Experimental enzyme kinetics; Linear plots and Enyzme Inhibition

Question 1

What steps would a typical enzyme kinetics experiment take?

Answer 1

Aim: to find the properties of a newly discovered enzyme by measuring K_M and V_max

• Measure initial rates v_o for a series of different [S] values
• Plot the Michaelis-Menten curve for the enzyme
• Find where vo = 0.5 V_max at this point, [S] = K_M

Question 2

How does K_m and V_max tell us about enzymes properties?

Answer 2

• V_max indicates the catalytic rate when 100% of the enzyme is occupied by substrate (saturated with substrate)
• higher V_max means a faster reaction, better catalysis Vmax = k2 [E]total (Vmax is a conditional constant)
• K_M indicates how well the substrate fits the catalytic site
• K_M is the concentration of S that gives an initial rate of 0.5 V_max
  
  • low K_M means good recognition, substrate binds well
  • high K_M means poor recognition, substrate binds poorly
• An enzyme with more than one substrate has a different K_{M value} for each
  
  • to measure K_M(A), keep [B] constant and vary [A]
  • to measure K_M(B), keep [A] constant and vary [B]

Question 3

What happens if V_max(A)max(B)?

If K_M(A)M(B)?

Answer 3

Question 4

How do you measure K_m and V_max?

Answer 4

• The Michaelis-Menten equation is a hyperbolic curve when vo is plotted vs. [S]
  
  • it approaches Vmax gradually
• when estimating by eye, the tendency is to underestimate V_max, and then 0.5V_max and K_M are too low
• Real experimental data often shows scatter due to measurement errors
  
  • difficult to plot a curve when points are scattered; easier to fit them to a line
• Linear transformation convert the Michaelis- Menten equation into a straight line form
  
  • Lineweaver-Burk method
• Slopes and intercepts of the straight line give better estimates of Vmax and KM

Question 5

What is the Lineweaver-Burk or double reciprocal plot?

Answer 5

• Take reciprocals of both sides of the Michaelis-Menten equation
• Use 1/v₀ for y and 1/[S] for x
• Result is a straight line equation
  
  • Slope = K_M/V_max
  • y-intercept = 1/V_max
  • x-intercept = -1/K_M ( = -y intercept/slope)
• Obtain K_M by extending the graph onto the negative x axis
• Lineweaver-Burk plots are widely used; they work well unless there are significant errors in the data

Question 6

What is the summary of Lineweaver-Burk method for enzyme kinetic analysis?

Answer 6

Hints: You can easily derive the intercepts from the axes

Write V_max in place of v_o , and write -K_M in place of [S]

slope = - y intercept / x intercept

Question 7

What are inactivators?

Answer 7

Inactivators and inhibitors both interfere with enzyme catalysis

• Inactivators: usually react with enzymes irreversibly
• Inactivation results from covalent chemical reaction between inactivator and enzyme
• Often irreversible– reaction destroys catalytic activity and“uses up” the enzyme
• Simple stoichiometry relationship between inactivator and enzyme

2 μmol inactivator + 3 μmol enzyme = 1 μmol active enzyme left

• Many inactivators are higly toxic- nerve gases inactivate enzyme acetylcholinesterase, interfering with nervous impulses

Question 8

What are inhibitors?

Answer 8

• Inhibitors: usually bind to enzymes reversibly
• Bound inhibitor decreases enzyme activity without destroying the catalytic function of enzyme molecule
• Enzyme activity is restored if inhibitor concentration is reduced
• non-covalent bonding- inhibitor binds to site on the enzyme by non-covalent forces, similar to substrate
• Degree of inhibition governed by binding equilibrium, not simple stoichiometry

Question 9

What is competitive and non-competitive inhibition?

Answer 9

• Presence of inhibitor may affect different stages of the catalytic reaction, gives different modes of inhibition
  
  • competitive inhibition affects ability to bind substrate
  • non-competitive inhibition affects catalytic rate (also mixed inhibition)

Question 10

What do inhibitors regulate? How does it benefit us?

