• 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

Investigating the influence of pH on the activity of potato catalase

Extracts from this document...

Introduction

Biology Coursework. Investigating the influence of pH on the activity of POTATO CATALASE Conclusion. Enzymes are globular proteins which can be denatured at extreme pH levels The active site is a region of the protein/enzyme such as a cleft or crevice in the tertiary structure of the enzyme, often constituting less than 5% of the surface area. The active site can be negatively charged to attract for example a positively charged amino group (basic side chains) and vice versa, namely positively charged to attract for example a carboxylic group. As the hydrogen ion concentration decreases the pH number increases and a solution becomes more alkaline / basic. Amino acids are amphoteric having both a basic (amine) group and acidic (carboxylic acid) group). In water, the carboxylic acid partially dissociates becoming negatively charged and similarly the amine group accepts a hydrogen ion becoming positively charged. In this state, an amino acid becomes a zwitterion which enables amino acids to act as buffers. When the hydrogen ion concentration of a solution of an amino acid increases, protons are attracted to the negatively charged carboxyl group and therefore the overall charge becomes positive due to the amine group also being positively charged. Alternatively, reducing the hydrogen ion concentration of the solution means the amine group donates hydrogen atoms (ions??) making the overall charge negative. Therefore, changing the hydrogen ion concentration of a solution, namely the pH value, alters the charge of amino acids, which in turn interact with any charges on the R groups (of the substrate??). This leads to changes in the ionic charges and eventually causes ionic bonds to break. All enzymes have a 3D tertiary structure. The function of a protein or an enzyme depends on its particular characteristics, in particular the shape of the protein / enzyme, which is determined by its bonds. If the bonds are altered or broken, the shape of the protein / enzyme changes, the characteristics change and the protein / enzyme function changes. ...read more.

Middle

time due to the formula: Percentage error = error o X 100 exact value To overcome this, a smaller surface area should be used and therefore, I believe a suitable improvement is to use a cylinder of potato, 1cm in length. A number of discs in one test tube have the ability to overlap or become stuck on the inner surface of the test tube and therefore, decrease the surface area, which decreases the rate of reaction. The potato cylinders or discs will already be accurately cut with a pre-determined sized cork borer and therefore, have the same circumference length. Error may occur in the accuracy of the length of the potato cylinders and therefore, the mass of the cylinders will change. A higher mass of potato will increase the rate of reaction, as there will be a larger surface area and therefore, a higher enzyme concentration. Another modification would be to weigh the 1cm cylinders so that they are all the same mass. The most accurate method of calculating the weight of the potato cylinder would be by difference, i.e. the weight of the conical flask should be recorded, the potato chip added and the new weight recorded. To calculate the weight of the potato chip, the weight of the flask should be subtracted from the weight of the flask and potato chip. This eliminates any error. Another source of error due to the method is that osmosis will have occurred to a greater extent on the discs that remained in the water in the petri dish for a longer period of time causing the potato chips to have a larger mass, as the water molecules move from a higher to a lower, concentration of water. The 1mm discs have a much larger surface area to volume ratio, compared to a 1cm cylinder, therefore osmosis will be at a greater rate due to Fick's Law, which states that increasing the surface area, increases osmosis. ...read more.

Conclusion

Another possibility is that the scientist may have begun cutting the disc with the scalpel but decided this length was under 1mm, so cut a longer disc leaving a dent in the surface. Also, when transferring the discs to the test tube the scientist may have pinched the discs too tightly with the tweezers. These features would both increase the surface area, which would have caused an increase in the rate of reaction and a possible reason as to why an unusually high bubble count was recorded. An improvement would therefore be to use larger discs e.g. using a cylinder of 1cm in length. Therefore using only one disc, eliminates the error of miss-counting and overlapping. In addition, the percentage error of cutting 1mm length would decrease as if the measurement was 0.1mm in length this would cause a smaller percentage error on the 1cm cylinder than on the 1mm disc. Another random error may have occurred from rinsing the equipment. The water residue remaining in the test tube may have been sufficient to dilute the hydrogen peroxide buffered solution causing a decrease in rate of reaction in the two unusually low results. The anomaly in the experiment at pH5 with a bubble count of 80, may have been due to the equipment not being washed thoroughly enough causing the substrate and buffered solution to remain in the tube, increasing the hydrogen peroxide concentration which in turn increases the rate of reaction because there are more substrate molecules to be converted into products, producing a larger volume of oxygen. An improvement would be to ensure the equipment was washed with the same degree of thoroughness or to use new equipment for each experiment. However, a more economical way would be to rinse and thoroughly dry the equipment in order to remove remaining residue. In conclusion the anomalous results could have been caused by any number of combinations of these factors. I predict that the modifications and improvements I have suggested, would produce a more accurate and reliable set of results. ...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

    The Effect of pH on Catalase

    4 star(s)

    cm3 s-1 Rate is measured by: Volume of O2 produced Time The results show that the catalytic reaction is affected by the pH and the data that will be obtained in the full experiment will be able to show the effect of pH on the enzyme.

  2. Marked by a teacher

    effects of substrate concentration on the activity of the enzyme catalase.

    4 star(s)

    in a table which looks like this :- Percentage Concentration % H�O� Volume Of H�O�(cm�) Volume of Water (cm�) Volume Of Catalase (cm�) Time Taken for 5cm� Gas to be Produced(1st)

  1. To investigate the effect of enzyme concentration on the activity of catalase in potato ...

    Timing the amount of time it took to produce 5ml of oxygen was very accurate and does not need to be changed. We measures the (min: sec: 1/100th sec) and this is accurate enough it is dependant on your own reaction times.

  2. Diffusion in Agar Block

    Graph 1 and 2 show that Time Taken for decolourization is directly proportional to the Surface Area: Volume ratio BUT is inversely proportional to Volume: Surface Area ratio. Graph 1 shows that the block with the lowest Volume: Surface Area ratio takes the highest time to lose its colour.

  1. Influence of pH on the activity of potato Catalase

    This will affect the averages that is the product form the results, as in some cases there is only 1 result the average will not be entirely true as it have no other readings to compare against. Over the first pH groups, 3 to 7, the time taken shown on

  2. Factors affecting the activity of potato catalase on hydrogen peroxide.

    By mixing the concentration of H2O2 with water this will decrease the rate of reaction This is because the enzymes found in the potato chip which are the catalyse produce water and oxygen and as I mentioned in the formula above because water molecules are now in there as well,

  1. Report on a practical - Investigating catalase activity.

    This depression is called the active site. The reaction then takes place and the molecules of product then leave the active site, freeing it for another substrate molecule. The active site of a particular enzyme has a specific shape into which only one kind of substrate will fit.

  2. Investigate the effect of enzyme concentration on the activity of catalase in potato tuber ...

    Which means patterns and trends could be affected. This is because if the bubbles come up all at once then this will cause a faster rate of reaction. If the bubbles come up very slow then this will cause a slower rate of reaction. A way of improving this is to make sure that the collecting tube is

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