Enzyme Kinetics Flashcards
Michaelis-menten kinetics assume that:
A) at the start of a reaction there is essentially no product formed
B) the concentration of enzyme-substrate complex stays constants throughout the reaction
C) the concentration of enzyme is far less than the concentration of the substrate
D) all of the above
E)none of the above
D) all of the above
Glucokinase (in liver) and hexokinase (in most cell types) both catalyze the phosphorylation of glucose following its transport into the cytosol from the bloodstream. The Km of hexokinase for glucose is 20 microliters, and the Km of glucokinase for glucose is 10 mM. Which one of the following statements regarding these enzymes is most accurate when the cytosolic glucose concentration is 5 mM? (Assume Michael is-menten kinetics)
A) hexokinase is operating at near maximal velocity
B) both hoxokinase and glucokinase are operating are near maximal velocity
C) glucokinase is operating at near maximal velocity
D) neither hexokinase nor glucokinase is operating near maximal level
E) glucokinase is inactive and hexokinase is active
A) hexokinase is operating at near maximal velocity
It an enzyme-catalyzed reaction with Km of 3.5 mM has a velocity of 5mM/min at a substrate concentration of 0.5 mM, what is the Vmax?
A) 0.625 mM/min
B)15 mM/min
C)17.5 mM/min
D)35 mM/min
E)40 mM/min
E)40 mM/min
Substituting the vales into the Michelis-me ten equation (v=Vmax [S]/(Km+[S])), then 5=Vmax(0.5)/(3.5+0.5), or Vmax= (5)(4)/(0.5). Thus Vmax =49mM/min
Your roommate tried to bake a cake, but accidentally created an inhibitor for chymotrypsin. Treatment of chymotrypsin with the inhibitor decreases the Vmax of chymotrypsin. However, the Vmax seems to continue to decrease the longer the inhibitor is allowed to interact with chymotrypsin. After inhibition, you purify chymotrypsin using gel filtration chromatography, but even with free inhibitor removed you find that chymotrypsin still has lowered Vmax. These results likely mean that:
A) Your roommate has created a competitive inhibitor for chymotrypsin
B) Your roommate has created a non-competitive inhibitor for chymotrypsin
C)chymotrypsin does not interact with this inhibitor
D)your roomate has created an irreversible inhibitor to chymotrypsin
E)this inhibitor can only bind at low pH
D)your roomate has created an irreversible inhibitor to chymotrypsin
Describe a simple enzyme catalyzed reaction in terms of rate constants, substrate/ product concentrations, and use of the Michaelis-Menten equation to calculate Km and Vmax
A unimolecular reaction has a velocity that is dependent on the concentration of only one substrate
A bimolecular reaction has a velocity dependent on two substrates
E+S<—>ES—->E+P
Initial reaction velocity only depends on the rate determining step
V= Vmax [S]/ Km +[S]
Hyperbolic curve
Km= substrate concentration when enzyme works at half of its maximum speed.
What is the catalytic rate constant (kcat)?
Catalytic rate constant or turnover number, when the enzyme is saturated with S
It determines how quickly and enzyme can act
What is catalytic efficiency and how is it measured?
Catalytic efficiency reflect both the enzyme rate (kcat) and the substrate affinity and is meassured by kcat/Km
You have discovered a new enzyme, and mutation of Asp 37 to Ala greatly reduces the kcat but does not affect the Km of the enzyme. Based on this, Asp 37 may be:
A) important for binding the substrate
B) not involved in substrate binding or catalysis
C) important for the catalytic machinist of this enzyme
D) needed for both substrate binding and catalysis
E) none of the above
C) important for the catalytic machinist of this enzyme
Describe enzyme inhibition and interpret graphical kinetic data
Irreversible enzyme inhibitors (often poisons): prevent catalysis.
Ex: diisopropylflurophosphate
Reversible enzyme inhibitors: bind to enzymes non-covalently. They are classified by where they bind and their effects on enzyme kinetics
Competitive inhibitors: appear to increase Km but Vmax stays the same. Large excess of substrate can overwhelm the inhibitor
Non-competitive inhibitors: bind to the allosteric site and decrease the Vmax, Km stays the same. The substrate and the inhibitor are not mutually exclusive, but the enzyme/substrate/inhibitor complex is less active
Mixed inhibition: similar to non-competitive inhibition, but binding to distant site modifies both Vmax (reduced) and Km (increased)
A competitive inhibitor will:
A) increase the Vmax of en enzyme
B) decrease the Vmax of an enzyme
C) increase the Km of an enzyme
D) decrease the Km of an enzyme
E) increase both the Vmax and Km of an enzyme
C) increase the Km of an enzyme
Explain the various short and longer-term mechanisms used to regulate enzyme activity in the body
Allosteric enzymes after have multiple subunits that show cooperativity: analogous to homoglobin vs myoglobin
Enzymes can have positive or negative allosteric
Affect both Vmax and Km
Allosteric effectors may inhibit or activate enzymes and are used to control metabolic pathways.
