S******9
发帖数: 2837 | 1 A man is known hetrozygous carrier of mutation that causes
hemochromatosis.(autosomal recessive disease).Suppose that 1% of general
population consist of homozygote for this mutation.If the man mates with
sombody from general population what will be 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
B
q=1/10
p=1-1/10=9/10
female: 2pq=9/50
female transfer this gene to kids: 9/50x1/2
male transfer this gene to kids: 1/10x1/2
so:( 9/50) x (1/2) x( 1/10) x (1/2)
is this the right way to calculate?
Thanks | d****y
发帖数: 2180 | 2 The answer is right.
But male transfer this gene to kids is: 1/2 (not 1/10x1/2).
Because he is a known heterozygous carrier. He had the mutated gene already.
Your calculation will get 0.0045, not 0.045. | S******9
发帖数: 2837 | | n***h
发帖数: 364 | 4 SUMO2009 and daisyy:
Do you mean the female is a hetrozygous carrier , not homozygote? | d****y
发帖数: 2180 | 5 Female is from general population.
The probability of heterozygous carrier in the general population is: 9/50
Carriers are asymptomatic. People with homozygous mutated genes will show
disease.
The question stem said the male marry someone in the general population,
the chance for him to marry a homozygous disease female is low,because
disease people are only 1% in the population. 99% of people are symptom-
free.
So most likely the female he married is either normal or a carrier.
【在 n***h 的大作中提到】
: SUMO2009 and daisyy:
: Do you mean the female is a hetrozygous carrier , not homozygote?
| n***h
发帖数: 364 | 6 Thanks.
In NBME and UW, there also have this kind of questions, most of cases, we
only need to assume their heter. , not homo.
【在 d****y 的大作中提到】
: Female is from general population.
: The probability of heterozygous carrier in the general population is: 9/50
: Carriers are asymptomatic. People with homozygous mutated genes will show
: disease.
: The question stem said the male marry someone in the general population,
: the chance for him to marry a homozygous disease female is low,because
: disease people are only 1% in the population. 99% of people are symptom-
: free.
: So most likely the female he married is either normal or a carrier.
| w********g
发帖数: 107 | 7 as your calculation:
( 9/50) x (1/2) x( 1/10) x (1/2)= 0.0045
but the correct answer in this question is (B) 0.045 =( 9/50) x (1/2) x( 1/
10) x (1/2)
Actually, 0.045 is the best answer here but not the perfect one.
I saw another version of this question as below:
A man is a known heterozygous carrier of a mutation that causes
hemochromatosis (AR). 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 best estimate for the probability that he
and his mate will produce a child who is an affected homozygote?
A) .025
B) .05
C) .09
D) .10
E) .25
One must first determine the probability that the man's mate will also be a
heterozygous carrier. This is done by applying the Hardy-Weinberg rule, p2+
2pq+q2 = 1 where p2 is the frequency of homozygous dominant individuals, 2pq
is the frequency of heterozygous individuals, and q2 is the frequency of
homozygous recessive individuals. If the frequency of affected homozygous is
1% (given), then the gene frequency of the hemochromatosis mutation is the
square root of 1%, or 0.10. Then the estimate frequency of heterozygous
carriers in the population, 2pq, is 2x0.9x0.1 = 0.18 (where p is 1-0.1= 0.9)
(notice that we do not use the 2q approximation in this case coz p is not
approximately equal to 1). The probability that two heterozygous carriers
will produce an affected offspring is 0.25, so the probability that the man
mates with a carrier and the probability that they will in turn produce an
affected offspring is obtained by multiplying the two probabilities together:
0.25x0.18 = 0.045. Additionally, there is a small chance (0.01) that the
person he mates with will be a homozygote, and thus there is a chance (0.01/
2 = 0.005) that the offspring of this mating will be affected, since the man
has a 1 in 2 chance of transmitting the hemochromatosis gene. The correct
answer then is 0.045 + 0.005 = 0.05.
Correct ans is B)
【在 S******9 的大作中提到】
: A man is known hetrozygous carrier of mutation that causes
: hemochromatosis.(autosomal recessive disease).Suppose that 1% of general
: population consist of homozygote for this mutation.If the man mates with
: sombody from general population what will be 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
| s********o
发帖数: 3319 | 8 where is this Q from? if it is from NBME/CD, please label it in the title.
【在 S******9 的大作中提到】
: A man is known hetrozygous carrier of mutation that causes
: hemochromatosis.(autosomal recessive disease).Suppose that 1% of general
: population consist of homozygote for this mutation.If the man mates with
: sombody from general population what will be 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
| n***h
发帖数: 364 | 9 yes, agree with you. if there is an other chocie of 0.05 in LZ's question,
we should choose it, not 0.045, that means we need also consider the
possibilty of homo. of the female, even though it is very low.
of
【在 w********g 的大作中提到】
: as your calculation:
: ( 9/50) x (1/2) x( 1/10) x (1/2)= 0.0045
: but the correct answer in this question is (B) 0.045 =( 9/50) x (1/2) x( 1/
: 10) x (1/2)
: Actually, 0.045 is the best answer here but not the perfect one.
: I saw another version of this question as below:
: A man is a known heterozygous carrier of a mutation that causes
: hemochromatosis (AR). 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 best estimate for the probability that he
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