namf
Forum Elite
Topics: 80
Posts: 312
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Hey!
Saw this question on the Utah site (http://umed.med.utah.edu/testfiles/frame_template...)
withouth any explanation for the answer. Help me out please!
An X-linked recessive disease has a gene frequency of 0.002 (1/500) in a population. The penetrance of the disease gene is 50%. What proportion of the male population is expected to be affected with this disease?
I'll post their answer just as soon as I hear back, but it's not 1/2000. What I got on second thought was saying 1/2 of boys born to carrier females x carrier females at 1/250 x 1/2 penetrance--is that how you'd do it to get 1/1000??
Thanks!!
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| mash
Forum Fanatic
Topics: 147
Posts: 1,326
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1/1000
gene frequency is proportion of chromosomes dat contain a sp. allele.
in XLR diseases only X ch carries the defective gene...dat means q^2 =1/500
if penetrance is 50% then no. of males affected wud be 1/500*1/2 = 1/1000
___________________
I hear and I forget. I see and I remember. I do and I understand.
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| mash
Forum Fanatic
Topics: 147
Posts: 1,326
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wats the ans?
___________________
I hear and I forget. I see and I remember. I do and I understand.
--Confucius
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| Malaysian
Forum Guru
Topics: 28
Posts: 778
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I got 1/1000 too.
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