Question

The following table gives you genotype frequencies at two blood loci in humans

A) Estimate the allele frequencies at each locus

B) Test for Hardy-Weinberg equilibrium at each locus

C) Estimate D and D'

D) Test for significant gametic disequilibrium

	MM	MN	NN
SS	91	32	5
Ss	147	78	17
ss	82	75	7

Answer 1

(A) MM7 (91+147+82)/3 = 106.67 (apbook) MN =(32+28+75) /+61-67(abbrox) NN (5+17++)/3 = 25 SS +(91+32+5)/3 = 42.67(abbrox) So=(147+787 17)/3=80.67abbrop 007 (82+ 75+7)/3 = 54,67 (abbay) Allele frequency, 5106.67 +(61.67/2) = 0.71 (abbrox) 193.34 napprox) N => $(61.67))+ 25% = 0.29 Cabbarox) 193.34 J S => (4267+ (50-67)2)) = 0.47 (abbrox) © = {(80167/2) + 54.67} = 0.53(apbox). 178.01 (B) Both locus - M2+2MN+MP -> (071)2 + 2X0.71X0-29+ (029)2 = 161) s+ 250782 = (0-47)²+2X0-47 +0-53+ (0.53)=1.7791671) Hence Locus MN is in HWE but locus So is not in HWE tok is higher than 4 © Haplotype - frequency - Value MS (A,B,) (0.71X0.41) = 0.33 Cab prox) Mo (A,B) 22 (0.74 X 0.53) = 0.38 (abbrox :) NS (A2B) 821 (0.29 X0.47): 0.14 Cabbrox) No (123) X32 (0:29 x 0.53) = 0.15 Calolbox.) Allele Frequency M(41) B = X4+242 =>(0-33+0.38) = 0.71 :N (A2) B2=221+222 (0.14 +0.15) -0.29 SCB) 4, = R15+X, (0.33+0.14) -0.52 = (B2) 92 = 212 +222 + (0.38+0-15) =0.53 Now, as we know, Rep= },9, +D or D = - 4, Or, D = {0-33-(0.71X6-52) are D = 40.33-0.36928 As, DCO, use Dmin. ore D =-0.04 (approx.). amin is higher of 8,92 & P291 ~.92 = (0.7180.53) = 0.3736 124, = (0.29 X0.53) = 8.1508 Now, D'al/mint = -0.04 Domin's 10.373 foll) (approx). течеh (0) Gametic disequilibrata D2/06,9, 1292) um m, g2=8f6-6.09970070)(0:52) 60.24)(0.53)} = 0.03(ab boy)