Hardy Weinberg Question (EK)

[KungFuPanda123]

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Question: About 0.25% of the members of a large population in Hardy-Weinberg equilibrium have CF. What proportion of this population are carriers for the CFTR mutation?

Choices:
A) 99.7%
B) 0.25%
C) 10%
D) 1%

My answer: C
Correct answer: D

Background info: CF (Cystic Fibrosis) is due to an autosomal recessive allele.

How I set up for my answer:
I know that p^2 + 2pq + q^2 = 1
I know that q^2 = 0.0025
so q = 0.05
so since p + q = 1... then p = 0.95
The proportion of carriers is 2pq, so 2*(0.95)*(0.05) = 0.095
0.095 * 100% = 9.5%

The closest answer to 9.5% is 10%, so I answered C.

Can anyone explain what is wrong with my logic?

 

[StarFall]

Math/logic seems right. EK may just be wrong. I got the same calcs.

 

[Mwooster]

Solid calculations. EK seems to be wrong there. Check the EK Board to see if this question is under their "errata"

 

[NextStepTutor_1]

This all looks correct to me too. Unless there's specific info from that passage that makes it incorrect, 9.5% is correct.

 

[shefv]

I also calculated it to be 9.5%