# Hard-Weinberg Question AQA Biology watch

1. Hi, I can't seem to get this question right...any help?

http://filestore.aqa.org.uk/resource...-74022-SQP.PDF
QUESTION 6

Use the Hardy–Weinberg equation to calculate the frequency of mosquitoes heterozygous for the KDR gene in this population in 2003

The population at 2003 reads 20%

I get 0.49 but the answer is 0.32 (this is a specimen paper so the MS could be wrong.)
2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

Just quoting in Danny Dorito so she can move the thread if needed
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3. (Original post by grahammnmr)
Hi, I can't seem to get this question right...any help?

http://filestore.aqa.org.uk/resource...-74022-SQP.PDF
QUESTION 6

Use the Hardy–Weinberg equation to calculate the frequency of mosquitoes heterozygous for the KDR gene in this population in 2003

The population at 2003 reads 20%

I get 0.49 but the answer is 0.32 (this is a specimen paper so the MS could be wrong.)
The MS is 100% correct haha, the hardy weinberg equation is p+q=1 right? The mosquitoes have either an KDR+ or KDR- allele, and 20% of the alleles are KDR-, so that means 80% are KDR+. Do you see why? If we make KDR- = p, and KDR+ = q, then 0.2 + q = 1, therefore q = 1 - 0.2 = 0.8 = 80%.

Now we have the proportion of alleles, you simply do 2 x 0.2 x 0.8 = 0.32. Bingo, there's your answer.

The reason you do 2 x 0.2 x 0.8 is because there are 4 combinations that the alleles can be in; they can be homozygous dominant (pp), heterozygous (pq or qp) and homozygous recessive (qq). We're only interested in the heterozygous combination, and pq+qp = 2pq = 2 x p x q = 2 x 0.2 x 0.8

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