Simple Method for Vancomycin Dosage Adjustment Based on Serum Creatinine Variation

Abstract

When a patient's serum creatinine changes and we need to know how to adjust the vancomycin dosage based on this new value of serum creatinine, the method described below can give a quick approximation of the needed dosage change.

For example, a patient in our hospital was on vancomycin 1000 mg every 12 hours. His serum creatinine was 1.0 mg/dL on May 25, 2010, with a measured vancomycin serum level of 8.0 mg/L (a rather low level). The next day, his serum creatinine was measured to be 1.4 mg/dL, and it was time to give him his next dose. Approximately what dosage change could be quickly calculated while waiting for the patient's measured vancomycin serum level?

On May 25, 2010, using Bayesian pharmacokinetics (based on measured level of 8.0 mg/L), we obtained patient's specific parameters to be V_d = 57.209 L, k_e = 0.09061193 hours ⁻¹ (corresponding to t_1/2 = 7.648 hours).

We can use the equation

k_{2} = \frac{k_{1} + 0.0044 (r - 1)}{r}

as derived in the note below to calculate the new elimination coefficient.

Let r = (1.4/1.0) = 1.4, and k₁ = 0.09061193 hours ⁻¹, then k₂ = 0.06597995 hours ⁻¹ (corresponding to t_1/2 = 10.503 hours). Thus, the patient's new specific parameters are now approximately V_d = 57.209 L, k_e = 0.06597995 hours ⁻¹, and a new dosage could be quickly calculated based on these values.

Just how good an approximation are these specific parameters? After a measured vancomycin serum level came back (value of 19.100 mg/L at 4:15 P.M. on May 26, 2010), using Bayesian pharmacokinetics, we obtained patient's new specific parameters to be V_d = 56.878 L, k_e = 0.0621301 hours⁻¹ (corresponding to t_1/2 = 11.154 hours), which give a predicting concentration value of 19.099 mg/L; the approximated values above of V_d = 57.209 L, k_e = 0.06597995 hours⁻¹, give a predicting concentration value of 18.339 mg/L (relative error = 3.98%).

Here is another example. Another patient in our hospital was also on vancomycin 1000 mg every 12 hours. Her serum creatinine was 1.40 on May 25, 2010, with a measured vancomycin serum level of 11.90 mg/L obtained at 3:00 A.M. Her vancomycin specific parameters calculated from Bayesian pharmacokinetics were V_d = 67.094 L, k_e = 0.0556358 hours ⁻¹ (corresponding to t_1/2 = 12.456 hours). On May 28, 2010, her serum creatinine was measured at 3:45 A.M. to be 1.10. What are her approximate vancomycin specific parameters? What should her dosage be?

Answer: r = (1.1/1.4) = 0.786, k₁ = 0.05563584 hours ⁻¹, so $k_{2} = \frac{k_{1} + 0.0044 (r - 1)}{r} = 0.06960925$ hours ⁻¹ (corresponding to t_1/2 = 9.956 hours). Thus, a possible dosage could be 750 mg every 8 hours for the patient's trough level to be between 15 and 20 (based on V_d = 67.094 L, k_e = 0.06960925 hours ⁻¹).

At 11:20 A.M., a later time on the same day, May 28, 2010, a vancomycin serum level was measured to be 11.50 mg/L. Using Bayesian pharmacokinetics, this patient's new specific parameters were calculated to be V_d = 66.990 L, k_e = 0.0652726 hours ⁻¹ (corresponding to t_1/2 = 10.617 hours). Thus, a possible dosage could also be 750 mg every 8 hours for the patient's trough level to be between 15 and 20, agreeing with the above approximate dosage. We also note that the approximate values (above) of V_d = 67.094 L, k_e = 0.06960925 hours ⁻¹ give a predicting concentration value of 10.364 mg/L at 11:20 A.M. on May 28, 2010 (relative error = 9.88%).

Note: Vancomycin half-life can be approximated by

t_{1 / 2} = \frac{0.693}{k_{e}}

where

R_e = 0.00083 (CrCl) + 0.0044

with

C r C l = (F e m a l e) \frac{(140 - A g e) W e i g h t}{72 S C r}

where Female = 0.85 if female, Female = 1 if male.

Thus,

t_{1 / 2} = \frac{a (S C r)}{b (W e i g h t) + c (S C r)}

where

a = 0.693

b = 0.00083 (F e m a l e) \frac{(140 - A g e)}{72}

c = 0.0044

ie,

t_{1 / 2} = \frac{a x}{b + c x}

where x is defined to be

x = \frac{S C r}{W e i g h t} .

And so we have

x = \frac{b t_{1 / 2}}{a - c t_{1 / 2}} .

Let

r = \frac{S c r_{2}}{S C r_{1}} = \frac{x_{2}}{x_{1}}

be the ratio between two serum creatinine values, then

x_{2} = r x_{1} = r \frac{b t_{1}}{a - c t_{1}} .

Since

t_{2} = \frac{a x_{2}}{b + c x_{2}},

substitute for x₂ from the equation above, we have

t_{2} = \frac{a x_{2}}{b + c x_{2}} = \frac{a r t_{1}}{a + c (r - 1) t_{1}}

Since

k_{e} = \frac{0.693}{t_{1 / 2}},

we have a particular simple equation in approximating k_e

k_{2} = \frac{k_{2} + 0.0044 (r - 1)}{r}

where $r = \frac{S C r_{2}}{S C r_{1}}$ , k₁ = k_e,1, k₂ = k_e,2.