Question

Alpha thalassemia is a blood condition resulting from abnormal hemoglobin production, due to loss of function of copies of either HBA1 or HBA2 genes. These genes are very similar, and for simplicity’s sake we talk about having four copies of the HBA gene (two copies of each).

HBA1 and HBA2 genes are located next to each other, and therefore are described within the same allele. For example, αα indicates that both HBA1 and HBA2 copies on an allele are functioning, α- indicates an allele with one of the two genes functioning, and -- indicates an allele with neither gene functioning.

There are multiple versions of alpha thalassemia, which depend on how many of the four HBA copies are functioning. This is outlined in the table below, in order of severity (most severe -> least severe):

Phenotype	Genotype
α-thalassemia major	Loss of all 4 α-globin genes
Hemoglobin H (HbH) disease	Loss of 3 α-globin genes
α-thalassemia trait	Loss of 2 α-globin genes in cis (--/αα) or trans (-α/-α)
α-thalassemia silent carrier	Loss of 1 α-globin gene (-α/αα)

A man who is a silent carrier has a child with a woman who has α-thalassemia trait.

i)              Assuming the woman has mutations in cis, what are the possible offspring phenotypes for this child? Give the probability for each possible phenotype.

ii)             Assuming the woman has mutations in trans, what are the possible offspring phenotypes for this child? Give the probability for each possible phenotype.

Answer 1

The genotype of the man is αα/-α. So the possible gametes for him are αα and -α

i)

If the woman has α-thalassemia trait in the cis form, she is --/αα genotype. since the alleles are so closely linked that they are considered as a single allele, her gametes would be -- and αα

So, the possible offspring are

gametes	αα	-α
αα	αα/αα (1/4 normal)	αα/-α (1/4 silent carrier)
--	αα/ -- (1/4 cis α-thalassemia trait)	-α/-- (1/4 Hemoglobin H (HbH) disease)

so the offspring would have the phenotype/ genotype with probabilities as shown in the Punnet square above.

ii)

If the woman has α-thalassemia trait in the trans- form, she is -α/-α genotype. her gametes would be -α genotype only.

So, the possible offspring with a man having αα and -α gametes would be

gametes	αα	-α
-α	αα/-α (1/2 silent carrier)	-α/-α (1/2 trans-α-thalassemia trait)

So, if she had trans-α-thalassemia trait, the offspring would have a probability of 0.5 for being a silent carrier and 0.5 for having trans-α-thalassemia trait.