Can a female get color-blindness from her father’s side?

The quick answer is that yes, a female can get a copy of the gene that leads to colorblindness from her father. In fact, if her father is colorblind she will most certainly inherit a copy of the colorblindness gene.

However, to be colorblind, a woman needs to get two copies of the gene that leads to colorblindness – one from their mom and one from their dad. What this means is that if there is no history of colorblindness in your family, then odds are that your daughter will not be colorblind herself, but will end up being a carrier for colorblindness. As a carrier, she'll have a 50% chance of passing the gene on to her kids.

Sons, on the other hand, can only get the gene that causes colorblindness from their mother. This is because the gene that causes colorblindness is found on the X chromosome. As you probably know, boys have an X and a Y chromosome while girls have two X chromosomes – this is usually how one becomes male or female. Because sons only have one X chromosome, they only need one copy of the colorblindness gene to be colorblind.

The version of the gene that causes colorblindness is also recessive. What this means is that if you have one “good” copy of the gene and one “colorblind” copy, then you won't be colorblind. This means that girls need to inherit two copies of the colorblindness gene to be colorblind – one from each parent.

So now that we have some background information, let’s look specifically at your situation. To make the explanation simpler, we’ll use Punnett squares.

Punnett squares are really just a way to organize all the genes and to figure out how likely a combination of genes is. For your current situation, if we assume that you do not carry a colorblindness gene, then we would say you are XX. (This just means you have 2 “normal” X chromosomes, at least as far as colorblindness is concerned.)

We would say your husband is X^cY meaning that he has the copy of the gene that causes colorblindness on his X chromosome and a Y chromosome. To make the Punnett square, we'll put your husband’s two possible contributions, X^c and Y on the top and your two, X and X on the left side like this:

In each square, we just combine the genes that come together in that square. This’ll give the following Punnett square:

As you can see, all of your daughters, XX^c, end up carrying the colorblindness gene. None of your sons carry it because they can only get an X from you. However, since each daughter also carries a good copy of the gene, they are not colorblind themselves but are carriers.

Now if your daughters had children with men who weren’t colorblind, their Punnett square would look like this:

As you can see, there would be a 1 in 4 or 25% chance for a daughter who is a carrier (X^cX), a 25% chance for a colorblind son (X^cY), and a 50% chance for a child without the colorblind gene (XX, XY). And if your daughter had children with a colorblind man? Then it would look like this:

Now there would be a 50% chance for a colorblind child (X^cX^c, X^cY), a 25% chance for a daughter who is a carrier (X^cX), and a 25% chance for a colorblind-free son (XY). I hope this helps. X-linked genetics can get a bit complex but I always found that the Punnett square helped me a lot