3Cooke and Ryder (1971) studied the of nestlings of Ross\'s goose, a small Arcti

Question

3Cooke and Ryder (1971) studied the of nestlings of Ross's goose, a small Arctic nesting goose. Goslings (baby geese) exist in two color morphs, grey or yellow. Cooke and Ryder reported that a population of geese at Karrack Lake, Canada included 263 yellow goslings and 413 grey goslings (676 total). They assumed that color is controlled by two alleles at a single locus. a) Calculate the frequencies of all three possible genotypes, assuming that grey is dominant, and that the population is in Hardy-Weinberg equilibrium. b. If 303 grey goslings survive to adulthood, but only 150 yellow ones do, calculate the fitness of the yellow phenotype relative to the grey one. c) Calculate the mean fitness ("w-bar") after the selection against some of the yellow goslings. d. Calculate the new allelic frequencies after the selection process.

Explanation / Answer

Answer:

3. a. For both of these calculations, p = frequency of dominant allele, and q = frequency of recessive allele.

If grey is dominant:

q2 = 263 / 676 = 0.389

q = (0.389) = 0.624 = frequency of yellow allele

p = 1 - q = 0.376 = frequency of grey allele

Predicted frequency of homozygous greys = 0.376 * 0.376 = 0.141

Predicted frequency of heterozygous greys = 2 * 0.376 * 0.624 = 0.469

Frequency of homozygous yellows = 0.389.

b. Let G be the grey allele and g be the yellow allele.

We’ve already figured out that p = freq (G) = 0.376 and q = freq (g) = 0.624.

Survival rate of grey goslings = 303/413 = 0.734

Survival rate of yellow goslings = 150/263 = 0.570

We could just use these as estimates of fitness, but remember that life is easiest if fitnesses are normalized so that the highest fitness value gets a value of 1.0, so let

wGG = 0.734 / 0.734 = 1.0

wGg = 0.734 / 0.734 = 1.0

wgg = 0.570 / 0.734 = 0.777

c. p2wGG + 2pqwGg + q2wgg = w-bar

(0.376 * 0.376 * 1) + (2 * 0.376 * 0.624 * 1) + (0.624 * 0.624 * 0.777) = w-bar

w-bar = 0.913

d. You get the effects of selection by dividing the above equation through by w-bar.

So: New frequency of GG geonotype = (0.376 * 0.376 * 1) / 0.913 = 0.155

New frequency of Gg genotype = (2 * 0.376 * 0.624 * 1) / 0.913 = 0.514

New frequency of gg genotype = (0.624 * 0.624 * 0.777) / 0.913 = 0.331

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