[istrydummy]

Eight tubes were incubated for 5 minutes at different temperatures. The assay was initiated by addition to each tube 2ml samples of the enzyme. After 5 minutes, a 1 ml sample was withdrawn from each of the incubation mixtures and mixed with 5 ml of borate buffer (ph9). The absorption of the samples were then measured at 410 nm. The absorption values were converted into nanomoles PNP produced per ml of reaction mixture using the conversion factor. The results:

Temp 0C          enzyme activity (nanomoles PNP /ml reaction mixture)
20                        26.0
25                        34.0
30                        44.5
and so on.

Given that each incubation mixture had a total volume of 10 ml ( 1 ml PNGP +6.9 ml buffer +0.1 ml cystiene + 2 ml enzyme) and that the incubation time was 5 minutes ( time between adding the enxyme and removing the sample), convert the reults above into ' Initial reaction velovities' expressed as nanomoles PNP produced per minute.

My method:  10 ml X  26.0= 260    therfore 260/5minutes = 52 nanomoles/minute
Is this right? and is there a dilution factor involved?

Also when wanting to plot a graph of initial reaction velocity against incubation temperature, does the temp need to be converted to kelvin?

Thanks for the consideration.

Billy

[Dan]

My method:  10 ml X  26.0= 260    therfore 260/5minutes = 52 nanomoles/minute
Is this right?

I think it's fine.

so for a reaction time of 5 min:

Amount of PNP = 10 mL x 26.0 nmol/mL = 260 nmol

So

Rate of PNP formation = 260 nmol / 5 min = 52 nmol/min

I agree.

Also when wanting to plot a graph of initial reaction velocity against incubation temperature, does the temp need to be converted to kelvin? [/color]

I don't think so, ^oC = K + constant, so the only difference in the plots would be the intercept on the temperature axis. It depends what you're going to use your graph for - a conversion at some point may be required.

[istrydummy]

Hey,
Another question if i may.

I have to produce a graph of log10 'initial reaction activity' as the ordinate against the reciprocal of the absolute temperature as abscissa.
Then i have to determine the activation energy given that slope of the plot is equal to -Ea/2.303R, where R = 8.314J/Mol/K

Here is the table: Just imagine the coloured rows along side each other. 1/Temperature (K)                               (4.d.p) 3.4130 x10¯³ 3.3557 x10¯³ 3.3003 x10¯³ 3.4675 x10¯³ 3.1949 x10¯³ 3.1447 x10¯³ 3.0960 x10¯³ 2.9155 x10¯³ Initial reaction velocity (nmole/minute) 52 68 89 115 147 125 60 10 Log10 of Initial reaction velocity (4.d.p) 1.7160 1.8325 1.9495 2.0607 2.1673 2.0969 1.7782 1.0000

The probelm: When i plot the graph of Log10 of Initial reaction velocity against 1/Temperature (K)  , i dont get  a straight line which the  slope  of it would be  equal to -Ea/2.303R where R is the Universal Gas Constant.

Questions: 1) Have I calculated the log 10 part correctly? if yes, then why doesnt my graph give a straight line.
                 2) How do you actually calculate the Activation energy from the graph if the graph was drawn correctly? Do you take the gradient of the line ?

Thanks for considering.


PS: In my initial post there is a mention of adding 1 ml sample to 5 ml buffer...does that have anything to do with the end result? Is it a dilution factor of some sorts?

[istrydummy]

Can someone lend a hand please?

[sjb]

Questions: 1) Have I calculated the log 10 part correctly? if yes, then why doesnt my graph give a straight line.
                 2) How do you actually calculate the Activation energy from the graph if the graph was drawn correctly? Do you take the gradient of the line ?

Thanks for considering.


PS: In my initial post there is a mention of adding 1 ml sample to 5 ml buffer...does that have anything to do with the end result? Is it a dilution factor of some sorts?

Given that it's an enzyme catalysed reaction, I don't really think that the plot would give a straight line, but I'm not sure how to actually get Ea, sorry.

[istrydummy]

Questions: 1) Have I calculated the log 10 part correctly? if yes, then why doesnt my graph give a straight line.
                 2) How do you actually calculate the Activation energy from the graph if the graph was drawn correctly? Do you take the gradient of the line ?

Thanks for considering.


PS: In my initial post there is a mention of adding 1 ml sample to 5 ml buffer...does that have anything to do with the end result? Is it a dilution factor of some sorts?

Given that it's an enzyme catalysed reaction, I don't really think that the plot would give a straight line, but I'm not sure how to actually get Ea, sorry.

After doing some research iv noticed some people instead of doing the Log of the initial reaction velocity against 1/T( in kelvin), they have done ln k on the Y-axis. Can you please explain why? . My sheet says to " produce a graph of log10 'initial reaction velocity' as the ordinate against the reciprocal of the absolute temperature as abscissa

[krees]

straight line.
                 2) How do you actually calculate the Activation energy from the graph if the graph was drawn correctly? Do you take the gradient of the line ?

Questions: 1) Have I calculated the log 10 part correctly? if yes, then why doesnt my graph give a
Thanks for considering.


PS: In my initial post there is a mention of adding 1 ml sample to 5 ml buffer...does that have anything to do with the end result? Is it a dilution factor of some sorts?

Given that it's an enzyme catalysed reaction, I don't really think that the plot would give a straight line, but I'm not sure how to actually get Ea, sorry.

After doing some research iv noticed some people instead of doing the Log of the initial reaction velocity against 1/T( in kelvin), they have done ln k on the Y-axis. Can you please explain why? . My sheet says to " produce a graph of log10 'initial reaction velocity' as the ordinate against the reciprocal of the absolute temperature as abscissa

I am doing this exact same experiment right now and i would appreciate some advice.

thanks kacey