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rizwansardar
Forum Newbie Topics: 1 Posts: 14 
I am having problem in understanding two questions of Kaplan genetics , since i am not having explanation for these can any one help me in giving explanation of these questions , which are given bellow: 1)If the incidence of Cystic fibrosis is 1/2500 among population of europeans, what is the predicted incidence of heterozygous carriers of a cystic fibrosis mutation in this population? A. 1/25 B. 1/50 C. 2/2,500 D. 1/2500 e. (1/2500)2 remember this 2 in option E is whole square. 2) A man is known heterozygous carrier of mutation that causes hemochromatosis ( autosomal recessive disease ).Suppose that 1% of the general population consists of homogyzotes for this mutation.If the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is an affected homozygote? A. 0.025 B. 0.045 C. 0.09 D. 0.10 E. 0.25 Please if some one can help me out , I will be obliged!  
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Huntseatchick
Forum Junior Topics: 11 Posts: 52 
I don't know if I can do the whole thing, but I'll get it started ... p2 + 2pq + q2 = 1 where: p2 = DD 2pq = Dd q2 = dd Cystic fibrosis = dd = q2 = 1/2500, and the square root of that is 0.02 so q = 0.02 Someone else take it from here  
meg
Forum Guru Topics: 62 Posts: 806 
"rizwansardar" wrote: 1)If the incidence of Cystic fibrosis is 1/2500 among population of europeans, what is the predicted incidence of heterozygous carriers of a cystic fibrosis mutation in this population? A. 1/25 B. 1/50 C. 2/2,500 D. 1/2500 e. (1/2500)2 remember this 2 in option E is whole square.! Since the incidence of this AUTOSOMAL RECESSIVE disease is 1/2500, q2 is 1/2500 so q= 1/50 p = 49/50 (since by Hardy Weinberg p+q=1) a heterozygous carrier would be someone with 2pq so heterozygous carrier is 2 X (49/50) X (1/50) since 49/50 is almost equal to 1, we can just say 2 X (1/50)= 1/25 "rizwansardar" wrote: 2) A man is known heterozygous carrier of mutation that causes hemochromatosis ( autosomal recessive disease ).Suppose that 1% of the general population consists of homogyzotes for this mutation.If the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is an affected homozygote? A. 0.025 B. 0.045 C. 0.09 D. 0.10 E. 0.25 Please if some one can help me out , I will be obliged! I think the answer for second question is A here, the man is a heterozygous, so he has a 50% (1/2) chance of giving the abnormal gene the homozygosity in the population is 1% (1/100), meaning q2= 1/100, so q= 1/10. So a heterozygous carrier will be 2pq= 1/5 So the probability will be 1/2 X 1/5 X 1/4 (since 1/4 will be affected homozygotes) = 1/40 = 0.025  
dxtxpx
Forum Guru Topics: 259 Posts: 1233 
thanks for explanation  
rizwansardar
Forum Newbie Topics: 1 Posts: 14 
Thank You you made my life easier!  
Topics: Posts: 
thanks a lot meg  
rizwansardar
Forum Newbie Topics: 1 Posts: 14 
This is for all of you guys who participated in my Question of genetics , i.e Question no 2 concerning HETEROZYGOTE INDIVDUAL HAVING HEMOCHROMATOSIS. One of my friend also solved this question and gave me explanation of that question which i think its right , read this and after that requires your coments: ONE MUST 1ST DETERMINE THE PROBABILITY THAT THE MAN'S MATE WILL ALSO B A HETEROZYGOUS CARRIER.THIS IS DONE BY APPLYING THE HARDYWEINBERG RULE. If the frequency of affected homozygotes is 1%, then the gene frequency of the Hemochromatosis mutation is the square root of 1% or 0.1 then the estimated freq. Of Heterozygous carriers in the population, 2pq is 2x0.9x0.1=0.18(notice that we don’t use the 2q Approximation in this case b/c p is not approximately equal to 1) the probability that the heterozygous Carriers will produce an affected offspring is 1/4, so the probability that the man mates with a carrier and the probability that they will in turn produce an affected offspring’s obtained by multiplying the 2 probabilities together i.e. 0.18x1/4=0.045  
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