Protein-protein interactions - lecture notes

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1 Protein-protein interactions - lecture notes arold P. Erickson December, 2002 Cell Biology, Duke University Medical Center h.erickson@cellbio.duke.edu Assigned reading: Reference #4 below, defining the hot spot, and this nice review: DeLano, W.L Unraveling hot spots in binding interfaces: progress and challenges. Curr Opin Struct Biol. 12: Reading list (recommended) 1. Chothia C, Janin J Principles of protein-protein recognition. Nature 256(5520): The original description of the protein-protein bond as proportional to surface area of the interface. 2. Amit AG, Mariuzza RA, Phillips SE, Poljak RJ Three-dimensional structure of an antigen-antibody complex at 2.8 A resolution. Science 233(4765): The first atomic structure of an antibody bound to a protein antigen. The interface is a lock-in-key structure confirming many of the principles of Chothia and Janin 3. Cunningham BC, Wells JA Comparison of a structural and a functional epitope. Journal of Molecular Biology 234: The first detailed chemical analysis of the growth hormone receptor interface by alanine scanning mutagenesis. 4. Clackson T, Wells JA A hot spot of binding energy in a hormone-receptor interface. Science 267: The followup study mutating the other side of the interface. 5. Atwell S, Ultsch M, De Vos AM, Wells JA Structural plasticity in a remodeled protein-protein interface. Science 278: Explores how an interface can change during evolution 6. DeLano WL, Ultsch M, de Vos AM, Wells JA Convergent solutions to binding at a protein-protein interface. Science 287(5456): An intriguing study of a protein patch that is optimized for binding multiple partners. 7. Kortemme T, Baker D A simple physical model for binding energy hot spots in protein-protein complexes. Proc Natl Acad Sci U S A 99(22):

2 8. Guerois R, Nielsen JE, Serrano L Predicting changes in the stability of proteins and protein complexes: a study of more than 1000 mutations. J Mol Biol 320(2): References 7 and 8 are two recent papers that are able to predict which amino acids contribute substantially to the binding energy (which are hot spot aa s). 9. Erickson P Cooperativity in protein-protein association: the structure and stability of the actin filament. Journal of Molecular Biology 206: Analysis of cooperativity in protein-protein interactions. Northrup S, Erickson P Kinetics of protein-protein association explained by Brownian dynamics computer simulation. Proceedings of the National Academy of Sciences of the United States of America 89: ow can proteins associate so fast?

3 Chapter 2. Basic thermodynamics of protein-protein association. Although many enzymes function alone, the majority of protein molecules associate with other protein molecules to achieve their functions. Protein subunits are held together by interfaces that we will refer to as a protein-protein bond. For some extracellular proteins the association interface is stabilized by covalent bonds (disulfide bonds are common; crosslinks formed by transglutaminase or lysyl oxidase are more specialized). We will ignore these covalent stabilizations and focus on the primary protein-protein bond, which is the basis for virtually all protein associations in the cytoplasm. Examples include oligomers like hemoglobin tetramers, self assembly of larger polymers like actin and microtubules, association of motor molecules with their cytoskeletal tracks, and association of growth factors with their receptors. This basic protein-protein bond is non-covalent. Bond formation is freely reversible, and the reaction can be characterized by an equilibrium constant and associated free energy. Consider the association of two subunits to form a dimer. We will use G to describe one subunit (think of it as a growth factor) and R for the other (the receptor, which is typically present in far fewer copies than G). [G][R] G + R < > G-R K D = - = 1/K A [G-R] Note that the units of K D are M. This is convenient because it expresses the strength of a reaction in units that can be compared directly with the concentration of the reactants. On the other hand K A is perhaps more intuitive for discussing bond strengths because a stronger bond means a bigger K A. One can rearrange the equation to solve for a useful ratio the number of liganded receptors [GR], over the number of unliganded receptors [R]: [G-R] [G] = [R] K D This relationship is exact, but it is complicated because [G] is the concentration of G that remains free at equilibrium. What we really want is a relationship in terms of [G o ], where [G] = [G o ] [G-R]. This can be substituted and the equation solved, but it is a quadratic equation, a bit messy. The analysis is simplified if we assume that G is present in great excess, [G o ] >> [R o ]. In this case the amount of G that is lost in the complex is negligible, and [G] ~ [G o ]. Then we have a simple relation for the fraction of R in complex [G-R] [G o ] = [R] K D Now we have three regimes: If [G o ] << K D, there will be relatively little complex; most of the R will remain unliganded. If [G o ] = K D, the R will be 50% free and 50% in liganded.

