In a certain reaction, a Kcat of 60 min-1 was observed, with a Vmax of 10 mMs-1. What is [E]?
In a certain reaction, a Kcat of 60 min-1 was observed, with a Vmax of 10...
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.
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
)* (Vmax). 4. When [S] = KM, Vo=( A) 0.5 B) KM C) 0.75 D) kcat E) [S] 5. The overall transformation shown in the following reaction: E s. Es p + E For the reaction, the steady state assumption A) ES breakdown occurs at the same rate as ES formation B) [P]>>[E] C) implies that ki-k-1 D) implies that k., and k2 are such that the [ES] = k1[ES] E) [S] = [P]
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...
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?
Estimation of Vmax and K by Inspection Although graphical methods are available for accurate determination of the Vmax and K of an enzyme-catalyzed reaction (see Box 6-1), sometimes these quantities can be quickly estimated by inspecting values of Vat increasing (Sl. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained. V (pm/min) V (m/min) 112 [S] (M) 2.5 x 10 4.0 x 100 1 x 10-5 2 x 10-5 [S(M) 4 x 10-5...
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?
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2
Kcat =30.0 s-1 Km= 0.005 M Operating at 1/4 Vmax What is [S] ? The solutions manuel doesn't explain the problem well. Whete does the 0.33 km come from? For part 2 : Plug in [S]= 1/2 Km, 2 Km, and 10 Km Where does the 1.5 Km come from? Answer (a) Here we want to find the value of [S] when Vo = 0.25 V max. The Michaelis-Menten equation is V = Vmax[S]/(Km + [SD so V = Vmax...
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?