L15. GRAM STAINING

L15. Gram staining

1. Introduction

Gram staining is a method of differentiating bacterial spieces into two large groups, gram positive and gram negarive. This differentation is based by the chemical and physical properties the their cell walls by detecting a peptidoglycan, which is present in a thick layer in gram -  positive bacteria.

Gram - negative: color pink
Gram - positive: color purple

2. Material

• 1 slide
• 1 cover slip
• Tongs
• Needle
• Gram stain
• Decolorize reagent : 96% ethanol
• Microscope
• Yogurt

3. Objectives

• Differentiate yogurt bacteria.
• Relate the staining procedure with the structure of the cells.

4. Procedure

First of all, prepare a heat-fixed sample of the bacteria to be stained and cover the smear with crystal violet for an exposure of 1 min.

After, rinse with distilled water and apply iodine solution for 1 min and again, rinse the sample with distilled water.

Decolorize using ethanol. Drop by drop until the purpule stops flowing. Wash immediately with disitilled water.

Cover the sample  with the safranin stain for an exposure time of 45 seconds and rinse the shample with distilled water.

Finally, gently dry the slide with paper.

L14. EPIDERMIS CELLS

L14. Epidermis cells

1. Material

• 1 slide
• 1 cover slip
• Distilled water
• 10% salt water
• Scissors
• Needle

2. Objectives

• Identify the shape of epidermis cells
• Identify and explore the parts of stoma
• Measure dimensions of the entire cell and the stoma

3. Procedure

PLANTS CELLS OBSERVATION:

First of all cut the stalk of the leek and in the place of the cut, pull out the transparent part of the epidermis using forceps.

After, using the brush, place the peel onto the slide containing a drop of the water and take a cover slip and place it gently on the peel with the aid of a needle.
View it in the microscope.

After, describe the change in the shape of the cells and finally, draw a diagram with the part of a stome.

SALT TREATMENT:

First of all, prepare a 10% of salt solution and put the salt with a dropper on the left part of the slide.

Then, place a piece of cellulose paper in the opposite part of the cover slip, and let the dissolution to go through your sample.

4. Questions

1. What is the major function of a cell membrane?

2. What is the major function of the cell wall?

3. How does salt affect the cells shape? And the stomes?

L13. ANIMAL CELLS vs PLANT CELLS

L13. Animal cells vs Plant cells

1. Objectives

1. Identify the major components of cells.
2. Differentiate between animal and plant cells.
3. Measure dimensions of the entire cell and the nucleus.

2. Material

• Forceps
• Dropper
• Needle
• 2 watch glass
• Toothpick
• 2 slides
• 2 cover slips
• Distilled water
• Methylene blue
• Iodine
• Onion
• Glycerine

3. Procedure

Plant cells

First of all pour some distilled water into a watch glass and peel off the leaf from half a piece of onion and using forceps, pull out a piece of transparent onion peel from the leaf.

After put the epidermis in the watch glass containing distilled water and take a few drops of safranin solution  in a dropper and transfer into another watch glass.

Using a bursh 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.

Take a peel from the safranin solution and place it in the watch glass containing distilled water and take a few drops of glycerine in a dropper and pour 2-3 drops at the center of a dry glass slide.

Using a brush, place the peel onto the slide containing glycerine and take a cover slip and place it gently on the peel with the aid of a needle.

Remove  the extra glycerine using a cellulose paper.

And finally, view it in the microscope.

Animal cells

Fist of all, gently scrape the inner side of the cheek using a toothpick, which will collect some cheek cells and place the cells on a glass slide that has water on it.

Mix the water and the cheek cells using a needle and spread them.

After, dry the sample under the light to fix the sample on the slide and take a few drops of methylene blue solution using a dropper and add this to the mixture on the slide.

After 2-3 minutes remove any excess water and stain from the slide using cellulose paper and take clean cover slip and lower it carefully on the mixture with the aid of a needle.

Using a top of the needle, press the cover slip gently to spread the epithelial cells.

Finally, remove any extra liquid around the cover slip using cellulose paper.

4. Augments

1. Cell - 5,5 cm = 55000um

NA = 40x · 10x = 400

400 = 55000/MR  ---- MR = 137,5um

2. Nucleous - 0,7 cm = 7000um

400 = 7000/MR ---- MR = 17,5um

L12. DNA EXTRACTION

L12. DNA extraction

1. Introduction

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

These nucleotides are joined to one another in a chain by covalent bonds between the sugar of one nucleotide and the phosphate of the next. Most DNA molecules consist of two strands coiled around each other to form a diuble helix. Hydrogen bonds bind the nitrogenous bases of the two separate strands.

