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Old 07-12-2011
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Genetics Autosomal dominant disorder and normal offspring

A man and woman are both affected by an autosomal dominant disorder that has 80% penetrance. They are both heterozygotes for the disease-causing mutation. What is the probability that they will produce phenotypically normal offspring?

A) 20%
B) 25%
C) 40%
D) 60%
E) 80%
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phenotypically- 40%?
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Default Ans.

i choose answer A.20%
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If you work out a punnett square then the chance of a normal child is 25%.
But Im confused about the 80% penetrance. Does that mean you subtract 20% (100-80) of 25% ? .. so 25% x 0.2 = 5%. Therefore the probability to have a normal child will be 20%.
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CCCCCCCCCC
Husband: Aa, Mom: Aa
mating results: AA, Aa Aa aa
offspings' chance of having disease = 75 % x 80% =0.6
chance of being normal = 1- 0.6 =0.4
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Quote:
Originally Posted by usluipek View Post
CCCCCCCCCC
Husband: Aa, Mom: Aa
mating results: AA, Aa Aa aa
offspings' chance of having disease = 75 % x 80% =0.6
chance of being normal = 1- 0.6 =0.4
So, in this case the correct answer is C)
but offspings' chance of having disease = 75 %!!!...
phenotypically chance of having disease = 75 % x 80% =0.6
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chances of offspring to carry abnormal genes is 75%

if the abnormal gene does not express itself, then there is no disease manifestations.

so when you say disease, it depends on whether you mean genotype or phenotype
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Old 07-12-2011
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80% x 1/4 ( which represent the normal phenotype ) = 20% ...
Im not good at this,, i just try...

----- A ------ a
A---A A ----- Aa
a---A a------ aa

This disease is DOMINANT which means just 1 alele is require to express the disease so 3/4 are Sicks 1/4 is normal...

Please if im wrong explain it... =)

So According to my plots Answer is A
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Quote:
Originally Posted by rulz View Post
80% x 1/4 ( which represent the normal phenotype ) = 20% ...
Im not good at this,, i just try...

----- A ------ a
A---A A ----- Aa
a---A a------ aa

This disease is DOMINANT which means just 1 alele is require to express the disease so 3/4 are Sicks 1/4 is normal...

Please if im wrong explain it... =)

So According to my plots Answer is A
The correct answer is this case is C)
If both parent are heterozygotes, there is a 75% chance that their offspring will receive one or two copies of the disease-causing gene. With 80% penetrance, the probability that the offspring will be affected (phenotypically) is 0.75 * 0.8 = 60%. The probability that the offspring will be phenotypically normal is 1 - 0.6 = 0.4 or 40%
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Thanks Bebix....=)
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Quote:
Originally Posted by bebix View Post
The correct answer is this case is C)
If both parent are heterozygotes, there is a 75% chance that their offspring will receive one or two copies of the disease-causing gene. With 80% penetrance, the probability that the offspring will be affected (phenotypically) is 0.75 * 0.8 = 60%. The probability that the offspring will be phenotypically normal is 1 - 0.6 = 0.4 or 40%
Thanks for the explanation! But the question asks for the probability of producing a phenotypically normal offspring. Yes there is a 75% chance of transferring the gene with 80% penetrance but the 25% chance of not transferring the gene will also result in a phenotypically normal offspring. Dont you have to take that 25% in consideration ?
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Quote:
Originally Posted by Hope2Pass View Post
Thanks for the explanation! But the question asks for the probability of producing a phenotypically normal offspring. Yes there is a 75% chance of transferring the gene with 80% penetrance but the 25% chance of not transferring the gene will also result in a phenotypically normal offspring. Dont you have to take that 25% in consideration ?
well, actually you are taking the 75% in consideration (which is 1-.25)...
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two ways to answer this question:
Aa x Aa = AA Aa Aa aa
80% penetrance

a)
0.75 * 0.80 = 60% affected (phenotypically)
100-60 = 40% "normal"

b)
0.25 (aa) * [0.75 * 0.2]
where, [0.75 * 0.2] = phenotypically normal
= 40%
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