Thursday, 19 November 2009

3rd year practicals - exam paper - Question 5!!

Hello Guys,

Today wego with the Question 5.

Check this out:)

Question 5 Part a) In this part you are asked to calculate the Vo (initial rate mmol/min/ml) of the Alkaline phosphatase. All details of the assay and obtained adosrbance are listed. To get the Vo we need: adsorbance (A410nm = 0.754), time (5min), volume of the reaction (3ml) and extinction coefficient (17500 1/Mcm). We know that:

A = e c l
where: A - adsorbance
e - extinction coefficient (1/Mcm)
c - concentration (M)
l - path length (cm)

To get the product concentration (c) we need to convert the formula:

c = A / (e x l)

Now we substitute the values:

c = 0.754 / (17500 x 1) = 4.3 x 10^-5 M = 4.3 x 10^-5 mM/ml

We know that assay was done for 5min in 3ml and 1/100 dilution of the enzyme was used. We need to inculde that data in our calculations. We have:

Vo = (c x 3ml x 100) / 5min = (4.3 x 10^-5 x 3ml x 100) / 5min = 2.58 x 10^-3 mmol/min/ml

In the Part b) you are asked to prepare Michelis - Menten and Lineweaver - Burk plots using the data provided in the Table:

To plot the Michaelis-Menten plot you use data from the Table, but to prepare the Lineweaver-Burk plot you need to calculate the reciprocals of the [S] (substrate concentration) and Vo (initial rate). After you have them you construct two plots (remember to scale as accurate as possible, in that way your results will be very accurate).

Part c) Remember that for the MM plot you read out your Vmax and Km values directly from the plot but for the second graph (LB plot) you need to do simple calculations to get the Vmax and Km. For example from the MM plot we can conclude that approximately Vmax = 5.1 mmol/min and that Km = 1.6mM. From the LB plot we have (approximately:):

y-intercept = 1/Vmax = 0.15
Vmax = 1/0.15 = 6.7mmol/min

x-intercept = -1/Km = -0.35
Km = -1/-0.35 = 2.8mM

Remember that values for Vmax and Km obtained from the MM plot are less accurate that from the LB plot. Why is that? Because linear model is easier to describe than the expotential one. I mean that it is always easier to find a linear equation that will fit to our data than the expotential one:)

Part d) Calculations again:) I will do it fast this time (if you need further assistance email me at or look at the previous Questions from this exam paper).

c1 = 300mM = 300 000 000 = 3 x 10^8 nM
dill. factor = c1/c2 = 300 000 000 / 30 x nM / nM = 3 x 10^7 = 10 000 000 = 10^7(dilution factor no units)
c1 / dlution factor = c3 = 300 000 000nM / 15 = 20 000 000nM = 20mM

That would be it:) Tomorrow the final question:)



1 comment:

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