Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 8th Lesson Viruses Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 8th Lesson Viruses

Very Short Answer Questions

Question 1.
What is the shape of T₄ phage? What is its genetic material?
Answer:
Tad pole-shape. Its Genetic material is double-stranded DNA.

Question 2.
What are virulent phages? Give an example.
Answer:
T-even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages Eg: Bacteriophage.

Question 3.
What is lysozyme and what is its function?
Answer:
Lysozyme is an viral enzyme, which breaks or lysed the bacterial cell wall to release their newly produced phage particles.

Question 4.
Define ‘lysis’ and ‘burst size’ with reference to viruses and their effects on host cells.
Answer:
The cycle in which the host cell wall ruptures to release the virions is called lysis. The number of newly synthesized bacteriophage particles released from a single affected host cell is called burst size.

Question 5.
What is a prophage?
Answer:
The phage DNA which is incorporated into bacterial DNA is called prophage.

Question 6.
What are temperate phages? Give an example.
Answer:
The phage DNA gets integrated into the circular bacterial DNA, becomes part of it and remains latent. Such phages are called temperate phages. Eg : A [Lambda] phage.

Question 7.
Mention the difference between virulent phages and temperate phages.
Answer:

Question 8.
What is the shape of Tmv? What is its genetic material?
Answer:
The shape of Tmv is rod and genetic material is single stranded RNA consisting of 6500 nucleotides.

Short Answer Questions

Question 1.
What is ICTV? How are viruses named?
Answer:
International Committee on Taxonomy of Viruses [ICTV] regulates the norms of classification and nomenclature of viruses. The ICTV has only three hierarchial levels, The family, Genus and Species. The family names end with the suffix viridae’ while the Genus name ends with virus and the Species names are common English expressions . Viruses are named after the disease they cause.
Eg : Polio virus.

Question 2.
Explain the chemical structure of viruses.
Ans. All viruses consist of two basic components, a core of nucleic acid that forms the genome and the surrounding coat of protein known as capsid. The capsid gives shape to virus and also protects the genome. Capsid is made up of protein sub-units called capsomeres.

The nucleic acid may be either a double stranded DNA (ds DNA) or single stranded DNA [ss DNA]. In general, viruses that infect plants have ss RNA and viruses that infect animals have ds DNA. Most viruses have a single nucleic acid molecule but a few have more than one. Eg: HIV has two identical RNA molecules.

Question 3.
Explain the structure of TMV.
Answer:

1. It is a rod shaped virus. It is about’ 300 nm long and 18 or 19 nm in diameter with a molecular weight of 39 × 10⁶ daltons.
2. The capsid is made up of 2,130 sub-units called capsomeres.
3. Capsomeres are arranged in a helical manner around a central core of 4 nm.
4. Each protein sub-unit is made up of a single polypeptide chain with 158 amino acids.
5. Single stranded RNA is present inside the capsid and is spirally coiled.
6. RNA of TMV consists of 6,500 nucleotides.

Question 4.
Explain the structure of T-even Bacteriophages.
Answer:

1. The viruses which attack bacteria are called Bacteriophages. They were discovered by Twort (1915).
2. Felix’d’ Herelle (1917) coined the term Bacteriophage.
3. Bacteriophages are tadpole-shaped with a large head and a tail.
4. The head is hexagonal and is capped by hexagonal pyramid, measures about 65 × 95 nm.
5. The head is formed with several cap-someres, each of which is a single protein.
6. The head protein forms a semipermeable membrane enclosing the folded double stranded DNA which is 1000 times longer than the phage.
7. The tail is composed of several parts present around central core.
8. The tail core is 95 nm long and 8 nm in diameter. It is surrounded by tail sheath composed of 144 sub-units which are arranged in 24 rings of 6 each.
9. The head and tail are joined by collar whose function is not known.
10. At the tip of the tail, hexagonal tail plate is present with six tail pins and tail fibres.
11. The tail fibres help in attachment of the phage on to the host cell.

Question 5.
Explain the lytic cycle with reference to certain viruses.
Answer:
T-even Bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five step process involving (a) attachment (b) penetration (c) biosynthesis (d) maturation and (e) release

a) Attachment:
The tail fibres of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration:
The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA from the bacteriophage head passes through the tail Core through the plasma membrane and enters the bacterial cell. The capsid remains outside the bacterial cell and is called ghost. Thus the phage particle functions like hypodermic syringe and injects its DNA into the bacterial cell.

c) Biosynthesis:
When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes, and capsid proteins are synthesized using the cellular machinery of the host cell.

d) Maturation:
Bacteriophage DNA and capsids are assembled into complete virions. This period of time between the infection by a virus and the appearance of mature virions is called Eclipse period.

e) Release:
The plasma membrane of the host cell gets dissolved or lysed due to viral enzyme, lysozyme. The bacterial cell wall breaks, releasing the newly produced phage particles or virions.

Long Answer Questions

Question 1.
Write about the discovery and structural organization of viruses.
Answer:
In 1892, the Russian Pathologist Dmitri Iwanowski, while studying Tobacco mosaic disease, he filtered the sap of diseased tobacco leaf through filters and is injected into a healthy plant. He found the symptoms of Mosaic disease in it. Finally he reported that, a filterable agent was responsible for the disease. Later Martinus Beijerinck repeated Iwanowski’s experiments and concluded that, the disease causing agent was a contagious living fluid’ (contagium vivium fluidum).

W.M Stanley (1935) purified the sap and announced that the virus causing mosaic disease in tobacco could be crystallized and was named TMV. Fralnke conrael (1956) confirmed that the genetid material of TMV is RNA.

Structure:-
Viruses range in size from 300 nm as in TMV to 20 nm as in Parvoviruses. All viruses consist of two basic components, a core of nucleic acid that forms the genome and the surrounding coat of protein known as capsid. The capsid gives shape to the virus and provides a protective covering for the genome. It is made up of protein sub-units called, capsomeres. A virus contains its genetic information in either a double stranded (ds) DNA or single stranded (ss) DNA. In general, phytophages have ss RNA and zoophages have ds DNA. Bacteriophages are usually ds DNA viruses.

Shape:-
a) Helical viruses resemble long rods.
Eg: Rabies Virus, Tobacco Mosaic Virus.

b) Polyhedral viruses. They resemble polyhedral shape (many sided)
Eg : Herpes simplex and polio viruses.

c) Enveloped viruses: The capsid is covered by an envelope which are roughly spherical.
Eg : Influenza virus.

d) Complex viruses: Viruses which infect bacteria have complicated structures.
Eg : Bacteriophages have polyhedral symmetry in the head and helical symmetry in the tail sheath.

Question 2.
Describe the process of multiplication of viruses.
Answer:
T-even Bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five step process involving (a) attachment (b) penetration (c) biosynthesis (d) maturation and (e) release

a) Attachment:
The tail fibres of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration:
The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA from the bacteriophage head passes through the tail core through the plasma membrane and enters the bacterial cell. The capsid remains outside the bacterial cell and is called ghost. Thus the phage particle functions like hypodermic syringe and injects its DNA into the bacterial cell.

c) Biosynthesis:
When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes, and capsid proteins are synthesized using the cellular machinery of the host cell.

d) Maturation:
Bacteriophage DNA and capsids are assembled into complete virions. This period of time between the infection by a virus and the appearance of mature virions is called Eclipse period.

e) Release:-
The plasma membrane of the host cell gets dissolved hr lysed due to viral enzyme, lysozyme. The bacterial cell wall breaks, releasing the newly produced phage particles or varions.

Some Bacteriophages such as λ(Lambda) phages do not cause lysis and death of host cell when they multiply. Instead, the phage DNA, upon penetration into an E.coli cell gets integrated in to the circular bacterial DNA, becomes part of it and remains latent. Such phages are called temperate phages. The inserted phage DNA is now called prophages.

The prophages replicate along with the bacterial genetic material. The prophage remains latent with in the progny cells. In some rare spontaneous events, or when the host cell is exposed to UV light or some chemicals the phage DNA separates from the bacterial genetic material leading to the initiation of the lytic cycle. This lysogenic cycle facilitates the phenomenon of transduction.

Intext Questions

Question 1.
When discussing the multiplication of viruses, virologists prefer to call the process as replication, rather than reproduction. Why?
Answer:
Viruses must invade a host cell and take over the hosts metabolic machinery. So it is called replication.

Question 2.
In dealing with Public Health, the approach to deal with bacterial diseases is treatment. Can you guess the nature of the general Public Health approach to viral diseases? What example do you cite to support your answer?
Answer:
“Prevention is better than cure” – in AIDS.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 7th Lesson Bacteria Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 7th Lesson Bacteria

Very Short Answer Questions

Question 1.
Write briefly on the occurrence of microorganisms.
Answer:
Microorganisms are omnipresent in vast numbers and are described as ubiquitous. Among them, Bacteria are found in soil, water, air and on or inside living organisms. They also occur in a variety of foods, can withstand extreme cold, heat and drought conditions.

Question 2.
Define Microbiology.
Answer:
Microbiology is a branch of biology that deals with the scientific study of microorganisms. That are too small to be seen with the naked eye.

Question 3.
Name the Bacteria which is a common inhabitant of human intestine. How is it used in biotechnology?
Answer:
Escherichia coli (E.coli). It is used in biotechnology for the production of Insulin hormone.

Question 4.
What are Pleomorphic Bacteria? Give an example.
Answer:
The Bacteria which change their shape depending up on the type of environment and nutrients available are called Pleomorphic Bacteria.
Eg : Acetobacter.

Question 5.
What is sex pilus? What is its function?
Answer:
Sex pilus is a conjugation apparatus that pulls two cells together prior to DNA transfer. It help in binding two conjugants and also help in passing F plasmid from F⁺ cell to F^– cell.

Question 6.
What is a genophore?
Answer:
Genophore is the main genetic material of Bacteria which is circular, double stranded DNA without histone proteins. It is also called Bacterial chromosome.

Question 7.
What is a plasmid? What is its significance?
Answer:
“Plasmid is a small self-replicating circular, double stranded DNA molecule present in a Bacteria in addition to main genetic material”. They are used as agents (vectors) in modem Genetic Engineering techniques.

Question 8.
What is conjugation? Who discovered it and in which organism?
Answer:
Transfer of DNA from one Bacterium (Donor) to the recipient Bacterium through direct contact is called conjugation. It was first observed in Escherichia coli by Lederberg and Tatum (1946).

Question 9.
What is transformation? Who discovered it and in which organism?
Answer:
Uptake of naked DNA fragments from the surrounding environment and their incorporation into the recipient cell is called transformation. It was discovered by Frederick Griffith in Streptococcus pneumoniae (1928).

Question 10.
What is transduction? Who discovered it and in which organism?
Answer:
The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction. It was observed by Lederberg and Zinder in Salmonella typhimurium (1951).

Short Answer Questions

Question 1.
How are bacteria classified on the basis of Morphology?
Answer:
Bacteria are classified into four types based on their shape. They are
I) Cocci : Spherical Bacteria.
Based on the number of cells and their arrangement, the Cocci are classified into
(a) Monococcus : A single cell
(b) Diplococcus : A pair of cells
(c) Tetracocci : A group of four cells
(d) Streptococcus : A linear chain of cells
(e) Staphylococci : A bunch of cells
(f) Sarcina : A group of 8 cells, arranged in cubes

II) Bacilli :
Rod-shaped Bacteria : The Bacillus forms exist as
a) Monobacillus : A single rod-shaped ceil.
b) Diplobacilli : A pair of rod-shaped cells.
c) Streptobacilli : A chain of rod-shaped cells.

III) Spirillum Helical rod-shaped cell.
It may be a distinct helical shape (spirillum) or Slender long and cork-screw shaped (Spirochete).

IV) Vibrio : Comma-shaped Bacteria.

Question 2.
How are Bacteria classified on the basis of number and distribution of flagella?
Answer:
Bacteria are classified into four types based on the number and arrangement of flagella.
They are
1) Monotrichous : A single polar flagellum is present at one end of Bacterial cell.
2) Lophotrichous : A tuft of flagella at one pole of the cell.
3) Amphitrichous : A single flagellum at each end of the cell.
4) Peritrichous : Flagella distributed over the entire cell.

Question 3.
What are the nutritional groups of Bacteria based on their source of .energy and carbon?
Answer:
1) Photoautotrophs :
They are photosynthetic i.e., they capture light energy and transform it into chemical energy and obtain carbon from atmospheric CO₂.
Eg: Chlorobium, Chromatium,

2) Chemoautotrophs :
They derive energy from the oxidation of inorganic substances and carbon from atmospheric CO₂.
Eg: Nitrosomonas, Nitrobacter, Beggiaotoaand Methanogens.

3) Photoheterotrophs :
They capture light energy and carbon from other organic sources.
Eg : Rhodospirillum and Rhodopseudomonas.

4) Chemoheterotrophs:
They derive both carbon and energy from organic compounds.
Eg: Xanthomonas, Salmonella.

Question 4.
Explain the conjugation in Bacteria.
Answer:
Transfer of Genetic material between two live Bacteria is called conjugation. It was first observed in 1946 by Lederberg and Tatum in Escherichia coli.

In E.coli, a small circular DNA strand occurs in the cytoplasm in addition to nucleoid called on F plasmid. The cell with F plasmid is called F⁺ cell and without F plasmid is called F^– cell. The F⁺ cell or Donor cell produces the sex pilus that makes contact with the recipient cell or F^– cell. During conjugation, F⁺ and F^– cells bind with each other with the help of sex pilus forms a bridge between them. The F plasmid replicates and the replicated DNA passes through bridge to the F^– cell. The F^– cell becomes F⁺ cell as it receives the F plasmid. After conjugation, the two cells separate from each other.

Long Answer Questions

Question 1.
Explain different methods of sexual reproduction in Bacteria.
Answer:
True sexual reproduction is absent in Bacteria. However, the exchange of genetic material takes place in three ways. They are 1) Conjugation 2) Transformation 3) Transduction

1) Conjugation:-
Transfer of Genetic material between two live Bacteria is called conjugation. It was first observed in 1946 by Lederberg and Tatum in Escherichia coli.

In E.coli, a small circular DNA strand occurs in the cytoplasm in addition to nucleoid called on F plasmid. The cell with F plasmid is called F⁺ cell and without F plasmid is called F^– cell. The F⁺ cell or Donor cell produces the sex pilus that makes contact with the recipient cell or F cell. During conjugation, F⁺ and F^– cells bind with each other with the help of sex pilus forms a bridge between them. The F plasmid replicates and the replicated DNA passes through bridge to the F^– cell. The F^– cell becomes F⁺ cell as it receives the F plasmid. After conjugation, the two cells separate from each other.

2) Transformation:-
It is uptake of naked DNA fragments from the surrounding environment and their incorporation into the recipient chromosome in a heritable form is known as transformation. It was discovered by Frederick Griffith (1928) in Streptococcus pneumoniae.

3) Transduction:-
The transfer of genetic material from one Bacterium to another through Bacteriophage is known as transduction. It was discovered by Lederberg and Zinder in Salmonella typhimurium.

Question 2.
“Bacteria are friends and foes of man” – Discuss.
Answer:
Many Bacteria are directly or indirectly beneficial and harmful to humans. Hence they are considered both as friends and foes of man.
1) Beneficial activities:-
Some Bacteria are important to man in the field of agriculture, industry, medicine and biotechnology.

I) Agriculture :
Bacteria play a significant role in maintaining the fertility of the soil.

They are :
a) Ammonifying Bacteria:
These Bacteria convert the proteins, and amino acids of the dead bodies into Ammonia. This process is called Ammonification.
Eg: Bacillus
b) Ammonia is oxidised to nitrates, by nitrifying bacteria, This process is called Nitrification Eg: Nitrosomonas, Nitrobacter.
c) Symbiotic Bacteria like Rhizobium, non-symbiotic Bacteria like Clostridium and photosynthetic Bacteria like Rhodospirillum, Rhodomicrobium and Chlorobacterium fix the atmospheric gaseous nitrogen and enrich the soil.
d) Bacillus thuringiensis is used as an agent against insect pests. Hence popularly used as a bioinsecticide.