4 If [G o ] >> K D, most of the R will be be liganded. K Note that it is the concentration of the species in excess, G, relative to K D, that determines the relative amount of the dilute species that is complexed or free. Of course, the the top equation is completely general, and with some additional mathematics the concentrations of reactants can be determined for any conditions. The dissociation constant, K D, is especially useful because it allows a quick comparison with the molar concentration of the reactants. The reaction can also be described by by an association constant, K A = 1/K D. For most of the lecture, and in the chapter on cooperativity, we will use the association constant, K A. The advantage of discussing association in terms of K A is that a stronger association corresponds to a larger K A. The equilibrium constant is related to the free energy by this standard equation: 1. RT ln K A = DG A (Note that favorable association, large K A, requires a negative free energy of association.) This free energy describes all of the chemical and energetic factors involved in the association reaction. It is extremely useful to break this term down into two opposing energies, one favoring association and one opposing it. (The importance of explicitely separating these two terms has been recognized since the work of Doty and Myers on insulin dimerization in 1953; it has been rediscovered about every ten years since; it was a crucial point in the analysis of Chothia and Janin, 1975; and it forms the basis for the analysis of cooperativity (Erickson, 1989). The two terms are the intrinsic bond energy, which includes all the chemical forces acting across the subunit interface, and the intrinsic subunit entropy (expressed as a free energy). 2. DG A = DG bond + DG s ; DG bond = DG A DG s The term DG bond is the intrinsic bond energy. All the chemical forces intrinsic to the protein-protein interface are included in the term DG bond. It is a negative number because it favors the association; the stronger the bond, the larger its absolute value. The Chothia and Janin analysis discussed next is directed at estimating this intrinsic bond energy and determining all of the chemical forces that contribute to it. The term DG s is the intrinsic subunit entropy, expressed in units of free energy. Think of this as the free energy required to immobilize a subunit in a dimer or polymer, independent of the type or number of bonds formed. Free energy is required because entropy is lost when the subunit is immobilized in the polymer. Before dimer formation each subunit has three degrees of translational and three degrees of rotational freedom. In the dimer one subunit still has its six degrees of freedom, but the other one has lost its independent translation and rotation. The question is, how much free energy does it take to immobilize a subunit, to compensate for the three translational and three rational degrees of freedom lost in forming the association? The best current value is DG s = +6. An essential part of this estimate is to realize that the bonded subunit is not completely immobilized, but retains rotational and translational vibrational modes corresponding to about ± 1 Å (see Erickson, 1989, Appendix 1, for details of the calculation). Importantly, the intrinsic entropy depends very little on the size and shape of the protein subunit (these

5 dimensions enter the calculation as a logarithm), so this single value can be used for all protein association reactions. It should be noted here that the theoretical calculation actually gave a value of 11. The value of 7 was the largest value that could give a reasonable fit to experimental measures of actin assembly (Erickson, 1989). An experimental study of protein-protein pairs with interfaces determined by X-ray diffraction gave a value DG s = 6.2 (orton and Lewis, 1992). We will use the value DG s = 6 for the remainder of the discussion. Note that DG bond is a negative free energy, indicating that it favors association. DG s is a positive number, opposing association. A useful way to think about this is as an entropy tax. The tax is a flat rate, not regressive. The entropy tax must be paid once for any association, and the tax is the same regardles of the size of the subunit and the strength of the bond. Another way to think about the equation 2 is that the intrinsic bond energy must be sufficient to achieve the observed K A, and it must also pay the entropy tax of 6. Let's put this in perspective and illustrate the calculations with some numbers. A typical modest protein-protein association (as for actin assembly or an antibody will have a K D = 10 6 M (so K A = 10 6 M 1 ). In eq. 1, R has the value 2 cal/deg.mol, and T is the absolute temperature = 300 degrees. 2a. DG A = RT ln K A = 600 ln (10 6 ) = 8,300 cal/mol = b. DG bond = DG A DG s = = 14.3 Thus the net free energy of association, 8.3, is very similar in magnitude to the intrinsic entropy term, 6 ; and the intrinsic bond energy is the sum of the two, The intrinsic bond energy must be sufficient to compensate for the entropic energy loss, and to produce the favorable association constant. Let s consider a higher affinity interaction, K A = 10 9 M -1 (this would be typical for a growth factor binding its receptor). 2c. RT ln K A = d. G bond = = It is interesting to consider that the 1,000-fold difference in K A is achieved by only a 29% increase in intrinsic bond energy. That is because the bond energy enters the K A as an exponential. The primary utility of DG bond is that components contributing to bond energy are simply additive. If one used DG A instead, mistakes in additivity would occur because the entropy tax is not properly handled (although many people, including the Wells group, get by with it because the error can be smaller than the accuracy of measurement). Probably the most important application of DG bond is in the calculation of cooperativity, considered later.

6 Chapter 3. The nature of the protein-protein bond, à la Chothia and Janin. In the 1960 s and early 1970 s X-ray crystal structures were obtained for several protein dimers and oligomers. This permitted one to look at the types of contacts or chemical bonds formed across the interface, and the information was staggeringly complex. There were numerous van der Waals contacts, hydrogen bonds and ionic bonds made across the interface. In these early times authors would present tables listing the specific van der Waals, hydrogen and ionic contacts made on each side of the interface. Sometimes estimates were given for the magnitued of each of these bonds: ionic bonds ~5-10, -bonds ~2-7, van der Waals bonds per atom contact. If one simply tried to add up all the contributions the total was enormous, so it was clear that the interactions had to be moderated. But overall, trying to make sense of the many interactions was impossibly complex. A first crucial observation of Chothia and Janin (1975) (which was actually based on work of F. Richards several years earlier) was that the interface between subunits looks like the interior of a protein molecule - in particular it consists of closely packed atoms with no spaces. This means that the two surfaces must fit together very snugly. Water was completely excluded, and there were van der Waals contacts between the surface atoms across the interface. It was also clear that many hydrogen bonds and ionic bonds were made across the interface. C & J then pondered the question of how each of the many interactions would contribute to the free energy of the protein-protein bond. C & J achieved an enormous simplification that now seems well accepted as the first order approximation they concluded that the intrinsic bond energy was simply proprotional to the area of the interface. -bonds, ionic and van der Waals interactions could be ignored. The key to this analysis was to realize that when the protein-protein bond is broken, the interface surfaces are not placed in a vacuum, but are exposed to the solvent. The bonds that were formed across the interface when the subunits were together are largely replaced with ones to solvent molecules when the subunits are apart. + + Na + Cl N O=C N O O O=C The diagram illustrates schematically a protein-protein interface. We should first emphasize that the interface is not a contact of a few atoms, but it covers an extensive area of each subunit. Typically 10-20% of each protein s surface area will be in the interface. Second, the interface is very snug, placing the atoms in contact and leaving no

