6.2 Enzyme Kinetics: Kinetic Parameters Used to Compare Enzymes

Michaelis Constant (Km)

Km : represents a relationship b/w various rate constants

• = the [S] needed to get to 50% max velocity

  • units: M

• affinity of enzyme for substrate

• Km is a constant

• often near physiological [substrate]

• DOESNT tell us abt catalysis

lower Km = better substrate for that enzyme

• means the enzyme has high affinity for that substrate & binds it easily, thus little is needed to hit half-speed

Turnover Number (kcat)

kcat = represents the ability of a single enzyme to catalyze product when its saturated with substrate

• going from ES to P + E

• kcat is a constant

  • units: s^-1

• a normalized value of Vmax (max rate), which depends on [enzyme]

  • note: Vmax isn’t a constant, thus is influenced by kcat

• units: per second

  

higher kcat = enzyme converts substrate to product faster, thus is better

→ THUS, the best enzyme would have a LOWER Km (higher affinity for substrate, thus less [s] is needed to be half-speed) & HIGHER kcat (faster conversion of substrate to product)

Specificity Constant

Specificity constant = a larger value can indicate catalytic perfection of an enzyme

• used to determine how good an enzyme is at converting sub→ product & can be used to compare enzymes

• a 2nd order rate constant

  • units: M^-1 s^-1

• has a max limit: upper limit of diffusion

  • some enzymes reach this, so they’re rxn immediately occurs

larger specificity constant = a better enzyme, bc kcat is increasing & Km is decreasing → THUS approaching catalytic perfection

Specificity Constant: Derivation of the Equation

1. take M-M equation and replace Vmax with what it equals bc it contains kcat

2. if [S] « Km (meaning [S] is less than the [S] needed to reach 50% max velocity) then V0 equals:

• bc if [S] is a lot less than Km, the denom is basically equal to Km → thus we can approximate it as equal to Km

• this is a pseudo 2nd order rate equation: bc its constant x [ ] x [ ]

  • 1st order would be constant x [ ]

3. final specificity constant equation

Q. The enzyme fumarase has a Km of 25.0µmol/L and a kcat of 900 s^-1 for malate. Is this close to catalytic perfection?

• malate is the substrate, and its asking for the specificity constant

  • have to convert 25µmol/L → M (÷ by 1million or x by 10^-6)

1. calc specify constant

2. is the specific constant near the upper limit of diffusion ( ~10^8 to 10^9)