pGLO Transformation Lab

pGLO Transformation Lab

Purpose

In this experiment, we aimed to gain an understanding of the process of DNA transformation. We were tasked with discovering what conditions were required to successfully insert the pGLO plasmid into E. coli bacteria to produce a sufficient transgenic organism, in addition to figuring out the purpose of the arabinose sugar’s purpose in the petri plate.

Introduction

DNA naturally contains plasmids, small circular pieces of DNA, that are necessary for survival. The transformation of plasmids allows bacteria to be antibiotic resistant. The bacteria used is E. Coli. E. Coli has an inducible operon system, meaning that a protein is needed to make the repressor inactive, turning on protein synthesis. The protein used is arabinose. The protein causes the gene for the green fluorescent protein (GFP) to be turned on. The pGLO plasmid has the genes for GFP and antibiotic resistance to ampicillin.

Methods

  There are several components used in the creation of the four different plates examined: agar, LB nutrient broth, ampicillin, arabinose sugar, calcium chloride, and pGLO. Agar is the medium off of which the bacteria grow, and the LB nutrient broth is the “food” needed for their growth. Ampicillin is an antibiotic that kills E. coli cells. When arabinose sugar is added, the gene for Green Fluorescent Protein (GFP) is switched on in transformed cells. Calcium chloride is the transformation solution used to allow the pGLO gene into the E. coli cells.

First, label one micro test tube -pGLO and the other +pGLO. Transfer 250 microliters of calcium chloride into each of the two test tubes.

Above: the four tubes--the blue and purple are the LB broth and transformation solution, the green and yellow are the +pGLO and -pGLO suspensions

Place these two tubes on ice.

Above: immersion of the sterile loop into a tube with a colony of E. coli bacteria

Pick up a single colony--a small group of circular cells--of E. coli from its plate with a sterile loop. Immerse the sterile loop into the transformation solution at the bottom of the +pGLO tube. Spin the loop between your fingers until the colony has been dispersed throughout the solution. Return this tube to the tube rack on ice. Repeat with the -pGLO tube with a new sterile loop.

Above: the sample plate of E. coli bacteria colonies

Using a new sterile loop, immerse the loop into the pGLO plasmid DNA stock tube, withdrawing a loopful. A see-through film of plasmid solution should cover the ring. Then, mix the pGLO plasmid DNA into the suspension in the +pGLO tube in the same way done with the E. coli colony. Close the tube and return it to the rack on ice. Then, close the -pGLO tube, but DO NOT add the pGLO plasmid DNA to it.

Above: the pGLO plasmid DNA stock tube

Below: the sterile loop being immersed in the pGLO plasmid DNA

Let the rack rest on ice for 10 minutes. In the meantime, label the four agar plates on the bottom as +pGLO (LB/amp), +pGLO (LB/amp/ara), -pGLO (LB/amp), and -pGLO (LB).

Above: the -pGLO and +pGLO suspensions on ice

Above: labels on the four agar plates

After the 10 minutes, transfer the rack into the warm water bath set at 42°C for 50 seconds. Quickly transfer the rack back to the ice and leave for 2 minutes.

Above: the rack in the warm water bath

Below: returning the tubes to the ice

Open one of the tubes. Add 250 microliters of the LB nutrient broth to it and reclose it. Repeat with the other tube. Then, leave both tubes for 10 minutes at room temperature.

Use a pipet to transfer 100 microliters of the suspensions onto each of the four nutrient agar plates.

Below: transfer of the suspensions to each of the four agar plates

Using four new sterile loops, one for each plate, spread the suspension evenly around the surface of each plate. Quickly skate the loop in a zigzag pattern across the agar surface, and do not press too deep.

Above & below: skating the sterile loop across the surface of the agar plate

Stack up the plates and tape them together. Label with the group name and leave the stack upside down in the incubator for a day.

Below: the stack of the plates for incubation

Data

The agar plate with +pGLO, LB nutrient broth, ampicillin, and arabinose sugar was the only one of the four that exhibited transformation of E. coli genes to take on GFP. Its bacteria clearly glowed a bright green under the UV light. Additionally, the bacteria on this plate were able to resist the effects of the ampicillin antibiotic.

Graphs & Charts

Graphs and charts were not necessary to the analysis of this lab.

