BIOCHEM. If correct will give a thumbs up. 2) Recall that the active site of HIV

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BIOCHEM. If correct will give a thumbs up.

2) Recall that the active site of HIV protease contains two aspartic acids. The pKa values for these acids are tuned in the protein such that one is deprotonated and the other is protonated at pH~5, where the enzyme is maximally active. 0 0 c) Inhibitors of HIV protease are particularly effective at reducing viral load because they limit the ability of the protease to process a single polypeptide containing all of the viral proteins into their constituent pieces. Chronic treatment with a protease inhibitor can provide selective pressure for mutations in the protease that help it evade the inhibitors. However, sometimes these so-called 'escape mutants' pay a price by exhibiting a lower overall enzymatic efficiency towards their native substrates. In your undergraduate project, you are tasked with studying one such mutant that has become resistant to inhibitors. You find that the efficiency of this mutant enzyme towards its normal substrate has been reduced by 10-fold. What is the most straightforward way to determine if the reduction in efficiency is due to a change in affinity for the substrate or a new deficiency in catalytic ability? Outline your experimental strategy in no more than 3 sentences and explain how your results would address this question.

Explanation / Answer

Ans. Experimental strategy:

I. Prepare a Michaelis- Menten or Lineweaver- Berk plot (better to choose LB plot for more accurate determination of Km and Vmax) at different [substrate] for the normal and mutant HIV protease enzyme in separate sets (two sets- one “normal E + [S]” and the other set as “mutant E + [S]”) of experiment. Compare the Vmax and Km for the two enzymes.

II. If Vmax remains the same but only Km of the mutant enzyme is increased, it means that the mutant enzyme has lower affinity for the substrate.

# If mutation increase only affects the affinity of enzyme for the substrate (measured in terms of Km), the Vmax would remain unaffected.

III. If Vmax and/ or Km of the mutant enzyme is affected (compared to that of normal enzyme), the mutation has affected the catalytic activity of the enzyme.   

# If the mutation affects catalytic activity of the enzyme, Vmax and/or Km are likely to be affected.

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