LECTURE NOTES INTRODUCTION

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1 LECTURE TES ITRDUCTI This course will explore the organic chemistry of three major classes of molecules famous for their roles in biological systems. ucleic acids, DA and RA, are essential for the transmission of genetic information. We will discuss the organic chemistry related to the biosyntheses of nucleic acids, and the organic chemistry that takes place between small organic molecules and nucleic acids leading to the initiation or treatment of cancer. Proteins are required for the maintenance of cellular structures and for the catalysis of biochemical reactions, and enzyme-catalyzed reactions and their cofactors will be discussed. Finally, natural products are small organic molecules with enormous diversity of structure. Despite this diversity, they are synthesized in cells by pathways conceptually related to those of the above macromolecules. The biosynthetic pathways that lead to the production of natural products will be examined in terms of the principles of organic chemistry.

2 STRUCTURE AD CEMISTRY F PRTEIS PRTEI STRUCTURE AMI ACIDS Amino acids are the building blocks of proteins and have the following general structure: 3 R Amino acids are linked via peptide (amide) bonds in order to form peptides. Proteins are simply long chains of amino acids, or polypeptides. The stereochemistry at the central α-carbon is always as depicted in natural amino acids (usually the S configuration, but see cysteine). These are called the L- isomers by convention, since plane-polarized light undergoes levorotation as it passes through a solution of S amino acids. ote that glycine is the only amino acid that has hydrogen as the R substituent and hence is an achiral molecule. owever, the two hydrogens at the α-carbon of glycine are not identical but rather are enantiotopic. If one of these hydrogens is substituted with a hydrogen isotope such as deuterium, the resulting molecule is chiral. nce other amino acids are linked to glycine, a chiral environment will be created in which the enantiotopic relationship of the two hydrogens will become important. In such an environment, the two protons are not identical and can therefore undergo different chemistry. The amino acids can be grouped into sets of hydrophobic, polar, and charged side chains. Each amino acid has both a three-letter and a one-letter abbreviation.

3 ydrophobic Amino Acids 3 C 3 3 C 3 C 3 3 C 3 C 3 3 C 3 C 3 alanine/ala/a valine/val/v leucine/leu/l isoleucine/ile/i C 3 S phenylalanine/phe/f tyrosine/tyr/y proline/pro/p tryptophan/trp/w methionine/t/m Polar Amino Acids C 3 glycine/gly/g serine/ser/s threonine/thr/t 3 S cysteine/cys/c asparagine/asn/ glutamine/gln/q Charged Amino Acids aspartate/asp/d lysine/lys/k histidine/is/ glutamate/glu/e arginine/arg/r

4 For a polypeptide chain to come together to form one unique conformation (the folded structure) is one of the miracles of the life sciences. It is an enormously complex problem to fold a polypeptide chain because of the large number of rotatable bonds in the polypeptide chain. ne common feature that is apparent in the hundreds of known crystal structures of proteins is that hydrophobic side chains tend to pack in the middle of the protein. What is truly amazing is the jigsaw puzzle network of these packing interactions. There is a beautiful fit of these residues in the protein, with very little space left. CFRMATIAL AALYSIS The three dimensional structures of the amino acids can be determined using the principles of conformational analysis. We will analyze each of the natural amino acids by using four very simple model compounds: ethane, butane, pentane, and propene. Ethane Consider ethane. In this system, there is only one important conformational concept, that of the staggering of bonds. The following ewman projection of ethane depicts the staggered conformation. Rotation about the carbon-carbon bond by 60 o produces the eclipsed conformation kcal / mol staggered eclipsed Microwave spectroscopy can be used to examine the energetics of this process. The conversion from the staggered to the eclipsed conformation is endothermic (requires the input of energy) by 3.0 kilocalories per mole (kcal/mol), which is a rather substantial energy requirement. In terms of an equilibrium ratio between the two conformations, such an energy difference results in a 160 to 1 ratio of the staggered to the eclipsed form in a population of molecules at room temperature. The energy difference has been the subject of some controversy, but is thought to arise from a favorable interaction

