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# Cellular Process - Essay Example

Summary
The Michaelis Menten plot usually consists of a plot for slope against the concentration of the substrate, thereby acting as the basis for procedure followed in the calculating Km. Consequently, in this case, there is need to plot the absorbance of the solution vs time for the…

## Extract of sample "Cellular Process"

Cellular process The Michaelis Menten plot usually consists of a plot for slope against the concentration ofthe substrate, thereby acting as the basis for procedure followed in the calculating Km. Consequently, in this case, there is need to plot the absorbance of the solution vs time for the each of substrate concentration. The plots gives 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 enzyme, the enzyme conservation law applies leading to the determination of the concentration complex (Raju & Madala, 2005, p. 188):
Where
Kr = constant for substrate binding
Kcat = conversion to product
Kf = binding to the enzyme
The equation above leads to the Michaelis & Menten of finding the velocity
At maximum concentration of substrate, both the Vmax for the inhibited reaction and for the unhibited reaction should always be equal (Ochs, 2014, p. 56). This brought about my surprise for 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 substrate leading to low turnover number, which is the Kcat. From the equation below, low turnover number leads to low Vmax in comparison to the unhibited reaction (Beard & Qian, 2005, p. 88).
Vmax = Kcat[E0]
Consequently, solving this issue will require for the reduction of competitive  inhibitor I in order to meet a turnover number equal to that of uninhibited reaction, at maximum concentration (Panesar et al, 2008, p. 78).
Bibliography
Rao, D. G. (2010). Introduction to biochemical engineering. New Delhi, Tata McGraw-Hill. 72
Raju, S. M., & Madala, B. (2005). Illustrated medical biochemistry. New Delhi, Jaypee Brothers Medical Publishers. 185
Ochs, R. S. (2014). Biochemistry. Burlington, Mass, Jones & Bartlett Learning.
Beard, D. A., & Qian, H. (2008). Chemical biophysics quantitative analysis of cellular systems. Cambridge, Cambridge University Press. http://proxy2.hec.ca/login?url=http://library.books24x7.com/library.asp?B̂&bookid=26546.
Panesar, P. S., Marwaha, S. S., & Chopra, H. K. (2008). Enzymes in food processing: fundamentals and potential applications. [S.l.], I K International Publish.
Berger, M. P. F., & Wong, w.-k. (2009). An Introduction to Optimal Designs for Social and Biomedical Research. Chichester, John Wiley & Sons. http://public.eblib.com/EBLPublic/PublicView.do?ptiID=454378. Read More
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