5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration...
7. a) In an enzyme catalyzed reaction which follows the Michaelis-Menten kinetics. The substrate concentration (Km, Michaelis constant) needed to reach 50% of the maximum reaction velocity (Vmax) is 20 μΜ. What substrate concentration is required to obtain at least 75% of the maximum reaction velocity? Show the work to get full points. (5 points) b) You want to load 10 μg of protein in 15 μL into one of the 10% polyacrylamide gel well. The protein needs to be...
An enzyme catalyzed reaction is studied at substrate concentration that is equal to 8Km. What will be the value of V0 in terms of fraction of Vmax?
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
What is the rate of an enzyme catalyzed reaction if the Vmax is 100µmol S→P/min and the Km is 7 mM and the substrate concentration is 11mM? Is the enzyme working at Vmax? What if the substrate concentration is raised to 25mM?
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?
The initial rate, V, of an enzyme catalyzed reaction varies with substrate concentration as follows: 106 x Initial rate, Ms SJ, M 0.020 0.585 0.004 0.495 0.002 0.392 0.001 0.312 0.250 0.00066 Determine Vmax and Km for this reaction
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
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.
At high (saturating) substrate concentrations, the rate of an enzyme-catalyzed reaction approaches Vmax. How close does the reaction rate actually get to Vmax? Determine how high (i.e. how many times Km) the substrate concentration must be for the reaction rate to be: a. 98% Vmax (show your work) (2) b. 99% Vmax (answer only) (1) c. 99.9% Vmax (answer only) (1)