This includes questions from Class 12 Biology chapters.
Q 1: Microsporogenesis means
(1) Formation of microspores from pollen mother cell (PMC) through mitosis
(2) Formation of microspores from primary endosperm cell (PEC) through meiosis
(3) Formation of microspores from pollen mother cell (PMC) through meiosis
(4) Formation of microspores from primary endosperm cell (PEC) through mitosis
✅ (3) Formation of microspores from pollen mother cell (PMC) through meiosis.
▶️ Microsporogenesis is the formation of microspores. These are small haploid cells that give rise to male gametophytes (pollen grains) in plants. This process takes place in the anthers of the flower's stamen.
▶️ The pollen mother cell (PMC) within the anther, undergoes meiosis to give haploid microspores.
▶️ During microsporogenesis, each PMC gives rise to 4 haploid microspores. They develop to mature pollen grains, each containing a male gametophyte.
Q 2: In human, optimum temperature for sperm production is
(1) 25-30° C
(2) 40-50° C
(3) 33-35° C
(4) 20-22° C
✅ (3) 33-35° C
▶️ Testes are situated outside the abdominal cavity within a pouch called scrotum.
▶️ The low temperature (2-2.5° C less than the body temperature) of scrotum helps for proper functioning of testes and for spermatogenesis.
Q 3: Which one is correct about Lactational amenorrhea?
(1) More effective with no side effects
(2) Less effective with no side effects
(3) More effective with higher side effects
(4) Less effective with higher side effects
✅ (2) Less effective with no side effects.
▶️ Lactational amenorrhea is the absence of menstrual cycle and ovulation due to intense lactation after parturition.
▶️ Fully breastfeeding increases lactation. This method helps to prevent conception.
▶️ This is effective up to 6 months following parturition. But the chances of failure are high.
Q 4: A heterozygous round seeded plant is crossed to recessive wrinkled seeded plant. The progeny would be
(1) 303 rounded: 301 wrinkled
(2) 301 rounded: 100 wrinkled
(3) 20 rounded: 99 wrinkled
(4) 99 rounded: 301 wrinkled
✅ (1) 303 rounded: 301 wrinkled.
▶️ When a heterozygous round-seeded plant (Rr) is crossed with a recessive wrinkled-seeded plant (rr), the possible genotypes of the offspring are: 50% Rr (round-seeded) & 50% rr (wrinkled-seeded). i.e., 1:1.
▶️ Here, 303 rounded: 301 wrinkled ≈ 1:1.
Q 5: Out of a population of 800 individuals in F2 generation of a cross between round yellow & wrinkled green pea plants, what would be number of yellow and wrinkled seeds?
(1) 350
(2) 400
(3) 200
(4) 150
✅ (4) 150
▶️ We know that dihybrid F2 phenotypic ratio = 9:3:3:1. i.e.,
Round Yellow (RrYY or RrYy): 9/16 of the offspring
Wrinkled Yellow (rrYY or rrYy): 3/16 of the offspring
Round Green (Rryy): 3/16 of the offspring
Wrinkled Green (rryy): 1/16 of the offspring
▶️ Here, the population has 800 individuals.
Round Yellow: (9/16) x 800 = 450
Wrinkled Yellow: (3/16) x 800 = 150
Round Green: (3/16) x 800 = 150
Wrinkled Green: (1/16) x 800 = 50
▶️ So, the number of yellow and wrinkled seeds would be 150.
Q 6: RNA differs from DNA in the type of
(1) Purines & pyrimidines
(2) Sugars & purines
(3) Sugars & pyrimidines
(4) Pyrimidine
✅ (3) Sugars & pyrimidines
▶️ RNA (ribonucleic acid) and DNA (deoxyribonucleic acid) differ in the type of sugars and one of the pyrimidines.
▶️ Sugar in DNA: Deoxyribose.
▶️ Sugar in RNA: Ribose.
▶️ Pyrimidines in DNA: Thymine and Cytosine.
▶️ Pyrimidines in RNA: Uracil and Cytosine.
Q 7: Which is correct statement?
