Red-green colorblindness is the most common form of colorblindness and is caused

Question

Red-green colorblindness is the most common form of colorblindness and is caused by two closely linked genes located near the tip of the human X chromosome. Red blindness, or protanopia, and green blindness, or deutanopia, are recessive conditions caused by mutant alleles p and d, respectively. Between these two genes lies the locus for glucose-6-phosphate dehydrogenase (G6PD), an important enzyme found in erythrocytes. A deficient form of this enzyme is caused by the recessive allele (g).

A woman, who is phenotypically normal, marries a man who has deutanopia and protanopia and is G6PD deficient.

The couple have a daughter who is green colorblind like her father, and they have a son who is red colorblind and G6PD deficient.

(a) What is the genotype of the father?
(b) What is the most likely genotypic arrangement of the genes in the mother?
(c) The woman then has an illegitimate son with a completely normal man. The son is protanopic, deutanopic, and G6PD deficient. How is this possible?

Explanation / Answer

a) the man has deutanopia, which is caused by the recessive allele d, protanopia, which is caused by the recessive allele p, and is G6PD deficient which is caused by the recessive allele g. Therefore the only possible way for him to exhibit all three recessive traits is to be homozygous recessive for all three, meaning his genotype would be ppddgg

b) Between the two children, they exhibit all three recessive phenotypes, meaning that the mother must be a carrier for all three of the recessive genes. Therefore she must be heterozygous and her genotype would thus be PpDdGg

c) If both the mother and the illegitimate father are heterozygous for all three genes, then their offspring could potentially be homozygous recessive and therefore be protanopic, deutanopic, and G6PD deficient

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