• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18

An Investigation into the Rate of Breakdown Of Hydrogen Peroxide By Potato Tissue

Extracts from this document...

Introduction

An Investigation into the Rate of Breakdown Of Hydrogen Peroxide By Potato Tissue Initial Observations: - 1) Hydrogen peroxide solution is toxic. 2) Hydrogen peroxide slowly beaks down to oxygen and water. 3) All living cells contain the enzyme catalase. 4) Catalase/living cells cause hydrogen peroxide to break down quickly. Hypothesis: - I predict that the greater the surface area the more oxygen will be produced in the first five minutes of collecting oxygen. Each enzyme has an active site, which is an area of a particular shape into which only one type of molecule will fit. For this reason enzymes are specific to their substrate. The enzyme catalase splits hydrogen peroxide as follows: - hydrogen peroxide oxygen + water 2H O O + 2H O The above reaction will only occur when the hydrogen peroxide molecule collides with a catalase molecule. The more collisions there are, the larger the volume of oxygen produced. If the temperature is increased there will be more successful collisions because the molecules have more energy so they move faster and collide more frequency. The collision will also have more energy so more catalase will be successful so therefore there will be more oxygen produced. It is for this reason that temperature in my investigation must be controlled. If the temperature is not controlled then the oxygen being produced has not just been produced from the surface area but also from the temperature. If the temperature is changed as well as the surface area then the oxygen is being produced by two variables. Enzymes Enzymes are proteins. A protein is made up of a series of amino acids, which are unique to that protein. This is known as the proteins primary structure. If the enzyme is deformed in any way, then it will not be the right shape and will not function. Each enzyme has its own specific shape, with a gap at a different position. ...read more.

Middle

3.0 cm� 2.4 cm� 3 minutes 3.6 cm� 3.0 cm� 3.5 minutes 3.9 cm� 3.5 cm� 4 minutes 4.2 cm� 4.0 cm� 4.5 minutes 4.4 cm� 4.2 cm� 5 minutes 4.9 cm� 4.8 cm� Graph 2 Length of potato = 4.5cm Surface Area = 2.78cm� Hydrogen Peroxide = 12cm� Oxygen Production Time 1st attempt 30s 0.5 cm� 1 min 1.2 cm� 1.5 minutes 1.8 cm� 2 minutes 2.4 cm� 2.5 minutes 3.0 cm� 3 minutes 3.5 cm� 3.5 minutes 4.2 cm� 4 minutes 4.5 cm� 4.5 minutes 4.7 cm� 5 minutes 5.0 cm� Unfortunately due to a mistake I made, instead of changing just the surface area, I changed the mass as well. This caused some of the results to be inaccurate, as two variables were changed instead of one. When the final experiment is done, the mass of the potato will be kept the same, but the surface area will be changed only. Hopefully this should make the results more accurate. Results of Final Investigation: Graph 3 Pieces of potato= 1x 5cm Surface Area = 8.25cm� Hydrogen Peroxide = 12cm� Oxygen Production Time 1st attempt 2nd attempt 30s 0.5 cm� 0.5 cm� 1 min 1.0 cm� 1.0 cm� 1.5 minutes 2.0 cm� 1.2 cm� 2 minutes 2.4 cm� 1.8 cm� 2.5 minutes 3.0 cm� 2.4 cm� 3 minutes 3.8 cm� 3.0 cm� 3.5 minutes 4.5 cm� 3.6 cm� 4 minutes 5.0 cm� 4.0 cm� 4.5 minutes 5.8 cm� 4.8 cm� 5 minutes 6.2 cm� 5.4 cm� Mean 3.42 cm� 2.77cm� Total Average 3.1 cm� Graph 4 Pieces of potato= 2x 2.5 cm Surface Area = 8.6 cm� Hydrogen Peroxide = 12cm� Oxygen Production Time 1st attempt 2nd attempt 30s 0.5 cm� 0.5 cm� 1 min 1.0 cm� 1.2 cm� 1.5 minutes 1.0 cm� 2.0 cm� 2 minutes 1.8 cm� 2.6 cm� 2.5 minutes 2.0 cm� 3.2 cm� 3 minutes 2.6 cm� 3.8 cm� 3.5 minutes 3.8 cm� 4.2 cm� 4 minutes 4.2 cm� 5.0 cm� 4.5 minutes 5.0 cm� 5.9 cm� 5 minutes ...read more.

