Exam: Lab Content (Notes from Assignments) Flashcards

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1
Q

What is the name of the cell line used in Lab 1? How do these cells grow in a culture flask? What lab space is required to work with these cells? How would you describe their ability to proliferate?

A
  • Ptk2 cells
  • adherent
  • CL1
  • very efficient
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2
Q

What was the experimental question in Lab 1? How did you stress/destress the cells? Why is the unstressed cell sample a negative control?

A
  • Q: How do Ptk2 cells respond to oxidative stress? How does the cell shape change?
  • To stress cells, I used 15 microliters (μL) of hydrogen peroxide (H2O2)
  • For unstressed cells, 30 μL of distilled water was added (H2O)
  • It is a baseline to compare with treatment cells
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3
Q

How does the viability assay distinguish between living and dead cells?

A
  • Living: appear clear or grey due to intact cell membranes
  • Dead: stained with blue dye, represented with compromised cell structures
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4
Q

Describe how well you were bale to see the cell shape and colour when you switched the microscope setting between brightfield (O) and phase contrast (Ph)?

A
  • Brightfield: it is hard to see cells/their shape
  • Phase Contrast: Ph1 able to see more cells and their structure due to high contrast between dye and actual cells
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5
Q

How do you calculate the concentration of cells? What formula do you use?

A

cells per mL = # cells / # squares x dilution factor (stock/dilutation x stock) x 10^4

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6
Q

Calculate the concentration of living cells in a tube. There are 28 live cells within 4 hemocytometer grid. You removed 20μL of cells and mixed it with 20μL Trypan Blue.

A

28 cells / 4 grids x (20μL + 20μL / 20μL) x 10^4

= 7 x 2 x 10^4
= 1.4 x 10^5 cells/mL

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7
Q

Determine how to dilute those cells before seeding them into new flasks:

The ideal cell concentration for seeding should be 1x10 cells/ml . Based on your calculations above, (you calculated 1.4 x 10^5 cells/mL), your live cell concentration is too high! Ack! Therefore, you must dilute your cells before putting them into new flasks
You are starting with 12 ml of cells in media in the collection tube.

Use the dilution formula (C1V2 = C2V2) to determine how much media you will need to add to your tube of cells to create a final concentration of 1x10^5 cells/ml?

A

C1 = 1.4 x 10^5 cells/mL
V1 = 12 mL
C2 = 1x10^5 cells/mL
V2 = ??

V2 = (C1 x V1) / C2
V2 = (1.4 x 10^5 cells/mL x 12 mL) / 1x10^5 cells/mL
V2 = 16.8 mL

16.8 mL - 12 mL = 4.8 mL

You will need to add 4.8 mL of media.

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8
Q

You prepare your samples and begin to load your gel. Oh no! You notice that the protein sample is not sinking into the well! Instead, it floats in the running buffer. What could have gone wrong with your sample buffer? Explain.

A

Not enough glycerol was added to the sample buffer since glycerol adds density to proteins allowing them to sink instead of float.

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8
Q

Hemoglobin is a tetramer composed of four polypeptides; two a-chains (each ~16 kDa) and two B-chains (each ~16 kDa). The four subunits are stabilized entirely by non-covalent bonds (i.e. no disulfide bonds).

What is the molecular weight of tetrameric hemoglobin?

A

4 polypeptides x 16 kDA length = 64 kDA

The molecular weight of tetrameric hemoglobin is 64 kDA.

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9
Q

Imagine that you would like to use gel electrophoresis to observe the size of hemoglobin in its quaternary structure. Suggest three procedural things that you would need to do (or not do) to see the size of the full tetramer.

A
  1. Avoid boiling the sample
    - Boiling the sample could denature the protein and disrupt the non-covalent bonds holding the tetramer together.
  2. Use a non-denaturing gel like Native-PAGE instead of SDS-PAGE.
    - A non-denaturing gel, like Native-PAGE, will preserve the quaternary structure of the protein during electrophoresis.
  3. Use a suitable molecular weight marker
    - The marker should have a range that includes the molecular weight of the tetrameric hemoglobin (~64 kDa) for accurate comparison.
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10
Q

If the inhibitory peptide on pepsinogen was only ~100 Da (i.e. 0.1 kDa) in size, and you used only your naked eye, would you see a difference between the inactive and active form on a gel?Why or why not?

A
  • You would NOT be able to see the difference.
  • The difference between the inactive and active form is only 0.1 kDA so we could guess its location but it’s so small that it likely wouldn’t present itself.
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11
Q

What is a Ponceau stain? What does it do? Why is it important?

A
  • stains the proteins on the nitrocellulose membrane temporarily
  • allows visualization of both control and treatment proteins
  • allows us to confirm that cell lysate proteins successfully transferred to nitrocellulose membrane
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12
Q

What is a milk incubation? What does it do? Why is it important?

A
  • coats the nitrocellulose membrane with casein protein
  • helps reduce the non-specific binding of antibodies to the membrane
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13
Q

What are the TBS-Tween Washes? What does it do? Why is it important?

A
  • washes the nitrocellulose membrane removing (non-specific) excess antibodies
  • this is important to do so that when luminescence rxns are present the sp. spots are dark not the entire membrane
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14
Q

What is Luminol/Oxidizing Reagent Incubation? What does it do? Why is it important?

