Begin 01/27/09 #4
Closed last lecture with a working definition of enthalpy. A second thermodynamic property, entropy, also changes during chemical reactions or changes of state. The change in entropy is denoted by ΔS, and a working definition can be taken as the degree of disorder. Disorder always increases for real-world spontaneous processes if order is the only thermodynamic property of the system that changes; so the condition for ΔS that would be required for a spontaneous reaction is: ΔS > 0, if entropy is the only thermodynamic property that changes.
Thus, changes in both ΔH and ΔS impose conditions on whether a reaction is spontaneous and these quantities have been combined to define a function called Gibbs free energy, given by the symbol ΔG. The change ΔG is a sufficient criterion for identifying spontaneous processes. The relation between ΔG, ΔH and ΔS is given by the equation:
ΔG = ΔH - TΔS
where T is the temperature in oK. Since ΔH is usually negative and ΔS is usually positive for spontaneous reactions, we conclude that overall ΔG will be negative for spontaneous reactions. Hence the criteria for assessing direction of our model reaction
are:
ΔG < 0, the forward direction is spontaneous
ΔG > 0, the forward direction is not spontaneous- work must be put into the system to drive it in the forward direction
ΔG = 0, the system is in equilibrium; i.e., there will be no net transfer from the left hand to right hand side or in the opposite direction.
Having defined ∆G, you can see why either H or S can drive a reaction, since it is only the net change in free energy that matters. Thus, for reactions which are dominated by the change in H, ∆G will still be negative if the absolute value of H > than the absolute value of TS, and the reaction will be spontaneous even if S is negative. S will be negative when products result in restricting degrees of freedom present in the starting compounds: an example, is the reaction we are describing, in which two molecules couple to form a single molecule. While the reaction will have negative entropy, if negative enthalpy change is large enough, the enthalpy change will still drive the reaction forward. Another situation in which S is negative is the formation of a cyclic compound from a linear compound.
There is a very simple relation between ΔG and the ratio of the concentrations of products to the concentration of reactants when a reaction is reversible. That is relationship is defined by the equilibrium constant, K, which is the ratio of the (multiplicative) product of the concentrations of the compounds on the right hand side of the equation, (by convention) the reaction products, to the product of the concentrations of the reactants, which by convention are the compounds on the left hand side. For the current example,
K = (A-B)(H2O)/(A)(B), where (A-B), (A), (B) and (H2O) are usually given in units of moles/L. K is always unitless, and that is one way in which you can check to see that you have set up the equilibrium equation correctly. Since this reaction takes place in solution, the water concentration is effectively constant in the aqueous medium, [H2O] can be assumed to remain unchanged at 55.6 M.
The free energy change ΔG for reversible reactions is related to K by the expression
ΔG = ΔGo + RT ln K
where T is in oK and R is the gas constant (R = 1.98 cal/mole-oK = 0.00198 kcal/mole-oK)
and ΔGo is defined as the net free energy for the formation of products: which is given by the expression
ΣGoproducts - ΣGoreactants = ΔGo
where ΣGoproducts is the free energy of formation of products from elements and ΣGoreactants is the free energy of formation of the reactants from elements at standard pressure and temperature. Pstd always = 1 atm, Tstd = 298 K (25 oC) for biochemists or 273 K (0 oC) for physical chemists. If you are doing serious calculations, you have to be aware of the different conventions for biochemistry and physical chemistry.
If the system is at equilibrium, ΔG = 0, and the expression: ΔG = ΔGo + RT ln K reduces to
0 = ΔGo + RT ln K, or after rearranging,ΔGo = -RT ln K
One way of looking at this expression is that ΔGo represents a difference in stability between reactants and products and from the relation with K you can see that the change in G will determine the end point of the reaction. Hence, in a dynamic situation, where the system is allowed to come to equilibrium, ΔGo determines the ratio of the products to reactants, that is, the extent to which the reaction will take place from left to right, if you start with a set of arbitrary concentrations, which was the objective of this exercise. Commonly, the superscript “o” of ΔGo is dropped, and the relationship is simply written:
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ΔG = -RT ln K
On the next overhead, I have calculated some thermodynamic quantities for the model reaction, which is the reversible formation/hydrolysis of a dinucleotide. For the sake of time, I will discuss only the results and leave you can go to the web site to work through the calculations.If you do have questions, we can fix that on Thursday. The first calculation is determining the equilibrium constant K, given an experimental value for∆G and the temperature. Since ∆G = +6 kcal/mole (> 0), you see immediately that the forward direction is not spontaneous,which means that the phosphodiester bond is unstable with respect to hydrolysis. That implies that DNA is not stable with respect to hydrolysis. We will take the standard temperature as 25 oC = 298 oK. Then K, equilibrium constant calculated for the forward reaction is ~3.8 x 10-5. In the next calculation on the overhead, the value calculated for K is used to determine the extent to which the dinucleotide will hydrolyze. We will start with a concentration of 1 x 10-3 M. An important point to make about this calculation is the approximation used to express the final concentration of the monomeric units A and B. We know from the equilibrium constant of ~ 10-5, that the equilibrium concentration of dinucleotide at equilibrium, which I have called x, will be very small and virtually all of the dinucleotide will hydrolyze. Hence the final concentrations of the reactants (l.h.s) can be approximated by 1 x 10-3 M, instead of using the exact expression (1 x 10-3 – x) M. This avoids the need to solve a quadratic equation to determine x. As you can see, there will be virtually no dinucleotide left.
The reason why this situation is not immediately threatening to life as we know it is that, although the hydrolysis is energetically favorable, it is extremely slow - and that introduces another important consideration in chemical interactions, which is kinetics, which deals with reaction rates.
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But before moving on to kinetics, though, we will go back to coupling problem and pose the question of how the forward reaction - the coupling of nucleotides to make a polymer - can be pushed by the input of energy. Remember, this is equivalent to the extension of DNA in the process of replication, so it is clear that nature developed a way to do this. In the case of DNA replication, this is accomplished by a cascade of reactions called group transfers. Conveniently, we can explain the meaning of group transfer reactions by continuing with the illustration of the dinucleotide synthesis. The key to the entire process is ATP, the acronym for adenosine-5-triphosphate, which you may recognize from biology or biochemistry as the “energy source” of biochemical systems that is supplied to cells by oxidative phosphorylation. Because of energetically unfavorable electronic interactions between the negatively charged oxygens of the triphosphate unit (remember the phosphates are deprotonated at physiologic pH), a large quantity of internal energy is released on hydrolysis of ATP as the next overhead shows:
ATP + H2O ADP + Pi where ADP = adenosine diphosphate and Pi = an inorganic phosphate group. For this reaction, ΔG = -7 kcal/mole. Because of the large negative ΔG for the hydrolysis reaction, the bond to the terminal phosphate is called a high-energy bond, and ATP is sometimes written as ADP~P to emphasize this.
The first stage in the coupling reaction is the transfer of a high-energy bond to a deoxynucleotide that is, a dN5′-monophosphate to generate a dNDP:
ATP + p-dN ADP + dNDP
This is followed by a second transfer of a high-energy phosphate to the 5′-nucleoside diphosphate by another ATP molecule:
ATP + dNDP ADP + dNTP
This last transfer can occur because ΔG is slightly negative for the transfer of the high-energy bond from ATP to the deoxynucleoside 5′-phosphates. At this point, 2 reactions called “group transfers” have been accomplished,with the “group” being the phosphate.
