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Phage Practice Problems for M315
Phage titer determination (Ex. 23):
1.
For each of the dilutions, 1 ml phage solution was mixed with E. coli
cells and plated with soft agar:
Dilutions
10
2
10
3
10
4
10
5
10
6
Plaques on plate
Complete lysis
Complete lysis
Complete lysis
>500
45
2.
Original phage stock titer: _____________ (what is the unit used?)
Burst size determination (Ex. 24):
1.
What does the term “one-step” really mean?
A) one generation of phage infection cycle; B) one growth phase in the phage cycle; C) one dilution step
needed
2.
Why are the following steps critical in the burst size determination?
A) MOI <<1
B) The initial 5 min incubation with high cell-phage density
C) The first 100-fold dilution to the DIL tube
D) The use of chloroform
3.
The One-Step growth curve of the T4 phage
In the ADS tube, 0.1 ml above phage stock was mixed with 2 ml E. coli
cells (3.0 x 10
8
cfu/ml) to a final
volume of 2.1 ml. At 5 min, 0.1 ml cell-phage mixture was transferred from ADS to the DIL tube. At various time points, TSB-1 and TSB-2 tubes were prepared. 0.1 ml sample was taken from the TSB-2 tube and assayed for plaques.
A) What is the number of E. coli
cells in the ADS tube?
B) What is the number of phage number in the ADS tube?
C) What is the M.O.I value of this infection? Why is it important to keep MOI<<1?
D) What is phage titer (pfu/ml) in the ADS tube
at Time 0? What is the concentration of the infected E. coli
cells
?
E) Assuming the total dilution factor from ADS to the final plaque assay plate is 10
7
, what is the theoretical pfu/ml value for each of the following time points?
Time (min)
Avg # Plaques on plate
Theoretical phage titers as it would be in the ADS tube (pfu/ml)
15
1.1
25
1.5
30
9
35
81
40
155
45
243
50
239
What is the peak titer?_______________________
What is the phage burst size?__________________
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Related Questions
DILUTION
COLONY COUNT
CFU/mL
1:106
155000
1:107
15500
1:108
1550
1:109
155
Given these values how would I fill in the rest of this serial dilution table? Also, what would be a 1:1 CFU/mL value based on this table?
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Time point (min)
Absorbance of culture at 660nm
Approximate cell concentration
Approximate # cells in 1mL extract
0
0.298
1.49 x 108 cells/mL
1.49 x 108 cells
10
0.316
1.58 x 108 cells/mL
1.58 x 108 cells
20
0.374
1.87 x 108 cells/mL
1.87 x 108 cells
30
0.429
2.145 x 108 cells/mL
2.145 x 108 cells
40
0.512
2.56 x 108 cells/mL
2.56 x 108 cells
50
0.544
2.72 x 108 cells/mL
2.72 x 108 cells
60
0.607
3.035 x 108 cells/mL
3.035 x 108 cells
a. Using these data, prepare a growth curve of this strain ofEscherichia coli (E. coli).b. Estimate the doubling time for this strain of E. Coli. Clearly showhow you estimated this value from the empirical data presented.
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A. A
Aa A
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AaBbCcDdF
AaBbCcDdF
AaBbCcDd
A
Normal
No Spacing
Heading 1
Heading 2
=y updates, fixes, and improvements, choose Check for Updates.
1. You have found a putative virus which is able to infect a bacterium causing increased
mortality and resistance to all antibacterial agents. You have systematically purified the
sample but have an unknown concentration of viruses. You perform the following serial
dilutions.
0.5ml
1ml
5ml
0.1ml
2.5ml
1ml
15ml
1
3
6
7
Phage Buffer
100.5ml
9ml
14.5ml
99.9ml 25ml
5ml 985ml
Tube Dilution in each Test Tube (Show Tube Dilution in each Test Tube (Show your
your work)
#
#
work)
4
1
6.
