How does Kcat influence the velocity of MM enzymes ?A. Linear relationshipB. Inverse relationshipC. No effectD. Only effects velocity if total enzyme concentration is lowv =Vmax [S]KM + [S]Turnover ratekcat=Vmax[ET]Speed(substrates product/sec)

Approach to solving the question: Detailed explanation:Answer: A. Linear relationshipThe turnover…

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter12: Chemical Kinetics

Section: Chapter Questions

Problem 22Q

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List two ways that enzyme catalysis of a reaction is superior to normal conditions.
Reaction Velocity (mm/sec) Ma no !! L [1 1 L IM 3M FO Substrate Concentration (M) 1. IN Which statement is true? KM AN ALM 4V The graph above displays the kinetic profiles of the same enzyme catalyzing two different substrates. The red curve displays the higher turnover number (kcat). The red curve displays higher affinity between substrate and enzyme. The blue curve displays competitive inhibition. The red curve displays higher cooperativity,
The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics. Product formed (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 1000 The Km for this enzyme is approximately: 4 mM. 2 mM. 1 mM. 6 mM. 1000 mM.
An enzyme catalyzes a reaction with a Km of 5.00 mM and a Vmat of 2.25 mM s. Calculate the reaction velocity, to, for each substrate concentration. [S] = 2.00 mM mM s- [S) = 5.00 mM mM -s- [S] = 12.5 mM Uo : mM s Enter numeric value [S] = 5.00 mM Uo : mM s- [S] = 12.5 mM
The steady state assumption, as applied to enzyme kinetics, implies: A. Km = Ks. B. the enzyme is regulated. C. the ES complex is formed and broken down at equivalent rates. D. the Km is equivalent to the cellular substrate concentration. E. the maximum velocity occurs when the enzyme is saturated. O A O B O C O D O E
Reversible Enzyme Kinetics: Consider the following reversible reaction: k, E + S ES P +E k_2 Show that the product will be greatly favored, i.e. [P]/[S] >> 1 only if kık2 >> k.ık.2. State all the assumptions.
What should be taken in consideration when designing an enzyme kinetics experiment so that the reaction will not be too slow or too fast? M - amount of substrate B - amount of enzyme |- temperature O - pH
For an enzyme -catalyzed reaction, KM=10.0 mmol dm-3, vmax=0.250 mmol dm²³ s-'; [E]o= 2.3 x 10-6 mmol dm-3. Calculate the catalytic efficiency (=kb/KM) of the enzyme. 0.920 dm³ mol-l s-1 3.1 x 107 dm³ mol-1 s-1 1.1 x107 dm³ mol-l s-1 0.250 dm³ mol-l s-1
For an enzyme that follows Michaelis –Menten kinetics, k1=1X106 M-1 sec-1, k-1= 2x103 sec-1 and k2=2X 102 sec-1. a. What is the Km for the enzyme ? If the enzyme concentration is 10 nanomoles in 1 ml, what is Vmax for the enzyme ? What is the catalytic efficiency for this enzyme?
Below which is false about the Michaelis constant Km in Michaelis-Menten kinetics? k₂ k₁ E+SESE + P k₁ k₂ A higher K indicates a higher affinity between the substrate and the enzyme. Km is effectively equal to the Dissociation Constant (k_₁/k₁) for ES. When [Substrate] = Km, Vo = 1/2 V max When [Substrate] = Km [ES] = [Efree] = [Etotal]
When an enzyme is added to a solution (1.0 mL) containing its substrate at a concentration of 100 times the Km, it catalyzes the conversion of 75 mol of the substrate into the product in 3.0 min in the 1.0 mL solution. The concentration of enzyme in the solution is 0.20 M. What is the enzyme's turnover number (kcat)? Note that the unit of kcat is min-1.
What is the turnover number in molecules/min of a catalase enzyme molecule that cleaves 8.4 x 107 H₂O₂ molecules in 15 minutes?

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How does Kcat influence the velocity of MM enzymes ? A. Linear relationship B. Inverse relationship C. No effect D. Only effects velocity if total enzyme concentration is low v = Vmax [S] KM + [S] Turnover rate kcat = Vmax [ET] Speed (substrates product/sec)