Q.65 The frequencies for autosomal alleles A and a are p = 0.5 and q = 0.5,
respectively, where A is dominant over a. Under the assumption of random
mating, the mating frequency among dominant parents is _________.
(Rounded off to two decimal places)
The mating frequency among dominant parents under random mating is the square of the dominant phenotype frequency, which is 0.75 for these allele frequencies.
Genotype Frequencies
In Hardy-Weinberg equilibrium with allele frequencies p = 0.5 (A) and q = 0.5 (a), genotype frequencies are p² for AA, 2pq for Aa, and q² for aa. This yields AA = (0.5)² = 0.25, Aa = 2(0.5)(0.5) = 0.50, and aa = 0.25. Since A dominates, the dominant phenotype (AA or Aa) frequency is p² + 2pq = 0.75.
Dominant Phenotype Proportion
The recessive phenotype frequency equals q² = 0.25, so the dominant phenotype covers the remaining 0.75 of the population. Random mating means individuals pair proportional to their frequencies in the population.
Mating Frequency Calculation
Under random mating, the frequency of matings between two dominant-phenotype parents equals (dominant frequency)² = (0.75)² = 0.5625, or 0.56 when rounded to two decimal places. This follows from panmixia, where mating probabilities are products of individual frequencies.
Introduction
In population genetics, understanding autosomal alleles A and a frequencies p=0.5 q=0.5 dominant parents mating frequency under random mating is crucial for exams like CSIR NET Life Sciences. This scenario applies Hardy-Weinberg equilibrium to predict mating probabilities among phenotypes.
Hardy-Weinberg Basics
Hardy-Weinberg principle assumes random mating, no selection, and equal allele frequencies across generations. For p=0.5 (dominant A) and q=0.5 (recessive a), genotypes are AA=0.25, Aa=0.50, aa=0.25. Dominant phenotype (A_) frequency is 0.75.
Step-by-Step Solution
- Compute dominant phenotype frequency: p² + 2pq = 0.25 + 0.50 = 0.75.
- Random mating frequency between dominant parents: (0.75)² = 0.5625.
- Round to two decimal places: 0.56.
No options provided, but common distractors include:
- p²=0.25 (AA only)
- 2pq=0.50 (heterozygotes)
- 0.75 (phenotype frequency)
The correct value accounts for all dominant individuals (AA × AA, AA × Aa, Aa × Aa).
CSIR NET Relevance
This matches CSIR NET Q.65 (2023-Feb), testing equilibrium applications in genetics. Maximum heterozygosity occurs at p=q=0.5, emphasizing why this setup is common in problems.
