How is turnover number mathematically related to Vmax and total enzyme concentration?
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How is turnover number mathematically related to Vmax and total enzyme concentration?
Determine the value of the turnover number of the enzyme catalase, given that Rmax (Vmax) for catalase is 41 mmol · L-1.5-1 and [E]t equals 3.6 nmol · L-1. Catalase has a single active site. turnover number: 8-1
help with these please The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number = - -=k2 (Rmax is often written as...
Vmax is a poor metric for comparing enzyme efficiencies because: it is independent of the total enzyme concentration b. it varies as a function of the Km value of the substrate it changes as a result of using competitive inhibitor d. the total ezymes concentration affects the value of Vmax it requires measuring enzyme velocity
Calculate turnover number if Vmax = 0.681 mMmin-1 assuming that 1 x 10^-4 mol of enzyme was used The book says the answer is 20.43, but I got 1.61. I used Vmax/Km (Km=0.421), but evidently that's wrong. Any help would be great! Thank you!
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 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?
Need help with number 13! I already asked about number 12. The inverse velocity and inverse substrate concentration relationship for an enzyme-catalyzed reaction is given below V Vmax Vmax S For the hydration of CO2 catalyzed by carbonic anhydrase, it was determined experimentally that (dm s mol 4023.9+ 39.934 at a total enzyme IS] concentration of 2.32 × 10-y mol-dm- What is the value of the Michaelis constant KM for this enzymatic reaction? (B). 9.92x103 mol dm3 (D). 100.8 mol...
1. The turnover number for an enzyme is known to be 5000 min. From the following set of data, calculate the Km, Vmax and the amount of enzyme present in this experiment. Use excel to obtain the lineweaver burk plot. Substrate concentration (MM) 1 Initial velocity (umol/min) 167 250 334 376 498 499 100 1,000
Does increasing enzyme concentration changes the value of Vmax, Km and Kcat? Why?
The maximum catalytic rate of an enzyme is defined as the turnover number (also called kcat)and is the maximum number of reactions a single active site can carry out per second for a given concentration of enzyme. i.Would a competitive inhibitor reduce the turnover number for an enzyme?Why? ii.Would an irreversible inhibitor reduce the turnover number for an enzyme?Why?