Bio 201 Set 5 Notes PDF

BIO 201 Set 5 Notes Bioenergetics and Free Energy  Contents (5.1) The Importance of Energy (5.2) Phototrophs and Chemotrophs (5.3) Oxidation and Reduction Reactions (5.4) Flow of Energy through the Biosphere (5.5) Bioenergetics (5.6) Laws of Thermodynamics (5.7) Free Energy (...

BIO 201 Set 5 Notes Bioenergetics and Free Energy  Contents (5.1) The Importance of Energy (5.2) Phototrophs and Chemotrophs (5.3) Oxidation and Reduction Reactions (5.4) Flow of Energy through the Biosphere (5.5) Bioenergetics (5.6) Laws of Thermodynamics (5.7) Free Energy (5.8) Equilibrium and ΔG (5.9) Limitations of DG (5.10) Standard Free Energy Change is DGo’ (5.11) Calculating DG’ (5.12) The Steady State  5.1 – The Importance of Energy Every cell has four essential needs: 1) Construction of molecular building blocks. (Covered in Set 3 notes) 2) Chemical catalysts (i.e. enzymes) (Covered in Set 6 notes) 3) Information to guide activities (i.e. DNA, RNA, etc.) (Covered in Sets 2 and 3) 4) Energy to drive rxns and processes essential to life.  5.1 – The Importance of Energy All living systems require an ongoing supply of energy. Energy can be thought of as the capacity to cause specific chemical or physical change. Physical and chemical changes that occur in the cell can be classified in 6 groups: Making energy-rich molecules from simpler starting molecules. (A) SYNTHETIC WORK: Changes in chemical bonds. Example: The work of biosynthesis results in the formation of new chemical bonds and the synthesis of new molecules.  5.1 – The Importance of Energy (B) MECHANICAL WORK: Change in location or orientation of a cell or a sub-cellular structure. - Mechanical work involves a physical change in the position or orientation of Deadlifting 10 lbs?? a cell or some part of it. Examples:  5.1 – The Importance of Energy (C) CONCENTRATION WORK: Moving molecules or ions across a membrane against a concentration gradient. Requires ATP! Example: Active transport of amino acids into the cell. Concentration of [amino acids] is sometimes low outside of the cell and requires active transport into the cell even though the cell already has a high concentration of [amino acids].  5.1 – The Importance of Energy (D) ELECTRICAL WORK: Moving ions across a membrane against an electrochemical gradient. Example: In the case of mitochondria or chloroplasts the difference is essential in the production of ATP. Why does there need to be a H+ gradient in the mitochondria (as shown in the diagram)?  5.1 – The Importance of Energy (D) ELECTRICAL WORK: Moving ions across a membrane against an electrochemical gradient. 3 Na+ out 2 K+ in Example: To maintain this electrochemical gradient, 3 Na+ ions have to pumped out of the nerve cell (axon) and 2 K+ ions have to be pumped into the nerve cell (axon). This active transport process requires a membrane protein (Na+/K+ ATPase) and requires the hydrolysis of an ATP molecule (for energy).  5.1 – The Importance of Energy (E) HEAT: An increase in temperature that is useful to warm- blooded animals. Living organisms do not use heat as a form of energy, but rather it is a by-product of metabolic processes (for example, oxidation of glucose). Producing heat is very important for homeotherms. ~67% (depends on the person) of your metabolic energy (i.e. glucose oxidation) is being used just to maintain your body temperature at 37oC. Q1. What is a homeotherm? Vasoconstriction (constricting of blood vessels) is another way to conserve heat if you are cold.  5.1 – The Importance of Energy (E) HEAT: An increase in temperature that is useful to warm- blooded animals. Rare example: The skunk cabbage (not a warm-blooded animal!) (S. foetidus) generates heat to melt overlying snow so that it can get a head-start on growing while other plants are still dormant underneath the snow.  5.1 – The Importance of Energy (F) BIOLUMINESCENCE: Production of light. Bioluminescence, the production of light, is important in a number of organisms such as fireflies, certain jellyfish and luminous toadstools. Bioluminescence is generated by the reaction of ATP with luminescent compounds. Don’t memorize this chemical reaction! In Bio 202, you will genetically engineer bacteria to express green fluorescent protein (GFP).  