O Focus
glish (United States)
W
30
DII
DD
000
000
F4
80
F7
F8
F9
F10
F3
F5
F6
$4
&
4
5
6
8.
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Confluence cells added onto 100-wells
plate, 2 ml trypsin added onto cells and 6 ml
media added to dilute trypsin, 1 ml was used
for counting, and average number of Cells
counted was 24(DF 2)
find out the the number of cells/ml and total
number of cells in 7 ml?
How many cells you need for 100 wells,
assuming Number of cells/well is 10 to the
power 4 cells?
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Average plaques for bacteriophage A,B,C are 137,36,25. PFU/ml is average plaques multiply by the volume dilution multiply by the dilution factor. Show your working for each bacteriophage.
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Biomedical Engincering Dept.
Biocompatibility Hello sir, I know it takes precious
Lab. Sheet-Level time. Please, I have a report on this
Semester 1
subject. Only 5 papers within a
Experiment Title: Cytotoxicity Test discussion solution. I want to solve
the text of the keyboard. Thank you,
Experiment no. 2
sir
1 12:34 PMA
Objectives: Study the eytotoxicity and how it is measured.
1- Background
Cytotoxicity can lead healthy living cells to three potential
cellular fates.
1. Necrosis (accidental cell death): Rapid loss of membrane integrity and cell
lysis
2. Apoptosis (programmed cell death): slower, more orderly, and genetically
controlled
3. Cytostasis (a decrease in cell viability): cells remain alive but fail to
actively grow and divide
Importance of measuring cytotoxicity two major reasons
1. Either you want specific cells to die and look for an adequate
compound/condition (cancer and immunotherapy)
2. Or you want to exclude cytoloxicity in specific cells (chemicals and drugs)…
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Helping tags: Biology, bacteria, lag phase
WILL UPVOTE, just pls help me answer the following questions and explain them clearly. Thank
you.
TOPIC: Bacterial Growth Curve
1. How may the following growth and culture conditions affect the length of the lag
phase? EXPLAIN YOUR ANSWERS.
a) inoculum is from an old culture
b) shifting cells from rich culture medium to a poorer one
c) exponentially growing culture is transferred into the same medium under the
same growth conditions
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and the minimum exposure time
1. Determine the minimum inhibitory concentration
given the following data:
Concentration of
ethanol
Number of colonies of E. coli on Nutrient Agar plates
Exposure time (in minutes)
10
5
15
Distilled water
+++
+++
+++
40%
+++
++
+
70%
95%
Minimum inhibitory concentration:
Minimum exposure time:
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Case Study: You are asked to inoculate lactose media (broth) with E. coli that was growing exponentially in glucose media (broth). You
monitor the growth of these cells in the lactose broth using a viable plate count technique and plot the Growth Curve for E. coli growing c
lactose. After a day of monitoring the cells, you note that the culture entered into lag phase growth initially and is now currently growing
exponentially. In addition, you find that the cultures doubling time is 30 minutes. You know that the doubling time of E. coli in glucose
media is 20 minutes. You did not change the oxygen content or temperature at Which you were growing the cells, E. coli is still aerobically
respiring at 37°C.
Question: If the lac operon is 'on' what does this imply about the cell?
O The cell is dying and the lac operon is going to induce apoptosis. Apoptosis is truly cell death and the cells will enter into the death phase of the Growth
Curve.
O Lactose is being catabolized and glucose is no…
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Vivid outline and Standard operating procedure of Viral hemagglutination inhibition test.
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Apple puree was analyzed for petulin by HPLC-MS-MS after SPE clean-up. The procedure was 10.0g of puree + 10μl of a 10μg/ml solution of isotopically labeled petulin as internal standard were treated with 10.0ml of pectinase and acetic acid, centrifuged, and filtered. Four ml of the filtrate was passed through a SPE cartridge. The petulin was eluted with 2.0ml of ethyl acetate. The sample was evaporated to dryness and the residue dissolved in 1.0ml of the mobile phase. The analyte signal was 127 and the internal standard 197. Calculate the concentration of petulin in the sample in μg/g (RRF=1).
