Question

You are studying green iguanas from the Caribbean and want to determine if a specific island's iguana population is in Hardy-Weinberg equilibrium with respect to the two M and m alleles segregating at the Mgene locus. In a screen of 1075 individuals from this population, 287 were homozygous MM, 665 were heterozygous Mm, and 123 were homozygous mm. Perform a chi- square test to determine if this population is or is not in Hardy-Weinberg equilibrium. (obs - exp) Class Obs Exp (obs - exp)|(obs - exp)2 exp MM Mm mm Chi-square: x = (Observed – Expected)? Expected Degrees of freedom (df) = 3 - 1 -1 = 1

Answer 1

Answer-

X2 = 76.45

P value is less than 0.01

The population is not in Hardy- Weinberg Equilibrium.

Explanation is given below:

Total number of genotype, m=287 mm = 665 mm = 123 Total (N) = 1075 Mm Allelie frequency of M = p - 2 Nmm Mmm 2N . Qx 287)+665 2x1075 - tas 1239 = 10.576 2150 QX123) +665 2x1075 - 911 +0.424 Allelie frequency of m, a = 2 2mm + Nem AN - 2150 we know, p+q=1 » (P+4) 2 = + p2 + 204 + q² = 4 (Mm) (Mm) (mm) Expected genotype frequency of MM = p2 -(0-576)* = [0-332 mm = 209 = 2X0.57680.424 = 0.488 mm=q² = (0.424)² = [0,18] Total population = 1075 Expected number of mm = 0.382x1075 = 357 mm = 0.488 x 1075 = 525 mm = 0.18 8 1075 = 193

obs (ol CoE) O-E) I CO-E2 Fup (E) class MM 287 357 -70 4900 13.73 Mm 665 525 140 19600 37.33 mm 123 193 25.39 -70 4900 lo75 loz5 Total x² = 76.45 ro L Chi-square, x² = 76.45 Total phenotypes = (because MM, Mm will show same. phenotype) i. Regree of freedom =2-1=1 Predlue = P<0.01 Conclusion Pualue is less than 0.05 (P 20.01) So, the hypothesis is rejected. This means the deviation of expected from observed is consi- dered significant statistically to reject Muell hypothesis The population is not in Hardy-Weinberg Equilibrium. Cobrewed and expected numbers are different)