Question

The likelihood of an individual in a population carrying two specific alleles of a human DNA marker, each of which has a frequency of 0.2, will be

• A. 0.4
• B. 0.32
• C. 0.16
• D. 0.08

Hint:

To find the frequency of a specific heterozygous genotype carrying two alleles with frequencies \( p \) and \( q \), always use the term \( 2pq \) from the Hardy-Weinberg expansion.
For \( p = 0.2 \) and \( q = 0.2 \), the value is \( 2 \times 0.2 \times 0.2 = 0.08 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Under the Hardy-Weinberg principle, the genotype frequencies in a large, randomly mating population remain constant from generation to generation in the absence of evolutionary forces.
For a gene locus with multiple alleles, the expected frequency of any heterozygous genotype can be calculated from the individual allele frequencies.
Key Formula or Approach:
For two specific alleles \( A_i \) and \( A_j \) at a locus with frequencies \( p_i \) and \( p_j \), respectively, the expected frequency of the heterozygous genotype \( A_i A_j \) carrying both alleles is calculated as:
\[ \text{Frequency} = 2 \cdot p_i \cdot p_j \]

Step 2: Detailed Explanation:
Let us identify the parameters given in the problem:
We have two specific, distinct alleles of a DNA marker.
The frequency of the first allele (\( p \)) is \( 0.2 \).
The frequency of the second allele (\( q \)) is \( 0.2 \).
Under random mating conditions (Hardy-Weinberg equilibrium), the probability of an individual inheriting the first allele from one parent and the second allele from the other parent is \( p \times q \).
Similarly, the probability of inheriting the second allele from the first parent and the first allele from the second parent is \( q \times p \).
Therefore, the total expected frequency of the heterozygous genotype carrying both specific alleles is:
\[ \text{Heterozygote Frequency} = 2pq \]
Substituting the given allele frequencies into the formula:
\[ \text{Frequency} = 2 \times 0.2 \times 0.2 \]
\[ \text{Frequency} = 2 \times 0.04 = 0.08 \]
Therefore, the likelihood of an individual carrying both specific alleles is 0.08 (or 8%).

Step 3: Final Answer:
The likelihood of an individual carrying both specific alleles is 0.08.