Chickens that carry both the alleles for rose comb (R) and pea comb (P) have walnut combs, whereas chickens that lack both of these alleles (that is, rr pp) have single combs. From the information about interactions between these two genes, determine the phenotypes and proportions expected from the following crosses: a. RR Pp × rr Pp;
b. rr PP × Rr Pp;
c. Rr Pp × Rr pp;
d. Rr pp × rr pp

Explanation:

Available data:

So, genotypes and phenotypes would be like

a) Cross: RR Pp × rr Pp

   Phenotypes:  Walnut  x  Pea

   Gametes)   RP, Rp, RP, Rp

                       rP, rp, rP, rp

   Punnett square)   RP       RP       Rp      Rp

                       rP    RrPP   RrPP    RrPp    RrPp

                       rP    RrPP   RrPP    RrPp    RrPp

                       rp    RrPp   RrPp    Rrpp     Rrpp

                       rp    RrPp   RrPp    Rrpp     Rrpp

F1) 12/16 = 3/4 R-P-, Walnut-combed animals

     4/16 = 1/4 Rrpp, Rose-combed animals

b. Cross: rr PP × Rr Pp

  Phenotype: Pea x Walnut

   Gametes:  rP, rP, rP, rP

                     RP, Rp, rP, rp

Punnett square)    RP         rP          Rp        rp

                    rP    RrPP     rrPP       RrPp     rrPp

                     rP    RrPP     rrPP       RrPp    rrPp

                    rP    RrPP     rrPP       RrPp     rrPp

                    rP    RrPP     rrPP       RrPp     rrPp

F1) 4/16 = 1/4 = 25% RrPP, Walnut-combed animals

    4/16 = 1/4 = 25% rrPP, Pea-combed animals

    4/16 = 1/4 = 25% RrPp, Walnut-combed animals

    4/16 = 1/4 = 25%  rrPp, Pea-combed animals

    1/2 = 50%  Walnut-combed animals, R-P-

    1/2 = 50% Pea-combed animals, rrP-

c. Cross: Rr Pp × Rr pp

   Phenotype:  Walnut  x   Rose

   Gametes: RP, Rp, rP, rp

                    Rp, rp, Rp, rp

    Punnett square)   RP          Rp           rP         rp

                      Rp     RRPp     RRpp      RrPp     Rrpp

                      Rp     RRPp     RRpp      RrPp     Rrpp

                      rp      RrPp       Rrpp      rrPp       rrpp

                      rp      RrPp       Rrpp      rrPp       rrpp

F1) 2/16 = 1/8 RRPp, walnut

    2/16 = 1/8 RRpp, Rose

    4/16 = 2/8 RrPp, Walnut

    4/16 = 2/8 Rrpp, Rose

    2/16 = 1/8 rrPp, Pea

    2/16 = 1/8 rrpp, single

    6/16 = 3/8 walnut, R-P-

     6/16 = 3/8 rose, R-pp

     2/16 = 1/8 Pea, rrP-

     2/16 = 1/8 Single, rrpp

d. Cross: Rr pp × rr pp

Phenotype: Rose   x   Single

Gametes) Rp, rp, Rp, rp

                rp, rp, rp, rp

Punnett square)   Rp       Rp        rp        rp

                    rp    Rrpp    Rrpp     rrpp    rrpp

                    rp    Rrpp    Rrpp     rrpp    rrpp

                    rp    Rrpp    Rrpp     rrpp    rrpp

                    rp    Rrpp    Rrpp     rrpp    rrpp

F1) 8/16 = 1/2 = 50% Rrpp, Rose-combed animals

    8/16 = 1/2 = 50% rrpp, single-combed animals

the answer is c) abandoned properties that are contaminated with hazardous materials.

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2. a 7. d. 12. d

3. c 8. b. 13. c

4. d 9. c. 14. d

5. c 10. d. 15. a

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