Genotype

The mother’s genotype must be heterozygous, that is, Ff. This is due to the fact that the son does not have the condition, implying that he got the recessive gene f from his mother, where f is any gene that does not code for freckles. As a result, the boy is homozygous (ff) for genes that do not create freckles. Given this, it is evident that the father provided a gene that did not code for freckles, and so he may be heterozygous or homozygous recessive. The reason behind the above assumption is that the freckle gene is dominant and the son lacks them (freckles) hence he must have inherited a recessive gene from the father as well.

One of the following assumptions is true to the son’s genotype proving the heterozygous state of the mother’s genotype:

if the father is heterozygous (Ff) and the mother heterozygous (Ff), the outcome would be FF, 2 Ff, and ff where the son is ff;

if the father is homozygous for the recessive gene (ff) and the mother heterozygous, the outcome would be 2 Ff and 2 ff where the son is one of the ff.

Where:

FF is homozygous for dominant gene;

Ff is heterozygous.

And lastly, ff is homozygous for recessive gene.

For an offspring to be homozygous recessive both parents have heterozygous or one of them heterozygous recessive and the other heterozygous (Sun et al., 2015). In any case, one of the parents is homozygous dominant for any trait; the probability that an offspring will lack the trait is zero. This is because in the presence of a dominant gene the trait will be expressed and hence this nullifies any possibility of the father or the mother being homozygous dominant.

Reference

Sun, Z. et al. (2015). Genotype-phenotype correlation of xeroderma pigmentosum in a Chinese Han population. British Journal of Dermatology, 172(4), 1096-1102.