Module 6

  1. How do enzymes offer rate acceleration/enhancement?
    By lowering activation energy (EA)
  2. Why did nature evolve enzyme? What do enzymes afford and how do they affect Gibbs Free Energy (ΔG)?
    • Biological reactions cannot simply rely/depend on spontaneity, some reactions require huge EA, and would take an eternity to go to completion
    • Enzymes afford reaction rate acceleration by lowering the barrier to activation (EA), they have NO EFFECT on Gibbs Free Energy (ΔG)
  3. Where do the rate enhancing reactions provided by enzymes take place?
    Catalysis takes place in a region called active site
  4. Define active site and state the expected residues)
    Region of an enzyme that contains catalytic residues (from CDESTY HKR category)
  5. Label both figures and state the expected ΔG (explain your choice)
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  6. What is the difference between ΔG and the activation energy?
    • The ΔG is a difference in a free energy of the products and reactants 
    • The activation energy is a measure of energy required to break covalent bonds in reactants
  7. What is the difference between an enzyme catalyzed reaction versus a non-catalyzed reaction?
    A catalyzed reaction ensures LOWERING of the activation energy
  8. The apex or peak of the curve is where the ________ state lies, it is extremely short-lived and _______
    • transition state
    • unstable
  9. Define substrate
    Substrate: a molecule upon which the enzyme acts. Enzymes catalyze chemical reactions involving the substrate(s). In the case of a single substrate, the substrate bonds with the enzyme active site, and an enzyme-substrate complex is formed.
  10. How does the rate of the reaction relate to the amount of substrate concentration, [S]?
    • At low [S] (red circle), rate is directly proportional to [S]
    • At higher [S], curve levels off and asymptomatically approaches the Vmax of the reaction

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  11. Define Km
    • Km (Michaelis constant): the substrate concentration at 1/2 the maximum velocity
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  12. If Km = [S] at half the maximum velocity, then V0 = ____
    In other words, what can Km be estimated as?
    • 1/2Vmax
    • estimated as 1/2Vmax
  13. When [S] = Km, exactly half of the ______ _____ have been filled with _______
    • active sites 
    • substrates
  14. If Km value is large, enzyme binds substrate ______. This means there is a large(r) probability that enzyme-substrate complex can ________. If the Km value is small, enzyme binds substrate more ______; small(er) probability that enzyme-substrate complex will ______
    • weakly
    • dissociate
    • strongly
    • dissociate
  15. What is kcat? State the equation too
    • The turnover number (kcat): The turnover number defines the number of substrate molecules formed into products per unit of time 
    • Kcat = Vmax/[E]

    **[E]: Enzyme concentration
  16. What would be the kcat of an enzyme that has a Vmax of 60,000 M/s and an enzyme concentration of 0.1M?
    (60,000 M/s)/0.1M = 600,000s
  17. How does kcat relate to the rate of the reaction?
    • Higher the kcat, the FASTER the reaction
    • Lower the kcat, the SLOWER the reaction
  18. The reciprocal of turnover number can provide us with the time it takes to?
    Catalyze a single reaction
  19. Why evolve enzyme inhibtors? (2)
    • Enzyme function needs to be modulated in the cell (NATURAL REGULATION)
    • Enzyme function needs to be halted or blocked (DRUGS)
  20. Define Irreversible inhibition
    Modification of the enzyme (active site) via covalent binding aka suicide inhibition
  21. Reversible inhibition
    Modification of the enzyme (active site) via non-covalent binding
  22. What are the 4 types of reversible inhibition?
    • Competitive
    • Non-competitive 
    • Uncompetitive
    • Mixed inhibition
  23. Define Competitive and Non-competitive inhibitors
    Competitive inhibitors: inhibitor (similar structure to substrate) binds to active site reversibly; prevents binding of substrate. For example, Tamiflu binding to viral neuraminidase or Statins such as Lipitor bind to HMG-CoA reductase

