👉 Download Biology Question Paper PDF - SET 2
PART A: BOTANY
I. Answer any 3 questions from 1 – 5. Each carries 1 score. (3 x 1 = 3)
✅ Rosie.
✅ Gause’s ‘Competitive Exclusion Principle’.
✅ Hind II
✅ Stratification.
✅ Tapetum
II. Answer any 9 questions from 6 – 16. Each carries 2 scores. (9 x 2 = 18)
✅ Answer
Bt. Cotton is genetically modified to produce a natural insecticide, the Bt toxin, derived from the bacterium Bacillus thuringiensis. This makes the cotton plant resistant to certain pests, reducing the need for chemical pesticides.5’ → GGAATTCC → 3’
3’ → CCTTAAGG → 5’
She wishes to add a new segment of DNA into it.
a. Identify the technology she planned.
b. Suggest the specific enzyme to make a cut in the DNA with above sequence.
✅ Answer
a. Recombinant DNA technology.b. A restriction enzyme called EcoRI.
✅ Answer
ADA deficiency can be treated through enzyme replacement therapy or gene therapy.In enzyme replacement therapy, patients receive regular injections of functional ADA enzyme.
In gene therapy, a functional ADA gene is introduced into the patient's cells, restoring the production of ADA enzyme.
a. Differentiate Natality and Mortality.
b. Differentiate Immigration & Emigration.
✅ Answer
(a) Natality (B): It is the number of births in a population during a given period.Mortality (D): It is the number of deaths in a population during a given period.
(b) Immigration (I): It is the number of individuals of the same species that have come into the habitat from elsewhere during a given time period.
Emigration (E): It is the number of individuals of the population who left the habitat and gone elsewhere during a given time period.
✅ Answer
a. Construct a food chain with the listed organisms.
b. Point out trophic level of each organism in the constructed food chain.
✅ Answer
a. Plants → Tadpole → Fish → Kingfisherb. Trophic levels of each organism:
Plants: Trophic Level 1 (Producers)
Tadpole: Trophic Level 2 (Primary Consumers)
Fish: Trophic Level 3 (Secondary Consumers)
Kingfisher: Trophic Level 4 (Tertiary Consumers)
✅ Answer
Case No. 1: MutualismCase No. 2: Competition
Case No. 3: Predation/Parasitism
Case No. 4: Commensalism
a. Expand PCR.
b. Name the steps A, B, C in the process.
✅ Answer
(a) Polymerase Chain Reaction.(b) Steps:
A= Denaturation
B= Annealing
C= Extension
a. Ovary develops into seed.
b. In flowering plants, zygote is formed inside the microsporangium.
c. Ovules develop into fruit.
d. Primary endosperm cell (PEC) develops into embryo.
✅ Answer
a. Ovary develops into fruit.b. In flowering plants, zygote is formed inside the ovule.
c. Ovaries develop into fruit.
d. Zygote develops into embryo.
✅ Answer
Exponential Growth: • Population increases at a constant rate.
• Unrestricted growth under ideal conditions.
• Shown by a J-shaped curve.
Logistic Growth:
• Population growth slows as it approaches carrying capacity.
• Limited resources cause a plateau in growth.
• Shown by an S-shaped curve.
✅ Answer
III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 x 3 = 9)
(b) Distinguish between mutualism and parasitism. Give one example for each.
✅ Answer
(a) The amount of living material in a trophic level at a given time is called standing crop. It is measured as the biomass (mass of living organisms) or the number in a unit area.(b) Mutualism and parasitism.
• Mutualism: It’s a type of symbiotic relationship where both species benefit. E.g., bees and flowers. Bees get nectar from flowers, and in return, they help in pollination.
• Parasitism: It’s a type of symbiotic relationship where one species (the parasite) benefits at the expense of the other (the host). E.g., Ticks feed on the dog’s blood.
✅ Answer
In 1983, Eli Lilly company prepared two DNA sequences corresponding to A & B chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A & B were combined by creating disulfide bonds to form human insulin (Humulin). • The human insulin gene is isolated from a human cell.
• The insulin gene is inserted into a plasmid of E. coli to form a recombinant DNA.
• Recombinant DNA plasmid is introduced into E. coli.
• E. coli multiply and produce insulin.
• Insulin is extracted and purified for medical use.
(i) Maize (ii) Vallisneria (iii) Rose
a) Group them based on pollinating agents.
b) Mention one adaptation of each flower related with the agents of pollination.
✅ Answer
a) Grouping based on pollinating agents: • Maize: Wind Pollinated (Anemophily)
• Vallisneria: Water Pollinated (Hydrophily)
• Rose: Insect Pollinated (Entomophily)
b) Adaptations related to pollination:
• Maize: It has long and feathery stigmas to catch pollen grains from the wind.
• Vallisneria: It has long stalked female flowers that reach the water surface to catch pollen grains.
• Rose: It has bright colour and sweet fragrance to attract insects.
b. How can you make fragments of DNA for electrophoresis?
c. Explain separation of DNA fragments using electrophoresis.
