W9.4_Enzymes

Question 1

What are enzymes? What are the major classes of enzymes? Describe the variability in specificity in enzymes.

Answer 1

• Proteins, large molecules, only small part of 3D structure is used as active site, may have allosteric sites, chiral to show stereospecificity, may need cofactors(coenzymes)/metal ions/organic molecules
• Major classes: oxidoreductases (oxidation/reduction), transferases (group transfer), hydrolases (hydrolysis), lyases (add or remove groups to form double bonds), isomerases (isomerisation), ligases (ligation of substrates to harm ATP hydrolysis)
• Specificity: varies (some to sequences/some to enantiomers)

Question 2

Explain the equilibrium in enzymatic reactions with the use of K value. How does enzymes affect a chemical equation through its rate?

Answer 2

• ∵ Enzymatic reaction is a chemical reaction
• ∴ Tendency to reach equilibrium is driven by ∆G (should be -ve) = -RTlnK (K still applies in eqm rules)
• K=k(f)/k(b) -> enzymes are catalysed by K^2 fold in both directions
• Enzymes don’t change value of K/position of eqm, but speed up the rate of reaching eqm by increasing kf & kb equally

Question 3

Explain the properties of transition states in enzymatic reactions and how ∆G affects the direction of the chemical reaction.

Answer 3

• S‡: transition state (enzymes help bring substrates into correct orientation to form it)
• Too unstable/high energy state ∴ transient (can go back/forward)
• ∆G decides which is more favourable (-ve: go forward, +ve: go back)
• Net energy change (∆G) is same regardless of presence of enzymes

Question 4

Explain the structure and features of active sites in enzymes.

Answer 4

• Active site: induced fit model (X lock and key model)
• Shape of active site is moulded after binding substrates to fit more precisely
• Only few amino acids involved in 3D cleft, water usually excluded, weak and reversible interactions dominantly
• Other features: product inhibition, denaturation, enzyme rates affected by environment

Question 5

Describe the different orders of reaction. Explain the general trend of rate of reaction in a typical chemical reaction.

Answer 5

• 0th order reaction: V = k (unit: Ms-1) -> rate independent of [reactants]
• 1st order reaction: V = k[A] (unit: s-1) -> rate varies with [species]x1
• 2nd order reaction: V = k[A][B] (unit: M-1s-1) -> rate varies with [species]x2
• Rate (dPdt): slope of tangent to the curve
• Rate drops off as substrate is used up
• Rate can also be sped up with substrate

Question 6

Regarding the step by step enzymatic reactions, explain the k(cat) value.

Answer 6

• Enzymes and substrates combined to form new physical complex (ES/Michaelis complex) -> change to product/dissociate back
• ∵ Initially all enzymes are in the form of ES so [P] is negligible
• ∴ k(-2) is ignored and k(2) = k(cat) (catalysis) -> initially V(0) = k(cat)[ES]
  (1st order expression)

Question 7

Explain the Michaelis-Menten equation and the each of the values in the equation.

Answer 7

• Michaelis-Menten equation: V(0)=V(max)[S(0)]/(K(M)+[S(0)])
• V(max): maximum velocity/rate (mols^-1)
• V(max)=k(cat)[E(0)] (enzyme concentration)
• K(M): Michaelis constant (M or molL^-1), [S(0)] required to reach half-maximal velocity K(M)=(k^-1+k(cat))/k(1)

Question 8

Explain the dissociation constant from the Michaelis-Menten equation and the assumptions made.

Answer 8

• When k(-1)»>k(cat), K(M) = dissociation constant for ES complex (high KM -> weak binding/less catalytic)
• Assumptions: use initial V(0), [S(0)]»>[E(0)], only single enzyme forms the product, negligible product formation without enzyme, no cooperativity/allosteric modulation/inhibition

Question 9

In different cases for different values of S(0) and K(M), explain how V(0) would be calculated.

Answer 9

• When [S(0)]«<K(M) -> V(0)=V(max)[S(0)]/K(M) (∴ Rate ∝ [S(0)], 1st order kinetics)
• When [S(0)]»>K(M) -> V(0)=V(max) (∴ Rate is maximum, 0th order kinetics)
• When [S(0)] = K(M) -> V(0)=V(max)/2 (∴ K(M) is defined as [S(0)] which gives V(max)/2)

Question 10

Explain the turnover number and catalytic efficiency derived from the Michaelis-Menten equation.

Answer 10

• k(cat): turnover number (number of substrate molecules processed per second per active site in V(max))
• Theoretically, enzymes can achieve rate enhancement of 108-1012 fold
• Realistically, much lower catalytic efficiency (k(cat)/K(M)) is observed
• ∵ [S]«<K(M) in normal cells -> low chance of substrates hitting enzymes to change to product
  ∴ Reaction becomes limited by diffusion rate (≈109M^-1s^-1)