Review Question - QID 107025

The probability of having a daughter with red-green color blindness from two phenotypically normal parents is closest to 0.

Since males only have one sex X chromosome, the incidence of 1/100 is the frequency of the mutant allele (q), and the frequency of the normal allele is 99/100 (p + q = 1). Since 99/100 is close to 1, we will use 1 to simplify the calculations. Therefore, the carrier frequency (2pq) is 2 x 1/100 x 1 = 1/50. The frequency of the normal individuals is (99/100)^2 which is ~98/100. In order for the daughter to have red-green color blindness, she must inherit two mutant alleles, one from the mother and one from the father. It is possible that the mother can be a carrier, but since the father has normal vision, he has the normal allele; therefore, it is impossible to have a daughter with red-green color blindness.

Deeb and Motulsky review red-green color vision defects which typically presents in males and does not result in other ophthalmologic abnormalities. Some individuals with mild red-green color vision defects may not be aware of their disorder until they are tested. This disorder is most common in those of Northern European descents with up to 8% of males having defects in red-green vision, and the disorder is less common in males of African (3-4%) and Asian (3%) descent.

Deeb and Motulsky discuss the four subclasses of red-green color vision defects. Normal individuals are trichromatic meaning they have three classes of cones - blue, green, and red. The two severe subclasses of the disorder are protanopia, only blue and green cones are functional, and deuteranopia, only blue and red cones are functional. The two mild subclasses of the disorder are protanomaly, blue and green cones are functional plus mutated red cones resulting in red weakness, and deuteranomaly, blue and red cones are functional plus mutated green cones resulting in green weakness.

Illustration A shows the distribution of red blue and green cones in a normal individual (left) and a color-blind individual (right).

Incorrect answers:
Answer 1: 1/100 is the frequency of the mutant allele (q).
Answer 2: 99/100 is the frequency of the normal allele (p).
Answer 3: 1/50 is the carrier frequency (2pq).
Answer 4: 98/100 is the frequency of the normal individuals (p^2).