5.2 – Phototrophs and Chemotrophs Q2. List examples of photoautotrophs in nature. PHOTOTROPHS: Capture light “photo” energy from the sun and transform it into chemical energy (like glucose) for nourishment “-troph”. Photoautotrophs: Use solar energy to produce all the carbon compounds they need from CO2 (inorganic carbon source). “auto” = self – use CO2 as carbon source.  5.2 – Phototrophs and Chemotrophs Photoheterotrophs: Harvest solar energy “photo” for SOME cellular activities but still rely on an intake of organic molecules as a source of carbon (other “hetero” sources of carbon are needed). They can not use CO2 as their SOLE carbon source.  5.2 – Phototrophs and Chemotrophs CHEMOTROPHS: Obtain energy by oxidizing chemical bonds in molecules (organic or inorganic). Chemoautotrophs: Oxidize inorganic compounds such as H2S, H2 or inorganic ions for energy, and use CO2 as a carbon source.  5.3 – Oxidation and Reduction Reactions OXIDATION: Is the removal of electrons from a substance. During oxidation, Each carbon loses H atoms or gains bonds to oxygen. Oxidation reactions RELEASE ENERGY. Example: glucose oxidation C6H12O6 + 6O2 → 6CO2 + 6H2O + energy Recall that when cells are oxidizing glucose for energy, the glucose is first broken down via glycolysis (10- step process). During this process, glucose is broken down in many enzymatic steps, and the electrons are LOST from glucose (and its intermediates). Highly reduced Oxidized Q3. In glycolysis, glucose is highly reduced and pyruvate (the last product of glycolysis) is oxidized. Through 10 enzymatic steps in glycolysis, glucose is continuously OXIDIZED (losing electrons). Where do the electrons from glucose go? Why is there a RELEASE of energy?  5.3 – Oxidation and Reduction Reactions Glucose (6C) 2x Pyruvate (3C) Glucose is the most reduced molecule in Pyruvate is the most oxidized molecule in the glycolytic pathway. Note that all of the glycolytic pathway. Note that 2 out of glucose’s carbons are bonded to 3 of pyruvate’s carbons contain 2 or more hydrogens and have single bonds with bonds to oxygen with no bonds to oxygen. hydrogen.  5.3 – Oxidation and Reduction Reactions REDUCTION: The ADDITION of electrons to a substance through addition of hydrogen atoms (and a loss of oxygen atoms). Requires an INPUT of Energy. Example: carbon dioxide reduction energy + 6CO2 + 6H2O → C6H12O6 + 6O2 (opposite of glucose oxidation) Q4. Where does this INPUT of energy occur for the reduction of carbon dioxide?  5.3 – Oxidation and Reduction Reactions REDOX Reactions (oxidation/reduction reactions): In any redox reaction, one compound is reduced and one compound is oxidized. (i.e. one compound has to gain the electrons and one compound has to lose the electrons). Glyceraldehyde-3-phosphate dehydrogenase Glyceraldehyde-3-phosphate (G3P) NAD+ NADH + H+ + Pi 1,3-bisphosphoglycerate (1,3-BPG) Q5. Fill in the blanks: In glycolysis, the 6th reaction is catalyzed by an enzyme called glyceraldehyde-3-phosphate dehydrogenase. The substrate G3P is ____(a)_______ to 1,3- BPG. The substrate NAD+ is ______(b)______ to NADH + H+.  5.4 – Flow of Energy through the Biosphere Cyclic Flow of Matter - Matter (things made of atoms) cycles between phototrophs and chemotrophs. - Matter enters the chemotrophic part of the biosphere as ENERGY-RICH compounds (glucose, fatty acids) and leave the chemotrophic part of the biosphere as ENERGY-POOR compounds (CO2, H2O, NO3-).  5.5 – Bioenergetics Bioenergetics - Energy flow is governed by the principles/laws of thermodynamics. - Thermodynamics concerns the laws governing energy transactions that accompany most physical and chemical processes. - BIOENERGETICS (applied thermodynamics) applies principles of thermodynamics to the biological world. Definitions: - Energy can be defined as the ability to cause chemical or physical changes. - Though energy is distributed throughout the universe, the energy under consideration in any particular case is called a SYSTEM and the rest of the universe is called the SURROUDINGS. - The boundary between the system and surroundings may be real or hypothetical.  