Comparing Gene Expression in Lung Cancer Cells and Normal Lung Cells Using Microarrays

Introduction:

a) All cells (except for gametes) contain the same DNA. What makes cells unique is how they express this DNA - sometimes the difference between a healthy cell and a diseased cell is a simple change in gene expression. Microarray analysis allows us to compare gene expression by first creating a gene chip, in which specific DNA sequences are put into individual spots on the chip. Meanwhile, cDNA (complementary DNA) is created from mRNA found in the healthy and diseased cells. The cDNA is then added to the chip and the fragments bind to their complementary strands of DNA that were placed there originally. By analyzing the chip, we can see what sequences are being expressed based on where the cDNA binds. For instance, let's say that the cDNA from the diseased tissue is labeled red and the cDNA from the healthy tissue is labeled green. When we analyze the results, spots that are green indicate that those particular sequences are only expressed in healthy tissue. Spots that are red indicate sequences that are only expressed in diseased tissue. Spots that are yellow (red light + green light = yellow light) indicate sequences which are expressed in both and spots that don't turn any color are not expressed in either.
b) The purpose of this experiment is to analyze gene expression in lung cancer cells versus healthy lung cells by using microarrays. Potentially, by gaining more knowledge about gene expression in this way, we could develop ways to alter cancer cells or cure them somehow. The more we understand about what makes a cancer cell different, the closer we get to solving the problem.
c) Most of the techniques or materials were explained in part "a". The microarray chip serves to isolate known DNA sequences, so that when we apply the cDNA to the chip, we can determine which sequences are active. The red and green labels show us which sequences are specifically expressed in either, both, or neither of the two cell types. We will use pipetting techniques to transfer the cDNA to the miniature microarray.
d) For my hypothesis, I think genes which are involved in typical cell functions (such as protein synthesis or cell respiration) will be expressed in both the healthy and the cancer cells. I think genes which are involved in replication will likely be turned off in the cancer cells, since they are known to divide uncontrollably. After that,  I will have to wait and see how the experiment turns out. The controls of this experiment are the DNA sequences on the microarray. We already know where specific sequences are located on the chip. The variables are the cDNA sequences, which are unknown (until they bond to the DNA on the chip of course).

Results:

The microarray results are shown above. Pink = expressed in lung cancer cells only (#1,5). Blue = expressed in healthy lung cells only (#3,6). Purple = expressed in both (#2). Clear = expressed in neither (#4).

Solving the Mystery using Restriction Enzymes and Gel Electrophoresis

Introduction:
a) Restriction enzymes are a specific type of protein which can be found in bacteria. They act as a defense against viruses by cutting up viral DNA at specific locations called palindromes (these specific sites can also be called restriction sites). Gel electrophoresis is a process used to analyze DNA - the DNA is placed in a gel and an electric current is sent through. DNA is negatively charged, thus it will be pulled toward the positive node. However, the smaller, lighter pieces of DNA can move faster and travel farther along the gel. If this cut-up DNA is run on a gel, we can determine the length of each fragment of DNA based on how far they traveled along the gel. The different fragments can be seen as dark bands along the gel. This is called Restriction Fragment Length Polymorphism (RFLP).
b) The purpose of this experiment is to analyze different suspects' DNA via gel electrophoresis and compare their DNA to the crime scene's. The suspect whose DNA matches the crime scene's (meaning the bands are the same) is the culprit.
c) The purpose of the restriction enzymes is to cut up the suspects' DNA - each person has a unique genetic code, thus each person's DNA will be cut at different locations and will result in unique DNA fragment lengths. A loading buffer containing blue dyes is added to the DNA samples before injection into the gel - this allows us to monitor the process of gel electrophoresis. The gel electrophoresis and RFLP will then show a unique banding pattern of each person's DNA fragment lengths on a gel.  We can then analyze these bands to determine which DNA matches the DNA of the crime scene.
d) I cannot predict which person will be the culprit since I have not seen the results yet. However, the control is the crime scene DNA and the variables are the suspects' DNA.

Procedure:
Basically explained above...

Results:
It turns out Suspect #3 was the criminal! The DNA bands from Suspect #3 clearly matched the DNA bands of the Crime Scene - the evidence is incriminating.

