Respuesta :
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% )
The Huntington's disease.
As per the question, the disease is a rare inherited disease that is caused by the breakdown of the nerve cells of the brain. The disease has a broad impact on functional abilities and usually results in movement, thinking, and other disorders.
Thus the answer is the chance of the disease to a child is about 50% as it starts at age of 30 to 40. The chances of six children are 5.4%.
- The disease is an inherited aspect of the body genes and the disorder s created is due to the diseases allele is heterozygous
- The results shows that the chances of the first child to have this disease will be 1/2 that is 50% and the chances if any six people out of seven that have this disease is about 5.46%
Learn more about Huntington's disease.
brainly.com/question/8935079.
