Bio 232 Quizzes, HW, and Lab Analyses
Plasmid A is pGEX Plasmid B is pCR-LDAP Plasmid C is pLux Plasmid D is pUC19
Based on the gel electrophoresis data and the plasmid maps, assign which plasmid DNA corresponds to pUC19, pCR-LDAP, pGEX and pLux.
Plasmid A which is Plux will emit light due to E.coli. Plasmid B which is PCR-LDAP will be white colonies because of inhibited lacZ gene. Plasmid C which is PUC19 will be blue because of activated lacZ gene. Plasmid D which is pGEX will produce recombinant proteins when with E.coli.
Based on the identity of each plasmid you determined with RE mapping, can you predict the phenotypes of bacterial colonies transformed with each of these plasmids, and plated on LB/Amp/X-Gal media?
Though our protein does not seem that pure due to the weak the intensity of the protein ladder, the ladder is still visible and is the only ladder present compared to other lanes, and is therefore relatively pure. Also, the recombinant protein elution worked fairly well because there were not that many gaps or pores in the protein ladder, therefore further proving the purity of the protein because eluting means that unnecessary protein is washed out.
Based on the intensity of the protein bands in your gel, how pure is your protein? How well did the elution of the recombinant protein from the agarose matrix work?
We have been working with the unknown recombinant protein GST-SidC with the Legionella ubiquitin protein since the protein band size is nearest the total MW of the recombinant protein that had Legionella ubiquitin. This protein band was around 150 kDa which is close to the expected MW of the protein is 135 kDa.
Based on your results, identify the unknown protein you have been working with.
The pellet would have larger molecules such as carbohydrates from the spinach. The supernatant contains the chlorophyll as it is the supernatant that is analyzed on the spectrophotometer for the chlorophyll concentration.
Briefly discuss the fractionation that occurred during centrifugation. What is in the pellet and what is in the supernatant after centrifuging the extracts?
Microwells with HIV antigens must be tested with the person's serum by a serial dilution. If HIV antibodies are present, then they will be in the wells bound. Then, microwells with unbound proteins are washed and discarded. The anti-HIV would then be added and incubated, then washed Color development solution will then be added and will have to be incubated After incubation, color intensity will indicate if the person has HIV or not
Describe an ELISA procedure that you would use to determine if a person was infected with the human immunodeficiency virus, the causative factor of AIDS.
Yes, they do. As expected, the negative control group had no colonies. Plasmid B, pCR-LDAP, was white and had colonies. Plasmid C, pLux, is glow in the dark. Plasmid D, pUC19, is blue. Clearly, each plasmid all matches their phenotype as expected.
Do the identities of the plasmids B, C, and D that you assigned based on RE mapping match the phenotypes of the transformed E.coli?
Yes they do. Based on the RE mapping, plasmid B was expected to be pCR-LDAP, plasmid C was expected to be pLux, and plasmid D was expected to be pUC19, which also all agree with the phenotypic identities of each plasmid.
Do the phenotypes you observed for each transformation agree with the predictions you made based on the RE mapping results
Yes, there are observable protein bands on the gel. The predominate GST known band has the kDa of 26 in lane 1. The GST-unknown predominant protein is around 150 kDa in lane 7.
Do you observe any protein bands on your gel? What is the size of the major (predominant) protein band on your gel (See Appendix for the Protein Marker information)?
The calculated protein as shown in the table in the results section correlates with the size exclusion activity because protein concentration increased until well 3, with a concentration of 5.51, which means that this well has the most hemoglobin protein, and that the large hemoglobin proteins particles eluted from the column first. The gradual decrease in the concentrations as the wells go on show that there is less and less protein mean that Vitamin B12 are the ones eluting last since they are smaller in particle size.
Does your calculated protein amount corelates with the size exclusion activity?
(10mg/mL)*(0.0001%) = 0.00001 mg/mL
Estimate the lowest concentration of IgG that you detected in your assay. (Hint: The serum concentration of IgG is about 10mg per ml and the concentration of serum in Well #1 is 1%.)
