Problems from Previous Class

Transcription

1 1 Problems from Previous lass 1. What is K m? What are the units of K m? 2. What is V max? What are the units of V max? 3. Write down the Michaelis-Menten equation. 4. What order of reaction is the reaction A B? What would be the units of the forward and reverse rate constants? 5. What order of reaction is the reaction A + B? What would be the units of the forward and reverse rate constants? 6. For an enzyme which obeys Michaelis-Menten kinetics, how does the rate of the reaction, v, change with [S] when [S] << K m? 7. For the same enzyme at very high [S], how does v change with [S]?

2 2 8. If the data obtained from an enzyme-mediated reaction are plotted as a Lineweaver-Burk plot, how would we obtain the values for K m and V max? 9. ow would a competitive inhibitor change the data points on a Lineweaver-Burk plot? 10. ow would a noncompetitive inhibitor change the data points on a Lineweaver-Burk plot? 11. Define k cat. 12. If V max = 1 mmol/l/sec, [E total ] = 1 µm and K m = 5 µm, has this enzyme achieved kinetic perfection?

3 3 ENZYMES 3 Mechanisms verview We shall examine: 1. Forces governing ES interactions 2. atalytic mechanisms

4 4 1. FRES GVERNING ENZYME SUBSTRATE INTERATINS Three types of noncovalent bonds may be formed between E and S. a. Electrostatic Bonding (Salt bridges) ppositely charged groups interact. The force of electrostatic interaction is given by oulomb's Law. where F qq 1 2 = 2 rd q 1 and q 2 are the charges upon the two groups r is the distance between them (optimum r 2.8Å) D is the dielectric constant of the medium. D of a vacuum is 1 and of water is 80 hence water can reduce the electrostatic interaction between 2 species.

5 5 b. ydrogen Bonding In a hydrogen bond, a hydrogen atom is shared by two other atoms bond bond N N 2.88 Å 3.04 Å donor acceptor donor acceptor -bonds energies range from 3-7 kcal/mol. -bonds are directional - are strongest when donor and acceptor are colinear. Strong Weak

6 6 e.g. ribonuclease R Uridine ring of substrate N N N ß a Threonine side chain of enzyme 2 Serine side chain of enzyme

7 7 Amino acid side chains of peptide main chain can form a variety of -bonds. 1. Side chains of Tryptophan and arginine serve as -bond donors only. 2. Side chains of asparagine, glutamine, serine and threonine can serve as both -bond acceptors and donors. 3. -bonding by lysine, aspartate, glutamate, tyrosine and histidine vary with p. e.g. 2 -acceptor -donor protonated form of glutamate or aspartate 2 -acceptor Ionized form

8 Tryptophan Arginine 8 2 N 2 N N 2 N N N 2 Asparagine glutamine serine threonine 2 N 2 N 2 N 2 N N 2 N 2 lysine aspartate tyrosine histidine 2 N 2 N 2 N 2 N N N 2 N 2

9 9 c. Van Der Waals Interactions Repulsive Attractive vdw contact distance 4 Distance Electronic charge around an atom changes with time and at any specific time the charge distribution is not perfectly symmetric. This transient electronic charge asymmetry alters the electron distribution around neighboring atoms allowing for electronic attraction. This attraction increases as the atoms come closer until the van der Waal's contact distance is reached. If the atoms then approach further, their electron clouds would overlap resulting in strong repulsion. van der Waal s ontact Radii Atom r N S 1.85 P 1.9

10 10 The van der Waal's bond energy for a pair of atoms is only 1 kcal/mol thus a given vdw bond counts for very little. owever, when numerous S atoms approach numerous E atoms vdw forces become significant. vdw forces fade rapidly when the distance between atoms is 1 Å > the vdw contact distance. vdw interactions between E and S are best when the shapes of E and S permit close contact. Thus while a single vdw bond shows no specificity, complementary shapes in E and S afford for many vdw interactions both attractive and repulsive.

11 11 Effects of Water on Interactions Water is polar and nonlinear. 105 o 0.99Å -ve region +ve region ence 2 can weaken electrostatic and -bonding interactions by surrounding a molecule or competing for electrostatic interactions.

12 12 Exam-type questions 1. Which of the following statements about enzyme(e):substrate(s) interactions is true: A. Van der Waal's bond energies between atoms are more significant than hydrogen bonding interactions. B. Electrostatic interactions between charged groups are unimportant in E/S interactions.. Polar solvents increase electrostatic interactions and hydrogen bonding between E and S. D. ydrogen bonding between atoms is weakest when donor and acceptor are co linear. E. Glutamate can act as both a hydrogen bond acceptor and donor.

13 13 2. Repulsive Van der Waal's forces between two molecules A. Decrease when the distance between atoms falls below the Van der Waal's contact distance. B. Involve disruption of hydrogen bonds.. Are highly specific. D. Increase when the electron clouds of atoms overlap. E. Involve disruption of salt bridges.

14 14 2. ATALYTI MEANISMS LYSZYME 3 2 N ß-glycosidic bond 2 N 3 3 LS dissolves certain bacteria by cleaving polysaccharide component of cell wall. PS component made of 2 types of sugar N-acetylglucosamine (NAG) and N- acetyl-muramic acid (NAM). NAG and NAM are linked by β-glycosidic linkages between -1 of sugar and -4 of the other. NAM NAG Lysozyme hydrolyzes the glycosidic bond between -1 of NAM and -4 of NAG. The other bond (between -1 of NAG and -4 of NAM) is not cleaved.

15 Structure of LS 15 LS = 129 amino acid polypeptide chain of MW 14.6 kda. LS is cross-linked by 4 internal disulfide bridges. omplete amino acid sequences known. Dark residues indicate those amino acids that form the binding pocket.

16 3-D structure of LS is known - 45 x 30 x 30 Å. Interior is almost entirely non-polar 16

17 LS contains 4 alpha-helices and 5 ß-sheets. 17

18 Locations of the four S-S (black atoms) bonds shown in two images below 18

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21 a 21

22 22 Where is the active site? nce a protein crystal has been obtained, it is possible to soak the crystal in substrate in order to produce the enzyme/substrate complex. Unfortunately, hexa NAG is rapidly cleaved by crystalline LS and is thus useless in studies attempting to identify the substrate binding site. tri-nag is hydrolyzed very slowly by LS - penetrates LS crystal and binds to active site. tri-nag is a potent competitive inhibitor of LS. Trp 62 N 3 2 A N N B 2 2 N N 3 Main chain Asp 101 Trp 63 N Main hain 59

23 23 X-ray crystallography shows tri-nag binds to LS in a cleft at the surface occupying 1/2 of cleft. 2 A N 3 Asp N B 2 Trp 62 N 2 N Trp 63 N 3 N Main hain 59 Main chain 107 Binding of tri-nag must be mediated by -bonding and vdw bonding as tri-nag lacks ionic groups. Binding studies indicate that residue of tri-nag makes the largest contribution to the affinity of binding. The crystallographic data show that residue makes a large number of -bonds and vdw interactions.

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26 Mode of binding of EXA-NAG (shown in black) to lysozyme. The locations of the sugar residues A, B and (left) are those observed in the tri-nag-lysozyme complex, whereas those of residues D, E and F (right) are inferred by model building. The two shaded amino acids directly participate in catalysis. 26

27 Enzymic Mechanism 1. ow does S bind? tri-nag data suggest room in cleft for 3 more NAG residues - these residues (D, E and F) have been fitted into cleft by model building. Residue D fits only if distorted. 27 Binding affinity (- G ') kcal/mol Keq 1000 B F A E D Keq A B D E F The contribution to binding of each of the 6 NAG residues to total hexa- NAG binding has been determined from careful measurements of binding to LS. The striking finding that residue D makes a negative contribution to binding affinity is consistent with the view that D is bound in a distorted form.

28 xygen A B Distortion of the D-ring of the substrate of lysozyme into a half-chair form: (A) a sugar residue in the normal chair form; (B) on binding to lysozyme, the ring oxygen atom and -5 of residue D move so that -1, -2, -5 and are in the same plane as shown in part. This is supported additionally by the increased affinity of a lactone derivative of tetra-nag for LS where the D residue has halfchair conformation. This conformation is structurally similar to the proposed distorted form of D. NAG N 3 o-planar atoms NAG 3 2 N 3 In lactone NAG, -1, - 2, -4, -5 and are coplanar. arbonium ion derivative of tetra-nag Lactone analog of tetra-nag

