Chapter 4, Problem 11CONQ

A type A woman is the daughter of a type O father and a type A mother. If she has children with a type AB man, what are the following probabilities?

A. A type AB child

B. A type O child

C. The first three children with type AB

D. A family composed of two children with type B blood and one child with type AB

Summary Introduction

To review:

The probability of following children of type Awoman with type AB man.

A. AB child of A type.

B. O child with A type.

C. First three children have AB type.

D. Family having two children one with B and one with AB blood type.

Introduction:

The blood group also known as the blood type. The classification of the blood is on the basis of absence or presence of inherited antigenic substances or the antibodies at the surface of the RBC’s (red blood cell). These antigenic substances can be carbohydrates, glycoproteins, or proteinson the basis of blood grouping system.

Explanation of Solution

The ABO blood group system finds its use when the blood group of an individual is to be identified. The individual might have type AB, type O, type B, or type A blood. The blood type of the offspring is determined by the pair of alleles it receives from both the parents. In order to find out the probabilities of the children, a cross is made between the type A and type AB blood group type.

A. AB child of A type:

The allele of type A is represented by I^A Iⁱ, whereas type AB is represented by alleles I^A I^B.

Alleles	IA	Ii
IA	IAIA	IA Ii
IB	IA IB	IB Ii

The above Punnett square shows the offspring obtained after the cross made between type A and type AB blood. Out of 4 children, only one offspring has I^AI^B alleles, which represents the AB blood type.

So, the probability of the AB type children is calculated as:

$$$$$$$\begin{array}{l}\text{formula of probability =}\frac{\text{Number}\text{of}\text{favourable}\text{outcome}}{\text{Total}\text{number}\text{of}\text{outcome}}\\ \text{probability of offsprings =}\frac{1}{\text{4}}\end{array}$

Hence, the 0.25 probability of having AB child.

B. O child with A type:

The alleles for the O blood group depicted by ii alleles. The cross between the type A and AB parent show that there is no type O child produced.

The probability of O type is calculated as:

$\begin{array}{l}\text{formula of probability =}\frac{\text{Number}\text{of}\text{favourable}\text{outcome}}{\text{Total}\text{number}\text{of}\text{outcome}}\\ \text{probability of offsprings =}\frac{\text{0}}{\text{4}}\\ \text{probability of offsprings = 0}\end{array}$

Hence, 0 probability of having child with blood O type.

C. First three children have AB type:

From the table, it is determined that the only one offspring has I^AI^B alleles, the probability of this is 0.04. So, the probability of first three offspring having AB type is:

$\begin{array}{l}\text{formula of probability =}\frac{\text{Number}\text{of}\text{favourable}\text{outcome}}{\text{Total}\text{number}\text{of}\text{outcome}}\\ \text{probability of offsprings =}\frac{1}{\text{4}}\text{×}\frac{1}{\text{4}}\text{×}\frac{1}{\text{4}}\\ \text{probability of offsprings = }\frac{1}{\text{64}}\end{array}$

Hence, $\frac{1}{\text{64}}$ is the probability of first three child.

D. Family having 2 children 1 with B and 1 with AB blood type:

The probability of family having two children one with B and one with AB blood type is calculated by using binomial expansion equation because the probability of two offprings can easily be depicted by the binomial expansion equation as shown below:

$\begin{array}{c}\text{n = 3, p = }\frac{\text{1}}{\text{4}}\text{, q = }\frac{\text{1}}{\text{4}}\text{, x = 2}\\ \text{P = }\frac{\text{n!}}{\text{x!}\left(\text{n-x}\right)\text{!}}{\text{p}}^{\text{x}}{\text{q}}^{\text{(n-x)}}\\ \text{ = }\frac{\text{3!}}{\text{2!}\left(\text{3-2}\right)\text{!}}{\text{p}}^{\text{2}}{\text{q}}^{\text{3-2}}\\ \text{ = }\frac{\text{3!}}{\text{2!}\left(\text{1}\right)\text{!}}{\text{p}}^{\text{2}}{\text{q}}^{\text{1}}\end{array}$

$\begin{array}{c}\text{= }\frac{\text{3×2×1}}{\left(\text{2×1}\right)\left(\text{1}\right)}\text{0}{\text{.25}}^{\text{2}}\text{0}{\text{.25}}^{\text{1}}\\ \text{= 3 × 0}\text{.0156}\\ \text{= 0}\text{.0468}\\ \text{ = 0}\text{.047 or 4}\text{.7\%}\end{array}$

Where, P = probability of offspring, n= no of first three children, p = number of B type, q = number of AB type, and x= children number having different blood types.

Hence, the probability of 2 children with one B type and one AB type is 0.047.

Conclusion

Therefore, it can be concluded that the probabilities for children with different blood type can be seen in the table below:

A. The probability of AB child of A type is $\frac{1}{\text{4}}$.

B. The probability of O child of A type is 0.

C. The probability of three children having AB type is $\frac{1}{\text{64}}$.

D. The probability of having two children one with B and one with AB blood type is 0.047.