1 Protein expression and purification • I. The molecular principles for understanding proteins Jozef Hritz, Lubomír Janda, Blanka Pekárová and Radka Dopitová Kód předmětu: C8980 I. The molecular principles for understanding proteins – Jozef Hritz Theory can complement the experiment: •Theory explains experiments Properties of existing protein mutants, organic molecules • Simulation suggests Design of drugs, enzymes new experiments Stock market prices less experiments better chance of success knowledge new ideas For which problems is understanding/theory/computational simulations useful ? I. The molecular principles for understanding proteins – Jozef Hritz Example: Mutations within E2 protein of alpha viruses Gardner C.L.; Hritz J.; Sun C.; Song T.Y.; Rogers M.B.; Vanlandingham D.L.; Higgs S.; Ghedin E.; Klimstra W.B.; Ryman K.D. Deliberate Attenuation of Chikungunya Virus by Adaptation to Heparan Sulfate- Dependent Infectivity: A Model for Rational Arboviral Vaccine Design. PLoS Negl. Trop. Dis. 2014, 8, e2719 I. The molecular principles for understanding proteins – Jozef Hritz Theoretical models allow also understanding of situations where: -experiments are very expensive -experiments are almost impossible -possibility for fantasy. How the situation would look like if e.g. water is neutral … For which problems is understanding/theory/computational simulations useful ? The constituents of proteins, the amino acids General structure of an amino acid The zwitterionic form of an amino acid 1.2. The amino acids 5 I. The molecular principles for understanding proteins – Jozef Hritz 1.2.2. Clasification of the amino acids in terms of polarity Non-polar side chain Ala, Gly, Ile, Leu, Met, Phe, Pro, Trp, Val Polar, uncharged side chain Asn, Cys, Gln, Ser, Thr, Tyr Polar charged side chain Arg, Asp, Glu, His, Lys 1.2.1. The variety of amino acids 1.2. The amino acids 6 I. The molecular principles for understanding proteins – Jozef Hritz 1.2.3.2. Ionization 1.2. The amino acids 1.2.3. General properties of the amino acids Amino Acid Symbol pK1 (COOH) pK2 (NH2) pK R Group Glycine Gly 2,4 9,8 Alanine Ala 2,4 9,9 Valine Val 2,2 9,7 Leucine Leu 2,3 9,7 Isoleucine Ile 2,3 9,8 Serine Ser 2,2 9,2 Threonine Thr 2,1 9,1 Cysteine Cys 1,9 10,8 8,3 Methionine Met 2,1 9,3 Aspartic Acid Asp 2 9,9 3,9 Glutamic Acid Glu 2,1 9,5 4,1 Asparagine Asn 2,1 8,8 Glutamine Gln 2,2 9,1 Arginine Arg 1,8 9 12,5 Lysine Lys 2,2 9,2 10,8 Histidine His 1,8 9,2 6 Phenylalanine Phe 2,2 9,2 Tyrosine Tyr 2,2 9,1 10,1 Tryptophan Trp 2,4 9,4 Proline Pro 2 10,6 7 http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/proteins.htm I. The molecular principles for understanding proteins – Jozef Hritz pH= log10 1 aH+ æ è ç ç ö ø ÷ ÷ @ log10 1 H+éë ùû æ è ç ç ö ø ÷ ÷ 1.3.1. The peptide bond 1.3. The primary structure of proteins Resonance stabilization of the peptide bond 8 I. The molecular principles for understanding proteins – Jozef Hritz I. The molecular principles for understanding proteins – Jozef Hritz Partial double bond character of peptide bond 1.3.1. The peptide bond 1.3. The primary structure of proteins The cis and trans forms of the peptide bond Structure of a tetrapeptide showing the alphacarbon atoms, side chains, and N and C-termini. 10 I. The molecular principles for understanding proteins – Jozef Hritz I. The molecular principles for understanding proteins – Jozef Hritz 2.8 Å2.8 Å ω ω I. The molecular principles for understanding proteins – Jozef Hritz Combination rules for Lennard-Jones potential sij = sii +s jj 2 eij = eiiejj Eij LJ = 4eij sij rij æ è çç ö ø ÷÷ 12 - sij rij æ è çç ö ø ÷÷ 6é ë ê ê ù û ú ú Calculate LJ potential energy between Cα..Cα and Cα..O at the distance 2.8 Å I. The molecular principles for understanding proteins – Jozef Hritz 2.8 Å2.8 Å ω ω I. The molecular principles for understanding proteins – Jozef Hritz What is general principle for stable – equilibrium state? 1 2 Condition for equilibrium state Ep->min What is the potential energy of a 1kg potato in the height 1m above the ground? 15 Constant Value Avogadro's Number NA = 6.022 × 1023 mol-1 Faraday Constant F = 96 485.33 C mol-1 Molar Gas Constant R = 8.314 J mol-1 K-1 Coulomb's Constant ke = 8.987 × 109 N m2 C-2 Speed of Light (Vacuum) c = 299 792 458 m s-1 Boltzmann Constant kb = 1.38 × 10-23 J K-1 Charge on a Proton/Electron e = 1.602 × 10-19 C Standard acceleration of gravity g = 9.8 m s-2 Planck's Constant h = 6.6 × 10-34 m2 kg / s Useful Physical/Chemical Constants Ep = mgh I. The molecular principles for understanding proteins – Jozef Hritz The object is molecule of nitrogen (N2) 10 points Question 1: What is the potential energy [in Joules] of an object in the height 10000 m above see level. The potential energy at the see level is considered to be zero. Molecular mass of nitrogen atom is 14 Da 8 points J = kg m2 s-2 6 points The weight of 1 mol of N2 molecules is 28 grams 4 points Please solve a problem. 