Question

In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rrP_). If two recessive genes are present at the first and the second locus (rrpp), a single comb is produced. Give genotypes for comb shape of the parents in the crosses listed below. a. Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring. b. Rose crossed with pea produces 20 walnut offspring. c. Pea crossed with single produces 1 single offspring. d. Rose crossed with pea produces 2 walnut, 1 single, and 1 pea offspring. e. Rose crossed with single produces 31 rose offspring. f. Rose crossed with single produces 10 rose and 11 single offspring.

Answer 1

Answer and Explanation:

Available data:

• Comb shape is determined by genes at two loci (R, r and P, p).

• The walnut comb genotype is R_P_.

• The rose comb genotype is R_pp.

• The pea comb genotype is rrP_.

• The single genotype is rrpp.

a. Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring:

Parental) RrPp x rrpp

Gametes) RP Rp rP rp rp rp rp rp

Punnet 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 phenotype: 25% walnut, 25% rose, 25% pea, and 25% single.

F1 genotype: 4/16 RrPp, 1/16 Rrpp, 4/16 rrPp, 4/16 rrpp.

b. Rose crossed with pea produces 20 walnut offspring.

Parental) RRpp x rrPP

Gametes) Rp Rp Rp Rp rP rP rP rP

Punnet 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 phenotype: 100% walnut.

F1 genotype: 16/16 RrPp.

c. Pea crossed with single produces 1 single offspring.

This is not possible, because the pea genotype involves at least one dominant allele P. There are two possible crosses: rrPp x rrpp, which must produce half of the progeny pea and the other half single, or rrPP x rrpp which produce a whole pea progeny with no single offspring.

Parental) rrPp x rrpp

Gametes) rP rp rP rp rp rp rp rp

Punnet 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 phenotype: 50% pea, and 50% single.

F1 genotype: 8/16 rrPp, 8/16 rrpp.

d. Rose crossed with pea produces 2 walnut, 1 single, and 1 pea offspring.

This is not possible, because having one of the parents with a rose phenotype involves at least one R allele, which means that there must be rose phenotype in the progeny.

Parental) Rrpp x rrPp

Gametes) Rp Rp rp rp rP rP rp rp

Punnet 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 phenotype: 25% walnut, 25% rose, 25% pea, and 25% single.

F1 genotype: 4/16 RrPp, 1/16 Rrpp, 4/16 rrPp, 4/16 rrpp.

e. Rose crossed with single produces 31 rose offspring.

Parental) RRpp x rrpp

Gametes) Rp Rp Rp Rp rp rp rp rp

Punnet 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 phenotype: 100% rose (31 individuals equal 100% of the progeny).

F1 genotype: 16/16 Rrpp.

f. Rose crossed with single produces 10 rose and 11 single offspring.

Parental) Rrpp x rrpp

Gametes) Rp Rp rp rp rP rP rp rp

Punnet 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 phenotype: 50% rose, 50% single.

F1 genotype: 8/16 Rrpp, 8/16 rrpp.