II) Industry:
Industrially, Bacteria are employed in a number of processes. They are :
a) Clostridium butyricum and C. felcinium are used in retting of sunhemp and flax respectively to extract the fibre.
b) Some Bacteria are used in tanning industry.
c) Bacteria such as Bacillus megaterium and Micrococcus are used in curing of tobacco and tea respectively.
d) Some Bacterm are used in fermentation process, Eg: Lactobacillus.
e) Bacteria like Methanococcus and Methanobacillus ferment the dung anaerobically and produce methane that is commonly referred to as gobar gas.
f) In various chemical industries, Bacteria are extensively used for the production of chemicals. Few, examples are given below.

III) Medicine:
a) Corynebacterium glutamicum produces lysine, an essential amino acid.
b) A large number of antibiotics are produced by Bacteria. Species of Streptomyces and Bacillus produce important antibiotics as mentioned below.

IV) Biotechnology:
a) With the help of recombinant DNA technology, it was made possible to produce insulin hormone from Escherichia coli.
b) Some Bacteria store large quantities of proteins inside their cells. These are used as a source of single cell protein (SCP).
Eg : Brevibatiterium.
c) Agrobacterium tumefaciens is used as vector in genetic engineering.

2) Harmful activities :
A few saprophytic and all parasitic Bacteria carry on some processes which are harmful to man.

i) Spoilage of food materials :
Bacteria grow on different types of food materials and render them unsuitable for human consumption. Some of these Bacteria produce powerful toxins while growing on the food materials. For instance, Clostridium botulinum releases a very potential toxin, ‘botulin’ which causes ’botulism’, a type of food poisoning.

ii) Plant diseases :
A number of species of Bacteria are reported to cause different plant diseases as mentioned below.

Some important crop diseases caused by Bacteria

Some Bacteria cause Human diseases. They are :

Intext Questions

Question 1.
Many people believe that bacteria do little more than cause human illness and infectious diseases. How does the information in the chapter help you correct that misconception?
Answer:
Bacteria are directly or indirectly beneficial to Humans. Hence they are considered both as “friends and foes of man”.

Question 2.
An organism is described as a peritrichous bacillus. How might you translate this bacteriological language into a description of the organism.
Answer:
It is Rod shaped Bacteria.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 6th Lesson Plant Growth and Development

Very Short Answer Questions

Question 1.
Define plasticity. Give an example.
Answer:
Ability of plants to follow different pathways in response to the environment or phases of life to form different kinds of structures in called plasticity.
Ex : Heterophily.

Question 2.
What is the disease that formed the basis for the identification of gibberellins in plants? Name the causative fungus of the disease.
Answer:
Bakane (foolish seedling) disease in rice seedlings. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 3.
What is apical dominance? Name the growth hormone that causes it.
Answer:
Growing Apical bud inhibits, the growth of Axillary buds is called Apical dominance. It is caused by Auxins.

Question 4.
What is meant by bolting ? Which hormone causes bolting?
Answer:
Sudden elongation of internodes prior to flowering is called bolting. It is caused by Gibberellins.

Question 5.
Define respiratory climactic. Name the PGR associated with it.
Answer:
The rise in the rate of respiration during the ripening of fruits is known as respiratory climatic. It is associated by ethylene.

Question 6.
What is Ethephon? Write its role in agricultural practices.
Answer:
It is an Ethylene releasing chemical formulation. It hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits. It promotes female flowers in cucumbers, there by increasing the yield.

Question 7.
Which of the PGRs is called stress hormone and why?
Answer:
ABC [Abscisic acid] is called stress hormone. ABA stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 8.
What do you understand by vernalisation?
Answer:
The method of inducing flowering quantitatively or qualitatively on exposure to low temperature is called vernalisation. It prevents precocious reproductive development late in the growing season, and enables the plant to have sufficient time to reach maturity. It specially refers to the promotion of flowering. It also stimulates a subsequent photoperiodic flowering response in biennials, [cabbage, carrot]

Question 9.
Define the terms quiescence and dormancy.
Answer:
The condition of a seed when it is unable to germinate only because favourable external condition normally required for growth are not present is called quiescence.

The condition of a seed when it fails to germinate because of internal conditions even though external conditions are suitable is called dormancy.

Short Answer Questions

Question 1.
Write a note on agricultural / horticultural applications of auxins.
Answer:

1. IBA, NAA and IAA help to initiate rooting in stem cuttings, widely used for plant propagation in horticulture.
2. Auxins like 2,4-D, 2,4,§-T acts as herbicides and kills broad leaved dicot weeds to prepare weed – free lawns.
3. Auxins stimulates fruit growth. Ex : Tomatoes.
4. Auxins induces flowering in pineapple.
5. Prevents pre harvest fruit drop.

Question 2.
Write the physiological responses of Gibberellins in plants.
Answer:

1. Gibberellins delay senescence. Thus fruits can be left on the tree longer so as to extend the market period. .
2. Spraying of Gibberellins on sugarcane crop, increases the length of the stem, thus increasing the yield as much as 20 tonns per acre.
3. GA hastens the maturity period of conifers thus leading to early seed production.
4. GA also promotes bolting in cabbages, beet etc.,
5. They also produce parthenocarpic fruits in grapes and tomato.
6. Gibberellins favour the formation of male flowers in cucurbita.

Question 3.
Write any four physiological effects of cytokinins in plants.
Answer:

1. Cytokinins induces cell division.
2. They help to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
3. Cytokinins help to overcome apical dominance.
4. They promote nutrient mobilisation which helps in the delay of senescence.
5. Cytokinins help in the opening of stomata by increasing the concentration of k+ ions in guard cells.

Question 4.
What are the physiological prodesses that are regulated by ethylene in plants?
Answer:

1. Ethylene promotes the ripening of fruits.
2. Ethylene promotes the senescence and abscission of leaves and flowers.
3. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds and sprouting of potato tubers.
4. Ethylene promotes rapid intemode/petiole elongation in deep water rice plants.
5. It also promotes root growth and root hair formation, thus helping plants to increase their absorption surface.
6. Ethylene is used to initiate flowering (mango) and for synchronising fruit set in pineapples.
7. It promotes female flowers in cucumbers, thereby increasing the yield.

Question 5.
Write short notes on seed dormancy.
Answer:
The inability of seed to germinate or grow is called dormancy. It may be due to either external factors or internal factors.

Internal factors:
1) Immature embryo :
The embryo has not reached morphological maturity to germinate.
Ex : Ranunculus.

2) Hard seed coat:
In fabacea members, seeds have hard seed coats which prevent uptake of oxygen or water. It can be broken by scarification in which the hard seed coat is ruptured or weakened.

3) Chemicals :
Seeds of some plants (tomato) contain chemical compounds which inhibit their germination.

External factors:
1) Low temperature treatment :
Many seeds (polygonum) with not germinate until they have been exposed to low temperatures in moist conditions in the presence of oxygen for weeks to months. The practice of layering the seeds during winter in moist sand and peat is called stratification or prechilling.

2) Seeds of many domestic plants may be limited only by lack of moisture or warm temperature.

Question 6.
Which one of the plant growth regulators would you use if you are asked to.
a) Induce rooting in a twig
Answer:
Auxins

b) Quickly ripen a fruit
Answer:
Ethylene

c) Delay leaf senescence
Answer:
Cytokinins

d) Induce growth in axillary buds
Answer:
Cytokinins

e) ‘Bolt’ a rosette plant
Answer:
Gibberellins

f) Induce immediate stomatal closure in leaves
Answer:
Abscisic acid

g) Overcome apical dominance
Answer:
Cytokinins

h) Kill dicotyledonous weeds
Answer:
Auxins – 2, 4 – D

Question 7.
Describe briefly, a) Sigmoid growth curve, b) Absolute and relative growth rates.
Answer:

a) Sigmoid Growth Curve : It consists of 3 phases namely
1) lag phase
2) log pase
3) stationary phase.

1) Lag phase :
Growth is slow

2) Log phase :
Growth increases rapidly at an exponential rate

3) Stationary phase :
Limited nutrient supply, slows down the growth leading to a stationary phase. If we plot the parameter of growth against time, we get a typical sigmoid or S-curve.

b) Absolute and relative growth rates :
Measurement and comparision of the total growth per unit time is called absolute growth rate. The growth of the given system per unit time expressed on a common basis, e.g., per unit initial parameter is called relative growth rate.

Long Answer Question

Question 1.
List five natural plant growth regulators. Write a note on discovery, physiological functions and agricultural / horticultural applications of any one of them.
Answer:
Auxins, Gibberellins, Cytokinins, Abscisic acid, Ethylene

Auxins:
Discovery :
Observation of Charles Darwin and his son Francis Darwin, that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light. It was concluded that the tip of the coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile. Auxin was isolated by F.W. Went from the tips of coleoptiles of oat seedlings.

Physiological functions :

1. Auxins help of initiate rooting in stem cuttings.
2. Auxins promote flowering in pineapples.
3. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
4. Auxins promote apical dominance.
5. Auxins also induce parthenocarpy in tomatoes.
6. They are widely used as herbicides 2,4-D widely used to kill dicotyledonous weeds.
7. Auxins also controls zylem differentiation and help in cell division.

Agricultural / Horticultural applications ;

1. Auxins help to initiate rooting in stem cuttings, widely used for plant propagation in Horticulture.
2. Auxins, 2, 4-D widely used as herbicide ie, kills dicotyledonous weeds to prepare a weed free-lawns.

Intext Questions

Question 1.
Fill in the blanks with appropriate word/words.
a) The phase in which growth is most rapid is ………… .
b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more ………… in the apical bud than in the lateral ones.
c) In addition to ouxin, a …………. must be supplied to the culture medium to obtain a good callus in plant tissue culture.
d) ………….. of vegetative plants are the sites of photoperiodic perception.
Answer:
a) LOg phase or exponential phase
b) Auxins
c) Cytokinin
d)Shootapices of plants

Question 2.
Gibberellins promote the formation of Male flowers on genetically Modified (dwarf) plants in cannabis whereas Ethylene promotes formation of Female flowers on genetically Modified (dwarf) plants.

Question 3.
Classify the following plants into long day plants(LDP), short day plants(SDP) and Day Neutral plants(DNP). Xanthium, Spinach, Henbane (Hyoscyamus), Rice, Strawberry, Bryophyllum, niger, Sunflower, Tomato, Maize.
Answer:
Long day plants – Henbane, Spinach,
Short day plants – Xanthium, Sunflower, Rice, Tomato
Day nutral plants : Maize, Bryophyllum

Question 4.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers. Which plant growth regulator can be applied to achieve this?
Answer:
Ethyline.

Question 5.
Where are the following hormones synthesized in plants?
a) IAA b) Gibberellins c) Cytokinins.
Answer:
a) IAA – growing apices of the stems and roots.
b) Gibberellins : Fungal pathogen, Gibberella fujikuroi.
c) Cytokinins – roots apices, shoot…, young fruits.

Question 6.
Light plays an important role in the life of all organisms. Name any three physiological processes in plants which are influenced by light.
Answer:
Growth, differentiation and development.

Question 7.
Growth is one of the characteristics of all living organisms. Do unicellular organisms also grow? If so, what all the parameters?
Answer:
Yes, unicellar organisms also grow, some parameters are increase in fresh weight, length, area, volume and cell number.

Question 8.
Rice seedlings infected with fungus Gibberella fujikuroi are called Foolish seedlings. What is the reason?
Answer:
The fungus cause foolish seeding disease in rice which show, plant grow. Very tall, become pale, produce less tillers and less yield.

Question 9.
Why isn’t any one parameter good enough to demonstrate growth throughout the life of a flowing plant?
Answer:
One parameter is not good enough to demonstrate growth because various parts show growth… is estimated in different parameters Eg: Growth of the pollentube is measured interms of the length, increase in surface are dexoles growth in a dorsiventral leaf.

Question 10.
‘Both growth and differentiation in higher plants are open’ Comment.
Answer:
Both growth and diffentitation in higher plants are open because, cells/tissues arising out of the same meristem have different structures of maturity and their location. Eg. : Cells away from root apical meristem differentiate as root cap cells, while those pushed to the periphery nature as epidermis.

Question 11.
Both a short day plant and a long day plant can produce flowers simutaeneously in a given place’. Explain.
Answer:
5 me plants require the exposure to light for a period exceeding critical duration and some require a light for a period less than critical duration for flowing.

Question 12.
Would a defoliated plant respond to photoperiodic cycle? Why?
Answer:
No, because some hormonal substance migraterwom leaver to shoot apices for induce flowing.

Question 13.
What would be expected to happen if
a) GAj is applied to rice seedlings.
b) Dividing cells stop differentiating.
c) A rotten fruit gets mixed with unripe fruits.
d) You forget to add cytokinnin to culture medium.
Answer:
a) Appearance of disease symptoms (Bakane disease) in uninfected rice seedings.
b) Growth of the plant body in stopped.
c) Unripe fruits became ripened due to ethylene.
d) Initiation of heafy shoots is inhibited.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 5th Lesson Respiration in Plants

Very Short Answer Questions

Question 1.
Different substrates get oxidized during respiration. How does respiratory quotient (RQ) indicate which type of substrate (i.e) carbohydrate, fat or protein is getting oxidized?
RQ = A/B. What do A & B stand for?
What type of substrates have RQ of 1, <1; > 1?
Answer:

carbohydrates have RQ = 1
Fats have RQ value = < 1
Organic acids have RQ = > 1

Question 2.
What is the specific role of F₀ – F₁ particles in Respiration?
Answer:
The F₁ headpiece is a peripheral membrane protein complex and contains the site for the synthesis of ATP from ADP and inorganic phosphate. F₀ is an integral membrane protein complex that forms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F₁ component for the production of ATP

Question 3.
When does anaerobic respiration occur in man and yeast?
Answer:
When oxygen is inadequate for cellular respiration, anaerobic respiration occurs.

Question 4.
What is the common pathway for aerobic and anaerobic respirations? Where does it take place?
Answer:
Glycolysis. It occurs in cytoplasm of the cell.

Question 5.
What cellular organic substances are never used as respiratory substrates?
Answer:
Pure proteins or fats are never used as respiratory substrates.

Question 6.
Why is the respiratory quotient (RQ) of fats less than that of carbohydrates?
Answer:
During the oxidation of fats, the amount of CO₂ evolved is less than the amount of O₂ consumed so respiratory quotient value is less than one. Where as in oxidation of carbohydrates, the amount of CO₂ evolved is equal to the amount of O₂ consumed. So the respiratory quotient value is one.

Question 7.
What is meant by Amphibolic pathway?
Answer:
Respiratory pathway involves both anabolism and catabolism so it is considered as amphibolic pathway. In this, the breakdown and the synthesis of fatty acids and proteins took place.

Question 8.
Name the mobile electron carriers of the respiratory electron transport chain in the inner mitochondrial membrane.
Answer:
Ubiquinone and cytochrome – C.

Question 9.
What is the final acceptor of electrons in aerobic respiration? From which complex does it receive electrons?
Answer:
Oxygen. It receives electrons from complex IV

Question 10.
Do you know of any step in Kreb’s cycle where there is a substrate level phosphorylation? Explain.
Answer:
In Kreb’s cycle during the conversion of succinyl – CoA to succinic acid a molecule of GTP is synthesised. This step is a substrate level phosphorylation.

Short Answer Questions

Question 1.
Why is the respiratory pathway referred to as amphibolic pathway? Explain.
Answer:
Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, j they would be degraded to acetyl CoA and enter the pathway Glycerol would enter the pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate.

Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. Rut when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one.

Question 2.
Write about two ATP yielding reactions of glycolysis.
Answer:
1) 1, 3 BPGA (bisphosphoglyceric acid) looses phosphate group in the presence of phosphoglycerokinase to form 3-phosphoglyceric acid. ADP accepts phosphate group and gets converted to ATP.

2) Phosphoenol pyruvic acid undergoes dephosphorylation in the presence of pyruvic kinase results in the formation of pyruvic acid. ADP accepts phosphate group and gets converted to ATP.

