Question

1) A total of 6147 North American Caucasians were blood typed for the MN locus, which is determined by two codominant alleles, L^M and L^N. The following data were obtained:

Blood type                     Number

       M                               1654

     MN                              3215

       N                               1278

Carry out a chi-square test to determine whether this population is in Hardy–Weinberg equilibrium at the MN locus.

2) Tay–Sachs disease is an autosomal recessive disorder. Among Ashkenazi Jews, the frequency of Tay–Sachs disease is 1 in 2400. If the Ashkenazi population is mating randomly for the Tay–Sachs gene, what proportion of the population consists of heterozygous carriers of the Tay–Sachs allele?

3) Color blindness in humans is an X-linked recessive trait. In most populations, we saw that approximately 10% of the men are colorblind, however in a unique region on a remote island the presence of color blind men is closer to 15% (6 points)

a. If mating is random for the color-blind locus, what is the frequency of the color-blind allele in this unique population?

b. What proportion of the women in this population is expected to be colorblind?

c. What proportion of the women in the population are expected to be heterozygous carriers of the color-blindness allele?

Answer 1

Answer:

1).

1654= MM

3215 = MN

1278 NN

Total M alleles = 2*1654 +3215 = 6523

Total N alleles = 2*1278 + 3215 = 5771

Total alleles =12294

Frequency of allele, M = 6523 / 12294 = 0.53

Frequency of allele, N = 5771 / 12294 = 0.47

Total population = 6147

Expected MM individuals = 0.53*0.53*6147 = 1726.69

Expected MN individuals = 2* 0.53 * 0.47 * 6147 = 3062.44

Expected NN individuals = 0.47*0.47*6147 = 1357.87

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
MM	1654	1726.69	-72.69	5283.84	3.0601
MN	3215	3062.44	152.56	23274.55	7.6000
NN	1278	1357.87	-79.87	6379.22	4.6980
Total	6147	6147			15.3581

X^2 = 15.36

Degrees of freedom = no. of phenotypes – 1

Df = 3-1=2

P value = 5.99

The X^2 value of 15.36 is greater than the p value of 5.99. Hence, null hypothesis is rejected and the population is not in Hardy-Weinberg equilibrium.

2).

Frequency of Tay-sachs disease = q^2 = 1/2400 = 0.00042

Frequency of recessive allele = q = SQRT of 0.00042 = 0.02

Frequency of dominant allele = p = 1-0.02 = 0.98

Proportion of the population consists of heterozygous carriers of the Tay–Sachs allele = 2pq = 2 * 0.98 * 0.02 = 0.0392

No. of heterozygotes = 0.0392 * 2400 = 94

3). a). Frequency of colorblind male = q = 15% = 0.15

Frequency of colorblind allele = q = 0.15 (as males have only one X-chromosome).

b). Frequency of colorblind woman = qq = 0.15 * 0.15 = 0.0225

c). Frequency of normal allele = p =1-0.15 = 0.8

Frequency of heterozygous female = 2pq = 2* 0.85 * 0.15 = 0.255