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1- A man is a known heterozygous carrier of a mutation that causes hemochromatosis
( autosomal recessive disease ) .. 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 probability that he
and his mate will produce a child who is an affected homozygote?
A- .0025
B- .045
C- .09
D- .10
E- .25

2- The incidence of Duchenne muscular dystrophy in North America is
about 1/3,000 males. On the basis for this figure what is the gene
frequency of this X- Linked recessive mutation?
A- 1/3,000
B- 2/3,000
C- (1/3,000)2
D- 2 x % (1/3,000)


for qn 1

in the population, its given p2=0.01, p=0.1, so q=0.9, carrier frq is 2pq=0.18

the qn stem gives us that man is a carrier.

Now these 2 carriers mate and the chance of them having an affected child is 0.25. So, the prob is 0.18 X 0.25=0.045.
so answer is B...

but if there was a choice of 0.05, then u need to calculate further.....

there can an additional chance that the woman can be homozygote, which is given in qn stem as 0.01. So when a carrier man and homozygote woman mate, 1/2 is the chance that they will produce a child as a carrier. so its 0.01 X 1/2=0.005

now final answer is 0.045+0.005=0.05

this qn was discussed sometime ago..


for q2..its A

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