The deoxynucleoside triphosphate can react with the free 3-OH of a d-nucleotide to make a dinucleotide or to extend the growing end of DNA by one additional nucleotide:
p-dN′ + p3-dN p-dN′-p-dN + p-pΔG = +0.5 kcal/mole
where p-p is a diphosphate molecule. Although this reaction is not thermodynamically favorable, the diphosphate immediately hydrolyzes in aqueous solution:
p-p + H2O 2PiΔG = -7 kcal/mole
as a result of the large negative ΔG and favorable kinetics, so the diphosphate does not stay around to run the coupling reaction in reverse. The overall free energy change for this series of reactions is the algebraic sum of the ΔGs, which is
ΔG = (+0.5 - 7.0)kcal/mole = -6.5 kcal/mole
The hydrolysis of the diphosphate is said to be “coupled” to the esterification reaction, and since the net ∆G < 0, the coupled system runs in the forward direction. This is exactly the same process that takes place at the growing end of a DNA strand during replication and explains the need for deoxynucleoside 5-triphosphates in order for replication to proceed spontaneously (that is, without the input of energy from some source other than the synthons themselves).
We have thus illustrated two concepts here: (1) group transfer as a means of using energy stored in ATP to drive a reaction and (2) the coupling of reactions as the mechanism to harness this energy. In our example, the dNTP synthesis was accomplished by group transfer and the formation of the dinucleotide or extension of DNA by one additional nucleotide was driven by coupled reactions.
Remember, we had noted in passing that hydrolysis of DNA, although thermodynamically favorable,is not a problem because the kinetics of hydrolysis was slow. More often than not, however, slow kinetics in biochemistry is a disadvantage and highly undesirable. As a result, biochemical systems have evolved schemes to control kinetics – and the key systems involved are (for the most part) specialized proteins called enzymes. The mechanisms of kinetic control can be illustrated by further elaborating the reaction we just described, but now we want to run our original example in reverse – we wish to add water to a phosphodiester bond to split the bond to two deoxynucleotides or to remove a deoxynucleotide from the end of DNA. This last reaction represents the activity of an enzyme called an exonuclease. The hydrolysis reaction, in three dimensions, can be represented by the scheme at the top of the next overhead, which is a 3-D picture of the process:
[OH, trigonal bipyramidal transition state phosphodiester hydrolysis + energy coordinate diagram]
The transient pentacoordinate transition state is high-energy as a result of unfavorable steric and electronic interactions. A profile of the free energy of the system as it progresses along the reaction coordinate from starting state to products is represented at the bottom of the overhead:
[OH, trigonal bipyramidal transition state phosphodiester hydrolysis + energy coordinate diagram]
G
ΔG‡
Dimer ΔG
2 x monomer
Rreaction coordinate
Therefore, in order to reach the final state, which is favorable because there is a net decrease of energy, it is first necessary to overcome the energy barrier to the transition state: the activation energy, G‡. In any ensemble of molecules, there is a distribution of energies, dependent on temperature. At any given instant, some molecules in the ensemble may have sufficient vibrational energy to propel themselves over the energy barrier of the transition state. The higher the barrier, the fewer the molecules can cross, and the rate of reaction, which is a function or the number of molecules crossing the barrier/time, will be slow.
In the reaction we are discussing, the barrier created by the activation energy is high because of electronic and steric repulsion between all the electronegative oxygen atoms crowded together in the transition state structure, and the hydrolysis will be slow. It is, however possible to affect the rate if the transition state can be lowered in energy. Enzymes are designed to accomplish this objective. They are highly specialized molecules, geared to deal selectively with specific classes of structurally congruent reactants – meaning that enzymes have high selectivity towards specific structural features. On the next overhead is a schematic of a protein that catalyzes the hydrolysis of the phosphodiester linkage of the nucleotide at the 3′ terminal of a DNA strand, which makes the enzyme a 3′ → 5′ exonuclease. For the hydrolysis of the phosphodiester linkage, the exonuclease has a cavity called an active site into which the phosphodiester (called the substrate) can fit (this is the origin of selectivity) and which is engineered to dissipate the electron density on the oxygens by hydrogen bonding, or bonding to metal ions A and B, as illustrated by the overhead:
[OH from J. Am. Chem. Soc.1997, 119, 12691-12692]
