Chapter 4.5, Problem 129E

(a)

To determine

To explain: The reason that Beth’s brother is albino although her parents are not albino.

Answer to Problem 129E

Solution: The child will be an albino $\left(aa\right)$ only when both the parents have allele $a$ in their set of genes. Thus, both of Beth’s parents have type $Aa$.

Explanation of Solution

It is provided in the question that Beth has non-albino parents but an albino brother. Therefore, the possible set of alleles for each parent are either $AA\text{ or }Aa\text{or }aA$. Also, it is provided that the albino brother has the set of alleles $aa$, each of which is obtained from every parent. Hence, the possible set of alleles of an offspring each of which is obtained from every parent can be depicted in the form of tables as follows:

		Alleles inherited from Parent
	Alleles	$A$	$A$
Alleles inherited from Parent.	$A$	$AA$	$AA$
	$a$	$Aa$	$Aa$

		Alleles inherited from Parent.
	Alleles	$A$	$a$
Alleles inherited from Parent.	$A$	$AA$	$aA$
	$A$	$AA$	$aA$

		Alleles inherited from Parent.
	Alleles	$A$	$a$
Alleles inherited from Parent.	$A$	$AA$	$Aa$
	$a$	$Aa$	$aa$

With both the parents as non-albino, that is, genes $AA$ for parent 1 and $AA$ for parent 2, the child will be a non-albino with set of genes $AA$. Also, here, $aA\text{and }Aa$ are considered as the same genetic set. It is observed from the tables that for a child to be albino (with set of genes $aa$ ), both the parents must have an allele $a$ in their set of genes. Hence, both of Beth’s parents have the set of genes $Aa$.

(b)

To determine

The genetic type of Beth’s parents’ child.

Answer to Problem 129E

Solution: Beth’s parents’ child could have the genetic type as $aa,Aa,\text{or }AA$.

Explanation of Solution

It is provided in the question that Beth’s parents are not albino, which means that each of them has a set of genes $AA\text{ or }Aa\text{or }aA$. The genetic type set of the offspring can be represented in the form of a table as follow:

		Alleles inherited from Parent.
	Alleles	$A$	$a$
Alleles inherited from Parent.	$A$	$AA$	$Aa$
	$a$	$Aa$	$aa$

Therefore, Beth’s parents’ child could have either of the set of genes $aa,Aa,AA$. Each type’s inheritance by the child is an equally likely event as the probability of occurrence of each event is equal and independent of each other.

To determine

To find: Probability of each type of genes inherited by the offspring.

Answer to Problem 129E

Solution: The probabilities for each set of genes are:

$\begin{array}{l}P\left(AA\right)=0.25\\ P\left(Aa\right)=0.5\\ P\left(aa\right)=0.25\end{array}$

Explanation of Solution

Calculation: It is provided in the question that the inheritance of the alleles by the child is an independent event. The probability for each of them is 0.5. Thus, the probabilities can be written as follows:

$P\left(A\right)=0.5$

And

$P\left(a\right)=0.5$

The probabilities for each set of genes inherited can be calculated as follows:

$\begin{array}{c}P\left(AA\right)=P\left(A\right)\times P\left(A\right)\\ =0.5\times 0.5\\ =0.25\end{array}$

$\begin{array}{c}P\left(Aa\right)=\left[P\left(A\right)\times P\left(a\right)\right]+\left[P\left(a\right)\times P\left(A\right)\right]\\ =\left[0.5\times 0.5\right]+\left[0.5\times 0.5\right]\\ =0.25+0.25\\ =0.5\end{array}$

$\begin{array}{c}P\left(aa\right)=P\left(a\right)\times P\left(a\right)\\ =0.5\times 0.5\\ =0.25\end{array}$

Hence, the obtained probabilities are

$\begin{array}{l}P\left(AA\right)=0.25\\ P\left(Aa\right)=0.5\\ P\left(aa\right)=0.25\end{array}$

(c)

To determine

To find: The conditional probabilities for Beth’s genetic type provided the fact that Beth is nonalbino.

Answer to Problem 129E

Solution: The conditional probabilities obtained are:

$P\left(AA|\text{not }aa\right)=\frac{1}{3}$

$P\left(Aa|\text{not }aa\right)=\frac{2}{3}$

Explanation of Solution

Calculation: The possible cases for types of genes of Beth based on the genes inherited from her parents are presented in the form of table in part (b) as follows:

		Alleles inherited from Parent.
	Alleles	$A$	$a$
Alleles inherited from Parent.	$A$	$AA$	$Aa$
	$a$	$Aa$	$aa$

Since, it is provided that Beth is not albino, therefore, the possible genetic types of Beth are either $AA\text{or }Aa$. Here, the type $aa$ will not be considered as this type is of albino people. Thus, after ruling out the type $aa$, a total of 3 genetic types remain. The conditional probabilities possible are calculated as follows:

$\begin{array}{c}P\left(\text{Beth's genetic type is }AA|\text{She is not albino}\right)=P\left(AA|\text{not }aa\right)\\ =\frac{1}{3}\end{array}$

And

$\begin{array}{c}P\left(\text{Beth's genetic type is }Aa|S\text{he is not albino}\right)=P\left(Aa|\text{not }aa\right)\\ =2\times \frac{1}{3}\\ =\frac{2}{3}\end{array}$

Hence, the obtained probabilities are:

$P\left(AA|\text{not }aa\right)=\frac{1}{3}$

$P\left(Aa|\text{not }aa\right)=\frac{2}{3}$