Saturday, 31 January 2015

P12: DNA Extraction.

  • Introduction:
Deoxyribonucleic acid (DNA)  is a nuclic acid that encodes the genetic instructions used in the development and fuctioning of all known living organsims and many viruses.
Nucleic acids are biopolymers formed by simple units called nucleotids. Each nucleotide is composed of a nitrogen-containing nucleobase (G, T, C, A) as well as a monosaccharide (deoxyribose) and a phosphate group.
Most DNA molecules consist of two strands coiled around each other to form a double helix. The two strands run in opposite directions to each other and are therefore anti-parallel. Moreover the bases of the two opposite strands unit according to base pairing rules : A-T and G-C.
  • Materials:
  1. 1L Erlenmeyer flask.
  2. 100mL beaker.
  3. 10mL graduated cylinder.
  4. Small funnel.
  5. Glass strining rod.
  6. 10mL pipet.
  7. Knife.
  8. Safety goggles.
  9. Cheesecloth.
  10. Kiwi.
  11. Pineaple juice (1mL/5mL).
  12. Distilled water.
  13. 90% Ethanol ice-cold.
  14. 7mL DNA buffer.
  15. 50mL dish soap.
  16. 15g NaCl.
  17. 900mL tap water.
  • Procedure:
Put the ethanol in the freezer-you will need it really cold later.
Prepare the buffer in a 0,5L beaker: Add 450mL of tap water, 25mL of dish soap and 7g NaCl. Stir the mixture.
  1. Pell the kiwi and chop it to small pieces. Place the pieces of the kiwi in one 600mL beaker and smash with a fork until it becomes a juice puree.
  2. Add 8mL of buffer to the mortar.
  3. Mash the kiwi puree carefully for 1 minutewithout creating many bubbles.
  4. Filter the mixture: put the funnel on top of the graduated cylinder. Place the cheesecloth on top of the funnel. 
  5. Add beaker contain carefully on top of the cheesecloth to fill the graduated cylinder. The juice will drain through the cheesecloth but the chucks of kiwi will not pass through into the graduated cylinder.
  6. Add the pineaple juice to the green juice (you will need about 1mL of pineaple juice to 5mL of the green mixture DNA solution). This step will help us to obtain a purer solution DNA. Pineaple juice contains an enzyme that breaks down proteins.
  7. Tilt the graduated cylinder and pour in an equal amount of ethanol with an automatic pipet. Put the ethanol through the sides of the graduated cylinder very carefully.You will need about equal volumes of DNA solution to ethanol.
  8. Place the graduated cylinder so that it is eye level. Using the stirring rod, collect DNA at the boundary of ethanol and kiwi juice. Do not stir the kiwi juice; only stir in the above ethanol layer!!
  9. The DNA Precipitate looks like long, white and thin fibers.
  10. Gently remove the stirring rod and examine what DNA looks like.
  • Solution: 



  • Questions:
  1. What did the DNA looks like? The DNA looks like long, small white and thin fibers.
  2. Why do you mash the kiwi? Where it is located inside the cells? Because you want to liberate the DNA that is located inside the nucleus.
  3. Explain what is the function of every compund of the buffer (soap ans salt) The salt breaks the nucleus and the cell and the soap takes away the proteins.
  4. DNA is soluble in water, but not in ethanol. What does this fact have to do with our method of extraction? This means that we can only see the DNA in the part of the ethanol beucause if it touches the water it will dissolve.

Sunday, 25 January 2015

P11: Cytochrome C Comparison Lab

  • Introduction:
To compare the relatedness between organisms by examinig the amino acid sequence in the protein, Cytochrome C.
Cytochrome C is associated with the inmer membrane of the mitochondrion. It is a small protein from eucariote cell. It is a Hemeprotein; iron group.
Function: produce energy, is a part of a electron transport chain (ATP).
  • Procedure:
We compare the mammals: horse, donkey, whale; beardds: penguin, chicken; insect: moth and fungui: wheat, yeast.
  • Conclutions:



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 H2O2 --------------------->   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?




