The mother has a second homozygous blood group and normal color vision, although

The mother has a second homozygous blood group and normal color vision, although her father was color blind. husband with the fourth blood group and color blind. to determine what is the likelihood of a healthy boy with a second blood group in this family. If the gene for color blindness is linked to the X chromosome and is recessive.

Since the father of a woman with normal vision was color blind, it can be argued that she is a carrier of the recessive gene for color blindness. When you marry a color-blind person, there is a 50% probability that a color-blind boy will be born.

The fourth blood group is determined by genes A and B. The second group is genome A. With a 50% probability, children will have a fourth group and with a 50% probability – a second.

The probability of the birth of a boy who does not suffer from color blindness with the fourth blood group will be equal to the product of the probabilities of the manifestation of these two phenotypes.

0.5 * 0.5 = 0.25,

or as a percentage: 0.25 * 100% = 25%.

With a probability of 25%, this couple will have a boy with good color vision and a second blood group.