P10 CATALASA

INTRODUCTION

Catalase is an enzyme that has the function of destroying the peroxide of hydrogen, which is a product of biological oxidation, to be become in water and oxygen.

The peroxide of hydrogen protects us from microorganisms, like anaerobic microorganisms. But because of the peroxide of hydrogen is toxic, the catalase has to act.

OBJECTIVE

• To know the quantity of catalase that there is in meat, in this experiment a liver, or in a vegetable, in this case a potato. 
• To observe the reaction of catalase and what happens if we boil the meat and potato.

MATERIAL

• Peroxide of hydrogen ( H₂O₂)
• Distilled water
• Bunsen burner
• A piece of a liver and a piece of potato
• A beaker (250 mL)
• 4 beakers (100 mL)
• A knife, scissors
• Latex gloves 
• 2 Test tubes with cork stopper.

PROCEDURE
We have done three different experiments.

PROCEDURE OF THE FIRST

• We cut into pieces of liver and potato.
• We take a 250 mL Beaker and we add the pieces, then we put in distilled water until it recovers all the pieces.
• Finally, we boil the beaker in a Bunsen burner. We observe the reaction.

RESULTS:  It has been a reaction, with effervescence when the water has boiled. 

CONCLUSIONS: The catalase of this tissues has denaturated, because of the change in the temperature, it has left with a primary structure and with no function. 


PROCEDURE OF THE SECOND 

We take four 100mL Beakers and then we put inside:

• In the first: we put a piece of the liver with 5 mL of H₂O₂.
• In the second: we put a piece of the potato with 5 mL of H₂O₂.
• In the third: we put a piece of the boiled potato with 5 ml of H₂O₂.
• In the fourth: we put a piece of the boiled liver with 5 ml of H₂O₂.

RESULTS: in the beakers that have been boiled , there is no action by the catalase, no effervescence. 

CONCLUSIONS: in the tissues that have been denatured, they don't have a reaction with the oxygenate water, because the catalase has lost the function. 


PROCEDURE OF THE THIRD 

• ·   First we take two test tubes with a cork stopper, with two perforations  in each of them, one to put a thermometer and the other to let the oxygen pass. And we put them in a beaker.
• ·   We add in a tube a piece of potato and in the other a piece of the liver.
• ·   Finally, we pour distilled water.
• ·   We close the test tubes and we observe the reactions. 

RESULTS: in the test tube with a piece of liver, the temperature has increased 10 degrees, but in the test with a piece of a potato, the temperature has almost increased nothing. 

CONCLUSIONS : the tissues from the animals have more catalose than the vegetables. In the vegetables the catalose produces a little efervescence, but the exothermic reaction is not as in the animals. 

QUESTIONS 

1.    What happens when the tissues are boiled? 
The tissues will lose the quaternary, the tertiary and secondary structure,
so the proteins are left with a primary structure, without function. That is 
called denaturation and it’s caused because of changes in thetemperature,
in the pH...

2.    What happens with the temperature?

The reaction of catalase is exothermal. In animals the temperature in the previous experiment (the third), in the liver has increased ten degrees and in the potato almost nothing; when it acts with oxygenate water. 

3.    Why the peroxide of hydrogen (oxygenated water) is a good disinfectant?

The oxygenate water is used to cure wounds because is a strong oxidant; it acts on the microorganisms, as anaerobes or aerobics, it destroys the extern membranes of them, therefore, their existence. The oxygen gets free when it acts with organic material and it produces effervescence.

4.    Do a graphic with the results of the third experiment.

5. Which tissue has more catalase? The liver has more catalase
because it has produced more effervescence and temperature has
increased much more than in the potato, the exothermic reaction is
stronger too. 

 

P9 BIURET TEST

OBJECTIVES

We want to observe the quantity of proteins of eggs and milk, and of its different types and parts.

MATERIAL

·         Tube rack

·         5 test tubes

·         5 beakers 250 mL

·         Pipette

·         Distilled water

·         An egg

·         Milk: skimmed and whole milk

·         Soy juice

·         CuCo₄

·         NaOH

·         Weighing machine

·         Spatula

·         Glass stirring

·         Dropper

PROCEDURE

I will divide this experiment in two parts, to make it strictly.