Discussion

Overall, three of our four petri plates sustained growth overnight - both of the +pGLO plates and the -pGLO/plain LB plate. There was a total absence of any surviving E. coli on the -pGLO/ampicillin plate; since this bacteria was completely unaltered, we can safely say that E. coli is not naturally resistant to ampicillin. The only plate to successfully express the gene for GFP (i.e. glow under UV light) was the +pGLO/ampicillin/arabinose plate; none of the others were able to fluoresce under the UV light.

Conclusion

In order to determine that E. Coli did not naturally glow a -pGLO LB plate was made. So there would not be a possibility of bacteria naturally glowing without the insertion of a plasmid. A second plate was made, –pGLO LB/AMP, This control proved that bacteria that did not have the plasmid that contained the ampicillin resistant gene could not grow in an environment with ampicillin.

Chapter 16-17 Remediation (Kayla Ruiz)

Chapter 16-17 Quest Remediation

Missed Concepts and Terms:

• The translation of mRNA into polypeptides
• Regarding this topic, I misnomered the two steps of translation and the specificity between its components. First, aminoacyl-tRNA synthetase must match tRNA and an amino acid, then there must be a correct match between the tRNA anticodon and an mRNA codon. It does not truly matter whether the ribosome is specific to either of the mRNA or tRNA components; ribosomes only facilitate their coupling. 

• The structure and process of tRNA
• I forgot what happened at the jutting 3' end of the tRNA molecule. This is the site where amino acids attach during translation.

• Types of mutations
• While the term "point mutation" generally applies to any mutation that alters a single base pair, a missense mutation is the right name for the given mutation in the quest. The codon would still code for an amino acid, just not the correct amino acid. 

• Difference between eukaryotic and prokaryotic codons
• In this question, I probably confused the difference between eukaryotic and prokaryotic codons for the differences in the bases of RNA and DNA. The genetic code is nearly universal because it is used by the simplest bacteria and the most complex animal species. There are differences between codons, but the proteins to which they correspond are the same in all organisms. The language of codons is seen as "glowing" genes of jellyfish are transferred to other organisms.

• Addition of amino acids to the polypeptide chain
• With this, I forgot the three steps in the addition of an amino acid to the growing chain in tRNA. There is codon recognition with the tRNA in the A site, then a peptide bond forms between it and the tRNA in the P site. Finally, there is translocation: the tRNA is "kicked out" of the P site into the E site.
• In a different but related question, I should have identified the A site as the site where the codon is being read in the ribosome.

• What causes termination of transcription
• I forgot the difference between termination of transcription in prokaryotes and eukaryotes. In the former, the polymerase stops transcription at the end of the terminator and the mRNA can be translated without further modification. In eukaryotes, RNA polymerase II transcribes the polyadenylation sequence.

• Semi-conservative process
• Watson and Crick's semiconservative model states that when a double helix replicates, each daughter molecule will have one old strand from the parent and one newly made strand. This was examined in an experiment in which older strands were marked with a heavy isotope, newer strands with a light isotope, and bands of replicated DNA were examined. If DNA were replicated in a conservative manner, the older DNA strands would rejoin. It would be seen as two separate bands of DNA. 

• The purpose of telomerase
• This enzyme catalyzes the lengthening of telomeres in germ cells. It does not necessarily cause their shortening, however.

• The purpose of DNA polymerase
• This enzyme catalyzes the elongation of new DNA at the replication fork. It adds nucleotides only to the free 3' end of the growing strand so a new DNA strand can only elongate in the 5' to 3' direction.

• The purpose of telomeres
• A telomere is defined as the nucleotide sequences at the end of DNA molecules that postpone the erosion of genes.

Chapter 16 & 17 Portfolio

Chapter 16 & 17 Portfolio

Explain the structure of DNA and nucleotides. Include a model of your explanation.

DNA is a structure of two strands of nucleotides whose twisting shape accounts for the name double helix. Each nucleotide has three components: a deoxyribose sugar, a phosphate group, and one of the four type of of nitrogenous bases (adenine, thymine, guanine, and cytosine). The two strands of nucleotides are held together by hydrogen bonds between nitrogenous bases, specifically a bond between a pyrimidine and a purine. There is a specific pairing between bases: adenine goes with thymine, cytosine with guanine. The curving ridge of DNA is called the sugar-phosphate backbone. The two strands of DNA also run in antiparallel fashion so that the 5’ end of one faces the 3’ end of the other, and vice versa.

