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Intro Bio Lecture 8

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1 Direction of reactionsFlow of in in E. coli Lec. 8

Macromolecules Polysaccharides Lipids Nucleic Acids Proteins

monomers

intermediates

biosynthetic pathway

glucose

Each arrow = a specific chemical reaction 2 In order for E. coli to double in 1 hour, 3 giant problems

1) Chemical reactions must be catalyzed to go fast. Very fast. Solution: enzymes

2) Chemical reactions must go in the right direction Solution: metabolism to produce and harness energy

3) Design enzymes and metabolism and remember this information Solution: DNA 3 Major concerns about the cell’s essential chemical reactions

1)Speed • We need it to go fast enough to have the cell double in one generation 2) Direction • We need it to go in the direction we want

Proteins are used to solve both of these problems.

4 Example: Biosynthesis of a

3 glucose’s One 18- fatty acid (18 C’s) Free energy change: ~ 300 kcal per mole of glucose used is REQUIRED

So: 3 glucose 18-carbon fatty acid

So getting a reaction to go in the direction you want is a major problem (but to be discussed second)

5 Free energy difference determines the direction of a chemical reaction atoms

transition state

energy catalyzed transition state ree f

free A + B } energy C + D difference

E + F 6 Change in free energy (DG) Energy that can do work For: A + B C + D, left-to-right direction as indicated:

Consider the quantity called the change in free energy associated with a chemical reaction, or: Δ G

• Δ G < 0 (negative): A AND B WILL TEND TO PRODUCE C AND D (i.e., tends to the right).

• Δ G > 0 (positive): C AND D WILL TEND TO PRODUCE A AND B. (i.e., tends to the left)

• Δ G = 0: THE REACTION WILL BE AT EQUILIBRIUM: NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY 7 The change in free energy (DG) A + B C + D products DG = DGo+ RTln [C][D] [A][B] reactants

• A, B, C and D = concentrations of the reactants and the products AT THE MOMENT being considered. (i.e., these A, B, C, D’s here are not the equilibrium concentrations)

• R = universal gas constant = 1.98 CAL / °K MOLE ( so R =~2)

• T = ABSOLUTE TEMP ( oK ) 0oC = 273oK; Room temp = 25o C = 298o K or ~ 300o K

• ln = NATURAL LOG

• D Go = a CONSTANT: a quantity related to the INTRINSIC properties of A, B, C, and D, . . . . to be explained 8 The change in free energy change (DG)

Also abbreviated form: DG = DGo+ RTlnQ (Q for “quotient”)

Where Q = ([C][D]/[A][B])

Qualitative term: Quantitative term: Josiah Willard Gibbs What molecules How much of each is (1839 - 1903) are in play. This depends present, at the moment on what A, B, C, and D under consideration are.

9 Standard free energy change (DGo)

STANDARD FREE ENERGY CHANGE of a reaction. If all the reactants and all the products are present at 1 unit concentration, then:

DG = DGo + RTln(Q) = DGo + RTln([1][1] / [1][1]) = DGo + RTln(1) = DGo + RT x 0, or DG = DGo, when all components are at 1 ….. a special case (when all components are at 1) “1” usually means 1 M 10 Standard free energy change (DGo)

So DG and DGo are quite different, and not to be confused with each other.

DGo allows us to compare all reactions under the same standard reaction conditions that we all agree to, independent of concentrations.

So it allows a comparison of the stabilities of the bonds in the reactants vs. the products.

It is useful.

AND,

It is easily measured. 11 DG and DGo at equilibrium

• Because at equilibrium, DG = DGo + RTln(Q) = 0

[C]eq [D]eq and only at equilibrium: Q = K = eq [A]eq [B]eq (a second special case).

