Consider an X-linked recessive trait that is present in 5% of the males in a population....
Consider an X-linked recessive trait that is present in 5% of the males in a population. Assume Hardy-Weinberg equilibrium and a 1:1 sex ratio in the population. What is the frequency of the trait in the females of that population?
Refer to question 7. What is the frequency of carrier females in that population?
** I understand how to do part one but not sure how to do part two. The correct answer is 0.0950**
Homework Answers
Answer:
The frequency of colorblind men = q = 5% = 0.05
The colorblind allele frequency = Colorblind phenotype frequency in male (as males have only one X-chromosome)
The frequency of colorblind allele =q= 0.05
The frequency of normal allele = 1-q = 1-0.05 = 0.95
The frequency of colorblind females = qq = 0.05 * 0.05 = 0.0025
The frequency of the carrier females = 2pq = 2*0.95*0.05 = 0.095
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