Genetics question

Question

If the father's blood group: AB M S and the mother's blood group: O N s

What's the possible blood group of children?

If the father's blood group: A MN s and mother's blood group: O MN s

What's the possible blood group of children?

If the father's blood group: O M Ss and mother's blood group: A N Ss

What's the possible blood group of children?

If the father's blood group: AB MN Ss and mother's blood group: AB MN S

What's the possible blood group of children?

Answer 1

This question is looking at three different blood type marker systems, so each of them need to be considered independently.

First, the in the ABO system, alleles for A and B are codominant, meaning that the corresponding antigen will be expressed on the red blood cells whether a person has one or two copies of the allele. The O allele does not code for a marker.

The possible genotypes and phenotypes are:

Genptype Phenotype

AA A

AO A

BB B

BO B

OO O

AB AB

So, for bloodtypes A and B, there are two possible genotypes. Since the question asks for all possible phenotypes of the children, assume the parents are heterozygous (AO for A, and BO for B)

The MN system is also codominant, but with just two alleles. This is more straigh forward.

Genotype Phenotype

MM M

NN N

MN MN

Lastly, the S system is a simple Mendelian dominant/recessive type of inheritance.

Genotype Phenotype

SS S

Ss S

ss s

The S phenotype has two possible genotypes, some assume heterozygosity (Ss). The s phenotype is recessive, so the genotype must be ss.

Now with those genotypes and corresponding phenotypes, we can address the questions.

Keep in mind that since we are looking for all [possible phenotypes for the children, we should assume that a parent with the dominat phenotype is heterozygous unless we are told that parent is homozygous.

1, The father is AB M S, so assume his genotype is AB MM Ss. Actually, the AB and MM are certain. The mother's phenotype is O N s, so it is certain that her genotype is OO NN ss.

Each child must inherit O N s from the mother since she is homozygous for all three traits.

The children must inherit M from the father, but can inherit either A or B for the ABO system, and either S or s for the S system. That means there are four possible combinations from dad:

A M S

A M s

B M S

B M s

Now combine those with the O N S from mom and there are four possible outcomes for the children:

Genotype Phenotype

AO MN Ss A MN S

AO MN ss A MN s

BO MN Ss B MN S

BO MN ss B MN s

2, Dad is A MN ss, again assume heterozygosity for A (AO genotype),. He is heterozygous for MN and Homozygous for s, so he can contribute four possible gametes: A M s, A N s, O M s, or O N s. Mon is Homozygous for O and s, and heterozygous for MN, so she can contribute either O M s or O N s. Putting those together yields eight combinations. Let's put this into a Punnett square:

O M s O N s

A M s AO MM ss (A M s) AO MN ss (A MN s)

A N s AO MN ss (A MN s} AO NN ss (A N s)

O M s OO MM ss (O M s) OO MN ss (O MN s)

O N s OO MN ss (O MN s) OO NN ss (O N s)

From this you can see that there are two possible phenotype for the ABO marker, three possible phenotypes for the MN system, and only one possibility for s. That gives six possible phenotype for the children: A M s, A MN s, A N s, O M s, O MN s, and O N s.

3. In this case dad is homozygous for O and M, and we are told he is heterozygous for Ss. So he can have either O M S, or O m s gametes. Mom is homozygous for N, and heterozygous for the other two traits (AO and Ss), so she has four possible genotypes for her gametes: A N S, A N s, O N S, and O N s.

Since the question is asking for phenotypes, let's take a shortcut now that we've already done a rigorous analysis.

For the ABO systemthe children must inherit O from dad and either A or O from mom, and the children can be phenotypically either A or O. They must inherit M from dad and N from mom, so the children must be MN. Both parents are heterozygous for S, so the children can be either S or s phenotypically. That gives two possiblilities for ABO (either A or O, two possiblities for S (either S or s) and all must be MN. Putting that together gives four possible phenotypes for the children:

A MN S

A MN s

O MN S

O MN s

4. This one might look a little more complicated, but it really is not if each marker is considered separately. We know that each parent is heterozygous for each trait (or in other words, this is a trihybrid cross)

AB x AB can yield A, B or AB blood types (three possiblitities0

MN x MN can yield M, N, or MN blood types (three possibliities)

Ss x Ss can yiels S or s blood types (two possibilities)

So there are 3 x 3 x 2 = 12 possible outcmes as follows:

A M S

A M s

A N S

A N s

B M S

B M s

B N S

B N s

AB M S

AB M s

AB N S

AB N s

Hope this helps.