Analysis flow cytometry.pdf

EXERCISES TOPIC 12 EXERCISE 1 You need to detect the expression of surface markers, CD73, CD90, CD105, and CD31 on adipose-derived stem cells (ASCs). Well known that CD73, CD90, and CD105 are highly expressed on ASC, while CD31 is negative. This table is the setting of flow cytometer we used in the...

EXERCISES TOPIC 12 EXERCISE 1 You need to detect the expression of surface markers, CD73, CD90, CD105, and CD31 on adipose-derived stem cells (ASCs). Well known that CD73, CD90, and CD105 are highly expressed on ASC, while CD31 is negative. This table is the setting of flow cytometer we used in the lab. 1A. SELECT THE APPROPRIATED ANTIBODY AND THE CORRESPONDING DETECTOR FOR EACH SURFACE ANTIGEN. EXPLAIN WHY? 2A. ANALYZE THE SPILLOVER SOURCE (HTTPS://APP.FLUOROFINDER.COM/) EXERCISE 2 Folder named” Viability”. There are 7 samples in total, and each sample is stained with FVS570 (fixable viability dye, https://www.bdbiosciences.com/en-dk/products/reagents/flow-cytometryreagents/research-reagents/single-color-antibodies-ruo/fixable-viability-stain-570.564995 ). So, what is the percentage of the dead cell for each sample, and analyze at least three of them? EXERCISE 3 Folder named” dilution”. The human blood samples are stained with different concentrations of CD25 conjugated with Mouse Anti-Human CD25(https://www.bdbiosciences.com/content/bdb/paths/generatetds-document.dk.562442.pdf), can you choose which one is the optimal Conc. to stain the lymphocytes. Tips: try to calculate the stain index 170 MFI: median fluorescence intensity - Set a forward scatter/side scatter gate on the cells of interest Create fluorescence histograms gated on forward/side scatter Create gates on negative and positive populations Create statistics view with median fluorescent intensity and SD (robust Standard Deviation) Calculate stain index according to the formula Create a plot similar to the example Example EXERCISE 4 Folder named” analysis”. ASCs are analyzed for the expression of CD201, CD36, and Stro-1. And this table is the antibody list and applied detector. So, try to analyze the expression of each surface marker and the proportion of CD201pos CD36 pos and Stro-1neg of alive stable singlets. Fluorochrome Marker Detector BV421 CD201 FL6 PB450 BV605 CD36 FL8 Violet610 Alexa Fluor 647 Stro-1 FL3 APC-H FVS570 FL10 PE Comments: if you can do this, you are able to basically analyze all the flow data. EXERCISE 5 A researcher would like to stain lymphocytes in human blood samples in order to perform a flow cytometric analysis. A mouse anti-human CD25 antibody is available, but the optimal concentration has to be determined. For that purpose, the researchers performed a titration of the antibody using eight different antibody concentrations (0.05, 0.1, 0.015, 0.2, 0.25, 0.5, 1, and 2 µg/ml), which yield the following data: 171 5A. EXPLAIN WHAT IS THE STAIN INDEX (SI) IN THE Y-AXIS AND OUTLINE HOW TO DETERMINE IT The stain index (SI) is a parameter used to quantify the signal-to-noise ratio in flow cytometry experiments. - The "signal" refers to the fluorescence intensity produced by the antibody binding to its target antigen on the cells of interest. - The "noise" represents background fluorescence or nonspecific binding that can interfere with accurate detection of the signal. The Stain Index (SI) is a method used to assess the relative intensity or brightness of a fluorochrome on a flow cytometry instrument. It is calculated based on the difference between the mean fluorescence intensity (MFI) of the positive and negative populations, divided by two times the standard deviation of the negative population. Higher SI values indicate better separation between signal and noise, meaning the signal is clearer and easier to distinguish from background noise. Researchers calculate the SI for different antibody concentrations used in their staining protocol. 5B. BASED ON THE DATA OF THE FIGURE, INDICATE WHAT WOULD BE THE OPTIMAL ANTIBODY CONCENTRATION TO STAIN THE LYMPHOCYTES Determining Optimal Antibody Concentration: - The optimal antibody concentration is typically associated with the highest SI value, indicating the best balance between signal intensity and background noise. - By identifying the antibody concentration that yields the highest SI, researchers can determine the most effective concentration for their staining protocol, ensuring accurate and reliable detection of their target cells. Based on these SI values, we can see that the highest SI value is 32, which gives an antibody concentration of 0.25 µg/ml. Therefore, the optimal antibody concentration to stain the lymphocytes would be 0.25 µg/ml. This concentration provides the best balance between signal intensity and background noise, ensuring accurate and reliable detection of the lymphocyte population in the flow cytometric analysis. 