star1 Forum Guru

Topics: 93 Posts: 826
| | 01/20/06 - 03:09 PM  
 
   
 
|   #1 |
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)
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| roopashri Forum Elite
Topics: 16 Posts: 190
| | 01/20/06 - 03:58 PM  
 
   
 
|   #2 |
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..
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| roopashri Forum Elite
Topics: 16 Posts: 190
| | 01/20/06 - 04:00 PM  
 
   
 
|   #3 |
for q2..its A
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