PETER PAZMANY SEMMELWEIS UNIVERSITYCATHOLIC UNIVERSITY Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework** Consortium leader PETER PAZMANY CATHOLIC UNIVERSITY Consortium members SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund *** **Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben * * * A p ro j e kt a z Eu ró p a i U n i ó t á m o g a t á sá v a l , a z E u r ó p a i S z o c i á l i s A l a p t á rsf i n a n sz í ro z á sá v a l v a l ó su l m e g . INTRODUCTION TO BIOPHYSICS (Bevezetés a biofizikába) MULTIPLE EQUILIBRIA (Többszörös egyensúlyok) GYÖRFFY DÁNIEL, ZÁVODSZKY PÉTER . Multiple equilibria occur when small moleculesbind to large molecules with multiple bindingsites, such as hormones to receptors, substrates to enzymes, antigens to antibodies, or oxygen to haemoglobin Law of mass action . Multiple equilibria are based on the law of mass action . Let us consider a reaction Ka plusP.A . P -A where P is a macromolecule, say a protein withone binding site, A is an unbound ligand, PA is acomplex of the protein and the ligand and Ka is the association constant Ligand binding of a protein with one binding site . The law of mass action for this reaction is [ P -A] Ka = [ P ][ A] where [P], [A] and [P-A] are the concentration ofthe protein, the unbound ligand and the complex, respectively . Let us introduce a new quantity, ., as the ratio of the number of moles of bound A and the total number of moles of protein P, that is .= [ P -A] [ P -A].[ P ] . We can combine the equations for . and Ka to eliminate all quantities with a P in them . This is because [P-A] and [P] always appear asratios K =[ P -A] a [ P ][ A] and after rearrangement [ P -A ] K a [ A]= [ P ] . For . .= [ P -A] [ P ].[ P -A] and dividing by [P] both the numerator and denominator on the right hand side we get .= [ P -A]/[ P ] 1.[ P -A ]/[ P ] . So K [ A].= 1. Ka [ A] a . This means that the amount of bound ligandper protein molecule depends only on theconcentration of the ligand and not on theconcentration of protein . Let us note, however, that it is the unbound concentration of A which is important Introduction to biophysics: Multiple equilibria Proteins with several binding sites . In most cases, proteins have several sites for binding . Now, let us consider proteins having severalidentical subunits . We also assume that these are independentbinding sites, i.e. site 1 does not sensewhether site 2 is occupied Protein with four subunits and binding sites . Let . denote again the ratio of the totalnumber of moles of the bound ligand and thetotal number of moles of the protein . The total number of moles of bound ligand isthe sum of the number of moles of ligandbound to the different binding sites [P1 -A][P 4 -A] .=.1......4 =..... [ P ].[ P -4·A ][ P ].[ P -4·A] where [ P -4·A]=[ P1 -A]=[P2 -A]=[P3 -A]=[P4 -A] 13 [ Pi -A] Ka [ A]= [ P ] where i=1,2 ,3,4 we get K [ A] K [ A] 4 K [ A] a aa .= .....= 1. Ka [ A] 1. Ka [ A] 1.Ka [ A] . In general for a protein with n equivalent, independent binding sites nK [ A] .= a 1. K[ A] a . Let us note that the amount of A bound to P is dependent only on n, Ka and the concentration of unbound A, [A] . Now the question arises how we can evaluate n and Ka . We can get these if we measure how much A isbound to P . Let us consider an equilibrium dialysis experiment Dialysis experiment . Place a solution of the protein, P, in the bagand the bag in a solution of the ligand, A . A can pass through the pores of the bag but P cannot . What do we know with precision? – The mass of the protein, mP, and that of the ligand, mA – The molecular weight of the protein, MP, and that of the ligand, MA – The number of moles of the protein, nP=mp/Mp, and that of the ligand, nA=mA/MA . Let us assume that [ A ]inside =[ A]outside . Now we only need to measure [A]out . We can do this by spectrophotometry or radioactive labelling . Based on these measurements, we can calculate . molesof bound [ A] total A-free A n A -[ A ]V .= == total molesof P total P nP where V is the total volume of the solution Scatchard plot . The Scatchard plot is a plot helping us toobtain n and K a . Let us set out from the expression nK [ A].= 1. Ka [ A] a . Performing rearrangements we obtain ..1.K a [ A ].= n K a [ A ] so ... K a [ A ]= n K a [ A ] and .=n K a [ A ]-. K a [ A ] Introduction to biophysics: Multiple equilibria . And finally we get the linear relationship . =nK a-. Ka [ A] . Let us plot ./[A] against . . The slope of the curve will be -Ka . The . intercept is n: at ./[ A]=0, nKa =n . so n=. . Let us consider an example Scatchard plot . In this example: -slope= Ka =106 M-1 and n=4 Introduction to biophysics: Multiple equilibria . But we encounter a problem: – Scatchard plots are never linear in reality . There are four reasons for the non-linearity ofScatchard plots: – Binding site heterogeneity: more than one class ofbinding sites with different Ka 's – Donnan potential: if A is charged, [A].