BSc Course ‘Biomedical Science’

St. George’s Hospital Medical School

DEGREE-SPECIFIC HANDBOOK

Containing:

Materials for Lecture Course ‘Chemistry in Life Sciences’,

Plan for Tutorials Accompanying the

‘Chemistry in Life Sciences’

Degree-Specific Timetable.

First term, 2002.

This Booklet Contains:

Page

1. Lecture Course ‘Chemistry In Life Sciences’ ………..……………….. 3

2. Course Overview (List of lectures etc.) ……………..………………….. 4

3. Tutorials ‘Chemistry in Life Sciences’ ….……………………………….. 29

4. Student Groupings for Tutorials, first term………..………………………………………………………………. .30

5. CLS Tutorial study material……………………………………………….32

6. Tutorials MCBHD course ………………………………………………52

7. Laboratory Practical Course……………………………………………….54

8. Overview on Term 1 Assessment ………………………………………. 54

9. Total Biomed Degree-Specific CFP Timetable …………………………54

Lecture course: CHEMISTRY IN LIFE SCIENCES

Unit Organiser: Prof BM Austen(email sghk200:sghms.ac.uk

AIMS OF THE COURSE

• To provide you with a basic understanding the chemical principles that underlie life science. This involves the chemistry of important molecules involved in life processes, in terms of structure, stereochemistry and reactivity.

• To familiarise you with the chemical and physical principles of solution chemistry and spectroscopy used for analysis in the laboratory in Biomedical Science.

• To enable you to understand the basis of covalent and non-covalent forces important for the three-dimensional structure and function of proteins, nucleic acids and other biologically relevant macromolecules.

• To provide you with the background knowledge that is required for the understanding of the principles underlying the methods frequently applied in the molecular life sciences.

LEARNING OBJECTIVES

After having completed this course, you should be able to:

• Appreciate the chemistry of important molecules involved in life processes in terms of structure, stereochemistry and reactivity.

• Know how to interpret instructions for making up solutions used in biochemical research and understand how spectroscopy is used in analyses in the laboratory in Biomedical sciences.

• Appreciate the thermodynamic bases of reactivity, and the spectral properties of biological molecules, as far as they are important in Biochemistry.

• Describe the most important properties of proteins and nucleic acids, and the relevance of these properties for biomedical laboratory methods used for isolation and analysis of these macromolecules.

RECOMMENDED READING MATERIALS:

1. Biomedical Sciences Explained (series editor: CJ Pallister)

JD Blackstock: ‘Biochemistry’, Butterworth & Heinemann, 1998.

2. McMurray: Principles of Organic Chemistry (1997?)

3. K. Wilson & J. Walker: Principles and Techniques of Practical Biochemistry.

(5th edition, Cambrigde University Press, 2000).

In addition, websites containing useful diagrammatic material and self-directed tutorials are given in this text as hyperlinks. This material will soon be available on the intranet; on my webpages at sghms.ac.uk/depts/ndu, under the tab teaching. When opened on a computer in the library, simply click on a hyperlink, and the relevant web page will be opened. However, if you face difficulties in accessing these websites, you can also find similar information in textbooks. t

COURSE OVERVIEW

Teaching will be via a lecture course on Thursdays,, and by biweekly tutorials. See the lecture outline below for further course materials, venues and times. The complete timetable for the first term Biomedical Sciences course can be found at the end of this booklet.

BSc in Biomedical Science Course, first term 2002.

Chemistry in Life Sciences (CLS) course

List of lectures:

Lecture 1: page

Thurs Oct 3^rd 15:15 The Chemistry of biological solutions ………………..………… 6

Prof. B. Austen

Lecture 2:

Thurs Oct 10^th 15:15 Protein Structure ………………………………………………….. 10

Prof. B. Austen

Lecture 3:

Thurs Oct 17^th 15:15 Protein Isolation …………………………………………………… 17

Prof. B. Austen

Lecture 4:

Thurs Oct 24^th 15:15 Protein Analysis…………………………………………………. … 20

Prof B Austen

Lecture 5:

Thurs Oct 31st 15:15 Carbohydrates and Stereochemistry

Prof B Austen ….…...…………………………..23

Lecture 6:

Thurs Nov 7^th 15:15 Properties and reactions of nucleic acids ……………………………27

Prof. M. Clemens

Lecture 7:

Thurs Nov 14^th 15:15 Spectroscopy ………………………………………………………26

Dr. M. Roberts

Lecture 8:

Thurs Nov 21^st 15:15 Tracer methods in Biological Sciences ………………………… 28

Dr. U. Bommer

Lecture 1 CHEMISTRY OF BIOLOGICAL SOLUTIONS.

THURS OCT 3rd, 15:15 Teaching Room 10

Prof Brian Austen

LEARNING INTENTIONS

To equip you with background chemical knowledge of laboratory methods used to make up and store solutions with predetermined content and pH suitable for performing biological reactions or analyses. At the end of this teaching, students should be able to prepare and pH solutions from details given in the format normally found in scientific publications.

SUMMARY

The concept of molecular weight will be revised, and the use of this concept in making solutions of predetermined concentration from information recorded in scientific publications will be described. The Bronsted-Lowry definitions of acids and bases will be reviewed, and the differential titration behaviour of weak and strong acids and bases described. The term ‘pK_a‘ will be defined, and the suitability of common buffers for specific purposes will be rationalised. The calibration and use of pH meters will be described. The behaviour of certain buffers at different temperatures will be explained. The chemical rationale for solubility of solutes in water or organic solvents will be explained, and the use of alcohols and other water-mixable solvents for introduction of insoluble substances into aqueous media discussed. Advice will be given on how to interpret concentrations of active biological substances given in activity units. Storage conditions for typical biological reagents will be rationalised.

CONTENT

Water is the solvent of choice for biological material because of its unique properties. These are relatively high melting point and boiling point, high heat of vaporisation, and a high dielectric constant giving the capacity to store high amounts of electrical energy, e.g. in polarised membranes. This is due to the molecular structure of water, in which two pairs of unpaired electrons on oxygen allows it to make hydrogen bonds with hydrogens on adjacent molecules. Thus each water molecule will make and remake hydrogen bonds with adjacent molecules allowing high fluidity. Water, then is the chosen solvent for preparing solutions in which to handle biological material in the laboratory, including living cells and purified macromolecules such as proteins.

Concentrations: The are different ways of describing the concentration of a substance in a solution. The most common ones are:

1. Molarity, (moles per litre final solution; abbreviated: ‘M’);

3. Percent (% (w/v) – g solid substance per 100 ml solution, or % (v/v) – ml liquid per 100 ml solution). (Please refer to p. 28 of your practical booklet for definitions of moles,

concentrations and other measures.)

In science, the usual way of describing solutions is molar (M), as the different substances react with each other based on their molar ratio. It is therefore important that you are familiar with the definitions and recalculations between molar and percent.

It is important to note that both definitions refer to the volume of the final solution. For example, to make up 100 ml of 0.1 M sodium hydroxide, adding 100 ml of water to 0.4 g of sodium hydroxide would be inaccurate as it would not take into account any volume of the solute or volume changes the sodium hydroxide causes the solvent. It would be accurate to make up the volume of the solution to 100 ml in a volumetric flask after the solute had dissolved in a smaller volume.

One mole of a substance is defined as the amount of substance in g, which contains as many elementary entities as 12 g of pure carbon C¹² or 1 g of H¹.

An entity is a single particle that can be described exactly, and can be an atom, a molecule, or an ion of an element or compound. For a compound that exists in a well-defined molecular structure such as ethanol, C₂H₅OH, the molar mass is readily calculated by summing atomic weights of all atoms in the molecule. For other substances, such as salts, which can exist in many ionic states with a variety of counter-ions in infinite arrays and variable degrees of water of hydration, this may be more problematical. Conveniently, most manufacturers record the molar masses (as Formula weights (FW) of the common laboratory chemicals they sell on the outside of the storage cartons.

Concentrations required may be much SMALLER than moles/l. These are abbreviated as follows: Millimolar (mM): 10^-3moles/l,

Micromolar (M): 10^-6moles/l,

Nanomolar (nM): 10^-9moles/l,

Picomolar (pM): 10^-12 moles/l,

Femtomolar (fM): 10^-15 moles/l.

Buffers are normally composed of a weak acid/base combined with conjugate strong base/acid. Polyprotic bases and acids, i.e. those with several groups that can ionise, also act as buffers.

Biological solutions are often buffered to avoid wide changes in pH during handling.

The Bronsted-Lowry definition of an acid is a substance that will donate a proton, and the definition of a base is a substance that accepts a proton.

Acids differ in their proton-donating ability, strong acids, such as HCl react almost completely with water, whereas weak acids such as acetic acid react only slightly.

The strength of any acid is given by its acidity constant K_a

HA + H₂O  A^- + H₃O⁺

K_a = [H₃O⁺][A^-]

[HA]

As the spread of values of equilibrium constants is so great, it is more convenient to use the value pK_a where pK_a = -log K_a.

The stronger the acid, the smaller the value of its pK_a.

Examples for ‘acids’ (pK_a in brackets): Ethanol (16), Water (15.74), Acetic acid (4.76),

Nitric acid (-1.3), HCl (-7.0).

A strong acid is one that loses its proton most readily, so its conjugate base does not hold the proton tightly, so it is weak.

An acid will donate its proton to the conjugate base of any acid with a larger pK_a

e.g.; CH₃----COOH + ^-O-----H  CH₃ ---- COO^- + H --- O — H

pK_a : 4.76 pK_a : 15.74

The events occurring during these interactions can be followed by a plot of pH during the stepwise addition of one component to a fixed component of the other component. This is called a titration curve.

The changes occur according to the Henderson-Hasselbalch equation:

_PH = pK_a + log10[base]

[acid]

Fig.: The curves shows the titration a weak acids with a strong base. The points of inflections centre on points at which the addition of strong base (or acid) makes relatively little difference to pH. The points of inflection occur when half of the equivalent of titrant has been added, and are at the pKa values of the weak components. Thus an adequate definition of pK_a is "the pH at which the acid (or base) is half ionised”, i.e. when [base] = [acid]. The buffer zone of the weak acid or base is defined as the pH region within pK_a -1 to pK_a +1.

Many biological compounds contain more than one ionisable group, e.g. amino acids, proteins or nucleic acids, which act as acids or bases. Amino Acids exist as zwitter ions; i.e. they have both negative and positive charges, at pHs between the pK_as of each group. The lowest pK_a is termed pK_a1, the second lowest pK_a2, and so on. For compounds that contain both basic and acidic ionisable groups, the pH at which these neutralise each other is known as the isoelectric point. It is also the pH at which the ionisable substance remains stationary in an electric field

Frequently used buffer systems

Compounds chosen as buffers have pK_as in the region of the pH chosen for handling the biological material, often pH 7.4, and are compatible with the material under study.

Popular buffers are:

• phosphate-NaOH, (You will use this buffer in the practical.),

• Tris-HCl (Tris[hydroxymethyl]methane),

• Hepes-NaOH (N-[2-hydroxyethyl]piperazine-N'-[2-ethanesulphonic acid]).

The ionisation of Tris changes more with temperature than that of Hepes, because it has a higher heat of ionisation, which means, it is less easy to buffer at different temperatures.

Other Examples: The carbonate-hydrogencarbonate system acts as a buffer in the blood to neutralise acids that are produced during metabolism. Although the pK_a value of carbonic acid is low at 3.8, it can buffer at pH 7, because it is in equilibrium with large volumes of CO₂ in the lung and dissolved in tissues.

Phosphate has three ionisations: H₃PO_{4 } H₂PO₄ ^- + H⁺ pK_{a 1} 2.14

H₂PO4^{- } HPO₄^- + H⁺ pK_a2 7.20

HPO₄^{-- } PO₄^-- + H⁺ pK_a3 12.40

Thus, it is a useful buffer at pH 7.

Substances that are frequently added to biological buffers:

1. To prevent inactivation of functional macromolecules that are sensitive to the presence of heavy metals, the chelating agent EDTA can be added.

2. To retain proteins in a reduced state in the presence of air, the reducing agents dithiothreitol or 2-mercaptoethanol are often added.

3. Addition of sodium azide or thiomersal prevents the growth of microorganisms, which is important for long term storage.

4. To prevent the potential degradation of proteins, protease inhibitors may be added.

Lecture 2: Protein Structure

THURS OCT10th 15:15, Lecture Theatre F.

Prof Brian Austen

LEARNING INTENTIONS

To enable you to understand the facets of protein structure that underlie dictate their function and direct how they may be isolated from biological sources.

SUMMARY

The chemical structures of amino acids and peptide bonds will be reviewed. The side chains of amino acids will be classified into acidic, basic, polar, non-polar, aromatic, sulphur-containing, The meaning of molecular weight of proteins will be explained. The common types of secondary structure and tertiary structure present in many proteins will be revised. The structural basis of active sites and binding sites will be explained

By the end of this lecture, you should understand how the structural diversity amongst proteins form the basis for the application of the following techniques for isolation, to be described in the following lecture:

CONTENT

The chemistry of amino acids is dominated by the amino and carboxyl groups.

Amines.

The chemistry of primary amines -NH₂ is dominated by a lone pair of electrons on nitrogen. They take up a proton H⁺ to form the positively-charged -NH₃⁺ group which react with acids to form acid-base complexes. Primary amines react with aldehydes and ketones to yield imines; secondary amines give eneamines, both important intermediates in metabolic reactions.

Carboxylates.

The carbonyl carbon in carboxylic acids has sp² hybridisation, and carboxyl groups -COOH are planar. They dissociate to give carboxylate anions –COO^- and H₃O^+. The acidity of carboxyls (and basicity of amines) are described by pK_a values. Substitution of the carboxyl by strong electron withdrawing groups increase acidity.

Amino Acids

Amino acids are difunctional, containing both a basic amine group and an acidic carboxyl group. The twenty alpha amino acids found in proteins possess both amino and carboxyl groups attached to the alpha-carbon atom. Nineteen of the amiono acids contain primary amines, while proline contains a secondary amine, whose nitrogen and alpha carbon are part of a pyrrolidine ring. Alpha amino acids differ in the nature of the substituent (side-chain) of the alpha carbon atom. These side chains may be hydrophilic, or hydrophobic, alkyl or aromatic, acidic or basic, or sulphur containing. http://www.ultranet.com/~jkimball/BiologyPages/A/AminoAcids.html

Amino acids are the building blocks (monomers) of proteins. 20 different amino acids are used to biosynthesize proteins. The shape and other properties of each protein is dictated by the precise sequence of amino acids in it.

Each amino acid consists of an alpha carbon atom to which is attached:

• a hydrogen atom

• an amino group NH₂ or NH (hence "amino" acid)

• a carboxyl group (-COOH). This gives up a proton and is thus an acid (hence amino "acid")

• one of 20 different "R" groups.

In the table on the following page, the structures of the amino acids occurring in proteins are listed with their linear formulas. For each amino acid, both the three-letter and single-letter codes are given. If you view the table in Word on a computer with the browzer on and internet connected, clicking on the amino acid name will show its three-dimensional structure over the web.

Physicochemical properties of amino acid side chains:

Alanine and valine have side chains that are hydrophobic, and to occupy internal positions in three-dimensional structure of proteins, where they are shielded away from water.

Glutamic acid and aspartic acids. Free side-chain carboxyl groups make these acidic and hydrophilic, and capable of forming electrostatic interactions with basic groups contained elsewhere in the same protein, in ligands, substrates or other proteins.

Glutamine and asparagine. The side-chain amide groups have potential for hydrogen bonding and so make Asn and Gln hydrophilic. Carbohydrate can be covalently linked ("N-linked) to the -NH of Asn, when this residue occurs next-but-one to a Threonine or Serine..

Cysteine and methionine. Oxidation of the sulfhydryl (-SH) groups of Cysteine (Cys) link 2 Cysteines together with a disulphide bond (S-S) in a protein and form a covalent bond which is important in dictating the three-dimensional structure of proteins. In contrast, methionine contains an S-methyl group, which is hydrophobic, but cannot form a disulphide.

Tryptophan contains the bulky aromatic indole group as a side chain. It is hydrophobic and contributes strongly to the UV absorption of proteins at 280nm, the wavelength commonly used ina spectrometer to measure protein concentration..

Histidine contains an imidazole side chain, which carries a partial positive charge at pH 7 (pK_a6.5). It is basic and hydrophilic, and often makes interactions with acidic groups in ligands.

Structures of Amino Acids; click on the amino acid name to open your browser and view the 3D structure of the molecule

Name Abbr. Linear structure formula

======================================================

Alanine ala a CH3-CH(NH2)-COOH

Arginine arg r HN=C(NH2)-NH-(CH2)3-CH(NH2)-COOH

Asparagine asn n H2N-CO-CH2-CH(NH2)-COOH

Aspartic acid asp d HOOC-CH2-CH(NH2)-COOH

Cysteine cys c HS-CH2-CH(NH2)-COOH

Glutamine gln q H2N-CO-(CH2)2-CH(NH2)-COOH

Glutamic acid glu e HOOC-(CH2)2-CH(NH2)-COOH

Glycine gly g NH2-CH2-COOH

Histidine his h NH-CH=N-CH=C-CH2-CH(NH2)-COOH

|__________|

Isoleucine ile i CH3-CH2-CH(CH3)-CH(NH2)-COOH

Leucine leu l (CH3)2-CH-CH2-CH(NH2)-COOH

Lysine lys k H2N-(CH2)4-CH(NH2)-COOH

Methionine met m CH3-S-(CH2)2-CH(NH2)-COOH

Phenylalanine phe f Ph-CH2-CH(NH2)-COOH

Proline pro p NH-(CH2)3-CH-COOH

|_________|

Serine ser s HO-CH2-CH(NH2)-COOH

Threonine thr t CH3-CH(OH)-CH(NH2)-COOH

Tryptophan trp w Ph-NH-CH=C-CH2-CH(NH2)-COOH

|_______|

Tyrosine tyr y HO-p-Ph-CH2-CH(NH2)-COOH

Valine val v (CH3)2-CH-CH(NH2)-COOH

(Continued from page 13)

Leucine and Isoleucine contain alkyl side chains. The amino acids are hydrophobic, and often located in internally in the three-dimensional structure of proteins, away from water.

The basic amino acids contain an amino group (lysine) or a guanidino group (arginine), which carry positive charges at pH 7, with a pK_a of 10.2 for the Lys amino group, and pKa of 11 for the Arg guanidino group. These amino acids are strongly basic and hydrophilic and make electrostatic interactions with acidic groups on ligands or with acidic residues in the same or other proteins.

Phenylalanine contains a phenyl group side chain, which makes the amino acid very hydrophobic so it generally occupies an internal position in the three-dimensional structure. It contributes weakly to the UV adsorption of proteins at 280nm.

Proline causes kinks in the polypeptide chain.

Serine and Threonine. The potential for hydrogen bonding to the side chain hydroxyl makes these residues hydrophilic. Carbohydrate can be covalently linked ("O-linked") to the –OH, or a phosphate group can be added via an ester bond formation. The latter is important for protein phosphorylation, which is often involved in activity alteration of the protein.

Tyrosine contains the phenolic side chain, which makes it hydrophobic, but the -OH group can form hydrogen bonds, so it is more hydrophilic than Phe. The aromatic side chain contributes to the UV adsorption of proteins.

Overall charge.

The overall charge of the protein is determined by the number of positive and negatively charged groups that are exposed to the water environment, on the surface of the folded protein. The isoelectric point of a protein is the pH at which the overall charge of the protein is neutral, whwere the negative charge of the acidic groups cancel out the positive changes of the basic groups.. So the protein would remain motionless in an electrical field during electrophoresis at that pH

Stereochemistry of amino acids

All amino acids, except glycine, are chiral compounds by virtue of the assymmetric centre of the on the tetrahedral alpha carbon atom which has four different substituents except in glycine (two protons). In consequence they exist as pairs of enantiomers, each with the property of rotating plane polarised light. These are designated L or D. The stereochemistry is best examined by viewing a molecular model of an amino acid e.g. on the following web site:

http://www.ultranet.com/~jkimball/BiologyPages/E/Enantiomers.html

There are two enantiomers of every amino acid except glycine, whilst threonine and isoleucine have an additional stereogenic carbon. Only one of the four possible stereoisomeric forms of each of these exist in proteins. Common amino acid configuration is referred to as L because of their configurational similarity to L sugars (see below). Enzymes can be used to resolve racemic mixtures of amino acids.

Peptide bonds

Amino acids are linked together into peptides and proteins via condensation of alpha-amino and alpha carboxyl groups, with the elimination of water, into a peptide bond, which is an example of an amide bond. Read more on:

http://www.iacr.bbsrc.ac.uk/notebook/courses/guide/aa.htm .

Amides are completely non-basic because of the delocalisation of the nitrogen lone pair electrons through orbital overlap with the carbonyl group. Since protonation loses the resonance energy, the unprotonated form is favoured. This overlap results in a planar structure of the amide bond, which can be seen in the simplest amide, formamide. There is a strong barrier to rotation around the peptide bond. Most peptide bonds are trans configuration, but sometimes cis bonds occur. Because of their basicity, amines are readily purified away from amides during their chemical synthesis, by extraction from an organic layer into water.

Structural organisation of proteins

Primary structure is a description of the linear sequence of amino acids in proteins. A single letter code is used to describe the sequence of a protein, with the amino acid bearing the free amino group (N-terminus) written at the left, and the one with the free carboxyl (C-terminus) at the right.

Secondary structure is the local arrangement of atoms in space within a short local sections of polypeptide chains. This is often a repeating conformation held together by hydrogen bonding from the peptide groups, and the helix -sheet and -turn are the local folding patterns that are commonly seen in most proteins. To see a helical protein structure, click on: http://www.paccd.cc.ca.us/instadmn/physcidv/chem_dp/chemweb/protein/intro.htm

Tertiary structure is the total three-dimensional arrangement of all the atoms of a protein subunit.

Quaternary structure is the arrangement of protein subunits in the fully functional protein. (This applies only to multimeric proteins.).

Refer also to:

http://www.ultranet.com/~jkimball/BiologyPages/P/PrimaryStructure.html

The forces involved in maintaining the overall conformation of proteins are the following:

1. Electrostatic interactions (salt bridges) between basic side chains (lysine, arginine or histidine) and acidic side chains (aspartic and glutamic);

2. Van der Waal's forces, which are intermolecular forces between adjacent groups due to opposite electrical dipoles induced by the movement of electrons,

3. Hydrogen bonds in which there is electron sharing between a hydrogen atom covalently bonded to an oxygen or nitrogen atom, and a nearby oxygen

4. Covalent (disulphide) bridges between cysteine residues

5. Hydrophobic forces. This is the energy resulting in the entropy factors of keeping hydrophobic resides together away from surrounding water. (This is best investigated in the tutorial on the web site on:

http://www.iacr.bbsrc.ac.uk/notebook/courses/guide/prot.htm)

Lecture 3. PROTEIN ISOLATION

Thurs 17^th Oct, 15:15 . Professor Brian Austen. Lecture Theatre F

Aims

To help students understand how the techniques commonly used to isolate proteins from complex mixtures in biological sources, rely on their structural features. In the Practical laboratory course you will obtain experince in isolation of proteins by ion-exchange chromatography, affinity chromatography and electrophoresis.

Learning Objectives

To understand the principles of the following isolation techniques:

• Salting out

• Organic precipitation

• Ion exchange chromatography

• Size exclusion chromatography

• Affinity chromatography

• Electrophoresis

• Reverse-phase chromatography

Molecular weights of proteins vary between 3000 and several millions. The units used are daltons where one dalton is the mass of a hydrogen atom. A thousand dalton is referred to as a kilodalton (kDa). The main factor determining the molecular weight of proteins is the length of each polypeptide chain and the number of polypeptide chains in the functional protein (quaternary structure).

1. Bulk separation techniques

Prior to column chromatography, initial fractionations of protein mixtures are often made by "salting out". Proteins differ in their solubilities at high ionic strength, depending on their isoelectric points and overall charge distribution. A mixture of proteins is fractionated by the addition of saturated solutions of ammonium sulphate. Final concentrations of between 20%(w/v) and 60%(w/v) ammonium sulphate are use, and separating off the precipitates by centrifugation.

Similarly addition of organic solvents mixable with water such as ethanol may be performed. Organic solvents act by decreasing the dielectric constant of water which depresses ionisation of charged amino acid side chains, and so decrease solubility of proteins.

Proteins may also be selectively precipitated from solutions by altering the pH. Proteins are least soluble at pH values close to their isoelectric points, the pHs at which the acidic charges are equal in number to the basic charges on the protein, so the overall charge on the protein will be neutral.

2. Size exclusion chromatography (or gel permeation chromatography) is a technique that separates native proteins on the basis of their molecular weights. The stationary phase or matrix of this type of chromatography is composed of porous beads containing cross-linked molecular network comprised of a polysaccharide, such as agarose or dextran, or of polyacrylamide. Proteins are dissolved in the mobile phase, which is generally a buffer at a pH, at which the proteins of interest are stable. The matrix is placed in a cartridge or column through which the mixture of proteins pass from one end to the other. The smaller proteins permeate into the cavities of the molecular network, and are retarded as the larger proteins pass through unhindered (and emerge first). The solution flowing through the matrix (or eluate) is kept at constant flow by a pump and is collected in a series of tubes in a fraction collector. The eluate is generally allowed to flow through a silica cell in the light path of a UV source in a spectrophotometer, set to measure absorbance at 280 nm, so that the position of elution of proteins can be monitored. The procedure is shown diagrammatically at the following website: http://wine1.sb.fsu.edu/bch5425/lect31/lect31.htm

Proteins elute from these columns at elution volumes that are proportional to the log of their molecular weights. Thus, molecular weights of an unknown protein can be estimated by calibrating the column with the elution volumes of proteins of known molecular weight.

2. Ion-exchange chromatography. You will gain first hand experience in the practical course. This is a technique that separates proteins on the basis of overall surface charge, which depends on the relative numbers of aspartic and glutamic acidic amino acids and lysine, arginine, and histidine basic residues disposed on the surface of the folded protein. The technique requires beads of a matrix of insoluble polymer containing charged groups The polymer can be chemically modifed cellulose or chemically crosslinked dextrans, which are known by their trade name Sephadex .The beads are packed as a suspension in buffer into a column or cartridge. Proteins dissolved in the same buffer are loaded onto the column, and will attach themselves by ion-pairing between their charged side chains and the charged groups on the matrix. Proteins are eluted in order of their affinity for the matrix by a series of buffers of increasing ionic strength. The ions in the buffers compete out the binding sites for the proteins on the matrix. Alternatively, the proteins may be eluted by changing the pH of the eluting buffer. Matrices with different types of charged groups are available for use. The procedure is shown in the diagram on: http://ntri.tamuk.edu/fplc/ion.html

4. Affinity chromatography. This is a technique capable of achieving a high degree of purification, A high-affinity ligand for the protein of interest is linked covalently to a dextran or Sephadex, and a mixture of proteins are applied. Only the protein with high affinity for the ligand will bind, other proteins are eluted away with excess buffered solution. Then the required protein is eluted by application of a solution of the ligand to the column, or extreme change in pH of the eluting buffer. If an enzyme is being isolated, the ligand may be a specific inhibitor. You will gain first hand experience in the practical course. A diagram is shown here:

5. Another less frequently used technique for proteins is reverse-phase chromatography. In this, the stationary phase (column matrix) is apolar, consisting of alkyl or phenolic groups attached covalently to silica, and proteins applied in water with as salts with trifluoracetic acid are eluted off by an apolar solvent which is miscible with water, such as acetonitrile. See on:

http://www1.amershambiosciences.com/aptrix/upp00919.nsf/Content/LC-tech+Reversed+Phase

6. Electrophoresis. Analytical electrophoretic techniques and their clinical application will be

shown to you in the practical course. See also on http://www.kendricklabs.com/

LECTURE 4 PROTEIN ANALYSIS

Thurs, Oct 24th, 15:15 Lecture Theatre F

Prof BM Austen

AIMS

To understand methods in use for identification of the protein components present in a mixture. Also as an essential prerequisite for finding out how proteins function, methods used for determination of the chemical composition of a purified protein should be known and understood, in particular modern methods for finding out the linear preimary sequence of a protein should be known.

CONTENT

Analysis of whole proteins

The method most commonly used to resolve individual proteins form a mixture of a large number of proteins is elecrophoresis through polyacrylamide. Acrylamide is induced to polymerise and cross-link by adding initiators of free radical formation and a cross-linker, which contains two acrylamide groupings. The ratio of bis-acrylamide to monoacrylamide determines the pore size of the gel that is formed.

As proteins are poly-ionic they migrate across the gel placed in an electric field according to their overall charge and the pH of the buffer in the gel. You will gain experience of this technique in the practical course.

SDS-gel electrophoresis. Proteins move according to their molecular weight, when an anionic detergent, sodium dodecyl-sulphate (SDS) is present in the electrophoresis buffer and in the sample. This has the effect of coating the proteins with anionic sulphate groups, so all proteins are negatively charged and migrate towards the anode at pH 8. There is an inverse relationship of the mobility of the proteins and the log of their molecular weight. The disadvantage of this method is that the proteins are denatured by the SDS. It is therefore mainly applied for analytical and not preparative purposes.

Iso-electric focussing (IEF). If buffers making a gradient of pH throughout the gel are used, proteins migrate to the pH of their isoelectric point/

Two-dimensional (2D) electrophoresis. A combination of isoelectric focussing and SDS-gel electrophoresis used at a 90 degree angle to each other give rise to a two-dimensional separation capable of resolving many hundreds of proteins in one experiment. See the technique on http://www-lecb.ncifcrf.gov/phosphoDB/2d-description.html

Spectral properties of proteins

The ultraviolet region of the electromagnetic spectrum of greatest use to the biochemist lies between wavelengths from 200 to 400nm. In this region, conjugated systems, i.e. ones with an extended series of overlapping  orbitals as found in aromatic ring systems, absorb energy. The energy promotes the electron to jump into the  non-bonding orbital, so it is called a   ^* transition. Aromatic residues tyrosine and tryptophan by these transitions contribute most to the UV spectrum of proteins, which absorb maximally at 280 nm. This provides a useful method for measuring protein concentration, and is used along with other colourimetric assays. At shorter wavelengths, the delocalised  electrons in the peptide bonds absorb, giving a  * transition at 215 nm, and an n  * transition at 200nm. This can also be used for quantitation of proteins, but many substances found in common biological buffers interfere at this low wavelength.

Modification of proteins

Some side chains in proteins are susceptible to modification by chemical reagents, which enable investigators to probe the role of these groups in function. In this way, the groups present at a binding site in a protein may be identified in the active site of an enzyme. Amino groups may be modified by acyl anhydrides such as maleic anhydride, tyrosines by nitration, or cysteines by iodoacetic acid. Some side chains become unusually susceptible to modification due to their configuration within an active site; e.g. the serine hydroxyl in the active site of trypsin becomes reactive to nucleophilic reagents by delocalisation of its proton on the hydroxyl grouphttp://stingray.bio.cmu.edu/~web/bc1/trypsin/trypsin.htm,

or the aspartic acid in the active site of pepsin.http://www.rcsb.org/pdb/molecules/pdb12_3.html

Some side chains in proteins are modified by metabolism after translation, e.g. Thr and Ser residues may be phosphorylated (addition of a phosphate group on the hydroxyl), and Asn, Ser and Thr residues may be glycosylated (carbohydrate added). These changes often modify the function or turn-over rates of proteins in their natural environment.

Protein Sequencing.

It is important to determine the linear sequence of amino acid residues in proteins or peptides.

http://www.ultranet.com/~jkimball/BiologyPages/P/PrimaryStructure.html

For sequence analysis, large proteins are broken down to small peptides. First, they are denatured in high concentrations of urea or a strong detergent. Disulphides are broken by reduction with thiol, and then alkylated by reaction with iodoacetate. The protein is then digested by a specific endoprotease that cleaves only at certain types of residues in the protein (e.g. trypsin cleaves at basic residues (K and R) and chymotrypsin at aromatics (Y and W) and leucine (L)). Alternatively, the chain can be cleaved by cyanogen bromide, (CNBr) which reacts with the electrophilic sulphur in the side chain of methionine in a protein, forming homoserine and breaking the chain.

Pure protein  denatured and S alkylated:

+ trypsin  peptides ending in K or R;

+ chymotrypsin  peptides ending in W,Y, or L; + CNBr  peptides ending with homoserine .

The molecular weights of the peptides formed are measured by mass spectrometry,

http://www.users.globalnet.co.uk/~freya01/maspec.html .

The results may give enough information to identify the gene from which the protein is derived; this process is called finger printing. Identification can be done using gene data banks available through the Internet (e.g. at http://cbrg.inf.ethz.ch/). This procedure is known as proteomics.

There are modern versions of mass spectrometers, which allow the determination of the complete sequence of a peptide, basing on the fragmentation of the peptide produced by bombarding with high-energy particles.

If the sequence of the coding DNA is not in the database, the protein is unknown. The peptides are then separated from each other by reverse-phase high-pressure liquid chromatography (HPLC), and their sequence are analysed further.

To see the amino acid composition of peptides, the amino acids in peptides are released by treatment with strong acid at 110^oC, and the free amino acids identified by chromatography. The N-terminal amino acid is determined by reaction with a fluorescent compound, dansyl chloride, which is identified by chromatography. Before the use of mass spectrometers, the sequence from the N-terminus was determined by Edman degradation, which derivatises the N-terminal amino acid residue as a phenyl isothiohydantoic acid. Acid-catalysed cyclisation then expels the chain-shortened peptide and forms a thiazolinone from the N-terminal amino acid released. The thiazolinone rearranges in aqueous acid to a PTH-amino acid, identified on HPLC after extraction into ethyl acetate. The cycle of reactions, which is normally automated in a sequencer is repeated and will identify the next N-terminal amino acid, and may be repeated until the yields get too low. N-terminal and C-terminal amino acids are also released by treatment with exopeptidases. Peptides can be broken down sequentially from the carboxy-terminus by ion bombardment in a mass spectrometer, and the molecular mass differences of the range of residual fragments detected is used to work out the sequence. http://www.biotech.iastate.edu/facilities/protein/nsequence494.html

Lecture 5: CARBOHYDRATES AND STEREOCHEMISTRY

Thurs, Oct 31^st 15:15 Lecture Theatre F

Prof Brian Austen

AIMS

The electronic properties of carbonyl groups, the main types of functional groups they are part of, and the main classes of reactions they undergo, will be described. You will acquire an understanding of carbohydrate structure and its importance in Biochemistry. The essential stereochemical features and biological format of carbohydrates will be described, as will chemical reactions that can be used to modify carbohydrates.

CONTENTS

The carbonyl group

One of the most important functional groups in biologically important molecules, the carbonyl group, contains sp² hybrid  bonds, a -bond, and corresponding anti-bonding orbitals. The carbonyl oxygen also has two non-bonding pairs of electrons that occupy its remaining two orbitals. A number of functional groups contain carbonyls; these include aldehydes and ketones, which cannot stabilise a negative charge, and carboxylic acids, acid halides, acid anhydrides, esters, lactones, amides and lactams, which can.

Carbon-oxygen bonds are polarised because of the high electro-negativity of oxygen compared to carbon. The carbonyl oxygen carries a partial negative charge, and reacts with electrophiles; the carbon carries a partial positive charge and reacts with nucleophiles.

Examples: Nucleophilic addition of hydrogen cyanide to a ketone is summarised here:

Nuleophilic addition of an alcohol to produce an acetal:

Nucleophilic addition of amines; imine formation.

Carbonyls also undergo alpha-substitution reactions and condensation reactions, which occur in metabolic reactions in living cells, e.g. the aldol reaction, catalysed by aldolase, that occurs in the course of glycolysis.

Carbohydrates

Carbohydrates are a broad class of polyhydroxylated aldehydes and ketones, commonly called sugars. Simple sugars are monosaccharides that cannot easily be hydrolysed into small molecules. Complex carbohydrates are made up of two or more sugars. Monosaccharides are classified into aldoses or ketoses, according to the nature of the carbonyl group.

Sugars are examples of chiral molecules that can exist in a number of stereoisomeric forms. Molecules that are not identical to their mirror images exist in two enantiomeric forms (http://qlink.queensu.ca/~7slb/chem210/enant.html - Enantiomers ), which are chiral.

(http://lab1.chem.queensu.ca/chem210/french/page3.html ).

