phenotypically- 40%?

 

Ans.

i choose answer A.20%

 

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%.

 

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

 

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

 

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%

 

Thanks Bebix....=)

 

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%