Question

A rare genetic disorder resulting from homozygosity for a recessive allele (\( r \)) occurs in 2 out of every 10,000 individuals in a population. Assuming that (i) the disorder is not lethal, (ii) the disorder does not impact reproductive success, (iii) no new mutations are introduced in the population, and (iv) the population follows Hardy-Weinberg equilibrium, the percentage (%) of the carriers in the population that pass the \( r \) allele to offspring is .......... (rounded off to 1 decimal)

Hint:

The carrier frequency in Hardy-Weinberg equilibrium is calculated using \( 2pq \), where \( p \) and \( q \) are the frequencies of the dominant and recessive alleles, respectively.

Answer 1

Step 1: Understanding Hardy-Weinberg equilibrium.
According to Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype (\( rr \)) is given by \( q^2 \), where \( q \) is the frequency of the recessive allele \( r \). The given information tells us that \( q^2 = \frac{2}{10,000} = 0.0002 \).
Step 2: Finding the frequency of allele \( r \).
To find \( q \), take the square root of \( q^2 \): \[ q = \sqrt{0.0002} \approx 0.01414 \] Step 3: Finding the frequency of the dominant allele.
Since \( p + q = 1 \), where \( p \) is the frequency of the dominant allele \( R \), we can calculate \( p \): \[ p = 1 - q = 1 - 0.01414 = 0.98586 \] Step 4: Finding the carrier frequency.
The frequency of the carriers (heterozygous \( Rr \)) is \( 2pq \): \[ 2pq = 2 \times 0.98586 \times 0.01414 \approx 0.0278 \] So, the percentage of carriers is: \[ \text{Percentage of carriers} = 0.0278 \times 100 \approx 2.78% \] Final Answer: \[ \boxed{2.8%} \]