Bio Lab Midterm

P20
Which micopipettor would be the best choice to measure a volume of 19µL most accurately?
135µL
You have been asked to meaure a volume of 0.135 mL. How much is this in microliters?
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500 mg or 0.5 g
A volume of 500µL will have a mass of approximately
0.250mL
You have been asked to measure the following volumes into well A1 in a microliter plate; 0.2mL of reagent A, 10µL of reagent B, 35 µL of reagent C, and 5µL of reagent D. What is the total volume of reagent in well A! in mL?
rinse your eyes immediately for at least 5 minutes and tell your instructor
If a chemical in the lab gets splashed into your eye, you should
False
T or F: Return all unused chemicals into their original containers at the end of each labratory.
T
T or F: Never remove chemicals or equipment from the lab.
a green fluorescence protein
pGlo plasmids, when taken up by a bacteria will code for
DNA
Genetic transformation occurs when a cell takes up and expresses a new piece or foreign ____________, often a circular plasmid.
salt
Hydrophobic Interaction Chromatography (HIC) allow for elution of certain proteins by using various concentrations of __________ buffer to load samples on the column and wash them off.
study of biological processes
observation of cell movement
use of GFP as a visual marker
What are 3 real-world links for the study/use of genetic GFP?
speed up the conversion of hydrogen peroxide to water and oxygen
In enzyme Microarray assay, the catalase was used to:
(actual-expected)/expected x 100
% error =
1000
1 mL = _____ µL
1000
1 L = _____ mL
hydrogen peroxide
___________ is produced in aerobic respiration and is toxic to cells, so the body must work quickly to remove it by turning it into water and oxygen.
Catalase
___________ is the enzyme found in all living things used to quickly convert hydrogen peroxide into water and oxygen.
it denatures them
what effect does heat have on enzymes?
true
T or F: Equilibrium constant of reactions are not altered by catalysts.
allosteric
Some enzymes are regulated by intracellular metabolites. These are called ___________ enzymes.
substrate
Reactant molecule in an enzyme catalyzed reaction is called the _________
true
T or F: Most aerobic life forms are capable of enzymatic peroxide detoxification.x
Any remaining H2O2 converts iodide (I-) to iodine, resulting in a brownish-red color. H2O2 will decrease in the incubation reaction, so the color will decrease.
Why does the brownish red color decrease in the incubation reaction in Lab 2?
detect the presence of the enzyme catalase in bacteria.
The catalase test (Lab 2) is used to
Bradford Assay uses Coomassie Blue dye which binds to the side chains of specific amino acids and shifts the absorbance from 470nm (reddish-brown) to 595nm (blue).
The intensity of the blue correlates with concentration of protein.
How does testing the absorbance of our milk samples (Lab 3) tell us the protein concentration?
phosphate buffered saline
What did we dilute the milk with in lab 3?
a series of protein standards of known concentrations
What did we compare the diluted milk samples with in lab 3 to determine their protein concentration?
measure the absorbence of all cuvettes, calculate the concentration of known protein standards, and use that to find the concentrations of the samples
After visually comparing our diluted milk samples to our series of protein standards in lab 3, what did we do?
A single-celled organism would be the best recipient for a genetic transformation, because it contains only one cell which needs to take up the new gene.
To genetically transform an entire organism, you must insert the new gene(s) into every cell in the organism. Which organism is better suited for total genetic transformation-one composed of many cells, or one composed of a single cell?
An organism which reproduces quickly. Fast production of offspring or new progeny will allow you to quickly assess if the new trait has been passed on.
Scientists often want to know if the genetically transformed organism can pass its new traits on to its offspring and future generations. To get this information, which would be a better candidate for your investigation, an organism in which each new generation develops and reproduces quickly, or one which does this more slowly?
The organism should not produce any toxins or compounds which could make people sick. The organism should grow vigorously in the lab environment, but should not be able to survive outside the laboratory. The organism should not be able to infect plants or animals.
Safety is another important consideration in choosing an experimental organism. What traits or characteristics should the organism have (or not have) to be sure it will not harm you or the environment?
