Introduction to Biochemistry

Aim: to understand life on a molecular level

3 principle areas: structure/function relationships metabolism-chemical reactions genetics

Biochemistry is interdisciplinary and is moreexperimental than theoretical

biochemist needs to understand: techniques based on physics basic chemical elements & structure of

biological compounds including:-stoichiometry-mechanisms of reactions-thermodynamics

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

Many important biological molecules arepolymers : joining of prefabricated monomers monomers of a given polymer have limited

diversity monomers polymerized by identical

mechanism to form covalent bonds

monomer polymerglucose cellulose-homopolymer

Polysaccharides serve asstructural component & storeenergy

nucleotides nucleic acids-heteropolymers(4 types) RNA & DNA- storage &

transmission of genetic info

amino acids proteins or peptides-hetero(20) Greater structural & functional

diversityLipids -chemically diverse group

-low solubility in aqueous solution andamphipathic nature results in formation ofparticular structures-major structural element in cell membranes

Range of Object Sizes of interest to Biochemistsand Techniques used to study them

Covalent vs Non-covalent bonds

some covalent bond energies (kJ/mole):

C-H 415C-C 345C C 615C C 810C-O 350O2 499O-H 463

This is the energy required to break one mole ofchemical bonds of a particular type in the gas state.

In many cases, it is not very dependent on themolecule.

Conjugated double bonds are an exception. Thesebonds are more stable than those moleculescontaining isolated C-C and C C bonds.

Example: For benzene, it takes ~ 158 kJ/mole moreenergy than expected to dissociate due to resonanceenergy which stabilizes the molecule.

Non covalent interactions ~ 10-100 times weaker than covalent essential that they can be broken and reformed consider Energy available in a typical biological

system (aqueous environment) to break/reformnon covalent interactions

RT = the thermal energy of the “bath”surrounding any molecule;

R = ideal gas constant (8.314 J/mol K)T = 273 + 37 = 310 K

~ 2.5 kJ/mole

(not usually enough to break covalent bonds)

THUS, biochemical function has evolved to centeraround “weak” or non covalent interactions

Weak interactions are fundamentally electrostaticin nature:

depend on the forces that electrical charges ordipoles exert on one another

Non covalent interactions

consider Energy available in a typical biologicalsystem (aqueous environment) to break/reform

Weak interactions are fundamentally electrostaticin nature:

Classification of Non Covalent Interactions1. Charge-charge (ion-ion)

Coulomb’s Law: F = k q1 q2 / r2 In vacuum

q = magnitude and sign of charger = distance between charges k = proportionality constant (depends on units)

if charges are repulsive (++ or - -), F is positiveif charges are attractive (+ -), F is negative

in some medium : F = k q1 q2 / r2

= dielectric constant of the medium reflects the effect the medium has on separating

the charges the medium shields the charges from each other

Relevant ’s: vacuum 1Hydrocarbon (organic solvent) 1-10Water 80

NOTE:Force between 2 charges in non-polar environment; for example, inside a biological membrane or inside aprotein (low ), can be much larger than in water

we are more concerned with changes in energy thatoccur during interactions, integration yields...

energy of interaction: U(r) = k q1 q2 / r

the energy required to separate 2 charged particlesuntil r reaches infinity

when working in common units:U(r) = 1389 q1 q2 / r (kJ/mole)

For: q in integral multiples of the electronic charge; r in angstroms

eg. two electrons in vacuum separated by 1 have: U = + 1389 kJ/mole (repulsion)

Moving on from: 1. charge-charge interactionsto: 2. ion-dipoleand: 3. dipole-dipole

what do we mean by dipole?molecules that do not carry a net charge and havean asymmetric internal distribution of charge arecalled polar molecules and have a permanentdipole moment ()

eg. water O

H H

dipole moment- the product of the magnitude ofthe charge and the distanceseparating the charges