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Case Study: You are asked to inoculate lactose media (broth) with E. coli that was growing exponentially in glucose media (broth). You
monitor the growth of these cells in the lactose broth using a viable plate count technique and plot the Growth Curve for E. coli growing on
lactose. After a day of monitoring the cells, you note that the culture entered into lag phase growth initially and is now currently growing
exponentially. In addition, you find that the cultures doubling time is 30 minutes. You know that the doubling time of E. coli in glucose
media is 20 minutes. You did not change the oxygen content or temperature at which you were growing the cells, E. coli is still aerobically
respiring at 37°C.
Question: If you want to grow E. coli as quick as possible would you grow it on lactose or glucose?
O Mixing the two carbon sources would provide the most optimal growth conditions.
O Lactose because of the generation time being faster.
O Lactose and an anaerobic environment.
O Glucose…
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a. Explain whether or not any of the methods in fi gure 2.9 could beused to determine the total number of cells present in a patient’s specimen.b. After performing the streak plate method on a bacterial specimen, theculture was incubated for 48 hours at 37°C. Upon viewing the plate, therewas heavy growth (with no isolated colonies) in the fi rst quadrant, but nogrowth was apparent in the remaining quadrants. Please discuss errors in the procedure that could have produced this result.
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Case Study: You are asked to inoculate lactose media (broth) with E. coli that was growing exponentially in glucose media (broth). You
monitor the growth of these cells in the lactose broth using a viable plate count technique and plot the Growth Curve for E. coli growing on
lactose. After a day of monitoring the cells, you note that the culture entered into lag phase growth initially and is now currently growing
exponentially. In addition, you find that the cultures doubling time is 30 minutes. You know that the doubling time of E. coli in glucose
media is 20 minutes. You did not change the oxygen content or temperature at which you were growing the cells, E. coli is still aerobically
respiring at 37°C.
Question: In order to determine the doubling time of E. coli a viable plate countris performed. This means that you are truly counting
O Dead cells only via a microscopic count
O Live cells only via membrane filtration or serial dilution
O All cells via membrane filtration or serial…
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6
2
4
3
5
LABORATORY REPORT FORM
EXERCISE 25
ASSAY OF ANTIMICROBIAL AGENTS: DISK-DIFFUSION METHODS
What is the purpose of this exercise?
Evaluate
1.
For each of the disinfectants or antiseptics tested, determine the active ingredient. What class of inhibitory
agent does it belong to and what is its mode of action?
CHEMICAL AGENT
Lysol
Sanitizer
Foaming
Handwash
Detergent.
Desinfecting
Hand
Sanitizer
ACTIVE INGREDIENT(S)
Alhyl
Dinethylaenzy/a.
chloroxylenol 0.3%
Alkyl, Dimethyl benzyl
ammonium chlorides
ethy benzyl A. C.
Ethyl Alcohol 70%
1/Cleansing
s Pray
Hydrogen
Peroxide
Benzalkonium
Lidocaine HC)
An Septic
CLASS OF
AGENT
MODE OF ACTION
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ISOLATE #2
Results for "Isolate #2":
***Please fill out the table****
1. Blood agar:
2. Nitrate Test:
Test
Results
Interpretation
Blood agar (type of hemolysis)
Nitrate test
3. Gelatinase Activity:
Gelatinase activity
Resistant or Susceptible?
Novobiocin Resistance
Zone Diameter:
4. Novobiocin Resistance:
5. Coagulase Test:
Coagulase (Sure-vue test)
2. Based on your results, what is the possible identification (Genus and species) of "Isolate #2"?
nhes
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Think about minimal inhibitory concentration (MIC) assays. A set of tubes, all containing the same amount of bacterial cells, have decreasing amount of antibiotic X added to them. Tube 1 had 0mg/ml, Tube 2 has 100mg/ml, tube 3 has 50mg/ml, 4=25, 5=12.5, 6=6.25, and tube 7 has 3.125. If bacteria grew in tubes 1, 4, 5, 6, 7, what is the MIC of this antibiotic for this bacteria?
