Question

A colorblind female is crossed with a normal male. What are the parental genotypes? Diagram the F1 results from the cross

Answer 1

Can you list the Genotypes?

Answer 2

Colorblindness is X-linked recessive. So for a female to be colorblind, both of her X chromosomes must have the colorblind allele: \[X ^{b}\] A normal male has just one X chromosome with the normal dominant allele:\[X ^{B}\] So the parental genotypes are:
\[X ^{b}X ^{b}\] for the mother, and \[X ^{B}Y\] for the father.

The F1 results: All daughters will be carriers, and all sons will be colorblind.

Answer 3

Here's the diagram of the F1 offspring.

Answer 4

Colorblindness is an \(X\) linked disorder. As all females have the alleles \(X~X\) and males have \(X~Y\). The super subscripts will if \(b\) is the colour blind allele, so the female will have both of these superscripts. If the male is normal, and sice he only has one \(X\) chromosome he will only have a dominant \(B\) allele superscript.