Essential Knowledge 1.B.1 (3D GameLab)

One conserved core biological process is the use of DNA and RNA as the carriers of genetic information for all living organisms. With this, organisms share dependence on the transfer of information through transcription, translation, and DNA replication. The influence of DNA is observed particularly in nucleotides; all living organisms share the same four adenosine, cytosine, thymine, and guanine nucleotides, though their sequences (and thus, the amino acids and proteins for which they code) may differ extremely. A specific example of how this conserved core biological process works between organisms is the transformation of recombinant DNA so that the coded characteristics of one organism may be taken up by a different organism. This supports the idea of common ancestry because over time, and over the millions of years of evolution, we can understand how small phenotypic changes corresponding to small changes in gene sequence were passed over many generations.

Another conserved core biological process is the presence of mitochondria and/or chloroplasts in all eukaryotic cells. Both are organelles which support the processing of energy. The endosymbiont theory states that a prokaryotic cell that consumed oxygen and a photosynthetic prokaryotic cell were both ingested by the ancestor of the eukaryotic cell. There came to be an endosymbiotic relationship, or an equally beneficial relationship, between the host cell and its bacteria. The occurrence of this proposed relationship over millions of years supports the idea of common ancestry in that all eukaryotic cells came from this common ancestor cell with these organelles. Even the organelles evolved from prokaryotic bacterium to take on new functions.

Another conserved core biological process is the conservation of metabolic pathways across all currently recognized domains. One important example is the performance of glycolysis of cellular respiration across all organisms. Similarly, organisms that rely on photosynthesis have both light and dark reactions within their chloroplasts. Such similarities between organisms support the idea of common ancestry in that all eukaryotic cells depend on the same processes, a level even deeper than that of common organelles.

1.A.2 Continued

One example of an evolutionary change that is partially related to a change in the environment is the emergence of dog breeds over hundreds of years. It has been impacted by both environmental as well as human factors. From the environmental standpoint, the regions from which certain breeds emerged played some part into what physical characteristics canines adapted. For example, the dense, thick fur of the Siberian Husky allowed it to endure the extremely cold and harsh climates of the Siberian Arctic. The Mexican Hairless Dog is a breed whose lack of hair allows it to be comfortable in the Central American heat. The role of humans has played a very large impact on the evolutionary change in these animals. When we bred ancient dogs for certain tasks, they developed characteristics suited to those tasks over many, many generations. For example, the muscular, compact body of the Portuguese Water Dog allowed it to herd fish into nets, retrieve tackle, and serve as courier from boat to land. Our actions in this respect were artificial selection. While dog breeds will continue to change and emerge, similar developments may be seen in similar species in the future. Domestication of foxes in recent years has notably resulted in foxes with dog-like traits such as less erect tails.

Essential Knowledge 1.A.2 (3D GameLab)

Peppered Moth Simulation

21.
 Light Forest -> 77% light moths and 23% dark moths at end
 Dark Forest ->  33% light moths and 67% dark moths at end

22.
In a light forest, a light-colored moth would have a much greater chance at survival because it can blend easily into its surroundings. The tan color of the bark was the same as if not similar to the color of the light-colored moths in that simulation. A light-colored moth would be in stark contrast to the light background, and predators would be able to pick it out much more quickly. In the same way, in the dark forest, a dark-colored moth would be much harder for a predator to find than a light-colored moth that would stick out so easily from its surroundings.

23.
Natural selection dictates that in a situation with variation in traits, differential reproduction, and heredity, the phenotype that allows the organisms that best chance at survival will continue to exist and be carried on in the next generation. In this example, the light or dark color that allows the greatest chance at survival from predators in the light or dark environment, respectively, will be passed on by the moths of that color to the offspring.

24.
If there were no predators in this case, the colors of the moths might still change, though not nearly as dramatically. Moths of both light and dark colors would continue to proliferate. The colors within the moth population might only change depending on the nature of the colors' genotype; if the relationship between the dark and light genes was incomplete dominance, there may be an intermediate color (perhaps gray) that would appear.

Restriction Mapping of DNA Plasmid Lab

Restriction Mapping of DNA Plasmid Lab

Purpose

This lab explores the use of restriction mapping by the use of restriction enzymes followed by gel electrophoresis in characterizing a DNA sequence. Restriction enzymes are used to digest DNA samples, and because each restriction enzyme is known to correspond to a certain sequence, their use gives clues as to what nucleotides are at those sites. They are also used in the creation of recombinant DNA; the needed sequence is isolated using restriction enzymes before insertion into a vector and connection with DNA ligase. The band lengths of DNA that are produced by the restriction enzymes and measured by means of electrophoresis can help determine the location of restriction sites to create a map of the given DNA.

