The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of K for this enzyme?

When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?

From a kinetics experiment, the Vmax was determined to be 450µM∙min-1.  For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole.  Assume a reaction volume of 700µL.  Calculate the kcat  (in sec-1) for the enzyme.

How can you find Kcat if you are only given Vmax, Km and [E] ? I tried using Kcat=Vmax/[E] but that didn't work. How do you know if [E] is the same as [E]total, and if it isn't how do you find it from this information: The Vmax for a particular enzyme is 10 nmols/L/s. The Km for its substrate is 5 microM. If the enzyme concentration is 10 nM, what is the kcat? a.80 nmoles/L/s b.8000 nmoles/L/s c.2 nmoles/L/s d.50 nmoles/L/s

The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.

82.16

Only 15-14

The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM

The following question focuses on how the parameters regulating enzyme function might change, and how these might appear graphically on a Michaelis-Menten plot and a Lineweaver-Burke plot. Carbonic anhydrase is an enzyme that will convert CO2 and water into HCO3. CO2 + H20 > H+ + HCO3 There are many different isoforms of this enzyme. (see for instance http://en.wikipedia.org/wiki/Carbonic_anhydrase 1 Assume that one variant has a Km of 10 µM and a different variant has a Km of 100 µM. Draw on the same graph a typical Michaelis-Menton plot showing the alteration in the rate of carbonic anhydrase as the CO2 level is varied for the two different variants of enzyme, assuming the concentration of the enzyme (10 mM) in the test tube is kept constant. Assume that you have equal amounts of the two different variants of carbonic anhydrase in a number of test tubes and that the Vmax for both enzymes are the same. Be sure to label the axes. For the same conditions as above, draw a…

The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)