Thermodynamics typical questions

Thermodynamics Glossary of biochemical terms

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Glossary of biochemical terms

Biochemistry is a science of living cells. A living cell synthesizes complex molecules. These molecules may be shuttled out of the cell or from one compartment to another. To accomplish these activities, energy is required. Bioenergetics is a quantitative analysis of how cells gain and use energy. The objective of this chapter is to introduce you to the concepts of thermodynamics as applied to biochemistry.

1. First Law of Thermodynamics: The law states that total energy of a system and its surroundings remains constant. Energy is neither created nor destroyed but may be used to do useful work on the system or surroudings. For example, energy produced by the digestion and breakdown of proteins may be used by the organism to synthesize other molecules, be used to keep the temperature of the organism constant etc. This law is represented as:

E= q- w ----- (1)

where delta E is the change in internal energy of the system.

q is positive number when heat is absorbed by the system from its surroundings and a negative number when heat iis absorbed by the surroundings from the system.

w is positive when work done by the system and negative when the work is done by the surroundings. Since the concept of w generally applies to gaseous systems where partial pressure or volume of the system may vary, it is generally ignored for reaction in solution e.g cellular reactions. Under these conditions, delta E is essentially change in heat content of the system. Therfore E = H (the change in enthalpy of the system).

2. Second Law of Thermodynamics: Direction of Reactions:

The first law is essentially keeps track of energy produced and spent and these must balance. The second law helps us understand why certain processes can proceed in a reversible manner whereas others essentilly are unidirectional. For example, it is poosible to convert water to ice and vice versa but if we burn a piece of cloth to carbon dioxide and water, it is not reversible. This concept can be understood if we realize that not all the energy or heat content of the system (delta H) in solution may not be availavle for useful work i.e. some of the energy may be lost in non-productive manner. Thus delta H may be defined as:

H = G + T S------ (2)

where G is the free energy change in the system and is the energy available to do useful work (convert a substrate to a product) and delta S is change in the entropy of the system (entropy is defined as degree of randomness of the components of the system). These are conditions of constant pressure, volume, and absolute temperature in degrees Kelvin (T= 273 + degree centigrade). This equation is generally written as:

G = H - T S ------- (3)

It is clear and logical to think that if the free energy of the product is less than that of the reactants, then the reaction will be favored in the forward direction. However, if G is positive then the backward reaction will be favored. Negative values of G can be achieved if dH is negative and dS is positive. Such reactions are considered as Enthalpy Driven and/or Entropy Driven. However, depending upon the magnitude, it is possible to have negative value for G even if H is positive or S is negative. An example is melting of ice to water. The H for this system is positive whereas dS is highly positive which compensates for the positive dH. On the other hand when water freezes to ice, S is negative (less randomness in the structure of water) but positive TS value is compensated for by negative enthalpy change. It must be remembered that favorable reactions must have a negative value for dG. Such reactions are called Exergonic. Reactions with positive G are not favored in the forward direction and are called Endergonic.

Since reactions are driven by overall free energy changes, let us see if thermodynamics can explain the cellular processes. Considering a cellular process: S ----> P, the free energy of substrate and product can be described as:

Gs = G0s+ RT ln {S} and

Gp = G0p + RT ln {P}

G0p and G0s are standard free energy changes and represent the free energy of product and substrate when they are present at 1M concentrations. For biochemical reactions, standard free energy changes are normalized to pH 7 because many of the biochemical reactions involve {H+} as one of the components of the reaction and it is simpler not to have to include {H+} as one of the components of the reaction. Thus G0 are represented as G0'. Therefore, change in free energy of a reaction can be represented by:

G= (Gp-Gs)= (Gp0'-Gs0' ) + RT( ln {P}- ln {S})

G= G0' + RT ln {P}/{S} (1)

When the reaction reaches equilibrium, there is no net change in the concentrations of substrate and product; therefore no useful work is done by the system and G =0. Under these conditions:

0=G0' + RT ln {P}e/{S}e or G0' = - RT ln {P}e/{S}e = -RT ln Keq or G0' = -RT ln Keq

Thus knowing standard free energy change for a reaction, one can calculate equilibrium constant. For these calculations, R is 1.98 cal/mole/degree Kelvin and T is (273 + Degree centigrade) (in recent literature, the heat energy is represented by Joules or KiloJoules; it is easy to convert calories or kilocalories to Joules: 1 cal = 4.184 joules). Based on these calculations, one can predict the magnitude of Keq. Thus

Go' = 0 Keq = 1

Go' is negative Keq > 1

Go' is positive Keq < 1

Keq values also indicate whether the reaction is favored in a particular direction. Keq =1 means that the reaction is equally favored in both directions. Keq greater than one means that the reaction is favored in the forward direction whereas Keq of less than one means that the reaction is favored in the backward direction.

