1. Enzyme Kinetics vo=vo= V max [S] K m + [S] KmKm V max & E1 E2 E3 1st order zero order CompetitiveNon-competitive Uncompetitive Direct plot Double reciprocal Bi-substrate reaction also follows M-M equation, but one of the substrate should be saturated when estimate the other Affinity with substrate Maximum velocity Inhibition Activity Observe v o change under various [S], resulted plots yield V max and K m k 3 [E t ] k cat Turn over number k cat / K m Activity Unit 1 m mole min Specific Activity unit mg Significance Juang RH (2004) BCbasics

2. Significance of Enzyme Kinetics v o = V max [S] K m + [S] Obtain V max and K m [S] = Low → High[S] = Fixed concentration zero order 1st order E3 E2 E1 Proportional to enzyme concentration v 0 = V max × K = k 3 [Et] × K Juang RH (2004) BCbasics

3. K m = [S] K m + [S] = 2 [S] V max 2 = V max [S] K m + [S] K m : Affinity with Substrate If v o = V max 2 S2 S1 S3 S1 S2 S3 V max 1/2 When using different substrate Affinity changes KmKm v o = V max [S] K m + [S] Juang RH (2004) BCbasics

4. K m : Hexokinase Example Glucose + ATP → Glc-6-P + ADP 123456123456 GlucoseAlloseMannose Substrate number K m = 8 8,000 5 mM CHO H-C-OH HO-C-H H-C-OH H-C-OH H 2 -C-OH CHO H-C-OH H-C-OH H-C-OH H-C-OH H 2 -C-OH CHO HO-C-H HO-C-H H-C-OH H-C-OH H 2 -C-OH Juang RH (2004) BCbasics

5. Turn Over Number, k cat v o = V max [S] K m + [S] (vo)(vo) E + S ES E + P k2k2 k1k1 k3k3 When substrate excess, k 3 = k cat, turn over number (t.o.n) When [substrate] is low = k 3 [E][S] K m + [S] = KmKm k3k3 [E][S] Omit the [substrate] Substrate specificity Start from M-M eq. Second order (IV) Juang RH (2004) BCbasics

6. Turn Over Numbers of Enzymes Catalase H 2 O 2 Carbonic anhydrase HCO 3 - Acetylcholinesterase Acetylcholine 40,000,000 400,000 140,000 b-Lactamase Benzylpenicillin 2,000 Fumarase Fumarate 800 RecA protein (ATPase) ATP 0.4 EnzymesSubstrate k cat (s -1 ) The number of product transformed from substrate by one enzyme molecule in one second Adapted from Nelson & Cox (2000) Lehninger Principles of Biochemistry (3e) p.263

7. Chymotrypsin Has Distinct k cat / K m to Different Substrates O R O H 3 C–C–N–C–C–O–CH 3 H H = – = –– –HGlycine k cat / K m 1.3 ╳ 10 -1 –CH 2 –CH 2 –CH 3 Norvaline 3.6 ╳ 10 2 –CH 2 –CH 2 –CH 2 –CH 3 Norleucine 3.0 ╳ 10 3 –CH 2 – Phenylalanine 1.0 ╳ 10 5 (M -1 s -1 ) R = Adapted from Mathews et al (2000) Biochemistry (3e) p.379

8. Enzyme Activity Unit Reaction time (min) Product [P] 0 10 20 30 40 Slope tan S → P m mole v o = [P] / min Unit = Activity Units Protein (mg) t m mole / min y x yxyx = tan Juang RH (2004) BCbasics Specific Activity =