Activitat enzimàtica de la CATALASA en diferents teixits animals i vegetals

En aquest experiment valorarem de forma quantitativa l'activitat enzimàtica de la Catalasa del fetge de pollastre. Dividirem la pràctica en dues parts. En la primera observarem la diferència d'activitat de la catalasa en diferents teixits animals i vegetals. Per aquesta part utilitzarem la pastanaga, la patata, el tomàquet i el fetge i el cor de pollastre. En la segona part veurem la influència de determinats factors en l'activitat enzimàtica...

La catalasa és un enzim present en els peroxisomes de les cèl·lules animals i vegetals que s'encarrega d'eliminar l'aigua oxigenada que es forma en algunes reaccions del metabolisme. La reacció química és la següent.
H2O2 --> 1/2 H2O + O2

Material:
Teixits animals; fetge i cor de pollastre
Teixits vegetals: patata, pastanaga, tomàquet
Tub d'assaig de coll ampla
Vas precipitats
Termòmetre
Gradeta
Aigua destil·lada
Aigua oxigenada al 3%
Pinces
Bisturí
Tisores

Protocol:
Experiment 1: Quin teixit presenta més activitat de la catalasa?
1- Tallarem tres trossos de patata, tomàquet, i pastanaga que pesin més o menys el mateix (mateixa mida 1cm3)
2- També tallarem dos trossos, un de fetge i un de cor de la mateixa mida i pes (1,2 g la patata, 0,9 g tomàquet, 1,1 g pastanaga, 0,9 g fetge i 1,2 g cor.) .
3- Ho posarem en cinc tubs amb cada teixit en un tub + 5 ml d'aigua destil·lada en cada.
4- Després posarem 2 ml d'aigua oxigenada i marcarem l'alçada que assoleixen les bombolles en cada tub. Mesurarem aquesta alçada en mm.

(imatge agafada de http://pbsciencelab.blogspot.com.es/)


Preguntes:
Variable dependent i independent?
Alçada de bombolles és la variable dependent i els diferents teixits la independent.
Problema que es vol investigar?
Quin teixit presenta més activitat de la catalasa?
Explicació dels resultats:
Hem pogut veure que el cor és el teixit que més activitat enzimàtica. Els teixits animals, comparat amb els teixits vegetals presenta més activitat.

Experiment 2: Com afecten diferents factors en l'activitat de la catalasa?
Agafarem el teixit del fetge i el posarem en 5 tubs d'assaig sota condicions diferents
1er tub: tros de teixit a temperatura ambient (a 19 graus 1 g)
2n tub: tros de fetge amb 10 ml d'HCl al 10% (0,9 g)
3er tub: fetge congelat (a 2 graus 1,2 g)
4t tub: fetge bullit. (1g)
5è tub: fetge submergit en una dissolucio saturada de NaCl (0,8 g)

Afegirem 2 ml d'aigua oxigenada i anotarem l'alçada de les bombolles (controlarem el temps que seran 10 minuts, el pes i la quantitat de dissolució.

PREGUNTES
Variable dependent i independent?
La variable dependent és l'alçada de les  bombolles i la independent les diferents condicions a les que hem sotmés el fetge.

Problema que es vol investigar?
Com afecten diferents condicions d'un teixit animal (fetge) en l'activitat de la catalasa sobre aquest?
Explicació dels resultats:
Hem pogut observar que el tub amb més activitat de la catalasa ha sigut el de NaCl, seguidament el congelat (perquè s'estava descongelant) i després el de temperatura ambient. Els dos que han presentat menys activitat han sigut amb àcid clorhídric i el bullit perquè estaven fora del pH òptim i de la Tª òptima. 

Quina és la funció de la catalasa en els teixits animals i vegetals? On es troba aquest enzim?
És un enzim que es troba als peroxisomas de les cèl·lules eucariotes. La funció de la catalasa és descomposar l'aigua oxigenada en aigua i oxigen ja que en el metabolisme es produeix aquesta aigua oxigenada que és una molècula tòxica.

Per què quan ens fem una ferida ens posem aigua oxigenada?
L'existència de catalasa en els teixits animals, s'aprofita per utilitzar l'aigua oxigenada com a desinfectant quan es tira sobre una ferida. Com molts dels bacteris patògens són anaerobis (no poden viure amb oxigen), moren amb el despreniment d'oxigen que es produeix quan la catalasa dels teixits actua sobre l'aigua oxigenada.