7 holes and no (or few) buried water molecules. The beak on the left subunit fitting into the pocket on the right one illustrates the snug, lock-in-key fit. Third, - and ionic bonds are made across the interface. The key to the simplification of C & J was to recognize that when the subunits are separated they are not in a vacuum. The numbers presented above for the energies of -, ionic and van der Waals bonds are those for making the bonds in a vacuum. Protein molecules are in an aqueous solution, and this provides the simplification. Consider first the ionic bond. When the subunits are together the + charge on one subunit neutralizes the charge on the other. When the subunits are separated these charges are not bare, but are neutralized by ions in the solution. Similarly for -bonds: When the subunits are separated both the -bond acceptor (N ) and donor (O C) form -bonds to water molecules. C & J made the approximation that the ionic and -bonds to solvent were similar in magnitude to those across the interface, and thus their net contribution to the intrinsic bond energy would be negligible. Similarly, the van der Waals interactions made across the interface would be replaced with van der Waals interactions with water. Clearly, the nature of the protein-protein bond is enormously simplified if we simply set all these chemical contributions to zero. The hydrophobic bond. What then provides the free energy for protein association? C & J conclude that the main source of free energy is the "hydrophobic bond," which is actually a complicated entropic effect of the solvent. The protein surface is not really hydrophobic as mentioned above water forms favorable van der Waals and hydrogen bonds with the surface. The driving force comes from the water molecules, which have a strong ydrophobic surface O O O O O Bulk water bonds, water molecules very mobile ydrophobic surface - ~3 -bonds, but surface water frozen

8 tendency to hydrogen bond to each other. The diagram illustrates the basic concept and why it is an entropic effect. In bulk water each molecule can form a maximum of 4 hydrogen bonds with neighbors. It actually achieves an average of 3.4 bonds per molecule. As discussed above, if a molecule is immobilized when it forms a bond it must lose entropy, but in water the molecules remain highly mobile. They are constantly rotating around the single bonds, and also breaking bonds and reforming them to new partners. The molecules in the bulk water therefore retain a high state of entropy, which is favorable. At the hydrophobic protein surface there is an interesting constraint. The water molecules next to this surface can t make hydrogen bonds to the protein (except to the hydrophilic aa s that provide donors or acceptors), but they can still make bonds to neighboring water molecules. But here they have to pay a price either lose a couple of hydrogen bonds, or orient themselves in a position that will place their hydrogen bonding sites facing the solvent. The water molecules next to the surface actually achieve about 3 hydrogen bonds each, but in order to do this they are much more restrained than those in the bulk solution. These water molecules are partially "frozen," they have less mobility than molecules in the bulk water. In thermodynamic terms their entropy is decreased, and this costs energy. When the two subunits come together to make the interface this layer of "frozen" water is released from each surface. The water molecules can now make -bonds in all directions; they have more freedom, i.e., increased entropy and a lower (more favorable) free energy. It is important to realize that the hydrophobic interaction is not a positive attraction of the hydrophobic surfaces. Their interaction (van der Waals) with the water was approximately as favorable as that with each other. It is the water that drives the subunits together, because the water molecules have a much stronger interaction with each other than with the subunit interface. C & J then go on to estimate the magnitude of the hydrophobic interaction, and argue that it is simply proportional to the area of protein surface that is removed from contact with water. Studies of model compounds have given a value of for each 1 Å 2 hydrophobic surface removed from contact with water. C & J calculate the surface area removed from contact with water by rolling a ball about the size of a water molecule over the surface of the interface (all this in the computer). This interface area (counting the surface of each subunit) is on the order of 1,000-1,500 Å 2 for hemoglobin, insulin dimer and trypsin-pti (see Table 1), so they estimate a hydrophobic free energy of about favoring these associations. These energies are larger than the measured intrinsic bond energy, which means that the hydrophobic bond energy is more than enough to account for the observed bonding. Complimentarity Saying that the net contribution of ionic bonds is zero does not mean that they are ignored. These bonds play a crucial role in protein association, namely in determining the specificity of the interaction. Protein association is not promiscuous: e.g., pancreatic trypsin inhibitor associates very strongly with trypsin, but not at all with most proteins. The basis for the high specificity is that the interfaces must be highly complimentary. This complimentarity comprises three features. (The classic C&J theory is given first;

