Male proband with short stature married the same girl. Three sons were born

Male proband with short stature married the same girl. Three sons were born from this marriage, one of them is of normal height. A son of normal height married a short girl. They have two children – a son of short stature and a daughter of normal stature. Make a pedigree, establish the nature of inheritance of the trait (dominant or recessive, linked or not linked to the sex, determine the genotypes of all family members.

Since parents with short stature give birth to a child of normal height, it should be concluded that the gene for short stature dominates over the gene for normal growth. It should also be concluded that both parents are heterozygous for the growth gene.

If the growth gene was linked to the X chromosome, then all F1 descendants would be short. If the growth gene were linked to the H – chromosome, then it would be inherited only through the male line, but girls are also stunted in the task. Therefore, the growth gene is not linked to the sex chromosomes.

Let’s designate the gene that causes low growth as K, then the gene for normal growth will be designated as k. Parents have genotypes Kk and produce two types of germ cells – K and k. Their offspring include the following possible options:

homozygous children with short stature (KK) – 25%;

homozygous children with normal height (kk) – 25%;

heterozygous children with short stature (Kk) – 50%.

People with normal growth have a homozygous kk genotype, because homozygosity is a prerequisite for the realization of a recessive trait in a phenotype.

Since a child of normal height appears in the marriage of a son with normal growth, it should be concluded that the short girl with whom he is marrying is heterozygous. The girl, therefore, has the Kk genotype and produces two types of eggs – k and K. Their theoretically possible offspring includes the following variants:

children – heterozygotes with short stature (Kk) – 50%;

children are homozygotes with normal height (kk) – 50%.

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