Let's break down the genetics described: - **Buckskin (B) is dominant to chestnut (b).** - **Trotting (T) is dominant to pacing (t).** Horse 1: **Homozygous for buckskin fur (BB)** and **heterozygous for trotting gait (Tt)** Horse 2: **Heterozygous for buckskin fur (Bb)** and **is a pacer (tt)** Let's write their genotypes: - Horse 1: BB Tt - Horse 2: Bb tt Now, let's find the probability for each trait in the offspring: **Buckskin Fur:** - BB x Bb → Offspring: 50% BB, 50% Bb (all buckskin, since both BB and Bb are buckskin) **Trotting Gait:** - Tt x tt → Offspring: 50% Tt (trotter), 50% tt (pacer) **Probability of Buckskin Trotter:** - All offspring will be buckskin (100%) - 50% will be trotters (Tt) So, the probability of being both buckskin **and** a trotter is: 1 (buckskin) × 0.5 (trotter) = **0.5 or 50%** **The answer is:** - **50%**

Let's analyze the problem step by step: ### Genetic Traits: - **Color:** - Buckskin (B) is **dominant** over chestnut (b). - **Gait:** - Trotting (T) is **dominant** over pacing (t). ### Parental Genotypes: - **Parent 1:** - Homozygous for buckskin: **BB** - Heterozygous for gait: **Tt** - Genotype: **BB Tt** - **Parent 2:** - Heterozygous for buckskin: **Bb** - Pacer: **tt** - Genotype: **Bb tt** --- ### Step 1: Determine possible gametes for each parent: - **Parent 1 (BB Tt):** - Gametes: **B T** or **B t** (since it's homozygous BB, all gametes carry B) - **Parent 2 (Bb tt):** - Gametes: **B t** or **b t** --- ### Step 2: Possible offspring genotypes: Create a Punnett square for each trait: | | **B T** | **B t** | |----------------|---------|---------| | **B t** (Parent 2) | **BB Tt** | **BB tt** | | **b t** (Parent 2) | **Bb Tt** | **Bb tt** | ### Step 3: Determine phenotype probabilities: - **Color:** - All **BB** or **Bb** offspring will be **buckskin** (since buckskin is dominant). - Offspring genotypes: - **BB Tt**: buckskin - **BB tt**: buckskin - **Bb Tt**: buckskin - **Bb tt**: buckskin - **Gait:** - **Tt**: trotter - **tt**: pacer ### Step 4: Calculate probability of each combination: - **Buckskin:** - All offspring are buckskin (since all are either BB or Bb). - **Gait (Tt vs. tt):** - From the genotypes: - **BB Tt**: trotters - **BB tt**: pacers - **Bb Tt**: trotters - **Bb tt**: pacers - **Probabilities for gait:** - Tt (trotter): 2 out of 4 genotypes → 50% - tt (pacer): 2 out of 4 genotypes → 50% ### **Final calculation:** - **Probability of foal being buckskin:** 100% - **Probability of foal being a trotter:** 50% **Therefore, the probability that the foal is both buckskin and a trotter is:** \[ \boxed{50\%} \] **Answer: 50%**