# Punnett Square

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#### cartman1980

##### Full Member

(Note, this isn't directly asked in TBR, but a slight variation of what i cam up w/ in trying to identify flaws in my understanding of Punnett Squares)

If Rh-negative mother has a baby w/ heterozygous Rh-positive father, what are the chances that the baby will have erythroblastocis fetalis? Assume, second pregnancy, and Rh antigen to be passed as dominant trait

Possible genotypes for Mom = A(r)A(r), A(r)a
Possible genotypes for Dad = Aa

First off, is this correct representation? (assuming r is super-script, and lower case implies non-carrier of Rh antigen)

Punnett square could look like:

A(r)A(r) A(r)a
----------------------------|------------------------|
Aa | AA(r), AA(r), aA(r), aA(r) | AA(r), Aa, aA(r), aa |

total combinations = 8
EF combinations = 6
% = 6/8 = 75%

#### cartman1980

##### Full Member
anyone? (think ferris bueller)

#### gunj122

##### Full Member
5+ Year Member
well first off, if the Rh antigen is passed on in a dominant fashion, shouldn't the only possible genotypes for the mom be "aa" and for the dad "Aa"?
so, doing a simple punnet square gives a 1/2 chance of having an Rh-positive phenotype.

in order to develop erythroblastocis fetalis, the mother needs to have two babies, both with Rh-positive. the second baby will be the result of the antibody-antigen reaction. therefore, two babies with Rh-positive phenotypes gives a 1/4 probability.

that's my way of doing the problem. not 100% sure but it makes sense.

#### ilovemcat

##### Full Member
Removed
(Note, this isn't directly asked in TBR, but a slight variation of what i cam up w/ in trying to identify flaws in my understanding of Punnett Squares)

If Rh-negative mother has a baby w/ heterozygous Rh-positive father, what are the chances that the baby will have erythroblastocis fetalis? Assume, second pregnancy, and Rh antigen to be passed as dominant trait

Possible genotypes for Mom = A(r)A(r), A(r)a
Possible genotypes for Dad = Aa

First off, is this correct representation? (assuming r is super-script, and lower case implies non-carrier of Rh antigen)

Punnett square could look like:

A(r)A(r) A(r)a
----------------------------|------------------------|
Aa | AA(r), AA(r), aA(r), aA(r) | AA(r), Aa, aA(r), aa |

total combinations = 8
EF combinations = 6
% = 6/8 = 75%

Okay, well first, realize that since this is her second pregnancy, the mother (who is Rh- negative), is now a carrier of Rh+ Antibodies. (An Rh- negative individual only develops antibodies after exposure of Rh+ blood; Unlike ABO blood types that develop antibodies throughout it's life cycle.) If the mother has an Rh+ Baby, it's very likely that the mothers antigens will attack the babies blood (since the mothers antibodies could diffuse into the babies bloodstream). This life-threatening condition is known as erythroblastocis fetalis.

However, before this could happen in the first place, the baby must have Rh+ blood type to begin with. So let's see if that's a possibility here:

The father has: Rh positive blood type (which we'll label as Aa), since we're also told he's heterozygous. The fact that the father has Rh blood type (but is heterozygous), must mean Rh positive blood is a dominant trait. (So even without being told Rh+ is a dominant trait, you should be able to figure this out.)

The mother has: Rh negative blood type. Since Rh+ is a dominant trait, the only way she could have Rh- blood type is if she carried two ressecive alleles (aa).

Now, to become Rh positive, the baby only needs 1 A (dominant allele) from his father's gamete, and 1 a (recessive allele) from his mother's gamete. Together, these gametes would fertilize into a zygote (producing the Aa genotype).

Of the father's 2 allele's, there's a 50% (1/2) chance the baby will get a dominant (A) allele.
Of the mother's 2 allele's, there's a 100% (1) chance the baby will get a recessive (a) allele.

The chances the baby will get a dominant allele from his father (1/2) AND a recessive allele from his mother (1) is equal to the product of those two probabilities (Rule of Multiplication). Therefore the chances that the baby will be Rh+ and therefore be afflicted with this condition is 1/2 or 50%.

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#### cartman1980

##### Full Member
The father has: Rh positive blood type (which we'll label as Aa), since we're also told he's heterozygous. The fact that the father has Rh blood type (but is heterozygous), must mean Rh positive blood is a dominant trait. (So even without being told Rh+ is a dominant trait, you should be able to figure this out.)

The mother has: Rh negative blood type. Since Rh+ is a dominant trait, the only way she could have Rh- blood type is if she carried two ressecive alleles (aa).

So this was the main problem with my understanding. I assumed because Rh is passed as dominant trait, and since mother is Rh-, she was either AA or Aa (resulting in dominant behaviour in kid). Guess i skipped a beat regarding the expression of Rh-. Mother can only express it if she doesn't have either of the dominant allele. Hence, mother = aa.

Thanks to both you guys.