Introduction

First, agarose gel is cast, liquid to gel with 6 wells to pipet the DNA into it. Once there is DNA in the wells it can go through electrophoresis, which uses the negative charges of DNA molecules to pull fragments through the gel. The wells are at the negative side of the chamber, and as the current runs through the chamber the dye and fragments will move through the gel toward the positive electrode. The speed of the DNA fragments’ movement is determined by their relative sizes (bigger fragments move slower), and since the electrophoresis is stopped before any of the bigger fragments can catch up to the smaller ones, the smaller fragments will move further through the gel.

The lambda/PstI DNA is what the other sites are being compared to. The others are single, double, and triple digests, which space out along the gel.

After the electricity moves through the gel and the bands move the gel is removed and one can compare the marker DNA to the other DNA in the chambers.

Methods

In the procedure of this lab, it should be noted that many steps were already performed by the teacher and teacher assistant in the casting of the agarose gel and setup of the electrophoresis chamber. As a lab group, we need merely load the samples into their respective wells.

The entirety of a sample would be drawn into a fresh pipet. It was made sure that the sample was expelled of air so air bubbles would not form around the well. The pipet would then be steadied over the well with two hands, deep enough to touch the bottom of the well and without pressure that would puncture the gel itself. The sample was then slowly injected into the well. This was done with lambda and the four DNA samples. The order of our samples was pMAP/PstI, pMAP/PstI/SspI, pMAP/PstI/HpaI, and pMAP/PsI/HpaI/SspI.

If we believed additional sample was needed for successful electrophoresis, more was injected into the well. The second well in our gel was also left empty.

Above: the gel before electrophoresis and the samples next to their corresponding wells

Above: loading a DNA sample into its well

The gel was carefully placed in the electrophoresis chamber and left until the longest DNA fragment was about 2 cm above the end of the gel.

Below: the gel in the electrophoresis chamber

Above: several gels in the electrophoresis chamber

After electrophoresis, the gel was taken by the teacher for staining.

Below: gels in the process of being dyed for analysis

Above: the gel in the middle of electrophoresis

Data

Above: the gel after electrophoresis

Above: the gel on the light box with and without an orange filter

Below: marked DNA fragments on the lightbox

Below: Estimations of the band lengths from the electrophoresis

Graphs & Charts

Graphs and charts were not necessary to the analysis of this lab. However, the lab document included these maps of the DNA fragments when cut with alternate combinations of restriction enzymes.

Discussion

Using the lambda/PstI lane for comparisons, we were able to approximate the lengths of the bands in each of the test lanes. The DNA cut with PstI only (lane 3) showed two resulting bands of lengths approximately 2838 bp and 900 bp; with PstI and HpaI together (lane 4), it showed three bands of lengths 2000, 1800, and 1000 bp; with PstI and SspI (lane 5), it showed three bands of lengths 2500, 900, and 650 bp; with all three (lane 6), it showed four bands of lengths 2000, 1200, 900, and 650 bp. The average total DNA length for each lane turned out to be 4335 bp, which we rounded to 4500 for simplicity’s sake. Using this, we converted the individual band lengths into usable estimations that added up to the rounded average for our plasmid map: the bands of lane 3 became 3500 and 1000 bp; lane 4’s became 2000, 1500, and 1000 bp; lane 5’s became 2700, 1000, and 800 bp; and lane 6’s became 1500, 1200, 1000, and 800 bp. Using these approximated lengths we determined the relative restriction sites on the plasmid as shown below.

Above the map, we have the original band lengths in black with reappearing band lengths highlighted in the color of the restriction enzyme associated with them, as well as the adjusted estimations in orange.

Conclusion

Due to rounding errors in the analysis of the gel, estimates of band lengths and restriction sites were not completely accurate. The appearance of the bands in the gel was also not completely clear, which contributed to errors in band measurement and plasmid mapping. The DNA cut with PstI only in particular had a total DNA length that was pretty far from the average (3738 bp vs 4335 bp), so that required some significant rounding. Despite all of the rounding and estimating, however, we have determined the relative locations of the restriction sites for the three enzymes with some level of accuracy.

References

The procedures and figures used were obtained from the document “Restriction Mapping of Plasmid DNA” from the Carolina Biological Supply Company.

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