The final step of the problem-based ideation process involves: a detailed situation analysis. the selection of one or more probl
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16. A Couple with the ability to taste PTC have two grown sons and one grown daughter. The sons have the ability to taste PTC. Their daughter is a PTC non-taster. She married with a PTC non-taster man and have two sons.
The ability to taste Phenylthiocarbamide<span> or PTC is inherited by autosomal dominant trait. The daughter of the first generation should be autosomal recessive(tt) as she was a non-taster. That means her parent would be heterozygotes(Tt) since they have the ability to taste PTC. Her brother could be homozygote dominant(TT) or heterozygote(Tt). The pedigree would be:
i. no-shade square(Tt) -----------------</span>no-shade circle(<span>Tt)
ii. </span>no-shade square(T?), no-shade square(T?), shaded circle(tt)------shaded square(tt)
iii. shaded square(tt) and shade square(tt),
17. She marries with non-taster too which means also a homozygote recessive. Since both parents are homozygote recessive, the children should be 100% homozygote recessive too.
The phenotype of the sons in generation III would be 100% non-taster.
18. The genotype of the female should be XX. To express the recessive trait, an individual need to have all recessive genes. If she has a recessive trait of an X-linked disease then both of the X genes should have a lowercase letter(recessive).
Answer:
C.
19. The genotype of a male should be XY. If he has a recessive trait of an X-linked disease then the X genes should have a lowercase letter(recessive). As there is only one X gene in male, to express recessive he only needs 1 recessive gene.
Answer:
E.Y
The ability to taste Phenylthiocarbamide<span> or PTC is inherited by autosomal dominant trait. The daughter of the first generation should be autosomal recessive(tt) as she was a non-taster. That means her parent would be heterozygotes(Tt) since they have the ability to taste PTC. Her brother could be homozygote dominant(TT) or heterozygote(Tt). The pedigree would be:
i. no-shade square(Tt) -----------------</span>no-shade circle(<span>Tt)
ii. </span>no-shade square(T?), no-shade square(T?), shaded circle(tt)------shaded square(tt)
iii. shaded square(tt) and shade square(tt),
17. She marries with non-taster too which means also a homozygote recessive. Since both parents are homozygote recessive, the children should be 100% homozygote recessive too.
The phenotype of the sons in generation III would be 100% non-taster.
18. The genotype of the female should be XX. To express the recessive trait, an individual need to have all recessive genes. If she has a recessive trait of an X-linked disease then both of the X genes should have a lowercase letter(recessive).
Answer:
C.
19. The genotype of a male should be XY. If he has a recessive trait of an X-linked disease then the X genes should have a lowercase letter(recessive). As there is only one X gene in male, to express recessive he only needs 1 recessive gene.
Answer:
E.
