Etiket Arşivleri: Quaternary structure

Enzymes

ENZYMES

Submitted to:

Dr. Gul Shahnaz

Submitted by:

Group 1 A
Adeel Afzal
Fatima Saleh
Fatima Zahid
Maida Shafiq
Najia Hafeez

CONTENTS

Chemistry
Classification
Mechanism of Enzyme Action
Enzyme Kinetics
Inhibition
Activation
Specificity

CHEMISTRY

Introduction

 Enzymes are biological catalysts that speed up the rate of the biochemical reaction.

 Most enzymes are three dimensional globular proteins (tertiary and quaternary structure).

 Some special RNA species also act as enzymes and are called Ribozymes e.g. hammerhead ribozyme.

STRUCTURE OF ENZYMES

 The active site of an enzyme is the region that binds substrates, co-factors and prosthetic groups and contains residue that helps to hold the substrate.

 Active sites generally occupy less than 5% of the total surface area of enzyme.

 Active site has a specific shape due to tertiary structure of protein.

 A change in the shape of protein affects the shape of active site and function of the enzyme.


Food System Lecture Note

Covalent bond : electron pairs are shared between atoms.

Non – Covalent bonds :

1)    Ionic bonds : result from the electrostatic attraction between two ionized groups of opposite charge.

2)    Hydrogen bounding : is an attraction between hydrogen atom covalently bounded to highly electronegative atom on a second electronegative atom

3)    Vander Waals Forces : are short range attractive forces. Vander Waals forces are weaker than ionic and hydrogen bonding.

Intermolecular force : is an attraction. The net attraction is the sum of pair of      interactions between the molecules.

4)    Hydrophobic Interactions : Molecules such as proteins have polar and nonpolar groups.

polar groups : have tendency to be solvated in water.

nonpolar groups : do not have tendency to contact with water.

Amino Acids : are the building blocks of proteins.

>>>        >>>

R Groups:

         1) Nonpolar or Hydrophobic R groups

         2) Unchanged , neutral R groups

         3) Polar negatively charged R groups

         4) Polar positive charged R groups

Properties of Amino Acids :

1)    They are white crystillane substances.

2)    They are much soluble in water than nonpolar ( organic ) solvent.

3)    Net charge , solubility , reactivity in chemical rxn and  acception behaviour of amino acids is related to the chemical nature of R groups. The solubility of proteins in water is related to essentially  to the distribution of the polar and nonpolar groups of the side chain

4)    Crystalline amino acids have relatively high melting points ( above 200 ). Ionic compounds exhibith high melting points because all the strong ionic attraction between the ionized groups.

5)    Aminoacids in neutral aqeous solutions have both positive and negative charges.

A dipolar ion of any amino acid act as an acid ( proton donor ) as a base ( proton accepter )

Any molecule containing both groups and being able to act as an acid or a base is called an amphoteric molecule.

PEPTIDE: small # of AA linked in defined sequence.

In proteins , the – carboxyl groups of one AA is joined to the -amino group of other amino acid by peptide bond.

Disulfide Bond :

-SH ( sulfydryl group ) is highly reactive.

interchain bond

Properties of Peptides :

1)    Usually high melting points

2)    – carboxyl groups and -amino groups in the peptide linkages can not ionize in the ph zone from 0 to 14

3)    Acid – Base properties of peptides are determined by free -amino group of the N-terminal residue , the free  -carboxyl group of C-terminal residue and those R groups of the residues which can ionize.

4)    Give the avarage elemental composition of proteins basis of the percent C,N,H,O …

Conformation of Polypeptide Chains:

         1) Primary Structure: involves the sequenced amino acids. Zig-zag shape with R groups protruding in opposite directions. NH group does not protonate between pH 0 to 14. The C-N link is not able to rotate freely ( very stable ).

CO-NH atoms involved in the peptide bond and two -C atoms are in the sample plane. A series of rigid planes are seperated by HCR groups.

The primary structured proteins are determined by breaking the peptide bond by hydrolysis reactions under the catalytic activity of enzymes or by acid hydrolysis.

2) Secondary Structure : H bonding between O and H of  amino group.

·         – helix

·         -plated sheet parallel , anti-parallel

helix structure : is governed by intramolecular H – bonding. -helix structure

is an ordered particularly stable structure. Side chains are located outside of the helix. Each peptide bond is engaged in the formation of H – bond , the helixal structure has hight stability and this structure resricts the H bonding with other molecules esp with water.

-plated sheet : a rises from the intermolecular H bonding. The side chains are located above and beneath the plane of the sheets , and their charges and long R groups have a little effect on the structure.

         3) Tertiary Structure : non-covalent interactions  if chains

·         electrostatic interactions

·         hydrogen bonding

·         vander waals interaction

·         interactional non-polar side chains

·         disulfide bonds

4) Quaternary Structure : non-covalent associate ion of polypeptide. chain subunits >> identical subunits , different subunits.

         Denaturation Of Proteins : ( change in the protein conformation ) Acid , alkali , concentrated soline  solutions , heat , radiation.

Protein denaturation is modification of secondary , tertiary or quaternary structure not accompained by the rupture of peptide bonds involved in the primary structure.

Pysical Agents :

·   Heat

·   Cold

·   Mechanical Treatment

·   Heigh pressure

·   Interfaces : adsorption at the interface >> irreversible denaturation

·   Irradiation

o   electromagnetic radiation

o   ultraviolet radiation is observed by aromatic a.a residues.

Chemical Agents :

·   Acid and alkalies

·   Metals

·   Organic solvents

·   Aq. solutions of organic compounds urea , Quanidine salt

·   Quanidine Hydrochloride and Ures  are widely used denaturating agents used for the preparation of partially unfolded proteins.

Determination of Sequence of Amino Acids:

  1. Break All Cys-S-S-Cys disulfide bonds :  mercaptoethonol urea used to cleaves all S-S bond

  2. Determine the amino acid composition : 6N HCI for abouth 20-24 hrs at 100 C all peptide bonds are broken down. Aminoacids >> molar ratios are determined by aminoacids analysis. Ninhydrin and fluorescomine are used for detection of a.a

  3. Identıfy N-terminal and C-terminal a.a :

  • Identification of N terminal a.a : Edman degradation and Sanger’s methot used.Edman reagent is Pheynlisothiocyanate and sanger’s method reagent is fluorodinitrobenzene.

  • Identification of C terminal a.a : Carboxypeptidase and hydrazine are used to determine C – terminal a.a because both enzymes do not react.

  1. Break polypeptides into fragments , determine the a.a composition of each fragment

  2. Determine a.a sequence of each fragment and determine sequence of polypeptide.

  3. Lorate S-S bond

Protein – Water Interactions: The amount water located inside the protein or strongly absorbed to specific surface sites  >>> 0.3 g /g dry protein.

The amount of water that occupies the first layer adjacent to protein >> 0.3g/g dry protein. Water interacts with protein through :

–       Their peptide bonds ( H – bonding , dipole-dipole )

–       Their a.a side chains ( ionized group , polar )

Water Solubility of Proteins :

·   Influence of pH: PI = isoelectric point

                                       At PI  # of negative = # of positive charges

                                       Net charge of protein is min. At pH value higher or lower than the isometric point >> protein carries a negative or positive electric charge  >> water molecules may interact with those charges so solubility of protein increases. (). Protein chains carrying the same electric charge have a tendency to repel each other to dissociate or unfold >> more groups capable interacting with water so solubility increases. () At PI there is minimum repulsion and no interactions with solvent , protein molecules interact with each other >>> precipitation . Hydrophobic interaction >>> aggregates >>> precipitation

·   Influence of İonic Strenght :  .At 0.5 -1.0 M >> solubility of protein increases >> salting in .  The ions react with charge of proteins and decrease the electrostatic attraction between the opposite charge of neigbouring molecules.

