Suppose that Rosie and Ross are both carriers for PKU (a rare autosomal recessive disease.) If they have eight children, what is the probability that all of eight children will have the same phenotype?

Answer:

1/65,536 = 0.001523 %

Explanation:

A: normal allele

a: recessive allele causing PKU

The probability that Rosie and Ross have one child with PKU will be >>  

Aa x Aa = 1/4 AA; 1/2 Aa; 1/4 aa

In consequence, the probability that Rosie and Ross have eight children with PKU will be >>  

(1/4)⁸ = 1/65,536 = 0.001523 %

If Rosie and Ross are carriers of PKU, if they intend having 8 children, the probability that they will all have the same phenotype is: [tex]\mathbf{\frac{1}{65,536}; 0.001523 \%}[/tex]

Given that Rosie and Ross are both carriers of the recessive disease, it means that for the disease to manifest itself, two recessive alleles must be present in an allelic pair.

Using P and p to denote the dominant and recessive alleles respectively, we have the following gene of both parents:

The cross between both parents has been provided using a Punnett square as shown in the attachment.

The cross shows that, for any cross between both parents, there is [tex]\frac{1}{4}[/tex] probability that the child would have the phenotype, pp, which means the child would have the are autosomal recessive disease.

Therefore, the probability for for 8 children having the same phenotype will be:

[tex]\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{65,536}[/tex]

This is also approximately: [tex]\frac{1}{65,536} \times 100 = 0.001523 \%[/tex]

Therefore, if Rosie and Ross are carriers of PKU, if they intend having 8 children, the probability that they will all have the same phenotype is: [tex]\mathbf{\frac{1}{65,536}; 0.001523 \%}[/tex]

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