Numerical Problems/ISC/Class 12
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Created by Titas Mallick
Biology Teacher • M.Sc. Botany • B.Ed. • CTET (CBSE) • CISCE Examiner
Numerical Problems - Hardy-Weinberg principle
In a population of 1000 individuals, 360 are homozygous dominant (AA), 480 are heterozygous (Aa), and 160 are homozygous recessive (aa). Calculate the frequencies of the dominant allele (p) and the recessive allele (q).
Solution:
Total number of alleles = 2 * Total individuals = 2 * 1000 = 2000
Number of A alleles:
Frequency of dominant allele (p) = (Total A alleles) / (Total alleles) = 1200 / 2000 = 0.6
Number of a alleles:
Frequency of recessive allele (q) = (Total a alleles) / (Total alleles) = 800 / 2000 = 0.4
Verification: p + q = 0.6 + 0.4 = 1.0
If the frequency of the recessive allele (q) for a certain trait in a population is 0.3, and the population is in Hardy-Weinberg equilibrium, what are the frequencies of the homozygous dominant, heterozygous, and homozygous recessive genotypes?
Solution:
Given: q = 0.3
Since p + q = 1, then p = 1 - q = 1 - 0.3 = 0.7
According to the Hardy-Weinberg principle (p^2 + 2pq + q^2 = 1):
Frequency of homozygous dominant (p^2): p^2 = (0.7)^2 = 0.49
Frequency of heterozygous (2pq): 2pq = 2 * 0.7 * 0.3 = 0.42
Frequency of homozygous recessive (q^2): q^2 = (0.3)^2 = 0.09
Verification: p^2 + 2pq + q^2 = 0.49 + 0.42 + 0.09 = 1.0
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