Energy coupling between two spontaneous reactions?

Energy coupling between two spontaneous reactions?

We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

This diagram describes energy coupling between a nonspontaneous reaction (the formation of glutamine from glutamic acid and ammonia) and a spontaneous reaction (the hydrolysis of ATP). I can see that if you add reaction 1 ($ce{Glu + NH3->Glutamine}$) and reaction 2 (ATP hydrolysis), then the overall ∆G will be negative, and this reaction will be spontaneous.

My question is this: why can't you add the reverse of reaction 1, which would on its own be spontaneous, to reaction 2? Would that not make an overall ∆G that is even more negative (more spontaneous)? This would be even more favorable than what we got from the energy coupling shown in the diagram. Yet it seems that energy coupling can produce the reaction presented in the diagram. How does the universe know to add the nonspontaneous reaction to the spontaneous reaction rather than the spontaneous reaction to another spontaneous reaction? It seems to me that the latter would be more probable.

Points to bear in mind

  1. That the biological coupling of an energetically favourable and unfavourable reaction (I would avoid using the term spontaneous†) is done through a composite reaction involving all the components of the two separate reactions. The reactants and products are the same as in the sum of the separate reactions - so it is valid to calculate the overall free energy change from the separate reactions - but we are talking about a different reaction or serious of reactions that occur on the enzyme. For example in the glutamate synthetase reaction that you mention there are the two reactions shown below. Gamma-glutamyl phosphate can be considered as an activated intermediate to which the free energy of hydrolysis of ATP has been transferred, rather than dissipated in heat, as would occur in the isolated reaction with inorganic phosphate as the product.

  1. Such coupled reactions are no different from non-coupleld thermodynamically favourable reactions in a cell in that they require an enzyme to catalyse them. They do not proceed spontaneously† because there is an activation energy barrier. If you regard this as “forcing” a reaction, then the answer to the question you pose in your comment to @VonBeche is 'yes'. However I would describe it as an enzyme having evolved to 'permit' or 'facilitate' a reaction. It does this by providing binding sites for the substrates and (often) a different reaction pathway that involves a lower activation energy.

So why no reaction Gln + ATP ➞ Glu + NH3 + ADP + Pi ?

This can be regarded in two ways.
(1) Why hasn't an enzyme evolved to do this sort of thing?
Either there is no selective pressure on an organism for an unneccesary enzyme to evolve, or (in this case) if the organism needs an enzyme to deaminate glutamine one evolves, but this enzyme (glutaminase) doesn't need to involve ATP as the reaction is already thermodynamically favourable.

(2) Why doesn't the glutamate synthetase enzyme catalyse this reaction
From your comment, this seems to be your concern. There are three binding sites on the enzyme, one for glutamate, one for ATP and one for ammonia; and I presume you are thinking why can't glutamine bind to the glutamate site (after all it leaves from there) and react with ATP to form γ-glutamyl phosphate. The answer to this is the chemistry of the reactions shown above. The mechanism of the glutamate synthetase reaction is complex, but at the simplest level the first stage is the nucleophilic attack of the negative acid oxygen (yellow) of glutamate on the γ-phosphate of ATP: In glutamine there is an amino group instead of the negative oxygen, so the reaction will not occur.

Furthermore, you should be aware that the enzyme is an active participant in the reaction, and the various substrates and products are oriented specifically in relation to residues at the active site of the enzyme to allow the reaction to proceed. To give a flavour of this I reproduce a portion of the summary of a paper by Liaw and Eisenberg on the mechanism of action of glutamine synthetase. I don't expect you (or anyone else) to digest this - it merely serves to illustrate that enzymes are highly sophisticated machines in which many components participate in producing a specific product.

Keep in mind that there needs to be an enzyme present to "couple" two reactions. You can imagine it like the energy from the ATP hydrolysis pushing the enzyme to a high energy state, and from that high energy state the enzyme can push the secondary reaction against the equilibrium.

You could do this, and there might even be examples (maybe for some reactions that are just barely exothermic), it's just that enzymes coupling two spontaneous reactions did not evolve that often as there's not much pressure to do that. A spontaneous reaction could just be sped up by a single enzyme, no need to build a big coupling enzyme for that. Coupling this to ATP hydrolysis would just lead to useless energy consumption and extinction.

Gibbs Free Energy and Its Formula | Thermodynamic System | Energy Management

After reading this article you will learn about the gibbs free energy and its formula,which is a process-initiating work obtainable from an isothermal,isobaric thermodynamic system.

Gibbs free energy is the measures of “useful” or process-initiating work obtainable from an isothermal, isobaric thermodynamic system. It is the maximum amount of non-expansion work that can be extracted from a closed system this maximum can be attained only in a completely reversible process. When a system changes from a well-defined initial state to a well-defined final state, the Gibbs free energy ∆G equals the work exchanged by the system with its surroundings. The free energy change (∆G) of a reaction determines its spontaneity. A reaction is spontaneous if ∆G is negative (if the free energy of the products is less than the free energy of the reactants).

Where ∆G= change in free energy, ∆G 0 = standard free energy change (with 1 M reac­tants and products, at pH 7), R = gas constant, T = absolute temperature

At equilibrium, ∆G equals zero. Solving for ∆G 0 ‘ yields the relationship at left.

K’eq, the ratio [C][D]/[A][B] at equilibrium, is called the equilibrium constant.

The standard free energy change (∆G 0 ‘) of a reaction may be positive and the actual free energy change (∆G) negative, depending on cellular concentrations of reactants and products. Many reactions for which ∆G 0 ‘ is positive are spontaneous because other reactions cause depletion of products or maintenance of high substrate concentrations.

An equilibrium constant greater than one, (more products than reactants at equilibrium) indicates a spontaneous reaction (negative ∆G°’). Free energy changes of coupled reactions are additive.

Examples of different types of coupling:

Some enzyme-catalyzed reactions can be expressed in two coupled half-reactions, one spontaneous and the other non-spontaneous. The free energy changes of the half-reactions may be added, to yield the free energy of the coupled reaction.

For example, in the reaction catalyzed by the Glycolysis enzyme Hexokinase, the two half-reactions are:

Pi + glucose <—> glucose-6-P+ H2O

ATP + glucose <—>ADP + glucose-6-P

Two separate enzyme-catalyzed reactions occurring in the same cellular compart­ment, one spontaneous and the other non-spontaneous may be coupled by a common inter­mediate (reactant or product). For example, reactions involving pyrophosphates,

Overall reaction A + ATP+ H2O <—> B + AMP + 2Pi ∆G 0 ‘ = (+15-33) = -18 kJ/mol

Pyrophosphate (PPi) is often the product of a reaction that needs a driving force. Its spontaneous hydrolysis, catalyzed by pyrophosphatase enzyme, drives the reaction for which PPi is a product.

Active transport of ions through membrane is coupled to a chemical reaction, e.g., hydrolysis or synthesis of ATP, while the transports of an ion say A + creates a potential dif­ference across the membrane.

The free energy change (electrochemical potential difference) associated with transport of an ion A + across a membrane from side 1 to side 2 is represented as:

Where R = gas constant, T = temperature, Z = charge on the ion, F = Faraday constant, and ∆Ψ = voltage across the membrane.

Each of our cells has an electric potential associated with it. This potential, or voltage, helps to control the migration of ions across the cell membranes. A major example of electri­cal work is in the operation of the nerves. The nerves when get stimulated, they generate an electrical impulse called an action potential which can communicate information to the brain, or carry a signal from brain to a muscle to initiate its movement.

Since free energy changes are additive, the spontaneous direction for the coupled reac­tion will depend on the relative magnitudes of ∆G for the ion flux (∆G varies with the ion gradient and voltage) and ∆G for the chemical reaction (∆G 0 ‘ is negative in the direction of ATP hydrolysis. The magnitude of ∆G also depends on the concentrations of ATP, ADP, and Pi).

Two examples of such coupling are:

The membrane transport requires energy or ATP. The spontaneous ATP hydrolysis (negative ∆G) is coupled to (drives) ion flux against a gradient (positive ∆G).

It takes place in mitochondria, which is also known as the power house of the cell and where the synthesis of ATP occurs. The spontaneous H + flux across a membrane (negative ∆G) is coupled to (drives) ATP synthesis (positive ∆G).


According to the popular but misleading definition of chemical coupling I have introduced in the previous section, metabolic pathways are a set of chemically coupled reactions. For that reason, it is not surprising that the confusion about the spontaneity of the so-called coupled reactions reappears when metabolic pathways are considered. To illustrate this point, I can quote: “Another common mechanism for coupling an unfavorable reaction to a favorable one is simply to arrange for one of the reactions to precede or follow the other” [ 4 ] or “Although any given enzymatic reaction in a sequence may have a characteristic positive free energy change, as long as the sum of all the free energy changes is negative, the pathway will proceed” [ 5 ]. These asseverations flagrantly disregard an important fact any overall reaction will proceed spontaneously if, and only if, all and every one of the elementary reactions are spontaneous by themselves [ 6 ]. Herein, I shall try to prove it by invoking the Second Principle of Thermodynamics.

An open thermodynamic system. The system is constituted by the reactants and enzymes involved into two sequential metabolic reactions. Each enzyme is isolated by a semi-permeable membrane (dashed line) that separates two compartments within the thermodynamic system. In contrast, all the reactants, which are able to cross the membrane, are freely distributed throughout the entire system. In addition, we assume that the chemical potential of every metabolite is kept constant through interchange with the environment (arrows). It must be noted that because both reactions are connected through the common intermediate B (which is diffusible), the concept of coupling, as it is defined in most texts, is fully applicable to the system sketched in this figure.

Energy coupling between two spontaneous reactions? - Biology

31 notecards = 8 pages ( 4 cards per page)

Biology chapter 8

The process of cellular respiration, which converts simple sugars such as glucose into CO2 and water is an example of ______.

A. A pathway in which the entropy of the system decreases
B. A catabolic pathway
C. A pathway that occurs in animal cells but not plant cells
D. A pathway that converts organic matter into energy
E. An endergonic pathway

Energy is observed in two basic forms potential and kinetic. Which of the following correctly matches the forms with a source of energy.

A. The motion of individual molecules: potential energy

B. The energy related to the height of the bird above the ground: Kinetic Energy

C. The covalent bonds of a sugar molecule: potential energy

D. The heat released from living organism: potential energy

E. The energy associated with a gradient of ions across a membrane: Kinetic Energy

The covalent bonds of a sugar molecule: potential energy.

Which of the following is true of metabolism and it’s entirety of all organisms?

A. Metabolism consists of all energy transformation reactions in an organism

b. Metabolism uses all of an organism’s resources

c. Metabolism manages the increase of entropy in an organism

d. Metabolism depends on a consistent supply of energy from food

Metabolism consists of all the energy transformation reactions in organism.

Which of the following is an example of potential rather than Kinetic energy?

a. A crawling beetle foraging for food

b. Water rushing over Niagara Falls

C. Light flashes emitted by a firefly

Most cells cannot harness heat perform work because ______.

a. Heeat is not a form of energy

b. Temperature is usually uniform throughout the cell

c. Heat can never be used to do work

d. He must maintain constant during work

Temperature is usually uniform throughout the cell

Which of the following statements about the combustion of glucose with oxygen to form water and carbon dioxide is correct?

a. The reverse reaction, making glucose from water and carbon dioxide, must be an Exergonic reaction

b. The entropy of the products is greater than the entropy of the reactants

c. The entropy of the universe decreases as a result of this reaction

d. This is the process of cellular respiration, an anabiolic pathway that releases free energy

e. The free energy lost in the combustion is less than the energy that appears as heat

The entropy of products is greater than the entropy of the reactants

Which of the following statements about equilibrium of chemical reaction is correct?

a. The equilibrium point is where the system has the highest free energy

b. Reactions can only go in the direction towards equilibrium

c. Most reactions in a living cell are close to equilibrium

d. Every action that is at equilibrium is not capable of doing any work

e. Equilibrium point of the reaction represents at least stable configuration for that reaction

A reaction that is at equilibrium is not capable of doing any work.

The mathematical expression for the expression for the change of a system is ΔG = ΔH - TΔS. Which of the following is correct?

a. H is the change in entropy, the energy available to do work

b. G is the change in free energy

c. S is the change enthalpy, a measure of randomness

d. T is the temperature in degrees Celsius

ΔG is the change in free energy

Which of the following statements about ATP is correct?

A. The energy release on hydrolysis of ATP is the result of breaking a high energy bond

b. The hydrolysis of ATP is an endergonic process

c. The hydrolysis of ATP can supply energy needed for catabolic pathways

d. Almost all of the free energy released on the hydrolysis of ATP is released as heat

e. The cycling between ATP and ADP + pi provides an energy coupling between catabolic and anabolic pathways

The cycling between ATP and ADP + Pi provides an energy coupling between catabolic and anabolic pathways

Why is ATP in important molecule in metabolism?

a. It’s terminal phosphate group contains a strong covalent bond that, when hydrolyzed, releases free energy

b. It’s terminal phosphate bond has higher energy than the other two phosphate bonds

c. It provides energy coupling between exergonic and endergonic reactions

d. It’s hydrolysis provides an input of free energy for exergonic reactions

It provides energy coupling between exergonic and endogonic reactions

Catabolic pathways _______.

b. Are spontaneous and do not need enzyme catalysis

C. Supply energy, primarily in the form of ATP, for the cells work

d. Combine molecules into more energy rich molecules

Supply energy, primarily in the form of ATP, for the cells work.

Enzymes are described as catalyst, which means that they ______.

a. Increase the free energy of the reactance to make the reaction go faster

b. Can alter the free energy change for a chemical reaction

c. Provide activation energy for the reactions they facilitate

e. Increase the rate of a reaction without being consumed by the reaction

Increase the rate of a reaction without being consumed by the reaction

Which of the following would be unlikely to contribute to the substrate specificity of an enzyme?

a. A hydrophobic group on the substrate interacts with several hydrophobic amino acids on the enzyme

b. The enzyme has an allosteric regulatory site

c. A similar shape exist between a pocket on the surface of the enzyme in the functional group of the substrate

The enzyme has a allosteric regulatory state

Which of the following is NOT a way in which an enzyme can speed up the reaction that is catalyzes?

a. Binding of the substrate to the active site can stretch bonds in the substrate that need to be broken

B. The enzyme binds a cofactor that interacts with the substrate to facilitate the reaction

c. The active site can provide heat from the environment that raises the energy content of the substrate

The active site can provide heat from the environment that raise the energy content of the substrate.

Which of the following is true when comparing and uncatalyzed reaction to the same reaction with a catalyst?

a. Catalyzed reaction will be slower

b. The catalase reaction will have the same as G

The catalyzed reaction will consume all of the catalyst

The catalyzed reaction will have the same ΔG

The binding of a compound to an enzyme is observed to slow down or stop the rate of the reaction catalyzed by the enzyme? increasing the substrate concentration reduces the inhibitory effects of this compound? which of the following could account for this observation?

A. The compound is a competitive inhibitor

b. The compound reduces disulfide bonds, causing the enzyme molecules to partially unfold

c. The compound is a negative allosteric regulator

The compound is a competitive inhibitor

Select the correct statement about chemical energy, a term used by biologists to referred to potential energy available for release in a chemical reaction.

a. Light energy is converted to a chemical energy during photosynthesis

b. When a glucose molecule is cannibal iced to CO2 and H2O, chemical energy is lost

C. A photosynthetic cell within a plant leaf produces chemical energy, stored within glucose molecules

Light energy is converted to chemical energy during photosynthesis.

Which statement about finding of enzymes and substrates is correct?

a. Substrate molecules that into the active site of an enzyme like a key fits into a lock

b. Substrate molecules binds to the active site of the enzyme only by weak bonds, such as hydrogen bonds or hydrophobic attraction

c. When substrate molecules bind to the active site of the enzyme, the enzyme undergoes a slight change in shape

When substrate molecules bind to an active site of the enzyme, the enzyme undergoes a slight change in shape.

Which of the following involves a decrease in entropy?

b. Reactions a separate monomers

c. Depolymerization reactions

Which of the following is the most correct interpretation of the figure.

A. Pi acts as a shuttle molecule to move energy from ATP to ADP

b. ATP is a molecule that acts as an intermediary to store energy for cellular work

c. Energy from ccatabolism be used directly for performing cellular work

d. ATP + pi are a set of molecules that store energy for catabolism

ATP is a molecule that acts as an intermediary to store energy for cellular work

How do cells use the ATP cycle shown in the figure?

a. Cells use the cycle to recycle energy released by ATP hydrolysis

b. Cells use the cycle to recycle ADP, phosphate, and the energy released by ATP hydrolysis

c. Cells use this primarily to generate heat

D. Cells use the cycle to recycle ADP and phosphate

Cells use the cycle to recycle ADP and phosphate

According to the induced fit hypothesis of enzyme catalysis, ______.

A. The binding of the substrate depends on the shape of the active site.

B. Active site creates of microenvironment idea for the reaction

C. Some enzymes change their structure when the activators bind to the enzyme

D. The binding of the substrate changes the shape of the enzyme’s active site

The binding of the substrate changes the shape of the enzyme’s active site

Which of the following is a statement of the first law of thermodynamics?

a. Energy cannot be created or destroyed

b. Energy cannot be transferred or transformed

c. Entropy of the universe is constant

Entropy of the universe is decreasing

Energy cannot be created or destroyed

Which of the following statements is a logical consequence of the second law of thermodynamics?

a. Every energy transformation Byselle decreases the entropy of the universe

b. Conversion of energy from one form to another is always accompanied by gain of free energy

c. Cells require constant input of energy to maintain their high level of organization

d. Without input of energy, organisms would tend towards decreasing entropy.

Cells require a constant input of energy to maintain their high level of organization.

