Calculations in the molecular biosciences Part 1: Characterization of biological molecules The next two chapters will explain the main types of calculations involved in studying proteins. In this chapter, we shall first of all deal with some general principles important in any calculations, and then concentrate on how to study the individual components of reactions. This includes the preparation and dilution of solutions and how spectrophotometry can be used to check concentrations in some cases. The behaviour of acids, bases, and buffers is described and this is followed by a discussion of the specific activities of proteins and the manner in which their purification is recorded. Chapter 4 will concentrate on the analysis of a number of processes and reactions of biological interest. At the end of this chapter, there are several problems for you to check your understanding of the material covered. If you wish to explore the topics covered here in more detail, the books mentioned in Chapter 1 can be consulted. I The golden rules for successful calculations ■H KEY CONCEPTS ■ Working out the strategy for a calculation ■ Setting out and explain the working H| ■ Making sure the number of significant figures and the units are correct Whenever you do a calculation, make sure that the answer to each of the following questions is yes'. This will not absolutely guarantee success but will go a very long way towards that goal. ■ Have I worked out the correct strategy for the calculation? ■ Is my working clearly set out and explained? ■ Is the value obtained reasonable? ■ Is it reported to the appropriate number of significant figures? ■ Is it quoted in the correct units? The aim of this chapter is to help you carry out calculations successfully in a number of topics covering the purification and characterization of proteins. 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES It is difficult to overestimate the importance of the need for this check. The average frequency of occurrence of amino acids in proteins is given in Chapter 1, Table 1.1. Leucine and alanine occur most frequently (9.6% and 7.8%, respectively); cysteine (1.5%) and tryptophan (1.2%) occur least frequently. These are only average figures and do not necessarily apply to any particular protein, for example the small protein metallothionein that plays a role in storing a number of heavy metal ions consists of about 30% cysteine residues. We can expand these points as follows: Have I worked out the correct strategy for the calculation? Before starting to write down your solution, make sure that you have worked out how you will proceed from the information supplied in the problem you are trying to solve to the point where you can calculate the final answer. If the path is not immediately obvious, it can be a good idea to think about what you can deduce from the information and then try to think of what might be required immediately before the final step. Working from both ends in this way can often help to define the strategy. 7s the working clearly set out and explained? It is very important to break down the calculation into steps and use words to explain. Do not just write down numbers without explanation. If you do, you will not convince anyone else reading your work that you know what you are doing. If you make a small slip in the first step and then this is (correctly) carried through subsequent steps, you should obtain credit for this; if you have not explained what you are doing and the final result is incorrect it is highly unlikely that you will obtain any credit at all. Is the value obtained reasonable? Sometimes it is not clear what sort of value might be expected, although it is usually possible to check that the outcome of the final calculation step is of the right order of magnitude (see Chapter 2, section 2.1). However, in many cases it is possible to think about whether the value is reasonable. For example, one would expect to find small integral values for the number of ligand-binding sites on a protein, e.g. one substrate binding site per polypeptide chain. Values of 0.0009 or 926 sites per polypeptide chain are very unlikely and suggest that a serious error has been made in the calculation, probably involving either the use of incorrect units or mistakes in handling powers of 10. A second example would be if you were asked to use some experimental chemical modification data to calculate the number of cysteine side chains in a protein of molecular mass 30 kDa. Before starting, one can make a useful estimate as follows. A 30 kDa protein will contain about 30 000/110 amino acids, i.e. 273 amino acids (see Chapter 2, section 2.1). There are 20 different kinds of amino acids in proteins, so if they occurred with equal frequency (which in practice they do not, but that is not important here), there would be about 14 of each type. Thus, we might expect to find about 14 cysteines in each molecule of the protein. In fact, cysteine is among the rarer amino acids so in most proteins of this size the actual value would be less than 14. The key point is that a value of, for example, 538 would clearly be incorrect. In addition, a value of 0.25 would be incorrect since the value should be either zero or an integral value such as 1,2, etc. Is the appropriate number of significant figures used? This has been dealt with in Chapter 2, section 2.2; remember to use the display in the calculator with caution when writing down the final answer. Is the answer quoted in the correct units? The subject of units of quantities is discussed in section 3.3. You should get into the habit of not only quoting the correct units of the final answer, but also quoting the units at the various stages of the calculation. This will act as a guide to the direction of the calculation process, and will also help to convince anyone else that you know what you are doing! 3.2 MAGNITUDES OF QUANTITIES 77 The key concepts at the start of the section can be checked at the end of section 3.3. SELF TEST Magnitudes of quantities KEY CONCEPTS ■ Appreciating the range of sizes of quantities used in describing biological systems ■ Performing calculations involving powers of 10 Molecular biology and biochemistry are quantitative sciences; the values of quantities we discuss range from the very large (numbers of molecules, Avogadro's number, frequency of radiation, etc.) to the very small (dimensions of cells, organelles and molecular complexes, concentrations of hormones, etc.); the range of dimensions of biological objects is illustrated in Fig. 3.1. This range of magnitudes means that you need to develop confidence in handling powers of 10 and the prefixes used for quantities (section 3.3). Fig. 3.1 shows the advantage of using a logarithmic scale. For example, if we wished to represent the range of dimensions from 10 nm to 100 m on linear scale graph paper with a grid where each 1 mm represents 10 nm, then to represent 100 m, we would need a piece of paper 104 km long (more than the distance from London to Los Angeles!). The key concepts at the start of the section can be checked at the end of section 3.3. Giant redwood Blue i whale 1 Human Longhorn j beetle Fire ant Fruit fly House Most Bacterial cell; dust animal and nucleus in mite plant cells eukaryotic cell I t 1 1 T 1 Virus Ribosome 100 m 10 m 1 m 100 mm 10 mm (1 cm) 1 mm 100 |im 10 urn 1 um 100 nm 10 nm Naked eye Light microscope Electron microscope Fig. 3.1 The dimensions of biological objects. The scale is logarithmic, with each division representing a factor of 10. The dimensions associated with each object are for guidance only as there will be a range of values in each case. Individual atoms and small molecules would occupy the zone between 0.1 nm (100 pm) and 1 nm. The lower arrow bars indicate the observation ranges for the naked eye, light microscope, and electron microscope. Highly sophisticated electron microscopes can resolve objects down to about 0.2 nm (200 pm) in size. 78 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES ■ Knowing the SI prefixes for units and interconvert them as appropriate ■ Using correct units in equations and graphs KEY CONCEPTS Units of quantities It is important to refer to a 1 M solution for calculation purposes, i.e. a Kd of 50 [lM should be expressed as 50 x 10-6 M (or 5 X lO"5 M). The absorption coefficient was previously known as the extinction coefficient; the latter term is still frequently used. Virtually all quantities we come across have units, for example mass (e.g. kg, kDa), length (e.g. mm, nm), concentration (e.g. g L"1, mol L_1 (or M)), time (e.g. ms), etc. It is essential to state these units in any description of a system or during calculations. The only exceptions arise where the quantity represents a ratio of two values, e.g. the absorbance (A) of a solution is defined in terms of the ratios of intensities of two light beams (incident on the solution and transmitted by the solution) and therefore does not have units. It should be pointed out that equilibrium constants (see Chapter 4, section 4.1) raise an apparent contradiction; strictly speaking they are dimensionless since all the terms in the definition refer to ratios to the standard state for that component. However, since for solutes, the values (of concentrations) are each expressed relative to a standard state of 1 M concentration (see Chapter 4, section 4.1), equilibrium constants are usually quoted with units of M to indicate this. Remember that in any equation, the units must balance on the two sides. For example, in spectrophotometry the Beer-Lambert law (eqn. 3.1; see section 3.6) states that: A = excxl 3.1 where A is the absorbance of a solution of concentration c in a cell of path length /. The absorption coefficient £ gives a measure of the ability of the solution to absorb light at the wavelength in question. The absorbance (A) does not have units (it is merely a ratio, expressed in logarithmic terms), c has units of M (molar), and I has units of cm. In order for the right-hand side of the equation overall to have no units, the units of £ must be M_1 cm-1. When plotting a graph, the units of the slope of the graph will be the units of the y-axis divided by the units of the x-axis. Thus, if you plot the concentration of product formed in a reaction (measured in mM units) on fhey-axis and the time (measured in min) on the x-axis, the units of the slope are mM min-1. 3.3.1 The Systeme Internationale (SI) system of units The SI system of units is based on the metre (m), kilogram (kg), second (s) quantities and units derived from these. Information on these units and units derived from them as well as the values of physical constants is included in Appendix 3.1 at the end of this chapter. It uses prefixes to bring numerical values of quantities into the 'easy to scale' range, i.e. between 0.1 and 100; thus 800 nm = 0.8 Ltm; 3.3 UNITS OF QUANTITIES 2000 pm = 2 mm, etc. The most important prefixes involve factors of 1000, so we can set up a scale as shown in Table 3.1: Table 3.1 The prefixes used in the SI system of units Prefix G, M, k, Base m, n, P. f, giga mega kilo unit milli micro nano pico femto Numerical 10' 106 103 1 io-3 10"6 io-» irr12 io-15 value We only need to go outside the range shown on very rare occasions; for interest 1012 is known as Tera (abbreviated'T') and 10~18 is known as atto (abbreviated as 'a'). It is very important indeed for you to be able to convert quantities between the different prefixed units rapidly and reliably, for example to show that 5000 nm = 5 |tm and that 0.012 timol = 12 nmol. Some intermediate prefixes are retained for convenience, e.g. c (centi, 10~2), as in cm; d (deci, 10_1)> as in dm (note 1 litre = 1 dm3). For measures of volume there would be an inconveniently large factor of 109 between 1 m3 and 1 mm3, hence the use of 1 dm3 and 1 cm3 (1 cm3 - 1000 mm3; 1 dm3 = 1000 cm3). One example of the use of the prefix T would be in denoting the frequency of radiation. Thus, red light of wavelength 700 nm has a frequency of 4.3 X 10" Hz (hertz) or 430 THz. The length of the C-N bond in the peptide bond unit is 0.132 nm. F.xpress this in terms of urn, pm and also in terms of Angstrom units (A), which are commonly used to describe protein structures (1 A = 10'* cm). STRATEGY This is a straightforward application of the interconversion of prefixes in the SI system of units. SOLUTION 1 nm= 10 'urn, so 0.132 nm= 1.32 x 10-1 pm = 132 pm = 1.32 A EXAMPLE / The Angstrom unit (A) is not an SI unit, but is widely used by structural biologists; most bond lengths are in the range 1-2 A (0.1-0.2 nm). WORKED E) The molecular mass of lysozyme is 14 300 Da, and its density is 1.4 gmL '.Assuming that the lysozyme molecule can be regarded as spherical, calculate its radius. Avogadro's number is 6.02 x 1023 mol-1. The volume of a sphere of radius r is given by (471/3)^. STRATEGY This problem involves calculating the mass of an individual molecule of lysozyme and hence its volume and radius. It is very important to keep track of the units and powers of 10. SOLUTION The mass of 1 mol lysozyme is 14 300 g (see section 3.4). Hence the mass of 1 molecule = 14 300/(6.02 x 1023) g = 2.375 x 10~20 g. Since volume = mass/density, the volume of 1 molecule = 1.697 x 10-20 mL (cm3). So (47t/3)r3 = 1.697 x 10~20, hence r3 = 4.05 x IO"21, therefore r= 1.59 x 10~7 cm, or 1.59 nm (note 1 cm = 10~2 m; 1 nm = 10~9 m). Thus, the radius of a lysozyme molecule = 1.59 nm. In this calculation, the cube root can be readily derived by using logarithms (see Chapter 2, section 2.3). 80 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES SELF TEST Check that you have mastered the key concepts at the start of sections 3.1 -3.3 by attempting the following questions. ST 3.1 A pipetting device is stated to be capable of delivering 2.5 mm3 portions of a solution. Express this volume in terms of mL and |_iL. ST 3.2 A Scatchard plot is often used to analyse binding data. In this plot, r/[L]frcc is plotted against r (r is a ratio and therefore has no units; [£]fr« is the concentration of free ligand and in the case in question has units of LiM). What are the units of the slope of the graph? ST 3.3 Calculate the radius of a bacterial ribosome which can be assumed to be spherical. The molecular mass of the ribosome is 2.7 MDa and the density is 1.4 g mL_I. Avogadro's number is 6.02 x 1023 mol-1. The volume of a sphere of radius r is given by (471/3)^. Answers ST 3.1 The volume delivered = 2.5 x lfr3 mL or 2.5 uL. ST 3.2 The units of the slope of the graph are jxM1. ST 3.3 The radius of the ribosome = 9.1 nm. Attempt Problem 3.1 at the end of the chapter. The concepts of moles and molarity KEY CONCEPTS Distinguishing clearly between moles and molarity Knowing the molecular mass of the solute, interconverting concentrations expressed in terms of mass/volume and in terms of molarity It is important to note that the term mole can refer to molecules, atoms, subatomic particles, photons, etc. One mole (abbreviated mol), which is the gram formula mass of a substance, contains Avogadro's number (6 x 1023, or more accurately 6.022 x 1023) of entities (which can be molecules, atoms, ions, etc.). Substances react with one another in terms of moles, not in terms of masses. Thus, in the hydrolysis of urea: CO (NH2)2 + H20 -» C02 + 2NH3 1 mol of urea (60 g) reacts with 1 mol water (18 g) to give 1 mol carbon dioxide (44 g) and 2 mol ammonia (2 x 17 g, i.e. 34 g). We can also refer to 1 mol of atoms (e.g. H atoms or CI atoms) or ions (e.g. H+ ions or electrons); in each case this will mean the Avogadro number of that species. 3.4 THE CONCEPTS OF MOLES AND MOLARITY 81 3.4.1 Molarity of solutions Virtually all biochemical reactions take place in solution. (A solution consists of one or more solutes dissolved in the solvent). A 1 molar (M or mol dm-3) solution contains 1 mol of the solute in 1 litre (L or dm3) of solution. number of moles = concentration (expressed in M) x volume (expressed in L) KEY RELATIONSHIP in A certain solution of NADH has a concentration of 0.6 mM. How many moles of NADH are present in 0.5 mL of this solution? STRATEGY This is an application of the relationship: number of moles = concentration x volume SOLUTION 1 L of the solution contains 0.6 mmol. Hence 0.5 mL contains 0.3 x 10-6 mol, i.e. 0.3 umol. WORKED EXAMPLE NADH (nicotinamide adenine dinucleotide, reduced form) is a biological reducing agent, which participates in redox reactions catalysed by dehydrogenases, such as lactate dehydrogenase (pyruvate + NADH + H+ ^ lactate+ NAD+). The structure of NADH is given If the molecular mass of the solute is X dalton (Da), then 1 mol = X g (e.g. for in Chapter l, section 1.8). glucose C6H1206 the gram formula mass would be 180 g). AIM solution therefore contains X g solute It' (for glucose, this would be 180 gL"1). With practice, you should be able to work out rapidly how many mol of a solute are present in a certain volume of solution of specified molarity. Most biochemical reactions involve small volumes of dilute solutions, so we are likely to be dealing with Limol, nmol or even pmol quantities of solute. We can readily work out the concentrations of solutions as follows: A solution containing 1 g solute of molecular mass X per litre would be (MX) M (for glucose this would be (1/180) M = 5.56 x lO"3 M). For biochemistry, g and L are large quantities; mg and mL (or even |ig and p.L) are much more likely to represent the scale of experiments undertaken and to take account of the high costs of many of the reagents involved. In terms of concentration, 1 pg pL 1 = 1 mg mL 1 = 1 g L ]. KEY RELATIONSHIP Q Thus, we can always calculate and obtain the molar concentration of a solution by dividing the mg mL 1 (or g L-1 or pg pL ') concentration by the molecular mass (in Da). Thus, a 0.2 mg mL1 solution of glucose = 0.2/180 M = 1.11 x 10"3 M (1.11 mM). 82 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES BSA is an abundant protein in the serum of cattle blood. It can bind a number of non-polar substances and serves to transport fatty acids in the blood stream. KED EXAMPLE A 0.15 mg mL 1 solution of bovine serum albumin (BSA) (molecular mass 66 000 Da) = 0.15/66 000 M = 2.273 LiM. The concentration of protein required for an nuclear magnetic resonance experiment is 0.5 mM. If the molecular mass of the protein is 27.5 kDa, what concentration is required in terms of mg mlr1? If the sample volume is 0.4 mL, what mass of protein is required? STRATEGY This is a relatively straightforward calculation, provided that you keep track of the units and powers of 10. SOLUTION AIM solution of protein contains 27 500 mg voir1. Thus a 0.5 mM solution contains 0.5 x 10~3 x 27 500 mg mlr' = 13.75 mg mlr1. In 0.4 mL there would be 13.75 X 0.4 mg, i.e. 5.5 mg. It should be noted that for large molecules such as proteins a 1 M solution is physically unrealistic; it would be impossible for instance to dissolve 27 500 mg (27.5 g) protein in buffer to give a 1 mL solution. Nevertheless, the concept of a 1 M solution represents a useful basis for calculations. In a number of cases, it is convenient to describe the concentration of a solute in a solution in terms of % weight/volume (abbreviated as w/v); this would represent the grams of solute per 100 mL solution. If two liquids are being mixed (e.g. ethanol and water) it can be convenient to express the concentration of ethanol in terms of percentage volume/volume (abbreviated as v/v). WORKED EXAMPLE The structure of acrylamide is given in Chapter 1, section 1.8. When acrylamide is polymerized and cross-linked it gives a gel which provides a convenient support medium for electrophoresis in which convection is minimized. The monomer (but not the polymer) is a powerful neurotoxin and must be handled with great care. Acrylamide, which can be polymerized to make a gel widely used in electrophoresis, is available as an aqueous solution at 30% (w/v). The molecular mass of acrylamide is 71.1 Da. What is the concentration of this solution in molar terms? STRATEGY This is again a relatively straightforward calculation provided the correct units are used throughout the calculation. SOLUTION A 30% (w/v) solution corresponds to 300 g L_I. AIM solution of acrylamide would be 71.1 g L_1. Hence, the concentration is 300/71.1 M = 4.22 M. A final note of caution: it is absolutely essential that you appreciate the distinction between the amount of a solute in a solution (measured in moles) and its concentration (measured in molar). You should be able to work out that, for example, 1 mL of a 1 mM solution contains 1 Ltmol and that 0.02 mL of a 0.7 LiM solution contains 14 pmol. 3.5 PREPARATION AND DILUTION OF STOCK SOLUTIONS 83 For some calculations it is appropriate to work in terms ofmolfor others it may be best to work in terms ofM. You must never slip from one to the other during a calculation without being aware of what you are doing. If you label the quantities with the correct units as you go along, you should avoid making mistakes. Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 3.4 How many moles of solute are present in 2.5 raL of a solution whose concentration is 1.35 mM? ST 3.5 A solution contains 2.1 pmol of a given protein in 0.5 LiL. What is the concentration of the solution? ST 3.6 A mass spectrometer requires 0.2 ng protein in a volume of 2.5 LiL. If the molecular mass of the protein is 34 kDa, what is the concentration of the solution in molar terms? ST 3.7 The molecular mass of a small peptide is 791 Da. What would be the concentration of a 5 mM solution of the peptide in terms of mg mL-1? Answers ST 3.4 The amount of solute = 3.375 x 10-6 mol or 3.375 Limol. ST 3.5 The concentration = 4.2 uM. ST 3.6 The concentration = 2.35 nM. ST 3.7 The concentration = 3.955 mg mtr1. Attempt Problems 3.2 to 3.8 at the end of the chapter. BX9 Preparation and dilution of stock solutions IhHB KEY CONCEPTS Working out how to prepare stock solutions of the required concentration ■BBBB ■ Calculating how to dilute these solutions appropriately for particular applications The accurate preparation and dilution of solutions is essential for accurate work in the molecular biosciences. The following sections should give you guidance as to how to carry out these operations successfully. 3.5.1 Stock solutions In experimental work, we often make use of'stock solutions' of compounds which have either been prepared to a high degree of accuracy in the laboratory or purchased from a reputable supplier. These stock solutions are usually much more 84 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES I Many biological compounds are stored at low temperatures because they are unstable at room temperature. When taking bottles from the fridge or freezer it is important to allow them to warm up to room temperature before opening, otherwise water vapour will condense on the inside of the bottle. The dissolution of many substances including urea and GdmCl is endothermic (i.e. takes in heat), so it is often necessary to warm solutions gently to aid dissolution. concentrated than our 'working solutions' (in actual experiments) and must be diluted appropriately before use. The preparation of stock solutions is usually carried out as follows. Suppose that you wished to prepare 5 mL of a 5 mM stock solution of NADH. You would need to know the molecular mass of NADH; inspection of the catalogue reveals that the disodium salt hydrate of NADH (which is the form you have purchased) has a molecular mass of 709.4 Da. A 1 M solution of NADH would be 709.4 g Lr1, i.e. 709.4 mg mL-1. Thus, a 5 mM solution would be 5 x 10-3 x 709.4 mg mL-1, i.e. 3.55 mg mLr1. If we wished to make up exactly 5 mL of this solution, we would have to weigh out 5 x 3.55 mg, i.e. 17.75 mg NADH on an appropriate balance. This could take some considerable time. It is much more likely that we require about 5 mL of the stock solution, in which case we can adjust the volume of solvent (water, buffer, etc.) so that the concentration is correct. Thus, if we weighed out 17.43 mg NADH, we would add 4.91 mL solvent to obtain the correct concentration. (In practice, we would probably use measurements of light absorbance to confirm the concentration of this stock solution, as explained in section 3.6. This would be particularly important in those cases where a substance may have absorbed moisture, which often happens to powders stored at low temperatures.) In some cases, particularly of very concentrated stock solutions, there can be significant volume changes when the solvent is added to the solute. For example, if we wished to prepare 10 mL of an 8 M solution of the denaturant guandinium chloride (GdmCl) of molecular mass 95.5 Da, we can readily calculate that we would require 95.5x8 mg mL"1, i.e. 764 mg mL1 GdmCl. For 10 mL, this would be 7.64 g GdmCl. If we were to put this into a graduated tube and add 10 mL of water or buffer, the final volume would be close to 14 mL and the solution would not be of the required concentration. Instead, we would carefully add the solvent, making sure that the GdmCl was dissolved and bringing the volume of the solution up to the 10 mL mark. It should be noted that 8 M is close to the solubility limit of GdmCl, so it takes some time to ensure that it is all dissolved. (Since GdmCl does not have a convenient absorption band, the concentration of the solution can be checked by density or refractive index measurements, by reference to standard tables of values). 3.5.2 Dilution of stock solutions Many procedures involve the preparation of working solutions from more concentrated stock solutions which have either been purchased or made up in the laboratory. It is very important to be able to do this accurately. Two basic points are: A. The amount of solute cannot be increased or decreased by any dilution procedure (matter cannot be created or destroyed!), and B. The amount of solute = the volume of solution multiplied by its concentration. 3.5 PREPARATION AND DILUTION OF STOCK SOLUTIONS Thus if: Cs = concentration of stock solution Vs = volume of stock solution taken and Cw = final concentration of working solution Vw = final volume of working solution Then we can derive the following important equations: KEY RELATIONSHIP concentration of stock x volume of stock = final concentration x final volume i.e.Csx Vs = CwxVw 3.2 from which it follows that: volume of stock = (final concentration x final volume)/concentration of stock Note the importance of relating the volume of the stock solution to the final (total) volume of the solution (eqn. 3.2), not to the volume of buffer or solvent The application of eqn. 3.3 is illustrated in the worked examples below. added to dilute the stock solution. WORKED EXAMPLE How much of a stock solution of ATP (4.0 mM) in a given buffer should be taken to give a concentration of 0.1 mM in a working solution of total volume 1.0 mL? STRATEGY This is an application of eqn. (3.3); it is important to ensure that the units are the same on both sides of the equation. SOLUTION In this case, Q = 4.0 mM, Cw = 0.1 mM, Vw= 1.0 mL Hence from eqn. (3.3) Vs = (0.1 x 1.0)/4.0 mL = 0.025 mL (25 uL) The remaining volume (0.975 mL or 975 |0.L) would be made up of the appropriate buffer. 86 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES / WORKED EXAMPLE Stock solutions of ATP and MgCL, are 4.0 and 20 mM, respectively, in a given buffer. What volumes of ATP, MgCl, and buffer would you add together to produce a working solution of final volume 2.0 mL, in which the concentrations of ATP and MgCL, were 0.5 and 1.5 mM, respectively? STRATEGY This calculation involves the application of eqn. (3.3) to each solute separately. Again it is important to ensure that the correct units are used on both sides of the equation. SOLUTION We must consider the components separately. For ATP, Q = 4.0 mM, Cw = 0.5 mM and Vw = 2.0 mL Hence from eqn. (2.3) Vs = (0.5 x 2.0)/4.0 mL = 0.25 mL For MgCl2, Cs = 20 mM, Cw = 1.5 mM and Vw = 2.0 mL Hence from eqn. (2.3) Vs = (1.5 x 2.0)/20 mL = 0.15 mL Thus, the working solution would consist of 0.25 mL ATP, 0.15 mL MgCl2 and (2.0 - (0.25 + 0.15)) mL, i.e. 1.6 mL buffer. ? SELF TEST Check that you have mastered the key concepts at the start of the section by attempting the following question. ST 3.8 You are provided with stock solutions of an enzyme (2.5 mg mL-1), its substrate (5 mM), and urea (8 M), all in a given buffer. You wish to make up a solution of volume 0.5 mL with the following final concentrations: enzyme, 0.1 mg mL-1; substrate 0.25 mM; urea, 2 M. What volumes of the stock solutions and buffer would you require? Answer ST 3.8 The volumes of the solutions required are: enzyme, 0.02 mL (20 u\L); substrate, 0.025 mL (25 uL); urea 0.125 mL (125 uL); buffer, 0.33 mL (330 ul). Attempt Problems 3.9 to 3.1 5 at the end of the chapter. In ST 3.8 remember to use eqn. 3.3 to calculate the volume of each stock solution required. The volume of buffer required to make up the total volume is calculated by subtracting the stock volumes added from the required volume of the working solution. ■EQl Absorbance measurements IHl KEY CONCEPTS HH ■ Using the Beer-Lambert law to calculate concentrations from absorbance data ■ Knowing the range of absorbance values where measurements are most HHHB reliable Measurements of the absorption of light by a solution (spectrophotometry) can be an extremely convenient way of determining the concentration of that solution, 3.6 ABSORBANCE MEASUREMENTS provided that (a) the solute has a suitable absorption band, and (b) you have access to a properly calibrated instrument (spectrophotometer). The fundamental equation governing spectrophotometry is the Beer-Lambert law (eqn. 3.1): A=excxl where A is the absorbance at a particular wavelength, e is the absorption coefficient (a measure of the degree of absorption of light at that wavelength), and I is the path length of the cuvette (or cell) that holds the solution of concentration c. From this equation we can see that at a given wavelength and at a constant path length, A is proportional to c. We have already seen this equation (section 3.3), where it was noted that the units of e will depend on the units of c and I, since overall, each side of the equation has no units. This is because A is defined by eqn. 3.4 as: where 70 is the intensity of the incident light, i.e. the light striking the cuvette, and 7, the intensity of the transmitted light, i.e. the light leaving the cuvette. WORKED EXAMPLE What percentage of the incident light is transmitted after passage through solutions which have absorbance values of (a) 0.05, (b) 0.3, (c) 1.0 and (d) 2.0? STRATEGY This is an application of eqn. 3.4 for each value of A by putting the value of A into the calculator and use the 10* function to obtain (/„/!,). The reciprocal of this value gives (IJI0). SOLUTION The value of (7,/70) at each value of A is multiplied by 100 to give the per cent of the incident light that is transmitted. Thus, the per cent of transmitted light is (a) 89.1%, (b) 50.1%, (c) 10% and (d) 1%. The logarithmic nature of A is not always readily appreciated because most modern instruments merely give a digital display of the absorbance. However, it is particularly important to note that you should always aim to measure absorbance values in the range 0.1-1.0. Despite what the manufacturers of the instrument may claim, absorbance values significantly above 1.0 are not usually reliable. You can bring absorbance values into the appropriate range either by an accurate dilution of a concentrated solution, or sometimes by using a cuvette of different path length. Measurements will be most accurately made when the absorbance is about 0.3. The absorbance at a given wavelength is often denoted by a subscript; thus Am0 is the absorbance at 280 nm. The absorption coefficient would be similarly abbreviated, e.g. e2g0. 88 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES If we know (or can look up) the absorption coefficient for a substance at a particular wavelength, we can use the Beer-Lambert law to determine the concentration of a solution, as illustrated below in the worked examples below. WORKED EXAMPLE When performing dilutions it is important to ensure that the correct pipettes are used and the solutions are completely mixed. The absorption coefficient of NADH at 340 nm is 6220 M-1 cnr1. You have made up a 2.5 mM stock solution of NADH (see section 3.5.1). How would you check its concentration spectrophotometrically? STRATEGY The key point is to make sure that the absorbance to be measured is in the range for reliable measurements (0.1-1.0). SOLUTION A 2.5 mM solution of NADH in a cuvette of 1 cm path length would, according to the Beer-Lambert law, have an absorbance at 340 nm (denoted as A340) of 6220 x 2.5 x 10~3 = 15.6. This is far too high to measure, so we need to perform an accurate dilution. If the solution were diluted 50-fold, e.g. by adding 0.02 mL to 0.98 mL buffer, the A340 should be 15.6/50 = 0.312, which is within the required range for measurement. The spectrophotometer should be adjusted so that the absorbance of buffer alone is set at zero. (Note that this dilution procedure would be appropriate for a cuvette of 1 mL capacity with a path length of 1 cm; if we were using a cuvette of 3 mL capacity, we would add 0.06 to 2.94 mL buffer.) / WORKED EXAMPLE A solution consists of a mixture of NADH and its oxidized form NAD+. NADH and NAD+ have equal absorbances at 260 nm (absorption coefficient 18 000 M"1 cm-1)- NADH absorbs at 340 nm (absorption