pGLO Transformation Lab

Introduction & Purpose:

For this lab, we were trying to genetically transform colonies of E.coli so that they would display a trait that allowed them to become bioluminescent (glow) as well as become antibiotic resistant. To do this we took a gene called Green Fluorescent Protein (GFP) that is originally found in jellyfish and inserted it into the dish of E.coli colonies. We also inserted a gene that allows bacteria to overpower antibiotics. Although the E.coli bacteria were never designed to become bioluminescent, they have the ability to transfer plasmids, causing them to share the new genes we inserted into them. The purpose of this experiment was to help us understand the procedure of genetic transformation. We were able to observe how different factors allowed the bacteria to transform. We learned how to insert new genes into the bacteria, and why the bacteria colonies were able to display the new traits.   

Procedure:

Analysis:

Through the lab it was determined that E. coli can be genetically modified to glow in the dark under UV light. Through the method used, it was determined that this could be done by inserting E.coli into an agar plate that contained quantities of the plasmid known as pGLO, along with ampicillin and arabinose. The plasmid pGLO contains the GFP gene which is what allowed the E. coli to glow in the dark by coding for a green fluorescent protein. It is also important to note the total amount of growth for each plate, especially the plates that had the least and most growth.

As predicted, the two control plates, those labeled -pGLO and -pGLO LB/amp exhibited the least amount of growth, with the -pGLO exhibiting no growth. The plates that exhibited the most amount of growth were labeled  +pGLO LB/amp and +pGLO LB/amp/ara, with the latter plate having the most growth out of the four plates. This may be due to the bacterias increased resistance to ampicillin.  

 
This would also explain why the control plates has less growth than those with the pGLO plasmid. The plate with the -pGLO had no growth because it had no LB broth or arabinose to stimulate growth. The plate labeled -pGLO LB/amp did not have the plasmid necessary for antibiotic resistance to ampicillin, and therefore the E. coli could not grow. However, the plate labeled +pGLO/amp/ara had both the plasmid necessary to encode for antibiotic resistance, as well as ambroise which will induce the bacterial cell to express the fluorescent trait and grow.      

 To make the environment within the the plate most suitable for cell growth it is then necessary to add the pGLO plasmid, the ampicillin, and the arabinose. It is plausible to believe that the E. coli cells will grow with the added presence of arabinose because this chemical is what begins to stimulate the E. coli growth with both the expression of the fluorescent trait, along with antibiotic resistance. For the transformation efficiency of the experiment it was determined that there were  transformants/μg. Because it is common for the transformation efficiency calculations to result in higher numbers it is possible that there might have been an error while conducting the experiment. It is plausible to believe that too little of LB broth or arabinose was added to the agar plate. It is also possible that too much ampicillin was added.       

While we obtained the results predicted, to obtain more accurate results, it would be suggested to take more accurate measurements of the time the plates were in ice or into the warm water, along with perhaps better, more detailed written observations of the transformation and control plates. It would also be beneficial to make sure that each plate contains very precise amount of LB broth and ampicillin to get the most accurate calculations of the efficiency of transformation by getting the desired growth rate in the E. coli. To improve this calculation, it is also possible to take more care in counting the bacterial colonies on the agar plates after the experiment by using a microscope and more concentrated UV lights.

Gummy Bear Lab

*This lab was done in collaboration with Amanda Diana & Ananya Bhattacharya

1.) 30 Pink: 10 White

Hypothesis: Since pink appears to be the dominant trait, having a pink:white, 3:1 ratio, the parents of these offspring must both be carriers for the recessive white.

Analysis:

As previously stated, this group of offspring has a 3:1 ratio, exhibiting simple Mendelian dominance. In order to get this ratio, the crossed parents must be two heterozygous pink gummy bears (Pw x Pw).

The result of crossing these two parents is three pink gummy offspring (one homozygous and two heterozygous) and one homozygous white gummy offspring, or a 3:1 ratio. Thus proving that this litter of gummy bears follows simple Mendelian dominance.