Which of the following types of reactions would decrease the entropy within a cell?

Which of the following is true for all Exergonic reactions?

a. The reaction goes only in forward direction, all reactants will be converted to products, but no products will be converted to reactants.

B. The reaction proceeds with the net release of free energy

c. The products have more total energy than the reactants

d. The net input of energy from the surrounding is required for the reactions to proceed

The Second Law of Thermodynamics

The second law of thermodynamics states that for any spontaneous process, the overall ΔS must be greater than or equal to zero yet, spontaneous chemical reactions can result in a negative change in entropy. This does not contradict the second law, however, since such a reaction must have a sufficiently large negative change in enthalpy (heat energy). The increase in temperature of the reaction surroundings results in a sufficiently large increase in entropy, such that the overall change in entropy is positive. That is, the ΔS of the surroundings increases enough because of the exothermicity of the reaction so that it overcompensates for the negative ΔS of the system. Since the overall ΔS = ΔSsurroundings + ΔSsystem, the overall change in entropy is still positive.


Thermodynamics is concerned with describing the changes in systems before and after a change. This usually involves a discussion about the energy transfers and its dispersion within the system. In nearly all practical cases, these analyses require that the system and its surroundings be completely described. For instance, when discussing the heating of a pot of water on the stove, the system may includes the stove, the pot, and the water and the environment or surroundings may include everything else. Biological organisms are what are called open systems energy is transferred between them and their surroundings.

1st Law of Thermodynamics

The first law of thermodynamics deals with the total amount of energy in the universe. It states that this total amount of energy is constant. In other words, there has always been, and always will be, exactly the same amount of energy in the universe. According to the first law of thermodynamics, energy may be transferred from place to place, but it cannot be created or destroyed. The transfers of energy take place around us all the time. Light bulbs transfer energy from an electrical power station into heat and photons of light. Gas stoves transfer energy stored in the bonds of chemical compounds into heat and light. Heat, by the way, is the amount of energy transferred from one system to another because of a temperature difference. Plants perform one of the most biologically useful energy transfers on earth: they transfer energy in the photons of sunlight into the chemical bonds of organic molecules. In every one of these cases energy is neither made nor destroyed and we must try to account for all of the energy when we examine some of these reactions.

1st Law and the Energy Story

The first law of thermodynamics is deceptively simple. Students often understand that energy cannot be created or destroyed. Yet, when describing an energy story of a process they (and their professors) often make the mistake of saying things such as "energy is produced from the transfer of electrons from atom A to atom B". While most of us will understand the point the speaker is trying to make, the wrong words are being used. Energy is not made or produced, it is simply transferred. To be consistent with the first law, when telling an energy story, make sure that you try to explicitly track all of the places that ALL of the energy in the system at the start of a process goes by the end of a process.

2nd Law of Thermodynamics

An important concept in physical systems is that of . Entropy is related to the with the ways in which energy can be distributed or dispersed within the particles of a system. The 2nd Law of Thermodynamics states that entropy is always increasing in an isolated system. This idea helps explain the directionality of natural phenomena. In general, the notion is that the directionality comes from the tendency for energy in a system to move towards a state of maximal dispersion. The 2nd law, therefore, means that in any transformation we should look for an overall increase in entropy (or dispersion of energy), somewhere. An idea that is associated with increased dispersion of energy in a system and/or its surroundings is that as dispersion increases the ability of the energy to be directed towards work decreases.

There will be many examples of where the entropy of a system decreases (things become more organized, rather than more random). To be consistent with the second law, however, we must try to find something else (likely a closely connected system in the surroundings) that must compensate for the "local" decrease in entropy with an equal or greater increase in entropy.

The entropy of a system can increase when:(a) it gains energy (b) a change of state occurs from solid to liquid to gas (c) mixing of substances occurs (d) the number of particles increases during a reaction.

Does the second law say that entropy is conserved?

Biological systems, on the surface, see to defy the Second Law of Thermodynamics. They don't. Why? Hint: You might want to employ the terms "isolated system" vs. "closed system" vs. "open system" in your answer.

A fine point. The figure below mentions order and disorder and shows that this is somehow related to a change in entropy (&DeltaS). It is common to describe entropy as a measure of disorder as a way to simplify the more concrete description relating entropy to the number of states in which energy can be dispersed in a system. While the idea of measuring disorder to define entropy has some flaws, it is sometimes a useful, if imperfect, proxy. Consider the figure above. Here order serves as a good proxy for approximating the number of ways to distribute energy in the system. Can you describe why this is the case?

What are Nonspontaneous Reactions

Nonspontaneous reactions refer to the chemical reactions that require an energy input to proceed. In nonspontaneous reactions, both enthalpy and entropy prefer the reactants. Thus, reactants are more stable than products. On that account, the chemical reaction is endergonic, absorbing heat. It decreases the entropy as well. The change of the Gibbs free energy over time in nonspontaneous reactions is shown in figure 2.

Figure 2: The Change of the Gibbs Free Energy/Time

The reaction between atmospheric nitrogen and oxygen is an example of a nonspontaneous reaction. It forms nitrogen monoxide. At the normal atmospheric pressure and temperature, this reaction is unfavourable. This means the reactants of the chemical reaction, i.e., nitrogen and oxygen gases, are more stable than the product: nitrogen monoxide. But, at very high temperatures such as when lightning, this reaction is favourable.

What is the Difference Between Coupled and Uncoupled Reaction?

Most chemical reactions that we know are non-spontaneous therefore, we need to couple them with some other reactions in order to make them progress. Thus, this new reaction type is called a coupled reaction while the previous non-spontaneous reaction type is called an uncoupled reaction. The key difference between coupled and uncoupled reaction is that coupled reactions show energy transferring from one side of the reaction to the other side whereas in uncoupled reactions there is no energy transfer taking place.

Below infographic tabulates more differences between coupled and uncoupled reaction.

Related Resources

The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses visit MIT OpenCourseWare at

PROFESSOR: OK, let's get started. Why doesn't everyone go ahead and take 10 more seconds on this clicker question. It should look familiar. You were pretty split on this question on Friday, so we're hoping after learning a little bit more about delta g of formation, we have at least one direction that wins out here.

OK, great. So now we're up to 85% of you, and hopefully in just a minute we'll be up to a 100%. But if delta g of formation here is less than zero, that means we're talking about a spontaneous reaction when we form the compound. So if it spontaneously forms the compound, that must mean that the compound is going to be stable relative to its elements. So that's kind of the last thing we went over in terms of topics on Friday, and we'll pick up right there today.

All right. Here we have our notes for today, so reminder exam 2 is on Wednesday -- I don't think I need to remind anyone of that. But just please, don't come to this room, make sure that you go to Walker to take the exam. And also if you have any questions or you feel like there's anything you don't understand, I have office hours today from 2 to 4 in my office. Your TAs have also all moved their office hours so they fall before the exam. So make sure you get any of your questions addressed before you're trying to sit there and figure it out in exam time.

All right, so today we're going to pick up where we left off on thermodynamics, so that was talking about free energy of formation. After that we're going to talk a little bit more about the effect of temperature on spontaneity. We touched upon this on Friday, but we're going to formalize exactly in which cases temperature can or can not affect whether a reaction is spontaneous or non-spontaneous.

Then we're going to look a little bit into thermodynamics and biological systems. Two examples that I wanted to talk about were ATP coupled reactions -- those are very important in biology. And also, thinking about the idea of hydrogen bonding, we're going to combine our thoughts on bond enthalpies, and also our ideas of bonding that we had from thinking about covalent and ionic bonds.

All right, so let's finish up first with free energy of formation. So as 85% of you just told me, when we have a case where delta g of formation is less than zero, what we're talking about here is a compound that is thermodynamically stable relative to its elements. So that means we know the inverse as well, which is when we're talking about a case where delta g is now greater than zero, the compound is going to be thermodynamically unstable relative to its elements.

So any time you're looking at delta g of formation, just remember to think about this is just the delta g of a reaction where you're forming a compound. So it should make sense that if it's negative, it's going to be spontaneous and you're going to have a stable compound here.

So we could look at any number of examples, really we could pick any compound to look at, but up on the screen here I'm just putting the compound, the formation of benzene, we've looked at benzene a lot. So benzene has a delta g of formation of 124 kilojoules per mole. Is benzene stable or unstable compared to its elements?

I can't really tell the difference between stable and unstable when you say it, so everyone start at the same time, go.

PROFESSOR: OK, excellent. Benzene is unstable compared to its elements. So that means the reverse reaction here is going to be stable -- or the reverse reaction is going to be spontaneous where we actually have the decomposition of benzene.

So when we're seeing this, when we're seeing that the reverse reaction is spontaneous, a question that might immediately come to us is well, why did we just spend all the time talking about benzene because clearly benzene's just going to break down, right. Why do we form benzene and not have it immediately decompose into its elements. Thermodynamically that is what should happen and it is what does happen, but the reality is that this reaction, this decomposition of benzene is actually very, very slow. It's so slow that, you know we use benzene all the time in organic reactions, we don't see it break down even when we heat it up, is because even though this is thermodynamically a non-spontaneous reaction to form benzene, it's very slow for the actual decomposition of benzene to occur.

