To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables, an objective function, and constraints based on the problem statement.

Let's define:
- $x_1$ as the number of servings of elk meat consumed.
- $x_2$ as the number of servings of bison meat consumed.

The objective is to minimize the total cost. The cost per serving of elk meat is $6, and the cost per serving of bison meat is $7. Therefore, the objective function can be represented as:
\[ \text{Minimize} \quad 6x_1 + 7x_2 \]

The constraints are based on the nutritional requirements:
- At least 30 units of iron must be consumed daily. Given that elk meat contains 5 units of iron per serving and bison meat contains 4 units of iron per serving, we have:
\[ 5x_1 + 4x_2 \geq 30 \]
- At least 40 units of zinc must be consumed daily. Since elk meat contains 3 units of zinc per serving and bison meat contains 4 units of zinc per serving, we have:
\[ 3x_1 + 4x_2 \geq 40 \]

Additionally, the number of servings cannot be negative, so we also have non-negativity constraints:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Now, let's represent this problem in the requested JSON format:

```json
{
    'sym_variables': [('x1', 'servings of elk meat'), ('x2', 'servings of bison meat')],
    'objective_function': '6*x1 + 7*x2',
    'constraints': ['5*x1 + 4*x2 >= 30', '3*x1 + 4*x2 >= 40', 'x1 >= 0', 'x2 >= 0']
}
```

Finally, here is the Gurobi code in Python to solve this linear programming problem:

```python
from gurobipy import *

# Create a new model
m = Model("Nutrition_Optimization")

# Define variables
x1 = m.addVar(name='x1', lb=0)
x2 = m.addVar(name='x2', lb=0)

# Set the objective function
m.setObjective(6*x1 + 7*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(5*x1 + 4*x2 >= 30, name='Iron_Requirement')
m.addConstr(3*x1 + 4*x2 >= 40, name='Zinc_Requirement')

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution found: x1 = {x1.x}, x2 = {x2.x}")
    print(f"Total cost: {6*x1.x + 7*x2.x}")
else:
    print("No optimal solution found")
```