5 between the C- bonding orbitals of one carbon and the C- antibonding orbitals on the adjacent carbon in the staggered conformation. This interaction lowers the energy of the staggered conformer by 3 kcal/mol relative to the eclipsed conformer. Since three pairs of eclipsed bonds produce this energy difference, each single eclipsed bond raises the energy by one kcal/mol. This energetic penalty applies not only to eclipsed C- bonds, as in ethane, but to most others as well, including eclipsed C-C or C-F σ bonds. Throughout the course, you will see other examples where this overlap of bonding and antibonding orbitals plays a crucial role in determining the structure and reactivity of a molecule. σ C- C C σ C- σ C- σ C- Molecules that have cylindrical symmetry about their σ bonds and can rotate very easily will avoid this energetic penalty at all costs and adopt a staggered conformation preferentially. In fact, there are now over 200,000 high-resolution X-ray structures in the Cambridge Structural Database of small organic molecules, and many of these contain methyl groups. f these molecules, only two of them have an eclipsed conformation. ature will always try hard to avoid an eclipsed interaction. Butane ow consider the conformations of butane. The conformation of larger molecules in this course will be drawn according to the following lattice skeleton, which we will call the template projection, in order to standardize our analyses:

6 In the case of butane, there is a new conformational issue with respect to the relationship of the two methyl groups. As in ethane, the central bond is a σ bond, which has almost free rotation, but in this case two different staggered relationships can exist: the anti and gauche conformations kcal / mol anti gauche ote that the anti conformation produces a symmetrical molecule. owever, if butane could be locked into the gauche conformation it would be chiral. f course, the rotation is so rapid that butane is, overall, an achiral molecule although at fractions of an instant it can adopt chiral shapes.

7 The energetics of this process are shown above. The conversion of the anti to the gauche conformation is endothermic; the anti conformation is favored by 0.9 kcal/mol. It is a less dramatic difference than that between the staggered and eclipsed forms; in this case, at a room temperature equilibrium the ratio of these two molecules would be on the order of 82% anti, 18% gauche. The origin of this destabilizing force is an electrostatic repulsion between electrons in the C- bonds of the methyl groups that begin to overlap slightly when the two methyls are close to each other in the gauche conformation. The 0.9-kcal/mol energy increase due to the gauche conformation can be applied to a variety of systems other than butane, as will be demonstrated shortly. Pentane Pentane has two rotatable bonds, not including the terminal methyl groups (note that these will simply adopt the staggered conformation about their C-C σ bonds). The conformation below is called the anti-anti conformation because it has an anti conformation about both central bonds. ote that the overall structure about each bond is similar to that seen in anti-butane. What are the energetic consequences of rotating these carbon-carbon bonds? Consider what happens when you rotate the indicated carbon-carbon bond by 120 o. This conformation of pentane is called anti-gauche, hopefully for obvious reasons. ne central σ bond still contains an anti butane-like conformation, but the other has been rotated to adopt a gauche conformation. The energetic consequence of this rotation is +0.9 kcal/mol. You can dissect out the two butane-like conformations; one stayed anti and therefore had no energy change, while the other converted to a gauche conformation. Recall that for butane, the gauche conformation is +0.9 kcal/mol more energetic than the anti is, and this quantity holds in this more complex system. A second 120 o rotation about the other C-C bond produces gauche-gauche pentane. This conformation is +1.8 kcal/mol more energetic relative to the anti-anti conformation, since