Select one:
a.None of the Above
b.3.125 mg/ml
c.25 mg/ml
d.0 mg/ml
e.50 mg/ml
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Please answer fast
Dilution Problem. A culture of Staphylococcus is diluted as follows:
(1) 20mL are added to 80mL of water.
(2) 10uL from (1) are then added to 9.99mL of water.
(3) A 10-2 dilution is made from tube # (2).
(4) 100uL from (3) are plated for a pour plate and incubated.
Growth Problem.A culture with approximately 2x103 cells/mL were incubated. After 3 hours, the number of cells had increased to 3x105.
a) How long was the generation time in minutes?b) How many generations have occurred
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for E-coli-
Describe results for Gram reaction, cell shape and arrangement (look up information)
What color changes were observed for each test? Any zones of inhibition (for antibiotics only)?
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Match the methods discussed to the description of what it would be used for.
HPLC-UV
Visual range standard curve
Affinity chomatography
UV analysis of protein fractions
Gel filtration
Ion Exchange
SDSPAGE…
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Q.6. You have a bacterial culture. From this culture suspension, six serial dilutions are made, and 1 ml is plated onto Plate Agar from the last dilution. After incubation, X colonies are counted on the plate. Calculate CFU/mL of the original Sample? (X=9)
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Plaque assay (Bacteriophage Titer) Answer questions 2 & 3
2. Why don't bacteriophages continue spreading over the entire plate until all of the E. coli cells on the plate are lysed?
3. Why is it important that bacterial cells should be in the exponential growth phase for this experiment?
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Diagram illustrating Miles & Misra technique for determining viable counts:
10ul
10-5
106
104
10-7
3. Set up controls of the donor and recipient cultures as follows:
• Spread plate 0.1 ml of the donor culture over the surface of each of
the 3 different selective media (i.e. nutrient agar + Ap; nutrient agar
+ Sm; nutrient agar + Rif)
• Similarly spread plate 0.1 ml of the recipient culture over the
surface of each of the 3 selective media
• What is the purpose of this step?
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A phagehunter performs a full plate titer using standard techniques (100 ul of each dilution added to 250 ul of Gordonia terrae cells) of a lysate obtained from an optimum webbed plate experiment. The phagehunter counts 72 plaques on the 10-7 dilution plate. Which of the following scenarios is the best choice for the phagehunter to do next?
a.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should adopt a phage from direct isolation.
b.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should go back to the pick-a-plaque experiment.
c.) The phagehunter has achieved a high enough titer lysate to move forward with characterization experiments.
d.) The phagehunter has not achieved a high enough titer lysate to move forward with characterization experiments, so they should try to make more optimum webbed plates.
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d-6f06-490d-b3d3-cb18767c100
Post-Lab: Streak Plate Isolation
Homework. Due in 12 hours
2/5 answered
Unanswered
Post-Lab: PIV 2.5
Homework Unanswered
How do colonies of E. coli differ from those of P. mirabilis?
inancwarnd
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IN YOUR OWN WORDS. Differentiate the following. (Not more than 10 sentences)
1. Aseptic Teachnique vs. Sterile Technique
2. Microbistatic Agent vs. Microbicidal Agent
3. Sterilization vs. Disinfection
4.How is radiation, autoclave and ultrasonic waves used in the hospitals or clinics as a means to inhibit the growthbof microbes?
5. How is filtration applied nowadays in relation to Covid-19?
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Z40 7 @
۱۲:۶۲
لا يوجد SIM و
السؤال
Roller compactor and chilsonators are
used in preparation of granules of aspirin.
false
true
حفظ الإجابة
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Give five (5) other examples of samples that are best prepared using Smear Preparation Technique (example: Human RBCs). Explain why.
