Try this pickup line, it always gets the girls:
I may be colorblind, but I can see you.
By not being colorblind and andswering all the questions correctly.
no
If the mother is a carrier of the colorblind gene (XcX) and the father has normal color vision, the probability of their child being colorblind is 50%. This is because the child has a 50% chance of inheriting the Xc chromosome from the mother and developing colorblindness.
There is no such assumption.
The probability of a colorblind child being born is 50%. This is because the male passes his Y chromosome to all his sons, and since he is colorblind, his sons will inherit the colorblind gene from him. The daughters will inherit their X chromosome from the mother and have a 50% chance of being carriers like her.
If Mary's mother is colorblind, and therefore carries the colorblind gene on one of her X chromosomes, then Mary would inherit that gene as well. If Mary's father is colorblind, he would have to pass on his X chromosome with the colorblind gene to Mary, making her colorblind too. If only Mary's mother is colorblind, Mary's father is likely not colorblind.
No, Taylor Lauter is not colorblind.
All dogs are colorblind.
Not necessarily. The allele for colorblindness is recessive. For a female, in order to be colorblind she must have to recessive alleles for colorblindness. Example: XcXc would be colorblind. XCXc would be a carrier for colorblindness, but not colorblind. For a male, because colorblindness is a sex-linked gene, he only needs one allele to be colorblind. Example: XcY is colorblind. XCY is not colorblind.
100% of all male offspring will be colorblind. 0% of all femal offspring will be colorblind.
Make a punnet square with the mother above, her genotype would be: X^B X^b, and the father to the left whose genotype is X^b Y.The probability of having a colorblind CHILD is 50%. The probability of them having a SON is 50%. Since we are asked what the probability of their SON being colorblind, it is 50% as well. The reason is because the chance of having a colorblind son, among sons only, (according to the punnet square) is 50%.
The woman with normal color vision likely has the genotype XX, where X is the normal vision allele. The father, being colorblind, has the genotype X^cY, where X^c represents the colorblind allele. Their daughter, who is colorblind, must have inherited one X^c from her father and one X from her mother, resulting in the genotype X^cX. Hence, the woman's phenotype is normal color vision, while the father's phenotype is colorblind, and the daughter's phenotype is colorblind.