• So: at equilibrium, o DG = DG + RTln(Keq) = 0 o • And so: DG = - RTln(Keq) • So just measure the Keq, • Plug in R and T • Get: DGo, the standard free energy change • Publish it. It gets written in a book. 12 E.g., let’s say for the reaction A + B C + D, Keq happens to be:

[C]eq[D]eq = 2.5 x 10-3 so, not very much product [A]eq[B]eq

o -3 Then DG = -RTlnKeq = -2 x 300 x ln(2.5 x 10 ) = -600 x -6 = +3600 3600 cal/mole (If we use R = 2 we are dealing with calories) Or: 3.6 kcal/mole 3.6 kcal/mole ABSORBED (positive number, +3.6)) So energy is required for the reaction in the left-to-right direction And indeed, very little product accumulates at equilibrium

(Keq = 0.0025) 13 Note: If for the reaction A + B C + D, ΔGo = +3.6

Then for the reaction C + D A + B, ΔGo = - 3.6 (Reverse the reaction: switch the sign)

And: For reactions of more than simple 1 to 1 stoichiometries: aA + bB <--> cC + dD, ΔG = ΔGo + RT ln [C]c[D]d [A]a[B]b

And RT = 2*300 = 600 cal/deg-mole at room temp., or RT= 0.002*300 = 0.6 kcal/deg-mole at room temp, more usefully 14 Some exceptions to the 1M standard condition: Exception #1: • 1) Water: 55 M (pure water) is considered the “unit” concentration in this case instead of 1M

The concentration of water rarely changes during the course of an aqueous reaction, since water is at such a high concentration.

• So when calculating DGo, instead of writing in “55” when water participates in a reaction (e.g., a hydrolysis) we write “1.”

• This is not cheating; we are in charge of what is a “standard” condition, and we all o agree to this: 55 M H20 is unit (“1”) concentration for the purpose of defining DG .

15 Some exceptions to the 1M standard condition: Exception #2: In the same way,

Hydrogen concentration, [H+]: 10-7 M is taken as unit concentration, by biochemists. since pH7 is maintained (buffered) in most parts of the cell despite a reaction that may produce acid or base.

This definition of the standard free energy change requires the designation ΔGo’

However, we will not bother. But it should be understood we are always talking about ΔGo’ in this course.

16 Summary

DG = DGo + RTln(Q)

This combination of one qualitative and one quantitative (driving) term tells the direction of a chemical reaction in any particular circumstance

o DG = - RTln(Keq)

The ΔGo for any reaction is a constant that can be looked up in a book.

17 Energy in a cell: (ATP) -- Hydrolysis of ATP: ATP + HOH à ADP + HPO4 ATP, a cellular small molecule that helps in the transfer of energy from a place where it is made to a place where it is needed.

Acid anhydride (new functional group: 2 acids joined)

adenine “base” ribose OH OH adenosine phosphate | | HO - P - OH HO - P - OH A-R: a nucleoside ||| ||| O O ATP: a nucleotide (a nucleoside triphosphate) Dehydration between 2 acids 18 The hydrolysis of ATP

ATP + HOH ßà ADP + Pi ______

(ADP)

The DGo of this reaction is about -7 kcal/mole. Energy is released in this reaction. This is an exergonic reaction under standard conditions Strongly to the right, towards hydrolysis, towards ADP 19 “High energy” bonds

• DGo of a least ~ -7 kcal/mole is released upon hydrolysis • Designated with a squiggle (~) often • ATP = A-P-P~P • Rationalized here by the relief of electrical repulsion upon hydrolysis:

_ _ _ _

DGo = -7 kcal/mole 20 Hydrolysis of ADP

_ AMP AMP _ _ _ _

ADP o (ADP) DG = -7 kcal/mole

_ A-R-P~P~P _

Not a high energy bond 21 What happens when we dissolve ATP in water?

ATP + HOH à ?

+ ATPase ATP + HOH ADP + Pi + heat (à7o­)

1M ATP, 1 ml water, 1 mmole ATP, @7 kcal/mole, 7000 cal/mole, 7 cal / mmole, 7 cal, @1 cal:1 ml water: 1o Cà7o­

22 How does the cell harness this energy? • The cell often uses the hydrolysis of ATP to release energy.

• The released energy is used to drive reactions that require energy.

• How does this work ??

23 An example: an endergonic reaction

Suppose: glucose + Pi glucose-6-phosphate 2- CH2OH CH2OPO3

+ Pi

o Glucose + Pi --> glucose-6-P + H2O; Δ G = +3.6 kcal/mole. -3 Keq= 2.5 x 10 o 1) ATP + H2O --> ADP + Pi Δ G = -7 kcal/mole o 2) Glucose + Pi --> G6P + H2O Δ G = +3.6 kcal/mole

ATP + H2O+ Glucose + Pi à ADP + Pi + G6P + H2O Δ Go = -3.4 kcal/mole overall Glucose + ATP à G6P + ADP Δ Go = -3.4 kcal/mole overall = net sum of the two considered reactions 24 An endergonic reaction, continued Our test tube of 1 M ATP in water:

Add glucose + PO4: à nothing

Enzymes needed … ATPase? Glucose phosphorylase?

