Lab Analysis study
7. For the PCR sample of your genomic DNA, list the contents of the tube and describe the role of each of the components found in the PCR reactions. Be sure to explain why D1S80 primers are needed to study the D1S80 locus. Be sure to explain why Taq Polymerase is used (and not human polymerase) in PCR and why DNA bases in the triphosphate form (dNTP's) are needed.
A PCR sample of my genomic DNA contains the segment of DNA to be copied, D1S80 primers, and a Master Mix composed of Taq polymerase, MgCl2, dNTPs, and 2X buffer. The template DNA is that which is being amplified in the PCR process. The D1S80 primers flank the D1S80 locus and are essential for initiation the elongation of DNA by providing Taq polymerase with the location for where to start adding nucleotide bases. The Master Mix does a combination of things: the Taq polymerase included in the Master Mix acts as a heat-tolerant enzyme to create new DNA strands by placing the new nucleotide bases obtained from the dNTPs in the location given to them by the D1S80 primers. The MgCl2 acts as a cofactor and catalyzes the Taq polymerase functions by removing phosphates from the dNTP, thereby enabling a faster reaction. The dNTPs provide the new nucleotide bases, and the 2X buffer provides the appropriate pH for the mixture so the functioning enzymes are not denatured and the reaction can take place ("Introductory Laboratory Experiments in Biology")
5. What is a 'Tris' solution? Why is it extensively used in cellular and molecular biology laboratories?
A Tris solution is composed of Trizma base—an organic compound chemically named Tris(hydroxymethyl)-aminomethane—dissolved in solution. It is extensively used in cellular and molecular biology laboratories as a component of buffer solutions and contains a primary amine that enables the reactions associated with typical amines.
7) What cellular components were visible in the BPAE cells? Which fluorophore was used to identify the location of each structure? Include a LABELED composite image exhibiting all three stains simultaneously.
A few cellular components were visible in the BPAE cells because of their fluorescence provided for by their binding fluorophore (each fluorophore was used to identify a location of a specific structure in the cell). Dapi binds to A-T regions in DNA, thus accumulating in the nuclei, making these regions appear blue because of the wavelength of light it emits. Texas Red binds to actin, outlining the cytoskeleton of the BPAE cells in red because of its emission of red light. Bodipy binds to microtubules such as tubulin, causing these areas to be green because the specific green-light wavelength it emits (Maas).
10. What is a primer dimer? Be sure to cite your source.
A primer dimer is the amplification of only short nonsense PCR products because of template DNA annealing to itself ("Multiple Primer Analyzer")
4. Report in a table for each of the objectives (4x 10x, 40x, & 100x), the corresponding N.A, total magnification, and the calculated theoretical resolution using Abbe's equation (show your work, indicate the wavelength of light chosen).
A stage micrometer is a type of microscope slide with a very small 2 mm ruler on the surface. It is different from an ocular reticle because it is a physical slide with units in mm, versus the ocular reticle is inside of the ocular and in reticle units. These two sources of measurement can be used together to calibrate the microscope by aligning the two and finding the value of the reticle spaces in terms of micrometers (finding a proportion) (Fellers)
6. The pGLO exercise demonstrates how the interaction between genotype and the environment determines phenotype. Give an example of how bacteria with the SAME genotype can have different phenotypes due to variation in environmental conditions.
An example of how bacteria with the same genotype can have different phenotypes due to variation in environmental conditions can be seen in the lab—specifically in analyzing (predicting) the genotypes and phenotypes of the bacteria on the +AMP plate and +AMP and +ARA plate. Both plates had transformed bacteria because of the ampicillin resistance granted by the expression of the pGLO plasmid genes (hence how they are referred to as "transformed" with the pGLO plasmid). The nontransformed bacteria on both plates were killed off by the ampicillin because of their lack of resistance, and thus it can be inferred that both plates had the same genotype (of only transformed cells). However, their phenotypes were different because of the +ARA presence on the one plate. Arabinose presence allows the inhibition of the araC operon which in turn allows the production of green fluorescent protein. The fluorescing phenotype of the bacteria cells that have a "transformed" genotype (which includes the gene that codes for GFP) is dependent upon the environmental condition of the presence of arabinose in the growth media (PGLO Bacterial Transformation Kit).
An individual has the genotype 24, 32 for the D1S80 locus. Are they homozygous or heterozygous? What does '24' indicate on a molecular level? What is the size of each allele (in bp)?
An individual with the genotype 24, 32 for the D1S80 locus would be heterozygous. The "24" and "32" indicate the number of short tandem repeats of organized nucleotide sequences. These tandem repeats vary in length (number of repeats) among individuals and are inherited from their parents. The formula for calculating size of an allele in bp is 16(number of repeats) +146. Thus, the allele with 24 repeats is 530bp and the allele with 32 repeats on the D1S80 locus is 658bp ("Exercise 5 and 6").
7. Using the digital images you acquired, describe the average length and width in reticle units and micrometers of a white blood cell. Show your calculations to convert these measurements into micrometers. Identify any cell structures and describe their function.
