Q-1. Force not acting in an enzyme substrate complex
a) Electrostatic
b) Covalent
c) Van der Waals
d) Hydrogen
Answer: Van der Waals
Explanation:
Mechanisms to facilitate catalysis:
Catalysis by proximity
Acid-Base catalysis
Catalysis by strain
Covalent catalysis
Force acting in an enzyme substrate complex:
Hydrogen and ionic binding (Electrostatic)
Hydrophobic interactions
Covalent bonding
Q-2. All are true about denatured proteins except:
a) Amino acid sequence remains intact
b) The biological function remains intact
c) The iso-electric pH change
d) Recovery after de-naturation is not possible
Answer: The biological function remains intact
Explanation:
Raising the temperature increases the rate of both un-catalyzed and enzyme-catalyzed reaction by increasing the kinetic energy and the collision frequency of the reacting molecules.
However, heat energy can also increase the kinetic energy of the enzyme to a point that exceeds the energy barrier for disrupting the non-covalent interactions that maintain its three dimensional structure. The polypeptide chain then begins to unfold, or denature, with an accompanying loss of catalytic activity.
Q-3. True about de-naturation of proteins is all except
a) Unfolding occurs
b) Disruption of secondary structure occurs
c) Sequence of amino acids remain the same
d) Biological activity is retained
Answer: Biological activity is retained
Explanation:
See above explanation
Q-4. Km value is defined as
a) Substrate concentration at Vmax/2
b) Substrate concentration of twice V max.
c) Substrate concentration of thrice V max
d) Substrate concentration of one third V max.
Answer: Substrate concentration at Vmax/2
Explanation:
The Michaelis constant Km is the substrate concentration at which v1 is half the maximal velocity (Vmax/2) attainable at a particular concentration of enzyme
Important points:
Km has the dimensions of concentration and is characteristic of an enzyme under a given set of reaction conditions.
Km does not depend on the concentration of enzyme, but can vary with pH. A non-competitive inhibitor decreases Vmax but does not alter Km.
Q-5. The Michaelis constant Km is: –
a) Numerically equal to 1/2 Vmax.
b) Dependent on the enzyme concentration.
c) Independent of pH
d) Numerically equal to the substrate concentration that gives half-maximal velocity
e) Increased in the presence of a non-competitive inhibitor.
Answer: Numerically equal to the substrate concentration that gives half-maximal velocity
Explanation:
See above explanation
Q-6. Km of an enzyme is –
a) Dissociation constant
b) The normal physiological substrate concentration
c) The substrate concentration at half maximum velocity
d) Numerically identical for all isozymes that catalyze a given reaction
Answer: The substrate concentration at half maximum velocity
Explanation:
See above explanation
- True about km
a) Half the substrate concentration at which velocity is maximum
b) Substrate concentration at which reaction rate is half the maximum
c) Michaelis constant
d) Dissociation constant of enzyme- substrate complex
e) Dissociation constant of enzyme product complex
Answer: b and c
Explanation:
See above explanation
Q-8. Hormone substrate concentration affects the velocity of enzymatic action, this is known as
a) Zimmerman reaction
b) Salkowski reaction
c) Michaelis Menten reaction
d) Lieberman – Burchard reaction
Answer: Michaelis Menten reaction
Explanation:
See above explanation
Q-9. Michaelis – Menten hypothesis states that –
a) Rate of enzymatic reaction is independent of substrate concentration
b) Rate of non-enzymatic reaction is proportional to substrate concentrate
c) Km. is the enzyme-substrate complex association constant
d) Enzyme substrate complex formation is essential in enzymatic reaction
Answer: Enzyme substrate complex formation is essential in enzymatic reaction
Explanation:
Michaelis – Menten hypothesis states that – Enzyme substrate complex formation is essential in enzymatic reaction.
Q-10. The type of enzyme inhibition (in which Succinate Dehydrogenase reaction is inhibited by malonate) is an example of: (AIIMS May 2006)
a) Non-competitive
b) Uncompetitive
c) Competitive
d) Allosteric
Answer: Competitive
Explanation:
Most frequently, in competitive inhibition the inhibitor (I) binds to the substrate-binding portion of the active site – thereby blocking access by the substrate.
The structures of most classic competitive inhibitors therefore tend to resemble the structures of a substrate, and thus are termed substrate analogs.
Important point:
Inhibition of the enzyme succinate dehydrogenase by malonate illustrates competitive inhibition by a substrate analog.
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Q-11. A competitive inhibitor of an enzyme:
a) Increase Km without affecting Vmax.
b) Decrease Km without affecting Vmax.
c) Increase Vmax without affecting Km.
d) Decrease Vmax without affecting Km.
e) Decreases both Vmax and Km
Answer: Increase Km without affecting Vmax.
Explanation:
A competitive inhibitor has no effect on Vmax but rises Km, the apparent Km for the substrate.
Simple Non-competitive Inhibitors Lower Vmax but do not affect Km.
Q-12. In competitive inhibition
a) Vmax unchanged
b) Apparent Km unchanged
c) Apparent Km decreased
d) Vmax decreased
e) Rate independent of substrate concentration
Answer: Vmax unchanged
Explanation:
See above explanation.
Q-13. In competitive inhibition the relation Km and Vmax is one of the following
a) Km and Vmax are the same
b) Km increases and Vmax is the same
c) Km decreases and Vmax increases
d) Km and Vmax decreases
Answer: Km increases and Vmax is the same
Explanation:
See above explanation.
Q-14. Km. increased but Vmax same enzyme is:
a) Competitive
b) Non competitive
c) Irreversible
d) Uncompetitive
Answer: Competitive
Explanation:
See above explanation.
Q-15. In non-competitive antagonism the true statement is
a) Km value decrease; V max normal
b) Km value decreased; V max decreased
c) Km value normal; V max decreased
d) Km value deceased; V max increased
Answer: Km value normal; V max decreased
Explanation:
See above explanation.
Q-16. Non-competitive inhibitor of an enzyme
a) Increases Km with no or little change in Vmax
b) Decreases Km
c) Decreases Vmax
d) Increases Vmax
Answer: Decreases Vmax
Explanation:
See above explanation.
- True about reversible non- competitive inhibitors
a) Lower Vmax
b) Lower km
c) Not affect km
d) Not affect Vmax
e) Affect both Vmax & km
Answer: a and c
Explanation:
See above explanation.
Q-18. Disopropyl phospho-fluoridate (DFP) reacts with serine proteases irreversibly and therefore is:
a) Allosteric inhibitor
b) Competitive inhibitor
c) Non-competitive inhibitor
d) A repressor
Answer: Non-competitive inhibitor
Explanation:
An enzyme that has been ‘poisoned” by an irreversible inhibitor such as a heavy metal atom or an acylating reagent remains inhibited even after removal of the remaining inhibitor from the surrounding medium.
- Enzyme activity is expresses as
a) Milli moles/lit
b) Mg/lit
c) Mg /dl
d) Micro-moles/min
e) Microgram/min
Answer: Micro-moles/min
Explanation:
An enzyme unit is defined as the enzyme activity that catalyses the conversion of 1 µmol substrate into product in one minute.
The SI unit of enzyme catalytic activity is the katal; however, it is less commonly used in practice.
1 katal = Amount of enzyme that catalyzes the conversion of 1 mole of substrate per second (1 katal = 1 mol/s).