A bacterium would be the best host organism. Bacteria are small, single-celled organisms which reproduce quickly and easily.
Based on the above considerations, which would be the best choice for a genetic transformation: a bacterium, earthworm, fish, or mouse? Describe your reasoning.
-Color of colonies, number of colonies, distribution of colonies on the plate.
Recall that the goal of genetic transformation is to change an organism’s traits (phenotype). Before any change in the phenotype of an organism can be detected, a thorough examination of its usual (pretransformation) phenotype must be made. Look at the colonies of E. coli on your starter plates. List all observable traits or characteristics that can be described.
Equal amounts of cells could be plated on two different LB nutrient agar plates, one which contains just LB nutrient agar and one which contains LB nutrient agar ampicillin. The growth of E. coli could be compared on the two plates. If ampicillin negatively affects the growth of E. coli, then there should be fewer colonies of bacteria on that plate. If ampicillin has no effect, there should be approximately equal numbers of colonies on both plates.
Describe how you could use two LB nutrient agar plates, some E. coli, and some ampicillin to determine how E. coli cells are affected by ampicillin.
Antibiotics usually kill bacteria (are bacteriocidic) or inhibit their growth (bacteriostatic). Thus, there should be few, if any, bacterial colonies present on the ampicillin plate. The presence of any colonies on the ampicillin plate would suggest that those bacteria are resistant to the antibiotic ampicillin
What would you expect your experimental results to indicate about the effect of ampicillin on the E. coli cells?
Bacteria which resemble the non-transformed will be found on the LB/(-) pGLO plate. These bacteria were removed from the starter plate, did not have any plasmid added to them, and were replated on an LB plate. Thus, they are virtually identical to the non-transformed starter .
On which of the plates would you expect to find bacteria most like the original untransformed E. coli colonies you initially observed? Explain your prediction.
The transformed cells are found on the LB/amp and LB/amp/ara plates. Genetically transformed cells have taken up the pGLO plasmid which expresses the ampicillin resistance gene—these cells can survive on the plates which contain ampicillin.
If there are any genetically transformed bacterial cells, on which plate(s) would they most likely be located? Explain your prediction.
The LB/amp (-) pGLO and the LB/amp (+) pGLO plates should be directly compared. Cells which were not treated with DNA (-pGLO) should not be expressing the ampicillin resistance gene and will not grow on the LB/amp plates. Cells which were treated with DNA (+pGLO) should contain the pGLO plasmid and should express the ampicillin resistance gene—the corresponding LB/amp plate will contain transformed bacterial colonies.
Which plates should be compared to determine if any genetic transformation has occurred? Why?
A control plate is a guide that is used to help you interpret the experimental results. In this experiment, both (-) pGLO plates are control plates. The LB/amp control plate can be compared to the LB/amp (+)pGLO plate. This comparison shows that genetic transformation produces bacterial colonies that can grow on ampicillin (due to the uptake of the pGLO plasmid and the expression of the ampicillin resistance gene). The (-) pGLO/LB control plate can be compared to any of the LB/amp plates to show that plasmid uptake is required for the growth in the presence of ampicillin. The (-) pGLO LB/amp plate shows that the starter culture does not grow on the LB/amp plate. Without this control one would not know if the colonies on the LB/amp (+) pGLO plate were really transformants.
What is meant by control plate? What purpose does a control serve?
The bacteria on the (+) pGLO LB/amp plate and the (-) pGLO LB plates should be whitish. The bacteria on the (+) pGLO LB/amp/ara plate should appear whitish when exposed to normal, room lighting, but fluoresce bright green upon exposure to the long-wave UV light.
What color are the bacteria?
There should be approximately ~ 75 bacterial colonies on the two (+) pGLO plates. The lawn of bacteria on the LB plate contains an even spread of bacteria and individual colonies can’t be counted.
Count how many bacterial colonies there are on each plate (the spots you see).
transformed traits and how you arrived at this analysis for each trait listed.-Bacteria are a whitish color;Colony size is similar both before and after transformation
Which of the traits that you originally observed for E. coli did not seem to become altered? In the space below list these non
The colonies on the LB/amp/ara plate fluoresce green under UV light; The transformed colonies can grow on ampicillin resistance
Of the E. coli traits you originally noted, which seem now to be significantly different after performing the transformation procedure? List those traits below and describe the changes that you observed.