As (faintly) seen, the white blood cell aligned with the ocular reticle was measured to be 11ru x 12ru (length by width). During the previous calibration of the ocular reticle, the conversion factor for micrometers to reticle units under the 4x, 10x, and 40x were found directly. From the 4x objective to the 10x objective (an increase in magnification of 2.5), there was a decrease in the ration of micrometers per 1 reticle unit by a factor of 2.5. In the same way, increasing from the 10x to 40x objective (an increase of magnification by a factors of 10), there was a decrease in the ratio of micrometers to reticle units by a factor of 10. In this way, we saw that an increase frpm the 40x objective to the 100x (an factor of 2.5) would result in a decrease in the proportion of 2.5 as well. Therefore the ratio of micrometers to reticle units for the 100x objective was found to be 1:1 (because the ratio for the 40x objective was 2.5:1). Thus, two simple proportions: 11𝑟𝑢 𝑥µm = 1𝑟𝑢 1µm and 12𝑟𝑢 𝑥µm = 1𝑟𝑢 1µm gave the converted dimensions of 11µm x 12 µm (the ration was 1:1 and thus the converted dimensions stayed the same numerically).
5. Predict the genotype and phenotype of the bacteria growing on LB only plates, +AMP plates, and +AMP and +ARA plates.
Bacterial cells that have been transformed with pGLO are referred to as "transformants", and those that have not been transformed with pGLO, "nontransformants" (aka nontransformed cells). The genotype of the bacteria growing on the LB only plates will be both transformed and nontransformed with the pGLO plasmid. Because no ampicillin is on this plate, both the nontransformed and transformed bacteria alike are able to grow on the LB only plate, and there is thus a very large amount of bacteria colonies that create a white "lawn" appearance of bacteria cells covering the plate anywhere there is available space. The phenotype of the bacteria growing on the LB only plates causes the bacteria to be less immediately visible because of the lack of green light being emitted (no fluorescence under the UV light). The genotype of the bacteria growing on the +AMP plates will show a complete transformation of all the cells because of the majority of the bacteria's resistance to ampicillin. Only previously transformed bacteria were able to survive because of their resistance (allowed by their genetic code for the production of proteins that form enzymes that breakdown the antibiotic), whereas those nontransformed bacteria were killed off by the antibiotic and thus not seen in the genotype. The phenotype of these bacterial cells will appear as dots of transformed cells (colonies). Lastly, the genotype of the bacteria growing on +AMP and +ARA plates will be of entirely transformed bacteria, and their phenotype will show fluorescently-glowing cells when shined under the UV light (PGLO Bacterial Transformation Kit).
2) Use the table below to describe the advantages and disadvantages of the following techniques. Technique Advantages Disadvantages Brightfield without Stain Brightfield with Stain Phase Contrast/Darkfield Fluorescence
Brightfield without Stain Microorganisms may remain alive for observations. One of the easiest methods for microscopy (short prep time). Not very effective for clear observations as there is very little contrast and many organelles and cell structures are too faint to be seen without a stain. Brightfield with Stain Fairly simple staining required for a substantial difference in contrast and thus clearer image of details. More cell structures visible. Specimen must be killed in order to stain them for better observation. Phase Contrast/Darkfield Specimen may remain alive. Staining procedures are not required—just adjustments in the condenser/diaphragm/illumination settings. Certain limitations as to magnification. Fluorescence Very sensitive and accurate observation technique. Only a few cells viewed at a time. Clear observations of cells structure. Specimen must be killed and stained. Specific stains must be used to allow fluorescence of different cell structures. Staining process takes the longest and the specimen may require multiple stains.
7. Explain the methods used to test this hypothesis as though you were submitting a paper for publication in a scientific journal. Use the 'Writing Scientific Papers' handout posted on LabClicker to help you write your methods section.
Data was obtained by recording the absorbance of dopachrome at 390.8 nm with either the addition or lack of sodium chloride in the enzyme-substrate solution to test the effects of ions on enzyme activity. This wavelength was found by running an absorbance spectrum with a LoggerPro-spectrophotometer function on a prepared blank (negative control)—consisting of the 30 µL sodium chloride, 150 µL tyrosinase, 900 µL potassium phosphate pH 7 buffer, and 150 µL deionized water—to determine the optimum wavelength at which to observe tyrosinase activity. An experimental cuvette was prepared with 900 µL potassium phosphate pH 7 buffer, 150 µL of 20 mM DOPA, 30 µL sodium chlorate and 150 µL tyrosinase enzyme (the potato homogenate extract being experimented with). A prepared ice bath was used to store the DOPA, potassium phosphate buffer, and tyrosinase before use to prevent the reaction from occurring with time, either by the endogenous compounds in the homogenate that can serve as the substrate, or by the spontaneous oxidization of the DOPA. The experimental cuvette was placed into the calibrated spectrophotometer and the reaction rate was recorded and analyzed in LoggerPro. A (positive) control cuvette without sodium chloride was then prepared with 900 µL potassium phosphate buffer, 150 µL (20mM) DOPA, 150 µL tyrosinase, and 30 µL deionized water (for volume control) and placed into the spectrophotometer for reaction rate analysis. A final cuvette was prepared with the same components of the buffer and run through the spectrophotometer for absorption v. time analysis. All three solutions were prepared a second time and analyzed using LoggerPro a second time for more conclusive results. All absorption curves were analyzed for the initial linear slopes (before enzyme saturation) by finding the best-fit linear regression line and recording the slope of each curve as its absorption rate (Maas).
