If 4% of a population has an autosomal recessive trait, what portion of the population are...

If 4% of a population has an autosomal recessive trait, what portion of the population are likely to be carriers? Assume that the population is large, and that the recessive trait doesn't affect fitness of the individual. List your answer as a percentage.

Expert Solution

According to Hardy-Weinberg equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1.
So p+q =1
p2+ 2pq+q2 =1
here p = frequency of dominant allele
q is the frequency of the recessive allele.
p2 frequency of homozygous dominant
q2 frequency of homozygous recessive
2pq frequency of heterozygous dominant or carrier.

Given that 4% of the population is an autosomal recessive trait. It means that the frequency of the autosomal recessive trait in the population is 4/100 = 0.04.

q2 = frequency of homozygous recessive

q2 = 0.04

q = 0.2.

p+q =1

So, p = 1- q = 1-0.2 = 0.8.

So the frequency of the dominant allele is 0.8.

Frequency of carrier = 2-p*q = 2*0.2*0.8 = 0.32.

Percentage of carrier population = frequency * 100 = 0.32 * 100 = 32 %.

So, 32% of the population is likely to be carrier.

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gladiator answered 2 years ago

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