Michaelis-Menten curve is a plot between substrate concentration and velocity of the reaction or enzyme activity.
VM is the maximum velocity at which the reaction can run beyond which, despite of increasing the substrate concentration, the enzyme gets saturated. Km is the substrate concentration at which the velocity of the reaction is half (Vm/2); it cannot be negative).
Competitive inhibition is a reversible inhibition wherein the inhibitor molecule is similar in shape and structure of the substrate allowing it to compete for the active site of the enzyme with the substrate. Hence, the maximum velocity of the enzyme(Vm) remains unchanged (function of enzyme is not interfered with). The substrates binding to the enzyme would change cause a decrease in the slope of the graph, it would need more substrate molecules to reach the same velocity as earlier. So, the Km shifts to a higher value.
please help to draw a Michaelis menten curve, Vmax of 4/sec and Km of -2.5 uM....
The equation that describes the above Michaelis-Menten curve: Vo TS]+K Vmax [S] Michaelis-Menten Equation Lineweaver and Burke manipulated the Michaelis-Menten equation to yield: Ko V I S Vmax [S] Lineweaver-Burke Equation Linewenver Burke Equation If you plot 1/ V. vs. 1/[S], you get the following Lineweaver-Burke plot: 1/V. Slope = km/Vmax Intercept = -1/KM -Intercept = 1/Vmax 1/[S] Which is easier to calculate values for Km and Vmax, using the linear (y=mx+b) Lineweaver-Burke Plot or the Michaelis-Menten curve?
1. Show, using the Michaelis-Menten equation, that when [S] >>> Km, vo = Vmax. Show, using the M-M equation that when [S] <<<Km, vo =[S][Et]kcat/Km. 2. What is Vmax? Provide both a mathematical and written description of Vmax? How can Vmax be experimentally altered? How can we use Vmax to determine the turnover number (kcat) of an enzyme-catalyzed reaction? What is the major challenge of determining Vmax from an Michaelis-Menten plot?
The kinetics of enzyme catalyzed reactions can be described the Michaelis-Menten equation and the Eadie-Hofstee equation as shown below: V0 = (-Km) V0 / [S] + Vmax a). Please derive the Eadie-Hofstee equation starting from the Michaelis-Menten equation. b). The Vmax and Km of the enzyme catalyzed reaction can be derived from a plot of V0 versus V0/[S]. Please draw one of these plots and explain how do you use it to derive Vmax and Km. c). Please draw a...
(I need help with part C, Drawing the expected Michaelis-Menten plot; Do NOT draw the Lineweaver-Burk plot. thanks!) 1. Michaelis-Menten kinetics- use the M-M equation to answer the following: a. An enzyme (5 µM) has a Vmax of 450 mM/min. What is kcat? b. When the substrate concentration is 50 mM, the initial velocity (V0) was measured to be 375 mM/min. Under the conditions described above, calculate the KM. c. Draw the expected Michaelis-Menten plot (label your axes and include...
b. For an enzyme that displays Michaelis-Menten kinetics, what is the initial velocity as a function of Vmax when: a. [S] Km b. S] 0.1 Km c. [S] 50Km c. What will be the initial velocity (yo) for an enzyme that has Km 2.5 [S]? Your answer will be a fraction of Vmax a.
HUUR UUU UUILUUITUL LUIUILINU 2 PV, 10 pul, posta 10. Applying the Michaelis-Menten Equation II An en- zyme catalyzes the reaction M = N. The enzyme is present at a concentration of 1 nm, and the Vmax is 2 um s-1. The Kn for substrate M is 4 um. (a) Calculate kcat. (b) What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2.0?
Please show how to calculate Km and Vmax for no inhibitor/low inhibitor given graph. Show how to solve for a.,a/a etc. Lineweaver Burk #4: No inhibitor, Low inhibitor 0.2 ▲ No inhibitor Low inhibitor 0.15 0.05 0.2 0.15 0.1 0.05 0.05 0.1 0.15 02 5 1/IS] in units of 1/mM Fill in the blanks. Show your work. No inhibitor Kmno Vmax,o- Vmax,w = ๙ Vmax,o Solve for ๙ inhibitor Krmkw =픕Km,o Solve for 픕 Hint treat, as a single number....
4. Basic concepts of Michaelis-Menten kinetics. The Michaelis-Menten equation is expression of the relationship between the initial velocity, Vo, of an enzymatic reaction and substrate concentration, [S]. There are three conditions that are useful for simplifying the Michaelis-Menten equation: [S] <<Km; [S] = Km; [S] >> Km. Match each condition with the statement(s) that describe it. TV, Vmox[S] Vo =Vmax m . V Vo - Vmax [S] Km +[S] V. (um/min) max [S] (mm) (a) Doubling [S] will almost double...
What is the velocity of a Michaelis-Menten enzyme reaction (in terms of vmax) when the concentration of substrate is 4 times the value of KM? Show your work.
Write the equations that describe the Michaelis-Menten and the Lineweaver-Burk double-reciprocal plots. Draw examples of each plot, demonstrating how Km and Vmax can be determined. On the same graphs, draw another plot where the same enzyme-catalyzed reaction is subjected to inhibition by a competitive inhibitor.