5. What is empty magnification? How does the use of immersion oil between the coverslip and the objective lens help to increase the NA and resolution?
Empty magnification is that which is above a level of contributing any useful magnification or resolving power. By immersing both the objective lens and specimen in an oil of high refractive index (through the process of oil immersion), the numerical aperture of the objective lens and thus the resolution of the microscope are thus increased. Increasing the refractive index allows more light rays to refract into the objective lens and thus improves the resolution, which depends on the refractive power of the lens (Spring, n.d.-b).
5. In section 4.6, you explored the effect of temperature on DOPA conversion by tyrosinase. Using your data, explain the effect of temperature on reaction rate. Include graphs and/or tables when discussing the results. Discuss whether or not this was expected given the enzyme's natural environment. Describe what was happening at the molecular level to all reaction components (enzyme, substrate, and product).
For our experiment, reaction rate decreased as temperature became extreme on either side of the scale of hot/cold. The highest rate of reaction was 0.01878 at room temperature. The normal environment of tyrosinase enzyme is the inside of a potato, usually stored at room temperature. It makes sense that the reaction would work best in a room temperature environment. At the molecular level, the DOPA substrate binds to the tyrosinase (enzyme)'s active sites and is catalyzed and converted to dopachrome (the product of the reaction). Enzyme activity is based on open active sites, and can go through multiple rounds of using its active sites until the substrate has been used up. Tyrosinase begins to denature at extreme temperatures. When the temperature is increased past a certain point, the bonds holding the protein together break, and with the enzyme denatured, the substrate has nothing to bind to. The reaction occurs at a much slower rate (Enzyme).
2. If you are offered a graduated cylinder, Erlenmeyer flask, and a beaker to measure a 60-mL volume, which would you select? Explain your selection.
If I was offered a graduated cylinder, Erlenmeyer flask, and a beaker to measure a 60-mL volume, I would choose the graduated cylinder because of the increased accuracy the smaller increments of measurement located on the graduated cylinder allow for. The Erlenmeyer flask has only one mark, while the beaker's marks are not accurate in their spacing
6. In section 4.7 you designed your own study of enzymatic activity. What was your hypothesis?
If salt is added to the enzyme-substrate solution, the resulting reaction rate will be decreased because of the effect of the ions destabilizing the ionic bonds holding the enzyme together
8) Explain how immunofluorescence works using an example from lab.
Immunofluorescence is when a fluorophore is coupled to an antibody created by the immune system. Direct and indirect immunofluorescence was observed by the conjugation of Bodipy to a primary (direct
4. Explain why arabinose must be present for bacteria to express the GFP gene.
In a bacterial cell that has been transformed with a pGLO plasmid, the GFP gene of the plasmid has been inserted into the arabinose operon of the bacterial cell. Because of its location, the GFP is prevented from transcription by the araC protein in the absence of arabinose. The GFP gene is only transcribed (and expressed) when arabinose is present in the growth media, causing the araC protein to allow transcription by RNA polymerase ("Introductory Laboratory Experiments in Biology").
2. What is Köhler illumination and why is important? What does a numerical aperture (N.A.) indicate? What is the relationship between resolving power and numerical aperture?
Köhler Illumination is a specific set of steps important for giving the optimal specimen illumination used for transmitted and reflected light optical microscopy. In the lab, we followed a set of steps for adjusting the diaphragm and condenser to optimize the condenser height and achieve Köhler illumination. The numerical aperture is a measure of an objective's resolving power for finer specimen at fixed object distances. Finding köhler illumination involves optimizing the shape of the beam of light that passes through the specimen. Changing the objective resolving power changes the adjustments that would need to be made to find the proper illumination for a specific specimen (Parry-Hil
6. What is metabolic acidosis? When might an individual experience metabolic acidosis? Use your knowledge of blood-buffer systems to explain why bicarbonate therapy might be used to treat this condition.
Metabolic acidosis is a condition of the body in which too much acid is accumulated. A person may experience metabolic acidosis due to kidney failure, a large buildup of body toxins, or the ingestion of certain drugs. A person with a certain rare complication due to diabetes may also experience metabolic acidosis. Metabolic acidosis typically shows up as nausea, vomiting, fast breathing, and many other symptoms. Bicarbonate therapy, which includes replacement of sodium bicarbonate to patients with sodium bicarbonate loss due to the combined symptoms, might be used to treat metabolic acidosis because of its ability to act as a buffer, effectively reducing the acidity of the patients' blood.