8 now a second gauche interaction has been introduced. A third rotation of 120 o about this same bond produces a gauche-gauche interaction again. owever, the energy of this conformation is approximately 4-5 kcal/mol greater in energy relative to the anti-anti conformation. This conformation has a new kind of effect, which is manifest only with pentane and not with butane. This particular gauche-gauche pentane has conspired, through the five carbon atoms, to place the two methyl groups particularly close to each other this is called the syn-pentane conformation, and it is avoided at all costs. anti-gauche gauche-gauche syn-pentane Cyclohexane ow consider cyclohexane. Recall that this molecule can undergo a chair-chair flip. In the case of 1,3-dimethylcyclohexane with the cis stereochemistry, there are two possible chair conformations. The energetic difference between these two can be estimated based on the principles outlined in the model systems above. When both groups are equatorial, the molecule has an anti-anti pentane conformation. From the point of view of the methyl groups, the same molecule, flipped into the diaxial conformation, has a synpentane interaction. ence, only the diequatorial conformation of 1,3-dimethylcyclohexane is observed. The previous model systems also allow an estimation of energy differences for cyclohexanes with single substituents, such as a single methyl group.

9 It is obvious that axial substituents are disfavored relative to equatorial ones, but now the energy differences between the two conformations can be estimated. ote that when the methyl is equatorial, there are two anti butane conformations, whereas after a ring flip there are two gauche interactions. The difference between these two conformations is thus +1.8 kcal/mol. The term for such an energy difference based on one substituent is the A value. It is a number assigned to a particular substituent on a cyclohexane ring, based on the energy difference between the axial and equatorial conformation. ence, the methyl group A value is +1.8 kcal/mole. This is the energetic cost of putting a methyl group in the axial position of cyclohexane, relative to the equatorial position. Finally, why is the chair conformation of cyclohexane is very stable despite the fact that it contains a series of apparent syn-pentane interactions? Cyclohexane does have the built-in syn-pentane geometry, but in this case, one of the hydrogens from each of the two terminal methyl (-C 3 ) groups has been replaced with a methylene (-C 2 R) unit. ence, the electrons from the C- bonds that would be repelling each other in the syn-pentane are instead forming bonds to the same carbon atom; they are part of a bond, which is a very stabilizing situation. ence, there are no true syn-pentane interactions in chair cyclohexane. Valine These principles of conformational analysis can be applied to amino acids as well. Peptide chemists have devised a nomenclature for the rotatable bonds along a polypeptide chain, as shown here: φ χ1 R ψ ω

10 Consider the amino acid valine, depicted below in the standard template projection. With valine and other amino acids, the carboxylic acid and amino groups will be considered here, to a first approximation, to be about the size of a methyl group. ote that in the case of peptides however, the actual steric effects seen in simple hydrocarbons tend to be magnified. This is because, in the case of the amino substituent, there is a bulky carbonyl group is attached to it. The same holds true for the carboxyl group. ence, these structures are actually more sterically bulky than what is observed in the simple hydrocarbon models. evertheless, the same conformational principles still apply. χ 1 Valine/Val/V The χ 1 bond of valine is rotatable, with three different possible positions for the hydrogen in the staggered conformation produced by a 120 o rotation. There are three possible conformations: one with two gauche and two anti interactions, while the other two have one anti and three gauche interactions. The first has the lowest energy, and 90-95% of the side chains will be in this conformation. I will call this the 180 o conformation, since there is an angle of 180 o between the two C- bonds that are attached to the carbon atoms of the χ 1 bond. The remaining molecules in the population have side chains that will be in one of the two 60 o orientations. If butane were the perfect model, the energy difference would be estimated at +0.9 kcal/mol, with about 82% of the structures in the 180 o conformation. owever, because the amine and carboxylic acid groups are large, the percentage of structures having this conformation is greater than 82%. Leucine In the case of leucine, the side chain is longer by one carbon atom and thus the conformation of a second C-C bond in the side chain (χ 2 ) must be considered.

