Question

There is an irregular mating population. If the frequency of an autosomal recessive lethal gene is 0.4, then the frequency of the carriers in a population of 200 individuals is

• A. 30
• B. 80
• C. 96
• D. 104

Answer 1

The correct option is C 96

• The gene is autosomal recessive lethal, the heterozygous individuals carrying one copy of the affected allele and one of wild type allele will serve as carriers.
• If the genotype of homozygous individuals for the recessive lethal gene is $qq$ and that of homozygous individuals for wild-type gene is $pp$ then the genotype of heterozygous individuals will be $pq$.
• According to the question, the frequency of a recessive lethal allele is (q = 0.4). Frequency of wild-type allele : $p=1-q,i.e.,1-0.4=0.6$
• Hardy Weinberg law states that the sum of all genotype frequencies is represented as the binomial expansion of the square of the sum of p and q. This sum is equal to one. $(p+q{)}^2$ = $p^2$ + $2pq$ + $q^2$ = $1$.
• Hence, the frequency of carrier genotype in total population $\left(2pq\right)$ = $2$ X $0.6$ X $0.4$ = $0.48$.
• And the number of carriers in the population of 200 is = $0.48$ X $200$ = $96$.

So, the correct answer is option C.