1. In lab, you investigated an enzyme common to many organisms that is responsible for pigmentation. Complete the table below concerning the enzymatic reaction. Molecule Role / Function Molecular Oxygen (O2) Dopachrome Tyrosinase DOPA Copper ions
Molecular Oxygen (O2) The enzyme tyrosinase catalyzed the hydroxylation of tyrosine into DOPA in the presence of oxygen. Dopachrome The orange/brown colored compound that is spontaneously produced from the conversion of dopaquinone to dopachrome—can be monitored using a spectrophotometer. Tyrosinase An enzyme that catalyzes the synthesis of pigment through a series of steps and thus turns into melanin. DOPA Rate limiting enzyme for the production of dopachrome and precursor to dopamine. Ends up being catalyzed into dopaquinone. Copper ions Universally important cofactor—especially in pigmentation of the skin.
7. Review the 'Notebook Guidelines' posted on LabClicker and evaluate the quality of your lab notes. Did you follow all of the guidelines? What will you do differently next week?
My lab group followed the guidelines in terms of our actions and using the proper technique for the procedure. However, the overall steps we followed veered off for a bit for our particular lab group because the pH meter given to us for the experiment turned out to not be working. The process of determining that it was not working took a few separate side steps, and it was a few minutes—during which we recalibrated the pH of our specific LabQuest computer system, determined it wasn't working, and switched our whole experiment to using a different lab table's pH meter—before we were able to get back to the preset guideline steps. In next week's procedure, we would proceed in the same way we did with the first lab, but would hopefully not be hindered by the same problems with the equipment as last time and thus improve the lab experience overall.
2) Explain how PCR amplicons are created, why they can vary in size, and how you are able to visualize the location of amplicons in a gel.
PCR amplicons are created as the products of amplification reactions. it can either be the source or product of the replicating event and tons of copies are made through this process. they can vary in size because they correspond to sizes of DNA, which naturally vary so there would be as much variation in amplicons as there are in length of DNA molecules. in a gel, the lines that are present are amplicons. it is the only thing you can see in the gel since they are what you are observing primarily
12. When you are given your PCR tube next week in lab, will you find millions of copies of all 46 of your chromosomes? Millions of copies of chromosome 1? Millions of copies of something else? Will you be able to actually see anything in your tube? Will you be able to figure out your genotype just by looking at the tube?
Q12. In this lab, we isolated all 46 of our chromosomes from our cheek cells, but only studied the D1S80 locus, therefore the PCR tube will only show the results of our test on the small D1S80 locus. We only have two copies of chromosome 1 (D1S80) which were studies in the lab, but each copy contains about 249 million nucleotide base pairs and 4.316 genes. Nothing will be immediately visible to us when given our PCR tubes next week that would enable the genotype to be found. We are going to have to pour gels and use Gel Electrophoresis to run the gels with PCR products to determine our actual genotype ("Exercise 5 and 6").
8. Which automatic pipette is the best to deliver 150 µL volume and why?
The 20-200µ automatic pipette is the best to deliver 150µL volume because it is the smallest pipette that can hold this volume. This allows for the best accuracy (accuracy decreases as unnecessarily large pipettes are used for small volumes).
3) During Procedure 3.2, you added FIX and PERM solutions to your sample of cheek cells. What does each solution contain and what is their purpose? What might happen if you forgot to add the PERM solution?
The FIX solution contains the toxic chemical paraformaldehyde, and is used to immobilize antigens and glue the targeted specimen into place; "fixing" the cell structure onto the slide for observation. The PERM solution contains a mild buffered detergent which permeates (creates holes in) the cell membrane to allow for the dye to have better access to inside structures—the detergent attacks the lipid bilayer (makes sense knowing the purpose of detergent). Without the addition of the PERM solution, the dye would not properly stain the specimen, resulting in a lack—or at least a lower degree—of fluorescence ("FIX & PERM")
1. How do you safely dispose of Tris solution? Pipette tips? Gloves?
The Tris solution was disposed carefully into the liquid chemical waste and the beaker was rinsed out with distilled water again into the liquid chemical waste to ensure complete disposal of the base. The automatic pipette tips were ejected using proper technique into the solid chemical waste container, and the gloves we wore during the experiment were taken off using the proper method and thrown into the designated glove bin.
2. In section 4.4, you monitored the activity of tyrosinase by measuring the change in absorbance over time. Using the data your group collected, describe how absorbance values changed over time. Why did the values change? What happened at the molecular level when absorption values no longer increased?
The absorbance values taken from the collected data on the activity of tyrosinase increased at a fairly-linear rate of 0.0157nm/s before rounding off at AROUND 50 seconds. From that point onward, the change in the absorbance began to decline to nearly, continuing as a horizontal line. The absorbance values changed due to the functioning enzyme, tyrosinase, acting on dopaquinone to produce dopachrome. In simplified terms, dopachrome production increased over time and the rate of production was monitored by the absorbance rate over time. The initial linear increase took place because the tyrosinase (enzyme) was acting on the DOPA to reach maximum absorbance. The absorbance values began to decrease because the DOPA molecules had absorbed the maximum amount of light possible. The dopaquinone was used up, leaving the enzyme with nothing to catalyze into the product, and thus with no active reaction, the absorbance values stopped changing (Boiret).