If the genetically transformed cells have acquired the ability to live in the presence of
The plasmid must express a gene for ampicillin resistance (the protein product of the bla gene codes for beta-lactamase, the protein that breaks down ampicillin).
the antibiotic ampicillin, then what can be inferred about the other genes on the plasmid that were involved in your transformation procedure?
The best way is to compare the control to the experimental plates. Cells that were not treated with the plasmid (LB/amp (-) pGLO and LB/amp/ara (-) pGLO plates) could not grow on ampicillin, whereas cells that were treated with the plasmid (LB/amp (+) pGLO and LB/amp/ara (+) pGLO plate) can grow on the LB/amp plate. Thus, the plasmid must confer resistance to ampicillin
From the results that you obtained, how could you prove that these changes that occurred were due to the procedure that you performed?
The plasmid sample did not fluoresce.
Recall what you observed when you shined the UV light source onto a sample of original pGLO plasmid DNA and describe your observations.
The pGLO plasmid DNA and the original bacteria can be eliminated from providing the fluorescent source.
Which of the two possible sources of the fluorescence can now be eliminated?
The source of fluorescence is probably from some protein that the plasmid encodes.
What does this observation indicate about the source of the fluorescence?
A successful experiment will be represented by the presence of colonies on the (+) pGLO LB/amp and (+) pGLO LB/amp/ara plates and the absence of colonies on the (-) pGLO LB/amp plate. Moreover, the colonies on the LB/amp/ara plate should fluoresce green. An unsuccessful experiment will show an absence of colonies on the (+) pGLO LB/amp and (+) pGLO LB/amp/ara plates. This could be a result of not adding a loopful of plasmid to the (+) pGLO tube or not adding a colony of bacteria to the (+) pGLO tube
Describe the evidence that indicates whether your attempt at performing a genetic transformation was successful or not successful.
Yes. The bacteria that did not receive the plasmid are growing on a plain LB plate.
Look again at your four plates. Do you observe some E. coli growing on the LB plates which do not contain ampicillin/arabinose?
No. You cannot tell if the bacteria are ampicillin resistant just by looking at them. Both types of bacteria (those that are ampicillin resistant and those that are ampicillin sensitive) look similar when cultured—think about the colonies on the LB starter plate and the colonies on the +pGLO LB/amp plate.
From your results, can you tell if these bacteria are ampicillin resistant by looking at them on the LB plate? Explain your answer.
The best test would be to take some of the bacteria growing on the LB plate and streak them on an LB/amp plate. If the bacteria are viable on the LB/amp plate, then they are resistant to ampicillin. If no bacterial colonies survive, then they were not ampicillin resistant (they were ampicillin sensitive).
How would you change the bacteria’s environment to best tell if they are ampicillin resistant?
The sugar arabinose in the agarose plate is needed to turn on the expression of the GFP gene. The UV light is necessary to cause the GFP protein within the bacteria to fluoresce.
What two factors must be present in the bacteria’s environment for you to see the green color? (Hint: one factor is in the plate and the other factor is in how you look at the bacteria).
The sugar arabinose turns on expression of the GFP gene by binding to a regulatory protein, araC, which sits on the PBAD promoter. When arabinose is present, it binds to araC, consequently changing the conformation of araC which facilitates transcription of the gene by RNA polymerase (see detailed description in Appendix D). When exposed to UV light, the electrons in GFP’s chromophere are excited to a higher energy state. When they drop down to a lower energy state they emit a longer wavelength of visible fluorrescent green light at 509 nm.
What do you think each of the two environmental factors you listed above are doing to cause the genetically transformed bacteria turn green?
Gene regulation allows for adaptation to differing conditions and prevents wasteful overproduction of unneeded proteins. Good examples of highly regulatable genes are the enzymes which break down carbohydrate food sources. If the sugar arabinose is present in the growth medium it is beneficial for bacteria to produce the enzymes necessary to catabolize the sugar source. Conversely, if arabinose is not present in the nutrient media, it would be very energetically wasteful to produce the enzymes to break down arabinose.