11 χ 1 Leucine/Leu/L χ 2 First, consider the effect of a rotation about χ 2 value by 120 o. The major effect of this rotation is that the methyl group is placed in a syn-pentane position relative to the carbonyl. This is a prohibitive interaction, and hence this conformation is not observed. What happens if χ 1 is rotated by 120 o (counterclockwise as drawn)? If χ 1 is rotated and χ 2 remains fixed, another syn-pentane interaction is created (you should make a plastic model to convince yourself of this fact). ence, this conformation is not observed either. owever, there is a 120 o rotation about χ 1 that is allowed it simply requires a simultaneous rotation about χ 2. The small hydrogen can now go back into that very crowded position and a syn-pentane interaction can be avoided. To a first approximation, this conformation is isoenergetic with the first one. The K eq for the two conformers related by that simultaneous rotation of both χ 1 and χ 2 is about one. Any other rotation about χ 2 is not allowed because it will create another syn-pentane interaction. ence there are only two lower energy conformations for the leucine side chain and these are approximately equally populated. ote also that leucine is a very large amino acid with a hydrophobic side chain. Therefore, leucine is usually found in the hydrophobic interior of a folded protein in water. Isoleucine Isoleucine is an isomer of leucine. Again, the starting conformation will be similar to that of valine, with the preferred χ 1 value. Isoleucine/Ile/I

12 The hydrogen will again be placed in the more crowded down position. What is the location of the methyl and ethyl groups? These are on a chiral carbon atom; hence, the absolute stereochemistry dictates the position of the methyl. The question then becomes, on the ethyl group, in which of the three different positions should the methyl be placed? It is placed in the more extended position to avoid a synpentane interaction. Therefore isoleucine has one preferred conformation, and 95% of isoleucine amino acid side chains adopt this single conformation. Isoleucine can thus be viewed as a rigid amino acid in spite of its seemingly rotatable bonds.

13 thionine The final amino acid we will consider is methionine. It has an unusually floppy side chain and therefore can access a very large number of conformations. S thionine/ t/m The first issue concerns the χ 1 value. To a first approximation, there will be an equal population of conformers with the methylene either to the right or to the left. ote again that it would be unfavorable to place the chain in the more hindered down position. The next substituent on the side chain is the sulfur atom. Again, this will not be placed in the very crowded position since this would produce a syn-pentane interaction. Therefore, as in the case of isoleucine, it is placed in the least crowded forward position. Finally, how are the three groups attached to sulfur arranged? ote that there are three two of them are lone pairs of electrons. By analogy to the previous discussion, the methyl, the largest of the three, is placed in the least crowded position between the two hydrogens. Consider now a rotation about χ 2. Placing the S-methyl group in the more hindered "up" position creates a gauche-butane-like interaction. owever, it turns out that the conformations with methionine up or in the first position drawn are almost isoenergetic. The reason is due to the nature of the C-S bond, which is a very long bond relative to a C-C bond. There is thus very little steric clash between the α- carbon and the thiomethyl. ow consider a rotation about χ 3. In one conformation, there is a gauche interaction between the terminal S-methyl and the rest of the chain. owever, since two very long C-S bonds separate them, the steric clash is minimized. ence, these two conformations, rotated about χ 3, are roughly isoenergetic. verall, there are very few steric interactions built into the methionine side chain. The unusually flexible methionine sidechain plays a special role in biology. ne illustrative example shows how nature uses this flexibility to allow a single host protein to recognize many different guest partners. When proteins are synthesized that function outside of the cell, the cell has to solve the problem of exporting that protein across the plasma membrane. It does so by equipping the protein with a signal peptide, which consists of about 20 hydrophobic amino acids. In the plasma membrane is a protein complex called the signal recognition particle (SRP), which binds the signal peptide and helps export the protein. owever, while there are many different signal peptide sequences, there is only one

