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A223 Biochemistry Notes (Problem 1 to 10)

by: Soh Yi Ling

A223 Biochemistry Notes (Problem 1 to 10) A223

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A223 Biochemistry
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This 70 page Study Guide was uploaded by Soh Yi Ling on Thursday March 3, 2016. The Study Guide belongs to A223 at Republic Polytechnic taught by in Spring 2013. Since its upload, it has received 20 views. For similar materials see A223 Biochemistry in Biochemistry at Republic Polytechnic.


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Date Created: 03/03/16
Problem 1 1) Element are substances that cannot be broken down and it is at their simplest form. 2) Molecule - two or more atoms chemically combined together. 3) Biomolecule - Molecule created by living things and most of the time the molecule is rather big in size. 4) Polymer - Made up of many same repeating monomers. 5) Macromolecule - A large molecule consist of many different molecule and it is larger than polymer in terms of size. 6) Monomer are basic unit of polymer. Many monomers come together to form polymer. Biomolecule (Polymer) Monomer Carbohydrate Monosaccharide Protein Amino acid Nucleic acid Nucleotide Lipid* Fatty acid and glycerol 7) Dehydration synthesis - Chemically combined two monomers together with removal of 1 molecule of water. 8) Hydrolysis - Introduce a molecule of water to separate two monomers from each other. 9) Carbon has 4 valence electrons which allow it to bond with a lot of different element or molecule. If we were to compare group 5 or 6, they cannot form 4 bonds like carbon. 10) Carbon is able to form large molecules because of catenation. 11) Catenation is the process in which carbon form long chain with the same kind of elements.(C-C-C-C-C-C) 12) Carbon only has two shells and this mean that it is able to attract other atom more easily. After that a bond is form and this bond is rather strong as carbon atom only has two shells. In addition the remaining electrons are used to form the other bond and hence it is able to form long chain like structure. 13) Because of 4 valence electrons and formation of strong bonds, it is able to form linear chain, ring structure and branch structure. 14) Most of biomolecules is formed by dehydration synthesis of monomers and in order to break the biomolecules into monomer we used hydrolysis to break the biomolecules. Diagram of Structure Name of Structure Ring structure Linear chain Branched structure Biomolecu Bond Name Bond Image le Polysacchari Glycosidic des bond Fat Ester bond Protein Amide bond /peptide bond Nucleic acid Phosphodiester bond Protein Carbohydrate Lipids Nucleic acid Use to form Serve as a Protect against Storing the entire catalyst building block dessication ‘recipe’ for the enzymes. for building (drying out). making of many kinds of protein Eg. (DNA) polymers. Used to form Storing energy Aid in the trans formation of membrane membranes. protein (receptor protein) Used in cell Act as a Storage of signalling structural energy (hormones) building material (Eg. Cell wall) Act as a Insulate against molecular “tag” cold and attach onto surface of cell and molecules (identification purpose) Absorb shock Involve in cell communication (Hormones endocrine signalling) Eg. Steroid hormone 15) Lipids are insoluble in water. Why carbon is able to be the backbone element of most of the biological molecules? 1) Carbon is unable to lose electrons to form ions and can only form covalent bond with 4 other non-metal atoms. 2) C-C is one of the strongest bonds compared to other elements bonds Eg.(Ge-Ge bond).This make it suitable to form long chain (Catenation). 3) Carbon and carbon triple bond is stronger than C-C bond and C=C bond. What are the factors that allow carbon to form many compounds? 1) The stability of certain carbon bonds, in particular of the C−H bond. 2) The existence of carbon in both sp and sp3 hybridizations so it can form either single or double bond. 3) The ability of carbon to form both chain and cyclic compounds (in which the chain of atoms is joined end to end to form a ring) based upon either carbon atoms alone or carbon atoms in combination with those of other nonmetals (e.g., oxygen, sulfur, nitrogen) and either upon single- or multiple-bond arrangements. 4) The capability of many carbon compounds to exist in isomeric forms (isomers are molecules with identical numbers of the same atoms bonded in different arrangements; such molecules have quite different properties). All but a very few carbon compounds are called organic compounds, and they are discussed in the article chemical compound. Why not silicon? 1) Silicon has the ability to form compounds that require extreme temperature to heat if silicon is used instead of carbon, it will be tough to burn human and there isn’t any need for temperature regulation within our body. 2) Our body do need to respire (O +2C  CO ).2ilicon has great affinity with oxygen (Si + 2O 2 SiO )4, it will be oxidised to form silicon dioxide (solid) making it hard for our respiratory system to remove it compared to CO gas.2 (Removal of waste gas from our body). Triglyceride has 3 fatty acid and 1 glycerol. Problem 2 1) In making of beer malt, water and yeast is needed in the making of beer. 2) Metabolism = Catabolism + Anabolism. 3) Anabolism - Forming of complex substances with simple substance. 4) Catabolism - Forming of simple substance by breaking down complex substances. 5) Metabolism - Breaking and forming of complex substances. - Is a maze of reactions in which biomolecules are broken down and converted to useful form. 6) Metabolic pathway is a series of chemical reaction and also involves sequential conversion of one compound to another. 7) From the above diagram, we could see that I is the metabolite/intermediate. 8) The difference between both of them is that for linear metabolic pathway is that you cannot get back the same product after the whole reaction whereas it is possible for cyclic metabolic pathway. 9) Malt contain starch, in order to make beer we must convert starch to glucose and maltose through hydrolysis of starch. After that from there yeast will make use of the glucose and glycolysis it and obtain energy. Next yeast will make use of pyruvic acid and ferment it into alcohol. 