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Old 03-21-2012
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Arrow NBME 12 question

Ok, I came across this HARDY-WEINBERG question in the 3rd section, where they ask you to figure out the carrier risk.
Man, I cant figure it out! I calculate the answer, but its not in the options list.

They have an incidence of CF in the general population to be 1/40,000

so i take a square root of it and get 1/200

trying to find the carrier rate (2pq)

I have 2 x 199/200 x 1/200

I get 0.0095 as a result!

Do they mean that this is kinda close to 1/100 or what? I have trouble with Hardy-Weinberg.
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Old 03-21-2012
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you are correct till 2pq = 200 ....... but have to take p as 1 so q = 1/100
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Originally Posted by Hitman View Post
you are correct till 2pq = 200 ....... but have to take p as 1 so q = 1/100
But why do u take p as a 1, when its 1-q
(p+q=1)
?
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Originally Posted by DocSikorski View Post
But why do u take p as a 1, when its 1-q
(p+q=1)
?
I know but its given in kaplan biochem ..... dont remember much but had noted down that to calculate carriers need to take p as 1 becoz p are normal people and are very large part of the population and q is a very small part so they approximate it .something like that dont know the exact detail ... just need to cram this up for solving them it i felt .....
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Default NBME form 12 block 3 #20

With these type of questions...I suggest that you write down what's given as you're reading the question

- autosomal recessive
- incidence of cystic fibrosis in the population: (1)/(40,000)
- what's the wife's risk for being a carrier?

this type of question is called "carrier prevalence" &/or "disease prevalence"

in this case, they are asking for carrier prevalence = "wife's risk"

this is the equation that you need to use for this type of question is called "carrier prevalence" or "disease prevalence"

2 + 2q + q^2

2q = carrier prevalence

q^2 = disease prevalence

they gave you disease prevalence in the question = incidence of cystic fibrosis in the population = q^2 = (1)/(40,000)

take the square root of [(1)/(40,000)] = 1/200 = q

what's the wife's risk for being a carrier? they want use to find "carrier prevalence"

"carrier prevalence" = 2q = (2)[1/200] = 1/100

ans. 1/100
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Quote:
Originally Posted by USMLE16 View Post
With these type of questions...I suggest that you write down what's given as you're reading the question

- autosomal recessive
- incidence of cystic fibrosis in the population: (1)/(40,000)
- what's the wife's risk for being a carrier?

this type of question is called "carrier prevalence" &/or "disease prevalence"

in this case, they are asking for carrier prevalence = "wife's risk"

this is the equation that you need to use for this type of question is called "carrier prevalence" or "disease prevalence"

2 + 2q + q^2

2q = carrier prevalence

q^2 = disease prevalence

they gave you disease prevalence in the question = incidence of cystic fibrosis in the population = q^2 = (1)/(40,000)

take the square root of [(1)/(40,000)] = 1/200 = q

what's the wife's risk for being a carrier? they want use to find "carrier prevalence"

"carrier prevalence" = 2q = (2)[1/200] = 1/100

ans. 1/100

But why do u take p as 1?
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Quote:
Originally Posted by DocSikorski View Post
But why do u take p as 1?
I think it's becoz value of P >>> q and it reaches approximately 1. So for simplicity p=1 is used.
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Originally Posted by terminator View Post
I think it's becoz value of P >>> q and it reaches approximately 1. So for simplicity p=1 is used.
Well no. For example if the prevalence is 1/25 in the general population, than you definitely cannot take p as 1.

So I guess In this case .0095 is rounded to .01 = 1/100...
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In that case, Yes you can't take p=1 becoz prevalence of the disease is very high vs in the question where it is very low. In any case, I would do it the old fashioned way. It's less complicated. lol
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