You can see the DNA backbone (trace on overhead) and also the trigonal bipyramidal transition state that we drew schematically on the overhead and the metal ions A and B which are most prominent structural feature of the exonuclease active site. In enzyme-substrate (dNMP) complex, metal ion B interacts directly with the 3' oxygen atom of the phosphodiester bond targeted for cleavage in the transition state, presumably stabilizing the developing negative charge on the oxyanion leaving group (the 3′-O of the penultimate nucleotide). Other important interactions which help to dissipate negative charge are between B and one of the equatorial phosphate oxygens of the trigonal bipyramid, and between metal ion A and the incoming OH-highlighted in blue and one of the equatorial oxygens of the phosphate. The next overhead shows how the energy profile of the reaction would be changed by decreasing G‡:
[OH, decrease in G‡ by enzyme-substrate complex]
G‡
G
Net G is unchanged, but G‡ is decreased and more molecules will pass over the activation energy barrier, so that the reaction rate increases. Since G is unchanged, the equilibrium is not affected – only the rate at which equilibrium is achieved. Thus all the relationships derived for equilibrium constant K and G remain valid. This observation is important, because it illustrates that enzymes do not perform magic by allowing unfavorable reactions to proceed- they simply speed up processes that are thermodynamically allowed (i.e., can’t make water run up hill). The necessity for the reacting species- the substrate of the enzyme to fit into the catalytic site = active site – in order for the enzyme to perform its function, brings up a very important point relevant to our consideration of the effects of chemical modification of DNA. Any change in the helix geometry will affect the manner in which DNA is processed by enzymes such as exonucleases shown here or endonucleases (which can perform the same hydrolysis at an internal phosphodiester bond) and the phosphodiester bond-forming polymerases either with regard to recognition or stability of complexation at the active site, and in general will interfere with normal function. There is not sufficient detail known concerning mechanisms of polymerases or proteins associated with recognition and repair of DNA damage to predict a priori the effect of any specific modification of DNA. However, this is an area where rapid progress will result from the continually improved capability for crystallographic and NMR structural determinations.
DNA REPLICATION
We will now apply some of what we learned to discuss how DNA is copied – the process of replication. This is the subject of chapters 15 and 18 in Genes IX, which cover the topic in much more detail than we have time to do. But at least a sketchy understanding of the mechanisms of replication is important to us because misincorporation of bases during replication of both normal and damaged DNA directly causes mutations, which, of course are contributory to cell transforming events. We will speak in terms of a general replication mechanism, which is applicable in outline to both prokaryotic and eukaryotic organisms. But remember that while the prokaryotic and eukaryotic processes are parallel in many aspects, the details may be different. And, especially in terms of drug development and interspecies extrapolation, the devil is in the details.
Replication is divided into three stages: (1) initiation, which involves recognition of a starting point, called an origin, (2) elongation, which is the polymerization of a daughter strand – the coupling reaction we just described in the section on thermodynamics – by a complex of enzymes and proteins called a replisome, and finally, (3) a termination process. We will first describe the elongation process. The enzymes that are actually responsible for synthesis of the new DNA are called polymerases. All prokaryotic and eukaryotic DNA polymerases share the same fundamental type of synthetic activity. They extend a DNA strand exclusively by adding deoxynucleoside-5′-triphosphates one at a time to the 3'-OH of the nascent strand. This is done as pictured in cartoon form on the overhead in the left panel:
[OH]
The overhead (left panel) also illustrates with a little graphics help from me the way in which the triphosphate group is used to supply energy for formation of the phosphodiester linkage, as we just described in the thermodynamics lecture, by coupling formation of the phosphodiester bond with the hydrolysis of the diphosphate to inorganic phosphates. Because the 5-phosphate of the newly added nucleotide is being coupled to the 3 end of the strand, the direction of addition is defined as 5→3 (strand grows 5→3 ). The reason for synthesis exclusively in the 5'→3' direction becomes clear in considering the manner in which organisms insure fidelity in base pairing. One way or another, the replisome (the entire agglomeration of enzymes and other proteins associated with the synthesis of new DNA) posses a 3'→5' exonucleolytic capability- i.e., the replisome can remove a newly added nucleotide in a hydrolysis reaction, also identical to the example we described. This capability is activated when a mismatch is detected at the active site of the polymerase, and is therefore called proofreading, which is shown in cartoon form in the right panel.
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