Question 3.
The net gain of ATP for the complete aerobic oxidation of glucose is 36. Explain.
Answer:
Balance sheet of ATP production in aerobic oxidation of Glucose.
1) Glycolysis :
1. ATP produced by substrate level phosphorylation
Bisphosphoglyceric acid to phosphoglyceric acid : 2 × 1 = 2ATP
phosphoenol pyruvic acid to pyruvic acid : 2 × 1 = 2ATP
ATP consumed : for the phosphorylation of glucose
and fructose-6-phosphate : -2 ATP
Net gain of ATP : +2 ATP

2. ATP from NADH generated in glycolysis :
G-3-P to BPGA (2NADH, each worth 2ATP) 2 × 2 = 4ATP
Total ATP gain from glycolysis in the presence of O₂ : ^(a)6 ATP

2) Oxidative decarboxylation of pyruvic acid
Pyruvic acid to acetyl COA
(2 NADH, each worth 3 ATP) : ^(b)2 × 1 = 6 ATP

3) Krebs cycle
1. ATP produced in substrate level phosphorylation:
Succinyl CoA to succinic acid : 2 × 1 = 2 ATP
2. ATP from NADH: Isocitric acid to Oxalosuccinic acid : 2 × 3 = 6 ATP
a-Ketoglutaric acid to succinyl CoA : 2 × 3 = 6 ATP
Malic acid to Oxalocaetic acid : 2 × 3 = 6 ATP
3. ATP from FADH2 : Succinic acid to fumaric acid : 2 × 2 = 4 ATP
Total ATP value of krebs cycle : ^(a)24 ATP
Net gain of ATP in aerobic respiration per mole glucose
(a + b + c) : 36 ATP

Question 4.
Define RQ. Write a short note on RQ.
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ cosumed in respiration is called the Respiratory Quotient (RQ).

The respiratory quotient depends upon the type of respiratory substrate used during respiration.
1) When carbohydrates are used as substrate, the RQ is 1, because equal amounts of CO₂ and O₂ are evolved and consumed.
Ex : C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + Energy
RQ = \(\frac{6CO_2}{6O_2}\) = 0.1

2) When fats are used in respiration the RQ is less than 1.
Ex : 2(C₅₁ H₉₈ O₆) + 145 O₂ → 1 O₂ CO₂ + 98 H₂O + Energy Tripalmitin
RQ = \(\frac{102CO_2}{145O_2}\) = 0.7

3) When protein are respiratory substrates the ratio would be about 0.9

Question 5.
Describe briefly the process of fermentation.
Answer:
In fermentation, the incomplete oxidation of glucose is achieved under anaerobic conditions by set of reactions where pyruvic acid is converted to CO₂ & Ethanol. The ezymes, pyruvic acid decarboxylase, and alcohol dehydrogenase catalyse these reactions. Two types of fermetations are :

1. Alcoholic fermentation
2. Lactic acid fermentation

1) Alcoholic fermentation :
This is most common type of fermentation process. In alcoholic fermentation, pyruvic acid is broken to ethyl alcohol & CO₂ in two steps.
i) The pyruvic acid is decarboxylated to acetaldehyde in the presence of enzyme pyruvic decarboxylase.

ii) Acetaldehyde is then reduced to ethyl alcohol by NADH⁺ + H⁺ produced in glycolysis acetaldehyde is final hydrogen acceptor. The reaction is catalysed by alcohol dehydrogenase.

2) Lactic acid fermentation:
Pyruvic acid formed at the end of glycolysis is converted to lactic acid by lactobacillus lactibacteria.

Question 6.
Explain various complexes involved in electron transport system of respiration.
Answer:
Electron flow through the mitochondrial electron transport (Aerobic respiration) is carried out by five enzyme complexes. These complexes are the integral proteins of the inner mitochondrial membrane. They are complex I, II, III, IV & V out of these first four complexes involved in Electron transport chain.
i) Complex I :
NADH – dehydrogenase (or) NADH-Q-reductase. This complex I transfers electrons from NADH to ubiquinone.

ii) Complex II :
Succinate dehydrogenase (or) ubiquinone.This complex trasfers electrons from succinate to ubiquinone via Fe – S – centres.

iii) Complex III :
Cytochrome ‘c’ Reductase or cyt-b-c, complex. This complex contains cytochrome ‘b’ and cytochrome ’c’ cyt-c(mobile carrier). The ultimate electron acceptor of complex-III. This complex works through Q-cycle mechanism.

iv) Complex IV :
Cytochrome -c- oxidase. It contains two heme proteins called cyta and cyta₃ and two copper proteins. This complex transfers electrons to oxygen.

v) Complex V :
ATP synthase or F₀ – F₁ complex. This complex involved for the production of ATP from ADP & inorganic phosphate.

Question 7.
Describe the structure of Complex-V and explain the process of oxidative phospho-rylation as explained by chemosmotic hypothesis.
Answer:

Complex – V is an ATP synthase. This complex-V consists of two major components, F₁ and F₀. The F₁ head piece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate. F₀ is an integral membrane protein complex that froms the channel through which protons cross the inner membrane. The passage of protons through the channel is coupled to the catalytic site of the F₁ component for the production of ATP For each ATP produced, 3H⁺ passes through F₀ from the intermembrane space to the matrix down the electrochemical proton gradient.

In photophosphorylation light energy is utilised for the production of proton gradient required for phosphorylation, but in respiration it is the energy of oxidation reduction utilized for the proton gradient. It is for this reason that the process is called oxidative phosphorylation. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that one molecule of FADH₂ produces 2 molecules of ATP.

Long Answer Questions

Question 1.
Give an account of glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Answer:
Glucose is broken down into 2 molecules of pyruvic acid is called glycolysis. It was given by gustav embden, Otto Mayerhof and J.Parnas so called EMP pathway. It occurs in the cytoplasm of the cell and takes place in all living orgnisms. In this, 4 ATP are formed of which two are utilised and 2 NADPH⁺ K⁺ are formed. A⁺ the end of glycolysis, 2PA, 2ATP and 2NADPH⁺ H⁺ are formed as end products. The ATP and NADPH⁺ H⁺ are utilised for fixation of CO₂.

Glycolysis occur in cytoplasm. Pyruvic acid, 2ATP, 2NADPH⁺H are the end products. In aerobic respiration, pyruvic acid, 2 NADPH + H⁺ are completely oxidised through TCA cycle, ETS pathway and produce 36 ATP molecules. In Anaerobic respiration, pyruvic acid is partially oxidised results in the formation of Ethyol alcohol and CO₂.

Question 2.
Explain the reactions of Krebs cycle.
Answer:

The acetyl CoA enters into the [mito chondrial matrix] a cyclic pathway tricarboxylic acid cycle, more commonly called krebs cycle after the scientist Hans Krebs who first elucidated it.

1) Condensation :
In this acetyl CoA condenses with oxaloacetic acid and water to yield citric acid in the presence of citrate synthetase and CoA is released.

2) Dehydration :
Citric acid looses water molecule to yield cisaconitic acid in the presence of aconitase.

3) Hydration :
A water molecule is added to cis aconic acid to yield isocitric acid in the presence of a conitase.

4) Oxidation I:
Isocitric acid undergoes oxidation in the presence of dehydrogenase to yield succinic acid

5) Decarboxylation :
Oxalosuccinic acid undergoes decarboxylation in the presence ofdecarboxylase to form a-keto glutaric acid.

6) Oxidation II, decarboxylation :
α – keto glutaric acid undergoes oxidation and decarboxylation in the presence of dehydrogenase and condenses with co.A to form succinyl co. A.

7) Cleavage :
Succinyl co.A splits into succinic acid and co.A in the presence of thiokinase to form succinic acid. The energy released is utilised to from ATP from ADP and PI.

8) Oxidation – III:
Succinic acid undergoes oxidation and forms Fumaric acid in the presence of succinic dehydrogenase.

9) Hydration :
A water molecule is alcohol to Fumaric acid in the presence of Fumarase to form Malic acid.

10) Oxidation IV :
Malic acid undergoes oxidation in the presence of malic dehydrogenase to form oxaloacetic acid.

In TCA cycle, for every 2 molecules of Acetyl co.A undergoing oxidation, 2 ATP, 8 NADPH+ H⁺, 2FADH₂ molecules are formed.

Intext Questions

Question 1.
Differentiate between
a) Respiration and Combustion
b) Glycolysis and Krebs cycle
c) Aerobic respiration and Fermentation
Answer:
a)

b)

c)

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
Substances which oxidises in respiration are called respiratory substrates. Ex: Glucose, Fats, Proteins and Organic acids. Among them glucose is the common respiratory substrate.

Question 3.
Give the schematic representation of glycolysis.
Answer:

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
Glycolysis = Cytoplasm
PAOD = Matiix (Mitochondria)
Krebs cycle = Matrix (Mitochondria)
Electron transport system = Matrix (Mitochondria)

Question 5.
Give the schematic representation of an overall view of Krebs cycle.
Answer:

Question 6.
Explain ETS
Answer:
NADPH + H⁺ or FADH₂ formed in glycolysis and Mitochondrial matrix are oxidised through the electron transport system and the electrons are passed on to O₂ resulting in the formation of H₂O. The metabolic pathway through which an electron passes from one carrier to another is called electron transport system.

NADPH + H⁺ is oxidised and releases electrons in the presence of NADH dehydrogenase (complex I), which are transferred to ubiquinone located with in the inner membrane. Ubiquinone also receives reducing equivalents via FADH₂ (Complex II). The reduced ubiquinone (Ubiquinol) is oxidised and releases electrons which are accepted by cytochrome c₁ Cytochrome_c (Complex III) cytochrome C is a small protein, acts as a mobile carrier for the transfer of electrons between complex III and TV to cytochromes a and a₃, finally reaches \(\frac{1}{2}\)O₂ along with 2H⁺, produce 1 H₂O molecule.

During this electron flow, 10H⁺ moves from matrix to inner mitochondrial membrane [4 H⁺ at complex I, 4H⁺ at complex III, 2 at complex IV]. As a result H⁺ concentration increases towards the inner membrane of mitochondria. So the H⁺ comes back to the matrix side through ATPage [F₀, F₁], involves in the synthesis ATP.

Question 7.
Distinguish between the following
a) Aerobic respiration and Anaerobic respiration
b) Glycolysis and Fermentation
c) Glycolysis and Citric acid cycle
Answer:
a)

b)

c)

Question 8.
What are the assumptions made during the calculations of net gain of ATP?
Answer:
It is possible to make calculations of the net gain of ATP for every glucose is oxidised.
These calculations can be made only on certain assumptions that:

1. There is a sequential, orderly pathway functioning with one substrate forming the next and with glycolysis, TCA cycle, ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesize any other compound.
4. Only glucose is being transferred – no other alternative substrates are entering in the pathway at any of the intermediary stages.

But these kind of assumption are not really valid in a living system all pathways work simultaneously and do not take place one after another.

Differences

Question 9.
Discuss “The respiratory pathway is an amplibolic pathway”.
Answer:
Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, they would be degraded to acetyl CoA and enter the pathway Glycerol would enter the 1 pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate.

Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. Rut when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one.

Question 10.
Define respiratory quotient (RQ). What is its value for fats?
Answer:
The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called respiratory quotient.
Respiratory quotient value of fats is less than one
2C₅₁ H₉₈ O₆ + 145 O₂ → 1 02 CO₂ + 98 H₂O + energy
(Tripalmitin) Respiratory quotient = \(\frac{102}{145}\) = 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
Synthesis of ATP from ADP and Pi coupled to electron transport from substrate to molecular oxygen is called oxidative phosphorylation.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough such that the energy is released can be coupled to ATP synthesis. All the energy entained in a substrate is not released free into the cell or in a single step. It is release in a series of slow step wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

13. Find the correct ascending sequence of the following, on the basis of energy released in respiratory oxidation.
a) 1 gm pf fat
b) 1 gm of protein
c) 1 gm of glucose
d) 0.5 gm of protein + 0.5 gm of glucose
Answer:
c → b → d → a

Question 14.
Name the products, respectively in aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast.
Answer:
In aerobic glycolysis :
Pyruvic acid, ATP, NADPH + H⁺

In skeletal muscle :
Lactic acid

In anaerobic fermentation :
CO₂ and Ethanol

Question 15.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Why?
Answer:
Glucose or fruit juice. They contain more sugars which oxidise in the presence of O₂ and releases more energy.

Question 16.
In a way green plants and cyanobacteria have synthesised all the food on earth. Comment.
Answer:
Only green plants and cyanobacteria can prepare their own food but not all food on earth.

Question 17.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Answer:
Red muscle fibres utilise fats or carbohydrates as their food. They can work for long time but with little force.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 4th Lesson Photosynthesis in Higher Plants Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 4th Lesson Photosynthesis in Higher Plants

Very Short Answer Questions

Question 1.
Name the processes which take place in the grana and stroma regions of chloroplasts.
Answer:
The processes that take place in the grana and stroma regions of chloroplasts are :
Trapping of light energy and is used for the synthesis of ATP and NADPH occur in the grana regions of chloroplast, whereas enzymatic reactions incorporate CO₂ into the plant, leading to the synthesis of sugar occurs in the stroma region of chloroplast.

Question 2.
Where does the photolysis of H₂O occur? What is its significance?
Answer:
Photolysis of water occurs in the grana of the chloroplast. It splits into protons, electrons and oxygen. This creates oxygen which is one of the net products of photosynthesis.

Question 3.
How many molecules of ATP and NADPH are needed to fix a molecule of CO₂ in C₃ plants? Where does this process occur?
Answer:
3 ATP and 2 NADPH are needed. It occurs in stroma of the chloroplast.

Question 4.
Mention the components of ATPase enzyme. What is their location? Which part of the enzyme shows conformational change?
Answer:
The ATPase enzyme consists of 2 parts:
One called F₀ is embedded in the membrane, that carries out facilitated diffusion of protons across the membrane. The other part is called F₁ which protrudes out into the stroma. F₁ particle of the ATPase shows confirmational change.

Question 5.
Distinguish between Action Spectrum and Absorption Spectrum.
Answer:

Question 6.
Of the basic raw materials of photosynthesis, what is reduced? What is oxidised?
Answer:
CO₂ is reduced. Water is oxidised.

Question 7.
Define the law of limiting factors proposed by Blackman.
Answer:
In a process participated by a number of seperate factors, the rate of the process is limited by the factor which is present in minimal value.

Question 8.
What is the primary acceptor of CO₂ in C₃ plants? What is first stable compound formed in Calvin Cycle?
Answer:
Primary acceptor of CO₂ in C₃ plants is RUBP. First stable compound formed is PGA (3C).

Question 9.
What is the primary acceptor of CO₂ in C₄ plants. What is the first compound formed as a result of primary carboxylation in the C₄ pathway?
Answer:
Primary acceptor of CO₂ in C₄ plants is PEPA.
First compound formed is = OAA (4C).

Short Answer Questions

Question 1.
Draw a neat labelled diagram of chloroplast.
Answer:

Question 2.
Tabulate any eight differences between C₃ and C₄ plants/cycles.
Answer:

Question 3.
Describe C₄ pathway.
Answer:
H.PKortschak and C.E Hartt found that in sugarcane the first products of photosynthesis were acids containing 4 – c atoms, rather than the 3c – acid. The same has been found true for many tropical plants.
Ex : Maize, Elensine, Sorghum, Amaranthus & Euphorbia etc. These plants are called C₄ plants. Hatch and Slack explained the manner of CO₂ fixation and reduction in such plants.

The leaves of C₄ plants show Kranz anatomy, means ‘wreath’ there vascular bundles are surrounded by two rings of cells, i) Bundle sheath cells 2) Mesophyll cells.

1. In C₄ – pathway, the primary acceptor of CO₂ is phospjioenol pyruvic acid (PEP) and which is converted into oxaloacetate in the presence of enzyme PEP carboxylase.
2. The oxaloacetate is converted into malate or aspartate depending upon the species.
3. The malate formed in mesophyll cells is transferred to the neighbouring bundle sheath cells via plasmodesmata between these cells.
4. In the bundle sheath cell the malate is now decarboxylated producing pyruvate and CO₂ by action of NADP specific malic enzyme.
5. In the bundle sheath cells the CO₂ arising from the decarboxylation of malate is now fixed again in exactly the same calvin pathway of C₃ plants.
6. Sucrose & starch are ultemately formed from 3 PGA in bundle sheath cells.
7. The pyruvate formed by decarboxylation of malate in the bundle sheath is transported bock to the mesophyll cells where it is converted into phosphoenol pyruvate by the action of enzyme pyruvate phosphate dikinase.

