CBSE Set Qa2 Biology Sample Test Papers For Class 12th for students online

Q 21. How does industrial melanism bring out the action of natural selection? (3 marks)
Ans21.The peppered moth, Biston betularia has two strains :
(i) a dull grey form, and
(ii) a black form: carbonaria
The occurrence of Carbonaria was rare before the Industrial Revolution because the dark form was more conspicuous on the tree trunks and was killed easily by the birds. However, a couple of centuries of Industrial Revolution offered better survivability to the dark form on the tree trunks
blackened with industrial soot due to its reduced conspicuousness. The grey form now stood out more conspicuously and was selectively predated by the birds. The frequency of dark form among moths increased manifolds. This phenomenon of 'Industrial Melanism' proves that the frequency of occurrence of a trait is dependent on its survivability and suitability against the factors of natural selection.

Q 22. How does intestinal juice contribute in the digestion of proteins? What provides alkaline pH in the small intestine? (3 marks)
Ans22. Intestinal juice contains the protease ‘Enteropeptidase’. Some of the inactive trypsinogen molecules of the pancreatic juice are first hydrolysed by enteropeptidase into an inactive peptide and active trypsin. Trypsin then hydrolyses the remaining trypsinogen molecules into trypsin. It also activates other pancreatic proteases, viz. chymotrypsin and carboxypeptidases. Trypsin hydrolyses basic proteins to peptides. Chymotrypsin hydrolyses casein and paracasein to peptides. Carboxypeptidases hydrolyse the terminal peptide bond of the peptide chain to release individual amino acids.
The bicarbonates of pancreatic and intestinal juices and bile from liver provide alkaline p_H.

Q 23. An mRNA strand has a series of codons out of which three are mentioned below.
(i) AUG, (ii) UUU and (iii) UAG
(a) What will these codons be translated into?
(b) What are the DNA codons that would have transcribed these RNA codons? (3 marks)
Ans23.
(a) (i) AUG and (ii) UUU will be translated. (iii) UAG is a stop codon which will not be translated.
(b)

Q 24. List any four symptoms shown by a Down's syndrome afflicted child. Explain the cause of this disorder. (3 marks)
Ans24. A child suffering with Down’s syndrome will have the following symptoms :
(i) A prominent forehead
(ii) A flattened nasal bridge
(iii) Habitually open mouth
(iv) Projecting lower lips
Down’s syndrome is caused by trisomy 21, i.e. an extra chromosome 21. Trisomy arises by non-disjunction during egg cell formation.

Q 25. Answer the following with reference to the anatomy of dicot stem :

(i) Where exactly are the cambial cells located in the vascular bundles?
(ii) What is the name given to such a bundle?
(iii) How are the xylem vessels arranged?
(iv) What type of cells constitute the pith? (3 marks)              jump to top
Ans.25.
(i) The cambial cells are located in the vascular bundles of the young dicotyledonous stem, between xylem and phloem.
(ii) It is called Fascicular Cambium.
(iii) Xylem vessels are arranged towards the center, protoxylem innermost and metaxylem next.
(iv) Parenchymatous cells constitute the pith.

Q 26. Explain any three chemical barriers that offer non-specific defence mechanism. (3 marks)
Ans26. The non-specific chemical defence to the body is provided by :
(i) Lysozyme: Lysozyme is an enzyme, which destroys cell wall of many bacteria, and thus renders them non-pathogenic. It is found on skin, tears, and saliva.
(ii) Oil and sweat: Glands secrete oils and waxes, which make the surface acidic. This discourages many microorganisms from establishing themselves on the skin.
(iii) Acidic gastric juice: Some bacteria who survive the salivary secretion are killed by the highly acidic gastric juice in the stomach.

Q 27. Trace the development of microsporocyte in the anther of a mature pollen-grain. (3 marks)
Ans27.The four microsporangia of the anthers have sporogenous cells into them. The sporogenous cells are characterised by their large size, abundant cytoplasm, and prominent nuclei.
These sporogenous cells or the microsporocytes undergo several mitotic divisions to increase in number and then function as Microspore Mother Cells (MMC). The MMCs are diploid cells. Each MMC divides meiotically to give a tetrad of four haploid microspores. The individual microspores get separated, enlarge and undergo a mitotic division to produce a large Vegetative cell (tube cell) and a small Generative cell. This entire structure representing the male gametophyte is called Pollen Grain. Normally pollen grain is shed at this stage.
In some plants, the generative cell undergoes mitosis to produce two male gametes before the pollen grains are shed.

Q 28. In the case of snapdragon (Antirrhinum majus) a plant with red flowers was crossed with another plant with white flowers. Trace the inheritance of flower colour up to the F₂ generation indicating the genotypesand phenotypes at each level. What special feature do you note in the genotypic ratios in F₂ generation? (5 marks )
Ans28.When homozygous red (RR) Antirrhinum majus is crossed with homozygous white (rr), the F₁ heterozygous (Rr) bears pink flowers. These heterozygous pink-flowered A. majus give homozygous red (RR), heterozygous pink (Rr) and homozygous white (rr) in a ratio of 1:2:1in their F₂ generation when underwent self pollination.
This is an example of ‘Incomplete Dominance’. The phenotypic and genotypic ratios are the same in case of Incomplete Dominance except for the fact that unlike in Complete Dominance, the homozygote and heterozygote for the dominant allele differ in their phenotypes.



Q 29. Name the hormone that regulates each of the following and mention the source of it:
(i) Urinary elimination of water.

(ii) Storage of glucose as glycogen.

(iii) Sodium and potassium ion metabolism.

(iv) Basal metabolic rate.

(v) Descent of testes into the scrotum 5 marks.


Ans29.

Q 30. Describe the C₄ pathway in a paddy plant. How is this pathway an adaptive advantage to the plant? (5 marks)
Ans30. Phosphoenolpyruvate (PEP) forms a 4-carbon compound oxaloaceticacid (OAA) by accepting CO₂ in outer mesophyll cells. OAA is converted to malic acid, which is transported to the cells surrounding the vascular bundle namely, the Bundle sheath cells. Malic acid is converted to
pyruvic acid in bundle sheath cells and CO₂ is released in the cytoplasm. Thus the bundle sheath cells maintain a high concentration of CO₂.
The pyruvic acid generated in the bundle sheath cells is transferred back to the mesophyll and is converted to PEP by the expenditure of an ATP molecule. This conversion generates AMP, which requires two additional ATPs to regenerate an ATP. The total additional requirement of ATPs per
six CO₂ molecules is thus 12 ATPs in comparison to C₃ pathway.
However, the plants having C₄ pathway save a lot of glucose from being wasted by photorespiration.
They can maintain a very high CO₂ concentration in bundle sheath cells where RuBPcarboxylase / oxygenase exists. It helps to avoid RUBISCO to go on the oxygenase mode and oxidise glucose even at high temperatures of tropics. Thus C₄ pathway is an adaptive advantage.