16 I. The molecular principles for understanding proteins – Jozef Hritz Ep = mgh g = 9.8 m s-2 h = 10000 m m = 28 g / 6.022 x 1023 = 4.6 x 10-23 g = 4.6 x 10-26 kg Ep = 4.6 x 10-26 kg 9.8 m s-2 10000 m = 45 x 10-22 kg m2 s-2 = 45 x 10-22 J Solution: 17 I. The molecular principles for understanding proteins – Jozef Hritz I. The molecular principles for understanding proteins – Jozef Hritz Why molecules of nitrogen do not fall to the ground if Ep->min? Boltzmann formula If the molecule can adopt two microstates A and B then the ration of their probabilities is: pB pA = e EB-EA( ) kT e – Euler number for T = 300K T*kb = 300 K * 1.38 × 10-23 J K-1 = 4.14 * 10-21 J How many times there is smaller probability to find molecule of nitrogen in the height 10000 m with respect to the ground level? p10km p0 = e - 4.5*10-21J 4.14*10-21J = 0.34 I. The molecular principles for understanding proteins – Jozef Hritz Sometimes it is simpler to calculate amounts per mol R*T = 300 K * 8.314 J mol-1 K-1 ≅ 2500 J mol-1 = 2.5 kJ mol-1 Ep = mgh = 0.028 kg * 9.8 m s-2 * 10000 m = 2744 J mol-1 p10km p0 = e - 2744Jmol-1 2500Jmol-1 = 0.33 pB pA = e EB-EA( ) RT I. The molecular principles for understanding proteins – Jozef Hritz What energy difference leads to the Boltzmann probability ratio of 0.01? I.e. it is large energy difference with respect to the thermal energy pB pA = e EB-EA( ) RT ln pB pA æ è ç ö ø ÷= EB -EA( ) RT -RTln pB pA æ è ç ö ø ÷= EB -EA( ) ln10 = 2.3 ln100 = ln102 = 2ln10 = 4.6 ln0.01= -ln100 = -4.6 for pB pA æ è ç ö ø ÷= 0.01 EB -EA( )= -2500J.mol-1 *(-4.6) =11500J.mol-1 =11.5kJ.mol-1 I. The molecular principles for understanding proteins – Jozef Hritz What energy difference leads to the Boltzmann probability ratio of 0.99? I.e. it is small energy difference with respect to the thermal energy -RTln pB pA æ è ç ö ø ÷= EB -EA( ) ln0.99 = -0.01 for pB pA æ è ç ö ø ÷= 0.99 EB -EA( )= -2500J.mol-1 *(-0.01) = 25J.mol-1 = 0.025kJ.mol-1 I. The molecular principles for understanding proteins – Jozef Hritz Pro All, but Pro: 20 kJ*mol-1 5 kJ*mol-1 cis trans The energy difference between cis and trans conformation of proline is Ecis - Etrans = 5 kJ.mol-1. Temperature is 27 oC. For simplicity: pcis+ptrans = 1 10 points Question 2: What is the probability of finding proline aminoacid in cis conformation? t = 27 oC corresponds to the T= 300 K 8 points pcis = pcis / (pcis+ptrans) = 1 / (1 + ptrans / pcis) 6 points (ptrans / pcis) = (pcis / ptrans)-1 4 point Please solve a problem. 23 I. The molecular principles for understanding proteins – Jozef Hritz Solution: 24 I. The molecular principles for understanding proteins – Jozef Hritz I. The molecular principles for understanding proteins – Jozef Hritz Pro All, but Pro: 20 kJ*mol-1 5 kJ*mol-1 cis trans ~10% of prolines are in cis conformations. Very slow conversion between trans and cis conformations due to high energy barrier is often rate limiting step in protein folding kinetics. Zoldak, G.; Aumüller, T.; Lucke, C.; Hritz, J.; Oostenbrink, C; Fischer, G.; Schmid, F.X. A library of fluorescent peptides for exploring the substrate specificities of prolyl isomerases. Biochemistry 2009, 48, 10423-10436 Pro All, but Pro: 20 kJ*mol-1 5 kJ*mol-1 cis trans 26 Proteins are composed from aminoacids with different physico-chemical properties Individual aminoacids are connected by peptide bond Two possible conformations around peptide bond: trans and cis. Proline is the only aminoacid where cis conformation is non-neglible (~12%) Trans and cis conformations are separated by quite high energy bariier that in case of prolines can be lowered by prolyl isomerases 1.4. The secondary structure of proteins 27 I. The molecular principles for understanding proteins – Jozef Hritz 1.4. The secondary structure of proteins I. The molecular principles for understanding proteins – Jozef Hritz 1.4. The secondary structure of proteins I. The molecular principles for understanding proteins – Jozef Hritz Energetically unfavorable conformational areas 1.4. The secondary structure of proteins I. The molecular principles for understanding proteins – Jozef Hritz Additional energetically unfavorable conformational areas because of the presence of carbonyl oxygen Glycine – less sterically restricted because of lacking the side-chain 1.4. The secondary structure of proteins I. The molecular principles for understanding proteins – Jozef Hritz Additional energetically unfavorable conformational areas because of the presence of Cβ Alanine