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 <em>Polydactyly trait</em>, 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 <em>blindness trait</em>, 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 <em>Polydactyly trait</em>, 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 <em>blindness trait,</em> 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
