Class 10 Science Chapter 8 Extra Questions (HOTS Level) – Heredity
Q1 (Assertion-Reason). Assertion (A): A recessive trait can be present in an organism without being expressed in its phenotype. Reason (R): A recessive allele is only expressed in the phenotype when present in the homozygous condition. Options: (a) Both A and R true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true. Answer: (a). A heterozygous individual (Aa) carries the recessive allele silently; it is masked by the dominant allele and only shows up phenotypically when paired with another recessive allele (aa).
Q2 (Reverse-engineering from a ratio). In guinea pigs, black fur (B) is dominant over white fur (b). A cross between two black guinea pigs produces 32 black and 11 white offspring. What are the likely genotypes of the parents, and why? Answer: The ratio 32:11 is close to 3:1, the classic monohybrid F2 ratio. Both parents are heterozygous (Bb x Bb) — each contributes either B or b, giving offspring in the ratio 1 BB : 2 Bb : 1 bb, i.e. 3 black : 1 white.
Q3 (Cross-prediction with a fresh trait pair). A plant with round seeds (R, dominant) and green pods (G, dominant) is crossed as RrGg x RrGg. Out of 160 offspring, how many are expected to show both recessive traits (wrinkled seeds, yellow pods)? Answer: A dihybrid cross RrGg x RrGg gives a 9:3:3:1 ratio; the double-recessive class (rrgg) is 1/16 of the total. 1/16 x 160 = 10 offspring.
Q4 (Pedigree/probability reasoning). Two carrier (heterozygous) parents for an autosomal recessive trait, neither showing the trait, have three children. What is the probability that all three children are unaffected? Answer: Each child independently has a 3/4 probability of being unaffected. Probability all three unaffected = (3/4)³ = 27/64 (~42%).
Q5 (General/abstract reasoning). Explain, in general terms, why a trait controlled by two independently assorting genes produces a 9:3:3:1 ratio in F2, rather than simply 3:1 for each trait combined. Answer: Each gene independently follows a 3:1 dominant:recessive segregation pattern (probability 3/4 dominant, 1/4 recessive). Since the two genes assort independently, the joint probabilities multiply: dominant-dominant = 3/4 x 3/4 = 9/16, each single-recessive combination = 3/4 x 1/4 = 3/16, and double-recessive = 1/4 x 1/4 = 1/16 — giving 9:3:3:1.
Q6 (Sex-determination, multi-part). A couple already has 3 daughters and is told the fourth child is “due” to be a son. Explain, using the biology of sex determination, whether this reasoning is scientifically correct, and state the probability the fourth child is a son. Answer: The reasoning is incorrect. Each pregnancy is an independent event: the sperm that fertilises the egg carries an X or Y chromosome with equal (1/2) probability regardless of earlier children — there is no “memory” effect. The probability the fourth child is a son remains 1/2 (50%).