1. How does the "field of view" change with magnification? Relate this to the rule that one should always use the low-power scanning objective first to locate a specimen. How does the "depth of focus" change as magnification increases? Are thick (e.g., Elodea) or thin (e.g., prepared Paramecium) samples of specimens preferred at high magnifications?
The field of view on a compound microscope is inversely proportional to the magnification of the object. As the magnification is increased, the field of view decreases in order to show a smaller portion of the slide in greater detail (because of the increased magnification). Scanning objectives range in power based on their magnification (a low-power scanning objective has a low magnification). When locating a specimen, it is best to begin with the lowest-powered objective because this allows for the largest "field of view" which thus enables the specimen to be found more easily without having to move the slide around too much ("Microscope Magnification"). Depth of focus refers to the thickness of the plane of focus and the relative ability to focus at increasing magnifications. When the magnification increases, depth of focus decreases (there is a thinner depth of plane that can be in focus), and thus thinner samples are preferred to enable the best images of the specimen (Spring, n.d.-a)
5) What cellular components were visible in your cheek cells? Include a LABELED drawing that shows both stains (label the cellular structures). Why were other areas in the cell black?
The image of the cheek cell stained with DAPI and Texas Red using fluorescent microscopy was drawn from a shared picture from another lab group in our lab section. The dimensions shared with me once again from another lab group in my lab section for cheek cells (that do not match this image) were approximately 24reticle units X 17 reticle units.
9. Explain the purpose of the negative control you prepared AND the positive control (that we prepared) for PCR.
The negative control served to provide a PCR reaction for observation without the presence of a DNA template (everything else was included). A negative control for PCR should give no amplicons, and thus if any visible bands appear, these observations would allow possible contaminants in the solution to be determined that could cause distortion of the results. The positive control in this experiment contained everything that our samples did plus a DNA template known to amplify correctly (prepared by the staff). The purpose of the positive control is to see whether the results of the experiment are correct ("Introductory Laboratory Experiments in Biology").
3. Explain how ampicillin allows you to determine which cells have been transformed.
The pGLO plasmid contains an ampicillin resistance gene which thus allows it to survive and grow in mediums containing ampicillin. Bacteria cells that survive and grow on a medium containing ampicillin are only those bacteria that have been transformed with pGLO plasmid (thus having the ampicillinresistance gene). Therefore, any cells present on the media containing ampicillin are those which have been transformed ("Introductory Laboratory Experiments in Biology").
3. In section 4.4, you determined the rate of the reaction by finding the best-fit linear regression line. Why is the slope of this line a good way to estimate the rate of the reaction? Why did you only use the initial linear region (first 30-60 seconds) to measure the slope?
The rate of a reaction is the change in whatever dependent variable is being tested versus the independent variable's change. Slope is defined as rise over run (is a measure of the steepness of the linear value gathered from the data). Thus, determining the slope of a reaction is the same as finding its rate. Finding the best-fit linear regression line of the reaction allowed us to find the most accurate linear slope over a chosen interval. The chosen interval for our purposes was the initial linear region (30-60 seconds) of the reaction for the purpose of observing the period of highest activity of the enzyme and its effects on absorbance vs. time. During this first initial period, the active sites of the enzyme are being filled, and once they're all filled up, the reaction stops increasing (all of the active sites are in use and although the enzyme is increasing the rate of the reaction versus having no enzyme, it can only speed it up to a certain point based on open active sites), and the values plateau off. Thus, finding the slope of the first 30-60 seconds allows for an accurate estimation of the rate of the initial increasing reaction without taking into account the rate as it slowed ("Fitting Lines").
6) You observed BPAE cells using fluorescence microscopy. What stains were used to prepare these cells? Which wavelengths of light were used to excite each stain? Which wavelengths of light were emitted by each stain? What is the purpose of the fluorescent filter cube turret?
The three fluorescent stains used to prepare cheek cells for observation were DAPI, Texas Red-Phalloidin, and Bodipy. DAPI was excited by a UV wavelength of 365 nm and emitted a slightly lower energy 420 nm (blueemission) wavelength. Texas Red-Phalloidin was excited by a (yellow-green light) wavelength of 550nm and emitted red light of 570 nm. Bodipy was excited by 495 nm blue light and emitted 525 nm green light. The fluorescent filter cube turret, using one or multiple sets of filter cubes to obtain precise wavelength band selection, plays a part in creating the black background by separating emitted fluorescent light from the dye in the specimen ("Flourescent Filter Combinations").
8. What are the necessary steps in a typical PCR program and what is the function of each? What would happen if you made a mistake programming the PCR machine and skipped the annealing stage?
The three steps in a typical PCR program are denaturation, annealing, and extension (also called elongation or polymerization). Denaturation is the first part of the PCR process that occurs at around 94˚C and entails the separation of the two double-helix DNA strands. This is done by making the sample hot enough to break the hydrogen bonds between base pairs of the DNA. Annealing occurs at about 54˚C and, because of the lowered temperature, functions to re-form the hydrogen bonds between base pairs, now between the short single-stranded DNA primers and their complementary regions on the DNA template strand. Extension/elongation/polymerization occurs at about 72˚C for the purpose of being at the ideal working temperature for the enzyme. In this step, Taq polymerase reads the template strand and adds complimentary base pairs from the substrate pool of dNTPS. Reading of the template is done from the 3' to 5' direction and addition of complementary base pairs is onto the 3' end of the elongating strand. If the annealing was skipped, the hydrogen bonds would not have been reformed between the DNA primers and template strand, and thus the template strand would not be complete (connected by hydrogen bonds to the DNA primers) for reading by the Taq polymerase and a new chain cannot be produced because there is no primer for the DNA polymerase to attach the dNTPs to (Bloom).