L.21 Mitosis in an onion root

On Monday 18th of May we did an experiment about the observation of mythosis. Mitosis is the process in which a eukaryotic cell nucleus splits in two, followed by division of the parent cell into two daughter cells. So the objective of this experiment is to view the different stages of a mitosis in the microscope.

MATERIALS
-Onion
-Orceine A and B
-Dropper
-Watch glass
-Beaker
-Forceps
-Bunsen burner
-Lighter 


PROCEDURE
1- A week ago we left an onion in a beaker with some water, (only the tip of the onion touched it) so its roots will grow so we can see the process of mitosis.
2- To start our experiment we took the onion and cut the tip of a root and put it in the watch glass.
3- Then with the dropper we took the orceine A and put some drops on the root and we took the watch glass with the wooden forceps and put it on the bunsen burner so the orceine and the root would heat. Some fumes began to evaporate. We had to be careful to not to burn the root so the watch glass could not be too hot, we should be able to touch it with our hand!!
4- After that we took the root with the forceps and put it on a slide and added a couple of drops of orceine B, we waited a couple of minutes.
5- Then with the scalpel we cut 3mm leaving the tip, and always knowing where the tip is. 
6- Finally we added a coverslip and used the squash method so we could observe the cells on the microscope.

Results and observations: Unfortunately, my group was able to only see the prophase so we asked our classmates if they found anything and they had more luck. 

 

L.20 The chloroplast and the photosynthesis

This Monday 11th of May we did an experiment with an algae. During the photosynthesis, plants and algae produce oxygen. The reasearch aspect of this science lab project is to assess how light intensity affects the rate at which photosynthesis occurs and the rate of oxygen production. This experiment is done to relate the light intensity with the photosynthesis process, to measure the rate of photosynthesis and to identify the products of the process and the variables that can affect it.

MATERIALS

-Algae 
-600 ml beaker
-Test tube
-Funnel
-Tap water 
-Light source
-Ruler

PROCEDURE

1- First we assigned the different distances to do the experiment and compare the results to each group.
2- We took the 600 ml beaker and placed 7 g of an algae under a clear funnel inside the beaker (the wide end goes over the algae like in the image). The funnel was raised off the bottom on pieces of blue-tack to allow unhampered diffusion of CO2 to Elodea. 
3-We didn't have sodium bicarbonate so we filled the beaker with tap water, the algae and the funnel should be completely under the water.
4- Then we filled a test tube with tap water and placed the thumb over the end of the test tube. We turned the test tube upside down taking care that no air enters and no water comes out and we put this test tube over the end of the funnel (the skinny part)
5- We marked the level of the water on the surface of the test tube with a marker pen.
6- Each group placed the preapartion close to a light source, each group placed the preparation in a different distance 5, 10, 20 and 25 cm, and one with no light source.
7- We also measured the temperature.
8- Finally we left this preparation for and hour and a half. After this time we measured the difference of gas accumulation on the top of the test tube.

Results and observations: everything will be explained in the questions below:

QUESTIONS

1- Identify the dependent and the independent variable of this experiment.
Dependent: gas production, Independent: distance (intensity of the light)
 
2-Using the data from your results prepare a graph and describe what happened to the amount of gas in the test tube.

 

My group had the 10 cm distance and our water level decreased about 0,4 mm in an hour and a half.
If you have more light intensity, the phtoshyntethic rate will be higher. We controled the temperature and the algae quantity because this could have and influence on the result. Also, we think that Laura's and Andrea's result is incorrect because the distance is higher so the water level decrease should be lower than the other. Maybe they measured it incorrectly.

3-How much gas was producted in the test tube after one hour? And an hour and a half?
We put the results in the graph.
 
4-Write the photosynthesis equation. Explain each part of the equation. Which subtances are produced by photosynthesis. Which gas is produced that we need in order to live? Plants take in carbon dioxide by diffusion through their stomata. Light energy enters the plant via leaves and water and nutrients enter through roots. The plant is then able to make glucose and oxygen. The glucose moves from the leaves to the plant and the oxygen diffuses out of the leaves. The gas that we need in order to live is oxygen.

