Biology, 30.03.2021 20:50, dpinzoner5952

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

Answers: 1

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answered: kdawg203

Explanation:

Available data:

There are two diallelic genes that code for chickens combDominant alleles R and P express rose comb and pea comb, respectivelyWhen they are together in a genotype, they express walnut combs, R-P-Recessive alleles are r and pGenotypes with only recessive alleles express single combs, rrpp

So, genotypes and phenotypes would be like

Walnut combs, RRPP, RrPP, RRPp, RrPp (One dominant allele of each gene)Rose comb, RRpp or Rrpp (only R gene with dominant allele)Pea comb, rrPP or rrPp (Only P gene with dominant allele)Single combs, rrpp (no dominant alleles at all)

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

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

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

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

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

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

answered: Guest

b is the best answer for this question.

answered: Guest

answer; i believe the correct answer to the above question is ( friction)

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Chickens that carry both the alleles for rose comb (R) and pea comb (P) have walnut combs, whereas c...