Hardy Weinberg Principle

Here is the trap that costs marks on Hardy-Weinberg questions: students write p² + q² = 1 and forget the 2pq heterozygote term entirely. They then solve for p or q, get a wrong number, pick a distractor, and lose 5 marks (1 lost + 4 forfeited by negative marking). The full equation is p² + 2pq + q² = 1, and dropping 2pq is the single most reliable way to get the wrong answer on this topic.

What the principle states. In an idealised, large population with no evolutionary forces acting, allele frequencies and genotype frequencies remain constant across generations (NCERT Class 12 Biology Chapter 6, page 138). The two equations are:

• Allele frequencies: p + q = 1 (where p = frequency of dominant allele A, q = frequency of recessive allele a)
• Genotype frequencies: p² + 2pq + q² = 1 (where p² = AA, 2pq = Aa, q² = aa)

The five conditions for equilibrium. Hardy-Weinberg holds only when ALL five conditions are met simultaneously: (1) no mutation, (2) no gene flow (migration), (3) no genetic drift (large population), (4) no natural selection, (5) random mating. Violation of any single condition pushes the population out of equilibrium — this is a common trap in NEET questions that ask "which factor disturbs/maintains Hardy-Weinberg equilibrium."

NEET application. Questions typically give you the frequency of one genotype (often the homozygous recessive aa = q²) and ask you to calculate carrier frequency (2pq) or dominant homozygote frequency (p²). The workflow: q² given → take square root → get q → compute p = 1 − q → plug into p², 2pq as needed. Every step requires you to remember that genotype frequencies have three terms, not two.

Watch-out. If a question asks what happens when one of the five conditions is violated, the answer is always "allele frequencies change" — the population evolves. Random mating alone does not cause evolution; it is the absence of all disturbing forces that maintains equilibrium.

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