Monday, 16 November 2009

3rd year practicals - exam paper - Question 2!!

Hello Again,

Today we will work with the Question 2 of the 3rd year practical - sample exam paper. I hope you are ready:)

Question 2 Part a) Everything you need to know is on the Figure below:

and here:

The main differences between DNA and RNA are listed in this Table:

Part b) In this question you are asked to explain role of the EDTA, sodium dodecylsulphate (SDS) and isopropanol in DNA extraction.

EDTA - ethylenediaminetetraacetic acid, is sequestering agent which is used to bind divalent metal ions like Mg2+, Ca2+ etc (see Figure below).

EDTA is used to "neutralize" divalent cations. In the case of the DNA isolation it is used because cations like Mg2+, Mn2+ (and others) are coffactors of enzymes that digest DNA. Simply saying, to prevent DNA degradation we inactivate the enzymes by removing the divalent ions that are necessary for their activity.

SDS - sodium dodecylsulphate is a strong detergent. As all detergents it contains a hydrophobic and hydrophilic part. See below:)

In DNA purification we use SDS for two main reasons: SDS will lyse the cells (by disrupting the cell membranes) and denaturate/precipitate proteins which in this process are dispensable. You can imagine that hydrophobic part of the SDS interacts with the hydrophobic regions of the protein what cause their precipitation.

Isopropanol - is a small molecular weight alcohol. See below:)

Isopropanol is used to precipitate the DNA because alcohols (as you probably now:) dehydrate. Just remember that alcohols form hydrogen bonds with water very easily what cause DNA solubility to drop down ==> result DNA precipitation. We can say that alcohols "bind" water molecules better than the DNA, so in the "absence" of the water molecules DNA precipitate.

Part c) In this part you are asked to calculate either DNA concentration knowing adsorbance of the sample or adsorbance of the sample knowing the DNA concentration. Look into the table:


These calculations are very easy: we know that if we masure DNA adsorbance in the cuvette of 1cm thick and the reading gives us 1.0, DNA concentration is equal to 50microg/ml. So we use simple proportion:

50microg/ml ==> 1.0 adsorbance
x microg/ml of DNA gives us for example (solution A diluted) 0.858

So to get the DNA concentration of diluted solution A we have to multiply the 50microg/ml by the adsorbance of the sample:) And we have:

50microg/ml x 0.858 = 42.9microg/ml

If the sample was diluted prior to adsorbance measurment you have to include the dilution factor which in case of the solution A is equal to 20:) It can not be simplier:) So neat sample A have concentration 20-times more so:

42.9microg/ml x 20 = 858microg/ml

For calculating the adsorbance from the concentration of the sample, we do opposite calculation:

50microg/ml ==> 1.0 adsorbance
75microg/ml (solution F) ==> x adsorbance

So to get the adsorbence we have to divide the concentration of the sample (in this case 75microg/ml) by the reference concentration (50microg/ml). And we have:

50microg/ml / 75microg/ml = 1.5 adosrbance

That is it:)

Part d) In this part you are asked to perform few simple calculations. What we have here:

solution c1 = 330kg/L and express it in microg/deciL. So we know that there is 1 000 000 microg in kg so we multiply the 330kg/L by milion:

c1 = 330kg/L * 1 000 000 = 330 000 000microg/L

In the next step we have to express the L in deciL (dL). We know that there deci is 1/10, so it means that there is 100dL in L. We have then:

c1 = 330 000 000microg/L = 330 000 000 microg/100dL = 3 300 000microg/dL

You are also asked to express the number in scientific writing, so:

c1 = 3 300 000microg/dL = 3.3microg/dL x 1 000 000 = 3.3 x 10^6 microg/dL

Further we have to calculate the dilution factor after diluting the c1 = 330kg/L solution to solution
c2 = 15mg/mL. First we have to express either c1 solution in mg/ml or c2 = kg/L. Let's choose the second option..

c2 = 15mg/ml, we know that there is 1 000 000mg in kg so we divide it by 1 000 000
c2 = 15mg/ml / 1 000 000 = 0.000015kg/ml
next we know that there is 1000 ml in L so we divide the ml by 1000
c2 = 0.000015kg/mL/1000 = 0.015kg/L

To get the dilution factor we divide c1 by c2 (starting solution by the final solution):

dilution factor = c1 / c2 = 330kg/L / 0.015kg/L = (units get reduced) = 330/0.015 = 22000 !!

Again the dilution factor do not have any units (because all of them get reduced).

In the next section of the part d) you  are asked to dilute the solution c1 = 330kg/L by dilution factor 66 to obtain solution c3. So you just have to divide the 330kg/L by 66:

c3 = c1 / 66 = 330 / 66 = 5kg/L

The last one is to express concentration of the solution c4 = 4mg/microL in ng/pL. We know that there is
1 000 000 of ng in mg so again we multiply the number by 1 000 000.

c4 = 4 x 1 000 000 ng/microL = 4 000 000ng/microL

We know also that there is 1 000 000 pL in microL. We have then:

c4 = 4 000 000ng / 1 000 000 pL = 4ng/pL

If you still have problems with convertion of the scientific prefixes visit this post Prefixes convertion - you will laugh:)

That is it:)

I hope you enjoy it:)

Maciek GGS Team

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