PGlo Transformation

Purpose

The purpose of this experiment was to test the idea that cells can take in genetic information. By using the glo gene, and putting these cells in different situations we were able to see what it takes for a cell to incorporate foreign DNA into their own DNA. This also helped us learn how viruses are able to take over an organism.  

Introduction 

In this experiment we did genetic transformation. Genes contain pieces of DNA that provides the information to make the proteins. Proteins gives an organism their trait. When changed caused by genes it is consider genetic transformation and many use in science. One application would be biotechnology in the fields of agriculture and medicine. For example, with bacteria by moving genes with the help of the plasmid, which the circular DNA. Plasmid DNA usually contains more than one trait that helps the survival of bacteria. Bacteria can transfer plasmids to adapt to new environment. Bacteria becomes resistance to antibiotics because of this transfer. Green fluorescent protein or GFP can by transformed by plasmids. It causes them to glow. When the transformation is done the bacteria will glow fluorescent in the dark. The PGLO plasmid codes for this gene that is resistant to antibiotic ampicillin. To control the rate of proteins that transferred into a cell caused by gene regulation system. Sugar arabinose can be active by GFP. This procedure takes place on antibiotic growth plate

Methods

First, we put in transformation liquid into
each tube; one labeled +pGlo and another
-pGlo.

Then we put them on ice .

Next, we transfered e coli to the tubes. 

We added plasmid DNA only to the
+pGlo tube.

Then we cold shocked the tubes for
10 minutes.

While we waited we labeled our petri dishes

Next we heat shocked the cells by putting
the tubes in a water bath, and back again into
the ice.

Then we added LB nutrient broth to the
mixtures, and mixed.

Lastly, we spreaded the solutions onto
the petri dishes, and let them
incubate for one day. 

Data and Graphs:

Discussion:

The LB/-PGLO plate had the most bacteria that resembled the original untransformed E. coli colonies we initially observed. This makes sense, if we take into consideration the fact that the bacteria were removed from the starter plate, did not have any plasmid added to them, and only had LB (this is broth, or in simpler terms, food) on the plate. Regulating these features makes it a control plate. The other control plate is the LB/amp/-PGLO plate, which had zero growth, because no plasmid was added. Plasmid actually has to be added in order for bacteria to multiply in the presence of ampicillin. The transformation plates include LB/amp/+PGLO and LB/amp/ara/+PGLO. First, both plates had the plasmid for the fluorescent gene added to the E. Coli spread on them. Because of the heat shock we gave the E. Coli, holes could be made in the cell membrane of the prokaryote cells. This means that the PGLO plasmid which expresses the ampicillin resistance gene is incorporated into the E. Coli. Thus the bacteria can survive on the plates that contain ampicillin. On the other hand, cells that were not treated with DNA did not express ampicillin resistance and did not grow on the LB/amp/-PGLO plate. The LB/amp/ara/+PGLO plate had about the same amount of growth as the other transformation plate, but the genes of the plasmid will only be expressed in the presence of arabinose, so it’s on this plate that the bacteria glowed under a UV light. The transformation efficiency (efficiency by which cells take in extracellular DNA and express its genes) for this plate is 406.25 transformants/microgram. This basically reflects how competent prokaryotic cells are at including new DNA. Our results are what they were supposed to be, and they support our belief that only the cells with DNA added to them would be genetically transformed. One way to improve this lab could be to let it go on for more than one day. That way we would try to calculate rate of reproduction in order to grasp how quickly plasmids can be generated in new cells.

Conclusion :

 We found that we were able to transform the DNA of this organism using plasmids. We concluded that E. Coli did not grow. Some bacteria was able to live in the surroundings with ampicillin  and glow under UV light in a environment with arabinose. Based on these  result our  hypothesis came true. 

Gel Electophoresis Lab

Background Information:

Gel Electrophoresis is a type of DNA technology that's used to separate nucleic acids or proteins that differ in size, charge, or other properties. The separation of the molecules depends on the rate of their movement in a polymeric gel in an electrical field. The distance traveled by a DNA molecule is inverse to its size. In other words, if the DNA molecule is larger in size, then it will travel a shorter distance because it has more resistance due to its size. On the contrary, smaller DNA molecule have less resistance so they travel faster down the gel. The DNA is always placed on the cathode side (negative) because DNA is also negative so when the power source is added the  the DNA will repel away from the cathode and towards the anode(positive) side. Restriction fragments can be used to cut up the DNA molecule into bands.
Purpose:
The purpose of this lab was to use restriction enzymes to sequence DNA. Using single, double, and triple digests, we tried to figure out the number of cut sights present in the DNA sequence for each enzyme as well as the position of those cuts relative to one another.

First our teacher cast an agarose gel with wells included.

Using a pipet we loaded the contents of a reaction tube (DNA with restriction enzymes) into a well in the gel.

We repeated this procedure with each reaction liquid into a different well.

When we finished loading it looked like this.

We put our gel in the electrophoresis chamber, and allowed the DNA to electrophorese until the bromphenol blue band was about 2cm from the end of the gel.

We removed the gel.

And we examined it on a light box. We assigned sizes to the lambda DNA size marker bands, then approximate sizes to the unknown DNA fragments, and determined the total size of digested DNA.

Data and Graphs

This shows the marker's lengths, so that we can
estimate the lengths of the DNA strands for the rest of the lanes.

Discussion: 

 Gel electrophoresis was used to determine the size of the unknown DNA sequence. The size of sequence was determined to be  about 5500 base pairs in length by comparing against a known lambda DNA sequence. When the unknown DNA sample was cut with restriction enzyme PstI, there were two bands at approximately 750 and 4700 base pairs.  The PstI had cut the unknown sequence twice. PstI and SspI there were three bands at approximately  750, 2140, and 2700 base pairs. This shows that the SspI sequence had only cut once. PstI and HpaI there are two bands at approximately 750 and 4700 base pairs.  Since there were only two bands, this shows that adding the HpaI did not cut the sequence. PstI, SspI and HpaI there are three bands at approximately 750, 2858 and 1093  base pairs. This is expected after since PstI cut twice, SspI cut once, and Hpal did not cut at all; giving a total of three cuts and three bands on the gel. Further investigations could look into the discrepancy between the bands in the lane containing PstI and SspI and the lane containing PstI, SspI, and Hpal.

Conclusion: Our results demonstrate how different restriction enzymes cut the plasmid into different sizes. All the different wells should have DNA fragments of the same size because we used the same DNA, but our variables were the restriction enzymes that digested the DNA at different locations. PstI only cut the plasmid into 2. PstI and SspI combined created a plasmid with 3 different sections. PstI and HpaI also created 3 different sections even if they one of them wasn't clearly visible on the gel. The last well with all the enzymes shows a combination of the fragments.

DNA Strawberry Extraction Lab

We began the extraction of DNA from out strawberries by squishing only the strawberries in the ziploc bag. We then added some buffer solution into the bag and mixed it well. The use of the buffer solution which consists of shampoo broke apart the cell membranes found in strawberries. Soap and shampoo molecules "have both properties of non-polar and polar at opposite ends of the molecule". This forces the polar heads and non-polar inside  of the plasma membrane to disassemble in order to join the shampoo molecules. Without a plasma membrane, the contents of the strawberry cells are released into the mixture. The salt in the buffer will allow the DNA to solidify.

 

 

 Info source: http://www.elmhurst.edu/~chm/vchembook/554soap.html

 We then filtered the contents in the bag so we could capture the DNA molecules.

Alcohol was added to the solution filtered. We poured the alcohol down the side of the test tube so that we could create a separate layer of the alcohol on top of the filtered solution. The addition of alcohol forces the DNA to become like a solid.

 The white, cloudy material at the top of the solution is our strawberry DNA.

