Chapter 2, Problem 45P

A man with Huntington disease (he is heterozygous HD HD+) and a normal woman have two children.

a.	What is the probability that only the second child has the disease?
b.	What is the probability that only one of the children has the disease?
c.	What is the probability that none of the children has the disease?
d.	Answer (a) through (c) assuming that the couple had 10 children.
e.	What is the probability that 4 of the 10 children in the family in (d) have the disease?

Summary Introduction

a.

To determine:

The probability of having the disease by the second child if a man is heterozygous for the Huntington disease mates with a normal woman.

Introduction:

Huntington’s disease is an inherited disorder which leads to the death of brain cells. This is an autosomal dominant disease mean a single copy of the defective gene may result in the disease.

Explanation of Solution

Only a copy of the defective gene can develop into Huntington’s disease because it is an autosomal dominant disease. The genotype of the male that is heterozygous for the disease is HD⁺HD. So, the possible gametes in males are HD⁺ and HD. The genotype of the normal female is HD⁺HD⁺ so; only one type of gamete will be formed in a female that is HD⁺.

The cross between a man who is heterozygous for the disorder and a normal female is as follows:

♂/ ♀	HD+	HD+
HD+	HD+HD+	HD+HD+
HD	HD+HD	HD+HD

By observing the given cross, it can be concluded that 50% of the children will have the disease, and the remaining 50% will be normal. Thus, the probability of having the disease by a second child will be 25%.

Summary Introduction

b.

To determine:

The probability of having the disease by one of the children.

Introduction:

Symptoms of Huntington’s disease include jerking, muscle problems and impaired posture. The symptoms of this disease develop after the age of 30, and another name for this disease is Huntington’s chorea.

Explanation of Solution

The cross between a man who is heterozygous for the disorder and a normal female is as follows:

♂/ ♀	HD+	HD+
HD+	HD+HD+	HD+HD+
HD	HD+HD	HD+HD

Two children are produced by the mating of heterozygous male and normal female. If the first child of the parents has a disease, then their second child will be normal and vice versa. Therefore, to calculate the probability of having a disease by one of the children multiplication and addition rule should be applied.

Thus, the formula for calculating the probability that only one child has the disease is as follows:

$\begin{array}{l}\left[\text{P}\left(\text{1st child has disease}\right)\right]\times \left[\text{P}\left(\text{2nd child normal}\right)\right]\\ +\left[\text{P}\left(\text{1st child normal}\right)\right]\times \left[\text{P}\left(\text{2nd child has disease}\right)\right]\end{array}$ ……(I)

By putting values of probabilities in equation (I):

$\begin{array}{rll}\text{Probability that one child has disease}& =& \left[\frac{\text{1}}{\text{2}}\times \frac{\text{1}}{\text{2}}\right]+\left[\frac{\text{1}}{\text{2}}\times \frac{\text{1}}{\text{2}}\right]\\ & =& \frac{\text{1}}{\text{4}}+\frac{\text{1}}{\text{4}}\\ & =& \frac{\text{1}}{\text{2}}\end{array}$

Thus, the probability that the only one has the disease is 50%.

Summary Introduction

c.

To describe:

The probability of having the disease by none of the children.

Introduction:

The cause of Huntington disease is the mutation in the HTT gene. This gene carries the information for producing a protein known as huntingtin. Huntingtin protein has an important role in nerve cells.

Explanation of Solution

The cross between a man who is heterozygous for the disorder and a normal female is as follows:

♂/ ♀	HD+	HD+
HD+	HD+HD+	HD+HD+
HD	HD+HD	HD+HD

Thus, the formula for calculating the probability that none of the children has the disease is as follows:

$\left[\text{P}\left(\text{1st child normal}\right)\right]\times \left[\text{P}\left(\text{2nd child normal}\right)\right]$ ……(I)

By putting values of probabilities in equation (I):

$\begin{array}{rll}\text{Probability that none of the child has disease}& =& \frac{\text{1}}{2}+\frac{\text{1}}{2}\\ & =& \frac{\text{1}}{4}\end{array}$

Thus, the probability that none of the children has the disease is 25%.

Summary Introduction

d.

To describe:

The probability of having the disease by the second child and the probability of having the disease by none of the children if it is assumed that the couple had 10 children.

Introduction:

There is no cure for Huntington disease, but patients can use medication that will help them to manage the symptoms of the disease.

Explanation of Solution

If it is assumed that the couple has 10 children, then the formula for calculating the probability of 2^nd child of the parents to have the disease is as follows:

$\text{P}\left(\text{2nd child has disease}\right)\times \text{P}\left(\text{rest 9 children are normal}\right)$ ……(I)

By putting values of probabilities in equation (I):

$\begin{array}{rll}\text{Probability that second child has disease}& =& \frac{\text{1}}{\text{2}}+{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{9}}\\ & =& {\left(\frac{\text{1}}{\text{2}}\right)}^{\text{10}}\\ & =& \frac{\text{1}}{\text{1024}}\text{ or 0}\text{.00097}\end{array}$

Thus, the probability that the second child has the disease is 0.00097.

Summary Introduction

e.

To describe:

The probability of having the disease of 4 children out of 10 children in the family.

Introduction:

When the symptoms of the disease appear in a patient that after 15-20 years, the complication of this disease like pneumonia and infection increases, which causes the death of the patients.

Explanation of Solution

A binomial expression that can be used to calculate the probability of having the disease by 4 children out of 10 children is as follows:

$\text{P}=\frac{\text{n!}}{\text{x!}\left(\text{n-x}\right)\text{!}}{\text{p}}^{\text{x}}{\text{q}}^{\text{n-x}}$

Thus, the probability of having the disease by 4 children out of 10 is as follows:

$\begin{array}{rll}\text{P}& =& \frac{\text{10!}}{\text{4!}\left(10-4\right)\text{!}}{\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^{\text{10-4}}\\ & =& \frac{105}{512}\\ & =& 0.20507\end{array}$

Thus, the probability of having the disease by 4 children out of 10 is 0.20507.