5.5 – Bioenergetics Open System Closed System An open system can have energy added to A closed system is SEALED from its it or removed from it. environment and can not take in energy nor release the energy. Organisms are open systems (in most cases) because organisms have the ability The system is completely blocked off from its to take-up energy from its surroundings and surroundings. also release its energy to its surroundings.  5.5 – Bioenergetics The State of a System A system is in a specific state if each of its variable properties is held at a specific constant value. Q6. Which 3 variables must be kept constant in order for a system to be kept at a specific “state”? In this situation, the total energy of the system has a unique value (held at these 3 variables). If the state of a system CHANGES, the total energy change can only be determined by looking at the initial and final states of the system. Biological Systems In biological systems, the 3 important variables are mostly constant during biological reactions.  5.5 – Bioenergetics Heat & Work Exchange of energy between a system and its surroundings occurs as either HEAT or WORK. HEAT is not a very useful source of energy for cells. Q7. Why? WORK is the use of energy to drive a process other than heat flow. Quantifying Energy change The units for quantifying the energy changes during chemical reactions are calories (cal). Calorie = the amount of energy required to raise 1 g of H2O by 1oC at 1 atm. 1 kcal = 1000 cal  5.6 – Laws of Thermodynamics 1st Law of Thermodynamics: Law of Conservation of Energy. “In every physical or chemical change, the total amount of energy in the universe remains constant.” In other words, energy may be converted from one form to another but CAN NOT be created or destroyed. Look at glucose oxidation as an example again. Here we have an OPEN SYSTEM with a cell exposed to the surroundings. C6H12O6 + 6O2 → 6CO2 + 6H2O + (energy) Glucose (from surroundings) work CO2, H2O (low energy heat (high energy compound) compounds released to surroundings) and HEAT (released to surroundings) ATP molecules Q8. Summarize how (energy stored - the 1st law of ability to do WORK) thermodynamics is obeyed in this open Cell “System” system of a cell.  5.6 – Laws of Thermodynamics CO2, H2O (low energy Glucose (from surroundings) compounds released to (high energy compound) surroundings) and HEAT (released to surroundings) ATP molecules (energy stored - ability to do WORK) The amount of energy stored in a system is known as the INTERNAL ENERGY. The internal energy is difficult to measure/quantify. This is because, there are millions of reactions that are occurring at any given time and the internal energy is constantly changing. Instead biologists/scientists primarily care about the change of energy that occurs for individual reactions inside the cell.  5.6 – Laws of Thermodynamics (equation 1) Δ! = !!"#$%&'! − !!"#$%#&%' This equation represents the change in energy for an individual chemical reaction. Change in Enthalpy In most biological rxns and processes, it is more interesting to look at the change in enthalpy (heat content). Enthalpy is represented by the symbol H (for heat). Enthalpy is related to the internal energy E by another term that combines both Pressure (P) and Volume (V) (equation 2) ! = ! + !" equation on formula sheet  5.6 – Laws of Thermodynamics Therefore, we can also say that the change in enthalpy is equal to the change in internal energy + the change in pressure and volume. (equation 3) Pressure and Volume In biological scenarios, most chemical reactions proceed without any changes to pressure or volume. Therefore equation 3 can be expressed as: 0 (equation 4) To summarize, the change in enthalpy is a very good approximation of the change in internal energy of the cell. Biologists are confident that the values of ΔH are valid