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#1




What probability of a child who is affected homozygote
A man is a known heterozygous carrier of a mutation that causes hemochromatosis (AR). Suppose that 1% of the general population consists of homozygotes for this mutation. If the man mates with somebody from the general population, what is the best estimate for the probability that he and his mate will produce a child who is an affected homozygote?
A) .025 B) .05 C) .09 D) .10 E) .25 I know its very challenging qn, plz try everyone. Thanks. 
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drcuddles (07162015), Usmle16Forall (09212016) 
#2




Guys... give me a guess....lol

#3




Tough question... I'm going to wait for your answer



#4




Quote:
If 1% of population are homozygotes(q^2) then q=0.1 the chance of being heterozygote will be 2pq=2x1x0.1=0.2(its the chance of being mate with heterozygote woman in this population) the chance of produce homozygote child if two parents were hetrozygotes is 0.25 so 0.25x0.2=0.05 
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#5




Could the answer be letter C?
1) p^2+2pq+q^2 = 1 also 2) p+q=1 q^2 = 0.01 (since the popn is 100%) so... q=0.1 using the second equation p+q=1 p= 1 0.1 = 0.9 then 2pq = 2(0.9)(0.1) = 0.18 mated with the man who is heterozygote 0.18 x 0.5 (since the man posses half of the recessive chromosome so 0.5) answer is 0.09 i dont know if this is right but if all of the questions of the USMLE would be like this i would be so DOOMED 
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aktorque (12082010) 
#7




Thank you very much for your effort. Fortunately, you won't see this kind of qn in the real exam, I put this qn for your practice. It's not an one minute qn. Don't worry bro. I'll put the correct answer with the fully explanation.

#8




Congrats to those guys who tried real hard to find the correct ans. You guys are so smart.
Here is the explanation: Correct ans is B) One must first determine the probability that the man's mate will also be a heterozygous carrier. This is done by applying the HardyWeinberg rule, p2+2pq+q2 = 1 where p2 is the frequency of homozygous dominant individuals, 2pq is the frequency of heterozygous individuals, and q2 is the frequency of homozygous recessive individuals. If the frequency of affected homozygous is 1% (given), then the gene frequency of the hemochromatosis mutation is the square root of 1%, or 0.10. Then the estimate frequency of heterozygous carriers in the population, 2pq, is 2x0.9x0.1 = 0.18 (where p is 10.1= 0.9) (notice that we do not use the 2q approximation in this case coz p is not approximately equal to 1). The probability that two heterozygous carriers will produce an affected offspring is 0.25, so the probability that the man mates with a carrier and the probability that they will in turn produce an affected offspring is obtained by multiplying the two probabilities together: 0.25x0.18 = 0.045. Additionally, there is a small chance (0.01) that the person he mates with will be a homozygote, and thus there is a chance (0.01/2 = 0.005) that the offspring of this mating will be affected, since the man has a 1 in 2 chance of transmitting the hemochromatosis gene. The correct answer then is 0.045 + 0.005 = 0.05. Tough ehh.... lol 
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#9




this is what I think !
1% of gen population is homozygous means q^2= 1/100
(we takeq^2 for affected people in AR diseases and only q in X linked recessive diseases) the man himself is a heterozygote i.e. 2pq, the chance that he mates with a heterozygote female is 99/100,more than him mating a homozygote female i.e. 1/100 because there are 99% heterozygotes in the population the probability that the two heterozygotes produce an affected child will be 25% or 1/4 or 0.25 Answer E is what I'll choose, I don't know I can be wrong, correct me if I am. 
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aktorque (12082010) 
#10




Quote:
Now remember that this man is heterozygous. So his children can only be homozygous diseased if he mates with
0.045 + 0.005 = 0.05 Looks simple though .... but took me like ages to figure that out... @aktorque Anyway...Thanks for the question.... now i think i will be able to mange Probability questions....
__________________
Knowing is not enough; we must apply. Willing is not enough; we must do. 
#11




Ahh hehehe nearly there oh well tnks for the info ^_^ glad to know how to solve this problem

#12




For me, the key was the line:

#13




Once you accept that, I think you're good to go, I couldn't make it work because I generally forget and use p=1 since most homozygote frequencies are much less than 1/100

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