Photo credit to the National Human Genome Research Institute

Develop a model which explains the major steps to replication, specifically a replication bubble.
First and foremost, the enzyme helicase unwinds the double helix structure of DNA and creates a replication bubble, which is essentially the active site for replication. Topoisomerase enzymes ahead of the helicase's replication fork relieve the twisting stress on the unwinding DNA to prevent the molecule from breaking. Single-stranded binding proteins keep the DNA strands steady and optimize them for replication, which begins with primers. Short sequences of RNA called primers attach to the origin of replication and kickstart the actual process of recreating DNA. The leading strand (on the 3' side of the primer) is continually elongated by DNA polymerase III, which adds free nucleotides to the growing DNA strand. On the 5' side of the original primer is the lagging strand, where more primers attach to the DNA and are elongated back towards the 3' end, synthesizing the DNA in pieces rather than consistently. Once all elongation is completed, DNA polymerase I removes the RNA primers from the new DNA strands and puts the correct nucleotides in their places. Finally, ligase joins the segments created by polymerase I and all of the lagging strand pieces, called Okazaki fragments, together with the leading strand to finalize the fully formed DNA molecule.

Figure 16-UN3 and 16-16b6 from Chapter 16: The Molecular Basis of Inheritance PowerPoint Lecture Presentation for Campbell Biology by Chris Romero and Erin Barley

Compare and contrast the difference between replication, transcription, and translation.

    Replication, transcription, and translation are similar in the sense that all 3 processes involve the "reading" of DNA or RNA to create new molecules. All 3 processes use enzymes and other proteins in order to make the desired product from the given DNA or the RNA created from it.
    The key differences in replication, transcription, and translation lie in their purposes. DNA replication occurs during the S phase in cell growth to double a cell's DNA content in order to prepare for mitosis or meiosis. Transcription and translation occur in conjunction with one another and are used in the process of gene expression. Transcription creates single-stranded RNA from DNA, and translation uses that RNA to build the proteins originally coded for in DNA.

Develop a model and explain how DNA is packaged into a chromosome.

  It is only in eukaryotic organisms that a linear strand of nucleotides, one long DNA molecule, is formed into a chromosome through association with large amounts of proteins called histones. One of the first steps in chromatin packaging is the development of nucleosomes. These are bunches of eight histones with 10 nm unfolded chromatin wrapped twice around them with unwound “string” in-between. The next level of packing creates chromatin 30 nm in thickness; interactions between nucleosomes cause the 10 nm fiber to fold. Then, 30 nm fiber loops around a scaffold composed of proteins to make a 300 nm fiber. The looped domains then fold mysteriously into a packaged chromosome of 700 nm width. In prokaryotic organisms such as bacteria, there are dense regions of DNA called the nucleoid where chromosomes are tightly coiled. Proteins cause the packing, but the process is much less complex.

Figure 16-21a from Chapter 16: The Molecular Basis of Inheritance PowerPoint Lecture Presentation for Campbell Biology by Chris Romero and Erin Barley

Compare and contrast the key terms gene expression, transcription, and translation.

  Gene expression is the broadest of the three terms. It is defined as the process by which DNA directs the synthesis of proteins and of RNA molecules involved in protein synthesis. In a larger context, the DNA inherited by an organism dictates which physical traits are going to be passed on to the generation under observation. Transcription and translation are the two steps of gene expression. Transcription is the process of synthesizing RNA from DNA, usually the mRNA that is so important to making proteins. The nucleotide “language” of DNA is simply translated into a slightly different one that makes up instructions understandable to protein-makers outside the nucleus. Translation is the synthesis of a protein using the instructions of mRNA. There is another change in language in this process as the nucleotide sequence of RNA is coded into a series of amino acids by ribosomes.

Photo credit to Bio-Social Methods Collaborative at University of Michigan, http://biosocialmethods.isr.umich.edu/epigenetics-tutorial/

Develop a model and explain the process of transcription and translation.

Explain how eukaryotic cells modify RNA after transcription and why it is necessary.

  One type of modification to RNA following transcription is the alteration of mRNA ends. A 5’ cap is a modified form of guanine added to the 5’ end of the strand. A poly-A tail is a series of 50-250 adenine nucleotides at the 3’ end. This type of modification serves three purposes: to help export mRNA from the nucleus, to help protect it from hydrolytic enzymes, and to help ribosomes attach to the 5’ end later on.

   Another type of modification to RNA is RNA splicing. Because there DNA sequences are so long but much shorter sequences are needed to code for the proteins, there are long stretches of RNA that serve little purpose in protein synthesis. Removing these sequences makes protein synthesis more efficient. The noncoding segments of nucleic acid are introns while the other regions are caused exons. Spliceosomes are large complexes of proteins and small RNAs that bind to sequences, release introns, and join the exons on either side of the introns. Below is a visual of how RNA splicing removes portions of sequences.

Photo credit to www.bio.utexas.edu and Pearson Education, Inc.

Develop a model which explains how point and frameshift mutations can impact a protein.
Point mutations are the smallest-scale mutation that DNA can experience, as only a single nucleotide pair is affected; in some cases, the mutation has no effect on the completed protein, as each amino acid has multiple similar combinations of 3-base codons to keep order in spite of mutations. The non-effective mutations are considered silent and usually occur as 2 paired nucleotides (usually in the 3rd position of the codon) switching sides of the DNA double helix, but producing a new codon for the originally coded amino acid. For example, if the original codon in a gene was GGC for glycine, and the third G-C pair flipped, the new codon would be GGG, which also codes for glycine. Other point mutations, however, are not silent; these are called missense and nonsense mutations. Missense mutations alter a codon so that it calls for a different amino acid, which frequently results in a dysfunctional protein. Nonsense mutations produce a stop codon instead of an amino acid codon, which would end the production of the protein prior to its completion, rendering it useless. Another type of mutation, called a frameshift mutation, is even more frequently devastating than point mutations. In an insertion, new base pairs are added to a DNA strand; a deletion is just the opposite. These are considered frameshift mutations because they move the 3-base "frame" that makes up amino acid codons - even if only one base pair is added or removed, the entire nucleotide sequence following it is affected. Proteins produced from frameshift mutations are even further from the originally coded protein as every amino acid in the sequence could be changed.

Figure 17.26 from Chapter 17: Gene Expression: From Gene to Protein PowerPoint Lecture Presentation for Campbell Biology by Nicole Tunbridge and Kathleen Fitzpatrick

Meiosis & Mitosis Lab

3A: Observing Mitosis in Plant and Animal Cells Using Prepared Slides of the Onion Root Tip and Whitefish Blastula

Purpose

For this experiment we were required to identify what stages of mitosis each of the onion root tip and whitefish blastula’s cells were in. We were expected to understand and identify each of the stages of mitosis of both organisms cells. And after the cells were identified they needed to be counted to estimate the length of time the cells spent in the each stage of the cell cycle.

Introduction

Mitosis is an essential process to all organisms whether they are plants, animals, bacteria, etc. because it allows organisms to grow with the use of identical somatic cells. This experiment observes the steps throughout the process of mitosis in samples of onion root tip and whitefish blastula. Onion root tip contains three different regions that can be easily identified in the provided slide. There is a region of maturation, elongation, and cell division. As cells are located more towards the terminal of the onion root tip, they progress to later stages of mitosis. The whitefish blastula samples lacked these distinct regions, but the different stages could still be readily identified.

Methods

The procedure for studying the progression of plant cells through mitosis using these samples was fairly simple. After placing each sample under a microscope, a 10X objective was used to find the areas of the dividing cells. A 40X sample was then used to identify cells in each of the five stages of mitosis (interphase, prophase, metaphase, anaphase, telophase) as described by biologists.

The tip of the needle under the microscope points to the cell undergoing each phase in the following pictures. Pictures were taken using a phone’s camera held up against the eye lens of the microscope. The phones had to be moved around and their lenses had to be zoomed in in order to show the cells clearly.

(Onion Root Tip)

Interphase

Prophase

Metaphase

Anaphase

Telophase

(Whitefish Blastula)

Interphase

Prophase

Metaphase

Anaphase

Interphase

In the second part of this experiment, onion root tip cells were examined to calculate the estimated time (in minutes) cells spend in each stage of mitosis. These amounts of time can be inferred from counting the number of cells in each stage because there will be more cells in a certain phase at one time if that phase takes longer to complete.

Two lab group members were needed to do the counting. One person would examine the cells under the microscope, identifying the phase of each cell out loud. The second person would then tally each cell under its respective phase of mitosis. The data collected from this follows.

Data

Graphs & Charts

Discussion

  The counted number of onion root tips cells in each phase can be subject to some error, however. The approximate number of cells depends on the ability of the person counting to correctly identify the stage and the ability of the person keeping track of the counting to keep up with them. As seen from the chart, data from three different fields of view were used in counting. This was done because one field of view would give an inaccurate idea as to how long cells throughout the sample would undergo each phase. For example, a field of view closer to the very tip of the sample would contain more cells in the latter stages, anaphase and telophase. Multiple fields of view spaced across the sample would give more accurate numbers.

