Question

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

Answer 1

The phenotypic outcomes for the four chicken comb crosses vary, depending on the alleles inherited from the parents. Each cross produces different ratios of comb types: a) 1:1 walnut to rose, b) 1:1 walnut to pea, c) 3:3:2 walnut to rose to single, and d) 1:1 rose to single combs.

Step-by-step explanation:

Understanding Chicken Comb Genetics

Genetics problems often concern the calculation of probability in inheritance patterns. Considering the genotypes given and the dominance of certain alleles, we can predict phenotypes and their proportions for the offspring of chicken crosses.

Crosses a to d

Cross d: Rr pp × rr pp will lead to offspring who are either Rr pp or rr pp, resulting in rose and single comb chickens at a 1:1 ratio.

Answer 2

Step-by-step explanation:

Available data:

• There are two diallelic genes that code for chickens comb
• Dominant alleles R and P express rose comb and pea comb, respectively
• When they are together in a genotype, they express walnut combs, R-P-
• Recessive alleles are r and p
• Genotypes 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

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