coefficient 6220 M"1 cm"'); NAD+ does not absorb at this wavelength. In a cuvette of 1 cm path length, the absorbance of the solution is 0.933 at 260 nm and 0.135 at 340 nm. What are the concentrations of NADH and NAD+ in the solution? STRATEGY We can use the A2m to determine the sum of [NADH] and [NAD+], since both absorb equally at this wavelength. The A340 can be used to determine the [NADH]. SOLUTION From the A2m, the sum of [NADH] and [NAD+] is 0.933/18 000 M = 51.8 uM. From the A340, [NADH] = 0.135/6220 M = 21.7 uM. Hence by difference [NAD+] = 30.lLtM. Spectrophotometry provides a convenient way of measuring the concentration of a protein solution. Virtually all proteins absorb radiation at 280 nm due to the aromatic amino acid side chains (principally tyrosine and tryptophan) they contain. The absorption coefficients of different proteins at 280 nm will vary because proteins differ in terms of the exact proportions of these amino acids they contain. For example, the absorption coefficient of BSA (a very commonly employed 3.6 ABSORBANCE MEASUREMENTS protein standard) at 280 nm (e280) is 0.66 (mg mL_1) 1 cnr1, whereas that of hen egg white lysozyme is 2.65 (mg mL"1)^1 cnr1, reflecting the greater proportion of aromatic amino acids in the latter. For proteins, the absorption coefficients are often referred to concentrations in terms of mg mlr1 (i.e. w/v) rather than in terms of molarity. If molarity were to be used as the basis for expressing concentrations, not only would the quoted absorption coefficients be inconveniently large (see worked example below), but in addition there also could be ambiguities about the molarity in the case of proteins with multiple polypeptide chains. Does the molarity refer to the individual polypeptide chains or to the intact protein? This should be clearly stated. WORKED EXAMPLE The molecular mass of lysozyme is 14.3 kDa. What is the molar absorption coefficient of the protein at 280 nm, given that the A2S0 of a 1 mg mlr1 solution in a cuvette of 1 cm path length is 2.65? STRATEGY This calculation relies on evaluating the properties of a 1 M solution of the protein, and scaling the absorbance accordingly. SOLUTION AIM solution of lysozyme would be 14 300 mg mlr1. Using the Beer-Lambert law, the A280of this solution in a 1-cm pathlength cuvette is 2.65 x 14 300 = 37 900. Thus, the absorption coefficient of lysozyme is 37 900 M"1 cm-1. The relationship between the absorption coefficient of a protein at 280 nm and the content of aromatic amino acids is discussed in Chapter 6, section 6.1.5. Many biochemical compounds, including proteins, nucleic acids and small molecules such as ATP and NADH, have convenient absorption bands. If at all possible, it is a good idea to use spectrophotometry to check the concentrations of solutions. The successful ligation of a DNA fragment (insert) into a cloning vector (DNA) requires a vector/insert ratio of 1/10 in the ligation reaction. Following the careful preparation of a 3.5 kilobase pair (kbp) vector and a 1.0 kbp insert, A260 measurements were made to determine the concentration of DNA in each preparation, using the relationship that a value of 1.0 for the A260 in a cuvette of 1 cm path length corresponds to 50 pg mL-1 DNA. If the A260 values of the vector and insert preparations are 0.80 and 0.60, respectively, calculate the volumes of the two preparations required for a ligation reaction which contains 20 fmol vector and 200 fmol insert. STRATEGY The DNA concentrations of vector and insert are worked out in terms of pg mL and hence mg mL-1. The molecular masses of the two samples of DNA can be estimated from the number of bases (see Chapter 2, section 2.1), and this can be used to express the concentrations in molar terms. From these values, the volumes of solution which contain the required number of moles can be calculated. WORKED EXAMPL 90 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES This worked example provides good practice in being able to interconvert the different prefixes for units in the SI system (see section 3.3.1). It also highlights the very small volumes of reagents used in many molecular biology techniques; great care should be taken in handling such small volumes. SOLUTION The DNA concentration of the vector is (0.80/1.0) x 50 ug mL"' = 40 Lig mL"1 (i.e. 0.04 mg mL"1). Similarly, the DNA concentration of the insert = 30 ug mL1 (i.e. 0.03 mgmL"1). The molecular mass of the vector can be estimated from the relationship (Chapter 2, section 2.1) that each nucleotide in a strand of DNA contributes 330 Da to the molecular mass. Thus, the molecular mass of the vector = 3500 x 2 x 330 Da = 2.31 x 106 Da (note that we have multiplied by 2 because there are 3500 base pairs in this sample of double-stranded DNA). Similarly the molecular mass of the insert = 1000 x 2 x 330 Da = 0.66 x 106 Da. Hence the concentration of the vector = concentration (in mg mL"')/molecular mass (in Da) = 0.04/(2.31 x 106) M = 1.73 x 10~8 M = 17.3 nM. The concentration of insert = 0.03/(0.66 x 106) M = 45.5 nM. The solution of vector contains 17.3 nmol L"1; this is equivalent to 17.3 fmol |iL~l. Thus, the volume of solution that contains 20 fmol = 20/17.3 uL = 1.16 uL. The solution of insert contains 45.5 nmol L"1; this is equivalent to 45.5 fmol u,L_1. Thus, the volume of solution that contains 200 fmol = 200/45.5 uL = 4.40 uL. SELF TEST In ST 3.11, it should be noted that a nitroanilide is a good substrate for a proteolytic enzyme (protease) as it contains an amide bond, which links the amino acid units in a protein (see Chapter 9, section 9.9.3). Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 3.9 A solution of JV-acetyltyrosine has an A280 of 0.265 in a cuvette of path length 0.5 cm. The absorption coefficient of the compound at this wavelength is 1280 M~' cm"'. What is the concentration of the solution? ST 3.10 A solution of bovine insulin, when diluted 20-fold into buffer, had an A,8, of 0.732 in a cuvette of path length 1 cm. If the absorption coefficient of insulin at 280 nm is 0.97 (mg mL"')"1 cm-1 and the molecular mass of the protein is 5734 Da, what is the concentration of the original solution expressed in terms of mg mL"1 and in terms of molarity? ST 3.11 4-Nitroaniline (4NA) is a product of the action of proteases on 4-nitroanilide substrates; the absorption coefficient of 4NA at 405 nm is 9500 M"' cm'. In a cuvette of 1 cm path length, the A405 of a solution of 4NA whose volume is 2.5 mL is 0.27. What is the concentration of the solution and how many moles of 4NA are present? Answers ST 3.9 The concentration = 0.414 mM or 414 uM. ST 3.10 The concentration = 15.1 mg mL"' or 2.63 mM. ST 3.11 The concentration = 2.84 x 10~5 M or 28.4 uM; there are 7.11 x 10"8 mol or 71.1 nmol present. Attempt Problems 3.16 and 3.1 7 at the end of the chapter. 3.7 ACIDS, BASES, AND BUFFERS Acids, bases, and buffers KEY CONCEPTS Using the Henderson-Hasselbalch equation, calculate the balance of an ionization equilibrium of known pK, at a given pH Understanding how buffers operate and how they can be prepared As mentioned in Chapter 2, section 2.3.1, the pH scale provides a convenient measure of the acidity of a solution; pH = -log [H+]. At 25°C (298K), pure water has a pH of 7. An acid can be defined as a substance which has a tendency to donate H+ ions and a base as a substance which has a tendency to accept H+ ions. Strong acids such as HC1 are completely dissociated (to give H+ and CI" ions), whereas weak acids (such as ethanoic acid) are only partially dissociated. In a similar manner, strong and weak bases will differ in terms of their readiness to accept H+ ions. Thus, NaOH is a strong base since it is completely dissociated to yield Na+ and OH" ions; the latter will readily accept H+ ions to form water. The strength of an acid HA (which dissociates to give H+ and A~ ions) can be expressed quantitatively in terms of its acid dissociation constant rva, which is defined in eqn. 3.5 as: Ka=([H+] [A"])/[HA] 3.5 By analogy with the definition of pH, we define pK, = -log Taking logarithms of both sides of eqn. 3.5, we obtain: K, logKa=log [H+]+log([A-]/[HA]) Rearranging the terms: -log [H+] = -logKa + log ([A-]/[HA]) which can be expressed as eqn. 3.6 (remembering that pH PKa = -logKa): = -log [H+] and that pH = pKa + log([A1/[HA]) 3.6 Eqn. 3.6 is known as the Henderson-Hasselbalch equation; it provides a simple way of visualizing the predominant species in the equilibrium as the pH varies with respect to the pKa; this is depicted in Fig. 3.2. The definition of pH means that as [H*] increases, pH decreases. Distilled water usually contains some dissolved CO, (i.e. carbonic acid) from the atmosphere, so that the observed pH is lower than 7, typically in the range 5.5-6.0. By analogy with the definition of pH, the lower the value of pK, of an acid, the stronger acid it is. 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES pHpJfa I 1 1 pH increasing t t t [A"] < [HA] [A] = [HA] [A] > [HA] Fig. 3.2 The effect of pH on the position of the HA H+ + A" equilibrium. When pH = pK,, exactly 50% of the HA has dissociated to give H+ and A". The exact ratio of [A-]:[HA] at any pH can be calculated using the Henderson-Hasselbalch equation (eqn. 3.6). y WORKED EXAMPLE What percentage of the acid is in the form of A" (a) when the pH is 1 unit below the pKa, (b) when the pH = pK,, and (c) when the pH is 1 unit above the pKa? STRATEGY This calculation applies eqn. 3.6; it is important to keep track of the signs and to use the 10' key on the calculator reliably. To calculate the percentage of each form we must remember that in each case, the sum of [A~] and [HA] is 100%. SOLUTION From the Henderson-Hasselbalch equation (eqn. 3.6), it is found that ([A"]/[HA]) is 0.1, 1.0 and 10 for (a), (b), and (c), respectively. This would give the percentages of A" as 9.1, 50, and 90.9%, respectively. The pK, values for some acids of biochemical interest are 4.76 for ethanoic acid m. c-r ■ ■ ■ (acetic acid), 7.2 for the second dissociation of phosphoric acid (H,POl ^ The structure ot I ns is given v ' rr \ i 4 in Chapter l, section 1.8. H+ + HP02,"), and 8.1 for Tris base (TrisH+ ^ Tris + H+). Since the structures and hence biological activities of many proteins are very sensitive to pH, we usually try to maintain the pH in the solution by the use of a buffer, which is a system designed to minimize the change in pH on addition of H+ or OH" ions. A buffer typically consists of a mixture of a weak acid (HA) and its salt (A-); the latter is usually supplied as its sodium salt, for example. (Note that a buffer can also be prepared from a mixture of a weak base and its acid.) We can explain the action of buffers by considering the equilibrium HA ^ H+ + A". If H+ ions are added to the system, they will combine with the A" ions to form HA and thus effectively be removed. When OH" ions are added, these will react with HA to generate H20 and A", again resisting the change in pH. Buffers will be most effective when there are reasonable concentrations of HA and A" present, and thus should be used within a narrow range (usually taken to be 1 pH unit on either side) of the relevant pK,. Thus, the phosphate buffer system based on H2P04" # H+ + HPO2" (i.e. using the salts NaH2P04 and Na2HP04) would be suitable for buffering in the range of pH between 6.2 and 8.2. For effective buffering the concentration of the buffer should be kept high. 3.7 ACIDS, BASES, AND BUFFERS 93 WORKED EXAMPLE The Tris buffer system (TrisH+ ^ Tris + H+, based on the base Tris and its hydrochloride TrisHCl) has a pK, of 8.1. What are the concentrations of the TrisH+ and Tris forms at pH 8.3, if the total concentration of Tris species is 50 mM? During a reaction, 5 mM H+ ions are formed. What is the new pH? STRATEGY This calculation depends on applying eqn. 3.6 to evaluate the concentrations of Tris and TrisH4 and then consider the effect of adding 5 mM H+ to the system. SOLUTION From the Henderson-Hasselbalch equation (eqn. 3.6), the ratio [Tris]/[TrisH+] = 1002 = 1.58. Thus, [Tris] = 61.2 mM and f TrisH'j=38.8 mM. On addition of 5 mM H+, [Tris] will be reduced by 5 mM to 56.2 mM and [TrisH+] will be raised by 5 mM to 43.8 mM. Application of the Henderson-Hasselbalch equation shows that the new pH would be 8.21. (Note if no buffer had been present, 5 mM H+ would give a pH of 2.3.) In the previous worked example, if the total concentration of Tris species is 20 mM, what would be the new pH on adding 5 mM H+ ions? STRATEGY This follows the same steps as the previous example, with the addition of 5 mM H+ having a proportionately greater effect. SOLUTION At pH 8.3, the concentrations of Tris and TrisH+ are 12.2 and 7.8 mM, respectively. On addition of 5 mM H+, the concentrations become 7.2 and 12.8 mM, respectively; the new pH is 7.85. Comparison with the previous example clearly demonstrates the beneficial effect of using a high concentration of buffer. WORKED EXAMPLE WORKED EXAMPLE J A phosphate buffer system can be made up by mixing appropriate quantities of the two salts NaH,P04 (supplied as the dihydrate NaH2P04.2H20; molecular mass 156 Da) and Na2HP04 (supplied as the anhydrous salt; molecular mass 142 Da). You wish to make up 1 L of a buffer of pH 7.5 with a total phosphate concentration of 0.1 M. Explain how you would make up 0.1 M stock solutions of the two salts in water and then in what proportion you would mix them to obtain the required pH. The appropriate pK, for the H,PO; ^ H+ + HP042- ionization is 7.2. STRATEGY This is a straightforward application of eqn. 3.6 to calculate the ratio of the two salts required for the buffer. SOLUTION QAM stock solutions of NaH2P04.2H20 and Na2HP04 can be made up by weighing out 15.6 (i.e. 156 x 0.1) g and 14.2 (i.e. 142 x 0.1) g of the salts, respectively, dissolving each in water and making up to 1 L in each case. From the Henderson-Hasselbalch equation (eqn. 3.6) the ratio of [HPO^J/tH^PO;;] at pH 7.5 is 1.995 (i.e. 10OJ!). Thus, to obtain 1 L of the required buffer, we need to mix 666 mL ((1.995/(1 + 1.995) x 1000) mL) of the 0.1 M Na2HP04 solution and 334 mL ((1/(1 + 1.995) x 1000) mL) of the 0.1 M NaH,PO.,.2H20 solution. 94 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES In practice, we could also make up the buffer in the last worked example (0.1 M sodium phosphate, pH 7.5) in a rather simpler way. The appropriate amount of the acidic form of the buffer (NaH2P04.2H20) is weighed out to make 1 L of the buffer, i.e.0.1xl56g=15.6g. This is then dissolved in about 900 mL water, and titrated to the required pH using concentrated (5 -10 M) NaOH. Finally, the volume is made up to the required mark (1 L) with water. One advantage of this second method is that if other components are added, e.g. NaCl, they could affect the pJCa of the buffer system, which would alter the ratio of the two solutions to be added in the first method. We would commonly use a titration method to make up Tris buffers. The required amount of the solid (Tris(hydroxymethyl)methylamine) is weighed out and dissolved in water before being titrated down to the required pH with the appropriate concentrated acid (HC1, ethanoic acid, etc.) before being made up to the required volume with water. Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 3.12 The pK, of the amino group in the side chain of lysine is 10.5. What percentage of lysine is in the unprotonated (-NH2) form at pH 8.2? ST 3.1 3 The pK, of a particular lysine side chain in a protein is 7.8. What percentage of this lysine side chain is in the unprotonated form at pH 8.2? ST 3.14 The pK, of ethanoic acid is 4.76. What concentrations of ethanoic acid and sodium ethanoate would be required to make up a buffer at pH 5.2, if the total ethanoic acid and ethanoate concentration is 0.2 M? Answers ST 3.12 0.5% of the lysine is unprotonated. ST 3.13 71.6% of the lysine side chain is unprotonated. ST 3.14 The concentrations are: 0.053 M ethanoic acid and 0.147 M sodium ethanoate. Attempt Problems 3.1 8 and 3.19 at the end of the chapter. SELF TEST In ST 3.14, the Henderson-Hasselbalch equation (eqn. 3.6) is used to calculate the ratio [ethanoate]/[ethanoic acid] at the stated pH. The total concentration of the two species is 0.2 M. |Ejy| Specific activities of proteins and enzymes MHMB KEY CONCEPTS ■ Understanding how to calculate the specific activity of a protein BMW ■ Expressing the result in the appropriate units A very important quantity when studying a protein is a measure of its biological activity, usually called an assay of its activity. In the case of a binding protein, we will want to know how much ligand it can bind and with what affinity. If the 3.8 SPECIFIC ACTIVITIES OF PROTEINS AND ENZYMES 95 protein was able to elicit some more complex response such as cell growth, we would need to measure the ability of a given quantity of the protein to cause cell proliferation in a suitable system. In the case of an enzyme, the activity will be measured in terms of how much substrate is converted to product in a given time. A very commonly used unit for enzyme activity is the amount that converts 1 Limol substrate to product in 1 min; in fact this is often abbreviated simply to 1 unit (U) or sometimes 1 International Unit (IU). In the SI system, the unit of enzyme activity (the katal) is defined as the amount which converts 1 mol substrate to product in 1 s. This is, in fact, a very large amount of activity and in practice we would often be dealing with nkatal amounts. An important property is the specific activity of the protein, which is defined as the amount of biological activity per given weight of the protein. For an enzyme, this might be measured in terms of Limol min-1 mg"' (U mg"1 or IU mg"1) or using the SI definition in terms of katal kg"1. Show that for an enzyme a specific activity of 1 katal kg 1 corresponds to 60 Limol min-' mg"1. STRATEGY The important point here is to keep a very careful track of the units. SOLUTION The specific activity of 1 katal kg"1 corresponds to 1 mol substrate s"1 kg-1, i.e. to 60 mol substrate min 1 kg-1. This is 60 x 106 u,mol substrate min-1 kg"1 and hence to 60 Limol substrate min"1 mg"1. Thus, despite the fact that both the katal and the kg are large and impractical amounts of activity and protein, the SI definition of specific activity gives numbers that are of similar orders of magnitude to the more widely used definition (U mg"1). The specific activity of a protein is essentially a measure of the quality and purity of the sample and will be discussed further in section 3.9. The activity of the enzyme lactate dehydrogenase can be measured by monitoring the reaction in the direction pyruvate + NADH —> lactate + NAD+. NADH absorbs radiation at 340 nm (absorption coefficient = 6220 M"1 cm-1); NAD+ does not absorb at this wavelength. In an assay, 25 LiL of a sample of enzyme (containing 5 |ig protein mL"1) was added to a mixture of pyruvate and NADH to give a total volume of 3 mL in a cuvette of 1-cm pathlength. The decrease in A340 was 0.14 min"1. What is the specific activity of the enzyme expressed in terms of |0.mol min"1 mg"1 and in SI units (katal kg"1)? STRATEGY This calculation depends on evaluating the amount of activity in terms of jimol min"1 and the amount of enzyme added. It is advisable to break down the calculation into small steps and set the working out clearly. WORKED EXAMPLE Strictly speaking the equation for the reaction catalysed by lactate dehydrogenase should include H* on the left-hand side. It is assumed that the reaction is carried out in a suitable buffer, so that H+ is often omitted. 96 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES SOLUTION The amount of enzyme added is 0.025 x 5 ug (note that 25 pL=0.025 mL), i.e. 0.125 u g. The rate of the reaction (change in A340 min"') can be converted to a concentration change min', i.e. 0.14/6220 M min"1 = 22.51 pM min"1. In a 3 mL assay mixture this corresponds to change of 22.51 x 3/1000 pmol min"1 = 0.0675 pmol min'. The specific activity is thus 0.0675 pmol min"1 per 0.125 ug, or 0.0675 x 1000/ 0.125 umol min"' mg"1 = 540 umol min"1 mg"1. In SI units this would be 540/60 katal kg"1, i.e. 9.0 katal kg"1. We shall see in Chapter 4, section 4.5 how the specific activity can be used to calculate the turnover number (kcM) of an enzyme, provided the molecular mass of the enzyme is known. ? SELF TEST Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 3.1 5 The rate of an enzyme-catalysed reaction as measured by the formation of product is 7.3 pM min"1 in a solution of volume 2 mL. What does this rate correspond to in terms of pmol min"1? If 0.15 ug enzyme had been added to the reaction, what is the the specific activity of the enzyme expressed as pmol min"1 mg"1 and as katal kg1? ST 3.1 6 The specific activity of a certain enzyme is 745 pmol min"1 mg"1. How much enzyme should be added to a reaction mixture of volume 1 mL to give a rate of reaction of 12 pM min"1? Answers ST 3.15 The rate corresponds to 0.0146 pmol min"1; the specific activity is 97.3 pmol min"1 mg"1 (or 1.62 katal kg"1). ST 3.16 The amount of enzyme to be added is 0.0161 pg. Attempt Problem 3.20 at the end of the chapter. Purification tables KEY CONCEPTS ■ Defining the terms specific activity, yield and purification factor I ■ Calculating these quantities to draw up a purification table A very important part of the study of proteins involves their purification. Only when purified protein is available can we determine its structure and investigate the basis of its biological activity. Traditionally, purification would have involved In ST 3.16 the rate should be calculated in terms of mol/min from the concentration change/min and the volume of solution. Knowing the specific activity of the enzyme allows you to work out how much enzyme is needed to give the required rate. 3.9 PURIFICATION TABLES initially identifying a suitable source in which the protein of interest was abundant, for example yeast would be an excellent source of enzymes of the glycolytic pathway. After breakage of the cells to provide a crude extract, various purification procedures would be applied in turn until pure, active protein was obtained. Recently, the trend has been to overexpress a recombinant version of the protein (often with some tag incorporated to facilitate purification) in a convenient host organism, such as Escherichia coli, yeast or insect cells, which can often be grown on a large scale using relatively simple growth media (see Chapter 5, section 5.4.2). Whatever method of purification is used, it is important to record the progress of the procedure in a suitable way; this involves drawing up a purification table, which will record the following information: ■ Step involved ■ Total protein ■ Total activity ■ Specific activity (i.e. the activity per unit weight of protein) ■ Yield (i.e. the recovery of activity expressed as a per cent of that in the initial extract) ■ Purification factor (the factor by which the specific activity has increased). It is essential for you to be able to calculate the specific activity, yield, and purification factor given data on the total protein and total activity. To illustrate the concepts involved, consider the following data from a purification of an enzyme. Step 1 refers to the crude extract. The total protein is 940 mg, and the total activity is 56 780 units (1 unit equals 1 Limol substrate consumed per min). Step 2 refers to the extract which has been subjected to ion-exchange chromatography on DEAE-cellulose. The total protein in the pooled peak fractions is 53 mg and the total activity is 47 640 units. We can then draw up our initial purification table (Table 3.2). Of course, in a real example it is likely that there will be several more steps before the protein is completely purified. Before undertaking a purification process, it is essential to devise a reliable assay system for the protein in question, this will usually involve a measurement of the biological activity (catalysis, binding, etc.); see Chapter 5, section 5.3. Table 3.2 Purification table for a protein Step Total protein Total activity (mg) (units) Specific activity (units mg"') Yield (%) Purification factor Crude extract 940 56 780 60.4 100 1.0 Ion-exchange 53 47 640 899 83.9 14.9 98 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES The specific activity is obtained by dividing the total activity by the total protein at that step (e.g. 56 780/940 = 60.4 units mg"'). The yield refers to the recovery of activity, not to the recovery of protein. After all, the purpose of the purification is to get rid of contaminating proteins and retain the protein of interest. In this case, the yield at step 2 is (47 640/56 780) x 100% = 83.9%. The purification factor refers to the factor by which the specific activity has increased, in this case 899/60.4 after step 2 = 14.9-fold. (Note in a multi-step purification, we can express this either in terms of the factor for each individual step or as a running total factor relative to the initial crude extract as 1.0). Some examples of protein purification experiments are described in Chapter 7. ? SELF TEST Check that you have mastered the key concepts at the start of the section by attempting the following question. ST 3.1 7 In a purification procedure for a certain enzyme, the total protein and total activity at each step was as follows: Step Total protein (mg) Total activity (units) Crude extract 53.5 780 Ion-exchange chromatography 8.7 525 Gel filtration 1.6 490 Calculate the specific activity, yield and purification factor at each step in the procedure. Answer ST 3.17 The values are: specific activity, 14.6, 60.3, 306.3 units mg-1; yield, 100, 67.3, 62.8%; purification factors (per step), 1.0, 4.1, 5.1). Attempt Problem 3.21 at the end of the chapter. 3.10 Problems Full solutions to odd-numbered problems are available to all in the student section of the Online Resource Centre at www.oxfordtextbooks.co.uk/orc/price/. Full solutions to even-numbered problems are available to lecturers only in the lecturer section of the Online Resource Centre. 3.1 Hair is composed of the protein cc-keratin, which is a-helical in structure. How many turns of cc-helix are produced per second in growing hair? Assume that the rate of hair growth is 18 cm year1 and that each turn of oc-helix corresponds to a length of 0.54 nm. 3.2 How many moles of GTP are present in 2 pi of a 15 pM solution? 3.3 GTP (sodium salt) has a molecular mass of 545 Da. What mass of GTP is present in 3 mL of a 15 pM solution? 3.4 A certain analytical technique is said to be capable of determining 100 pmol of the amino acid tyrosine (molecular mass 181 Da). What mass of tyrosine does this correspond to? 3.5 The concentration of haemoglobin in red blood cells is estimated to be 250 mg mL"'. Haemoglobin consists of four polypeptide chains each of molecular mass 16.5 kDa (16 500 Da); what is the molar concentration of haemoglobin chains in the red blood cell? 3.6 The molecular mass of chymotrypsin is 24 500 Da (24.5 kDa). What is the concentration of a 0.35 mg mL-1 solution expressed in molar terms? 3.7 A solution of aspartate aminotransferase is stated to be 13.6 pM in terms of the dimeric (two polypeptide chain) form of the enzyme (the molecular mass of each polypeptide chain is 44 kDa). What is the concentration of the enzyme in terms of mg mL"1? 3.8 The enzyme phosphoglycerate mutase from Saccharomyces cerevisiae (polypeptide molecular mass 27 kDa) is typically assayed at a concentration of 0.05 pg mL"1. What is the molar concentration of enzyme polypeptide chains in the assay? 3.9 Explain how you would make up 15 mL of a 10 M solution of urea in water. The molecular mass of urea is 60 Da. 3.10 For the preparation of a certain gel for electrophoresis, 20 mL of an 8% (w/v) acrylamide solution is required. What volume of a 30% (w/v) stock solution of acrylamide is required for this? 100 3. CALCULATIONS IN THE MOLECULAR BIOSCIENCES 3.11 A stock solution of a protein in buffer is 4.3 mg mL"1. How would you prepare 1 mL of a solution containing 0.2 mg mL"1 protein? 3.12 The concentration of glacial acetic (ethanoic) acid is stated to be 17.3 M. How would you prepare 200 mL of a 1 M solution of the acid in water? 3.13 IPTG (isopropyl-(3-D-thiogalactopyranoside) is widely used as an inducer for the expression of recombinant proteins in the bacterium E. coli. Its molecular mass is 238.3. How much IPTG is required to make up 5 mL of a 0.5 M stock solution in water? What volume of this solution should be added to 50 mL of bacterial culture to give a final concentration of 1.2 mM? 3.14 The four deoxyribonucleotide triphosphates (NTPs), i.e. dATP, dCTP, dGTP, and dTTP, are essential components of the polymerase chain reaction (PCR) used to amplify a selected section of DNA. The NTPs are often supplied as separate stock solutions, each at a concentration of 100 mM in sterile H20. How would you make up a working solution (in sterile H20) of final volume 1 mL which contains the four NTPs each at a concentration of 2 mM? 3.15 The concentration of a stock solution of NADH is 3 mg mL"1. What does this correspond to in molar terms? (The molecular mass of NADH (disodium salt) is 709 Da). What volume of this stock solution would you add to an assay mixture (of final volume 1.0 mL) to obtain a final concentration of 0.1 mM? 3.16 The absorbance of a solution of ATP was checked by performing two successive dilutions, the first 50-fold and the second 40-fold. The A26l) of this final solution was 0.47 in a cuvette of 1 cm path length. What is the concentration of the original solution of ATP? The absorption coefficient of ATP at 260 nm is 15 000 M"1 cm"1. 3.17 A solution of BSA is made up by weighing out 20 mg of the lyophilized (freeze dried) powder of the protein supplied and dissolving in 2 mL water. The concentration of the protein was checked by adding 0.04 mL of the solution to 0.96 mL water; the A280 of the resulting solution relative to a water blank was 0.237. The absorption coefficient for BSA is 0.66 (mg mL"1)"1 cm"'. Calculate the actual concentration of the original solution of BSA and comment on your answer. 3.18 What would be the change in pH on adding OH" ions to a final concentration of 20 mM to a 0.1 M sodium phosphate buffer, pH 7.0? The piC, for the H2P04" =± H+ + HP042" ionization is 7.2. 3.19 Explain how you would make up 1 L of a 50 mM Tris/HCl buffer at pH 7.5, containing 20 mM NaCl. The molecular masses of Tris(hydroxymethyl)methylamine and NaCl are 121.1 and 58.45 Da, respectively. 3.20 A sample of malate dehydrogenase (3.2 mg mir1) was diluted 500-fold for assay. 20 pL of this solution was added to an assay system of total volume 1 mL in a cuvette of 1 cm path length. The increase in A340 (corresponding to the conversion malate + NAD+ —> oxaloacetate + NADH + H+) was 0.23 min-1. What is the specific activity of the enzyme expressed in terms of umol min"1 mg"1 and in SI units? (The absorption coefficient of NADH is 6220 M"1 cm"1; NAD+ does not absorb at this wavelength.) 3.21 The enzyme adenylate kinase catalyses the interconversion of adenine nucleotides in the reaction AMP + ATP ^ 2ADP. The enzyme was purified from pig muscle by procedures including precipitation at low pH, elution from phosphocellulose by a pulse of AMP and gel filtration on Sephadex G-75. The following results were obtained starting from 1 kg muscle: Step Total protein (mg) Total activity (units)* Crude extract 72 500 413 000 LowpH 18 700 365 000 Phosphocellulose 290 223 000 Gel filtration 77 200 000 *1 unit of enzyme activity corresponds to 1 fimol substrate consumed per minute. Complete the purification table for this procedure. i Appendix Appendix 3.1 The SI units and physical constants SI units are based on the metre-kilogram-second system of measurement, and are very widely accepted among scientists. SI units for various physical quantities are listed below. Quantity SI unit Notes Amount of substance mole (mol) This quantity contains 1 Avogadro number of the basic units (atoms, molecules etc.) Electric charge coulomb (C) Length metre (m) A (Angstrom) = 1 x lfr1" m = 0.1 nm Mass kilogram (kg) Multiple units are based on g (e.g. mg) although kg is the basic unit Molecular mass dalton (Da) 1 Da = 1 atomic mass unit« mass of 1 hydrogen atom (1.66 x 10~24 g); more exactly it is 1/12 of the mass of the 12C isotope of carbon Temperature degree Kelvin (K) 0°C = 273.15K Time second (s) Volume cubic metre (m3) 1 litre (L) = 1 dm3; 1 mL = 1 cm3 Electric potential volt (V) (J C"') Energy joule (J) (1 m2 kg s~2) lcal = 4.18J Force newton (N) (1 m kg s-2) Frequency hertz (Hz) (1 s"1) Power watt (W) (Is"') Pressure pascal (Pa) (1 N nr2) latm= 101.325 kPa Radioactive decay becquerel (Bq) 1 s~' 1 disintegration per second; 1 curie (Ci) = 3.7x 1010 s"' Concentration mol kg-1 Moles per kg of solvent, also called molality Concentration is more usually quoted in terms of molarity (M), moles per L of solution For dilute aqueous solutions, molarity is very nearly equal to molality. The following prefixes are used in conjunction with the SI units to bring the numerical values of quantities into a convenient range (see Chapter 3, section 3.3.1). Prefix G, M, k, Base m, (X, n, p, giga mega kilo unit milli micro nano pico Numerical 10' 106 103 1 io-3 io-6 IO"9 10"12 value Physical constants used in this book are: Avogadro number (L) 6.022 xlO23 mol 1 Faraday constant (F) 9.648 x 10" C mol"1 Gas constant (R) 8.314 J K'1 mol-1 Planck constant (h) 6.626 x 10" 34Js Speed of light in a vacuum (c) 2.997 x 10s m s"1