- Joined
- Dec 22, 2014
- Messages
- 405
- Reaction score
- 333
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?
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?