Evolution
Tamil Nadu Board · Class 12 · Zoology
Most important questions from Evolution for Tamil Nadu Board Class 12 Zoology board exam 2026. MCQs, short answer, and long answer questions with marks.
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August Weismann disproved Lamarck's theory of inheritance of acquired characters by conducting an experiment. What was the key finding of his experiment?
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Cutting the tails of mice for 20 generations still produced mice with tails, proving somatoplasm changes are not inherited
Step 1: Lamarck proposed that characters acquired during an organism's lifetime can be inherited by offspring. Step 2: Weismann tested this by cutting the tails of mice for 20 consecutive generations. Step 3: All mice born across 20 generations still had tails, showing that the physical change (tail removal) was NOT inherited. Step 4: Weismann proposed his germplasm theory — only changes in the germplasm (reproductive cells) are inherited; changes in somatoplasm (body cells) are NOT transferred to the next generation. Step 5: Option D is wrong because Weismann's theory says germplasm changes C
Industrial melanism in the peppered moth (Biston betularia) is a classic example of which evolutionary concept?
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Natural selection
Step 1: Before industrialization, white-winged peppered moths were more common because they blended with light-coloured lichens on trees, escaping predators. Step 2: After industrialization, tree trunks turned dark due to soot and smoke, and lichens disappeared. Step 3: The white-winged moths became visible to predators on dark tree trunks, while dark-winged moths were better camouflaged. Step 4: Nature 'selected' the dark-winged moths by offering them a survival advantage — this is natural selection in action. Step 5: This is NOT genetic drift (which is random) or founder's effect (which invo
In a Hardy-Weinberg equilibrium population, the frequency of allele 'A' (p) is 0.6 and allele 'a' (q) is 0.4. What is the expected frequency of heterozygotes (Aa) in the population?
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0.48
Step 1: Recall the Hardy-Weinberg equation: (p + q)² = p² + 2pq + q² = 1. Step 2: The frequency of heterozygotes (Aa) is given by 2pq. Step 3: Substitute the values: p = 0.6 and q = 0.4. Step 4: 2pq = 2 × 0.6 × 0.4 = 2 × 0.24 = 0.48. Step 5: So the frequency of heterozygotes is 0.48 or 48%. The other options represent: p² = 0.36 (frequency of AA), q² = 0.16 (frequency of aa), and 0.24 = pq (without the factor of 2, which is a common mistake).
Which of the following is a post-zygotic isolating mechanism?
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Hybrid sterility
Step 1: Isolating mechanisms are divided into two categories: prezygotic (acting before fertilization) and post-zygotic (acting after fertilization). Step 2: Prezygotic mechanisms prevent mating or fertilization — these include ecological, seasonal, behavioural (ethological), morphological, physiological, and cytological isolation. Step 3: Post-zygotic mechanisms act AFTER fertilization and include hybrid inviability, hybrid sterility, and hybrid breakdown. Step 4: Hybrid sterility is a classic post-zygotic mechanism — the hybrid is formed but is sterile (e.g., a mule, which is a cross between
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