14 type of receptor. ow can a single receptor recognize these peptides based on their hydrophobicity rather than their precise sequence? This question was answered when the sequence of the receptor protein was determined. The amazing feature of this protein is that 40% of its amino acid residues are methionine! thionine is very hydrophobic, and undoubtedly all of these methionines project inwards; hence, the receptor protein can be viewed as a hydrophobic channel with very flexible side chains. Thus, when any hydrophobic peptide is inserted into it, the channel will be able to rotate its many methionine sidechains to accommodate the signal peptide. If the peptide is hydrophilic, it will be expelled from the hydrophobic channel. Propene The final model molecule for conformational analysis is propene (propylene). In particular we will consider the sp 2 -sp 3 bond similarly to the sp 3 -sp 3 bond in ethane. Eclipsed kcal / mol Staggered Imagine looking at a ewman projection down this bond. There are two conformations with respect to the C=C double bond: staggered and eclipsed, just as in ethane. It turns out that the eclipsed conformation is more stable by about +2.0 kcal/mol. Again, the source of the energy difference is still controversial, but it is generally believed to result from electron repulsion between the π orbital system of the double bond and two of the methyl C- σ-bonds in the staggered conformation. ote that in the eclipsed conformation, the C- bond facing the double bond is projected into the nodal plane of the π- bond, where these is no electron cloud to repel kcal / mol Repulsion between π-system and σ-orbitals ow consider a substituted form of propene.

15 + 3.5 kcal / mol In the eclipsed conformation, there are now two different positions that the methyl groups on the allylic (sp 3 ) carbon can occupy. The conformation on the right is reminiscent of a syn-pentane interaction or a 1,3-diaxial methyl-methyl interaction. This form is energetically unfavorable and is therefore not observed. Strain that exists in such an allylic moiety is called A 1,3 (or allylic) strain, since the sterically interacting substituents are on atoms 1 and 3. ow does this model relate to amino acids? Rather than looking at side chains, we will now focus more on the main polypeptide chain. Amides have a lone pair on the nitrogen that can delocalize into the adjacent carbonyl as illustrated in the resonance structure: R amide resonance + R ence, there is a certain similarity between the amide functionality and the trimethyl-substituted propylene. Both are, to a first approximation, isosteric. This substitution pattern is found in every single amide linkage in a polypeptide chain. Thus, in every amide linkage, the main chain rotatable bonds will adopt conformations that minimize A 1,3 strain. If the only important consideration in peptide chain conformation were allylic strain, minimizing it would produce one of two recurring motifs in peptides, the β conformation. This conformation places the hydrogen in the same plane as the carbonyl. This conformation repeats itself in the long extended strand. An important aspect of β-strands is that two β- strands, aligned in either parallel or antiparallel fashion, completely satisfy the hydrogen-bonding propensity of the amide carbonyl and the amide. ote that minor rotations away from the conformation minimizing A 1,3 strain are permissible and will produce additional peptide structures such as α-helices. There are additional considerations with respect to the amide bond resonance. In order for delocalization to occur, the amide moiety has to be planar. Therefore, the dihedral angle ω, which corresponds to the C- bond, has two allowed conformations of 0 o and 180 o, with the angle corresponding to the angle between the two largest groups on the carbon and nitrogen. These angles correspond to cis and trans stereochemistry, respectively. ote that the nomenclature is with respect to the two R groups. If they are opposed to each other, they are trans; if they are on the same side of the C-

16 bond, the stereochemistry is cis. owever, while the energetic difference between cis- and trans-2- butene is rather small about +1 kcal/mol in favor of the cis form the energetic difference between the cis- and trans-amide is very large. This is due to the fact that the R groups in this case are much larger and the steric clash is more pronounced. ence, the cis conformation is rarely found in polypeptide chains. ccasionally, however, the cis conformation is found in proline and glycine residues. It tends to be very common in proline residues, and much less so in glycine, though it is still observed. It is interesting that there are several disease states in which the cis conformation of glycine is important. For example, there is a fiber found in the brains of Alzheimer s patients called the amyloid fiber. It has been a contentious issue for years as to whether this is a cause or an effect of the disease, though there is evidence on the side of it being a cause. The one difference between the fibrous form of the protein found in patients and the non-fibrous natural form is that the glycine amides have rotated into a cis conformation. It is easily understood why proline readily adopts both cis and trans conformations. Proline is a unique amino acid in that the three atoms attached to nitrogen are all carbon atoms there is no hydrogen. owever, the methylene is still smaller than the methyne (-CR 2 ) bearing the acyl group. Therefore, the trans conformation is still the more common one, in a ratio of approximately 85:15 trans to cis conformers. In light of the propene model, proline is interesting in that one of the main chain bonds, whose dihedral angle is denoted Ψ, becomes fixed in a peptide chain in the conformation that minimizes the A 1,3 interaction. The hydrogen of this allylic carbon is usually found in the same plane as either the methylene or the methyne in a proline peptide. Thus, even though Ψ is a cylindrically symmetric and rotatable σ bond, it is essentially frozen in proline. ω ω +1.0 kcal / mol ψ ψ PRTEI FLDIG ne of the miracles of proteins is that although their polypeptide chains are floppy, they manage to find one single global conformation in the folded state of the protein. Four major forces act on these