Question 4.
Describe in brief photorespiration.
Answer:

1. Photorespiration is a light dependent release of CO₂ and uptake of O₂ by the green tissue of particularly C₃ plants. To understand photorespiration we must know first step in calvin cycle. In which RuBP combines to form 2 molecules of 3PGA that are catalysed by RuBiSCO.
2. RuBiSCO is the most abundant enzyme in the world. It has active site can bind to both CO₂ and O₂. RuBiSCO has a much greater affinity for CO₂ than O₂.
3. In C₃ plants some O₂ does bind to RuBiSCO and hence CO₂ fixation is decreased. Here the RuBP instead of being convert 2 molecules of PGA, binds with O₂ to form one molecule of phosphoglycerate and phosphoglycolate in pathway called photorespiration.
4. In photorespiration there is no synthesis of ATP or NADPH & sugars. Therefore, photorespiration is a wasteful process.
5. In C₄ plants photorespiration does not occur. This is because they have mechanism that increase the cone of CO₂ at the enzyme site.

Long Answer Questions

Question 1.
In the light of modem researches describe the process of electron transport, cyclic and non-cyclic photopnosphorylation.
Answer:
Electron transport and mechanism of ATP formation :
The synthesis of ATP from ADP and Pi in presence of light within the chloroplast is called photophosphorylation. It is of two types-viz., non – cyclic photophosphorylation and cyclic photophosphorylation.

Non – cyclic photophosphorylation:
In chloroplasts, the electrons continuously supplied from water, are transported to photosystem-II, to photosystem-I and finally to NADP+ which is reduced to NADPH. This transport of electrons is called non-cyclic electron transport. Formation of ATP in association with non-cyclic electron transport is called non-cyclic photophosphorylation.

Non – cyclic electron transport requires two large, multimolecular complexes called as photosystem -I and photosystem – II. The two complexes operate in series and are linked by electron carrier molecules. The photo systems and electron carriers are arranged in the thylacoid membranes of chloroplasts.

Photosystem-I consists of several molecules of chlorophyll a, some molecules of chlorophyll b and some β – carotene molecules. The reaction centre of PS -I consists of four to six molecules of chlorophyll a and is designated as P₇₀₀. Light harvesting complex I (chlorophyll – protein complexes) is associated with PS I. PS I is activated by red light of longer wavelength (> 682 nm) when PS I absorbs red light, the absorbed light energy is transferred to P₇₀₀.

It is first excited and then oxidized. The electron released from P₇₀₀ is transferred to a special chlorophyll a. The electron is then passed through quinone and Fe – S – proteins and finally to ferredoxin. From ferredoxin, electron is transferred to NADP⁺ which is reduced to NADPH. This reduction takes place in the stroma and catalysed by the enzyme ferredoxin NADP⁺ oxidoreductase.

Photosystem – II consists of several molecules of chlorophyll a, some molecules of chlorophyll b and some β – carotene molecules. The reaction centre of PS – II consists of four to six molecules of chlorophyll a and is designated as P₆₈₀. Light harvesting complex – II is associated with PS – II. PS – IT is activated by red light of shorter wavelength (< 682 nm). When PS – II absorbs red light, the absorbed light energy is transferred to P₆₈₀. It is first excited and then oxidised.

The electron released from P₆₈₀ is transferred to pheophytin (a colourless chi. a molecule which lacks magnesium ion). From pheophytin electron is quickly transferred to quinone and then to plastoquinone, which requires two electrons. Reduced plastoquinone picks up two protons from stroma and moves from PS – II to cytochrome b6 f complex. Electrons from reduced plastoquinone are transferred first to Fe – S – protein and then to cytochrome f. Protons released from oxidised plastoquinone diffuse into lumen of thylacoid membrane. From cyt. f the electron is transported to plastocyanin (a mobile protein containing copper ion). From plastocyanin electron moves to P₇₀₀ of PS – I, bringing it to reduced or normal state.

The electrons that reduce oxidised P₆₈₀ are supplied by a cluster of four manganese ions which are associated with oxygen evolving Complex (OEC) proteins. OEC splits water and releases electrons and oxygen. OEC is located on the lumen side of thylacoid membrane. Cl^– also bind to OEC.

The result of electron transport from water to PS – II and from PS – II to PS – I is accumulation of protons in the lumen of thylacoid membrane. The thylacoid membrane is impermeable to protons. Thylacoid membrane shows many proton channels called ATP synthase or coupling factors. A coupling factor consists of a basal, hydrophobic complex (CF₀) and apical, hydrophilic complex (CF₁). CF₀ contains the active site for ATP synthesis and CF₀ forms a proton channel across the membrane. When protons diffuse from thylacoid lumen into stroma through coupling factor, ATP is synthesised from ADP and Pi.

Cyclic photophosphorylation:
In chloroplasts, when sufficient amount of NADP⁺ is not available the electrons released from activated P₇₀₀ of PS – I return to P₇₀₀ through plastoquinone, cytochrome complex and plastocyanin. This electron transport is independent of PS – II and called cyclic electron transport. Formation of ATP associated with cyclic electron transport is called cyclic photophosphorylation.

Question 2.
Explain Calvin cycle.
Answer:
The plants in which the first stable product of photosynthesis is 3 – PGA are called C₃ plants. The path of carbon in C₃ plants was traced by Melvin Calvin, Benson and S.A. Bashaam (1945-1957).

The path of carbon in C₃ plants can be represented in the form of a cycle called reductive pentose phosphate pathway or Calvin – Benson cycle or Photosynthetic Carbon Reduction (PCR) cycle. It takes place in stroma of the chloroplast and includes three main parts, namely carboxylation, reduction and regeneration.

I) Carboxylation:
The carbon atom of carbondioxide is first fixed by RuBP, A6 – Carbon unstable compound is formed. It is spontaneously hydrolysed into two molecules of PGA. The enzyme RuBP ase (Rubisco) catalyses the reaction. The enzyme is light activated.

II) Reduction :
The second part of path of carbon is reduction of PGA to phospho- glyceraldehyde. This part includes the following reactions.

1. PGA is phosphorylated by phosphoglycero kinase to yield B is PGA. ATP donates the phosphate group.
2. B is PGA is reduced by phosphoglycero dehydrogenase to yield phospho- glyceraldehyde. NADPH donates the hydrogen atoms. Energy rich phosphate group is released.
   Some molecules of phosphoglyceradehyde are used to form starch.

III) Regeneration :
For the continuity of photosynthetic carbon reduction, the primary acceptor RuBP should be regenerated. Regeneration of RuBP takes place in three pathways.

First pathway:

1. Some molecules of PGAL (3c) are isomerised by triose phosphate isomerase to yield DHAP (3c).
2. Some molecules of DHAP and PGAL are condensed into fructose bis-phosphate (6c) by aldolase.
3. Fructose bis – phosphate is dephosphoiylated into fructose monophosphate (6c) by . phosphatase. The enzyme is light activated. Energy rich phosphate radical is released.
4. Fructose monophosphate reacts with PGAL to form xylulose monophosphate (5c)
   and Eiythrose monophosphate (4c). Transketolase catalyses the reaction.
5. Xylulose monophosphate is epimerised into Ribulose monophosphate (5c) by phosphopentose epimerase.
6. Ribulose monophosphate is phosphorylated by phosphoribulokinase to regenerate RuBP (5c). ATP donated the phosphate group. The enzyme is light activated.

Second Pathway:

1. Erythrose monophosphate (4c) condenses with DHAP (3c) to form Sedoheptulose bisphosphate (7c) Aldolase catalyses the reaction.
2. Sedoheptulose bis – phosphate is dephosphorylated byphosphatase into sedo-heptulose monophyosphate (7c). Energy rich phosphate radical is released. The enzyme is light – activated.
3. Sedoheptulose monophosphate (7c) reacts with PGAL (3c) to form xylulose monophosphate (5c) and ribose monophosphate (5c). Transketolase catalyses the reaction.
4. Xylulose monophosphate is epimerised by phosphopentose epimerase into Ribulose monophosphate (5c).
5. Ribulose monophosphate is phosphorylated by phosphoribulokin-ase to regenerate RuBP. ATP donates the energy rich phosphate group. The enzyme is light activated.

Third Pathway:

1. Ribose monophosphate (5c) is isomerised by phosphopentose isomerase into ribulose monophosphate.
2. Ribose monophosphate is phosphorylated by phosphoribulokinase to regenerate RuBP. ATP donates the energy rich phosphate group. The enzyme is light activated.

Intext Questions

Question 1.
By looking at a plant externally, can you tell whether a plant is C₃ or C₄? Why and how?
Answer:
C₄ plants have greater productivity of Biomass and can tolerate higher temperatures.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Answer:
Large bundle sheath cells are present around the vascular bundles in C₄ plants and the C₄ plants are identified.

Question 3.
Even though a very few cells in a C₄ plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
Answer:
In C₄ plants CO₂ fixation occur in two types of cells, mesophyll and bundle sheath cells so they are highly productive.

Question 4.
RUBisCO is an enzyme that acts as both as a carboxylase and oxygenase. Why do you think RUBisCO carries out more carboxylation in C₄ plants?
Answer:
The bundle sheath cells in C₄ plants are rich in RuBisCO but lack PEP case. Thus RuBisCO carries out more carboxylation.

Question 5.
Suppose there were plants that had a’ high concentration of chlorophyll b, but lacked chlorophyll a, would it carryout photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Answer:
Yes, they carry out photosynthesis. Chlorophyll b, xanthophylls and caroteriods also absorb light and protects the chlorophyll a from photpoxidation.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Answer:
If we kept the leaf in dark, photosynthesis does not occur so the leaf turns to yellow or pale green. Chlorophyll is the pigment more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Answer:
The leaves of the plant kept in sunny side get more light to absorb and is utilised for photosynthesis become dark green than the plants kept on shady side.

Question 8.
Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.
a) At which point/s (A, B or C) in the curve is light a limiting factor?
b) What could be the limiting factor/s in region A?
c) What do C and D represent on the curve?

Answer:
a) In Region ‘A’ Light is limiting factor where photosynthesis is minimum.
b) Light is the limiting factor in region A.
c) ’D’ shows the optimum point and gives the light intensity at which maximum photosynthesis is recorded. ‘C shows the rate of photosynthesis remains constant after this point even though the density of light changes.

Question 9.
Give comparison between the following :
a) C₃ and C₄ pathways
b) Cyclic and Non-cyclic photophosphorylation
c) Anatomy of leaf in C₃ and C₄ plants.
Answer:
a)

b)

c)

Question 10.
Cyanobacteria and some other photosynthetic bacteria do not have chloroplasts. How do they conduct photosynthesis?
Answer:
In Cyanobacteria, the peripheral cytoplasm contains the photosynthetic apparatus. Flattened membrane vesicles called thylakoids house the pigments and proteins that make up photosynthetic machinery. Each thylakoid is studded with phycobillisomes (Chlorophyll a) and accessory pigments (phycobilliproteins) trap photons is used in photosynthesis.

Question 11.
Why photorespiration does not occur in C₄ plants?
Answer:
C₄ plants have a mechanism that increases the concentration of CO₂ at the enzyme site. This takes place when C₄ acid from the Mesophyll cell is broken down in the bundle sheath cells to release CO₂, results in increasing intracellular concentration of CO₂. In turn RUBIsCO functions as carboxylase, minimising the oxygease activity.

Question 12.
Tomatoes, chillis and carrots are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Answer:
Carotene. Yes, it is a photosynthetic pigment.

Question 13.
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?
Answer:
No, it cannot carry out photosynthesis. If sugar solution is given to plant as supplement, it maintains its growth up to certain extent.

Question 14.
Photosynthetic organisms occur at different depths in ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesise under these conditions?
Answer:
Yes, green sulphur bacteria carry out photosynthesis even live in deeper part of the ocean. In deeper ocean, they derive energy from chemical reactions that do not need light, e.g. : Chemoautotrophs.

Question 15.
In tropical rain forests, the canopy is thick and shorter plants growing below it, receive filter light. How are they able to carry out photosynthesis?
Answer:
Due to thick canopy, small plants cannot get enough sunlight, so they adapt to live on the huge branches of large trees as Epiplytes and receive light for photosynthesis.

Question 16.
Why do you believe chloroplast and mitochondria to be semiautonomous organelle?
Answer:
The chloroplast and mitochondria are semiautnomous organelles because they have its own genetic material (DNA) Ribosomas and the ability to make its own proteins.

Question 17.
Is it correct to say. that photosynthesis occurs only in the leaves of plant? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis? Justify.
Answer:
Photosynthesis occurs in all green parts of the plant and green roots also (Taeniophyllum) because green parts consists of chloroplasts in them.

Question 18.
What can we conclude from the statement that the action and absorption spectrum of photosynthesis overlap? At which wavelength do they show peaks?
Answer:
Yes, they overlap. Rate of photosynthesis is high at red light and blue light. Light absorbed is also high during Red light and Blue light.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 3rd Lesson Enzymes

Very Short Answer Questions

Question 1.
How are prosthetic groups different from co-factors?
Answer:
Prosthetic groups are organic compounds that are tightly bound to the apoenzyme (protein part of the enzyme) whereas cofactors are non-protein parts of holoenzyme.

Question 2.
What is meant by feedback inhibition?
Answer:
The end product of a chain of enzyme-catalysed reactions inhibits the enzyme of the first reaction as a part of homostatic control of metabolism is called feedback inhibition.

Question 3.
W hy are ‘oxido reductases’ so named?
Answer:
Enzyme which catalyse oxidation and reduction of substrates usually involving hydrogen transfer are called oxido reductases.

Question 4.
Distinguish between apoenzyme and cofactor.
Answer:
The protein part of a holoenzyme is called apoenzyme the non-protein part of a holoenzyme is called cofactor.

Question 5.
What are competitive enzyme inhibitors? Mention one example.
Answer:
Substances which are closely resembles the substrate molecules and inhibits the activity of the enzyme are called competitive inhibitors.
Eg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.

Question 6.
What are non-competitive enzyme inhibitors? Mention one example.
Answer:
The inhibitor has no structural similarity with the substrate and bind to an enzyme of locations other than the active sites so that the globular structure of the enzyme is changed are called non-Competitive enzyme inhibitors.
Eg : Metal ions of Copper, Mercury.

Question 7.
What do the four digits of an enzyme code indicate?
Answer:
In enzyme code, the first digit indicates the major class of the enzyme the second digit and the third digit indicate sub class and sub-sub class respectively. The last digit of the code is the serial number of the enzyme.

Question 8.
Who proposed ‘Lock and Key hypotheses’ and ‘Induced fit hypothesis’?
Answer:
Lock and key hypothesis was proposed by Emil Fisher (1884). Induced fit hypothesis was proposed by Daniel E. Koshland (1973).

Question 9.
Define Michael’s constant?
Answer:
Substrate cencentration required to cause half the maximal reaction rate is termed as michalis menten constant (km).

Short Answer Questions

Question 1.
Write briefly about enzyme inhibitors.
Answer:
The chemicals that can shut off enzyme activity are called inhibitors. They are of 3 types.
1) Competitive inhibitors:
Substances which closely resemble the substrate molecules and inhibits the activity of the enzyme are called competitive inhibitors.
Eg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.

2) Non-competitive inhibitors :
The inhibitors which have no structural similarity with the substrate and bind to an enzyme at locations other than the active sites, so that the globular structure of the enzyme is changed are called non competitive inhibitors.
Eg: Metal ions of Copper, Mercury.

3) Feedback inhibitors :
The end product of a chain of enzyme catalysed reactions inhibit the enzyme of the first reaction as a part of homoeostatic control of metabolism are called feed back inhibitors.
Eg : During respiration (Glycolysis) accumalation of Glucose-6 Phosphate occurs, it inhibits the Hexokinase.

