Biology · Evolution · NEET
p is the frequency of the ALLELE A, not the genotype. p and q are allele frequencies (p for A, q for a) and they always add up to 1, so p + q = 1. The genotype frequencies are the squared/product terms: AA = p², Aa = 2pq, aa = q². So p = allele A, but p² = genotype AA. Mixing these up is the most common Hardy-Weinberg mistake in NEET.
They describe two different things. p + q = 1 adds up the two ALLELE frequencies (only A and a exist, so together they make the whole = 1). p² + 2pq + q² = 1 adds up the three GENOTYPE frequencies (AA + Aa + aa = whole population = 1). The second equation is simply the first one squared: (p+q)² = p² + 2pq + q² = 1² = 1.
2pq is the frequency of heterozygous individuals (Aa). The 2 is there because a heterozygote can form in two ways: allele A from the father and a from the mother, OR a from the father and A from the mother. Each way has probability pq, so together it is pq + pq = 2pq. NEET has directly asked which term stands for heterozygotes, and the answer is 2pq.
Genetic equilibrium means the gene pool is not changing. The gene pool is the total of all genes and their alleles in a population. If allele frequencies stay constant generation after generation, the population is in genetic equilibrium (also called Hardy-Weinberg equilibrium). This is the 'no evolution' baseline. NCERT states it clearly: the gene pool remains constant, and this is called genetic equilibrium.
It shows that evolution is happening. When the measured allele frequency differs from the expected value, that difference (and its direction) is a measure of evolutionary change. So Hardy-Weinberg is used as a null test: if frequencies stay put, no evolution; if they shift, one of the five disturbing factors is acting. This 'disturbance = evolution' link is exactly why NEET loves this topic.
In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by
A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?
Which of the following factors will not affect the Hardy-Weinberg equilibrium?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It states that allele frequencies in a population stay constant from generation to generation when no evolutionary force acts on it, keeping the gene pool in genetic equilibrium.
p² + 2pq + q² = 1, which is the binomial expansion of (p+q)². Here p² = AA, 2pq = Aa, q² = aa, and separately p + q = 1 for the two allele frequencies.
Gene migration (gene flow), genetic drift, mutation, genetic recombination and natural selection. Any of these changes allele frequencies and therefore signals that evolution is occurring.
p is the frequency of the allele A. p² is the frequency of the genotype AA (homozygous dominant). Always square the allele frequency to get the homozygous genotype frequency.
It acts as a 'no-evolution' baseline. If real allele frequencies differ from the values the equation predicts, that difference measures the amount and direction of evolutionary change.
