Question

inheritance of colorblindness

• females have two x chromosomes
• males have one x chromosome and one y chromosome
• colorblindness is coded for by a recessive allele (c) only found on the x chromosome
• in females (xx), a recessive colorblind allele on one x chromosome can be masked by a dominant allele on their other x chromosome
• the girl would have normal vision but would be a carrier (xcxc) of the trait
• in males (xy), a recessive colorblind allele on the xc chromosome always produces the colorblind trait because there is no corresponding allele on the y chromosome

genotype	phenotype
xcy	male - normal vision
xcxc	female - carrier
xcxc	female - colorblind
xcy	male - colorblind

1. the father has normal vision (xcy) and the mother has normal vision but she is a carrier for colorblindness (xcxc).

father (normal vision)

	xc	y
xc

phenotype probability (%)
normal vision:
colorblind:

1. the father is colorblind (xcy) and the mother is a carrier for colorblindness (xcxc).

father (colorblind)

	xc	y
xc

phenotype probability (%)
what is the probability that their child will be colorblind?

1. the father has normal color vision (xcy) and the mother is colorblind (xcxc).

father (normal vision)

phenotype probability (%)
if they have a daughter, what is the probability that she will be colorblind?
if they have a son, what is the probability that he will be colorblind?

1. the father is colorblind. the mother has normal color vision, and she is not a carrier of the colorblind trait.

father (colorblind)

phenotype probability (%)
what is the probability that their child will be colorblind?

Answer:

To solve these genetics problems related to colorblindness (an X - linked recessive trait), we analyze each problem using Punnett squares and the rules of X - linked inheritance.

Problem 11

• Father's genotype: \(X^{C}Y\) (normal vision, since he has one \(X\) with the dominant allele \(C\) and a \(Y\)).
• Mother's genotype: \(X^{C}X^{c}\) (normal vision but a carrier, as she has one dominant \(C\) and one recessive \(c\) on her \(X\) chromosomes).
• Punnett Square Setup:
• The father can contribute \(X^{C}\) or \(Y\) gametes.
• The mother can contribute \(X^{C}\) or \(X^{c}\) gametes.
• The Punnett square will have the following combinations:
• \(X^{C}\) (father) and \(X^{C}\) (mother) \(\to X^{C}X^{C}\) (female, normal)
• \(X^{C}\) (father) and \(X^{c}\) (mother) \(\to X^{C}X^{c}\) (female, carrier)
• \(Y\) (father) and \(X^{C}\) (mother) \(\to X^{C}Y\) (male, normal)
• \(Y\) (father) and \(X^{c}\) (mother) \(\to X^{c}Y\) (male, colorblind)
• Phenotype Probabilities:
• Total number of possible offspring genotypes: 4.
• Normal vision ( \(X^{C}X^{C}\), \(X^{C}X^{c}\), \(X^{C}Y\)): 3 out of 4. So the probability of normal vision is \(\frac{3}{4}\times100 = 75\%\).
• Colorblind ( \(X^{c}Y\)): 1 out of 4. So the probability of being colorblind is \(\frac{1}{4}\times100=25\%\).

Problem 12

• Father's genotype: \(X^{c}Y\) (colorblind, as he has the recessive \(c\) on his \(X\) and a \(Y\)).
• Mother's genotype: \(X^{C}X^{c}\) (carrier, with one \(C\) and one \(c\) on her \(X\) chromosomes).
• Punnett Square Setup:
• The father can contribute \(X^{c}\) or \(Y\) gametes.
• The mother can contribute \(X^{C}\) or \(X^{c}\) gametes.
• The possible offspring genotypes are:
• \(X^{C}X^{c}\) (female, carrier)
• \(X^{c}X^{c}\) (female, colorblind)
• \(X^{C}Y\) (male, normal)
• \(X^{c}Y\) (male, colorblind)
• Probability of Having a Colorblind Child:
• Out of 4 possible genotypes, the colorblind genotypes are \(X^{c}X^{c}\) and \(X^{c}Y\), which is 2 out of 4. So the probability is \(\frac{2}{4}\times100 = 50\%\).

Problem 13

• Father's genotype: \(X^{C}Y\) (normal vision).
• Mother's genotype: \(X^{c}X^{c}\) (colorblind, since she has two recessive \(c\) alleles on her \(X\) chromosomes).
• For a Daughter:
• The father can only contribute \(X^{C}\) (because if he contributes \(Y\), the offspring will be male).
• The mother can only contribute \(X^{c}\).
• The daughter's genotype will be \(X^{C}X^{c}\) (carrier, not colorblind). So the probability of a daughter being colorblind is \(0\%\).
• For a Son:
• The father contributes \(Y\), and the mother contributes \(X^{c}\).
• The son's genotype is \(X^{c}Y\) (colorblind). So the probability of a son being colorblind is \(100\%\).

Problem 14 (Incomplete Information, but General Approach)

• Father's genotype: \(X^{c}Y\) (colorblind).
• Mother's genotype: \(X^{C}X^{C}\) (normal vision, not a carrier, as she has two dominant \(C\) alleles).
• Punnett Square Setup:
• The father can contribute \(X^{c}\) or \(Y\) gametes.
• The mother can only contribute \(X^{C}\) gametes.
• The possible offspring genotypes:
• \(X^{C}X^{c}\) (female, carrier)
• \(X^{C}Y\) (male, normal)
• Probability of Having a Colorblind Child:
• None of the offspring genotypes result in colorblindness (since all offspring have at least one dominant \(C\) allele from the mother). So the probability is \(0\%\).

Final Answers

1. Normal vision: \(\boldsymbol{75\%}\), Colorblind: \(\boldsymbol{25\%}\)
2. Probability of colorblind child: \(\boldsymbol{50\%}\)
3. Probability of colorblind daughter: \(\boldsymbol{0\%}\), Probability of colorblind son: \(\boldsymbol{100\%}\)
4. Probability of colorblind child: \(\boldsymbol{0\%}\)