BCH210H1 Lecture Notes - Lecture 20: Enzyme Kinetics, Turnover Number, Enzyme Inhibitor
25 May 2018
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Lecture 20: Enzyme Inhibition
Understanding Km
• The kinetic activator constant
• Km is a constant derived from rate constants
• Km is, under true Michaelis-Menten conditions, an estimate of the dissociation constant of E from S
• Also affinity – small Km means tight binding; high Km means weak binding
Physiological consequence of Km
• Mitochondrial form of AD = low Km
- Responsible for carrying out majority of rxn
• Cytoplasmic form of AD = high Km
- Needs a lot of acetaldehyde to get enzyme
working (single AA mutation in mito AD → enzyme
inactive, rely solely on cyto AD)
Understanding Vmax
• The theoretical maximal velocity
• Vmax is a constant and a theoretical maximal rate of the reaction
- But it is NEVER achieved in reality (asymptotically)
• To reach Vmax, would require that ALL enzyme molecules are
tightly bound with substrate
• Vmax is asymptotically approached as substrate is increased
• Always at saturating conditions
Lineweaver-Burk (LB) Plot
• Took reciprocal plot of M-M equation
The turnover number defines activity of one enzyme molecule
• A measure of catalytic activity – how efficient is your enzyme
• Kcat, the turnover number is the number of substrate molecules converted to product per enzyme molecule per
unit of time, when E is saturated with substrate
• Larger the Kcat, the faster the enzyme works
• If the M-M model fits, k2(RLS) = kcat (under saturating condition) = Vmax/Etotal
- Vmax = kcat x enzyme total
• Values of Kcat range from less than 1/sec to many millions per sec
Question
• What value of [S] as a fraction of Km is required to obtain 20% Vmax?
- V0 = Vmax[S]/Km + [S]
- Set Vmax to 100, V0 to 20
- Km must be ¼ of [S] to get 20% Vmax
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Document Summary
Needs a lot of acetaldehyde to get enzyme working (single aa mutation in mito ad enzyme inactive, rely solely on cyto ad) Understanding vmax: the theoretical maximal velocity, vmax is a constant and a theoretical maximal rate of the reaction. But it is never achieved in reality (asymptotically: to reach vmax, would require that all enzyme molecules are tightly bound with substrate, vmax is asymptotically approached as substrate is increased, always at saturating conditions. Lineweaver-burk (lb) plot: took reciprocal plot of m-m equation. Larger the kcat, the faster the enzyme works. If the m-m model fits, k2(rls) = kcat (under saturating condition) = vmax/etotal. Vmax = kcat x enzyme total: values of kcat range from less than 1/sec to many millions per sec. Km must be of [s] to get 20% vmax. Inhibitor competes with substrate (inhibitor binds to active site)