Mitosis 101

Summary of Mitosis

Cell Communication

Introduction:
Yeast cells have two "sexes," called a-type and alpha-type. They locate each other via specific secreted factors. The exchange of factors causes them to mate. When these two cultures mix, the haploid cells (meaning they only have one complete set of chromosomes) become gamete cells (a mature sexual reproductive cell). The yeast cells halt asexual reproduction and grow into pear-shaped gametes called schmoos. When a-type and alpha-type come into contact with each other, they fuse, and the haploid nuclei form a diploid nucleus.

Purpose: The purpose of the lab was to analyze the cell signaling that occurs between the alpha-type and a- type yeast cells. The mating interaction of the two yeast types mixed together was the focus of the experiment. Our results for the mixed culture should support the idea of yeast cells mating with the use of special factors that attach to the opposite mating type. The characteristics of each yeast cell type were also individually analyzed.

Methods: 

1.We labeled 3 agar plates and 3 culture tubes as alpha-type, a-type, or mixed.

2. We added about 2 mL of sterile water to each tube and transferred a small amount of each yeast type into its designated tube.

To do this we used a toothpick to gather yeast cells and mixed the yest cells with the sterile solution by swishing the toothpick in the solution. For the mixed culture, we gathered a-type cells with a toothpick and placed it into the solution. Then with a different toothpick we gathered alpha-type cells and placed it in the same solution. Drops of yeast suspension were placed on their designated plates. We used different cotton swabs like in the picture below to spread the suspension on the agar plates.

 

 We then looked at the yeast under the microscope at different intervals of time to see what occurred in the life cycle. We captured images at 0 minutes,30 minutes, and 24 hours later. Unfortunately, the 0 and 30 minute images are unable to upload, but below are images of the alpha-type, a-type, mixed-type, and experiment yeast suspensions on the agar plates after 24 hours.

Mixture of alpha and a type saw the most growth on the agar plate. Best cell communication

alpha type yeast saw some growth too, but still not as dense as the mixed plate. Mating factors of the different yeast type can signal the opposite sex type, which impedes cell communication.

a type yeast cells grew a bit

we placed alpha and a type yeast on opposite ends of the plate (separated). Growth occured in the sides of each individual type but failed to develop towards the middle of the plate since the mating factors never or barely came in contact.



Data/graph

 

Discussion 

A large part of analyzing this lab is noticing the differences in the yeast over a period of time. In order from least amount of change to most amount  of change overnight it goes; the mixed separate dish, a-type, mixed, and then alpha-type. The reasons that the separate mixed was the least had to do with the fact that we started the yeast on opposite sides of each other. This makes it difficult for the yeast cells to sense the actual amount of cells in the dish, and therefore they do not shmoo as much as they should. Even when they do, it is a far distance across the dish for these little cells to get to. This slows down the overall process even more. The reasoning for the mixed being one of the highest producing ones is because it was mixed. This gives more opportunity for diversity within the cells and makes them able to produce faster. The reasons, I believe, that alpha was the largest change, was just due to its genetic make-up. It was obviously more adapted to move and produce quickly. This experiment proves how cells communicate with one another. After they got to a certain density, they stopped producing. We were also able to see how they changed their cytoskeleton to reach one another. This experiment truly shows how cells signal one another. 

Conclusion:
This lab proves that yeast cells communicate through pheromones, also known as chemical signals transmitted between organisms. Because a-type and alpha-type cultures changed into their gamete form, they detected a signal from the opposite type. Also based on the fact that alpha-type yeast changed the most, we can conclude that this type releases the most mating factor.

Plant Pigments and Photosynthesis

Plant Pigment Chromatography
Purpose: In this experiment, we were trying to separate and identify pigments as well as other molecules found in plant extracts. We needed to calculate the Rf constant, which represents the relationship between the distance moved by a pigment to the distance moved by the solvent. This helped us determine the factors involved in the separation of pigments.

