# Destroyer Bio Q.

Discussion in 'DAT Discussions' started by Golfguy, Jul 24, 2011.

1. ### Golfguy 5+ Year Member

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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 4%. What I don't understand is why q^2= .0004 (1/2500). Shouldn't q =.0004? I thought q was the frequency of the recessive allele, and q^2 was the phenotypic frequency. Can someone please explain this to me?

Thanks Much!

2. ### PooyaH 2+ Year Member

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When it says it's an autosomal RECESSIVE trait, it's talking about the genotype. And since CF is a recessive disorder, you have to be homozygous recessive to show it. But I think it's not a very well-worded problem. It shouldn't say the frequency of the "allele" in the beginning of the question. That's def misleading.

3. ### Hope30

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Regarding the question, I tried to solve it and have something close to 4%, but not sure it is right.

p^2= 1/2500 --> take the square root of 1/2500 will give you p=0.02
p + q = 1
1 - 0.02 = 0.98

p^2 + 2pq + q^2 = 1
The question asks for heterozygous, so use 2pq
2(0.98)(0.02) = 0.0392
0.0392*100% = 3.92% close to 4%

4. OP

### Golfguy 5+ Year Member

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Thanks guys. It seems as if it's just worded incorrectly then. It should read the prevalence of cystic fibrosis is......not allele prevalence.

I thought the frequency of an allele is p or q, not p^2 or q^2.
I thought the frequency of the phenotype is p^2, or q^2.

For example, if 80% of a population express the dominant phenotype, then p^2 + 2pq = .8, and q^2 = .2 for the recessive.

So, solving from the recessive (easier). We would get q = .44, 1-.44= .56 = p.

Does this look correct?

Oh, and Hope, you are correct.

5. ### PooyaH 2+ Year Member

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Yes you are correct. q and p refer to the frequency of the alleles. p^2, 2pq, and q^2 refer to the frequency of genotypes among a population.

6. OP

### Golfguy 5+ Year Member

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Thanks PooyaH!

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