Discussion:
a) Via gel electrophoresis, the fragments were separated by length along the gel - the shorter fragments traveled farthest toward the positive node while the longer fragments remained further behind. By comparing the bands, it was clear that the DNA bands of Suspect #3 exactly matched the bands of the Crime Scene.
b) Possible sources of error:

• Improper pipetting
• Contamination of DNA
• Not run on the gel long enough to accurately disperse DNA fragments

Creating biofuels from cellulose using enzyme cellobiase

Introduction:
a) Enzymes speed up the rate of chemical reactions without being used up by first binding to the reactants (substrate). The location of the binding is called the enzyme's active site. This bond lowers the activation energy (the energy needed to start the reaction), making the reaction occur faster. Enzymes' effectiveness can be affected by pH, temperature, and salinity. The reaction tends to occur faster if the concentration of enzymes or substrates is increased, but there is a point when the solution becomes saturated with either enzymes or substrates and cannot work any faster. The enzyme we will be using on this lab is cellobiase, which breaks down cellobiose - a sugar derivative of cellulose, a polysaccharide found in plant cell walls. The biofuel agency uses enzymes such as cellobiase to break down cellulose in plant matter and convert it to glucose. Glucose can then be converted to ethanol by microbial fermentation. Ethanol can then be used as an energy source to power certain motors and engines.
b) The purpose of this lab is to experience first hand in the classroom what biofuel companies do on a massive scale. Biofuel companies are constantly experimenting to find the most efficient way to make biofuels - and the method we will use in this experiment is only one method out of many.
c) We will use pipetting techniques to carefully and accurately transfer enzymes, buffers, and substrates to the solution. On Day #1, we will transfer a little bit of the solution at various time points to another test tube with a strong base to stop the reaction. The base will also turn p-nitrophenol (a product of the reaction) yellow, so we can judge how much product (including the glucose) has been created. On Day #2, we will have to carefully grind up mushrooms and add the mushroom extract to the reaction and measure how this affects the speed of the reaction.
d) I predict that, as time goes on, more product will be produced from the reaction. Also, I think the mushroom, which contains more enzymes, including cellobiase, will speed up the reaction rate. The variable of this experiment is the mushroom. The control is the amount of product produced under normal circumstances (Day #1).

Results/Observations:
The solution from Day #1 turned more yellow over time (the first cuvette was the lightest and the last cuvette, the darkest). The solution with the mushroom (Day #2) also turned darker over time, only faster.

Discussion:
a) The solution from Day #1 turned more yellow over time (the first cuvette was the lightest and the last cuvette, the darkest). This indicates that more product was produced over time. On Day #2, the mushroom turned darker faster, indicating that the reaction was occurring faster. This is probably because the mushrooms contain more enzymes, such as celliobiase, which can speed up the reaction to an extent. There is a limit to how many enzymes are effective, but in the case of Day #2, the extra enzymes did help speed up the reaction.
b) Possible sources of error...

• Incorrect measuring with the pipettes
• Contamination of the solution(s)
• Timing errors

Extracting and precipitating DNA from cheek cells

Introduction:
a)  All living things contains DNA in each of their cells. Eukaryotes contain a nucleus inside their cells, which houses the DNA. Prokaryotes do not have a nucleus, but their DNA moves freely throughout the cell. DNA is a double-stranded helix composed of sugar (deoxyribose), phosphates, and bases (thymine, adenine, guanine, and cytosine). Thymine pairs with adenine and guanine pairs with cytosine. Genes are particular segments of DNA which code for a specific protein. Proteins are what give your eyes color, carry out cell communication, and build muscles.
b)  The purpose of this lab is to precipitate (or anti-dissolve) DNA, so that we can see and observe it with the naked eye. In a professional setting, such as in a biotech lab, one can use the extracted DNA to map the genome, clone genes, compare DNA, test for genetic diseases, or do forensics. However, in the classroom, we will simply be admiring our DNA.
c)  We will use a "chewing on the insides of our cheeks" technique in order to loosen cheek cells. The saline solution (which contains 0.9% salt water) will be used to give the cells an isotonic solution to keep them from bursting or shriveling. The lysis buffer (which is a detergent) will serve to dissolve the cell membranes (which are made of phospholipids), releasing the DNA out into the open. DNase is an enzyme that lives in the cell in order to kill foreign DNA; however, we do not want the DNase to kill our own DNA now that it's been released, so we add the enzyme protease to destroy it. DNA is negatively charged, due to the negative phosphates. A negative charge makes DNA polar and hydrophillic, meaning it likes water and will dissolve in it. We do not want our DNA to dissolve - in fact, we want to anti-dissolve it or precipitate it - so, we add Na+ ions to give DNA a neutral charge. This makes DNA nonpolar and hydrophobic so it will not dissolve. The hot water bath serves to speed up the enzyme reactions and helps break open the cell membranes (lysis buffer works better with hot water). Finally, the cold ethanol (rubbing alcohol) helps with precipitation by forming a very cold layer on top of the hot water solution.