10) The below diagram shows the process of glycolysis of glucose. Detailed diagram of glycolysis 10 Steps in Glycolysis 1) In the beginning, 2 ATP molecules are needed to start the glycolysis. 2) This result in the glucose molecule will have 2 phosphate groups as it take it from ATP molecules. 3) Next the glucose molecule will split into 2 identical molecules (with 1 phosphate group on 1 molecule). 4) Next NAD will bring one phosphate group to the “split glucose” and result in it to have 2 phosphate group on them .Next the “split glucose” will reduce NAD to form NADH. 5) Lastly 2 molecules of ADP will go to one of the “split glucose” molecule to form 2 molecules of ATP. 6) At the end 2 molecules of pyruvic acid is formed, in total 4 molecules of ATP is generated and 2 molecules of NADH are formed. 7) The purpose of this process is to produce energy. 8) Fermentation process of pyruvic acid 9) The conversion of pyruvic acid to lactic acid need an enzyme called lactate dehydrogenase. (Lactic acid pathway) 10) The conversion of pyruvic acid to acetaldehyde need an enzyme called pyruvic decarboxylase. (Alcohol pathway) 11) After that conversion of acetaldehyde need an enzyme called alcohol dehydrogenase. (Alcohol pathway) 12) When our body is unable to produce sufficient energy to meet the demand, anaerobic respiration occurs and this will stimulate our body to produce lactate dehydrogenase. After that this enzyme will convert pyruvic acid into lactate (lactic acid). This process will convert NADH into NAD and NAD is + needed for glycolysis of glucose to produce energy. Lactic Acid Formation Uses of Fermentation 1) ATP is needed to make the glucose unstable as the transfer of phosphate group will make glucose unstable and hence easier to break down. 2) If an enzyme name end with decarboxylase this means than the reaction that the enzyme catalyse produces CO . 2 3) The reason why glucose undergoes so many steps in order to provide energy is to ensure that is no wastage of energy and energy is released when there it is needed. In addition it is to control the release of energy. Problem 3 1) Aerobic respiration will produce 38 ATP molecules. 2 molecules from glycolysis,2 from kreb’s cycle and 34 from oxidative phosphorylation. 2) On other hand anaerobic respiration will produce 2 ATP molecules all molecules is from glycolysis. 3) Pyruvate acid is produced from the glycolysis of glucose. 4) Glycolysis can go on with or without oxygen. 5) Aerobic respiration need oxygen during the process on the other hand the anaerobic respiration does not need any oxygen. 6) In order to determine whether the process is aerobic respiration or anaerobic respiration, we first determine whether the process take in oxygen or not. 7) Kreb’s cycle is a circular metabolic pathway and this is valid provided there is constant concentration of Acetyl coA. Kreb’s Cycle Diagram Detailed Diagram of Kreb’s Cycle Detailed Diagram of Oxidative Phosphorylation 8) Oxaloacetic acid is the substrate in the Kreb’s cycle.(With reference to the circular metabolic pathway) 9) In the diagram (Kreb’s cycle) pyruvate acid is converted to the acetyl coenzyme through a process pyruvate oxidation. This result in 2 moles of NADH and 2 moles of CO 2roduced. 10) After that acetyl coenzyme is produced, it will react with oxaloacetic acid to form citric acid. 11) Next the citric acid will form oxaloacetic acid. It will result in 6 NADH, 4 CO 2nd 2 FADH . 2 12) NADH= 3ATP molecules 13) FADH =2 ATP molecules 14) The reason is that the NADH will be bonded to 3 complex proteins. First one the first complex protein, followed by third complex proteins and lastly the fourth protein complex. This result in NADH to producing 3 ATP molecules. 15) On the other hand, FADH wi2l only be bonded to two complex proteins and this would result in FADH to 2 produce 2 ATP molecules. CoQ-Complex (1-3) Cyt C(complex 3-4) 16) At the beginning of the electron chain process, 1 molecule of NADH will first lose 2 electrons and hydrogen ions. The losing of electrons will provide energy for the first complex protein to pump out hydrogen ions to the intermembrane layer.Next the electrons will be transported to complex protein lll and energy is released.This energy is used to pump protons into the intermembrane layer and after that the electrons will move to complex lV.Energy (from electron) will be released at complex lV after that this energy will also pump out protons into the intermembrane layer and this result in the concentration of the hydrogen ions in the intermembrane layer to be high and the protons want to diffuse back into the matrix due to the concentration gradient difference between matrix and intermembrane layer.The protons will diffuse back into the matrix through ATP synthase. *Electrons will bind to other substance and result in change in energy level of the electrons. This will result in the release of energy and this energy is used to pump protons into the intermembrane space. 17) When hydrogen ions diffuse into ATP synthase, it will cause the axle found in the protein to turn and when the axle turn it will cause ADP and phosphate group to come together closer which stimulate the ADP to merge with phosphate group to form ATP. 18) The H ions will react with 2 electrons and 1 oxygen atom to form water molecule. 19) The remaining NAD molecules will then be used for glycolysis and Kreb’s cycle. 20) Glycolysis occurs in cytoplasm whereas oxidative phosphorylation and electron transport chain occur along the inner membrane of mitochondria. 21) Oxidative phosphorylation is somehow identical to electron transport chain. As oxidative phosphorylation is the process in which electrons from NADH and FADH is 2 being passed down through a series of complex proteins. 22) From the above diagram we can see that a single molecule of NADH is capable of producing 1 hydrogen ion and 2 electrons. No Parameter Aerobic Anaerobic . Respiration Respiration 1. Requirement for Yes No oxygen (Yes/ No?) 2. Breakdown of Complete Incomplete Glucose (Incomplete/ complete) and its final product 3. Amount of ATP 38 2 (Glycolysis) produced per molecule of glucose 4. Location in the Mitochondria Cytoplasm cell where it occurs 23) From the above table we could see that anaerobic respiration is not a complete breakdown as the final product of it can either be alcohol or lactic acid. These two compounds can be further breakdown to provide energy. 24) Glycolysis and Kreb’s cycle will produce 4 ATP molecules, 10 NADH and 2FADH 2 25) Glycolysis will produce 2 ATP molecules and 2 NADH. 26) Right before Kreb’s cycle, pyruvate acid will be converted to Acetyl CoA (Acetyl coenzyme A) and this will happen through a process call pyruvate oxidation. After which it will released 1NADH molecules and C2 . (Only for 1 molecules of pyruvate acid). 27) Kreb’s cycle will produce 6 NADH,2 FADH2,CO2and 2 ATP molecules. 28) Prokaryotic cell is bacteria, they are unicellular organism and they got no distinct organelles. 29) Prokaryotic cell will only produce 36 ATP molecules because 2 of the ATP molecules are being used to transport NADH into the matrix. 30) Eukaryotic cell will produce 38 ATP molecules. 31) The above number may vary as some of the energy will be loss due to release of heat energy. 32) The reason why pyruvate is being converted into Acetyl CoA, is to ensure that no pyruvate acid will escape from the matrix. As we know pyruvate acid is created at cytoplasm and they will diffuse into mitochondria’s matrix. After a period of time it will cause the concentration of pyruvate acid to be higher in the matrix than the cytoplasm and pyruvate acid tend to diffuse out. In order to prevent the leakage of the pyruvic acid, matrix must convert pyruvate acid into Acetyl CoA to maintain the concentration gradient of pyruvate acid and ensure there is constant supply of pyruvate acid diffuse into the matrix to support the Kreb’s cycle. 33) Lipids and protein can provide energy provided there is oxygen. 34) For lipids, it wills first being converted into glucose then the whole process will goes on. 35) Whereas for protein it will be directly converted into Acetyl CoA and after that it will go through the Kreb’s cycle and oxidative phosphorylation. 36) The substrate of kerb cycle is CoA. 37) Glycolysis occurs in cytosol. 38) Electron transport chain occurs in the inner membrane of mitochondria. 39) Glycolysis occurs at cytoplasm of the cell and after that pyruvate acid diffuse into the matrix of mitochondria. 40) Kreb’s cycle occurs at the matrix of mitochondria and electron transport chain occurs along the inner membrane of mitochondria. Production FADH ,2ADH , CO and2ATP molecules FADH 2 NADH CO 2 ATP molecul molecul molecul molecul e e e e Glycolysis 0 2 0 2 Formation of 0 2 2 0 Acetyl CoA (oxidative phosphorylati on) Kreb’s cycle 2 6 4 2 Total 2 10 6 4 1 NADH = 3 ATP molecules 1 FADH 2 2 ATP molecules Detailed Diagram of Electron Transport Dystem (ETS) Blue bold arrow with yellow colour thingy passing through this is the path of electrons movement during electron transport chain/system. 1) Cyt c is the substrate of complex lll and it transport electrons from complex lll to complex lV. It is a peripheral membrane protein. 2) Q is an electron carrier and it is known as ubiquinone it carry electrons to complex lll from complex l and complex ll. 3) In complex lll Q cycle occurs, it tells us the sequential oxidation and reduction of lipid loving electron carrier (Q). 4) First Q coenzyme will carry 2 electron and hydrogen 3+ atoms to Fe -S (complex lll) and from there it will lose one of the electrons and all of the hydrogen atoms. The hydrogen atoms will also be pumped out of membrane to the surrounding and it will form back Q. The first electron will go to Fe-S to Cyt c1 then to Cyt C where the electron will be carried by Cyt C to complex lV. As shown in the below diagram. After that the second electron will bond to Fe -S then to Cyt b after which the electron will bond back to Q (containing one electrons) to from back QH .T2e whole process will go on again. A step by step diagram showing the Q cycle at complex lll. Problem 4 1) ΔG = free energy of products – free energy of reactants 2) ΔG = ΔH – TΔS Delta G = the net change in the gibbs free energy Delta H = the enthalpy change or the change in heat energy. T =Temperature in kelvin. Delta S = the net change in entropy. 3) Transition state is the intermediate stage at which reactants becoming products .In the transition state it is possible for it form either reactants or products .It is tough to isolate this state because at transition state, the molecules have the highest amount of energy and it is rather unstable. 4) Delta G is the net gibbs free energy level and this energy is the amount of useful energy that can be used to fuel certain reaction that needs energy. 5) If the delta G is negative, it means that the reaction is spontaneous and the vice versa applies. 6) Eactis the amount of energy needed to start a reaction. 7) In order to form interaction with other molecules, the molecules must collide with the other molecules with sufficient amount of energy and proper orientation. 8) Graph A is an endergonic reaction because the amount of energy possess by reactants is less than the amount of energy possess by the product. 9) Graph B is an exergonic reaction because the amount of energy possess by reactants is more than the amount of energy possess by the product. 10) Endergonic reaction requires energy whereas exergonic reaction releases energy. 11) ATP molecule is a nucleotide (a class of biomolecule).As ATP contains adenine, nitrogenous base and 3 phosphate groups. 12) Coupled reaction contain of both endergonic and exergonic reactions. In the coupled reaction, exergonic reaction produces energy to fuel the endergonic reaction. 13) Coupled reaction can also be reaction that produce energy and the energy can be used to fuel reaction that requires energy. 14) In coupled reaction, a spontaneous reaction can be used to drive non-spontaneous reaction. 15) The bond between the phosphate groups is called phosphoanhydride bond. 16) Catabolism most of the time is exergonic reaction and on the other hand anabolism is endergonic reaction. As we need energy to form large molecules hence making anabolism reaction endergonic. Classes of Enzymes 1) Kinase - A class of enzyme that transfer phosphate group to another molecule. 2) Isomerase - A class of enzyme that convert one isomer to another 3) Dehydrogenase - A class of enzymes that transfer hydrogen atom to other molecules through oxidation. 17) Energetically favourable mean that the input of the energy is less than or equal to the output energy and most of the time energetically reaction tend to be a spontaneous reaction. 