Conc. of salts > 1.0 M >>> solubility of proteins decreases >>> salting out. Mostly the water molecules interact with ions . Not enough water molecule is avaible for the interaction with protein so the protein –protein interaction occurs >>> aggregation >>> precipitation.

·   Influence of Temperature : ( at constant  and pH ) . Solubility of protein with T  in 0 – 40 -50 C . The molecular motion becomes sufficent to distrupt the bonds involved the 2 dary 3 ery structures >>> partial denaturation >>> solubility . Above  40 – 50 C , the complete denaturation of protein can occur. Through hydrohobic interaction between the protein molecules >>> aggregation >> precipitation >>> decrease solubility.

·   Influence of Nonaqueous Solvents: The electrostatic forces of repulsion among protein molecules >>> contributes to the aggregation.

General Information:

Complex macromolecules  %50 more of dry  weight of living organisms.

>>> Homoproteins composed of all a.a

>>>Heteroproteins : a.a + prosthetic groups such as nucleoproteins , lipoproteins , phospoproteins , hemoproteins

According to the conformatios : fibrous proteins and globular proteins.

According to their functions : Structure proteins protein with biological activity , food proteins >>> are those that are palatable , digestible , nontoxic and available economically for humans, essentiall a.a such as isoleurine , leucine , lysine , methionine

Surface Tension :

Liquids have tendency to have the minimum surface to maintain minimum free energy.Surface tension is the energy or work required to increase the surface area of a liquid by unit amount . As the temperature –and hence the intensity of molecular motion increases , intermolecular forces become less effective. Less work is required to extend the surface of a liquid surface tension decreases with increase temperature.

1.    Cohesive Forces : intermolecular forces between like molecules.

2.    Adhesive Forces : intermolecular forces between unlike molecules.

Substances that reduce surface tensioned water and allow it to spread more easiyl are known as wetting agents

Ex: water >> have concave meniscus  because adhesive forces > cohesive forces

mercury >> have convey meniscus because adhesive forces < cohesive forces

air – water >>> surface tension  ,  oil – water >>> interfacial tension

The interfacial tension of a binary system lies between the surface tensions of the two pure liquid at the same temperature.

The presence of interface between two liquid phases indicates that the intermolecular forces between the molecule in the liquids are not balanced.

Adsorption from solution :

Adsorption is the phenomenon whereby dissolved molecules accumulate at the surface in greater concentrations than conc. which obtained when the molecules are randomly distributed through out the system.

Two types of surface active materials :

·         small  molecule surfactant  ( are considered  to adsorp reversibly )

·         macromolecules such as protein  ( are considered  to adsorp irreversibly )

Gibbs adsorption equation :

Materials that adsorp strongly at the interface and cause substantial lowering of the surface tension at low conc. are called surfactant ( small molecules + protein )

Surface properties of proteins:

·   a soluble protein

·   diffusion towards interface

·   adsorption >>> once a portion of protein contacts the interface than the nonpolar a.a residues orient toward the nonaqueous phase.

·   rearrangement >> the remainder of the protein adsorbs spontenously

Absorption and unfolding in achieved easily in a short time by flexible proteins >>> good emulsifiers . Globular proteins need a long time for unfolding after adsorption even adsorption might be slow with globular proteins.

