Genetics: What is the probability that the third child will be affected

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SUMMARY

The probability that the third child will be a normal daughter, given that the first child has an autosomal recessive disorder and both parents are carriers, is calculated as ½ (probability of being a daughter) multiplied by ¾ (probability of avoiding the bad allele), resulting in a 37.5% chance. The confusion arose from miscalculating the probability of being affected, which was initially stated as 12.5%. The correct interpretation of the genetic inheritance pattern clarifies that the third child has a 75% chance of being normal, but this option was not provided in the answer choices.

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Homework Statement
About 1% individuals in a population suffer from a genetic disorder. The cause was traced to such individuals being homozygous recessive for a single locus with two alleles. The elder of the two children of a family(where both the parents are normal) suffers from the disorder, while the younger one is normal. What is the probability that the third child will be a normal daughter?
(a) 49.5%
(b)12.5%
(c)37.5%
(d)25%
Relevant Equations
Ans: (c)
pg-363, sum-334

Attempt:

The disorder is autosomal recessive.
The probability = Probability of being a daughter x Probability of an affected child being born = ½×¼ = 1/8 x100 =12.5%
 
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SanjuktaGhosh said:
Problem Statement: About 1% individuals in a population suffer from a genetic disorder. The cause was traced to such individuals being homozygous recessive for a single locus with two alleles. The elder of the two children of a family(where both the parents are normal) suffers from the disorder, while the younger one is normal. What is the probability that the third child will be a normal daughter?
(a) 49.5%
(b)12.5%
(c)37.5%
(d)25%
Relevant Equations: Ans: (c)

pg-363, sum-334

Attempt:

The disorder is autosomal recessive.
The probability = Probability of being a daughter x Probability of an affected child being born = ½×¼ = 1/8 x100 =12.5%

It looks like you've calculated the probability of an affected daughter.
 
PeroK said:
It looks like you've calculated the probability of an affected daughter.
Oh! My bad.
P= ½×¾ x 100=37.5%

Thanks :)
 
I looked at this thread out of curiosity, and I find myself confused.

Since the first child has the disorder, and both parents do not, this means each parent carries the bad allele on just one of the two chromosomes which are home to the relevant gene. For a third child to not have the disorder, means the child must avoid receiving the bad allele from both parents. A child will get the bad allele from the mother half the time, and from the father half the time, and from both (1/2)*(1/2) = 1/4 the time. Thus the third child avoids inheriting the bad allele form both parents 1-(1/4) = 3/4 of the time. What really confuses me is that 75% is not listed among the four answer choices.
 
Buzz Bloom said:
I looked at this thread out of curiosity...
Thanks.
What really confuses me is that 75% is not listed among the four answer choices.

''What is the probability that the third child will be a normal daughter?''

Actually, the question is about a normal daughter, so its ½ × ¾.
 
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SanjuktaGhosh said:
Actually, the question is about a normal daughter, so its ½ × ¾.
Hi Sanjukta:

Thanks for the correction. One of these days I will regain my ability to read accurately.

Regards,
Buzz
 
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Buzz Bloom said:
Hi Sanjukta:

Thanks for the correction. One of these days I will regain my ability to read accurately.

Regards,
Buzz
Me too. I saw the answer and thought what?? Then I re-read the op.
 

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