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INTERPRETATION OF RESULIS
NEGATIVE: Two lines appear. One colored line should be in the control region (C), and another
apparent colored or faded color line adjacent should be in the test region (T). This negative result indicates
that the drug concentration is below the detectable level.
POSITIVE: One colored line appears in the control region (C). No line appears in the test region (T).
This positive result indicates that the drug concentration is above the detectable level.
INVALID: Control line fails to appear. Insufficient specimen volume or incorrect procedural techniques
are the most likely reasons for control line failure. Review the procedure and repeat the test using a new
test panel. If the problem persists, discontinue using the lot immediately and contact your local distrībutor.
READING OF RESULTS
INTERPRETATION
SONTROL
TEST
SAMPLE
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An adult male patient (52 yr, 75kg) whose serum creatinine is 2.4mg% is to be given gentamicin sulfate for a confirmed gram-negative infection. The usual dose of gentamicin in adult patients with normal renal function is 1 mg/kg every 8 hrs by multiple IV bolus injections. Gentamicin sulfate (Garamycin®) is available in 2-mL vials containing 40 mg gentamicin sulfate per mL. Calculate:
a. The creatinine clearance in this patient by the Cockcroft and Gault method and
b. The appropriate dosage regimen of gentamicin sulfate for this patient in mg and mL.
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Please answer fast
1. a. Give an example of lysogenic conversion.
b. An example of lytic phage and temperature phage in E. coli.
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Case Study #6: A 20 year old male college student develops a fever after gall bladder
surgery and has a pus exudate around the incision. You receive a cotton swab of the
exudate which you streak on a blood agar plate. Bacterial colonies on the plate show B
hemolysis, and microscopic examination reveals Gram + cocci occurring singly, in pairs,
and in short chains. Further testing reveals that the isolate is catalase negative and
phenol red fermentation broth with mannitol turns yellow.
1. What is the name of the most likely bacterium causing this infection?
2. What is the most likely place that this bacterium originated to cause this
infection?
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LABORATORY REPORT: Aseptic Technique & Inoculation of Bacteria
Guide Questions.
1. Why is direct flaming preferred when disinfecting loops and needles?
2. Why is it important to flame the entirety of the loop and not just the tip? What consequences can be seen when this process is not done correctly?
3. What is the difference between quadrant streak method A from method B?
4. Why do we use a pencil instead of a pen when labelling culture tubes or plates? Are there other alternatives in labelling?
5. Why is an inoculating needle used in Subculturing Techniques? Can a loop be used instead?
NOTE: Please try to answer all of the question asked, i promised to give you a good ratings
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Lung Cancer Patient
Whats the use of the given medications below:
● Cisplatin100 mg/m2 IV q4 weeks● Trametinib Dimethyl Sulfoxide 2mg/tab 1 tablet OD● Roflumilast 500 mcg PO● Spiriva Respimat: 5 mcg (2 actuations; 2.5 mcg/actuation)● Budesonide 1 neb q6hr● Hydrochlorothiazide 25 mg tab PO OD● Levothyroxine 175 mcg PO OD● Hydralazine 50 mg tab PO TID● Metformin 500 mg tab PO BID
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Escherichia
coli
Mycobacterium
phlei
Bacillus
Micrococcus
subtilis
luteus
A
B
Figure 1: Agar Slant Cultures of Bacteria
(Gary E. Kaiser, Ph.D.- Professor of Microbiology)
. Observe and describe the colour of the slant cultures (A-D) in fig 1. (
. Define the following terms: pure culture, sterile medium, inoculum, aseptic
technique, and colony.
What is the name of the cultures that you used
. Where they gram negative or positive
Define the following terms: psychrophile, mesophile, thermophile, and
hyperthermophile.