(1) DNA and histone are positively charged.
(2) DNA and histone are negatively charged.
(3) DNA is negatively charged and histone is positively charged.
(4) DNA is positively charged and histone is negatively charged.
✅ (3) DNA is negatively charged and histone is positively charged.
▶️ In eukaryotes, there is a set of positively charged, basic proteins called histones.
▶️ Histones are rich in positively charged basic amino acid residues lysines and arginines.
▶️ 8 histones form histone octamer.
▶️ Negatively charged DNA is wrapped around histone octamer to give nucleosome.
Q 8: Homologous organs fundamentally show
(1) Similar function & origin, but different structure
(2) Similar function & structure, but different origin
(3) Similar origin, but different functions & structure
(4) Similar structure & origin, but different functions
✅ (4) Similar structure & origin, but different functions.
▶️ Homologous organs are the organs having fundamentally similar structure and origin but different functions. This phenomenon is called Homology. E.g., Human hand, Whale’s flippers, Bat’s wing & Cheetah’s foot.
▶️ Analogous organs are the organs having similar function but different structure & origin. This phenomenon is called Analogy. E.g. Wings of butterfly and wings of birds.
Q 9: What is true about Interferons?
(1) They are a group of hormone substances
(2) They are released by cells against viruses
(3) They are antibodies
(4) They are Viral proteins
✅ (2) They are released by cells against viruses.
▶️ Interferons are proteins released by cells in response to the presence of viruses.
▶️ They belong to the large class of proteins known as cytokines. They act as Cytokine barriers in immune system.
▶️ They play a crucial role in the immune response by signaling neighboring cells to heighten their antiviral defenses.
▶️ Interferons are not hormones, antibodies, or viral proteins.
Q 10: Acquired immunity is
(1) Non- specific to pathogen and based on memory
(2) Pathogen specific but not based on memory
(3) Pathogen specific and based on memory
(4) Non-specific to pathogen and not based on memory
✅ (3) Pathogen specific and based on memory.
▶️ Acquired immunity is specific to particular pathogens and involves memory cells that "remember" the pathogens encountered before. This allows the immune system to mount a quicker and more effective response upon subsequent exposures to the same pathogen.
Q 11: Choose the correct combination.
(1) Aspergillus niger - Citric acid
(2) Trichoderma polysporum - Butyric acid
(3) Streptococcus - Pectinase
(4) Monascus purpureus - Lactic acid
✅ (1) Aspergillus niger - Citric acid.
▶️ Trichoderma polysporum – Cyclosporine A
▶️ Clostridium butylicum – Butyric acid
▶️ Streptococcus – Streptokinase
▶️ Monascus purpureus – Statins
▶️ Lactobacillus – Lactic acid
Q 12: Restriction enzymes belong to a class of enzymes called
(1) Nucleases
(2) Ligases
(3) Hydrolases
(4) Endonucleases
✅ (1) Nucleases (4) Endonucleases
Options 1 and 4 can be considered as right answers.
▶️ Restriction Enzymes (‘molecular scissors’) are the enzymes that cut DNA at specific sites into fragments.
▶️ They belong to a class of enzymes called nucleases.
▶️ Restriction enzymes 2 types:
🔹1. Exonucleases: They remove nucleotides from the ends of the DNA.
🔹2. Endonucleases: They cut at specific positions within the DNA. E.g. EcoRI.
Q 13: Hind II cuts DNA molecules by recognizing a specific recognition sequence of
(1) 4 base pairs
(2) 6 base pairs
(3) 8 base pairs
(4) 10 base pairs
✅ (2) 6 base pairs
▶️ Hind II is the first restriction endonuclease. It cuts DNA molecules by recognizing a specific sequence of 6 base pairs.
▶️ Restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA. It is a sequence of base pairs that read the same on the two strands in 5' → 3' direction and in 3' → 5' direction.
▶️ E.g. Palindromic sequence for EcoRI:
5' — GAATTC — 3'
3' — CTTAAG — 5'
For Hind II:
5’ — AAGCTT — 3’
3' — TTCGAA — 5'
Q 14: Pick out the incorrect statement.
