Question

Consider a population of 425 diploid giant Sequoia trees. In this population, you observe the following genotypic counts: 100 homozygous dominant genotypes, 250 heterozygous genotypes, and 75 homozygous recessive genotypes. What is the allele frequency of the recessive allele in this population (use two decimal places and the usual rounding conventions)

Answer 1

Answer:

q = 0.42

Explanation:

This question is an example of Hardy-Weinberg question and there are two equations necessary to carry out this question;

p + q = 1

p² + 2pq + q²  = 1

where;

p = the frequency of the dominant allele

q =  the frequency of the recessive allele

p² = the frequency of individuals with homozygous dominant genotype

2pq  = the frequency of individuals with heterozygous genotype

q² = frequency of individuals with the homozygous recessive genotype

Since the total  population = 425

q² = $\frac{individuals with recessive genotype}{Total Population}$

= $\frac{75}{425}$

q² = 0.1765            

To find q; we need to square root both side to           eliminate the square from  q².

∴ $\sqrt{q^2}=\sqrt{0.1765}$

q = 0.4201

q = 0.42       (to two decimal places)