At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes. At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision. A heterozygous brown-eyed woman who is a carrier of colorblindness has children with a blue-eyed man who is not colorblind. What is the probability that their first child will be a blue-eyed female who has normal color vision? (Enter the probability as a percent. Enter the number only without the percent sign. For example, enter 100% as 100 and enter 12.5% as 12.5)

Question

Fill in the blank(s) with the appropriate word(s).

Answer 1

25

Clarify Question
• What is the key concept addressed by the question?
        o This question addresses both X-linked and autosomal inheritance.
• What type of thinking is required?
        o You need to analyze the information given, using logic, to dissect the problem and determine the answer.
• What key words does the question contain?
        o Autosomal refers to genes on a non-sex chromosome. X-linked genes are on the X chromosome, like the colorblindness gene here. Dominant alleles are expressed when heterozygous, but recessive alleles are expressed only when homozygous or on the only X chromosome in males (hemizygous). 

Gather Content
• What do you already know about X-linked and autosomal inheritance?
        o X-linked genes have different patterns of gene expression because males have only the one X chromosome.
        o The segregation of an allele on the X will not affect the segregation of an allele on an autosome. 

Consider Possibilities
• What other information is related to the question? Which information is most useful?
        o Punnett squares are a useful way to determine both genotype and phenotype ratios. 

Choose Answer
• Given what you now know, what information and/or problem solving approach is most likely to produce the correct answer?
        o Try drawing a Punnett square. Write the possible gametes on the top and the side. You will have to create your own abbreviations for the gene alleles.
        o Let's call the colorblindness gene C, and the eye color gene B.
        o The mother is a carrier for colorblindness (C/c) and heterozygous for brown eyes (B/b). Her gametes are C B, C b, c B, and c b.
        o The father is not colorblind (C/Y) and is blue-eyed (b/b). His gametes are C b and Y b.
        o This creates 8 combinations in the offspring.
        o Consider just the 1/2 of the offspring which are female. The females are CC Bb, CC bb, Cc Bb, and Cc bb. Half of these females are blue-eyed, and ALL will have normal color vision.
        o So 1/2 (female) × 1/2 (blue-eyed) × 1 (noncolorblind) = 1/4. or 25%
        o You can also solve this with a simpler approach:
        o 1/2 the offspring will always be female. Those females will ALL receive the C allele on the one X from their father. (A daughter always receives her father's X chromosome, not the Y.) Of those females, half will get the blue-eyes allele from their mother (and guaranteed that allele from the homozygous recessive father).
        o So 1/2 × 1 × 1/2 = 1/4 or 25%.

Reflect on Process
• Did your problem-solving process lead you to the correct answer? If not, where did the process break down or lead you astray? How can you revise your approach to produce a more desirable result?
        o This question required you to analyze the information given, using logic, to dissect the problem and determine the answer.
        o Did you recognize that a daughter will always receive the X from her father?
        o Did you try drawing a Punnett square, or at least writing out the cross, to help answer this question?