    Non-competitive inhibitors: binds to enzyme AWAY from active site; binding alters enzyme shape, thus affecting catalysis
  24. Define Uncompetitive and Mixed inhibition
    • Uncompetive inhibition: binds ONLY to enzyme-substrate complex
    • Mixed inhibition: inhibitor binds to enzyme in presence or absence of substrate. However, it should be noted that sometimes there is a preference
  25. Define Uncompetitive and Mixed inhibition
    • Uncompetitive inhibition: binds ONLY to enzyme-substrate complex
    • Mixed inhibition: inhibitor binds to enzyme in presence or absence of substrate. However, ti should be noted that sometimes there is a preference
  26. Generally explain what is happening in this reaction?
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    • This is irreversible inhibition, an example of the mechanism and mode of action of a common non-steroidal anti-inflammatory drug (NSAID)
    • Aspirin acetylates a serine residue on cyclooxygenase (COX-1)
    • Acetylserine is formed
    • Leads to irreversible inhibition of enzyme
  27. How is the type of inhibition best determined?
    Through use of what are known as double-reciprocal plots aka Lineweaver-Burk or a double-reciprocal plot of non-competitive inhibition
  28. What are 4 factors of non-compeitive inhibition
    • Inhibitor doesn't interfere with the binding of substrate to active site since the inhibitor doesn't occupy same site
    • Inhibitor (I) and Substrate (S) are different shapes
    • Vmax is altered
    • Binding to inhibitor alters enzyme shape affecting catalysis
  29. How to determine the Km value from plot?
    Extrapolate line (use a ruler!), make best estimate of x-intercept, solve algebraically
  30. What does this plot tell you about the Km and Vmax?
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    • Km is unaffected
    • Vmax is reduced
  31. Label the graph and state the type of inhibition. How are Km and Vmax affected? Can the inhibitors be catalyzed by the enzyme?
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    • Km increased
    • Vmax unaffected
    • Inhibitors CANNOT be catalyzed by the enzyme
  32. What kind of inhibition is displayed? How are Km and Vmax affected?
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    • Mixed inhibition Lineweaver Burk
    • Km increased (in the presence of inhibitor)
    • Vmax decreased
  33. Depict a Lineweaver-Burk Plot for uncompetitive inhibition. How are Km and Vmax affected?
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    • Km reduced (in presence of inhibition)
    • Vmax decreased
  34. Where do uncompetitive inhibitors bind? Uncompetitive binding of inhibitor stimulates binding of _______! What does this cause?
    • Bind to a site other than the active site
    • substrate!
    • Causes lowering of Km in the presence of inhibitor
  35. Fill out the table for each of the 4 types of inhibition 
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  36. Draw and describe the following situations:
    No inhibition
    Irreversible inhibition
    Competitive inhibition
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    • No inhibition: substrate binds normally
    • Irreversible inhibition: imagine a deprotonated serine side chain in covalent engagement with the inhibitor, I.
    • Competitive Inhibition: Inhibitor competes with normal substrate for binding to the active site
  37. Draw and describe the following situations: 
    Non-competitive inhibition
    Uncompetitive Inhibition:
    Mixed inhibition:
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    • Non-competitive inhibition: Inhibitor binds to an allosteric site AWAY from enzyme altering enzyme's shape and function
    • Uncompetitive inhibition: Inhibitor binds AWAY from active site. Binding of inhibitor stimulates enhanced binding of substrate lowering Km
    • Mixed inhibition: Inhibitor can bind to enzyme alone AWAY from active site OR inhibitor can target the enzyme-substrate complex. In both cases, inhibitor binds AWAY
  38. We have been with the following data for an enzyme known as butyrylcholinesterase. Based on the double-reciprocal plot, please determine: 
    The type of inhibition 
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    • Solution: Obtain reciprocal values and ALWAYS label and show units 
    • x-axis: 1/[S] (mM)-1 
    • y-axis: 1/V0 (min/mM)
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  39. Given the information from the previous card, estimate the Km and the Vmax
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    • **The Vmax value came from 1/.0036 (1 divided by the y-intercept (estimated) of the line drawn for the "no inhibitor"
    • **The Km value came from 1/2.1 (1 divided by the estimated x-intercept of the line with the inhibitor)
    • **Do NOT worry if your kinetic parameter values are off from the problem set key! Your answers are being graded on organization 
    • Do the Km and Vmax values make sense?
  40. PT I: In the table below, you have been provided with enzyme kinetics data investigating the rate of the enzyme, ATP sulfurylase, in the presence and absencce of the inhibitor, chlorate
    Find your x and y values and graph 
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  41. PT II: Determine Km and Vmax in both the presence and absence of an inhibitor 
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Author
chikeokjr
ID
340664
Card Set
Module 6
Description
Module 6
Updated