Introduction:  Paper chromatography separates the components of cell extract. Different molecules and pigments move up the paper at varied rates due to differences in solubility, molecular mass, and hydrogen bonds (1). Chlorophyll a, the primary pigment that absorbs light, absorbs mostly violet and blue light for photosynthesis. Chlorophyll b is an accessory pigment that broadens the absorption spectrum on the spinach leaf by absorbing different wavelengths than chlorophyll a. Chlorophylls contain oxygen and nitrogen and have a greater affinity for the paper (3). Carotenoids are also accessory pigments that absorb violet and blue-green light. Their function is to perform photoprotection to dissipate excessive light energy that could damage chlorophyll pigments. They are very soluble and form no hydrogen bonds with the paper. Xanthophyll is a division of the carotenoid group that has a similar structure to carotenes but contain oxygen atoms and create hydrogen bonds with the paper. (2).

Methods:
We got a 50mL graduated cylinder that had 1 cm of solvent in the bottom. Then we cut a piece of filter paper long enough to reach the solvent and made sure the end was cut into a point.
We used a coin to crush leaf cells about 1.5 cm above the point of the paper. By rubbing the coin against the leaf, we were able to extract pigments.

We put the filter paper in the cylinder so the pointed end was barely immersed in the solvent, and stopped the cylinder.

When the solvent moved about 1cm from the top of the paper, we removed the paper and marked the location of the solvent and the bottom of each pigment band. Next we measured the distance each pigment moved from the bottom of the pigment origin to the bottom of each band.

 

Data and Graphs

Discussion:

In this lab we found out that the solubility, size of particles, and their attractiveness to the paper are all involved in the separation of pigments.  The different solubility’s of the pigments would change the Rf values. The reaction center contains chlorophyll a. The other pigments collect different light waves and transfer the energy to chlorophyll a.  Xanthophylls went furthest up the paper. We examined that the closer the rf factor to each other the distance  of the pigment traveled is closer to the distance traveled by the solvent. The separation of pigment in chromatography allowed us to look at the different pigments there in the plant. We can tell if a plant will reflect the color that showed and doesn’t absorb as much from the light and wavelengths. The orange, yellow, and green light will be somehow reflected from the Spinach leaves. To find the RF you take the distance pigment migrated divided by the distance solvent. The RF for band 1 is .252 mm and RF band 2 is .326. Band 3 is .467 and RF band is .585. The larger the RF is, then the more distance that was traveled by the pigment. Pigments like the carotene have the highest RF factor since they are the least polar and travelled the most. The chlorophyll pigments are extremely polar and have a high affinity for the paper which slows them down.

 

  
Conclusion: The 5 different pigment bands seen on the paper demonstrate that a mixture of these pigments is needed for photosynthesis to occur. The light spectrum of each pigment provides the spinach leaves with a large source of light that can be used to power the light reactions of photosynthesis. The first line of pigment from the bottom appears to be chlorophyll b since it is the most polar and most soluble due to its carbonyl group. This high polarity inhibits this pigments ability to move up the paper. Chlorophyll a is the next most polar which makes it the 2^nd pigment line. The less polar pigments are able to travel up the paper since they are less likely to create interactions with the paper. Errors that could’ve have occurred might be calculation errors and measurements.

Sources:

1.     http://www.phschool.com/science/biology_place/labbench/lab4/concepts1.html

2.     http://en.wikipedia.org/wiki/Xanthophyll

3.     Lab Introduction

Photosynthesis:
Purpose: We were trying to test the hypothesis that light and chloroplasts are required for light reactions to occur. To do this we had to measure the transmittance and absorbance from four cuvettes, each of with contained a different mixture. By measuring transmittance/absorbance, we were able to determine photosynthetic rates. We also took data at different times to see how light intensity affects the rate of photosynthesis.

