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Consider the trypsin binding-pocket specificity structure scenario and critical AAs interactions ...

Consider the trypsin binding-pocket specificity structure scenario and critical AAs interactions within: G226-D189-G216 (see

Consider the trypsin binding-pocket specificity structure scenario and critical AAs interactions within: G226-D189-G216 (see L3, slide 26): a single nucleotide polymorphism within the D189 codon resulted in a first nucleotide Guanine replacement by Cytosine. What is the consequence of this mutation relative to binding pocket- substrate specificity? (3 pts) 5. You work at the infamous enzymology lab; a colleague asks you to evaluate her enzyme activity data test based on the efficacy of two inhibitors, each at the [10 mM] concentration. She shows you the Lineweaver-Burk plot (below): two inhibitors are indicated by "+A" or "+B", respectively. E represents the enzyme activity with no inhibition. (Note: material from Monday Lecture you should have enough time to complete this problem before the HW is due) 6. 0.35 0.3 E 025 0.2 0.15 0.1 006 0.05 3 21 1/IS], 10M What type of enzyme inhibition is depicted in the a) +A data set? b) +B data set? (1.5 pts each) c) Since you were able to identify two types of inhibitors, she asks you which of the two inhibitors is likely to be more efficient at LOW substrate concentrations? Please explain your reasoning. (2pts)
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a) +A data set = Competitive inhibition,

b) +B data set = non competitive inhibition.

The standard Line-Weaver Burk Plot follows the following equation,

1 Km V Vmax S Vmar

Where, the, Km = Michaelis–Menten constant, signifies enzyme's affinity towards the substrate and Vmax = Maximum velocity.

When you add a specific inhibitor, the inhibitor will change the Km or Vmax or both of the reaction and the apparent Km or Vmax will be higher or lower of the standard enzymatic reaction condition.

Now, from the Enzyme kinetics we know that, for a competitive inhibitor, apparent Km will increase, {K'm > Km}. So, the value of -(1/Km) will come towards zero in the graph but there will be no change in the Vmax as you see in the graph for +A.

But for Non competitive inhibition there will be no change in Km but the apparent Vmax will decrease, so the value of 1/Vmax will go away from the zero as you see in the graph for +B.

c) Competitive inhibitor.

When you see the graph you see that after adding the competitive inhibitor the apparent Km increases. That means the enzyme's affinity towards the substrate decreasing. But how is it possible? because these are the same enzyme we used before. This happen because the competitive inhibitors bind to the active site of the enzyme instead of the substrate. As we can not count the inhibitor's binding to the enzyme for enzyme kinetics calculation it seems like the enzyme's affinity is decreasing.

Now, if you use more substrate (increase substrate concentration) along with the competitive inhibitor, the inhibitors have to compete with more the substrate, so the change of Km will be less. Now, if the substrate concentration is much much more than the concentration of the inhibitor, then there will be almost no effect of the inhibitor as the high substrate concentration will out compete the inhibitors for binding towards the enzyme's active site.

So, the competitive inhibitors will act efficiently at less substrate condition.

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