Question

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some populations along the coast of British Columbia. Kermit Ritland and his colleagues determined that white coat color in these bears results from a recessive mutation (G) caused by a single nucleotide replacement in which guanine substitutes for adenine at the melanocortin 1 receptor locus (mcr1), the same locus responsible for red hair in humans (K. Ritland, C. Newton, and H. D. Marshall. 2001. Current Biology 11:1468–1472). The wild-type allele at this locus (A) encodes black or brown color. Ritland and his colleagues collected samples from bears on three islands and determined their genotypes at the mcr1 locus. (Section 25.2) Genotype Number AA 42 AG 24 GG 21 a. What are the frequencies of the A and G alleles in these bears? b. Give the genotypic frequencies expected if the population is in Hardy–Weinberg equilibrium. c. Use a chi-square test to compare the number of observed genotypes with the number expected under Hardy–Weinberg equilibrium. Is this population in Hardy–Weinberg equilibrium? Note that DF =1. Show Chi Square value, DF, P value, and interpretation.

Answer 1

Answer and Explanation:

a) Frequencies of A and G lalleles are as follows:

f(A) =( 84+24)/174 = 0.62

f(B) = (42+24)/174 =0.38

b) Expected genotype frequencies:

f(AA) = (0.62) (0.62) = 0.384

f (AG)= 2(0.62) (0.38) =0.471

f(GG) = (0.38) (0.38) = 0.144

c) Genotype Observed Expected O-E (O-E)2 (O-E)2/E

AA 42 33 9 81 2.45

AG 24 41 17 289 7.05

GG 21 13 8 64 4.92

Chi squared = 14.42

The number of degrees of freedom is the number of genotypes minusthe number of alleles= 3-2 =1

The p value is much less than 0.05, therefore we reject the hypothesis that these genotype frequencies may be expected from HArdy Weinberg equilibrum