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vaieri [72.5K]
3 years ago
7

Two independently segregating genes determine kernel color in wheat. Red color results from at least one dominant allele in both

genes. Kernels on rrbb plants are white, and the lack of a dominant allele in either the R locus or the B locus (but not both) results in brown kernel color. If you mate a plant that is true-breeding for red color kernels with a plant that is true-breeding for white color kernels:a.What is the expected phenotype of the F1 plants?b. What are the expected phenotypic classes in the F2 progeny and their relative proportions?
Biology
1 answer:
Lera25 [3.4K]3 years ago
7 0

Answer:

a) The expected phenotype of the F1 plants is 100% RrBb, red kernels.

b) The expected phenotypic classes in the F2 are: 9:3:3:1

   9/16 R-B-, 3/16 rrB-, 3/16 R-bb,  1/16 rrbb

   Proportions 9:6:1.

9/16 Red kernel (R-B-), 6/16 Brown kernel (rrB- + R-bb), 1/16 White kernel (rrbb)    

Explanation:

<u>Available data:</u>

  • red kernel: R-B-
  • brown kernel: R-bb or rrB-
  • white kernel: rrbb

1º Cross)  RRBB      x      rrbb

F1)             100% RrBb (red kernels)

2ºCross)   RrBb       x       RrBb

Gametes)  RB                   RB

                  Rb                   Rb

                  rB                    rB

                  rb                    rb

Punnet Square)      RB           Rb         rB         rb

                 RB      RRBB      RRBb     RrBB     RrBb

                 Rb      RRBb      RRbb     RrBB     Rrbb

                 rB       RrBB       RrBb      rrBB      rrBb

                 rb       RrBb       Rrbb       rrBb     rrbb

F2)  Phenotypic classes:

       <em>9/16 R-B-</em>

<em>        3/16 rrB-</em>

<em>        3/16 R-bb</em>

<em>        1/16 rrbb</em>

      Phenotypic proportions:

      <em>9/16 Red kernel (R-B-)</em>

<em>       6/16 Brown kernel (rrB- + R-bb)</em>

<em>       1/16 White kernel (rrbb)         </em>                      

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