Sickle cell anemia is a homozygous recessive genotype caused by a mu.pdf

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Sickle cell anemia is a homozygous recessive genotype caused by a mutation in an autosomal gene for hemoglobin. James does not have sickle cell anemia, neither do his parent, but he has a sister with the disease. Natasha does not have sickle cell anemia, neither do his parents or two brothers, but has a sister with the disease. If James and Natasha get married, what is the probability that their child will have sicle cell anemia? Solution Given, I. Sickle cell anemia (SCA) is homozygous recessive genetic disorder. That is, an individual is affected only when it has recessive (mutated) allele at both the loci. Let the normal, dominant allele be HbA, and mutated, recessive allele be HbS. II. James parents are NOT affected with sickle cell anemia. James is also NOT affected. His sister is affected. Now, III. For James’ to be affected with sickle cell anemia, she must be HbS HbS. Since each parent contributes one allele to the child, the parents must have been carrying the recessive allele. Also, given that parents are not affected, they must be the carrier. So, genotype of parent = HbA HbS = heterozygous, carrier of SCA. IV. gametes produced by male parent = HbA, HbS gametes produced by female parent = HbA, HbS (Male x female) parents = [HbA, HbS] x [HbA, HbS] = HbA HbA, HbAHbS , HbAHbS,HbS HbS So, Genotype ration = Normal (HbA HbA) : Carrier (HbAHbS) : Affected (HbS HbS) :: 1: 2: 1 Probability of a carrier child = 2/4 = ½ ; i.e. 2 out of 4 progeny are carrier. Probability of an affected child = ¼ ; i.e. 1 out of 4 progeny is affected. Thus, probability of James being a carrier = 1/2 V.Similarly,probability of Natasha being a carrier = 1/2 Now, 1. Probability of each parents being carrier, p = 1/2 Probability that both parents are carrier = p (James being carrier) x p (Natasha being carrier) = ½ x ½ = 1/4 2. Probability of getting an affected child from carrier parent = ¼ ; [see, IV] 3. For an affected child to appear in the family, both James and Natasha must be carrier simultaneously. So, Overall probability of affected child = p (both parents are carrier) x p (affected child from carrier parents) = (1/4) x (1/4) = 1/16 Thus, probability of a child being SCA affected = 1/16.

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3. For an affected child to appear in the family, both James and Natasha must be carrier
simultaneously.
So,
Overall probability of affected child = p (both parents are carrier) x p (affected child from carrier
parents)
= (1/4) x (1/4) = 1/16
Thus, probability of a child being SCA affected = 1/16

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