 

INVESTIGATION

-Which is the origin of the oxygen that we breathe?

 The trees and plants that are around us and other organisms that do the photosynthesis.

-Where are the lungs of our planet?

Phytoplankton need two things for photosynthesis and thus their survival: energy from the sun and nutrients from the water. Phytoplankton absorb both across their cell walls.
In the process of photosynthesis, phytoplankton release oxygen into the water. Half of the world's oxygen is produced via phytoplankton photosynthesis. The other half is produced via photosynthesis on land by trees, shrubs, grasses, and other plants.


 

19. Cell organelles

 

 

L.18 Life in a drop of water

We observed pond water and we found an eucariotic unicelular organism with two flagels.

L17. Gram staining

This was the second experiment that we did on Monday 2nd of March. Gram staining is a method of differentiating bacterial species into two large groups: Gram positive and Gram negative. This differentiation is based by the chemical and physical properties of their cell walls by detecting a peptidoglycan, which is presents in a thick layer in gram-positive bacteria. The objectives of this experiment were to differentiate yogurt bacteria and realte the staining procedure with the structure of the cells.

MATERIALS

-Hot plate
-1 slide
-1 coverslip
-Tongs
-Needle
-Gram stain (crystal violet, iodine and safranin)
-Ethanol
-Microscope
-Yogurt

PROCEDURE 

 1- First we prepared a heat fixed sample of the bacteria by spreading somre yogurt on a slide and drying it on the hot plate.
2-Then we covered the smear with crystal violet and waited for 1 min. After that we rinsed it with distilled water.
3-We applied iodine solution for another 1 min and again rinsed it with distilled water.
4-Then we decolorized using ethaol. Drop by drop until the purple stops flowing and washed immediately with distilled water.
5-Lastly we covered the sample with safranin stain for and exposure time of 45 seconds and rinsed the sample with tap water.
6-Finally we dried the under part of the slide with paper and viewed it on the microscope.

Results and observations: We saw some bacteria red and other purple. Why?

Gram Positive Cell Wall:

Gram-positive bacteria have a thick cell wall which is made up of peptidoglycan (50-90% of cell wall), which stains purple.  Peptidoglycan is mainly a polysaccharide composed of two subunits.  The thick peptidoglycan layer of Gram-positive organisms allows these organisms to retain the crystal violet-iodine complex and stains the cells as  purple.

Gram Negative Cell Wall:

Gram-negative bacteria have a thinner layer of peptidoglycan (10% of the cell wall) and lose the crystal violet-iodine complex during decolorization with the alcohol rinse, but retain the counter stain Safranin, thus appearing reddish or pink. They also have an additional outer membrane which contains lipids,  which is separated from the cell wall by means of  periplasmic space.

L16. Epidermis cells

On Monday 2nd of March we did two experiments using the microscope, and now i'm going to explain the first one. The objective of this experiment was to identify the shape of epidermis cells, and to identify and explore the part of the stoma and see how it changes its shape when we add salt water.The pores open to facilitate uptake of carbon dioxide and close to limit the loss of water

MATERIALS

-Slide
-Cover slip 
-Tap water
-10% salt water
-Forceps
-Dropeper -Scissors
-Needle
-Leek

PROCEDURE

1-First we cut the stalk of the leek and pulled out the transparent part of the epidermis using forceps.
2-Then we placed the peel into the slide containing a drop of tap water (so the cells don't die) .
3-Next we took a cover slip and placed it gently on the peel with the aid of a needle.
4-We viewed it in the microscope and took pictures of it.
5-Then we prepared a 10% salt solution and put the solution with a dropper on the left part of the slide (so it touched the cover slip) and placed a piece of cellulose paper in the opposite side of the cover slip to let the dissolution go through the sample.
6-Finally we looked through the microscope once more and took more pictures.


Results and observations:  
When we first saw the cell we noticed the characteristic shape of a plant cell, a geometric one, a squareand the cell wall. Then we looked closer at it and saw the stomas: they were open. Stoma opens when the guard cells are turgid, when the water potential of the cells adjacent to the guard cells are higher than that in the cell sap of the guard cells and the water molecules from the adjacent cells move into the guard cells by osmosis. The opening of the stoma is an advantage because it allows gaseous exchange to take place.
Then we added salt water and took a second look. Now the stomas were closed because the adjacent cells were hypertonic and the guard cells hypotonic so the water molecules moved out of the guard cells into the adjacent cells by osmosis. When this happens, the guard cells become plasmolysed which in turn causes the stoma to close.