So this is just another case, and I'll keep saying this, and when Professor Drennan starts talking about kinetics, she'll keep repeating this as well. What we want to keep in mind is that delta g tells us whether a reaction will happen or whether it won't happen. It tells us absolutely nothing about how long it takes for that reaction to happen. It tells us nothing about the rate of the reaction. We'll keep seeing examples of this, so hopefully no one will be confused by the time we do get to kinetics.

So if we're talking about calculating delta g for any reaction, now we actually have several ways to do it. The first way is very analogous to thinking about delta h for the reaction. We can just look up a table where we have delta g as a formation. So we can take the sum of the delta g of formation of the products, and subtract from it the delta g of formation of the reactants.

We also have another way if maybe we don't have that information available to us. We can also take a look at using this reaction or this equation right here, which is telling us that delta g of a reaction is equal to the change in enthalpy minus t delta s. So that tends to be very helpful, especially when we want to take into consideration temperature.

So let's take into consideration temperature. We've done this a little bit so far, but let's really take a look at some reactions where temperature's going to make a big difference. So the reaction we're going to look at here is the decomposition of sodium bicarbonate, or sodium bicarb here, and it decomposes it into sodium carbonate, plus c o 2, carbon dioxide, and water. So we can think about calculating that delta g of this reaction. I'll tell you that the change in enthalpy, this is actually an endothermic reaction. It requires heat. It's plus 135 . 6 kilojoules per mole.

So let's go to a clicker question and I want you to pick out from several choices which of the changes in entropy seem reasonable to you here. So what would you predict the delta s for this reaction to be? All right, let's do 10 more seconds on that.

OK, so we've got the majority, but not everyone. So let's take a look at why this is the correct answer, that it should be plus 0.334 . If we're going from 2 moles of solid, to 1 mole of solid, plus 2 moles of gas, are we increasing or decreasing the disorder?

PROFESSOR: We're increasing the disorder. If we're getting to a more disordered state, then we're going to have a positive change in entropy. We're going to increase the disorder. The only one with a positive delta s is this choice here. So if we switch back to our notes, we can see that, in fact, so what we see is that it's 0.334 kilojoules per k per mole is our delta s, so we can go ahead and calculate our delta g for the reaction, so we're just plugging in our delta h. So our delta h is 135 . 6. And we're talking about to start with let's talk about room temperature. So 298 k times 0.334 .

So what we end up having for the delta g of our reaction is that its 36 . 1 kilojoules per mole. So this is at room temperature. So is our reaction spontaneous or non-spontaneous at room temperature?

PROFESSOR: Non-spontaneous. All right, but let's take a look at a different temperature. For example, let's look at baking temperature. So if we think about baking cookies, we maybe bake them at 350 degrees fahrenheit, so that would be 450 kelvin -- our ovens are usually set to fahrenheit and not kelvin. So if we think about this reaction that this temperature, first of all let me point out why we would be talking about baking cookies for this particular reaction here. Does anyone know what another name for sodium bicarb is?

PROFESSOR: Yeah, it's just baking soda. So this is the reaction that causes your cookies or your cakes to actually rise. So we're producing gas here, and when we're producing that gas when this sodium bicarb decomposes, we're forming these pockets of gas in our baked goods. So if we bake our cookies at room temperature, obviously, they don't bake and they also don't rise because the sodium bicarb, clearly it was non-spontaneous at room temperature, this reaction just doesn't happen.

But if we take a look at baking temperature now, we're going to plug in this new temperature, which is 450 k, and plug this into our delta g reaction. What we find is now our delta g for the reaction is negative 14 . 7 kilojoules per mole. So in this case, we are dealing with a spontaneous reaction, which is good, because this means that when we put our cookies in and we turn it to 350 fahrenheit or 450 k, the baking soda will decompose and we'll got our cookies to rise a little bit.

All right, so one thing that I want you to notice when we were talking about the case with the decomposition of sodium bicarb is that the delta h for that reaction and the delta s, they both had the same sign. And something that you can keep in mind in general is any time that both delta h and delta s have the same sign, it's actually possible to switch from spontaneous to non-spontaneous or vice versa just by changing the temperature of the reaction.

And we can think about this graphically. If we assume that delta h and delta s are independent of temperature, which is a good assumption to a first approximation, in this case we find that the delta g of the reaction is the linear function of the temperature. So that means we can go ahead and graph what we saw for the case of baking soda.

So our first point here was that we saw at room temperature, about 298 k, the delta g was positive, it was 36 kilojoules per mole. We also saw that once we heated it up to baking temperature, we actually had the delta g now at negative 15 kilojoules per mole. So since this is linear, we can actually draw a straight line right through here, and we can think about the fact that we have this temperature here where anything below this temperature has a positive delta g, and anything above this temperature is going to have a negative delta g.

And let's actually think about the fact that this is a line and see what our slope and our y-intercept is going to mean. We can say that delta g is equal to negative delta s t plus delta h, if we want to put it in the formation of the equation for line. So what this tells us is that our slope is going to be negative delta s here, and if we think about our y-intercept, that's going to be the change in enthalpy for the reaction.

So again, what I want to point out is that any time we're at a temperature that's lower than this change in temperature here, we're going to find that delta g is greater than zero, and we're going to be dealing with a non-spontaneous reaction.

However, if we move our temperature up and up and up so we go across the graph this way, eventually we'll hit a point where if we're above that temperature, we'll find that delta g is less than zero, and now we have a spontaneous reaction. So we can actually think about what this temperature is. We can call this T star. This is the temperature at which our reaction switches, whether it's spontaneous or non-spontaneous. And if you're thinking about trying to get a reaction to go, it's very important to be able to calculate what this temperature is. A lot of times in organic chemistry laboratories, they need to heat up reactions -- part of why they do that is kinetics, but the other part is sometimes they need to make a reaction go from being non-spontaneous to spontaneous.

So let's think about how to calculate T star or this change in temperature. So we're talking about this threshold temperature, so we're talking about where delta g is going to be equal to zero, because if we set delta g equal to zero, we know that anything on one side of that temperature is going to be spontaneous, and the other side is going to be non-spontaneous.

So if we do this, we can just rewrite our reaction, delta g equals delta h minus t delta s, and let's plug in our zero for delta g there, and now rearrange our reaction so that we're talking about this threshold temperature. So that T star is just going to be equal to the change in enthalpy divided by the change in entropy. So it's very easy for us to calculate, so let's go ahead and do this for the case with baking soda. And for baking soda what we saw was that delta h was 136 kilojoules per mole, and the change in entropy was 0.334 kilojoules per k mole. And that means if we do that simple division, then what we end up as our temperature star is 406 kelvin.

So basically, what this tells us is if we tried to bake our cookies below 406 kelvin they would not rise, if we bake them above 406 kelvin they will rise because this reaction is now spontaneous.

All right, so this was a case where we had seen that the delta h and the delta s both had a positive value. But let's take a look at what happens when now delta h and delta s are both negative. So we can think about this just by plotting it on our graph again -- we don't have actual values, but we can think about what the sign should be. So if we talk about our zero point where temperature is absolute zero, if we have a positive, or excuse me, if we have a negative delta h now and we're at temperature equals zero, what is delta g going to be? Negative or positive? Yeah, it's going to be negative. So if we have negative delta h and we're at temperature of zero, our s term completely falls out, so we're definitely going to start with a negative delta g.

As we increase the temperature higher and higher, at some point that delta s term is going to become greater than our delta h term, so at some point we're going to flip to where the reaction has now a positive delta g.

So you can draw this into your graphs in your notes where we're going to start, in this case if delta g is negative where delta g starts negative, and then when it hits that threshold temperature, anything above that temperature is going to be positive.

So if we had actual numbers we could plug those into our graph, but we should be able to understand what the general trend is going to be even in a hypothetical case where we're just dealing with is it positive or is it negative.

So what we see is that below this temperature, we have a spontaneous reaction, and above this temperature we have a non-spontaneous reaction. That was flipped from the case where we had delta h and delta s both be positive.

All right, so let's actually summarize what all of our four different scenarios could be if we're dealing with delta h's and delta s's. So to start with, why don't you tell me what you think if we have a reaction where we have a negative delta h and we have a positive delta s, do you think that this will be a reaction that's never spontaneous, always spontaneous, or will this be one of these cases where the spontaneity depends on temperature?

All right, let's take 10 more seconds on this. OK, great. So most of you got it. So let's go back to the slides and think a little bit about why this is. So, this case is always spontaneous, because in this case we have a negative delta h, which contributes to a negative delta g, and a positive delta s, but since the equation says minus t delta s, that means a positive delta s is also going to contribute to a negative delta g. So regardless of what the temperature is, we're going to have a spontaneous reaction. So we say this is always spontaneous -- delta g is less than zero at all temperatures.

