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

Q21. Human skin colour is believed to be controlled by at least three separate genes :
What do you call this type of inheritance?
i) Suppose the genes are designated as A, B, and C and darkness is dominant over fairness, what shall be the genotypes of the darkest, fairest and intermediate skin colour?
ii) What will be the total number of allele combinations in the gametes of a person heterozygous,all the three genes.
Ans21.(i) This inheritance is called quantitative inheritance and individual genes show incomplete dominance.
(ii) The genotype of the darkest will be AABBCC, that of the intermediate
will be AaBbCc; and that of the fairest will be aabbcc.
(iii) The number of possible allele combinations in the gametes is eight for
such a three-gene combination (2³).

Q22. Draw a labeled diagram of the vertical section of the human heart and show the pulmonary circulation in it by means of arrows.
Ans22.

Q23. Expand PEP. Where is it produced in C₄ plants? What is its role in the biosynthetic process?
Ans23.Phosphoenolpyruvate.

It is produced in the mesophyll cells in C₄ plants.

Its function is to fix CO₂ by getting carboxylated. The carboxylation of PEP
produces Oxaloacetate, which gets converted to malate. Malate is transported
across to the Bundle sheath cells where malate is degraded to release CO₂ and Pyruvate. Pyruvate is returned to the mesophyll cells and CO₂ is fixed in
Calvin cycle. Its role is to fix CO₂ out of the Calvin cycle in the mesophyll
cell and thereby separate the fixation of CO₂ by Rubisco from uptake part.

Q24. What type of striated muscles constitutes the muscles of the eyeball? How is this specific type of muscle advantageous?
Ans24.Six striated ocular muscles attatch the sclera of the eye-ball to the
eye-socket. The ocular muscles move the eye ball for looking at different
directions.

Q25. The base sequence on one of the strands of DNA is TACTAGGAT.
(i) Write the base sequence of its complimentary strand.
(ii) What is the distance maintained between the two consecutive paired bases in the DNA molecule?
(iii) Who contributed the base complementarity rule?
Ans25.(i) The base sequence will be : ATGATCCTA.
(ii) The distance between two consecutive base pairs in DNA helix is 0.34 nm.
(iii) Chargaff proposed the base complementarity rule.

Q26. Name the parasympathetic nerve from the brain that innervates most of the visceral organs. List any four functions of this nerve.
Ans26.The Vagus nerve innervates most of the visceral organs.
Its major four functions are :
(i) Gastric and pancreatic secretion,
(ii) Cardiac slowing,
(iii) Gastrointestinal movements,
(iv) Respiratory reflexes.

Q27. Sometimes natural selection is not operative if the abnormal allele is somewhat advantageous to the individual. Explain this mechanism with a suitable example.
Ans27. Sometimes natural selection is not operative if the abnormal allele
is advantageous to the individual because the abnormality offers better
survivability due to its advantages over the normality.

Sickle cell anaemia is a good example of this phenomenon, where a defective
form of haemoglobin is found in the RBCs. Individual homozygous for this
trait die at an early age. Even those homozygous for this trait have
sickle-shaped RBCs, which are highly inefficient in binding oxygen. However,
natural selection has not eliminated this trait in those geographical locations in the world, where there is a prevalence of malaria. This has happened because a sickle shaped RBC effectively kills the malaria parasite, which harbours in it. Thus individuals homozygous for Sickle-cell trait have a better survivability in the event of malaria epidemics and are favoured during the natural selection despite of their abnormal trait.

Q28. What develops into a microspore mother cell in a flower? Trace the development of the cell into a pollen grain, which is ready for germination. Draw a labeled figure of a mature pollen grain.
Ans28. 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
with two sets of chromosomes. 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.

Q29. What envelops the mammalian ovum preventing the entry of sperm into it easily? How does the sperm gain the entry eventually? What is the significance of the point of entry of the sperm?
Ans29. The cellular layer of follicle cells, called Corona Radiata; and non-cellular layer, called Zona Pellucida prevent the entry of sperm into the Ovum easily. In order for fertilisation to take place, the Corona Radiata must be penetrated or dissolved. This is accomplished by enzymatic hydrolysis of the material cementing the cells of Corona Radiata together. The sperm releases Sperm Lysin from the Acrosome, which consists of mainly Hyaluronidase enzyme that dissolves the follicle-cells away.

At the point of entry of sperm, the cortical granules below the Zona Pellucida forms a thick covering called fertilization membrane. The fertilisation membrane prevents Polyspermy.

Q30. Who demonstrated the semi conservative replication of DNA? Explain the experiment in detail.
Ans30.Meselson and Stahl showed that the DNA replication is semiconservative, that is after the replication one of the DNA strands is of daughter generation and the other is of parental generation.

They grew Escherichia coli in a medium containing nitrogen salts labelled
with stable, heavy isotope ¹⁵N. ¹⁵N got incorporated into both strands of
DNA and this DNA was heavier than the DNA obtained from E. coli grown in ¹⁴N medium. The heavier DNA could be separated from the lighter by centrifugation.
E.Coli cells grown in ¹⁵N medium were transferred to ¹⁴N medium and the DNA was isolated after one generation-time when one bacterial cell has divided into two. This DNA had its density intermediate between that of ¹⁵N DNA and ¹⁴DNA. This was because, during one generation, new DNA molecules with parental ¹⁵N strand and a complementary ¹⁴N new strand were formed and this had an intermediate density between the DNA having ¹⁵N in both the strands and the DNA having ¹⁴N in both the strands.

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