Procedure:
*Our video was accidentally deleted by another group - however I will briefly go over the procedure here.

See Introduction part "c" for details on each step of the procedure. However, I will note that after the DNA had precipitated entirely, we extracted it with a pipette and put it into a necklace and wore it around school!

Results/Observations:
I observed that almost instantly after we added the cold ethanol to the hot water solution, the DNA began precipitating between the two layers. It formed a stringy, white substance that was very easy to see and very cool! It definitely looked like DNA, with the long strands all tangled together. It was very interesting.

Discussion:
a)  I did not get very much DNA, however, it was definitely there and very easy to see. The procedure largely explains how the DNA was extracted and there is not anything to analyze or explain really about the results. All I can say is that it worked!
b)  Possible sources of error could be...

• Food or other contents from the mouth could have gotten into the solution
• Insufficient chewing on the cheeks may not have extracted enough cheek cells
• Some DNA may have been destroyed in the process by a hypo/hypertonic solution (if the saline solution did not work) or DNase (if the protease did not kill it in time)

Making yogurt out of pasteurized milk and yogurt bacteria while testing Koch's postulates

Introduction:
a) Koch's postulates are... 1. Isolate a microbe that can be found in all sick people (in this case the "sick people" is yogurt) but not in healthy people ("healthy people" = milk). 2. Culture that microbe. 3. Inoculate the healthy people with the microbe. 4. Culture the microbe from the new sick people and it should be the same microbe as the original one. The microbe, or bacteria, that we will be using in this experiment is yogurt bacteria. Bacteria are prokaryotes, meaning they do not have a nucleus to hold their DNA. Bacteria can be both good and bad. E. Coli are bacteria that live in our intestines and help us digest food, which is very beneficial to us. Other bacteria however can cause disease and antibiotics can be taken to kill these harmful bacteria. Yogurt bacteria are a good kind of bacteria - they help with digestion and promote general good health. Yogurt bacteria break down milk sugar (lactose) into pyruvic acid, and then, using enzymes, the pyruvic acid is broken down into lactic acid. The lactic acid causes the milk to be more acidic, thus denaturing the milk protein casein into a more solid form - yogurt!
b) The purpose of this experiment is to make yogurt out of milk and yogurt bacteria. We will also be testing Koch's postulates using microbial techniques (extracting and culturing bacteria and identifying if the bacteria are the same).
c) We will be culturing bacteria using specific techniques. We will be using sterile inoculating loops to transfer yogurt bacteria from the petri dish to the test tubes. Precautions will be taken to avoid contamination, such as keeping the lab area clean and not touching the inoculating loops to any potentially contaminating surfaces. A vortex will be used to thoroughly mix the contents of each test tube and an incubator will encourage the production of yogurt overnight.
d) The controls of this experiment are the milk only tube (negative control) and the milk + yogurt bacteria tube (positive control). The variables are the milk + yogurt + ampicillin tube, and the milk + E. Coli tube. I think the negative control will do nothing and the positive control will produce yogurt (as should happen). The first variable should not produce yogurt because the ampicillin will kill the yogurt bacteria which are necessary to denature the milk into yogurt. The second variable will probably result in some of the milk being digested by the E. Coli, because that's what E. Coli do - digest food.

Procedure:
See video posted.

Results/Observations:

Discussion:

a) Tube #1, the negative control which contained only milk, obviously did not turn into yogurt because it did not have the yogurt bacteria in it. The sour, tainted milk smell is the result of unrefrigerated milk gone bad - it did not have the yogurt bacteria which helps preserve freshness. Tube #2, the positive control, turned into yogurt, as expected. The acidity of 4 can be attributed to the lactic acid, which was created by the breakdown of lactose and pyruvic acid (see Introduction part "a"). Tube #3 did not produce yogurt and turned out sour because the ampicillin killed all the yogurt bacteria, which are essential in the creation of yogurt. Tube #4, which served to test whether or not any type of bacteria can create yogurt out of milk, did nothing - the milk turned rotten and stayed in liquid form. This proved that the E. Coli does not produce yogurt out of milk and clearly, yogurt bacteria is the necessary ingredient for this reaction to take place.
b) Some possible sources of error include...

• Outside bacteria getting into the solution
• Not enough yogurt bacteria, ampicillin, and/or E. Coli being added to the appropriate tube
• Milk improperly pasteurized