BCHM 4115 - Lectures 18-21

In noncompetitive inhibition (not mixed), both v_max and k_m ...

• Vmax decreases

  Km stays the same

• Vmax stays the same

  Km stays the same

• Vmax stays the same

  Km decreases

• Vmax stays the same

  Km increases

Would a high k_m value be considered favorable? Why or why not?

Answer explanation

No, a higher k_m value would NOT be considered favorable because k_m , the Michaelis constant, reflects the enzyme's affinity for its substrate. A lower k_m ​ indicates higher substrate affinity, meaning the enzyme can reach half of its maximum catalytic activity V_max at a relatively low substrate concentration, compared to a high km which indicates that a lot more substrate is needed to reach this same threshold. This suggests the enzyme is efficient at binding and processing the substrate, even when substrate levels are limited.

Why is the induced fit model a better representation of substrate binding?

Answer explanation

The induced fit model is a more accurate representation of enzyme-substrate interactions because it accounts for a conformational change in the enzyme upon substrate binding. This dynamic adjustment allows the enzyme to bind to the substrate more precisely and catalyze the reaction more effectively. In contrast, the lock and key model treats the enzyme as a rigid structure, which does not reflect the flexibility observed in enzymes during substrate recognition.

RNA molecules that are catalytic are called _.

Draw the enzyme-substrate reaction pathway. Include all molecular species (E, S, ES, P), the reversible and irreversible steps, and label the corresponding rate constants (k₁, k₋₁, k₂)

Answer explanation

What is the name of the two scientists that won the Nobel Prize in Chemistry for CRISPR DNA editing?

Answer explanation

Jennifer Doudna and Emmanuelle Charpentier

What type of enzyme inhibition does this Linweaver-Burk plot exhibit and why?

Answer explanation

This Lineweaver-Burk plot depicts competitive inhibition. This is because the V_max remains unchanged, while the enzyme's affinity for the substrate decreases, as indicated by an increased K_m. This makes sense because in competitive inhibition, the inhibitor competes with the substrate for binding at the enzyme’s active site, effectively reducing the enzyme’s binding affinity for the substrate.

What are the three amino acids present within the active site of tripsin (a digestive protease).

Answer explanation

Aspartate (102)

Histidine (57)

Serine (195)

Explain why the two transition state analogs depicted at the bottom of the graphic are effective at preventing racemization between the L- and D-enantiomers.

Answer explanation

The two transition state analogs shown at the bottom are effective at preventing racemization between the L- and D-enantiomers because they closely mimic the structure of the planar transition state. Their planarity allows them to fit tightly into the enzyme’s active site, binding even more strongly than the natural substrate. As transition state analog inhibitors, they block the enzyme’s catalytic activity by occupying the active site, thereby preventing the interconversion between L-proline and D-proline.

Tags

NGSS.HS-PS1-4

Why is the velocity starting to level off as the reaction is approaching its V_max?

Answer explanation

As a reaction approaches its Vmax, it indicates that the enzyme is becoming saturated with substrate. At this point, even as substrate concentration increases, the rate of ES complex formation plateaus because all of the enzyme’s active sites are already occupied with substrate.

Tags

NGSS.HS-PS1-5