Question

monohybrid sex - linkage problems

1. what is the genotype of a normal male?
2. what is the genotype of a normal female?
3. what is the genotype of an affected male?
4. what is the genotype of an affected female?
5. what is the genotype of a carrier female?
6. a mother, who is a carrier for color blindness, marries a normal male.

a. what is moms genotype? what is dads genotype?
b. what is the chance that any of their sons would be color - blind?
c. what is the chance that any of their daughters would be carriers?

1. the mom in problem #6 has a mother who is normal. if so, what would her fathers genotype have to be?
2. is he color - blind or not?
3. a mother, who is a carrier for male pattern baldness, marries a bald male.

a. what is moms genotype? what is dads genotype?
b. what is the chance that any of their sons would be bald?
c. what is the chance that any of their daughters would be carriers?
d. what is the chance that any of their daughters would be bald?

1. who always passes a sex - linked trait onto a daughter?
2. cross a father who has hemophilia with a normal mother.

a. what is moms genotype? what is dads genotype?
b. what is the chance that any of their sons would have hemophilia?
c. what is the chance that any of their daughters would be carriers?
d. what is the chance that any of their daughters would have hemophilia?

1. a mother, who is not bald, passes male pattern baldness down to her son. her husband is not bald.

a. what is moms genotype? what is dads genotype?
b. where did the mom get the defective \x\ from?
c. what is the chance that any of her sons would be bald?
d. what is the chance that any son would also pass it down to their sons?

Explanation:

Step1: Recall sex - linked inheritance basics

Sex - linked traits are carried on the X or Y chromosomes. For X - linked traits, males have one X and one Y chromosome (XY), and females have two X chromosomes (XX).

Step2: Determine normal male genotype

A normal male for an X - linked trait has the genotype $X^NY$.

Step3: Determine normal female genotype

A normal female has two normal X chromosomes, genotype $X^NX^N$.

Step4: Determine affected male genotype

An affected male for an X - linked recessive trait has the genotype $X^nY$, where $n$ represents the mutant allele.

Step5: Determine affected female genotype

An affected female for an X - linked recessive trait has two mutant X chromosomes, genotype $X^nX^n$.

Step6: Determine carrier female genotype

A carrier female has one normal and one mutant X chromosome, genotype $X^NX^n$.

Step7: Problem 6 - A

Mom is a carrier for color blindness, so her genotype is $X^NX^n$. Dad is normal, so his genotype is $X^NY$.

Step8: Problem 6 - B

For sons, mom can pass either $X^N$ or $X^n$ and dad passes Y. The chance of getting $X^n$ from mom (and thus being color - blind) is $\frac{1}{2}$ or 50%.

Step9: Problem 6 - C

For daughters, dad passes $X^N$. Mom can pass either $X^N$ or $X^n$. The chance of being a carrier ($X^NX^n$) is $\frac{1}{2}$ or 50%.

Step10: Problem 7

If mom is a carrier ($X^NX^n$) and her mom is normal ($X^NX^N$), her father must be affected ($X^nY$) to pass the mutant allele.

Step11: Problem 9 - A

Mom is a carrier for male - pattern baldness, so her genotype is $X^BX^b$ (where $B$ is the normal allele and $b$ is the baldness allele). Dad is bald, so his genotype is $X^bY$.

Step12: Problem 9 - B

For sons, mom can pass either $X^B$ or $X^b$ and dad passes Y. The chance of getting $X^b$ from mom (and thus being bald) is $\frac{1}{2}$ or 50%.

Step13: Problem 9 - C

For daughters, dad passes $X^b$. Mom can pass either $X^B$ or $X^b$. The chance of being a carrier ($X^BX^b$) is $\frac{1}{2}$ or 50%.

Step14: Problem 9 - D

For daughters to be bald, they need $X^bX^b$. The chance is $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ or 25%.

Step15: Problem 10

A father always passes a sex - linked trait onto a daughter because he gives her his X chromosome.

Step16: Problem 11 - A

Mom is normal, so her genotype is $X^HX^H$. Dad has hemophilia, so his genotype is $X^hY$.

Step17: Problem 11 - B

For sons, mom passes $X^H$ and dad passes Y. The chance of having hemophilia ($X^hY$) is 0.

Step18: Problem 11 - C

For daughters, dad passes $X^h$ and mom passes $X^H$. All daughters are carriers ($X^HX^h$), so the chance is 100%.

Step19: Problem 11 - D

The chance of daughters having hemophilia ($X^hX^h$) is 0.

Step20: Problem 15 - A

Mom passes male - pattern baldness to her son and is not bald, so her genotype is $X^BX^b$. Dad is not bald, so his genotype is $X^BY$.

Step21: Problem 15 - B

Mom got the defective "X" from her father.

Step22: Problem 15 - C

For sons, mom can pass either $X^B$ or $X^b$ and dad passes Y. The chance of being bald ($X^bY$) is $\frac{1}{2}$ or 50%.

Step23: Problem 15 - D

If a son is a carrier ($X^bY$), when he has sons, he has a 50% chance of passing the $X^b$ to his sons.

Answer:

1. $X^NY$
2. $X^NY$
3. $X^NX^N$
4. $X^nX^n$
5. $X^NX^n$
6. A. Mom: $X^NX^n$, Dad: $X^NY$

B. 50%
C. 50%

1. Her father's genotype is $X^nY$
2. Her father is color - blind
3. A. Mom: $X^BX^b$, Dad: $X^bY$

B. 50%
C. 50%
D. 25%

1. Father
2. A. Mom: $X^HX^H$, Dad: $X^hY$

B. 0
C. 100%
D. 0

1. A. Mom: $X^BX^b$, Dad: $X^BY$

B. Her father
C. 50%
D. 50%