Answer:
Part A
[tex]0.108[/tex]
Part B
B.1
Fifty Percent
B.2
[tex]16.4[/tex] %
Explanation:
Part A
Given -
Both the parents are heterozygous for the sickle cell anemia
Let the genotype of the parents be
Ss
where S is the allele for presence of sickle cell anemia
and s is the allele absence of sickle cell anemia
If the above two parents are crossed, following offspring are produced
Ss * Ss
SS, Ss, Ss, ss
The probability of children with sickle cell anemia is equal to
[tex]\frac{3}{4}[/tex]
The probability of children with sickle cell anemia is equal to
[tex]\frac{1}{4}[/tex]
Probability that three of the children have sickle cell anemia and four of the children are healthy
[tex]\frac{7!}{3! * 4!} * \frac{3}{4}^4 * \frac{1}{4}^3\\ = 35 * \frac{81}{256 *64} \\= 0.108[/tex]
Part B
B.1
Hh * hh
Hh, Hh, hh, hh
The first child of the parent has a probability of fifty percent to have Huntington\'s disease
B.2
[tex]\frac{7!}{2!*5!} * \frac{1}{2}^2 * \frac{1}{2}^5[/tex]
[tex]= 16.4[/tex]