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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
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B
80
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C
96
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D
104
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Solution

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 $$(2pq)$$ =  $$2$$ 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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