The two strands run in opposite directions to each other and are therefore anti-patallel. Moreover the bases of the two opposite strands unit according to base pairing rules: A-T and C-G.

Within cells, DNA is organized into structures called chromosomes.

2. Objectives

• Study DNA structure
• Understand the process of extracting DNA from a tissue.

3. Material

• 1L Erlenmeyer flask
• 100mL beaker
• 10mL graduated cylinder
• Small funnel
• Glass stirring rod
• 10mL Pipet
• Knife
• Safety goggles
• Cheesecloth
• Kiwi
• Pineapple juice
• Distilled water
• 90% Ethanol ice-cold
• 7mL DNA buffer
• 50mL dish soap
• 15g NaCl
• 900mL tap water

4. Procedure

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

First of all peel the kiwi and chop it to small pieces and place the pieces of the kiwi in one 600mL beaker and smash with a fork until it becomes a juice puree.

Add 8mL of buffer tot he mortar and mash the kiwi puree carefully for 1 minute without creating many bubbles. Filter the mixture: put the funnel on top of the graduated cylinder. Place the cheesecloth on top of the funnel.

After, add beaker containq carefully on top of the cheesecloth to fill the graduated cylinder. The juice will drain through the chessecloth but the chucks of kiwi will not pass through into the graduated cylinder.

Add the pineapple juice to the green juice. This step will help us to obtain a purer solution of DNA, Pineapple juice contains an enzyme that breaks down proteins.

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.

Place the graduated cylinder so that it is eye level. Using the tirring rod, collect DNA at the boundary of ethanol and kiwi juice. Do not stir the kiwi juice.

The DNA precipitate looks like long, white and thin fibers,

Finally, gently remove the stirring rod and examine what the DNA looks like.

5. Observations

We can observe the fibers of the DNA.

6. Questions

1. What did the DNA looks like?

DNA was placed in the top part floating in ethanol was.

2. Why do you mash the DNA? Where it is located inside the cells?

The crush to extract the liquid from the kiwi, the DNA is in the nucleus of cells.

3. DNA is soluble in water, but not in ethanol. What does this fact have to do with our method of extraction?

DNA because the floats in ethanol therefore we can observe as it does not dissolve.

L11. PROTEIN AND EVOLUTION

L11. Protein and evolution

1. Introduction

Cytochrome C is a small protein from eukariotic cell associated wirh the inner membrane of the mitochondrion.
Is a hemprotein and produce energy. Is an essential component of the electron transport chain.

Genes are made of DNA and are inherited from parent to offspring. Some DNA sequences code for mRNA 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 usuable cytochrome C are unlikely to survive.

2. Objectives

To compare the relatedness between organisms by examining the amino acid sequence in the protein, Cytochrome C.

3. Procedure

First of all compare the amino acid sequence of Cytochrome C in various organisms. Mark the amino acids which are different. Use the exemple to show you how.
Count and record the total number of differences ans share your data with the rest of the class and complete Table 1.
After, make a branching tree or cladogram using the data in Table 1.

4. Conclusions

As far from each other and are more differences further evolutionarily aminoacids. Are grouped by groups of animals.


5. Questions

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

0


2. Make a branching tree, or cladogram for chickens, penguins, and turkeys.

chicken- turkey : 0
turkey - penguin: 3







3.

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

i. Horse and zebra: 1-2
ii. Donkey and zebra: 1-2

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

If they can reproduce and the offspring is festile.

5. List three other things used to determine how organisms are related to each other.

Comparing the organs, anatomic prove, embrions..

6. Explian why more closely related organisms have more similar Cytochrome C.

Evolutionarily not so long ago parted hence have not so many mutations.

7. Other data, including other genes, suggest that fugi are more closely related to animals than plants. What are some reasons that the Cytochrome C data suggest that fugi, plants, and animals are equally distantly related?

Have a very high name mutations plants therefore are further from the yeast. When it exceeds the number 40 that organism dies.

L.10 PROTEIN DENATURATION

L10. Protein denaturation

1.Introduction

Denaturation is a process in which protein 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 or compounds such as a strong acid or base, a concentrated inorganic salt or organic solvent.

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

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

In very few cases denaturation is reversible and protein can recuperate their native state when the denaturating factor is removed. This process is called renaturation.

2.Objectives

• Study the relation between the structure and the function of protein.
• Understand how temperature, pH and salinity affect to the protein structure.

3. Material

• 2x150mL beaker
• 4 test tubes
• Test tube rack
• 10mL pipet
• knife
• Glass marking pen
• potato
• distilled water
• hydrogen peroxide
• NaCl
• HCl

4.Procedure

In this experiment we are going to test the catalase activity in different enviroment situations. We are going to measure the rate of enzyme activity under various conditions, such as different pH values and temperatures. We will measure catalase activity by observing the oxygen gas bubbles when H2O2 is destroyed.If lots of bubbles are produced, it means the reactions is gappening quickly and the catalase enzyme is very active.

