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Old 08-10-2011
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Default genetics-Hardy weinberg question

Q. A man is a known heterozygous carrier of disease Y, which is an autosomal recessive condition. Disease Y has a prevalence of 4% in the population. What are the man's chances of mating with a heterozygous carrier female and having a child affected with disease Y?


A. 2/25
B. 3/50
C. 1/100
D. 3/100
E. 1/400


Explanation must !!!!


Thanks in advance !!!!
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Old 08-10-2011
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I agree

in AR allele freq q2=4%, q=0.2
carrier freq=2*0.2=0.4
chance of child gettin disaes in AR= 1/4
mother s bcum a carrier =0.4
=1/4*0.4
1/10,
YA ME TOO getting 1/10.
its nt in ans choice...

Last edited by surez18; 08-10-2011 at 06:36 AM.
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Old 08-10-2011
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Default

Quote:
Originally Posted by surez18 View Post
in AR allele freq q2=4%, q=0.2
carrier freq=2*0.2=0.4
chance of child gettin disaes in AR= 1/4
mother s bcum a carrier =0.4
=1/4*0.4
1/100
I calculated by the same method..but the answer comes to be 1/10..
Please check the calculations..
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Old 08-10-2011
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Correct Answer ans A

i can explain tis
q2=4%
q=0.2 (1/5)
in AR p+q=1
p=1-0.2
p=0.8
carrier freq=2pq=2*0.2*0.8
chance of child gettin disease in AR =1/4
mother chance f being carrier=2*0.2*0.8
==2*0.2*0.8*1/4
=2/25.

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Old 08-10-2011
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Man Aa
Female = probability Aa =
p^2 + 2pq + q^2 = 1
q^2 = 0.04 = q = 0.2 AND p = 0.8 (p is only to 1 when q->0 )
2pq = 2*0.2*0.8 = 0.32 = 32/100 = 8/25

autosomal recessive condition = aa = probability of 1/4
1/4*8/25 = 8/100 = 2/25
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