3. Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was...
Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, PNPP, had been converted to PNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1 st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5...
3. Acid Phosphatase was assayed at [S] =2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, PNPP, had been converted to pNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5 minutes?...
1. An enzyme with a Km of 1x10 % M was assayed using an initial substrate concentration of 3x10-- M. After 2 min, 5 percent of the substrate was converted. How much substrate will be converted after 10 min. 30 min. 60 min? How long must the reaction be run to achieve 99% conversion? (Assume that the enzyme follows Michaelis-Menten kinetics.)
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22 micromoles X liters^-1 Xmin^-1). What velocity would be observed in the presence of 2X10^-4 M substrate and 5X10^-4M of a. a competitive inhibitor b. a noncompetitive inhibitor c. an uncompetitive inhibiter Ki in all three cases is 3X10^-4M. What is the degree of inhibition in all three cases?
Do you mind checking my work please? I feel like there's
something wrong with question 2 and 3 but I can't quite put my
finger on it. If it is wrong, could you please explain it in
detail? Also, need help filling out the table for the mass of the
enzyme. Thank you for your help!
1. Math/Calculation: What is the concentration of substrate in the reaction cocktail? C 50 mM pNPP 50mM -0.05 moVL pNPP Vi-500 μ1-0.0005 L Reaction...
3. Why is an allosteric enzyme more sensitive to substrate concentration around Km values than a Michaelis-Menten enzyme with the same Vmax? 4. Explain how pH and temperature influence enzyme activity. ( A Lineweaver-Burk (double reciprocal) plot was used to compare the effects of three different reversible inhibitors (A, B and C) on an enzyme. The plot of 1/V vs 1/[S] for the enzyme with no inhibitor is shown in a solid black line. The plot of 1/V vs 1/[S]...
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
The results for enzyme activity analysis are presented in Table 5 (below). In this case, do not use the graph to determine: (1.5 pts each) (Hint- use M-M equation) a) Vmax; b) Km; c) initial velocity at [S] 1 x 10-1 M; d) the amount of product formed during the first 5 minutes at [S] 2x10 3M; at a [S of 2x10 6M? e) what is Km and Vmax if the free [E is increased 3-fold? 3. v (umol/min 0.25...
3. Use the below tables to complete lab calculations on the worksheet on pages 2-6 of this document. You will submit the worksheet via dropbox on Canvas by you assigned lab time the week of March 30th through April 3rd. A document with sample calculations of different concentrations is provided to you on canvas. Enzyme Kinetics Lab Buffer volume (mL) Enzyme Volume (ml) Substrate Volume (ml) TOTAL VOLUME (mL) 0.04 0.5 1.5 0.04 0.25 1.5 0.04 0.1 1.5 0.04 0.05...