9 modern experimental work with alanine scanning mutations have identified important exceptions, which will be discussed next under hot spot ). a) Ionic complimentarity. The original formulation of C & J suggested that all possible salt bridges must be made across the interface. If there is a minus charge group on one subunit, there must be a plus charge opposite on the other side to neutralize it. Otherwise the subunit association would require breaking an ionic bond to solvent (2-6 ) and replacing it with nothing. Burying a charge might be very expensive in free energy and could strongly destabilize the protein-protein interaction. b) ydrogen bond complimentarity. The argument for bonds is exactly analogous to that for ionic bonds. Any hydrogen bond donor on one subunit must find a hydrogen bond acceptor on the opposite. Losing a hydrogen bond would cost 0.5-6, depending on the chemical nature and charge of the buried group (see Fersht et al., 1985 for this important experimental analysis). c) Steric complimentarity. When separated, the subunit interfaces are covered with 2 O, and all the atoms on the protein surface make van der Waals contacts with water molecules. The van der Waals bonds are less than 1 kcal per atom, but there art lots of atoms. In order not to lose the energy of the van der Waals interactions when the bond is made, contacts with solvent must be replaced by contacts with the other subunit. This means that the subunits have to fit together very snugly. Separating a pair of atoms even 1 Å will eliminate the van der Waals energy. This is indicated in the above figure by the projecting beak of one subunit fitting into a complementary pocket in the other. ow does the "hot spot" of binding affinity reconcile with the Chothia Janin theory. Our understanding of the hot spot will be based on reading the paper by Delano and Wells (2000). The noted below briefly address the background to this paper and recent progress in understanding hot spot residues. The simplest interpretation of the Chothia Janin theory is that we can ignore the chemical nature of the amino acids at the interface and simply add up the water accessible surface area buried. Several groups have now attacked this question directly by systematically mutating to ala each of the residues in the interface. Mutation to ala essentially removes the side chain, leaving only the beta carbon (mutation to gly would remove everything, but gly could also introduce changes in the backbone folding). The mutation causes a change in the DG of binding, referred to a DDG. If the side chain was contributing to the binding, changing it to ala reduces the binding affinity, producing a positive DDG. Fersht and colleagues have studied the barnase-barstar interaction, which has a very efficient interface (65%, i.e., good complemetarity - efficiency is indicated in Table 1 and will be addressed in the final section, after cooperativity). Cunningham and Wells have studied the G-GR (growth hormone - growth hormone receptor) interaction, which has one of the poorest interfaces known (30% efficient). Both groups came to the same conclusion. Some aa s, although making contact in the interface, contribute very little to the bonding energy, and most of the bonding energy is attributed to a number of key aa s. The Wells group identified 31 residues of G that made contact with GR, but

10 found that only 8 of them led to significantly reduced binding when changed to ala. Importantly, these 8 aa's were clustered in the center of the interface, in what they call a "hot spot" of binding interaction. (In current terminology a hot spot aa is one that decreases DG by 2. They are typically clustered near the center of the interface (Bogan and Thorn, 1998).) aa's in the periphery, although they make contact across the interface, contributed less to the interaction (some even destabilized the binding, Clackson, 1998). From the prior description of the C-J theory one might expect any -bond or ionic aa in the interface to be a hot spot aa, because changing the aa on one side to ala would leave its partner on the other side unpaired. As stated above it should be expensive to bury a charge group or -bond acceptor. owever, the Wells study found that this was true sometimes, but was not generally true. Several aa s making important looking - or ionic bonds across the interface of G-GR could be mutated to ala with minimal effect on the binding. A possible explanation is that the interface has some plasticity, and if a -bond acceptor is removed, the donor side chain shifts and finds another acceptor (see Altwell...Wells, 1997 for demonstration of this kind of plasticity). A rather dismal and frustrating conclusion of the Wells study was that there was no way to tell from looking at the x-ray structure which aa s would be hot spot ones and which ones would have little effect when changed to ala. It was perhaps comforting to find that some prominent aa s behaved as predicted by the C-J picture. In particular, W104 and W169 of GR, which are right in the center of the interface and bury the most surface area, turned out to cause a major loss of binding when changed to ala. The DDG was >4.5, the largest value observed for any change. owever, in general one could not predict from the structure which aa s would be hot spot and which neutral. One interesting point is that when one calculates the area of the hot spot, it predicts the binding energy if the entire hot spot is contributing the full hydrophobic energy of Å 2. Three recent studies have considerably brightened the dismal picture. Bogan and Thorn did the first survey of all interfaces that had been subjected to ala scanning, and came to a number of predictive conclusions. Perhaps most interesting they found that trp, arg and tyr are found in hot spots (DDG > 2 ) much more frequently than their abundance in proteins or interfaces in general. Similarly val, leu and ser were almost never hot spot residues. Surprisingly ile, although structurally very similar to leu, is a frequent hot spot residue. Two studies have now developed algorithms that can actually predict with fairly good accuracy which aa s will be hot spot aa s, and which ones neutral (Guerois, Nielsen, and Serrano. 2002; Kortemme and Baker 2002). The prediction uses a computer analysis of the atomic structure of the interface. It is based on calculations of bonding energy across the interface, with explicit terms for several types of -bonds, ionic interactions, vdw contacts and a term for desolvation of the aa s. The different interactions are given variable weight and trained to best match the DDG s of a large set of ala mutations of various protein-protein interfaces. The trained algorithim correctly predicted 64% of the hot spot aa s in G:GR, and 100% in EL:D1.3 and barnase:barstar (Kortemme). Thus it is still true that the G:GR interface is difficult to interpret, but for the more efficient interfaces the algorithm is pretty good at determining whether any particular aa is neutral or hot spot.

