Question

Another classification of blood group antigens is known as MN. Individuals are either homozygous for M (MM) or N (NN), or they express both antigens (MN). You are studying the distribution of alleles in a population of people. You determine that 90 people are MM, 60 are MN, and 50 are NN. Assign symbols for the allelic frequency of the M and N alleles in the population. Determine the frequency of each allele. Based on the allelic frequencies, determine (out of 200 individuals) the number of individuals in the population that are expected for each genotype. Test, by chi square, whether the population is in Hardy-Weinberg equilibrium. Show your work and circle your answer.

Answer 1

Answer:28.12

Explanation:

GENOTYPE FREQUENCIES:

MM (p2) = 90/200 = 0.45

MN (2pq) = 60/200 = 0.30

NN (q2) = 50/200 = 0.25

ALLELE FREQUENCIES:

Freq of M = p = p2 + 1/2 (2pq) = 0.45 + 1/2 (0.30) = 0.45 + 0.15 = 0.60

Freq of N = q = 1-p = 1 - 0.60 = 0.40.

EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):

MM (p2) = (0.60)2 = 0.36

MN (2pq) = 2 (0.60)(0.40) = 0.48

NN (q2) = (0.40)2 = 0.16

EXPECTED NUMBER OF INDIVIDUALS of EACH GENOTYPE:

# MM = 0.36 X 200 = 72

# MN = 0.48 X 200 = 96

# NN = 0.16 X 200 =32

CHI - SQUARE (X2):

X2 = Σ(O - E)2 / E

X2 = (90-72)2 /72 + (60-96)2 /96+ (50-32)2 /32

= (18)2 /72 + (-36)2 /96 + (18)2 /32

= 4.5 + 13.5 + 10.12

= 28.12

X2 (calculated) < X2 (table) [3.841, 1 df, 28.12 ls].

Therefore, conclude that there is no statistically significant difference between what you observed and what you expected under Hardy-Weinberg. That is, you fail to reject the null hypothesis and conclude that the population is in HWE.