Add these enzymes: ATP à ADP + Pi, Glucose + Pi à G6P But just get 7 kcal/mole as heat again.

Energy absorbed mostly by raising the temperature of all the water around 7o (25oà32o). Just get warmer glucose. But: 25 Directly coupled reactions Answer: Add Hexokinase (instead of ATPase and glucose phosphorylase)

Glucose + AR-P-P-P ßà glucose-6-P + AR-P-P

o -3.4 kcal/mole Glucose +ATPßà glucose-6-P04 + ADP, DG =

A directly coupled reaction. A new reaction, ATP is not simply hydrolyzed.

What a good idea!

Direct coupling of reactions is one of two ways the cell solves the problem of getting a reaction to go in the desired direction. The second: later. 26 ATP: the energy “currency” of the cell Flow of glucose in E. coli

Macromolecules Polysaccharides Lipids Nucleic Acids Proteins ATP ATP ATP monomers ATP ATP

ATP ATP ATP ATP intermediates

biosynthetic pathway

glucose

Each arrow = a specific chemical reaction 27 Where does this ATP come from?

• Glucose is the only carbon source in our minimal medium. • Need to make ATP from glucose; that path TAKES energy. • But need only to regenerate ATP from ADP:

coupled to drive endergonic reactions ATP . . . ATP Glucose ATP ADP biosynthetic pathway to ATP

regeneration of ATP from ADP?

Just need to put that P back onto ADP i 28 Regeneration of ATP from ADP

Two solutions:

• 1)

• 2) Catabolism of organic compounds (e.g., glucose) Metabolic breakdown of an

29 Glucose catabolism overview/preview

1- (6C à 3C)

2- KREBS CYCLE (3Cà 1C, CO2 release)

3- ELECTRON TRANSPORT CHAIN ( uptake, water release)

Glycolysis, in detail, as: A) Basic mechanism of energy metabolism (getting energy by glucose breakdown.)

B) An example of a . 30 dihydroxyacetone phosphate

fructose-1,6- glucose-6- fructose-6- diphosphate phosphate phosphate glyceraldehyde-3-phosphate Glycolytic pathway, Glycolysis glucose 1,3-diphosphoglyceric acid

phosphoenol-pyruvic 2-phosphoglyceric 3-phosphoglyceric acid acid acid

yeast lactic acetaldehyde ethanol acid pyruvic acid oxidative pathways Handout 8A 31 32 The firstThe 5 stepsfirst 5 ofsteps glycolysis of glycolysis dihydroxyacetone pnosphate

isomeriz- ation again 5

PO3

gyceraldehyde +ATP +ATP Aldolase -3-phosphate Glucose G-6-P Fructose-6-P Fructose- +HOH, phosphorylation phosphorylationPhosphorylation isomerizationIsomerization Phosphorylation 1,6-di-P -HOH 2 moles of G3P 1 2 3 4 for every mole of glucose. And: have spent 2 ATPs dihydroxyacetone phosphate

fructose-1,6- glucose-6- fructose-6- diphosphate phosphate phosphate glyceraldehyde-3-phosphate Glycolytic pathway Glycolysis glucose 1,3-diphosphoglyceric acid

phosphoenol-pyruvic 2-phosphoglyceric 3-phosphoglyceric acid acid acid

yeast lactic acetaldehyde ethanol acid pyruvic acid oxidative pathways Handout 8A front 33 Oxidation = loss of electrons Acid anhydride functional group (if add HOH get 2 acids)

glyceraldehyde-3- phosphate 1,3,-diphospho- glyceric acid

Abbreviations for the phosphate group Need an acceptor for these electrons: an oxidizing agent 34 Nicotinamide adenine dinucleotide (NAD)