[A]. Most in out biological molecules are charged – Debye-Hückel effect: if P and A are charged, thenwhen A binds to P there is an electrostatic interaction in addition to the normal binding site – Binding site cooperativity: binding sites are not independent Binding site heterogeneity . Let us suppose that there are two classes ofbinding sites: – Class 1 with Ka1, n1 – Class 2 with Ka2, n2 . Let us consider a protein with two classes ofbinding site, each having four actual sites Binding site heterogeneity . Class 2 binding sites might be adventitious binding sites which frequently exhibit non­specific weak binding . These class 2 sites affect the total binding . If the binding sites are still independent, then .=.1..2 ......8 n1 K a1 [ A ] n2 K a2 [ A]= 1.K a1 [ A]. 1.K a2 [ A] . Now, let us plot ./[A] vs. . again Scatchard plot for binding site heterogeneity . The initial slope is -n1 K a1 2 .n2 K a2 2 initial slope= n1 K a1.n2 K a2 . If Ka1>>Ka2, then slope.-K a1 . But if Ka1 is only one order of magnitude greater than Ka2, this will give a 10% error . The final slope is -.n1 .n2 . K a1 K a2 final slope= n1 K a1.n2 K a2 . If Ka1>>Ka2, then -.n1.n2 . slope. K a2 n1 . The first intercept is .n1 K a1.n2 K a2 .2 first intercept= 2n1 K a1 2 .n2 K a2 . If Ka1>>Ka2, then intercept.n1 . The second intercept is second intercept =n1 .n2 where n1 and n2 must be integers . Thus we have four unknowns but we also have four equations . Thus, we can calculate n1, n2, Ka1 and Ka2 Donnan effects . Let us consider the system shown below . The protein on the left-hand side releases fiveNa+ ions: P.Na5 . P- 5.5 Na. Donnan effect Concentrations of ion species . Let us recall that we added NaCl to side B to avoid generating a large Donnan potential .c. c- .A =.c. c- .B so .5a .x.. x .=. b- x.2 . Thus x is from this equation b2 x= 5a .2b . Thus the Donnan potential is RT c. A RT .5a. x . ..= ln = ln zF c. BzF b- x . How does this Donnan potential influence thebinding of small molecule A which has twopositive charges? . Since A2. [ ]«[ Na.] to a first approximation, we can neglect thecontribution of A2+ to the Donnan potential . But A2+ still feels the Donnan potential. . If we ignored the Donnan potential, it wouldappear that the protein binds one A2+ molecule although it does not . We can set the pH of the solution to theisoelectric point of protein to avoid this effect . Let us consider the case when the protein is uncharged (is at its isoelectric point) Ka int P0 .Az . P -Az where z is the charge on A . There is no electrostatic interaction so K is a int the intrinsic association constant 0 .Gint =-RT ln K a int . Now, let us compare the case when the proteinis charged Ka obs PZ .Az . P -A where Z and z are the charge on the protein andon A, respectively . is the observed association constant Ka obs which includes electrostatic interactions 0 .Gobs=-RT K aobs . The difference between these standard free energies is equal to the electrical work necessary to bring the charges together .Gobs 0 -.Gint 0 =..r . zeN where r is the radius of the protein . Substituting the expression with theassociation constants into the equation above, we get K aobs RT ln =-..r. zeN K a int . Thus after rearrangement -zeN ..r ./ RT K aobs = K aint e Ze e-. r ..r.= Dr RT so it is plausible to write the equation above inthe following form -2 wZ z K aobs = K aint e where Ne2e-. r w = Z Dr RT . For the binding of a small molecule A withcharge z to a protein P with charge Z number of moles of bound A .= totalnumberofmolesof protein P may be written in terms of an observedassociation coefficient Ka obs . is affected by the charge interactions and Ka obs may be expressed in terms of the intrinsicassociation constant K a int . The term -2 wZz e has to do with bringing the charged molecule tothe protein surface from infinity . Let us consider the following example P .4 A2. so the charge, z, on A is 2 . If we begin at the isoelectric point, Z =0, initially K aobs = K aint for the first A2+ . However, the second A2+ to bind will see a charge of +2 on the protein . How does this affect Ka obs . Let us note that this was calculated using avalue of w=0.1 (w is always on the order of0.1) . It is dimensionless and varies with the proteinradius and the ionic strength Ne2e-. r w = Z Dr RT and from the Debye-Hückel theory 8 . Ne2 I .2 = 1000 Dk BT . Let us note that Ka obs decreases by more than an order of magnitude from K in this example a int . Even beginning at the isoelectric point, the Scatchard plot is badly curved if we do not take the Debye-Hückel theory into account . Let us plot ./[A] as a function of . ./[A] vs. . . When we are binding more than one chargedmolecule to a protein we get a curved plot . Instead: -2 wZz .