Michaelis Menten Equation and Estimation Of Vmax and Tmax

1. PRESENTED BY:
Deepak A. Thakre
M.Pharm I Year
Industrial Pharmacy
Department
Shri Sadashivrao Patil Shikshan Sanstha's
SMT. KISHORITAI BHOYAR COLLEGE OF PHARMACY,
KAMPTEE
GUIDED BY :
Dr. Jayshree B. Taksande
HOD Pharmaceutics
Department
TOPIC : 1. The Michaelis-Menten equation &
2. Estimating Km and Vmax
1

2.  If metabolism is the only pathway of elimination, the rate of metabolism or elimination is defined by
the Michaelis-Menten equation.
 The (Non- linear) kinetics of capacity-limited or saturable processes is best described by Michaelis-Menten
equation:
−
𝒅𝒄
𝒅𝒕
=
𝑽𝒎𝒂𝒙𝑪
𝐊𝐦+𝐂
------- (1)
Where,
–dc/dt = rate of decline of drug concentration with time,
Vmax = theoretical maximum rate of the process,
Km = Michaelis constant,
C= drug Concentration .
 Three situations can now be considered depending upon the values of Km and C:
A] When Km = C
 Under this situation, the equation (1) reduces to i.e. −
𝑑𝑐
𝑑𝑡
=
𝑉𝑚𝑎𝑥
2
-------(2)
i.e. the rate of process is equal to one-half its maximum rate.
2

3. 3
A plot of Michaelis-Menten equation (elimination rate dC/dt versus concentration C). Initially, the rate
increases linearly (first-order) with concentration, becomes mixed-order at higher concentration and then
reaches maximum (Vmax) beyond which it proceeds at a constant rate (zero-order).

4. Michaelis-Menten equation is generally used to explain the kinetics of in-vitro, few
enzyme catalyzed in-vivo and in-situ processes.
4

5. 5
B] Km>> C
When the concentrations are low, i.e. Km> C, then Km +C is approximately
equal to Km,
−
𝒅𝒄
𝒅𝒕
=
𝐕𝐦𝐚𝐱𝐂
𝐊𝐦
------------ equation .(3)
The above equation is identical to the one that describes first-order elimination
of a drug where Vmax/Km = KE. This means that the drug concentration in the
body that results from usual dosage regimens of most drugs is well below the
Km of the elimination process with certain exceptions such as phenytoin and
alcohol.
Because both Vmax and Km are constants, the metabolism rate is proportional
to the drug concentration and is constant (i.e., first-order process).

6. C] C>> Km
When the concentrations are high, i.e. C > Km, then
𝒅𝒄
𝒅𝒕
=
𝑽𝒎𝒂𝒙𝑪
𝑪
= 𝑽𝒎𝒂𝒙 ------- equation (4)
Equation 4 gives the zero order kinetics.
Therefore, it was concluded that at high plasma concentrations, first order kinetics were not seen.
The above equation is identical to the one that describes a zero-order process i.e. the
rate process occurs at a constant rate Vmax and is independent of drug concentration.
 When given in therapeutic doses, most drugs produce plasma drug concentrations well below
the Km for all carrier mediated enzyme systems affecting the pharmacokinetics of the drug.
 Hence most of the drugs at normal therapeutic concentrations follow 1st order rate processes.
Some of the drugs like phenytoin and salicylate saturate the hepatic mixed function oxidases at
higher therapeutic doses.
 With these drugs, elimination kinetics is 1st order at low doses and mixed at high doses and
shows zero-order at very high therapeutic doses.
6

7. If a single IV bolus injection of drug D0 is given at t=0, the drug concentration, Ct in the
plasma at any time may be calculated by integrated form of Eq 1 is given by
𝑪𝒐−𝑪𝒕
𝒕
= 𝑽𝒎𝒂𝒙 −
𝑲𝒎𝑪
𝒕
𝑰𝒏
𝑪𝒐
𝒕
----equation (5)
Where C0 is the concentration at time t=0.
Alternatively the amount of the drug in the body after an IV bolus injection may be
calculated by the following relationship.
𝑫𝒐−𝑫𝒕
𝒕
= 𝑽𝒎𝒂𝒙 −
𝑲𝒎𝑪
𝒕
𝑰𝒏
𝑫𝒐
𝒕
------equation (6)
Where Do is the amount of the drug in the body at t=0.
…Equation (5)
…Equation (6)
Equation 6 can be used to study the decline of the drug in the body after the administration
of different therapeutic doses. Here, the Km and Vmax of the drug are unknown.
7