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a. What is the total dilution of Tube #4? Express the answer in exponential format.
b. You plated 1 mL of the Tube #4, After incubating, you counted 500 colonies on the plate. What is the concentration of Tube #0? include units.
c. How could you change the experiment in part B to get a plate in the countable range? Be specific about any dilution factors and/or plated volumes you would change.
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Related Questions
- DILUTION COLONY COUNT CFU/mL 1:106 155000 1:107 15500 1:108 1550 1:109 155 Given these values how would I fill in the rest of this serial dilution table? Also, what would be a 1:1 CFU/mL value based on this table?arrow_forwardTime point (min) Absorbance of culture at 660nm Approximate cell concentration Approximate # cells in 1mL extract 0 0.298 1.49 x 108 cells/mL 1.49 x 108 cells 10 0.316 1.58 x 108 cells/mL 1.58 x 108 cells 20 0.374 1.87 x 108 cells/mL 1.87 x 108 cells 30 0.429 2.145 x 108 cells/mL 2.145 x 108 cells 40 0.512 2.56 x 108 cells/mL 2.56 x 108 cells 50 0.544 2.72 x 108 cells/mL 2.72 x 108 cells 60 0.607 3.035 x 108 cells/mL 3.035 x 108 cells a. Using these data, prepare a growth curve of this strain ofEscherichia coli (E. coli).b. Estimate the doubling time for this strain of E. Coli. Clearly showhow you estimated this value from the empirical data presented.arrow_forwardA. A Aa A 三,三,行,E AaBbCcDdE Aa AaBbCcDdF AaBbCcDdF AaBbCcDd A Normal No Spacing Heading 1 Heading 2 =y updates, fixes, and improvements, choose Check for Updates. 1. You have found a putative virus which is able to infect a bacterium causing increased mortality and resistance to all antibacterial agents. You have systematically purified the sample but have an unknown concentration of viruses. You perform the following serial dilutions. 0.5ml 1ml 5ml 0.1ml 2.5ml 1ml 15ml 1 3 6 7 Phage Buffer 100.5ml 9ml 14.5ml 99.9ml 25ml 5ml 985ml Tube Dilution in each Test Tube (Show Tube Dilution in each Test Tube (Show your your work) # # work) 4 1 6. O Focus glish (United States) W 30 DII DD 000 000 F4 80 F7 F8 F9 F10 F3 F5 F6 $4 & 4 5 6 8.arrow_forward
- Confluence cells added onto 100-wells plate, 2 ml trypsin added onto cells and 6 ml media added to dilute trypsin, 1 ml was used for counting, and average number of Cells counted was 24(DF 2) find out the the number of cells/ml and total number of cells in 7 ml? How many cells you need for 100 wells, assuming Number of cells/well is 10 to the power 4 cells?arrow_forwardAverage plaques for bacteriophage A,B,C are 137,36,25. PFU/ml is average plaques multiply by the volume dilution multiply by the dilution factor. Show your working for each bacteriophage.arrow_forwardBiomedical Engincering Dept. Biocompatibility Hello sir, I know it takes precious Lab. Sheet-Level time. Please, I have a report on this Semester 1 subject. Only 5 papers within a Experiment Title: Cytotoxicity Test discussion solution. I want to solve the text of the keyboard. Thank you, Experiment no. 2 sir 1 12:34 PMA Objectives: Study the eytotoxicity and how it is measured. 1- Background Cytotoxicity can lead healthy living cells to three potential cellular fates. 1. Necrosis (accidental cell death): Rapid loss of membrane integrity and cell lysis 2. Apoptosis (programmed cell death): slower, more orderly, and genetically controlled 3. Cytostasis (a decrease in cell viability): cells remain alive but fail to actively grow and divide Importance of measuring cytotoxicity two major reasons 1. Either you want specific cells to die and look for an adequate compound/condition (cancer and immunotherapy) 2. Or you want to exclude cytoloxicity in specific cells (chemicals and drugs)…arrow_forward
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