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The Cellular Process - Essay Example

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This essay "The Cellular Process" focuses on the Michaelis Menten plot usually consists of a plot for slope against the concentration of the substrate, thereby acting as the basis for the procedure followed in calculating Km. There is a need to plot the absorbance of the solution…
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The Cellular Process
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Cellular process The Michaelis Menten plot usually consists of a plot for slope against the concentration of the substrate, thereby acting as the basis for the procedure followed in calculating Km. Consequently, in this case, there is a need to plot the absorbance of the solution vs time for each of substrate concentration. The plots give a straight line which helps in determining the slope for each. The determined slope gives the reaction rate for the concentration therein. The units for the concentration rate derived from the plots are Delta A/time (Rao, 2010, p.72). Another important step is plotting the slope, which is the concentration rate, against the substrate concentration.

Mathematically the above derivation relied on the direction from the following In the presence of an enzyme, the enzyme conservation law applies to lead to the determination of the concentration complex (Raju & Madala, 2005, p. 188): WhereKr = constant for substrate binding can = conversion to products = binding to the enzyme equation above leads to the Michaelis & Menten of finding the velocity At a maximum concentration of substrate, both the Vmax for the inhibited reaction and for the uninhibited reaction should always be equal (Ochs, 2014, p. 56). This brought about my surprise at the observation, which indicated different Vmax for the two reactions.

This issue may have arisen from the action of the inhibitor. The low Vmax for the inhibited reaction means that more inhibitor I was supplied. The presence of an inhibitor slows the rate of binding of the enzyme to the substrate leading to a low turnover number, which is the Kcat. From the equation below, a low turnover number leads to low Vmax in comparison to the uninhibited reaction (Beard & Qian, 2005, p. 88). Vmax = Kcat[E0]Consequently, solving this issue will require the reduction of competitive inhibitor I in order to meet a turnover number equal to that of uninhibited reaction, at a maximum concentration (Panesar et al, 2008, p. 78).

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