11 Protein interfaces are preformed and rigid lock in key vs induced fit As a final very important point we need to address the question of the rigidity of protein molecules and how this relates to the interface. First, rigidity. In a study of microtubules and actin Joe oward s group concluded that the microtubule s Young s modulus would be ~1.2GPa, similar to Plexiglas and rigid plastics. (Gittes et al. J. Cell Biol. 120:923(1993). It is reasonable to conclude that this value extends to tubulin-tubulin interfaces, the tubulin molecules themselves, and to globular proteins in general. This is very rigid, completely different from the description of soft wet materials sometimes applied to biological molecules. In this picture the two interfaces that come together to make the protein-protein bond are preformed and rigid, and they come together as a lock in key. This picture has in general been confirmed for many cases where atomic structures have been solved for the complex and the separate subunits. The atomic structure of the interface is virtually superimposable on the corresponding structure of the separate subunits. There are three exceptions to this general rule, that are interpreted as induced fit i.e., one or both binding partner shows significant changes in atomic structure when bound in the interface. See Wilson and Stanfield, 1993 for details. a) Surface side chains can move. This is fairly common, and seen even in complexes that are largely described as lock in key. b) Surface loops can move. This is an especially important aspect of induced fit that has been demonstrated in several structures. Loops are less constrained than the body of a globular domain and can move substantially and change shape to fit a binding interface. c) Domains can move at hinges. Many proteins form two or more independently folding globular domains. Each domain is rigid, but they can be connected by a flexible hinge. Movements at the hinge are common (e.g. IgG molecules have hinges between the Ig domains). In summary, induced fit does occur in antibody antigen interactions and other proteinprotein associations, but lock in key association of rigid, preformed interfaces is a much more general picture of protein-protein associations. Summary description of the protein-protein interface from Janin and Chothia, 1990: The protein-protein recognition sites discussed here [proteaseprotease inhibitors and antigen-antibody] have very similar structural properties: 34 ± 7 close-packed contact residues bury 1,600 ± 350 Å 2 of surface (these are total from two sides of the interface) and form 8 13 intermolecular hydrogen bonds. The conformational changes that occur on association involve only the rotation of certain side chains and small movements in some sections of the main chain.

12 An important problem that can be addressed with the knowledge of the proteinprotein interface. A particularly valuable probe for microtubule studies is tubulin labeled with biotin, a small molecule of ~200 MW that (a) can be attached covalently to reactive lysines or other groups on the surface of the protein, and (b) can be subsequently labeled with antibodies for localization. What would this biotin molecule do to tubulin assembly if it were attached to an amino acid (a lys) in the middle of the tubulin-tubulin interface. Tubulin has a MW of 50,000. ow could tubulin be labeled to minimize reaction at the interface? Answer (1) Inserting a biotin molecule in the interface would disrupt it completely, since it would block the contact of complementary amino acids. (2) The solution is to assemble the Mt s first. Then the biotin will not be able to penetrate the interface but will label aa's on the outer surface of the tubulin subunits.

13 Ch 4. Cooperative association and efficiency of bonds. What happens to K D if you double the intrinsic bond energy? A hypotherical example would be to compare binding a bivalent Ab to binding of the monovalent Fab. Specifically, let's assume we have a EL dimer, with the two EL subunits held together so they can each bind the Fab of the IgG. Assume that the monovalent Fab binds with K A Fab = 4.5 x 10-7 M -1 What is K A IgG? To make the calculation easy we will introduce two assumptions, neither of which are really valid. 1. Assume that the Ab is rigid. (actually the Fab fragments on an IgG have considerable rotational flexibility). 2. Assume that the crosslinked EL dimer is also rigid, and that the crosslinking presents the EL epitopes so that the rigid IgG can bind both of them without strain or distortion. Essentially the problem asks what happens to K A if we double the intrinsic bond energy. Is K A also doubled? Do you square it? Both of these quick guesses have been proposed for certain cases, but they don't work. We need to consider the intrinsic bond energy to do the calculation. Most important, we have to pay close attention to the intrinsic subunit entropy.

14 1. Calculate DG bond Fab = RTlnK A 6 = = Calculate DG bond IgG. This is simply 2 x DG bond Fab = Calculate K A IgG : RTln K A IgG = DG bond IgG + 6 = = K A IgG = exp (25,400) / RT = 2.4 x M. This is an enormous enhancement in binding affinity, from 4.5 x 10 7 to 10 19, from doubling the intrinsic bond energy. The key to this effect, and to this calculation, is to realize that the entropy tax is paid only once, while the intrinsic bond energy is fully counted twice. Since the IgG is assumed to be rigid it is fully immobilized when the first Fab binds. The second Fab then binds "for free" and all of its energy goes to increasing K. Of course this is an oversimplification, since it depends critically on the assumptions of (a) rigidity of the Ab, and (b) the perfect fit to the EL dimer. In a real Ab the second Fab will still have substantial rotational entropy after the first one has bound. This will have to be compensated when the second head binds (but it should be much smaller than 7 ). Also an IgG binding to a virus will probably have to strain a bit to make the bivalent attachment. Nevertheless, experimental measurements of Abs binding to viruses show 1,000 X greater affinity for IgG compared to Fab. Analysis of cooperativity in G-GR interaction (and comments on the "efficiency" of hydrophobic bonding. Table I summarizes data on interface area and actual bond energy for a number of proteins, with a particular interest in the question - ow does the actual bond energy compare to the Chothia-Janin prediction based on Å 2? It is comforting to see that actual bond energies are all less than that predicted from interface area. This suggests that some imperfections in the interface may decrease the bond energy from its full potential. We can calculate a "bond efficiency" which is the ratio of the actual bond energy to the maximum hydrophobic energy available from the interface area. We see that the first four proteins vary in efficiency from 70 to 50%. The G-GR complex presents an interesting case. The formation of the single G- GR through site 1 is well characterized from a mutant G that can't bond the second GR (see Clackson 1998). The area of this interface is relatively large, giving this interface an exceptionally low efficiency, 29%. The addition of the second GR to the already-formed G- GR is also well characterized. This involves two interfaces, the site 2 interface between G and the second GR, and a GR-GR interface at the bottom of the second FN-III domain. Adding up the interface areas we find that the efficiency of this whole complex is also 30%. We will therefore assume that each of the individual interfaces, site 1, site 2 and the GR-GR, are 30% efficient. We can now use the reverse of the cooperativity argument to estimate the affinity of GR for GR through the interface in the second FN-III domain. This is a very important question, because if this affinity were anywhere near significant it would result in receptor dimerization, and signalling, in the absence of G. owever, we see that this affinity is ridiculously low, only 2 M -1 meaning you would have to have receptors at 0.5 M before this association became important. The affinity of GR for site 2 on G is also very low, so this association is insignificant for an isolated pair of G and GR. Yet once the G-GR has formed through site 1, these two very week affinity patches provide a cooperativity that makes the second GR bind with very high affinity, forming a strong receptor dimer.