Sunday, 11 January 2015

SEVENTH EXPERIMENT: Protein Identification

  • Introduction:
Biuret's test is a chemical test used for detecting the presence of peptide bonds. Polypeptides as proteins, are chains of amino acids link together by peptide bonds.
The biuret reaction can be used to assess the concentration of proteins because peptide bonds occurs with the same  frequency per amino acid peptide.
  • Materials:
-7 × 250mL beaker.
-6 test tubes.
-Test tube rack.
-6 × 10mL Pipet.
-Mortar.
-Glass marking pen.
-Gloves.
-Goggles.
-Milk.
-Rice Milk.
-Egg.
-Yogurt.
-Potato.
-Distilled water.
-NaOH 20%.
-10 drops of CuSO4.
  • Procedure:
We are going to determine protein protein concentration in some foods: milk, rice milk, yogurt, potato and egg. Be careful, aa we are going to test separately the egg white and the yolk. Put gloves and use goggles, remember that NaOH is caustic!!
First of all we are going to dilute the protein.
1.  Add 100mL of distilled water to each 250mL beaker. Label them with M(milk), R(rice milk) and EW(egg white), EY(yolk) and P(potato).
2. Separate the egg white and the yolk in another beaker.
3. Smash the potato in a mortar and add some amount of the of the smashed potato to the P beaker.
Prepare the samples:
4. Add 10mL of a dispersions of each food (M, S, EY, EW and Y) to the indicate beaker. Calculate the final concentration. All the groups will use the same dispersion from the beakers.
5. Prepare 6 test tubes (clean and dry) and label M, S, EW, EY, Y and P. Add 2mL of the every food dilution of each beaker.
6. Add 2mL of 20%NaOH dissolution.
7. Shake gently and add 5 drops of CuSO4 in each tube. Allow the mixture to stand for 5 minuts.
8. Note any colour change. Remember that proteins will turn solution pink or purple.
9. Compare the test tubes.
  • Questions:
1. Wich food has proteins? Egg white, milk, yogurt, potato.
2. Wich food has more proteins? Why? Egg white, because they are animal food and is rich in proteins.
3. Do you find any difference between soy rice and cow milk? Yes, rices milk doesn't have proteins, however cow milk does.
4. Is there any difference among milk and yogurt? Why? Yes, because yogurt has more proteins than milk.

Tuesday, 2 December 2014

SIXTH EXPERIMENT: Saponification.

  • Materials:
  1. 900 mL beaker.
  2. Spoon.
  3. Clock Glass.
  4. Bascule.
  5. Rod.
  6. 32 g NaOH.
  7. 90 mL H2O.
  8. 270 mL recycled oil.
  • Procedure:
We take 32 grams of NaOH in 90 mL water and mix to dissolve you, then let throwing the oil slowly while saponification happens.

  • Results:

CH3-(CH2)7-CH=CH-(CH2)7-COOH + NaOH --->  CH3-(CH2)7-CH=CH-(CH2)7-COONa + H2O





Monday, 1 December 2014

FIFTH EXPERIMENT: Lipids Properties

  • Introduction:
These compounds are characterized as natural substances taht do not mix with water but dissolve in organic solvents.
Lipids are made in general of a long chains of hydrocarbons with relatively little oxygen.As a result of this, they tend to be non-polar, meaning they do not dissolve in polar solvents such as water.
  • Materials:
  1. Test tube rack.
  2. 250ml beaker.
  3. Water.
  4. 6 test tube.
  5. Cellulose paper.
  6. Dropper.
  7. Scissors.
  8. Glass rod.
  9. Olive oil.
  10. Soap (detergent)
  11. Milk with different fat content: Full-cream, semi-skimmed and skimmed milk.
  12. Petroleum ether.
  13. Ethanol.
  14. Sudan III.
  • Procedure:
Solubility of some lipids:
  1. Clean and dry three test tube. Label as W (water), ethanol (E) and PE (ether)
  2. Add 3 drops of oleic acid to 3 small test tubes.
  3. Add 1 mL of water in teh first test tube (W).
  4. Add 1 mL of ethanol in the second test tube (E).
  5. Add 1 mL of petroleum ether in the third test tube (PE).
  6. Shake carefully each test tube and record solubility and observations in your worksheet.
 
 
  • Results:
The oil and ethanol don't dissol in water and ether can dissol in water.

