The color blindness gene is recessive and X linked. It is written as Xb. A woman who is color blind has a son with a man with normal vision. What is the genotype of the son? Use a Punnett square to answer the question.

Genotype of the son = XᵇY.

Explanation:

The trait is X linked recessive that means in female if both the alleles are Xᵇ then only female will show traits of color blindness whereas in males since they only have one X chromosome even if one Xᵇ will be present, he will show color blindness traits. But in the given example, woman is color blind while man has normal vision. It simply means that genotype of woman is XᵇXᵇ while that of man is XY.

The Punnett square diagram depicting the cross is attached as under:

It may be noted that their son will be born with a genotype XᵇY so as already explained he will be color blind.

This enables the differential expression of genes. While the sigma promoter is canonically placed upstream of genes at position -10 and -35 across organisms, they differ slightly in sequence. This allows for the binding of sigma factors of RNA polymerase by different affinities across different gene promoters. Expression of genes will, therefore, be high where the sigma factors bind with high affinity and vice versa.