A molecule cannot be chiral, if it contains a plane of symmetry. The most common (but not only) cause of chirality in organic chemistry is the presence of a carbon atom bonded to four different groups. Such carbons are termed asymmetric centres. Chirality is the property of the whole molecule, wheras a stereogenic centre is the cause. Examples are:

Enantiomers are classified either as R or S. (http://www.sci.ouc.bc.ca/chem/nomenclature/nom-920.htm). When more than one asymmetric centre is present, the R & S nomenclatures are used after the numbering of the relevant stereogenic centre:

(http://www.sci.ouc.bc.ca/chem/nomenclature/nom-930.htm.) .

Diastereoisomers are stereoisomers that are not mirror images of each other. They have the opposite configuration at some (one or more) stereogenic centres, but have the same configuration at others. Enantiomers, by contrast, have opposite configurations at all stereogenic centres. For example, the amino acid threonine has two asymmetric centres:

(http://qlink.queensu.ca/~7slb/chem210/diastereo.html - dstereo) and 2² stereoisomers. Only the 2S,3R isomer occurs naturally in plants and animals. Cholesterol has 8 stereocentres, giving 2⁸ (=256) stereoisomers. Only one of these is produced in nature. Molecules that have a number of asymmetric carbon atoms and possess a plane of symmetry and are known as meso-compounds.

Classifications of carbohydrate sugars

The stereochemical classifications of carbohydrate sugars are based on that of glyceraldehyde (http://www.jonmaber.demon.co.uk/monosaccharide/). Carbohydrates are classified as either D or L depending on configuration of the highest-numbered asymmetric carbon atom, the asymmetric carbon at the lowest position in the Fisher representation.

Alcohols undergo rapid nucleophilic additions with ketones or aldehydes; as both of these groups are found in one sugar molecule, hemi-acetals are formed from sugars by cyclisation; the forms are either pyranose or furanose.

Hexoses form six-membered rings that can be represented by Haworth projections, in which the hemiacetal ring is viewed flat with the oxygen at the far right corner. The cyclisation produces yet another stereogenic centre, to produce two diastereoisomers called anomers. Chair or boat structures of sugars represent their actual conformations.

Treatment of an aldose or ketose with sodium borohydride reduces it to a polyalcohol called an alditol. Sugars are easily oxidised to corresponding monocarboxylic acids.

Treatment of a monosaccharide hemiacetal with an alcohol and an acid catalyst yields an acetal with the anomeric OH replaced by OR, a glycoside. If the alcohol is itself a sugar, the product is a disaccharide, e.g. cellobiose or maltose.

Monosaccharides play important roles in metabolism, for example in the series of reactions called glycolysis, which generates ATP and smaller high-energy compounds.

Polysaccharides are carbohydrates in which tens, hundreds, or even thousands, of simple sugars are linked together. Examples include starch and glycogen. Glycosaminoglycans are components of biological fluids composed of repeating units of specific disaccharide structures. Other important carbohydrates are components of cell surfaces, such as the blood group substances.. Carbohydrates are linked to some proteins, e.g. thyroglobulin. The linkages are of two types: N-glycosidic, where an N-acetyl glucosamine is linked to asparagine or O-glycosidic, where galactose is linked to serine or threonine.

Lecture 6: PROPERTIES AND REACTIONS OF NUCLEIC ACIDS

Thurs Nov 7^th 15:15 Lecture Theatre F.

Prof Mike Clemens

AIMS

To provide a basic understanding of the structures of DNA and RNA, the forces involved in determining these structures, and the specific properties of DNA and RNA that result from such structures. The lecture will also describe how these properties allow us to separate different molecular species of nucleic acids, to detect individual sequences by the process of hybridisation and to obtain information about the linear order of the four nucleotides, which constitute these sequences.

TOPICS COVERED:

• Principles of primary and secondary structure of DNA and RNA;

• differences in structure between DNA and RNA;

• summary of the different types of RNA in cells;

• techniques used to separate nucleic acids from other molecules and from each other;

• molecular hybridisation as a means of detecting specific DNA or RNA species;

• principles of DNA sequencing;

• chemical and enzymatic methods for the synthesis of nucleic acids.

LEARNING OBJECTIVES:

After this lecture, and additional reading, you should be able to do the following:-

1. Describe the essential aspects of DNA and RNA structure and the differences between them;

2. Describe the types of RNA found in cells;

3. Understand the ways in which nucleic acids can be purified;

4. Define what is meant by nucleic acid hybridisation and describe how it is used

to detect individual DNA or RNA species;

2. Describe the basis of the most commonly used method for sequencing DNA.

Lecture 7: APPLICATIONS OF SPECTROSCOPIC METHODS IN

BIOLOGICAL SCIENCES

Thurs, Nov 14^th 15:15 Lecture Theatre F

Dr M Roberts

AIMS:

The components of atoms and the molecules from which they are made can be stimulated to higher energy states by electromagnetic radiation. This radiation is absorbed by molecules at specific wavelengths determined by the separation of these energy states. Each molecule will therefore have its own characteristic radiation absorption pattern that can be used for its quantitative and qualitative analysis. For example protein and nucleic acid are biological macromolecules with distinct ultraviolet (UV) absorption spectra. If the absorption is known, then the concentrations of either of these molecules can be calculated. Different types of protein or nucleic acid can be further distinguished by more sophisticated spectroscopic techniques, such as circular dichroism (CD) and nuclear magnetic resonance (NMR). The principles of absorption of electromagnetic radiation by biological macromolecules will be described, and how these can be used in analytical procedures in spectrophotometers in the Biomedical laboratory.

TOPICS TO BE COVERED:

• The electromagnetic spectrum and the radiation wavelengths used in biomolecular analysis.

• The electronic structures of biomolecules that give rise to UV absorption.

• The basics of the UV spectrophotometer.

• The significance of Beer’s Law.

• The mechanisms of fluoresence and luminesence.

• The use of CD and NMR in biomolecular structure analysis and the levels of information these techniques provide.

LEARNING OBJECTIVES:

This lecture should enable you to understand:

1. how the different types of absorption spectra arise in biomolecules;

2. the basic principles of operation of the UV and NMR spectrometers;

3. what circular dichroism means;

4. how to calculate the concentration of protein or nucleic acid in solution from the UV absorption spectra;

5. which spectroscopic techniques you would use to understand more about the 3D structures of biomolecules;

6. the mechanisms of the emission of light in the forms of fluoresence and luminesence;

7. how spectroscopic methods can be used to monitor interactions between biomolecules.

Lecture 8: TRACER METHODS IN BIOLOGICAL SCIENCES

Thurs Nov 21^st 15:15 Lecture Theatre F

Dr Uli Bommer

AIMS

The ‘visualisation’ of classes of molecules or of specific macromolecules is a central theme in the Biomolecular Sciences. A plethora of techniques have been developed to achieve this goal, however there is a distinct number of principles underlying these ‘labelling methods’. This lecture will introduce you to some of these principles, many of which you will come across in various forms during your practical classes throughout the curriculum and thereafter.

TOPICS TO BE COVERED:

• Overview on principal methods to ‘visualise’ groups of macromolecules or specific molecules.

• Direct ‘staining’ of classes of molecules,

• Radioactive labelling of biomacromolecules;

• Non-radioactive tracer methods;

• Antibody-mediated labelling techniques.

• Labeling of specific molecules by ‘helper reactions’, e.g. enzyme-linked assays.

LEARNING OBJECTIVES:

At the end of this session, you should be able to:

1. name at least one staining method each to visualise proteins and nucleic acids;

2. describe the underlying principles of radioactive tracer methods and how they can be used to quantify reactions and visualise reaction products;

3. name alternative techniques which have been developed to replace the use of hazardous radioactive precursor molecules.

4. reason why antibodies are very versatile reagents, suitable to visualise and quantify specific proteins in a variety of assay methods;

5. give examples of the use of enzymes in ‘helper reactions’ to visualise specific molecules.

Chemistry in Life Sciences Tutorials

Aims:

• Supporting you to understand the main concepts presented to you during the lectures of the ‘Chemistry in Life Sciences’ course. This involves discussion of the chemical properties of the various classes of biological (macro)molecules and of the underlying principles of methods frequently applied in the Biomedical Sciences.

• Complementing the lectures by discussing questions that remained open to you during your self-directed learning and by introducing certain aspects of the subject which are difficult to cover in a normal lecture format. Make actively use of this possibility by coming forward with your own questions.

• Encouraging you to develop your scientific curiosity and your skills in basic calculations and in writing scientific essays.

Format of the tutorials:

Each student group will have four tutorial sessions, numbered 1 through 4. Each session will cover the material of the preceding lectures according to the following topics:

Tutorial 1: Biological solutions – calculations and concentrations.

Tutorial 2: pH and buffers

Tutorial 3: Protein structure and spectroscopy.

Tutorial 4: Properties of nucleic acids.

Assessment:

The tutorials are an important part of the course, your attendance and participation will be noted. Ensure that you prepare for the tutorial in advance by looking through the material, and working out the exercises. You will be asked for the answers in the tutorials. Moreover, each of you will be expected to hand in one essay. Essay writing is an important skill to acquire in biomedical sciences as it is often required to present scientific facts and research in a clear, logical format that is lucid to the reader. Know your material before you start writing, and plan your essay carefully; subheadings and diagrams are useful to help you to communicate, and word-processed or neatly-hand written presentations will always be easier to read. Try aand cover your material as succinctly as possible, avoiding lengthy incoherent rambling. Choose a title from the following, and hand in your essay to your tutor for mrking on the third tutorial

1. What are the chemical and biological priciples to apply when choosing a buffer for protein isolation.

2. Explain the chemical principles behind size-exclusion chromatography used as a method for protein purification.

3. What are meant by the primary, secondary, tertiary and quaternary structures of proteins. What are the forces and bonding responsible for protein folding.

4. What are the chemical principles of nucleic acid hybridisation, and how is it used in clinical diagnosis.

Tutors: Christina Sidera (email csidera@sghms.ac.uk) Groups 1 &4;

Gwyn Davies (jdavies@sghms.ac.uk)Groups 2 and 5;

Mike Roberts (sgbm400@sghms.ac.uk) Groups 3 & 6

Student Groupings for Tutorials

CLS Tutorial Studies

Theme 1; Tutorial 1; Solution chemistry.

Background on units

Science is concerned with counting and measurements, and measurements require units, so units are obviously important to scientists. (They are also important in everday life: think how easy it is to imagine situations where an error in the use of units could prove to be very expensive, or even fatal!) Hence it's necessary to become familiar with the units used in any particular discipline, and often to be able quickly to convert from one unit to another. In general, the units used in a laboratory are those that are most convenient, but may be different from those given on the label of a bottle or in an experimental protocol. For example, it's easier both for me to say and you to understand "you'll need five micromoles of X" than "you'll need five times ten to the minus six moles of X". The following units, prefixes and abbreviations are commonly used and should be learned.

^§ 1Å = 10^-10 m ^*A useful definition of a mole is M_r in g. For example, 342 g of sucrose is equivalent to 1 mole of sucrose.

^† It is extremely important to distinguish between the terms mole and molar. Molar is a concentration term.

^¶ Mole per 1000 g solvent (Another concentration term - but not used as commonly as molar.).

^‡ Parts per hundred. When used for solutions the units must be defined: % (w/v) means g solute/100 ml solution and % (v/v) means ml solute/100 ml solution

^# A Curie is equivalent to 2.22 x 10¹² dpm and a Becquerel to 1 dps. (1Ci = 37GBq)

Repeated usage of the various common units soon makes conversion among them familiar and easy; yet it's still important always to pause and check "Does this make sense ?" and "Is this a reasonable number ?". Usually it's sensible and worthwhile to check the result of a calculation by working backwards from your answer to ensure that it gives the original information correctly. A few examples of typical conversions among units are given below.

(Note that throughout this booklet the term "log" is used for "logarithm to the base 10", with "ln" denoting "logarithm to the base e".)

Examples

1. Express 5 µmol as (a) mol (b) pmol (c) mmol

(a) By definition 1 µmol = 10^-6 mol

Therefore 5 µmol = 5 x 10^-6 mol

(b) By definition 1 µmol / 1 pmol = (10^-6 mol) / (10 ^-12 mol) = 10⁶

Therefore 1 µmol = 10⁶ pmol and 5 µmol = 5 x 10⁶ pmol

(c) By definition 1 µmol / 1 mmol = (10^-6 mol) / (10 ^-3 mol) = 10^-3

Therefore 1 µmol = 10^-3 mmol and 5 µmol = 5 x 10^-3 mmol

2. Express 1 litre in terms of (a) m³ (b) mm³

(a) By definition 1 litre = (0.1 m)³ = 10^-3 m³

(b) By definition 1 m = 10³ mm and so 1 m³ = (10³ mm )³ = 10⁹ mm³

Therefore 1 litre (=10^-3 m³⁾ = (10^-3)(10⁹) mm³ = 10⁶ mm³

3. Express 1 µm³ in terms of litre

By definition 1 µm = 10^-6 m and so 1 µm³ = 10^-18 m³

Also 10^-3 m³ = 1 litre

Therefore 1 µm³ = (10^-18 / 10^-3) litre = 10^-15 litre

4. Express 3 mM in terms of (a) mol/litre (b) µmol/ml (c) nmol/100 µl

Remember that concentration is a ratio of different units (amount/volume) and both must be changed as required.

(a) 3 mM = 3 mmol/litre = 3 x 10^-3 mol/litre (Only change required is mmol to mol)

(b) 3 mM = 3 mmol/litre = 3 x 10³ µmol/10³ ml = 3 µmol/ml (Both units changed)

(c) 3 mM = 3 mmol/litre = 3 x 10⁶ nmol/10⁶ µl = 3 nmol/µl (Both units changed)

or 300 nmol/100µl (Both x 100)

5. Calculate the number of mol of ethanol in

(a) 5 ml of a 10 mM solution

(b) 5 ml of a 10% (v/v) solution

(a) 10 mmol/litre = 10 x 10^-3 mol/10³ ml = 10^-5 mol/ml or 5 x 10^-5 mol in 5 ml

(b) 10% (v/v) = 10 ml/100ml = 0.1 ml/ml or 0.5 ml in 5 ml

More information required to go further! Need to know specfic gravity of ethanol is 0.8 and its M_r = 46.

1 ml weighs 0.8 g, so that 0.5 ml = 0.4 g = (0.4/46) mol = 8.7 x 10^-3 mol

6. Calculate the number of molecules of glucose inside an erythrocyte, assuming it's a disk with internal diameter 8 µm, & thickness 1 µm, containing 5 mM glucose. (Avogadro's number is 6 x 10²³)

Volume = _R²H = _(4)²(1) µm³ = 50 µm³ = 50 x 10^-15 litre (see question 3 above).

5 mmol/litre = (50 x 10^-15)(5 x10^-3) mol in (50 x 10^-15) litre = 2.5 x 10^-16 mol or

(6 x 10²³ x 2.5 x 10^-16) molecules = 1.5 x 10⁸ molecules

Exercises

Use scientific notation whenever appropriate.

(E.g., write 2.3 x 10 ^-9 instead of 0.0000000023)

1. Calulate the number of mols of ATP in:

(a) 0.5 ml of 10 mM-ATP solution

(b) 500 µl of 0.01 M-ATP solution

(c) 50 ml of 100 µM-ATP solution

4. 2 ml of 5 x 10^-5 M-ATP solution

2. Complete the equations:

(a) 3 mol = ............... µmol (b) 3 mmol = ............... nmol

(c) 3 µm = ............... mm (d) 3 nm = ............... µm (e) 3 mg = ............... pg (f) 3 litre = ............... µl

(g) 3 nl = ............... ml (h) 3 litre = ............... cm³

(i) 3 m³ = ............... litre (j) 3 µl = ............... mm³

3. Express the following concentrations in molar (M = mole/litre) units.

(a) 20 mM (b) 0.3 µM (c) 400 nM (d) 0.05 mM (e) 60 pM

4. The M_r of glycine ( H₂N- CH₂-COOH) is 75.

Given that H = 1, C = 12, N = 14, O = 16, P = 31, draw corresponding structures and calulate the M_r values of the following compounds.

(a) Alanine (C₃H₇O₂N)

2. Glucose (C₆H₁₂O₆) [Give ring structure.]

Use scientific notation whenever appropriate.

(E.g., write 2.3 x 10 ^-9 instead of 0.0000000023)

5. Complete the equations:

6. Express the following in molar (M = mol/litre) units:

(a) 25 mM (b) 150 µmol/ml (c ) 0.006 mM (d) 90 µM

(e) 20 nmol/litre

7. Express 5 M in units of:

(a) mmol / ml (b) µmol / µl (c) nmol / nl

(d) mol / ml (e) mmol / µl (f) µmol / nl

(g) mmol / litre (h) µmol / ml (i) nmol / µl

3. Calculate the molarity of a solution containing

3. 360 mg glucose/ml (M_r glucose = 180)

(b) 300 µg glycine/ml (M_r glycine = 75)

(c) 356 mg alanine/ml (M_r alanine = 89).

Preparation of solutions; Background information

(a) The amount of solute required to prepare a solution of given concentration & volume is calculated according to the formula:

Number of g = (M_r of solute)(molarity of solution)(volume in litres)

where M_{r is the relative molecular mass ( or "molecular weight") of the solute}

E.g., to prepare 250 ml of 200 mM-glucose we require

(180)(0.2)(0.25) = 9 g ofglucose

(b) If the concentration is given in terms of % :

For solids: number of g = 10(w/v % value)( volume in litres)

For liquids: number of ml = 10(v/v % value)(volume in litres)

E.g., to prepare 250 ml of 10 % (w/v) trichloroacetic acid we require

(10)(10)(0.25) = 25 g of TCA.

To prepare 50 ml of 2% (v/v) ethanol we require

(10)(2)(0.05) = 1ml of ethanol.

Dilutions

For dilution in general use the formula: C₁V₁ = C₂V₂

where V₁ is the volume of the initial solution, concentration C₁, necessary to give volume V₂ of the required solution with concentration C₂. It does not matter what units are used for C and V, provided they are the same throughout. Usually we want to find the value of V_1,

so that we calculate:

V₁ = C₂V₂ / C₁

For example, prepare 250 ml of a solution containing 0.01 M-glucose and 0.15 M-NaCl, given stock solutions of 2M glucose and 3M NaCl.

For glucose

C₁ = 2M, C₂ = 0.01M and V₂= 250 ml, so that V₁ = (0.01 x 250) / 2 = 1.25 ml

For NaCl

C₁ = 3M, C₂= 0.15M and V₂ = 250ml, so that V₁ = (0.15 x 250) / 3 = 12.5 ml

Hence, pipette 1.25 ml of 2M-glucose, then 12.5 ml of 3M-NaCl and make the mixture up to 250 ml with water.

Exercises.

1. Given the molecular weights CaCl₂ = 111.0 and NaCl = 58.44, calculate the amount of

(a) CaCl₂ required to prepare 1000ml of a 2.5M solution.;

(b) NaCl required to prepare 1000ml of a 5M solution;

(c) CaCl₂ required to prepare 50ml of a 1 M solution;

(d) NaCl required to prepare 100ml of a 10mM solution.

2 Calculate the following:

(a) the amount of TCA required to prepare a 500ml of a 10% (w/v) solution;

(b) the amount of SDS required to prepare a 100ml of a 5% (w/v) solution.

3. Calculate the volumes of the following stock solutions required to make the indicated solutions.

(a) 500ml of 50mM CaCl₂ from a 1M solution;

(b) 10ml of 1%(v/v) solution from a 10% solution of TCA.

4. Calculate the number of mols of glucose in

(a) 50 ml of 100 µM glucose solution

(b) 0.002 litres of 2.5 mM glucose solution.

5. Express 2 M in units of: (a) mol/ml (b) pmol/ml (c ) nmol/µl (d) mmol/100µl.

SAMPLING

Many experimental procedures involve successive sampling and dilution steps before the final measurements can be made. Initially an extract is often made of some cells, then proteins may be precipitated by the addition of acid, or lipids extracted with organic solvents, different phases separated, further samples taken, reagents added to produce a coloured derivative, and so on. The various samples are not uniform and the volumes involved vary at each stage in the procedure; but the aim is usually the same: from the measurement of the amount or concentration of "X" in the final solution, calculate the amount or concentration of "X" in the orginal cell extract. Such procedures may be summarised diagrammatically as shown below.

The original solution has volume Vo and contains "X" at concentration C_o. The final solution has volume V_f and concentration C_f. The successive samples taken have volumes S₁, S₂.........S_f and the volumes at the successive steps are V₁, V₂.......V_f. ( Note that these volumes are the TOTAL volumes at each stage, the sum of the sample plus whatever else is added to it, or it is added to.) Successive application of the equation C₁V₁ = C₂V₂ shows that the value of C_o is readily calculated from C_f according to the equation:

C_o = C_f [(V₁)(V₂).......(V_f )] / [(S₁)(S₂)......(S_f)]

Examples

(1) 5 ml of 10% TCA were added to 3 ml of a tissue homogenate and the precipitated protein was sedimented by centrifugation. Then 2 ml of the supernatant solution were mixed with 3ml of a solution containing reagents that convert Pi to a coloured derivative. Spectrophotometric analysis of a 3 ml sample of the latter mixture showed that it contained 1.8 µmol Pi. i

(a) What was the concentration of Pi in the original homogenate ?

(b) How many µmol of Pi were there in the 3 ml of homogenate ?

Vo = 3 ml: Co = ?

S1 = 3 ml; V1 = (3+5) = 8 ml; S2 = 2 ml; V2 = (2+3) = 5 ml; S3 = 3 ml; V3 = 3 ml

C3 = (1.8/3) µmol/ml or 0.6 mM

(a) Substituting these values into the equation gives: Co = 0.6 [(8)(5)(3)] / [(3)(2)(3)] = 4 mM

2. CoVo = 4 (µmol/ml) x 3 (ml) = 12 µmol

Exercises

1. A protein has a M_r of 66 000. Calculate:

(a) the molarity of a solution containing 66 mg of the protein per ml

(b) the number of protein molecules in 1 ml of the solution.

[Assume Avogadro’s number = 6 x 10²³]

2 (a) Equal volumes of solution A and water were mixed and then a 0.1 ml sample of the mixture was shown to contain 0.2 mg of solute P.

What was the concentration of P in solution A ?

(b) Enough acid was added to 0.5 ml of solution A to dilute the latter 10-fold and then a 2 ml sample of the mixture was shown to contain solute R at a concentration of 2 mM.

How many µmol of R were there in the original 0.5 ml sample of solution A ?

(c) Protein was preciptated from 2 ml of cell homogenate by addition of 3 ml of acid. Lipid was extracted from a1ml sample of the supernatant solution by mixing it with 10 ml of chloroform/methanol. After separation of the two phases, the volume of the upper, aqueous, phase was 1.5 ml and it contained solute S at a concentration of 2 mM.

What was the concentration of S in the cell homogenate ?

Tutorial 2 Theme 2; pH, pK_a and Buffers

The basic concepts involved in this topic are given in the first CLS lecture.. The relevant equations are listed below, with square brackets denoting concentration.

(1) pH = -log[H⁺] when [H⁺] is in molar (mol/litre) units.

(Similarly, pOH = -log[OH^-], pK_a = -logK_a and pK_w = -logK_w)

(2) pK_w = (pH + pOH) = 14

(3) pH = pK_a + log[conjugate base]/[conjugate acid]

Henderson-Hasselbalch equation for buffer solutions - i.e., solutions of weak acids or weak bases with their salts.

NOTE THAT pH = pK_a WHEN [CONJUGATE BASE] = [CONJUGATE ACID],

i.e., WHEN THE WEAK ACID OR BASE IS HALF DISSOCIATED

(4) pH = 0.5(pK_a - logC)

Solution of a weak acid, or a salt of a strong acid and a weak base, concentration C mol/litre

Examples

1 (a) What is the pH of a solution with [H⁺] = 40 nM ?

(b) What is the H⁺ concentration, in nmol/litre, in a solution at pH 8 ?

(a) [H⁺] = 40 x 10^-9 mol/litre = 4 x 10^-8 mol/litre

Therefore pH = -log (4 x 10^-8 ) = 7.4

(b) log[H⁺] = - 8, therefore [H⁺] = 10^-8 M or 10 x 10^-9 M or 10 nM

2 Given that the pK_a of formic acid (methanoic acid) is 3.8

(a) What is the pH of a solution containing 20 mM H.COOH and 40 mM H.COOK ?

(b) What are the concentrations of H.COOH and H.COONa in a

60 mM (formic acid/sodium formate) buffer at pH 3.5 ?

(a) pH = pK_a + log[H.COOK]/[H.COOH] = 3.8 + log[2] = 3.8 + 0.3 = 4.1

(b) pH = pK_a+log[H.COONa]/[H.COOH] so log[H.COONa]/[H.COOH] = (3.5 - 3.8) = - 0.3 and [H.COONa]/[H.COOH] = 10^-0.3 = 0.5 Since [H.COOH] + [H.COONa] = 60 mM, [H.COOH] = 40 mM and [H.COONa] = 20 mM

3 Given that the pKa of CH_3.COOH is 4.8, how many g of NaOH would be required to convert 60 ml of 100 mM CH_3.COOH to a buffer solution at pH 5.3? [H = 1, Na = 23, O = 16]

pH = pK_a + log[CH_3.COO^-]/[CH₃,COOH], therefore [CH_3.COO^-]/[CH₃,COOH] = 10^0.5 = 3.16

60 ml of 100mM contains (0.06 x 100) mmol = 6 mmol acid

Hence require (3.16)(6) = 19 mmol NaOH or (19 x 10^-3 x 40) g = 0.76 g NaOH

4. Given that the pK_a values of the terminal -NH₂ and -COOH groups are 8.0 & 3.5, respectively, whilst the Lys side chain has a pK_a value of 10.5, calculate the average net charge per molecule of the tripeptide Ala-Lys-Ala in solution at pH (a) 10.5 and (b) 3.5.

Remember that when the pH equals the pK_a value, 50% of the groups are dissociated. When the pH is 1 unit above the pK_a value, 90% are dissociated; 2 units above, 99% dissociated.. Conversely, when the pH is 1 unit below the pK_a value, only 10% are dissociated; 2 units below, 1% dissociated.

Exercises

1. Calculate (to 1 decimal place) the pH of the following solutions.

(a) 0.1 mM-HCl (b) 3 µM-HCl (c) 6 µM-KOH (d) 3 µM-NaOH

3. A mixture of equal volumes of (b) and (c).

2 Calculate (to 2 significant figures) the

(a) [H⁺] of blood plasma (pH = 7.4) (b) [OH^-] of saliva (pH = 6.6)

(c) [H⁺] of tomato juice (pH = 4.3) (d) [H⁺] of gastric juice (pH = 1.5)

3. [OH^-] of pancreatic juice (pH = 7.9)

3. Given the pH of blood plasma is 7.4, calculate the value (to 2 significant figures) of the ratios

(a) [HCO₃^-] / [H₂CO₃] and (b) [HPO₄^2-] / [H₂PO₄^-] in blood plasma.

(pK_a values for H₂CO₃ and H₂PO₄^- are 3.8 and 6.8, respectively.)

Tutorial 3. Theme 3; Protein Structure and Spectrophotometry.

Exercises

1. (a) What is the average net charge per molecule of the each of following amino acids in solution at pH 7 ? (Arg, Asp, Glu, Gly, Leu, Lys, Met, Val)

(b) If they were constituents of a globular protein in aqueous solution at pH 7, which would you expect to have their side chains inside and which outside ?

(c) An ion-exchange column consisting of polystyrene beads containing sulphonic acid residues interacts with molecules via both the negative charges of the acid residues and the strongly hydophobic backbone of the resin. Predict which member of each pair of amino acids listed below would be eluted first from such a column by a pH7 buffer solution.

(i) Asp & Lys (ii) Arg & Met (iii) Glu & Val (iv) Gly & Leu

Spectrophotometry

Background information

The equations needed for calculations involving measurements of light absorption are:

where "I_o" and "I" are the intensities of the incident and emerging light, respectively; "A" is the absorption (often called 'absorbance' ), "E" the extinction coefficient, "C" the concentration of the solution and "L" the light path length, measured in cm. "C" is usually measured in molar units, in which case "E" is the "Molar extinction coefficient", with units of (litre mol^-1 cm^-1). [Note that "A" is the logarithm of a ratio of identical units and so has no units, although its values are usually given as "absorption (or absorbance) units".

Usually these absorption measurements are made to determine the concentration of a solute in solution, when either the solute itself, or an easily formed derivative of it, absorbs light at a convenient wavelength. This method requires either knowledge of the extinction coefficient of the absorbing species, or the construction of a standard calibration curve

Examples

1. The absorption at 260 nm of an ATP solution in a 1 cm light-path cuvette is 0.390. Given _ME₂₆₀ = 13 x 10³ litre.mol^-1.cm^-1 what is the concentration of the ATP solution ?

C= (A/E) mol/litre = (0.390)/(13 x 10³) = 0.03 x 10^-3 M or 30 µM

2. Bovine serum albumin (BSA) was used as a standard protein in a spectophotometric method for assaying protein, which involved the formation of a coloured derivative absorbing at 590 nm. Two samples, A & B, of a solution of unknown protein concentration were assayed at the same time, giving the following results.

Calculate the protein concentration in the "unknown" solution.

When the results are plotted a curve is obtained. Instead of calculating an extinction coefficient, this standard curve must be used - the unknown values being simply read off directly, as indicated by the dashes lines on the graph.

Sample A contained 35 µg protein in 50 µl

Sample B contained 69 µg protein in 100 µl,

Hence the concentration of the unknown was about 70 µg/100 µl or 0.7 mg/ml.

Exercises.

Unless it is indicated otherwise, assume that all absorbance values refer to measurements with a 1 cm light path.)

1. (a) The absorbance at 260 nm of a solution of cytosine is 0.666.

Given that _ME₂₆₀ = 5.55 x 10³ litre mol^-1 cm^-1 for cytosine, calculate the concentration of the solution.

(b) A 0.2 ml sample was taken from 6 ml of GTP solution and mixed with 9.8 ml water; then 0.5 ml from the latter solution was mixed with 4.5 ml water. 3 ml of the latter solution had an absorbance (at 260 nm) of 0.540. How many µmol of GTP were there in the original 6 ml of GTP solution ? (_ME₂₆₀ = 7.2 x 10³ litre mol^-1 cm^-1 for GTP)

2. Solutions of ATP were prepared as shown below and their absorbances at 260 nm measured, giving the values indicated.

Plot absorbance against concentration and calculate (to 3 significant figures) the value of the molar extinction coefficient for ATP from the best-fit line.

3. A spectrophotometric assay for protein consists of measurement of the absorption at 750 nm of a blue derivative which is produced by reaction with "reagents I & II". Bovine serum albumin (BSA) is commonly used as a "standard protein" to calibrate the method with the formation of a "standard curve". In this example the method was used according to the protocol given below to determine the protein content of two solutions (A & B). Each measurement was made in duplicate, giving the absorption values recorded below.

* Solution B diluted 10-fold with water

Draw the standard curve and use it to calculate the concentration of protein in A and B.

2. Qualitative analysis revealed that the ATP in solution A had been partly hydrolysed, releasing inorganic phosphate (Pi). To estimate the extent of the hydrolysis, solution A was analysed quantitatively for (a) (ATP + ADP + AMP) and (b) Pi.

(a) 100 µl A diluted by addition of 100 µl water to give B. Then 100 µl of B diluted with water to give10 ml of C. Duplicate measurements of A₂₆₀ for C gave values of 0.635 and 0.665.

(_ME₂₆₀ for adenosine phosphates is 13 x 10³ litre mol^-1cm^-1)

(b) Pi was converted to a phosphomolybdate complex, which absorbed at 680 nm when reduced:

2. Calculate the percent hydrolysis of the ATP.

Tutorial 4.Theme 4; NUCLEIC ACIDS

The genetic information contained in DNA consists of a linear sequence of successive words, known as codons. Each codon is a specific sequence of three nucleotides (three nucleotide pairs in double-stranded DNA), and each codes for a single amino acid unit in a protein. Where necessary, assume that (a) the average molecular weight of nucleotide pair is 660; (b) each nucleotide pair contributes 0.34nm to the length of DNA; and (c) the average molecular weight of an amino acid in a protein is 110.

1 (a) The partial composition (in mole-fraction units) of one of the strands of double- helical DNA molecule is: [T] = 0.35; [C] = 0.25. What can be said about

(i) [A] and [G] of the same strand

(ii) [A], [C], [G] and [T] of the complementary strand ?

(b) In the DNA of certain bacterial cells, 13% of the nucleotides are adenine. What are the percentages of the other nucleotides ?

2. A protein synthesised in rat liver has 192 amino acid residues. What is the minimum number of DNA base pairs required to code for this protein ? It is in fact encoded by a gene of molecular weight 9.5 x 10⁵. Suggest possible explanations for this.

3. The molecular weight of an E.Coli DNA molecule is very great, about 2.5 x 10⁹.

(a) Given the assumptions noted above, calculate the length of an E.coli DNA molecule. Compare the length of the DNA molecule with the actual cell dimensions. How does it fit in?

(b) Assuming that the average protein in E.coli consists of a chain of 400 amino acids, what is the maximum number of proteins that can be coded by an E.coli DNA molecule?

MCBHD tutorials

Aims:

• Supporting you to learn the main concepts in Biomolecular Sciences and their importance to describe basic life processes, both in health and disease. This involves discussion of the cellular structures, of the main groups of molecules involved, and of the most important processes underlying the functioning and interaction of cells within the body;

• Complementing the lectures by discussing questions that remained open to you during your self-directed learning and by introducing certain aspects of the subject which are difficult to cover in a normal lecture format. Make actively use of this possibility by coming forward with your own questions.

• Encouraging you to develop your scientific curiosity and your skills in literature research, in writing essays and in presenting a scientific subject.

Format of the tutorials:

Each student group will have four tutorial sessions, numbered 1 through 4. Each session will cover the material of the two preceding lectures according to the following topics:

Tutorial 1: Cells - their basic structure and functions, and how they make up our body.

Tutorial 2: Molecules that are involved in the structure and functioning of the cell.

Tutorial 3: Enzymes – biocatalysts that facilitate processes in our body.

Tutorial 4: Nucleic acids, macromolecules providing the chemical basis of inheritance.

Yours tutors will discuss further details with you.

Assessment:

The tutorials are an important part of the course, your attendance and participation will be noted. Moreover, each of you will be expected to hand in one essay and to give one brief presentation. Your tutors will give to you the questions for the essays and presentations in the first tutorial.

You will receive an overall mark made up from the mark of your written essay and your presentation. Your attendance and participation in the discussions will also form a component of the overall mark, which will count one point towards the end-of-term exam (see page 31).

Molecular and Cellular Basis of Health and Disease (MCBHD) course

Allocation of students in groups are as for the CLS tutorials (see above).

Tutors are Prof M Clemens (ext 5762) (with Dr Jeffries (ext 5760))(Groups 1 &4), Dr U Bommer(ext 5754)(Groups 2 & 5), Dr S O'Dell (email (sodell@sghms.ac.uk)Groups 3 & 6)

Recommended textbooks:

• Alberts et al.: Essential Cell Biology, 1998.

• Meisenberg and Simmons: Principles of Medical Biochemistry, 1998.

• Dow, Lindsay, Morrison: Biochemistry. Molecules, Cells and the Body, 1996.

Biomedical Sciences Laboratory Pratical Course.

The lab Practical Course will run on Tuesday afternoons in lab 5 Hunter Wing level 5. A seperate handbook is available for this course from the AVA

Overview on Term 1 Assessment

Summative assessment.

The assessments will be combined to give you a mark out of 15. This mark will contribute to an overall mark for the year out of 100. You will need at least 45 marks out of the 100 to pass the first year and proceed to the second year of the course.

The assessment of your work during this term will take the form of:

1. In-course assessment:

• Biomedical Practical Course (write-ups and viva) 3 marks

• Overall mark for Chemistry tutorials 1 mark

• Overall mark for MCBHD tutorials 1 mark

2. Formal end-of-term examination: 10 marks

total: 15 marks.

The end-of-term examination will consist of a CFP-based exam, and a degree-specific exam, based on the “Chemistry in Life Sciences “ and the Practical course, to be taken on the afternoon of Wed, Dec 6^th. The latter will consist of a written paper with multiple choice questions, short answer questions, and an essay. The practical examination for the anatomy course will be taken on Thurs Dec 7^th.

Degree-specific time table of the Common Foundation Module; It is important the Biomeds follow this time-table, as you are not expected to do the more clinical aspects of the CFP, nor CBLs

M – MBBS, B – Biomedical Sciences, P – Physiotherapy, D – Diagnostic Radiotherapy, T – Therapeutic Radiotherapy, N - Nursing