Recall that this plot is used for Michaelis –Menten equation. It helps to calculate the constants much easier (via graph)
From Michaelis–Menten equation.
V = Vmax*[S] / (Km + [S])
Where
Vmax = max rate velocity (M/s)
[S] = substrate concentration (M)
Km = Michaelis–Menten constant [M]
V = reaction rate [M/s]
Now… mathematical manipulation
Inverse:
1 / [ V = Vmax*[S] / (Km + [S]) ]
1/V = (Km + [S]) / (Vmax*[S])
Separate species
1/V = (Km) / (Vmax*[S]) + [S] / (Vmax*[S])
1/V = Km / Vmax*[S] + 1/ Vmax
Now, note that:
1/V = Km / Vmax*[S] + 1/ Vmax
1/V = y-axis
Km/Vmax = slope
1/[S] = x-axis
1/Vmax = y-intercept
-1/Km = x-intecept ( when y = 0 )
Now… Plot the your DATA:
x-axis = 1/[S]
y-axis = 1/V
from:
y-axis --> 2.9701 = 1/Vmax
Vmax = 0.3366
slope = Km/Vmax = 1229.4
Km/0.3366 = 1229.4
Km = 1229.4*0.3366
Km = 413.81
21. For the L-8 graph below, what is the Km and Vmax for this enzyme? ◆...
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. 5 4.5 4 a) What is the KM? KM: v. (mM/s) 3.5 3 2.5 2 1.5 1 0.5 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the inhibitor (write them below) and draw an appropriate curve...
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 KM: 3mm V. (mM/s) 1.5 1 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the inhibitor (write them below) and draw an appropriate curve...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration was 6 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? Express your answer to three significant figures. kcat kcat = 2.73×104 min−1 Part B Calculate kcat/KM for the enzyme reaction.
How do I calculate the apparent vmax? 16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 M/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 1.5 KM: 3mm ve (MM/s) 1 0.5 0 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the...
3. The Michaelis-Menten Graph also shows the theoretical maximum rate of the enzyme (Vmax), the point where the enzyme is working at its maximum rate (Vmax/2), and amount of substrate needed to bind half of the active sites (Km). Label these points on the graph. Vmax represents: Vm Vmax/2 represents: Reaction velocity v Vmax 2 Km represents: Kim Substrate concentration (5)
You perform a series of enzyme activity assays and then graph the data using a Lineweaver-Burk plot. You determine the X-intercept is at -0.02 mM-1 and the Y-intercept is at 5.0 (mM/sec)-1. Calculate the Vmax and Km for this enzyme. A. Vmax = 0.20 mM/sec; Km = 50.0 mM B. Vmax = 0.20 mM/sec; Km = ‒50.0 mM C. Vmax = 5.0 mM/sec; Km = 0.02 mM D. Vmax = 5.0 mM/sec; Km = ‒0.02 mM
To determine the kinetic characteristics of an enzyme you used 1 nmol/L of enzyme in a series of assays where you measured the rate of reactions as you varied the concentration of substrate in each assay (Table A). Estimate from a Michaelis-Menten plot approximate values for Vmax, KM, Kcat, and the specificity constant for this enzyme and substrate. (The only information that is given is this paragraph and the table below). Table 1: [S] (μM) v (μmol/L/min) 0 0 5...
An enzyme has a Km for substrate of 10 mM and Vmax of 5 mol L-1 sec-1 at a total enzyme concentration of 1 nM. At [S] = 10 mM, kcat is: A) 2500 per M per sec. B) 5000 per M per sec. C) 1250 per M per sec. D) 2500 per sec. E) 5000 per sec.
At an enzyme concentration of 1microM the Vmax is 100microM/min and the Km is 20nm. At an enzyme concentration of 2microM, what would the V0 be at a substrate concentration of 40mM?
- Vmax is a kinetic property associated with an enzyme. Describe what occurs when an enzyme reaches its Vmax. - Km is also another intrinsic property of an enzyme. Practically, what does the Km tell about how an enzyme interacts with its substrate?