Final answer:
In a cross between a heterozygous carrier female for color blindness (X/Xc) and a color blind male (XcY), there is a 50% chance their male offspring will be color blind and a 50% chance their female offspring will be carriers. No daughter will suffer from color blindness phenotypically. This occurs because color blindness is an X-linked recessive trait predominantly affecting males.
Step-by-step explanation:
To analyze the phenotype and genotype percentages of offspring from a cross between a heterozygous carrier female for color blindness and a color blind male, we need to use a Punnett square to predict the genetic outcomes. We denote the normal vision allele as 'X' and the color blindness allele as 'Xc'. The female is represented as X/Xc (heterozygous carrier), and the male, who is suffering from color blindness, is represented as XcY. The Punnett square will look like this:
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- Father's alleles: Xc, Y
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- Mother's alleles: X, Xc
When we cross these, we get the following genotype combinations:
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- X Xc (Carrier female)
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- XX (Normal vision female)
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- Xc Y (Color blind male)
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- XY (Normal vision male)
Based on these combinations, we can determine the phenotype percentages:
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- 50% chance of male offspring being color blind
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- 50% chance of male offspring having normal vision
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- 50% chance of female offspring being carriers
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- 50% chance of female offspring having normal vision
This implies that there is a 50% chance that a son will suffer from color blindness, while there is a 0% chance for a daughter to suffer from color blindness, but a 50% chance she will be a carrier of the gene like her mother.