  Judging by the calculated numbers of minutes for each phase, the amount of time in the phase generally shortens as mitosis progresses. It takes more than half of the time phrase for interphase to occur while it takes only 2% of the time for anaphase to occur. From this, it can be inferred that it takes longer for the nondividing cell to prepare for splitting than for a cell to actually split.

Conclusion

As seen visually from the number of onion root tip cells in each phase, the greatest number of cells were counted under interphase, and this signified that this phase took the longest amount of time to complete. Assuming that the entire cycle would take a 24-hour day, based on the number of minutes in a day (1,440), it would take 58.1% of the time or 836.64 minutes of the day for a cell to undergo interphase. In order from greatest to least amount of time, the stages are interphase, prophase, metaphase, telophase, and anaphase.

References

• The procedure and chart included here were obtained from the "Lab Three: Mitosis and Meiosis" document from CollegeBoard.

3B: Simulation of Meiosis

Purpose

To better understand meiosis and crossing over within meiosis lengths of beads were provided to simulate chromosomes. As instructed the beads were placed into a structure replicating the chromosomes in the different stages of meiosis I and meiosis II. In part two, crossing over was further examined when we counted the Asci in order to count the centromere distance (map units) between genes.

Introduction

Unlike mitosis, meiosis goes through two nuclear divisions ending with four different daughter cells increasing genetic variation. The differences in cells is a result from crossing over. Meaning chromatids of homologous pairs exchange physical parts of genes. Crossing over occurs during prophase I. Meiosis I reduces chromosome numbers from diploid to haploid and separates homologous pairs. While meiosis II separates the sister chromatids and ends with the four new cells.

Methods

Special meiosis simulation kits were used in imitating the different stages of meiosis. Two strands of one color represented one chromosome, and two pairs of the two colors represented homologous chromosomes. The creation of each stage had specific instructions.

Interphase includes DNA replication. The formation of identical chromatids was simulated by bringing the magnetic centromeres of strands of the same color together so that they connected.

Crossing over, which is unique to meiosis I, takes place in prophase I. This was simulated by popping off beads from one strand of each chromosome and swapping those with its homologous chromosome.

In metaphase I, the chromosomes line up in the center of the cell.

Also unique to meiosis I is the way in which whole chromosomes are pulled to opposite poles during anaphase I in the cell rather than single chromatids.

Formation of the first two daughter cells from cytokinesis and later re-development of the nuclear envelope takes place in telophase I. Note that there is no interphase during the two stages of meiosis because DNA is being further divided rather than replicated.

Prophase II includes movement of the two daughter cells to different areas within the organism so their further process of meiosis take place farther apart.

In metaphase II, the chromosomes are aligned in the center of each daughter cell.

In anaphase II, the chromatids of the chromosome within each daughter cell move to separate poles.

Meiosis ends after telophase II when the nuclear envelope forms, cytokinesis occurs, and the two daughter cells each split to form four daughter cells.

Data

Graphs & Charts

Graphs and charts were not necessary to this portion of the lab.

Discussion

Taken from this experiment were three major differences of mitosis and meiosis. In prophase I of meiosis each pair of chromosomes come together in a process called synapsis. A second difference happens during anaphase I, homologous chromosomes separate and eat the spindle fiber towards the opposite sides of the cell. DNA replication does not occur during interphase II, renamed interkinesis. This is the final difference of mitosis and meiosis. There are also a few differences in meiosis I and II; meiosis I starts with one parent cell that has 46 chromosomes and meiosis II starts with two parent cells that each have 23 chromosomes.

The oogenesis is the female gamine division although it produces four cells ,only one cell has the correct amount of organelles while the other three do not have organelles and therefore are at a disadvantage. The spermatogenesis is the male gamete division and the four haploid daughter cells are fully equipped for fertilization. Resulting in meiosis being important for sexual reproduction, it produces gametes with 23 chromosomes allowing different daughter cells with 23 gamete chromosomes to create an organism.

Conclusion

We learned about the differences between mitosis and meiosis, that crossing over only occurs during meiosis and all of the phases of each of them.

References

• The procedure and chart included here were obtained from the "Lab Three: Mitosis and Meiosis" document from CollegeBoard.