Which one of the following equations is based on Hardy-Weinberg princ…

The correct answer is Option (1) → \( p^2 + 2pq + q^2 = 1 \)

The Hardy-Weinberg principle says that allele frequencies in a population stay constant from generation to generation. Take one gene with two alleles, A and a, with frequencies p and q. Then AA has the frequency p², aa has q², and Aa has 2pq. The three genotypes make up the whole population, so their frequencies add up to 1. This is the expansion of \( (p+q)^2 \).

(1) \( p^2 + 2pq + q^2 = 1 \) — Correct. This is the equation itself.

(2) \( p^2 + pq + q^2 = 1 \) — Incorrect. The heterozygote Aa arises in two ways (A from the father and a from the mother, or the reverse), so its frequency is 2pq. Check with \( p = q = 0.5 \):

    \( \displaystyle 0.25 + 0.25 + 0.25 = 0.75 \ne 1 \)

(3) \( p + 2pq + q = 1 \) — Incorrect. The first and third terms are the allele frequencies, and \( p + q = 1 \) already. The extra 2pq would make the total more than 1.

(4) \( p^2 + 2pq + q^2 = 0 \) — Incorrect. The three genotype frequencies cannot add up to zero, because every individual belongs to one of the three genotypes. The total is 1.

Note: A similar question was also asked in CUET (UG) 2026 on 29-05-2026 (Slot 2).

NCERT Extract:
Hardy-Weinberg Principle – “… The frequency of AA individuals in a population is simply p2. This is simply stated in another ways, i.e., the probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e., p2. Similarly of aa is q2, of Aa 2pq. Hence, p2+2pq+q2=1. This is a binomial expansion of (p+q)2.”