Sickle cell disease (sickle cell anemia) is a relatively common disorder in indi

Question

Sickle cell disease (sickle cell anemia) is a relatively common disorder in individuals of African descent and affects approximately 1 in 500 African-Americans. Due to a mutation in an autosomal gene for hemoglobin, the homozygous recessive genotype leads to red blood cells that are relatively stiff and sticky, and deform into a sickle shape as they lose oxygen. This leads to problems in the spleen, and anemia.

There are hundreds of different hemoglobin alleles. Individuals heterozygous for the sickle cell allele and one non-sickle cell allele usually do not experience major complications or even noticeable symptoms. A genetic test is available to diagnose carriers.

James does not have sickle cell disease, neither does his mother or father, but his sister does. His wife Natasha does not have sickle cell disease, and neither of her parents does either. Natasha also has a sister who has sickle cell anemia and two brothers without the disease.

QUESTION: If James and Natasha have a child, what is the probability that the child would be a homozygote for the recessive allele and have sickle cell disease? You may draw the Punnet square but please solve the problem using probability calculations, then explain your reasoning in 2-5 complete sentences

Explanation / Answer

James does not have sickle cell disease, neither does his mother or father, but his sister does. So, both parents of James are carriers of the recessive allele for sickle cell disease. Let us define "A" as being the dominant normal allele and "a" as the recessive abnormal. Both parents are heterozygous (Aa)

The Punnet square shows the below combinations of allele with equal probabilities.

Given, James does not have sickle cell disease, so James does not have 'aa' recessive allele. So there are chnaces of AA, Aa and Aa with equal probabilities.

So, Chances of James having normal homozygous (AA) allele = 1/3

Chances of James having heterozygous (Aa) allele = 2/3

Similarly,

Chances of Natasha having normal homozygous (AA) allele = 1/3

Chances of Natasha having heterozygous (Aa) allele = 2/3

Now, we have the followng cases -

Case 1 - Both James and Natasha having normal homozygous (AA) allele

Probability of the case = (1/3) * (1/3) = (1/9)

In this case, the child will not have recessive (aa) allele.

Case 2 - James have normal homozygous (AA) allele and Natasha have heterozygous allele (Aa)

Probability of the case = (1/3) * (2/3) = (2/9)

In this case also, the child will not have recessive (aa) allele.

Case 3 - Natasha have normal homozygous (AA) allele and James have heterozygous allele (Aa)

This is similar to case 2 and the child will not have recessive (aa) allele.

Case 4 - Both James and Natasha having normal heterozygous (Aa) allele

Probability of the case = (2/3) * (2/3) = (4/9)

In this case, the child will have recessive (aa) allele with probability 1/4 (as there is only one chance of aa allele out of 4 combinations)

So, the probability that the child will have recessive (aa) allele = (4/9) * (1/4) = 1/9

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