Experiment 1

Analytical Methods for amino Acid Separation and Identification

Separation and identification of amino acids are operations that must be performed frequently by biochemists. The 20 amino acids present in proteins have similar structures. However, each amino acid is unique in polarity and ionic characteristics. In this experiment, we will use a combination of ion exchange chromatography and paper chromatography to separate and identify the components of an unknown amino acid mixture.



I. Amino acids (Revise Chapter 4.1 of Garrett & Grisham – your Biochem-I lecture textbook).


Twenty amino acids are the fundamental building blocks of proteins. Amide bond linkages between a-amino acids construct all proteins found in nature. The amino acids isolated from proteins material all have common structural characteristics. The general formula for an amino acid is:




        R -C -COO-




The distinctive physical, chemical and biological properties associated with an amino acid are the result of the R group. There are 20 major amino acids that differ in their R-group.  The R-group can be hydrophobic or polar, aromatic or aliphatic, charged or uncharged.  The different R-groups are responsible for amino acids having different polarities, solubilities and chromatographic behavior (see below).


The structure and biological function of a protein depend on its amino acid composition. It is a matter of basic importance to understand practical methods used for the separation and identification of the 20 common amino acids.


II. Acid-Base Chemistry of the Amino Acids (Revise Chapter 4.2 of Garrett & Grisham – your Biochem-I lecture textbook).


Amino acids are amphiprotic because they contain both an acidic group and basic group.  The COOH group is acidic with a pKa value of 1.7-2.4.  Thus at pH values below this, the group exists as COOH while at higher pH values, the group exists as COO-.  The NH2 group is basic with a pKa of 9-10.5, so below this it exists as NH3+ while above this pH it exists as NH2.  At neutral pH values, both groups are ionized and the amino acid exists in a dipolar form with no net charge.  This form is called a zwitterion.  The pH at which all the amino acid molecules are in this form is the isoionic point (pI) of the amino acid where (for amino acids with non-ionizable side chain chains)


pI   =  pK1 + pK2


























III. Ion Exchange Chromatography of Amino Acids (Revise Chapter 4.6 of Garrett & Grisham – your Biochem-I lecture textbook and
Chapter 3D and 3E of your lab textbook – Boyer’s 3rd Ed).


Different types of ion exchange resins (a) Cation exchanger (b) Anion exchanger.


































Operation of a cation exchange column for the separation of the amino acids Asp, Ser and Lys.















































Separation of amino acids on a cation exchange column





























Chromatographic separation of a mixture of amino acids on ion exchange columns using Amberlite IR-120 (similar to Dowex-50)




























IV. Paper chromatography of amino acids


Paper chromatography can separate different amino acids based on their varying solubilities in two different solvents.  In this method, a sample of an amino acid (or mixture of amino acids) is applied as a small spot near one edge of a piece of chromatography paper.  The edge of the paper is then placed in a shallow layer of solvent mixture in a chromatography tank.



The solvent mixture contains several components, one of which is usually water and another of which is a more non-polar solvent.  As the solvent mixture moves up the paper by capillary action, the water in the mixture binds to the hydrophilic paper (cellulose) and creates a liquid stationary phase of many small water droplets.  The non-polar solvent continues to move up the paper forming a liquid mobile phase.  Since amino acids have different R-groups, they also have different degrees of solubility in water vs. the non-polar solvent.  An amino acid with a polar R-group will be more soluble in water than in the non-polar solvent, so it will dissolve more in the stationary water phase and will move up the paper only slightly.  An amino acid with a hydrophobic R-group will be more soluble in the mobile non-polar solvent than in water, so it will continue to move up the paper.  Different amino acids will move different distances up the paper depending upon their relative solubilities in the two solvents, allowing for separation of amino acid mixtures.


The movement of amino acids can be defined by a quantity known as Rf value, which measures the movement of an amino acid compared to the movement of the solvent.  At the start of the chromatography, the amino acid is spotted at what is called the origin.  The chromatography is then performed, and the procedure is stopped before the solvent runs all the way up the paper.  The level to which the solvent has risen is called the solvent front.  The Rf value of an amino acid is the ratio of the distance traveled by the amino acid from the origin to the distance traveled by the solvent from the origin.


Since Rf value for an amino acid is constant for a given chromatography system, an unknown amino acid can be identified by comparing its Rf value to those of known amino acids.


Certain technical aspects are important when performing paper chromatography.  First, it is necessary to keep the applied amino acid spot very small.  The spot tends to spread out as it moves up the paper, so starting with a big spot will produce a large smear by the end of the procedure, making it difficult to measure an accurate Rf value.  Second, the chromatogram paper must be kept very clean.  Fingerprints or other types of contamination will interfere with the chromatography and give poor results.  Finally, since amino acids are colorless, something must be done to detect the amino acids at the completion of the chromatography.  One of the simplest methods for this involves spraying the paper with ninhydrin.  When heated, ninhydrin reacts with amino acids to produce a blue-purple color (yellow in the case of proline), making the amino acids spots visible for analysis.




In this experiment, paper chromatography will be performed using an unknown amino acid along with known standards.  Through a comparison of Rf values, the unknown amino acid will be identified.


V. Procedures


A. Ion Exchange Chromatography


Column Preparation

Clamp a glass column or buret to a ring stand with two clamps, making sure that the column is vertical. With the stopcock close, pour 2-3ml of citrate buffer, pH 3.0, into the column. With a long glass rod, put a small plug of glass wool into the bottom of the column. Press any air bubbles out of the glass wool. This glass wool barrier prevents the ion exchange resin from running out the column. Use enough glass wool to fill the hole in the column; otherwise the flow rate will be greatly diminished. Obtain about 20-25ml of cation exchange resin from you TA in a small graduated beaker and carefully pour the resin into the column. You might need a funnel to accomplish this task. Swirl the resin slur as you are pouring it into the column. Allow the resin to settle into 20-25cm in height. When all the resin beads have settled, open the stopcock and allow the buffer to drain from the columns until the buffer meniscus is just above the top of the resin. DO NOT ALLOW THE COLUMN TO RUN DRY.

Application of Sample and Collection of Fractions.

Prepare 30 test tubes for fraction collection by numbering the tubes and marking 1ml level on each tube with a sharpie. Using a pipette, carefully apply 0.5ml of your unknown amino acid mixture to the top of your resin. Open the stopcock and begin to collect you fractions in test tube #1. When the top level of the buffer has just entered the resin, turn off the stopcock and add 1ml of citrate buffer (pH=3.0) to the column to wash away any amino acid from the inside wall of the column. Allow the buffer to enter the resin and continue to collect fraction in tube #1 until 1ml is collected; then begin to collect in tube #2.

When the buffer has all entered the resin, turn off the stopcock and carefully add 10ml of buffer to the top of the column. Continue to collect 1ml fractions. If an acidic amino acid is present in the unknown mixture, it will be eluted during this procedure. To test the fraction for the presence of an amino acid, place two drops from each fraction onto a strip of chromatographic paper (3x10cm).  Allow the spots to air-dry, spray them in the hood with the ninhydrin solution. Place the paper strip in the oven (T=110oC) for 10min. Red-purple spots indicate the presence of n amino acid.

If an amino acid has been completely removed from the column (no colored spot for fraction #10), change the eluting buffer by draining all citrate buffer from the column (continue to collect fraction) and add 10ml of 0.05M citrate buffer, pH=6.0, to the top of the column. Collect fractions until the top of the buffer has reached the top of the resin. Test each fraction as before for amino acid content or check the pH change with the use of the pHmeter (you TAs will instruct you on the use of the pH meters). Neutral amino acids in the unknown mixture should have bee eluted during this wash. Now wash the column with 10ml of Tris, pH=9.0. Collect the fractions until the entire buffer has entered the resin. Spot two drops of each fraction onto chromatographic paper and neutralize each spot by applying two drop of glacial acetic acid. Allow the paper to dry, spray with ninhydrin and place in the oven as before. Determine which fractions contain amino acids and save them for part B.




