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+--- Thread: q5 - super6818 (/showthread.php?tid=801420)
q5 - super6818 - ArchivalUser - 02-28-2015
A couple wish to know their risk of having a child with an autosomal recessive disorder. The man's brother is affected with the disorder, whereas his wife has no family history of the condition. The carrier frequency in the general population is estimated to be 1 in 100. The risk that they would have an affected child is:
A 1/4
B 1/400
C 1/600
D 1/800
E 1/1200
0 - ArchivalUser - 02-28-2015
I'm getting .0066 so clearly I'm not doing it right. Anyone able to break it down?
0 - ArchivalUser - 02-28-2015
b.
0 - ArchivalUser - 02-28-2015
cc
0 - ArchivalUser - 02-28-2015
answer should be ee
0 - ArchivalUser - 02-28-2015
can someone plzzz explain this one ???
0 - ArchivalUser - 02-28-2015
is it b ?? .. can someone with help me with this one ?
0 - ArchivalUser - 02-28-2015
C.
chance the man has a mutant allele is 2/3. chance he will pass it on is 1/2
chance the child will get this mutant allele is 2/3 * 1/2 = 1/3
chance the mother has this mutant allele is 1/100. chance she will pass it on is 1/2.
chance the child will get this mutant allele is 1/100 * 1/2 = 1/200
and then you do 1/3 * 1/200 = 1/600
0 - ArchivalUser - 02-28-2015
carrier frequency 2pq=1/100 so,q=1/200 ,so chance of maternal being carrier q=1/200,chance to pass a allele 1/2.in father side 2/3*1/2=1/3 ,so total probability=1/200*1/3*1/2=1/1200
0 - ArchivalUser - 02-28-2015
how do u know that the father is 2/3 chance of mutant allele?