Question

For a particular gene associated with a degenerative neurological disorder, you surveyed genotype frequencies in a population and found that 16 people had genotype AA, 97 had genotype Aa, 15 had genotype aa.

a) Calculate the allele frequencies.

b) Calculate the expected genotype frequencies under Hardy-Weinberg Equilibrium. Use the chi-square test to determine whether the observed number of people with different genotypes and expected number of people with different genotypes are significantly different (at an alpha level of 0.05). What will the allele frequencies be in the next generation?

For the Chi-square calculations associated with Exercise 2, the degrees of freedom will be one.You get this number with the following formulat: df = k - 1 - m

k is the number of classes (the three genotypes)

m is the number of independent allele frequencies estimated from the data. Since there are only two alleles, the frequency of the second allele is not independent of the first allele.

Chi-square formula: X² = ∑ (observed – expected)²/expected

DF	P = 0.1	P = 0.05	P = 0.01	P= 0.001
1	2.71	3.84	6.64	10.83
2	4.6	5.99	9.21	13.82
3	6.25	7.82	11.34	16.27

Answer 1

Answer:

16 AA = 32 A alleles.

97 Aa = 97 A and 97 a alleles

15 aa = 30 a alleles

Total alleles = 256

Total A alleles = 32+97= 129

Frequency of A allele = 129/256= 0.504

Total a alleles = 97+30=127

Frequency of a allele = 127/256= 0.496

Expected number of each Genotype:

Total progeny= 128

No. of AA = 0.504*0.504*128= 32.51

No. of Aa = 2*0.504*0.496*128= 64

No. of aa = 0.496*0.496*128= 31.49

Genotypes	Observed	Expected	(O-E)^2	(O-E)^2/E
AA	16	32.51	272.58	8.39
Aa	97	64	1089	17.02
aa	15	31.49	271.92	8.64
Total	128	128		34.05

Chi-square value= 34.05

Degrees of freedom = 3-1=2

Critical value= 3.84

The chi-square value of 34.05 is greater than the critical value of 3.84. hence, null hypothesis is rejected frequency of second allele is not idependent of first allele.