orblind marries a normal man. What is the expected proportion of color-blind chi
ID: 151767 • Letter: O
Question
orblind marries a normal man. What is the expected proportion of color-blind children? Give 9. Red-green colorblindness is a sex-linked trait. A normal woman whose father was red- both phenotypes and genotypes, including gender 10. A woman and a man have a red-green colorblind daughter. What can you say about the genotypes of the parents? 11. In cats, coat color is an X-linked trait. Black coat color is determined by allele XB, and yellow is generated by the Xb allele. Black and yellow are co-dominant; heterozygous cats have a tortoise shell coat. Suppose a female yellow cat is bred to a black tom. What will the phenotype ratio of kittens be? (Include genders in your ratios.) 12. What is the expected phenotypic ratio of a cross between a tortoise shell female and a black male?Explanation / Answer
Answer:
9). Based on the information, the woman is carrier for red-green clolorblindness.
XC = Normal ; Xc = red-green clolorblindness
XCXc (woman) x (normal man) XCY--------Parents
XC
Y
XC
XCXC (normal daughter)
XCY (normal son)
Xc
XCXc (normal but carrier daughter)
XcY (Colorblind son)
Only 25% of the children are colorblind.
10). To produce red-green colorblind daughter, father must be color-blind and mother either colorblind or carrier for that gene.
Father genotype = XcY
Mother genotype = XcXc or XCXc
11).
Female yellow cat (XbXb) x (black tom) XBY----P1
XB
Y
Xb
XBXb (totoise shell female cat)
XbY (yellow female cat)
Tortoise shell : Yellow = 1:1
12).
XBXb (totoise shell female cat) x (black male) XBY—P1
XB
Y
XB
XBXB (black female)
XBY (black male)
Xb
XBXb(tortoise female)
XbY (yellow male)
Black : Tortoise : Yellow = 1:2:1
XC
Y
XC
XCXC (normal daughter)
XCY (normal son)
Xc
XCXc (normal but carrier daughter)
XcY (Colorblind son)
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