All right. So let's look at the reverse case here where delta h is now greater than zero, and delta s is less than zero. Is this always, never, or sometimes spontaneous?

PROFESSOR: Nope. So, this is going to be never spontaneous in this case. So for the same reason, because delta h is greater than zero, that contributes to a positive delta g, and delta s being negative also contributes to a positive delta g. So we'll say again, that this delta g is greater than zero at all temperatures.

All right, so now we have a case where delta h is greater than zero and delta s is greater than zero as well. Is this going to be always, never, or sometimes spontaneous?

PROFESSOR: Sometimes, good. So, even before thinking, you can just remember if the signs are the same, we can have a dependence on temperature here. So what we find is that this is sometimes spontaneous. So we can think about when this happens. Would it be when the actual temperature is greater or less than that threshold temperature?

PROFESSOR: All right, so I heard a lot of people say greater than. It's greater than, because when it's greater than that threshold temperature, that means that the delta s term is the one that's going to actually sort of take over, it's going to be at some point greater than the delta h term. The delta h is going to be making the delta g positive. That means it would make it non-spontaneous, but once delta s gets large enough, that's going to override, and now we're going to have a negative delta g only when the temperature is above that threshold temperature.

So similarly, when we see that delta h is negative and delta s is negative, this is also going to be sometimes spontaneous, and specifically, it will be spontaneous or delta g will be less than 0 when the temperature is less than that threshold temperature.

All right, so you should be able to look at any situation where you have or you figure out or you calculate the enthalpy and the enthropy change in the reaction, and you should right away, before doing any calculations, be able to know is this always spontaneous, is this never spontaneous, or is this something where I'm going to need to really take temperature into consideration. And if I do, you can actually calculate what that temperature is going to be where you flip from spontaneous to non-spontaneous or vice versa.

All right, so shifting gears a little bit, let's take a look at a few examples where it's important to think about thermodynamics and biological systems. So there's two things I particularly want to focus on. So the first is the idea of ATP coupled reactions. So this is really important, because when we're talking about biological reactions, a lot of the reactions that take place in our body are actually non-spontaneous, so energetically we figure that the delta g for this reaction is actually positive. So I just show a schematic biological reaction here where we have some molecule that's broken up into two building blocks. So let's say, for example, this has a delta g that's greater than zero. We need to think about how it is that our body can actually make this reaction happen.

And the way that it does this is that it takes the hydrolysis of ATP, and the hydrolysis of ATP is what's called that it's coupled to this non-spontaneous reaction. And when you have a spontaneous reaction coupled to a non-spontaneous reaction, you can potentially drive the reaction in the forward direction. Remember the hydrolysis of ATP is spontaneous, so what we're doing in this case is we're taking a spontaneous reaction where we give off energy, and we're coupling it to a non-spontaneous reaction that requires energy.

So what we will hope is when we add up the total energy changes between these two reactions, if the sum of those two is a negative delta g total, now we can have this whole process move in the forward direction.

So let's take a look at thinking about what we can do in terms of a coupled reaction. The first thing we need to do if we're thinking about using the hydrolysis of ATP is actually calculate what the delta g for the reaction of ATP hydrolysis is, and I picked 310 kelvin because that's the temperature of our bodies. So again, this reaction is taking adenosine triphosphate that has three phosphate groups and hydrolyzing it, so reacting it with water to form ADP plus phosphate plus acid here.

So I'll tell you that the delta h of this reaction is negative 24 kilojoules per mole. This is what we actually calculated in class on Friday. And that the delta s is plus 22 joules per k, per mole, so we can go ahead and calculate what this delta g is here. So the delta g of this overall reaction is just going to be, of course, delta h minus t delta s, and our delta h is minus 24 kilojoules per mole, and we'll subtract 310 kelvin times 0 . 022 kilojoules over k mole. So if we do this, what we end up getting is a delta g of negative 31 kilojoules per mole of ATP.

All right, so again, this is good, we got a negative number, that's what we were expecting. So that means that we will have energy available to use if we hydrolyize ATP and couple it to another reaction here. So what we saw for our delta g, negative 31 kilojoules per mole.

All right, so let's talk about a reaction that is a ATP coupled reaction, and there's just tons of examples of this in our body. One I picked because it has to do with glucose, which we have talked so much about, is the conversion of glucose to glucose 6 p. So 6 p -- p just stands for a phosphate group here, so what we're doing is taking, if we numbered the glucose carbons, this would be carbon number six, and we're putting a phosphate group on carbon number six.

So first of all, why would our body need to do this reaction? It turns out that glucose, we know that we use glucose for energy. Glucose is somewhat apolar, so it can actually move in and out of our cells, because our cell walls, those are very greasy. So what we're actually doing here is we're putting a charged molecule onto the glucose. There's two negative charges in a phosphate group, this is going to make sure that our glucose is now it's very polar molecule. And now that we have a very polar molecule, we're not going to be able to have the glucose move in and out of the cell. So we keep it in our cell by putting this phosphate group on it. That's why we want to do this reaction, but let's think about the energy of actually doing it. And it turns out that this requires energy. The delta g for this reaction is 17 kilojoules per mole.

All right, so that could be a problem if our body had not come up with a way to solve it, which it has, and what it actually does is it couples this reaction here with the conversion of ATP into ADP. And we just calculated that that has a delta g of negative 31 kilojoules per mole. So this means we can think about delta g total, and by total we mean of this overall process here. So that's just equal to 17, and we're adding it to the other delta g, which is negative 31. So we get an overall delta g for this process of negative 14 kilojoules per mole.

All right, so that's one really simple example just to illustrate to you how these ATP coupled reactions work. Some ATP coupled reactions require one molar equivalent of ATP, some require a lot more. But essentially, the idea is the same. We have this reaction that's energetically unfavorable that we couple with an energetically favorable reaction. So, quite literally, this is what we mean when we talk about ATP as being the energy currency for the cell. We spend some ATP in order to get these non-spontaneous reactions to go.

All right, so I also want to talk to a little bit about hydrogen bonding. This also is a topic that deals with the thermodynamics, and it also is related to the ideas that we talked about before in terms of thinking about different types of bonds.

So if we talk about a hydrogen bond, a hydrogen bond, first I just want to be really clear, is not a covalent bond. So this h x bond here is a covalent bond, this is not a hydrogen bond. But a hydrogen bond can form when you have a partial positive on a hydrogen, and you have what's called a hydrogen bond donor atom, which has a partial negative on it, and also has a lone pair. Typically this y atom will be on a separate molecule. And what happens is you have that Coulombic attraction between the partial positive on the hydrogen and the partial negative on this hydrogen bond donor atom. So what you form between them is a hydrogen bond, and you'll notice that I drew it as a dashed line. H bonds or hydrogen bonds are drawn either as dashed lines or as dotted lines, and they're done so to differentiate them from covalent bonds, which we draw with this straight line here.

So let's think about what can form hydrogen bonds. First of all, there's a requirement for what this h x bond can be. It has to be a really electronegative atom in this x here, because we need to we need to form that partial positive on the hydrogen, which means that the atom that it's covalentally bonded to needs to be pulling away some of that electron density from the hydrogen, such that we have a delta positive on a hydrogen atom. Similarly, the same atoms are what can be the y here, so it can either be a nitrogen or an oxygen or a fluorine. So these are the only hydrogen bond donors that can form h bonds.

And we can think about the reason for this and the reason is pretty straightforward. We need to have a small atom, but it also needs to be really electronegative. This should make sense because we need to have this partial negative charge here to have that Coulomb attraction. And the other requirement is that we definitely need to have that lone pair of electrons so it can actually interact in this hydrogen bond here.

So let's take a look at an example of a hydrogen bond, and that's between two water molecules. So let's go ahead and re-draw our water molecules -- they're just kind of randomly oriented there. But let's re-draw them as if they were going to have a hydrogen bond between them. And one thing I want to point out about hydrogen bonds is that they're strongest when you actually have all three atoms in a straight line like this, because it keeps the dipoles the strongest, so the partial positive here and the partial negative here can interact.

So let's re-draw our water molecules like that. So we have our first water molecule here, and let's say we're going to be talking about this hydrogen in terms of hydrogen bonding. Obviously, we have four hydrogens up there that can take place or take part in h bonding, but we're just going to focus on this one here. So that means we want our straight line to be something like this to our second oxygen. So let's draw the hydrogens on this water molecule as well.

All right, so thinking about what the Lewis structure of water is, how many lone pairs do we have on this oxygen here?

PROFESSOR: Two lone pairs, great. So let's draw those in, and let's draw these lone pairs in as well. All right, and we're going to be talking about these two atoms in terms of our hydrogen bond, so does this have a delta plus or a delta negative on it? Delta plus. So this is a polar bond here, this o h bond, so we have a delta plus on our hydrogen, and we have a delta minus on our oxygen. So these also have delta plusses, of course, and this also has a delta minus.

