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Discussion.docx-In this experiment we performed a
Discussion.docx-In this experiment we performed a
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Discussion.docx-In this experiment we performed a
Page 1
In this experiment we performed a connuous assay.
This type of assay basically means that as
soon as the enzyme is mixed with the substrate the reacon starts and the substrate is used up and
turned into product (65 Lab Manual).
To perform this assay you must first blank the spectrophotometer
using a cuvee filled with the appropriate phosphate soluon.
The next step is to add the enzyme,
cover with Para film, quickly invert it and then place it in the spectrophotometer.
spectrophotometer is then recorded in 10 second increments.
In order to keep all trials consistent with
3mL, all three trials used 2.7 mL of the para-nitrophenyl phosphate and then variable amounts of
enzyme and water.
In the full enzyme, 100 microliters were used.
In the half enzyme 50 microliters
were used.
In the inhibited trial, water was used with the inhibitor.
We collected the data for all three trials and recorded the appropriate absorbance values.
using the change in absorbance divided by the me in seconds you can calculate the V
This is also
equivalent to the slope when using the Absorbance vs. Time graphs.
The next step was to convert this
rao into absorbance per minute which can be done by mulplying the value by 60.
However, we want
in terms of micromoles per minute so the next step was to mulply by 0.003L.
Once V
was obtained it was important to produce a Michaelis-Menten Plot. This plot is the
substrate concentraon vs. the V
This plot is important because it can demonstrate the differences in
rate of reacons between the trials.
The uninhibited enzyme was clearly the fastest reacon which
verified with inial predicons.
The inhibited enzyme was not the slowest however.
The half
uninhibited enzyme was actually the slowest of the three trials. This means that even if an enzyme is
inhibited, it will sll occur at a faster rate if the enzyme is there in a higher concentraon.
This leads to
an important conclusion that concentraon is the dominant factor.
Next we needed to create a Leinweaver Bruk Plot which was important for determining the V
and K
In a Leinweaver-Bruk plot, you must plot the inverse of the substrate concentraon
verses the inverse of V
This is important because the y-intercept then will produce the inverse of V
To do this we first created the plot using excels.
Then using the trend-line feature we were able to
determine the equaon of the line for each trial.
Using the equaon of the line we could use the y-
intercept value to obtain the V
for the three trials.
Next we could use the equaons of the line to also
find the slope which is equal to the V
To solve for K
you then mulply V
by the slope.
calculaons can all be found in the lab report.
The important reason for creang this plot as well is that
the Michaelis-Menten Plot cannot give the V
because it only approaches the V
From the informaon from the graph and also the tables on page 74 one can infer that the
uninhibited enzyme had the greatest V
as well as Km value.
The half enzyme had the lowest Vmax
and Km. The Vmax means the maximum velocity at which the reacon can occur so in the case of the
fully uninhibited enzyme having the highest Vmax value means of the three trials this trial had the
highest potenal for reacon rate.
The Km is an esmate of the dissociaon constant of the enzyme
and substrate.
Having a high Km means there is low affinity and weak enzyme to substrate binding.
half enzyme had the lowest Km which means it had the ghtest binding which is why the graph
resembles a more hyperbolic curve.
The inhibited trial had the middle values.
This means that it had
neither the highest potenal nor the ghtest/weakest binding.

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