BIOCHEM Exam 2 Lecture 8

increase the rate at which a reaction approaches equilibrium

promote chemical reactions without undergoing permanent change themselves

enzymes are biological catalysts

• most enzymes are proteins but not all
• -ribozymes (RNA)

the specific area of the enzyme where the reactions occur

a 3D cleft or crevice in the enzyme that contains catalytic residues

active sites have many more amino acids than participate in the reaction --> proper folding

• geometric complementarity
• electronic complementarity
• specificity depends on arrangement of atoms in active site
• substrate binds in active site through weak interactions

???  enzyme and substrate fit together like lock and key....?

• change occurs in substrate
• results in some sort of strain in the enzyme and the enzyme changes conformation

• apoenzyme = enzyme w/o cofactor
• holoenzyme = enzyme w/ cofactor 

apoenzyme + cofactor --> holoenzyme

the transition state is an activate form of a molecule that has undergone a partial chem rxn

the activation energy delta G ++ is the input of energy need to reach the transition state

• enzymes work because they lower delta G ++
• -stabilize transition state
• -provide alternate mechanism or rxn pathway from reactants to product

• the energy derived from enzyme - substrate reactions
• activation energy is lowered due to the binding energy released from formation of enzyme substrate complex
• holds substrates in the proper orientation to react

• enzymes do not effect delta g
• enzymes do not effect keq

S + E --> E + P

• the study of rates of chemical reactions
• -rate can be defined as decrease in A over time or increase in P over time (for A-->P)
• -rate constant k

• single reactant, single product
• A-->P
• V=k[A] (units s-1)

• Two reactants, single product
• A+B-->P
• V+k[A][B]
• k has units of mol x s^-1

• E+S-->E+P
• [E] is constant
• velocity depends on S and is hyperbolic

• assumes enzymes reversibly associate to make an enzyme substrate [ES] complex
• the saturation curve on v vs. [S] curves suggested that an ES exists
• E+S-->ES, k1 is the rate constant for forming, k-1 is constant for dissociation
• assuming rapid eq, k1=k-1, product is formed in second step E+S-->ES-->E+P

• 1. steady state assumption
• -assumes that [ES] reaches steady state value quickly, that it is formed from E+S as quickly as it is broken down to P
• 2. initial velocity: [P] is negligible at early times in the rxn
• -experimentally can measure velocity before P has a chance to accumulate E+P<-->E+S becomes significant, v= v0 = k2[ES]

• rate eqtn for one substrate enzyme catalyzed rxn
• v0 = Vmax[S]
•         km+[S]
• vmax determined from asymptote
• km determined from [S] at 1/2 max

• double reciprocal plot of 1/vo as a function of 1/[S]
• 1/vo = (Km/Vmax)*(1/[S])+(1/Vmax)
• in the form y = mx + b
• x intercept = -1/km
• y intercept = 1 /vmax

• michaelis constant
• concentration of substrate at 1/2 maximal rxn velocity
• units = mass/vol, M
• ratio of rates of diss. (ES break) to rates of assoc. (ES form)
• km = k-1+k2
•             k1
• the constant for a given enzyme and substrate

• the velocity of the reaction when the enzyme is fully saturated
• vmax =k2[E_T] where k2 is rate of ES-->P
• theoretical max, not measurable
• units of conc/time (M/sec)

• the rate of substrate converted to product when the enzyme is fully saturated with enzyme 
• k_cat = V_max
•            [E_T]
• units s^-1

• kcat/km
• allows the comparison of different enzymes or different substrates of the same enzyme
• depends on: affitinity of the enzyme for a given substrate, rate of catalysis w/ that substrate
• measure of catalytic efficiency
• can be used to compare substrate preference for an enzyme
• kcat/km near 10⁸ or 10⁹ M^-1sec^-1 near limit of diffusion

• interfering with substrate binding
• interfering with substrate turnover
• reversible inhibitors associate and dissociate rapidly from enzyme 
• irreversible inhibitors dissociate very slowly, it at all

• compete with substrate for active site
• usually resembles substrate but isn't reactive
• competitive inhibitors can be overcome by increasing the amount of substrate so that [S] >>[I]

• can only bind to ES complex
• does not affect enzyme substrate binding 
• cannot overcome inhibition by increasing [S]

• can bind to E or ES
• inhibitor binds at a distinct site other than active site 
• does not interfere with substrate binding

• -no change in vmax
• -km increases

• vmax decreases
• -1/v intercept does up as inhibitor binds ES complex

• KM appears to decrease
• -inhibitor decrease [ES] beacuse ESI is unproductive
• -appears to need less S to reach 1/2 max velocity

• vmax decreases
• -1/v intercept increases in value

• Km may not change
• -no change if inhibitor has no impact on substrate binding