## **Core Concept**
The problem involves understanding the kinetics of glucose transport via a specific transporter, characterized by its Michaelis constant (Km) and maximum velocity (Vmax). The Km value indicates the glucose concentration at which the transport rate is half of Vmax, reflecting the transporter's affinity for glucose. The Vmax value represents the maximum rate of glucose transport when the transporter is fully saturated with glucose.

## **Why the Correct Answer is Right**
To calculate the rate of glucose transport, we use the Michaelis-Menten equation:
[ V = frac{V_{max} cdot [S]}{K_m + [S]} ]
where:
- (V) is the rate of glucose transport,
- (V_{max} = 36) micromoles glucose/sec/mg of transporter,
- (K_m = 45) mM,
- ([S] = 15) mM (glucose concentration in peripheral blood).

Substituting the given values:
[ V = frac{36 cdot 15}{45 + 15} ]
[ V = frac{540}{60} ]
[ V = 9 ]
So, the rate of glucose transport is 9 micromoles glucose/sec/mg of transporter.

## **Why Each Wrong Option is Incorrect**
- **Option A:** Without calculating, we cannot assume its correctness or incorrectness directly, but let's assume it's not 9.
- **Option B:** Similarly, let's assume it's not the correct calculation result.
- **Option C:** This option suggests a calculation of (frac{36 cdot 15}{45 + 15} = frac{540}{60} = 9), which matches our calculation.
- **Option D:** This would imply a different calculation or incorrect substitution into the Michaelis-Menten equation.

## **Clinical Pearl / High-Yield Fact**
The Michaelis-Menten kinetics is crucial for understanding how transporters and enzymes work in the body, especially in pharmacology and physiology. For glucose transporters, like GLUT4 in muscle and adipose tissue, insulin stimulates the translocation of these transporters to the cell surface, increasing glucose uptake. Remembering that a high Km means lower affinity (more substrate needed to achieve half Vmax) and a high Vmax means a higher maximum transport rate can help in quickly assessing how changes in substrate concentration affect transport rates.

## **Correct Answer:** C. 9