Question

Two parents come to see you for genetic counseling. Dad is sickle cell carrier (sickle cell...

Two parents come to see you for genetic counseling. Dad is sickle cell carrier (sickle cell is a recessive genetic disease) and has a normal cholesterol. Mom is also a carrier for sickle cell and has struggled with high cholesterol, and a test by her doctor recently showed that she is heterozygous for hypercholesterolemia (hypercholesterolemia is a genetic disease that shows incomplete dominance). What are the odds of them producing a child with sickle cell disease?-------A child who is a carrier for the sickle cell trait?--------A child with heterozygous hypercholesterolemia?-------A child with homozygous hypercholesterolemia?----------What are the odds of producing a child heterozygous for both?--------

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Answer #1

Let the normal allele be "S" and sickle cell allele be "s" . Also, normal allele for cholesterol be C1 and hypercholesterolemia allele be C2.

Note that S is dominant over s allele, but C1 and C2 alleles show incomplete dominance, which means that heterozygous condition has an intermediate phenotype.

Dad

(sickle cell carrier, normal cholesterol)

Genotype of Dad : SsC1C1

Mom

(sickle cell carrier, heterozygous for (hypercholesterolemia)

Genotype of Mom: SsC1C2

Thus, the required cross is SsC1C1 × SsC1C2.

Let us set up punnet squares for each gene as follows:

Ss × Ss

Cross for

Sickle cell

S s
S SS (normal) Ss (carrier)
s Ss (carrier) ss (affected)

For cholesterol gene, C1C1 × C1C2

C1 C2
C1 C1C1 (normal) C1C2 (heterozygous forhypercholesterolemia)

Ans (a) :

The odds of producing a sickle cell child (ss) is 1/4 or 0.25.

Ans (b) :

Probability of child who is carrier for sickle cell trait (Ss) is 1/2 or 0.5.

Ans (c) :

Probability of a child with heterozygous hypercholesterolemia (C1C2) is 1/2 or 0.5.

Ans (d) :

Probability of a child with homozygous hypercholesterolemia (C2C2) is zero.

Ans (e) :

The odds of producing a child heterozygous for both (SsC1C2) can be determined by multiplying the individual probability of heterozygous condition for each trait.

= p(Ss) × p(C1C2)

=(1/2) × (1/2)

= 1/4 or 0.25.

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