µ = qx where q = fractional (partial) charge x = distance between thecharges (q- and q+) in themolecule

eg. Carbon monoxide q+ q-C O

is a vector, in your book directed towards thepartial + charge (most books directed to - charge)for larger molecules, have to sum all individual µ’sthrough-out the molecule to get the overall dipolemomentµx = qi xi (sum of all in x-direction)

similarly for y and z components

overall

ii

Units of (Coulomb-meter) or Debye (D)

1 Debye = 3.34 X 10-30 Coulomb-meter

2. Ion-dipole interaction:consider charge Q at a distance r from the centerof a polar molecule with dipole moment, ,subtending an angle to the line joining the twomolecules

Q+ -----------------------------------------------r

q+

q–

U(r) = – ( Q cos ) / r2

•Consider cos as an orientation term•Important to note that U is inversely proportional to r2

where for charge-charge interactions it was inversely proportional to r.

Simplified example:Na+ ion near a water molecule (m = 1.85 D) requires96 kJ/mole at 300o K to pull them apart

3. Dipole-dipole interaction:

1---------------------------------2

r12

12

U(r) = 1389 12 (cos 12 - 3cos1cos2) / r123

where: 12 = angle between 1and2

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

nointeraction

attraction repulsion

The magnitudes of dipole moments can be substantial.

HCl 1.04 DUrea 4.56 DPeptide bond 3.70 D

Peptide bond dipole momentis parallel to N-H bond(N is q-)

C i-1 N

C Ci

O Ri

HTotal

for amino acid: glycine (Ri is H, simplest) 16.7 Dglycylglycine 28.6 D

for large protein:hemoglobin hundreds

QUESTION?for the same r, which would have the largest interactionenergy? A. a charge-charge interaction

B. an ion-dipole interactionC. a dipole-dipole interaction

NEXT,even if molecules have no permanent dipole moments,there are forces between them! (induced dipoles)

So far we have seen weak interactions that areelectrostatic in nature, now we will consider

Polarization interactions (also weak)- arising fromdipole moments induced in atoms and molecules bythe electric fields of nearby charges or permanentdipoles

4. Charge-induced dipole : proportional to 1/r4

5. Dipole-induced dipole : proportional to 1/r5

these are even shorter range than permanent dipoleinteractions

other non covalent interactions include:6. Dispersion forces (London forces)- -all atoms attract each other-results from instantaneous fluctuations incharge distributions -depends on the polarizability of the molecule (iethe deformability of the electron cloud)

U(r) = A/r6 where A > 0

-this force occurs between all atoms. It is small butis summed over all species in a structure.

7. Steric repulsion (van der Waals repulsion)-all atoms repel at short distances-occurs when outer electronic orbitals overlap

U(r) = + B/r12 where B > 0

Non covalent Interaction Energyof two approaching particles

Van der Waalsradii

Ener

gy o

f Int

erac

tion

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-A B ……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

anefficient path for moving charges-the proton can easily move between nuclei

O-H …. O O …. H-O

Hydrogen bonding in ice and in water

a: ice, space filling;b: ice, skeletalc: 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 200-800 kJ/mole

•H-bonds 25

•ion-ion 20-250

•ion-dipole 15

•dipole-dipole 0.3-2

•dispersion 2

•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 EquilibriaAcids 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, get partial dissociation with equilibrium between weak acid and conjugate base

Weak Acid Conjugate base + proton

H3PO4 H2PO4– + H+

phosphoric acid Dihydrogen phosphate ion

H2PO4– HPO4

2– + H+

monohydrogen phosphate ion

HPO4 2– PO4 3– + H+

phosphate ionthe 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

H2O + H2O H3O+ + OH–

often see written as:

H2O 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:

Kw = [H+][OH–] = 1 x 10 –14 M2

[H+] and [OH-] do not vary independentlyIf [H+] is high, [OH-] must be ______?