Question 2.
Explain different types of co-factors.
Answer:
Cofactors are two types.
1) Metal ion cofactor :
Metallic cations get tightly attached to the apoenzyme are called metallo enzymes.
Eg : Cu⁺² cytochrome oxidase

2) Organic cofactors :
They are two types.
a) Coenzyme :
They are small organic molecules which are loosely associated with the apoenzyme.
Eg : Thiamine pyrophosphate, Vitamin B.

b) Prosthetic group :
They are the organic cofactors which are tightly bounded to the apoenzyme.
Eg : Haeme is the prosthetic group of enzyme peroxidase.

Question 3.
Explain the mechanism of enzyme action.
Answer:
The substrate ‘S’ has to bind the enzyme of its active site with in a given cleft. The substrate has to diffuse towards the active site leads to the formation of ES complex. This is called transition state structure. Very Soon, after the expected bond breaking /making is completed, the product is released from the active site. In other words, the structure of substrate gets transformed into the structure of products.

When we represent this pictorially through a graph by taking potential energy on y-axis and progress of reaction on x-axis, we notice the energy level difference between ‘S’ and ‘P’. If ‘P’ is at lower level than S, it is an exothermic reaction. One need flot supply energy to form the product. However, whether it is an exothermic or an endothermic reaction (Energy requiring reaction) the ‘S’ has to go through a much higher energy state or transition state.

The difference between average energy content of ‘S’ and that of this transition state is called activation energy. –

Each enzyme [E] has a substrate [S] binding site in its molecule so that a highly reactive enzyme substrate complex [ES] is produced. This complex is short lived and dissociates into its products [P] and the unchanged enzyme with an intermediate formation of the enzyme product complex [EP].
E+S → ES → EP → E+P

Formation of ES complex has been explained with lock and key hypothesis by Emil Fisher and much later with induced fit hypothesis by Daniel E. Koshland.
The catalytic cycle of an enzyme action ‘is

1. The substrate binds to the active site of the enzyme.
2. The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
3. The active site of the enzyme breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate.

Intext Questions

Question 1.
Explain how pH effects enzyme activity with the help of a graphical representation.
Answer:

Enzymes generally function in a narrow range of pH. Each enzyme shows its highest activity at a particular pH called optimum pH.
Activity declines both below and above the optimum level.

Question 2.
Explain the importance of [ES] complex formation.
Answer:
Each enzyme [E] has a substrate [S] binding site in its molecule so that a highly reactive enzyme-substrate complex [ES] is produced. This complex is short-lived and dissociates into its products [P] and the unchanged enzyme with an intermediate formation of the enzyme product complex [EP].
E+S → ES → EP → E+P

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 2nd Lesson Mineral Nutrition

Very Short Answer Questions

Question 1.
Define hydroponics.
Answer:
The technique of growing plants in a specified nutrieht solution in complete absence of soil is known as hydroponics.

Question 2.
How do you categorize a particular essential element as a macro or micronutrient?
Answer:
The elements which are present in large amounts i.e., in excess of 10m mole kg^-1 of dry matter are called Macronutrients. The elements which are needed in small amounts i.e., less than 10m mole kg^-1 of dry matter are called micronutrients.

Question 3.
Give two examples of essential elements that act as activators for enzymes.
Answer:
Molebdinum, Magnesium and Zn⁺².

Question 4.
Name the. essential mineral element that play an important role in photolysis of water.
Answer:
Calcium and Manganese.

Question 5.
Out of the 17 essential elements, which elements are called non-mineral essential elements?
Answer:
Carbon, Oxygen, Hydrogen and Nitrogen.

Question 6.
Name two amino acids in which sulphur is present.
Answer:
Cysteine and Methionine.

Question 7.
When is an essential element said to be deficient?
Answer:
The element said to be deficient when present below the critical concentration [The Cone, of the essential elements below which plant growth is retarded is termed as critical concentration].

Question 8.
Name two elements whose symptoms of deficiency first appear in younger leaves.
Answer:
Sulphur and Calcium.

Question 9.
Explain the role of the pink colour pigment in the root nodule of legume plants. What is it called?
Answer:
The enzyme nitrogenase is highly sensitive to the molecular oxygen. It is protected from oxygen by pink colour pigment cailed an oxygen scavenger, leg haemoglobin.

Question 10.
Which element is regarded as the 17^th essential element? Name the disease caused by its deficiency.
Answer:
Nickel. It’s deficiency cause Mouse ear in pecan. [Slightly wrinkled leaves round or blunt leaflets].

Question 11.
Name the essential elements present in nitrogenase enzyme. What type of essential elements are they?
Answer:
Nitrogenase enzyme contains Mo and Fe elements. They are Micronutrients.

Question 12.
Write the balanced equation of nitrogen fixation.
Answer:
N₂ + 8H⁺ + 8e^– + 16 ATP → 2NH₃ + H₂ + 16ADP + 16 Pi

Question 13.
Name any two essential elements and the deficiency diseases caused by them.
Answer:
Chlorosis occurs due to the deficiency of N, K, Mg, S, Fe, Mn, Zn and Mo.
Neurosis occurs due to the deficiency of Ca, Mg, Cu, K.

Short Answer Questions

Question 1.
Explain the steps involved in the formation of root nodule.
Answer:

1. Roots of legumes release sugars, amino acids which attracted Rhizobium. They get attached to epidermal and root hair cells of the host.
2. The root hair curl and the bacteria invade the root hair.
3. An infection thread is produced, carrying the bacteria into cortex of the root.
4. Bacteria initiate nodule formation in the cortex of the root. Then the bacteria released from the thread into the cortical cells of the host and stimulate the host cells to divide. Thus leads to the differentiation of specialized nitrogen fixing cells.
5. The nodule thus formed establishes a direct vascular connection with the host for exchange of nutrients.

Question 2.
Write in brief, how plants synthesize amino acids.
Answer:
Amino acids are synthesized into two ways. They are
1) Reductive Amination :
In this ammonia reacts with a-ketoglutaric acid and forms glutamic acid.

2) Transamination :
In this, transfer of an amino group from an amino acid to the keto group of a keto acid. Glutamic acid is the main amino acid from which the transfer of NH₂ takes place and other amino acids are formed in the presence of transaminase.

Question 3.
Explain in brief, how plants absorb essential elements.
Answer:
The process of absorption can be demarcated into two main phases. In the 1^st phase, there is an initial uptake of ions into the free space or outer space of cells – the apoplast. It is a passive process. In the second phase of uptake, the ions are taken in slowly into the inner space – the symplast.

The passive movement of ions into the apoplast from the cell along the concentration gradient usually occurs through ion-channels. The entry or exit of ions to and from the symplast against the concentration gradient requires metabolic energy which is an active process.

Question 4.
Explain the Nitrogen cycle, giving relevant examples.
Answer:
The cyclic movement of nitrogen from the atmosphere to soil and from soil back into the atmosphere through plants, animals and micro-organisms is termed as nitrogen cycle. Nitrogen cycle is a continuous pathway and it involves five steps.

1. Nitrogen Fixation
2. Nitrogen Assimilation
3. Ammonification
4. Nitrification
5. Denitrification

1) Nitrogen Fixation :
The gaseous dinitrogen is fixed into inorganic nitrogenous substances is called nitrogen fixation. It occurs by two methods.
a) Abiological or physical method,
b) Biological method – Dia2otrophy.

a) Abiological or physical method :

1. It occurs in atmosphere. Due to thunders and lightening, dinitrogen is converted into nitric acid which is further oxidised to nitrogen dioxide.
2. These oxides dissolves in rain water and reaches soil as nitrous and nitric acids.
3. These acids react with alkali radicles of soil and form nitrates.
4. The soluble nitrates are directly absorbed by plants. The reactions of physical nitrogen fixation are as follows.
   1) N₂ + O₂ → 2NO
   2) 2NO + O₂ → 2NO₂
   3) 2NO₂ + H₂O → HNO₂ + HNO₃
   4) HNO₃ + Ca or K salts → Ca or K nitrates.
   5) A biological nitrogen fixation is carried put on industrial scale by Haber Bosch process at 0°C with 1000°C bars pressure.

b) Biological method – Diazotrophy :
1) Conversion of dinitrogen into nitrogenous compounds by prokaryotes is called biological nitrogen fixation. Such microbes are called as diazotrophs or nitrogen fixers.
Eg : Free-living bacteria – Azotobacter, Clostridium.
Symbiotic bacteria – Rhizobium in the root nodules of Fabaceae members.
Blue green algae – Nostoc and Anabaena.

2) Nitrogen Assimilation :

1. The process of absorbing nitrates, Ammonia to produce organic nitrogen constituents is called nitrogen assimilation. It is the second step of nitrogen cycle.
2. Nitrates and Ammonia formed in the first step are absorbed by plants and converted and constitute the organic nitrogen.
   3) When plants are eaten by animals, this Organic nitrogen is passed on into the animal body.

3) Ammonification :

1. The process of conversion of organic nitrogenous compounds from the dead bodies of plants and animals into ammonia is called Ammonification. It is mineralization process.
2. The bacteria responsible for this are called Ammonifying bacteria.
   Eg: Bacillus ramosus, B. vulgaris, B. mycoides.

4) Nitrification :

1. This is fourth step of nitrogen cycle. It is an oxidative and an exergonic process.
2. The conversion of ammonia into nitrites and nitrates by bacteria is called nitrification. Sucfybacteria are called nitrifying bacteria. It occurs in two steps.

A) In the step, Ammonia is converted into nitrites by bacteria like Nitrosomonas and Nitrococcus.
2NH₃ + 3O₂ → 2NO₂^– + 2H⁺ + 2H₂O

B) In the next step, the nitrites are further oxidised to nitrates by Nitrobacter.
2NO₂^– + O₂ → 2NO₃^–

5) Denitrification :
This is final step in nitrogen cycle. It is also exergonic process released around 11,000 cal of energy. Conversion of nitrates of soil into molecular dinitrogen is called denitrification. Denitrification occurs in four steps
NO₃^– → NO₂^– → NO → N₂

This process is brought about by denitrifying bacteria like Thiobacillus denitrificans, Pseudomonas denitrificans and Micrococcus denitrificans.

Intext Questions

Question 1.
Who should be credited for initiation of Hydroponics?
Answer:
Julius von sachs (1860).

Question 2.
Are all the essential elements required by plants mineral elements? Explain.
Answer:
No. based upon the criteria of essentiality, only a few elements have been found to be essential for plant growth and metabolism. They are micro elements and microelements.

Question 3.
Which essential element is needed to activate the enzymes required for CO₂ fixatin?
Answer:
Mg⁺².

Question 4.
Name a cation and anion that maintain osmotic balance in cells?
Answer:
K⁺ and Cl^–

Question 5.
Which element is required for the formation of mitotic spidle?
Answer:
Calcium

Question 6.
What is the role of sulphur in plant life?
Answer:
Sulphur is present in two amino acids, cysteine and methionine. It is the main constituent of several coenzymes, vitamins (Thiamine, biotin, coenzyme A) and ferredoxin. It forms disulphide bridges which help in stabilizing protein structure.

Question 7.
Which micro element is required in more quantify than die other micronutrients?
Answer:
Iron.

Question 8.
Which element is necessary for the synthesis of the chief photosynthetic pigment without being its structural component?
Answer:
Iron, Magnesium

Question 9.
Which micronutrient necessary for photolysis of water is absorbed by plants in anionic form?
Answer:
Chlorine (Cl^–)

Question 10.
Which enzyme is activated by the 17^th essential element?
Answer:
Nickel acts as an activator for urease.

Question 11.
When is an element considered to be toxic?
Answer:
Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic.

Question 12.
Which element when supplied in excess leads to appearence of brown spots surrounded by chlorotic venis?
Answer:
Magnesium.

Question 13.
Name an anaerobic, free living, photo-hetrotrophic nitrogen fixing bacterium.
Answer:
Rhodospirillum.

Question 14.
Which microorganism produces nitrogen-fixing nodules in Alnus?
Answer:
Frankia.

Question 15.
When the cross section of root nodules of ground nut plants are observed under microscope, they appear pinkish, why?
Answer:
This is due to the presence of leguminous haemoglobin or leg-haemoglobin which is a pink coloured pigment and is also called oxygen scavenger.

Question 16.
Apart from the cortical cells, which other cells are stimulated to divide by the bacteroids inside the root nodules?
Answer:
Inner cortex and pericycle cells.

Question 17.
What is the ratio of electrons and protons required for the fixation of atmospheric molecular nitrogen through biological mode?
Ans. 8H⁺ + 8e^– = 1 : 1.

Question 18.
What acts as oxygen scavenger in the legume-root nodule combination?
Answer:
Leg-hemoglobin.

Question 19.
In what why does aspargine differ from aspartic acid?
Answer:
Aspargine contains more nitrogen than aspartic acid.

Question 20.
Through which tissue the amino acids are transported inside the plant body?
Answer:
Xylem vessels.

Question 21.
Plants like the Picture and Venesfly trap have special nutritional adaptations. Name the essential element and its source for which they show such adaptations.
Answer:
Nitrogen.

Question 22.
Excess ‘Mn’ in soils leads to deficiency of Ca, Mg and Fe. Justify.
Answer:
Manganese competes with iron and Magnesium for uptake and with magnesium for binding with enzymes. Mn also inhibits calcium translocation in the shoot apex. Therefore excess of Mn may infact, induce deficiencies of Ca, Mg and Fe.

Question 23.
What acts as a reservoir of essential elements for plants? By what process is it formed?
Answer:
Soil, it is formed due to weathering and breakdown of Rocks.

Question 24.
Nitrogen fixation is shown by prokaryotes only. Why not by eukaryotes?
Answer:
In prokaryotes, nitrogenase enzyme which is a Mo-Fe protein is present and is capable of nitrogen reduction. It is absent in Eukaryotes.

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 1st Lesson Transport in Plants

Very Short Answer Questions

Question 1.
What are porins? What role do they play in diffusion?
Answer:
Porins are proteins that form huge pores in the outer membranes of the Plastids, mitochondria and some bacteria, allowing molecules upto the size of small proteins to pass through.

Question 2.
Define water potential? What is the value of water potential of pure water?
Answer:
Water potential is defined as the chemical potential of water and a measure of energy for reaction or movement. The value of water potential of pure water is zero.

Question 3.
Differentiate osmosis from diffusion.
Answer:

Question 4.
What are apoplast and symplast?
Answer:

Question 5.
How does ffuttation differ from transniration?
Answer:

Question 6.
What are the physical properties of water responsible for the ascent of sap through xylem in plants?
Answer:
1) Cohesion :
Mutual attraction between water molecules.

2) Adhesion :
Attraction of water molecules to polar surface (such as the surface of trachery elements).

3) Transpiration pull:
Driving force for upward movement of water.

Question 7.
With reference to transportation of food within a plant cells, what are the sources and sink?
Answer:
Source is the part of the plant which synthesise the food. Sink is the part of the plant which utilizes depending on the season or the plants needs.

Question 8.
Does transpiration occurs at night? Give an example.
Answer:
Yes. In succulent plants, stomata opens during the night and remain closed during the day time. (Scotoactive stomata) Eg: Bryophyllum, Cacti.

Question 9.
Compare the pH of guard cells during the opening and closing of stomata.
Answer:
Increase in the pH of guard cells leads to the opening of stomata. Decrease in the pH of guard cells leads to closure of stomata.

Question 10.
In the wake of transpirational loss, why do the C₄ plants are of more efficient than C₃ plant?
Answer:
C₄ plants minimise water loss and fixes more CO₂ than C₃ plants.

Question 11.
What is meant by transport saturation how does it influence facilitated diffusion.
Answer:
Transport rate reaches a maximum when all of the protein transporters are being used is called transport saturation. It allows cell to select substances for uptake.

Question 12.
How does ABA bring about the closure of stomata under water stress conditions?
Answer:
Under water stress condition, ABA, a natural anti-transpirant drive the K⁺ ion put of the guard cells making them close.

Question 13.
Compare imbibing capacities of pea and wheat seeds.
Answer:
Proteinaceous pea seeds swell more on imbibition than starchy wheat seeds, because proteins have very high imbibing capacities compared to carbohydrates.

