10. In humans, the condition for normal blood clotting dominates the condition f

ID: 3478599 • Letter: 1

Question

10. In humans, the condition for normal blood clotting dominates the condition for nonclotting (hemophilia). Both alleles are linked to the X chromosomes. A male a carrier is a woman who has an allele for normal blood clotting and an allele for hemophilia.) What are the chances that if they have a male child he will be normal for blood clotting? hemophiliac marries a woman who is a carrier for this condition. (In this respect, In humans, the condition for normal vision dominates color blindness. Both alleles are linked to the X chromosome. A color-blind male marries a color-blind female. If they have a daughter, what are the chances that she will have normal vision? 11. 12. A red-green color-blind man (sex-linked gene) marries a normal woman. From this union one color-blind daughter is born. Give the genotypes of these three people. 13. A man who is hemophilic (sex-inked gene) marries a woman whose father was hemophilic. What percentage of their boys will be hemophilic? What percentage of their girls will be hemophilic? 14. After eight years of married life, during which time she had frequent intercourse with her husband but had failed to become pregnant, Mrs. X met and fell in love with Mr. Y. During the ensuing five years, three children were born. In the meantime, the persons involved had tried to come to an understanding and wished to determine which of the two men was the father of each child. The blood of those involved was examined with the following results: ABO Group Blood of Husband Lover Wife First Child Second Child OO Third Child Determine if possible who is the father of each child. AO AO If a man with blood type B, who has one parent with blood type O, marries a wor with blood type AB, what percentage of their children are likely to have each of t following blood types? 5. AB =

Explanation / Answer

First question-

The hemophilia trait presents in X chromosome of sexual chromosome. We will represent the hemophilic X chromosome as X’.

Male hemophilic will have 44A + X’Y and Carrier woman has 44A + X’X. the female is carrier because X chromosome dominates over X’ but in male it is the only X chromosome; therefore, the male is hemophilic. Here, A represents autosomal chromosome.

Now cross between 44A + X’Y and 44A + X’X

44A + X’Y × 44A + X’X

Gametes

Female 22A+ X’

Female 22A+ X

Male 22A+ X’

44A + X’X’ (hemophilic girl)

44A + X’X (carrier girl)

Male 22A+ Y

44A + X’Y (hemophilic boy)

44A + XY (Normal boy)

Here we can see that the children of the parents are 2 hemophilic, 1 carrier, and 1 normal.

Therefore, the chances of having a normal child= N÷ Nt

Here N is the numbers of individuals of desired trait, and Nt is the total individuals in the generation.

The chances of having a normal child= 1÷ 4

The chances of having a normal child= 0.25 or 25%