17 polypeptides to achieve this folded state. The first is hydrogen bonding. This is an essential element in the previously described β-sheet structures, formed by hydrogen bonding between two strands. In the case of α-helices in peptide chains, the main element that holds helix together are hydrogen bonds between the amide and carbonyl. Recall that with any amide moiety (other than proline) there is an on one side and a carbonyl with the oxygen lone pairs on the other side. These groups would like to form hydrogen bonds, with the group donating a proton and the carbonyl donating the lone pair electrons. Proteins fold in a way to maximize this propensity for hydrogen bonding. elices and sheets are common motifs for accomplishing this. Another common folding motif in polypeptide chains are turns. These form when a carbonyl and an that are 10 atoms apart share a hydrogen bond. In a turn, the main chain comes in one direction and exits in another it is a way to change the directionality of the chain. The second important effect is electrostatic in nature: the formation of salt bridges between negatively and positively charged groups. Salt bridges often form between a positively charged arginine residue and a nearby negative charge, such as a glutamate. Such salt bridges are very common, but they

18 are probably most energetically critical in the rare cases in which they are found in the interior core of the protein. This core usually has a low dielectric constant and is nonpolar; hence, any highly charged polar residues must be neutralized. Salt bridges are also found on the surface of proteins, but since the dielectric constant in the surrounding water is high, this electrostatic interaction is of less consequence to the overall energy of the protein. The third important component for protein folding is the disulfide bond. These are formed when two thiols are oxidized to release two electrons and two protons, and form a bond between the two sulfur atoms. S S + 2 e -, S If cysteines are close in space, they will form such a bond. Clearly, this bond is much stronger than hydrogen bonds and imposes a major constraint on protein structure. ne important aspect of disulfide bonds is that the lowest energy conformation is that with a 90 o angle between the S-R bonds. ote that the inside of a cell is a reducing environment. Disulfide bonds are therefore not observed frequently in intracellular proteins. owever, proteins that are secreted from the cell, such as hormones, often have many such bonds. The fourth force is the hydrophobic effect. There is no way to do justice to such a very complicated phenomenon in a brief description, but it can be illustrated in the following manner. Consider a polypeptide chain, as it goes from an unfolded to a folded state. ne of the problems with the unfolded state is that there are hydrophobic amino acids like phenylalanine, valine, and leucine in an aqueous milieu surrounded and solvated by water molecules. Upon folding, these hydrophobic side chains pack together inside the protein and solvate each other. The result is that water molecules previously solvating the side chains are now released into the bulk media. Consider the energetics of this process. The unfolded state will be called the first state, and the folded the second state. The energetics of the process can be determined by monitoring the changes in entropy (the amount of disorder) and enthalpy (the amount of heat ) as the protein goes from the first to the second state. A hypothetical phenylalanine side chain can be used as a model for the protein. In the unfolded state, the phenylalanine is solvated by water. owever, the phenylalanine side chain is hydrophobic and cannot directly hydrogen bond to water. Rather, the water molecules form extensive hydrogen bonds with each other, thereby creating a lattice of hydrogen-bonded water molecules around the phenylalanine. The water molecules become very ordered, and form a three dimensional ice-like

