A man with hemophilia and color blindness married a healthy woman who was not a carrier of these genes. What is the probability that a child from his daughter’s marriage with a healthy man a) will have one of these diseases 2) will have both anomalies?
It is known that both the hemophilia gene and the color blindness gene are inherited in the X chromosome and are recessive genes.
Let’s designate the gene that causes the development of hemophilia in humans as g, and the gene for color blindness as d. The genes that determine the absence of these diseases will be denoted as G and D, respectively.
The chromosome containing the hemophilia gene will be designated as X g, and the chromosome with the color blindness gene as X d. Chromosomes with genes for normal development of blood coagulation and vision will be designated respectively – X G, X D.
A man with color blindness and hemophilia has only one X chromosome, therefore his genotype will be X g d Y. He produces the following two types of sperm: X g d and Y.
The woman is healthy and has the genotype X GD X GD. It produces eggs of the same type – X GD.
The offspring of this married couple will be represented by the following options:
healthy girls, heterozygous for both traits (X GD X g d) – 50%;
healthy boys (X GD Y) – 50%.
Answer: a) 0%; b) 0%. All offspring will be great.
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