how do you work out rate of reaction?
I increased the concentration of my substrate (casein) from 1-5%
at 1% my substrate was broken down in 1 min and 9 seconds
at 2% - 2 mins and 47 seconds
at 3%- 3 mins and 50 seconds
at 4% - 4 mins and 45 seconds
at 5% - 5 mins and 40 seconds
how do I change this into rate of reaction?
my question was to find the effects of substrate conc on rate of enzyme catalysed reaction, I know that as the substrate is increased the rate of reaction should increase to a point then level off, but using my results how do I show this?
how to work out rate of reaction Watch
- Thread Starter
- 28-01-2015 21:43
- 28-01-2015 22:48
Im not 100% sure, but I think you could do amount broken down per second. That usually happens when you're working in moles so you'd do the total amount in moles divided by seconds taken to break it down, and that would give you the number of moles broken down per second. Not sure if you could do it the same with percentage though
- 29-01-2015 15:22
You would first need to plot a graph with your results and generally, if you want to calculate the rate of reaction you would calculate the rate of reaction at the start of the experiment - so the initial rate of reaction as generally this would be the fastest.
- 06-02-2015 19:38
"Rate of reaction" means the speed of a reaction, and rate is measured as a unit change per time, usually. However, in this case, there isn't a measurable change in the dependent variable to divide by time because the dependent variable in your experiment was the time taken to break down the substrate, so you might have to calculate rate as 1 divided by the time taken to break down the substrate (in seconds) (so rate = 1/time). This might give you quite a small number, though, so if you have to plot a graph, it might be advisable to multiply all your figures by a power of ten (e.g.: x10^3) so that you're not trying to plot minuscule decimal numbers (make sure to put rate x10^3 or whatever figure you use on the axis of the graph so that it stays accurate).
Hope that helps, good luck! C: