Question

A heterozygous brown-eyed human female who is color blind marries a homozygous brown-eyed male who is not color-blind (remember, color-blindness is sex-linked).  Assume that eye color is an autosomal trait and that brown is dominant over blue.

a. Write out genotypes of parents and the Punnett squares for the possible offspring below first (Use the table feature to help and clearly label parental and offspring genotypes):

*What is the probability that any of the offspring produced have the following traits?  (please provide possible genotypes and show mathematical work for full credit)

b. carrier of color blindness?

c. blue-eyed normal vision female?

d. How many copies of the colorblind/normal vision allele do females carry? How many do males carry? Please explain any differences and define any terms.

Be complete and specific in your responses.

Answer 1

# Let's assume XcXc is for colour blind female; XcY is colour blind male. BB is for brown eye and bb is for blue eye (Brown eye is dominant).

Genotype for heterozygous brown eyed colur blind female is Bb XcXc.

Genotype for homozygous brown eyed normal male is BB XY.

Bb XcXc X  BB XY

The possible gametes are BXc, bXc, BX,BY

a. Punnette square

	BX	BY
BXc	BB XcX	BBXcY
bXc	BbXcX	BbXcY

BB XcX ; Brown eyed carrier female

BBXcY ; Brown eyed colour blind male

BbXcX ; Brown eyed carrier female

BbXcY ; Brown eyed colour blind male

b. 50% of the female offspring will be carrier.

c. There is no blue eyed female.

d. For a normal vision, Male should not carry any colour blind allele. Even a single colour blind allele causes colour blind vision in a male. But, if female contains one color bind allele, they will show normal vision but they will act as a carrier. If they have colour blind then, both the allele should be colour blind.