PART 1

1.    We take 5 beakers and we indicate which substance we are going to add on it, and we put 100 mL of distilled water in every beaker.

2.    After that, we add 10 mL of one of those substances (soy milk, skimmed milk, whole milk, egg yolk, the egg white) in every beaker, with the help of a pipette.

3.    Then we remove it with the help of a glass stirring.

PART 2

1.    We take five test tubes and we indicate again which substance we are going to add on it, the content of each beaker.

2.    We add in every test tube 2 mL of one of the beakers, different in each tube.

3.    We put 2 mL of Na OH dissolution 20% (20 g of NaOH and 80 g of distilled water).

4.    Finally, we add 5 drops of CuCO4 1% dissolution (1 g of CuCO4 and 99 g of distilled water).

RESULTS  

If we observe the results are:

1.    Egg yolk. (colour dark purple)

2.    Soy juice. (colour purple)

3.    The white egg.  (colour dark blue)

4.    The whole milk.(colour blue)

5.    The skimmed milk. (colour blue- grey)

CONCLUSIONS

Biuret Test is used to reveal the quantity of proteins that a substance has, depending on the result colour.

The substance that has more quantity of proteins is which is close to the colour black, then the colour purple, blue. The substance with the less quantity of proteins is close to the colour grey. 

In the results is already arranged from which has more proteins to which has minus quantity of proteins.

P8 SOAP

P8 SOAP

OBJECTIVE

Understand the saponification reaction and produce soap.

MATERIAL

·         600mL Beaker

·         Glass road

·         Goggles

·         Gloves

·         Clock glass

·         Spatula

·         Bunsen burner

·         NaOH

·         Water (always distilled)

·         Oil

PROCEDURE

We have made three experiments, that they are differenced in the quantities of oil and distilled water.

First experiment

MATERIAL: 500 mL of oil, 183 g of distilled water, 63 g NaOH, 35 mL of an essence, spatula and an Erlenmeyer.

Procedure:  

1. Firstly, we have to weight out the necessary quantity of NaOH and then, we dissolve it with the distilled water.
2. Secondly, we add it in an Erlenmeyer and we mixed with oil.
3. Finally, we add the essence (it can be any essence).

Second experiment

MATERIAL: 50 mL of oil, 20 g NaOH, 80 g of distilled water.

1.     We make dissolution from NaOH and water. 
2.     From this mixture we only take 50 mL.
3.     Then we add the oil and we mix it. 
4.     Finally, we have to heat it. We can also add an essence.

Third experiment

MATERIAL:  20 g NaOH, 80 g distilled water, 2mL of oil.

Procedure:

1. We dissolve NaOH with distilled water.
2. In a test tube we have put 2mL of that dissolution and 2mL of Oil.
3. We have made two test tubes like this.
4. We can add an essence.
5. We hit it in a “Baño María” (Water Bath).

QUESTIONS 

• What is the soap?  Soaps are carboxylate salts with very long hydrocarbon chains.
• Which is the reaction that we have done? Soap can be made from the base hydrolysis of a fat or an oil. This hydrolysis is called saponification.
• Which capes we have produced? 2 layers. Above soap and the below layer is oil and glycerin that have’nt reacted yet.
• Formulate the reaction of it.

P6 STARCH

OBJECTIVES 

Reveal the presence of starch in different foods and in a plant and know why it appears in any of them.

MATERIAL 

·        A piece of Frankfurt

·        A piece of jam

·        A leaf

·        A piece of a potato

·        A watch glass

·        A knife

·        Wire gauze

·        A lab burner

·        Lugol.

·        A dropper.

·        Ethanol

PROCEDURE

THE LEAF’S PROCEDURE

·        We have to heat the leaf for 2 minutes.

·        Then we take the leaf and we put it inside a test tube with ethanol.

·        We come back to heat it for 10 minutes.

·        Then we bring it out from the tube and we put it in a watch glass.

·        We add it lugol.

THE FOOD’S PROCEDURE

·        We cut a part from a potato, from Frankfurt and from jam.

·        Then we put them all in a different watch glass.

·        And we add lugol to reveal the starch presence.