Lipids identification:
      Translucent Mark:
  1. Cut two pieces (10x10cm) of cellulose paper.
  2. Put 1 drop of water in the first squared piece. You will see a transluced spot. Wait for a while and observe what is happening.
  3. Put 1 Drop of olive oil in the second suqred piece of cellulose papaer. You will see a transluced spot. Wait for while and observe what is happening. Has the spot dissapered? Why?
  • Results:
The oil is transluced because doesn't evaporate, water can evaporate.

Sudann III dye: Be careful Sudan red cxan stain clothes!
Sudan III is a red fat-soluble dye that is utilized in the identification of the presence of lipids, triglycerids and lipoproteins in liquids.
  1. Take the W test tube of the first experiment and add 2 drops of Sudan III.
  2. Prepare four test tubes: 3 with milk with different fat content (M1, M2, M3) and soda (S). Add two drops of Suddan III and observe the results.
  • Results:
The proof of Sudan III doesn't work possibly because the solution was too diluted.

Permament emulsion:
  1. Take a 250 mL beaker and put 100 mL water.
  2. Add 1  mL of olive oil. With a glass rod stir the mixture vigorously and let it stand for a few minutes.
  3. Make note of what is happening.
  4. Add 2 drops of soap and stir the mixture again. Let it stand for a few minutes and notice the differences between both mixtures.












  • Results:
The oil is translucent because it doesn't evaporate, however the water if it evaporates, leaving different stains.
  • Questions:
  1. From your observation, wich compunds can dissolev lipids? Ether
  2. Do the oil and water mix? What can you conclude about the polarity of the oil if you know that water is polar? No, the oil non-polar.
  3. Why is liquid the olive oil at room temperature? And why not the lard? Because is the faty acid insaturated. The lard is a faty acid saturated.
  4. Why does the lipid leave a translucent spot on paper? Because the lipid does not evaporate.
  5. Wich type of milk contains more lipids? Why? Milk full-cream, because they take away the semi-skimmed milk fats.
  6. Did the oil and water mix when you added the soap?No, but a monolayer and a micelles separated oil from water formed.
  7. What did the soap do to the fat? Separated in micelles.
  8. Can you think about process and locations were compounds like the soap would be important to an animal? Acids biliars.

    

Sunday, 2 November 2014

FOURTH EXPERIMENT: Saccharides Properties


  • Introduction:
Saccharides are organic molecules consisting of C, H and O atoms. Usually the empirical formula is CnH2nOn . Are divided into three groups:

-Monosaccharides: formed by a linear carbon chain, are the building blocks of oligo and polysaccharides.
-Oligosaccharides: small polymer containing between 2 and 10 monosaccharides.
-Ploysaccharides: big polymers with more than 10 monosaccharides.
Saccharides yield 4,2 Kcal/gr.

  • Materials:
  1. Test tube rack.
  2. 10 ml pipet.
  3. Water.
  4. 5 test tubes.
  5. 1 dropper.
  6. 1 Spatula.
  7. Lactose.
  8. Maltose.
  9. Glucose.
  10. Sucrose.
  11. Starch.
  12. Lugol's iodine.
  13. Distilled Water.
  • Procedure:
Complete the properties' table that you have below from the 5 saccharides you have at the lab:


Glucose (G)
Maltose (M)
Sucrose (SU)
Lactose (L)
Starch (S)
Flavour
(sweet/no sweet)
sweet
sweet
Very sweet
No sweet
No sweet
Cystalline from
crystalline
no cystalline
cystalline
no crystalline
no crystalline
Colour
(white/cream)
white
white
white
white
cream
Solubility
(soluble/insoluble)
soluble
soluble
soluble
soluble
Insoluble
Lugol's iodine (+/-)
-
-
-
-
+

                                            
1.      In the first part of the experiment we are going to test some physical propierties of the saccharides you have in the lab: flavour, crystal structure and colour. How?
a)      Flavour: put  a small amount of each saccharide in your han and taste it! Is it sweet or not?
b)      Crystals: Observe a small amount of each  saccharide on a clock glass under magnification.
c)      Colour: White, trasparent or creamy.
To test solubility:
2.      Clean and  dry 5 test tubes and label them  "G, M, L, SU, S".
3.      Put 5ml of water in each test tube.
4.      With the aid of a spatula, put a small amount of eavh saccharide inside the labelled test tube and test tube if they are soluble or insoluble.
5.      Observe if each saccharide forms a mixture called dissolution or a colloidal suspension.

