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 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. 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. Genotype Number AA 42 AG 24 GG 21 What are the frequencies of the A and G alleles in these bears? Give the genotypic frequencies expected if the population is in Hardy-Weinberg equilibrium. Use a chi-square test to compare the number of observed genotypes with number expected under Hardy-Weinberg equilibrium. Is this population in Hardy-Weinberg equilibrium?

Answer 1

Answer and Explanation:

The number of observed individuals:

• AA 42

• AG 24

• GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

• f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

• f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

• F (AA)= 0.62 ² = 0.3844

• F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

• F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

• AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

• AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

• BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.