QUESTIONS

1- What is the major function of a cell membrane?
The membrane is selectively permeable to ions and organic molecules and controls the movement of substances in and out of cells. Also it protects the cell from its surroundings.

2- What is the major function of the cell wall?
It surrounds the cell membrane and provides structural support and protection to it. Also it acts as a filtering mechanism and as a pressure vessel, preventing over-expansion when water enters the cell.

3-How does salt affect the cells shapes? And the stomes?
I explained it earlier.

 

L13. ANIMAL CELLS vs PLANT CELLS

Last Monday we compared animal and plant cells with the mycroscope. The objectives were to identify the major components of cells, differentiate between animal and plant celss and to measure dimensons of the entire cell and the nucleus.

First we peeled off a leaf from an onion. dyed it with safranin stain (red) and then viewed it in the mycroscope.

Then we did some calculations to know the real size of the cell and its nucleous.

After that, we extracted a cell from our cheeks. 

L12. DNA Extraction

On Monday 26th of January we did a new experiment about DNA extaction. The objective of this experiment was to study the DNA structure and to understand the process of extracting DNA from a tissue.

MATERIALS

- 600 ml Beaker
- 10 ml graduated cylinder
- Small funnel
- Glass stirring rod
- 10 mL pipet
-Safety goggles
-Cheesecloth
-Automatic pipet
-Kiwi
-Pineapple juice
-Distilled water
-90 % ethanol (ice cold)
- 8 ml DNA buffer (25 ml dish soap, 7 g NaCl, 450 ml tap water)

PROCEDURE

First of all we prepared the buffer in a 600 ml beaker. We put 450 mL of tap water, 25 ml of dish soap and 7g of NaCl and stirred the mixture careful because we don't want any foam or bubbles to form! 

1-We peeled the kiwi and chopped it to small pieces. We placed them in one 600 ml beaker and smashed them with the pestle to become a juice puree. 
2- We added 8 ml of the buffer to the mortar.
3- We mashed the kiwi puree carefully for 1 minute without creating many bubbles.
4- Then we filtered the mixture: we put the funnel on top of the graduated cylinder. We placed the cheesecloth on top of the funnel and added all the contents of the mortar carefully on top of the cheesecloth to fill the graduated cylinder. The juice drained through the cheesecloth but the chunks of kiwi didn't pass through.
6- We added the pineapple juice to the green juice (4 ml because we had 20 ml of green mixture DNA solution). This helped to obtain a purer solution of DNA because pineapple juice contains an enzyme that breaksdown proteins.
7- Next, we tilted the graduated cylinder and poured in an equal amout of ice-cold ethanol with an automatic pipet. We put the ethanol through the sides of the graduated cylinder to form a clear layer on top of the DNA solution.
8- We placed the graduated cylinder to eye level and using the stirring rod we stirred only the ethanol and DNA came up. 

Results and observations:
The DNA looks like long, small, white and thin fibers.  

QUESTIONS
1- What did the DNA look like?
 The DNA looks like long, small, white and thin fibers.  

2-Why do you mash the cells? Where it is located inside the cells?
Because you want to libreate the DNA that is located inside the nucleus.

3- Explain what is the function of every compound in the buffer
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 has to do with the method of extraction?  
This means that we can only see the DNA in the part of the ethanol because if it touches the water it will dissolve.

On Monday 2nd of February we repeated this experiment but we extracted the DNA from our own cells from our mouths. We did the exact same procedure but instead of mashing the kiwi we took mineral water with salt and we rinsed our mouth with it. At the end we observed the DNA in a microscope. 

L.11 Cytochrome C Comparison Lab

PROTEINS AND EVOLUTION

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. 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. The cytochrome C is a small protein found loosely associated with the inner membrane of the mitochondrion. It is found in eucariotic cells and has an hemeprotein. It is essential to the electrone transport chain and it is involved in using energy in the cell (ATP).

The purpose of this practice is to compare the relatedness between organisms by examining the amino acid sequence in the protein.

