Outline of first two lectures: size scales- biochemical events depend on the structure of large molecules and assemblies of molecules weak interactions- the structures depend on weak interactions (ie weaker than those usually considered in organic chemistry) ionic equilibria thermodynamics in biology- 2nd law vs order in biological systems

Range of Object Sizes of interest to Biochemists and Techniques used to study them

eg. waterOH overall  ii ii

Units of  (Coulomb-meter) or Debye (D) 1 Debye = 3.34 X Coulomb-meter 2. Ion-dipole interaction: consider charge Q at a distance r from the center of a polar molecule with dipole moment, , subtending an angle  to the line joining the two molecules Q r  q+ q–  U(r) = – ( Q  cos  ) /  r 2 Consider cos  as an orientation term Important to note that U is inversely proportional to r 2 where for charge-charge interactions it was inversely proportional to r. Simplified example: Na + ion near a water molecule (m = 1.85 D) requires 96 kJ/mole at 300 o K to pull them apart

3. Dipole-dipole interaction:   2 r 12 11 22 U(r) = 1389  1  2 (cos  cos  1 cos  2 ) / r 12 3 where:  12 = angle between  1  and  2 dipole-dipole interaction energy depends on the inverse cube of the distance between the two dipoles and their orientations consider the following dipole orientations: repulsion attraction no interaction attractionrepulsion

The magnitudes of dipole moments can be substantial. HCl1.04 D Urea4.56 D Peptide bond3.70 D Peptide bond dipole moment is parallel to N-H bond (N is q-) C i-1 N C C i O R i H Total  for amino acid:glycine (R i is H, simplest)16.7 D glycylglycine28.6 D for large protein:hemoglobinhundreds QUESTION? for the same r, which would have the largest interaction energy? A. a charge-charge interaction B. an ion-dipole interaction C. a dipole-dipole interaction NEXT, even if molecules have no permanent dipole moments, there are forces between them! (induced dipoles)

Non covalent Interaction Energy of two approaching particles Van der Waals radii Energy of Interaction

One last important non covalent interaction : Hydrogen bonds- stabilize and specify the structures of proteins and DNA determine the structure of liquid water an H atom that interacts simultaneously with two other atoms is said to form H bond takes place between: an acidic hydrogen H atom that is covalently bound to a donor group; like -O-H or N-H depends upon electronegativity of donor and a pair of non-bonded electrons on an acceptor group; like O=C- or N B+H-AB …… H-A length of H-bond is nearly the same in all species approximately 0.3 nm defined as the distance between the center of the H- donor atom (A) to the center of the acceptor atom (B) H-bonds are directional the H-donor atom (A) tends to point directly at acceptor e- pair on B maximum stability when B …… H-A are co-linear bent H-bonds have reduced stability

H-bonds are extremely important in dictating protein structure. Both  -helical and  -sheet structures are stabilized by intramolecular H-bonds. The nature of H-bonds is electrostatic -attraction of one proton to two nuclei provides an efficient path for moving charges -the proton can easily move between nuclei O-H …. OO …. H-O

Hydrogen bonding in ice and in water a: ice, space filling; b: ice, skeletal c: water, simulation

Hydration of ions in solution As salt dissolves, non-covalent interaction between ions and water produces a hydration shell. The energy gained helps overcome ion-ion interactions that stabilize the crystal.

Amphipathic molecules are both hydrophobic and hydrophilic Hydrophobic: the molecular property of being unable to engage in attractive interactions with water molecules. Hydrophobic substances are nonionic and nonpolar; they are nonwettable and do not readily dissolve in water. Hydrophilic: the ability of an atom or a molecule to engage in attractive interactions with water molecules. Substances that are ionic or can engage in hydrogen bonding are hydrophilic. Hydrophilic substances are either soluble in water or, at least, wettable.

SUMMARY: Several weak interactions are present within or between biomolecules. The energy sums up to an impressive total. The energy is minimized for a particular conformation. One central problem of modern Biochemistry is to predict what the 3-D structure of a protein will be given its primary sequence. SUMMARY Typical magnitudes of Interaction Energies covalent kJ/mole H-bonds25 ion-ion ion-dipole15 dipole-dipole0.3-2 dispersion2

Water’s tendency to form H-bonds makes it unique causing boiling point, melting point, heat of vaporization to be anomalously high Each water molecule can be H-bond donor and acceptor simultaneously Dipolar nature can reduce effective electrostatic force between two interacting ions: the orientation of water dipoles between the two charges acts as counterfield Hydrophilic molecules tend to form H-bonds with water and readily dissolve Ions in aqueous solutions become hydrated as water forms hydration shells around them (energetically favorable, energy is released) Hydrophobic molecules have limited solubility in water clathrate structures form the ‘caged’ water structure is ordered which decreases the entropy (randomness) of the mixture hydrophobic molecules tend to aggregate in water with a single “cage” surrounding them Important Properties of Water Amphipathic molecules tend to form ordered structures in water: monolayers, micelles, bilayers

