Sunday, 15 November 2009

3rd year practicals - exam paper !!

Hi ya all students,

Today in the 3rd year practical section you will not find lab report tutorial but I think you will be interested as well:) In the next few posts you will find all the sample exam paper anwsers:) Everyday (since today) one question will be solved!

So lets begin:

Question 1) Part a) In this question you have to calculate different values using the data provided in the Table. This is easy busy. Like here:

To calculate the Specific Enzyme Activity (IU/mg) you have to divide the Total Enzyme Activity (IU) by the Total amount of the protein (mg). It means that you have to express how much enzyme activity is there for each mg of protein present in the extract. Actually as you look at the unit of the Specific Enzyme Activity (IU/mg) it tells everything:).
To calculate the Purification Factor you divide the Specific Enzyme Activity (of each purification method) (IU/mg) by Specific Activity of Crude extract (IU/mg).
To get the Yield (%) you express in % how much of the protein is there after purification in relation to protein content of the Crude Extract.

Part b) In this part you are asked to explain what Specific Enzyme Activity and Total Enzyme Activity are? The first one is already explained in part A (no need to do that again).
Total Enzyme Activity is (as the name suggest) total activity of the sample you have obtained. So if you examined activity of let's say 1ml sample taken from your extract and it gaves you for example 5.0IU. And you have in total 40ml of your extract/purification your total enzyme activity is 40x5.0 = 200IU.

Part c) In this part you are asked to discuss 2 factors that can influence the enzyme activity in the assay. There is a lot of them, for example:
- buffer (wrong pH, salt concentration (ionic strength), detergents, etc),
- temperature (to low, to high),
- purity of the enzyme (low purity ==> low activity),
- inhibitors,
- enzyme handling,
- other molecules essential for enzyme activity (like coffactors, ions etc),
- and more:)

Lets say you choose buffer and temperature. In you discussion you should say that each enzyme works in the optimal pH (different for different enzyme) and if the buffer is not pH properly then to high or to low pH will affect our assay (it may even cause lost of enzyme activity). It goes similar with the temperature. Again as you imagine there is temperature range/optimum in which each enzyme works. Wrong temperature used during the assay will affect our results.

Part d) In this question you have to perform simple calculations (baby ones:). First you have to express 40IU/ml as mIU/microL. So, we know that there is 1000 mIU in IU yeah? So we multiply our number by 1000, what gives us this:
40IU/ml x 1000 = 40 000mIU/ml
And we also know that there is 1000 microL in ml. We have:

40 000mIU/1000microL yeah? = 40mIU/microL

Done:) If you still have problems with the refixes convertion go here Prefixes Convertion - you will laugh

Next one is even easier. You have to dilute the solution c1 = 40IU/ml by factor 20:) So you just divide the concentration of the solution c1 by dilution factor 20:) And you have:

40IU/ml / 20 = 2IU/ml
That is it:)
Next, you have to dilute solution c3 = 15mg/ml to solution c4 = 10microg/ml and caluclate the dilution factor. So to get the dilution factor you have to simply calculate how much more solution c3 is concentrated than the solution c4. To get that you divide the c3 / c4 but first you have to express either c3 in microg/ml or c4 in mg/m. Let's say we choose the first option so:

c3 = 15mg/ml = as we know there is 1000microg in mg = 15 x 1000microg/ml

So now:

dilution factor = c3 / c4 = 15 000microg/ml / 10microg/ml
dilution factor = c3 / c4 = 1500microg/ml x ml/microg
dilution factor = 1500 and the unit:) There is no unit:) Because all units are reduced.

I hope you find these answers clear and straight. Tomorrow there we will KILL Question 2:)

Maciek GGS Team

1 comment:

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