BIOL0280 First Midterm Examination 2013

Is the molecule shown D- or L-alanine? D-alanine 2. [2 points] What kind of reaction is peptide bond formation (eg. redox, cleavage, hydrolysis, etc)? a condensation reaction 3. [2 points] In a highly basic solution, pH = 13, the dominant form of glycine is: A) NH —CH —COOH. 2 2 – B) NH —CH —COO . 2 2 + – C) NH —CH —COO . 2 3 D) NH +—CH —COOH. 3 2 + – E) NH —CH —COO . 3 2 Circle the correct answer. 4. [6 points] Draw the structure of Met–Leu-Arg in the ionic form that predominates at pH 7. Draw all atoms (including hydrogens) as well as any charges as necessary. O O O H H H N CH C N CH C N CH C O 3 CH2 CH2 CH2 CH2 CH CH2 H C CH S 3 3 CH2 CH3 NH C NH2+ NH2 5. [2 points] Why are peptide bonds planar? The partial double bond character makes the peptide bond planar.

Which of the following refers to particularly stable arrangements of amino acid residues in a protein that give rise to recurring patterns? A) Primary structure B) Secondary structure C) Tertiary structure D) Quaternary structure E) None of the above Circle the correct answer. 7. [2 points] A D-amino acid would interrupt an α helix made of L-amino acids. Another naturally occurring hindrance to the formation of an α helix is the presence of: A) a negatively charged Asp residue. B) two Ala residues side by side. C) a nonpolar residue near the carboxyl terminus. D) a positively charged Lys residue. E) a negatively charged Glu residue. Circle the correct answer. 8. [2 points] Which two amino acids are most commonly found in β-turns? Pro and Gly 9. [4 points] What types of amino acids are typically found in the interior of a water-soluble globular protein and why are these amino acids located in the interior? Hydrophobic amino acid residues cluster away from the surface in globular proteins, so much of the protein’s interior is a tightly packed combination of hydrocarbon and aromatic ring R groups with very few water molecules. 10. [2 points] List two (or more) ways to denature proteins High or low pH, detergents, temperature, chaotropic agents 11. [2 points] In a mixture of the five proteins listed below, which should elute first in size- exclusion (gel- filtration) chromatography? A) cytochrome c M = 13,000 r B) immunoglobulin G M = 145,000 r C) ribonuclease A M = 13,700 r D) RNA polymerase M = 450,000 r E) serum albumin M = 68,500 r Circle the correct answer.

In one sentence, define the term ‘ligand’ A ligand is a molecule that is bound reversibly by a protein. 13. [2 points] In the binding of oxygen to myoglobin, the relationship between the concentration of oxygen and the fraction of binding sites occupied can best be described as: A) linear with a negative slope. B) hyperbolic. C) linear with a positive slope. D) random. E) sigmoidal. Circle the correct answer. 14. [2 points] Carbon monoxide (CO) is toxic to humans because: A) it binds to myoglobin and causes it to denature. B) it is rapidly converted to toxic CO . 2 C) it binds to the globin portion of hemoglobin and prevents the binding of O . 2 D) it binds to the Fe in hemoglobin and prevents the binding of O . 2 E) it binds to the heme portion of hemoglobin and causes heme to unbind from hemoglobin. Circle the correct answer. 15. [2 points] Which of the following is not correct concerning 2,3-bisphosphoglycerate (BPG)? A) It binds at a distance from the heme groups of hemoglobin. B) It binds with lower affinity to fetal hemoglobin than to adult hemoglobin. C) It increases the affinity of hemoglobin for oxygen. D) It is an allosteric modulator. E) It is normally found associated with the hemoglobin extracted from red blood cells. Circle the correct answer. 16. [2 points] Which one of the following statements is true of enzyme catalysts? A) Their catalytic activity is independent of pH. B) They are generally equally active on D and L isomers of a given substrate. C) They can increase the equilibrium constant for a given reaction by a thousand-fold or more. D) They can increase the reaction rate for a given reaction by a thousand-fold or more. E) To be effective, they must be present at the same concentration as their substrate. Circle the correct answer.