A
  • it is the substrate for HRP to bind to
  • cataylzes a rxn that releases light that we can capture on film, intensity allows us to know about the protein expression while location of the light tells us about the nuclear weight of the protein.
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15
Q

Imagine that you completed the procedure, for lab 4 - western blotting, but after examining the developed film, you find that there are NO BANDS on the blot for tubulin. Your Ponceau stain indicated that there were cell lysate proteins on the nitrocellulose. Suggest two procedural (mis)steps that could generate this result.

A
  1. Using expired reagents
  2. Protein did not transfer correctly from the gel to the nitrocellulose membrane (incorrect transfer or insufficient protein)
15
Q

For lab 4, do you expect there to be a difference in the band intensity for tubulin when comparing your treatment and negative control samples?

When you look at your blot, do the tubulin bands look the same or different?

Based on the above questions what assumption can you make about your experimental results?

A
  • There should NOT be a difference in band intensity when comparing T and (-) C samples.
  • This a procedural check so we’d expect the bands to be the same.
  • The assumption can be made that ___________

—> down regulation of proteins?

16
Q

What was your protein of interest for lab 4? What did you expect to see in your blot? Why?

A
  • Protein of interest was p53
  • You’d expect to see thick an noticeable bands
  • Why? due to up regulation of proteins
17
Q

Based on your blot for lab 4, what is the ~MW of your protein of interest? was the expression of your protein of interest up-regulated, down-regulated, or unaffected by the stressor?

Consider the purpose of your experiment. Write one sentence to summarize your results.

A
  • ~MW of your POI: 68 kRa
  • Up-regulated due to stressor
  • purpose: _____________________
18
Q

Because you LOVE cell biology, you do some extra reading, and you find out that both Hsp27 and p53 are activated by phosphorylation in response to stress!

a) Assuming that the antibody we used today binds to both the phosphorylated and the non-phosphorylated form of the protein, do you think you could tell if your protein of interest was phosphorylated? (Hint: a phosphate has a molecular weight of ~0.1 kDa!). Why or why not?

b) Suggest a hypothetical primary antibody that could be used in your experiment to determine if your protein of interest was phosphorylated in your samples. (Provide the full name using accepted nomenclature.)

A

a) No, since the protein or sample is so small, you need something to show it’s location (ex: use western blot to illuminate where bands of a protien are)

b) Mouse anti-phospho-p53, conjugated to HRP

19
Q

A completed western blot is the product of many steps.

a) Name one thing that could have gone wrong in Lab 2 (Bradford!) that would cause a poor result today.

b) Name one thing that could have gone wrong in Lab 3 (PAGE!) that would cause a poor result today.

c) Name one thing that could have gone wrong during today’s procedure that would cause a poor result.

A

a) Improper calculations of dilutions causing error in concentrations of protiens

b) Improperly inputting sample protein concentrations into wells

c) Not doing a milk bath, this would cause an error in the light rxn, highlighting all proteins not the specific ones you are aiming to indentify.

20
Q

Compare Western Blotting and Immunofluorescence on the following:

i. Cell intact or lysed?
ii. What reporter molecule was used?
iii. Mircoscope needed to visualize the reporter molecule?
iv. Can you see sub-cellular localization of sp. cell structure?
v. Requires washes for removing unbound antibodies?

A

Western Blotting
- Lysed
- HRP
- No
- No
- Yes

Immunofluorescence
- Intact
- FITC
- Yes
- Yes
- Yes

21
Q

Regarding these q’s think back to lab 5.

i. What does the methanol do?
ii. Full name of antibody?
iii. Indirect or direct visualization?
iv. Protien probed for?
v. Structure?
vi. Colour of structure?
vii. Colour of nucleus?
viii. Antibody attached to DNA in cells?
ix. Cells alive when viewed by microscope?

A
  • Fixes & permeabilizes
  • Mouse anti-tubulin labelled with FITC
  • Direct
  • Tubulin
  • Microtubules
  • Mircotubules = Green
  • Nucleus = Blue
  • No DNA attached
  • No, cells not alive
22
Q

In lab 5, imagine that you saw NO glowing in the microscope with the correct filter. Sadness! If you wanted to make it more likely for the procedure to be a success, would you leave the cells in antibody for longer amount of time or a shorter amount of time?

A

You would want to leave the cells in the antibody for a LONGER amount of time.

23
Q

Situation A: Direct visualization for immunofluorescence.

A primary antibody against actin was raised in a giraffe. It glows red when viewed under the appropriate filter of a fluorescence microscope.

What is the name of this antibody?

A

Giraffe anti-actin labelled with Cy3.5

24
Q

Situation B: Indirect visualization for immunofluorescence.

A primary antibody against actin was raised in an elephant.

What is the name of this primary antibody?

A

Elephant anti-actin

25
Q

Situation B: Indirect visualization for immunofluorescence.

A secondary antibody (that recognizes the primary above in Situation B, Part I) was raised in a sloth. It glows green when viewed under the appropriate filter of a fluorescence microscope,

What is the name of this secondary antibody?

A

Sloth anti-elephant labelled FITC

26
Q

Situation C: Indirect visualization for western blotting.

Now you want to do a western blot for actin! You are provided with the same actin primary antibody as in Situation B, Part I.

Suggest the name of a hypothetical Secondary antibody that could work with that antibody for western blotting.

A

Sloth anti-elephant labelled with HRP.