What advantage would there be for an organism to be able to turn on or off particular genes in response to certain conditions?
A macromolecule which consists of chains of amino acids.
What is a protein ?
Antibodies, digestive enzymes, hair proteins, hormones, and hemoglobin are all examples of proteins.
List three examples of proteins found in your body.
Genes contain the genetic code which determine the amino acid composition of a protein. There is a unique gene for each protein within all of the cells of the body.
Explain the relationship between genes and proteins.
The duplication and propagation of a cell or organism.
Using your own words, define or describe cloning.
A library of bacterial cells contains a diverse mixture of bacterial types which contains a diverse mixture of genes. A single colony of bacteria originates from an individual clone which only contains a single gene.
Describe how bacterial cells in a “library” are different from the cells of a single colony.
One can isolate a single fluorescent green colony of bacteria and grow large amounts of the bacteria in a liquid growth media. Bacteria in liquid media can be concentrated by centrifugation. After the bacterial cells are lysed to release the cancer curing protein, the protein can be isolated by passage through a chro- matography column which has an affinity for the cancer-curing protein.
Describe how you might recover the cancercuring protein from the bacterial cells.
A bacterial colony is a large group or cluster of bacterial cells that originated from a single, clonal cell.
What is a bacterial colony?
ultraviolet (UV) light—The green protein fluoresces, and is thus visible, when exposed to the UV light.: incubator—The incubator provides a warm temperature which enables the bacteria to grow.; shaking incubator—A shaking incubator provides a warm temperature and provides aeration which oxygenates the bacterial cultures. Increased oxy- gen accelerates the growth rate of bacteria.
How are UV light, incubator, shaking incubator items helpful in this cloning experiment?
If a green colony was streaked onto an LB/amp plate, the resulting colonies would be white. This plate does not contain arabinose which is needed to induce expression of the GFP gene and generate green fluorescent colonies. If a white colony was streaked onto an LB/amp/ara plate, the resulting colonies would be green. This plate contains arabinose which induces expression of the GFP gene and generates green fluorescent colonies.
Can you predict what would happen if you took one of the green colonies from the LB/amp/ara plate and streaked it onto an LB/amp plate? Conversely, what would hap pen if you took a white colony from the LB/amp plate and streaked it onto an LB/amp/ara plate? Explain your answer.
The green colony seeded into the (+) tube fluoresces green because the arabi- nose in the liquid culture media continues to induce expression of the GFP gene, which results in a green culture. The white colony seeded into the (+) tube fluo- resces green because the arabinose in the liquid culture media turns on expres- sion of the GFP gene which was previously off on the LB/amp plate (which lacks arabinose), which results in a green culture.
Can you explain why both liquid cultures fluoresce green?
The supernatant contains the bacterial growth media and does not contain the desired GFP.
Why did you discard the supernatant in this part of the protein purification procedure?
The white culture of bacteria does not contain the GFP and is not needed for the subsequent purification step.
Why did you discard the white liquid from the “x” tube but keep the green one?
When a bacterial cell freezes, the volume of cytoplasm expands. The expansion puts pressure on the weakened cell wall, which then ruptures from the pressure.
Can you explain why the bacterial cells’ outer cell wall ruptures when the cells are frozen? What happens to an unopened soft drink when it freezes?
The bacteria need to be ruptured in order to release the GFP, which can then be purified using column chromatography.
What was the purpose of rupturing or lysing the bacteria?
-The pellet should be a whitish or pale green color. The supernatant should fluo- resce bright green. The fluorescent green color of the supernatant indicates that the green protein was released from the bacteria and remained in the super- natant. The much lighter color of the bacterial pellet suggests that the GFP was released from the bacteria upon lysis.
What color was the pellet in this step of the experiment? What color was the supernatant? What does this tell you?
The pellet contains unwanted bacterial debris—bacterial cell walls, membranes, and chromosomal DNA. The pellet contains little, if any, GFP and can be discarded.
Why did you discard the pellet in this part of the protein purification procedure?
