Does increasing enzyme concentration changes the value of Vmax, Km and Kcat? Why?
Does increasing enzyme concentration changes the value of Vmax, Km and Kcat? Why?
4. What does enzyme kinetics study? What is Vo, km, Vmax, Kcat, respectively? If you plot Vo versus (substrate), or 1/Vo versus 1/[substrate], how the curves would look like, and how to get Vmax and Km values?
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.
Kcat= 30.0sec^-1 Km= 0.0050M Total enzyme Concentration= 1uM What is the max rate of the enzyme?
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?
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
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...
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.
A. Prove that Km equals the substrate concentration at ½ Vmax . (Show solution) B. Why do structural analogs if the transition-state intermediate of an enzyme inhibit the enzyme competitively and with low Ki values (binds tightly)?
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km of 0.0030 M have an initial velocity that is 25% of the maximum velocity? a. 0.0010M b 0.0090 M c. 0.0015 M d. 0.0060 M e.Cannot be determined
Find Km, Vmax, Ki and Kcat values. Find Vmax, Km and Ki values according to the graph 350 y = 51.843x + 99.634 300 y = 45.555x + 93.809 250 y = 33.086x + 88.634 200 No In 150 100 10 pl • 20 ul ........ Linea -50 Linea -100 Benze Linea 1/[S] mM Benze