Question

Red-green colorblindness is an X-linked recessive mutation. A colorblind female is crossed with a normal male. a. What are the parental genotypes? b. Diagram the F1 results from this cross c. Diagram the F2 results. (Show your Punnett Square).

Answer 1

The sub questions sort of guide you through the problem. The first step is to decide what notation should be used to represent the different alleles.

Because they are sex chromosomes, the woman has two X chromosomes whereas the man has one X and one Y chromosome. So Mom we know is XX and Dad is XY. The Y chromosome is 'blank' - it carries a couple genes, but never carries the gene for color blindness. The X chromosomes do carry genes. So all you need is a notation to distinguish between the X chromosomes that do carry the allele for color blindness and those that do not carry the allele for color blindness.

If that wasn't helpful enough let me know.

Answer 2

It was I understand the first part, but I am confuse about the F1 diagram. Can you help please?

Answer 3

OK - so what genotypes do you have for the parents? You will need those before you can set up your Punnett square and get your F1 results...

Answer 4

Xa Xa for mom XY for dad

Answer 5

Nice. Where of course Xa stands for the X chromosome with the color blind allele on it (Mom's case) and just plain X stands for the X chromosome without the color blind allele on it (Dad's case).

To set up the Punnett square, the idea is that either one of of Mom's two X chromosomes could wind up in a specific child. Similarly, any child of Dad's has a random chance of getting either his X chromosome or his Y chromosome.

So your square should be set up like this, with Mom's chromosomes on the top and Dad's on the side:

Answer 6

Ok got it, do I do the same for F2? Or do I change mom and dad position on the punnet square?

Answer 7

first diagram would look like this