Q.32 The Hardy–Weinberg equation can be used to test :

1. Whether evolution is occurring at a population level.
2. To determine the genotypes of offspring in a genetic cross.
3. Whether evolution is occurring at an individual level.
4. To determine the phenotypes of offspring in a crossing experiment.

   Hardy-Weinberg Equation Tests Evolution at Population Level

   The Hardy-Weinberg equation determines if evolution is occurring at a population level by comparing observed vs. expected genotype frequencies, making the first option correct.

   Introduction

   The Hardy-Weinberg equation serves as a null model to test whether evolution is occurring at the population level. By comparing observed genotype frequencies against expected equilibrium values (p² + 2pq + q² = 1), deviations indicate evolutionary forces at work. This population genetics cornerstone is essential for GATE Life Sciences preparation.

   Option Analysis

   H-W Assumptions: Infinite population, random mating, no selection/mutation/migration/drift.

   • Whether evolution is occurring at a population level: Correct. Primary application: χ² test compares observed (O) vs. expected (E) genotypes. Significant deviation (p<0.05) indicates evolution via selection, drift, etc.

   • To determine the genotypes of offspring in a genetic cross: Incorrect. Punnett squares/Mendelian ratios predict individual crosses; H-W analyzes population allele frequencies, not single matings.

   • Whether evolution is occurring at an individual level: Incorrect. Evolution operates on populations (allele frequency change), not individuals. Individuals don’t evolve; populations do.

   • To determine the phenotypes of offspring in a crossing experiment: Incorrect. Phenotypic ratios from test crosses/chi-square; H-W specifically tests genotypic frequencies under equilibrium.

   Hardy-Weinberg Applications

   Equilibrium Test Protocol:

   text

   1. Count genotypes → Calculate p, q (p + q = 1)
2. Expected: p²(AA) + 2pq(Aa) + q²(aa) = 1
3. χ² = Σ(O-E)²/E → df = #genotypes-1
4. p-value <0.05 = Evolution occurring[web:161]


   Example: Sickle cell (HbA/HbS)

   • Observed: AA=0.81, AS=0.18, SS=0.01

   • Expected: p²=0.81, 2pq=0.18, q²=0.01 (equilibrium)

   • Malaria regions show heterozygote advantage deviation.

   Force	Effect on H-W	Detection
   Selection	Excess homozygotes	χ² significant
   Drift	Random allele change	Small populations
   Migration	Introduces alleles	Frequency shifts
   Non-random mating	Excess homozygotes	Genotype deviation

   Exam Relevance

   GATE Signature Question: “Population in H-W equilibrium?” Calculate p,q from recessive frequency (q=√SS), verify p²+2pq+q²=1. Key: Tests population evolution, not individual inheritance.

   Mnemonic: “POPulation genetics tests POPulation EVOlution with H-W EQUAtion.”