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Sagot :
To solve this problem, we'll use the Michaelis-Menten equation, which describes the relationship between dosing rate and steady-state concentration for drugs that follow a Michaelis-Menten kinetics. The equation is given by:
[tex]\[ DR = \frac{V_{\max} \cdot C_{ss}}{K_m + C_{ss}} \][/tex]
where:
- [tex]\(DR\)[/tex] is the dosing rate
- [tex]\(V_{\max}\)[/tex] is the maximum rate of drug metabolism
- [tex]\(K_m\)[/tex] is the Michaelis constant
- [tex]\(C_{ss}\)[/tex] is the steady-state concentration of the drug in the blood
We are provided with two sets of dosing rates and their corresponding steady-state concentrations:
1. Dosing rate ([tex]\(DR_1\)[/tex]) = 600 mg/day, Steady-state concentration ([tex]\(C_{ss1}\)[/tex]) = 9.8 mg/L
2. Dosing rate ([tex]\(DR_2\)[/tex]) = 1200 mg/day, Steady-state concentration ([tex]\(C_{ss2}\)[/tex]) = 28.6 mg/L
First, we need to determine [tex]\(K_m\)[/tex] and [tex]\(V_{\max}\)[/tex] from these two equations. Using the Michaelis-Menten equation for both sets of data, we set up two equations:
[tex]\[ 600 = \frac{V_{\max} \cdot 9.8}{K_m + 9.8} \][/tex]
[tex]\[ 1200 = \frac{V_{\max} \cdot 28.6}{K_m + 28.6} \][/tex]
By solving these two equations simultaneously, we determine the values of [tex]\(K_m\)[/tex] and [tex]\(V_{\max}\)[/tex]. The values obtained are:
[tex]\[ K_m \approx 31.14 \text{ mg/L} \][/tex]
[tex]\[ V_{\max} \approx 2507 \text{ mg/day} \][/tex]
Next, we need to determine the dosing rate ([tex]\(DR\)[/tex]) that will achieve a target steady-state concentration ([tex]\(C_{ss}\)[/tex]) of 15 mg/L. We use the Michaelis-Menten equation again with these newly found constants:
[tex]\[ DR = \frac{V_{\max} \cdot C_{ss}}{K_m + C_{ss}} \][/tex]
Substitute [tex]\(V_{\max} \approx 2507 \text{ mg/day}\)[/tex], [tex]\(K_m \approx 31.14 \text{ mg/L}\)[/tex], and [tex]\(C_{ss} = 15 \text{ mg/L}\)[/tex]:
[tex]\[ DR = \frac{2507 \cdot 15}{31.14 + 15} \][/tex]
[tex]\[ DR \approx \frac{37605}{46.14} \][/tex]
[tex]\[ DR \approx 815 \text{ mg/day} \][/tex]
Thus, to achieve a steady-state concentration of 15 mg/L, the dosing rate should be approximately 815 mg/day.
In summary:
- [tex]\(K_m \approx 31.14 \text{ mg/L}\)[/tex]
- [tex]\(V_{\max} \approx 2507 \text{ mg/day}\)[/tex]
- The dosing rate to achieve a steady-state concentration of 15 mg/L is approximately 815 mg/day.
[tex]\[ DR = \frac{V_{\max} \cdot C_{ss}}{K_m + C_{ss}} \][/tex]
where:
- [tex]\(DR\)[/tex] is the dosing rate
- [tex]\(V_{\max}\)[/tex] is the maximum rate of drug metabolism
- [tex]\(K_m\)[/tex] is the Michaelis constant
- [tex]\(C_{ss}\)[/tex] is the steady-state concentration of the drug in the blood
We are provided with two sets of dosing rates and their corresponding steady-state concentrations:
1. Dosing rate ([tex]\(DR_1\)[/tex]) = 600 mg/day, Steady-state concentration ([tex]\(C_{ss1}\)[/tex]) = 9.8 mg/L
2. Dosing rate ([tex]\(DR_2\)[/tex]) = 1200 mg/day, Steady-state concentration ([tex]\(C_{ss2}\)[/tex]) = 28.6 mg/L
First, we need to determine [tex]\(K_m\)[/tex] and [tex]\(V_{\max}\)[/tex] from these two equations. Using the Michaelis-Menten equation for both sets of data, we set up two equations:
[tex]\[ 600 = \frac{V_{\max} \cdot 9.8}{K_m + 9.8} \][/tex]
[tex]\[ 1200 = \frac{V_{\max} \cdot 28.6}{K_m + 28.6} \][/tex]
By solving these two equations simultaneously, we determine the values of [tex]\(K_m\)[/tex] and [tex]\(V_{\max}\)[/tex]. The values obtained are:
[tex]\[ K_m \approx 31.14 \text{ mg/L} \][/tex]
[tex]\[ V_{\max} \approx 2507 \text{ mg/day} \][/tex]
Next, we need to determine the dosing rate ([tex]\(DR\)[/tex]) that will achieve a target steady-state concentration ([tex]\(C_{ss}\)[/tex]) of 15 mg/L. We use the Michaelis-Menten equation again with these newly found constants:
[tex]\[ DR = \frac{V_{\max} \cdot C_{ss}}{K_m + C_{ss}} \][/tex]
Substitute [tex]\(V_{\max} \approx 2507 \text{ mg/day}\)[/tex], [tex]\(K_m \approx 31.14 \text{ mg/L}\)[/tex], and [tex]\(C_{ss} = 15 \text{ mg/L}\)[/tex]:
[tex]\[ DR = \frac{2507 \cdot 15}{31.14 + 15} \][/tex]
[tex]\[ DR \approx \frac{37605}{46.14} \][/tex]
[tex]\[ DR \approx 815 \text{ mg/day} \][/tex]
Thus, to achieve a steady-state concentration of 15 mg/L, the dosing rate should be approximately 815 mg/day.
In summary:
- [tex]\(K_m \approx 31.14 \text{ mg/L}\)[/tex]
- [tex]\(V_{\max} \approx 2507 \text{ mg/day}\)[/tex]
- The dosing rate to achieve a steady-state concentration of 15 mg/L is approximately 815 mg/day.
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