estimates of ΔE.  5.6 – Laws of Thermodynamics (equation 5) equation on formula sheet This equation represents the change in enthalpy of an individual chemical reaction that occurs in a cell. This is a valid estimation of the change in internal energy. The Meaning of ΔH If ΔH < 0 then the reaction is EXOTHERMIC. Example (combustion of propane): C3H8 + 5O2 à 3CO2 + 4H2O + energy If ΔH > 0 then the reaction is ENDOTHERMIC. In endothermic reactions, energy is absorbed. (e.g. the melting of an ice cube.)  5.6 – Laws of Thermodynamics The 2nd Law of Thermodynamics The 2nd law of thermodynamics is the law of thermodynamic spontaneity. “In every physical or chemical change, the universe tends toward greater disorder or randomness (entropy).” Entropy (S) Entropy (S) is a measure of randomness or disorder. Entropy increases when a system becomes LESS ORDERED (e.g. when ice melts solid to liquid or the explosion of TNT). Don’t memorize this exact chemical rxn. Trinitrotoluene (TNT) Many molecules of CO2, H2O and N2 Very ordered Very disordered  5.6 – Laws of Thermodynamics Entropy (S) Entropy decreases when a system becomes MORE ORDERED (e.g. when ice forms from water or synthesis of glucose from H2O and CO2.) Entropy Change as a Measure of Thermodynamic Spontaneity (equation 6) equation on formula sheet ΔSuniverse is positive for EVERY spontaneous process or reaction (increases the entropy of the universe). This rule is never violated for any reaction or process. Easy example: The oxidation of propane: C3H8 + 5O2 à 3CO2 + 4H2O + energy There are 6 molecules on the reactant side and 7 molecules on the product side. The products are MORE disordered than the reactants (more molecules of product than reactant). Therefore, ΔSsystem > 0 and ΔSuniverse > 0 as well.  5.6 – Laws of Thermodynamics Entropy Change as a Measure of Thermodynamic Spontaneity Another easy example The oxidation of glucose: C6H12O6 + 6O2 → 6CO2 + 6H2O + (energy) There are 7 molecules on the reactant side and 12 molecules on the product side. The products are MORE disordered than the reactants (more molecules of product than reactant). Therefore, ΔSsystem > 0 and ΔSuniverse > 0 as well. Harder Example The freezing of ice H2O (l) à H2O (s) Q9. Is Δssystem positive or negative for the freezing of liquid water into ice? Freezer (system)  5.6 – Laws of Thermodynamics ΔSuniverse is positive for EVERY spontaneous process or reaction (increases the entropy of the universe). This rule is never violated for any reaction or process. Surroundings (Universe) Freezer (system) Even though in the isolated system (your freezer), the entropy is DECREASING, this is not violating the laws of thermodynamics because only the isolated SYSTEM is decreasing in entropy. We need to take into account the SURROUNDINGS as well. How is your freezer powered? You plug-it in and electricity provides the energy for the freezer to stay cold. If your electricity is provided by a hydroelectric dam, the water is increasing in entropy as it falls down the dam. This represents an increase in entropy of the surroundings (or the universe).  5.7 – Free Energy It is much more convenient to have a term/value that can measure the spontaneity of the SYSTEM alone, and that is using FREE ENERGY (Gibbs’ Free Energy) represented by G. A measure of spontaneity for a SYSTEM ALONE (no surroundings) is called FREE ENERGY (G). (equation 7) equation on formula sheet ΔG is related to the enthalpy and entropy of a reaction. (equation 8) equation on formula sheet T = temperature of the system in degrees Kelvin. oK = oC + 273 equation on formula sheet  5.7 – Free Energy Free energy is a readily measurable indicator of spontaneity. Every spontaneous reaction is characterized by a DECREASE in the free energy of a system. If ΔG < 0, the reaction is thermodynamically spontaneous. EXERGONIC & ENDERGONIC REACTIONS Exergonic rxns are energy-yielding and occur spontaneously (D G < 0) Endergonic rxns are energy-requiring and do NOT occur spontaneously under the conditions specified. (DG > 0) The values of BOTH DH and -TDS influence the value of DG  5.7 – Free Energy FREE ENERGY: Biological Example Consider the oxidation of glucose (a highly exergonic process) C6H12O6 + 6O2 → 6CO2 + 6H2O + energy Under STANDARD conditions (25oC, 1 atm, 1M of each reactant and product) DH = -673 kcal/mol glucose and DS = +0.0436 kcal/mol oK Q10. Calculate DG for the oxidation of glucose.  5.7 – Free Energy FREE ENERGY: Biological Example Consider the reverse reaction for the formation of glucose (endergonic process) 6CO2 + 6H2O + energy → C6H12O6 + 6O2 Under STANDARD conditions (25oC, 1 atm, 1M of each reactant and product) DH = +673 kcal/mol glucose and DS = -0.0436 kcal/mol oK Compared Q11. Calculate DG for the formation of glucose.  5.7 – Free Energy The Meaning of Spontaneity The term spontaneous tells us that a reaction CAN take place, not that it WILL. Whether an exergonic reaction will proceed depends on a favorable negative ΔG but also the availability of other factors (i.e. enzymes). Usually an input of activation energy (Ea) is also required as well.  5.8 – Equilibrium and ΔG The equilibrium constant (Keq) is the ratio of the product concentrations to reactant concentrations at equilibrium. At equilibrium there is NO net change in the concentrations of reactants or products. Consider the very simple reaction: The equilibrium constant would be expressed as:  5.8 – Equilibrium and ΔG If you know the equilibrium constant (Keq) for a reaction, you can tell whether a particular mixture of products and reactants is in equilibrium. The tendency toward equilibrium provides the driving force for every chemical reaction.  5.8 – Equilibrium and ΔG Biological Example The conversion of glucose-6-phosphate to fructose-6-phosphate occurs in glycolysis. The equilibrium constant (Keq) for the reaction shown to the left is 0.5 (at 25oC). (NOTE: pressure is not specified) Enzyme: Phosphoglucose isomerase This means that at equilibrium at 25oC, there is always 2X more glucose-6- phosphate compared to fructose-6- phosphate.  5.8 – Equilibrium and ΔG Biological Example Concentration Ratio of Products-to- Reactants Q12. According to Le Chatalier’s Principle, if the concentration ratio of (fructose-6-phosphate:glucose-6- phosphate) was LESS than Keq in a cell, which way would the reaction proceed? Q13. According to Le Chatalier’s Principle, if the concentration ratio of Enzyme: Phosphoglucose isomerase (fructose-6-phosphate:glucose-6- phosphate) was MORE than Keq in a cell, which way would the reaction proceed?  5.8 – Equilibrium and ΔG Let’s go back to our original example of reactant A forms product B. Let’s say that the equilibrium constant Keq for this reaction is equal to 1. (i.e. at equilibrium, A and B are in exactly equal concentrations.) In the diagram, the lowest free energy occurs when the reaction is at equilibrium. Hence again, the driving force for every chemical rxn is the tendency towards equilibrium. Q14. If the concentration ratio of [B]/[A] in a cell was 100, what direction would the reaction proceed and would the reaction proceed spontaneously?  5.8 – Equilibrium and ΔG There is another equation we can use to calculate the free energy of a chemical reaction if the reactants and products are NOT at equilibrium: [.]# $ Δ" = −RT ln )!" + +, ln % & These equations will not be on the DG is free energy change in cal/mol formula sheet because you will not (NOTE UNITS) need to use them on the exam. You R is the gas constant (1.987 cal/mol oK) will be given another one that is similar to it (see later slides). T is the temperature in kelvins (oK) Keq is equilibrium constant at standard temperature of 298 K (25oC) ln is the natural log  5.8 – Equilibrium and ΔG Biological Example: Q15. Assume that the concentrations of [glucose-6-phosphate] in a cell is 10 μM and the concentration of [fructose-6-phosphate] in a cell is 1 μM at 25oC. The Keq is 0.5 at 25oC (note this reaction does not have a specified pressure). Calculate the DG of the conversion of glucose-6-phosphate to fructose-6-phosphate at these non-equilibrium conditions. What direction will the reaction proceed?  