2. The M-N blood locus was examined in a sample of 900 Australian Aborigines. Th

ID: 256704 • Letter: 2

Question

2. The M-N blood locus was examined in a sample of 900 Australian Aborigines. There were 30 individuals with blood type MM, 230 with type MN, and 640 with type NN. (1 point) A. Calculate the frequencies of the M and N alleles in this population. B. Calculate the expected genotype frequencies and the expected number of individuals of each genotype under the assumptions that the population is in Hardy-Weinberg equilibrium. (2 points) C. Compare the observed and expected number of individuals of each genotype by calculating the chi- squared statistic for each genotype. [Note: a chi-squared value greater than 3.84 constitutes statistical evidence that the population is not in Hardy-Weinberg equilibrium; that is, when the chi-squared statistic is greater than 3.84, the disagreement between observed and expected genotype frequencies is so great that we must reject the hypothesis that the population is in Hardy-Weinberg equilibrium.) Does the chi-squared statistic provide any evidence that this population is not in Hardy-Weinberg equilibrium for the MN blood locus? (2 points)

Explanation / Answer

Answer:

A).

30 MM= 60 M alleles

230 MN = 230 M & 230 N alleles

640 NN = 1280 N alleles

Total alleles =1800

Total M alleles = 60+230 = 290

Frequency of allele, M = 290/1800 = 0.16

Total N alleles = 230 + 1280 = 1510

Frequency of allele, N = 1510/1800 = 0.84

B).

Genotype, NN frequency = 0.84 * 0.84 = 0.7056

Expected NN individuals = 0.7056 * 900 = 635

Genotype, MM frequency = 0.16 * 0.16 = 0.0256

Expected MM individuals = 0.0256 * 900 = 23

Genotype, MN frequency = 2 * 0.84 * 0.16 = 0.2688

Expected MN individuals = 0.2688 * 900 = 242

C).

Phenotype

Observed(O)

Expected (E)

O-E

(O-E)2

(O-E)2/E

49.00

2.13

640

635

25.00

0.04

230

242

-12

144.00

0.60

900

2.76

Chi-square value = 2.76