A 45-year-old man with a history of generalized tonic-clonic seizures is being managed in … — USMLE Step 3

A $45$ -year-old man with a history of generalized tonic-clonic seizures is being managed in the neurology clinic with oral phenytoin. His seizure diary indicates incomplete control. His medication history and laboratory results show the following steady-state serum phenytoin concentrations based on his daily maintenance dose:

- At a dose of $300 \text{ mg/day}$ , his steady-state concentration was $5 \text{ mg/L}$ .
- His dose was then increased to $400 \text{ mg/day}$ . After $4$ weeks of adherence, his steady-state concentration was $10 \text{ mg/L}$ .

His seizures are less frequent but still occasionally occur, and he reports no symptoms of toxicity. The neurologist decides to target a steady-state serum concentration of $15 \text{ mg/L}$ to optimize efficacy. Phenytoin exhibits capacity-limited (Michaelis-Menten) elimination kinetics.

Assuming $100\%$ bioavailability and that no salt factor adjustments are needed ( $S = 1.0$ ), what is the required new daily maintenance dose in $\text{mg/day}$ to achieve this exact target steady-state concentration?

Answer: 450 mg/day

Answer

450

The correct answer is

450

. Phenytoin undergoes capacity-limited (Michaelis-Menten) elimination, meaning the enzymes responsible for its metabolism become saturated within the therapeutic range. Consequently, the relationship between the daily maintenance dose (

R

) and the steady-state concentration (

C_{ss}

) is non-linear and is described by the equation

R = \frac{V_{max} \times C_{ss}}{K_m + C_{ss}}

. By setting up a system of two equations using the patient's two known dosing states (

300 \text{ mg/day}

yields

5 \text{ mg/L}

, and

400 \text{ mg/day}

yields

10 \text{ mg/L}

), we calculate the patient's specific Michaelis-Menten constant (

K_m = 5 \text{ mg/L}

) and maximum metabolic rate (

V_{max} = 600 \text{ mg/day}

). Using these calculated parameters, targeting a steady-state concentration of

15 \text{ mg/L}

requires a new dosing rate of

\frac{600 \times 15}{5 + 15} = 450 \text{ mg/day}

Step-by-Step Solution

Set up the Michaelis-Menten equation for the first dosing state.

300 = \frac{V_{max} \times 5}{K_m + 5} \Rightarrow 300K_m + 1500 = 5V_{max} \Rightarrow 60K_m + 300 = V_{max}

To establish the first of two simultaneous equations needed to solve for the pharmacokinetic constants

V_{max}

and

K_m

Set up the Michaelis-Menten equation for the second dosing state.

400 = \frac{V_{max} \times 10}{K_m + 10} \Rightarrow 400K_m + 4000 = 10V_{max} \Rightarrow 40K_m + 400 = V_{max}

To establish the second equation required to solve the system for the two unknown variables.

Solve the system of equations for the Michaelis-Menten constant (

K_m

60K_m + 300 = 40K_m + 400 \Rightarrow 20K_m = 100 \Rightarrow K_m = 5 \text{ mg/L}

By setting the two expressions for

V_{max}

equal to each other, we can isolate and solve for

K_m

, the concentration at which the elimination rate is half of

V_{max}

Calculate the maximum elimination rate (

V_{max}

V_{max} = 60(5) + 300 = 600 \text{ mg/day}

Substituting the value of

K_m

back into either of the rearranged equations yields the patient's individual maximum metabolic capacity for the drug.

Calculate the new maintenance dose required for the target steady-state concentration of

15 \text{ mg/L}

R_{new} = \frac{600 \times 15}{5 + 15} = \frac{9000}{20} = 450 \text{ mg/day}

Using the patient-specific

V_{max}

and

K_m

parameters, the Michaelis-Menten equation predicts the exact dosing rate needed to achieve the new target steady-state concentration.

Key Concept

Phenytoin exhibits capacity-limited (Michaelis-Menten) non-linear pharmacokinetics, requiring the calculation of patient-specific

V_{max}

and

K_m

to accurately adjust maintenance doses.

Alternative Method

The problem can also be solved algebraically using the Ludden linear transformation of the Michaelis-Menten equation:

R = V_{max} - K_m \times \frac{R}{C_{ss}}

. Substituting the given values produces two linear equations:

300 = V_{max} - 5K_m \times (300/5)

and

400 = V_{max} - 10K_m \times (400/10)

, which simplify to

300 = V_{max} - 60K_m

and

400 = V_{max} - 40K_m

. Subtracting the equations yields

100 = 20K_m

, so

K_m = 5 \text{ mg/L}

. Then

V_{max} = 300 + 60(5) = 600 \text{ mg/day}

. To find the new dose

R

, substitute the target

C_{ss} = 15

back into the linear equation:

R = 600 - 5 \times \frac{R}{15}

, solving this gives

R + \frac{R}{3} = 600 \Rightarrow \frac{4R}{3} = 600 \Rightarrow R = 450 \text{ mg/day}

Estimated Time:3m 0s

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Step-by-Step Solution

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