Question

The probability 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. 0.02
• B. 0.04
• C. 0.08
• D. 0.16

Hint:

To find the probability of a heterozygous carrier of two specific alleles, always use the formula \(2pq\).
Here, \(2 \times 0.2 \times 0.2 = 0.08\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a population in Hardy-Weinberg equilibrium, the genotype frequencies for a multi-allelic locus are determined by the respective frequencies of the individual alleles.
Key Formula or Approach:
For a genetic marker with two distinct, specific alleles \(A_1\) and \(A_2\), having frequencies \(p\) and \(q\), respectively, the probability of an individual carrying both of these specific alleles (which means having the heterozygous genotype \(A_1A_2\)) is calculated as: \[ f(A_1A_2) = 2pq \]

Step 2: Detailed Explanation:
Let the two specific alleles be designated as \(A_1\) and \(A_2\).
The frequencies of these alleles are given as: \[ p = f(A_1) = 0.2 \] \[ q = f(A_2) = 0.2 \] An individual carrying both of these specific alleles must be heterozygous (\(A_1A_2\)) for this pair.
According to the Hardy-Weinberg law, the frequency of heterozygotes for two specific alleles is: \[ \text{Probability} = 2pq \] Substitute the given values into the formula: \[ \text{Probability} = 2 \times (0.2) \times (0.2) \] \[ \text{Probability} = 2 \times 0.04 = 0.08 \] Therefore, the probability of an individual carrying these two specific alleles in the population is 0.08.

Step 3: Final Answer:
The calculated probability is 0.08, which corresponds to option (C).