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Author7 Posts
  #1

I am having problem in understanding two questions of Kaplan genetics , since i am not having explanation for these can any one help me in giving explanation of these questions , which are given bellow:

1)If the incidence of Cystic fibrosis is 1/2500 among population of europeans, what is the predicted incidence of heterozygous carriers of a cystic fibrosis mutation in this population?

A. 1/25
B. 1/50
C. 2/2,500
D. 1/2500
e. (1/2500)2

remember this 2 in option E is whole square.

2) A man is known heterozygous carrier of mutation that causes hemochromatosis ( autosomal recessive disease ).Suppose that 1% of the general population consists of homogyzotes 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. 0.025
B. 0.045
C. 0.09
D. 0.10
E. 0.25

Please if some one can help me out , I will be obliged!




  #2

I don't know if I can do the whole thing, but I'll get it started ...

p2 + 2pq + q2 = 1

where:

p2 = DD
2pq = Dd
q2 = dd

Cystic fibrosis = dd = q2 = 1/2500, and the square root of that is 0.02 so

q = 0.02

Someone else take it from here


  #3

"rizwansardar" wrote:

1)If the incidence of Cystic fibrosis is 1/2500 among population of europeans, what is the predicted incidence of heterozygous carriers of a cystic fibrosis mutation in this population?

A. 1/25
B. 1/50
C. 2/2,500
D. 1/2500
e. (1/2500)2

remember this 2 in option E is whole square.!


Since the incidence of this AUTOSOMAL RECESSIVE disease is 1/2500,
q2 is 1/2500
so q= 1/50
p = 49/50 (since by Hardy Weinberg p+q=1)
a heterozygous carrier would be someone with 2pq
so heterozygous carrier is 2 X (49/50) X (1/50)
since 49/50 is almost equal to 1, we can just say 2 X (1/50)= 1/25



"rizwansardar" wrote:
2) A man is known heterozygous carrier of mutation that causes hemochromatosis ( autosomal recessive disease ).Suppose that 1% of the general population consists of homogyzotes 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. 0.025
B. 0.045
C. 0.09
D. 0.10
E. 0.25

Please if some one can help me out , I will be obliged!


I think the answer for second question is A
here, the man is a heterozygous, so he has a 50% (1/2) chance of giving the abnormal gene
the homozygosity in the population is 1% (1/100), meaning q2= 1/100, so q= 1/10. So a heterozygous carrier will be 2pq= 1/5

So the probability will be 1/2 X 1/5 X 1/4 (since 1/4 will be affected homozygotes) = 1/40 = 0.025


  #4

thanks for explanation


  #5

Thank You

you made my life easier!


  #6

thanks a lot meg


  #7

This is for all of you guys who participated in my Question of genetics , i.e Question no 2 concerning HETEROZYGOTE INDIVDUAL HAVING HEMOCHROMATOSIS.

One of my friend also solved this question and gave me explanation of that question which i think its right , read this and after that requires your coments:


ONE MUST 1ST DETERMINE THE PROBABILITY THAT THE MAN'S
MATE WILL ALSO B A HETEROZYGOUS CARRIER.THIS IS DONE
BY APPLYING THE HARDY-WEINBERG RULE. If the frequency
of affected homozygotes is 1%, then the gene frequency
of the Hemochromatosis mutation is the square root of
1% or 0.1 then the estimated freq. Of Heterozygous carriers in the population, 2pq is 2x0.9x0.1=0.18(notice that we don’t use the 2q
Approximation in this case b/c p is not approximately
equal to 1) the probability that the heterozygous
Carriers will produce an affected offspring is 1/4, so
the probability that the man mates with a carrier and
the probability that they will in turn produce an
affected offspring’s obtained by multiplying the 2
probabilities together i.e. 0.18x1/4=0.045






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