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How do we predict the outcome of chemical reactions inside of a cell? 9/5

How do we predict the outcome of chemical reactions inside of a cell? 9/5

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## How do we predict the outcome of chemical reactions inside of a cell? 9/5

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**How do we predict the outcome of chemical reactions inside**of a cell? 9/5 • What happens to chemicals thrown into a container called a “cell”? Can we predict reactions under non-standard conditions? YES! • What is an Equilibrium Constant? • What is a Standard Free Energy? • How do we use ΔG’o ? • What happens to ΔG in a cell where conditions are not standard? • Sample calculations • PubMed search examples vs Google Examples**What happens when we put chemicals in a solution?If they can**they automatically chemically convert!If they can means a RXN is thermodynamically feasible!However, SPEED of the reaction cannot be predicted. • Consider Glucose-6-P Fructose-6-Pi • If we put in 1M G-6-P and F-6-P what happens? • Assume: Temp=25oC pH=7.0 • Eventually reaction comes to an equilibrium! • At EQ: 2 G-6-P for every 1 F-6-P molecules! • Keq= [Product]/[reactants]= [1]/[2]=0.5 • This says the system would eventually stabilize with more G-6-P than F-6-P!**What happens when we put chemicals in a solution?If they can**they automatically chemically convert!If they can means a RXN is thermodynamically feasible! • Consider Glucose-1-Pi Glucose-6-Pi • If we put in 0.2M G-1-P what happens? • Assume: Temp=25C pH=7.0 • Eventually reaction comes to an equilibrium! • At EQ: 1 G-1-P for every 19 G-6-P molecules! • Keq= [Product]/[reactants]= 19/1=19 • This says the system would eventually stabilize with more G-6-P than G-1-P! • Why isn’t water listed in these Keq equations? Because it is present at about 55M on both the reactant and product sides of the equation!**Standard Free Energy (ΔGo’ ) is an immutable constant for**any chemical reaction that can occur! Tables Exist! • Assume reactants and products are initially present at molar concentrations of 1 M and pH=7.0 • Assume: R is the gas constant: 1.987 cal/mole K • Assume: T is temperature in Kelvin (usually 298 K) • K’eq: Equilibrium Constant-ratio of products and reactants when the reaction comes to its normal equilibrium (‘) means at standard T. • ΔGo’=-2.303RT log keq OR!! ΔGo’=-RT ln keq • -2.303: mathematical constant converting log and ln • The more negative (-) ΔGo’ more energy is released in achieving Keq! • The more positive the ΔGo’ the more difficult the reaction is and the more energy is req. enter the system to achieve Keq!**We can use the observed Keq to calculate the energy change**needed to reach Keq under special conditions! • Consider G-1-P G-6-P • Standard free energy of the reaction equals: • ΔGo’=-2.303RT log keq • OR ΔGo’=-2.303RT log [G-6-P]/[G-1-P] • OR ΔGo’=-2.303RT log [19]/[1]=-2.303(1.987)(298)(1.28) • ΔGo’= -1,740 cal/mol OR -1.74 Cal/mol • ΔGo’= negative value so reaction can and can proceed! • However the SIZE of ΔGo’ does NOT matter for determining the RXN rate! • What if Keq was 190/1? Products favored!!! • ΔGo’= -(1364)(log 190)= -3.1 Cal/mol • What if Keq was 0.19/1?Products not favored!!!! • ΔGo’= -(1364)(log 0.19)= +0.98 Cal/mol**Remember that natural log (ln) is equivalent to the 10-based**log times 2.303! The book uses ln (natural log), but many calculators lack this function now you can convert the two methods.**Lets consider a couple reactions in glycogen breakdown**occurring when you exercise a skeletal muscle cell?Glycogen is broken into G-1-P by glycogen phosphorylaseG-1-P G-6-P F-6-P…this is fed into glycolysis! • Glucose-6-P Fructose-6-P ΔGo’= +0.40 Cal/mol • Why won’t this RXN go with regards to ΔGo’? • BUT WAIT!!!!G-1-P G-6-6 ΔGo’=-1.74Cal/mol • This is Spontaneous? Why do you know? • We can link the two reactions and also their two opposing ΔGo’ values together if they occur in the same place at the same time (in the cytosol)! • Net ΔGo’= -1.74 + 0.40 = -1.34 Cal/mol • Would this work if one of the reactions occurred in the mitochonrial matrix? • The math: Note substition of (2.3)log for ln in book!! • ΔGo’= -2.3RT log Keq or ΔGo’= -2.3RT log 0.5= +400 cal/mol**Hydrolysis means removal