In one or two sentences, describe the lock-and-key and the transition state model as proposed by Emil Fischer and Linus Pauling, respectively. Lock-and-key: Enzymes are structurally complementary to their substrates Transition state model: Enzyme active sites are complementary not to the substrate but to the transition state conformation. 2 points each. 18. [2 points] Michaelis and Menten assumed that the overall reaction for an enzyme- catalyzed reaction could be written as k k 1 2 E + S ES E + P k-1 Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: A) k ([E ] – [ES]). 1 t B) k ([E ] – [ES])[S]. 1 t C) k [ES]. 2 D) k [ES] + k [ES]. -1 2 E) k-1 [ES]. Circle the correct answer. 19. [2 points] The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V Substrate added 0 (mmol/min) (mmol/L) ————————————— 217 0.8 325 2 433 4 488 6 647 1,000 ————————————— The K for this enzyme is approximately: m A) 1 mM. B) 1,000 mM. C) 2 mM. D) 4 mM. E) 6 mM. Circle the correct answer.

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by 1/V = K /( V [S]) + 1/V . 0 m max max To determine K from a double-reciprocal plot, you would: m A) multiply the reciprocal of the x-axis intercept by -1. B) multiply the reciprocal of the y-axis intercept by -1. C) take the reciprocal of the x-axis intercept. D) take the reciprocal of the y-axis intercept. E) take the x-axis intercept where V = 1/2 V . 0 max Circle the correct answer. 21. [2 points] The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the: A) dissociation constant. B) half-saturation constant. C) maximum velocity. D) Michaelis-Menten number. E) turnover number. Circle the correct answer. 22. [2 points] Which three amino acids (residues) form the catalytic triad of chymotrypsin? Ser, His, Asp 23. [4 points] Describe the functional contribution of the hydrophobic pocket for the enzymatic reaction catalyzed by chymotrypsin. The hydrophobic pocket accommodates aromatic residues (F, Y, W) and defines substrate binding and enzymatic specificity. 24. [2 points] Indicate with numbers the correct order of steps in the chymotrypsin reaction mechanism (must be all correct to receive credit): ____6_____ release of product 2 ____3_____ release of product 1 ____2_____ ester formation ____5_____ ester hydrolysis ____ 1_____ substrate binding ____4_____ water deprotonation

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that: A) a Glu residue on the enzyme is involved in the reaction. B) a His residue on the enzyme is involved in the reaction. C) the enzyme has a metallic cofactor. D) the enzyme is found in gastric secretions. E) the reaction relies on specific acid-base catalysis. Circle the correct answer. 26. [2 points] A good transition-state analog: A) binds covalently to the enzyme. B) binds to the enzyme more tightly than the substrate. C) binds very weakly to the enzyme. D) is too unstable to isolate. E) must be almost identical to the substrate. Circle the correct answer. 27. [2 points] What is the effect of a double bond on fatty acid structure? Most double bonds in fatty acids are in the cis configuration. This results in a rigid bend in the hydrocarbon chain. 28. [4 points] What is the (molecular) difference between triglycerides and phosphoglycerides and what are their most common functions in cells? Triglycerides: glycerol esterified with three fatty acids Phosphoglycerides: 3-phosphoglycerate esterified with two fatty acids Triglycerides: storage, phosphoglycerides: membrane components 29. [2 points] Which of these is a general feature of the lipid bilayer in all biological membranes? A) Individual lipid molecules in one face (monolayer) of the bilayer readily diffuse (flip-flop) to the other monolayer. B) Polar, but uncharged, compounds readily diffuse across the bilayer. C) Individual lipid molecules are free to diffuse laterally in the surface of the bilayer. D) The bilayer is stabilized by covalent bonds between neighboring phospholipid molecules. E) The polar head groups face inward toward the inside of the bilayer. Circle the correct answer.