Protein chromatography is a technique which can be used to separate or purify proteins from other molecules. This lab used hydrophobic interaction chro- matography to purify GFP based upon its hydrophobic properties.
Briefly describe hydrophobic interaction chromatography and identify its purpose in this lab.
Equilibration Buffer—This buffer prepares the column for the application of GFP. Equilibration buffer raises the salt concentration of the column to match that of the bacterial GFP lysate.; Binding Buffer—This buffer raises the salt concentration of GFP which causes a conformational change in GFP, exposing the hydrophobic regions.; Wash Buffer—Wash buffer functions to wash away less hydrophobic, contaminating proteins from the column.; TE (elution) Buffer—This buffer functions to remove GFP from the column.
Based on your results, explain the roles or functions of the equilibrium, te, binding, and washbuffers. Hint: how does the name of the buffer relate to its function?
Binding buffer >>Equilibration buffer>>Wash buffer>>Elution buffer;Binding buffer has the highest concentration of salt because it is needed to raise the salt concentration of the GFP lysate. The hydrophobic patches of proteins are exposed in high salt buffer. TE elution buffer has the lowest salt concentration because it causes GFP to elute from the column. The hydrophobic patches of proteins re-orient to the interior, and the hydrophilic regions are exposed in low salt buffer.
Which buffers have the highest salt content and which have the least? How can you tell?
If tube 3 fluoresces green, the student was successful in purifying GFP. If GFP is not present in tube 3, examine the column—application of an incorrect buffer would prevent the elution. Alternatively, if the student did not start with a bright green culture, then tube 3 will not be extremely bright.
Were you successful in isolating and purifying GFP from the cloned bacterial cells? Identify the evidence you have to support your answer.
5
If the GAATTC palindrome is repeated four times on the same piece of linear DNA, and the restriction enzyme that recognizes that base sequence is present, how many DNA fragments will be produced?
Random sized fragments will be produced.
If the GAATTC palindrome repeats are randomly spaced along the DNA strand, then what can you say about the size of the fragments that will be produced when the DNA is digested with a restriction enzyme that recognizes that sequence?
The DNA in the uncut lambda DNA tube, which gets no enzyme, should remain intact as a single band. No restriction enzymes are added to that tube, thus no digestion will occur, and no fragments will be produced. This tube (L) is the control tube.
In which tube do you expect no changes to occur, that is, no DNA fragments produced? What is missing in that tube that leads you to that decision?
The P tube contains the restriction enzyme PstI. There should be an enzymatic digestion that occurs in that tube, which results in the production of restriction fragments. The L tube only contains DNA and no enzyme and so should not produce any restriction fragments. Restriction enzymes digest DNA at specific sites. If lambda DNA contains restriction sites for the enzyme PstI, then the DNA should be cut into fragments. With no added enzyme, no digestion should occur.
Compare tube P to tube L; What do you expect to happen in the P tube compared to the L tube. Why do you expect this difference?
Most likely a restriction enzyme was inadvertently added to the L tube. This could have happened by accidently adding an enzyme to the tube, or by not changing pipet tips, which could result in enzyme being carried over between tubes.
If the DNA in tube L becomes fragmented at the conclusion of the reaction, what can you conclude?
No. The DNA still appears colorless.
Is there any visible change to the DNA after adding restriction enzymes?
3
If a DNA molecule has two restriction sites, A and B, for a specific restriction enzyme, how many fragments would be produced, if it is cut by that enzyme?
6.
Draw a DNA molecule that has five randomly spaced restriction sites for a specific palindrome. How many fragments would be produced if they were each cut by a restriction enzyme?
The large fragments would be towards the top of the gel because it is more difficult for the larger pieces to be strained through the gel.
In electrophoresis, where would the larger fragments, those with the greater number of base pairs, be located; toward the top of the gel or the bottom? Why?
There would still be only 4 bands present.
Suppose you had 500 pieces of each of the four fragments, how would the gel appear?
slower
The larger the DNA fragment, the _______ it migrates through an agarose gel.
The DNA is a colorless solution
What color was the DNA before you added loading dye?
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