Usefulness of G0'

(1) Used for calculating Keq for a variety of reactions as dGo' values can be calculated.

(2) Can predict the direction of a reaction when reactant and product concentrations are equal or are 1 M each.

(3) Go' values for two reactions having a common intermediate are additive.

Let us see if thermodynamics can explain how some physiological reactions in the cell. Consider the physiological reaction:

Glucose + ATP ------- > Glucose--6--Phosphate + ADP G0' = -- 4 K cal/mole

This is a combination of two reactions;

Glucose + phosphate ------- > Glucose--6--Phosphate G0' = +3.3 Kcal/mole (a)

ATP -------- > ADP + Phosphate G0' = --7.3 Kcal/mole (b)

Reaction a would not occur without the participation of second reaction. Since reactions a and b have one of reactants and products in common, the standard free energies can be added, giving the overall energetics of the overall reaction. In addition to the negative standard free energy change, the reaction is further favored by continuous siphoning of the product, glucose-6-phosphate, in subsequent reactions of the glycolytic cycle. This gives a G value which is more negative than -4 K cal/mole. Another clear example of usefulness of additive aspects of G0' is the reaction:

D-Fructose-1,6,-bisphosphate --------- > Dihydroxyacetone-phosphate+ D-Glyceraldehyde--3--phosphate G0' = +5.77 Kcal/mole

The only reason this reaction proceeds in glycolysis is because under physiological conditions in the cell, the concentrations of products are much smallet than the reactant and when these concentrations are substituted in equation 1, the value of G comes out to be --0.32 Kcal/mole.

Some typical questions that may be asked

1. Which reaction as written below is more endergonic?

A. AMP + ppi --> ATP

B. AMP + 2pi ---> ATP

answer B

2. The free energy change for a reaction at equilibrium is

A. less than 1.

B. zero.

C. greater than 1.

D. equal to -RT In Keg.

E. can be utilized to do work in the cell.

answer B

3. Which of the intermediates listed below has a standard free energy

of hydrolysis more negative than that of ATP + water --> ADP + Pi?

A. D-glucose I-phosphate

B. D-glucose 6-phosphate

C. 3-phospho-D-glycerate

D. 2-phospho-D-glycerate

E. 2-phosphoenolpyruvate

answer E

4. The phosphorylation of glucose by ATP (ATP + glucose --> ADP + glucose

6-phosphate) has a standard free energy change of -4.5 Kcal

per mole. The equilibrium constant for the reaction

A. cannot be determined from the data given.

B. is less than one.

C. is greater than one.

D. changes with the concentration of enzyme.

E. depends on the concentration of ATP.

answer C

BioWeb

Frequently Asked Questions (FAQs)

Enzyme Problems Set1.

What is the direction of the following reactions when the reactants are present at equimolar amounts?

a. ATP + Creatine ----------------> Creatine-P + ADP

b. ATP + Glycerol ----------------> Glycerol-3-P + ADP

c. ATP + Pyruvate -----------------> Phosphoenolpyruvate + ADP

The standard free energies of hydrolysis of ATP, creatine-P, glycerol-3-P and phosphoenolpyruvate are -7.3, -10.3, -2.2 and -14.8 Kcal/mole respectively.

To solve such problems, remember that the direction of reaction can only be predicted by the sign of free energy change i.e deltaG. However, since the concentrations of reactants and products are equal, deltaG is equal to deltaG0' applying the equation:

deltaG = delta G0' + RT log {P1}{P2}/{S1}{S2}

The ratio {P1}{P2}/{S1}{S2} is equal to 1 and log1=0. Therefore, if deltaG0' comes out to be negative, the reaction will be exergonic and proceed to the right. If deltaG0' is positive then the reaction as written is endergonic and proceeds in the opposite direction.

