Question

For each of these problems, be sure to justify your answer(s) by including an allele key, parental cross (genotypes), phenotypes, and the resulting Punnett square representing all possible mating of eggs and sperm.

1. Tom and his wife Julie are both carriers for cystic fibrosis, an autosomal recessive disease. What is the chance that their children will have cystic fibrosis and carriers?

2. In cattle, a dominant “Dexter” gene, combined with its recessive allele, results in abnormally short legs, while the homozygous dominant condition results in such severe deformities that the fetal calf nearly always dies in utero. If a cow and bull, each heterozygous for the Dexter gene, are mated, what is the chance that they will produce a calf with normal legs?

3. The Martin family knows of three relatives who had kinky hair disease, a sex-linked recessive disorder in which a child does not grow, experiences brain degeneration, and dies by the age of two years. Affected children have peculiar while stubby hair, from which the disorder takes its name. Joan Martin is hesitant about having children because her two sisters have each had sons who died from kinky hair disease.

(a.) If Joan is a carrier, what is the chance that a son of hers would inherit kinky hair disease?

4. “Hairy ears” is an unusual, sex-linked Y-linked trait. If Kyle, with hairy ears, marries Heather, with normal ears, what are the expected genotypes and phenotypes of their children? What is the probability the Kyle and Heather will have hairy-eared daughters? Hairy eared sons?

5. A man whose father was a hemophiliac, but whose own blood clotting time is normal, marries a normal woman with no record of hemophilia in her ancestry. What is the chance of hemophilia in their children?

6.   If a husband and wife have a heterozygous girl for colorblindness, a normal boy, a colorblind girl, and a colorblind boy, what would be the genotypes of the parents?

Answer 1

Ans: 1) As given in the question that Tom and Julie are both carriers of cystic fibrosis which is an autosomal recessive disease. So Let's assume that Tom and Julie both have Cc genotype as they are carriers so at the time of mating they both produce two type of gamete C and c so punnett square for this cross is:

Female(gametes) Male(gametes)	C	c
C	CC	Cc
c	Cc	cc

So from the cross we can say that 25% of their children will have cystic fibrosis and 50% of them will be the carrier of cystic fibrosis.

2) As given in the question homozygous dominant condition nearly kill the fetal calf and Heterozygous condition will produce calfs with abnormal legs this means that only calfs having homozygous recessive condition will have normal legs.

So bull and cow both are heterozygous for dexter gene are mated together so they both will produce two type of Gamete one with Dominant allele and one with recessive allele. So the punnett diagram for this cross is:

Female (Gamete) Male (Gamete)	D	d
D	DD	Dd
d	Dd	dd

So from the punnett diagram we can say that there will be only 25% chances that they will produce a calf with normal legs.

3) As given in the question that Joan is a carrier of the disorder which is a X linked recessive disorder. So the possible genotype of Joan is Xx where x is the caustive agent for this disease. Her husband possible genotype will be XY. So Joan will produce two gametes X and x whereas her husband will produce X and Y gametes. So mating will result in the following type of progeny:

Female (Gamete) Male (Gamete)	X	x
X	XX	Xx
Y	XY	xY

So from this punnett square we can say that her having a son has 0.50 chances and from these 0.50 half will die from the disorder so the probability is 0.50*0.50= 0.25 or 25%.

4) As given in the question Hairy ears is a Y linked trait. Kyle has hairy ear so his possible genotype is XY and Heather his wife has possible genotype of XX. So the expected genotypes of their children are XX and XY and possible phenotypes are hairy ears and normal ears.

Since this trait is only Y linked so the possibility of having hairy eared daughter is none and having hairy-eared son is 100%.

5) Hemophilia is a X-linked recessive disorder. As given in the question the man's father had hemophilia but his blood clotting time is normal which means that he does not have recessive X chromosome and he married with normal woman with no record of hemophilia in her ancestry. So they both does not contains any recessive x gene for hemophilia so their children will have 0% chances of hemophilia.

6) From the question we can say that their child had red green color blindness which is a x linked recessive disorder. They have Heterozygous girl for colorblindness which means she has a genotype of Xx, a normal boy who has a genotype of XY, a colorblind girl who has a genotype of xx and a colorblind boy who has a genotype of xY.

So from this data we can say that the mother's genotype is Xx and the father's genotype is xY. For confirmation let's make a punnett square:

Female (Gamete) Male (Gamete)	X	x
x	Xx (Heterozygous girl for colorblindness)	xx (colorblind daughter
Y	XY (Normal son)	xY (colorblind son)