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. When answering the given question, consider that the answers include the proportions of only some of the possible phenotypes; other phenotypes are also expected to occur but are not included. If Bill and Martha marry, what proportions of children with specific phenotypes would they be expected to produce? 1/4 color‑blind girls with normal fingers, 1/4 boys with normal vision and polydactyly 1/8 color‑blind girls with polydactyly, 1/8 boys with normal vision and normal fingers 1/8 color‑blind girls with normal fingers, 1/4 boys with normal vision and polydactyly 1/4 girls with normal vision and polydactyly, 1/8 boys with normal vision and polydactyly

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

Bill´s parents cross:

(mother) pp XB Xb -      x      P- X-Y (father)

                       (Bill) Pp Xb Y

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 XbPp XB XbPp XB XbPp XbXb    Pp XbXb

p Xbpp XB Xbpp XB Xbpp Xb Xbpp XbXb

P Y          Pp XBY           Pp XB Y    Pp Xb YPp Xb Y

pY           pp XB Y    pp XB Y     pp Xb Ypp 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

well

explanation:

the answer is c. it will give up it's valence electron to the other atom.