of phosphate (PO4 or Pi). To**calculate the ΔG’o for the phosphorylation (adding Pi) to ADP ATP, just reverse the sign, so this reaction requires +7.3 Cal/mol.Kinase:Adds a phosphate to substratePhosphatase: removes a phosphate from a substrate**The close proximity of the negative charges on the Gamma,**Beta, and Alpha-phosphate groups means they want to separate, when they do energy is released! • ATP ADP + Pi ΔGo’=-7.3 Cal/mol**ATP hydrolysis releases energy allowing coupling of**chemically unfavorable to favorable reactions. This lets cells do things they otherwise could not! • How does the first step for glycolysis in glucose phosphorylation occur? Glucose + Pi G-6-P Glucose + Pi>G-6-P +H2O • ΔGo’ = +3.3 Cal/mol energy required! • ATP > ADP + Pi • ΔGo’ = -7.3 Cal/mole energy released! • Net ΔG = (-7.3) + (+3.3)= -4 • ATP+Glucose> G-6-P +ADP • ΔGo’= -4 kcal/mole • Neg. ΔGo’ so reaction occurs • The enzyme hexokinase speeds the reaction up! • Remember why water is ignored in equation?**Lets look at what happens inside a red blood cell when ATP**is actually hydrolyzed to ADP and Pi! • Are the concentrations in the cell 1 M? • Biologically, this condition is extremely RARE! ΔG’= ΔGo’+2.3RT log[product C][product D]/[reactant A][reactant B] This lets you see what happens in NON-IDEAL conditions! • The Reaction: ATP + H2O ADP + Pi ΔGo’= -7.3 Cal/mol • The Cell: Human Red Blood Cell • Cells maintain intracellular concentrations that seldom exist at Keq! • This imbalance is based on the balance between use and production of these compounds!**Try to be aware of the fact that all cells are not alike**with respect to the amounts of intracellular compounds found inside. This is a central element of cell biology and cell differentiation!**The reality of concentrations inside the red blood cell:**things are not ideal/not 1 M! • [ATP]= 2.25 mM [ADP]=0.25 mM and [Pi]= 1.65 mM • Convert to M for equation to work! • ATP=0.00225M ADP=0.00025M Pi=0.00165M • ΔG’= -7300 + 2.3RT log [ADP][Pi]/[ATP] • ΔG’= -7300 + 1360log [0.00025][0.001650]/[0.002250] ΔG’= -7300 + 1360(-3.74)=-7300 cal/mol-5100cal/mol=-12,400cal • ΔG’= -12.4 Cal/mol • Therefore the reaction yields even MORE energy than would be expected based in just the ΔGo’ !**These problems are due next Wednesday: If you show that you**tried to work out the math for the questions you will get most of the 10 possible points. Show all steps and calculations used for your math! • 1) Calculate the ΔG0’ from the given Keq values. Assume standard R, T and pH values for these important enzyme catalyzed reactions: • A) Glutamate + Oxaloacetate Aspartate + alpha-ketoglutarate • K’eq= 6.8 Aspartate transaminase: Important for amino acid synthesis! • B) Dihydroxyacetone phosphate glyceraldehyde-3-phosphate K’eq=0.0475 Triose phosphate isomerase: Important in glycolysis! • C) Fructose-6-Pi + ATP Fructose-1,6-Pi + ADP • K’eq= 254 Phosphofructokinase: Important in glycolysis!**2) Calculate the K’eq for these problems given standard**conditions and the given ΔGo’ . • A) G-6-P + H2O Glucose + Pi • ΔGo’ = -3.3 Cal/mol Glucose-6-phosphatase: Gluconeogenesis • B) Lactose + H2O Glucose + Galactose • ΔGo’ = -3.8 Cal/mol Lactase: Digestion of milk sugar • C) Malate Fumarate + H2O • ΔGo’ = +0.75 Cal/mol Fumarase: Citric Acid Cycle • 3) Calculate the K’eq for the sum of this coupled reaction using the Table of ΔGo’ values in your notes. • RXN #1 G-1-P G-6-P • RXN #2 G-6-P F-6-P • Net RXN G-1-P F-6-P**4) Calculate the ΔGo’ for these two ATP-Coupled reactions**using the values given in the table found in Friday’s notes.A) Phosphocreatine + ADP ATP + CreatineImportant for weight lifters hoping to maintain strength!B)ATP+FructoseADP+Fructose-6-Pi Important for diabetic control, fruit consumption and glycolysis • 5) Calculate the physiological ΔG’ for this reaction in a normal brain cell and one that is unable to generate ATP when the oxygen supply is deficient (hypoxia): • RXN: Phosphocreatine + ADP Creatine + ATP Assume T= 25 C • Normal Cell:Phosphocreatine= 4.7 mM, Creatine = 1.0 mM, ADP=0.20 mM and ATP=2.6 mM • Hypoxic Cell: Phosphocreatine= 4.7 mM, Creatine = 1.0 mM, ADP=1.30 mM and ATP=1.3 mM • Are the two values the same or different? Which yields more energy to the cell? • Phosphocreatine is a P-reserve to maintain ATP in emergencies.