Suite H. HYDROPHOBICITY - People 10-10-11.pdf · Suite H. HYDROPHOBICITY INTRODUCTION. The...

© P.S.Phillips October 10, 2011 EXPERIMENT H:1

Suite H. HYDROPHOBICITYINTRODUCTION. The hydrophobic effect accounts for the behavior of non-ionic species in water. Polar species dissolve in water because they can hydrogen bond to water (enthalpically favorable) and then dispersal (entropically favorable). This is balanced against the breaking of water’s hydrogen bonds (enthalpically unfavorable), the break up of the lattice of the polar compound (if solid) and the loss of entropy caused by the binding of water to solute (solvation). For small polar molecules ∆G is favorable, but for larger molecules, where the fraction of polar functional groups is often small, the molecules are not so soluble. If the fraction of polar groups is high (e.g. sugars and some proteins) the molecule will usually be soluble. From these observations, one may deduce that the hydrocarbon chains are responsible for the lack of solubility: they do not hydrogen bond and the entropy of mixing is insufficient to overcome the disruption of hydrogen bonding. In fact, this is not true. Although hydrocarbons do not H-bond, the van der Waals forces are quite large and they should be moderately soluble. This is where the hydrophobic effect comes in.

Water cannot H-bond to hydrocarbons (note that it just cannot bond, it is not repelled ), but it can with itself. To make sure there are no dangling (i.e. unused) H-bonds, which is energetically unfavorable, the water molecules have to take on a geometry (on average) to minimize this. It’s a little difficult to envision, but basically the normal and H-bonds (which are interchangeable in water) organize

themselves so that they form a hollow polygon (see the figure). The hydrocarbon is located in the cavity. This shell is labile, but under high pressures, clathrates can form. The best known is methane clathrate, a white crystalline solid

found on the bottom of the deep ocean. It has the amusing property of being flammable. The net effect of hydrophobicity is that water becomes highly structured in the vicinity of non-polar molecules so ∆S becomes negative, and since ∆H is small in the absence of polar groups, ∆G becomes negative and non-polar species do not dissolve.

This has a major impact on the structure of proteins which

have hydrophilic (polar) and hydrophobic (non-polar) regions. They must fold in such a way that the hydrophilic sections are inside and the polar sections are on the outside. It is possible to change the nature of hydrogen bonding in water thereby reducing the hydrophobic effect allowing the protein to unfold (denature).

We will do two experiments here. One where we study we study the solubility of n-butanol and n-pentanol in water as a function of temperature. And the second the where we study the solubility of toluene (a generic non-polar species) in water in the presence of species that change H-bonding in water (co-solvents).

Part 1. FREE ENERGY OF TRANSFER. THEORY. For a two phase system, (here we have water and alcohol) the chemical potential of any give species must be the same if the two phases are in equilibrium (see the appendix for details). That is

1 2phase phaseA Am m=

but (see appendix)

ln for any species in solution.oA A ART Xm m» +

If we consider the transfer (partitioning) of a hydrocarbon, HC, into water, W, then

lno o HCHC W

W

XRT

Xm m- =-

or, if the hydrocarbon is only slightly soluble (but that’s not the case here) then

Wln as otransfer HC HCG RT X X XD =- (1)

∆Gtransfer is maximum when the solution is saturated so all we need do is determine the saturation point (and its concentration) as a function of temperature and we can get ∆So, ∆Ho and ∆Go for the process.

( .)

( .)ln HC sato

transferW sat

XG RT

XD =- (2)

and

/ and

(1/ )

o oo otransfer transfer

P P

G G TS H

T T

ö ö¶D ¶D÷ ÷÷ ÷=D =D÷ ÷÷ ÷¶ ¶÷ ÷ø ø (3)

EXPERIMENT H:2 © P.S.Phillips October 10, 2011

PROCEDURE. RTFM! Rather than titrate in alcohol to saturation (which takes too long because of the equilibrium times required) we will prepare a series of samples and observe which ones are cloudy (saturated and two phases) and which ones are clear (unsaturated and one phase).

The usual way to proceed is to prepare a series of mixtures solutions in the supplied tubes, say, as follows (label them!) : 0.30 to 11.10g of butanol in 0.40g steps in 10.00g of e-pure water. 0.40-4.00g of n-pentanol in 0.20g steps again in 10.00g of e-pure water. As usual, the amount doesn’t have to be exact, just accurately know. (This is roughly equivalent to doing a titration in 0.2mL steps.) Shake the tubes thoroughly and put them in the rack in the shaker or other bath as directed bath. We then repeat the process with n-pentanol as described below.

Start at 25C and work your way up to 65C in 5C increments. Also, make up an ice-water bath for a 0C point. Ask the prof. about temperatures between 0 and 25 C. Wait at least 15min at each temperature for equilibrium to be achieved. Be sure to vigorously shake the sample a few times. Using the ultrasonic cleaner helps as well. To find the saturation point, look for a pair of adjacent (in mole fraction) tubes where one is cloudy and one is clear. If the cloudy tube is only faintly so (and little other evidence of a second phase), then take that is the saturation point. If it’s fairly cloudy then average the mole fraction for the two tubes. This means your maximum error in X is half the increment.

Aggghhh! He’s torturing us; death by a million samples. No… I just outlined it because it makes things a little clearer. We are going to use a really neat trick to find the cloud point, which goes as follows: the method above is called a linear search and unfortunately requires the preparation of many samples and limits the accuracy to about 3% for 19 samples, i.e. 5% composition intervals. The search time is of order of the number of samples, O(#samples). However, it is possible to get and accuracy of <1% with nine or less samples in a similar time. This is achieved by use the semi-numerical method of a bisection search - search time is O(log #samples).

Bisection is the division of a given curve, figure, or interval into halves. A simple bisection procedure for iteratively converging on a solution which is known to lie inside some interval [a,b] proceeds by evaluating the function in question at the midpoint of the original interval i.e. at m=(a+b)/2 and testing to see in which of the subintervals [a, m] and [m,b] solution lies. The procedure is then repeated with the appropriate subinterval as often as

needed to locate the solution to the desired accuracy. The method is very robust, if there is a solution, this method will find it with an absolute error of, at most |b-a|/2n after n steps. However, if there is more than one solution, it will only find one of them without more information.

Note that are not many restrictions on the nature of a, b and x. a and b just need to be distinguishable (e.g. by a sign change). Nor do they need to be continuous, a sorted list will suffice, in which case it’s called a binary search. This method one of many similar search algorithms. In this case it’s a variant of the divide and conquer algorithm. Interested students can look at Wikipedia, NIST or the Wolfram site, all of which have fairly accessible material.

In our case a and b are the concentrations of our bracketing samples, our “equation solution” is the opalescence concentration and the bracketing test is “one phase” or “two phase”. Ideally we want “”cloudy” (the cloud point). In practice it will be cloudy with signs of two phases. We know what the 0% and 100% samples look like so firstly, you would normally make up a roughly 10% and 90% solution and shake the sample. This defines the direction of the search (the sign change): if the 90% solution separates then the search direction is low to high, if the 10% solution separates the direction is reversed. Weighing on a taring balance is efficient and avoids problems with volume changes due to non-ideality. We can then convert weight % to mol% (or molality).

This is a really elegant approach to the problem, but there is one difficulty – what if there is more than one cloud point! Clearly that can occur, we can have a dilute solution of alcohol in water or a dilute solution of water in alcohol. However, we are interested in studying the hydrophobic, interaction. That is, how the alcohol disrupts the water, so we are only interested in solutions with small mole% of alcohol. It’s interesting to consider how many cloud points one might observe with a two component system at constant pressure.

So prepare a 10% and 50% sample, equilibrate it at the desired temperature for 15 min then check it. Shake it in an ultrasonic bath to ensure complete dissolution. If the 50% sample is a two phase and the 10% solution one phase then you know the opalescence point lies between 50% and 10%. You then bisect the concentrations; make up a [(50+10)/2]% = 30% solution and equilibrate. If that’s clear then you bisect again to between[ (50+30)/2]% = 40% and so on to an interval of 1% or less. If the 50% and the 10% are both two phase then the cloud point is between 0 and 10% and you bisect accordingly. Record the flanking concentrations, with a note of the degree of separation in the


cloudy one, and take the average. If a sample is only faintly cloudy (two phases not immediately obvious), you can take that as the transition point.

Note that the bisection intervals need not be exact as long as there are no gaps in the search. You should keep each sample, as they may be reusable at another temperature.

The whole process requires only seven samples or less to reach a 1% absolute accuracy as opposed to 3% accuracy mentioned for the linear search method.

As initially described, you still need to vary the temperature. Make up a beaker with ice and water in it. That’s your 0 C bath. Start the heated bath at 25C and work your way up to about 70C in 7C increments. Wait at least 15 min for each sample to equilibrate (to save time you could make up the next two bracketing samples needed, then only use the one you need). Shake the tubes vigorously and regularly. Watch for pressure build up in the tubes at the higher temperatures. Crack the tops open to release the pressure occasionally. For the samples close to the cloud point sonicate them for five minutes then return them to the bath.

Do pentanol/water and butanol/water systems in parallel.

CALCULATIONS. 1) Calculate ∆So, ∆Ho and ∆Go for transfer, and their errors, using equations (2) and (3). 2) The two alcohols differ by one methylene group. Calculate the hydrophobicity increment (∆G difference) for hydrocarbon chains. See if butanol and pentanol are 4 and 5 times this value, respectively. 3) Compare your data with the literature values and comment.

QUESTIONS. 1) There are many reasons to store solutions of biomolecules in the cold. Is hydrophobicity one of them? 2) We could use ∆Go= ∆Ho-T∆So to get, why didn’t we ? (Hint look at the error). 3) At this point, you know what the search direction is. Make some arguments to show that you could predict it for similar solvents. 4) Can you think of other measures one could use for the composition?

Part 2. THE EFFECT OF CO-SOLVENTS INTRODUCTION. Here we look at the transfer of toluene (our model for the hydrophobic core of a protein) into aqueous solutions of “co-solvent” by UV spectroscopy. We will use two protein co-solvents; guanidine chloride – which denatures (increase solubility of) proteins, and

sodium chloride which crystallizes (decreases solubility of) proteins.

A co-solvent will denature a protein if it decreases the hydrophobic effect, thus allowing the hydrophobic core of the protein to be exposed with a reduced entropy penalty. On the other hand, if we use sodium chloride the hydrophobic effect is increased so solubility will be decreased and the protein will precipitate out.

The exact mechanism by which hydrophobicity is changed is unclear, but we just wish to demonstrate the effect.

THEORY. As before, except we want the change in chemical potential (for saturated solutions) of toluene, A, into pure water, W, vs. toluene into co-solvent, S.

, , ,lnoA W A W A WRT Xm m= +

, , ,ln .oA S A S A SRT Xm m= +

,, ,

,ln A Wo

transfer A W A SA S

XG RT

Xm mD = - =-

It’s more convenient to use the concentration scale here. Since toluene is only slightly soluble and it’s a constant volume system, we get

,

,ln A Wo

transferA S

cG RT

cD =-

We can get the concentration, cS, of toluene in the aqueous phase from the *p p® transition of toluene at 268nm. cS, is just the concentration or pure water. So we get

lno Wtransfer

S

AG RT

AD =-

where AW is the absorbance in pure water and AS that in the co-solvent.

PROCEDURE. This experiment should be done concurrently with part 2 or 3. Make up saturated solutions of toluene (just shake one mL with the solution) in 0-6M guanidine chloride and in 0-5M NaCl, both in 1M increments. Equilibrate the solutions at 25C for at least an hour. Shake the solutions every 15min (or use the shaker bath). Zapping it with the ultrasonic cleaner a few times my help, but don’t overdo it or you’ll disperse some of the toluene into the aqueous phase. Record the UV spectrum of the aqueous phase from 260-280nm. Use matched quartz cells with toluene free solutions as the blank.


CALCULATIONS. Using the absorbance data, calculate otransferGD for each concentration and plot it vs.

concentration.

QUESTIONS. Suppose each sodium ion has a solvation shell of six waters – we will call this “bound “water. How much “free” water is there in a 5M NaCl solution (ignore the chloride ion)? Now consider a 1mM solution of a model protein of molar mass 100kDa (say poly glycine). Would there be enough water to hydrogen bond this molecule completely (assume 2 H-bonding sites per base). Is it possible that we the effect is not due to hydrophobicity changes, but a loss of H-bonding?

Part 3. PARTITIONING INTRODUCTION. In the last two sections we looked at the mutual solubility of two liquids. Here we will look at how a solute distributes between two immiscible solvents, and the effect of pH.

You’ve all heard the expression “oil and water” don’t mix. To be more specific, oil and water are immiscible and form a two phase system. The reason for this is the hydrophobic effect, discussed above.

You’ve also heard the phrase “like dissolves like”. Again, to be more specific, polar materials dissolve in polar solvents and non-polar materials dissolve in non-polar solvents. There are of course many exceptions, acetone dissolves freely in hexane and water, although these two solvents are immiscible. This then begs the question, what happens if you mix acetone, water and hexane? The answer is that the acetone will dissolve in both; it will partition itself between the two solvents. The degree of partitioning will relate to the polarity of the solute and the relative polarity of the two solvents. This apparently mundane observation is rather important: we exploit it in the chemistry laboratory (and industrially) to extract the non-polar species from aqueous solution (by shaking with a non-polar solvent and decanting the non-polar solvent off). Similarly, we can extract polar species using water. In environmental science it’s important to know how various chemicals (a.k.a. pollutants) distribute themselves between water (highly polar), mud (polar) and fish (partly non-polar). In pharmacy and biochemistry it’s important to know how a drug or bioactive species distributes itself between the cell membranes or fat (both non-polar) and water (90% of most mammals).

Here we will explore the partition coefficient of a simple species, an acid-base indicator, between water (the most important polar solvent) and octanol (the canonical non-polar species – although in food science olive oil is usually

used). We will also examine the effect of pH on the partition coefficient.

THEORY. The partition coefficient for a solute X, is defined as

octanol

water

[X]

[X]owK =

So Kow tends to be large for non-polar species. Kow is, of course, an equilibrium constant, albeit for a physical process, rather than a reaction. The corresponding free energy

lnoowG RT KD =-

is just the free energy change when one mole of X is transferred from water to octanol.

One widely overlooked fact is that, for acids and bases (and amphiphiles), Kow depends on pH. For instance, acetic acid (CH3COOH) is quite soluble in non polar solvents. It’s also soluble in water because of H-bonding and also because the conjugate base (CH3COO-) is very polar. The degree of dissociation depends on the pH. For a sufficiently high pH, acetic acid will cease to partition into the non-polar phase because it’s completely dissociated. This means that partition coefficients for organic acids and bases vary with pH. The tabulated values (which are for arbitrary concentrations of the acid or base) are completely useless for environmental or biochemical work where the aqueous phase is nearly always buffered to near neutral.

The system is a parallel (as opposed to sequential) equilibrium:

HA( ) H ( ) A ( )

HA( )

a

ow

Kaq aq aq

K

oct

+ −←→ +

↑↓

which is described by two equations

[H ][A ]

[HA]aK+ +

=

and

octanol

water

[HA]

[HA]owK =

as before, where X is now HA. It’s easy to see that

octanol[HA]

[H ][A ]a

owK

K + +=

This can be rearranged, as shown later, in terms of the mass of the acid used.


BASIC PROCEDURE. This experiment should be done concurrently with part 1 or 2. The simplest way to measure partition coefficients is by mixing the material with water, shaking the solution with octanol, then titrating the aqueous layer to find how much is left (or the octanol layer to find out how much crossed). However, this cannot be easily done as a function of pH as buffers (which are often organic acids) interfere with the titration. Here we will use UV-Vis spectroscopy to measure the concentrations. This is a simple experiment so some of the experimental details will be left for you to work out.

There are basically three steps: a) Preparation of stock solutions. b) Preparation of calibration standards. c) The partitioning experiment itself.

Make sure all flasks and their caps are clean and dry. Note that octanol is quite oily, it will pipette slowly. It is also quite smelly, not too unpleasant, but persistent. You should wear gloves and work in the fume hood.

Stock Solutions. Weigh out accurately about 20mg of bromophenol blue (an acid-base indicator) and make up a 200mL aqueous indicator stock. Take 10mL of this stock and dilute with pH 7 buffer to 100mL (10µg/mL). Take another 10mL and dilute to 100mL with pH 4 buffer (prepare as indicated on the bottle). Also, make up a similar stock in pH 2 buffer. Also, do pH 5 if directed to do so. These will be your working solutions.

Calibration Solutions. You need to prepare a range of indicator solutions in water, from the stock indicator solution. About (but accurately known) solutions of 0, 2, 4, 6, 8 µg/mL should do. I’ll leave you to work out the details (but use a one stage dilution). Once you have the standards, run their UV-Vis spectra in the range 320-700nm to get the calibration curve (discussed later). Check the absorbances are below 2 units. Save the spectra on a USB drive as directed.

Partition run. Take the 100mL of your working solution (pH 7, 4, 5, and 2) and shake it with 5mL of octanol in a separatory funnel. (Remember to safely vent it. If you haven’t used one before get the instructor to show you.) Let the solution settle and drain off the aqueous layer into a beaker (record it’s pH if directed to do so. This serves as a check as some of the buffer may partition as well.) Next filter the solutions with the 45µm syringe filter into a 1cm cuvette and take their UV spectra. Use the corresponding original buffer as a blank).

Interpreting the UV spectra. The indicator is the sodium salt of an organic acid (pKa 3.85) and so exists in two states;

the neutral state, HI, absorbing at 440nm (coloured yellow; denoted by subscript HI) and its conjugate base, I-, absorbing at 580nm (coloured purple; denoted by subscript I). To get a calibration curve simply plot amplitude of the 580nm peak vs. concentration (in µg/mL). You may need to do some baseline corrections to get the correct amplitudes. The concentrations of the test samples can then be just read off this graph.

CALCULATIONS. We are studying a distribution between octanol (denoted by subscript, o) and water (denoted by subscript, w). We can define the partition coefficient many ways, however it’s only useful to define them in terms of the same species or for total concentrations (Ctotal). As charged species (i.e. I-) do not partition into non-polar solvents we can try:

1 2 o o

w w

HI HIow ow

HI total

C CK K

C C= =

2owK is the usual definition of partition coefficient, but is pH dependant as and

w wHI IC C are pH dependant. On the other hand

1owK should be pH independent (why?) so is, perhaps, more relevant. In the literature you will see the symbol owD , but there is some confusion as to which of the two K’s this represents.

The calibration is done at pH 7 where only the charged species is present so we can get the extinction coefficient for that species and hence the concentration of I- in the aqueous phase of all the samples.

wI I IA Ce= l

where the cell length, l, is 1cm.

In the aqueous phase the concentration of HI and I- are related by the Henderson-HasselBach equation (see the appendix) so

log w

w

Ia

HI

CpH pK

C= +

As the molar mass of HI and I- differ by only one, so we can write

where m is the mass of the species in the sample. It’s convenient to define

w wI HIR m m= so

- logapH pK R=

The mass of the indicator can be determined by

log w

w

Ia

HI

mpH pK

m= +


conservation of mass:

w w oHI I HI totalm m m m+ + =

so (1 1/ )o wHI total Im m m R= - +

Hence (since C=m/V)

1o

w

HI oow

HI w

m VK

m V=

(1 )w

total w

I o

Rm VR

m V

æ ö÷ç ÷ç= - + ÷ç ÷ç ÷çè ø

similarly

2

11 1wI w

owtotal o

m VK

m R V

æ öæ ö÷ç ÷ç ÷= - +ç ÷ç ÷ç ÷ç ÷ç è øè ø

We have and wtotal Im m from the spectra, so we can use

these equations to get 2owK and

1owK at the three pH’s.

Note that there is an interplay between wIm and R that

can result in –ve partition coefficients if these numbers are just a fraction off. This is easily resolved by noting that the result is not –ve, but zero within experimental error.

DISCUSSION. 1) Tabulate your results. Are the partition coefficients consistent with the spectra and the Henderson-Hasselbach equation? 2) Explain in terms of the UV spectra why you see the colors that you do.

QUESTIONS. 1) Synthetic membranes can be made with bilayers of phosphatidyl choline (see fig.1). Would you expect the following species to be able to penetrate the membrane.; water, any anion, alcohol, oxygen. Briefly explain why.

Figure 1. Schematic of a cross-section of part of a bilayer in aqueous solution. The sphere is the polar head. The tails are non-polar chains.

2) How would expect the following compounds to partition between octanol and water (i.e. Kow<1, Kow>1 or other). Briefly explain your answers. Phthalic acid, mercury, ethanol, sodium chloride.

3) How do detergents affect partitioning (see Exp.M). 4) Name one or more instrumental methods that are based on partitioning and very briefly describe them.

APPENDIX. The Henderson-Hasselbach equation relates the concentration ratio of an acid and it’s conjugate base in a solution of given pH. Consider an acid HI;

HI(aq)]H+(aq)+I-(aq)

My preference is to write H+(aq) rather than the hydronium ion. H3O+. (or 25H O+ which is probably closer to the truth) ) so for a dilute solution

+ -[H ][I ][HI]aK =

but pKa = -log Ka and pH = -log [H+]

therefore

-[I ]

pH = pK + log[HI]a

This equation can be used to determine the extent of dissociation of an acid in a buffered solution. Why such a simply derived equation is named after two people, is a mystery. You may wish to delve into the literature to try and discover why.

APPENDIX. Pressure Dependence of Free Energy. We have, by definition, and using the chain rule

G H TS

dG dH TdS SdT

º -\ = - -

but H U PV

dH dU PdV VdP

º +\ = + +

so dG dU PdV Vdp TdS SdT= + + - -

but U q wº +

For a reversible change in a closed system of constant composition (no reactions), with no non-expansion work, this becomes

revU q PdV= -

but since revq TdS= , substituting back into the expression for dG we get


dG TdS PdV PdV VdP Tds SdT= - + + - -

hence we get dG VdP SdT= - (1)

so at constant T T

GV

T

ö¶ ÷ =÷÷ø¶

Integrating between P1 and P2 assuming P scales inversely with V (e.g. PV=nRT)

22 1

1( ) ( ) ln

PG P G P RT

P

æ ö÷ç ÷= + ç ÷ç ÷çè ø (2)

Solids and liquids are incompressible so the volume change with pressure is tiny, so the change is negligible (1-2 J), but for gases it’s quite large.

If we measure changes in G with respect to some reference state, they become ∆G’s. To simplify things further, we usually make P1 the reference state, so P1 become Po and ∆G(P1) becomes ∆Go , so we get for some pressure P

( ) lnoo

PG P G RT

P

æ ö÷çD =D + ÷ç ÷÷çè ø (3)

Chemical Potential. Gibbs Free Energy is a function of composition

1 2( , ,... )mn n n , temperature, T and pressure, P, i.e.

1 2( , , , ,... )mG f T P n n n=

V is considered a function of P and concentration is function of V and n so these two variables are not needed. Using the slope rule, we get for m components

, , ,... , , ,...1 2 1 2 , , 1

...... (4)P n n T n n T P all nk i

m

iii

G G GdG dT dP dn

T P n¹=

öö ö¶ ¶ ¶ ÷÷ ÷ ÷= + +÷ ÷ ÷÷ ÷ ÷ø ø¶ ¶ ¶ ÷øå

We define the last term as the chemical potential, µ, the variation of G with composition i.e.

, , k i

ii T P all n

G

nm

¹

ö¶ ÷÷º ÷÷¶ ø

so (4) becomes

, , ,... , , ,...1 2 1 2 1

.......P n n T n n

m

i ii

G GdG dT dP dn

T Pm

=

ö ö¶ ¶÷ ÷= + +÷ ÷÷ ÷ø ø¶ ¶ å

Also, for 1 mol, µ is just G so (2) can be written

( ) lnoo

PP RT

Pm m

æ ö÷ç= + ÷ç ÷÷çè ø (5)

It’s a little difficult to see why we would introduce chemical

potential; after all, we can only measure changes in it, which, at constant T and P, is ∆G. There are a couple of reasons, but the simplest is that it easier to talk about. Saying ∆Gvap is +ve is the same as saying the chemical potential of a liquid is lower than the vapor. The latter is somehow clearer and more intuitive (things roll downhill). Another example is, that at equilibrium the chemical potential of multiple phases have the same chemical potential. That statement captures the situation more easily than some description using free energy. It also allows us to introduce non-ideality more easily (chemical potential is also known as the partial molar free energy). Lastly, (4) is the not whole equation, there are other terms we won’t discuss here.

Solutions. The chemical potential of a substance is the same throughout a sample at equilibrium, regardless of how many phases are present. This is very important because it allows us to say something about complicated systems. In particular, for a liquid/vapor system, if we know the chemical potential of the vapor, we know the chemical potential of the liquid. This is important because we can calculate (or measure) a lot about gases, even non-ideal ones, but we know very little about liquids and, currently, can calculate diddly-squat about them, but if we have the chemical potential of the vapor, we know everything we need to know (thermodynamically) about the liquid.