Short Answer Questions

Question 1.
Define and explain water potential
Answer:
“The measure of the relative tendency of water to move from one area to another is called water potential”. It is denoted by the Greek symbol Psi or Ψ and is expressed in pascals (Pa). It has two components namely solute potential and pressure potential.

a) Solute potential:
If some solute is added to pure water, the solution has fewer free water molecules and the concentration of water decreases reducing its water potential. The magnitude of this lowering due to dissolution of a solute is called solute potential. It is denoted as Ψ_s. It is always negative.

b) Pressure potential :
When water enters a plant ceil due to diffusion, causing a pressure to build up against the cell wall. This makes the cell turgid. The magnitude of increase in water potential in such turgid cell is called pressure potential It is usually positive and is denoted as Ψ_p.
Ψ = Ψ_S + Ψ_P

Question 2.
Write short note on facilitated diffusion.
Answer:
Facilitated diffusion :
The diffusion rate depends on the size of the substances. Obviously smaller substances diffuse faster. The diffusion of any substance across a membrane also depends on its solubility in lipids, which also move faster. Substances that have a hydrophilic moiety find it difficult to move through the membrane. Their movement has to be facilitated by membrane proteins without utilising metabolic energy and there must be concentration gradient.

This is called facilitated diffusion. It cannot cause net transport of molecules from low to high concentration, this would require input of energy. Facilitated diffusion is very specific, it allows cell to select substances for uptake. It is sensitive to inhibitors which react with protein side chains.

Some proteins allow diffusion only if two types of molecules move together. In asymport both molecules cross the membrane in the same direction, in an antiport, they move in opposite directions. When a molecule moves across a membrane independent of other molecules, the process is called uniport.

Question 3.
What is meant by plasmolysis? How is it practically useful to us?
Answer:
Plasmolysis is the process in plant cells where the cytoplasm pulls away from the cell wall due to the loss of water through osmosis. This occurs when cell is placed in a hypotonic solution, water moves out causes the shrinkage of protoplast leading to the seperation of plasma membrane from the cell wall in the comers called incipient plasmolysis. The salting of pickles and preserving of fish and meat in salt are good examples of practical applications.

Question 4.
How does ascent of sap occur in tall trees?
Answer:
Upward movement of water through xylem against gravitational force is called ascent of sap. The transpiration driven ascent of xylem sap depends on (a) Cohesion – mutual – attraction between water molecules (b) Adhesion – attraction of water molecules to polar surfaces (c) transpiration pull – driving force for upward movement of water. These properties give water high tensile strength and high capillarity.

In plants capillarity is aided by the small diameter of the tracheary elements. As water evaporates through the stomata, Since the thin film of water over the cells is continuous, it results in pulling of water, molecule by molecule into the leaf from the xylem. Also, because of lower concentration of water vapour in the atmosphere, water diffuses into the surrounding air. This creats transpiration pull. The forces generated by the transpiration can create pressure sufficient to lift a xylem sized column of water over 130 metres high.

Question 5.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
Glucose is prepared at the source is converted to sucrose, then moved into the companion cells and then into the living phloem sieve tube cells by active transport. Water in the adjacent xylem moves into the phloem by osmosis. As osmotic pressure builds up, the phloem sap moves to the cells which will use the sugar converting it into energy, starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem.

To explain this, Munch conducted one experiment. In this, he took two osmometers A and B. In ‘B’ he took concentrated sugar solution and in A, he took pure water. These two bulbs A and B are connected by a tube, ‘C’. He then, placed A and B bulbs in X and Y water tubs which are connected by tube ‘Z’. Due to osmosis, water moves from A to B, B to A through ‘C and then A-X and finally X-Y through Z occurs until the solution becomes isotonic. In this he compared ‘B’ bulb as source, A’ bulb as sink, ‘C’ tube as phloem, X apd Y tubs are xylem. .

Question 6.
Transpiration is necessary evil’. Explain.
Answer:
Transpiration has both beneficial and harmful effects. They are

Beneficial effects :

1. It helps in passive absorption of water.
2. It also helps in passive absorption of mineral salts by Mass flow mechanism.
3. It is the main force for ascent of sap.
4. It regulates the temperature of plant body and provides cooling effect.

Harmful effects :

1. Excessive transpiration makes the cells flaccid which retards growth.
2. Excessive transpiration leads to closure of stomata thus obstructing gaseous exchange.
   Hence transpiration is considered to be a necessary evil.

Question 7.
Transpiration and photosynthesis – a compromise. Explain.
Answer:
Transpiration has more than one purpose; it

1. Creates transpiration pull for absorption and transportation in plants
2. Supplies water for photosynthesis
3. Transports minerals from the soil to all parts of the plant
4. Cools leaf surface
5. Maintains the shape arid structure of the plants by keeping the cells turgid.

An actively photosynthesising plant has an insatiable need for water. Photosynthesis is limited by available water which can be swiftly depleted by transpiration. C₄ photo- synthetic system is one of the strategies for maximising the availability of CO₂ and minimizing water loss. C₄ plants are twice efficient than C₃ plants in fixing carbon and also water loss.

Question 8.
Explain the mechanism of opening and closing of stomata.
Answer:

Mechanism of opening and closing of stomata :
Levitt (1974) proposed K⁺ pump theory to explain the opening and closing of stomata. According to this, K⁺ ions are accumulated into guard cells from the subsidiary cells in the presence of light. This coupled efflux of protons leads to increase in pH of the guard cells. K⁺ ion accumulation is associated with influx of Cl^– ions to decrease the water potential of guard cells. Water enters into guard cells making them turgid. The outer walls of the guard cells are thin, expand out- wardly with the help of microfibrils in the cell walls of guard cells resulting in opening of stomata.

At night, the K⁺ and Cl^– ions move out of the guard cells due to which the water potential of guard cells increases and water stands moving out of guard cells leading to the closure of stomata.

Under water stress conditions ABA (Abscisic acid), a natural antitranspirant drives the K⁺ ions out of guard cell making them close.

Intext Questions

Question 1.
Differentiate uphill and down till transport.
Answer:

Question 2.
Compare facilitated diffusion and simple diffusion.
Answer:

Question 3.
What happens when two solutions of different concentrations are separated by an egg membrane? State the reason.
Answer:
If two solutions are separated by an egg membrane, the solution moves from low concentrated solution to a high concentrated solution through egg membrane (semipermeable membrane) until the two solutions become equally concentrated. It is called osmosis.

Question 4.
In general in a plant which path of water movement is more and why?
Answer:
Transpiration pull. Generally due to transpiration, tension is created in mesophyll cells which withdraws water from stem → root → Root hair and finally from the soil. Thus water goes upwards in the form of a continuous column.

Question 5.
Why pinus seeds fail to germinate in the absence of mycorrhizae?
Answer:
Pinus plants have an obligate association with the mycorrhizae. The fungus only provides water and minerals. That is why seeds of pinus cannot germinate in the absence of mycorrhiza.

Question 6.
Why do stomata close under water stress conditions?
Answer:
In water stress conditions, ABA (Abscisie acid), a natural antitranspirants drives thfe K⁺ ions out of guard cells making the stomata close.

Question 7.
How are stomata distributed in a typical monocot plant?
Answer:
Stomata are equally distributed on both the surfaces in monocot also plant.

Question 8.
In what form the sugars are transported through phloem?
Answer:
Sucrose, other sugars, hormones and amino acids.

Question 9.
Why does the root endodermis transport ions in one direction only?
Answer:
The walls of the endodermis are suberised (casparian strips) and is impervious to water. So water is directed to wall regions that are not suberised into the cells proper through membranes. So because of suberin, it has the ability to transport water in one direction only.

Question 10.
If a ring of bark is removed from an actively growing plant, what will happen and why?
Answer:
In the absence of downward movement of food, the portion of the bark above the ring on the stem becomes swollen after a few weeks. This shows that, phloem is the tissue responsible for translocation of food.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids

Very Short Answer Questions

Question 1.
Although phenoxide ion has more resonating structures than carboxylate ion carboxylic acid is a stronger acid than phenol. Why?
Answer:

• Phenoxide ion has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom.
• The negative charge is delocalized over two electronegative oxygen atoms in carboxylate ion whereas in phenoxide ion the negative charge less effectively delocalised over one oxygen atom and less electronegative carbon atoms.
  

Question 2.
Write equations showing the conversion of
i) Acetic acid to Acetyl chloride
ii) Benzoic acid to Benzamide
Answer:
i) Acetic acid reacts with PCl₃ (or) PCl₅ (or) SOCl₂ to form acetyl chloride.

ii) Benzoic acid reacts with ammonia to form benzamide.

Question 3.
List the reagents needed to reduce carboxylic acid to alcohol.
Answer:
The reagents required to reduce carboxylic acid to alcohol are

1. LiAlH₄/Ether (or) B₂H₆
2. H₃O⁺

Question 4.
Compare the acidic strength of acetic acid, Chloroacetic acid, benzoic acid and Phenol. (IPE 2014) (Mar. ’14)
Answer:

Question 5.
Explain the position of electrophilic substitution in benzoic acid.
Answer:
Benzoic acid undergo electrophilic substitution reactions in which carboxyl group acts as a deactivating and meta directing group.

Question 6.
Write the mechanism of esterification.
Answer:
Mechanism of esterification of carboxylic acids : The esterification of carboxylic acids with alcohols is a kind of nucleophilic acyl substitution. Protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol. Proton transfer in the tetrahedral intermediate converts the hydroxyl group into – ⁺OH₂ group, which, being a better leaving group, is. eliminated as neutral water molecule. The protonated ester so formed finally loses a proton to give the ester.

Question 7.
What is Etard reaction. Give equation.
Answer:
Etard Reaction : Benzaldehyde is prepared by the oxidation of Toluene with chromyl chloride.

Question 8.
What is Gater man – Koch formylation reaction. Give equation.
Answer:
Gates man – Koch reaction:
Benzene or its derivatives treated with CO and HCl in presence of AlCl₃ gives Benzaldehyde.

Question 9.
An organic acid with molecular formula C₈H₈O₂ on decarboxylation forms Toluene. Identify the organic acid.
Answer:
The organic acid is phenyl acetic acid.

Question 10.
What is decarboxylation ? Give equation.
Answer:
Decarboxylation: Carboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with soda lime (a mixutre of NaOH & CaO In ratio 3: 1).

• This reaction is called decarboxylation.
  

Short Answer Questions

Question 1.
What is Tollens reagent ? Explain its reaction with Aldehydes.
Answer:
Tollens Reagent: Freshly prepared ammonical silver nitrate solution is called Tollens reagent.

• On warming an aldehyde with Tollens reagent a bright silver mirror is produced due to formation of silver metal.
  R – CHO + 2 [Ag (NH₃)₂]⁺ + 3OH^– → RCOO^– + 2Ag + 2H₂O + 4NH₃

Question 2.
Describe the following:
i) Cross aldol condensation
ii) Decarboxylation
Answer:
i) Cross Aldol Condensation: When aldol condensation is carried out between two different aldehydes and (or) ketones, it is called cross aldol condensation.

• If both the reactants contain α-hydrogen atoms, it gives a mixture of four products.
  

Ketones can also be used as one component in the cross aldol reactions.

ii) DecarboxylationCarboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with soda lime (a mixture of NaOH & CaO in ratio 3:1).

• This reaction is called decarboxylation
  

Question 3.
Explain Nucleophilic addition reaction mechanism of aldehydes and ketones.
Answer:
Nucelophilic addition reaction mechanism:

Nucleophile attack on the electrophilic carbon atom of the polar carbonyl group to form an intermediate this intermediate capture a proton to give the product. Aldehydes are more reactive towards the nucleophilic addition reaction due to steric and electronic reasons. In ketones two large substituents sterically hinder the approach of nucleophile towards carbonyl carbon. Aldehydes having only one alkyl group the attack of nucleophile is easy. Electronically two alkyl groups in ketones reduce the positive charge density on carbonyl carbon by + I effect, than aldehydes having only one alkyl group.

Question 4.
Explain the role of electron withdrawing and electron releasing groups on the acidity of carboxylic acids.
Answer:

• Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive effect.
• Electron donating groups decreases the acidity of carboxylic acids by destabilising the conjugate base.
  

Eg : Cl^– is a electron withdrawing group acidic strength order in case of chloro acetic acids CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH

Question 5.
Write any two methods for the preparation of Aldehydes.
Answer:
Preparation of Aldehydes:
From Hydrocarbons:
a) Hydration of alkynes in the presence of H₂SO₄ and HgSO₄ give aldehydes and ketones.

b) Reduction Acetyl Chloride (Rosen mund’s reduction)

c) From nitriles (Stephen reaction)

Nitriles are selectively reduced by di-isobutyl aluminium hydride (DIBAL – H) to imines which hydrolysis give aldehydes.

Question 6.
Write any two methods for the preparation of Ketones.
Answer:

Acyl chlorides react with dialkyl cadmium give ketones.

Question 7.
Explain the following terms. Give an example of the reaction in each ease.
i) Cyanohydrin
ii) Acetal
iii) Semicarbazone
iv) Aldol
v) Hemiacetal
vi) Oxime (IPE 2016 (TS))
Answer:
i) Cyanohyclrin
Aldehydes and ketones react with hydrogen cyanide (HCN) forms addition products called (or) known as cyanohydrins.

ii) Acetal

• In the presence of dry HCl gas, an aldehyde reacts with two equivalents of a monohydric alcohol forms gem-dialkoxy compounds are known as acetals.
• In acetal two alkoxy groups are present on the terminal C-atom.
  

iii) Semicarbazone
Aldehydes/ketones react with semicarbazide forms certain compounds called as semicarbazones.
For example:

iv) Aldol
When an aldehyde ((or) ketone) having atleast one α-hydrogen atom undergo a reaction in the presence of dilute alkali as catalyst to form aldol (or) β – hydroxy aldehydes ((or) ketals in case of ketones), the reaction is called aldol condensation.
For example:

v) Hemlacetal: In the presence of dry HCl gas an aldehyde reacts with one molecule of a monohydric alcohol forms gem-alkoxy alcohols. These are known as hemiacetals.
For example:

vi) Oxime : In weak acidic medium, an aldehyde ketone reacts with hydroxylamine forms products which are known as oxims.

Question 8.
Explain Clemenson’s reduction and Wolf Kishner reduction reactions.
Answer:
i) Clemenson’s reduction: In this reaction carbonyl group is reduced to CH₂ hence aldehydes and ketones are directly converted into alkanes. The reducing agent in this reaction is zinc amalgam and concentrated HCl.

ii) Wolf Kishner reduction : In this reaction also aldehydes and ketones are reducted to alkanes. Addition of hydrazine followed by heating in the presence of KOH in glycol gives alkanes.

Question 9.
What is haloform reaction? Give équation.
Answer:
Haloform reaction: (Oxidation of methyl ketones)
Aldehydes and ketones having at least one (CH₃CO) group react with halogens in presence of NaOH or sodium hypohalite to form haloform (CHX₃) and salt of carboxylic acids having one carbon atom less than that of carbonyl compound.

Question 10.
What is Cannizaro reaction? Give equation. (IPE – 2015 (AP), B.M.P)
Answer:
Cannizaro reaction : Aldehydes having no α – hydrogen undergo self oxidation and self reduction reaction on heating with alkali. In this reaction alcohol and salt of carboxylic acid are formed.

Question 11.
What is HVZ reaction? Give equation. (T.S. Mar.’18) (IPE 2016 (TS))
Answer:
Carboxylic acids having α-hydrogens are halogenated at the α-position on treatment with chlorine or bromine in presence of small amount of red phosphorus to give α-halo carboxylic acids.
This reaction is named as Hell-Volhard — Zelinsky (HVZ) reactions.

Question 12.
Write any three methods for the preparatión of Carboxylic acid.
Answer:
Preparation of carboxylic acids:
i) From primary alcohols : Primary alcohols on oxidation with acidified KMnO₄ (or) K₂Cr₂O₇ (or) Chromium trioxide (CrO₃) gives carboxylic acids.

ii) From Alkyl benzenes : Aromatic carboxylic acids are prepared by the oxidation of alkyl benzenes with KMnO₄ or CrO₃, any alkyl group irrespective of chain length is oxidised to -COOH.

iii) From Nitrites and Amides: Nitnies on hydrolysis in the presence of H⁺ or OH^– gives amides which on hydrolysis give acids.

