Question

Red-green color blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and normal color vision. Her mother is normal in all respects, but her father is color blind and polydactylous. Bill is color blind and polydactylous. His mother has normal color vision and normal fingers and toes. If Bill and Martha marry, what types and proportions of children can they produce?
The following two genotypes are crossed: Aa Bb Cc X+ Xr × Aa BB cc X+Y, where a, b, and c represent alleles of autosomal genes and X+ and Xr represent X-linked alleles in an organism with XX-XY sex determination. What is the probability of obtaining genotype aa Bb Cc X+ X+ in the progeny?

Answer 1

Answer:

1) Children type and proportion:

• 2/16 individuals are expected to be normal girls for both traits, X⁺X⁻Pp  

• 2/16 individuals are expected to be normal boys for both traits, X⁺YPp  

• 2/16 individuals are expected to be girls who are normal for vision and have polydactyly, X⁺X⁻pp  

• 2/16 individuals are expected to be boys who are normal for vision and have polydactyly, X⁺Ypp

• 2/16 individuals are expected to be girls who are normal for fingers and toes and have colorblindness, X⁻X⁻Pp  

• 2/16 individuals are expected to be boys who are normal for fingers and toes and have colorblindness, X⁻YPp

• 2/16 individuals are expected to be girls that are color-blinded and have polydactyly, X⁻X⁻pp  

• 2/16 individuals are expected to be boys that are color-blinded and have polydactyly, X⁻Ypp

2) The probability of getting genotype aa Bb Cc X+ X+ = 1/64

Explanation:

Due to technical problems, you will find the complete explanation in the attached files