- Created by: phoebeb1799
- Created on: 27-09-19 22:27
Hydrogen ions, H+ (more correctly termed hydroxonium ions, H3O+, though we will call them H+ for simplicity) take part in many biological reactions, so the H+ concentration (represented as [H+]) is very important. This concentration is usually expressed as pH, defined as
pH = –log10 [H+]
where [H+] is expressed in moles per litre (M).
It is sometimes easier to calculate the concentration of OH– ions rather than that of H+ ions. You can then calculate the [H+], as follows:
[H+] × [OH–] = 10–14.
where [OH–] is also expressed in moles per litre (M).
Strong acids and bases
A ‘strong’ acid or base is one that is always fully ionised, such as HCl (which is really a mixture of H+ and Cl–) and NaOH (= Na+ and OH–). Note that ‘strong’ does not mean concentrated; you can have a dilute solution of a strong acid. The following examples show how to calculate the pH of a solution of a strong acid or base:
Calculate the pH of 1 M HCl
[H+] = 1 M. pH = –log10 1 = 0.0. Calculate the pH of 0.005 M HCl
[H+] = 0.005 M pH = –log10 0.005 = 2.3. Calculate the pH of 0.001 M NaOH
[OH–] = 0.001 M [H+] = 10–14 / 0.001 = 10–11 M, pH = 11.0. Calculate the pH of 10 M NaOH
[OH–] = 10 M [H+] = 10–14 / 10 = 10–15 M, pH = 15.0.
Note that the pH scale is not confined to the range 0–14; it can be negative or over 14.
Weak acids and bases
Most of the acids and bases that we encounter in biology are ‘weak’ (i.e., not always fully ionised). For example, lactic acid (CH3—CHOH—COOH) is a weak acid, and ammonia (NH3) is a weak base. Their degree of ionisation depends on the pH of the solution in which they find themselves. In biology, we often deal with a low concentration of the weak acid (or base) in a solution whose pH is held approximately constant by the presence of a relatively concentrated buffer (see below). In such situations, the weak acid or base itself, being dilute, has little effect on the pH of the solution, but the pH of the solution affects the ionisation of the weak acid or base.
A moderately concentrated solution of a weak acid or weak base, at a pH near its pKa, tends to resist the changes in pH that would otherwise occur when small amounts of other acids or bases were added, and is thus said to be a pH buffer. Since pH affects enzyme activity, we need to control the pH in enzyme assays, so we always add a buffer. The buffer chosen is a weak acid or weak base with a pKa within roughly 1 unit of the required pH. Thus, ethanoic acid (= acetic acid; CH3—COOH), which ionises to form ethanoate (acetate; CH3—COO–) with a pKa of 4.7, can be used as a buffer within the pH range approximately 3.7–5.7. Phosphoric acid (H3PO4) can ionise three times
H3PO4 H2PO4– + H+ H2PO4– HPO42– + H+ HPO42– PO43– + H+
pKa = 2.2 pKa = 7.2 pKa = 12.4
so phosphate buffer is useful within the approximate pH ranges 1.2–3.2, 6.2–8.2 and 11.4–13.4. In enzyme assays, buffers are usually used at concentrations in the range 20–200 mM.
Relative molecular mass (Mr) and molecular weight
The Mr of a compound is calculated by adding together the atomic masses of all the atoms present (H = 1, C = 12, N = 14, O = 16, Na = 23, P = 31, S = 32, etc.). For example, the Mr of NaOH is 23 + 18 + 1 = 40, the Mr of ethanol (CH3—CH2OH) is 48, and the Mr of the enzyme peroxidase is approximately 40,000. Since the Mr is relative, it is dimensionless.
You will also come across the term molecular weight (MWt). This is given a unit — the dalton (Da), which is the same as g mol–1. Thus, ethanol has a molecular weight of 48 Da, and peroxidase has a molecular weight of 40 kDa. Numerically, molecular weight is equal to Mr.
Quantity and concentration: mole versus molar
Wherever possible, quantities of biochemicals are expressed in moles rather than grams. This is because we are more interested in how many molecules we have rather than in their weight. A mole (abbreviation: mol) is a certain number of molecules, and is thus a unit of quantity. [The ‘certain number’ is Avogadro’s number, or about 6.023 × 1023, although you rarely need to use this figure.]
If you calculate the relative molecular mass (Mr) of a compound, and then weigh out that many grams, you have 1 mol. Thus, 1 mol of NaOH is 40 grams, 1 mol of ethanol is 48 grams, and 1 mol of the enzyme peroxidase is 40 kg.