✅ Answer
(a) a is largest DNA fragments and e is smallest DNA fragments.(b) DNA is fragmented using restriction endonucleases that cut DNA at specific sequences.
(c) DNA fragments are loaded into agarose gel, and an electric current is applied. DNA is negatively charged, so it moves towards the anode. Smaller fragments move faster while larger fragments move slower and travel less distance. This results in separation of DNA fragments based on size.
PART B: ZOOLOGY
I. Answer any 3 questions from 1 – 5. Each carries 1 score. (3 x 1 = 3)
✅ Genetic drift
✅ Spermiogenesis
(a) 10th day (b) 14th day
(c) 1-5th day (d) 15-28th day
✅ (b) 14th day
✅ Edward Wilson
✅ Propionibacterium sharmani
II. Answer any 9 questions from 6 – 16. Each carries 2 scores. (9 x 2 = 18)
b) Mention any two salient features of human genome.
✅ Answer
(a) BAC: Bacterial Artificial ChromosomeYAC: Yeast Artificial Chromosome
(b) Salient features of human genome:
• Human genome contains 3164.7 million nucleotide bases.
• Total number of genes= about 30,000.
• Average gene consists of 3000 bases, but sizes vary.
• 99.9% nucleotide bases are same in all people. Only 0.1% difference makes every individual unique.
(b) Write any two symptoms of this disorder.
✅ Answer
(a) Turner’s syndrome. Genetic constitution: 44A + X0(b) Sterile, Ovaries are rudimentary. Lack of secondary sexual characters. Dwarf. Mentally retarded (Any 2).
(b) Name any two microbes used as biofertilizers.
✅ Answer
(a) Yes. Methanogens are the microbes that can be used for biogas production.(b) Rhizobium, Azotobacter, Cyanobacterium, Mycorrhiza etc. (any 2)
✅ Answer
b. Mention its importance.
✅ Answer
(a) An antibody. It is produced by B-lymphocytes (B-cells).(b) These are the proteins to fight the pathogens. i.e., gives immunity.
(b) Who experimentally disproved this theory?
✅ Answer
(a) It states that life came out of decaying and rotting matter like straw, mud etc.(b) Louis Pasteur disproved this theory.
a. Blastocyst becomes embedded in the endometrium.
b. Mammary glands produce milk towards the end of pregnancy.
c. Cap-like structure at the tip of sperm head.
d. Yellow-coloured structure developed from ruptured Graafian follicle after ovulation.
✅ Answer
(a) Implantation(b) Lactation
(c) Acrosome
(d) Corpus luteum
✅ Answer
(i) GIFT: Gamete Intra Fallopian Transfer(ii) ICSI: Intra Cytoplasmic Sperm Injection
✅ Answer
Divergent evolution: It is the evolution by which related species become less similar to survive and adapt in different environmental condition.Convergent evolution: It is the evolution by which unrelated species become more similar to survive and adapt in similar environmental condition.
✅ Answer
Salient features of genetic code: • Triplet code. 61 codons code for amino acids. UAA, UAG & UGA are stop codons (Termination codons).
• Genetic code is universal.
• No punctuations b/w adjacent codons.
• An amino acid is coded by many codons. So the code is degenerate.
b. Mention its significance.
✅ Answer
(a) Test cross. It is the cross between F1 hybrid with a recessive parent.(b) It is used to find out the unknown genotype of a character.
III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 x 3 = 9)
a. Mention any three such warning signs.
b. Mention any three ways for prevention and control of drug/alcohol abuse.
✅ Answer
(a) Warning signs: • Drop in academic performance and absence from school.
• Lack of interest in personal hygiene.
• Withdrawal and isolation.
• Depression, fatigue, aggressive and rebellious behaviour.
• Change in sleeping and eating habits.
(b) Prevention and control:
• Avoid undue peer pressure.
• Education and counselling.
• Seeking help from parents and peers.
• Looking for danger signs.
• Seeking professional and medical help.
a. Define linkage and recombination.
b. Mention his any two conclusions from this experiment.
✅ Answer
(a) Linkage is the physical association of genes on a chromosome.Recombination is the generation of non-parental gene combinations.
(b) He found that
• When two genes were situated on same chromosome, proportion of parental gene combinations was much higher than the non-parental type. It is due to linkage.
• Genes of white eye & yellow body were very tightly linked and showed only 1.3% recombination.
• Genes of white eye & miniature wing were loosely linked and showed 37.2% recombination.
• Tightly linked genes show low recombination. Loosely linked genes show high recombination.
(b) Biodiversity (species richness) is highest in tropics. Give any two reasons.
(c) Define the term hotspots.
✅ Answer
(a) Extinction of the parasites when the host is extinct.In co-evolved plant-pollinator mutualism, extinction of one causes the extinction of the other.
(b) Biodiversity (species richness) is highest in tropics because
• Tropics had more evolutionary time.
• Relatively constant environment (less seasonal).
• They receive more solar energy which contributes to greater productivity.
(c) Hotspots are the regions with very high species richness, high degree of endemism but most threatened.
(b) Name the initiator codon and the amino acid coded by it.
✅ Answer
(a) Central Dogma of Molecular Biology: