5.5 Explain the effect of each type of inhibitor on the apparent kinetic parameters: (1) Inhibitor binds only free enzyme does Km increase, decrease, not change? does Vmax increase, decrease, not change? (2) Inhibitor binds only ES complex does Km increase, decrease, not change? does Vmax increase, decrease, not change? (3)Inhibitor binds E and ES equally does Km increase, decrease, not change? does Vmax increase, decrease, not change?

5.5 Explain the effect of each type of inhibitor on the apparent kinetic parameters: (1) Inhibitor binds only free enzyme does Km increase, decrease, not change? does Vmax increase, decrease, not change? (2) Inhibitor binds only ES complex does Km increase, decrease, not change? does Vmax increase, decrease, not change? (3)Inhibitor binds E and ES equally does Km increase, decrease, not change? does Vmax increase, decrease, not change?

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter18: Glycolysis

Section: Chapter Questions

Problem 17P

See similar textbooks

Similar questions

3.18: An enzyme E binds a substrate S and a cofactor C. The equilibrium dissociation constantKd,S of the enzyme-substrate complex ES is 1 μM, for EC it is 10 μM. When the cofactor Cis present, K’d,S is decreased to 0.1 μM. What is the value for the dissociation constant K’d,C of the enzyme-cofactor complex in the presence of substrate S? Calculate the interactionenergy ΔΔGint for cofactor and substrate binding.
NOTE: the enzyme-inhibitor complex requires 450 joule/mol to dissociateJOULE/MOL
For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K M
The following kinetic data were obtained for an enzyme in the absence of an inhibitor (1) and in the presence of two different inhibitors (2) and (3) at 5mM concentration. [S], mM (mmol/mL/sec) (mmol/mL/sec) (mmol/mL/sec) (1) (2) (3) 1 12 4.3 5.5 2 20 8 9 3 29 14 13 8 35 21 16 12 40 26 18 Determine the Vmax and KM of the enzyme Determine the type of inhibition.
From your Lineweaver-Burk plot,the vlaues are: Km Vmax Uninhibited 0.09 mmol/L 3.02 min/mmol Inhibited 6.22 mmol/L 9.98 min/mmol By describing the potential changes in the kinetic parameters, identify and justify the type of inhibitor that was inhibiting the acid phosphatase activity.
Velocity (mmol/minute) [S], (mM) No inhibitor Inhibitor 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8 The kinetics of an enzyme are measured as a function of substrate in the presence and the in absence of 2mM inhibitor (I). What are the values of Vmax and KM in the absence of inhibitor? In its presence? In its presence? What is the type of inhibition?
Calculate α' for an inhibitor with KI' = 10.0 nmol L-1 when 100 nmol L-1 of inhibitor is present.
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?
What is the impact of the lower value Vmax on the affinity for enzyme for substrate? And what is impact of the lower V max on the amount of product formed ? If the lower value of black resulting in the new plot (red curve) is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible ? And why?
5. 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 equilibrium or the steady state? Justify your answer. b) What is kcat for this reaction? Justify your answer. c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1, and each enzyme has two active sites. d) What substrate concentration would be required for the reaction in (c) to reach half of Vmax. Justify your answer mathematically. e) A second Michaelis-Menten enzyme has k1 = 4.2 ⅹ 107 M-1 s-1, k-1 = 6.1 ⅹ 104 s-1, and k2 = 5.3 ⅹ 102 s-1. Which enzyme is most efficient? 6. A pharmaceutical company is trying to develop a
An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?
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.