15 Table 1. Examples of protein-protein bonds. Calculation of solvent-accessible surface area buried in interface, and relation to actual bond arold P. Erickson, October 10, 1993; revised Oct. 25, 1995 (note: this table uses 7 for DG S, not 6) Proteinprotein pair (a) Area in interface Å 2 trypsin PTI = 1,390 barnasebarstar = 1,590 insulin dimer = 1,130 EL mabd1.3 (Fab) = 1,438 (b) Max Avail. ydropho bic bond energy 25 cal/mol x area (c) K A experimental M -1 (d) DG bond = RTlnK A M = = = x = 17.6 (e) Bond efficiency d/b 25/35 = 71% 26/40 = 65% 14/28 = 50% 17.6/36 = 49% G GR1 1,230 (x2) = 2,460 (G GR1) \ GR2 900 (x2) 500 (x2) = 2, x = x = /61.5 = 32% 20.4/70 = 29% G GR2 GR1 GR2 900 (x2) = 1, (x2) = 1, (5 x 10 4 ) back calculated (2 M -1 ) back calculated ( 13.5 ) back calculated ( 7.5 ) back calculated 30% assumed 30% assumed

16 Conclusions. 1. Protein-protein bonds are always weaker than the maximum available hydrophobic bond energy. They achieve from 70% to 30% of the available energy 2. The growth hormone receptor bonds appear to be especially inefficient, primarily because the areas reported buried in the interface are much higher than for other proteins. Cunningham and Wells demonstrated that a small cluster of residues in the center of the interface contributes most of the binding energy. This cluster apparently contributes at near 100% efficiency, while the peripheral residues contribute near 0%. 3. Cooperative bonding of GR2 to GR1 G complex is 10 9 stronger than GR1 to unliganded GR2.

17 Chapter 4a. ow cooperativity makes possible the actin filament. (This chapter still under construction based on Erickson JMB 1989.) The following section addresses the question of how to make a cytoskeletal filament, like actin. The typical actin filament can be 1-10 µm long, which at 5 nm per subunit pair (in the double helix) 500 5,000 subunit pairs long. There are two requirements that we can specify for the association constants of the filaments. 1. The filament must be very stable to fragmentation. Consider fragmentation and reannealing as a reversible reaction, characterized by an association constant K A frag. In order to achieve the desired stability, K A frag must be on the order of M -1 (see box at end for details). 2. The filaments must be assembled from a pool of subunits near micromolar concentration. In the actual actin filaments this is achieved by having the K A for addition of subunits to the end ~ 2 x 10 5 M -1. Thus the actual actin filament has two association constants that are orders of magnitude apart. It is highly stable to fragmentation in the middle, yet subunits are very weakly attached at the ends and easily removed. 4a-1. Isodesmic assembly. Let s try first to see what we could achieve with a onesubunit thick filameent, in which subunits are connected by a single type of bond, all identical. This is called isodesmic assembly. Although the diagram K end A on the left indicates separate reactions for fragmentation and association onto the end, if all the bonds are identical, fragmentation and end assembly K end D must be identical. For isodesmic assembly K A frag = K A end = K A. K D frag K A frag It should be clear that the average filament length will be increased by increasing either K A or C t, the total concentration of subunits. In fact there is a simple formula (Romberg et al, 2001, JBC 276:11743) Nav = 1/2 (1 + (1 + 8K A C t ) 1/2 ) Where N av is the weight average number of subunits, K A is the association constant and C t is the total concentration of subunits. Let s assume that we have 10 µm subunits (actin), and see what average filament length is obtained for different values of K A.