Fraction 3 has the most hemoglobin and fraction 8 has the most vitamin B12 based on the red color intensity in fraction 3 and the pink color intensity in fraction 8.
Examine the 10 fractions that you collected. Which tube contains the "peak" fraction for hemoglobin and vitamin B12? The peak fractions contain the highest concentration of protein/vitamin and will be the most intense in color.
Look at lab report 5
Examine your plates and record your observations in the table below:
Graph lab 13
Graph the volume of O2 consumption (mL) as a function of time (min) for each seed sample.
Non-germinating seeds does not consume much oxygen because they are not active and do not need oxygen to live and grow like the germinating seeds. They use very little respiration.
How did the non-germinating seeds affect the rates of respiration in the seeds, and why?
This extraction extracts chlorophyll pigments from plant cell chloroplasts while alkaline lysis inserts DNA through plasmids into a prokaryotic cell.
How is this extraction method different than alkaline lysis?
Myosin, hemoglobin, myoglobin will be the order of passing since the more daltons, the bigger the particles, and bigger particles elute first.
If the following mix of molecules were purified using size exclusion chromatography, what would be the order in which the molecules pass through the opening in the bottom of the column? Mixture containing hemoglobin, 65,000 daltons; myoglobin, 17,000 daltons; myosin, 180,000 daltons
For the germinating sample, the rate of oxygen consumption is 0.0124 mL/min and for the nongerminating sample, the rate of oxygen consumptions 0.0052 mL/min.
Using these data, determine the oxygen consumption rates (mL/min) measured for each seed sample.
In our sample, there was not that much of a difference between the treated and control samples but the pH treated extract would have caused denaturation and therefore should've caused more of a difference only in structure but not in the concentration. The 45 degree Celsius treatment is not that much different from room temperature and therefore would not cause much denaturation, thus having not much difference from the control.
Was there a difference in protein concentration of control vs treated extract samples? If so, why?
No Lysis Buffer in plasmid DNA extraction cannot be used because the P2 buffer denatures the cells and will therefore denature the protein and its protein-protein interactions. The protein can therefore cannot be purified if the interactions are ruined.
We used sonicator to disrupt the cells prior to protein purification via affinity chromatography. Could have we used the same Lysis Buffer (P2) we did in plasmid DNA extraction, in this procedure?
The 5-min incubation period denatures the protein. It is important to incubate the protein because with the protein sample buffer before the analysis so that the secondary structures can be denatured before the gel loading.
What changes does the protein sample undergo during a 5-min incubation at 90˚C with Protein sample buffer (think about the charge, shape and confirmation of protein molecules)? Why is it important to incubate the protein sample with the protein sample buffer prior to the SDS-PAGE analysis?
The calculated chlorophyll content for extract A is 0.94917 mg and 0.4666mg for B. Measuring chlorophyll to analyze the health of a plant so that unwanted organisms such as algae can be tracked. A metabolite similar to chlorophyll is a carotenoid.
What was the calculated Chl content? What was the purpose for measuring Chl content? What is another general metabolite that we could measure to use in a similar fashion?
PE 1=2 (0.150-0.0179)/0.1976= 1.337 mg/mL PE 2=2 (0.161-0.0179)/0.1976= 1.448 mg/mL PE 3=2(0.161-0.0179)/0.1976= 1.448 mg/mL
What was the protein concentration of your extract? Remember step 7, you used diluted extract for treatments. Consider dilution factor while calculating concentration. What units are used here?
The hemoglobin exited first because it is bigger in particle size and the bigger particle elutes first.
Which molecule, hemoglobin, or vitamin B12, exited the column first? Would this molecule be the larger or smaller of the two?
The highest concentration is 685.45 ng/ul in plasmid A and the lowest is 462.14 ng/ul in plasmid D. This high concentration number means that the copy number is high which is further supported by the high vector size of PGEX and low concentration means that copy number is low which is also supported by the low vector size of pUC19.
