2. Color-blindness is an X-linked recessive disorder. Mike is color-blind. His wife, Meg, is homozygous for normal color vision allele. Using Punnett squares, derive and compare the genotypic and phenotypic ratios for the offspring of this marriage. a. What is Mikes's genotype?

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Related questions

. A recessive allele on the X chromosome is responsible for red-green color blindness in humans. This means that females have two copies (since human females are XX) and males have one copy (since human males are XY). This is called a sex-linked trait. A red-green color blind male is genotype Xn Y. A normal vision male is genotype XN Y. For this problem, a normal vision woman whose father is color-blind marries a color-blind male. N= not color blind allele, n = color blind allele. Y = Y chromosome.

a. What is the genotype of the color-blind father? Why?

b. What is the genotype of the color-blind husband? Why?

c. What is the genotype of the normal vision woman? Why?

d. What is the probability that this couple will have a son that is color-blind?

e. What is the probability that this couple will have a daughter that is color-blind?

The frequency of the X-linked recessive allele (c) responsible for color-blindness is 0.10, and the normal allele (C) is at a frequency of 0.90 in a random mating population. A color-blind woman marries a man with normal vision from this population. What is the probability that their first daughter will be color-blind? (2 points)

Color blindness is an X-linked recessive trait. A color blind man marries a homozygous normal female. They have two children, both with what is known as Turner Syndrome (monosomy X or X0 or X_). One child has normal vision and one is colorblind.

Did nondisjunction occur in the mother or father for the child with normal vision? Explain.