Ionic Equilibria Acids and Bases (proton donors and acceptors) REVIEW: strong acids- essentially complete dissociation HCl H + + Cl - strong base- essentially complete ionization to yield OH - ions; proton acceptors weak acids and bases - important in biochemistry, do not completely ionize at physiological pH, getpartial dissociation with equilibrium between weak acid and conjugate base Weak AcidConjugate base + proton H 3 PO 4 H 2 PO 4 – + H + phosphoric acidDihydrogen phosphate ion H 2 PO 4 – HPO 4 2– + H + monohydrogen phosphate ion HPO 4 2– PO 4 3– + H + phosphate ion the stronger the acid, the weaker the conjugate base (the conjugate base does not have a strong tendency to accept a proton and reform the acid) -which of the above is the strongest acid?

water has a slight tendency to ionize it can act as acid and base H 2 O + H 2 O H 3 O + + OH – often see written as: H 2 O H + + OH – but proton never exists as free ion in solution, it is always associated with other water molecules equilibrium can be expressed as ion product of water: K w = [H + ][OH – ] = 1 x 10 –14 M 2 [H + ] and [OH - ] do not vary independently If [H + ] is high, [OH - ] must be ______? For pure water at 25° C [H + ] = [OH – ] = 1 x 10 –7 (neutral) to simplify, define pH = – log [H + ] for neutral solution: pH = –log (1 x 10 –7 ) = 7 physiological pH range = Equilibria involving water and pH

Weak Acid and Base Equilibria Many weak acids are found in biological systems, for eg: catalytic proteins have ionizable side chains whose state of ionization is dependent on pH (catalytic activity occurs only at a certain pH) to describe weak acid strength consider: dissociation constant, K a, (equilibruim constant) HAH + + A – where K a = [products]/[reactants] = [H + ] [A – ] / [HA] (large K a, greater tendency to dissociate, stronger acid) also definepK a = – log K a small pK a, stronger acid H 3 PO 4 H 2 PO 4 – + H + pK a = 2.14 H 2 PO 4 – HPO H + pK a = 6.86 HPO 4 2– PO 4 3– + H + pK a = 12.4 Environment influences pK a values: -hydration of proton favors dissociation -electrostatic attraction between proton and conjugate base opposes dissociation -identical groups in different local environments (different regions in a protein) will dissociate to differing degrees -consider high dielectric constant vs. low dielectric constant media

Buffers and the Henderson-Hasselbalch Equation -many biological processes generate or use H + - the pH of the medium would change dramatically if it were not controlled (leading to unwanted effects) --biological reactions occur in a buffered medium where pH changes slightly upon addition of acid or base -most biologically relevant experiments are run in buffers how do buffered solutions maintain pH under varying conditions? to calculate the pH of a solution when acid/base ratio of weak acid is varied: Henderson-Hasselbalch equation comes from: K a = [H + ] [A – ] / [HA] take (– log) of each side and rearrange, yields: pH = pK a + log ( [A – ] / [HA] ) some examples using HH equation: what is the pH of a buffer that contains the following? 1 M acetic acid and 0.5 M sodium acetate

Titration example (similar one in text:) Consider the titration of a 2 M formic acid solution with NaOH. 1. What is the pH of a 2 M formic acid solution? use K a = [H + ] [A – ] / [HA] HCOOH H + + HCOO – let x = [H + ] = [HCOO – ] thenK a = 1.78 x 10 –4 = x 2 / (2 – x) for an exact answer, need the quadratic equation but since formic acid is a weak acid (Ka is small), x <<< [HCOOH] and equation becomes K a = 1.78 x 10 –4 = x 2 / 2 so x = [H + ] = [HCOO – ] = and pH = Now start the titration. As NaOH is added, what happens? NaOH is a strong base --- completely dissociates OH – is in equilibrium with H +, K w = [H + ] [OH – ] = 10 –14, K w is a very small number so virtually all [OH – ] added reacts with [H + ] to form water

Titration continued: - to satisfy the equilibrium relationship given by K a K a = [H + ] [HCOO – ] / [HCOOH] = 1.78 x more HCOOH dissociates to replace the reacted [H + ] and -applying HH, see that [HCOO – ] / [HCOOH] will increase pH = pK a + log ( [HCOO – ] / [HCOOH] ) -leading to a slow increase in pH as the titration proceeds _______________________________________________ consider midpoint of titration where half of the HCOOH has been neutralized by the NaOH [HCOO – ] / [HCOOH] = 1 HH becomes: pH = pK a + log 1 = pK a = 3.75 for HCOOH Titration curve: - within 1 pH unit of pK a over most of curve - so pK a defines the range where buffering capacity is maximum - curve is reversible

