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

Solution

## The correct option is C 96The 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 $$(2pq)$$ =  $$2$$ 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.Biology

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