B. Paper chromatography


Obtain a piece of chromatography paper, handling it only by the upper edge or using gloves.  Mark lightly in pencil an origin line 2cm from the bottom edge of the paper.  Along the origin, spot 8 known amino acids and 1 unknown using capillary tubes and 2% amino acid solutions. Keep the spots small.  To ensure that enough amino acid is applied so that it can be detected at the end of the chromatography, make 4-6 applications to each spot, allowing the spot to dry between applications.


After all the spots have dried, place the paper in a chromatography tank containing the solvent mixture (acetonitrile and 0.1M ammonium acetate, 60:40, pH=4.0). The solvent level should be below the origin.  Allow the solvent to rise to 1cm from the top of the paper (or as time permits), remove the paper, and dry it in a hood.  Spray the paper lightly with ninhydrin in the hood, and develop the color in a 110oC oven for 10 minutes.  Note the color of each amino spot, and measure the Rf values.  Handle the chromatogram only with gloves!




Day 1 – Prepare the ion exchange column. Run the first ion exchange elution with citrate (pH=3.0).

Day 2 – Run the second (citrate, pH=6.0) and third (Tris, pH=9.0) ion exchange elutions. Paper chromatography of the mixture.


VI. For the lab report


A. Introduction - Include a one paragraph introduction explaining the purpose and general outline of the experiment.


B. Procedures - Include any procedure modification (if any).


C. Results - Include all data


1.      Scan the spot test results from ion chromatography and paste then in you report

2.      Scan your paper chromatogram

3.   Calculate Rf values.


D.        Discussion


1.         Using each of the two techniques, identify your amino acid and defend your choice.

2.         Draw the form or forms of the amino acid present at each point ion exchage fraction.

3.         Compare the two techniques as to accuracy, speed, difficulty, and amount of sample required. What are the advantages and disadvantages of each technique?


Questions for the Lab Exam


1.      Write the predominant ionic structure of Lys that would exist in aqueous solution at the following pH values: pH2, pH5 and pH11.

2.      Draw all possible ionization states of glutamic

3.      Predict the relative order of Rf values for the amino acids in the following mixture: S, K, L, V and A. Assume that the separation was carry out with the solvent system described in this experiment

4.      Calculate the net charge of the following amino acids at the given pH:


A@ pH=5, D@pH=4.0, K@8, S@6 and H@9


5.      Predict the order of elution of the following mixture of amino acids from a cation-exchange column: E, A, S, P and R. Assume that the procedure ws the same as in this experiment. Explain your answer.

6.      For each pair of amino acids listed below, predict which one would be first to elute from a cation-exchange columns. Assume the buffer pH begins at 3.25 and it is increased in a stepwise fashion


a) D and E b) S and V c) K and F d) F and A e) L and V f) G and A


7.         What are the major sources of error in the paper chromatography?

8.         Would phenylalanine have the same Rf value if a bigger piece of paper was used and the solvent was allowed to run twice as far?  Would phenylalanine have the same Rf value if the solvent was run the same distance, but the composition of the solvent mixture was changed?

9.         Write the reaction between ninhydrin and an amino acid.  Why does proline produce a yellow spot?




The amino acids in your unknowns are within the list of eight amino acids shown below.


Amino Acids             pKa Carboxyl                     pKa Amino                           pKa side chain

Proline                        2.00                                       10.60

Phenylalanine            2.58                                        9.24

Serine                        2.21                                         9.15

Threonine                  2.63                                        10.43

Aspartic acid            2.09                                         9.82                                       3.86

Glutamic acid            2.19                                         9.67                                       4.25

Lysine                         2.18                                         8.95                                       10.53

Histidine                    1.82                                          9.17                                       6.0






Experiment 2

Spectroscopic Analysis of Proteins

 The purpose of this experiment is to familiarize yourself with the theory and practice of spectrophotometry.   In this experiment, spectrophotometric techniques will be studied using a variety of proteins.

I. Spectroscopy

 Spectrophotometry is widely used in biochemistry.   Many biochemical compounds absorb light in the ultraviolet (200-400 nm), visible (400-700 nm), or near infrared (700-900 nm) regions of the spectrum.   Even if a particular compound does not absorb light itself, it can often be reacted with another compound to produce a light-absorbing substance.   Thus spectrophotometry allows for the qualitative and quantitative determination of biochemical compounds.   In addition, such techniques are often simple, fast, and clean.   Because of their sensitivity, these methods are frequently employed by biochemists.

 When white light is passed through a solution containing a colored compound, certain wavelengths of light are absorbed.   Which wavelengths (energies) of light are absorbed depends upon the chemical structure of the compound.   The absorption of a particular wavelength of light indicates the absorption of photons possessing particular energies, and the absorption of these photons increases various types of molecular energy (electronic, rotational, vibrational, etc.) of the compound.   Those wavelengths of light that are not absorbed by the compound are reflected or transmitted, and are responsible for the appearance of the compound.   Since different types of compounds have characteristic wavelengths at which they absorb light, it is possible to measure the absorbance of a substance at many different wavelengths to obtain its absorption spectrum.   A compound can often be qualitatively identified in this manner.

 Two basic laws govern quantitative spectroscopy.   The first is Lambert's law, which states that the proportion of incident light absorbed by a medium is independent of its intensity.   Mathematically,

I = IOe-al

where Io is the intensity of the incident light, I is the intensity of the transmitted light, a is the absorption coefficient of the medium, and l is the length of the light path through the medium in cm.   Converting to logarithmic form,

ln IO/I = al

Converting to logarithms to the base 10,

log IO/I = Kl,  where K = a/2.303

The term log IO/I is termed absorbance (A) or optical density (OD).   The term I/IO is referred to as the transmittance, so A = -log T where T is the transmittance.

 The second important law is Beer's law, which takes into account the effect of concentration.  Beer's law states that the light absorption is proportional to the number of molecules of absorbing substance through which the light passes.  We can then write

K = Ec

where E is the extinction coefficient (or absorption coefficient) and c is the concentration. Therefore,

log IO/I = Ecl = A

If concentration is measured as molarity, then E = e the value of the absorbance of a 1 M solution with a light path of 1 cm.   The value of E is constant only for a particular wavelength of light and a particular solvent system.   If A vs. c is plotted for a particular wavelength, then a straight line is obtained, the slope of which is El.   This is called a Beer's law plot, and it allows for the determination of concentrations. Thus the behavior of one light-absorbing substance is relatively simple, and if the values of E and l are known, the compound can be quantified from its absorbance.

 When dealing with a mixture of light-absorbing substances, often the behavior is more complicated.   The presence of a second component may cause a change in the light-absorbing properties of the first component.   The absorbance of the solution may not be equal to the sum of the absorbances of separate solutions of the two components, and there is no easy way to deal quantitatively with such a system.   In other cases, however, the two components do not interfere with each other, the absorbance of the mixture is additive, and Beer's law applies.   For each component at a given wavelength, A = Ecl. E and l are often combined into a single constant, k, called the extinction constant.   Since the absorbance of the mixture is additive,

A = k1 c1 + k2 c2

where A is the absorbance at the chosen wavelength, k1 is the extinction constant for the first component at that wavelength, c1 is the concentration of the first component, k2 is the extinction constant of the second component at that wavelength, and c2 is the concentration of the second component.

  All spectrophotometers have the same basic components.   1)   A light source must emit a steady amount of light in the required wavelength ranges.   A tungsten lamp provides for visible wavelengths, while a hydrogen lamp emits UV light.   2)   To limit the light to a certain wavelength, filters or a prism are used.   In most spectrophotometers, the light resulting from a particular wavelength setting has a maximum energy at this wavelength, but wavelengths slightly longer or shorter than the chosen one are not totally excluded.   For instance, when set at 500 nm, a spectrophotometer might emit light, 95% of which falls between 495 and 505 nm.   3)   To vary the intensity of the light, a slit in the light path can be adjusted.   4)   The light then passes through the sample which is held in a tube or cuvette.   In order to keep the light path constant, cuvettes are made very precisely, and sets of cuvettes are designed to be identical.   Nevertheless, it is desirable to measure the absorbance of plain solvent or buffer in all cuvettes that are to be used and to correct for variations as necessary.   Glass cuvettes are employed for visible wavelengths, but quartz cuvettes must be used for UV wavelengths.   5)   To detect the transmitted light, some sort of photocell or photomultiplier tube is required.

 There are several general considerations when using a spectrophotometer.   The wavelength at which measurements are made must be carefully chosen.   Light at that wavelength must be appreciably absorbed by the substance under analysis since accurate measurements will not be possible if the transmitted light is of too strong an intensity.   Similarly, the substance must not absorb too much light, or else the intensity of the transmitted light will be too weak for a precise determination.   The best compromise between too much and too little light comes in the region between 15 and 65% transmittance.   In addition, the wavelength of light chosen must fall in a region of the absorption spectrum of the substance where the absorbance is not changing rapidly with wavelength.   This is because spectrophotometers cannot isolate a single wavelength of light, but instead isolate a narrow band of wavelengths.   However, if all wavelengths in the band are absorbed to nearly the same extent, then the situation is almost the same as measuring absorbance at a single wavelength.

II. Protein Analysis

 The preceding discussion applies to both inorganic and biochemical spectrophotometry.   However, in biochemistry, only a few important compounds are highly colored and so can be studied directly.   Many biochemical molecules absorb UV light, but the amount of absorption is often too small for an accurate analysis if one is dealing with a limited amount of the compound to be analyzed.   To circumvent this difficulty, various reactions have been developed in which a particular type of biochemical compound is converted into a highly colored substance.   In performing such quantitative determinations, a series of solutions of the compound (or a similar one) are made, the concentrations of which are known.   Under defined conditions, the compound in these solutions is reacted with an excess of the color-forming reagents.   The absorbances of the solutions are measured, and a standard Beer's law plot  showing the variation of absorbance with concentration can be drawn.   In addition, a blank is prepared which contains all of the color-forming reagents, but none of the compound being assayed.   The absorbance of the blank serves as a control.   Then, the color-forming reaction can be performed with the sample where the concentration of the compound is unknown, and a quantitative determination can be made.

 Proteins in particular are a biochemical compound that must often be measured.   Proteins absorb UV light at 280 nm due to the presence of aromatic amino acids, allowing for a direct determination of protein.   Most pure protein solutions containing 1 mg/mL of protein have an absorbance of about 1.0 when the light path is 1 cm.   This method is simple, rapid, and allows for full recovery of the protein.   However, many other biochemicals absorb near this wavelength, making an accurate quantitation difficult.   Furthermore, different proteins absorb to different extents depending upon their aromatic amino acid content. To avoid these problems, the Biuret test was developed.  Compounds containing 2 or more peptide bonds take on a purple color when treated with dilute copper sulfate in an alkaline solution.   This reaction is reproducible from protein to protein, but it requires a relatively large amount of protein (1-20 mg).   The Folin-Ciocalteu (Lowry) test depends upon the reaction of protein with alkaline copper (as in the Biuret test) and the reduction of phosphomolybdate-phosphotungstate salts by tyrosine and tryptophan residues to yield a blue-green color.   This test is very sensitive, requiring as little as 5 g of protein.   However, variations from protein to protein can be substantial.

 In this experiment, proteins will be analyzed by spectroscopic methods.  Bovine serum albumin (BSA) and lysozyme, two colorless proteins, will be analyzed using absorbance at 280 nm, the Biuret test,  and the Lowry test.

III. Procedures

A. Absorbance at 280 nm
Fill two quartz cuvettes with deionized water.  Zero the spectrophotometer at 280 nm using one cuvette. Measure the absorbance of the second cuvette and make sure the absorbance is very close to zero so that the cuvettes are matched.
Using a stock solution of bovine serum albumin (500 g/mL in water) prepare 5.0 mL of BSA solutions that are 100, 200, 300, 400 and 500 g/mL.    Use deionized water for all dilutions.  Make a similar set of solutions for lysozyme
(500 g/mL in water).
Measure the absorbance of the protein solutions at 280 nm using a water blank.  Also measure the absorbance of the unknown.

 B. Biuret Test
Using a stock solution of bovine serum albumin, (10 mg/mL), prepare 1.0 mL of solutions that are 2, 4, 6, 8, and 10 mg/mL.   Make a similar set of solutions for lysozyme, (10 mg/mL).   Include a blank which is 1.0 mL of deionized water.  Also include the unknown.  To each solution add 4.0 mL of biuret reagent, mix, and let stand at room temperature for 30 minutes.   Read the absorbance at 550 nm.

 C. Lowry Test
Using a stock solution of bovine serum albumin (500 g/mL) prepare a series of 6 solutions where the protein concentration varies from 50-300 g of protein in 1.0 mL.   Make a similar set of solutions for lysozyme (500 g/mL).   Include a water blank and the unknown.  Make up an appropriate amount of alkaline copper reagent with the proportions of 1 mL of 1 percent CuSO4, 1 mL of 2 percent sodium tartrate, and 98 mL of 2 percent Na2CO3 in 0.1 N NaOH.   Add 6 mL of the alkaline copper reagent to each protein solution.   Mix and let stand for ten minutes.   Then add 0.3 mL of Folin-Ciocalteu reagent to each tube, mix immediately, and let stand for 30 minutes.   Read the absorbance at 500 nm.

IV. Schedule:

 Day 1 - Absorbance at 280 nm, Biuret Test and Lowry Test

V. For the lab report

A. Introduction - Include a one paragraph introduction.

B. Procedures - Include all procedures as actually done in the lab.

C. Results - Include all data

  1.    Show sample calculations for making all solutions.
  2.    Make a table for the data at 280 nm. Make the two Beer's law plots.
  3.    Make a table for the Biuret test data. Make the two Beer’s law plots.
  4.    Make a table for the Lowry test data.  Make the two Beer’s law plots.
  5.    Using the appropriate plot or plots determine the concentration of your unknown.

D. Discussion

1. Indicate which plot or plots you are using for your unknown and why.
2. If you did not know the identity of your protein, could you identify it from these tests?  Why or why not?
3. What are the advantages and disadvantages of these three different methods for measuring protein?  Compare these different methods as to accuracy, speed, sensitivity, and limitations of the techniques.

VI. Questions for the lab exam

 1.    What characteristics should a Beer's Law plot have?   Should BSA and lysozyme at 280 nm produce good Beer's Law plots?   Explain why or why not?

 2. What kind of a Beer's Law plot would be obtained at 280 nm if  lysozyme solutions between 0.5 mg/mL and 2.0 mg/mL were used?

 3.    Why is the timing important during the Biuret test and the Lowry test?

 4.    What is the purpose of the alkaline copper reagent in the Lowry test?

 5.    How did BSA and lysozyme differ in their reaction in the Biuret and Lowry tests?

 6.    Can you identify a protein using any of these methods?   Why or why not?

 7.    In the Biuret test, what would be the effect of using a blank that was just deionized water without the reagent added?

 8.    What kind of a Beer's Law plot would be obtained if protein solutions between 50 and 300 g/mL were run in the Biuret test?

 9.    What are the major sources of error in this experiment?

 10.    Describe the chemistry of the Biuret test and the Lowry test.

 11. Is one of the six Beer’s law plots most accurate for determining the concentration of the unknown?  Explain.

Experiment 3

Protein Molecular Weight Determination

The purpose of this experiment is to determine the molecular weight of a protein using gel filtration and SDS-gel electrophoresis.

I. Gel Filtration

 Gel filtration is a chromatographic technique that separates different molecules on the basis of size.  It is commonly used during protein purification to remove unwanted proteins from the protein being purified.  It can also be used to determine the molecular weight of a protein.

 In gel filtration, a dextran, polyacrylamide, or agarose gel is suspended in buffer and packed in a glass or plastic column.  The sample to be analyzed is applied to the top of the column and is allowed to run down into the gel.  A continuous supply of buffer is then provided at the top of the column, and, as the buffer runs through the column, the components in the sample are carried down the gel and separated.  The buffer is collected at the bottom of the column in fractions of constant volume (i.e. 1.0 mL), and all the fractions are analyzed for the presence of the various components in the sample.  The separation of the components is caused by cross-linking in the gel which creates pores.  Small molecules can penetrate the pores and so are slowed down and retained as they pass down the column.  Large molecules cannot penetrate the pores and so run down the column quickly.  Gels with different degrees of cross-linking (and therefore different sized pores) are commercially available to separate molecules in different molecular weight ranges.  In this experiment, Sephadex G-75 will be used.  This gel is a dextran capable of separating proteins with molecular weights between 3000 and 70,000.

 For a Sephadex column, the total volume, Vt, is equal to the sum of the volume of the gel matrix, the volume inside the gel matrix, and the volume outside the matrix.  The total volume is also , in most cases, equal to the amount of the buffer required to run a substance through the column (also known as eluting a substance) when the substance is small enough to completely penetrate the pores of the gel.  Such a substance is said to be completely included by the gel.  For Sephadex G-75, compounds with molecular weights less than 3000 are completely included.  The volume outside the gel matrix is known as the void volume, Vo.  This is the volume required to elute a substance so large that it cannot penetrate the pores at all.  Such a substance is said to be completely excluded by the gel.  For Sephadex G-75, proteins with molecular weights greater than 70,000 are completely excluded.  Compounds with intermediate molecular sizes that can partially penetrate the pores elute between the void volume and the total volume, and are said to be partially included by the gel.  The volume of buffer required to elute any given substance is known as the elution volume, Ve, of the compound.  Thus on Sephadex G-75, a protein with a molecular weight of 60,000 will be less included than a protein with a molecular weight of 30,000.  The larger protein will have a smaller elution volume and run through the column more quickly than the smaller protein.

 During protein purification, a mixture of many proteins can be subjected to gel filtration, and all proteins that have molecular weights different from the one being purified can be separated out.  Thus gel filtration is a powerful technique for purifying a protein.  Gel filtration can also be used to determine the molecular weight of a protein.  To do this, several proteins with known molecular weights are run on the column and their elution volumes determined.  If the elution volumes are then plotted against the log molecular weight of the corresponding proteins, a straight line is obtained for the separation range of the gel being used.  If the elution volume of a protein of unknown molecular weight is then found, it can be compared to the calibration curve and the molecular weight determined.

 Gel filtration has many advantages as a biochemical technique.  It is relatively simple to perform, and the mild conditions used tend to prevent denaturation of proteins, unlike some other techniques.  The protein that runs off the column can be collected and used for further analysis, so no protein is consumed in gel filtration.  However, there are also disadvantages as well.  The column must be carefully prepared to obtain optimal separation.  Any cracks or discontinuities in the column will interfere.  The size of the sample and the rate of buffer flow must be strictly controlled.  If a column is run several times, each run must be done under the exact same conditions in order to compare the different runs.  finally, some substances stick to Sephadex and do not elute properly.

 In this experiment, a Sephadex column will be prepared.  Five substances of known molecular weight will be used to construct a calibration curve.  Then another protein will be run on the column, its molecular weight determined, and the results compared to the known size of the protein.

II. SDS-gel electrophoresis
 The second method used to find the molecular weight of a protein will be
SDS-gel electrophoresis.  When a charged protein is placed in an electric field, it will migrate toward the oppositely charged region, and this is the basis of electrophoresis.  In most electrophoresis methods, the molecules being analyzed are placed on a solid support and then allowed to migrate.  For proteins, a polyacrylamide gel support is commonly used.  The proteins are applied to the gel, and the gel is contained in an electrophoresis cell, which in turn is connected to a power supply which creates a positive electrode and a negative electrode in the cell.  Buffer is used to complete the circuit in the cell between the gel and the electrode wires.  The buffer in the cell and contained in the gel is important, since its pH determines the charge on the protein molecules.

 Usually the determining factor in the separation of the molecules is their charge.  The more highly charged the molecule, the faster and farther it will move during electrophoresis.  With proteins, however, a second effect is seen, namely the size of the protein.  As a protein moves through the gel, it must overcome frictional forces which oppose its movement.  The larger the protein, the greater the frictional force.  Thus in most gels, the exact rate of movement of a particular protein depends on both its charge and its size.

 One type of electrophoresis is SDS-gel electrophoresis.  In this method, the proteins to be separated are denatured (usually in urea) and then mixed with the detergent SDS (sodium dodecyl sulfate).  SDS binds along the length of the protein, obscuring the protein’s own charges and giving all proteins the same negative charge per unit length.  Thus charge is essentially removed as a factor in the separation and size alone becomes important.  All proteins will move toward the positive electrode, but large proteins will move more slowly than small proteins.  The distance moved is inversely proportional to the log of the molecular weight. It is therefore possible to run several proteins of known molecular weight in an SDS-gel electrophoresis procedure, measure their migration distances, and construct a calibration curve.  The distance moved by a protein of unknown molecular weight can be compared to the standards and its size determined.

 Some proteins are colored and can be seen directly on a gel, but most are colorless.  To visualize most proteins, a staining procedure is needed.  Coomassie blue is a general protein stain, causing the protein to be come visible as blue bands within the gel.  Silver stain can detect very small amounts of proteins, causing them to turn brown-black.

III. Procedures

A. Gel filtration

  You will be supplied with one liter of buffer (0.05 M KCI, 0.05 M tris, pH8.0).  You will also be supplied with a Sephadex G-75 slurry.  Both the buffer and the Sephadex have been autoclaved to prevent microbial growth. De-gas the gel under a vacuum for 10 minutes to remove air bubbles.  Clean a chromatography column, making sure it runs freely but does not leak.  Pour about 2 inches of buffer into the column, clamp it off, and add the slurry by gently pouring it down the side of the column.  Fill the column, and let the gel settle for 20 minutes.  At that time, open the column and let the excess buffer run out.  Add any remaining slurry and allow the gel to settle.  Never permit the top of the column to dry out.  When adding buffer, do it gently with a pipet to avoid disturbing the top of the gel.

  To calibrate the column, allow the buffer to run down until the top of the gel is just dry.  Do not allow the gel to dry out.  Gently pipet 1.0 mL of calibration mixture onto the top of the column, being sure not to disturb the gel.  The first calibration mixture contains blue dextran (MW = 2,000,000), and myoglobin (MW = 16,900).  Let the calibration mixture run into the gel.  Connect the column to the fraction collector and begin to collect fractions of 1.0 mL.  Wash the calibration mixture completely into the gel with one mL of buffer.  Add more buffer to fill the column and maintain a constant supply.  Collect fractions until all the color is off the column.  Then wash the column with buffer.  Add 2.0 mL of deionized water to all fractions.  Measure the absorbance of the blue dextran fractions at 660 nm.  Measure the absorbance of the myoglobin fractions at 410 nm.

  Repeat the above procedure with the second calibration mixture containing bovine serum albumin (MW = 66,000), carbonic anhydrase (MW = 29,000) and aprotinin (MW = 6500).  Be sure to exactly duplicate the conditions of the first calibration run.  Measure the absorbance of appropriate fractions at 280 nm.

  Repeat the above procedure using a sample of cytochrome c.  Measure absorbance at 410 nm.  Determine the molecular weight of this protein.  Compare the result you obtain to the actual molecular weight of 12,400.

 B. SDS-gel electrophoresis

Remove the pre-cast polyacrylamide gel from the plastic package and rinse with deionized water.  Wash the sample wells with running buffer (0.24 M tris, 0.19 M glycine, 0.1% SDS, pH = 8.3), making sure no air bubbles remain.  Place the gel in the electrophoresis cell, add buffer, and check for leaks.

Heat both the protein standards and the cytochrome c sample in a boiling water bath for one minute.  The standards include the following proteins conjugated to various dyes.