= nK aint e[ A] -2 wZ z 1. Ka int e[ A] . From this equation we get .-2 wZz =nK aint- K aint . [ A] e . Now, let us plot . -2 wZz [ A] e as a function of . to get a linear relationship ./[A]e-2wZz vs. . . This type of analysis may be applied to proteintitration, that is the binding of protons toproteins . Protons are small charged particles, andproteins have specific binding sites for them . These are the acidic and basic side chains of amino acids . They can be classified by their chemical character and characterized by their pK a int . All negative charges on the surface of a protein are due to ionized acidic groups . All positive charges on the protein surface aredue to ionized basic groups . The titration curve for a protein can begenerated from the amino acid composition . Deviations from this behaviour: – When the protein has a large positive or negativecharge (due to gain or loss of protons) it begins tobind anions or cations from the solution – At very low or very high pH, the protein denaturesexposing buried groups and changing the radius ofthe protein, thus changing w Cooperativity . Let us suppose that the binding sites are notindependent of each other, that is there iscommunication between binding sites . As an example, let us compare haemoglobinand myoglobin Comparison of haemoglobin and myoglobin Comparison of saturation curves of myoglobin and haemoglobin . Its simple to explain the saturation curve for myoglobin nK a [O2 ] .= 1. K[O2 ] a . In the case of myoglobin n=1 . Now, let us switch to the use of partialpressure instead of molar concentration of oxygen pO2 =. [O2 ] where ß is a constant . So K 'pO2 .= a 1. Ka' pO2 where K a'= Ka /. . The experimental curve fits the calculatedcurve for myoglobin almost perfectly . For haemoglobin, things do not work this well . We could try lots of different equations of theform n ni K ai [O2 ] .=. i =11.K ai [O2 ] but this will always give a curve with decreasingslope, it will not give the sigmoidal curve . This can also not be due to a Donnan or Debye-Hückel effect because O2 is not charged . Something new is going on at molecular level –this is the cooperativity . How does cooperativity work at molecular level? . Let us consider two kinds of subunits, . and ß . Binding sites on . subunits are stronger . O2 binds to an . subunit which induces a conformational change and thus an increasedaffinity of ß to O2 (positive cooperativity) . As a consequence, it allows haemoglobin to release O2 at higher pO2 than myoglobin . Haemoglobin dumps O2 into the tissues over a very narrow range of pO2 . This keeps the pO2 more constant throughout the body . Now, let us devise a model for extreme positive cooperativity . Let us consider a protein with four subunitswith four hidden binding sites . A high concentration of A is necessary to bindthe first O2 to the first site Positive cooperativity . This model predicts only two kinds of largemolecule – With no O2 is bound – With 4 O2's are bound . There is a negligible amount of the forms 1,2 and 3 O2's bound . The model predicts that we will have onlyextremes of saturation: 0% and 100% Ka P .4 A. P -A4 where [ P -A4 ]Ka =4 [ P ][ A] . The saturation is molesof bound A 4 [ P -A4 ] .= moles of total P =[ P ].[ P -A4 ] and after substitutions and rearrangement 4 K [ A]4 .= a 4 1. K[ A] a . This equation gives an S-shaped curve . Let us recall that for independent binding sites 4 K [ A] .= a 1. K[ A] a . What if we do not have extreme positive cooperativity? . More realistically, we cannot neglect the other partially saturated species . There is an intermediate case Hill plot . The Hill equation is nK [ A]x .= ax1. K[ A] a where n is the number of binding sites and x is the cooperativity and 1. x.n . If x=n an extreme cooperativity exists . If x=1 there is no cooperativity, so the binding sites areindependent of each other . Hill did not prove this equation, he just showedthat it works pretty well . Subsequently, it has been shown why it workswell . Now, we would like to know how we get x . Let us take the Hill equation and solve it for k[A]x, then take the logarithm of both sides k [ A]x =. n-. . log . .=log k .x log [ A ] n-. . Hill plot is a plot of . log .. n-. vs. log [ A] Hill plot . For independent binding sites, we get a slopeof 1 . For extreme positive cooperativity, we get aslope of n (in this case, 4) . For intermediate positive cooperativity, the plot is straight in the centre with a slope of x . It tapers off at the ends to a slope of 1 . The cooperativity, x, is the slope of the curveat the centre (1886-1977)