Which plasmid had the highest and the lowest DNA concentration (in lab 2)? How does it relate to the type of the plasmid vector and its copy number?
Before adding the HCl, row A, which is the chicken sera appeared to react a little in wells A2 and A4, and row D which is rabbit, reacted very strongly with wells D1, D2, D3 being high intensity, and D4 being low intensity. After HCl, more interactions showed with colors of A2 and A4 being medium intensity, A3 being low intensity, B1-B5 and B6 being low intensity, all of row C being low intensity, and D1, D2, D3 being high intensity, D4 being medium intensity, and D5 being low intensity. Since a goat-anti-rabbit IgG peridoxase was used, it makes sense that it binds to the rabbit serum the strongest since an IgG from a rabbit would cause the goat to produce antibodies that will recognize the rabbit IgG molecules. The goat-anti-rabit IgG must have had lower interactions with the other sera as reflected in the color intensities because other serum like the chicken's may have an antigen that can bind with the immunoglobin IgG.
Which sera samples reacted with the antibody and which did not. Explain why a difference was observed.
Since fraction 1 served as the blanks due to being just column buffers, extra blanks were not necessary.
Why did we not include a blank in the protein measurement assay?
chemically inert, is always solid or liquid adsorbed on solid surface, should allow free flow of mobile phase
in case of column chromatography, stationary phase
MgCl2
maintain ionic strength
**know cell cycle images phases
metaphase, anaphase, telophase, prophase, telophase
size, shape, net charge
migration of particles in an electric field is dependent on the following particle properties
peptide
primary structure of proteins is a stretch of amino acids linked together by these type of bonds
denatures DNA and changes the osmotic pressure so cells rupture, denatures plasma membrane, denatures lipid molecules
role of using NaOH in the lysis buffer
**know in a diagram
secondary antibody conjugate, enzyme
true
the development of blue color from orange reddish color is due to binding of coomassie dye with basic amino acids
repress the gene expression of protein of interest
the function of gene expression of protein of interest
true
the higher the bis acrylamide concentration while making polyacrylamide the slower the proteins will migrate
Glycerol
used in extraction of organelles
they are signals for molecules to stimulate or stop cell division to instruct cell differentiation or to initiate death
what are cell cycle regulators
radiation, pH, and solvents
which of the following factors affect protein denaturation
only way of reproduction for single cells; cell division is crucial for growth and/or replacement of damaged or dead cells
why is cell division for both single and multicellular organisms
beta mercaptoethanol
The limitation of BCA assay is that it is incompatible with some components of the buffer. Which of the following components affect outcome of BCA assay.
PVPP
avoid browning from phenolics
Selection of transformed cells that grow on plate
Ampicillin which is added to LB media while making LB-Amp-X-gal plates helps with
The cells treated with acetic acid are dead. The cells treated with glucose release high level of CO2 due to glycolysis
Based on Lab 13 yeast experiment, why do yeast cells with glucose produce more CO2 compared to the cells treated with acetic acid?
beta-mercaptoethanol
During lab 6, while making extraction buffer, you used a reducing agent to keep proteins from oxidizing, during extraction. The name of this chemical is
antigen, antibody, proteins, glycoprotiens
ELISA tests can be used to identify which of the following substances
LacZ protein is produced but not doing its function because there were no lactose to break down.
Place lac promoter, lac operator and lacZ gene to their respective places in the simulation. Let the simulation proceed for 30-60 sec.
SDS-PAGE is composed of a stacking gel and a resolving gel which are used to enhance protein separation for sharper bands. While the stacking gel has a low acrylamide percentage and pH value, the resolving gel has a higher acrylamide percentage and pH value. The stacking gel allows a sample to get concentrated and to enter into the resolving gel, and then the resolving gel allows the sample to run and separate proteins based on their molecular weight.