5C. WHEN STAINING CELLS WITH ANTIBODIES IN FLOW CYTOMETRY, FC-RECEPTORS CAN BE A PROBLEM. HOW CAN YOU FIND OUT WHETHER YOU HAVE A PROBLEM OR NOT? IF YOU DO HAVE A PROBLEM WITH FC RECEPTORS, WHAT CAN YOU DO? Isotype Control Staining: Include an isotype control antibody staining as a negative control alongside your primary antibody staining. Isotype control antibodies are antibodies that have the same immunglobulin class and subclass and fluorochrome as the primary antibody but lack specificity for the antigen of interest. 172 They are used to assess non-specific binding and background fluorescence. If the isotype control antibody binds to the antigen, it indicates non-specific binding. This means that the observed fluorescence signal in your experimental samples may not be solely due to the specific interaction between your primary antibody and the antigen of interest, but rather to non-specific interactions between the isotype control antibody and components on the cell surface. This nonspecific binding can lead to false-positive results. If I do have a problem with Fc receptors: To prevent unspecific binding of the antibody Fc-region to Fc-receptor you can block the FC-receptors with blocking agent a protein like bovine serum albumin or milk. EXERCISE 6 In an experiment with cultured keratinocytes you want to assess the expression of actin and envoplakin by means of immunostaining and fluorescence microscopy. For both markers, the primary antibodies are readily available (mouse monoclonal anti-actin IgG and goat monoclonal anti-envoplakin IgG). There is also a suitable Alexa-fluor 488 anti-mouse IgG secondary antibody in your stock. However, you have two labelled anti-goat IgG secondary antibodies to choose from: Alexa-fluor 555 and Alexa-fluor 633. After analyzing the chart below, which shows the emission spectra of the dyes, you decide that the optimal election for envoplakin staining is the Alexa-fluor 633 antibody. Assuming that the available microscope contains the light sources and filter combinations that are needed for proper visualization of the indicated dyes: 6A. EXPLAIN WHY AF 633 IS A BETTER OPTION THAN AF 555 FOR VISUALIZATION OF ENVOPLAKIN. In fluorescence microscopy, the emission spectrum of the fluorophore should ideally be well-separated from the emission spectra of other fluorophores used in the experiment. This prevents spectral overlap and allows for clear visualization of each specific target. Looking at the emission spectra chart, we can observe that the emission of AF 448 (emission peak at around 500 nm) overlaps into the emission spectra of AF 555 (emission peak at around 565 nm), while the emission of AF 448 does not overlap much into the emission spectra of AF 633 (emission peak at around 650 nm). 173 The optimal choice would be to select the secondary antibody (Alexa-fluor 633) with the emission spectrum that is furthest away from the emission spectrum of the secondary antibody used for actin visualization (Alexa Fluor 488). By selecting AF 633 for envoplakin staining, which emits light at a longer wavelength compared to AF 555, we minimize the risk of spectral overlap between the two channels used for actin and envoplakin visualization. This helps to improve the specificity and clarity of the fluorescence signal, ensuring accurate detection and visualization of envoplakin in the keratinocytes without interference from the actin signal. 6B. FOR COUNTERSTAINING OF THE CELL NUCLEI, YOU ARE CONSIDERING EITHER DAPI DYE (EXCITATION 358 / EMISSION 461 NM) OR PROPIDIUM IODIDE (EXCITATION 493 / EMISSION 636 NM). WHICH ONE WOULD YOU CHOOSE? JUSTIFY YOUR ANSWER. DAPI dye has an emission peak at approximately 461 nm, which falls within the blue spectrum. Propidium iodide (PI) has an emission peak at approximately 636 nm, falling within the red spectrum. Considering that the secondary antibody used for actin visualization is Alexa Fluor 488, with an emission peak around 500 nm, and for envoplakin visualization is Alexa Fluor 633, with an emission peak around 660 nm, it's essential to choose a counterstain with minimal spectral overlap with these fluorophores to avoid signal bleed-through and maintain imaging specificity. Given that DAPI emits in the blue spectrum, which is far from the emission peaks of both Alexa Fluor 488 and Alexa Fluor 633, it would be a suitable choice for counterstaining of the cell nuclei in this case. DAPI is less likely to cause spectral overlap with the emission signals of the other fluorophores, allowing for clear visualization of the nuclei without interference from the actin and envoplakin signals. SUMMARY Method Cellular or molecular Quantitative or qualitative 174