Question

The frequency of a particular lethal recessive allele in a population is 0.02. Given this information, calculate the percentage of individuals who are carriers of the lethal recessive allele.

Answer 1

Answer:

approximately 4%

Explanation:

Given,  frequency of recessive allele (p )= 0.02

so, the frequency of dominant allele is (q )= 1-p

so q = 1-p

 p = 0.02

 q = 1- 0.02

 q = 0.98

Once we know p and q, which factor of the Hardy-Weinberg equation. we can calculate percentage of individuals who are carriers of lethal recessive allele.

so

Result : percentage of individuals who are carriers of the lethal recessive allele is  4%.

Answer 2

Carrier individuals are those who carry the recessive lethal allele in their heter0zyg0us genotype. The percentage of individuals who are carriers of the lethal recessive allele is 4%.

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The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,  

• p is the frequency of the dominant allele,  

• q is the frequency of the recessive allele.

The genotypic frequencies after one generation are  

• p² (H0m0zyg0us dominant genotypic frequency),  

• 2pq (Heter0zyg0us genotypic frequency),  

• q² (H0m0zyg0us recessive genotypic frequency).  

*****If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation. *****

The addition of the allelic frequencies equals 1

p + q = 1

The sum of genotypic frequencies equals 1

p² + 2pq + q² = 1

According to this information, if we know the frequency of the recessive allele, we can calculate the frequency of the dominant allele and the genotypic frequencies.

Let us name the dominant allele M, and the recessive one m.

• f(M) = p = frequency of the dominant allele

• f(m) = q = frequency of the recessive allele

• F(MM) = p² = frequency of the h0m0zyg0us dominant genotype

• F(Mm) = 2pq = frequency of the heter0zyg0us genotype

• F(mm) = q² = frequency of the h0m0zyg0us recessive genotype

We know that f(m) = q = 0.02.

By clearing the following equation we can get the p value.

p + q = 1

p + 0.02 = 1

p = 1 - 0.02

p = 0.98

Now we need to calculate the heter0zyg0us genotypic frequency ⇒ carrier individuals are heter0zyg0us because they carry the recessive allele.

2pq = 2 x 0.98 x 0.02 = 0.0392 ≅ 0.04

This is the frequency of heterozygous individuals. If we want to check is our results are ok, we need to sum all genotypic frequencies and the result should be one.

If q = 0.02,

q² = 0.02² = 0.0004 ⇒ These individuals are death

If p = 0.98,

p² = 0.98² = 0.96

Now we can make the addition by replacing the terms,

p² + 2pq + q² = 1

0.96 + 0.04 + 0.0004 ≅ 1  ⇒  Our calculations are ok.

So, the percentage of individuals who are carriers of the lethal recessive allele is 4%.

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