Question

If an autosomal recessive disorder, which shows Hardy-Weinberg equilibrium, has an incidence of 1 in 6400, then the frequency of a carrier is approximately -

• A. 1 in 20
• B. 1 in 40
• C. 1 in 80
• D. 1 in 160

Hint:

For rare autosomal recessive genetic disorders where $q$ is very small, the dominant allele frequency $p$ is extremely close to $1$. Consequently, the carrier frequency can be quickly estimated as simply $2q$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Hardy-Weinberg principle states that allele and genotype frequencies in a large, randomly mating population remain constant from generation to generation in the absence of evolutionary forces.
For a bi-allelic locus, the frequencies of the dominant allele ($p$) and recessive allele ($q$) sum to $1$.
Key Formula or Approach:
The Hardy-Weinberg equation representing genotypic frequencies is:
\[ p^2 + 2pq + q^2 = 1 \]
where:
- $p^2$ is the frequency of homozygous dominant individuals.
- $2pq$ is the frequency of heterozygous carriers.
- $q^2$ is the frequency of homozygous recessive affected individuals.

Step 2: Detailed Explanation:
We are given the incidence of the autosomal recessive disorder ($q^2$):
\[ q^2 = \frac{1}{6400} \]
Taking the square root of both sides to find the recessive allele frequency ($q$):
\[ q = \sqrt{\frac{1}{6400}} = \frac{1}{80} \]
Next, calculate the dominant allele frequency ($p$):
\[ p = 1 - q = 1 - \frac{1}{80} = \frac{79}{80} \approx 1 \]
Now, calculate the frequency of heterozygous carriers ($2pq$):
\[ 2pq \approx 2 \times 1 \times \frac{1}{80} = \frac{2}{80} = \frac{1}{40} \]
Thus, approximately 1 in 40 individuals in this population is a carrier for the recessive disorder.

Step 3: Final Answer:
The frequency of the carrier is approximately 1 in 40, which corresponds to option (B).