Let’s consider a pure (denoted by a superscript *) solvent, A, in equilibrium with its vapor in a closed system (no air present), then

* *( ) ( )A Avapor liquidm m=

Since the liquid is pure, and if we have one mole, it’s in its standard state (the pressure is not one bar, it’s whatever its vapor pressure is, but we have shown elsewhere that this effect is negligible for liquids and solids so we can use the standard state), so

*( ) ( )oA Aliquid vaporm m=

However, the chemical potential of the vapor is pressure dependent, (5). so if P* is the vapor pressure of the pure liquid we get

** *( , ) ( , ) lno oA A o

Pvapor P vapor P RT

Pm m

æ ö÷ç ÷ç= + ÷ç ÷ç ÷è ø (6)

The standard state for the vapor being the vapor at one bar pressure.

Now let’s consider a solution of solvent A and a single non-


volatile solute B (i.e. it has no vapor pressure).

* ( ) ( )A Avapor solutionm m=

since the chemical potential of A must be the same in both phases. Therefore from (5)

( ) ( , ) lno oA A o

Psolution vapor P RT

Pm m

æ ö÷ç= + ÷ç ÷÷çè ø (7)

where P is now the vapor pressure over the solution.

From (6)

**( ) ( , ) ln lno

A A o oP P

solution vapor P RT RTP P

m mæ ö æ ö÷ ÷ç ç÷ ÷ç ç= - +÷ ÷ç ç÷ ÷ç ç÷ ÷è ø è ø

so *

( ) ( ) lnoA A

Psolution liquid RT

Pm m

æ ö÷ç ÷ç= + ÷ç ÷ç ÷è ø

but Raoult's Law for an ideal solution is

*A A AP X P=

( )( ) ( ) lnoA A Asolution solvent RT Xm m= +

That is, the solute lowers the vapor pressure of the solvent (since XA is <1). Note that the identity of B is irrelevant; vapor pressure is a colligative property. For a real solution, we change X to activity.

We can now rearrange this equation to give us the chemical potential change for converting one mole of solvent to a solution of concentration XA i.e. o

transferGD Raoult’s Law. Raoult’s Law is empirical and only applies to cases where the solution is dilute or the solute and solvent are very similar. This problem can be side-stepped, in the usual way, by stating “for dilute solutions.....”, however, that defeats the purpose of studying non-ideal cases. The more useful way is to define the activity aA(“apparent “ or thermodynamic concentration) as follows

( ) ( ) lnoA A Asolution liquid RT am m= +

The activity, aA, is purely empirical, but the equation is exact. For dilute solutions of non-polar solutes A Aa X→ . For dilute solutions of ionic compounds, the activity can be calculated by Debye-Huckel theory or one of its extensions.

Some Nomenclature. There are some evil forces at work that refer to the hydrophobic effect as hydrophobic forces, or worse bonds.

However, there are some sources of confusion. The first being intermolecular vs. intramolecular. Intermolecular refers to “between molecules” and intramolecular “within a molecule” (and is mainly responsible for protein folding). The next is the distinction between forces, bonds and effects. A bond involves sharing of electrons and is directional. Forces are usually electrostatic in nature (ionic bonds aren’t really bonds, they are an electrostatic force). They may or may not be directional. Hydrogen bonds, π-stacking etc. are electrostatic forces, which we collectively call intermolecular forces (although they can be intramolecular forces. π-stacking is almost exclusively intramolecular.) On the other hand the hydrophobic effect is just that an effect (on entropy). It is caused by H-bonding, but is not bonding, or even a force. To avoid confusion were call all the conditions that organize the structure of molecules, covalent and ionic bond’s excepted, intermolecular interactions.

© P.S.Phillips October 10, 2011 Permeability:1

Suite P. PotpourriThis is a suite of three miscellaneous experiments, but they are united by a central theme of modeling (see appendix). All experiments demonstrate a foundational principle with fairly obvious applications, but are also useful for demonstrating modeling, as opposed to theory. The osmosis/permeability demonstrates two very important processes in biology – it is a model for passive transport through membranes and simple cells and diffusion. This experiment is short and not terribly exciting. The wine experiment is straightforward and is about buffering in complex solutions and how to model them. The experimental is easy: modeling not so much. The final one is about glow in the dark stuff. This is a ROB experiment where you develop the methodology. It’s straight forward, we are pretty sure it works, and you will be given help with the fluorimeter. It demonstrates phosphorescence, which is not that well understood. Your data will, in principle, form a model for the process. We then may be able to develop a theory.

Experiment 1 Osmosis and Permeability. Be sure to read the question section before you start; they effect the procedure.

INTRODUCTION. Roughly speaking, a permeable solid is one that permits the passage of materials through it. A semi-permeable material is one that allows the passage of some materials, but not others. Such discrimination is made of the basis of molecular polarity or size. This is widely exploited in chromatographic methods to analyze materials based on size or polarity and in dialysis, which separates low molecular weight components from solution. The latter is used to remove waste product from blood (without removing proteins), alcohol from wine (why?) or smoke taint from wine. A good understanding of permeability is necessary to understand transport across bio-membranes and fluid movement in rocks. An important feature in understanding permeability is osmosis, which is what drives the materials across a membrane, and also diffusion, which describes motion in fluids. This experiment demonstrates osmosis and diffusion, but the focus is on permeability.

METHOD. Diffusion, permeability, osmosis are often simply illustrated in first year biology labs. by dropping a piece of sealed dialysis tubing, full of sucrose solution, into a beaker of water and watching it get fat – this models what happens when you drop a cell into water (it destroys them).

We are going to exploit this to get a quantitative value for the permeability of the dialysis tubing to water. We could use salmon fry in salt water but it would get vetoed. No wait; salmon are adapted to survive that. Besides they are inhomogeneous. This is a short experiment. Note I’m leaving a lot of details out deliberately.

A schematic of diffusion across a semi-permeable membrane (SPM) is shown below

SPM

Water [A]

Figure 1. Schematic of the experiment. The right hand side represents the dialysis bag. There will be a net transfer of water to the solution of A, in the bag. Species A does not cross the membrane.

Here we use a semi-permeable bag (made by clipping the ends of dialysis tubing). Water will enter the dialysis tubing due to the concentration gradient of 10% solute, changing the weight of the tube. The mass change will follow some kind of exponential curve.

Do the experiment with 10%, 20%, 30% and 40% sucrose. (Uh oh! What does 40% mean? 40g/100g solution or 40g/100g water. Doesn’t matter, except that’s why you should never use % or ppm as a concentration unit –it’s ambiguous, just make sure it matches what you find in the CRC tables for density of sucrose solutions).

THEORY. Consider the expression for general first order kinetics (of a species S)

SSr

dnk n

dt= − (1)

Note the use of number of moles not concentration. In inhomogeneous systems (i.e. real ones, particularly ones separated by a membrane, e.g. cells) this expression must be used. This is a very important point and commonly missed. If your reaction is not in a single container, then start all calculations with (1). (In homogenous systems [S]=nc/V so the V cancels on both sides giving the usual expression).

Now we need to introduce a new definition; flux, J. Flux is the amount of material passing through unit area per unit

Permeability:2 © P.S.Phillips October 10, 2011

time so

SS

1 dnJ

A dt= (4)

where A is the area (not to be confused with a species A). By Fick's first Law of diffusion we have

SS S

dcJ D

dz= (5)

Where DS is the diffusion coefficient and z the distance. Thus

SS

[S]dn dAD

dt dz= − (6)

Now, membranes are pretty thin, say l, so we can approximate to D[S]/Dz using simple differences

S SS

[S]([S ] [S ])i

o idn AD

ADdt z l

D= − = −

D (7)

Where the subscripts ‘i’ and ‘o’ denote in and out respectively. Now we can see that our rate constant is sort of related to the diffusion constant. This makes sense, but we are not there yet. The concentration gradient is the gradient within the membrane, not across it! That is the concentrations need to be modified by the partition coefficient, K of species S between the solution and the membrane.

S S S ([S ] [S ])io i

dn AD K

dt l= − − (8)

This illustrates another important point; diffusion coefficients actually don’t vary much in solution for small species so the transport of species across membranes is mainly dictated by the partition coefficient, that’s why these rather mundane looking constants are so important in real systems.

The constant DK/l is called the permeance, P, and represents; well you tell me. (Hint, look at the units. Note that the permeability is DK, but we don’t have l so we settle for permeance, but some texts confuse the two). It also turns out to be relatively easy to measure (compared to D and K). So we get

S([S ] [S ])i

o idn

APdt

= − − (9)

Replacing iSn with Vi[Si] we finally get

[S ]([S ] [S ])i i

o id V

APdt

= − − (10)

which is in terms of readily measurable parameters.

Equation (10) looks like what we would have expected from simple kinetic arguments based on (1), except for the bug-a-boo of Vi. (Which is why I laboured through this derivation).

In general, concentrations are easy to work with, but not as easy as mass. So using []=n/V and dropping the subscript S (In this case note that it’s water that crosses the membrane, not the sucrose – the sucrose just provides the concentration gradient) and using i and o to represent inside (the bag) and outside respectively we get

i o i

o i

dn n nAP

dt V V

= − −

(11)

but n=m/MW (mass over molecular weight) so

i o i

o i

dm m mAP

dt V V

= − −

(12)

where (just as a reminder) mi is the mass inside, mo the mass outside and Vi and Vo are the respective volumes.

Next, we recognize that the m/V terms are just densities and our observed mass is

, , ,i total i water i sucrose tube clipsm m m m += + +

So

i totaldm dm

dt dt=

Now, if we restrict ourselves to short times and make sure that Vo >>Vi , then both densities will be constant providing that sucrose does not cross the membrane so

( ) constant, totalo i

dmAP Q

dtρ ρ= − − = (13)

Integrating (13) we get

( )( )total o im t AP tρ ρ= − − + R (14)

i.e. a plot of total mass vs. time a straight line (it’s a zero order process) that dead ends when the bag is empty. The slope is ( )o iAP ρ ρ− − so you can get P. The R, of course, is just the mass of the empty bag + clips.

This derivation illustrates, very nicely, the dilemma of applied thermodynamics. The physical chemistry concepts needed for real systems (diffusion, permeability etc.) are quite simple. However, the maths is much worse than in traditional thermodynamics and is full of pitfalls; this is not even a complicated system.

QUESTIONS 1) One parameter you need is the MW cutoff for the dialysis tubing. What does this mean and what value does it have in


this case. 3) Find a permeability value for a biological system and compare it with that of the dialysis tubing.

4) Does our constant density approximation at short time seem reasonable in light of the experimental results?

5) There is a serious flaw in the design of this experiment for use with sucrose. What is it? How would you demonstrate (if possible) that this flaw exists?

6) Following up on the question above. I tried this experiment with a strong solution of polyvinyl alcohol, MW about 30000, and also starch of a similar MW. They are extremely viscous solutions. Would you expect the experiment to work properly?

7) Plot the initial rate of permeation vs. % concentration of the sucrose solutions. Would you expect this value to change or be constant. Explain. Support your answer with numbers.

8) Define diffusion, effusion, permeation (or permeability), percolation, porosity, osmosis, partitioning, convection and (ion) conduction. Be sure the definitions distinguish each process clearly. Incorporate comments on their relationships, if any. And don’t use Wikapedia!

9) How would you test the approximation that sucrose does not diffuse. Hint use sugar in the tube, and work out how to analyse for sucrose. Do the experiment if time permits.

10) Plot the permeability vs. sucrose concentration and comment. Given an explanation of the data if needed.

11) Analyze and present the data. Answer the questions. In addition write up this experiment for a second year lab, just the procedure. I’ve left a lot of details out.

Experiment 2. Buffer Capacity of a Wine INTRODUCTION. (Draft) Grapes are very unusual in that the principle acid in them is tartaric acid. In fact, they are the only fruit that contains significant amounts of

tartaric acid. They also have the highest concentration

of sugar of any fruit. The high sugar levels gives them a propensity to

ferment producing

alcohol or what we call wine, although some people refer to

it as “nectar of the gods”. (Note the small “g” and the “s”. Monotheistic religions tend to disapprove of alcohol.) The inter play between the sweetness of the sugar, the tartness of the acids, and the fieriness of the alcohol (ok that’s probably just tolerated – it’s the inebriating effect) can make a pleasant drink that plays a significant role in our social history.

The flavor is also influenced by the grape skin (red vs. white wines), which give them individual flavors and determine which wines are best suited to drink with various food. This is not just snobbery, red wine will overpower chicken, but try a gewürztraminer with a light curry. Also, wines from a given grape can vary from year to year and from vine to vine. (The Dirty Laundry in Peachland has three or four kinds of gewürztraminer). However, given that in double blind tests so called wine experts can’t distinguish between red wine and white wine dyed red, ones choice comes down to three things, how much is it, how much of a hangover do you get, and whether you like the taste.

A final comment is that wine is potable, unlike most of the water in the Mediterranean regions. This probably contributes to it’s popularity.

Given the influence of the balance of the acid, sugar, and alcohol (in white wine in particular) we are going to model the acidity of wine. Our model will be tartaric acid, malic acid, sugar and alcohol.

The malic acid occurs in northern grown grapes making them too acid. To raise the pH the malic acid is removed by adding calcium carbonate (calcium malate is insoluble) or by converting it to weaker lactic acid (there are other methods). The latter, called malolactic fermentation, produces biogenic amines, which are responsible for the mild allergies some people have to red wine and some chardonnays. If I could find out which wineries did this I would avoid their wines. Note that there is an interplay between alcohol and sugar. Low sugar and a long fermentation would leave little sugar and all the acid. High sugar and long fermentation gives you high alcohol (14%) and some sugar, but in my experience it taste pretty foul. (It’s interesting to take a dry wine – high acidity – and add sugar, alcohol, and change the pH –with chalk or food grade NaOH, and see how it effects the taste. Sugar of course sweetens it, alcohol can make it “chemically”, and most interesting, raising the pH makes it flat and bland. Unfortunately, we have a bureaucratic ethics committee that prevents us from doing this experiment.)

Probably the most important determinant in the basic flavor of a white wine is the pH so we will model our wines


using mixtures of tartaric acid and malic acid and their potassium salts. Potassium is the most prevalent metal ion in grape juice. Sugar doesn’t seem to affect pH, but we need to quickly verify that, but alcohol does, although the mechanism is not understood. We will look at that briefly.

So here’s what we will do. Will measure the pH of an unbuffered solution of tartaric acid and see how it changes the pH. We will get a titration curve for tartaric acid and make sure we can find its two pKa’s. Then we’ll mix in some malic acid and see what that does to the titration curve. Then, we add in some potassium hydrogen tartrate (Make it from tartaric acid and KOH then add in more tartaric acid and malic acid.) We will use quantities commensurate with a typical must (the crushed grape mix used for making a wine). Finally we will titrate a red wine. We will use red wine because it adds some interest when using a pH meter to do titrations. We will do the titrations over a wide range because we want to get some insights into pH changes in moderately complex systems. To be specific we want to look at buffering.

Buffering is important in biological systems so that minor changes (say increased CO2 in the blood when exercising) doesn’t cause wild pH fluctuations. It’s similarly important in environmental systems. A poorly buffered lake cannot tolerate much acid rain without its pH plummeting and killing all the fish. In wine, buffering effects the palette. A poorly buffered wine will shift pH in the mouth, affecting its flavor. There are other important effects as well, but they are beyond the scope of this discussion. How little a system changes pH when acid is added is called the buffer capacity.

One of the problems about buffers is that if you look them up in a biochemistry text the authors will waive their arms and refer you to a physical chemistry text. If you look in a physical chemistry text, you will rarely find an index entry for buffers. Sometimes you will find an entry under acid-base equilibria. Even then, you might find they refer you to a biochemistry text. Anyway, if you want some background you will have to hunt around.

Let’s do some experiment design. Our auto-titrators are tied up so we want to minimize our titrations. We don’t want to measure a whole titration curve (pH 3-13) every time so let’s look at what we need. You may need to fill in some experimental blanks.

We want to see if there are significant impacts on the activity of the acids by alcohol and sugar to see if we need them in our model system. If there are then we need to get our corrected pKa’s but for now we can do as follows. Make up 500mL 7.0g/L solution of tartaric acid. Split it in five (use a 100mL pipette – that’ll be good enough) samples. Take

two samples and pipette (graduated pipette) or titrate in three 6mL aliquots of water to one sample, measuring the pH at each stage (four pH measures) – this is your control. Now repeat with alcohol. Plot the data against each other. Any deviation from a straight line of slope one means the alcohol affects the pKa. To test the effect of sugar just add 10g to 100mL of the acid solution in a measuring cylinder (you’ll have one sample for a spare), then measure the pH, and record the volume. then add another 10g and get the pH and volume again. 20g of sucrose That’ll give you two pH’s measures. Show me your data. The question then becomes how much and how do you calculate the pKa from the pH. We’ll worry about what to do with the data when we see it.

Next we need to get the composition of our wine sample so we need to do HPLC to get the tartaric and malic acid concentration. We also need to get the mono-cation concentration. No wait: you are doing this course to avoid analytical chemistry. If you really want to try this, you can, I have some papers, we should have the columns. Instead, we’ll assume it’s a typical wine and just do the titration curve up to about pH11 (it should be about pH3.4 to start. and get it’s shape and measure the buffering capacity.

Now we just make up our model wine to compare the real wine with. Take 7.00g/L tartaric acid, 5.00g/L malic acid and 1400ppm of potassium ions (as KOH) and do the titration curve up to about pH11. Get it’s shape and measure the buffering capacity. The curve is broad so you don’t have to use small increments.

Now how are we going to interpret this data? For the influence of sucrose and alcohol samples simply set up an ICE box to get the pH at the various concentrations (remember an ICE box works with moles not concentrations). Compare your results with the calculated results. Plot graphs as needed. Tables are good to.

The alcohol should have a small effect. Redo the calculations assuming that the alcohol doesn’t affect the acid concentration. Alternatively assume the alcohol dilutes the water, so five mole% water will shift the hydrogen ion concentration by 5% (you’ll need to show me how I did that calculation). If that doesn’t work, speculate how alcohol may influence the activity of the acid. Hydrogen bonding is an obvious choice, but sucrose does that as well, but you should find that it does not change the pH.

Non-polar solutes tend not to change the activity of other species in solution, so we should not see an effect. However, the concentrations used here will affect the activity of water re: osmotic pressure so there may be an indirect effect. We could measure the pKa using a conductiometric titration, but I suspect for high sucrose concentrations we


would see an effect (not the effect mentioned above). Why do you think I suspect that?

Buffering capacity is important. It tells us if the wine is likely to shift in taste as it ages (as the acids change). Also, the mouth is alkaline, a poorly buffered wine will shift in taste with each mouthful if it’s poorly buffered. The buffer capacity is simply a measure of how well a solution resists pH changes when acid or base is added. Typically for a wine it’s the amount of KOH that needs to be added to raise the pH by one point (and is bizarrely expressed in equivalents of tartaric acid). We will take a physical chemistry approach. It’s simply the differential of titration curve. Close to zero means a high buffer capacity in that region. The bigger that region, the better the buffering. The width of the region as defined by some convenient parameter is the buffer capacity. See the glycine experiment on how to differentiate data.

If you want a simple reminder of the effects of buffering take 50mL of water stick a pH electrode in, measure, and then and add 1mLof 0.1M KOH, it should shift the pH about 3 units. Repeat the experiment with pH 7 buffer. There should be little effect.

PROCEDURE. This is a simple pH titration. First, you will need to calibrate the pH meter, using the provided standard buffers - pH 2.00, pH 7.00 and pH 9.00 or 10.00 (You may need help with this unless you have a pre-calibrated meter; ask the instructor).

All titrations are done the same way: take a 25.0mL aliquot of the sample and put it into a beaker, along with the pH electrode (do not add indicator), and titrate in, while stirring, the 0.100M KOH. (A little warning you may need 0.25M KOH to get to pH 9), let me know). Add the KOH in 1.00mL increments (about, but accurately known). You may have to let the pH meter stabilize between readings. You should be able to get stable values to +0.01 pH units). Make sure you collect data beyond the end-points to get a good “baseline” (see fig.1). Don’t forget your 0mL pH reading.

Figure 1. A typical titration curve and its differential for a monoprotic acid. The weaker the acid (or base) the less well defined the inflection point becomes. A

diprotic acid will be two superimposed curves.

Firstly we need to do a titration of an unbuffered solution for comparison. Simply titrate 0.10M HCl with the NaOH and plot the data. Next we will do a simple buffer solution to see what difference that makes. We will use a mixture of 0.100M acetic acid with 0.100M sodium acetate and titrate that with 0.100M NaOH.

Finally, we will titrate a red wine with the 0.100M NaOH. Buffer capacity is more important in white wines, but that’s for drinking. Red is only suitable for dyeing cloth and chemistry experiments.

CALCULATIONS.

1) Determine the end-point and hence the two pKa‘s of the acid directly from a plot of pH vs. VHCl. (Use the Henderson-Hasselbach equation – look it up). Also refer to the figure 2 below.

Figure 2. Titration of a diprotic acid. The two pKa’s are determined from the pH at the ½ equivalence points (e.g. ½V2 and pK2 on the diagram). Note ½V1 (not marked) is ½ way between 0 and V1. ½V2 is ½ way between the two equivalence points – this is the buffer zone.

2) The buffer capacity can be calculated from the inverse of the differential of the titration curve. Typical results are shown in figure 1. To get the buffer capacity at a certain pH use your titration data to find the volume corresponding to the desired pH. Then use that volume to read the value of the derivative off your differential graph. The buffer capacity (for the purposes of this experiment anyway) is the reciprocal of that value. Get the buffer capacity of the wine at the pKa of acetic acid (when wine goes off it generates acetic acid), pH 3.4 (a typical wine pH) and pH 6.8 (pH of


the mouth).

MODELING. We can compare the shapes of the titration curves to see if our model is reasonable. They will not match because the compositions are different, but the overall shape should be the same. To test this properly we need to be able to match our compositions. However, we are just after insights, we can skip the chemistry altogether and simulate the titration curves. This is the physical chemistry bit. ICE will not work so we have to start from the ground up. Diprotic acids dissociate as follows:

1 2- 2-2H A H + HA H + A

K K+ +

where + - + 2-

2

H HA H A1 2

H A HA

a a a a

K Ka a −

= =

a is the activity which we will approximate to concentrations and ignore the fact that pH electrodes actually give the activity. We will have another two equations for the other acid.

1 2- 2-2H M H + HM H + M

K K+ +

+ - + 2-

2

H HM H M1 2

H M HM

a a a a

K Ka a −

= =

Next we invoke mass balance - 2-

2 2- 2-

2 2

[H A] = [H A]+ [HA ] + [A ] and

[H M] = [H M]+ [HM ] + [M ]

total

total

Now we get to the bit that may be new to you, charge balance - 2-

- 2-

[K ]+ [H ] = [OH ]+ [HA ] + 2[A ] and

[K ]+ [H ] = [OH ]+ [HM ] + 2[M ]

+ + −

+ + −

Note the potassium ion because we need potassium salts to model our wine. We can ignore the hydroxide below pH 7 (why?)

Since we make up our own solutions we have the total amount of acid and potassium (we include the potassium counter ions in with the total acid). We now have seven equations and seven variables. If you rearrange this you end up with cubics in [H+]. This is where Maple (or whatever, I haven’t had any success with MatLab or MathCad though). comes in it can solve this kind of thing easily, but there are two problems. The first is that Maple is quite general so it will give you all three solutions for the [H+]. You have to make sure you get the real +ve root only. You then have to get Maple to generate all the concentrations for given starting values, in a nice table or an array. To do that you have to put

everything in a loop. Finally, you have to get Maple to plot it, although it might be easier to cut and paste your table into Excel.

SAFETY NOTES. HCl, KOH and NaOH solutions are a little corrosive. Use with caution. Wine is toxic if ingested in large amounts. The symptoms are too well known to bother describing here. Stealing the wine stock bottle is even more hazardous. Symptoms include loss of dignity, credibility, marks, and external organs.

Experiment 3. Luminescence. Basically we will take three samples. Illuminate them at a high frequency (blue light), cut that off, that’s the tricky bit, and then observe the sample at a lower frequency in a fluorimeter. Sample one is a strip of “Glow in the Dark” paint. (In an alternate reality you would have to prepare the paint – copper doped zinc sulfide; we just want to do proof of principle.) Mount it a 45o and get the light intensity as a function of time. You may need to “prime” the paint with a light bulb rather than the fluorimeter. You’ll have to check that out. The decay may consist of two parts so you will have to do some kind of stripping as described in suite K. Repeat the experiment with the supplied europium salt. (In another alternate reality you would have to prepare the salt.) Then repeat the experiment with a terbium EDTA complex which you will make from terbium chloride and EDTA. I suggest an small excess of EDTA because Tb3+ alone phosphoresces, albeit much more weakly. Furthermore is TbCl3.xH2O, you’ll need a work around for the “x”. Write it up and tell me what’s going on in the experiment and about phosphorescence in general. Remember, I don’t know what’s going on so a clear and informative report earns lots of brownie points.

Why did we do a phosphorescence experiment instead of fluorescence experiment? Fluorescence has more scientific applications.

SAFETY NOTES. None of the materials have noted toxicity, but it would be a poor idea to eat them or slather yourself with them.

APPENDIX. Modeling, Theories, Laws,… A theory a set of statements or principles, often

expressed as equations or laws, and occasionally postulates, devised to explain a group of facts or phenomena. To be accepted as a theory, as opposed to cacodoxy, it must have


been be repeatedly tested and can be used to make predictions about natural phenomena.

A model is schematic description of a system or phenomenon that accounts for its known or inferred properties, and may be used for further study of its characteristics. A model may be a simplified version of the system. This makes studying it more tractable and enables you to deduce the critical elements. A model can be a physical system or software.

A simulation is done entirely with a computer. The input information is the theoretical equations and maybe some constraints. There is no reference to real data, although it may be compared with data at the end. This is useful for spectral analysis where the theory and constraints are well known and the solutions tend to be unique. Some believe that computers can substitute for an experiment. They can be a useful as a starting point, but eventually you have to get your hands dirty.

There are fuzzy areas. Generating a straight line is a simulation. A least squares fit to a straight line is a model – the model being that the data is linear. An equation that is part of a theory could of course be of the form y=mx+c. Anyway, a model is a route to a theory, but there is no real requirement for prediction, or laws, or for it even to be a physical entity. They tend to be used for multipart or complicated systems where the interrelationships are not always known. They are useful for eliminating or determining those relationships. The use of computer models without experimental verification is just plain stupid. Simulations are ok for well-defined systems, but given that the early models for the weather led to the discovery of chaos theory. You can see how much you can become unstuck. Anyway, they are an important and a powerful tool if handled with care.

The models here are just simplifications of real systems. The osmosis model tends to breakdown because of something called active transport (there are pumps in the membranes). The wine model can be extended successfully to much more realistic versions of wines, but in the end a vineyard is just a farm with a marketing manager: they are not interested in computer models. The phosphoresce experiment is a little simpler. We analyze the data using a model based on kinetics and all that implies. There is, a priori, no reason to believe that photochemical data will conform to kinetic equations.


3

© P.S.Phillips October 10, 2011 EXPERIMENT G:1

Exp T. MYOGLOBIN TRANSITIONS INTRODUCTION. There are four levels in the hierarchy of protein structure that are recognized. They are: primary – the linear sequence of amino acids; secondary – the regular, recurring orientation of the amino acids in a peptide chain due to H-bonds; tertiary- the complete 3-D shape of a peptide due to weak dipole-dipole interactions, p-stacking and van der Waal’s forces; and quaternary – the spatial relationships between different polypeptides or subunits.

Figure 1. 3-D ribbon diagram showing the structure of myoglobin, with extensive a- helices and haem binding site, undergoing a transition to a b-pleated sheet.

Myoglobin is a metalloprotein that acts as a temporary storage of oxygen needed for aerobic metabolism in muscle. Like hemoglobin the oxygen binds at a heme site containing iron. Unlike hemoglobin it is a monomer (it has no quaternary structure); hemoglobin is a tetramer and can carry four dioxygen molecules. Here will investigate changes in the secondary and tertiary structure of myoglobin. Common types of secondary structures include a-helices, b-pleated sheets, random coils and b-turns, usually in various amounts. Myoglobin is unusual in that it consists almost entirely of a-helices. Myoglobin is characterized as a globular protein; its tertiary (3-D) structure consists of eight a-helices which fold in such a manner that most of the hydrophilic groups are on the outside of the protein, facing the aqueous environment (what else would a hydrophilic group do?).

The hydrophobic groups are, as expected, mainly inside the protein. The hydrophobic effect plays a large role in maintaining the stability of the folded protein. Anything that disrupts the hydrophobic effect will change the protein structure. Here we will primarily look at the effect of

temperature. Remember, ∆G=∆H-T∆S and since the hydrophobic effect is an entropic one, it’s clear that temperature will affect it. We will also look the effects of SDS, guanidinium·HCl and pH.

METHOD.

By using model compounds, amino acids, and proteins to classify the IR spectra of proteins, biochemists have found the most useful vibrations to be the C=O stretch (amide I, 1655cm-1), N-H bend and C-N stretch (amide II, ~1550cm-1) of the polypeptide backbone. The variation in absorbance spectra of proteins with different secondary structure (and to some extent tertiary structure) is due to the characteristic hydrogen- bonding patterns of the C=O and N-H bonds. For a protein that is predominantly a −helical, IR spectra show amide I peaks centered around 1650 cm-1. Proteins mostly composed of b-sheet conformation show maximum intensity around 1640 cm-1. Random chain proteins have an amide I peak around 1643 cm , and denatured and or aggregated proteins show absorbencies around 1610–1628 cm-1.

One problem of IR spectroscopy is the strong absorbance of water in the amide I region. However, if very short path lengths are employed, spectra can be obtained in water. We can also shift the water absorbance by making use of the isotope effect. If we measure the spectra in deuterium oxide there is a ~400 cm-1 shift of solvent vibrations to lower energy. Furthermore, the use of D2O allows greater distinction between a-helices and random coils as the latter has a larger shift upon deuteration. Amide I peaks shift about 5-10cm-1, but the amide II peaks shift about 100cm-1. The tertiary structure will also influence the position of these vibrations: a-helices are around 1650cm-1 when buried inside the protein and shift down to 1635cm-1 when the helix is solvated; aggregated proteins often display intermolecular anti-parallel b-sheet structure with distinct sharp bands showing up at 1615 and 1685cm-1.

We will use IR spectroscopy to monitor these peaks and how they change as a function of temperature, pH, and the effect of H-bond disruptors. We will also compare myoglobin with two other proteins: chymotrypsin and lysozyme.

PROCEDURE. We will use FTIR coupled with either an ATR attachment or a short path-length IR cell with calcium fluoride windows and a 25mm spacer. The


comparison of the different denaturants and protein solutions will be carried out using the ATR, while the temperature denaturation of myoglobin will be performed in the short-path cell. We will use 10-20mL sample of about 60mg/mL of the appropriate protein in deuterium oxide. The protein stock solutions must be stored on ice. Solutions of myoglobin in 160mM SDS, 3M guanidine·HCl or pH 2 buffer made with H2O or D2O, may also be supplied along with solutions of other proteins. These samples need only be 0.1mL.

For the ATR samples, the experiment is very simple. Place 10-20mL of a blank on the ATR crystal and measure the background using 25-50 scans and a 4cm-1 resolution (or as instructed).

Repeat with 10-20mL of your sample. Collect all your data as absorbance spectra (rather than the more familiar transmission spectra). This will enable you to compare peak intensities directly and calculate difference spectra.

Clean the crystal with a small amount of methanol after every run and be sure to allow the crystal to dry thoroughly before running your next sample. Measure IR spectra of myoglobin in 3M guanidinium·HCl, 160mM SDS and at pH 2. Remember to take blanks of the appropriate solvent. Measure the IR spectra of each of the other two protein solutions provided.

When all the ATR runs are complete, remove the ATR attachment (ask for help) and install the cell holder in the sample compartment. For the variable temperature run, fill the cell with D2O buffer first and measure a background at room temperature. Use this room temperature blank for all subsequent temperature runs.

Fill the cell with myoglobin sample and take a room temperature spectrum to compare with the other spectra.

For the temperature series, Place the cell inside a zip-lock bag and submerge the bag in the water bath. Try to keep the cell from getting wet as the water vapour from evaporation in the FTIR will cause problems. The cell should stay in the bath for at least 15 min to let the sample equilibrate (see the appendix). The temperature range should be 30C to 80C (see the safety note) – choose equally spaced temperatures (about every 8-9 degrees) but make sure that you will have enough time (at about 20 minutes per temperature point) to get to 80C.

Remember to save all your spectra in csv format. Bring a USB drive so you can take it home.

CALCULATIONS. This consists of two parts: determining the transition temperatures for myoglobin and identifying the main contributors to the secondary structure of each protein.

Overlay all the temperature spectra on a single plot from 1300-1800 cm-1. This will allow you to identify the positions of largest change.

You should also try difference spectroscopy. Subtract your room temperature myoglobin spectrum from the other myoglobin spectra and plot an overlay of the subtracted spectra from 1300-1800 cm-1. This should highlight the appearance of the intermolecular antiparallel b-sheets. The a-helices will probably appear as –ve peaks, but the quantitative relationship is retained.

To find the transition temperature, plot the absorbance at selected frequencies vs. incubation temperature. The amide I peak is an obvious frequency to try, but you should try a few others and compare the results. Does the transition temperature depend on the peak chosen?

For other spectra you should try to identify the main changes in the tertiary structure. Calculating the second derivative of the spectra may help, but then the quantitative relationships become unclear. Find literature values if possible.

Tabulate your results.

QUESTIONS/DISCUSSION. The questions and discussion are somewhat interlocked so read this whole section before proceeding.

How do we know that the transitions we see are due to the unfolding of helices? Well it’s historically the other way round; we know there are helices because they unfold – there is a phase transition! Nowadays we can get good structures of hydrated proteins from X-ray crystallography, but even then you need to know that there are helical structures to solve the diffraction patterns. What you will do to complete this experiment is to explore the enzyme structures using modeling programs and use them to study the folding/unfolding based on the structures. The programs interfaces, and there, capabilities are all different so that is discussed in another document provided on-line.

Play with the various programs (note the plural: playing with different methods of data analysis is part of the course). Print up a couple of pictures of your compounds that make the structural differences clear.

That was fun wasn't it? The PDB ProtienWorkshop (and


other programs) have some nice visualization features. Conformation Type and Hydrophobicity are the two that I find of most interest.

Hydrophobicity: To a first approximation the proteins will fold (if sufficiently hydrated) to a structure where the hydrophobic regions are on the inside and the hydrophilic (least hydrophobic) are on the outside. Do the structures you see conform with those expectations? Cross check with other programs. The ones showing surface hydrophobicity may help.

Conformation: The conformation types are listed in PDB-WS are Turn, Helix, Coil, Strand. The elements I'm familiar with are Helix, Random Coil, Chain and b-pleated sheet, ribbon. So help me out here what's what? Be sure to explain things in terms of structures and intermolecular interactions (H-bonding, sulphur bridges, Pi stacking, hydrophobicity etc.)

Also, some of the programs (Ramplot and Swissplot) have options for Ramachandran plots which seems a neat way of sorting out the different structural types. So what are Ramachandran plots (Wikipedia is as good as any place to start)? Apply them to the species you studied.

At room temperature the more ordered helical structure exists in preference or the less ordered b-sheet, or more to the point a random coil. This would appear to violate the Second Law, but it doesn’t; explain. (i.e. explain the transitions between helices and whatever in terms of enthalpy and entropy, as well as intermolecular interactions.

Rationalize the spectral changes you saw in terms of structural changes (and accompanying changes in intermolecular interactions).

See if you can find an X-ray structure of myglobin above it’s transition temperature (i.e. above 90C). Print it out.

SAFETY NOTES. The maximum safe temperature for domestic water is 50C. Water at 80C will definitely scald you, so be careful with the water bath above 50C. Make sure you do not use a plastic water bath. pH 2 buffer is no fun in the eyes. The rest of the stuff is pretty benign, but don’t eat it, it’s too expensive.

APPENDIX.

INFLECTION POINTS. To determine the inflection point of a sigmoidal curve dy/dx vs. x. (here y is absorbance and x temperature). Setup Excel to do a simple derivative using 1 1/ ( ) / ( )i i i idy dx y y x x Typical results are shown in figure 2. The inflection point is the maximum of the derivative.

Figure 2. Figure 2. Typical sigmoidal curve and its differential. (It’s for a electrochemical titration in this case).

EQUILIBRATION TIMES. The experiment may seem strange; we spend 15 min equilibrating a tiny sample, and then stick it into a room temperature FTIR sample compartment. You are able to do this because t h e t r a n s i t i o n s a r e s l o w . In some cases the transition is actually irreversible e.g. boiled eggs. Many proteins are destroyed by heat, pH changes or H-bond disruptors. This is called denaturation and is a result of aggregation, polymerization or folding. Some show “real” phase transitions (i.e. reversible changes). However, you should note that such changes are intra-molecular not intermolecular, that is, they are changes in molecular structure, not macroscopic structure. So even though the proteins are in solution, these phase changes are essentially solid-solid (or in the case of membranes, gel-gel) transitions, which are very slow. Long equilibrium times are required, >15min, even for tiny samples. You are not just trying to bring the sample up to temperature – you have to wait for the transition (in either direction).

REFERENCES 1. F. Meersman, L. Smeller, K. Heremans Biophys. J. 82, 2635-2644 (2002).

USING THE PROTEIN DATABASE (Draft. Report any problems to me.)

Sign onto www.rcsb.org with you web browser. The home page is shown below. As you can see it is very cluttered there are only two points of interest – see comments to right.

This is what you get lots and lots of stuff (336 structures). You need to narrow it down.

Comment [PSP1]: This button leads you to a list of important proteins and their stories.

Comment [PSP2]: This the search bar. Type the name of the molecule here. Watch you spelling. You can also use the 4 letter code.

Comment [PSP3]: Click on items in the list. For us organism and method are good lists to use to narrow stuff down.

I selected humans and a 1-2 resolution X-ray structure. You still get tons of stuff and need to scroll through items at the bottom of the page until you get the one you want. Here we have two or three examples with the same information. Plea ignorance (i.e. say you are a chemist) and select one. In this case exceptional ignorance as somewhere it decided to insert hemoglobin into the myoglobin list. Better go find that 101 button at the top.

Anyway lets click on 2HHb, or more specifically the the title next to it.

Comment [PSP4]: Yay! Our

first Windows 7 pop up. This one is insidious as it can pop up under the window not on top, so don’t touch the keyboard or mouse once you hit that title bar. Just click run. Do not check the box or it won’t work. This is for IE9 under Windows 7. Other OS’s/browsers may behave differently.

A picture at last. I’ve scrolled down so we can see the bottom menus. These give some interesting views, especially under the surface options. We’ll get back to that later.

There are other viewing options so let’s scroll back up to see them.

Comment [PSP5]: Here’s our other options. Simple and other viewers are simple. Protein Workshop gives the best pictures. Jmol is the default one.

The Workshop option has a couple of wrinkles. Now wait for Java to load. It’s called Java because you have to go to coffee while it loads. I think it should be called C2 (crippled computing) although this applet seems to be well programmed. Anyway after you’ve done this a few times you will want to try off-line programs such as Jmol, Pymol, Ramplot, and SwissView are some. I can put them on disk or they maybe emailable, or you can find them online. To be able to use them we have to download the data files To do that find the download menu.

Comment [PSP6]: After dealing with the popup below click here to launch the workshop.

Comment [PSP7]: Click on save. It’s a little java snippet that can be delete later. There maybe other popups when you first run the workshop because it installs itself on your computer

Comment [PSP8]: The download option is a dropdown menu. Save the files using the PDB file (Text option) then hit save when the popup appears at the bottom.

Before we go on and discuss the viewing programs (Jmol, Pymol, Ramplot, and SwissView) in detail here’s a few pointers.

The online Jmol is mainly a viewer but it has some dropdown menus at the bottom that make it a very interesting viewer. I love the way it refers to the standard text book view for proteins as cartoons.

The offline Jmol is just the simple viewer in the options list. If that’s what you want go for it.

Kiosk (the “other” viewer) makes a good screen display for open house.

The Protein Workshop is a supped up viewer. I think it give the best pictures. There are a few bells and whistles I have checked out yet. When you first run it it installs itself on your computer.

I my humble opinion Pymol is garbage.

RamaPlot (Ramchandran Plot Explorer) is possibly the most interesting. It’s old and the visualization is not fancy but it seems to give stuff other programs don’t, or at least the kind of stuff relevant to this course

SwissView (Swiss-PDB Viewer) Lots of options but no fancy pictures.

Qutemol is interesting if you want to see the importance of lighting and shading when rendering 3D molecules.

For other options see

http://en.bio-soft.net/3d.html

http://www.pdb.org/pdb/static.do?p=software/software_links/molecular_graphics.html

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