Practice Questions - Week 10 - Enzymology (CATCH UP) Flashcards
T/F: The Michaelis constant determines the Vmax of an enzymatic reaction
False
(The Michaelis Menten constant tells us about the concentration of substrate at which the rate of reaction is half the maximum rate, or the velocity is half of the Vmax that can be achieved under the given conditions. It is denoted by the symbol Km)
Competitive inhibitors have this effect:
- Modifying the Km value
- Changing the value for Vmax
- Interfering with substrate binding
- Changes the Km and interferes with substrate binding
- All of the above are correct
4.
The value of Vmax changes in
a) competitive inhibition
b) noncompetitive inhibition
c) both forms of inhibition
d) neither form of inhibition
b
Which of the following is true?
a) The E-S complex often dissociates with no reaction taking place.
b) The E-S complex must form before a reaction can take place
c) Once the E-S complex forms, it can go on to form product or dissociate to E + S
d) All of these
d
The initial rate of an enzymatic reaction is usually determined in order to assure that
a) the enzyme is active
b) there is no reverse reaction of product to the enzyme-substrate complex
c) the substrate is not used up
d) the experiment can be completed quickly
b
The Michaelis constant is
a) related to the molecular weight of the enzyme
b) a measure of the resistance of the enzyme to denaturation
c) a refection of the percentage of polar amino acids in the enzyme
d) a measure of how tightly the substrate is bound to the enzyme
d
An important step in elucidating the behaviour of an enzyme is
a) obtaining a crystalline sample of the enzyme
b) insuring that metal ions are always excluded from the enzyme
c) determine the active site residues
d) none of these
d
(Enzyme structures unfold (denature) when heated or exposed to chemical denaturants and this disruption to the structure typically causes a loss of activity. Protein folding is key to whether a globular protein or a membrane protein can do its job correctly)
The amino acids in the active site can be involved in all of these processes, except:
a) Binding of the substrate
b) Becoming part of the product of the reaction
c) The actual chemical mechanism for the reaction
d) Binding of some necessary cofactor
e) All of these can be functions of the amino acids in the active site
b
What effect is seen on a Lineweaver-Burk graph when a competitive inhibitor is added
a) the y-intercept is changed, but does not change the slope
b) the slope of the line is changed but not the y intercept
c) both the slope and y intercept is changed
d) neither the y intercept or slop of the line is changed
b
Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water.
The best explanation is that:
a) The larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.
b) The —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is attached to C.
c) Water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme.
d) Water normally will not reach the active site because it is hydrophobic.
a - The larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.
The data shown below were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics:
What value will be the best estimate for the Km from this kinetic study?
a) 2 mM
b) 1 mM
c) 4 mM
d) 6 mM
a)
This one is easy!
The velocity at 1000 mmol/L is near maximal – that is ~Vmax.
Km is the [substrate] equal to ½ Vmax.
In your head divide 647 by 2 = ~325.
The V value at 325 equates to a substrate concentration of 2 mM.
Hence, the Km is approx 2 mM.
a
To determine Km from a Lineweaver-Burk or double-reciprocal plot, you need to multiply the reciprocal of the x-axis intercept by -1.
d) is twice the rate observed when the concentration of substrate is equal to the Km
Covalent
(The role of serine at the active site of serine proteases is to act as a covalent catalyst, while the histidine residue serves as a acid-base catalyst)
a