2.) 15 Pink: 30 Pink-White: 16 White

Hypothesis: Here we have an interesting batch of offspring. ¼ of the offspring are pink, ¼ are white, and ½ are pink and white. This appears to be a 1:2:1 (pink: pink-white: white) ratio. The parents of these offspring most likely are carrying a gene for white and pink, and these genes can both be expressed simultaneously.

Analysis: Because approximately half of the offspring are showing both traits, it can be assumed that pigment trait is co-dominant. This means that heterozygous genotypes will code to show both yellow and white, as opposed to complete dominance, where in heterozygous genotypes the more dominant gene surfaces.The parents for these offspring are infact carrying a gene for white and pink, as is shown below.

This also supports the hypothesis because the ratio of possible offspring here are also 1:2:1.

3.) 20 Pink: 18 White: 14 Green: 19 Pink-Green

Hypothesis:

This particular set of offspring is interesting in that four different phenotypes are exhibited. Since the offspring exhibits four different possible pigmentations it is likely to believe that the mode of inheritance may be multiple alleles, similar to the mode of inheritance for blood types.

Analysis:

Due to the evidence shown in this punnett square, we can conclude that the mode of inheritance is multiple allele. All possible phenotypes are demonstrated in the genotypes of the potential offspring shown in the punnett square. There is also a 1:1:1:1 ratio demonstrated in the Punnett Square equivalent ratio for the phenotypes of the offspring, supporting the idea that the dominance for this set of gummy bear genes is multiple alleles.

4.) 38 White

Hypothesis: In this grouping, all 38 gummy bears were white. Since all of the offspring share a phenotype, we can conclude that the parents must be homozygous white. The entire litter would not be all white, since white has proven to be a recessive trait in every other litter thus far, unless both parents had a genotype of ww.

Analysis: The offspring have a ratio of 1, meaning all offspring are white. These offspring are displaying simple dominance. To yield theses results, the parents of this litter of bears must have been homozygous white (ww x ww).

By crossing these two bears we see that all the offspring will share a genotype and phenotype. (ww, white).

5.) 15 White: 15 Pink: 30 Orange

Hypothesis: In this grouping, there are 15 white / yellow gummy bears, 15  pink gummy bears, and 30 orange gummy bears. The ratio of the offspring is 1:2:1 (Red:Orange:White). From these results we can assume that both parents have a genotype of heterozygous orange. We know this because heterozygous parents will always display a 1:2:1 ratio when expressing incomplete dominance.

Analysis: The offspring have a ratio of 1:2:1, meaning the amount of orange offspring is double the amount of white and pink offspring. This litter is displaying incomplete dominance, in which the parents are heterozygous orange gummy bears (as displayed in the punnet square below, with PW = orange).

In this graph, we are comparing the phenotype frequency for each gummy bear phenotype.

Mitosis / Meiosis Lab

3.A.1: Observing Mitosis in Plant & Animal Cells Using Onion Root Tip & Whitefish Blastula

Introduction / Purpose:
In this lab we learned about the processes of Mitosis in plant and animal cells. By using a microscope we were able to get a clear view of the cells in an onion root tip and a whitefish blastula. In these cells we observed the different stages of mitosis and were able to get a visual understanding of how cells divide. By seeing all the stages of mitosis, we learned what happens in each stage.

Procedure:
Examine prepared slides of either onion root tip or whitefish blastula. Locate the meristematic region of the onion or locate the blastula, with the 10X objective and then use the 40X objective to study individual cells. For convenience in discussion, biologists have described certain stages of the continuous mitotic cell cycle, as outlined on this page and the next. Identify one cell that clearly represents each phase. Sketch and label the cell in the boxes provided.

Image Credit goes to our fellow Early Bird Biology classmates

Analysis:

The purpose of this section of the experiment was to observe and gain experience in how an onion root tip cell looks when it is undergoing each phase of the cellular cycle. The observation of these cells helped us see how each cell looks like during each phase of the cycle, and how plant and animals cells may be different when dividing since a plant cell must accommodate for the presence of a cell wall.