So because these two can interact and they're lined up to do so, what we can do is we can draw a hydrogen bond right in between this hydrogen and this oxygen atom here.

So again, I pointed out that, in fact, hydrogen bonds are not as strong as covalent bonds. Since they're not as strong, if you remember thinking about the relationship between bond strength and bond length, would you expect a hydrogen bond to be longer or shorter than a covalent bond?

PROFESSOR: Good, OK, it should be longer here. So this is our longer bond because it's our weaker bond. They're not held together as tightly. All right, so this is an example of an intermolecular hydrogen bond, it's between two different molecules. Hopefully, you can also see that if we're focusing on any single one water molecule, we could also form hydrogen bonds between other water molecules with either of these two hydrogens as well, and also with our other lone pair here. So any water molecule can actually form four different hydrogen bonds.

All right, so let's talk a little bit about these hydrogen bonds in terms of their bond enthalpies here. I said that they were weaker than covalent bonds, so let's look at a few comparisons here. So we can look at an h o hydrogen bond versus an h o covalent bond. And what we find is that the bond enthalpies in kilojoules per mole, it's 20 kilojoules per mole for a hydrogen bond versus 463 kilojoules per mole for a covalent bond. So we're not talking about just a little bit weaker, we're talking about much, much weaker for the hydrogen bonds here.

Similarly, if we look at nitrogen, for nitrogen hydrogen bonds, if we have an o h bound to a nitrogen, that's 29 kilojoules per mole. If it's an n h hydrogen bound to a nitrogen, that's 14 kilojoules per mole, and if we're talking about an h n covalent bond, now we're talking about 388 kilojoules per mole.

All right, so again what we see in these cases is that the hydrogen bonds are much, much weaker. They tend to be as low as 5% of what the covalent bond is, so this is actually much weaker than any kinds of bonds we have within molecules, but it's also the strongest type of bond that we can have between two different molecules. And one thing that you find as in the case of water, is when you have lots of hydrogen bonds between molecules it changes the property of, for example, water here. It makes the boiling point much, much higher than you might expect if you looked at the other properties, because you have to actually break apart all of these individual hydrogen bonds, and even though it's not that much energy for one individual hydrogen bond, once you get a huge number of hydrogen bonds, you're talking about huge energies here.

So let's look at some examples of where we see hydrogen bonds in biological processes. First thinking about proteins, they're absolutely hugely important in proteins. And in proteins, we're talking about a molecule that's so large that you don't just see h bonds in between two different molecules, what you actually see is what's called intramolecular hydrogen bonding, so hydrogen bonding within the protein molecule itself. And hydrogen bonding is incredibly important, it, in fact, is the shape of a protein, which I'm just showing an example here, histone deacetylase, and the histone deacetylase molecule actually has just countless hydrogen bonds in here, and the hydrogen bonds largely govern the shape of the molecule.

So, if you're thinking about any protein that has a shape, so for example, we see these ribbon structures here, they're called alpha helices, you can see some arrows, which are just like a sheet structure or beta sheets, that actual shape is stabilized and formed in the first place by just many, many hydrogen bonds within the protein being in a correct orientation that allows for this particular shape to be stabilized. And we know that shape is so important when we're talking about proteins, and that's because the actual shape of the protein governs the interaction with other proteins or other small molecules.

The shape of the protein is very important in terms of positioning the specific atoms that are involved in the chemistry that the enzyme carries out being in the correct position. So hydrogen bonds are very important in terms of proteins. We can also think of them in terms of sugars or polysaccharides. This is a nice dramatic example, thinking about the structure of a protein is a very microscopic way of thinking about hydrogen bonding.

If we want to take it to the macroscopic level, we can talk about h bonding in trees. So if we're talking about trees, trees are made up -- a major part of a tree is the polysaccharide cellulose. So that's shown right here. Cellulose is a chain of glucose molecules linked together. It can be as many as 1,000 or more glucose molecules linked together covalentally. But what actually happens to those individual chains is that if you look at this picture here, all those dotted lines are actually hydrogen bonds. So the individual molecules are in a much larger chain of hydrogen bonded molecules and they're incredibly rigid. You have so many hydrogen bonds that now it takes a huge amount of energy to break all of these.

So that accounts for why wood is such a hard, solid material. And also, I mean for any plant, and tree is the most dramatic, the structure of a tree, the macroscopic structure, can be explained by hydrogen bonding. It's hydrogen bonds that keep all of those molecules together in the forest and in trees.

All right. So, that's proteins, that thinking about sugars. We can also talk about the importance of hydrogen bonding in DNA. So if you think about the characteristic DNA double helix, hydrogen bonds, we have in a double helix we have two strands of DNA that you can see are intertwined here. And those two strands are actually held together by hydrogen bonding. So specifically, it's hydrogen bonding between what are called complimentary bases within the DNA.

So, for example, if we look at guanine and cytosine here, you can see that there's several places, when they're lined up like this, where we can have hydrogen bonds. So in terms of thinking about how these two molecules are lined up, how many h bonds would you expect between these two bases? Yeah, so it's lined up that there are three that can form here. So we have an o h hydrogen bond, an h n, and then an h o hydrogen bond that can form.

So the other side of complimentary based pairs are AT base pairs, and if you look at the way these are lined up here, how many hydrogen bonds can form between A and T base pairs? Yup, so it it's two that we can form here. We have one between the h o, one between the n h, and you see that we've lost this nitrogen, this n h group here, and first of all this would be too far apart anyway for hydrogen bond, but we can't have a hydrogen bond form when we have a carbon h. Remember, it has to either be a nitrogen or an oxygen or a fluorine because those are the atoms that are going to pull away enough electron density from the hydrogen to give it a partial positive charge that it needs.

So in terms of thinking about DNA, this is a really neat case to consider the actual thermodynamics or the bond enthalpies of the hydrogen bonds, because, of course, does not always stay in it's double helix -- when we're talking about transcription, we actually need to unzip or separate this helix into its two strands, so each individual strand can be copied.

So it's really important that hydrogen bonds are strong enough to hold the DNA double strand together, but that they're not so strong that when you actually pull the hydrogen bonds apart to open up the double strand, that you actually, you don't want to break all of a covalent bonds in DNA as well.

All right, so that's all we're going to say in terms of thinking about thermodynamics and biological systems. Hold on a sec, we have plenty of time left. I want to go over a few clicker questions. I didn't want to give you too long set of notes, because we're switching over to Professor Drennan's going to start lecturing on Friday, so I thought maybe I shouldn't have to have her finish up with my notes.

So, Professor Drennan will continue talking about thermodynamics and thinking about equilibrium, and then she'll transition that into talking about kinetics. So that's actually what's going to start happening after the exam on Wednesday. But let's take a look, since we do you have a little bit of time left, at a few clicker questions just to make sure everyone has caught what we've gone over so far. And a few reviews for the exam, and I will have this clicker question or one of the ones following, be a clicker question quiz, so make sure you are still answering these questions here.

So the first one covers what we went over today, so I want you to tell me and try to not look back in your notes for this, for a reaction where you have a positive enthalpy and a negative entropy change, would you expect this reaction to be never, always, or sometimes spontaneous? So let's take 10 seconds on that, that should be pretty fast.

OK, excellent. 90% is very good. So this should never be spontaneous, because we both have the entropy and the enthalpy term contributing to a positive delta g.

All right, let's shift gears to a couple of questions that are review for the exam. So hopefully, these will all be very straightforward for you now. So I want you to think about bond lengths here, and tell me which molecule contains the shorter nitrogen oxygen bond. So, we're comparing n o minus 1, and n o 2 minus 1. So if you wanted to get started on this problem, what's the first thing you should do? Yeah, draw some Lewis structures. So that might be a good place to start in thinking about this.

So I'll do that up here as well, but don't look if you want to try it on your own. Remember, the quiz points come for answering, not for getting it right. So try doing it on your own here.

All right, let's go ahead and take 10 more seconds on this one. OK, hold on. Don't show the correct answer. You know we're going to re-poll and give you a little bit more time to see if we come to a consensus here. So let's take another 30 seconds or so on this while I finish drawing these up here.

OK, good. So we've got 81% have it correct that in the n o minus 1 bond here, this is going to be a double bond.


PROFESSOR: No. It would be a double bond here. So, if you follow just the Lewis structure rules, and you go ahead, you'll find that we end up having four bonding electrons available. So you can just go ahead and plug those in, and you end up with this many left. If we had a triple bond, then we would end up having more bonding -- or we would end up using more electrons than we have available for bonding here.

All right, so we have a double bond in the case of n o. What is this bond that we have here? Yup, so it's actually a 1 . 5 bond. What is this bond that we have here? STUDENT: [INAUDIBLE]

PROFESSOR: All right, so we have two 1 . 5 bonds in this case, how come these are 1. 5 and not, for example, a double bond?

PROFESSOR: Great, so that's the key. Even though we have double bonds in each of these structures, in this case here, we have resonance, so it's turns out to be, in fact, two 1 . 5 bonds. So since this is the double bond, it's going to be the stronger bond. Since it's the stronger bond, it's also going to be the shorter bond. So make sure you can make those relationships between bond strength and bond length and thinking about if you have resonance or you don't have resonance in a particular situation.