First of all, we prepare 30mL of H2O2 10% in a beaker using a pipet, 30mL of HCl 10% in a beaker and 30mL of NaCl 50% in a beaker.

After, peel a fresh potato tuber and cut the tissue in five cubes of 1cm3. Weigh them and equal the mass.
Next label 5 test tubes (1,2,3,4,5)

Immerse 10 minutes your piece of potato inside HCl beaker and 10 minutes another piece of potato inside NaOH beaker. After, boil another piece of potato and with a mortar, mash up the thrid piece of potato.

Prepare 5 test tubes as indicated below:
1. Raw potato 
2. Boiled potato
3. Potato with HCl treatment
4. Potato with NaCl treatment
5. Mashed up potato

Next, add 5 mL H2O2 10% in each test tube and with a glass-marking pen mark the height of the bubbles. Measure it with a ruler.
Finally, compare the results of the 5 test tubes.

5. Observations

We can observed that the mashed up potato have more activity of the bubbles than other potato.


6. Conclusions

mashed > raw > NaCl (activity) > HCl (no activity) > boiled (no activity)







7. Questions

1. How did the temperature of potato affect the activity of catalase?

The temperature affect the enzyme X of the catalase.

2. How did the change of the pH of the potato affect the activity of catalase?

When the pH changes for acid the catalase no activity while pH changes for a basic the catalase activity.

3. In which potato treatment was catalase the most active?  Why do you think this was?

The mashed up potato, because when we mashed up the potato break the enzyme X.

L9. PROTEIN IDENTIFICATION

L9. Prtoein identification

1. Introduction

Biuret's test is a chemical test used for detecting the presence of peptide blond's. Polypeptids as protein, are chains of amino acids links togerther by peptide bonds.

A peptide bond can be broken by hydrolysis. In organisms, protein molecules called enzymes facilitate the process.
The solution to be tested in treated with a strong base followed by a few drops of copper(II) sulphate.

Only peptids with a chain of at least 3 amino acids give a significant measurable colour shift with these reagents. Protein in the alkaline environment reduce Cu2+ to Cu+, which forms a coordination complex with protein, leading to a blue to pink-purpule colour change.


2. Objectives

• Identify peptide bonds
• Compare protein concentration in different foods.

3. Hypotesis

the hypotesis is getting to identify food that contains protein and not.

4. Material

• 7x250mL
• 6 test tubes
• test tube rack
• 6x10mL pipet
• mortar
• glass marking pen
• gloves
• goggles
• milk
• rice milk
• egg
• yogurt
• potatoe
• distilled water
• NaOH 20%
• 10 drops of CuSO4

5. Procedure

We are going to determine protein concentration in some foods: milk, rice milk, yogurt, potatoe and egg. We are going to test separately the egg white and the yolk.

First of all we are going to dilute the protein. We add 100mL of ditillated water to each 250 mL beaker. Label them with M(milk),R(rice milk),EW(egg white),EY(yolk), Y(yogurt) and P(potato).

After, we separate the egg white and the yolk in another beaker and smash the potatoe in a mortar and add some amount of the smashed potato to the P beaker.

Now we prepare the samples:

First of all, we add 10mL of dispersion of each food(M,R,EY,EW and Y) to the indicate beaker. Calculate the final concentration. All the groups will use the same dispersion from the beaker.

After, we prepare 6 test tubes and label M,R,EW,EY,Y and P. Add 2mL of the every food dilution of each beaker.

We add 2mL of 20%NaOH dissolution and shake gently and add 5 drops of CuS in each tube. Allow the mixture to stand for 5 minutes. 

Finally, we note any colour change. Remember that proteins will turn solution pink or purrple. Compare the test tubes.

6. Observations

To dilute the food we like food you contain protein canvian color, and react with a stronger or less lilac, segons the amount of protein they contain.

7. Conclusions

Has arrived at the conclusion that food animals are you contain a high level of proteins, such as egg white, milk, yogurt, potato.

8. Questions

1. Which food has protein?

Egg white, milk, yogurt, potato, utimately, the animal food.

2. Which food has more protein? why?

1st - Egg white, 2n- yogurt, 3rd- cow milk. Because the animal food is rich in protain.

3. Do you find any difference between rice and cow milk?

Rice milk dosen't have protain however cow milk does.

4. Is there any difference among milk and yogurt? why?

Yes, yogurt has more protain than milk.