Hemophilia is caused by an X-linked recessive gene. The condition clotting time.
ID: 254293 • Letter: H
Question
Hemophilia is caused by an X-linked recessive gene. The condition clotting time. John and Sue have a son and a daughter. Be careful o. Sue has normal clotting time, but her father han results in slower than normal bMood hemophilia. John has normal cloting time a. What is the genotype for Sue- b. What is the chance the daughter is a carrier? and John What is the chance the son has hemophilia? d. What is the chance the son is a carrier? c. What is the chance the child is normal (not have hemophilia)y 10. John and Sue later have another son, Bill. Bill has hemophilia. He marries Donna. Donna has normal clotting time, but her father also has hemophilia. (seems to be a lot of that going around) They also have a son and a daughter. Be careful agair a. What is the genotype for Donna b. What is the chance the daughter has hemophilia? c. What is the chance the son has hemophilia? and Bill - Her father is color blind. Mike has normal vision and so do his parents, but he has 11. Sally has normal vision. a brother who is color blind. a. What is the genotype for Sally - b. What is the probability their first child will be a boy? c. Sally and Mike have a daughter, what is the chance that she is colorblind? d. What is the probability that a child will be a carrier? e. If they have twenty children, how many of them would you expect to be colorblind boys? and Mike A man and woman of normal phenotype have a son with Duchenne muscular dystrophy, a sex-linked, recessive disorder. They are planning to have another child and ask you for genctic counseling. *What is the genotype of the parents? a. What is the likelihood that another child will have this disease, if it is a boy? b. What is the likelihood that another child will have this disease, if it is a girl? 12. c. What is the overall chance of any child in their family to have the disease?Explanation / Answer
9.
Answer: Let X be the X-chromosome carrying hemophilia gene and asterisk (*) indicate the presence of mutation.
a) Sue = XX and John = XY
Their children can have genotypes: XX, XY, XX, XY
girl boy girl* boy*
b) probability that their child, the girl is a carrier is: 1/2 = 50% (of the two possible genotype that the girl can have, one carries the mutation)
c) probability that the son has hemophilia is: 1/2 = 50% (of two possible boy genotypes, one has the mutation)
d) The son CANNOT be a carrier. He would contain only one copy of X-chromoom, and so it can only be either mutated or normal. There is no carrier condition in males. Therefore, the probability is zero.
e) Probability that their child is normal is: 2/4 = 1/2 = 50% (of the four different genotype their children can have, two contains the mutation)
10.
Answer:
a) Bill = XY and Donna = XX
Their children can have genotypes: XX, XX, YX, YX
girl** girl* boy* boy
b) chance daughter has hemophilia (for female to have hemophilia, both X-chromoomes must contain mutation) is: 1/2 =50% (their girls will either have hemophilia or be a carrier with equal probalility)
c) chance the boy has hemophila is: 1/2 = 50% (of the two genotype boys can have, one is hemophilic and other, normal)
11.
Answer: Let X be the defective X-chromoome carrying coloblidness gene
a) Sally = XX and Mike XY
Their kids can have: XX, XY, XX, XY
girl boy girl* boy*
b) probability their first child is a boy: 1/2 = 50%
c) Sally and Mike cannot have a colorblind daughter. They can only have carrier daughters since Mike has a normal X-chromoome. So, answer is zero.
d) probability a child will be a carrier is: 1/4 = 25%.
e) chance of a colorblind boy is 1/4 when one child i born. So, when 20 kids are born, the chance of colorblind boys among them is: 20 x 1/4 = 5. Five of the twenty would be expected to be colorblind boys.
12.
Answer: Let X be the X-chromoome with the mutation.
The Man and Woman has normal phenotypes, but son has the disease. Therefore, the mother must be a carrier.
Man = XY and Woman = XX
Thier kids can have: XX, XX, YX, YX
girl girl* boy boy*
a) if it is a boy, the probability the next child will have the disease is: 1/2 = 50%.
b) They will not have a girl with the disease. They can only have a carrier or a healthy daughter. Therefore, the probability is zero.
c) Chance of any child in that family having the disease is: 1/4 = 25%
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