This includes questions from Class 12 Biology chapters.
Q 1: Intine is made up of
(1) Sporopollenin & Suberin
(2) Cellulose & Pectin
(3) Sporopollenin & Pectin
(4) Cellulose & Suberin
✅ (2) Cellulose & Pectin
▶️ A pollen grain has a two-layered wall: exine & intine.
1. Exine: Hard outer layer. Made up of sporopollenin (highly resistant organic material). It can withstand high temperature and strong acids and alkali. Enzymes cannot degrade sporopollenin. Pollen grains are preserved as fossils due to the presence of sporopollenin. Exine has apertures called germ pores where sporopollenin is absent.
2. Intine: Inner wall. It is a thin and continuous layer made up of cellulose and pectin.
Q 2: For producing 1000 sperms and 250 eggs they require
(1) 250 primary spermatocytes & 250 primary oocytes
(2) 1000 spermatids & 250 secondary oocytes
(3) 500 secondary spermatocytes & 250 secondary oocytes
(4) All the above
✅ (4) All the above
▶️ 250 primary spermatocytes produce 1000 sperms (since each primary spermatocyte forms 4 sperms), and 250 primary oocytes develop into 250 eggs (since each primary oocyte develops into 1 egg).
▶️ 1000 spermatids mature into 1000 sperms, (since each spermatid forms 1 sperm), and 250 secondary oocytes develop into 250 eggs (since each secondary oocyte develops into 1 egg).
▶️ 500 secondary spermatocytes produce 1000 sperms (since each secondary spermatocyte forms 2 sperms), and 250 secondary oocytes develop into 250 eggs.
Q 3: The stage seen closest to the lumen of seminiferous tubule is
(1) Spermatogonia
(2) Spermatids
(3) Spermatozoa
(4) Secondary spermatocytes
✅ (3) Spermatozoa
▶️ During spermatogenesis within the seminiferous tubules, cells develop from the outer region towards the lumen.
▶️ Spermatogonia and primary spermatocytes are found towards the outer region of the tubule, while spermatids and spermatozoa are found closer to the lumen.
▶️ The mature spermatozoa are released into the lumen when they are fully developed.
Q 4: Progestogen–oestrogen combinations are used in
(1) Oral contraceptives and injectables
(2) Injectables and barriers
(3) Barriers and oral contraceptives
(4) Oral contraceptives and Lippes loop
✅ (1) Oral contraceptives and injectables
▶️ Oral contraceptives: Oral administration of progestogens or progestogen-oestrogen combinations in the form of tablets (pills). They inhibit ovulation and implantation and thicken cervical mucus to prevent entry of sperms.
▶️ Injectables & Implants: Progestogens or Progestogens-oestrogen combination are used by females as injections or implants under skin. Their mode of action is like that of pills and their effective periods are much longer.
Q 5: Two pink flowered snap dragon plants (Rr) are self-pollinated. The probability of the offspring to have white flowers are
(1) 25%
(2) 50%
(3) 75%
(4) 0%
✅ (1) 25%
▶️ This is the case of Incomplete dominance.
▶️ Phenotypic ratio= 1 Red: 2 Pink: 1 White.
i.e., 25% Red, 50% Pink and 25% White.
▶️ Genotypic ratio= 1 (RR): 2 (Rr): 1(rr)
Q 6: A blue fowl obtained from mating between black and white fowls, is self-crossed. The F2 ratio is
(1) 1 black: 2 white: 1 blue
(2) 2 black: 1 white: 1 blue
(3) 1 black: 2 blue: 1 white
(4) 1 black: 1 white: 1 blue
✅ (3) 1 black: 2 blue: 1 white
▶️ This is another example for Incomplete dominance.
▶️ Blue fowl is a result of incomplete dominance between black and white color genes.
▶️ When self-crossed, the offspring would follow a 1:2:1 ratio, with 1 black (BB), 2 blue (Bb), and 1 white (bb).
Q 7: What happened when heat killed S cells along with live R cells were injected into the mice?
(1) Mice survived and showed live S cells
(2) Mice died and showed live S cells
(3) Mice survived and showed live R cells
(4) Mice died and showed live R cells
✅ (2) Mice died and showed live S cells
▶️ When heat-killed S cells (virulent) and live R cells (non-virulent) were injected into the mice, the mice died and showed live S cells. This is because the DNA from the heat-killed S cells was taken up by the live R cells in a process called transformation, changing them into virulent cells.
▶️ This experiment was part of Griffith’s experiment which provided the first evidence of bacterial transformation.
Q 8: Which statement is incorrect about transcription?