+2H●

-2H●

“niacin”

or NADred or “NADH2” or NAD or NADH + H+

35 fructose-1,6- glucose-6- fructose-6- diphosphate phosphate phosphate Loose end #2 = NAD glyceraldehyde-3-phosphate Loose end #1 was ATPs Glycolytic pathway

glucose 1,3-diphosphoglyceric acid { Top carbon has been oxidized phosphoenol-pyruvic 2-phosphoglyceric 3-phosphoglyceric acid acid acid { -2 ATP + 4 ATP E. coli, yeast muscle + 2 ATP acetaldehyde ethanol lactic pyruvic acid acid “glycolysis” ends here, at pyruvate 36 dihydroxyacetone phosphate

fructose-1,6- glucose-6- fructose-6- diphosphate phosphate phosphate glyceraldehyde-3-phosphate Glycolytic pathway

glucose 1,3-diphosphoglyceric acid ATP debt paid in full phosphoenol-pyruvic 2-phosphoglyceric 3-phosphoglyceric acid E. coli, acid acid muscle - 2 ATP yeast lactic acetaldehyde ethanol + 4 ATP acid pyruvic acid oxidative pathways + 2 ATP “glycolysis” ends here Handout 8A 37 Net reaction:

1 glucose + 2 ADP + 2 Pi + 2 NAD 2 pyruvate + 2 ATP + 2 NADH2 DGo = -18 kcal/mole

So overall reaction goes essentially completely to the right.

38 Energy levels in glycolysis kcal/mole

glucose

2 pyruvates 39 Energy levels in glycolysis kcal/mole

DG = DGo + RTln[products] [reactants] glucose

pull

pull

2 pyruvates 40 The second way the cell gets a reaction to go in the desired direction:

1) First way was: a directly coupled reaction (i.e., a different reaction) . One of two ways the cell solves the problem of getting a reaction to go in the desired direction o Glucose + ATP ßà glucose-6-P04 + ADP, DG = -3.4 kcal/mole

2) The second way: • Removal of the product of an energetically unfavorable reaction • Uses a favorable downstream reaction • “Pulls” the unfavorable reaction • Operates on the second term of the DG equation. • DG = DGo + RTln([products]/[reactants]) 41 Regeneration of NAD 42 • So glucose à pyruvic acid

• ADP à ATP, as long as we have plenty of glucose • Are we all set?

• No…. What about the NAD?.. We left it burdened with those electrons.

• Soon all of the NAD will be in the form of NADH2 • Very soon

• Glycolysis will screech to a halt !!

• Need an oxidizing agent in plentiful supply to keep taking those electron off the NADH2, to regenerate NAD so we can continue to run glucose through the glycolytic pathway. Oxidizing agents around for NAD:

1) Oxygen Defer (and not always present, actually)

2) Pyruvate, our end-product of glycolysis In E. coli, humans:

Pyruvate à lactate, NADH2 à NAD, coupled

In Yeast: Pyruvate à ethanol + CO 2 43 fructose-1,6- glucose-6- fructose-6- diphosphate phosphate phosphate glyceraldehyde-3-phosphate 44 Glycolytic pathway

glucose 1,3-diphosphoglyceric acid

phosphoenol-pyruvic 2-phosphoglyceric 3-phosphoglyceric acid E. coli, acid acid muscle

yeast

lactic acetaldehyde ethanol acid pyruvic acid oxidative pathways 44 Overview:

Glucose

GAL-3-P 1,3-Di-PGA

ATP ATP Glucose NAD NADH2 Biosynthetic pathway to NAD biosynthetic pathway to NAD

Lactate Pyruvate

excreted 45 Glycolytic pathway 46

glucose 1,3-diphosphoglyceric acid

phosphoenol-pyruvic 2-phosphoglyceric 3-phosphoglyceric acid E. coli, acid acid muscle

yeast

lactic acetaldehyde ethanol acid pyruvic acid oxidative pathways H yeast C=O

E. coli | humans CH3 Alcoholic fermentation Acetaldehyde detail 46 Fermentation goes all the way to the right Efficiency: glucose--> 2 lactates, without considering the couplings for the formation of ATP's (no energy harnessing): Δ Go = -45 kcal/mole kcal/mole Out of this comes 2 ATPs, worth 14 kcal/mol. 31 kcal/mole output as heat. So the efficiency is about 14/45 = ~30% Not bad at all.