Answer 10

• Inhibitors can regulate enzyme activity in the cell
  
  • More economical for cell to make and destroy a small inhibitor than a large enzyme
• Many drugs are enzyme inhibitors – focus of most drug company research
  
  • Acetyl salicylic acid (aspirin) inhibits cyclo-oxygenase enzymes that make prostaglandins
    
    • prostaglandins affect inflammatory response, blood pressure and blood clotting, and intestinal action
    • research found replacements with fewer side effects; COX2 inhibitors - new pain killers, e.g. Celebrex
  • Statins (e.g. Lipitor) inhibit a key liver enzyme involved in cholesterol biosynthesis (HMG CoA reductase), thus lowering blood cholesterol levels and heart attack risk

Question 11

Competitive inhibition arises when inhibitors can only bind to what?

Answer 11

Competitive inhibition arises when the inhibitor can only bind to unoccupied enzyme E

• inhibitor binding is governed by equilibrium constant K_i
• Formation of EI complex means less E available to bind substrate
• Inhibitor and substrate compete for available enzyme – high [S] can overcome competitive inhibitor
• S and I often share same binding site and may resemble one another in terms of chemical structure

Question 12

Enzyme behaviour in presence of competitive inhibitor:

effect of increasing [S] when [I] is held constant

Answer 12

• No effect on V_max , but apparent K_M is increased: K_M’ = K_M (1 + [I]/K_i)
• Inhibition factor = 1 + [I] / K_i (increases as [I] increases)
  
  • K_i is a characteristic constant for each inhibitor (like K_M)
• If [I] is set equal to K_i, K_M’ = K_M (1 + 1) = 2 K_M
  
  • K_i is the concentration of inhibitor [I] that causes K_M to double

Question 13

Competitive inhibition can be easily seen using what?

Answer 13

Competitive inhibition can be easily seen using a Lineweaver-Burk plot

• different lines represent different [I]
• no effect on V_max , so all graph lines have same y intercept (y-int = 1/V_max)
• K_M’ increases as [I] increases, so x intercept gets smaller (x-int = - 1/K_M’)
• slope increases as [I] increases; L-B slope = K_M‘/V_max

Question 14

When does non-competitive inhibition arise?

Answer 14

• Non-competitive inhibition arises when inhibitor can bind to both E and ES
• inhibitor binding is governed by equilibrium constant K_i
• Formation of EI and EIS means less ES to undergo catalysis, but substrate can still bind to EI without yielding product
• Inhibitor binding site is different from substrate binding site
• Bound inhibitor may disorganize the catalytic component of enzyme
• If EI and EIS steps each have a different K_i, get mixed inhibition

Question 15

Enzyme behaviour in the presence of non-competitive inhibitor:

effect of increasing [S] when [I] is held constant

Answer 15

• V’_max decreases as [I] increases: V’_max = V_max / (1 + [I] / K_i)
• K_M is unchanged (mixed inhibition may show a small effect)
• If [I] is set equal to K_i, inhibitation factor = (1+1) =2

K_i is the concentration of inhibitor [I] that causes V_max to halve

Question 16

Non-competitive inhibition can be easily seen using what?

Answer 16

• different lines represent different [I]
• no effect on K_M, so all graph lines have same x intercept (x-int = -1/K_M)
• V’_max decreases as [I] increases, so y intercept gets bigger (y int = 1/ V’_max)
• Slope increases as [I] increases;

L-B slope = K_M/ V’_max

• For mixed inhibition, lines meet above the x-axis

Question 17

What is the summary of inhibition effects?

Answer 17

• K_M is increased by multiplying by (1 + [I] / K_i)
• V_max is decreased by dividing by (1 + [I] / K_i)
• For mixed inhibition, both V_max and K_M change