Greetings, The question states: 160/1000 members exhibit a recessivetrait that d

Question

Greetings, The question states: 160/1000 members exhibit a recessivetrait that does not affect viability of an individual. How manyindividuals in this population are heterozygous carriers of thegene that causes this trait? The answer is: 480. I understand the p+q=1 and p^2 + 2pq+ q^2=1 equations. But, Idon't understand why the solution took the square root of160/1000? May someone show me the step by step calculation to thisproblem (no calculators allowed on the test I am taking, so if youcould approximate answers by paper, that would be great). Thanks and I will rate! Greetings, The question states: 160/1000 members exhibit a recessivetrait that does not affect viability of an individual. How manyindividuals in this population are heterozygous carriers of thegene that causes this trait? The answer is: 480. I understand the p+q=1 and p^2 + 2pq+ q^2=1 equations. But, Idon't understand why the solution took the square root of160/1000? May someone show me the step by step calculation to thisproblem (no calculators allowed on the test I am taking, so if youcould approximate answers by paper, that would be great). Thanks and I will rate!

Explanation / Answer

If 160/1000 members have the recessive trait with genotype qq, thenthe frequency of this genotype is .16 ( or 160/1000). Thisfrequency is q^2 in the hardy weinberg equilibrium. So we know the following q^2 = .16. So q= .4 Since p+q = 1 and q = .4 then p + .4 = 1 so p = .6 The carrier frequency is represented by 2pq in the equlibrium. So 2pq = 2*.4*.6 = .48 Multiply this number by 1000 to get the total number of carriers inthe population: .48 * 1000 = 480.

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