5.9 – Limitations of DG DG is a thermodynamic parameter that tells us whether a reaction is thermodynamically possible as written. It also tells us how MUCH free energy would be liberated if a reaction took place. Q16. How much free energy is liberated in the previous example (on the previous slide)? DG does not tell us anything about the RATE or MECHANISM (i.e. there could be multiple steps to convert reactant A to product B) of the reaction.  5.10 – Standard Free Energy Change is DGo’ Because DG may vary under different conditions, we must identify the conditions for which a given measurement of DG is made. Biochemists have agreed on conditions to define the STANDARD STATE. These conditions are 25oC (298oK), 1 atm pressure, and all products and reactants at a concentration of 1 M. Biochemists also specify a standard pH of 7.0 (b/c most rxns take place at pH 7). The concentration of [H+] and [OH-] is 10-7 M, which means that the standard concentration of 1.0 M does not apply to H+ or OH- ions. Values of DG and Keq are written with a “prime” (i.e. DG’ and K’eq) to signify that these values were determined or calculated at pH = 7.0. The standard free energy change (DGo’) has a superscript “ o ” (knot) which refers to standard conditions of temperature, pressure and concentration. The prime (’) refers to the standard H+/OH- ions at a concentration of 10-7 M.  5.10 – Standard Free Energy Change is DGo’ Let’s use the equation from before to calculate free energy change: If the DG and Keq was calculated at standard temp., pressure, concentration and pH = 7, we can now change these terms in the equation to: DGo’ and K’eq If the concentration of all of the reactants and products is equal to 1.0 M, the second term in this equation contains ln (1) which equals to zero. 0  5.10 – Standard Free Energy Change is DGo’ Thus, we can get the following equation: equation on formula sheet We can simplify this further by knowing that RT = 592 cal/mol under standard conditions. equation on formula sheet DGo′ values are convenient They are easily determined from the equilibrium constant and provide a uniform convention for reporting free energy changes But, DGo′ is an arbitrary standard, referring to impossible conditions for most biological systems  5.10 – Standard Free Energy Change is DGo’ What does all of this mean?? equation on formula sheet Knowing the DGo′ can tell us whether a reaction will proceed in the forward direction (towards the products) under STANDARD CONDITIONS or if the reaction will proceed in the reverse direction (towards the reactants) under STANDARD CONDITIONS. If K′eq > 1.0 then DGo′ will be negative and the reaction can proceed to the right (towards the products) under standard conditions. If K′eq < 1.0 then DGo′ will be positive and the reaction will tend toward the left (toward the reactants) under standard conditions.  5.11 – Calculating DG’ For real life situations, DG′ is the most useful measure of thermodynamic spontaneity DG′ provides a measure of how far from equilibrium a reaction is, under the conditions in a cell. Under the special case where DG′ = 0, the reaction is in equilibrium; however, reactions in living cells are rarely in equilibrium. We will use this equation if you want to determine DG′ where the concentration of the products and reactants are NOT all 1.0 M, you need to use the following equation: equation on formula sheet  5.11 – Calculating DG’ This table is a great summary of everything we just learned about DGo’ and DG’  5.12 – The Steady State Note: This isn’t the same as the “steady state assumption” in the Michaelis- Menten derivation in Set 6 Notes. Steady State - At equilibrium the forward and reverse rates of a reaction are the same, so there is NO net flow of matter, nor energy produced. - Living cells are characterized by continuous reactions and maintain themselves in states far from equilibrium. - Cells maintain a steady state in which reactants, products and intermediates are kept at levels far from equilibrium. Q17. Why is it so important that cells keep themselves in the “steady state”?