In one sentence, define integral membrane proteins. Integral membrane proteins are tightly associated (and span) the membrane and can only be removed by disrupting hydrophobic interactions with detergents or organic solvents. 31. [2 points] Movement of water across membranes is facilitated by proteins called: A) annexins. B) aquaporins. C) hydropermeases. D) selectins. E) transportins. Circle the correct answer. 32. [2 points] Glucose transport into erythrocytes (not into intestinal epithelial cells) is an example of A) primary active transport B) secondary active transport C) facilitated symport D) facilitated uniport E) none of the above Circle the correct answer. 33. [2 points] The molecule shown to the right is a(n) A) aldopentose B) ketopentose C) aldohexose D) ketohexose E) aldotetrose Circle the correct answer. 34. [2 points] When two carbohydrates are epimers: A) they differ only in the configuration around one chiral carbon atom. B) they differ only in the configuration of the anomeric carbon. C) one is a pyranose, the other a furanose. D) one is an aldose, the other a ketose. E) they differ only in the number of carbon atoms. Circle the correct answer.

Which of the following pairs is interconverted in the process of mutarotation? A) α-D-glucose and β-D-fructose B) α-D-glucose and β-D-galactose C) α-D-glucose and β-D-glucosamine D) α-D-glucose and β-D-glucose E) α-D-glucose and β-L-glucose Circle the correct answer. 36. [4 points] a) Explain how it is possible that a polysaccharide molecule, such as glycogen, may have only one reducing end, and yet have many nonreducing ends. The molecule is branched, with each branch ending in a nonreducing end. 2 points for the word branched b) Explain how homopolysaccharides of glucose can have either helical or planar structures. Helical confirmation results from α1-4 linkages between the glucose monomers and planar structures result from β1-4 linkages between the glucose monomers. Hydrogen bonding stabilizes these structures 2 points for discussion of α/β linkages or 1 point for discussion of H bonds 37. [2 points] Penicillin and related drugs inhibit the enzyme ; A) β-lactamase B) transpeptidase C) chymotrypsin D) lysozyme E) hexokinase Circle the correct answer. 38. [2 points] All of the following contribute to the large, negative, free-energy change upon hydrolysis of ATP except: A) electrostatic repulsion in the reactant. B) low activation energy of forward reaction. C) stabilization of products by extra resonance forms. D) stabilization of products by ionization. Circle the correct answer.

Biological oxidation-reduction reactions always involve: A) direct participation of oxygen. B) formation of water. C) transfer of electron(s). D) mitochondria. E) transfer of proton(s). Circle the correct answer. 40. [2 points] The first five reactions of glycolysis (the preparatory phase) transform each glucose molecule that enters into: A) One molecule of fructose-1,6-bisphosphate B) One molecule of dihydroxyacetone phosphate and one molecule of glyceraldehyde-3- phosphate C) Two molecules of glyceraldehyde-3-phosphate D) Two molecules of pyruvate E) Two molecules of dihydroxyacetone phosphate Circle the correct answer. 41. [2 points] Transfer of a high-energy phosphoryl group to ADP, resulting in ATP, occurs when: A) 1,3-bisphosphoglycerate → 3-phosphoglycerate B) phosphoenolpyruvate (PEP) → 2-phosphoglycerate C) 3-phosphoglycerate → 2-phosphoglycerate D) Both A and B E) Both A and C Circle the correct answer. 42. [2 points] The main function of the pentose phosphate pathway is to: A) give the cell an alternative pathway should glycolysis fail. B) supply pentoses and NADPH. C) make use of pyruvate produced in glycolysis. D) supply ATP. E) supply NADH. Circle the correct answer. 43. [2 points] During strenuous exercise, the NADH formed in the glyceraldehyde 3- + phosphate dehydrogenase reaction in skeletal muscle must be reoxidized to NAD if glycolysis is to continue. The most important reaction involved in the reoxidation of NADH is: A) dihydroxyacetone phosphate → glycerol 3-phosphate B) glucose 6-phosphate → fructose 6-phosphate C) pyruvate → lactate D) isocitrate → α-ketoglutarate E) oxaloacetate → malate Circle the correct answer.