"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
|
| dxtxpx
Forum Guru
Topics: 259
Posts: 1,233
|
thanks for explanation
|
| rizwansardar
Forum Newbie
Topics: 1
Posts: 14
|
Thank You
you made my life easier!
|
|
|
thanks a lot meg
|
| rizwansardar
Forum Newbie
Topics: 1
Posts: 14
|
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
|
|
| |
| | | | | | 
