Questions: Quantitative Genetics model

(Questions. Answers.)

Question 1: This question aims to show that a diploid population can be polymorphic at a locus (i.e., have genetic variance) and yet have no additive genetic variance (i.e., have a narrow sense heritability of zero).

(a) Derive the average fitness of individuals that inherit an A allele from one parent at a locus with heterozygote advantage and an allele frequency p equal to the equilibrium expectation of p = (WAa - Waa)/(2 WAa - WAA - Waa). Assume that the other allele is drawn randomly from the population at large. [Hint: The fitness will thus be WAA times p, the chance that the other allele is A, plus WAa times q, the chance that the other allele is a.]

(b) Derive the average fitness of individuals that inherit an a allele from one parent at a locus with heterozygote advantage and an allele frequency p equal to the equilibrium expectation of p = (WAa - Waa)/(2 WAa - WAA - Waa). Assume that the other allele is drawn randomly from the population at large.

(c) Show that the results from (a) and (b) are identical, indicating that the allele inherited from a parent has no effect on the average fitness of offspring. For this locus, the breeding value of all individuals is thus the same and h2 = 0.

Question 2: A study of Darwin's finches on the Galapagos islands by Boag and Grant (1978) found that the regression between offspring and mid-parent in bill depth was 0.82 (+/- 0.15 S.E.).

(a) Assuming that the phenotypic variance remains the same from one generation to the next, this regression coefficient will equal the correlation between offspring and mid-parents. From this, can you estimate the broad-sense or narrow-sense heritability? What is it?

(b) In 1977, there was a severe drought during which many of the birds died. Birds with large beaks were more likely to survive, because they were able to crack and eat some of the larger seeds on the island. Before the drought, 642 birds measured had an average bill depth of 9.42 mm. Later in that same generation, 85 surviving birds had an average bill depth of 9.96 mm. Using the information here and in part (a), what is the expected average bill depth among the offspring of these survivors?

(c) Name two factors that could cause the average bill depth of the offspring to differ from the above calculations (there are several possible answers).

Answer 1:

(a) The average fitness of individuals that inherit an A allele equals WAA p + WAa q. Plugging in q = 1 - p and the equilibrium value of p, this can be shown to equal (WAa2 - WAA Waa)/(2 WAa - WAA - Waa).

(b) The average fitness of individuals that inherit an a allele equals WAa p + Waa q. Plugging in q = 1 - p and the equilibrium value of p, this can be shown to equal (WAa2 - WAA Waa)/(2 WAa - WAA - Waa).

(c) Both results are the same (you might need to factor your equations to get them into the above form).

Answer 2:

(a) Narrow-sense heritability = h2 = 0.82.

(b) Selection differential is 9.96 - 9.42 = 0.54 mm. The expected response to selection is h2 S = (0.82) (0.54) = 0.44. Therefore, the expected average bill depth among the offspring of these survivors is 9.42 + 0.44 = 9.86 mm.

(c) Name two factors that could cause the average bill depth of the offspring to differ from the above calculations (there are several possible answers).

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