b) Recessive mutations in the B-glabin gene can give rise to pthallassemia or si
ID: 258033 • Letter: B
Question
b) Recessive mutations in the B-glabin gene can give rise to pthallassemia or sickle cell anemia the p-globip.gene is on chromosome. Recessive mutations of the phenylalanine bydrexxlase (PAH) gene on chromosome 12 gives rise to phenylketonuria (PKU). A couple is thinking of having children. The wife is a carrier for sickle cell anemia and her father has PKU. The husband is a carrier for p-thalassemia: his mother has PKU A. What is the probability that the couple has a son who is a carrier for sickle cell anemia and has PKU B. What is the probability that the couple has a daughter who does not have any of the diseases and is NOT a carrier for any of the diseases? for PKU? C. What is the probability that the couple has a child who is a carrier for sickle cell and a carrier D. Suppose that the wife's father has a deletion in the PAH gene, while the husband's mother has a point mutation in the PAH gene. You amplify a region of the PAH gene from the wife, husband, wife's father, and husband's mother. What would you expect to see on the gel?Explanation / Answer
PKU is a recessive disease and, if the parents have a recessive disease, then children would be carrier of that disease. Therefore, wife and husband both is carrier for PKU disease.
Now in the cross between both wife and husband-
Wife = A’A +B’B +XX x A’A +B’B+XY = Husband
The A’ is showing carrier for Sickle cell anemia and B’ is for PKU. The X and Y are sex-determining chromosomes.
Now children-
Gametes
A’+B’+X
A’+B+Y
A+B’+X
A+B’+Y
A’+B’+Y
A’+B+X
A’+B’+X
A’A’+B’B’+XX
A’A+B’B+XY
AA’+B’B’+XX
A’A+B’B’+XY
A’A’+B’B’+XY
A’A’+B’B+XX
A’+B+Y
A’A+B’B+XY
AA’+BB’+XY
A’A’+BB+XY
A+B’+X
AA’+B’B’+XX
A’A+B’B+XY
AA+B’B’+XX
AA+B’B’+XY
AA’+B’B’+XY
A’A+B’B+XX
A+B’+Y
A’A+B’B’+XY
AA+B’B’+XY
A’A+B’B+XY
A’+B’+Y
A’A’+B’B’+XY
A’A+B’B’+XY
A’A’+B’B+XY
A’+B+X
A’A’+B’B+XX
A’A’+BB+XY
A’A+BB’+XX
A’A+B’B+XY
A’A’+BB’+XY
A’A’+BB+XX
Here we can see that 4 sons (Bold letter) have sickle cell trait with PKU disease. Total children are 27. Therefore, the probability is 4/27.
Now, we can calculate probability for other children.
Gametes
A’+B’+X
A’+B+Y
A+B’+X
A+B’+Y
A’+B’+Y
A’+B+X
A’+B’+X
A’A’+B’B’+XX
A’A+B’B+XY
AA’+B’B’+XX
A’A+B’B’+XY
A’A’+B’B’+XY
A’A’+B’B+XX
A’+B+Y
A’A+B’B+XY
AA’+BB’+XY
A’A’+BB+XY
A+B’+X
AA’+B’B’+XX
A’A+B’B+XY
AA+B’B’+XX
AA+B’B’+XY
AA’+B’B’+XY
A’A+B’B+XX
A+B’+Y
A’A+B’B’+XY
AA+B’B’+XY
A’A+B’B+XY
A’+B’+Y
A’A’+B’B’+XY
A’A+B’B’+XY
A’A’+B’B+XY
A’+B+X
A’A’+B’B+XX
A’A’+BB+XY
A’A+BB’+XX
A’A+B’B+XY
A’A’+BB’+XY
A’A’+BB+XX
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