Class NEET 2026 ALL Q #2060
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is
(A) 0.99
(B) 0.01
(C) 0.02
(D) 0.1
Correct Answer: B

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Detailed Solution

Step 1: Identify the Hardy-Weinberg Principle

The Hardy-Weinberg principle states that for a gene with two alleles, A and a, with frequencies $p$ and $q$ respectively, the genotype frequencies in a population at equilibrium are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.

Step 2: Assign Variables

Given that the frequency of allele A is $p = 0.1$. The frequency of the homozygous dominant genotype (AA) is represented by $p^2$.

Step 3: Calculate the Frequency

Substitute the value of $p$ into the expression for the AA genotype frequency: $$p^2 = (0.1)^2$$ $$p^2 = 0.01$$

Final Answer: 0.01

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to utilize the Hardy-Weinberg mathematical model to solve a specific population genetics problem.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific algorithmic process (squaring the allele frequency) to derive the genotype frequency.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. This question tests the application of evolutionary biology principles, which is a core requirement for the NEET Biology syllabus under the unit of Genetics and Evolution.