The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V for anenzyme-catalyzed, single-substrate reaction E + SES →E + P. The model can be more readily understood whencomparing three conditions: [S] << Km. [S] = Km, and [S] >> Km.Match each statement with the condition that it describes.Note that "rate" refers to initial velocity V, where steady state conditions are assumed. [Etotal] refers to the total enzymeconcentration and [Efree] refers to the concentration of free enzyme.[S] << KmIncreasing [Ewall will increase Km.Almost all active sites are empty.[S] = KmAnswer Bank[Enel is about equal to [Exotal].[ES] is much higher than [Efree][S] >> KmReaction rate is independent of [S].Not true for any of theseconditionsThe rate is half of the maximum rate.

Enzymes are protein molecules that increase the rate of reaction by decreasing the activation…

Answered: The Michaelis-Menten equation models…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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At what substrate concentration would an enzyme with a kcat of 33.0 s-1 and a Km of 0.0046 M operate at one-tenth of its maximum rate? Assume the enzyme obeys Michaeli s-Menten kinetics.
An enzyme with an unknown reaction mechanism has been isolated and is being investigated in the lab. Based on the information given below, propose one or more specific catalytic strategies (covalent, acid-base, and/or metal ion catalysis) used by the enzyme and predict the role of each catalytic residue in the enzyme's reaction cycle. Explain your reasoning. (a) Mapping of the active site with substrate analog showed that it contains a critical Ser residue. (b) Studies conducted in buffers of varying pH showed that the enzyme is sensitive to pH changes and has two catalytically important residues with pKa values of ~10.5 and ~4.0. (c) Substitution of the Ser residue with Ala resulted in a five-fold increase in the KM but had no effect on the kcat. (d) The reaction rate was insensitive to the addition of metal chelators such as EDTA and EGTA.
(i) Which graph indicates an enzymatic reaction without inhibitor?(ii) Which type of inhibitor is it? Briefly explain.(iii) Which graph indicates the highest concentration of inhibitor?(iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers.
Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places
For an enzyme-catalyzed reaction, the velocity was determined at two different concentrations of the substrate. Estimate the value of Vmax. [S] (MM) 10 20 88 nmol/min 79 nmol/min 67 nmol/min V(nmol/min) 51 nmol/min 27 Not enough information is given to form a reasonable estimation. 48
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/T
Using the appropriate graph and table above, explain what the R48C mutation appears to be doing to the enzyme’s function. Discuss the kinetic parameter changes and their meaning in this context, not the structure of the enzyme, which was not given to you.
The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.
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.
a particular enzyme catalyzes a single reactant S to a single product P, following michaelis-menten kinetics rp=(VmaxCs) / (Km + Cs) 1. A reaction with this enzyme is carried out at very low substrate concentrations. Draw and label a curve on the plot that describes the reaction kinetics under those conditions.
The typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURK
The Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo = Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches/Vm Vmax 2α' Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. = Apparent, or observed, Km is equivalent to the [S] at which Vo max. apparent Km =

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