SDS-PAGE is composed of two gels and what are they? and what is the purpose of having two gels?
The buffer ensures that the mobile phase will move through the column, resulting in completion of the chromatographic separation procedure.
While performing SEC in lab, we added extra buffer (1 ml at a time for a total of 3 ml) towards later steps in the protocol. Why did we applied this higher amount of buffer?
O2
The terminal electron acceptor for cellular respiration is
ion exchange chromatography
The type of chromatography that uses electrostatic interaction between the charged proteins is called
allows antigens in serum to bind to the well surface of the plate
While working on ELISA technique similar to your lab protocol, why it is important to incubate the serum in the plate?
The calcium chloride will be attracted to the negative phosphate groups in the phospholipid bilayer to neutralize this charge and allow plasmids to get into the cell.
Why is calcium chloride added to the cell membranes?
protease inhibitor
abolish function of protein degrading enzymes
carotenoid
Apart from Chlorophylls and which other pigment is involved in transferring electrons in antenna complex
Lactose was broken down by LacZ as lactose was added into the system.
Add lactose to the system by clicking Auto on the Lactose Injector (top left corner). Let the simulation proceed for 30-60 sec.
Lactose binds to the LacI gene blocking the lac operator hindering the LacZ production so some LacZ genes are being produced which then breaks down some lactose, but not enough LacZ is produced to break down all of the lactose fast enough because LacI is also being produced hence blocking the lacZ production until a lactose binds to it.
Add lactose to the system generated in the previous simulation by clicking Auto on the Lactose Injector (top left corner). Let the simulation proceed for 30-60 sec. Record your observations.
In the germinating seeds respirometer, it is a cellular respiration process so O2 consumes and moves downward. The yeast respirometer is a fermentation process produces CO2 gas which is less dense than the liquid and therefore the bubble moves up.
Based on the demonstration of germinating seeds respirometer, the colored bubble in the pipette moves downwards whereas the colored bubble in the yeast respirometer moves upwards. Why?
(picture was in HW 4)
Based on the gel images, please select the options that best describe the agarose gels. The gel image shows Ladder, uncut plasmid DNA and the plasmid DNA that is digested with restriction enzyme
The concentration of the chemical solution is more accurate in the serial dilutions method because first of all, the values are to the decimal points which shows more accuracy. Second, the dilution factor is known so the concentration is easier to be accurately found as the concentration gets diluted in a uniform manner. Additionally, solution gets patchy at low concentrations cell suspension-wise.
Based on what you learned from the video and the description, is the serial dilutions method more likely to produce accurate results for determining the concentration of chemical solutions or cell suspensions? Why?
WT is (31/4)*4*10^4 = 3.1 x 10^5 cells/mL UV-S is (77/4)*4*10^4 = 7.7 x 10^5 cells/mL
Calculate the cell concentration from the total cell counts for each yeast strain by using the following formula: (Total # 𝑐𝑒𝑙𝑙𝑠 / # 𝑠𝑞𝑢𝑎𝑟𝑒𝑠) × dilution factor × volume factor = #𝑐𝑒𝑙𝑙𝑠 / 𝑚𝐿
0.025% final concentration and 12.5 uL & 3 mM and 300 uL
Calculate the components of the buffer with given stock solutions to make 10ml of protein extraction buffer (HW 6)
(table were in HW 3)
Calculate the volume of water to be added to each reaction, to bring the total volume to 50 μL
The dye is red, therefore cationic, when present in the Bradford Protein Assay Reagent and is blue, therefore anionic, when bound to the protein.
Coomassie G-250 can exist in three forms: cationic (protonated), neutral and anionic (unprotonated). Which form of the dye is present in the Bradford Protein Assay Reagent? Which form is the dye in when bound to the protein?
The further away from the root tip, the less division the cells have and are therefore bigger and more irregular in shape.
Do you see any change in the cells as you move away from root tip?
