L.15 ANIMAL CELLS vs PLANT CELLS

INTRODUCTION:

We compared the animal cells and the plant cells. Animal cells are similar to plant cells in that they are both eukaryotic cells and have similar organelles. Animal cells are generally smaller than plant cells. While animal cells come in various sizes and tend to have irregular shapes, plant cells are more similar in size and are typically rectangular or cube shaped. A plant cell also contains structures not found in an animal cell. Some of these include a cell wall, a large vacuole, and plastids. Plastids, such as chloroplsts, assist in storing and harvesting needed substances for the plant. Animal cells also contain structures such as centrioles, lysosomes, cilia, and flagella that are not typically found in plant cells.

Our Objectives:

- Identify the major components of cells. 

- Differentiate between animal and plant cells.

- Measure dimensions of the entire cell and the nucleus.

MATERIALS:

- Toothpick
- 2 Slides
- 2 Cover slips
- Distilled water
- Methylene blue
- Iodine ( safranin)
- Onion
- Glycerine
- Watch glass
- Droper
- Needle
- Brush
- Cellulose paper
- Microscope

PROCEDURE:

- Plant cells observation: 
1. Pour some destilled water into a watch glass.
2. Peel of the leaf from half a piece of onion and using forceps, pull out a piece of transparent onion peel (epidermis) from the leaf.
3. Put the epidermisin the watch glass containing distilled water.
4. Take a few drops of safranin in a droppper and transfer into another watch glass.
5. Using a brush (or a needle), transfer the peel  into the watch glass containing the dye. Let this remain in the safranin solution for 30 seconds, so that the peel is stained.
6. Take the peel from the odine solution and place it in the watch glass containing distilled water.
7. Take a few drops of glycerine in a dropper and pour 2-3 drops at the center of a dry glass slide.
8. Using the brush, place the peel onto the slide containing glycerine.
9. Take a coverslip and place it gently on the peel with tha aid of a needle.
10. Remove the extra glycerine using cellulose paper.
11. View it in the microscope.

- Cheek cells observation:
1. Gently scrape the inner side of the cheek using a toothpick, which will collect some cheek cells.
2. Place the cells on a glass slide that has water on it.
3. Mix the water and the cheek cells using a needle and spread them.
4. Dry the sample under the light to fix the sample on the slide.
5. Take a few drops of methylene blue solution using a dropper and add this to the mixture on the slide.
6. After 2-3 minutes remove any excess water and stain from the slide using cellulose paper.
7. Take clean cover slip and lower it carefully on the mixture with the aid of a needle.
8. Using the top of the needle, press the cover slip gently to the spread the epithelial cells.
9. Remove any extra liquid around the cover slip using cellulose paper.

RESULTS AND CONCLUSIONS:

Plant cells: 


MR= MA/ NA
cell:
MR= 6,9/400= 0'017cm
0'17cmx10000= 172'5 microns

nucleous: 
400= 0'7x10000/X
X= 7000/400= 17'5 microns



 Eduard cheek cells:

400= 1'5x10000 microns/ X 
X= 1'5x10000 microns/ 400= 37'5

400= 0'3x10000 microns/ X
X=  0'3x10000microns/400= 7'5 microns

L.12 DNA EXTRACTION

INTRODUCTION:

Desoxyribonucleic acid (DNA)  is a nucleic acid that encodes the genetic instructions used in the development and functioning of all known living organisms and many viruses.

Nucleic acids are biopolymers formed by simple units called nucleotides. Each nucleotide is composed of a nitrogen-containing nuclease (G, T, C, A) as well as a monosaccharide (desoxyribose) and a phosphate group.

This nucleotides are joined to one another in a chain by covalent bonds between the sugar of the nucleotide and the phosphate of the next.

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.

Within cells, DNA is organized into structures called chromosomes.

MATERIALS:

- 1L Erlenmeyer flask.

- 100mL beaker.

- 10mL graduated cylinder.

- Small funnel.

- Glass stirring rod.

- 10mL pipet.

- Knife.

- Safety goggles.

