Thursday, 5 November 2009

3rd year biochemistry practicals - Alkaline phosphatase

Hi all students,

This week you have been performing the biochemical analysis of the biocatalyst. An alkaline phosphatase was used as an example.

Practical was split into two different parts. In the first one you were measuring how much of the product was produced by different concentration of enzyme. At the end of that part you had to decide which enzyme concentration is the best for our assay (correct concentration was that which gave you less thatn 10% of the product after 5min of raction). This is because you want to measure the initial rate of enzyme activity. You want to work in the conditions were not significant concentration of product is produced beacause otherwise it may affect your results (some reaction product may be inhibitors of the enzyme). If not enough enzyme is used then detection of enzyme activity becomes too complex or just belowe the detection threshold.

In the second part you were working with the chosen enzyme concentration to estimate the initial enzyme rate for different substrate concentration. In this case there where more substrate was used more product should be produced what indicates that enzyme worked with higher speed. Then you had to use these velocity values to calculate the Km (Michealis Constant which tells you how much of the substrate you need to get half of the Vmax of the enzyme) and Vmax (which is simply the maximum substrate to product conversion velocity, a characteristic feature for each enzyme).



So lets beggin the calculations.

1. First you have to orginise your Part A results into a small, cool table:

Then we have to calculate the rate of the reaction, vo at each concentration of enzyme used.

Rate (vo) = x micromol product produced/min/assay

To estimate the product concentration we use the super easy Lambert-Beer Law which says that adsorbance of the substance x is proportional to the concentration and the path length.

A = e x c x l
where A - adsorbance
          c - substance concentration [mol/l = M]
          l - path length [1cm]
          e - molar extinction coefficient (constant, characteristic for each substance) = 18.600 [L/(mol x cm)]

From that formula we extract the c which gives you:

c = A / (e x l)

For the 1/5 enzyme dilution we have:
c = 1.581/ (18600 x 1) = 0.000085 mol/L

Now we convert the mol/L to micromol/ml. There is 10^6 (1 000 000) micromols in the mol so we multiply our result by milion and there is 1000ml in L so we have to divide the result by a 1000 as well so we have

0.000085 x 1000000/1000 = 0.000085 x 1000 = 0.085micromol/ml

Now we use our calculated concentration to estimate how much of the micromoles of the product was produced per minut, per volume used in the assay. So we have to divide our result by 5 (because there was 5min incubation) and multiply by 3 (beacause reaction took place in the 3ml volume). So we have

(0.085 micromole/ml x 3) / 5 = 0.051 micromol/min/ml

In the next step we have to calculate enzyme concentration that was used for each dilution. So for the first 1/5 dilution we have: we know that neat enzyme concentration is 1mg/ml. So in 1/5 dilution we have 5 times less which is 0.2mg/ml. And we have taken 0.5ml of the dilution to the reaction which gives us 0.2 x 0.5 = 0.1mg of the enzyme. Now to calculate the concentration of the enzyme in the reaction we have to remember that volume of the reaction was 3ml, so 0.1mg of the enzyme in the 3ml gives us 0.1mg / 3ml = 0.033mg/ml.
We construct next table:

Using this table we plot the graph (see point 2 in you lab manual) product concentration produced per min per assay volume (y-axis) against the enzyme concentration [E] (x-axis) and answer the two question which you find in the manual.

3. In the next step we need to calculate the %Conversion of the substrate to the product to check which enzyme concentration is appropriate for our assay (remember that % conversion should be less than 10%). We use the similar formula as previously. We want to know how much of the product was produced in our 3ml volume reaction in 5min. So

micromol [product] formed in 5min = (A/18600) x 1000 x 3

So for 1/5 dilution in my case we have:

micromol [product] formed in 5min = (1.581/18600) x 1000 x 3 = 0.255micromol

Now we have to express that amount as the % of the initial amount of the substrate used. So we know from the protocol that stock concentration of the substrate was 20mM = 20mmol/L = 20micromol/ml. We used 0.5ml and used that in 3ml of reaction. So we took 20micromol/ml x 0.5ml = 10micromol.

%Conversion = 100 x (micromol [product] formed in 5min) / (micromol substrate present initially)

So for the 1/5 dilution we have:

%Conversion = 100 x 0.255 / 10 = 2.55%

We perform analogous calculation for other dilutions and we answer questions from the Section 3.

We put our results in the Table:

Now we go to the Part B of the report. First we fill in the table. To calculate the Vo we use the previous formule:

Vo = (Adsorbance - blank) / (18600 x 5)

To plot plot the Michealis-Menten graph we will use (Adsorbance - blank) / 5 (where 5 is our 5min incubation). This value is proportional to the Vo, as more adsorbance is present there where Vo rate is higher:).

So you should get a plot which look like that:

Then using the last table and the Michealis-Menten plot we try to estimate the Vmax and Km values. You have to remember that using the Michelis-Menten plot it is very hard to estimate those values. Usually mathematical methods are used to describe (to get the equation) how the curve is behaving. For our purpose it is ok if you roughly estimate those values. Next plot the Lineweaver-Burk plot is more accurate. So what you do is what you see in the next graph:

You gather the Vmax by dividing the Delta A/min value by extinction coefficient (18600, Vmax unit is micromol/min) and Km data directly from MM plot and you go to the next step which is the Lineweaver-Burk plot. To construct the LB plot we will need to recalculate some of the date we have used for the MM plot.
We construct a small table again:

In this table we calculate the reciprocal values of the initial substrate concentration and (Adsorbance - blank) / min which we had calculated in the last table. Remember that the (Adsorbance - blank) / min is proportional to the Vo. When you plot the 1/[S] versus 1/(Adsorbance - blank .....  you should get a straight line plot like this.

When you look at the plot you can see that the value where the line cross the x-axis gives you -1/Km which in this case is:

-1/Km = -1.5 ==> Km = 0.66 mM

And the value where the line crosses the y-axis gives you the 1/Vmax which in this case is:

1/Vmax = 5 ==> Vmax = 0.2

Remember that we have used (Adsorbance - blank.....) value insetad of the Vo so you have to divide the obtained result by 18600 (exticntion coefficient). At this stage you have Km and Vmax estimated from the MM plot and the Km, Vmax values calculated from the LB plot. Put them into the nice table and answer the question from the manual:)

Uffff that was a long report:)

See you next week:)


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