Description of Module Subject Name Paper Name 12 Module Name/Title 01 Electrophoresis: Basic Concept
1. Objectives 1.1 To understand basic concept of electrophoresis 1.2 To explain what determines charge or net charge on macromolecule such as protein 1.3 To understand why buffers are used in electrophoresis and how are these selected 1.4 To know requirements for performing electrophoresis experiment 2. Basic Concept Electrophoresis is defined as movement of charged particles in an electric field leading to their separation. In other words, it is separation technique which is based on differences in mobility of charged particles in an electric field. It is essential to understand factors which determine mobility of particles and these are being explained with the help of following statements. 2.1 When voltage (potential difference) is applied across the electrodes, it results in generation of potential gradient (E) which is equal to applied voltage (V) divided by distance between the electrode (d). Thus, E= V/d. It is required to be noted that d will vary from equipment to equipment and thus for the same applied voltage, potential gradient will vary. 2.2 If charge on molecule is q coulombs, then force exerted on particle is equal to Eq newtons. This force causes the charged particles to move towards electrode. Particles carrying positive charge are referred as cations and will migrate towards cathode whereas particles carrying negative charge are called as anions and these will move towards anode. Thus, cathode is negatively charged and anode is positively charged. It is also essential to be noted that particles under movement experiences frictional resistance which retards movement of charged particles. The frictional force is dependent on size, shape,
and hydrodynamic volume of particle as well as on pore size of support matrix used for separation and viscosity of medium (buffer) used in electrophoresis. Thus, velocity v is function of E and frictional coefficient ƒ and is defined by following equation. v = E / ƒ = Eq / ƒ 2.3 Electrophoretic mobility µ of an ion is defined as ratio of velocity to potential gradient and is expressed as µ = v / E = q / ƒ 2.4 Molecules with identical or similar charges can be also separated if these molecules experiences different frictional force. Thus, separation of particles differing in charge as well as of similar charge is feasible in electrophoresis. In fact, separation of charged particles based on differences in frictional forces is routinely used in laboratories. 2.5 In electrophoresis, electric field must be switched off before ions reach respective electrode to allow their separation. Failure to do so will result in their neutralization at respective electrodes, a process called electrolysis. 2.6 During electrophoresis, heat is generated in supporting medium and is given by following equation W =I 2 R, where W is power generated in Watts, I is current which increases with increase in voltage (V/I= R) and R is resistance. Power generated is dissipated as heat. Heating of medium (support) will cause following problems. Sample and buffer ions will diffuse resulting in broadening of bands Viscosity of buffer will decrease and thus, resistance of medium will also decrease. Thus, as per Ohm s law, it will result in increase in current at a given voltage. Heat can denature proteins and enzymes which may lose activity. If during electrophoresis, constant voltage is used, current will continue to slowly increase with time because of heat generated. More is the value of current, more is heat generated. Thus, this increased amount of heat will further increase current. As such amount of heat will be continuously increasing with time. Some workers prefer to run electrophoresis at constant current. Thus it should be clear that higher is the current, higher is heat generated and higher is
velocity. In order to minimize heat generation, low voltage could be an option. However, this will end in long separation time which ultimately will result in diffusion of separated bands. Affects of heat can be minimized if support is maintained at low temperature. This is achievable if electrophoresis is performed at low temperature or cold water is circulated through buffers in electrode chambers or support material is layered over surface with good heat conducting properties or through Peltier cooling. It is also repeated here that force exerted on charged particle is dependent on voltage, not on current. It is therefore advised to perform electrophoresis on constant voltage which when applied should not result in generation of excess heat and also should not lead to diffusion of separated bands. 3. 0 Charge or Net Charge on Macromolecule For the students of biochemistry or life-science subject, electrophoretic separation of protein is more frequently encountered and therefore this macromolecule is discussed here. Now we will explain how charge on protein is contributed. To make this clear, it is essential to understand concept of ph, buffer, dissociation constant, pka and different amino acids which acts as monomeric unit in protein. ph is minus logarithmic of hydrogen ion activity (concentration). ie ph = -log [H+] Water is weak electrolyte. It ionizes to small extent to give [H + ] and [OH - ]. dissociation and ionic product are summarized in following equations. Its Concentration of H+ and OH- ions from water is equal and is equal to 10-7 mol 2 dm -3. Thus, ph of water is 7. For alkaline solution, ph is greater than 7 while for acidic solution, ph is lower than 7. Also, it
may be noted that ph values for strong acids will be less than 0 and that of strong bases greater than 14. For weak acids such as acetic acid, extent of dissociation and concentration of acetic acid will determine ph of solution. Above equation is referred as Henderson Hasselbalch equation. This equation can be used for calculating extent of dissociation of group at a given ph. pka which is defined as negative logarithmic of dissociation constant is constant for a given group. Thus by changing ph, extent of dissociation can be changed and consequently the charge contribution. Carboxylic group on dissociation will give one negative charge whereas un-dissociated will be charge-less. Contrary to it, amino group in undissociated state will carry one positive charge while dissociated state will be charge-less. Proteins are polymers of amino acids which are linked together by peptide bond. Amino acids contain amino group and carboxylic groups. Carboxylic group of one amino acid is linked with amino group of other amino acid and thus the groups involved in peptide bond formation will not contribute to charge. However, α-amino group of first amino acid and α- carboxylic group of last amino acid in protein or peptide will contribute to charge on molecule. Different amino acids contain different side chains and some of them do contain groups which can contribute to charge on protein. For example side chains of aspartic acids and glutamic acid have carboxylic group while that of lysine has amino group. Similarly imidazole of histidine can contribute to charge. Thus (i) pka of α-amino, ε-amino, α-carboxylic, ß- carboxylic, γ- carboxylic and imidazole, (ii) number of these groups in protein and (iii) ph will determine net charge on protein. It is again emphasized that dissociation of carboxylic group will lead to one negative charge while dissociation of amino group will abolish one positive charge. pka values of groups in protein is given table.
Group pka Group pka α-amino (all amino acids) 8.80 to 10.6 Guanidino (Arg) 12.48 (Different values for different amino acids) ε-amino (Lys) 10.53 Imidazole (His) 6.00 α-carboxylic (all amino acids) 1.83 to 2.83 Sufhydryl (Cys) 8.18 (Different values for different amino acids) ß- carboxylic (Asp) 3.65 Phenolic (Tyr) 10.07 γ- carboxylic (Glu) 4.25 ph of solution and number of dissociable groups in protein and their pka values will decide the charge on protein molecule. Extent of dissociation can be determined by use of Henderson Hasselbalch equation. When ph is equal to pka, there will be 50% dissociation. When ph is one unit higher than pka, there will be about 90% dissociation. When ph is two units higher than pka, there will be about 99%.When ph is one unit lower than pka, there will be about 10% dissociation. When ph is two units higher than pka, there will be about 1% dissociation. Isoelectric point (pi) of protein is ph at which net charge on protein is zero and at this ph, protein in electric field will not migrate. At ph value higher than pi, protein will have net negative charge and move towards anode. However, at ph value lower than pi, protein will have net positive charge and move towards cathode. Most of proteins have PI in the range of 6.0 to 6.8 and therefore at alkaline ph, such proteins will have negative charge while at ph lower than 6.0, the net charge will be positive. As the ph is increased from say from 2.0 to 10.0, net positive charge will gradually decrease and will reach zero and then acquire negative charge. Also, it is made clear that net charge on protein will continue to increase as one move away from pi. From these discussions, it is also clear that ph is required to maintained constant during the electrophoresis so that electrophoretic mobility remains dependent only on composition of protein. ph can be maintained by using buffer during electrophoresis. Let us know more about buffers. 4.0 Buffers.
Buffers are solutions which resist change in ph on addition of acid or alkali. Contrary to it, if acid or alkali is added to water, it will bring large changes in ph. Buffers are mixtures of weak acids and salts or weak base and its salts. For example, mixture of acetic acid and sodium acetate will constitute buffer. Molar ratio of sodium acetate and acetic acid will determine ph of buffer. Addition of acid or alkali to buffer brings the changes in ratio of sodium acetate and acetic acid. If you see Henderson Hasselbalch equation, log of ratio of dissociated state (eg. CH3COO - ) to undissociated state (eg. CH 3COOH) will determine ph of buffer. Because of log of ratio, buffer resists change in ph. Alternate way of explaining how buffer functions requires consideration of dissociation of acetic acid as given below CH3COOH CH3COO - + H + When acid is added, the equilibrium is shifted to the left, in accordance with Le Chatelier's principle. Thus, hydrogen ion concentration increases by less than the amount expected for the quantity of strong acid added. Similarly, when alkali is added to the mixture, the hydrogen ion concentration decreases by less than the amount expected for the quantity of alkali added. Buffers exhibits better buffering in ph range of one unit either side of pka values of weak acid. Buffering is very poor or negligible if ph of buffer is more than one unit away from pka value. The pka value of most common buffer is given in Table Group pka Group pka Phosphate 1.97, 6.82, 12.5 Acetate 4.76 Citrate 3.09, 4.75, 5.41 Tricine 8.05 Boric acid 9.23, 12.74, 13.80 Tris (hydroxymethyl) 8.06 amino methane Carbonic acid 6.35, 10.33 4.1 Preparation of buffer. Preparation of Buffer is illustrated with the help of preparation of 0.1 M acetate buffer of ph value 5.76. Use Henderson Hasselbalch equation in determining ratio of sodium acetate and acetic acid. pka of acetic acid is 4.76. ph-pka= 5.76-4.76 = 1.0 = logarithmic of ratio of sodium acetate and acetic acid molar concentrations. Antilog of 1.0 = 10 = ratio of sodium acetate and acetic acid molar concentrations. Mixing of 100 ml of 0.1 M sodium acetate ( 8.203 g CH 3COONa is dissolved in water in total volume of 1000 ml) with 10 ml of 0.1 M acetic acid ( 6.005 g CH 3COO H diluted with water in total volume of 1000 ml) will make acetate buffer of ph 5.76 and of 0.1 M molarity. Check ph of buffer in ph meter calibrated with standard buffer. Adjust the ph with 0.1 M CH3COOH or 0.1 M CH3COONa if required.
Alternately, 90.90 mmoles of sodium acetate (7.456g ) and 9.09 mmoles of acetic acid (0.5458g) can be dissolved in 900 ml water. ph is checked and adjusted with HCl or NaOH if required. Total volume is then made to 1000 ml with water. 5.0 Equipment required for Electrophoresis Performance of electrophoresis requires two basic equipments; power supply and an electrophoresis assembly. Power supply can be used in constant voltage as well as constant current mode. It has meter to display voltage and current and two slots for connecting lids. Power supply to electrophoresis assembly was through lids. The colour of lid is usually colour-coded. Red lid and red marked slot for lid is positively charged and is required to be connected to anode. Similarly black colourlid is connected to cathode. Assembly is different for Horizontal Gel Electrophoresis and Vertical Gel Electrophoresis. In Vertical Gel Electrophoresis, buffer chamber containing cathode is connected to buffer chamber containing anode through support gel. The assembly as well as support gel is kept in vertical position.
On the other hand, horizontal electrophoresis is being usually done in one buffer system and support gel is submerged in buffer during electrophoresis. Agarose gel electrophoresis is done in horizontal electrophoresis Vertical Gel Electrophoresis Unit
Horizontal Electrophoresis Unit