# #160 on DESTROYER BIO

Discussion in 'DAT Discussions' started by yomikepump, Jun 2, 2008.

1. ### yomikepump 2+ Year Member

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This question says that the ALLELE frequency for cystic fibrosis(autosomal recessive) is 1/2500. Shouldnt that translate into just q. the solutions say that that translates into q^2. i think that q = 1/2500 since it is the allele not the trait. Am i right? is this an error or am i wrong?

2. ### userah 7+ Year Member

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no it's q^2 because it's a autosomal recessive meaning you need to be homozygous recessive; thus q^2.

hope that helps?

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### yomikepump 2+ Year Member

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I know if it is autosomal reccessive then you would have to have q^2 in order to have the TRAIT. the question says ALLELE frequency which can only be just q since it is recesseive. right?

4. ### userah 7+ Year Member

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maybe bad choice of wording? The way I approached this problem was that if they're giving statistics of people in a population, the only way they have such a condition is if they're homozygous recessive/dominant so i'd have to calculate p or q from that. I never concentrated on the word 'allele'

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### yomikepump 2+ Year Member

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well thanks for all the help man. o and congrats on ur scores i saw ur posted scores, very nice job i hope i can do the same im takin the test june 30 and still dont know hardly any equations for gchem or any ochem. i hope i can get done in time.

6. ### userah 7+ Year Member

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just make sure to know destroyer cold. especially for orgo. it was really helpful.

7. ### arpitpatel86 2+ Year Member

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ok i had the same problem with this one....hopefully some1 can clearify it....if u look at number 12 on the bio section it says "A dominant allele has a gene frequency of 0.3. What percentage of heterozygotes is present in thies population assuming Hardy-Weinsberg equilibrium?
The solution says"

p= dominant allel frequency=0.3
q= recessive allele frequency=0.7 (1-0.3)
Then they went on to do 2pq which gave the the correcet answer.

so for number 160 they said "the frequency for the allele for cystic fibrosis , an autosomal recessive trait is (1/2500)amonth white Americans. What is teh heterozygote frequency"\

so if q= recessive allele frequency shouldnt that equal (1/2500) like in number 12? 1/2500 equal 0.0004 so just like they gave the dominant allele frequency(0.3) in the problem of number 12 and said that =P why doesnt the recessive allele q= equal just 0.0004 and not q^2=0.0004. these two question seem to be contradicting each other because if 160 is correct then why didnt p get squared in the problem 12?

8. ### DentalDeac 2+ Year Member

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I think think what people are struggling with is the definition of allele. Using the pea example, you have two alleles, round (R) and wrinkled (r). When they are discussing alleles and the frequency of the alleles in a population, they are talking about the phenotype (use hardy-weindberg). When they ask for the gene frequency they are giving you p or q. Does that make sense?

9. ### Glycogen 2+ Year Member

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If I understood ur qs right,you are confused with the fact that why they did not simply have q=1/2500
I don't have destroyer right in front of me so I don't know the qs but whenever they are mentioning frequency,you've got to know that it is dealing with p^2+2pq+q^2=1
So,in this case,when you are given 1/2500 you have to get the root b/c q^=1/2500 which is 1/50=.02 now you have ur q and you can simply calculate ur p which is .8 and 2pq which is .032
Feel free to correct me if I am wrong b/c this is also my understanding of these sort of problem!

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### yomikepump 2+ Year Member

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I totally understand where ur comin from that is why i posted this thread in the first place. I get the explanations and exactly how to do the problem but i still think it is a typo cuz allele means 1/2 of a gene which is either p or q. when the problem says 1/2500 show the trait it means that it has to be q^2 for autosomal recessive.

q= allele, i dont think there is any way that q^2 could mean allele frequency. it is just the frequency of homozygous recessive people.

11. ### War Eagle 5+ Year Member

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yomikepump, you're right. It must be a typo. The only way 1/2500 would be q^2 is if that were the number of people that have cystic fibrosis. Allele frequecy is p + q = 1. Check out Campbell's pg 458 for a similar example with PKU. They say that the OCCURANCE of PKU is 1/10,000 so to find q, they take the sqare root. OCCURANCE is q^2, ALLELE FREQUENCY is just q. If this is your first time through the destryer you'll find that there are other mistakes, so don't be too surprised. It is an awesome resource, but not even the orgoman himself can write a book with 700+ questions without errors.

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12. ### Mamona 5+ Year Member

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If you have Campbell biology go to page 458 there is a paragraph about population genetics and human health and explain this type of problems. By overlooking seems that there is not typo in Destroyer's question......

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