2. Consider the reaction: ATP + Pyruvate -------------> ADP + Phosphoenolpyruvate

This can be written as two half-reactions:

ATP --------------> ADP + P deltaG0' = -7.3 Kcal/mole

Pyruvate + P ------> Phosphoenolpyruvate deltaG0' = + 14.8 Kcal/mole

These two reactions can be added, because of a common intermediate, P. So final reaction has deltaG0' of

7.5 Kcal/mole. Therefore,

deltaG0' = -RTlog Keq

7.5x1000 cal/mole = -2cal/mole/degK x 298deg K log Keq

Calculate Keq as 3.16x10-6

Keq = {PEP}{ADP}/{ATP}{Pyr}

3.16x10-6 = 1/10 x {PEP}/{Pyr}

If you do the math now, the ratio

{pyr}/{PEP} will be 3.16x104.

3. The Michaelis-Menten equation can be arranged in many ways. One way to do it is:

v = VmaxS/Km + S

v(Km + S ) = Vmax S

vKm + vS = Vmax S

vS = Vmax S - vKm........divide this equation by S

v = Vmax - vKm/S

A plot of v vs v/S should be linear with slope equal to -Km and intercept on Y-axis equal to Vmax.

Glucose Homeostsis Problem

You can do this problem two ways:

1. Use equation for two substrates: v= Vmax/(1+ Km1/S1 + Km2/S2)

For brain metabolism KmATP/ATP = .1/1.0 and KmGlu/Glucose = .01/5

v = Vmax/ (1+.01 + .002) or v ~Vmax; you can do the same for liver

2. You could also use simpler MM equation. In this case [ATP]/Km ATP= 1/.1 = 10 so the enzyme is saturated with ATP. Therefore, the reaction rate will only depend on glucose concentration. Use equation: v = Vmax S/ Km + S. Substitute values for Km for enzymes in the liver and brain and determine the relationship between v and Vmax.

Glossary of biochemical terms

Frequently asked questions

This site will display FAQs from NJMS, NJDS, and GSBS students of UMDNJ as well as questions from students in other institutions. Therefore, it is likely that these questions may not be in a sequential orders of lectures.

FAQs from uiuc

You said that Eukaryotes are monocistronic because only one protein product can be made per mRNA template. Does this mean one kind of protein and that multiple copies of this type of protein can be made or do you mean one protein period?

I mean that only one particular protein but many copies of this protein can be (and normally are) madeI

Can proteins be renatured like DNA?

Yes, proteins can be renatured like DNA if they are returned to favorable conditions.

Can you please tell me how an enzyme localizes substrates to an active site?

When an enzyme binds to a substrate it does so because the enzyme has an active site that allows the substrate to fit - once in the region of the active site more specific interactions may occur that allow the enzyme to hold onto the substrate long enough for the reaction to proceed. These interactions may involve a conformational change that "engulfs" the substrate - like your baseball glove closing in on the ball as you make the catch. Covalent interactions or weak interactions may also aide in the retention of the substrate. Unfortunately, most enzymes do not have the luxury of a "traffic cop" that actually participate in guiding the substrate into the active site. Much of the enzyme-substrate interaction is dependent upon the concentration of both and anything else in solution that aides in crowding them together.

How is it that an enzyme orients a substrate such that it is able to fit in the active site? Is it a matter of probability that the substrate will approach the enzyme in the correct orientation or are there some type of molecular/atomic interactions which sort of coerce the substrate into the correct position?

Good question. I don't know that this concept is fully understood as yet. I believe that there are some enzymes that undergo a conformational change to aide in substrate binding and to some extent charge is important - but "probability" has a role as well and there is a higher probability of a productive

interaction occurring with a higher substrate concentration.

What exactly are domains? Are they the same thing as active sites?

Domains are localized folding patterns within protein molecules that may have specialized functions. Enzymatic active sites are one example of such a specialized function, but not the only one. There are other examples such as fitting proteins into membranes or forming complexes with other molecules.

I was curious as to how enzymes work by means of concentration. That is, how do enzymes increase the effective concentration of substrates in a reaction?

Enzymes increase the effective "local" concentration of the substrate molecules simply by virtue of the fact that they hold the substrate molecules together in the same place which is the active site. Therefore, the reaction doesn't have to rely on a random collision between substrate molecules in solution, as the enzyme essentially "pulls" the substrate molecules out of solution and holds them in one place and in the proper orientation for the reaction to occur between them.

An enzyme helps catalyze a reaction that can take place only may be too slow. But I am having difficulty defining if these rxns are spontaneous or not and if an endothermic reaction could ever be spontaneous. Also, the book said that the active site of an enzyme can change, but your notes, I think, said this never happens - only the substrate changes.

