I understand sex linkage when there's only 2 alleles for each parent, but what if there is one trait that is X-linked and the other trait is not? IM SO CONFUSED please help, thank you!
-
Let's look at colorblindness. A normal male would have the genotype: X^CY the C is for normal color vision. A colorblind male would be X^cY the c is for the colorblind trait. A normal female would be
X^CX^C, a carrier female (normal color vision but containing the colorblind trait) X^CX^c and a colorblind female X^cX^c. So you see you have to show the sex of the parent since the gene is on the sex chromosome. It is not on the Y chromosome that is why the males either are colorblind or not. Females contain 2 X chromosomes so they have 3 possibilities.
Now let's say you want to show another trait like being tall (T) or short (t) along with being colorblind.
Here you don't need to show the sex since it is not found on the X chromosome.
Let's mate a colorblind male who is heterozygous tall with a normal female who is short.
The Genotypes would be like this: X^cYTt X X^CX^Ctt
Each parent has the genes for color eyesight and height. This is a dihybrid cross. So you need to figure out the gametes according to the law of independent assortment.
The father can produce the following gametes: X^cT, X^ct, YT and Yt (4 gametes)
The mother can produce the following gametes: X^Ct. They are all the same.
Now put the 4 from the male on one side and the one from the female on the side.
You will have a rectangular box divided into 4 parts each part will contain the combination of the male and female gamete possibilities.
X^CX^C, a carrier female (normal color vision but containing the colorblind trait) X^CX^c and a colorblind female X^cX^c. So you see you have to show the sex of the parent since the gene is on the sex chromosome. It is not on the Y chromosome that is why the males either are colorblind or not. Females contain 2 X chromosomes so they have 3 possibilities.
Now let's say you want to show another trait like being tall (T) or short (t) along with being colorblind.
Here you don't need to show the sex since it is not found on the X chromosome.
Let's mate a colorblind male who is heterozygous tall with a normal female who is short.
The Genotypes would be like this: X^cYTt X X^CX^Ctt
Each parent has the genes for color eyesight and height. This is a dihybrid cross. So you need to figure out the gametes according to the law of independent assortment.
The father can produce the following gametes: X^cT, X^ct, YT and Yt (4 gametes)
The mother can produce the following gametes: X^Ct. They are all the same.
Now put the 4 from the male on one side and the one from the female on the side.
You will have a rectangular box divided into 4 parts each part will contain the combination of the male and female gamete possibilities.