A brown-eyed fair-haired man, heterozygous for 1 trait, and a blue-eyed dark-haired woman

A brown-eyed fair-haired man, heterozygous for 1 trait, and a blue-eyed dark-haired woman, heterozygous for 2 traits, entered into marriage. Identify the genotypes of the married couple, as well as the genotypes and phenotypes of the children. Establish the probability of birth of children with a genotype homozygous for one trait.

Let the gene that determines the development of brown eyes in a person be M, then the gene for blue eyes will be m.

Let the gene that causes a person to develop light hair be n, then the gene for dark human hair will be N.

A fair-haired brown-eyed man is heterozygous for the gene for eye color – Mmnn. It produces two types of spermatozoa – Mn, mn.

A blue-eyed woman with dark hair, heterozygous for the hair color gene – mmNn. This woman also produces two types of eggs – mN, mn.

Probable offspring born to a given family will include the following options:

children with dark hair and brown eyes (MmNn) – 25%;

children with dark hair and blue eyes (mmNn) – 25%;

children with blond hair and brown eyes (Mmnn) – 25%;

children with blond hair and blue eyes (mmnn) – 25%.

Genotypes homozygous for one trait – mmNn and Mmnn – 50%.

Answer: 50%.