Conclusion

This was done by adding up the 10 results recorded at the correct times and dividing by 10. This was done for both attempts and for every size or length of potato, for both attempts. Then the two means were added together and divided by two to give a total mean. This was repeated for a further three times to draw graph 7. These means were then joined together to see if the larger the surface area, the more oxygen produced. What can be seen from the graph is that there is an anomalous result. This was the first part of the investigation being done. The surface area for the anomalous result is 8.25cm�, or one piece of potato at 5cm in length. Apart from this one anomalous result, the hypothesis made at the beginning of the investigation, "I predict that the greater the surface area the more oxygen will be produced in the first five minutes of collecting oxygen." has been proved on graph 7. This shows us that as the surface area increases so does the amount of oxygen produced. This is because when there is a large surface area, then there is more room for collisions to happen and they therefore happen very quickly and continually. On graph 8 it also shows that the prediction has been proved but on a smaller scale with the pieces of potato on the x-axis and not the surface area. If I had to redo the investigation experiment, then I would change my method so that the delivery tube was placed inside the measuring cylinder after the potato was cut and the hydrogen peroxide measured out and placed inside the boiling tube. The delivery tube would be the last thing I would place in position before the potato and bung were put in the right place. When I do redo the investigation, I will then record my results accurately and make sure that they are taken at the right time reading the numbers correctly, from the scale on the measuring cylinder. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Life Processes & Cells section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Life Processes & Cells essays

  1. Marked by a teacher

    Investigating the effects of surface area on the rate of enzyme reactions.

    4 star(s)

    If the graph does eventually level out, it is due to there not being enough substrate to bind with all the free active sites there are. A higher concentration of enzymes after this "maximum point" will do nothing for the amount of gas evolved.

  2. Marked by a teacher

    The aim of this investigation is to find out what effect pH has on ...

    4 star(s)

    This is because slowly and gradually the enzyme will start to denature, which means it will change shape; it also means that the active site of the enzyme where the substrate fits into and breaks down will change its structure, disallowing the substrate to break down.

  1. An investigation into the effect of substrate concentration on the activity of the enzyme ...

    Volume of Water (cm�) Average at 60secs Average at 120secs Average at180secs Average at 270secs Average at 360secs Average time taken to reach 10 cm� of gas. (seconds) 10 0 5.3 9.5 - - - 126 8 2 5.1 9 - - - 134 6 4 4.6 8.3 9.6 -

  2. Osmosis investigation

    This provided a good size potato chip so I chose to keep the size when doing my actual results. If I were to have a larger cork borer, for example size 6, then the potato would not be able to have the right amount of chips taken from it so

  1. The endeavour of this investigation is to ascertain if there is any effect of ...

    No evidence of repetition Use of ratios makes calculations more difficult Not a lot detail regarding the experiment so reliability is low 'Biological science' (published by C.U.P in 1990) This set of results uses length as the measuring constituent. However the length is to 1 d.p so the accuracy is

  2. For my coursework I will be performing an investigation into an experiment using hydrogen ...

    * 1x - Water bath - to heat the hydrogen peroxide to the right temperature * 1x - plastic measuring cylinder (mm units) - hold the water which will measure the oxygen level * 1x - Plastic tub - to hold the water inside the upside down measuring cylinder * 1x - Electronic scales (accurate to )

  1. Decomposition of Hydrogen peroxide

    In order to make my results as precise as possible, I will use the stop clock to help get the exact time and readings and I will use a measuring cylinder that has accurate markings. I will measure all my readings on the same scale - ml.

  2. Hydrogen peroxide will breakdown to oxygen and water in the presence of Catalase.

    The greater the number of discs, the greater the enzyme concentration. Apparatus - i) A manometer ii) 30ml hydrogen peroxide iii) Manometer fluid iv) 6 boiling tubes v) Tongs vi) A test tube rack vii) A potato viii) A petri dish ix)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work