Since 2 ATPs ARE produced, taking them into account, for the reaction: Glucose + 2 ADP + 2 Pi à 2 lactate + 2 ATP ΔGo = -31 kcal/mole (45-14) Very favorable. All the way to the right. Keep bringing in glucose, keep spewing out lactate, Make all the ATP you want. 47 Characteristics of metabolic pathways

Aside from its role in energy metabolism:

Glycolysis is a good example of a metabolic pathway.

Two common characteristics of a metabolic pathway, in general:

1) Each step = a small chemically reasonable change

2) The overall DGo is substantial and negative.

48 Energy yield But all this spewing of lactate turns out to be wasteful. Using oxygen as an oxidizing agent glucose could be completely oxidized, to: … CO2 That is, burned.

How much energy released then? Glucose + 6 O2 à 6 CO2 + 6 H2O DGo = -686 kcal/mole ! Compared to -45 for glucose à 2 lactates (both w/o ATP production considered)

Complete oxidation of glucose, Much more ATP But nature’s solution is a bit complicated. The fate of pyruvate is now different 49 From glycolysis: acetyl-CoA 50 pyruvic acid Scores: per glucose 2 NADH 2 ATP pyruvic acid 2 NADH

2 CO2

Krebs cycle Tricarboxylic acid cycle TCA cycle

a keto glutaric acid

succinic Handout 9A acid 50 Acetyl-OCoA O||

CH3 –C –OH + co-enzyme A à acetyl~CoA acetic acid (acetate)

coA acetate group

pantothenic acid (vitamin B5) 51 Per glucose FromB glycolysis: acetyl-coA pyruvate Input 2 oxaloacetates 2 NADH 2 ATP pyruvic acid 2 NADH 2 NADH 2 CO2 2 NADH 2 CO2 citric acid 2 CO2 oxaloacetic acid 6 CO2 Krebs Cycle isocitric acid malic acid

fumaric acid a keto glutaric acid

52 GTP is energetically equivalent to ATP

GTP + ADP à GDP + ATP

ΔGo = ~0

G= guanine (instead of adenine in ATP)

53 acetyl-CoA B Per 54glucose 2 oxaloacetate 2 NADH 2 ATP pyruvic acid 2 NADH 2 NADH 2 NADH 2 FADH2 citric acid 2 NADH oxaloacetic acid 2 CO2 2 CO2 Krebs cycle 2 CO2 isocitric acid 6 CO malic acid 2

fumaric acid a keto glutaric acid

succinic acid 54 FAD = flavin adenine dinucleotide

Business end (flavin) ~ Vitamin B2 ribose

adenine -

ribose FAD + 2H. à FADH 2 55 D acetyl-CoA Per 56glucose oxaloacetate

pyruvic acid 2 NADH 2 ATP 2 NADH 2 ATP 2 NADH 2 NADH citric acid 2 FADH2 oxaloacetic acid 2 CO2 2 CO2 Krebs cycle 2 CO2 isocitric acid malic acid Note label is in OA after one turn of cycle, half the time fumaric acid on top, half on bottom. So a keto glutaric acid no CO2 from Ac-CoA after succinic just one turn. (CO2 in first acid turn is from OA). Succinic dehydrogenase 56 E

Per glucose57 2 NADH 2 ATP pyruvic acid 2 NADH 2ATP 2 NADH 2 NADH 2 FADH2 citric acid 2 NADH oxaloacetic acid 2 CO2 2 CO2 Krebs Cycle 2 CO isocitric acid 2 malic acid

fumaric acid a keto glutaric acid

succinic acid 57 Glucose + 6 O2 à 6 CO2 + 6 H2O :

By glycolysis plus one turn of the Krebs Cycle:

1 glucose (6C) à 2 pyruvate (3C) à 6 CO2 √

2 X 5 NADH2 and 2 X 1 FADH2 produced per glucose 4 ATPs per glucose (2 from GLYCOL., 2 from KC as GTP)

NADH2 and FADH2 still must be reoxidized …. No oxygen yet to be consumed No water produced yet

Paltry increase in ATP so far 58 Oxidation of NADH by O2

NADH2 + 1/2 O2 --> NAD + H2O ΔGo = -53 kcal/mole

If directly coupled to ADP à ATP (7 kcal cost), 46 kcal/mole waste, and heat

So the electrons on NADH (and FADH2) are not passed directly to oxygen, but to intermediate carriers,

Each transfer step involves a smaller packet of free negative energy change (release) 59