Free Characteristics of Enzymes MCQs with Answers

31. Which treatment permanently destroys the activity of an enzyme?

• A. Strong heating well above its optimum
• B. Cooling the solution to 0 degrees Celsius
• C. Diluting the enzyme solution
• D. Adding more substrate

Explanation: Strong heat breaks the weak bonds holding the tertiary structure, so the active site is permanently distorted. Cold only slows the molecules down and activity returns on warming, which is why refrigeration preserves without sterilising.

32. Sulfa drugs control bacterial infections because they

• A. denature every bacterial protein at once
• B. dissolve the bacterial membrane
• C. raise the temperature inside the bacterium
• D. resemble PABA and competitively block the bacterial enzyme that uses it

Explanation: The drug mimics para-aminobenzoic acid, a substrate bacteria need to make folic acid, so it occupies the enzyme's active site and starves the pathway. Humans obtain folate from food, which is why the drug harms bacteria selectively.

33. Many modern medicines are designed as competitive inhibitors. To work this way, the drug molecule must

• A. bind permanently to the enzyme by covalent bonds
• B. denature the whole protein
• C. mimic the shape of the enzyme's natural substrate
• D. block the synthesis of the enzyme

Explanation: A competitive inhibitor succeeds only if it fits the active site, so medicinal chemists build molecules shaped like the substrate. Permanent covalent binding would make the drug an irreversible inhibitor instead, a different mechanism.

34. An irreversible inhibitor differs from a reversible one in that it

• A. can be removed simply by dilution
• B. binds only at the allosteric site
• C. attaches by weak hydrogen bonds
• D. forms permanent covalent bonds with the enzyme

Explanation: Irreversible poisons chemically modify amino acids at or near the active site, so the enzyme is permanently disabled and the cell must make new molecules. Reversible inhibitors bind weakly and fall off, so their effect depends on concentration.

35. In feedback inhibition, the end product of a metabolic pathway usually binds to

• A. the substrate of the first reaction
• B. an allosteric site on the first enzyme of the pathway
• C. the active site of every enzyme in the cell
• D. the DNA that codes for the pathway

Explanation: The end product docks at a regulatory site on the pathway's first committed enzyme, changing its shape and shutting production until the product is used up. Binding the substrate or the DNA would be too slow and too blunt for minute-to-minute control.

36. Penicillin kills susceptible bacteria by inhibiting the enzyme that builds the

• A. bacterial DNA
• B. bacterial ribosomes
• C. bacterial cell wall
• D. bacterial flagellum

Explanation: Penicillin blocks the transpeptidase that crosslinks the peptidoglycan wall, so a growing bacterium bursts under its own osmotic pressure. Human cells have no wall at all, which is why the drug is selectively toxic.

37. Which change would best restore the rate of an enzyme controlled reaction in the presence of a competitive inhibitor?

• A. Increasing the concentration of substrate
• B. Increasing the concentration of inhibitor
• C. Raising the temperature above the optimum
• D. Lowering the pH well below the optimum

Explanation: Because substrate and inhibitor compete for the same active site, flooding the mixture with substrate outcompetes the inhibitor and the normal maximum rate returns. This is the classic test that distinguishes competitive from non competitive inhibition.

38. Aspirin relieves pain and fever by irreversibly blocking the enzyme cyclooxygenase. In enzyme terminology, aspirin acts as

• A. a coenzyme
• B. an activator
• C. a substrate
• D. an enzyme inhibitor

Explanation: By disabling cyclooxygenase, aspirin stops the making of prostaglandins, the chemicals that promote pain, fever and inflammation. It is not consumed as a substrate nor does it assist the enzyme, so it can only be classed as an inhibitor.

39. Which enzyme is involved in DNA replication?

• A. RNA polymerase
• B. DNA ligase
• C. Peptidase
• D. Lipase

Explanation: DNA ligase joins Okazaki fragments by forming phosphodiester bonds between adjacent nucleotides.

40. The enzyme carbonic anhydrase is found in

• A. Stomach
• B. Red blood cells
• C. Pancreas
• D. Liver

Explanation: Carbonic anhydrase in RBCs catalyzes CO2 + H2O ↔ H2CO3, facilitating CO2 transport in blood.