Question

(HELP PLEASE)(punnet squares)

A male is colorblind and his wife has normal vision. What percent chance do the sons have of


being colorblind and what percentage do their daughters have of being colorblind? Show your


answer using a Punnett square.

Answer 1

In the case of a colorblind male and a non-colorblind female having offspring, there is a 50% chance for their sons to be colorblind while their daughters have a 0% chance of being colorblind.

Step-by-step explanation:

In genetics, color blindness is a sex-linked trait, meaning it is associated with an individual's sex. This trait is predominantly carried on the X chromosome. To illustrate how it is inherited, let's denote color blindness as 'b' and normal vision as 'B'. Since males only have one X chromosome, the male in question would have an XbY configuration for this trait. The female with normal vision would have an XX configuration.

The possible offsprings from their genetic combination can be represented in a Punnett Square. The resulting Punnett Square would be as follows: XbY x XX=in XbX, XX, XbY, XY. Correspondingly, the offsprings will be 50% colorblind males (XbY), 50% normal males (XY), 50% carrier females (XbX), and 50% normal females (XX).

Therefore, the chances of the offspring being color blind are: sons 50% (since they can be XbY or XY) and daughters 0% (since they can be carrier (XbX) but won't be colorblind).

Learn more about Punnett Square