Question:
A large population is showing random mating and not undergoing any evolutionary change. If frequency of allele A is 0.6 , then find out the percentage of homozygous recessive and heterozygous organisms respectively?
A large population is showing random mating and not undergoing any evolutionary change. If frequency of allele A is 0.6 , then find out the percentage of homozygous recessive and heterozygous organisms respectively?
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Use p + q = 1 and the Hardy-Weinberg equation.
Updated On: Nov 13, 2026
\(36\%,24\%\)
\(48\%,16\%\)
\(16\%,36\%\)
\(16\%,48\%\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Population is in Hardy-Weinberg equilibrium, so \(p^2+2pq+q^2=1\).
Step 2: Calculation
Given \(p=0.6\), so \(q=1-0.6=0.4\).
Homozygous recessive \(=q^2=0.16=16\%\).
Heterozygous \(=2pq=2\times0.6\times0.4=0.48=48\%\).
Step 3: Compare the options
So the answer is 16% and 48%.
Final Answer:
Homozygous recessive is 16% and heterozygous is 48%. \[ \boxed{\text{(D) }16\%,48\%} \]
Population is in Hardy-Weinberg equilibrium, so \(p^2+2pq+q^2=1\).
Step 2: Calculation
Given \(p=0.6\), so \(q=1-0.6=0.4\).
Homozygous recessive \(=q^2=0.16=16\%\).
Heterozygous \(=2pq=2\times0.6\times0.4=0.48=48\%\).
Step 3: Compare the options
So the answer is 16% and 48%.
Final Answer:
Homozygous recessive is 16% and heterozygous is 48%. \[ \boxed{\text{(D) }16\%,48\%} \]
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