METHOD

First, we compared the amino acid sequence of Cytochrome C in various organisms:
-horse, donkey, whale (mammals)
-penguin, chicken (birds)
-snake (reptile)
-moth (insect)
-wheat (plant)
-yeast (fungi)

1- We marked the amino acids which were different.
2-Then we counted and recorded the total number of differences.
3- We shared the data with the rest of the class to complete Table 1.


After that, we made a cladogram (branching tree)
1- The two most closely related species had the fewest differences in amino acid sequence. We placed the two most closely related species on the two shortest branches of the tree.
2-Then we placed the next two closest species on the next shortest branches.
3- And we continued until all the species had been placed.

RESULTS AND OBSERVATIONS

CONCLUSIONS

There are 0 differences between chickens and turkeys in their Cytochrome C amino acid sequence.

We think that the horse and the zebra have 1 or 2 differences, like the donkey and zebra.
To make this prediction we used this information: if they can reproduce and if the offspring will be fertile or not, we compared organs, compared embrios...

More closely related organisms have more similar Cytochrome C because evolutionarily, it hasn't been that long since they separated. If the species are close, it means that less time has passed since they separated (there are less accumulated mutations). They have a common ancestor.

Other data, including genes, suggests that fungi are more closely related to animals than plants. But Cytochrome C data suggests that fungi, plants and animals are equally distantly related because if there are more than 40 genes, there are too many mutations to see it clearly.

L9. Protein identification

On december 22nd we did an experiment about proteins. The objectives are to identify the peptide bonds and to compare the protein concentration in different foods.

Biuret's test
The Biuret's test is a chemical test used for detecting the presence of peptide bonds (the bonds that link together amino acids that form chains --> proteins). A peptide bond can be broken by hydrolysis. The intensity of the colour of the Biuret's test is directly proportional to the protein concentration. The solution to be tested in treated with a strong base followed by a few drop sof copper II sulphate. If the solution turns purple protein is present. Only peptides with a chain of at least 3 amino acids give a significant measurable colour shift with these reagents (polypeptides).


MATERIALS

-250 ml beakers
-Test tube rack
-6 test tubes
-6 x 10 ml pipet
-Mortar
-Glass marking pen
-Gloves
-Goggles
-Milk
-Rice milk
-Egg (white and yolk)
-Yogurt
-Potato
-Distilled water
-NaOH 20%
-10 drops of CuSO4

PROCEDURE

First of all we had to dilute the protein:
1- We added 100 ml of distilled water to each 250 mL beaker, and with the glass marking pen we labeled them as M (milk), R (rice milk), EW (egg white, EY (egg yolk), Y (yogurt) and P (potato).
2- Then we added 10 ml of a dispersion of each food in the belonging beakers, we separated the egg white and yolk in two different beakers and mashed the potato.

Then we prepared the samples:
3- We cleaned and dried the test tubes and labeled them like the beakers. Each group used the dispersions from the same beakers. Then we added 2 mL of every food dilution of the beaker to each corresponding test tube.
4- After that we added 2 mL of 20%NaOH dissolution to each test tube, Andrea did this because NaOH is caustic so she had to put gloves and goggles on.
5-We shook gently and added 5 drops of CuSO4 in each test tube. Then we allowed the mixture to stand for 5 minutes.
6-Finally we saw the colour change (to pink or purple) amd we compared the test tubes.

Results and observations:
Milk - positive
Rice milk - negative
Egg white - positive
Egg yolk - negative
Yogurt - positive (it has less water, the protein is more concentrated)
Potato - positive, but our teacher said that it had to be negative because it had starch.

We saw that all the foods of animal origin gave a positive result, they all have proteins. Rice milk is negative because it has starch like the potato (but strangely the potato was positive). Also, we ordered the food from more concentration of protein to less according to the colour: 1- egg white, 2- yogurt and potato, 3- milk.

QUESTIONS

1- Which food has proteins?
Eggwhite, milk yogurt and potato.

2- Which food has more proteins? Why?
Eggwhite because it's animal food and it's rich in proteins. Then the yogurt and potato and finally the milk.

3-Do you find any difference between rice milk and cow milk?
Rice milk doesn't have proteins however cow milk does as i had said before.

4- Is there any difference among milk and yogurt? why?
Yes, yogurt has more protein than milk because the protein is more concentrated than in the milk, it has less water.