(1) It is the process of copying genetic information from one strand of the DNA into RNA
(2) Adenine pairs with uracil instead of thymine
(3) An initiation factor (σ factor) present in RNA polymerase initiates transcription
(4) It takes place in 3’→5’ direction
✅ (4) It takes place in 3’→5’ direction
▶️ Transcription takes place in 5’→3’ direction.
▶️ The DNA-dependent RNA polymerase enzyme catalyzes the polymerization only in 5’→3’ direction.
▶️ The DNA strand in 3’→5’ polarity acts as template. RNA is built from this.
▶️ 5’→3’ strand acts as coding strand. This is copied to RNA.
Q 9: In a population, Hardy-Weinberg equilibrium allele frequency of ‘B’ is 0.85. Frequency of ‘Bb’ individuals is
(1) 0.7225
(2) 0.255
(3) 0.0225
(4) 0.12
✅ (2) 0.255
▶️ Frequency of ‘Bb’ individuals can be calculated using the Hardy-Weinberg equilibrium equation, p2 + 2pq + q2 = 1.
Here, ‘p’ is the frequency of the ‘B’ allele and ‘q’ is the frequency of the ‘b’ allele.
▶️ Given that p = 0.85
Hence q = 1 - p = 0.15.
▶️ The frequency of ‘Bb’ individuals is 2pq. So, 2pq = 2 x 0.85 x 0.15 = 0.255.
Q 10: Which of the following is not true about HIV?
(1) It is a retrovirus having RNA genome.
(2) It mainly spreads through unprotected sexual contact.
(3) It can change its genome to DNA in presence of DNA polymerase.
(4) It uses macrophages as ‘HIV factory’.
✅ (3) It can change its genome to DNA in presence of DNA polymerase.
▶️ HIV does not use DNA polymerase to change its genome from RNA to DNA. Instead, it uses an enzyme called reverse transcriptase.
▶️ Reverse transcriptase is unique to retroviruses like HIV and allows the virus to create a DNA copy of its RNA genome, which can then be integrated into the host cell’s DNA. This is a key step in the HIV life cycle and allows the virus to replicate within host cells.
Q 11: Which is not the property of cancer cells?
(1) Uncontrolled multiplication
(2) Metastasis
(3) Contact inhibition
(4) Starving normal cells
✅ (3) Contact inhibition
▶️ Metastasis is the ability of cancer cells to invade nearby tissues and spread to distant regions of the body.
▶️ Cancer cells starve normal cells by consuming nutrients and oxygen at a higher rate. It leads to nutrient and oxygen deprivation for normal cells.
▶️ Contact inhibition is a regulatory mechanism of normal cells. If a cell has contact inhibition, it will stop dividing when it comes into contact with another cell. Cancer cells do not have this property. They continue to grow and divide even when in contact with other cells, leading to the formation of multi-layered, disorganized masses of cells.
Q 12: Which is correct statement?
(1) Lipases are used in detergents to remove all types of stains from the laundry.
(2) Cyclosporine A is used as immunosuppressive agent which increases the power of immune system.
(3) Statin is produced by yeast and used to inhibit enzymes responsible for synthesis of cholesterol.
(4) A bacterium, Aspergillus niger, is used to produce citric acid.
✅ (3) Statin is produced by yeast and used to inhibit enzymes responsible for synthesis of cholesterol.
▶️ Statin is produced by a yeast called Monascus purpureus.
▶️ Lipases are used in detergents, but they specifically break down fat (lipid) stains.
▶️ Cyclosporine A is an immunosuppressive agent, but it decreases the power of the immune system, not increases. It is used to prevent organ rejection after transplantation.
▶️ Aspergillus niger is a fungus, not a bacterium, and it is used to produce citric acid.
Q 13: During gel electrophoresis, separated DNA fragments can be seen as bright orange coloured bands when they are
(1) Stained with ethidium bromide and exposed to IR radiation.
(2) Stained with methylene blue and exposed to UV radiation.
(3) Stained with methylene blue and exposed to IR radiation.
(4) Stained with ethidium bromide and exposed to UV radiation.
✅ (4) Stained with ethidium bromide and exposed to UV radiation.
▶️ Gel electrophoresis is a lab method used to separate mixtures of DNA, RNA, or proteins based on their size and charge.
▶️ DNA samples are placed in a gel and an electric field is applied. Smaller fragments move faster and travel further than larger ones. This results in the fragments being spread out across the gel, allowing for analysis and identification.
▶️ Separated DNA fragments can be seen as bright orange coloured bands when they are stained with ethidium bromide and exposed to UV radiation.
