Biology · Evolution · NEET
In a diploid organism with two alleles at a locus, p is the frequency of the dominant allele A and q is the frequency of the recessive allele a. They are ALLELE (gene) frequencies, not people. Because these are the only two alleles, they must add up to 1: p + q = 1. NCERT says the sum total of all allelic frequencies is 1.
p² = frequency of homozygous dominant individuals (AA). q² = frequency of homozygous recessive individuals (aa). 2pq = frequency of heterozygous individuals (Aa). The '2' in 2pq is there because a heterozygote can form two ways (A from mother and a from father, or a from mother and A from father). NCERT: 'AA is p², aa is q², Aa is 2pq, hence p²+2pq+q²=1.'
Because the only people you can spot by looking are the homozygous recessive ones (aa) — they SHOW the recessive trait (like a disease). That count gives you q² directly. Take its square root to get q, then p = 1 − q. Once you have p and q you can find every genotype frequency. Heterozygotes (Aa) look like AA, so you can never count them directly — you must calculate 2pq.
AA = p². Here p (frequency of A) = 0.1, so AA = (0.1)² = 0.01. Students wrongly pick 0.10 by forgetting to square. This is exactly the ReNEET 2026 question. Also note q = 1 − 0.1 = 0.9, aa = q² = 0.81, and Aa = 2pq = 2 × 0.1 × 0.9 = 0.18. Check: 0.01 + 0.18 + 0.81 = 1.
The 1 in 100 showing the recessive disease are aa, so q² = 1/100 = 0.01. Then q = √0.01 = 0.1 and p = 1 − 0.1 = 0.9. Carriers are heterozygotes = 2pq = 2 × 0.9 × 0.1 = 0.18, i.e. 18 in 100. So carriers are far more common than affected people — a favourite NEET twist.
A population of diploid organisms is at Hardy–Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is ________.
In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by
Try the real previous-year questions from this chapter — each with the answer and a full solution.
p + q = 1 and p² + 2pq + q² = 1. The first relates the two allele frequencies; the second relates the three genotype frequencies. Every NEET numerical is solved by combining these.
Only q² (the homozygous recessive aa), because those individuals actually show the recessive trait and can be counted. From q² you get q, then p, then everything else.
No. Hardy-Weinberg deals in frequencies (proportions between 0 and 1), not counts. If they give you numbers of people, convert to a fraction of the total first.
Yes — two Aa (dominant-looking) parents can have an aa child. It does not break the equation; heterozygotes (2pq) carry the hidden recessive allele, which is why carriers are often more common than affected individuals.
By NCERT convention p is the frequency of allele A (dominant) and q of allele a (recessive), but the maths works either way as long as you stay consistent. Read the question to see which allele's frequency is given.
