Question

A population at a bi-allelic locus has 23 A/A, 15 A/C, 12 C/C individuals. After randomly mating, if in the next generation, there are 100 individuals, the expected number of A alleles across this population is:

Select one:

a. 122

b. 61

c. 39

d. 78

e. 100

Answer 1

Answer-

According to the given question-

Here we have AA genotype = 23 , AC genotype = 15 and CC genotype = 12 .

Total = 50 .

According to Hardy Weinberg equilibrium

p+ q= 1 and p2 + q2 + 2pq = 1

p = dominant allele frequency.

q= recessive allele frequency

p2 = frequency of individual having dominant genotype

q2 = frequency of individual having recessive genotype.

2pq = frequency of individual having heterozygous genotype.

So q = recessive allele frequency = 2* 12+ 15 / 2*50

q= 0.39

q2 = 0.15

So p= 1- 0.39 = 0.61

p2 = 0.37.

So 2pq = 1- p2 +q2 = 1- 0.37- 0.15 = 0.48.

We know that the frequency remain same for both p and q in other generation.

So frequency of A allele will be = p2+pq which is equal to frequency of AA + half the frequency of AC i.e. 0.37 + 0.48/2 = 0.37 + 0.24 = 0.61

So for 100 individual = 0.61*100.= 61.

Hence the correct answer for the given question- is (B) 61.