Q 14: In order to induce bacterial uptake of plasmids, the bacteria are made competent by first treating with
(1) Sodium chloride
(2) Potassium chloride
(3) Magnesium chloride
(4) Calcium chloride
✅ (4) Calcium chloride
▶️ Since DNA is a hydrophilic molecule, it cannot pass through cell membranes. So bacterial cells are made ‘competent’ to take up alien DNA or plasmid as follows:
▶️ Treat bacterial cells with a specific concentration of a divalent cation (e.g. calcium) → DNA enters the bacterium through pores in cell wall → Incubate the cells with recombinant DNA on ice → Place them briefly at 42°C (heat shock) → Put them back on ice → Bacteria take up recombinant DNA.
Q 15: Which strategy is used to prevent the nematode infection in the roots of tobacco plant?
(1) Use of agrochemicals
(2) RNA interference
(3) Use of Bt toxin genes
(4) Gene mutation
✅ (2) RNA interference
▶️ A nematode Meloidogyne incognita infects the roots of tobacco plants causing a reduction in yield.
▶️ It can be prevented by RNA interference (RNAi) strategy.
▶️ RNAi is a method of cellular defense in all eukaryotic organisms. It prevents translation of a specific mRNA (silencing) due to a complementary dsRNA molecule.
Q 16: The genetic defect, adenosine deaminase (ADA) deficiency may be cured permanently by
(1) Administering adenosine deaminase through injection.
(2) Enzyme replacement therapy.
(3) Introducing gene from marrow cells producing ADA into the cells at early embryonic stages.
(4) Injecting the functional ADA cDNA into the lymphocytes of the patient.
✅ (3) Introducing gene from marrow cells producing ADA into the cells at early embryonic stages.
▶️ Gene therapy for ADA deficiency: Collect lymphocytes from the patient’s blood and grow in a culture → Introduce a functional ADA cDNA into lymphocytes (using a retroviral vector) → return them to the patient.
▶️ This should be periodically repeated as lymphocytes are not immortal.
▶️ If ADA gene from marrow cells is introduced into cells at early embryonic stages, it could be a permanent cure.
Q 17: In the equation dN/dt=rN(K-N/K), what do K & N denote?
(1) K = Population density at a time, N= Carrying capacity
(2) K = Intrinsic rate of natural increase, N= Carrying capacity
(3) K = Carrying capacity, N= Population density at a time
(4) K = Intrinsic rate of natural increase, N= Population density at t= 0
✅ (3) K = Carrying capacity, N= Population density at a time
▶️ A population with limited resources shows initially a lag phase, phases of acceleration & deceleration and finally an asymptote. This type of population growth is called Verhulst-Pearl Logistic Growth. It is described by the equation dN/dt=rN(K-N/K).
▶️ Where N = Population density at time t
r = Intrinsic rate of natural increase
K = Carrying capacity
Q 18: In a particular climatic condition, decomposition rate is slower if
(1) Detritus is rich in nitrogen, sugars & lignin
(2) Detritus is rich in lignin & chitin
(3) Detritus is rich in chitin, nitrogen & sugars
(4) Detritus is rich in nitrogen & humus
✅ (2) Detritus is rich in lignin & chitin
▶️ Chemical composition of detritus:
o Decomposition is slow in detritus rich in lignin & chitin.
o It is quicker in detritus rich in nitrogen and water-soluble substances like sugars.
▶️ Climatic factors (temperature & soil moisture):
o Warm and moist environment favour decomposition.
o Low temperature & anaerobiosis inhibit decomposition resulting in buildup of organic materials.
Q 19: Regarding climatic factors that influence decomposition, select the correct statement.
(1) Cool and moist environment favour decomposition.
(2) Warm and moist environment inhibit decomposition.
(3) Low temperature & anaerobiosis inhibit decomposition.
(4) High temperature & anaerobiosis inhibit decomposition.
✅ (3) Low temperature & anaerobiosis inhibit decomposition.
▶️ Warm and moist environment favour decomposition.
▶️ Low temperature & anaerobiosis inhibit decomposition.
Q 20: Rivet Popper Hypothesis is proposed by
(1) Paul Ehrlich
(2) Humboldt
(3) David Tilman
(4) Robert May
✅ (1) Paul Ehrlich
▶️ According to the study of Alexander von Humboldt in South American jungles, within a region, species richness increases with increasing explored area, but only up to a limit.
▶️ According to David Tilman, plots with more species shows less year-to-year variation in total biomass.
▶️ According to Robert May’s Global estimate, about 7 million species would have on earth.