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Please see the attached image.
![The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate Vo for an
enzyme-catalyzed, single-substrate reaction E + S = ES → E + P. The model can be more readily understood when
comparing 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 Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme
concentration and [Efree] refers to the concentration of free enzyme.
[S] << Km
[S]:
= Km
[S] >> Km
Not true for any of these
conditions
[ES] is much lower than [Efree].
[Efree] is about equal to [Eotal].
Almost all active sites will
Increasing [Etotal] will increase
be filled.
Adding more S will not increase
The rate is half of the
Km.
the rate.
maximum rate.
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- Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved residues in slightly different positions can change the catalytic rate of reaction.Using the ActiveModel for aldose reductase, describe the structure of the TIM barrel motif and the structure and location of the active site.Based on the definition of kcat, substitute a value that can be measured and yet still represents the value associated with the original concentration of the R. What would the rate or velocity of the reaction be equal to under these circumstances? How can cells increase Vmax? What variable that we could change would directly impact Vmax? Would the value of KM be affected by the ways you determined that Vma,x could be increased? What does this indicate about KM? Thinking about how catalysts work, about the Michaelis-Menten Equation, and the definition of kcat, what specifically does the enzyme change in the reaction mechanism to increase the rate? If an enzyme follows the 2 step mechanism proposed by Michaelis-Menten, what do you know about this enzyme? Be very specific and comprehensive. Please answer very soon will give rating surely
- In 1966, Ferdinand showed that a random-order ternary-complex mechanism for a two-substrate enzyme-catalysed reaction can lead to sigmoidal kinetics being observation in the absence of cooperative binding. Discuss this scenario so that it is clear why a plot of Vo versus [AXo] at constant [Bo] will be sigmoidal.The following reaction sequence consists of two different substrates catalyzed by an enzyme:let's assume he described his reactions.;E + S1: ES1ES1 + S2: ES1S2ES1S2 → P + Ea.Derive the reaction velocity equation with Michaelis-Menten acceptance.b. Derives the rapid equality of S1 substrate concentration, rather than S2 substrate concentrationsimplify for reaction cards where it is higher.Compare and contrast Bound Fraction equation in ligand binding and Michaelis-Menten equation in enzyme kinetics, including their double-reciprocal forms. Discuss what Km is important for and what Vmax (or kcat) is important for? Under what (substrate) conditions is Km more important than Vmax, and under what (substrate) conditions is Vmax more important than Km? Based on the discussions in question 2, explain what type of inhibitors works best under (a) high substrate concentration and (b) low substrate concentration.
- In the scheme for enzymatic catalysed reaction proposed by Michaelis and Menten, the steps involve reversible formation of enzyme-substrate (ES) complex followed by conversion of the complex to the product (P). a) Derive the rate equation for enzymatic process. State an assumption made in this derivation. Terbitkan persamaan kadar tindak balas bagi proses enzim. b) By showing appropriate reaction mechanisms and rate equations, explain how enzyme catalytic reactions may be affected by competitive inhibitionThe Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate ?0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<?m[S]<<Km, [S]=?m[S]=Km, and [S]>>?m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity ?0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
- If a 0.1 M solution of glucose 1- phosphate at 25 °C is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Calculate Keq and ΔG′° for this reaction.An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?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.