v (mm/sec) 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Estimate from the M-M plot below: Vmax, Km, Keat, and enzyme efficiency 1.0 [ET] =2.0 mM 2.0 3.0 [S](MM) 4.0 5.0 6.0 7.0
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![v (mm/sec)
5.0
4.0
3.0
2.0
1.0
0.0
0.0
1.0
0
Estimate from the M-M plot below:
Vmax, Km, Keat, and enzyme efficiency
2.0
[ET] =2.0 mM
3.0
[S](MM)
SmartArt
4.0 5.0
6.0
7.0
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- Use the Michaelis-Menten equation to complete the enzyme kinetic data set, whenKm is known to have a value of 1 mmol L−1 [S] (mmol L−1) Vo (μmol L−1 min−1) 0.5 50 1.0 2.0 3.0 10.0calculate the reaction velocity at saturating substrate concentrations. Your numerical answer is assumed to be in units of M sec-1. [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMThe Km value of an enzyme-catalyzed reaction and its Vmax is 70 mmol/min. What is the rate of the reaction of the reaction when the substrate concentration is 7 × 10−2 mmol/min? a. 35 mmol/min b. 50 mmol/min c. 60 mmol/min d. 70 mmol/min
- An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the initial velocity (V0) at a substrate concentration of 15.2 uM? Report your answer to three significant figures in units of uM/s.Given the following set of kinetic data, what is the KM of the enzyme? Numerical answer, only (assuming the same units as seen in the table). [S] (micro M) vo (Micro mol/sec) 1 49 2 96 8 249 50 498 100 656 1000 778 5000 999If 10 µM enzyme was used to obtain the data in the summary plot below, k-cat = ________________ sec-1 and efficiency = _________________ sec-1 M-1 for this enzyme. (Enter numeric answers to the nearest integer; Do NOT write unts.)
- An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the kcat value? Report your answer to three significant figures in units of 1/s.What percentage of max is obtained when the substrate is present at 80% of the Km? Use two digits in your answer.An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the kcat value? Report your answer to three significant figures in units of 1/s. Do NOT include the units in your answer.
- The Lineweaver-Burke plots of a reaction without inhibitor and one with non-competitive inhibitor will have the same 1. Vmax2. Km3. Km/Vmax4. 1/VmaxAn enzyme has a KM = 10 mM and Vmax = 100 mmol/min. Identify the substrate concentration (in mM) in which the velocity will near Vmax when there is a 10-fold decrease in KM.Calculate for the water activity of the soy sauce in 80.3% ERH. Suppose the soy sauce is subjected to a storage area with RH of 95.9%. What will be the water activity of the soy sauce?