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Homework for lab 4: Complete exercise 4 page 76-77-78 (question 6 only for page 78). Design your own experiment to examine th


5. Design a table to collect your data and write a brief procedure. Keep in mind that someone else should be able to come alo
. . EXERCISE 5: CONDUCTING YOUR OWN EXPERIMENT Objectives Perform the experiment your team designed. Present and interpret th
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HYPOTHESIS

Enzymes possess a defined three-dimensional structure. Any change in this structure causes a change in the enzyme’s activity. The pH of the reaction mixture modifies this structure and therefore, the activity. Every enzyme has an optimum pH where it shows maximum activity. Significant differences from this pH cause changes in the enzyme’s three-dimensional structure that reduce its activity. Catechol oxidase enzyme has an optimum pH of 7.

Independent Variable: pH range

pH5: 5mL ± 0.5mL

pH7: 5mL ± 0.5mL

pH9: 5mL ± 0.5mL

Dependent Variable:

Enzyme activity and reaction rate depends on the intensity of the colour produced (visible reaction)

Controlled Variables:

Source of enzyme: The potatoes are relatively the same size, same concentration

Temperature of variables: the temperature of the potato, pH, and all other variables and materials will remain at a constant room temperature of 25°C.

Amount of enzyme (potato): 10 drops of the potato solution will be used for each test tube

Amount of pH: 5mL of the selected pH will be used for each test tube

Extraction of Catechol Oxidase

Catechol oxidase may be extracted from bananas or potatoes. If you use potatoes, peel and chunk them, then blend at a high speed using 700 ml of cold, distilled water. Filter this potato juice through cheesecloth and refrigerate.

Experimental Details

We have buffer solutions with pH values 5,7 and 9. Label three test tubes with these pH values. Fill each tube to one-fourth capacity with the respective buffer. Add 10 drops of catechol oxidase extract into each of these tubes, followed by 10 drops of catechol. Shake the tubes and note down the color of each tube on a scale from 0 to 5 where 0 indicates no color and 5 indicates a dark, brown color. Continue shaking the tubes and note the color every five minutes for the next 20 minutes.

Result Interpretation (OUTCOME EXPECTED)

Use the data you obtain for the 20 minute reading to plot a graph. On the X-axis, indicate the pH of the buffers. On the Y-axis, indicate the color intensities ranging from 0 to 5. For every pH value, mark the color intensity and join these markings to obtain the final graph. Locate the peak of this plot and identify the optimum pH for the catechol oxidase reaction. Provided you have performed the experiment correctly, the optimum pH value will be 7. At pH 7, the enzyme is most active and rapidly catalyzes the oxidation of catechol to give the dark, brown color.

pH solution INTENSITY OF COLOUR PRODUCED INTENSITY RANGE
5 light, brown colour 2
7 dark, brown color 5
9 slightly dark brown colour 4

8 optimum pH 6 intensity range 4 2 3 5 7 9 pH of the buffer solution

Reason for the result: Catechol is oxidized by catechol oxidase in the presence of oxygen to form benzoquinone, which, on exposure to air, forms melanin. This enzyme is also known by other names, such as tyrosinase, diphenol oxidase and polyphenol oxidase. Potatoes, apples and bananas contain catechol oxidase which acts on colorless catechol and converts it to brown-colored melanin. The browning that occurs when you cut and expose these items to air is a result of this reaction.

Precautions

keep the catechol oxidase extract on a block of ice before you add it into each tube. Catechol is poisonous, so ensure you don’t allow it to contact your skin. Don’t pipette catechol solutions and if there is a spillage, wear gloves to clean it using paper towels.

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