Respuesta :
Answer:
B) 1/8 color blind girls with polydactyly, 1/8 boys with normal vision and normal fingers
Explanation:
Available data:
•Red–green color blindness is an X linked recessive trait in humans (expressed by Xb allele)
•Polydactyly (extra fingers and toes) is an autosomal dominant trait (Expressed by P allele)
•Martha has normal fingers and toes and normal color vision. (pp XB-)
•Her mother is normal in all respects (pp XB-)
•Her father is color blind and polydactylous (P- Xb Y)
•Bill is color blind and polydactylous. (P- - Xb Y)
•His mother has normal color vision and normal fingers and toes. (pp XB-)
Martha´s parents cross:
(mother) pp XB- x Pp Xb Y (father)
(Martha) pp XB Xb
- For the Polydactyly trait, Martha received one allele from her mother and one allele from her father. Her mother was normal, pp, and her father was Polydactylous. Martha is normal. As Polydactyly is a dominant trait, Martha must have received a recessive allele from both her parents. This means that her father was heterozygous for the trait.
- For the blindness trait, she also got an X chromosome form her mother and one from her father. Her father was blind so he gave Martha a Xb. Her mother was normal, and so Martha, so her mother gave her a XB
Bill´s parents cross:
(mother) pp XB Xb - x P- X-Y (father)
(Bill) Pp Xb Y
- For the Polydactyly trait, Bill received one allele from his mother and one allele from his father. His mother was normal, pp, and Bill is Polydactylous, which means his father gave him the P allele. This means that his father was Polydactylous too.
- For the blindness trait, he also got an X chromosome form her mother and Y chromosome from his father. Bill is blind so got a Xb from his mother, which means that his mother ws heterozygous for the trait.
Martha and Bill´s cross:
Parental) pp XB Xb x Pp Xb Y
Gametes) p XB , p XB , p Xb , p Xb
P Xb , p Xb , P Y , pY
Punnet Square)
p XB p XB p Xb p Xb
P Xb Pp XB Xb Pp XB Xb Pp XbXb Pp XbXb
p Xb pp XB Xb pp XB Xb pp Xb Xb pp XbXb
P Y Pp XBY Pp XB Y Pp Xb Y Pp Xb Y
pY pp XB Y pp XB Y pp Xb Y pp Xb Y
F1) 8/16 female
2/8 = ¼ polydactylous and normal-sighted females, Pp XB Xb
2/8 = ¼ polydactylous and blind females, Pp XbXb
2/8 = ¼ normal females, pp XB Xb
2/8 = ¼ normal fingers and toes and blind females, pp XbXb
8/16 male
2/8 = ¼ polydactylous and normal-sighted males, Pp XB Y
2/8 = ¼ polydactylous and blind males, Pp XbY
2/8 = ¼ normal males, pp XB Y
2/8 = ¼ normal fingers and toes and blind males, pp XbY
What proportions of children with specific phenotypes would they be expected to produce?
A) 1/4 color blind girls with normal fingers, 1/4 boys with normal vision and polydactyly
B) 1/8 color blind girls with polydactyly, 1/8 boys with normal vision and normal fingers
C) 1/8 color blind girls with normal fingers, 1/4 boys with normal vision and polydactyly
D) 1/4 girls with normal vision and polydactyly, 1/8 boys with normal vision and polydactyly
• 1/8 color blind girls with polydactyly (Pp XbXb)
Of the whole progeny, only two female individuals are color blind girls and polydactyous,
This is 2/16 = 1/8 Pp XbXb
• 1/8 boys with normal vision and normal fingers (pp XBY)
Of the whole progeny, only two male individuals have normal vision and normal fingers
This is 2/16 = 1/8 pp XBY