 Protein  Normal                    MW       MW with dye    Color
      ovalbumin                         45,000     51,000                yellow
      carbonic anhydrase           29,000     30,000                orange
      trypsin inhibitor                 20,100     23,000                green
      a-lactalbumin                   14,200     16,500                purple
      aprotinin                             6,500     10,500                blue

Both the protein standards and the cytochrome c sample are prepared in sample buffer (0.063 M tris, 2% SDS, 20% glycerol, 100mM dithiothreitol, 0.005% bromophenol blue, pH 7.5).  The purpose of the glycerol is to make the solution heavy so it will sink beneath the running buffer and stay in the sample well, the purpose of the dithiothreitol is to reduce any disulfide bonds in the proteins, and the purpose of the bromophenol blue is to migrate ahead of the proteins when the gel is run and so indicate how the electrophoresis is progressing.

Ten microliters of both the protein standards and the cytochrome c are then loaded into the sample wells using micropipets.  Sample buffer is loaded into any wells which do not contain a protein solution to ensure even running of the gel.  Fill the buffer chambers with running buffer, close the electrophoresis cell, and connect the power supply with the power OFF.  Run the gel at 125 volts for about 60-90 minutes.  The run is complete when the bromophenol blue tracking dye reaches the bottom of the gel.  Shut off the power, disconnect the power supply, and remove the gel from the electrophoresis cell.  Using a gel knife, separate the plates and remove the top plate.  Measure the migration distance of each protein, noting the color of each protein band.

Determine the molecular weight of cytochrome c and compare the result to the actual molecular weight.

IV. Schedule:

 Day 1 - Pour column & run first calibration mixture
 Day 2 - Run second calibration mixture and cytochrome c
 Day 3 - Electrophoresis

V. For the lab report

 A. Introduction - Include a one paragraph introduction explaining the purpose and general outline of the experiment.

 B. Procedures - Include all procedures as actually done in the lab.

 C. Results - Include all data

  1. Plot an elution profile of absorbance vs. fraction number for both calibration mixtures.  Determine the elution volume of each substance.
  2. Construct a calibration curve by plotting log MW vs. elution volume for all the calibration substances.
  3. Plot an elution profile for the cytochrome c sample.  Find the protein’s elution volume.  Using the calibration curve, determine the MW of the protein.
  4. Plot a calibration curve of migration distance vs log molecular weight for the protein standards in the electrophorresis procedure.
   Find the cytochrome c’s migration distance and determine its MW.

 D. Discussion

  1. Compare the experimental MW of cytochrome c in both methods to its known MW.  Explain any discrepancies.

  2. Compare the two methods as to accuracy, speed, sample size, sources of error, etc. What are the advantages and disadvantages of each method.?

VI. Questions for the lab exam

 1. What problems might arise if the Sephadex is not de-gassed prior to pouring the column.

 2. What problems might arise if the top of the gel is disturbed or runs dry?

 3. What is the effect if too large a volume of sample is loaded onto a Sephadex column?

 4. What is the result if too few milligrams of protein are loaded onto a Sephadex column?  What if too many milligrams are placed on the column?

 5. Lysozyme is one of the few proteins that does not run accurately on Sephadex.  Lysozyme tends to stick to the Sephadex and takes longer to elute than expected.  What characteristics of lysozyme might cause this problem?

 6. If a mixture contained two proteins whose molecular weight varied by only 3000, would you expect to be able to separate the two proteins by gel filtration?  Why or why not?  What could you do to maximize the separation?  Would these proteins more likely be separated by SDS-gel electrophoresis?  Explain.

 7. Why is gel filtration a particularly good technique to use when purifying an enzyme?  Why is SDS-gel electrophoresis not a good technique to use when purifying an enzyme?
 8. Would you expect a fibrous protein with a molecular weight of 30,000 to have the same elution volume as a globular protein with a molecular weight of 30,000?  Explain.

 9. Suppose a gel filtration during the purification of enzyme X shows several peaks of protein.  If you do not know the molecular weight of X, how could you tell which protein peak contains X?  If several bands of protein show up during SDS-gel electrophoresis, could you tell which band contains X?

 10. What problems might arise if too small a sample is loaded onto the SDS-gel?  What if too large a sample is loaded onto the gel?

 11. What might be the effect on the electrophoresis if insufficient SDS is present?

 12. A protein which normally has a molecular weight of 40,000 appears to have a molecular weight of 20,000 when run on an SDS-gel.  What is the most likely explanation of this result?

 13. Ten minutes after starting an electrophoresis run, the protein samples have not migrated into the gel.  What is the most likely cause of this?

  14. Cytochrome c and myoglobin are both composed of a single polypeptide chain.  Both can be seen directly on a gel because both contain heme and are therefore red.  While cytochrome c gives an accurate molecular weight in SDS-gel electrophoresis, myoglobin does not.  With myoglobin, the red color runs quickly through the gel with a molecular weight much smaller than the known molecular weight.  What is the most likely explanation of this result?

Experiment 5

Analysis of Plant Pigments

 The purpose of this experiment is to extract, separate, and identify various plant pigments.

I. Lipids

 Lipids are water-insoluble substances found in cells.  Lipids have several functions.  They are structural components of membranes, they store metabolic fuel, and they are protective components of cell walls and skin.  Other compounds classified as lipids include some vitamins, pigments, and hormones.

 Lipids are chemically diverse and many different classes of lipids exist, including fatty acids, triglycerides, phospholipids, glycolipids, sphingolipids, steroids, and terpenes.  Plant pigments also have lipid-like properties.  The yellow pigments (carotenes and xanthophylls) and the green pigments (chlorophylls) are unsaturated polyalcohols and hydrocarbons.  Typical pigments found in plants include chlorophyll a, chlorophyll b, $-carotene, lutein, neoxanthin, violaxanthin, and phaeophytin.

II. Extraction and separation of lipids

 Lipids can be difficult to handle experimentally because of their insolubility in water and because they can be labile in oxygen and light.  In general, lipids can be extracted from tissue by using organic solvents.  Non-lipid contaminants can be removed from the organic phase by washing with an aqueous salt solution.  This produces a solution of lipids in organic solvent.  This solution can now be analyzed in several ways.

 One good method for separating different lipids is thin-layer chromatography.  TLC is very similar to paper chromatography, except that instead of using paper, a plastic sheet coated with a thin layer of silica, cellulose, etc. is used.  After spotting the lipids at one end of the TLC sheet, the sheet is placed in a chromatography tank with a solvent which moves up the thin-layer by capillary action.  The mobile phase consists of the non-polar solvent components which move up the thin-layer.  The stationary phase consists of the polar solvent components that bind to the thin-layer or the solvent that was used to make the thin-layer and is still present within the thin-layer.  Lipids move with Rf values that depend upon a particular lipid’s relative solubility in the mobile and stationary phases.

 Another method for separating lipids is adsorption chromatography.  Here a column is packed with a solid adsorbent (alumina, silica, etc.)  The lipids are added at the top of the column, and then the column is developed by adding a solvent at the top.  As the solvent runs down the column, it carries the lipids with it.  How fast a particular lipid runs down the column depends on its relative affinity for the adsorbent vs. the solvent.


 An alternate method for separating lipids, including pigments, is high performance liquid chromatography (HPLC).  HPLC is a relatively new technique that is actually an extension of traditional column liquid chromatographic methods such as adsorption chromatography and gel filtration.  In HPLC, a column is filled with a solid support, and some liquid (buffer, organic solvent, etc.) is run through the column.  Different molecules can be separated, analyzed, and identified depending upon their interaction with the solid support and the liquid.  The advantages of HPLC (and its differences compared with traditional column chromatography) are several.  First, the solid stationary phase consists of very small particles, creating a large amount of surface area.  This results in better separations and resolution than can be achieved with classical column chromatographic methods.  Second, the solid support can withstand very high pressures (up to 5000 psi.), and running the column under high pressure speeds up elution and makes HPLC a rapid technique.  Other advantages include the ability to re-use the column a large number of times, improved reproducibility between runs, and automated equipment.

 The basic components of an HPLC system are shown in the link below.  A solvent reservoir holds the liquid phase.  The solvent is drawn through a filter to remove any particulate matter, and then through a pump which provides a constant, reproducible flow of solvent through the column.  An injection port is used to introduce the sample into the column, and the column itself is generally made of stainless steel or glass-Teflon tubing, with a diameter of 2-30 mm and a length of 5-100 cm, packed with the stationary phase.  An oven surrounding the column provides a constant temperature, if required.  After the column is a detector.  Photometric detectors monitor UV or visible absorbance as compounds elute from the column.  Also used are fluorescent detectors and differential refractometers (which monitor refractive index).  The detector is connected to a recorder or computer for data acquisition.  Following the detector is a fraction collector for collecting samples.

HPLC: User's guide (go over this link to review all concepts related to HPLC)

 Stationary phases often consist of small porous particles with an inert solid core covered with a thin porous outer shell of silica, alumina, copolymers, or silica with bonded substituent groups.  Alternatively, microporous particles are made entirely of silica, alumina, etc. and allow for greater resolution.  The most common supports have particles of 4-10 m in size. Mobile phases must be of high purity and include a wide variety of organic solvents, solvent mixtures, and aqueous solutions.  Materials are available to run adsorption chromatography, gel filtration, and other types of column chromatography as HPLC, each producing separations according to the theoretical basis of the type of chromatography being performed.  Normal phase chromatography uses a polar solid support and a nonpolar mobile phase solvent.   Reversed-phase HPLC employs a non-polar solid support with a more polar solvent and is better for polar biomolecules.

 The separated compounds eluting from the column are described by their retention time (tr), which is the time necessary for elution of a particular compound.  The recorder produces an elution profile of the compounds.  If a mixture of three compounds is run on an HPLC column, a typical profile would be that shown in the link above, where each substance’s tr could be measured from its peak.  Most HPLC systems are equipped with a computer for data handling, including quantifying data by peak integration.

 HPLC works best with very clean samples, so crude samples must be pre-treated to remove salts, metal ions, detergents, and particulates.  The concentration of the sample must be properly adjusted, and conditions must be optimized, including type of support, type of solvent, flow rate, etc.  When performed correctly, HPLC is a powerful tool for biomolecule separation.  Very small samples can be analyzed or, with a bigger column, large amounts of a biomolecule can be isolated.

IV. Procedures

 A. Extraction

  Two grams of spinach leaves, with the veins removed, are cut up and placed in 100 mL of boiling water for two minutes.  The water is then cooled and discarded.  The leaves are then extracted for about ten minutes with 25 mL of 90 percent methanol / 10 percent diethyl ether.  The extraction should consist of gentle swirling to avoid generating small bits of leaves.  Pour the organic mixture into a separatory funnel.  Add 25 mL of 70 percent methanol/30 percent diethyl ether to the leaves and repeat the extraction.  Combine the organic mixtures in the separatory funnel.  To the extract, add 10 mL of petroleum ether and 25 mL of saturated aqueous NaCl solution.  Shake to mix thoroughly and discard the lower layer, saving the green organic phase.  Wash the pigments twice with 25 mL of water.  Then place the organic phase in a small beaker and concentrate to 1 mL using a stream of nitrogen in the dark.

 B. Thin layer chromatography

  “Format” a cellulose TLC plate using soft pencil and only the very mildest of pressure.  Make a short (0.5 - 1.0 cm) horizontal mark on both sides of the plate 1 cm up from the bottom to mark the origin.  There should be 3.0+ cm left free in the middle of the plate.

  To pre-elute the plate, place 50 mL of acetone: petroleum ether (1:1, v/v) into a clean TLC developing chamber, insert the TLC plate which has already been formatted, cover, and let develop until the solvent front is about cm from the top.  Remove the plate and gently swing it in the air until it is dry.  Touch only the top of the plate and lay it down only onto a piece of aluminum foil.  The presence of fats, oils and greases from the human body, and other contamination, leads to smearing of the chromatogram and thus gives a poor separation.

  Load the sample onto the TLC plate using a capillary tube.  Insert the capillary into the extract and allow the green solution to rise up in the tube.  Using your finger, as you do with a pipet, stopper the tube and withdraw it from the solution.  “Spot” the origin by barely touching the plate in the area between the origin marks with the capillary tube.  Done properly, only a very small amount of solution will be pulled out of the tube and onto the cellulose of the plate.  It is necessary to touch the plate and remove the tube rapidly in order to keep the size of the loading spot to a minimum.  Repeat this across the plate to form a series of overlapping spots; these will begin to resemble a small streak.  Let the first loading dry and repeat several times until your origin is relatively dark green.  This “loads” enough pigment onto the TLC plate such that separated components can be visualized.

To develop the chromatogram, empty the TLC chamber into an “Organic Solvent” waste container and refill with 50 mL acetone: petroleum ether (1:9, v/v) as the developing solvent.  Place your TLC plate into the TLC tank, cover and let the separation develop.  Try to keep the tank and the pigments out of bright light as much as possible.  Light, in the presence of molecular oxygen (O2), degrades these pigments (photooxidation).  Stop the TLC run when the pigments have separated.  Air dry the chromatogram.  Draw a picture of your TLC plate, indicating the size and color of the spots.

To recover the pigments, prepare 6 TLC ‘spot removers’ using Pasteur pipets and glass wool.  Fold a small amount of glass wool, insert it into the top (large end) of the Pasteur pipet and push down about 0.5 cm, leaving the remaining ‘fluffy’ excess sticking up.  Trim off the excess, forming a small glass plug in the pipet.  Push the formed plug down to the tip constriction using a piece of wire, but do not force - just snug!

Label the 6 TLC ‘spot removers’ #1-6.  Prepare 6 small test tubes by putting them in a rack and labelling them #1-6.

Begin with the dark yellow spot at the top, which will be #1.  Hook up a piece of vacuum tubing to the vacuum, insert the tip (pointed end) of a Pasteur pipet ‘spot remover’ into the vacuum tubing, and using the larger open end of the pipet scrape the pigmented material off the TLC plate slowly.  You will note that the material is vacuumed into the pipet and is stopped by the glass wool plug.  Perform this ‘vacuuming’ in such a way as to collect the majority of each pigment.  Do not collect material from any overlapping areas.  Invert the pipet so that the glass wool plug is down, remove from the vacuum tubing and place (tip down) into the correspondingly labelled test tube.  Repeat for spots #2-6.

To elute the pigment from the cellulose, raise the ‘spot remover’ about 2 cm up from the bottom of the test tube and rinse the inside of the pipet twice with 1.0 mL of 100% acetone.  Discard the ‘spot remover’ in the broken glass receptacle.  Repeat for each of the 6 spots.


    The HPLC system for this experiment has the following features.  First, it uses a column in which the solid support consists of silica beads (10 m) linked to a hydrocarbon chain of 18 carbons, creating octadecylsilane (ODS).  This makes the column material non-polar, producing reversed phase HPLC.  Second, the instrument employs a single pump and therefore a single solvent system, unlike more complicated HPLC systems with multiple pumps that can use multiple solvents and solvent gradients.  The solvent used here is one part acetonitrile: water (9:1) and one part ethyl acetate.  Third, the detector is a dual wavelength detector that can monitor absorbance of the pigments coming off the column at two different wavelengths simultaneously.  The detector is set for 410 nm (to detect the gray phaeophytin) and 440nm (to detect the green and yellow pigments).  Fourth, sample loading is accomplished by using an autoinjector that automatically loads the sample onto the column.

  To run the pigment sample, first evaporate the sample in petroleum ether almost to dryness under nitrogen.  Add some 100% acetone and evaporate again.  Dissolve pigments in 90% aqueous acetone.  Draw some of the sample into a syringe and filter through a 0.45 micron filter.  Expel the sample into a clean glass vial.  Place the sample in the autosampler. With the help of the instructor, program the HPLC to load the sample.  As the column runs, data will be collected and displayed.

V For the lab report

 A. Include a one paragraph introduction.

 B. Include all procedures.

 C. Results - Include all data

  1. Draw a picture of your chromatogram showing the pigment bands you observed.  Draw a picture of your column showing the bands you observed.

  2. Include the HPLC chromatogram.

  3. Include the absorbance spectra of your samples.

  4. Include the known absorbance spectra of the various pigments.

 D. Discussion

  1. By comparing the spectrum of each of your pigment bands to the known spectra, determine what pigment or pigments are present in each pigment band obtained in the different techniques.  Be sure to explain how you made this determination by citing absorption peaks, band widths, color, polarity, chemical structure, etc.

  2. Explain the reasons for any poor separations that occured.

  3. Compare the methods as to speed, resolution, sample size, etc.

 VI. Questions for the lab exam

 1. Why are the spinach leaves placed in boiling water?

 2. Why is the organic phase washed with a salt solution?

 3. Why is the concentration done in the dark under nitrogen?

 4. What will be the result if the TLC chromatogram is overloaded?
  What if it is underloaded?

 5. Why does HPLC give better resolution of the pigments compared to other methods?

 6. Why does the pigment with the largest Rf value in TLC have the largest retention time in reversed-phase HPLC?

Experiment 6

Identification of a Monosaccharide

The purpose of this experiment is to identify an unknown monosaccharide using polarimetry and NMR spectroscospy.

I. Carbohydrates

Carbohydrates are an important class of biochemical compounds having many functions in metabolism.  The simplest carbohydrates are the monosaccharides which have the empirical formula (CH2O)n where n is greater than or equal to three.  The carbon skeleton is unbranched and each carbon except one has a hydroxyl group.  The remaining carbon contains a carbonyl oxygen.  Monosaccharides can be aldoses or ketoses, and they can belong to either the D or L family of stereoisomers.

Figure 1. The stereochemical relationship (Fisher Projections) among the D-aldoses with three to six carbon atoms.

Figure 2. The stereochemical relationship (Fisher Projections) among the D-ketoses with three to six carbon atoms.

Alcohols react with the carbonyl groups of aldehydes and ketones to form hemiacetals and hemiketals, respectively. Consequently, the linear form of most monosaccharides will undergo cyclization reaction reaction to form a structure of 5 or 6  membered rings.


Cyclic Sugar Have Two Anomeric Forms. The cyclization of a monosaccharide renders the former carbonyl asymmetric. The resulting pair of diastereomers are known as anomers and the hemiacetal or hemiketal carbon is referred as the anomeric carbon.

In this experiment we will determine the identity of a monosaccharide and we will evaluate the possibility of formation of anomers and their specific configuration using polarimetry and NMR spectroscopy.

II. Polarimetry

Polarimetry is commonly used to analyze chiral substances. Most biomolecules are chiral and hence rotate polarized light. Carbohydrates are particurlarly amenable for analysis using polarimetry.  The magnitude and direction of rotation of the plane of polarized light by an asymmetric compound is a specific physical property of the compound that may be used for its characterization.  When plane polarized light is passed through a solution of a pure asymmetric compound, the extent of rotation depends upon the concentration of the compound, the length of the light path through the solution, the wavelength of the light, and the temperature.  These variables are related by the following equation:

         [a]TD = [a] obs
                             l x c

where [a]T is the specific rotation of the compound at temperature T using the D line of the sodium spectrum as the light source, [a] obs is the observed rotation, I is the path length through the solution in decimeters (usually 2 dm), and c is the concentration of the compound in g/mL.

The instrument used to make these measurements is called a polarimeter.  A polarimeter takes light vibrating in all planes, isolates the light vibrating in a single plane, projects the light through a tube filled with a solution of an asymmetric compound, and measures the amount of rotation.  The polarimetry tubes and end plates are precision items and must be handled with care.



Required Reading: Chapter 5C (Boyer's 3rd ed.). Chapter 4.5 (Garrett & Grisham, 2nd Ed. Lecture Textbook). Usually most organic chemistry textbook contain excellent introduction to NMR spectroscopy, i. e. Organic Chemistry by  Solomons' (most FAU organic course use this textbook) NMR is covered extensible in Chapter 9.

Nuclear magnetic resonance spectrometry (NMR) is the only technique that provides details of the molecular structure and dynamics of a compound at the level of atomic solution in solution.

Modern NMR spectroscopy is frequently divided into several categories;
   1.High resolution mode on homogenous solutions.
   2.High power mode on highly relaxing nuclei which exhibit very broad lines, or polymers etc.
   3.The study of solids using e.g. Magic angle spinning techniques.
   4.NMR 3D imaging to resolutions of ~ 1 mm.

This experiment  is concerned entirely with the first category. The types of information accessible via high resolution NMR include;
   1.Functional group analysis (chemical shifts)
   2.Bonding connectivity and orientation (J coupling),
   3.Through space connectivity (Overhauser effect)
   4.Molecular Conformations, DNA, peptide and enzyme sequence and structure.
   5.Chemical dynamics (Lineshapes, relaxation phenomena).

The nuclei of all elements carry a charge. When the spins of the protons and neutrons comprising these nuclei are not paired, the overall spin of the charged nucleus generates a magnetic dipole along the spin axis, and the intrinsic magnitude of this dipole is a fundamental nuclear property called the nuclear magnetic moment, . The symmetry of the charge distribution in the nucleus is a function of its internal structure and if this is spherical (i.e. analogous to the symmetry of a 1s hydrogen orbital), it is said to have a corresponding spin angular momentum number of I=1/2, of which examples are 1H, 13C, 15N, 19F, 31P etc. Nuclei which have a non spherical charge distribution (analogous to e.g. a hydrogen 3d orbital) have higher spin numbers (e.g. 10B, 14N etc.) outside the scope of this particular experiment.  1H, 13C, 15N and 31P are biologically important nuclei that have net spins. 1H is the most common isotope studied by NMR.

NMR involves the absorption of radio waves by a molecule in a strong magnetic field, Bo . In quantum mechanical terms, the nuclear magnetic moment of a nucleus can align with an externally applied magnetic field of strength Bo in only 2I+1 ways, either re-inforcing or opposing Bo. The energetically preferred orientation has the magnetic moment aligned parallel with the applied field (spin +1/2) and is often given the notation a, whereas the higher energy anti parallel orientation (spin -1/2) is referred to as b. The rotational axis of the spinning nucleus cannot be orientated exactly parallel (or anti parallel) with the direction of the applied field Bo (defined in our coordinate system as about the z axis) but must precess about this field at an angle (for protons about 54š) with an angular velocity given by the expression;

 wo = gBo ...(1) (the Larmor frequency, in Hz)

The constant g is called the magnetogyric ratio and relates the magnetic moment m and the spin number I for any specific nucleus;

 g = 2pm/hI ...(2) (h is Planck's constant)

For a single nucleus with I=1/2 and positive g, only one transition is possible (D I=1, a single quantum transition) between the two energy levels;

NMR is all about how to interpret such transitions in terms of chemical structure. We will first consider the energy of a typical NMR transition. If angular velocity is related to frequency by wo = 2n, then

 n = gBo/2 ...(3)

It follows that proton NMR transitions (DI=1) have the following energy;

hn = DE = hgBo./2p ...(4)

For a proton g = 26.75 x 107 rad T-1 s-1 and Bo ~ 2T, DE = 6 x 10-26 J. The relative populations of the higher (n2) and lower (n1) energy levels at room temperature are given by the Boltzmann law;

 n2/n1 = e-DE/kT ~ 0.99999. ...(5)

For NMR, this means that the probability of observing a transition from n1 to n2 is only slightly greater than that for a downward transition, i.e. the overall probability of observing absorption of energy is quite small. This relationship also explains why a larger Bo favors sensitivity in NMR measurements, increasing as it does the difference between the two Boltzmann levels, and why NMR becomes more sensitive at lower temperatures.

Under the influence of Bo,  nuclei create a magnetic moment which would  be aligned with the external field,  is the most stable energetic state.  The magnetic moments can also be aligned against the external field, which is a higher energy state.  Radio waves have the proper energy to be absorbed, causing the nuclei to “flip” between these two alignments.  The exact energy required for a given nucleus to flip depends on both the energy of the radio waves and the strength of the magnetic field.  Modern NMR instruments use a radio frequency pulse 2-20 microseconds through the sample to excite all nuclei simultaneously to the higher energy state.  After the pulse, nuclei flip back to the original lower energy state (resonate), emitting energy of a particular frequency which is then detected.  The emitted frequencies are stored in a computer.  As the pulse is repeated many times, the data is compiled and produces an NMR spectrum which show which radio frequencies are emitted by the nuclei of that sample.  The exact frequency at which a given nucleus resonates depends upon the electronic configuration, the location of neighboring atoms, and other types of intramolecular and intermolecular forces.

The NMR instrument is composed of a strong magnet, a radio frequency transmitter, a radio frequency detector, a computer, and a sample holder which correctly positions the sample and spins it to ensure it is subjected to a homogeneous magnetic field.  Instruments with different magnetic field strengths are available, and are designated by the frequency at which protons resonate in their magnetic field, typically between 60 and 900 megahertz (MHz).  Thus a 500 MHz instrument has a stronger magnetic field than a 300 MHz instrument.  The stronger the magnetic field, the better the chemical shift dispersion.

Proton NMR spectroscopy looks at signals created by 1H nuclei in the sample.  Protons attached to different atoms within a molecule will require different energies to resonate and so will give different signals within the NMR spectrum.  The number of signals gives information about how many different types of protons are present.  The frequency at which each signal occurs indicates the different electronic environments of the protons.  The signal height (intensity) shows how many protons are producing that signal, and the splitting of one signal into several peaks is caused by the presence of other nearby protons.  All these characteristics, when put together, describe the molecular structure of the sample.

For example, the compound ethane

                CH3 - CH3

has six protons, all of which are structurally identical.  The NMR spectrum therefore shows just one signal.  In contrast, the compound propane

                CH3 - CH2 - CH3

has protons of two different types, six equivalent protons on the terminal carbons and two equivalent protons on the central carbon.  The NMR spectrum shows two signals with a relative intensity of 3:1.  Furthermore, the fact that each carbon has non-equivalent protons on the adjacent carbon produces interactions that cause each signal to be split into multiple peaks.  Protons that interact in this way are said to be ‘coupled’ and the resulting splitting of signals is termed spin-spin splitting.  The number of peaks into which a signal is split is given by the n + 1 rule, where n equals the number of adjacent, non-equivalent protons.  Thus the signal produced by the six protons on the terminal carbons of propane is split into three peaks since there are two non-equivalent protons on the central carbon.  Likewise, the signal caused by the two protons on the central carbon is split into seven peaks because there are six neighboring non-equivalent protons at the ends of the molecule.

The position of the signal in the spectrum (the radio frequency of the signal) depends upon the electronic environment of the proton.  For instance, a proton of a carbonyl group has relatively little electron density around it because of the electron-withdrawing effect of the oxygen.  Such a proton is termed ‘ deshielded’ and absorbs low energy radio frequencies.  In contrast, the methyl group protons of an alkane have a greater electron density, are said to be more “shielded”, and absorb higher energy radio frequencies.  The positions of the signals are called chemical shifts and are given in units of parts per million (ppm) of the total magnetic field.  The protons of tetramethylsilane are used as a standard and assigned a value of 0.0 ppm, creating the d scale.  The proton chemical shifts of most compounds then appear at 0-15 ppm.  This scale is used rather than frequency units because  values are independent of the strength of the magnetic field and so can be compared between different NMR instruments.  Other factors that can influence chemical shifts include hydrogen bond formation, temperature, solvent, and sample concentration.

 To obtain a proton NMR spectrum, the sample is dissolved in a solvent that either contains no hydrogens, or is deuterated (D2O).  A small amount of solution
(~ 0.5-0.8 mL) is needed at a concentration from the micromolar to the milimolar range.

When the proton NMR spectrum is determined for a monosaccharide, a relatively complex spectrum is obtained.  The large number of different types of protons in a monosaccharide molecule gives rise to many signals, and neighboring non-equivalent protons produce complicated splitting patterns.  Despite the close isomeric relationship of several sugars, a distinct spectrum results for each monosaccharide because the protons in the isomers are sterically non-equivalent. Thus if the NMR spectra are obtained for different monosaccharides, the spectrum of an unknown sugar can be compared and the unknown identified.

By way of contrast, a carbon 13 spectrum of a monosaccharide is relatively simple as it contains signals that correspond only to uncoupled carbons atoms in your sample (This topic will be discuss in the lab lecture. See also Solomon's Chapter 9). You will record the 13C spectrum of your sample and you will interpret the spectra with the aid of the 2D heteronuclear experiment HSQC (see below).

Nowadays, most structural determinations using NMR spectroscopy make use two dimensional NMR experiments (2D-NMR). The literature that regards the theory and practice of 2D NMR spectroscopy is extensive and beyond the level of this course. However, there are two experiments that are widely used and relatively easy to interpret: COSY (COrrelation SpectrospY) and HSQC (Hetero Single Quantum Coherence). COSY allows to determine which protons are coupled to each other. HSQC allows to determine which protons are correlated to which carbons. The principles and practical aspects of these 2D-NMR experiments will be discussed in detail in the lab lecture.

Links to relevant NMR sites (You are required to browse through these sites)
NMR in Biochemistry. an excert from your textbook on the role of NMR in biochemistry
NMR Spectroscopy: Principles and Applications. Softmore level NMR lecture by Henry Rzepa from the Imperial College in London
Basic 2D-NMR. An interactive web site that presents the basic general concepts behind 2D-NMR spectroscopy

IV. Procedures

A. Polarimetry

To calibrate the polarimeter, rinse the polarimeter tube twice with deionized water.  Then fill it completely with deionized water, close, wipe the outside, and place in the polarimeter.   Rotate the analyzer of the polarimeter until both halves of the circular field, as seen through the eyepiece, are equally illuminated.  Read the rotation from the circular vernier scale.  The reading should be close to zero since water does not rotate light.  Change the analyzer, bring back to make the field uniform, and read again. Take three readings and average.  This value is then used to correct all other readings.

Take a 10% sugar solution (one of the five knowns), rinse out the polarimeter tube, fill the tube, and read the rotation.  Make three readings and average.  Repeat for the other four knowns and one unknown.  Since these solutions are not fresh, they will contain the equilibrium mixtures of a and b forms for each sugar.  The known specific rotation for each sugar is shown below.

      [a] D20
    D-fructose -92.0
    D-galactose +80.2
    D-glucose +52.7
    D-xylose +18.8
    D-arabinose -108


You will be given a sample of your unknown (30mg/mL in D2O).  With the help of an instructor, record the following NMR spectraof you monosaccharide:
1D proton
1D 13C
2D 1H-13C HSQC

You will compare some of your spectra with the spectra of the five known sugars.

V. For the lab report

A. Include one or two pages of  introduction.

B. Include all data recorded.

C. Results - Include all data

1. Write out the of the five different sugars.

2. Calculate the specific rotation of each known sugar as well as the unknown.

3. Include the NMR spectrum of each known sugar as well as the spectrum of the unknown.

D. Discussion

  1. From the results of the polarimetry, identify your unknown.

  2. From the results of the NMR, identify your unknown.

  3. Compare the results of the two  methods.  Also compare the methods as to accuracy, speed, sample size, advantages, limitations, etc.

VI. Questions for the lab exam

 1. Glucose exists primarily as a ring form.  If glucose existed primarily as a straight-chain molecule, what differences, if any, would you observe in its oxidation by periodate?

 2. What would be the effect on the polarimetry results if fresh solutions of the sugars were prepared and used?  What additional source of error might there be using fresh sugar solutions?

 3. What are the major sources of error in each of these two methods?