I would expect plasmid pUC19, pGEX, pLux to have two fragments of increasing size from pUC19 to pLux. The plasmid pCR-LDAP should have 3 fragments, with one relatively bigger than the others.
How many fragments and which size you would expect for each plasmid after EcoRI enzyme
20
If a cell has 10 chromosomes, how many molecules will be seen during metaphase?
Ice repairs the damage to the cell well because there's less energy so phospholipids would be firm.
If heat shock is needed to open up the cells so the plasmids can enter, why do we place the cells on ice after heat shock for a few minutes?
I would expect about 100.67 colonies because a 906 number of particles divided by the 9 mL leads to this amount of colonies.
If the particles in this simulation were yeast cells, and you plated 1mL from container/tube 2 on the agar plates, how many colonies would you expect to form on the plate? Explain your reasoning.
2.4x10^8 cells/mL
If you are given a solution of E. coli and asked to perform serial dilution. You perform a dilution series of 10-1, 10-2, 10-3, 10-4 and 10-5 and plat 5 ul spots on the plates. You incubate plates for a day. You find that the dilution of 10-5 has nicely spread-out colonies that you can count. There are 12 colonies on the spot corresponding to 10-5 dilution. Based on this observation, what is the concentration of the E. coli solution given to you?
Only LB/Kan/ara will be the media with growth.
If you transformed bacteria with a plasmid carrying the kanamycin resistance gene and a gene for green fluorescent protein (GFP) under the control of an arabinose-dependent promoter, what media can you use to observe the bacteria grow? What media can you use to observe the bacteria glow? (Yes/No) (2 pts.)
200 µL = .200 mL6 x 10^8 * .200 mL = 1.2x10^4 cells
If you were given a liquid cell culture with a concentration of 6 x 10^8 cell/mL, and asked to transfer 200µL of this culture to 100mL of fresh media, how many cells would get transferred? Please, show your calculation.
Yes, bacterial growth will occur in the P+ tubes.
Imagine, you transformed competent E.coli cells with pUC19 plasmid (in P+ tube), and set up a negative control transformation, where competent E.coli cells were mixed with water instead of plasmid (in P- tube). You plated the cells from both tubes on two types of media: LB and LB/Amp. Would you expect formation of bacterial colonies on these plates for each transformation (Yes/No)? (1 pt.)
a. Volume occupied around the beads or stationary phase.
In Size exclusion chromatography, the void volume refers to
look at Lab Report 4
In the table below, record the size of the DNA bands you observe in each lane. Size of the DNA fragments in your samples can be determined by comparing level of their migration to that of the fragments of known size in 1Kb+ DNA Ladder (See Appendix 5).
On our slide, the right side, specifically the bottom right has the most active cell division as there was more frequency and variety in the different stages of the cell cycle.
In which area of root do you find the most active cell division process?
low metal binding capacity
One of the properties of Good buffers is they are relatively free from side effects and have
Antigen is immobilized to the surface of the multi-well plate and detected with the antigen specific antibody which is directly conjugated to HRP or other detection molecules in direct ELISA. Indirect ELISA uses a two-step process with the primary antibody specific antigen binding to the target and a labeled secondary antigen against the host species of the primary antibody binds to the primary antibody for detection. So, in this indirect ELISA, the antigen is not immobilized to the surface of the multi-well plate and has multiple steps for binding. **Lost points on this one**
Read the article about ELISA and discuss the advantages of direct ELISA over indirect ELISA.
DNA stain
SYBR Safe that was added while making the gel is an example of
UV-S has a higher concentration and it does agree with the OD600 readings in these cultures because exercise 1.1 values were 0.11 for WT and 0.51 for UV-S, hence showing a higher concentration for UV-S.
Which yeast strain has a higher cell concentration? Does this difference in concentrations agree with the difference in OD600 readings for these cell cultures?
true
T of F: size exclusion chormatography can be used for separation of viral particles
false
T or F: Addition of glutathione beads is important to isolate all the proteins from the lysate
false
T or F: Competent cells made by treatment with CaCl2 can be used for the transformation using electroporation
True
T or F: In size exclusion chromatography, the proteins with higher molecular weight are eluted first?