Question 13.
Explain Ring substitution reactions of aromatic carboxylic acids.
Answer:
Ring Substitution : Aromatic carboxylic acids undergo electrophilic substitution at meta position. -COOH group is electron with drawing group and deactivates the ring at ortho and para position.

Question 14.
How do you prepare the following compounds from acetic acid?
i) Acetyl chloride
ii) Acetamide
iii) Acetic anhydride
iv) Ethyl alcohol
Answer:
i) Acetyl chloride: Reaction with PCl₃ (or) PCl₅ (or) SOCl₂ : Carboxylic acids form acid chlorides with PCl₃, PCl₅, SOCl₂.

ii) Acetamide formation (Reaction with ammonia) : Carboxylic acids react with Ammonia and form ammonium salt which on heating gives amides.

ii) Acetic anhydride : Carboxylic acids on heating in the presence of dehydrating agents like conc H₂SO₄, P₂O₅ give anhydride.

iv) Ethyl alcohol : Acetic acid is reduced to ethyl alcohol by LiAlH₄.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 12(a) Alcohols, Phenols, and Ethers which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 12(a) Alcohols, Phenols, and Ethers

Very Short Answer Questions

Question 1.
Explain why propanol has higher boiling point than that hydrocarbon-butane.
Answer:
Propanol has higher boiling point (391K) than that hydrocarbon butane (309K).
Reason : In propanol strong intermolecular hydrogen bonding is present between the molecules. But in case of butane weak vander waal’s force of attractions are present.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explanation:

• Alcohols and water are both polar solvents. Alcohol is dissolves in water, due to formation of hydrogen bonding with water molecules.
• Hydro carbons are non polar and these donot form hydrogen bonds with water molecules. So alcohols are soluble in water where as hydrocarbons are not soluble in water.

Question 3.
Name the reagents used in the following reactions.

1. Oxidation of primary alcohol to carboxylic acid
2. Oxidation of primary alcohol to aldehyde.

Answer:

1. The reagents used for the oxidation of 1° – alcohols to carboxylic acid are acidified K₂Cr₂O₇ (or) Acidic/alkaline KMnO₄ (or) Neutral KMnO₄.
2. The reagents used for the oxidation of 1°- alcohols to aldehyde are pyridine chloro chromate (PCC) in CH₂Cl₂.

Question 4.
Write any one method for the preparation of ethyl alcohol.
Answer:
From Alkenes : Alcohols can be prepared from alkenes by hydration or hydroboration oxidation.

Question 5.
What is Williamsons synthesis ? Give example. (IPE – 2016 (AP) (TS), 2015 (AP))
Answer:
Williamsons synthesis : Reaction of alkyl halide with sodium alkoxide to give ether is called Williamson’s synthesis.

Question 6.
Write the equations for the following reactions.

1. Bromination of phenol to 2, 4, 6-tribromophenol
2. Benzyl alcohol to benzoic acid.

Answer:

1. Bromination of phenol to 2, 4, 6 tribromophenol.
   
2. Benzyl alcohol to benzoic acid converted as follows.
   

Question 7.
Give the reagents used for the preparation of phenol from chlorobenzene.
Answer:
Phenol is prepared from chlorobenzene as follows. Reagents required are

1. NaOH, 623K, 300 atm,
2. HCl.

Chemical reaction:

Question 8.
what is Esterification ? Give equation.
Answer:
Esterification : Alcohols react with carboxylic acids or acid halides or acid anhydrides to form esters.

Question 9.
What is Dehydration ? Give equation.
Answer:
Dehydration : Alcohols undergo dehydration in the presence of dehydrating agents like cone H₂SO₄, (or) H₃PO₄ etc and form alkenes. The relative ease of dehydration of alcohols follows the following order. Tertiary alcohol > Secondary alcohol > primary alcohol.

Question 10.
What is Reimer Tiemann reaction ? Give equation.
Answer:
Reimer-Tiemann reaction : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction.

Question 11.
What is Kolbe’s reaction ? Give equation.
Answer:
Kolbe’s reaction : Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophillic substitution with CO₂ to form salicylic acid.

Question 12.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
Case – I : When methoxy methane reacts with cold.dil. HI then methyl alcohol and methyl iodide are formed.
Mechanisms:

Case-II: When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:

Question 13.
Identify the reactant needed to form t-butylalcohol from acetone.
Answer:
When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butylalcohol.

Question 14.
Write the Oxidation reaction of phenol.
Answer:
Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.

Short Answer Questions

Question 1.
Give the equations for the preparation of phenol from Cumene. (TS Mar. ’17) (Mar. ’14)
Answer:
Phenol is prepared from Cumene as follows.

1. Oxidation of Cumene to Cumene hydroperoxide.
2. Cumene hydroperoxide on acidic hydrolysis to form phenol.
   

Question 2.
Explain the acidic nature of phenols and compare with that of alcohols. (AP Mar. ’17)Board Model Paper
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.

ii) Phenol reacts with aq.NaOH and forms sodium phenoxide

• In phenol hydroxyl group is attached to the Sp² hydridised carbon of benzene ring which acts as electron with drawing group. The formed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Comparison of acidic character of Phenol and Ethanol:

• The reaction of phenol with aq. NaOH indicates that phenols are stronger acids than alcohols.
• The hydroxyl group attached to an aromatic ring is more acidic than in hydroxyl group is attached to an alkyl group.
• Phenol forms stable phenoxide ion stabilised by resonance but ethoxide ion is not.
  

Question 3.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction of phenol: Phenol undergo reduction in presence of zinc dust to form benzene.

b) Oxidation of phenol : Phenol undergo oxidation with chromic acid and forms a conjugated diketone known as benzoquinone.

Question 4.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences +R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophilic substitution reactions.

In Anisole eletron density is more at O—and P—positions but not at M—position. So O—and P-products are mainly formed during electrophillic substitution reactions.

Question 5.
Explain whý phenol with bromine water forms 2,4, 6-tribromophenol while on reaction with bromine in CS₂ at low temperatures forms para—bromophenol as the major product.
Answer:
a) Phenol undergoes Bromination in presence of CS₂ to form p—bromophenol as major product.

b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).

Explanation : In bromination of phenol, the polarisation of Br₂ molecule takes place even in the absence of Lewis acid. This is due to the highly activating effect of -OH group attached to the benzene ring.

Question 6.
Explain the acidic nature of phenol.
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.

ii) Phenol reacts with aq.NaOH and forms sodium phenoxide.

• In phenol hydroxyl group is attached to the Sp² hydridised carbon of benzene ring which acts as electron with drawing group. The fõrmed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Question 7.
Explain the electrophilic substitution reaction of Anisole.
Answer:
Electrophilic substitution in Anisole:

In all the above reactions p – isomer is the major product.

Question 8.
Write equations of the below given reactions:
i) Alkylatlon of anisole
ii) Nitration of anisole
iii) Friedel—Crafts acetylation of anisole
Answer:
i) Friedel crafts alkylation of anisole:

ii) Nitration of anisole:

iii) Friedel – Crafts acetylation of anisole

Question 9.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
When alkenes undergo addition reaction with diborane to form tri alkyl boranes. These followed by the oxidation by alkaline H₂O₂ to form alcohols. This reaction is called as hydroboration-oxidation reaction.

Question 10.
Write any two methods for the preparation of phenol. (IPE 2014)
Answer:
Preparation of Phenol : Phenol can be prepared from halobenzene, benzene diazonium chloride and cumene etc.
i) From halobenzene

ii) from Benzene diazonium chloride

Question 11.
Write the structures of the following compounds.
i) 2 Methyl butan -1 – o1
ii) 2, 3 – diethyl phenol
iii) 1 – ethoxy propane
iv) Cyclohexyl methanol
Answer:

Students get through AP Inter 2nd Year Chemistry Important Questions 10th Lesson Chemistry In Everyday Life which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 10th Lesson Chemistry In Everyday Life

Very Short Answer Questions

Question 1.
What are drugs ?
Answer:
Drug : The chemicals of low molecular masses ranging from 100 to 500 U that react with macromolecular targets to produce biological response are called drugs.
E.g. : Morphine, Codeine, Heroin etc.,

Question 2.
When are the drugs called medicines ?
Answer:
When the biological response of a drug is therapeutic and useful then the chemical substances (drugs) are called medicines.

Question 3.
Define the term chemotherapy.
Answer:
Chemotherapy: The use of medicines (chemical substances) in the treatment of diseases is called chemotherapy.
In chemotherapy diagnosis, prevention and treatment of diseases are involved.

Question 4.
What are antagonists and agonists ?
Answer:

1. Antagonists : The drugs that bind to the receptor site and inhibit its natural function are called antagonists. ’
   • These are useful when blocking of message is required.
2. Agonists : The drugs that mimic the natural messenger by switching on receptors are called agonists.
   • These are useful when there is lack of natural chemical messenger.

Question 5.
What are antacids ? Give example. [IPE – 2014, 2016 (TS)] [Mar. 14]
Answer:
Antacids : Chemicals that remove the excess of acid in the stomach and maintain the pH to normal level are antacids.
E.g. : Omeprozole, Lansoprozole etc.

Question 6.
What are antihistamines ? Give example. [IPE 2014]
Answer:
Antihistamines : Chemicals that prevent the interaction of histamines with receptors of the stomach wall thus producing less amount of acid are antihistamines.
E.g. : Dimetapp, Terfenadine (Seldane).

Question 7.
What are tranquilizers ? Give example. [IPE 2015 (TS)]
Answer:
Tranquilizers : The drugs which are used in the management (or) treatment of psychoes and neuroses are called tranquilizers. ‘
E.g.: Luminal, Seconal, Barbituric acid etc.

Question 8.
What are analgesics ? How are they classified ?
Answer:
Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc.
Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics. E.g. : Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive
   properties. Their analgesic use is limited to mild aches and pains. .
   E.g.: Aspirin, Ibuprofen etc.

Question 9.
What are antimicrobials ?
Answer:
Antimicrobials : The chemical substances which destroy (or) prevent the development (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials.
E.g.: Lysozyme, Lactic acid etc.

Question 10.
What are antibiotics ? Give example. [A.P. Mar. 16]
Answer:
Antibiotics: The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
(Or)
The substance produced totally or partly by chemical synthesis which in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.
E.g.: Penicillin, Chloramphenicol etc.

Question 11.
What are antiseptics ? Give example. [A.P. Mar. 15]
Answer:
Antiseptics : The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g.: Dettol, Bithional etc.

Question 12.
What are disinfectants ? Give example. [A.P. Mar. 17]
Answer:
Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
i) These are applied to inanimate objects like floors, drainage systems etc.
E.g.:

1. 4% aqueous solution of formaldehyde called formalin is a disinfectant
2. 0.3 ppm chlorine aqueous solution is disinfectant.
3. SO₂ in very low concentration is disinfectant.

Question 13.
What is the difference between antiseptics and disinfectants ?
Answer:
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.

Question 14.
What are antifertility drugs ? Give example.
Answer:
Antifertility drugs: These are birth control pills and contain a mixture of synthetic estrogen and progesterone derivatives.
E.g. : Norethindrone, Ethynylestradiol (novestrol)

Question 15.
What are artificial sweetening agents ? Give example. [IPE – 2016 (A.P.), (T.S.)]
Answer:
The chemical substances which are used instead of sucrose (or) sugar are called artificial sweetening agents.
E.g. : Aspartame, Alitame, saccharin.
These decrease the calorific intake and at the same time several times sweeter than sucrose.

Question 16.
Why is the use of aspartame limited to cold foods and drinks ?
Answer:
Aspartame is unstable at cooking temperature so it’s use is limited to cold foods and soft drinks.

Question 17.
What problem does arise in using alitame as artificial sweetener ?
Answer:
While using alitame as artificial sweetener, the control of sweetness of food is difficult. Alitame is a high potency sweetner.

Question 18.
Why do soaps not work in hard water ?
Answer:
In hard water Ca, Mg-dissolved salts are present. Ca⁺², Mg⁺² ions form insoluble Ca, Mg, soaps respectively when sodium (or) potassium soaps are dissolved in hard water.

1. These insoluble soaps separate as scum in water and are useless as cleansing agent. ’ These are problematic to good washing because the ppt adheres into the fibres of cloth
   as gummy mass.
2. Hair washed with hard water looks dull.
3. Dye does not absorb evenly on cloth washed with soap using hard water due to this gummy mass.

Question 19.
What are synthetic detergents ?
Answer:
The cleansing agents which are having all the properties of soaps but donot contain any soap are called synthetic detergents.
Synthetic detergents can be used both in soft and hard water as they give foam even in hard water.
E.g.: Sodium dodecyl benzene sulphonate.

Question 20.
What is the difference between a soap and a synthetic detergent ?
Answer:

1. Generally soaps are sodium or potassium salts of long chain fattyacids.
2. Synthetic detergents are cleansing agents having all the properties of soaps and donot cpntain any soap.
3. Soaps donot work in hard water but synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 21.
What are food preservatives ? Give example. [A.P. Mar. 17 – IPE 2016 (T.S)]
Answer:
The chemical substances which prevent the spoilage of food due to microbial growth are called food preservatives.
E.g.: Sodium benzoate, Salt of sorbic acid etc.

Question 22.
How are synthetic detergents better than soaps ?
Answer:
Soaps donot work in hard water but synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 23.
Name a substance which can be used as an antiseptic as well as disinfectant.
Answer:
Phenol is used as antiseptic as well as disinfectant.

1. 0.2% phenol is antiseptic
2. 1% phenol is disinfectant.

Question 24.
Why chemicals are added to food ?
Answer:
Chemicals are added to food for i) preservation ii) enhancing their appeal iii) adding nutritive value in them.

Question 25.
Name different categories of food additives.
Answer:
The following are the categories of food additives.

1. Food colours
2. Flavours and sweetners
3. Fat Emulsifiers and stabilising agents
4. Anti oxidants
5. Flour improvers – antistaling agents and bleaches
6. Preservatives
7. Nutritional supplements such as minerals, vitamins and amino acids.

Question 26.
Why do we require artificial sweetening agents ?
Answer:

1. Artificial sweetening agents are very useful to diabetic persons.
2. These decrease the calorific intake and at the same time several times sweeter than sucrose.
3. These are harmless.

Question 27.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
The sweetening agent used in the preparation of sweets for a diabetic patient is saccharin (or) sucralose. It is stable at cooking temperature.

Question 28.
Name two most familiar antioxidants used as food additives.
Answer:
The most familiar antioxidants are butylated hydroxy toluene (BHT) and butylated hydroxy anisole (BHA).

Question 29.
What is saponification ?
Answer:
The process of formation of soaps containing sodium salts by heating esters of fatty acid with aq. NaOH solution is called saponification.

Question 30.
What are the main constituents of dettol ?
Answer:
Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.

Question 31.
What is tincture of iodine ? What is its use ?
Answer:
Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water.

Question 32.
What are enzymes and receptors ?
Answer:
Enzymes : The proteins which perform the role of biological catalysts in the body are called enzymes.
Receptors : The proteins which are crucial to communication system in the body are called receptors.

Question 33.
Which forces are involved in holding the drug to the active site of enzymes ?
Answer:
The forces involved in holding the drugs to the active site of enzymes are ionic bonds, vander waal’s forces, hydrogen bonds, dipole-dipole interactions etc.

Question 34.
What are enzyme inhibitors ?
Answer:
The drugs which inhibits the catalytic activity of enzymes and can block the binding site of the enzyme and prevent the binding of substrate are called enzyme inhibitors.

Question 35.
While antacids and antiallergic drugs interfere with the function of histamines why do not these interfere with the function of each other ?
Answer:
Antacids and antiallergic drugs donot interfere with the function of each other because they work on different receptors in the body.

Question 36.
What are antipyretics ? Give example.
Answer:
The medicines which reduce body temperature during fever are called antipyretics.
E.g.: Paracetamol

Short Answer Questions

Question 1.
What are analgesics ? How are they classified ? Give examples.
Answer:
Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc. Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g. : Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g. : Aspirin, Ibuprofen etc.

Question 2.
What are different types of microbial drugs ? Give one example for each.
Answer:
Antimicrobials : The chemical substances which destroy or prevent the development (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials.
E.g.: Lysozyme, Lactic acid etc.,
Different types of antimicrobial drugs are antibiotics, antiseptics, disinfectants.

