Refer to the below Enzyme activity data (Table 4) to determine the following: (1.5 pts each) a) V...
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...
Q2) Using the data in the below table, which were obtained for enzyme activity without presence of inhibitors. [15 points) IS (MM) 1.3 2.6 6.5 13.0 26.0 VomM/s) 2.50 4.00 6.30 7.60 9.00 A) Draw Lineweaver-Burk plot and determine the K. and Vmax for this uninhibited enzyme. (You may use hand, excel or other program to draw the graph and find the equation. Then copy and paste the graph here, you need to show axis labels) [6 points) B) From...
For each of the following enzyme regulation mechanisms/situations below, list whether it will: A. Always decrease activity V B. Always increase activity V C. May increase or decrease activity V D. May decrease or possibly have minimal effect on activity V E. Will definitely have no significant effect on activity V 1. Competitive Inhibitor with [I] > Ki and [S] < Km 2. Noncompetitive inhibitor with [I] > Ki and [S] < Km 3. Phosphorylation of Enzyme 4. Product Inhibition...
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 ?
4. The following data were obtained from an enzyme kinetics experiment. Graph the data using a Lineweaver-Burk plot and determine, by inspection of the graph, the values for Km and Vmax. ISI (M) V (nmol/min) 0.20 0.26 0.33 1.00 1.43 1.67 2.08 3.33 5. You measured the kinetics of an enzyme activity as a function of substrate concentration (see Table). The enzyme concentration was maintained constant at a level of 1 M. [S] AM Vopmol/min 2.9 3.8 4.4 Plot the...
Q1: A marine microorganism contains an enzyme that hydrolyzes glucose-6- sulfate (S). The assay is based on the rate of glucose formation. The enzyme in a cell-free extract has kinetic constants of km = 6.7 x 10-4, Vmax = 300 nm/L/min in presence of 10-5M competitive inhibitor (Galactose-6-sulfate) and 2 x 10-5M substrate (Glucose-6-sulfate), velocity was 1.5 nmoles /L/min. a) Calculate Ki for Galactose-6-sulfate b) Calculate velocity in absence of the inhibitor
please help!! please please
"Looking at table 6-8 on Enzyme Reactions from the lecture,
which achieves an excellent specificity constant with a very
favorable kcat (insert answer here) , which with a
favorable Km (insert answer here) , and which with
moderate values for each (insert answer here) ?
(pick from acetyl cholinesterase, B-lactamase, and catalase)"
Question 2: "The units of Vmax are (insert answer
here) and the units of Km are (insert answer
here) (pick from M, M-1, M/s, s/M...
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. 5 4.5 4 a) What is the KM? KM: v. (mM/s) 3.5 3 2.5 2 1.5 1 0.5 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the inhibitor (write them below) and draw an appropriate curve...
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.(a)At what inhibitor concentration will 75% of the enzyme be bound to the inhibitor if there is no substrate present? (b) This enzyme has a Km of 4.0 × 10-5 M and a Vmax of 50 μM/s. At a substrate concentration of 3.0 × 10-4 M, calculate (i) the velocity of reaction in the presence of the inhibitor at 4.8 x 10-5 M (ii) the degree of...