18 K A = 10 6 N av = 5 (Moderate association (K D 10 times C t ) can t make isodesmic filaments). K A = 10 9 filaments) K A = filaments) N av = 45 (Strong association can make short to medium isodesmic N av = 5,000 (A super strong association can make long isodesmic These numbers suggest that we could indeed make filaments with the length and stability of actin by having a super strong K A = M -1. But if we did this, all subunits would be assembled into these stable filaments. There would be a miniscule pool of free subunits, and if this were sequestered it would take days for the stable filaments to disassemble. Thus there would be no rapid control over the assembly and disassembly of the filaments. In order to achieve filaments that are (1) highly stable to fragmentation, and (2) with subunits weakly attached at the ends (providing rapid assembly/disassembly controlled by the pool of monomers), we need a more complex filament. Actin achieves this by a two-stranded filament built from two types of bonds. Cooperativity is the basis for the solution. 4a-2. The amazing effect of cooperativity in the actin filament. If we look at the structure of the actin filament we see that it is actually two strands of subunits, which are connected within the strand by one type of bond (we will call it a longitudinal bond) and across the filament by a second type (we will call it a diagonal bond). Designate their K A and intrinsic bond energies as KA long = exp -1/RT (DG long Bond + 6) KA diag = exp -1/RT (DG diag Bond + 6) We don t know what these association constants are, because the dimers that would be formed by them are too weak to measure. But we can determine them indirectly from the association constants we do know, fragmentation and end association. The challenge now is to determine whether this two-stranded structure can permit us to build an actin filament with our specified parameters K A frag = M -1, and K A end = 2 x 10 5 M -1 by specifying reasonable values for K A long and K A diag

19 Consider first the association of a subunit onto the end, specifying our desired weak association K A end = 2 x 10 5 M -1 Note that this association involves forming one longitudinal and one diagonal bond Consider next the fragmentation reaction, specifying our desired super strong association K A frag = M -1 You will notice that fragmentation involves breaking two longitudinal bonds and one diagonal. The situation is now different from that of isodesmic assembly since fragmentation involves a different bonding from end dissociation. Specifically, taking a subunit off the end involves breaking one longitudinal and one diagonal bond, while fragmentation requires breaking these plus one additional longitudinal bond. The extra bond will enhance the association by cooperativity. But is it possible to find bond strengths that actually produce the 10 8 fold difference in these reactions? Your assignment is to make this work and find the strengths of the single bonds.

20 Problem set for arold Erickson structural biology A. Make an actin filament what are the bonds. 1. Find values of DG long Bond and DG diag Bond that will produce the specified values of K A frag and K A end. 2. What are the K A values for forming the two types of dimers, connected by a single longitdinal or diagonal bond? What does this mean relative to the ~µm cellular concentration of actin? 3. What is the lifetime of a subunit on the end of the filament, before it dissociates? 4. What is the lifetime of each internal position for fragmentation? If the actin filaments is 10,000 subunits long (5,000 pairs), what is the average time before the filament breaks somewhere? B. The reading 1. From the Delano paper, determine what is the efficiency (as calculated in Erickson s Table 1) of the Fc-III peptide binding to Fc? Interface area 650 Å 2 X 2 (this may be a problem. It is not REAL obvious from the paper that the 650 Å refers to one side only, need to multiply by 2 for the total let s see how it plays out) Max available hydrophobic bond energy K A experimental 1/25x10-9 M = 4 x 10 7 Intrinsic bond energy = RTlnK A 6 = / 32.5 = 51% They may use 7 for the entropic factor, that s ok. If they calculate using only 650 they will get 100% efficiency. Take something off for that and let them know that it is not that good. 2. Also from this paper, list two structural features that distinguished the favored binding patch on Fc from random patches of identical surface area? If you refer to solvent accessible surface fraction explain what this means. The favored interface had only 19% of its surface capable of hydrogen bonding, as opposed to 35% for the average interface.

21 It also had 49% of the solvent accessible surface fraction, as opposed to 35% for the average. This fraction is the fraction of the atom s surface area that is actually exposed, relative to the maximum possible (it is lower than the maximum because side chains are blocked by secondary and tertiary structure). ence in the favored interface side chains expose more of their surface area than in the average interface. Also, the polar (hydrogen bonding) sites are largely in clefts surrounded by hydrophobics. 3. List three structural features in which the G-Gbp hot spot differs from the neutral residues. The hot spot is generally better packed than the neutral area. The neutral area contains significantly more buried water molecules The hot spot is overall more hydrophobic, but with some buried -bonds. (If somebody comes up with something reasonable give them credit). Estimate of actin fragmentation - kinetics. If fragmentation/reannealing is a diffusion limited reaction, k 2 = 2x10 6 M -1 s -1. The value of K A frag was set equal to for several reasons (Erickson, 1989, Appendix 1), one of which is that it leads to a lifetime for breaking an internal bond of t F = 1/k -1 = 1/K D k 2 = 1/(10-13 )(2 x 10 6 ) = 5 x 10 7 s. Since the filament breaks whenever any pair of subunits breaks, we need to divide by 10,000 subunits to obtain the lifetime of the filament. The lifetime of the filament is 500 s. Cytoplasmic actin filaments are turning over on the order of minutes in the cell, so this should be appropriate. Obviously the actin filaments in muscle need to be further stabilized. So K Afrag = M -1 is the estimate from kinetics to produce reasonably stable actin filaments.