8. By rearranging the above Equation 6, time to decline a certain amount of the dose of a drug
can
be calculated by the following equation
𝑡 =
1
𝑉𝑚𝑎𝑥
𝐷𝑜 − 𝐷𝑡 + 𝐾𝑚 𝐼𝑛
𝐷𝑜
𝐷𝑡
-----equation (7)
Equation 7 explains an inverse relationship between the time for the dose to decline to a
certain amount of the drug in the body and Vmax.
Actually, Km can be said as the hybrid constant in enzyme kinetics that may represents both
forward as well as backward reaction.
It is equivalent to the concentration of the drug in the body at ½ Vmax.
The one compartment open model having capacity limited elimination pharmacokinetics
effectively explains the plasma drug concentration time profiles for a number of drugs.
8

9. 9
The parameters of capacity-limited processes like metabolism, renal tubular secretion
and biliary excretion can be easily defined by assuming one-compartment kinetics for
the drug and that elimination involves only a single capacity-limited process. The
parameters Km and Vmax can be assessed from the plasma concentration-time data
collected after i.v. bolus administration of a drug with nonlinear elimination
characteristics.
Rewriting equation
−
𝒅𝒄
𝒅𝒕
=
𝑽𝒎𝒂𝒙𝑪
𝐊𝐦+𝐂
---------equation (1)
Integration of above equation followed by conversion to log base 10 yields:
𝒍𝒐𝒈𝑪 = 𝒍𝒐𝐠𝐂𝐨 +
𝐂𝐨−𝐂
𝟐.𝟑𝟎𝟑𝐊𝐦
−
𝐕𝐦𝐚𝐱
𝟐.𝟑𝟎𝟑𝐊𝐦
-------(2)
Estimating Km and Vmax

10. 10
A semilog plot of C versus t yields a curve with a terminal linear portion having
slope –Vmax/2.303Km and when back extrapolated to time zero gives Y-
intercept log Co .The equation that describes this line is:
Fig: Semi-log plot of a drug given as i.v.
bolus with nonlinear elimination and
that fits one-compartment kinetics.
----equation (3)

11. 11
From equation 2 and 3 ( at low Plasma Concentration)
𝐿𝑜𝑔𝐶𝑜 = 𝑙𝑜𝑔𝐶𝑜 +
𝐶𝑜−𝐶
2.303𝐾𝑚
𝑙𝑜𝑔𝐶𝑜 − 𝑙𝑜𝑔𝐶𝑜 =
𝐶𝑜−𝐶
2.203𝐾𝑚
OR
𝑙𝑜𝑔
𝐶𝑜
𝐶𝑜
=
𝐶𝑜−𝐶
2.303𝐾𝑚
For this equation Km can be obtained.
Vmax can be Computed by Substituting the value of Km in the Slope value.

12. Estimating Km and Vmax
Method : Equation 1 gives the relationship of the drug biotransformation to the
concentration of the drug in the body. When an experiment is performed by
using different concentrations of the drug C, a series of reaction rates (V) may
be calculated for each concentration. Km and Vmax can then be determined by
using the special plots.
Hence,
V=
𝑽𝒎𝒂𝒙𝑪
𝐊𝐦+𝐂
------ equation (8)
Rearranging the above equation 8, [As Per Equation of Slope- 𝒚 = 𝒎𝒙 + 𝒄]
1
𝑉
=
𝐾𝑚
𝑉𝑚𝑎𝑥
1
𝐶
+
1
𝑉𝑚𝑎𝑥
-----equation (9)
Above Equation 9 gives the linear equation, when 1/V is plotted against 1/C,
the resultant intercept for the line is 1/Vmax and the slope is Km/Vmax
Fig: Plot of 1/V against 1/C for the determination of Km and Vmax
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