Question

There is a “sex-linked”  gene that produces a protein that allows us to properly distinguish between red and green.  There are two alleles of this gene: One allele produces the functional protein; the other allele produces a non-functional protein.  In order to see normally, an individual needs only one allele that produces the functional protein (i.e. it’s not necessary to be homozygous for the functional-protein-producing-allele in order to see normally).

1. On what chromosome is this gene found? X

2. The allele that produces the functional version of the protein would most appropriately be represented by a  (BOLDthe correct answer)

capital letter    or    lower-case letter

3. The allele that produces the non-functional version of the protein would most appropriately be represented by a  (BOLDthe correct answer)

capital letter    or    lower-case letter

4. When choosing letter symbols to represent a gene that’s found on the X-chromosome, the following convention is typically used:

i. What letter combination would you chose to represent the allele that produces the functionalcolor-distinguishing protein? XA

ii. What letter combination would you chose to represent the allele that produces the non-functionalcolor-distinguishing protein? Xa

5. How many alleles of this gene do women have?  Why?

6. How many alleles of this gene do men have?  Why?

7. Complete the following table:

Genotype	XEXE	XEXe	XeXe	XEY	XeY
Is this individual a man (M) or a woman (W)?
Does this person have normal vision (N) or is the person colorblind (CB)?

8. Suppose a man who is colorblind mates with a woman who is homozygous for normal vision.  Construct a Punnett square to show the expected genotypes and phenotypes of this couple’s offspring.  Use the letter symbols from part g.

                  i. If this couple has a son, what’s the probability that he will be colorblind?

                  ii. If this couple has a daughter, what’s the probability that she will be colorblind?

Suppose that a woman who is a carrier (heterozygous) for the colorblind trait mates with a man who has normal vision.  Construct a Punnett square to show the expected genotypes and phenotypes of this couple’s offspring.  Use the letter symbols from part g.

                  i. If this couple has a son, what’s the probability that he will be colorblind?

                  ii. If this couple has a daughter, what’s the probability that she will be colorblind?

Is it possible for a colorblind woman to have a son with normal vision?  Explain your answer;  you should used a Punnett square on scratch paper to help you justify your answer.

Please answer the highlight question...

Answer 1

a. X chromosome. Because it is a sex-linked trait

b. Capital letter. It is the dominant allele of the gene. One copy of the dominant allele is sufficient to exhibit the dominant trait

c. Low case letter. It is the recessive allele of the gene. Two copies of the recessive allele are required to exhibit the recessive trait

d.

i) X^C

ii) X^c

e. A woman has two X chromosomes. So, a woman has two alleles of the gene

f. A man has one X chromosome and one Y chromosome. So, a man has only one allele of the gene

g.

i. Genotype - X^EX^E

The individual has two X chromosomes. So the individual is a woman (W).

The individual carries two copies of the dominant allele, hence the woman has the normal vision (N).

ii. Genotype - X^EX^e

The individual has two X chromosomes, So the individual is a woman (W).

One copy of the dominant allele is sufficient to have a normal vision. The woman is a carrier but has a normal vision (N).

iii. Genotype - X^eX^e

The individual has two X chromosomes. So the individual is a woman (W).

The woman has two copies of the recessive allele, so the woman is colorblind (C)

iv. Genotype - X^EY

The individual has one X chromosome and one Y chromosome. So the individual is a man (M).

Since the dominant allele of the gene is present, the man has a normal vision (N)

v. Genotype - X^eY

The individual has an X chromosome and a Y chromosome. So the individual is a man (M)

A man has one X chromosome, so one copy of the recessive allele is sufficient to cause the disease. The man is colorblind (C)

h)

Genotype of the colorblind man: X^eY

Genotype of the woman : X^EX^E

Parent cross : X^eY * X^EX^E

gametes of the father: X^e, Y

gametes of the mother: X^E

Offspring: X^EX^e (carrier daughter), X^EY (normal son)

i ) if the couple has a son, then he will be normal. A son inherits X chromosome from the mother. Since the mother is homozygous for normal vision,  the probability to have a colorblind son is 0.

ii) if the couple has a daughter, then she will be normal, but a carrier. The daughter will receive one copy of the mutated gene from the father and one copy of the normal allele from the mother. The probability to have a colorblind daughter is 0

I)

Genotype of the carrier woman: X^EX^e

Genotype of the normal man: X^EY

parent cross: X^EX^e * X^EY

gametes of the mother: X^E, X^e

gametes of the father: X^E, Y

offspring: X^EX^E (normal daughter), X^EY (normal son), X^EX^e (carrier daughter), X^eY (colorblind son)

i) Half the sons will be normal and half of the sons will be colorblind. So if the couple has a son, the probability to have colorblindness is 1/2

ii) All daughters will have normal vision. But Half of the daughters will be carriers (heterozygous) for colorblindness. And the other half will be homozygous normal. if the couple has a daughter, the probability to have colorblindness is 0

J)

Colorblindness is an X linked recessive trait. In a sex-linked trait, there is no male to male transmission. A son inherits an X chromosome from the mother and the Y chromosome from the father. In X linked recessive inheritance, an affected mother will always have an affected son inheriting one copy of the mutated gene (X chromosome) from the mother. In X linked recessive inheritance, All sons of an affected mother will be affected. For example, If the mother is colorblind and the father is normal, then all sons of the couple will be  colorblind and all daughters will be carriers

Genotype of the colorblind woman: X^eX^e

Genotype of the normal man: X^EY

Parent cross: X^eX^e * X^EY

gametes of the mother: X^e

gametes of the father: X^E, Y

offspring: X^EX^e (carrier daughter), X^eY (colorblind son)

All sons will be colorblind

All daughters will be carriers