Tuesday, 17 November 2009

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

Hi ya all,

It is time for another Question from the 3rd year practicals sample exam paper:)

Question 3, part a) In this one you are asked to explain principle of measuring the adsorbance of protein sample at A260 and A280. If you remember from the Question 2, A260nm is used to measure concentration of DNA. A280nm is specific for the proteins because at that wavelength aromatic aminoacids adsorb the light. If you measure adosrbance of you protein sample at 260nm and 280nm, you can see if your sample is mainly proteins or proteins and DNA (you estimate purity of the sample). So what you should get is low adsorbance at 260nm and high at 280nm.

You can find all aminoacids structures, three letter codes, one letter codes etc. here:

Way to remember aminoacids

or here

Way to remember aminoacids - chemist way

or here:)

Interactive aminoacid quiz

Part b) In here you have to use the data provided in the Table (see below) to prepare standard curve and estimate concentration of unknown samples.

Using the known concentration of BSA standard and corresponding adsorbance, we construct plot concentration against the adsorbance. Remember to label the x and y axis (with the units if applicable). Using the standard curve you estimate the concentration of the Unknown samples which correspond to particular adsorbance. In the case of dilutied samples do not forget to multiply obtained concentration by the dilution factor (for example Unknown A have to be multiplied by 20 as dilution factor is 20:). I used the trendline here and the equation to estimate the unknown samples concentrations. But you probably will have to do that on the paper, so remember to scale your plot as accurate as possible:)

Part c) I already answered that question. Look up here (Question 1 part a):

To calculate specific activity of the Unknown sample E we use the concentration estimated from part b). In my case is 20.66microg/ml, we also need Total Enzyme Activity (Activity of the sample) which is 540UI/ml.
So to get the Specific Enzyme Activity we have to divide Total Enzyme Activity (Activity of the sample) by the concentration of the sample:

Specific Enzyme Activity = 540UI/ml / 20.66microg/ml = 540/20.66 UI/ml x ml/microg
Specific Enzyme Activity = 26.14UI/microg

See also Question 1 part a from the same exam paper:

Part d) As usually part d of each question is about simple calculations. First we need to express solution
c1 = 150microg/ml in kg/L. We know that there is 1 000 000 microg in kg so we need to divide our value by 1 000 000.

c1 = 150kg / 1 000 000ml = 0.00015kg/ml

and we know that there is 1000ml in L so we need to multiply the number by 1000:

c1 = 0.00015 * 1000 kg/L = 0.15kg/L

Next you are asked to give the concentration of the solution c1 after dilution by factor of 30. So it can not be simplier:). You just divide the c1 by 30:

c2 = c1 / 30 = 150microg/ml / 30 = 5microg/ml

And finally you are asked to calculate the what is the dilution factor if you go from solution c1 = 150microg/ml to solution c3 = 15ng/ml. As you remember first we have to express the solution c1 in ng/ml or solution c3 in microg/ml to have the same units. Let's say we choose the first option. We know that there is 1000ng in microg, so we need to multiply our c1 concentration by 1000:

c1 = 150microg/ml * 1000 = 150 000 ng/ml

In the next step we need to calculate how much more solution c1 is concentrated that solution c3. To get that, which is actually our dilution factor we divide c1 / c3:

dilution factor = c1 / c3 = 150 000 / 15 ng/ml x ml/ng = 10 000

Remember that dilution factor do not have any units because all of the get reduced:)

Problems with convertion of scientific prefixes?? If yes go here Prefixes convetion - you will laugh:)

And that how it is done:)

Good luck:)


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