A farsighted father and a mother with normal vision gave birth to 4 children, all farsighted.
A farsighted father and a mother with normal vision gave birth to 4 children, all farsighted. Identify the dominant and recessive genes. Write down the crossing scheme, determine the genotypes and phenotypes of P and F1.
Based on the fact that all children show only one of the parental signs – hyperopia, 2 conclusions should be drawn:
Farsightedness is a dominant sign. Let’s designate it as B;
The father is homozygous for the gene for this trait, BB, and produces the same type of sperm B.
A mother with normal vision will be recorded as cc. The eggs produced by her body will be of the same type – b.
All children of this married couple will have the BB genotype, which will manifest itself in the development of vision pathology in them by the type of hyperopia.
Answer: the mother is homozygote for the gene for the recessive trait (bb), the father is homozygote for the dominant gene (BB), the children are hyperopic heterozygotes (Bb).