22 Ch 5. Kinetics of protein-protein interaction. A. Rigidity of protein-protein bond. ow much can you bend the protein subunits about the bond interface? Not very much. The diagram below shows the correctly docked interface on the left, and a bent one on the right. If the tilt pulls the subunits apart by even 1 Å, it will disrupt van der Waals and hydrogen bonds. Opening it any more would let water into the hydrophobic interface. It is a reasonable guess that this 1 Å split would be the maximum before the bond is disrupted. Across the 20 Å interface this would mean a 2.3 degree tilt. The take-home of this brief analysis is that protein subunits could rock back and forth by about 2 degrees and remain attached. Of course rotation about the plane of the interface would also be highly restricted, about 2 degrees. The protein-protein bond is quite rigid in all directions. 1 Å opening 2.3 deg tilt B. ow rigid are protein polymers? Like jello? No, much more rigid. Gittes and oward (J. Cell Biol , 1993) measured thermal fluctuations of actin and microtubules and reported that: The microtubule's Young's modulus is ~1.2 GPa, similar to Plexiglas and rigid plastics. This Young's modulus should apply to protein molecules in general and to their polymers - a very important concept. Proteins and their polymers are very rigid. C. Kinetics Let's approach kinetics from the point of a receptor binding its ligand (growth factor, GF). We assume that the receptors on the cell surface are present in very low concentration relative to the GF. There are two questions. 1. Lifetime of the empty receptor. If the receptor is empty, how long will it take to be occupied? This depends on the concentration of GF. In many cases the rate does not depend on the bond energy, but is limited simply by diffusion of GF. 2. Lifetime of the complex. Once a receptor has bound a GF, how long will the complex last before the GF dissociates? This is determined largely by the bond energy.

23 1. The lifetime of the empty receptor will be defined here as the reciprocal of the association rate: t E = 1/(k 2 [GF]). This is the average time it will take before an empty receptor is occupied (actually it is the time for 1/e of the receptors to be occupied, we will ignore this fine point). The lifetime of the empty receptor depends only on the concentration of GF and k 2. t E = 1/(k 2 [GF]) = 1/(( 2 x 10 6 )[GF]) The second order association rate, k 2, is a key parameter. In principal k 2 can vary considerably depending on the protein pairs, and would require experimental measurement for each case. In fact it turns out that a large number of protein-protein associations occur at the same rate, k 2 = 2 x 10 6 M -1 s -1. In the next section we will explain that this is the generic, diffusion limited rate constant for protein-protein interaction. There are some proteins that associate faster and some slower, so experimental measurements are important when they are available. But in the absence of experimental data, this is a very good guess. Assume: k 2 = 2 x 10 6 M -1 s -1 - for diffusion-limited protein-protein association (The basis for this assumption is discussed below.) t E = 1/(( 2 x 10 6 )[GF]) If the growth factor binding to its receptor is diffusion limited, here are some lifetimes to occupy a receptor for differenc concentrations of GF. [GF] t E GrFac 10-9 M 500 s 10-8 M 50 s (Actin) 10-6 M 0.5 s 2. The lifetime of the complex is t C = 1/k -1, the reciprocal of the rate of dissociation of GF from the receptor complex. t C is completely independent of the concentration of free GF but it depends very much on the bond energy (for associations with diffusion limited

24 k 2, K D is determined completly by k -1 ). It frequently happens that one does not know the kinetic constant, k -1, but the equilibrium binding constant, K D, is known. Now we can make a good guess, using the principle above that the association rate is diffusion limited, and equal to k 2 = 2 x 10 6 M -1 s -1, for many protein-protein associations. For these associations, the dissociation rate is determined directly by K D, and we can estimate k -1 directly from K D. t C = 1/k -1 = 1/K D k 2 = 1/K D (2 x 10 6 ). Again some examples, starting with the very strong complex of trypsin-trypsin inhibitor. K D t C Trp-TrIn x 10 6 s (58 days) GF-Rec s (8 min) actin s It is interesting now to consider the association-dissociation as a cyclic event. Consider a GF binding to its receptor. If the GF concentration is 10-9, equal to K D, the lifetime of the empty receptor is 500 s, and once a complex is formed its lifetime is also 500 s. This is another way of saying that when the concentration of P 1 is equal to K D, the receptor if 50% occupied. If the concentration of GF is now increased ten-fold to 10-8 M, the lifetime of the complex remains unchanged at 500 s, but the empty receptor now binds ligand in only 50 s. The receptor is ~90% occupied. 3. The diffusion-limited rate constant for protein-protein association. Koren and ammes (1976) surveyed a number of protein associations, and found that a large number of them had k 2 = x 10 6 M -1 s -1. To some in the field this rate seemed incredibly fast. The problem was a calculation by Smoluchowski (1917) that the diffusion-limited rate of encounter of smooth spheres was 7 x 10 9 M -1 s -1. According to this theory, if proteins behaved as smooth spheres, and they formed a complex every time they collided (without regard to orientation), they would form a complex at the rate k 2 = 7 x 10 9 M -1 s -1. This is very fast, but consider now how it will be affected by the extremely high steric specificity of the protein-protein bond. The correct bonding requires that each subunit be oriented rotationally to about 1/1000 of its possible rotations (this would correspond to a ± 1 Å rotation of a surface atom). A simple but incorrect argument would be that the kinetics of association should be slowed from the Smoluchowski limit by a factor of 1,000 for each subunit. That would mean that proteins should associate 1,000,000 times slower than the Smoluchowski rate. This would give a maximum, diffusion-limited rate of only 5 x 10 3 M -1 s -1. Actual proteins associate 1,000 times faster. ow? Northrup and Erickson (1992) resolved this question by using Brownian dynamics, a computer simulation treating the protein subunits as Brownian particles. It turns out the very slow rate, 5 x 10 3 M -1 s -1, would be appropriate if proteins behaved as spheres in a vacuum - where they would bounce apart and separate upon every