Question

Huntington\'s disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington\'s disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington\'s disease allele, and her test results show she is unaffected, hh. A.) What is the percent probability that the first child of this couple will have Huntington\'s disease? B.) What is the probability that any six of the seven children will have Huntington\'s disease?

Answer 1

Answer:

a) probability that the first child of this couple will have Huntington's disease is 1/2 = 50%

b) probability that any six of the seven children will have Huntington's disease is 0.05468  ( 5.468% )

Step-by-step explanation:

a)

Given that;

Hh ( diseased father) and hh (normal)

         H   h

h Hh hh

h Hh hh

therefore Hh ⇒ 2 and hh ⇒ 2

so here we can see that chances of getting Huntington's disease is 1 : 1

so probability that the first child of this couple will have Huntington's disease is 1/2 = 50%

b)

probability that 6 out of the 7 children will have Huntington's disease;

we use the expression

<P> = (n! / (S! × t!)) × p^S × q^t

where n! is total number of children factorial

S! is diseased number of children factorial

t! is healthy number of children factorial

p is probability of diseased children

q is probability of healthy children

so we substitute

<P> =  (7! / (6! × 1!)) × 1/2^6 × 1/2^1

= (7 × 6! / (6! × 1!)) × 1/2⁶ × 1/2

= 7 × 1/64 × 1/2

= 0.05468

probability that any six of the seven children will have Huntington's disease is 0.05468  ( 5.468% )