Final answer:
The woman's genotype is XCX as she inherited one colorblindness allele from her colorblind father and carries the trait, but has normal vision herself. Her sons have a 50% chance of being colorblind.
Step-by-step explanation:
The genotype of the woman whose father and first son are colorblind, and whose male partner has normal color vision, can be deduced based on the inheritance pattern of color blindness, which is an X-linked recessive trait. This means that the gene for colorblindness (denoted as C in this case) is found on the X chromosome. Since the woman has a colorblind father, she must have received one X chromosome with the colorblind gene from him. However, because she has normal vision, her other X chromosome must have the normal vision gene. Therefore, her genotype is XCX.
When she has a son, he receives one of his mother's X chromosomes and his father's Y chromosome. If he inherits the X chromosome carrying the colorblindness trait (XC), he will be colorblind since males have only one X chromosome and no second X chromosome to mask the recessive gene. This explains why her first son is colorblind. Given the 50% chance of inheriting the XC chromosome, it's expected that 50% of her sons would be colorblind. It's important to note, males cannot be carriers of this form of color blindness because they have only one X chromosome and will either be affected by color blindness or not, depending on whether that X chromosome carries the mutation.