How does the concentration of hydrogen peroxide affect how the enzymes in potatoes work?

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How does the concentration of hydrogen peroxide

 affect how the enzymes in potatoes work?

In this experiment I will see the affects of reacting different concentrations of hydrogen peroxide with the enzymes in potatoes. Potatoes contain the enzyme catalyse and this enzyme reacts with the hydrogen peroxide to form oxygen and water. The reaction can be seen by the formula:

2H²O²                                2H²O + O²

(hydrogen peroxide)                        (water + oxygen)

I will obtain sets of results by varying the concentration of hydrogen peroxide used; this will affect the rate of reaction and, in turn, oxygen given off within a certain time period will vary accordingly.

I will change the concentration of hydrogen peroxide by diluting it with distilled water.

It is vital that certain factors are kept constant throughout the investigation in order to carry out a ‘fair test’. The factors that must remain un-altered are:

  • Temperature. This has to be kept the same because enzymes work best at different temperatures and if the temperature is closer to their optimum temperature in one experiment compared to another this will cause the investigation to become unfair because as you get closer to the optimum temperature the rate of reaction increases. Also if the temperature gets too high then the enzymes can be killed and will therefore not react. We kept the temperature at around 20 degrees Celsius (room temp.) throughout the investigation.
  • Surface area of the potato chip. The surface area mustn’t be changed because a large surface area means that more of the catalyse molecules are accessible compared to smaller surface area where there are fewer catalyse molecules able to react with the surrounding hydrogen peroxide particles and as a result the rate of reaction is dramatically decreased. The noticeable differences in the rate of reaction will result in the investigation not being a fair test.
  • Size of the potato chip. The size of the potato chips must remain unchanged because if it did change it is likely that the larger chip will contain more catalyse than the smaller chip and consequently a wide range of results will be recorded. This is because if there is more catalyse for the hydrogen peroxide to react with then the rate of reaction will be increased and vice versa if there is less catalyse.
  • Same combined volume of H²O²and distilled water. I will have to keep the volume of the solution the same (20cm³) because if this was altered then the potato chip may only be partially submerged and therefore there would be less hydrogen peroxide for the catalyse to react with. Equally, if the volume of the solution was extremely high then there will be more particles for catalyse to react with.
  • Time given for the reaction to take place. If the reaction time was consistently changing then the results would be extremely inconclusive because obviously if the reaction is allowed more time then more gas will be produced and this is why it is essential that the reaction time is the same throughout the investigation. For each experiment I gave the reaction 2 minutes to take place.
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Method

  • Set the apparatus up as in the diagram.
  • Half-fill the tub with water.
  • Completely fill up a measuring cylinder with water and place it into the tub trying to prevent any water loss. Any water lost must be recorded.
  • Cut out all the potato chips that will be used for the experiment using a cork-borer and then use a scalpel to accurately measure the chips are all the same size. Record the size of the potato chip used so that future chips will also be that size. Also record the size of the cork-borer used ...

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