[S] << Km [S] = = Km [S] >> Km Not true for any of these conditions Almost all active sites are empty. [ES] is much higher than [Efree]. [Efree] is equal to [ES]. Reaction rate is independent of [S. [Efree] is about equal to [Etotai]- Increasing [Etotal] will lower Km-
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The 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.
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- Would you expect an “enzyme” designed to bind to its target substrate astightly as it binds the reaction transition state to show a rate enhancementover the uncatalyzed reaction? In other words, would such a protein actuallybe a catalyst? Explain why or why not.At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?If you graph the velocity of an enzymecatalyzed reaction vs. [S] for each of two substrates that are partof a random mechanism, would you expect to see the same shapecurve? Why or why not?
- Do you think the hydrolysis GTP and ATP are similar? Explain why or why not?An enzyme contains an active site aspartic acid with a pKa = 5.0, whichacts as a general acid catalyst. On the accompanying template, draw thecurve of enzyme activity (reaction rate) versus pH for the enzyme (assumethat the protein is stably folded between pH 2–12 and that the active siteAsp is the only ionizable residue involved in catalysis). Briefly explain theshape of your curve.In the scheme below which represents the mechanism of action for a large number of enzymes: A+B⟺AB⟶C The steady state approximation is reached when: d[AB]/dt≈0 k2≫k1 k−1≫k1 k−1=k1
- A potent inhibitor effectively inhibits an enzyme catalyzed reaction. What kind of a Ki value you would expect for a potent inhibitor?Consider the binding reaction L + R → LR, where L is a ligand and R is its receptor. When 1 × 10−3 M of L is added to a solution containing 5 × 10−2 M of R, 90 percent of the L binds to form LR. What is the Keq of this reaction? How will the Keq be affected by the addition of a protein that facilitates (catalyzes) this binding reaction? What is the dissociation equilibrium constant Kd?When enzyme solutions are heated, there is a progessive loss of catalytic activty over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 45 degrees Celsius lost 50% of its activity in 12 minutes, but when incubated at 45 degrees Celsius in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substrates.
- A histidine side chain is known to play an import-ant role in the catalytic mechanism of an enzyme; how-ever, it is not clear whether histidine is required in its pro-tonated (charged) or unprotonated (uncharged) state. Toanswer this question you measure enzyme activity over arange of pH, with the results shown in Figure Q2–1. Whichform of histidine is required for enzyme activity?If the higher value of KM resulting in the new plot ( red curb ) is due to the presence of an enzyme inhibitor is inhibitor reversible or irreversible? And why?An enzyme catalyzes a reaction in which substrate A is cleaved into two products, P and Q. In the catalytic mechanism, the enzyme converts A to an covalently-bound reaction intermediate X and product P, P then desorbs from the enzyme, and in a second chemical step, the enzyme converts the intermediate X in the EX complex to the final product Q (in EQ), which then desorbs from the enzyme E. You discover two inhibitors of this enzyme, I and J. I is a competitive inhibitor of the substrate A, and has nearly double the molecular weight of J. On the other hand, J is a mixed inhibitor of enzyme E, and its inhibitory effect on Km / Vmax (the slope effect from the double reciprocal plot) is greater than that of 1 / Vmax (the intercept effect in a double reciprocal plot). That is Kis < Kii . At low pH, the conversion of EX to EQ is greatly slowed, kcat is decreased, and the intercept effect of inhibitor J is elevated, that is, the value of Kii is diminished. When a high, fixed concentration…