Lugol's iodine Test:

6.      Finally, add 2 drops of Lugol's iodine to each test  tube and test if the reaction is positive or negative. Lugol's is a solutions of elemental iodine (I) and potassium iodine (KI) in waterr that is use to test a saccharide. The reaction is positive when iodine reacts by turning from yellow to a purple, dark-blue/black colour.
Finally we are going to test if some foods contain starch. Try with potato!
  1. Add 2 or 3 drops of lugol's solution to one piece of potato and observe what is happening.
  • Questions:
  1. Write the empirical formula of each saccharide that you have use. Show structures of the five saccharides. Classify each in one group: mono, oligo or polysaccharide.
    Glucose (mono): C6H12O6
    Maltose (disaccharide): C12H22O11 · H2O
    Sucrose (disaccharide): C12H22O11 · H2O
    Lactose (disaccharide): C12H22O11 · H2O
    Starch (polysaccharide): (C6H10O5)n
  2. Which of the monosaccharides are aldoses and wich ketoses? Aldoses: glucose, maltose, lactose, starch. Ketoses: sucrose.
  3. Which bond links monosaccharides? Oglicocidic.
  4. Which saccharide/s is/are insoluble? Is this property related to the structure of the molecule? Glucose, Maltose and Sucrose. Polysaccharides.
  5. Which saccharide/s is/are insoluble? Is this property related to the structure of the molecule? Starch, high molecular density.
  6. Wich saccharide has reacted with Lugol's iodine solution? Starch.
  7. Which kind of food contains starch? Flour, cereal, pasta, rice, legums, etc.
  8. Caculate the energy that comes from the saccharides. 23gr saccharides · 4,2 Kcal/gr = 96,6 Kcal.





Fheling's test: reducing sugar.
  • Introduction:
Fehling's solution is a chemical test used to different between reducing and non-reducing sugars. This test is based on the reaction of a functional group of sugar molecules with Fehling's reagent.
Fehling's A: is a blue aqueous solution of copper (II) sulphate.
Fehling's B: clear and colourless solution of potassium sodium tartrate and sodium hydroxide.
  • Procedure:
In this experiment you will first determine with sugars give a positive test Fehling's reagent and then, by testing the reaction of some organic molecules containing onlya single functional group, you should be able to deduce which functional group of sugar is reacting with Fehling's reagent:
  1. Take 5 test tubes and label: G, M, S, L, ST.
  2. Put 2 mL of distilled water inside each tuve.
  3. With different spatulas put a small amount of each sugar. Dissolve the sugar.
  4. Add 2mL of Fehling's A solution and then Fehling's B.
  5. Place each test-tuve in a boiling water bath (250mLbeaker on a hotplate stirrer).
  6. Observe what is happening.
Starch Hydrolysis:
Hydrolysis is the reaction of a compound with water. As you know, starch is a polymer, consisting of many units of α-D-glucose covalently linked together.
  1. Place 2mL of 1% starch in a test tuve and add 0.5mL of 3M HCl. Mix and place this mixture in a boiling water bath for 10 minutes.
  2. After 10 minutes, remove the tuve from the water bath and let it cool. Neutralize this solution with 1M NaOH and mix well.
  3. Transfer 8-10 drops of this solution to a small test tuve.
  4. Add 1mL of Fehling's A solution and 1mL of Fehling's B.
  5. Heat for a few minutes in a boiling water bath.
  6. Record your observations. Compare the results of this test with your results for unhydrolyzed starch in the step 1 of this experiment.
  7. You can test the absence of starch with iodine solution too!!
  •  Questions:
  1. From your observations and the structures of the sguras given above, indicate wich functional group in the sugar molecules reacts with Fehling's reagent. Glucose, Fructose, Maltose, Lactose
  2.  Compare the results you obtained for the Fehling's test of the starch and Fehling's test hidrolyzed starch. Explain results. The starch react and becomes blue, because it react with lugol's iodine.
  3. Would have you obyined a Fehling's positive test if you had hydrolyzed the sucrose (as you have done with starch9? Why? The sucrose has no free OH.
  4. What does "reducing suars" term mean? Are reacting positively reduces teh Fehling's test. Because it has a fre OH.