3.A.2: Time For Cell Replication

Introduction / Purpose:
For this experiment we used a microscope to further observe the cells of an onion root tip. Once again we were able to observe the process of mitosis and examine the differences between each stage. We were instructed to take a tally of how many cells were seen in each stage. Through our results were were able to get an understanding of which stages take the longest amount of time. 

Procedure:
1. Observe every cell in one high-power field of view and determine which phase of the cell cycle it is in. This is best done in pairs. The partner observing the slide calls out the phase of the each cell while the other partner records. Then switch so the recorder becomes the observer and vice versa. Count at least two full fields of view. If you have not counted at least 200 cells, then count a third field of view.

2. Record your data on Table 3.1

3. Calculate the percentage of cells in each phase, and record it in Table 3.1  

Analysis:

The purpose of the experiment was to try to estimate the amount of time a cell spends in each stage of the cell cycle by observing onion root tip cells under a microscope and counting the number of cells in each phase. The phase in which most cells were observed to be in would be the longest stage. The stage in which the least amount of cells were observed to be in would be the shortest stage. From the data we gathered, it was determined that the highest number of cells were currently in Interphase, thus making it the stage in which cells spend the majority of their time. From the rest of the data and the times recorded, it is plausible to believe that the cells spend the least amount of time in the Telophase stage of the cycle. The cycles from longest to shortest, as demonstrated by the data, were Interphase, Prophase, Metaphase, Anaphase, and Telophase. A way to improve the data and gather perhaps more accurate results would be to do a double count of the cells, or have a technique to more accurately count the number of cells in each phase.

3.B.1: Simulation of Meiosis

Introduction / Purpose:
In this experiment we created visual representations of each stage in meiosis. This allowed us to gain a better understanding of what occurs in each stage, and in what order the processes are completed. After this experiment we should be able to identify each stage and understand what occurs in that stage. 


Procedure:

  

 

Analysis:

Through this exercise, it was possible to observe the results of crossing over. It was helpful to observe a difference between Meiosis 1 and 2 during the exercise. Meiosis 2 separates the chromatins of the chromosome into two daughter cells all formed by the division of Meiosis 1. Several difference between Mitosis and Meiosis were also noted. Meiosis produces haploid cells, only occurs with gamete, and occurs in two staged. Mitosis produces diploid cells, occurs with somatic cells, and does not include a Mitosis 2.

3.B.2: Crossing Over During Meiosis

Introduction / Purpose:
In this experiment we viewed photos of hybrid Sordaria asci. We observed these photos and identified all the points of crossing over. It was difficult to see all the points at which crossing over occurred, but by studying the photos we were able to get a better understanding of the process. The purpose of this experiment was to help us understand a visual representation of crossing over.

Procedure:
1. Obtain two images of hybrid Sordaria asci containing both tan and black ascospores.

2. This is best done in pairs. Have one partner count off the amount of non-crossing over 4:4 asci while the other partner keeps a tally. Write down your results in Table 3.3

3. Next have one partner count off the number of crossing over asci while the other partner keeps a tally. Asci patterns can appear in a few different varieties (image in column 2 in Table 3.3). Write down your results in Table 3.3

4. Add number of non-crossing over 4:4 asci with the number of crossing over asci to get the total amount of asci. Write down your results in Table 3.3. To calculate the map unit (an arbitrary unit of measurement used to describe relative distances between linked genes) use the number of asci showing crossing over and divide it by the total amount of asci. Take this number and multiply it by 100. Write down your results in Table 3.3 and create a simple sketch in the box provided.

Analysis:

One of the main observations made from the experiment was that genes further away from the centromere are more likely to cross over to the other chromosome. This was seen as the sordaria sample  greater amount of crossing over occurring also had a larger distance separated the genes and centromere than the sordaria with less crossing over occurring. It may also be more likely that genes further from the centromeres are more likely to cross over because it is the ends of the chromatids that crossover and exchange genes.