- Cheesecloth.

- Kiwi.

- Pineapple juice (1mL/5mL).

- Distilled water.

- 90% Ethanol ice-cold.

- 7mL DNA buffer.

- 50mL dish soap.

- 15g NaCl.

- 900mL tap water.

PROCEDURE:

Put the ethanol in the freezer, you will need it really cold later!

Prepare the buffer in 0,5L beaker: Add 450mL of tap water, 25mL of dish soap and 7g NaCl. Stir the mixture.

1. Peel 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 pure.

2. Add 8mL of buffer to the mortar. 

3. Mash the kiwi puree carefully for 1 minute without 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 cheesecloth to fill the graduated cylinder. The juice will drain through the cheesecloth but the chucks of kiwi will not pass through in to the graduated cylinder.

6. Add the pineapple juice to the green juice ( you will need about 1mL of pineapple juice to 5mL of the green mixture DNA solution). This step will help us to obtain a purer solution of DNA . Pineapple juice contains an enzyme that breaks down the 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 the ethanol and kiwi juice; only the 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 the DNA looks like. 

QUESTIONS:

1.- What did the DNA looks like?
The DNA looks like white  and thin fibers.

2.- Why do you mash the kiwi? Where it's located inside the cells?
The DNA is located in the nucleus and we mashed it to liberate it. 

3.- Explain what is the function of every compound of the buffer (soap and salt). 
The salt can breaks the nucleus cell and we put soap to take 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?
We can see the DNA in the part of ethanol because, if we touch the water the DNA can dissolve.

L.11 CYTOCROM C COMPARISION LAB

INTRODUCTION:

Genes are made of DNA and are inherited from parent to offspring. Soma DNA sequences code form RNA which, in turn, codes for the amino acid sequence of proteins. Cytochrome C is a protein involved in using energy in the cell. Cytochrome C is found in most, if not all, known eukaryotes. Over time, random mutations in the DNA sequence occur. As a result, the amino acid sequence of Cytochrome C also changes. Cells without usable Cytochrome C are unlikely to survive. 

Cytochrome C is associated with the inter membrane of the mitochondrion. It is a small protein from eucaryote cell.  

the function is to  produce energy, is a part of a electron transport chain (ATP).

Our propose is to compare the relatedness between organism by examining the amino acids sequence in the protein, Cytochrome C. 

PROCEDURE:

- Compare the amino acid sequence of Cytochrome C in various organism. 

- Make a branching tree, or cladogram. 

RESULTS AND CONCLUSIONS:

QUESTIONS:

1.- How many Cytochrome C amino acid sequence differences are there between chickens and turkeys?

0 differences.

2.- Make a branching tree for chikens, penguins, and turkeys. 

3.- a) Predict the number of Cytochrome C amino acid sequence differences you would expect to see between 

- horse and zebra: 1 or 2

- donkey and zebra: 1 or 2

b) what other information did you use to make this prediction? 

They can reproduce, the offspring is fertile, comparing the organs or anatomic proofs, comparing embryos.

4.- Explain why more closely related organism have more similar Cytochrome C.

Not so long ago that there have been separated and many mutations. There are fewer differences in DNA.

5.- Other data including other genes, suggests that fungi are more closely related to animals than plants. What are some reasons that the Cytochrome C data suggests that fungi, plants, and animals are equally distantly related?

If you have more than 40, suffered a lot of mutations and can not be compared. If Compares other proteins, there are other changes and we can not draw more conclusions.

L.10 PROTEIN DENATURATION 1

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. 

If proteins in a living cell are denatured, this results in disruption of cell activity and possibly cell death.

Denatured proteins can exhibit a wide range of characteristics, from loss of solubility to communal aggregation. This last effect results from the bonding of the hydrophobic proteins to reduce the total area exposed to water.  

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.

POCEDURE:

- 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. Prepare 5 test tubes:

1.- Raw potato

2.- Boiled potato

3.- Potato with HCl 

4.- Potato with NaCl

5.- Mashed up potato

1. Add 5mL H₂O₂  10% in each test tube.
2. With a glass-marking pen mark the height of the height of the bubbles.
3. Compare the results of the test 5 test tubes.

