Answer:
Explanation:
From the given information:
Since the parent are normal; Let assume that their trait is Pp
The affected sister will be recessive in nature, so the trait will be pp
Now, the husband who is the groom will be a carrier with a probability of 2/3 since her sister is affected.
Also, the wife will be carrying a heterozygous trait since she is normal as well.
However, using Hardy-Weinberg assumption:
[tex]q^2 = \dfrac{1}{2500}= 0.0004[/tex]
[tex]q = \sqrt{0.0004}[/tex]
q = 0.02
Recall that:
[tex]p+q =1 \\ p = 1 - q \\p = 1 - 0.02 \\p = 0.98\\[/tex]
Thus, [tex]p^2 = (0.98)^2[/tex]
[tex]p^2 = 0.9604[/tex]
[tex]2pq = 2 \times 0.98 \times 0.022[/tex]
2pq = 0.0392
Finally, the required probability that there will be an affected offspring amongst the family is:
[tex]= \dfrac{1}{4} \times \dfrac{2}{3} \times 0.0392[/tex]
[tex]= \dfrac{1}{153}[/tex]