1. Antibiotics : The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
   (Or)
   The substance produced totally or partly by chemical synthesis in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.
2. Antiseptics: The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
   Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
3. Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
   These are applied to inanimate objects like floors, drainage systems etc.
   E.g. :
   • 4% aqueous solution of formaldehyde called formalin is a disinfectant
   • 0.3 ppm chlorine aqueous solution is disinfectant.
   • SO₂ in very low concentration is disinfectant.

Question 3.
Write notes on antiseptics and disinfectants. [T.S. Mar. 17]
Answer:
Antiseptics : The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g.: Dettol, Bithional etc.
Phenol is used as antiseptic as well as disinfectant. 0.2% phenol is antiseptic.
Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.
Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water.

Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
These are applied to inanimate objects like floors, drainage systems etc.
E.g.:

1. 4% aqueous solution of formaldehyde called formalin is a disinfectant.
2. 0.3 ppm chlorine aqueous solution is disinfectant.
3. SO₂ in very low concentration is disinfectant.

Phenol is used as antiseptic as well as disinfectant. 1% phenol is disinfectant.

Question 4.
How do antiseptics differ from disinfectants ? Does the same substance be used as both ? Give one example for each.
Answer:
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.
Phenol is used as antiseptic as well as disinfectant.
i) 0.2% phenol is antiseptic.
ii) 1% phenol is disinfectant.

1. Examples of Antiseptics :
   • Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.
   • Tincture of iodine (antiseptic) is a mixture of 2.3% Iodine solution in alcohol-water.
2.  Examples of disinfectants :
   • 4% aqueous solution of formaldehyde called formalin is a disinfectant.
   • 0.3 ppm chlorine aqueous solution is disinfectant.
   • SO₂ in very low concentration is disinfectant.

Question 5.
Explain the following terms with suitable examples.
i) Cationic detergents
ii) Anionic detergents
iii) Non-ionic detergents
Answer:
Synthetic detergents are classified into three types.
i) Cationic detergents : These are synthetic detergents.
a) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides (or) bromides as anions.

b) Cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Hence these are called cationic detergents.
E.g.: Cetyl trimethyl ammonium bromide
It is used in hair conditioners.

ii) Anionic Detergents : These are synthetic detergents.
a) Anionic detergents are sodium salts of sulphonated long chain alcohols (or) hydrocarbons.
b) Anionic detergents are formed by the treatment of long chain alcohols with cone. H₂SO₄ followed by the neutralisation with alkali.
E.g.: Sodium lauryl sulphate.

These are used for house hold work and in tooth pastes.

iii) Non-ionic detergents : These are synthetic detergents.
a) Non-ionic detergents donot contain any ion in their constitution..
b) The detergent formed by the reaction of stearic acid with poly ethylene glycol is an example of non ionic detergent.

Non-ionic detergents are used in liquid dish washing purpose.

Question 6.
What are biodegradable and non-bio degradable detergents ? Give one example for each.
Answer:
i) Biodegradable detergents :

• The detergents which are degraded (or) decomposed by micro organisms are called biodegradable detergents. Biodegradable detergents have less branching.
• These do not cause water pollution.
  E.g. : n-dodecyle benzene sulphonate, soap (non synthetic)

ii) Non Biodegradable detergents :

• The detergents which are not decomposed (Or) degraded by microbes (or) micro organisms are called non-biodegradable detergents. These have more branching.
• These cause water pollution.
  E.g.: ABS detergent.

Question 7.
What are broad spectrum and narrow spectrum antibiotics ? Give one example for each.
Answer:
The range of bacteria (or) other micro organisms that are effected by a certain antibiotic is expressed as its spectrum of action.
Broad spectrum antibiotics : Antibiotics which kill (or) inhibit a wide range of gram¬positive and gram-negative bacteria are called broad spectrum antibiotics.

Narrow spectrum antibiotics: Antibiotics which are effective mainly against gram-positive (or) gram-negative bacteria are called narrow spectrum antibiotics.
E.g.: Penicillin – G is a narrow spectrum antibiotic.

Limited spectrum antibiotics : Antibiotics which are effective mainly against a single organism (or) disease are called as limited spectrum antibiotics.

Question 8.
Name different types of soaps.
Answer:
The following are the different types of soaps.

1. Toilet soaps
2. Soaps that float in water
3. Medicated soaps
4. Shaving soaps
5. Laundry soaps
6. Soap powders and scouring soaps etc.

Question 9.
Write notes on antioxidants in food.
Answer:
Antioxidants :

• Antioxidants are important and necessary food additives.
• Antioxidants help in food preservation by retarding the action of oxygen on food.
• Antioxidants are more reactive towards oxygen than the food material which they protect.
• The most familiar antioxidants are Butylated hydroxy toluene (BHT) and Butylated hydroxy anisole (BHA).
• The addition of BHA to butter increases its shelf life from months to years.
• BHT and BHA along with citric acid are added to produce more antioxidant effect.
• SO₂ and sulphites are useful anti oxidants for wine and beer, sugar syrups and cut, peeled (or) dried fruits and vegetables.

Students get through AP Inter 2nd Year Chemistry Important Questions 9th Lesson Biomolecules which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 9th Lesson Biomolecules

Very Short Answer Questions

Question 1.
Define Carbohydrates.
Answer:
The compounds which are primarily produced by plants and form a very large group of naturally occuring organic compounds are called Carbohydrates.
Eg: Glucose, Fructose, Starch. Carbohydrates are the polyhydroxy aldehydes (or) ketones.

Question 2.
Name the different types of carbohydrates on the basis of their hydrolysis. Give one example for each.
Answer:
On the basis of the hydrolysis, carbohydrates are classified as

1. Monosaccharides, Eg : Glucose, fructose No saccharides
2. Oligosaccharides, Eg : Sucrose, maltose two monosaccharides
3. Polysaccharides, Eg : Starch, cellulose large number of monosaccharides.

Question 3.
How is Glucose prepared ? [IPE – 2014]
Answer:
Glucose is prepared by the hydrolysis of starch in presence of a little acid.

Question 4.
Why are sugars classified as reducing and non-reducing sugars ?
Answer:

• Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
  Eg: Glucose.
• Carbohydrates that doesnot reduce Fehling’s reagent, Tollen’s reagent are called non reducing sugars.
  Eg : Sucrose.

Question 5.
What do you understand by invert sugars ?
Answer:
During the hydrolysis of sucrose there is a change in the sign of rotation, from dextro (+) to laevo (-) and the product is named as invert sugar.

Question 6.
What do you mean by essential amino acids ? Give two examples for non essential amino acids ? [IPE 2014] [T.S. Mar. 16]
Answer:
Essential amino acids : The amino acids which cannot be synthesized in the body and must be obtained through diet are known as essential aminoacids.
Eg : valine, leucine etc.
Examples of non essential amino acids are Alanine, Glycine, Aspartic acid.

Question 7.
What is zwitter ion ? Give an example. [IPE 2015 (AP)]
Answer:
Zwitter ion: In aqueous solution of amino acids, the carboxyl group can lose a proton and amino acid can accept that proton to form a dipolar ion. This ion is called as zwitter ion.

Question 8.
What are reducing sugars ?
Answer:
Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
Eg: glucose.

Question 9.
What are proteins ? Give an example.
Answer:
Proteins : A poly peptide with more than hundred amino acid residues, having molecular mass higher than 10,000 units is called a protein.
Eg : keratin, myosin, insulin.

Question 10.
What are the components of a nucleic acid ?
Answer:

• Nucleic acids are long chain polymers of nucleotides i.e., poly nucleotides.
• Nucleic acids are constituted by pentose sugar, phosphoric acid, and nitrogenous hetero cyclic base (purine (or) pyrimidine).

Question 11.
What are essential and non – essential amino acids ? Give one example for each. [IPE – 2016 (TS)]
Answer:

• Essential amino acids : The amino acids that cannot be synthesised by the body and must be supplied in the diet are called essential amino acids.
  Eg : valine, leucine, phenyl alanine etc.
• Non-essential amino acids : Other amino acids synthesised by the tissues of the body are called non-essential amino acids..
  Eg: Glycine, alanine etc.

Question 12.
Differentiate between globular and fibrous proteins.
Answer:
Globular proteins

1. In this proteins the chains of poly peptides coil around to give a spherical shape.
2. hese are soluble in water.
3. Eg : Insulin

Fibrous proteins

1. In this proteins the poly peptides run parallel and are held together by hydrogens and disulphite – bonding.
2. These are insoluble in water.
3. Eg: keratin

Question 13.
Why are vitamin A and vitamin C essential to us ? Give their important sources.
Answer:
Vitamin A and Vitamin C are essential to us.
Explanation :

• Deficiency of vitamin A causes night blindness, redness in eyes, xerophthalnia.
• Deficiency of vitamin C causes pernicious anaemia (RBC deficient in haemoglobin).

Sources:

• Vitamin – A : Fish liver oil, carrots, butter and milk.
• Vitamin – C : Citrous fruits, amla, green leafy vegetables

Question 14.
What do you understand from the names (a) aldo pentose and (b) ketoheptose ?
Answer:
a) Aldo pentose : If a monosaccharide contains 5 carbon atoms with aldehyde group then it is known as aldo pentose.
b) Ketoheptose : If a monosaccharide contain seven carbons with a ketone group then it is called ketoheptose.

Question 15.
What are anomers ?
Answer:
Anomers : The two isomeric structures of a compound which differ in configuratiofi at C-l only are called Anomers.

Question 16.
What are amino acids ? Give two examples.
Answer:
The organic compounds which contain amino (-NH₂) functional group and carboxyl (-COOH) functional group are called amino acids.
Eg : Glycine, Alanine etc.

Question 17.
What are fibrous proteins ? Give examples.
Answer:
When the poly peptide chains run parallel and are held together by hydrogen and disulphide bonds then fibre – like structure is formed. These are called fibrous proteins. These are insoluble in water.
Eg : keratin, myosin.

Question 18.
What are globular proteins ? Give examples.
Answer:
When the chains of polypeptides coil around to give a spherical shape then globular proteins are formed. These are usually soluble in water.
Eg : insulin and albumins.

Short Answer Questions

Question 1.
Explain the denaturation of proteins.
Answer:
The phenomenon of disorganization of native protein structure is known as denaturation. Denaturation results in the loss of secondary, tertiary and quaternary structure of proteins. This involves a change in physical, chemical and biological properties of protein molecules.
Agents of denaturation:
Physical agents – Heat, violent shaking, X – rays, UV – radiation.
Chemical gents – Acids, alkalies, organic solvents, urea, salts of heavy metals.

Question 2.
What are enzymes ? Give examples ?
Answer:
The group of complex proteinoid organic compounds, elaborated by living organism which catalyse specific organic reactions are called enzymes.
Eg.: Lipases, Rennin, Maltase, Invertase etc. Practically all biological processes such as digestion, respiration etc., are carried on through the agency of enzymes.
Enzymes may be defined as biocatalysts synthesised by living cells.
The functional unit of enzyme is known as holo enzyme made up at apo enzyme (protein part) and co enzyme (non-protein part).

Question 3.
Write notes on vitamins.
Answer:
Vitamin is defined as an “accessory food factor which is essential for growth and healthy maintenance of the body.”
Classification : Vitamins are broadly classified into two major groups.
a) the fat soluble Eg : vitamin A, D, E and K.
b) water soluble Eg : vitamin B – complex and C.

Question 4.
What are harmones ? Give one example for each. [IPE – 2016 (TS)] [Mar. 14]

1. steroid hormones
2. Poly peptide hormones and
3. amino acid derivatives.

Answer:
Hormones: Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its function”.
Eg : testosterone, oestrogen.

1. Example for steroid hormones : Testosterone, oestrogen
2. Example for poly peptide hormones: Insulin
3. Example for Amino acid derivative : Thyroidal hormones thyroxine.

Question 5.
Write the importance of carbohydrates.
Answer:
Importance of carbohydrates:

• Carbohydrates are essential for life of plants.
• Carbohydrates are used as storage molecules as starch in plants.
• Cell wall of plants is made up of cellulose.
• Carbohydrates are also essential for life of animals. Carbohydrates are used as storage molecules as glycogen in animals.
• Carbohydrate source honey is used for a long time as an instant source of energy.

Question 6.
Give the sources of the following vitamins and name the diseases caused by their- deficiency [T.S. Mar. 17] [IPE AP & TS (Mar. 15) BMP, 2016 (AP]
(a) A
(b) D
(c) E and
(d) K
Answer:
Vitamin : A
Sources : Fish oils, liver, kidney
Deficiency diseases : Night blindness, Redness in eyes.

Vitamin : D
Sources : Fish oils, butter, milk, egg
Deficiency diseases : Rickets in children, osteomalacia in adults

Vitamin : E
Sources : Wheat germ oil, egg yolk, green vegetables
Deficiency diseases : Sterility

Vitamin : K
Sources : Green vegetables, Intestinal flora.
Deficiency diseases : Blood coagulation is prevented

Question 7.
Write notes on proteins. [IPE – 2016 (TS)]
Answer:
Proteins are polypeptide chains of amino acids.
Classification of Proteins : Proteins can be classified into two types on the basis of their molecular shape.
a) Fibrous Proteins : These are fibre like proteins, the polypeptide chains are parallel which are held together by hydrogen and disulphide bonds. These are insoluble in water.
Ex : Keratin present in hair, wool, silk etc., and myosin present in muscles.

b) Globular proteins: In these proteins, the polypeptide chains coil around to give a spherical shape. These are soluble in water.
Ex : insulin and albumin.

The structure of proteins is explained in four different levels
a) Primary structure
b) Secondary structure
c) Tertiary structure
d) Quaternary structure

Denaturation of proteins : A protein in a biological system with a specific structure and biological activity is called a native protein. The process in which a protein loses its activity when subjected to heating, change in pH, addition of reagents is called denaturation of protein. Denaturation may be reversible or irreversible.
Ex : Coagulation of egg white on boiling is an irresersible denaturation.
Renaturation is the reverse process of denaturation.

Question 8.
Write about polysaccharides with starch and cellulose as examples.
Answer:
Polysaccharides : The saccharides which on hydrolysis to form large number of monosaccharides are called polysaccharides.
Eg : Starch and cellulose

Starch :

• Starch is the most important dietary source for human beings.
• Vegetables, roots, cereals are important sources of starch.
• It is a polymer ofα – glucose.
• It is constituted by two components Amylose and amylopectin.

Amylose :

• It constitutes 15 – 20% of starch.
• Amylose is water soluble component.
• Amylose is a branched chain with 200 – 1000 α – D – glucose units held by C – 1 to C – 4 glycosidic linkage.

Amylopectin :

• Amylopectin constitutes 80 – 85% of starch.
• It is a branched chain polymer of a – glucose units in which chain is formed by C. – 1 to C – 4 glycosidic linkage whereas branching occurs by C – 1 to C – 6 glycosidic linkage.

Cellulose :

• Cellulose occurs in plants and it is the most abundant organic substance.
• It is a major constituent of cell wall of plant cells.
• Cellulose is a straight chain polysaccharide composed only of β – D – glucose units . which are joined by glycosidic linkage between C – 1 of one glucose and C – 4 of the next glucose.

Question 9.
Write notes on the functions of different hormones in the body. [IPE 2014]
Answer:
Functions of Hormones:

• Hormones help to maintain the balance of biological activities in the body.
• Insulin maintains the blood glucose level within the limit.
• Growth hormones and sex hormones play role in growth and development.
• Low level of thyroxine (produced from thyroid gland) causes hypothyroidism. High level of thyroxine causes hyper thyroidism.
• Gluco corticoids control the carbohydrate metabolism, modulates the inflammatory reactions.
• The mineralo corticoids control the level of excretion of water and salt by the kidney.
• Adrenal cortex does not function properly then results in Addison’s disease.
• Hormones released by gonads are responsible for development of secondary sex characters.
• Testosterone is responsible for development of secondary sex hormone produced in male.
• Estradiol is the main female sex hormone responsible for development of secondary female characterstics like control of menstrual cycle.
• Progesterone is responsible for preparing the uterus for implantation of fertiised egg.

Question 10.
Write the sources of vitamin and diseases due to vitamin deficIency. [AP Mar. 20]
Answer: