Changes in factors such as pH, temperature and concentration

Spectrophotometer  Pipettes – 100 µl, 200 µl and 1000 µl ADH (from yeast; 1 mg/ml)  10 mM Sodium Phosphate Buffer with 0. 1% (w/v) BSA, pH 7. 5 NAD+ (disodium salt, 250 mg/160 ml) 50 mM EtOH  50 mM Sodium Pyrophosphate buffer, pH 8. 8 deionised water (25 ml)  1. 5 ml plastic cuvettes Mixing rods, Parafilm  Timer METHOD  To follow the ADH reaction, we take advantage of the fact that the cofactor NADH has an absorption maximum at 340 nm, which is absent in the oxidised form (NAD+). NAD+ is added in excess to prevent it from affecting the reaction velocity.

For all your measurements, start the reaction by adding ADH and follow the change in absorbance, note time [min]. Keep the enzyme and NAD+ on ice until used.  The concentration of ethanol – the substrate – in the series of reaction mixtures is 0. 5 mM, 1 mM, 2. 5 mM, 5 mM, 10 mM and 20 mM.  To start the reaction – add the ADH to the cuvette, mix the cuvette contents and replace in spectrophotometer.  Record the data for a period of 2 – 3 min.  Remember the kinetics of an enzyme reaction proceed initially at a fast velocity which progressively slows as the reaction proceeds.

RESULTS: Table 1: Absorbance data obtained at 340 nm for alcohol dehydrogenase catalysed reactions in the presence of increasing concentrations of substrate (ethanol). Concentration of EtOH [mM] Time (min) 0. 5 1 2. 5 5 10 20 Absorbance at 340 nm (A340 nm) 0. 5 0. 056 0. 093 0. 149 0. 205 0. 242 0. 254 1 0. 105 0. 186 0. 291 0. 409 0. 478 0. 502 1. 5 0. 155 0. 248 0. 422 0. 546 0. 622 0. 682 2 0. 211 0. 316 0. 508 0. 657 0. 719 0. 768 2. 5 0. 260 0. 384 0. 576 0. 719 0. 769 0. 818 3 0. 316 0. 446 0. 608 0. 744 0. 794 0. 843.

Table 1 shows that the spectrophotometrically determined NADH absorbances can be used to determine the concentration of NADH using the Beer-Lambert Law which states that the absorbance of a solution increases with an increase in its substrate concentration under ideal conditions (Berg et al. 2007). Thus, between absorbances 0 & 1; molar absorptivity of NADH at 6220 M-1cm-1; length of light path at 1cm, then following equation can be used to determine the concentrations of NADH at varying concentrations of the substrate (ethanol) and at different times c=A / ?

Where c = molar concentration of solution, A = absorbance of solution, and ? = molar absorptivity of solute at a defined wavelength. Table 2: Concentration obtained at 340 nm for alcohol dehydrogenase catalysed reactions in the presence of increasing concentrations of substrate (ethanol). Concentration of EtOH [mM] Time (min) 0. 5 1 2. 5 5 10 20 Concentration of NADH at 340 nm (A340 nm) 0. 5 0. 009 0. 015 0. 024 0. 033 0. 039 0. 041 1 0. 016 0. 03 0. 046 0. 066 0. 077 0. 081 1. 5 0. 025 0. 04 0. 068 0. 088 0. 1 0. 11 2 0. 034 0. 051 0. 082 0. 105 0. 116 0. 124 2. 5 0. 042 0. 062 0. 093 0. 116 0. 124 0.

132 3 0. 051 0. 072 0. 098 0. 12 0. 128 0. 136 The velocity of the reaction is observed by analysing the progress curves (Fig. 1a—1c) which represents change in NADH concentration over time. These curves aid in obtaining the initial rate of a reaction—the rate at which the product concentration is zero. Figure 1(a): Progress curve of the reaction showing the change in concentration of [NADH] with time at 340 nm. This graph helps in determining the initial velocity of the reactions at different substrate [EtOH] concentrations by evaluating the slope of the curve between time t=0 and t=0. 5 minutes.

This graph needs to be extrapolated back to zero. Figure 1(b): Progress curve of the reaction showing the change in concentration of [NADH] with time at 340 nm – graph—Extrapolation back to zero. The slope of each curve which is the initial velocity of each of these reactions (in ? mol/mL/min units) can be determined using the formula: Slope = Y2 – Y1 / X2 – X1 Figure 1(c): Concentrations of NADH between t=0 and t=0. 5 minutes used to determine the velocity profile of each of the reactions at varying substrate (EtOH) concentration. Table 3: Reaction velocities (v) of ADH-catalysed reactions at different [EtOH] concentrations at 340 nm.

v(? mol/mL/min) 0. 018 0. 03 0. 048 0. 066 0. 078 0. 082 [EtOH] (mM) 0. 5 1 2. 5 5 10 20 Thus, we can infer that the initial velocity of an enzyme-catalysed reaction is dependent on the substrate concentration and this can be quantitatively expressed through the Michaelis-Menton Equation whose three critical assumptions can be hypothesized as follows: E + S ES E + P where E is the enzyme, S being the substrate, ES the enzyme-substrate complex, P the product and k1,k2,k3 are the rate constants of the reactions. This leads to Vmax = k3 [ES].

If the steady-state expression for formation and breakdown of the ES complex is expressed as Km= (k2+k3)/k1, then the rate expression obtained will be: v = Vmax. [S]/Km+ [S] where Vmax and Km are the constants of this rate equation and it is these parameters that are kinetically unique to each enzyme under specific pH and temperature conditions. A saturation curve that plots reaction velocities against substrate [EtOH] concentration at 340 nm will help in determining the rate constants of the Michaelis-Menton equation. Figure 2: Saturation curve displaying the reaction velocity dependence on substrate [EtOH] concentration.

From this saturation curve, the dependence of reaction velocity on ethanol concentration can be calculated based on the following three cases: (i) At [EtOH] << Km, [EtOH] in the denominator can be dropped. Thus the Michaelis-Menton Equation will be, v = Vmax. [EtOH]/Km Replacing Vmax/Km by a new constant K, V = K [EtOH] Interpretation: The reaction velocity depends upon the substrate ethanol concentration when the ethanol concentration is below the Km (first-order reaction). (ii) At [EtOH] << Km, Km can be dropped in the denominator. The Michaelis-Menton Equation becomes: v = Vmax. [EtOH]/[EtOH] or v = Vmax.

Interpretation: The reaction tends to be zero order when the ethanol concentration is above Km. (iii) At [EtOH] = Km, if we solve the Michaelis-Menton Equation, we will deduce that the initial velocity is half-maximal. Thus, we can conclude that …

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