P10: Protein Denaturation I

• Introduction:

Denaturation is a process in which proteins or nucleic acids lose the quaternary, tertiary and secondary structure that is present in their native state. Denaturation is the result of the application of some external stress (heat and pH change) or compounds such as a strong acid or base, aconcentrated inorganic salt or organic solvent. In very few cases denaturation is reversible and proteins can recuperate their native state when the denaturing factor is removed. This process is called renaturation.

• Materials:

-2x250mL beaker.
-4 test tubes.
-Test tube rack.
-10 mL pipet.
-Knife.
-Glass marking pen.
-Potato.
-Distilled water.
-Hydrogen Peroxide.
-NaCl.
-HCl.

• Catalase Activity:

Catalase is a common enzye found in nearly all-living organisms exposed to oxygen. It catalyzes the decomposition of hydrogen peroxide (H₂O₂) to water and oxygen. It is a very important enzyme in protecting the cell from oxidative damage  and preveting the accumulation of hydrogen peroxide.

                                             2 H₂O_{2 --------------------->}   2 H₂O + O₂ 

Catalase is a tetramer of four polypeptide chains, ecah over 500 amino acids long. It contains four Porphyrin Heme groups (iron groups) that allow the enzyme yo react with the hydrogen peroxide. The optimum pH for human catalase is aprox. 7, in other organisms vary between 4 and 11. The organelle that stores catalase in eukaryotic cells is the peroxisome, which also contains peroxidases.

• Procedure:

In this experiment we aregoing to test the catalase activity in different enviroment situations. we are measures the rate of enzyme activity under varios conditions, such as different pH values and temperature. We will measure catalase activity by observing the oxygen gas bubbles when H₂O₂  is destroyed. If lots of bubbles are produced, it means the reaction is happening quickly and the catalaseenzyme is very active.

1. Prepare 30mL of H₂O₂  10% in a beaker (use a pipet).
2. Prepare 30mL of HCl 10% in a beaker.
3. Prepare 30mL of NaCl 50% in a beaker.
4. Peel a fresh potato tuber and cut the tissue in five cubes of  1cm³. Weigh them and equal the mass.
5. Label 5 test tubes (1,2,3,4,5).
6. Immerse 10 minutes your piece of potato inside HCl beaker.
7. Immerse 10 minutes another piece of potato NaOH beaker.
8. Boil another piece of potato.
9. With a mortar, mash up the third piece of potato.
10. Add 5mL H₂O₂  10% in each test tube.
11. With a glass-marking pen mark the height of the height of the bubbles.
12. Compare the results of the test 5 test tubes.

• Conclutions:

Mash potato has more activity because proteins are more exposed and react more.

The raw potato is the second having more activity because it has not had any previous alteration.

Potato with NaCl practically no present activity.

 Boiled potato and potato with HCl no present activity because the proteins were denatured and above.

-Completed the table below with the important parts of this experiment:

Parts:	In this experiment this was...
Independent variable	Tratament of each potato.
Dependent variable	The height of the bubbles.
Experimental Group(s)	Boiled, with HCl, with NaCl and mashed up potato.
Control Groups	Raw potato.
Constants	Weight, same amount of H2O2 , time...

-Represent your results in a chart: Treatment vs bubble's height.

• Question:

1. How did the temperature of the potato affect the activity of catalase?
2. How did the change of the pH of the potato affect the activity of catalase?
3. In which potato treatment was catalase the most active? Why do you think this was?