# Destroyer Bio 160

Discussion in 'DAT Discussions' started by drtoothy, May 30, 2008.

1. ### drtoothy

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160. The frequency for the allele for cystic fibrosis, an autosomal recessive trait, is 1 in 2500 among white Americans. What is the heterozygote frequency?
A) 8%
B) 4%
C) 1%
D) 10%
E) 14%

The answer is B because he uses 1 in 2500 to represent q^2. But if the question says the frequency of the ALLELE (not genotype) is 1 in 2500, I would think that equals equal q. The answer I got using this, 12.5% is not one of the answer choices, but regardless, can someone clarify? Or is this a Destroyer mistake?

3. ### doc3232

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Can someone explain this?
My bio is rusty....

4. ### Zerconia2921 Bring your A-game!

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1 out of 2500, is the ratio of the frequency. So the frequency is 0.0004.
This is equal to = p^2

p^2 = .0004 same as 4x10E-4
p =2x10E-2 => same as.02

p+q = 1
q = 1-.02
q=.98

2(.98)(.02) =
= 0.039 ~ 0.04 x 100% = 4%

5. ### userah

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if it helps to clarify at all, if it's discussing population statistics and it's recessive, the only way to get recessive in a population is by being homozygous recessive. That's how you know it's q^2 and not q. Based on that you can figure out all the other details. let me know if you have any more questions =)

6. ### drtoothy

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So this question was bothering me and I looked it up in Campbell's Bio (7th ed - Pg. 456-466, if you're interested). Allele frequencies are different from genotype frequencies. The frequency of finding a particular allele in the gene pool is p or q. The frequency of finding a specific genotype can be predicted from that using p^2 + 2pq + q^2. I think it comes down to a matter of wording...correct me if I'm wrong, but I'm pretty sure that means it's a Destroyer error. His explanation and calculations would work only if he called it the frequency of the "genotype" of cystic fibrosis.

That make sense?