Answered

IDNLearn.com makes it easy to find answers and share knowledge with others. Get comprehensive and trustworthy answers to all your questions from our knowledgeable community members.

Theophylline was administered to a patient at dosing rates of [tex]600 \, \text{mg/day}[/tex] and [tex]1200 \, \text{mg/day}[/tex], and the respective steady-state concentrations were found to be [tex]9.8 \, \text{mg/L}[/tex] and [tex]28.6 \, \text{mg/L}[/tex].

Find [tex]K_m[/tex] and [tex]V_{\text{max}}[/tex]. Determine the dosing rate to achieve a [tex]C_{\text{Ss}}[/tex] of [tex]15 \, \text{mg/L}[/tex].

Answer:
[tex]K_m = 31.14 \, \text{mg/L}[/tex], [tex]V_{\text{max}} = 2507 \, \text{mg/day}[/tex], and [tex]DR = 815 \, \text{mg/day}[/tex].

Sagot :

GinnyAnsweridn
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.