CONCLUTIONS:

Important parts of the experiment:

Parts:	In this experiment this was...
Independent variable	Tratament of each potato: temperature, pH...
Dependent variable	The height of the bubbles.
Experimental Group(s)	Boiled, with HCl, with NaCl and mashed up potato.
Control Groups	Raw potato. (nº 1)
Constants	Weight, same amount of H2O2 , time...

L9. PROTEIN IDENTIFICATION

INTRODUCTION:

Biuret's test is a chemical test used for detecting the presence of peptide bonds. ( covalent bond formed between two peptides when the carboxyl group of the first one reacts with amino group of the second peptide). Polypeptides as proteins, are chains of amino acids link together by peptides bonds. 

A peptide bond can be broken by hydrolysis (the adding of water). In organims, protein molecules called enzymes facilitate the process.

The biuret reaction can be used to assess the concentration of proteins because peptide bonds accur with the same frequency per amino acid peptide. The intensity of the colour is directly proportional to the protein concentration. 

The solution to be tested in treated with a strong base followed by a few drops of copper (II) sulphate. If the solution turns purple, protein is present. Only peptides with a chain of at least 3 amino acids (two peptide bonds) give a significant measurable colour shift with these reagents.

Proteins in the alkaline environment reduce Cu2+ to Cu+, which forms a coordination complex with proteins, leading to a blue to pink-purple colour change. 

Our objectives:  

- Identify peptides bonds.

- Compare protein concentration in different foods.

MATERIALS:

- 7x250mL

- 6 test tubes

- test tube rack

- 6x10mL 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 concentration in some foods: milk, rice milk, yogurt, potato and egg. Be careful, as we are going to test separately the egg white and the yolk. 

First of all we are going to dilute the protein.

1. Add 100mL of distilled water to each 250mL beaker. Label them with (M, RM, EW, EY, Y, P).

2. Separate the egg white and the yolk in another beaker.

3. Smash the potato in a mortar and add some amount of the smashed potato to the P beaker. 

Prepare the samples: 

1. Add 10mL of a dispersion of each food (M, RM, EW, EY, Y, P) to the indicate beaker. Calculate the final concentration. All the groups will use the same dispersion from the beakers.

2. Prepare 6 test tubes and label them. Add 2mL of each test tube of the every food dilution of each beaker.

3. Add 2mL of 20% NaOH dissolution.

4. Shake gently and add 5 drops of CuSO4 in each tube. Allow the mixture to stand for 5 minutes.

5. Note any colour change. Remember that proteins will turn solution pink or purple.

6. Compare the test tubes. 

RESULTS AND CONCLUSIONS:

Milk: positive
Rice milk: negative
Egg white: positive
Egg yolk: negative
Yogurt: positive
Potato: positive --> It had starch so it had to be negative. 
The positive result, are the foods of animals, because all they have proteins. Rice milk have starch like potato so they are negative. 
We ordered from the more concentration to the less concentration thanks to the colour. 
1. Egg white.
2. Potato and yogurt.
3. Milk.

QUESTIONS:

1.- Which food has protein?
Egg white, milk, yogurt, potato, ultimately, the animal food.

2.- Which food has more protein? why?
1 Egg white

2 Yogurt 

3 Cow milk => Because the animal food is rich in protein.

3.- Do you find any difference between rice and cow milk?
Rice milk doesn't have protein however cow milk does.

4.- Is there any difference among milk and yogurt? why?
Yes, yogurt has more protein than milk.

L.8 SAPONIFICATION

INTRODUCTION:

It is a reaction of a fatty acid with a strong base, which gives rise to a salt of fatty acid, called soap. This reaction is very important because the fatty acids are insoluble in water, the soap molecules form groups (mecel·les) that disperse in water. That is, even though they are insoluble form colloidal dispercions. The difference is due to the fact that the carboxyl group (COOH) of the fatty acid is ionized and the little group (COO + strong base) are ionized and cause very many radicals.

