al₂ (Si₆O₁₈)
5. Sheet silicates : When SiO₄ units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si₂O₅)_n^2n-. These compounds appear in layer struc-tures. They can be cleaved.
   Ex : Kaolin Al₂ (OH)₄ Si₂O₅.
6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO₄ tetrahedron results in three dimensional lattice of formula SiO₂.
   Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na₈ [Al₆ Si₆ O₂₄] S₂), zeolites.

Question 8.
What are Silicones ? How are they obtained ?
Answer:

• Silicones are the organo silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – 0 – Si.
  Preparation : These are formed by the hydrolysis of chlorosilanes.
• Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH₃ – group).
  

Question 9.
Write a short note on Fullerene.
Answer:
Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – membered rings.
  
•  In C₆₀ 6 – membered rings can combine with 5 (or) 6 – mem-bered rings while 5 – membered rings only combine with 6 – membered rings.
• In Fullerens each carbon undergoes sp2 hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1,43A° and 1.38A° respectively.

Question 10.
Why SiO₂ does not dissolve in water ?
Answer:
Silica (SiO₂) is a non reactive compound in it’s normal state.

1. This non reactivity is due to very high Si – O bond enthalpy.
2. Silica is a giant molecule with 3-dimensional structure.
3. In Silica each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms
4. Hence SiO₂ is insoluble in water.
5. But slightly dissolves at high pressures when heated.

Question 11.
Why is diamond hard ?
Answer:
In diamond, each carbon undergoes sp3 hybridization. A carbon atom is bound to four carbon atoms, arranged in a tetrahedral symmetry, with single bonds. A three dimensional arrangement of the tetrahedral structures result in giant molecule. The bond energy is very high (348 kJ mol^-1). It is very difficult to break the bonds. So, diamond is hard.

Question 12.
What happens when the following are heated
a) CaCO₃
b) CaCO₃ and SiO₂
c) CaCO₃ and excess of coke.
Answer:
CaCO₃ up on heating gives Quick lime.


Quick lime (CO) with silica gives Calcium silicate.


Coke reacts with Quick lime and form Carbides.

Question 13.
Why does Na₂CO₃ solution turn into a suspension, when saturated with CO₂ gas.
Answer:
An aq. solution of Na₂CO₃ when saturated with CO₂, gives Sodium bicarbonate (NaHCO₃).
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃
NaHCO₃ is less soluble compared to Sodium carbonate, hence suspension is formed.

Question 14.
What happens when
a) CO₂ is passed through slaked lime
b) CaC₂ is heated with N₂.
Answer:
a) Slaked lime, Ca(OH)₂ is turned milky on passing CO₂ with the formation of insoluble calcium
carbonate Ca(OH)₂ + CO₂ → CaCO₃ + H₂O on passing more ‘CO₂‘, CaCO₃ is converted into Calcium bicarbonate.
CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂
b) CaC₂ on heating with N2 gives calcium cyanamide.

Question 15.
Write a note on the anomalous behaviour of carbon in the group -14.
Answer:
Carbon shows ariamalous behaviour in group – 14 elements. The following facts support that anamalous behaviour.
Except carbon all other elements of group -14 has available d-orbitals and can expand octet in valency shell.
Carbon occurs in free state but not the other elements of this group.

• Maximum covalency of carbon is four but for silicon is six.
• C – C bond energy is very high (348 kJ/Mde).
• Carbon can form multiple bonds with C, O, S, etc.
• Hydrocarbons are more stable thermally than silanes.

Long Answer Questions

Question 1.
What are Silicones ? How are they prepared ? Give one example. What are their uses?
Answer:

• Silicones are the organo silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – 0 – Si.
  Preparation : These are formed by the hydrolysis of chlorosilanes.
• Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH₃ – group).
  

Uses of Silicones:

• These are used in surgical and cosmetic plants.
• These are used in preparation of Silicons rubbers.
• These are used as sealoant, greases etc.
• These are used in preparing water proof clothes and papers.
• These are used as insulators.
• These are used in paints and enamels.

Question 2.
Explain the structure of Silica. How does it react with
a) NaOH and
b) HF.
Answer:

• Silica is a giant molecule with 3 – dimensional structure.
• In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
• Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
• Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
• In SiO₂ Silicon atom under goes sp³ hybridisation.
  

a) Silica reacts with NaOH and forms Sodium Silicate (Na₂SiO₃)
SiO₂ + 2 NaOH → Na₂SiO₃ + H₂O

b) SiO₂ is treated with HF to form SiF₄. Which on hydrolysis to form H₄SiO₄ and H₂SiF₆.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 4H₂O → H₄SiO₄ + 2H₂SiF₆

Question 3.
Write a note on the allotropy of carbon.
Answer:
The property of an element to exist in two or more physical forms due to difference in the arrangement of atoms is called Allotropy. Allotropes have more or less similar chemical properties but different physical properties.

Diamond, graphite and fullerenes are the crystalline allotropes of carbon.
Structure of Diamond : In diamond, each carbon atom bonded to four carbon atoms situated tetrahydrally around it.
In diamond, each carbon atom is in sp³ hybridisation and is linked to four carbon atoms by single covalent bonds.
C – C bond distance in diamond is 1.54 A°.
bond angle in diamond is 109° 28″.
Uses of Diamond :

1. Diamonds are used as precious stones for jewellery because of their ability to reflect light.
2. Diamonds are used for cutting glass and drilling rocks due to their remarkable hardness.

Structure of Graphite : Graphite consists of a series of layers in which hexagonal rings made up of carbon atoms.
In Graphite, each carbon atom undergo sp² hybridisation and forms three covalent bonds with three other carbon atoms.

The fourth electron present in the pure ‘p orbitaI which is unhybridised. The electron become Free Electron.
The C — C bond length in graphite is 1.42 A°.

The distance between the two layers in graphite is 3,4 A°.
These layers are held together by Vander Waal’s forces which are weak.
Graphite is a layer lattice structure.
Uses of Graphite:

1. Graphite is used as a lubricant.
2. It is used in the manufacturing of lead pencils.
3. It is used in the manufacturing of Electrodes and Refractory crucibles.

Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – membered rings.
• In C₆₀ 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
•  In Fullerens each carbon undergoes sp² hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1.43A° and 1.38A° respectively.
• Amorphous allotropes of carbon are coal, coke, animal charcoal wood charcoal, lamp black, carbon black, gas carbon, petroleum coke and sugar charcoal.

Question 4.
Write a note on
a) Silicates
b) Zeolites
c) Fullerenes.
Answer:
a) Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.

The silicates can be divided into six types. They are mentioned here.

1. Orthosilicate or Nesosilicates : Their general formula may be M₂¹¹ (SiO₄).
   Ex : Willemite Zn₂ (SiO₄).
2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO₇^-6 units. Pyrosilicates are rare.
   Ex : Thorveitite Ln₂ (Si₂O₇).
3. Chain silicates : They have the units (SiO₃)_n^2n-
   Ex : Spodumene LiAl (SiO₃)₂.
   Amphiboles are one type chain silicates. Generally double chains are formed in them.
4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
   Ex : Beryl Be₃Al₂ (Si₆O₁₈)
5. Sheet silicates : When SiO₄ units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si₂O₅)_n^2n-. These compounds appear in layer struc-tures. They can be cleaved.
   Ex : Kaolin Al₂ (OH)₄ Si₂O₅.
6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO₄ tetrahedron results in three dimensional lattice of formula SiO₂.
   Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na₈ [Al₆ Si₆ O₂₄] S₂), zeolites.

b) Zeolites : The three dimensional structure contain no metal ions. If some of the Si⁺⁴ in them is replaced by Al⁺³ and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si₂O₈]^-2n form zeolites. Zeolites act as ion ex-changes and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH₃, CO₂ and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca⁺² ions from hard water and replace them by Na⁺² ions.

Uses of Zeolites :

• Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
• Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

c) Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon. .
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – mem- bered rings.
  
• In C₆₀. 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
• In Fullerens each carbon undergoes sp² hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1.43A° and 1.38A° respectively.

Solved Problems

Question 1.
Select the member(s) of group 14 that

1. forms the most acidic dioxide
2. is commonly found in +2 oxidation state
3. used as semiconductor.

Solution:

1. Carbon
2. lead
3. Silicon and germanium

Question 2.
[SiF₆]^2- is known whereas [SiCl₆]^2- not. Give possible reasons.
Solution:
The main reasons are :

1. Six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.
2. Interaction between lone pair of chloride ion and Si⁴⁺ is not very strong.

Question 3.
Diamond is covalent, yet it has high melting point. Why ?
Solution:
Diamond has a three – dimensional network involving strong C – C bonds, which are very difficult to break and in turn has high melting point.

Question 4.
What are Silicones ?
Solution:
Simple Silicones consist of chains in which alkyl or phenyl groups occupy the remaining bonding positions on each silicon. They are hydrophobic (water repellant) in nature.