## Step 1: Define the symbolic representation of the problem
Let's denote the number of scoops of Gamma as $x_1$ and the number of scoops of Delta as $x_2$. The objective is to minimize the cost, which is $1.5x_1 + 2.5x_2$. The constraints are:
- $7x_1 + 12x_2 \geq 60$ (at least 60 grams of iron)
- $10x_1 + 9x_2 \geq 45$ (at least 45 grams of biotin)
- $x_1 \geq 0$ and $x_2 \geq 0$ (non-negativity constraints, as the number of scoops cannot be negative)

## 2: Express the problem in the required symbolic format
The symbolic variables are:
- $x_1$ for Gamma
- $x_2$ for Delta

The objective function is:
- $1.5x_1 + 2.5x_2$

The constraints are:
- $7x_1 + 12x_2 \geq 60$
- $10x_1 + 9x_2 \geq 45$
- $x_1 \geq 0$
- $x_2 \geq 0$

In the required format:
```json
{
    'sym_variables': [('x1', 'Gamma'), ('x2', 'Delta')], 
    'objective_function': '1.5*x1 + 2.5*x2', 
    'constraints': ['7*x1 + 12*x2 >= 60', '10*x1 + 9*x2 >= 45', 'x1 >= 0', 'x2 >= 0']
}
```

## 3: Implement the problem in Gurobi code
To solve this linear programming problem using Gurobi in Python, we will use the following code:

```python
import gurobi

def solve_optimization_problem():
    # Create a new model
    model = gurobi.Model()

    # Define the variables
    x1 = model.addVar(name="Gamma", lb=0)  # Number of scoops of Gamma
    x2 = model.addVar(name="Delta", lb=0)  # Number of scoops of Delta

    # Define the objective function
    model.setObjective(1.5 * x1 + 2.5 * x2, gurobi.GRB.MINIMIZE)

    # Add constraints
    model.addConstr(7 * x1 + 12 * x2 >= 60, name="Iron_Requirement")
    model.addConstr(10 * x1 + 9 * x2 >= 45, name="Biotin_Requirement")

    # Optimize the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print("Optimal solution found.")
        print(f"Number of scoops of Gamma: {x1.varValue}")
        print(f"Number of scoops of Delta: {x2.varValue}")
        print(f"Total cost: {model.objVal}")
    else:
        print("No optimal solution found.")

solve_optimization_problem()
```