Question

for the following problem please provide me a key for allele symbols, the genotype for each parent, a punnett square with the male and female symbols, and answers to the questions.
This is a dihybrid cross. in horses, black coloration is due to a dominant allele (B) and chestnut is due to a recessive alleles (b). the "trotting" style of running is due to a dominant allele (T), while the "pacing" style of running is due to a recessive allele (t). a black pacer who is heterozygous for both traits (mother) is mated to a chestnut trotter who is homozygous for both traits (father). what will the F generation phenotypes be? what are the phenotypic percentages of the F1 generation?

Answer 1

Allele/key
- B = black (dominant)
- b = chestnut (recessive)
- T = trotting (dominant)
- t = pacing (recessive)

Parent genotypes (interpretation)
- Mother (given as “heterozygous for both traits”): BbTt
(Note: a true “pacer” phenotype requires tt, so “black pacer who is heterozygous for both traits” is inconsistent. I assume the intended phrase was simply “heterozygous for both,” i.e. BbTt.)
- Father (chestnut trotter, homozygous for both): bbTT

Gametes
- Mother (BbTt) produces BT, Bt, bT, bt
- Father (bbTT) produces only bT

Punnett square (female on left, male on top)

Male: bT
Female
BT -> BbTT
Bt -> BbTt
bT -> bbTT
bt -> bbTt

Offspring genotypes and phenotypes
- BbTT (25%) → black trotter
- BbTt (25%) → black trotter
- bbTT (25%) → chestnut trotter
- bbTt (25%) → chestnut trotter

Phenotypic percentages (F1)
- 50% black trotters
- 50% chestnut trotters
- 0% pacers

(If you intended the mother to be a black pacer with genotype Bbtt, crossed to bbTT, the phenotypic result is the same: all offspring get a T from the father so all are trotters, and B vs b segregates 50:50 → 50% black trotters, 50% chestnut trotters.)