Question

a.)How many different types of gametes are possible from the genotype AABbDdEe? What is the probability of producing each gamete type?

b.)A woman with type O blood has children with a man with type AB blood. What blood types cannot be present in their children?

c.) A woman with type A blood has children with a man with type B blood. What blood types (if any) cannot be present in their children? (You must explore all possibilities to answer this question).

Answer 1

Solution (a) :-

Parental genotype is AABbDdEe

The gametes produces by this parents are

1. ABDE
2. ABDe
3. ABdE
4. ABde
5. AbDE
6. AbDe
7. AbdE
8. Abde

So, the answer is 8.

The probability of producing each gametes will be 1/8.

So, the answer is 1/8.

Solution (b):-

The parents are :- AB (IAIB) x O (ii)

Punnet square

The possible genotypes are IAi and IBi , i.e..., A or B blood groups. All other blood groups other than A and B will not be present in their Childrens.

So, the answer is O and AB Blood groups.

Solution (C):-

Women with "A" blood group x Man with " B" blood group

The possible combination are

1. IAIA x IBIB
2. IAIA x IBi
3. IAi x IBIB
4. IAi x IBi

So, the genotypes of progenies can be found out by creating a punnet square of the above combinations

1.IAIA x IBIB

	IA	IA
IB	IAIB	IAIB
IB	IAIB	IAIB

The possible genotype is IAIB, i.e..., AB blood group.

So, the children should not have A, B and O blood groups.

2. IAIA x IBi

	IA	IA
IB	IAIB	IAIB
i	IAi	IAi

The possible genotype is IAIB and IAi, i.e..., AB and A blood group.

So, the children should not have B and O blood groups.

3. IAi x IBIB

	IA	i
IB	IAIB	IBi
IB	IAIB	IBi

The possible genotype is IAIB and IBi, i.e..., AB and B blood group.

So, the children should not have A and O blood groups.

4. IAi x IBi

	IA	i
IB	IAIB	IBi
i	IAi	ii

The possible genotype is IAIB, IAi, IBi and ii, i.e..., AB, A, B and O Blood groups.

The children can have any of the blood groups.