The term ‘molar’ (abbreviation: M) is a unit of concentration: it describes a solution which contains 1 mole of a given substance per litre of solution.
Take great care to distinguish M (a unit of concentration) from mol (a unit of quantity). Remember, 1 M = 1 mol / litre.
Km and Vmax 1
The maximum rate (or maximum ‘velocity’) at which a given enzyme sample can theoretically act is called its Vmax. Thus, Vmax has the same units as enzyme activity. Ideally, an enzyme would work at its Vmax when the substrate concentration(s) were ‘infinitely’ high. Since infinite concentrations do not exist, an enzyme can never quite attain its Vmax, though it may reach a rate very close to it. This occurs when almost every active site is effectively occupied by a substrate molecule and the enzyme is said to be ‘saturated’ with substrate. Under these circumstances, when an enzyme sample is acting close to its Vmax, every enzyme molecule is spending almost all its time catalysing the reaction, and almost no time ‘searching’ the surrounding solution for its next substrate molecule.
If the substrate concentration is now progressively decreased, the enzyme has to spend an increasing proportion of its time waiting for its next substrate molecule, and therefore fewer and fewer reactions occur per minute. The substrate concentration at which the reaction rate is 50% of Vmax is called the Km. Different enzymes have different Km values. If the Km is low, i.e. the enzyme can still efficiently make ‘productive encounters’ with its substrate even in a dilute solution, we can suggest that the enzyme has a high affinity for that substrate. Conversely, a high Km suggests a low affinity.
Km and Vmax 2
At any given substrate concentration, the rate (v) of the reaction can be calculated using the Michaelis–Menten equation:
Vmax [S] v= ———— [S] + Km where [S] is the actual substrate concentration.
Concentrations of substrates and products can often be measured in a spectrophotometer, which records the proportion of light or ultraviolet radiation (of a given wavelength) that is absorbed by a sample of the solution in a cuvette. The key equation from which to calculate the substrate concentration is:
A = EcL
where A = absorbance (at a specified wavelength),
E = molar absorbance (= the absorbance of a 1 M solution in a 1-cm cuvette),
c = concentration of compound being measured,
L = path-length (in cm) of the light as it travels through the cuvette (almost always 1 cm).
Often in biological studies you will prepare a final solution (e.g. for an enzyme assay) by mixing together several more concentrated (‘stock’) solutions. What you should quote, when formally reporting such experiments, is the final concentration of each component (that is, in the case of an enzyme assay, the initial substrate concentrations actually ‘experienced’ by the enzyme).
^G° ́ values, also called Gibbs or standard free energies, are the free energies of reactions, at pH 7.0, with all reagents (other than H+) initially present at 1 M concentrations. Details of G° ́ can be found in text books. Its significance is that if the ^G° ́ is negative, the reaction is exergonic (sometimes described as ‘spontaneous’ or ‘down-hill’) in the direction written. The opposite of exergonic is endergonic (‘up-hill’). ‘Spontaneous’ does not mean that the reaction necessarily occurs at a measurable rate without an enzyme. Thus ^G° ́ says nothing about the rate of a reaction, but about the direction in which it will occur (if it occurs at all).
The equilibrium constant (K ́eq) of the reaction A + B <--> C + D is given by:
K ́eq = [C] [D] / [A] [B]
where [A], [B], [C] and [D] are the concentrations of the four reactants when at equilibrium.
Oxidation and reduction
You probably know that OXIDATION IS LOSS of electrons, and REDUCTION IS GAIN of electrons (mnemonic: ‘OIL-RIG’). When Cu+ is converted to Cu2+, it is losing an electron and thus being oxidised; when Fe3+ is converted to Fe2+, it is being reduced. However, you may find the ‘OIL-RIG’ rule difficult to apply to organic reactions. For example, if ethanol (CH3—CH2OH) gets metabolised to ethanal (CH3—CHO), has the ethanol been reduced or oxidised or neither? To help, you may find it useful to remember the following additional rules:
- loss of electrons, H atoms and gain of O atoms is oxidation
- gain of electrons, H atoms and loss of O atoms is reduction
The Primordial Biomolecules*
The most important ‘building blocks’ from which more complex cellular molecules are constructed, e.g. proteins, nucleic acids, polysaccharides and lipids. It also shows several closely related derivatives of them such as glucose 1-phosphate. You should aim to have a clear picture in your mind of these vital substances.
Optical isomers 1
If a compound contains an asymmetric centre, i.e. a carbon atom that is attached to four distinguishable groups, then there can be two isomers depending on the orientation of these four groups (see diagram of glyceraldehyde, below). One isomer is the mirror image of the other. The two isomers have the same name but are given different prefixes — in biology, usually D- and L-. [Chemists use a more generally applicable system, with the prefixes (R)- and (S)-.] The two isomers have the same water-solubility, melting point, density etc., and they react chemically in the same way with most other substances. However, solutions of the two isomers rotate polarised light in opposite directions, so they are called optical isomers. Crucially, enzymes can distinguish between the two optical isomers [this is because enzymes are built entirely from L-amino acids, not D-, so they can only fit together correctly with one of the two optical isomers of the substrate]. Therefore only one of the two isomers will participate in any given metabolic step.
The classic example is glyceraldehyde, whose middle carbon atom has the following four different groups attached to it: —H, —OH, —CHO and —CH2OH. The diagrams show D- and L- glyceraldehyde in ‘realistic’ and conventional representations. The wedge-shaped bonds are supposed to be coming out of the paper towards you; the dashed bonds are receding (from the central C atom) into the paper.
Optical isomers 2
The two isomers of glyceraldehyde were named D- and L- because they rotate polarised light right (Latin, dexter) and left (Latin, laevus) respectively. Other biologically important asymmetric compounds are also named D- or L-, but this is by reference to their structural similarity to D- and L- glyceraldehyde, not according to the direction they rotate polarised light.
Other biologically important asymmetric compounds include 3-phospho-D-glycerate, L-malate, D-lactate, all the L-amino acids, D-glucose, D-mannose, D-fructose, D-ribose and L-arabinose. Some of these are illustrated below; in each case the particular asymmetric carbon that is used in defining D- versus L- is shown in bold. Compare these structures to those of D- and L-glyceraldehyde.
In the case of glucose and most other sugars, there are several asymmetric carbons (not just one as in glyceraldehyde). In this situation, D- versus L- is decided by the orientation of the —H and the —OH at the last asymmetric C atom (i.e., usually the penultimate carbon: C-5 in glucose). Note that D- and L-glucose differ from each other at all four asymmetric centres, not just at C-5. L-Glucose is extremely rare in biology.
Naming phosphorylated compounds
An organic compound that carries a single phosphate ester group may be named with the suffix phosphate or with the prefix phospho. A numeral is used to indicate which position of the organic molecule carries the phosphate group. Thus, glyceraldehyde 3-phosphate is the same as 3- phosphoglyceraldehyde. In the case of phosphorylated nucleosides (i.e., nucleotides), it is conventional also to specify ‘mono’, to emphasise that there is only a single phosphate, as in adenosine 5 ́-monophosphate (= AMP). The 5 ́ (‘five-prime’, as opposed to 5) indicates that the phosphate is attached to position 5 of the ribose, not of adenine. If an organic molecule has two or more phosphate groups, you need to specify whether these groups are either:
- joined separately to different positions on the organic molecule (such compounds are called bisphosphates, trisphosphates, etc., e.g. fructose 1,6-bisphosphate, ribulose 1,5- bisphosphate, inositol 1,4,5-trisphosphate),
- attached to each other and joined to a single position on the organic molecule (such compounds
are called diphosphates, triphosphates, etc., e.g. guanosine 5 ́-diphosphate (GDP) and adenosine 5 ́-triphosphate (ATP)). The three phosphate groups in ATP are distinguished by the Greek letters a, b and y, starting with the phosphate nearest the organic molecule.
The two most central metabolic pathways*
These are the reactions by which aerobic cells respire (oxidise) glucose (C6H12O6) to 6 CO2. They also provide the starting materials for the biosynthesis of all other cell metabolites. You should aim to acquire a good working knowledge of these two pathways as TDC lecturers will refer to them repeatedly.
Properties of a high-polarity compound
- Hydrophilic; high solubility in water
- Lipophobic; low solubility in oils
- Do not readily pass through cell membranes unless a specific carrier protein is present
- Partition mainly into the aqueous phase when shaken with octanol + water
Properties of a low-polarity compound
- Hydrophobic; low solubility in water
- Lipophilic; high solubility in oils
- Readily pass through cell membranes even without the assistance of carrier proteins
- Partition mainly into the octanol phase when shaken with octanol + water