Question:

How is Hardy-Weinberg expression (\(p^{2} + 2pq + q^{2} = 1\)) derived?

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In numerical problems, always find \(q^{2}\) (frequency of recessive phenotype) first to calculate \(q\), then use \(p = 1 - q\) to find \(p\).
Updated On: Mar 29, 2026
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Solution and Explanation

Step 1: Understanding the Concept:
The Hardy-Weinberg principle states that allele frequencies in a population are stable and constant from generation to generation in the absence of evolutionary influences.
Step 2: Key Formula or Approach:
Let \(p\) be the frequency of dominant allele 'A' and \(q\) be the frequency of recessive allele 'a'.
Sum of frequencies: \(p + q = 1\).
Step 3: Detailed Explanation:
In a diploid organism, if the frequency of allele 'A' is \(p\) and allele 'a' is \(q\), the probability of an individual being:
1. Homozygous dominant (AA) is the product of individual probabilities: \(p \times p = p^{2}\).
2. Homozygous recessive (aa) is the product of individual probabilities: \(q \times q = q^{2}\).
3. Heterozygous (Aa or aA) can occur in two ways: \(p \times q + q \times p = 2pq\).
Since the total of all genotype frequencies must equal 1 (representing 100\% of the population):
\[ (p + q)^{2} = p^{2} + 2pq + q^{2} = 1 \]
Step 4: Final Answer:
The expression is a binomial expansion representing the total probability of all genotypes in a population where allele frequencies are \(p\) and \(q\).
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