Yes, I am afraid that the terminology is confusing. A "spontaneous" reaction is any reaction in which the products have less energy than the reactants (substrates). Therefore, all exergonic reactions are spontaneous because they all release energy and, hence, the products end up with less energy than the reactants. However, the fact that these reactions are "spontaneous" does not mean that they occur spontaneously (i.e. without an input of energy) - dumb huh! The reactants of a spontaneous reaction exist in energy troughs and require activation energy to push them out of these troughs so that the reaction can occur. Therefore, a "spontaneous reaction" is simply a reaction that is favorable because it releases energy - but it needs an initial input of energy to get it to go. The reason for this though is obvious. If the exergonic reactions in our cells could occur spontaneously (i.e. without an input of energy), they would occur all of the time in an unregulated fashion. All of the energy released would be lost as heat and we would go up in a puff of smoke. The need to provide activation energy prevents this from happening. When the cell receives a signal to "turn on" a particular reaction, the enzyme that catalyzes that reaction is produced and it is this enzyme that reduces the activation energy barrier so that sufficient energy can be obtained from within the cell for the reactants to reach the point where they can react.

Endergonic reactions are not spontaneous because they do not release energy - instead they require energy because the products have more energy than the reactants. Therefore, the energy needed to drive an endergonic reaction is a lot greater than the small amount of energy needed to overcome the activation energy barrier in exergonic reactions. The energy needed to drive an endergonic reaction is the enrgy needed to "push" the reactants almost all of the way to the products as the reaction is unfavorable and requires considerable energy input to occur.

One theory for a way in which enzymes catalyze reactions is the so-called "induced fit" hypothesis. This hypothesis proposes that the fit between the substrate and the enzyme is not precise. Therefore, when the substrate fits into the active site, it becomes strained which puts pressure on its bonds making them easier to break and, hence, "helping" the reaction to occur [remember that for most reactions to occur bonds have to be broken]. There is no evidence for this model but it seems likely that enzymes might work in this way.

What I said in lecture is that the enzyme remains unchanged by the reaction that it catalyzes. Enzymes are used over again and again within the cell so they have to remain unchanged. In other words, any change that occurs at the active site of an enzyme to promote a chemical reaction is only temporary - when the products of the reaction are released, the enzyme assumes its original conformation.

In non-competitive inhibition can the substrate still bind?

In competitive inhibition, the substrate and inhibitor compete for the same site on the enzyme. In non-competitive inhibition the inhibitor binds at a different site - but it may change the conformation of the active site and the substrate either cannot bind at all or cannot bind in a productive interaction, where product will be formed.

What is the difference between an allosteric regulator and a non-competitive inhibitor?

Both allosteric regulators and non-competitive inhibitors act in the same way in that they both effect enzyme activity by binding to the enzyme at a site which is NOT the active site. As a result of the binding of the allosteric regulator or non-competitive inhibitor, the enzyme undergoes a conformational

change which effects the binding of the substrate to the active site. However, there are two important differences between an allosteric regulator and a non-competitive inhibitor:

1. Allosteric regulators are always reversible. They regulate important pathways in cells so they have to be reversible. Non-competitive inhibitors, on the other hand, are usually irreversible.

2. Allosteric regulators can be positive as well as negative regulators. In other words, binding of some allosteric regulators can actually serve to INCREASE the rate of a particular reaction. Non-competitive inhibitors, on the other hand, always act to DECREASE the rate of a reaction - they cannot increase enzyme activity.

Why does feedback inhibition occur at the first committed step in a pathway?

Many pathways have a common beginning and then branch off to form specific products, this is advantageous because then less anzymes are required to get the job done. In order to inhibit one pathway specifically, you need to inhibit the "first committed step" - this means that this is the first step in the pathway that goes only to those specific products - including the one that is capable of feedback inhibiting the pathway. Many of these pathways are necessary - turning down most of them is not going to support life. The key is that metabolism involves an interwoven series of pathways and those that

are not necessary at the moment can be shut down to conserve energy - but the rest must remain functional.

Could you please explain what a truncated protein is?

A truncated protein is a protein whose synthesis has been terminated prematurely by the formation of a stop codon in the middle of the protein coding sequence. Many mutations can give rise to stop codons. Remember, the stop codons are specified by the codons UAA, UAG and UGA and there are many codons that normally encode amino acids that can give rise to one of these stop codons if they undergo a single base change.