19 structure. They form six-membered rings using three water molecules held together by three hydrogen bonds. This shell is probably several molecules thick and continues around the phenylalanine side chain. The formation of an ice-like water lattice has interesting consequences. The unfolded (state 1) form, with the lattice of water molecules, allows each of the negatively charged lone pairs on oxygen to feel a positively charged proton. This electrostatic effect lowers the enthalpy () of the water molecules. In other words, 1 is lower than 2. Therefore, Δ for the folding reaction is positive. This is counterintuitive; the folding of a protein increases the enthalpy of the system because water loses these very strong hydrogen bonds. By folding the protein, the water molecules are released into the bulk media, where they are moving around rapidly and cannot form stable hydrogen bonds. Due to this lattice of water molecules, state 1 is more ordered it has a lower entropy (S). ence, S 1 is smaller than S 2 and therefore ΔS for the folding process is positive. Recall the equation ΔG = Δ - T ΔS, where ΔG is the free energy of the process. A process is favored when there is a decrease in the free energy; hence, ΔG must be negative. If Δ and ΔS for protein folding are both positive, then ΔG can only be negative if T (temperature) is large enough to make the TΔS term larger than the Δ term. A simple experiment highlights these principles. Years ago a whole series of heat of transfer reactions were carried out using calorimetry to measure the change in energy and using temperature dependency to study the role of S and. In this experiment, water was used as the solvent in one case and carbon tetrachloride (CCl 4 ) in the other. thane will be a model for valine and phenylalanine since it is a very hydrophobic molecule. Water as a solvent is a model for the unfolded state, while carbon tetrachloride, being nonpolar, is a model for the folded state, mimicking the hydrophobic interior of the folded protein. This experimental system can be used to measure the change in ΔG calorimetrically. Varying the temperature and monitoring its effect on ΔG allows the calculation of Δ and ΔS according to the previous equation. The ΔG for the movement of methane from water to carbon tetrachloride is -2.5 kcal/mol: the reaction is exothermic (releases energy). This is expected intuitively since methane prefers to be in carbon tetrachloride rather than water. The surprising finding is that Δ for this reaction is +2.5 kcal/mol. This is an enthalpically endothermic process. In terms of the enthalpy, methane would rather be in water! Why? Because when methane is in water, the shells of water molecules form and there are very strong hydrogen bonds. This lowers the energy of the system. In order to balance the contribution of enthalpy and still give a negative ΔG, -T ΔS must be strongly negative. Experimentally, it was found to be 5.0 kcal/mol. ence, ΔS is a positive number, as expected. This is a very simple model for protein folding. This phenomenon can be observed in biological systems in a process called cold denaturation. Folded proteins are known that will unfold at lower temperatures. This demonstrates that entropy plays

20 an important role in the hydrophobic effect. With a positive ΔS, lowering the temperature decreases the favorable contribution to ΔG. The lower temperature shifts the equilibrium to the unfolded state. A fluorescent reagent can be used to visualize the cellular cytoskeleton under the microscope. It is the microtubule component of the cytoskeleton that forms a rigid lattice, allowing the cell to maintain its physical shape. Microtubules are made of two proteins, α-tubulin and β-tubulin. ot surprisingly, to make such a three-dimensional structure, these tubulin proteins fold together. If the temperature is lowered, it is possible to see the unfolding of the microtubules under the microscope the fluorescence disappears. This is a completely reversible process; increasing the temperature, thereby increasing the entropic component to folding, allows the cytoskeleton to refold and form the skeletal architecture. There is small organic molecule called taxol that it is one of the few major new anti-cancer drugs. It binds to tubulin and stabilizes its folded state. In the presence of taxol, at low temperature, the cytoskeleton persists. Taxol binds to the cytoskeleton and stabilizes it. This is a problem for cancer cells because the process of segregation of chromosomes during replication requires that the spindle apparatus, composed of microtubules, disassemble. When this process is prevented by taxol, replication is arrested and the cell cannot divide.