(1) Insulin consists of two polypeptide chains that are linked by hydrogen bonds.
(2) Insulin can be used to manage adult-onset diabetes.
(3) It is possible to produce human insulin using bacteria.
(4) In mammals, insulin is synthesized as a pro-hormone (pro-insulin).
✅ (1) Insulin consists of two polypeptide chains that are linked by hydrogen bonds.
▶️ Insulin consists of two short polypeptide chains (chain A & chain B) that are linked by disulphide bridges.
▶️ In mammals, insulin is synthesized as a pro-hormone (pro-insulin). It is processed to become mature and functional hormone.
▶️ The pro-insulin contains an extra stretch called C peptide. This is removed during maturation into insulin.
Q 15: First clinical gene therapy (1990) was given to a 4-year-old girl with
(1) Adenosine deaminase (ADA) deficiency
(2) Beta-thalassemia
(3) Phenylketonuria
(4) Hexosaminidase A (HexA) deficiency
✅ (1) Adenosine deaminase (ADA) deficiency
▶️ Adenosine deaminase (ADA) deficiency is caused due to the deletion of a gene of adenosine deaminase (an enzyme for the functioning of immune system).
▶️ 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 16: In population attributes, N represents
(1) Natality
(2) Mortality
(3) Population density
(4) Emigration
✅ (3) Population density
▶️ Population density (N) is the number of individuals of a species per unit area or volume.
▶️ 4 basic processes that fluctuate the population density:
1. Natality (B)
2. Mortality (D)
3. Immigration (I)
4. Emigration (E)
Q 17: In the given figure, K, L, M & N depict the sign of 
(1) K= +, L= +, M= –, N= –
(2) K= –, L= –, M= +, N= +
(3) K= +, L= –, M= +, N= –
(4) K= –, L= +, M= –, N= +
✅ (1) K= +, L= +, M= –, N= –
▶️ Natality & immigration increase the population density.
▶️ Mortality & emigration decrease the population density.
Q 18: The rate of formation of new organic matter by consumers is called
(1) Net primary productivity (NPP)
(2) Gross primary productivity (GPP)
(3) Secondary productivity
(4) Respiration loss (R)
✅ (3) Secondary productivity.
▶️ Gross primary productivity (GPP): It is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is used by plants in respiration.
▶️ Net primary productivity (NPP): It is the available biomass for the consumption to heterotrophs (herbivores & decomposers).
i.e., NPP is the Gross primary productivity minus respiration losses (R).
NPP = GPP – R ▶️ Secondary productivity: It is the rate of formation of new organic matter by consumers.
Q 19: Which of the following processes during decomposition is correctly described?
(1) Catabolism: Last step in the decomposition under fully anaerobic condition.
(2) Leaching: Water soluble inorganic nutrients rise to the top layers of soil.
(3) Fragmentation: Carried out by organisms such as earthworms.
(4) Humification: Leads to accumulation of humus which undergoes faster microbial action.
✅ (3) Fragmentation: Carried out by organisms such as earthworms.
Steps of decomposition:
▶️ Fragmentation: It is the breakdown of detritus into smaller particles by detritivores (e.g. earthworm).
▶️ Leaching: Water soluble inorganic nutrients go down into soil horizon and precipitate as unavailable salts.
▶️ Catabolism: Degradation of detritus into simpler inorganic substances by bacterial & fungal enzymes.
▶️ Humification: Accumulation of humus in soil.
▶️ Mineralization: Release of inorganic nutrients due to degradation of humus by some microbes.
Q 20: The relation between species richness and area is describes on a logarithmic scale by the equation
(1) Log S = log C – Z log A
(2) Log S = Z log A – log C
(3) Log S = log C + Z log A
(4) Log S = 2 log C + Z log A
✅ (3) Log S = log C + Z log A
▶️ 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.
▶️ Relation between species richness and area gives a rectangular hyperbola.
S= CAz
Where, S= Species richness
A= Area
C= Y-intercept
Z= slope of the line (regression co-efficient)