Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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When substrate [S] = Km, the velocity of an enzyme catalyzed reaction is about: *
0.1 Vmax.
0.2 Vmax.
0.5 Vmax.
0.9 Vmax
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- Two enzyme catalyzed reactions, RA (red) and Rp (blue), have kinetic profiles shown below. Report which reaction has the higher maximal rate and which reaction has the smaller Km. в A 1/ [S] Higher Vm [ Select ] Lower Km [ Select ] A/Tarrow_forwardFor an enzyme with a substrate Km of 0.001 M. a) What substrate concentration for an initial rate equal to one-half maximaly velocity? and b) for an initial rate equal to 0.9 maximal velocity> * a) [S} = 0.001 M, b) [S] = 0.009 M a) [S} = Km, b) [S] = 0.09 M a) [S] = 0.01 M, b) [S] = 0.9 M a) [S] = 1 M, b) [S] = 9 Marrow_forwardkcat is: a measure of the catalytic efficiency of the enzyme the rate constant for the reaction ES → E + P the rate constant for the reaction ES → E + S the [S] that half saturates the enzyme ½ Vmaxarrow_forward
- Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the kcat value for this reaction. Report your answer to three significant figures in units of 1/sec.arrow_forwardAt what substrate concentration would an enzyme with a Km of 0.005M operate at one quarter of its Vmax? What would be the Km for an enzyme if it is operating at 90% of its Vmax when the substrate concentration is 0.01Marrow_forwardWhich of the following statements are true about the relationships of [S], KM, and Vmax? (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.arrow_forward
- In Lineweaver Burk plot: O a. The slope of the plot is the Vmax/Km O b. The intercept on X-axis is 1/Km. O c. The intercept on y-axis is 1/Vmax O d. double-reciprocal plot of enzyme kinetics is generated, where 1/VO is plotted as a function [S]. O e. All answers are correctarrow_forwardThe KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →arrow_forwardFor an enzyme that has a Km of 25 mM, Vmax of 50 mM/s and kcat of 250 s-1, how long does a single reaction take? Select one: 50 ms 250 ms 4 ms 25 s 4 sarrow_forward
- Consider the following free energy diagram for an uncatalyzed and enzyme-catalyzed reaction. Select all the statements that are true. Without enzyme With enzyme A+B Time AB Oa. The reaction is now spontaneous due to the addition of enzyme b. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O C. The reaction is exergonic O d. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction e. The enzyme stabilizes the transition state for the reaction Released Energy pesarrow_forwardConsider the following free energy diagram for an uncatalyzed and enzyme-catalyzed reaction. Select all the statements that are true. Without enzyme With enzyme A+B Time AB O a. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O b. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction O c. The enzyme stabilizes the transition state for the reaction Od. The reaction is exergonic е. The reaction is now spontaneous due to the addition of enzyme Released Energyarrow_forwardThe following data have been obtained for two different initial enzyme concentrations for an enzyme-catalyzed reaction. [E]=0.015 g/l) (g/l-min) [S] (g/l) v([E]=0.00875 g/l) (g/l-min) 1.14 20.0 0.67 0.87 10.0 0.51 0.70 6.7 0.41 0.59 5.0 0.34 0.50 4.0 0.29 0.44 3.3 0.39 2.9 0.35 2.5 a. Find K b. Find V for [E]=0.015 g/l. c. Find V for [E]=0.00875 g/l. d. Find k2. Use Eddie-Hostee plot to find your answer. Compare your result using Hanes- Wolf plot.arrow_forward
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