For pure water at 25° C [H+] = [OH–] = 1 x 10 –7 (neutral)

to simplify, definepH = – log [H+]

for neutral solution: pH = –log (1 x 10 –7) = 7

physiological pH range = 6.5 - 8.0

Equilibria involving water and pH

Weak Acid and Base EquilibriaMany 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, Ka, (equilibruim constant)

HA H+ + A–

where Ka = [products]/[reactants] = [H+] [A–] / [HA] (large Ka , greater tendency to dissociate, stronger acid)

also define pKa = – log Ka

small pKa , stronger acid

H3PO4 H2PO4– + H+ pKa = 2.14

H2PO4– HPO4

2 + H+ pKa = 6.86

HPO4 2– PO4 3– + H+ pKa = 12.4

Environment influences pKa 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: Ka = [H+] [A–] / [HA]

take (– log) of each side and rearrange, yields:

pH = pKa + 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 Ka = [H+] [A–] / [HA]

HCOOH H+ + HCOO–

let x = [H+] = [HCOO–]

then Ka = 1.78 x 10 –4 = x2 / (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 Ka = 1.78 x 10 –4 = x2 / 2

so x = [H+] = [HCOO–] = 0.019 and pH = 1.7

2. Now start the titration. As NaOH is added, what happens?•NaOH is a strong base --- completely dissociates

•OH– is in equilibrium with H+ , Kw = [H+] [OH–] = 10–14 ,

•Kw 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 Ka

Ka = [H+] [HCOO–] / [HCOOH] = 1.78 x 10 -4

more HCOOH dissociates to replace the reacted [H+] and

-applying HH, see that [HCOO–] / [HCOOH] will increase

pH = pKa + 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 = pKa + log 1 = pKa = 3.75 for HCOOHTitration curve:- within 1 pH unit of pKa over most of curve

- so pKa defines the range where buffering capacity is maximum

- curve is reversible

Simple problem:-have one liter of a weak acid (pKa = 5.00) at 0.1 M-measure the initial pH of the solution, pH = 5.00-so it follows that initially,

[A–] = [HA] where pH = pKa

-add 100mL of 0.1M NaOH, following occurs

HA + OH– = A– + H2O 0.01moles

-so, 0.01 moles of HA reacted andnew [HA] = 0.1 – 0.01 = 0.09

new [A–] = 0.11

-use HH to get new pH = 5 + log (0.11 / 0.09) = 5.087_______________________________________________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 10 -3 = [OH– ]

use Kw = [OH–] [H+] = 1 x 10 -14

to get pH = 11.96_______________________________________________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 4.00-so to prevent the pH from drifting during the reaction, use weak acid with pKa close to 4.00 -- 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 = 3.75 + log [HCOO–] / [HCOOH]

[HCOO–] / [HCOOH] = 10 0.25 = 1.78

-so can make a formate buffer at pH 4.0 by using equal volumes of 0.1 M formic acid and 0.178 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 = 6.86 - 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 1 NH3+ – CH2 – COOH net charge +1

pH 6 NH3+ – CH2 – COO– net charge 0

zwitterion

pH 14 NH2 – CH2 – COO– net charge –1

pKa valuescarboxylate group 2.3amino group 9.6

can serve as good buffer in 2 different pH ranges______________________________________________use glycine to define an important propertyisoelectric point (pI) - pH at which an ampholyte or polyampholyte has a net charge of zero.

for glycine, pI is where:[NH3

+ – CH2 – COOH] = [NH2 – CH2 – COO– ]

can calculate pI by applying HH to both ionizing groupsand summing (see text) yields:

pI = {pK COOH + pK NH 3+ } / 2 = {2.3 + 9.6} / 2 = 5.95

pI is the simple average for two ionizable groups

polyampholytes are molecules that have more than 2 ionizable groups

lysine NH3+- C- (CH2)4 - NH3

+

COOHtitration 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 = 10.28•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 (Mi Zi2)

(sum over all small ions) M is molarityZ 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 solution2. 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