True
T or F: In the presence of excess substrate (glucose), yeast cells undergo anaerobic respiration even in the presence of O2
True
T or F: Size exclusion chromatography uses beads with pores as a stationary phase and protein mix as a liquid phase. Hence it is considered as a type of liquid chromatography
true
T or F: bradford assay with the coomassie dye is compatible with chelating agents
true
T or F: restriction enzyme EcoRI cuts DNA at specific sites which have palindromic DNA sequences
true
T or F: while performing agarose gel electrophoresis, the migration speed is inversely proportional to molecular weight of DNA
6
The amount of CO2 molecules produced during aerobic respiration are
12 uL
The concentration of DNA you extracted from E. coli culture, was 125ng/ul. How many ul you need to use in a restriction enzyme digestion reaction to get 1.5 ug total DNA. The volume of the digestion reaction is 20 ul.
LacZ protein production was stopped by the lacI protein as it binds onto the lac operator.
Turn off the lactose supply to the system by clicking Manual on the Lactose Injector. Add lacI promoter and lacI gene to the system by placing these components to their respected places. Allow the simulation proceed for 30-60 sec. to reach a new equilibrium. Record your observations.
X = (0.48 + 0.0185 ) / 2.1262 = 0.2344 mg/mL
Use the following standard curve generated using BSA standards. Calculate the concentration of the unknown sample if the absorbance of the unknown protein is "0.48"
since neither protein should come out in tube 1, it can act as a blank
While performing Bradford assay, we used 10ul of sample from each tube, but we did not include a blank beacuse
0.0294 (based on the slope)
Using the graph below, what would be the rate of respiration in the germinating seeds
Silver staining can be and fluorescent dyes are other methods that can be used. Coomassie blue is significantly faster, easy protocol to stain, and is affordable as opposed to these other two. Also, compatibility with downstream applications is more flexible in Coomassie blue as opposed to the other two methods.
We used InstantBlue (Coomassie dye) dye for visualization of proteins bands on the gel. What are the other two methods that can be used for visualization of proteins bands. What is the advantage of using Coomassie blue over these other methods.
Yes, we were able to identify prometaphase in our slide towards the most active cell division area on the right as we saw chromosomes not in the middle of the cell nor at the end and more so scrambled in the cell.
Were you able to identify prometaphase in your slide?
Glycolysis, the citric acid cycle, and electron transport chain are the three metabolic pathways.
What are the three metabolic pathways that contribute to cellular respiration
a foreign susbtance for the body which induces an immune responds, can be body's own cells which induces immune response, a substance that is required for production of antibodies in organism.
What is an antigen
SDS solubilizes the proteins and lipids that make up cell and nuclear membranes so the contents can freely float in solution.
What is the purpose of SDS in the lysis Buffer?
to make sure DNA visualized, to give density to DNA samples so they can sink in the wells, to monitor the migration of dye during electrophoresis
What is the purpose of mixing loading dye with the DNA samples to be run on agarose gel?
The purpose of SDS for protein sample buffers and SDS-PAGE is to denature proteins, for cell lysis, and to even out the charge. They separate the proteins according to their electrophoretic mobility.
What is the purpose of using SDS for protein sample buffers and SDS-PAGE?
The amount of protein in the sample will be underestimated if the protein concentration is higher than the standard curve. Dilution, for example serial dilution, of said protein and then re-assaying will lead to a more accurate protein concentration.
What problem are you likely to run into if the concentration of your Unknown protein is higher than the concentration of the standard protein dilutions used for building the standard curve? What solution to this problem can you offer?
excluded
When using size exclusion chromatography, a column is said to have an exclusion limit of 40,000 daltons, would hemoglobin (60,000 daltons) be fractionated or excluded from the column?
