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Question 87 - According to Hardy-Weinberg Equation, p x p + 2 x p x q + q x q = 1 where p = dominant allele frequency and q = recessive allele frequency. Let p + q = 1. Fraction of population has 2 copies of the p gene = p x p. Fraction of population has 2 copies of the q gene = q x q. Fraction of population has a copy of p gene and a copy of q gene = 2 x p x q. In a small town, the allele frequency is q = 0.2 for a recessive gene, the delta-32 mutation, that gives human protection from HIV infection. (a) Find the allele frequency a dominant gene, p. (b) What percent of the population has at least a copy of the gene that cause the population either immune to HIV or less susceptible to the disease?
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Answer 87 - (a) Let p + q = 1, then p = 1 - q = 1 - 0.2 = 0.8. (b) Percentage of population has 2 copies of the p gene = p x p x 100 = 0.8 x 0.8 x 100 = 64 %. Percentage of population has at least a copy of the q gene = (1 - Fraction of population has 2 copies of the p gene) x 100 = (1 - p x p) x 100 = (1 - 0.8 x 0.8) x 100 = (1 - 0.64) x 100 = 36 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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