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enzyme activity as a function of subtrate concentration

Posted: Wed Dec 30, 2009 9:46 am
by joshuadavis
I am conducting a scientific analysis based on the experiment at this link:

https://www.sciencebuddies.org/science- ... U&from=TSW

At the bottom of this page there is a variation to design and carry out an experiment to measure enzyme activity as a function of substrate concentration and graph results using a double-reciprocal plot.

Thr procedures are clear; however I don't understand how to collect the data and interpret it to fit the concepts and variables for the production of the plot.

Any answers?

Thanks,
Joshua Davis
Brunswick High School
12th Grade

Re: enzyme activity as a function of subtrate concentration

Posted: Wed Dec 30, 2009 12:20 pm
by deleted-71417
Hi Joshua,

You may find this site helpful in understanding your questions. I suggest particularly the sections starting with Michaelis-Menten Kinetics.

http://themedicalbiochemistrypage.org/e ... etics.html

Here is an experimental writeup discussing the issues:

http://www.science-projects.com/catalasekinetics.htm

The bibliography links to Wikipedia contributrs might also be helpful, but were temporarily unavailable when I tried to access them.

Best regards,

Barrett L. Tomlinson

Re: enzyme activity as a function of subtrate concentration

Posted: Wed Dec 30, 2009 4:23 pm
by joshuadavis
Thank you for the links Mr. Tomlinson! I am finding them to be very helpful in my research. I know the contributors are very busy people but if anybody has any time to look back at the link before that I posted as my project inspiration. Under the experimental procedure subject the effect of enzyme concentration on reaction rate. Self explanatory that its experimenting with the enzyme concentration. I understand in the experiment that the potato solution is the enzyme and the hydorgen peroxide is the substrate. If I were to follow those procedures yet using the hydrogen peroxide instead of the potato enzyme solution, would that give me my answer towards explaining enzyme activity as a function of substrate concentration?

And also Michaelis-Menten kinetics uses velocity. Is the reaction time mentioned in the experimental apparatus reaction velocity meaning that the saturation point is reached at the maximum time? It may seem obvious that the answer is yes but I don't want to go off my own assumption.
Joshua Davis
Brunswick High School
12th grade

Re: enzyme activity as a function of subtrate concentration

Posted: Thu Dec 31, 2009 2:06 am
by MelissaB
Hi,

Yes, what you would do is vary the concentration of hydrogen peroxide (you are correct that it is the substrate) in your solution. I don't think you can get it more concentrated, but you can certainly dilute it by adding distilled water.

Does that answer your question.

Re: enzyme activity as a function of subtrate concentration

Posted: Sun Jan 10, 2010 10:15 am
by joshuadavis
Yes that perfectly answers my question. I just have one more question. According to my research, it would be best to express the reaction time (in seconds) as moles per second. In order to do that would I multiply the reaction time by the moles in hydrogen peroxide (H2O2) for each data point up to the maximum reaction time?

Re: enzyme activity as a function of subtrate concentration

Posted: Sun Jan 10, 2010 10:49 am
by donnahardy2
Hi Joshua,

The experiment in the science buddies website is written to measure the time that it takes a filter paper disc to rise to the surface of the solution due to oxygen bubbles forming. The protocol suggests graphing time vs. concentration of hydrogen peroxide. This does not actually measure the moles of oxygen produced, so it would not be possible for you to calculate moles per second. If you did capture all of the oxygen produced, you could calculate the moles of oxygen based on the volume of gas produced. Did you follow the protocol on the website, or did you modify the procedure? Please let us know.

If you are using hydrogen peroxide from the pharmacy, it is usually 3%. The protocol suggests using 1%, which would be a 1:3 dilution (one parts hydrogen peroxide plus two parts water). 1 percent hydrogen peroxide is 1 gram of H202 per 100 ml, or 10 grams per liter, or:

10 gram H202/L x 1 M H202/36 grams = 0.0278 M or 27.8 mM/L.

Does this help? Let us know if you need more information.

Donna Hardy