2. In procedure 5.2 you transformed bacteria with the pGLO plasmid. What does it mean to 'transform' bacteria with pGLO? What are the main steps of a bacterial transformation? Will all of the bacteria be transformed? What happens to the majority of bacteria during the transformation process?
The transformation of bacteria with pGLO is the processes of causing a bacterial cell to take up the pGLO plasmid DNA by creating openings in the membrane where the plasmid can insert itself. Once the process is complete, a chemically competent bacterial cell capable of division is formed that expresses the pGLO genes. The main steps of bacterial transformation are to obtain chemically competent cells (E. coli for our purposes), insert the pGLO plasmid into these cells and "flick" the contents (to mix) and incubate on ice for 15 minutes, heat shock the cells by placing them in 42˚C water bath for 30 seconds and then back onto ice for 5 minutes, add LB broth for nutrients, and incubate the final transformation "cultures" in the 37 ˚C incubator for 40-60 minutes. Just because the cells are chemically competent does not mean that they will become transformed, which is what happens to a majority of cells—they are chemically competent but do not uptake the plasmid. A lot of the bacterial cells die in the heat shock process as well (PGLO Bacterial Transformation).
4) You observed your cheek cells using fluorescent stains. Which two stains were used to prepare these cells? Which wavelengths of light were used to excite each stain? Which wavelength of light were emitted? What is the relationship between excitation and emission wavelengths?
The two fluorescent stains used to prepare cheek cells for observation were DAPI and Texas Red-Phalloidin. DAPI was excited by a UV wavelength of 365 nm and emitted a slightly lower energy 420 nm (blue-emission) wavelength. Texas Red-Phalloidin was excited by a (yellow-green light) wavelength of 550nm and emitted red light of 570 nm. The relationship between excitation and emission wavelength for a fluorophore is that some specific wavelength of electromagnetic radiation is absorbed and some longer (lower energy) wavelength of radiation is emitted ("Triple Band Excitation").
8. Include a digital image of Elodea cells. Be sure to indicate and identify any cell structures and then describe their function. What are the dimensions of an average cell?
There were two options of cells to analyze depending on the focus. I chose the smaller cells. Because there was a lot of variation in cell size, I took the measurements of around 5 cells and found the average dimensions to be about 40ru x 9ru. Using the conversion factor for ru to µm for a 40x objective (2.5 µm = 1ru), the dimensions in terms of SI units were 100 µm x 22.5 µm
10. a) Ensures previously published work on topic is recognized b) Aa summary of the paper c) Describes the contents of the paper in one or two phrases d) Explains how the data were collected e) Describes what was discovered f) Discusses the implications of the findings g) Ensures recognition for the writer(s) h) Ensures those who helped/funded the research are recognized i) Ensures the article is correctly identified in abstracting and indexing services j) Explains the problem
Title: (c) Describes the contents of the paper in one or two phrases • Authors: (g) Ensures recognition for the writer(s) • Abstract: (b) A summary of the paper • Key words: (i) Ensures the article is correctly identified in abstracting and indexing services • Introduction: (j) Explains the problem • Methods: (d) Explains how the data were collected • Results: (e) Describes what was discovered • Discussion: (f) Discusses the implications of the findings • Acknowledgements: (h) Ensures those who helped/funded the research are recognized • References: (a) Ensures previously published work on the topic is recognized
6. Based upon the steps you performed to determine the size of the Paramecium specimen using the ocular reticle and the stage micrometer, list a set of brief instructions to determine the actual size (in SI units) of any object.
To begin, we used the ocular reticle to measure the dimensions of our individual paramecium specimen (in reticle units). Then we went through the steps of calibrating the ocular reticle. This included lining up the ocular reticle with the stage micrometer at each objective magnification and recording how many reticle units corresponded to 1 millimeter. From this measurement, with the knowledge of simple conversion factors such as "100 micrometers = 0.1 millimeter", we then completed a short dimensional analysis using the measurements of reticle units per 1 mm at each objective magnification to determine the respective conversion factors of however many micrometers per reticle unit. This resulting conversion factor (varying based on objective magnification), is the key to determining the actual size in SI units of any object using different objective magnifications.
4. In section 4.5, you monitored the effect of increasing substrate concentration on tyrosinase activity. How is tyrosinase activity affected by the amount of substrate? Include your graphs and/or tables when discussing the results.
