Question

p² + 2pq+q2 = 1 A total of 6129 North American Caucasians were blood typed for the MN locus, which is determined by two codominant alleles, ĽM and N. Blood type Number M 1787 MN 3039 N 1303 Calculate the observed frequency of each: 91) Observed Frequency of MM blood type=| 92) Observed Frequency of MN blood type = 93) Observed Frequency of NN blood type = Based on these numbers, what is p and q? 94) p= 95) = Assuming the population is in Hardy Weinberg equilibrium, what would you expect the frequencies to be? 96) Expected Frequency of MM blood type: 97) Expected Frequency of MN blood type: 98) Expected Frequency of NN blood type:

Answer 1

Ans:

91) The observed frequency of MM blood type = 0.2915 (Blood type = M with L^ML^M or MM genotype)

92) The observed frequency of MN blood type = 0.4960 (Blood type = MN with L^ML^N or MN genotype)

93) The observed frequency of NN blood type = 0.2125 (Blood type = N with L^NL^N or NN genotype)

94) p = 0.54

95) q = 0.46

96) Expected frequency of MM blood type = p² = 0.2916

97) Expected frequency of MN blood type = 2pq = 0.4968

98) Expected frequency of NN blood type = q² = 0.2116

All the calculations are given below-

Blood type Number lii] 1787 Do M MN 3039 N 1303 6129 Total. The observed frem equency of mm blood type Number of mm blood type Number of total blood tipe 1187 6129 20:2915 The observed freameney of MN blood type= G 3039 6129 = 0.4960 The observed fremuency of NN blood type= 1 303 6129 20:21 Scanned by TapScanner

of m allele (P). Frequency Number of my Number of 24 mm homozygotes 2x Total No. of individuals heterozygotes 2X 1787 + 3039 2 X 6129 0:54 MN 2x homozygotes 2x Total No. of individuals Number of MN Number of heterozigores Frequency of Hallele (9) 2X1303 + 3039 2x6129 = 0.46 Expected frequency of mm blood type = p² 0.542= 0.2916 Expected freemency of M blood type= 2pq = 2x0.5480.46=0.4968 Expected freauency of NN blood type= q2.0.462 0.2116 Scanned by TapScanner