Internal Assessment on Enzymes

Topics: Enzyme, Standard deviation, Catalysis Pages: 10 (1509 words) Published: July 18, 2011
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Enzymes are catalysts which speed up a reaction by providing an alternative reaction pathway of lower activation energy. But they do not undergo permanent changes and so they remain unchanged even at the end of the reaction. Many enzymes consist of proteins and non-proteins called the cofactor. The intra- and intermolecular bonds that hold proteins in their secondary and tertiary structures are disrupted by changes in temperature and pH. This affects shapes and so the catalytic activity of an enzyme is pH and temperature sensitive.

The basic mechanism by which enzymes catalyze chemical reactions begins with the binding of the substrate (or substrates) to the active site on the enzyme. The active site is the specific region of the enzyme which combines with the substrate. The binding of the substrate to the enzyme causes changes in the distribution of electrons in the chemical bonds of the substrate and ultimately causes the reactions that lead to the formation of products. The products are released from the enzyme surface to regenerate the enzyme for another reaction cycle.


Just as only the correctly shaped key opens a particular lock, so does only one enzyme reacts with a substrate. It is shown in the diagram above.


In the graph above, we can see that the x shows the optimal temperature (this is the right temperature for an enzyme to act at it fastest rate). We can also see that as the graph is making a progress and after it reaches the point x is gradually declines.

In this experiment I will show that as the temperature increases, enzymes start to denature. They need a suitable optimum temperature to react and after a certain temperature the enzyme denatures and the graph slops downwards.


• Test tube - 1
• Measuring cylinder – 1
• Water
• Stop watch
• Weighing balance
• Beaker - 1
• Tripod stand - 1
• Test tube holder
• Thermometer - 1
• Burning splint - 1
• Filter papers – 3 (to wipe the potato pieces after washing them) • Potato (cylindrical shape. Using pipette) 1 piece weighing 10 grams • Fire gauze

• First we peal and cut the potato with the help of a knife • We wipe them with the help of filter paper and keep them aside for the time being. • Mean while we place a tripod stand (at the top of it we have a fire gauze) and burn it with the help of a burning splint • Then we place a beaker with 100 ml of distilled water in it and place a potato in it • And heat till the temperature rises to 10 oC and count the number of bubbles for the next 10 minutes • Do the same with 20 oC and count the number of bubbles produced in the next 10 minutes • Repeat the same with 30 oC, 40 oC, 50 oC, 60 oC and 70 oC and record the number of bubbles produced.

The effect of different temperature on the number of oxygen bubbles produced in a potato during enzymatic reaction.

|Temperature / 0C |Number of bubbles produced / minute | | |1 |

|18 |1.732 |

Student’s T-Test 1 (Temperature 10 oC and 70 oC)

Step 1

Formulate the Ho and H1 hypothesis.

Ho states: there is no difference between the numbers of bubbles produced / minute. This means that the mean should be equal ((1 = (2)

H1 states: there is a difference between the numbers of bubbles produced / minute. Step 2

| | |The statistics for Temperature 10 oC and 70 oC...
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