All right, let's try one more clicker question here. We'll have this be the last one for you. So this is one that hopefully you'll all get right, because we've gone over this again and again both in class and recitation. So let's talk about the hybridization of the specific carbon and oxygen atoms in ATP, so I want you to go ahead and tell me what these hyrbridizations are for these two atoms.

OK, let's take 10 more seconds here, get those final answers in. OK, good, so 83% of you got this. Let's take a look at why. So carbon a is bonded to three things, but it is bonded to no lone pairs, so we need to have it be three hybrid orbitals, so it's s p 2. And oxygen is bonded to two atoms plus two lone paris. So we need four hybrid orbitals or s p 3.

All right, so this was your quiz question, so as long as you answered it, you got your quiz points for today. And you can get going a little bit early today and finish studying for this exam.

Energy coupling between two spontaneous reactions? - Biology

I. Life can metabolize
Living organisms carry out a complex series of chemical reactions. These reactions are collectively termed metabolism. Thus one of the basic properties of life: living organisms can metabolize. This is one of the unique properties that distinguishes living objects from inanimate ones. In this section of the course we will examine some of the basic principles governing the metabolism of cells.

II. Overview of Chemical Reactions

where A and B are the starting materials (reactants) and C is the result (product) of this reaction. The completion and direction of this reaction depends on the free energy (G) of the system. Free energy is the energy available to do work.

  1. these reactions release energy
  2. the products have less energy than reactants
  3. the energy that is released during the reaction can be lost as heat or trapped in other molecules such as ATP
  4. the free energy change that occurs during this reaction, symbolized by ΔG, is equal to the difference between the energy of the products and the energy state of the reactants. In an exergonic reaction, Δ G is negative
  5. exergonic reactions are typically breakdown (hydrolytic, catabolic) reactions, such as the hydrolysis of starch or cellular respiration and
  6. exergonic reactions are spontaneous that is, they don't need outside help (an external energy source) to occur.
  1. these reactions require energy
  2. the products have a greater energy content than the reactants
  3. there is a positive ΔG
  4. endergonic reactions are synthetic (buildup, anabolic) like photosynthesis and
  5. are not spontaneous (i.e., require an external energy source to occur).

Note that since the overall ΔG is negative, this reaction sequence is now spontaneous and will occur.

The actual "coupling" process is usually the result of ATP binding to one of the starting materials. This essentially "energizes the reaction" making it thermodynamically favorable.

C. Activation energy
Even if a spontaneous reaction or reaction sequence has a negative ΔG, it may only occur very slowly at the temperatures of life. For example: the conversion of bread (starch) to glucose takes about 1000 years if it sits in water. However, if you heat up the mix, it takes about 1 hour. But, eat some bread and our body converts it to glucose in minutes.

Similarly, rocks don't just roll down hill. Most just sit at the top waiting patiently, until we give them a little push. This push is analogous to what the biochemists call activation energy (symbolized Ea), which is the minimum energy needed to start a chemical reaction. Thus, reactions require a "push", or more appropriately, the necessary activation energy to proceed.

The Hill Model once again.

III. Molecules and energy
To better understand this process, let's study the energy distribution of a population of molecules by plotting the following graph:
# of molecules vs. energy/molecule

Our graph is a bell-shaped curve. We observe that some molecules have a lot of energy, some little, but most are happily average. For most reactions at life temperatures, only a fraction of the molecules have the necessary Ea - so the reaction proceeds slowly.

IV. What can be done to speed things up?

  1. Increase the concentration of the reactants - Not too feasible, especially in organisms where many cell components are present only at very low concentration.
  2. Heat - Works great for chemists, but not biologists. Why? 'cuz heat increases the energy of molecules, which can lead to thermal instability. Just ask a fried egg. Note we're not changing Ea, just the energy possessed by individual molecules.
  3. Lower the Activation Energy Requirement by adding a catalyst - This is molecular version of the "if you can't beat ‘em, join 'em" philosophy. A catalyst is a substance that speeds up the rate of a chemical reaction by lowering the activation energy requirement for a reaction. The catalyst is not "used up" during the process, it is recovered unchanged. Once the activation energy requirement has been lowered, most of the molecules will have the necessary energy to react. This is the approach taken by life.

V. Enzymes - the catalysts of life

  • are catalysts (speed up a reaction, but are not consumed in the process). Thus, enzymes are not destroyed in the reactions they are involved and can be used over and over.
  • are proteins, which in turn, are made of chains of amino acids linked together by peptide bonds. Remember that proteins have many different functions in organisms - enzymes, the catalytic proteins, are just one type. However, they are arguably the most important group of proteins, because without enzymes, no metabolism could occur
  • are mostly globular proteins (tertiary structure) and some have quaternary structure.
  • lower the activation energy
  • requirement for a reaction
  • end in "ase": ex., tyrosinASE
  • named by a standardized system of nomenclature. The official name is based upon: (a) the normal substrate and (b) the type of reaction. This system was developed by the International Union of Biochemists in 1956. Enzymes may also have a trivial or common name (such as catalase).
  • are usually specific, in vivo, for one reaction
  • have a unique 3-D shape that is determined by the kind and sequence of amino acids in the protein
  • were first studied by Jon Jakob Berzelius (1835). The Swede Arrehenius developed the concept of the enzyme-substrate complex (1889). The lock and key model was described in 1894.

B. Mechanism of Action.
Enzymes interact with its reactant(s), called a substrate, to form an enzyme-substrate complex. The substrate fits into an active site (catalytic region of the enzyme). The product is released and the enzyme is recovered unchanged. A brief equation: E + S → E-S complex → E + P. This reaction is reversible, depending upon the chemical equilibrium that is established.

Here's some examples: catalase is an enzyme that hydrolyzes hydrogen peroxide (substrate) to oxygen and water (products). Click here for Spuds McSaupe. In a lab during a past semester we studied the enzyme tyrosinase. It is found in many organisms and its main job is to oxidize (breakdown) tyrosine (one of the amino acids) and other phenolic molecules (with a benzene ring and attached hydroxyl group). Unlike many enzymes that are substrate-specific, tyrosinase isn't and works on a variety of phenolics. We took advantage of that fact and used a different substrate, L-DOPA (dihydroxyphenylalanine). Tyrosinase converts L-DOPA into the product, L-DOPA-quinone. Thus the reaction can be diagrammed:

L-DOPA + tyrosinase ↔ L-DOPA-tyrosinase complex ↔ L-DOPA quinone + tyrosinase

C. Kinetics of a enzymatic reaction.
Consider a typical enzyme-catalyzed reaction such as mixing catalase and hydrogen peroxide. Assuming a constant amount of enzyme, how will the concentration of the product, substrate and E-S complex change over time? At time "0", the amount of substrate is high, the product is low (zero), and the ES complex is zero. How do you predict the concentration of the product (oxygen and water), substrate (hydrogen peroxide) and ES complex to change as the reaction proceeds?

graph: concentration vs. time

D. Enzymes and Substrates.
An analogy (model) for the interaction of substrate and enzyme at the active site is the Lock-and-Key Model. The active site of the enzyme had a specific shape, like a lock, that is only recognized or opened by only a single key, like a substrate molecule. This explains why most enzymes are substrate-specific, that is, they act on only a single substrate (or sometimes a group of related substrates like tyrosinase). Now also know that the active site is not rigid like a metal lock, but that the enzyme shape is more flexible, like, say, a bean bag chair. The chair has a particular shape but it can change slightly. When you sit in it, it conforms to your body. Another good analogy is a handshake. This is called the induced fit model, the binding of the enzyme causes the active site to conform to the substrate.

E. So how do enzymes lower activation energy requirements?

  1. Enzymes increase the concentration of substrates in vicinity of the enzyme
  2. Enzymes orient substrates to the most favorable position for reaction (remember for any molecules to react they must first come in contact with one another
  3. Enzymes change conformation that may strain bonds of the substrate
  4. Enzymes provide a micro-environment that may be more favorable chemically (i.e., hydrophilic/hydrophobic or acid/etc.) for reaction.

VI. A Model
Let's use a model to help explain enzyme activity. Consider the game of musical chairs. The chair symbolize the enzymes and the seat represents the active site. The players are the substrates who will be converted into product (which happens when you sit in the seat). We will blindfold the players and allow them to wander randomly around the room (cell). [Note - the chair should also be moving, perhaps on rollers] When a player and chair collide, the player sits and is converted to product and then stands back up.

Let's play! We'll start with 5 chairs (enzymes) and 100 players (substrates) in a large room. Every minute we will count the number of people who have contacted and sat in a chair (turned into product). Then, we will plot [product] vs. time (min).

graph: product formed vs. time

The slope of this graph will give us the rate or velocity of the reaction product formed per minute. To give you an example of the speed of some enzyme catalyzed reactions check out the following table:

Speed of some select enzymes
Enzyme Reaction Rate (# molecules product formed per enzyme per minute)
carbonic anhydrase 36,000,000
catalase 5,600,000
β-amylase 1,100,000
succinate dehydrogenase 1150

Thus, it’s obvious that enzymes can really speed up the rate of a reaction!

VII. Factors that influence the rate of enzymatic reactions

A. Enzyme concentration.
For a constant amount of substrate - the more enzyme, the more active sites, the more product formed per unit time (the greater the reaction rate). Chair model: to simulate the effect of enzyme concentration on reaction rate, let's start with 2 chairs (enzymes) and 100 players (substrates) in a large room. Then every minute for an hour we'll count the number of people who found and sat in a chair. We can then plot the [product] vs. time Repeat with 5, 10, and 100 chairs (enzymes).

graph: product formed vs. time (for several enzyme concentrations)

We can now calculate the slope of each line to give us the reaction rate ([product formed]/minute) for each enzyme concentration. Now, we can plot reaction rate vs. enzyme concentration.

graph: reaction rate vs. enzyme concentration

Conclusion: The more chairs, the less time it will take any one person to encounter one. The more enzyme, the more active sites, the greater the reaction rate.

B. Substrate Concentration.
For a constant enzyme concentration - as the substrate concentration increases, the rate of reaction increases until a saturation level of substrate is reached at which point no matter how much more is added, there will be no additional increase in rate. Chair model: start with 10 chairs and 10 players. Every minute for one hour record the number of players (substrates) who find and sit in a chair (enzyme). Plot the product formed (number of people sitting vs. time). Repeat with 20, 50 and 100 players.
graph: product formed vs. time (for several substrate concentrations

Now, calculate the slope of each line to yield a reaction rate and then plot reaction rate vs. substrate concentration.

graph: reaction rate vs. substrate concentration

Conclusions: As the players increase, the likelihood of rapidly finding a chair increases. So the rate goes up. At low substrate concentrations, the rate-limiting step is finding a chair. At higher concentrations, the rate limiting step is the speed of conversion of enzyme into product.

(The next three paragraphs are not on the exam) Two measures of the effectiveness of an enzyme-catalyzed reaction are Vmax and Km. Vmax = maximum rate of reaction. Km = substrate concentration that yields half the maximum reaction rate. These can be found graphically. Km for many enzymes range from 1 mM to 0.1 uM.

Based on the relationship between reaction rate and substrate concentration in the graph above, Michaelis and Menton derived the following equation:

The beauty of this equation is that if we know the substrate concentration for a given enzyme we can calculate its predicted velocity. In practice it can be difficult to determine Km and Vmax from the graph. However, using the Michaelis-Menten equation, Lineweaver and Burk realized that if a graph of 1/V vs 1/[S] is plotted, the X intercept is the negative inverse of Km and the Y intercept yields the inverse of velocity. One advantage of making a "double reciprocal" plot is that the result is a straight line and that Km and Vmax are easily to determine.

C. Temperature
As temperature increases, so does the reaction rate because you heat the molecules and increase the probability of a collision. Each enzyme has a unique temperature optimum. For example, thermophilic bacteria can tolerate extremely hot temperatures in contrast to other organisms. The rate declines rapidly after the optimum because the enzyme denatures - looses it normal (native) shape, which in turn changes the shape of the active site making it less favorable for the reaction. Chair model: Imagine 10 chairs and 100 players. Play the game at room temperature. Repeat a 0 C and 30 and 100 C. Imagine everything moving slow at the cold temperatures. And, it's none too pleasant at 100. Hmmmm…the model doesn't work too well for this one!

D. pH.
Each enzyme has a unique temperature optimum. A pH greater or less than the optimum result in reduced rates of reaction. pH causes changes in the structure of the enzyme (affects pH sensitive bonds that hold the enzyme in its 3D shape), and hence, changes the shape of the active site. Thus, the reaction becomes more or less favorable for the reaction to occur.

E. Enzyme Inhibitors.
There are two groups of molecules that inhibit chemical reactions: competitive and non-competitive.

1. Competitive inhibitors.
These compete with the substrate for the active site. These inhibitors typically have a chemical structure similar to the normal substrate. Some examples: (a) succinic dehydrogenase converts succinic acid to fumaric acid. Malonic acid is a competitive inhibitor (b) ribulose bisphosphate carboxylase binds carbon dioxide to RuBP. Oxygen is a competitive inhibitor.

2. Non-competitive Inhibitors.
These molecules bind to a site on the enzyme other than the active site. Binding causes changes in the shape (conformation) of the enzyme, which in turn, changes the shape of the active site, and hence activity. Examples: (a) Hg and disulfide bonds (Mad Hatters and mercury)

F. Enzymatic control of metabolism

1. Allosteric effectors.
These bind to a site (allosteric site) other than active site and cause a change (slight) in the active site. These changes will enhance or inhibit the reaction. These are used to regulate rate of metabolic processes.

2. Availability of cofactors/coenzymes.
These are inorganic and organic substances, respectively, that are required by enzyme, in addition to the substrate, for activity. Vitamins are important coenzymes. Many metal ions are enzyme cofactors. Cofactors and coenzymes may be bound loosely or tightly to the enzyme.

3. Presence of activation factors.
Some enzymes require activation. For example, pepsinogen must be converted to pepsin. Many enzymes in photosynthesis are light activated.

4. Cooperativity
The binding of one substrate to the enzyme enhances the ability of the enzyme to bind to additional substrates.

5. Feedback inhibition - an end product inhibits an enzyme in its pathway

6. Multi-enzyme complexes and membrane compartmentalization.

VIII. The importance of enzymes (not tested directly on exam)

  • Serum glutamic oxaloacetic transaminase (SGOT) is found in tissues of high metabolic activity (heart, liver, pancreas, muscle) and is released following injury. Thus it is a good indicator for myocardial infarction (MI, heart attack) and liver disease (cirrhosis, hepatitis)
  • Serum glutamic pyruvic transaminase (SGPT) is found in liver, heart, muscle, kidney and is especially diagnostic for liver disease
  • Lactic dehydrogenase (LDH), is found in a wide variety of tissues, and indicates cell death and leakage. It is used to confirm a diagnosis of MI. There are five 5 isozymes: 1,2 = MI, 3=pulmonary, 5=liver
  • Creatine phosphokinase (CPK) is found in heart and skeletal muscle and is released upon injury to the myocardium and skeletal muscle
  • Acid phosphatase - which has a wide distribution, esp. liver, spleen, kidney, red blood cells, and prostate and
  • Alkaline phosphatase - is found in bone, liver, and placenta functions best pH 9, and indicates liver/bone disease.

B. Several human diseases are caused by a deficiency in a key enzyme in a biochemical pathway.
In other words, these conditions are the result of abnormal metabolic activity. Examples include phenylketonuria and albinism.

C. A Case Study.
Several years ago I had a complete blood work-up that included tests for SGOT and CPK. My CPK level was astronomically high, but the SGOT level was normal. My health was normal. How do you explain these results?

D. Some Take-Home-Lessons:
(1) Certain enzymes can serve as clinical markers for some human disorders (2) Some enzymes have a restricted distribution in an individual (i.e., found in a specific tissue, organ, cell and even organelle) (3) Enzyme deficiencies are inherited disorders (remember last semester) (4) A final, most excellent, conclusion - Properly functioning enzymes are crucial for survival.

VIII. Commercially important enzymes include (not tested directly on exam)

  • amylase - brewing - convert starch to fermentable sugar
  • invertase - artificial honey - convert sucrose to glucose & fructose
  • lactase - ice cream - prevent crystallization of lactose
  • naringinase - citrus - remove bitter tasting naringin
  • pectic enzymes - coffee - coat of bean
  • pectic enzymes - juice - improve juice yield
  • pectic enzymes - wine - clarification
  • proteases - brewing - aid filtration & clarification
  • rennet - cheese - coagulate casein
  • catalase - milk - destruction of H2O2
  • proteases - detergents - laundry aid
  • photography - recovery silver from spent film

IX. Some fun enzymes (not tested directly on exam)

A. Rennin
Breaks certain bonds in casein, the major protein in milk converting it to paracasein. Paracasein is insoluble and precipitates out of solution, clotting or curdling the milk. In vivo, the function of this enzyme is to slow the passage of milk proteins through the digestive tract to improve digestion. Rennin used in cheese making was traditionally obtained from the fourth stomach of a calf or other bovine (hoofed) animal, but now it is obtained primarily from microbial fermentations (trade name of rennilase).

B. Amylase.
Catalyzes the hydrolysis of starch to maltose. Starch is a polymer of glucose maltose is a disaccharide of glucose. Amylase occurs in the saliva where it serves to initiate the breakdown of dietary starch. It is also abundant in seeds where it is used for breaking down starch to be used for germination.

C. Bromelin.
Protein digesting enzyme in pineapple (papain from papaya). This is the reason why you can’t make jello from fresh pineapple.