29 29 2. Which bond is cleaved? Rate of hydrolysis of NAG oligomers increases markedly with increasing NAG residues. derivative rate of hydrolysis NAG 2 0 NAG 3 1 NAG 4 8 NAG 5 4,000 NAG 6 30,000 NAG 8 30,000 suggest 6 residues needed for optimal binding to cleft and catalysis

30 30 A B BINDING SITE NAG NAG NAG NAG NAG NAG NAM NAG NAM NAG NAM NAG NAG NAG NAG NAG NAG NAG D E F Excluded since tri-nag is stable Excluded since NAM is too big for site Inferred hydrolysis site leavage between A-B and B- excluded by stability of tri-nag. NAM cannot fit into site because of its lactyl side chain. As the PS is a NAG-NAM alternating polymer, this means NAM cannot occupy site E. The only remaining site is D-E. exa-nag is cleaved into tetra-nag and di- NAG supporting this theory. As the bond cleaved is NAM-NAG, these observations preclude cleavage between A-B, B-, -D and E-F.

31 31 3. Which groups are involved in catalysis? 2 18 studies indicate LS cleaves the 1 - bond rather than - 4 bond. D 1 4 E D E

32 32 At p 5, LS works optimally. At this p, aspartate can act as an -acceptor and glutamate as an -acceptor and donor. atalytic groups participating in cleavage appear to be aspartic 52 and glutamic 35. These can act as -bond acceptors and donors permitting abstraction or donation of a proton. rate of catalysis p LS remains active when all of its carboxyl groups are esterified except Glu 35 and Asp 52. In the presence of S, Asp 52 and Glu 35 cannot be esterified but following removal of S, these residues are esterified. Modification of Asp 52 E inactivation.

33 4. Proposed Scheme of atalysis 1. - group of glu 35 donates + to bond between -1 of D ring and the glycosidic oxygen atom thereby cleaving bond. 33 E NAG E NAG Glu D - Asp 52 Glu D - Asp 52 NAG 3 NAG 3

34 2. This creates a +ve charge on -1 of D - a carbonium ion. 3. The di-nag (E-F) diffuses away from enzyme. 4. arbonium ion intermediate reacts and - from solvent and tetra-nag diffuses away from LS. 5. Glu-35 is protonated and enzyme is ready for another round of catalysis. 34 Glu D - - Asp 52 Glu 35 1 D - Asp 52 NAG 3 NAG 3

35 35 The carbonium ion intermediate is stabilized by (i) Asp 52 is in the negatively charged carboxylate form and interacts with -1 of ring D. (ii) D is distorted into a half-chair conformation causing planarity of -1, -2, -5 and ring. This geometry is similar to that of the carbonium ion intermediate.

36 Exam-type questions Lysozyme: A. Interactions with hexa NAG indicate residue E is bound in a distorted form. B. ydrolyzes Tri NAG very slowly because of the inability of tri NAG to bind to lysozyme.. ydrolyzes the glycosidic bond between 1 of N acteylmuramic acid (NAM) and 4 of N acteylglucosamine (NAG). D. Interactions with NAG and NAM are largely electrostatic E. leaves NAG oligomers between residues E and F.

37 37 2. Polysaccharide hydrolysis by lysozyme is A. Mediated by cleavage of the 1 glycosidic oxygen bond of residue D. B. An example of covalent catalysis.. Enhanced by stabilization of a half-chair configuration of residue E due to interaction with Asp 52. D. Initiated by donation of a proton from Glu 35 to the bond between 4 on the E ring and its glycosidic oxygen. E. Inhibited by esterification of carboxyl groups in the presence of polysaccharide.

38 Summary of Lysozyme/enzyme Action E-S interactions include -bonding, vdw interactions and electrostatic interactions. Water can interfere markedly with -bonding and electrostatic reactions. 2. Lysozyme binds substrate in a surface cleft and binding is accounted for by a large number of -bonding and vdw interactions. 3. LS cleaves the glycosidic linkage between -1 and -4 of R-NAM-NAG- R. 4. The active site cleaves the hexameric binding substrate between residues D and E. 5. Glu-35 is 3Å from the hydrolyzed bond and donates a proton to the bond between -1 of the D ring and the glycosidic cleaving the bond. 6. The D-residue -1 atom transiently becomes a carbonium ion being stabilized by adjacent negatively charged Asp 52 of LS and the distortion of the D-ring of the tetrameric sugar. 7. The carbonium ion reacts with solvent - and Glu 35 is protonated. Tetra-NAG is released and the catalytic cycle begins again.