Our objective is prepare soap. 

MATERIALS:

- 600mL Beaker
- Spatula
-  Watch glass 
- Stirring rod
- 2 beakers 250mL
-electronic balance
- Gloves
- Safety goggles
- Hot plate

- 32g NaOH
- 90mL H2O
- 270mL Oil

PROCEDURE:

First we measured the grams of each substance with the balance, we took the beaker 250ml and we put 90 mL of water, and another beaker 250ml and we put 270 mL of oil. All this have led to a 600ml beaker, slowly pulling the oil and mixed, but be careful because the reaction is exothermic and it heats! Mix a good time, we put the 600mL on a hot plate for heat and this accelerate the reaction. Then we put a few drops of aroma and finally we put the mixture in a few evaporating dish. Now we have to wait for obtain the hard soap. 

Now when we have the flavor we wanted, we poured a box of porcelain. Now all that remains hard. And so how does the soap.

The reaction of saponification:

L.7 LIPIDS PROPERTIES

INTRODUCTION:

Lipids are heterogeneous group of compounds synthesized by organisms that are present in all biological tissues. Lipids are made in general of 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. 

In this experiment we test the solubility of different oils and how to identify lipids from a sample, Our objectives where: 

- Test the solubility of lipids.

- Identify lipids in liquids compounds

- Understand what are an emulsion and the effect of detergents.

MATERIAL:

- Test tube rack
- 250mL beaker
- Water
- 6 test tubes- Cellulose paper
- Dropper
- Scissors
- Glass rod

- Olive oil
- Soap
- Milk: full-cream, semi-skimmed and skimmed milk.
- Petroleum ether
- Ethanol
- Sudan III

PROCEDURE:

Solubility of some lipids:

- Clean and dry 3 test tubes. Label as W (water), E (ethanol) and PE (ether).

- Add 3 drops of oleic acid to 3 small test tubes.

- Add 1mL of water in the first test tubes (W).

- Add 1mL of ethanol in the second test tube (E).

- Add 1mL of petroleum ether in the third test tube (PE).

- Shake carefully each test tube and record solubility and observations.

Lipids identification: 

Translucent mark:

- Take 2 pieces of cellulose paper.

- Put 1 drop of water in the first piece. You will see a translucent spot. Wait for a while and observe what is happening.

- Put 1 drop of olive oil in the second piece of cellulose paper. You will see a translucent spot. Wait for a while and observe what is happening.

Sudan III dye:

Sudan III is a red fat soluble dye that is utilized in the identification of the presence of lipids, triglycerides and lipoproteins in liquids. 

- Take the W test tube of the first experiment and add 2 drops of Sudan III. 

- Prepare four test tubes: 3 with milk with different fat content, and soda. Add to drops of Sudan III and observe the results. 

Permanent emulsion:

- Take 250 mL beaker and put 100mL of water.

- Add 1 mL of olive oil. With a glass rod stir the mixture vigorously and let it stand for a few minutes.

- Make note of what is happening. 

- 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 AND CONCLUSIONS:

- Lipids are insoluble in water and soluble in ether (organic compound). In the second test tube, the oil and ethanol didn't mix because the oil formed micelles.
- Sudan III stains lipids, but in our experiment didn't work.
- Lipids leave a translucent spot in cellulose paper. 

QUESTIONS:

1.- From your observation, wich compounds can dissolve lipids? 

Ether can dissolve lipids. 

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 is insoluble in water, the oil is non-polar. 

3.- Why is liquid the olive oil at room temperature? And why not the lard?

Oil is the fat acid insaturated. Because the lard is a fat acid saturated.

4.- Why does the lipid leave a translucent spot on paper? 

Because the lipid don't evaporate.

5.- Wich type of milk contains more lipids? Why?

 Full-cream milk, because they take away the semi-skimmed milk fats. Contain all the lipids.

6.- Did the oil and water mix when you added the soap?

No, it forms micelles that can separated oil from water. 

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? 

Bile acids.