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

how do we predict the outcome of chemical reactions inside of a cell 9 5 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
How do we predict the outcome of chemical reactions inside of a cell? 9/5 PowerPoint Presentation
Download Presentation
How do we predict the outcome of chemical reactions inside of a cell? 9/5

play fullscreen
1 / 16
How do we predict the outcome of chemical reactions inside of a cell? 9/5
170 Views
Download Presentation
feng
Download Presentation

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

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. 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

  2. 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!

  3. 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!

  4. 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!

  5. 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

  6. 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.

  7. 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

  8. 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

  9. 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

  10. 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?

  11. 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!

  12. 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!

  13. 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’ !

  14. 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!

  15. 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

  16. 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+FructoseADP+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.