Tyrosinase is the active enzyme which aids in speeding up the reaction time until all of the active sites of the substrate have been filled. As our stock concentration was increased (by factors of two), our reaction rates similarly increased by factors close to two. For the first several concentration increases, the rates of reaction similarly increased by factors of two, but after a certain point, the effect of concentration began to be less important, and the increase in rate of reaction per increase in concentration wasn't as proportional. Tyrosinase is affected by the amount of substrate up to a certain point. If too much substrate (a limit differing depending on each specific enzyme) is present, the enzyme activity is inhibited because it cannot catalyze all of the substrate at the same rate it was previously doing so at (Maas)
1) Create a numbered list of the specimens (e.g., cheek cell, diatoms, etc.) you examined during Exercise 3. For each specimen, describe the type(s) of microscopy you used to examine the specimen.
We examined cheek cells using two different types of microscopy. First, we followed proper procedure to prepare a wet mount of Methylene blue-stained cheek cells for observation under a compound microscope. We used standard brightfield microscopy o view the cells—starting at 40X total magnification for focusing the specimen and working up to 400X total magnification for a closer examination of the cells. The brightfield method produced a bright background field of view that allowed for a sharp contrast with the stained cheek cells. We also examined a second prepared wet mount of cheek cells stained with fluorescent dyes (Dapi and Texas Red- Phalloidin) using fluorescence microscopy. This involved utilizing "flourophores" as stains (Dapi and Texas Red-Phalloidin) because these molecules have a fluorescent property that allows them to absorb light of some particular wavelength and then emit light of slightly longer wavelength. () Dapi and Texas Red-Phalloidin were used particularly because of their ability to bind to specific molecules within the cheek cells. Dapi, for example, emits a blue light upon excitation with UV light and intercalates into DNA, allowing for a clear visual of chromatin and chromosomes within cells. Rhodamine or Texas Red binds to actin in cells to allow for better visualization of actin filaments (Maas). We examined prepared FluoCells slides (permanent mounts of bovine pulmonary artery endothelial— BPAE—cells) using both brightfield and fluorescence microscopy with a research-grade Olympus microscope. Brightfield microscopy with the Olympus microscope was similar to that with the compound microscope; once again the brightfield illumination produced a bright background that contrasted with the stained BPAE cells. The BPAE cells were stained with three different flourescent dyes to allow greater contrast to several different structures. Texas Red-phalloidin, green-flourescence, and blue-flourescent DAPI allowed the actin, microtubules, and nuclei to have greater contrast in the brightfield setting. Flourescence microscopy involved viewing the BPAE cells at 400X total magnification at each of the three different excitation wavelengths to accumulate the final composite image of a BPAE cell (Maas). We examined prepared permament dry mounts of Diatom species (a kind of microscopic algae) using brightfield microscopy, darkfield microscopy, and phase contrast microscopy with the compound microscope. Standard procedure was used to obtain the proper brightfield illumination for the species under the 40X-400X total magnification settings. The diatom frustules were not stained, so although the brightfield method allowed for a clear image, it was faint. This is why we then switched to darkfield microscopy. With the use of a flat plate attached to the condenser and adjustments to the illumination intensity and condenser diaphragm, darkfield was achieved and the frustules were brightly illuminated in contrast with the dark background. The settings were changed a final time to observe the diatoms under phase contrast, where differences in refractive indexes among the specimen's components produce sharp contrasts in the image. This involved adjusting the condenser slider, diaphragm, and illumination intensity to obtain a clearly visible image (Maas).
9. Using the digital images you acquired, thoroughly compare and contrast the structure, size and number of red and white blood cells. In which way is their structure/size/number a reflection of their function? Be sure to cite any sources used to answer this question. Also include a digital image of the blood cells present in the permanent dry mount.
White blood cells and red blood cells differ in many ways, and these differences in structure, size, and number reveal many things about their unique functions. Not all of the structures can be seen with the digital images acquired in this lab, but comparison is still possible with what is visible. Red blood cells lack a nucleus but have a cytoplasm filled with hemoglobin to absorb and bind oxygen for transport (functions in the respiratory system). The lack of nucleus makes them appear hollow and disk-shaped. White blood cells, on the other hand, contain many things red blood cells lack, such as a nucleus (in this case multi-lobed) which functions to store important genetic information and plays a part in their more irregular shape. In addition, the cytoplasm of white blood cells (may) contain (in this case) pale red/blue cytoplasmic granules which aid in the lysis of certain cells. Chloroplasts: convert light energy into chemical energy via photosynthesis Central vacuole: holds materials and wasted/maintains pressure/provides structure and support Cytoplasm: provides support and suspension for internal structures and organelles/stores vital chemicals/ maintains shape and consistency of the cell Cell wall: protects/provides structure Cell membrane: semi-permeable to allow transport of ions and molecules/adhesion Cytoplasm containing hemoglobin: binds oxygen Pale red/blue cytoplasmic granules (barely visible): Often contain components that aid in the lysis of targeted cells Multi-lobed nucleus: contains important genetic information Cell membrane: semipermeable/regulates what enters/exits the cell Both red and white blood cells have a cell membrane whose functions differs based upon the function at hand, but both are semi-permeable and allow certain things to pass through (ions/molecules/etc.). Although I did not physically count every red blood cell in this image, about a quarter of the image contains around 100 red blood cells, meaning there are probably around 400/500 or so red blood cells total versus the one white blood cell just in this piece of the slide (the ratio of cells in normal blood makeup is about 40-45% red blood cells and only 1% white blood cells).In addition, the red blood cells are very obviously smaller in comparison (the white blood cell shown above was 11µm x 12µm and the average size of red blood cell was 7µm x 7µm). Their small size is because of their movement through the vessels, veins, and capillaries, and their larger number is due to their function of carrying oxygen (through the tiny vessels, veins, and capillaries) to all parts of the body. White blood cells are bigger and less numerous because their function is more specific (producing antibodies) and less frequently necessary (compared to needing red blood cells at all times to remain alive from consumption of oxygen), and they leave the capillaries to move to injury sites which explains their increased size ("Blood Basics").
3. A student, who has not learned properly how to prepare solutions, added 205.386 g of sucrose (FW = 342.31) to exactly 200.0 mL of dH2O and assumed that a 3.0 M solution had been prepared. The total solution volume increased to 220.0 mL after the sucrose was dissolved and the solution was thoroughly mixed. Determine the actual concentration of the sucrose solution prepared by the student. Show your work. What did the student do wrong?
With the way the student prepared the solution, the concentration of the sucrose solution tuned out to be 2.73M. This is because the student added 200mL of dH2O after obtaining the proper weight of sucrose instead of filling the container with dH2O until the total final volume was 220mL. The calculations for this experiment are to divide the measured weight of sucrose: 205.386g by its molar mass: 342.31g which yields 0.6 Moles of sucrose. This divided by the correct amount of volume, 0.20L results in a 3 Molar solution of sucrose, but, if the solution is made correctly and the final volume is 220mL, the 0.6 Moles of sucrose is divided instead by 0.22L which yields the incorrect concentration value of 2.73M that the student prepared.
4) What is the purpose of a molecular ladder in gel electrophoresis?
a molecular ladder helps determine the actual lenth of a molecule. it is a solution of DNA molecules of different, known lengths. this is often used as a reference ot estimate the size of unknown DNA molecules.
1. What is pGLO? Describe the structure of pGLO. Be sure to discuss the function/role of any genes or regions important for its role as a genetic vector. Give an example of how a plasmid like pGLO can be used by researchers. Be sure to cite your sources.
pGLO is a structurally-circular bacterial plasmid (extrachromosomal DNA source) that is separated from, and replicates independently of its bacterial chromosome. pGLO shares its genetic information with bacteria and contains an ampicillin resistant gene (bla), and a gene that codes for green fluorescent protein (GFP) that has been inserted into the arabinose operon that controls the transcription of the arabinose-digesting protein (araC). The location of the GFP gene on the arabinose operon causes the GFP gene to only be transcribed when arabinose is present in the growth media. pGLO contains an an origin for DNA replication and multiple cloning sites as well. pGLO can be used to determine whether incorporation of a gene of interest into genetic material on a culture of bacteria has occurred by platting the bacteria with LB, penicillin, and arabinose (to transcribe the GFP) and observing the resulting bacteria ("Introductory Laboratory Experiments in Biology").
4. Define the term pH. What is the H+ concentration of a solution with a pH of 3? What is the H+ concentration of a solution with a pH of 5? Of these two solutions, which of these solutions is more acidic? What is the difference in acidity between the two solutions?
pH is defined as an expression of the acidity of a solution on a logarithmic scale from 1-14 (1 being the most acidic value and 14 being the most basic). pH is a measure of the Hydrogen ion concentration measured with the equation -log10c, with "c" being the hydrogen ion concentration in moles per liter. The hydrogen ion concentration of a solution with a pH of 3 is 10-3. A solution with a pH of 5 would have a hydrogen ion concentration of 10-5. The solution with a pH of 3 would be more acidic in comparison to that with a pH of 5. The difference in acidity would be a value of 2. Because the pH scale is logarithmic, the difference between ion concentration levels is 10 each. The difference in acidity could also be described in terms of hydrogen ion concentrations, and would be a difference of 100 hydrogen ions in this case; in other words, a solution with pH 3 has 100 more hydrogen ions than that with a pH 5.
9) What are restriction enzymes? Give an example of how they are used by research scientists.
restriction enzymes are used to cut DNA into smaller fragments. the cuts are always made at specific nucleotide sequences. different restriction enzymes recognise and cut different DNA sequences. restriction enzymes are a basic tool for biotechnology research. they are used for DNA cloning and fingerprinting
3) What is the difference in function between loading dye and Syber-Safe DNA stain?
sybr safe is a highly sensitive gel stain for visualization of DNA in agarose gel and emits a green light. loading dye does not bind to DNA, instead it mixes with DNA and serves the purpose of helping you track how far your run has progressed.
1) Explain why you poured two gels in Procedure 6.1. Why was one gel 'thicker' than another?
two different gels were used during procedure 6.1 for different sized segments that are being observed. one gel is thicker than the other and the thicker gel is used for smaller segments so they can move through the gel nicely and not be washed out too quickly with bigger segments
