Q.91 Imagine a population of diploid species in Hardy–Weinberg equilibrium.
The population has two alleles for a gene which are ‘a’ and ‘A’. The number of
individuals with ‘aa’ genotype in this population is 1 in 10000. The frequency of
the allele ‘A’ in the population is ______ (up to two decimal places)
The frequency of allele ‘A’ is 0.99.
In Hardy-Weinberg equilibrium, the ‘aa’ genotype frequency equals q², where q is the frequency of allele ‘a’.
Problem Solution
Given 1 in 10000 individuals have genotype ‘aa’, the frequency of ‘aa’ is q² = 1/10000 = 0.0001.
Take the square root: q = √0.0001 = 0.01.
Since p + q = 1, the frequency of ‘A’ is p = 1 – 0.01 = 0.99.
Step-by-Step Derivation
Hardy-Weinberg equilibrium for two alleles states genotype frequencies as p² (AA), 2pq (Aa), and q² (aa), where p² + 2pq + q² = 1.
The recessive homozygote ‘aa’ frequency directly gives q², allowing direct calculation of q without needing other genotypes.
This holds in diploid populations under equilibrium assumptions (no selection, mutation, migration, drift, random mating).
In Hardy-Weinberg equilibrium for diploid populations with two alleles A and a, solving for A allele frequency when aa genotype frequency is 1 in 10000 yields precise results essential for CSIR NET Life Sciences. This keyphrase—hardy-weinberg equilibrium aa genotype frequency 1/10000 A allele—guides competitive exam preparation in population genetics.
Core Principles
The Hardy-Weinberg principle predicts genotype frequencies remain constant without evolutionary forces. For alleles A (frequency p) and a (frequency q), where p + q = 1, frequencies are AA: p², Aa: 2pq, aa: q².
Here, aa individuals (1/10000) mean q² = 0.0001, so q = 0.01 and p = 0.99.
Detailed Calculation
q² = 1/10000 = 0.0001
q = √0.0001 = 0.01
p = 1 – q = 0.99
This matches equilibrium expectations for rare recessives.
Exam Relevance
No options provided, but common CSIR NET traps include mistaking q for p or ignoring square root. Direct q² use simplifies recessive trait problems.
Verify via expected frequencies: AA ≈ 0.9801, Aa ≈ 0.0198, aa = 0.0001 (sums to 1).
1 Comment
Ritika Jangir
January 5, 2026Done