Introduction: Plants depend on light energy to fuel the light reactions of photosynthesis, which produces the reactants of the light independent reactions. The absorption of light occurs in a photosystem light-harvesting complexes which contain various pigments (discussed in lab 4A) that harvest light and send it to the reaction-center complex. Electrons are boosted to high energy levels and must be carried by the electron transport chain and another photosystem in order to return to a more stable condition (1). In this process, ATP and NADPH are produced. NADPH is produced by the reduction of NADP+.  DPIP replaces the electron acceptor NADP+ in dye-reduction. DPIP begins with a blue liquid and as it’s reduces it becomes colorless.

Methods:

We set up an incubation area that included a light and a water flask.

Then we prepared each cuvette. The first one included 1mL phosphate buffer, 4 mL H2O, and three drops of unboiled chloroplasts. This served as our blank, and we used it to calibrate the colorimeter (we measured the light transmittance through each of the other tubes as a percentage of light transmitted through this tube).

The other cuvettes all had the phosphate buffer, H2O an DPIP in them. Cuvette 2 was covered in foil and had 3 drops of unboiled chloroplasts in it. 3 had three drops of unboiled chloroplasts. 4 had 3 drops of boiled chloroplasts. 5 had no chloroplasts whatsoever.

After mixing cuvette 2, we removed the foil sleeve and put it into the colorimeter. Then we recorded % transmittance and absorbance at time 0. We replaced cuvette 2 in the sleeve and placed it in the incubation area. We took and recorded additional data at 5, 10, and 15 minutes. With each cuvette, we did the same steps, only no foil sleeves were involved.

Data and Graphs

 Discussion:

The graph shown above proves that there is an inverse relationship between transmittance and absorbable. As the amount of blue dye solution decreased, transmittance increased.

Each curvette had a specific purpose:

Cuvette 1 with no DPIP and chloroplasts was used to calibrate the colorimeter. The difference between unboiled and boiled chloroplasts is that boiling chloroplasts denatures them which changes the shape of their protein and might lead to changes in function. Boiling chloroplast reduces the efficiency of these chloroplasts which negatively impacts the rate of photosynthesis.

Cuvette 2 with unboiled chloroplast,DPIP, and no light (due to the aluminum foil wrapped around) supported the idea that light is needed for the rate of photosynthesis to increase. Cuvette 3 with unboiled chloroplasts, light, and DPIP demonstrates that perfect conditions can result in a quick rate of photosynthesis. The chloroplasts are functioning properly, light fuels the light reactions that produce products used in the dark reactions, and DPIP acts as a new electron carrier. Cuvette 4 had all the same elements of Cuvette 3 except that Cuvette 4 had denatured chloroplasts, which negatively impacted the rate of photsynthesis. Finally, Cuvette 5 was used to demonstrate that DPIP can't be reduced on its own. Chloroplasts are needed to reduce DPIP, even if light is present DPIP can't function without the light that excites electrons.

In this lab the DPIP is the electron acceptor in this experiment. The molecule that found in chloroplasts is DPIP and substitutes for the NADP molecules. The source of the electron that will reduce DPIP is the electrons that come from the photolysis water.  The amount of light transmittance is measured by a spectrophotometer. The effect of darkness will have no reaction take place. The effect of boiling the chloroplast on the subsequent reduction of DPIP will stop the reduction.   The difference in the percentage of transmittance between the live chloroplasts that were included in the light and those that were kept in the dark was the light energy. In the dark there isn’t a flow of electrons and photolysis water while the light does. In Cuvette one was our control and set to 100% transmittance. Cuvette two was light reaction work in dark.  In cuvette three was light reaction work in live chloroplast. Cuvette four boiled chloroplast.  Finally, in cuvette five shows us that chloroplast is needed in plants.

Conclusion: The spectrophotometer measured the light transmittance through the cuvettes and chloroplast solutions. The biggest change in transmittance (low to high) occurred in the cuvette with unboiled chloroplasts that was placed in the light. This proves that the rate of photosynthesis increased the most. Factors like light availability and denaturing of chloroplasts affected the rate of photosynthesis. Errors could have occurred from simple mistakes in calculations. The amount of time that the cuvettes spent in front of the light might not also be exact which could lead to skewed data.

Sources

1. Biology Book