RESULTS: WHE WE ADD IODINE SOLUTION TO THE FOOD SAMPLES (POTATOE, FRANKFURT SAUSAGE AND JAM), POTATOE AND FRANKFURT SAUSAGE STAIN TO DARK PURPLE OR BLACK. IT DOESN’T HAPENS WITH JAM.

CONCLUSION: IODINE SOLUTION IS AN INDICATOR OF STARCH PRESENCE. POTATOE AND FRANKFURT SAUSAGE CONTAIN STARCH.

THE PRESENCE OF STARCH IS AN EVIDENCE THAT PHOTOSYNTESIS HAS OCCURRED. GLUCOSE, THE PRODUCTE OF IT, HAS CONVERTED TO STARCH. 

QUESTIONS

Which is the origin of the starch that you can see in the leaf?  Photosyntesis. 
Explain the significance of boiling the leaf in water? Interrupt the photosyntesis.

Explain the significance of boiling the leaf in ethanol? Extract the clorophyll. 
Explain the significane of rinsing the leaf in water? To eliminate any remains of colorophyll.

P7 LIPIDS PROPERTIES

OBJECTIVES

Test the solubility of lipids. Identify lipids in liquid compounds and understand what are an emulsion and the effect of detergents.

MATERIAL

·         Test tube rack

·         250 mL beaker

·         Water

·         6 test tubes.

·         Cellulose paper

·         Dropper

·         Scissors

·         Glass rod

·         Olive oil

·         Soap (detergent)

·         Milk with different fat content: full-cream, semi-skimmed and skimmed milk.

·         Soy juice

·         Petroleum ether

·         Ethanol

·         Sudan III

PROCEDURE

We have made four different experiments to identify lipids.

Solubility of lipids:

1. Clean and dry three test tubes. Write on the first: W (water); in the second: Ethanol (E); and in the third: PE (petroleum ether).
2. Add three drops of acid oleic to the three test tubes.
3. Add 1 ml of water in the tube W; add 1 ml of ethanol in the tube E and 1ml of petroleum ether in the PE.
4. Shake each test tube and record the solubility of each tube.

RESULTS:

WATER+OIL------- INSOLUBLE

ETER+OIL   --------- SOLUBLE

PE+ OIL------------ SOLUBLE

CONCLUSION: Lipids are insolubles in wàter but are soluble in apolar orgànic dissolvents.

Translucent mark:

1.  Cut to pieces (10cmx10cm) of cellulose paper.
2.  Put one drop of water in the first piece, you will see a translucent spot, wait and observe what is happening.
3.  Put one drop of olive oil in the other, you will see a translucent spot. Is that going to disappear, why?

 CONCLUSION: Fats will leave a glossy, oily residue when wiped on paper. When light falls upon paper, a part of it is transmitted, a part is scattered, a part is absorbed, and a part is reflected

Sudan III dye

1. Take the W test tub and add two drops of Sudan III.
2. Prepare four test tubes: 3 with milk, each of different fat content and the fourth with soy juice, observe the results.

RESULTS:

1-      SKIMMED MILK: The solution stain soft orange

2-      Semi-skimmed milk: The solution stain medium orange

3-      Whole milk: The solution stain reddish- Orange

4-      Soy juice: It doesn’t change the colour.

•          CONCLUSION: Sudan III is not soluble in water; it is, however, soluble in lipids. If lipids are present the Sudan III will stain them reddish-orange.

EMULSION

1.  Take the beaker of 250 mL and put 100mL of water.
2.  Add 1 mL of oil.
3.  With a glass rod stir the mixture and left it for 5 minutes. Observe what is happening.
4.  Add two drops of soap and then notice the different of both mixtures.

1. RESULTS:
   WATER AND OIL FORM A TEMPORARY EMULSION
   WATER , SOAP AND OIL FORM A PERMANENT EMULSION
   CONCLUSION:  SOAP IS A TENSOACTIVE THAT ELIMINATE THE SURFACE TENSION BETWEEN THE WATER MOLECULES, BY ONE HAND, AND THE SURFACE TENSION OF OIL. THEN THE MIXTURE IS HOMOGENEUS (PERMANENT EMULSION).