In our lysis buffer, the resuspension buffer P1 was first used to resuspend the bacterial pellet before the Lysis SDS buffer. This chemical allows the solution to not have extreme pH changes.
Which chemical was used in our Lysis Buffer, which was not used in the video? What is the purpose of this chemical in the plasmid DNA extraction procedure
ampicillin
Which component is NOT part of the S.O.C. medium added to the transformed E.coli cells during the cell recovery period?
Proteinase K was added to the procedure video that was not in our lab exercise. This enzyme breaks down peptide bonds and digests contaminating proteins.
Which enzyme was used in the lysis step of the procedure in the video, but not in our lab exercise? What is the purpose of this enzyme in the DNA extraction procedure?
SDS
Which of the following chemicals from the protein loading dye is used to give a negative charge to the proteins in the lysate
immunodiffusion test
Which of the following methods is NOT used as a diagnostic tests in detection of HIV infection
ATP
Which one of the following have the highest amount of energy per molecule.
The conformation of the RFP enable them to separate the column chromatography because when it's unfolded, hydrophobic proteins can stick to resin, and hydrophilic folded proteins can pass through the columns.
Which physical property of the red fluorescent protein (RFP) protein is used to separate it by column chromatography in the example described in the article?
XL Blue-BL21
Which strain of bacteria was used for protein expression using the expression vector in this lab?
Germinating seeds have a higher respiration rate because the germinating seeds are living and require extra oxygen for growth since the seed germinates as its respiration rates increase, while non-germinating seeds are not living.
Why the germinating seeds have a higher rate of respiration compared to non-germinating seeds.
The fully folded RFP protein will have stronger interactions with the mobile phase because the unfolded proteins will expose these folded proteins along with the mobile phase and will pass through the column.
With which phase (stationary or mobile) will the fully folded RFP protein have a stronger interaction in the chromatography example? Why?
The unfolded RFP will have a stronger interaction with the stationary phase because the hydrophobic proteins unfolded will stick to the resin.
With which phase (stationary or mobile) will the unfolded RFP protein have a stronger interaction in the chromatography example? Why?
Two fragments (there were two ecoRI enzymes)
You are asked to do a restriction digestion of the following pRAK232 plasmid as shown below. You cut the plasmid similar to the one performed in the lab using EcoRI enzyme and separate the restriction digestion reaction on the agarose gel. How many DNA fragments will be seen on the gel.
three fragments (there were two bamHI enzymes)
You have a linear piece of DNA as seen in the figure below. If you use the BamHI enzyme to cut this piece of DNA how many fragments will you expect to see if you run your digest on agarose gel?
16 ng/uL
You isolated plasmid DNA from E. coli contain pUC19. When you checked the concentration of DNA using 1 ul on nanodrop machine, the setting was changed by the previous lab section. Now you are only able to see the absorbance at 260nm. If your absorbance reading was 0.32, what would be the concentration of your DNA sample in ng/ul.
50
You want to extract Anthocyanins from spinach leaves using the same method as we performed for chlorophyll extraction. You know that you need to dilute your extract in proper solvent before doing spectrophotometry. If you take 200ul of extract and added to 9.8 ml of solvent. You add 5 ml from this solvent to a cuvette and read absorbance. What would be the dilution factor you need to consider while calculating anthocyanin content.
SDS
detergent that denatures proteins
70% ethanol
what chemical is used DNA extraction process to remove excessive salts
an abundance of lack of cells therefore causing problems that may be minimal or life threatening
what happens if cell cycle regulators don't function properly
interphase, prophase, metaphase, anaphase, and telophase
what is the correct order in which cell cycle takes place
carbohydrate source
what is the purpose of having dextrose in the media while growign saccharomyces cerevisiae cultures
to check if cell is prepared for the cell division and to check if there are any mutations
what is the purpose of the G1/S checkpoint
*know which portion on antibody is responsible for binding with antigens in a diagam
which of the above portions of an antibody is responsible for binding with antigens
