Answer:
Pr= (3/16)
Explanation:
It was stated that both the man and the woman are both carriers for two autosomal recessive disorders, PKU (chromosome 12) and cystic fibrosis (chromosome 7).
∴ let Aa¹Bb° represent the traits in the Man
Where;
Aa¹ = carrier of PKU
Bb° = carrier of cystic fibrosis
Also, let Cc¹Dd° represent the traits in the woman
Where;
Cc¹ = carrier of PKU
Dd° = carrier of cystic fibrosis
Now, if Aa¹Bb° self-crossed, we'll have the F1 progeny as AB, Ab° , a¹B and a¹b°
also, if Cc¹Dd° self-crossed, we have CD, Cd° , c¹D and c¹d° as their F1 progeny
In the F2 generation, the dihybrid cross between the F1 generations will be:
AB, Ab° , a¹B, a¹b° × CD, Cd° , c¹D, c¹d°
ACBD, ACb°D, a¹CBD, a¹Cb°D
ACBd°, ACb°d°, a¹CBd°, a¹Cb°d°
Ac¹BD, Ac¹b°D, a¹c¹BD, a¹c¹b°D
Ac¹Bd°, Ac¹b°d°, a¹c¹Bd°, a¹c¹b°d°
Only (a¹CBD, Ac¹BD, a¹c¹BD) shows the probability that she will have PKU but not CF.