Simple problem: -have one liter of a weak acid (pK a = 5.00) at 0.1 M -measure the initial pH of the solution, pH = so it follows that initially, [A – ] = [HA] where pH = pK a -add 100mL of 0.1M NaOH, following occurs HA + OH – = A – + H 2 O 0.01moles -so, 0.01 moles of HA reacted and new [HA] = 0.1 – 0.01 = 0.09 new [A – ] = use HH to get new pH = 5 + log (0.11 / 0.09) = _______________________________________________ now consider,100mL of 0.1 M NaOH added to 1 L without the weak acid to see how well the weak acid buffers 0.01 moles OH – / 1.1L = 9.09 x = [OH – ] use K w = [OH – ] [H + ] = 1 x to get pH = _______________________________________________ what happens when 0.1 moles of base have been added? what happens when the next 1 mL of base is added? Known as overrunning the buffer

Sample Buffer Calculation (in text) -want to study a reaction at pH so to prevent the pH from drifting during the reaction, use weak acid with pK a close to formic acid (3.75) -can use a solution of weak acid and its conjugate base -ratio of formate ion to formic acid required can be calculated from the Henderson - Hasselbalch equation: 4.00 = log [HCOO – ] / [HCOOH] [HCOO – ] / [HCOOH] = = so can make a formate buffer at pH 4.0 by using equal volumes of 0.1 M formic acid and M sodium formate -Alternatively, exactly the same solution could be prepared by titrating a 0.1 M solution of formic acid to pH 4.00 with sodium hydroxide. _______________________________________________ some buffer systems controlling biological pH: 1. dihydrogen phosphate-monohydrogen phosphate pKa = involved in intracellular pH control where phosphate is abundant 2. carbonic acid-bicarbonate pKa = 6.37, blood pH control 3. Protein amino acid side chains with pKa near 7.0

Example of an ampholyte - molecule with both acidic and basic groups glycine: pH 1NH 3 + – CH 2 – COOH net charge +1 pH 6NH 3 + – CH 2 – COO – net charge 0 zwitterion pH 14 NH 2 – CH 2 – COO – net charge –1 pKa values carboxylate group2.3 amino group9.6 can serve as good buffer in 2 different pH ranges ______________________________________________ use glycine to define an important property isoelectric point (pI) - pH at which an ampholyte or polyampholyte has a net charge of zero. for glycine, pI is where: [NH 3 + – CH 2 – COOH] = [NH 2 – CH 2 – COO – ] can calculate pI by applying HH to both ionizing groups and summing (see text) yields: pI = {pK COOH + pK NH 3+ } / 2 = { } / 2 = 5.95 pI is the simple average for two ionizable groups

polyampholytes are molecules that have more than 2 ionizable groups lysineNH C- (CH 2 ) 4 - NH 3 + COOH titration of lysine shows 3 pKa’s: pH<2, exists in above form first pKa = 2.18, loss of carboxyl proton second at pH = 8.9 third at pH = need model compounds to decide which amino group loses a proton first _____________________________________________ to determine pI experimentally use electrophoresis (see end of Chapter 2) 1. Gel electrophoresis-electric field is applied to solution of ions, positively charged ions migrate to cathode and negatively charged to anode, at it’s pI an ampholyte does not move because net charge = zero 2. Isoelectric focusing- charged species move through a pH gradient, each resting at it’s own isoelectric point _____________________________________________ Macromolecules with multiples of either only negatively or only positively charged groups are called polyelectrolytes polylysine is a weak polyelectrolyte - pKa of each group influenced by ionization state of other groups

Solubility of macroions (polyelectrolytes and polyampholytes, including nucleic acids and proteins) depends on pH. For polyampholytes: high or low pH leads to greater solubility (due to – or + charges on proteins, respectively) At the isoelectric pH although net charge is zero, there are + and – charges and precipitation occurs due to: - charge-charge intermolecular interaction - van der Waals interaction to minimize the electrostatic interaction, small ions (salts) are added to serve as counterions, they screen the macroions from one another Ionic Strength = I = ½  (M i Z i 2 ) (sum over all small ions)M is molarity Z is charge Consider the following 2 processes that can take place for protein solutions: 1. Salting in: increasing ionic strength up to a point (relatively low I ), proteins go into solution 2. Salting out: at high salt, water that would normally solvate the protein goes to solvate the ions and protein solubility decreases. Most experiments use buffers with NaCl or KCl