Consider the reaction catalyzed by PFK. In the presence of ATP, which of thefollowing will be expected? Select all that apply.The T-state is stabilizedThe rate of the reaction is enhanced.The initial velocity plot would show a shift in the curve to the rightThe quaternary structure of PFK is stabilized to promote efficient binding of substrate

Phosphofructokinase-1 (PFK-1) is a glycolytic enzyme that catalyzes the transfer of a phosphoryl…

Answered: Consider the reaction catalyzed by PFK.…

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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when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
The following diagram shows reaction curves for aspartate transcarbamoylase (ATCase) with carbamoylphosphate and different concentrations of aspartate, in the absence of ATP (curve 1) and presence of ATP (curve 2). What do the shapes of the curves tell us about the ATCase enzyme? 2 جر [aspartate] It binds substrate through a sequential mechanism. It binds substte cooperatively. It obeys Michaelis-Menten kinetics. It binds substrate through a concerted mechanism.
An allosteric enzyme that follows the concerted mechanism has a T/R ratio of 500 in the absence of substrate. Suppose that a mutation reversed the ratio. How would this mutation affect the relation between the rate of the reaction and substrate concentration? The mutant enzyme would behave like an enzyme that obeys Michaelis Menton kinetics. The mutant enzyme would have a smaller vmax There would be no difference in the mutant enzyme in terms of substrate binding and catalysis. More than one answer is correct. The mutant enzyme would display cooperativity more than the wild type. MacBook Air 888 F5 F4 F3 F2 %23 2$ %
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.
During chymotrypsin-mediated catalysis, which of the following statements is true? Select any/all answers that apply. O A. A basic (positively-charged) side chain of the substrate polypeptide sits within the oxyanion hole of the enzyme. O B. A basic (positively-charged) side chain of the substrate polypeptide sits within the specificity pocket of the enzyme. O C. Ser195 hydrogen-bonds with the backbone carbonyl group of the substrate polypeptide within the oxyanion hole of the enzyme. O D. All three residues that comprise the catalytic triad of the enzyme interact directly with the substrate polypeptide chain. O E. Both covalent catalysis and acid-base catalysis occur.
Consider an enzyme-catalyzed reaction that follows Michaelis-Menten kinetics with KM =3.0 mmol·dm-3. What concentration of a competitive inhibitor with KI = 20 umol·dm-3 willreduce the rate of formation of product by 50% when the substrate concentration is held at 0.10mmol·dm-3.
Here’s a kinetic model for some enzyme E acting on substrate S. Give the equation for the change in the concentration of free enzyme (E) with time.
The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
.When saturated with substrate, an enzyme has a maximum initial rate of 110 umoles of substrate converted to product per second. At a substrate concentration of 100 µM, the same enzyme converts substrate to product at a rate of 0.010 mmoles/sec. Assuming that Michaelis- Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2 x 10 M.
Why does growth of ETPUM in Medium # require oxygen? ' ink about this in terms of howETPUM can generate a net gain in ATP by processing polyurethane. Remember that the degradationof polyurethane by polyurethanase does not expend ATP. In order to answer this question, address eachof the following questions in your answer:a. Is there a net gain or loss of ATP during the transport of the citrate?b. Consider the ATPs that can be generated via substrate-level phosphorylation. Will glycolysis beuseful for generating any ATPs during growth on polyurethane? How many ATPs can be generatedvia TCA? Is this enough to support growth (is there a net positive in the ATP tally)?c. Now consider how else ETPUM can generate ATPs (if not by substrate- level phosphorylation). Canthis process generate a net positive in the ATP tally?d. Now explain the importance of oxygen as relates to the ATP tally.
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?
The enzymatic activity of an enzyme with Kg = 2 mM that converts substrate S into product P is measured at an initial substrate concentration S, of 10 µM. After 5 min, the substrate concentration is halved. What is the rate constant k, the maximal velocity vmax and the concentration of product after 12 min?

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