AP Biology Lab Book Lab #2
Analysis of Results.
1.
Time Intervals (seconds)
Initial 0 to 10
Initial 10 to 30
Initial 30 to 60
Initial 60 to 90
Initial 90 to 120
Initial 120 to 180
Initial 180 to 360
Rates
-0.01
0.025
0.01
0.013
0.0067
0.01
0.002
2. When is the rate of the highest? Explain why.
The initial rate is the highest at 10-30 because there is more H2O2 making it easier to get a reaction.
3. When is the rate the lowest? For what reasons is the rate low.
The lowest initial rate the class had was from 0-10; however, it should be from 180-360 because there is now less H2O2 make it harder to get a reaction.
4. Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this to enzyme structure and chemistry.
If a reaction takes place and sulfuric acid is added to the catalase, you will get a less product. The sulfuric acid changes the ph of the enzyme causing it to denature. When the product is less or less action, you are inhibited. This relates to chemistry because lowering the pH takes away the H+ ions and taking away too many causes it to denature.
5. Predict the effect that lowering the temperature would have on the rate of enzyme activity. Explain your prediction
I predict that the lower the temperature of the enzyme activity the lower the kinetic energy which causes fewer collisions. I predict this because I know that the higher the kinetic energy the higher the temperature and the more collisions, so obviously lowering the temperature will have a opposite effect.
6. Design a controlled experiment to test the effect of the varying pH, temperature, or enzyme concentration.
First gather 4 cups of H2O2 with 10 mL each, and into each of the four cups add 1 mL of catalase enzyme of different concentrations. Let each 1mL of catalase react for 30 seconds before stopping it with 5 mL of H2SO4. This way you have more control of all the amounts and time but the concentration. Then do the drop by drop method to see what difference it made.
Written by sh1ppu
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