18) Spontaneous reactions are reactions that will occur with little or no input of energy. 19) Phosphorylation-Addition of phosphate group into the molecules. 20) Oxidative and substrate level phosphorylation are metabolic pathways. 21) Metabolism is the oxidation of substrate. Components of ATP molecule ATP consists of 3 phosphate groups, one ribose sugar and a base called adenine. Recharging ATP molecule Oxidation Phosphorylation 1) Oxidation phosphorylation is a process whereby ATP is formed by the transfer of electron from NADH and FADH 2 through electron carriers. 2) The movement of protons into the ATP synthase produce energy to fuel the formation of ATP molecule. From the below equation we can see that the formation of ATP from ADP need energy. The movement of protons through the ATP synthase produces energy to fuel the formation of ATP from ADP. Hence there is a coupling reaction as energy as shown above. 3) Proton diffusion= exergonic reaction 4) ATP synthesis from ADP= endergonic reaction ATP + H 2 <-> ADP + P + HI+ energy 5) An enzyme involved is ATP synthase. 6) Occur in mitochondria. 7) Coupling of proton gradient to ATP synthase. Substrate Level Phosphorylation 1) Phosphate group is transferred to ADP from a molecule that is more energetic than ATP and this kind of biosynthesis is called substrate level phosphorylation. 2) The formation of ATP is coupled to the removal of phosphate group from another molecule. 3) Can occur by cytoplasm and catalyse by various enzymes. Problem 5 Diamond structure Graphite structure 1) Diamond is hard and strong. Whereas graphite is soft and brittle. 2) Diamond is hard and strong because it is held by strong covalent bond and it has a rigid structure. 3) In each layer of carbon atom, the carbon is held by strong covalent bond and the layers between the carbon is held by weak van der Waals forces of attraction. It is this weak interactions that allow the layers of carbon to slide over easily and made the graphite soft and brittle. Structure of Amino Acid 4) An amino acid molecule consists of a carboxylic acid group, amino group, hydrogen atom and a side chain (R) that is bonded to one alpha carbon. 5) The chemical composition of side chain is not fixed and it varies. It is the side chain that is responsible for the properties of the amino acid molecules. 6) Some properties of side chain are acidic, basic, neutral, polar, non-polar, hydrophilic and hydrophobic. Protein 1) Protein is a class of biomolecules and its monomer is amino acid. 2) Proteins are important to us because protein is used to form enzymes, form hormones, form cell structure and form receptor protein. 3) In a protein (polypeptide) , there are more than 20 amino acid. Protein Folding First stage: Primary structure 1) The formation of peptide bond and this is formed by dehydration synthesis of two or more amino acid. 2) It is formed by combining amino group of one amino acid with carboxylic acid of another amino acid. 3) Peptide bond is one of the strongest bonds and this bond can only be broken through hydrolysis (including an enzyme). 4) In primary structure, peptide bond/amide bond is visible. 5) At the end of primary structure, a polypeptide chain is formed. Second stage: Secondary structure *Dotted lines represent hydrogen bonding 1) Alpha helix and beta-pleated can be seen at this stage. 2) Hydrogen bonding is responsible for the formation of alpha helix and beta-pleated sheet. 3) Only one polypeptide chain is involved in this stage. Third structure: Tertiary structure 1) Bonds like hydrophobic attraction, ionic bond .van der Waal attraction and disulphide bridge (only for cysteine amino acid) 2) Some of the protein is able to work and only protein that has one sub-unit is able to work when it has tertiary structure. 3) At this point of time there is only one polypeptide. 4) This stage occurs when there are attractions between the alpha helix and beta-pleated sheet. 5) 3-D structure of protein appears. Fourth reaction: Quaternary structure 1) Various polypeptide come together to form quaternary structure and this structure consisting of more than 1 polypeptide. 2) Protein at this stage is ready to function. Summary of Protein Folding 1) Different bond have different strength and it is this various kind of bond result in protein to have a special shape. If you have stronger bond, you will pull the protein closer together on the other hand if you have weak bond you will pull the protein further apart. If you don’t have any bonds at all, it will result in the protein to be linear in shape. 2) Water can also shape the protein, the R group of protein is responsible for the folding of protein and if the R group of the protein is hydrophobic, it will bond to another protein R group which is also hydrophobic. An example shown below: 3) The above diagram is showing the R group of amino acid. 4) Water also drives the formation of interaction. An example will be if an amino acid R group is hydrophobic and it is placed in aqueous environment it will tend to form interaction with another hydrophobic R group from another amino acid. Bonds: 1) Disulphide bond-It is form by oxidising the side chain and it is possible by the removal of hydrogen atom from sulphur at the side chain of amino acid (R).Disulphide bond will only form when there is oxidising condition. 2) When there is reducing condition, the disulphide bond will break and the sulphur (found in the side chain of amino acid) will gain back hydrogen and forming back 2 cysteine molecules. 3) Disulphide bond will determine the protein structure (tertiary structure). *The blue colour represents the hydrophobic residues. 4) Hydrophobic interaction is an interaction whereby the hydrophobic residues tend to seek out another hydrophobic residue and they form interaction. 5) They try to maintain a hydrophobic domain when they are in aqueous environment. 6) One of the bonds that are found in during tertiary structure and it is also one of the bonds that determine the shape of the protein. 7) It is also one of the bonds that hold the protein subunits together in the quaternary structure. 8) When protein is denatured, it will fold back into the original structure in order to function well provided the denaturation is removed. 9) Polar interactions, ionic interactions and other protein side chain amino acid can stabilise the protein sub-units together. Factors that may affect the denaturation of protein: 1) Salt concentration 2) Temperature 3) pH concentration 4) Reducing agent Folding of protein 1) In order to facilitate the folding of protein, there is a special kind of proteins in our body to ensure that protein is fold correctly and it is called molecular chaperones. 2) Gelatin is at primary structure. 3) Most of the time if the side chain have a ring structure if will be more insoluble. Converting bones to jelly: 1) Bones contains collagen 2) Jelly contains gelatin. Problem 6 Model on how enzymes work: 1)Induced fit model In induced fit model, the enzymes will alter its configuration site just to allow the substrate to bind with the enzymes and after the end of the reaction the enzymes will obtain its original configuration. 2)Lock and key model In lock and key model, only substrate with specific shape can only binds to the active site of the enzymes and the enzymes will not change its configuration state just to allow the substrate to fit onto it. Enzymes: 1) Enzymes are catalyst that alters the rate of reaction by lowering the activation energy needed to start the reaction. 2) Even if the reaction ends, the enzyme will remain unchanged. 3) Enzymes are protein in nature and they can be denatured exactly the same way protein does. 4) When enzyme is bind to a substrate, it will form enzyme-substrate complex. 5) Eais the activation energy and this energy refers to the amount of energy needed to start a reaction. 6) In order for a successful collision to occur, molecule must collide with each with sufficient amount of kinetic energy and proper orientation. 7) After that this energy is converted to potential energy and this required to change the reactants to transition state complex. 8) Enzyme is able to lower activation energy by weaken certain key bonds and hence activation energy is lowered. 9) From the above graph we can see that enzymes will also lower the free energy of the transition state complex. 10) However enzyme does not affect delta g of the reaction. 11) The shape of the active site is determined during protein folding. From the above equation: E - enzyme S - substrate ES - enzyme-substrate complex EP - enzyme binds to the newly formed product P - product The enzyme and the substrate will come together to form enzyme substrate complex and from there it will convert to the product. EP is the 2 intermediate product and from there EP complex will be gone forming enzyme and product. The enzyme remains unchanged and it is noted with E on the right hand side of the equation. Factors affecting enzyme activity: We can determine the enzyme activity be seeing the rate at which product is formed and the rate at which reactant is gone. 1) pH concentration 2) Temperature 3) Substrate and enzyme concentration 1)pH concentration *Noted that the graph is based on the enzymes that work best in room temperature and pressure. Also work best in neutral condition. Different enzymes got different optimum conditions. (Example: pepsin (stomach) works best in acidic condition and 37 C) Low temperature: From graph (a), we can see that at low temperature the enzyme activity is low because at low temperature the kinetic energy in the enzyme and substrate is low. This results in them to collide less often and when this happens the tendency to form enzyme substrate complex is low and eventually it will result in lesser product formed which lead to lower enzyme activity. Optimum temperature: From graph (a), we can see that at this temperature, the enzyme works the best as the temperature does lead to denaturation, the substrate and the enzymes have sufficient kinetic energy to collide with each other to form enzyme substrate complex. This results in the frequency of forming enzyme substrate complex is high and hence resulting in more products formed which lead to higher enzyme activity. High temperature: From graph (a), we can see that at high temperature, there isn’t any enzyme activity. The reason is that, the extreme temperature denatures the enzymes and when this happens the active site is being changed and it will be tough for the substrate binds to the enzymes. When this happens, there will be lesser enzyme substrate complex forms and eventually lesser product is formed. Hence no enzyme activity. Low pH: From the graph (b) we could see that at very low pH, there isn’t any enzyme activity as the low pH lead to the denaturation of the enzyme and result in the active site configuration being altered. This result in the substrate being unable to bind to the enzymes and hence no enzyme substrate complex is formed. When this occurs the number of product formed decrease and there will not be any enzyme activity. Optimum pH: From graph (b) we could see that the enzymes work best at pH 7.When enzymes are placed in that pH, there will not be any denaturation occurs and this mean that the substrate can bind to the enzymes to form enzyme substrate complex more easily. When there is more enzyme substrate complex formed, this result in more product being formed and hence leading to higher enzyme activity. High pH: From the graph (b) we could see that at very high pH, there isn’t any enzyme activity as the high pH lead to the denaturation of the enzyme and result in the active site configuration being altered. This result in the substrate being unable to bind to the enzymes and hence no enzyme substrate complex is formed. When this occurs the number of product formed decrease and there will not be any enzyme activity. Substrate concentration: If there is low substrate concentration, the tendency for the substrate to collide with the enzymes decrease and when this happens it will result in lower frequency of enzyme substrate complex to appear. When this happens, it will result in the formation of product decrease eventually lead to lower enzyme activity. Enzyme concentration: When the substrate concentration is high and the enzyme concentration is low, there will be limited active site and when there is limited active site, even if we increase the substrate concentration, the rate of reaction will still remain the same as it has reach the maximum and it is to form more enzyme substrate complex. (10 enzyme=10 enzyme substrate complex .It is impossible to form 11 enzyme substrate complex with only 10 enzyme at the same time). Progress Curve Progress curve 50 40 30 Conc of product formed (mg20l) 10 0 0 2 4 6 8 101214 Time (mins) Time [P] (minutes) (mg/ml) 0 0 2 11 4 22 6 33 8 39 10 40 12 40 The second graph is using this data plot the second graph above. In progress curve there is a straight line because at that period of time almost all the substrate is used up to form product and hence there is a straight line produce as there is no more substrate to be converted to products. Michaelis-Menten Curve The Michaelis Menten curve tells us the relationship between substrate concentration and the initial rate of reaction. Vmax(mg/ml)*–The maximum velocity of the reaction. *Depends on the graph and situation. Km (mg)* – The concentration of substrate needed in order for the reaction to reach half of the maximum velocity of the reaction. These 2 factors determine the enzyme activity. Higher Km means that the enzyme affinity for the substrate is low and on the other hand lower Km means that the enzyme affinity for the substrate is high. The reason why there is a straight line in M-M graph because all the active site of the enzymes have been taken even if we increase the concentration of the substrate, the rate of reaction will not increase as the enzymes are saturated and this means that there the number of enzymes substrate complex formed has reach maximum number and when this occur, the amount of product formed will remain the same throughout. As the enzymes is working at maximum speed. Vmaxis just an estimation of the maximum speed that a reaction can go. It is tough to determine the maximum speed of the reaction in the lab because during experiment high level of substrate is needed in order to determine the V max and it is impossible to obtain high concentration of substrate in lab. How to plot the Michaelis menten graph? 1) We must plot a progress curve for the enzyme at various concentrations. 2) In order to find the speed of the reaction we find the gradient of the reaction. The initial velocity is the steepest part of the progress curve. 3) Next we have to find the initial velocity of the reaction for all the concentration in which the same enzymes is placed in. 4) After finding all the initial velocity of individual reaction, together with concentration of substrate that is used, we can plot the M-M graph. Progress curve Michaelis-Menten graph Y-axis Concentration Initial rate of reaction of product (mg/min) formed(mg) X-axis Time (min) Substrate concentration (mg) Gradient Velocity of the reaction (mg/min) Measure Measure the Measure the rate of concentration reaction of enzyme of the product catalyzed reaction (SP) at formed against different substrate time concentration. M-M equation = To be used with the graph. v - Intial velocity of the reaction Vmax- Maximum velocity [S] - Substrate concentration Km- Concentration of substrate needed to reach half of Vmax Km is a constant that tells you how loosely or tightly an enzyme is bind to a substrate. Higher Km would mean that the enzyme binds loosely with the substrate and the vice versa applies. Problem 7 Enzyme inhibitor - They decrease enzyme activity by binding to the enzyme. 1) Irreversible inhibitor: Inhibitors that bind irreversibly with the enzyme and this is possible as they form strong bonds with the enzyme (Example: Covalent bond)  Alcohol  Acid and bases  Temperature  Heavy metal  Reducing agent Left side-Before the inhibitor bind to the enzymes. Right side- After the inhibitor binds to the enzyme. From the above diagram we can see that there is a covalent bond formed on the methyl group. 2) Reversible inhibitor: Inhibitor that binds reversibly with the enzyme and this is possible as they form weak interactions with enzymes.  Competitive inhibitor  Non-competitive inhibitor  Uncompetitive inhibitor From the above image, specific means that the only certain kind of inhibitor can bind to specific enzymes. Where non- specific means that the inhibitor can destroy or denature the enzyme without even binding to the enzyme. In fact non- specific can destroy all enzymes. Competitive inhibitor: 1) Inhibitor that binds reversibly with enzymes. 2) Bind at the active site, fight with substrate at the active site and it effect can be reduced by increasing the substrate concentration. 3) It increase Km value but does not affect Vmaxvalue. The reason is because the inhibitor is fighting for the active site with the substrate by doing so the affinity between the enzyme and the substrate decrease and hence this result in higher Km value. The V maxvalue remains the same as sufficient concentration of substrate is able to reduce the effect of the inhibitor. V max Km (Enzyme affinity) Remain the same Increases Non-competitive inhibitor: 1) Mixed inhibitor is inhibitor that can bind to the enzymes directly and also bind to the enzyme substrate complex. Non-competitive inhibitor is an example of mixed inhibitor. 2) Inhibitor that binds reversibly with the enzymes. 3) Does not compete for active site instead bind at another site found on the same enzymes. 4) Can bind to the enzymes or bind to enzyme substrate complex and after that it will change the enzyme active site and this result in the enzyme being unable to bind to substrate. 5) Excess substrate will not reduce the effect of this inhibitor. 6) Non-competitive inhibitor has same affinity for enzyme and enzyme substrate complex. 7) The enzyme affinity (Km) will not decrease because the substrate and inhibitor can bind at the same time and this means that the enzyme affinity for the substrate is not affect. In addition these 3 can come together to form enzyme substrate inhibitor complex. 8) The Vmax (mg/ml)*decreases because ESI complex is unable to produce product. When there is lesser product formed the Vmax value reduced. *Depends on situation. 9) Also the active site of the enzyme is being altered and this will result in the enzyme being unable to function normally and when this happens, the number of products formed per seconds is greatly reduced (rate of reaction) hence the Vmax is reduced. V Km (enzyme affinity) max Lower Remain the same Uncompetitive inhibitor: 1) Can only bind to enzyme substrate complex but not the enzyme. 2) This inhibitor is reversible. 3) This inhibitor can only bind to another site found on the same enzymes. 4) Km lower because if the concentration of ES complex will decrease as the inhibitor will bind to the complex and it result in the concentration of ES to decrease. According to Le chatelier principle, this will result in more E and S to bind together in order to form ES complex and replenish the loss of ES complex due to the presence of the inhibitor. Hence the enzyme affinity for substrate is higher as there is a stimulant to cause E and S to come together compared reaction without inhibitor. 5) Vmax is also lowered because it form ESI complex and when this is formed, no products are formed and in the reaction there will little product formed per minute (velocity) compared to reaction without inhibitor. V max Km (enzyme affinity) Lower Lower Allosteric site is a special site where the inhibitor binds to it and active site is not allosteric site. Allosteric site is applicable for non-competitive and uncompetitive inhibitor. Problem 9 Most of the time our body store excess glucose in the form of glycogen. Conversion of excess glucose to glycogen Enzymes involved: 1) Hexokinase 2) Phosphoglucomutase 3) UDP-glucose pyrophosphorylase 4) Glycogen synthase 5) Branching enzyme. Mechanism in converting glucose to glycogen 1) Excess glucose is being converted into glucose-6-phosphate with the assistance of ATP molecule and after that ADP is being produced. An enzyme called hexokinase facilitate this reaction. 2) Next glucose-6-phosphate will be converted to glucose-1- phosphate with the help of an enzyme called phosphoglucomutase and this conversion is reversible. 3) Next glucose-1-phosphate will react with UTP to form UDP- glucose and PPi. This reaction is being catalysed by an enzyme called UDP-glucose pyrophosphorylase. 4) Next the UDP-glucose will be attached to the chain of residue and an enzyme called glycogen synthase is involved in attaching UDP-glucose onto the chain. In this process, UDP and glycogen is produced. 5) After that a long linear chain will form and from there will be a branching enzyme to break the chain and the chain will be split into 2 (one the main chain and the other the side chain). 6) Next the side chain will bind to the main chain in a branching manner and it will lead to the formation of alpha 1,6 glycosidic bond.This is being catalyse by an enzyme called branching enzyme. Glucose + ATP + hexokinase  Glucose-6-phosphate + ADP Glucose-6-phosphate + phosphoglucomutase  Glucose-1- phosphate Glucose-1-phosphate + UDP + UDP-glucose pyrophosphorylase  UDP-glucose UDP-glucose + (glycogen)  (Glycogen) + UDP (uridine n n+1 diphosphate) Converting glycogen to glucose Enzymes involved: 1) Glycogen phosphorylase 2) Glycogen debranching enzymes 3) Phosphoglucomutase Mechanism in converting glycogen to glucose 1) Glucose phosphorylase break alpha 1-4 glycosidic bond and it will stop 4 residues away from the branching point (including the residue that is being branch out). 2) Glycogen debranching enzymes transfer three residues from the side chain to main chain. 3) Glycogen debranching enzymes break alpha 1-6 glycosidic bond. 4) After the 3 residues is brought to the main chain, glucose phosphorylase will continue it action from there. 5) Phosphoglucomutase convert glucose-1-phosphate to glucose-6-phosphate. 6) Glucose-6-phosphate is being converted to glucose with the help of an enzyme called glucose-6-phosphatase and this only happens in liver as only liver can produce this enzymes. Most of the time the monomer is form and cleaving at the non- reducing end Glycogen synthase cannot create residues from nothing.So a protein is needed to initiate this process and the protein is called glycogenin. Glycogenin is a complex protein consist of 2 subunit and one of the subunits will bring in 8 residues onto the other subunit. Problem 10 Lipids: 1) Commonly known as triglyceride. 2) Made up of 1 glycerol molecule and 3 fatty acid molecules. 3) 3 molecules of fatty acid and 1 molecule of glycerol undergo dehydration synthesis to produce 1 molecule of triglyceride and 3 molecules of water. Glycerol: Fatty acid: 1) Contain a hydrophobic region and a hydrophilic region. 2) The brown part of fatty acid molecule is carboxylic acid group and it is this component that is responsible for hydrophilic of fatty acid. 3) The chain of methyl group is responsible for the hydrophobic properties of fatty acid. Unsaturated and saturated fatty acid: 1) The term saturated is used to describe the number of double bonds in the fatty acid. 2) Saturated fat does not have any double bonds in it. Cis and trans-fat: 3) In natural, cis fat is most common form of fat. 4) If we keep on using the oil for deep frying, eventually the oil will become trans-fat due to the extreme heat and trans-fat is carcinogenic, or cancer-causing. 5) Omega-3 is called omega-3 because it is fat that have at least a single double bond, we count from the omega end -tail (methyl group) and the number indicate number of carbon from the tail to the double bond. 6) Most of the omega are cis-fats. Fats vs Glycogen Property glycogen fat Biomolecule type Carbohydra Lipids te Energy density of the 4 9 molecule (in kcal/g) What amount of the 25kg 11.111kg molecule (in kg) is needed to store 100,000 kcal of energy? Because of its solubility in 50kg 11.111kg water, 1 g of glycogen will usually bind about 2g of water when it is stored. Given this situation, what is the total weight (in kg) required now to store 100,000 kcal of energy? (N.B.: fats are hydrophobic and do not bind water molecules.) 1) From the above table we can see that fat has higher energy density compared to glycogen. 2) Fats have higher energy density compared to glycogen as it has the ability to store more energy in a fixed amount of mass compared to glycogen. 3) In addition, we need 25Kg of glycerol to store 100,000Kcal of energy and on the other hand we only need 11Kg to store 100,000Kcal of energy. Fatty acid synthesis: Formation of citrate: 1) Acetyl CoA will react with oxaloacetate to form citrate and leave the mitochondria into the cytoplasm. This reaction is catalyze by citrate synthase and CoASH is released. 2) This reaction will happen as Acetyl CoA and oxaloacetate cannot leave mitochondria. However citrate can leave mitochondria. Citrate breakdown: 3) After the citrate reaches the cytoplasm, it will be broken down in Acetyl CoA and OAA (oxaloacetate). This will happen with the help of citrate lyase and CoASH. OAAMalate: 4) The OAA will be converted to malate with the help of NADH.NADH will act as a reducing agent by introducing hydrogen atoms to OAA to form Malate. 5) NAD is generated and this is used in glycolysis. Malate Pyruvate: *Left side is Malate Right hand side is Pyruvate + 6) This process release CO 2nd converts NADP into NADPH. 7) NADPH can be used for the reduction of GSSG into 2 GSH. Acetyl CoA  Fatty acid: 8) Some of the acetyl CoA is converted to malonyl CoA. 9) Fatty acid synthase is a large complex polypeptide one of the site contains an amino acid called cysteine and the other contains an amino acid called phosphopantetheine. Both of these amino acid has a thiol group (-S-H) attached to it. 10) The thiol group on the site will only bind to the acetyl group and malonyl group of acetyl CoA and malonyl CoA. 11) Acetyl CoA will bind to the C part of the synthase and malonyl CoA will bind to the P part of the synthase. 12) After that acetyl CoA will move to the malonyl CoA and release carbon dioxide. 13) (-C=O) bond is removed and (-C-H) bond is formed and this is done by introducing hydrogen atom. 14) The growing chain will be transfer from the P site to C site. 15) Another malonyl group can bind to the P site and continue this reaction. 16) When the hydrocarbon chain of the new fatty acid is 16 carbon atoms long the bond joining the fatty acid to the pantetheine-SH carrier site is finally broken and the 16-C saturated fatty acid palmitate is released. Beta oxidation of fatty acid: 1 step: Activation of fatty acid 1) 2 ATP molecules were used to “activate” fatty acid. 2) Activate the fatty acid means adding of CoA group to the fatty acid. 3) This reaction is catalyse by an enzyme called fatty acyl CoA synthetase together with the help of CoASH group (not an enzyme) 4) Fatty acyl CoA synthetase is a thiokinase, it bring in thiol group (-SH) with the expense of ATP. 5) The end product is called fatty acyl CoA. Step 2:Transport fatty acyl CoA into the matrix of mitochondria 1) Fatty acyl CoA will react with carnitine and these reactions is catalyst by an enzyme called carnitine acetyltransferase I. 2) Carnitine acetyltransferase I will transfer canitine to the fatty acid chain. 3) After that the complex formed will move into the matrix of mitochondria. 4) After that when the complex enter the matrix, the complex will be broken into fatty acyl CoA and canitine. 5) These reaction will be catalyse by an enzyme called carnitine acetyltransferase II 6) The canitine will move out of the cytosol when it can be used. 7) Step 1 is need as fatty acyl CoA cannot move into the matrix directly. Step 3:Formation of FADH 2 1) Fatty acyl CoA will undergo oxidation when one hydrogen is lost from alpha carbon and the other one lose from beta carbon. + 2) The 2 hydrogen lost will be taken up by FAD to form FADH 2 3) After that a double bond is formed between alpha carbon and beta carbon. 4) After that the second compound (above diagram) hydration. 5) This will result in the third compound (above diagram) to gain a hydroxyl group on the beta carbon and hydrogen atom in alpha carbon. Step 4: Formation of acetyl CoA 1) The compound will undergo oxidation and 2 hydrogen atoms is removed from beta carbon. 2) This hydrogen will be used to reduce NAD to NADH + H . + 3) The second compound will undergo thiolysis and when this happens acetyl CoA will form together with the remaining chain of the fatty acid. 4) The remaining chain of fatty acid will undergo beta oxidation all over again. Thiolysis-Using of thiol group to cleave a compound into 2.Similar to hydrolysis but thiolysis uses thiol group instead. Calculation of energy released from 1 molecules of palmitic acid (16C) Energy released from breakdown of 1 palmitic acid (16C): Produ Numb No. of ATP Total number of cts er of produced after ATP formed produc products go ts through oxidative phosphorylation Acetyl- 8 8[(3x3)+1+(2x1)] 96 coA NADH 7 21 21 FADH 2 7 14 14 TOTAL ATP 131-2=129 ATPS produced: 1) 2 ATP molecules need to be deduct as these 2 molecules is needed for the activation of fatty acid. 2) 1 Palmitic acid will undergo 7 rounds of beta oxidation as during the seventh round of beta oxidation, 2 acetyl CoA is generated. Detailed step of beta oxidation: End Description Diagram Enzyme product Dehydrogenation b Acyl CoA Trans-Δ - y FAD: The first Dehydrogen enoyl-CoA step is the ase oxidation of the fatty acid by Acyl- CoA- Dehydrogenase. The enzyme catalyzes the formation of a double bond between the C-2 and C-3. Hydration: The next step is the hydration of the bond between Eenoyl CoA L-β- C-2 and C-3. The hydroxyacyl reaction Hydratase CoA is stereospecific, forming only the Lisomer. + Oxidation by NAD :The third step is 3- the oxidation of L- hydroxyacyl- β-hydroxyacyl CoA β-ketoacyl by NAD . This CoA CoA Dehydrogen converts the hydroxyl group ase into a keto group. Thiolysis: The final step is the An acetyl- CoA cleavage of β- molecule, ketoacyl CoA by and an acyl- thethiol group of β- CoA another molecule ketothiolase of Coenzyme A. molecule that is two The thiol is carbons inserted between shorter C-2 and C-3. *Acyl CoA is the remaining chain of the cleaved fatty acid. What happen if there is odd number of carbon in fatty acid? 1) Beta oxidation will still occur but the end product will be acetyl CoA and propionyl CoA. 2) Acetyl CoA will still undergo Kreb’s cycle. 3) From the above diagram we can see that propionyl CoA will be converted to succinyl CoA and succinyl CoA will undergo Kreb’s cycle. What happen to unsaturated fatty acid? 3 3  If the acyl CoA contains a cis-Δ bond, then cis-Δ -Enoyl CoA isomerase will convert the bond to a trans-Δ bond,2 which is a regular substrate. 4  If the acyl CoA contains a cis-Δ double bond, then its dehydrogenation yields a 2,4-dienoyl intermediate, which is not a substrate for enoyl CoA hydratase. However, the enzyme 2,4 Dienoyl CoA reductase reduces the3 intermediate, using NADPH, into trans-Δ -enoyl CoA. As in the above case, this compound is converted into a suitable intermediate by 3,2-Enoyl CoA isomerase. 2 2 Naming: Trans-Δ -enoyl-CoA- The symbol Δ designates the position of the double bond. Trans represent whether the fatty acid is in trans configuration For fatty acid the carbon that is attached to the CoA group is always the first carbon.


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