Cross a heterozygous running, heterozygous black mouse with a homozygous running, homozygous bl ouse. What is the probable phenotype ratio? \( \begin{array}{ll}\text { a. } \operatorname{RrBb} \times \mathrm{RRBB} & \text { b. }\end{array} \)

To solve this genetics problem, let's break down the cross step by step. ### **Given:** - **Parent 1:** Heterozygous running and heterozygous black mouse (Genotype: **RrBb**) - **Parent 2:** Homozygous running and homozygous black mouse (Genotype: **RRBB**) ### **Assumptions:** 1. **R** (Running) is dominant over **r** (Not Running). 2. **B** (Black) is dominant over **b** (Non-Black). 3. The two genes assort independently. ### **Step 1: Determine the Gametes** - **Parent 1 (RrBb)** can produce four types of gametes: - **RB** - **Rb** - **rB** - **rb** - **Parent 2 (RRBB)** can only produce one type of gamete: - **RB** ### **Step 2: Create the Punnett Square** Since Parent 2 can only contribute **RB**, each gamete from Parent 1 will combine with **RB** from Parent 2: | Parent1 Gametes | Parent2 Gametes (RB) | Offspring Genotype | |-----------------|----------------------|---------------------| | RB | RB | **RRBB** | | Rb | RB | **RRBb** | | rB | RB | **RrBB** | | rb | RB | **RrBb** | ### **Step 3: Determine the Phenotypes** - **RRBB:** Running, Black - **RRBb:** Running, Black - **RrBB:** Running, Black - **RrBb:** Running, Black All possible offspring genotypes result in the **Running, Black** phenotype because: - Each offspring inherits at least one dominant **R** allele (ensuring the Running trait). - Each offspring inherits at least one dominant **B** allele (ensuring the Black trait). ### **Conclusion:** All offspring from this cross will display the **Running, Black** phenotype. **Probable Phenotype Ratio:** - **100% Running, Black** All offspring exhibit both dominant traits; 100 % are running and black mice.