(a)
Interpretation:
V0 versus pH curve when the substrate concentration is much greater than the enzyme KM should be drawn.
Concept introduction:
KM is the Michaelis constant which is the substrate concentration when the reaction rate is half of the maximum rate achieved by a system. Michaelis constant is a measure of its binding affinity to a substrate. The higher the Michaelis constant, the lower affinity for its substrate. That means an enzyme with lower KM needs a higher concentration of substrate to achieve Vmax.
(b)
Interpretation:
V0 versus pH curve when the substrate concentration is much less than the enzyme KM should be drawn.
Concept introduction:
KM is the Michaelis constant which is the substrate concentration when the reaction rate is half of the maximum rate achieved by a system. Michaelis constant is a measure of its binding affinity to a substrate. The higher the Michaelis constant, lower affinity for its substrate. That means an enzyme with lower KM needs a higher concentration of substrate to achieve Vmax.
(c)
Interpretation:
The pH at which the velocity will equal half of the maximal velocity attainable under these conditions should be determined.
Concept introduction:
KM is the Michaelis constant which is the substrate concentration when the reaction rate is half of the maximum rate achieved by a system. Michaelis constant is a measure of its binding affinity to a substrate. The higher the Michaelis constant, lower affinity for its substrate. That means for an enzyme with lower KM needs a higher concentration of substrate to achieve Vmax.
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Biochemistry
- pls explainWhy was absorbance measured at 420 nm in the enzyme kinetics experiment? To monitor thea. decrease in catechol concentrationb. increase in quinone concentrationc. decrease in polyphenol oxidase concentrationd. polyphenol oxidase-catechol complex concentrationLineweaver-Burk plot is the most used linearization method of the MichaelisMenten equation, but the main drawback is a/an:a. overemphasis to low substrate concentration and less emphasis to high substrate concentration.b. overemphasis to high substrate concentration and less emphasis to low substrate concentration.c. exaggeration to both ends of the reciprocal of the substrate concentration.d. lack of independent variables to x- and y- coordinatesarrow_forwardCrymotripsin is an enzyme that degrade polypeptides The turnover number (k2) is measured to be k₂= 100- S a) Calculate Vmax for a batch process where the enzyme concentration is 0.01 mmol enzyme/L The enzyme reaction follows the Michaelis-Menten kinetic and the relative reaction rate (v/Vmax) has been measured as a function of urea concentration (S) 1 6 12 18 100 Substrate S (mm) Relative rate V/Vmax (-) 0.167 0.545 0.706 0.782 0.954 b) Calculate Km from the data in the tablearrow_forwardBIOCHEMISTRY QUESTION The initial rate (VO) data as a function of substrate concentration [S] for an enzyme (Enzyme1) that obeys Michaelis-Menten kinetics are shown in the table below. The total enzyme concentration is 2.5 nM. [S](μm) 16.0 8.0 4.0 2.0 1.0 v (μMsec ¹) 48.5 32.7 19.8 11.1 5.88 (a) Calculate KM for the enzyme, using the estimated slope (slope = (y2-y1)/(x2-x1)) from a Lineweaver-Burk plot. Show your work and don't forget the units! (b) When the enzyme concentration is 2.5 nM, a Lineweaver-Burk plot of this data gives a line with a y- intercept of 0.0106 μ M-1 sec. Calculate kcat for the enzyme. (Show your work and don't forget to include units). (c) A second enzyme (Enzyme2) also obeys Michaelis-Menten kinetics. Its kcat and KM are 800 sec-1 and 5 µM, respectively. Which is the more efficient enzyme? Briefly explain d) Is Enzyme 1 diffusion limited? Explain.arrow_forward
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- Activity, Enzyme Kinetics Biol 250, Spring 2022 The initial rate for an enzyme-catalyzed reaction has been determined at a number of substrate concentrations. Data are as follows: [S] (umol/L) V [(µmol/L) min1] 5 22 10 39 20 65 50 102 100 120 200 135 (a) Estimate Vmax and KM from a direct graph of v versus [S]. Do you find difficulties in getting clear answers? (b) Now use a Lineweaver-Burk plot to analyze the same data. Does this work better? (c) Finally, try an Eadie-Hofstee plot of the same data. (d) If the total enzyme concentration was 1 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? (e) Calculate kcat/KM for the enzyme reaction. Is this a fairly efficient enzyme?arrow_forwardSample pH vs Initial Velocity (AA450/60 seconds) a bo16 0.0014 0.0012 o.001 0.0008 0.0006 0.0004 0 0002 pH Sample what does it mean? Does it agree with what we know about enzyme kinetics and why or why nol? If there were anomalies, what were they and what are some possible reasons for this occurring? Was an optimal pH identifiable? Why or why not? Initial Velocity (AA450/60seconds)arrow_forwardReaction rate 0.35 0.30 0.25 0.20 0.15- 0.10 0.05 0.00 0 1000 2000 Enzyme total is 5mM 3000 4000 1. Substrate concentration 1. The above graph is an enzyme reaction, what is the vMax, KM, and Kcat of this enzymatic reaction? What does vMax, kM and Kcat mean, explain? 2. What are the three forms of regulation that follow the Michaelis Menten kinetics? What are the mechanisms by which these inhibitors can regulate the enzyme?arrow_forward
- . The steady-state kinetics of an enzyme are studied in the absence and pres- ence of an inhibitor (inhibitor A). The initial rate is given as a function of substrate concentration in the following table: v[(mmol/L)min '] [S] (mmol/L.) No inhibitor Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38 (a) What kind of inhibition (competitive, uncompetitive, or mixed) is involved? (b) Determine Vmax and KM in the absence and presence of inhibitor.arrow_forwardnot true about the Michaelis-Menten equation? The equation that gives the rate, v, of an the substrate concentration [S] is the Michaelis-Menten equation = Vmax[S]/(Km + [S]), where V, enzyme-catalyzed reaction for all values of max and Km are constants. Which of the following is a) for [S] << Km, V = Vmax applies to most enzymes, but allosteric enzymes have different kinetics when [S] = Km, then v = Vmax/2 gives the rate when the enzyme concentration, temperature, pH, and ionic strength are constant for very high values of [S], v approaches Vmax e) Which is correct about the constant Km in the Michaelis-Menten equation? also called the catalytic constant or turnover number equal to the number of product molecules produced per unit time when the enzyme is saturated with substrate it is the constant in the first order rate equation v = k[A] it is the constant in the second order rate equation v = equal to the substrate concentration at which the velocity or rate of a reaction is ½ the…arrow_forwardAn enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-arrow_forward
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