Little things in the lab

 

Stairs, alleys and labyrinths

Rajasthan, 2021

 

Perspectives

 

Flow cytometry I: Compensation

Put simply, as long as we don't find fluorochromes with absolute fidelity (i.e. very narrow emission peaks), we simply have to compensate raw readings from a flow cytometer. Let's start wth just two fluorochromes in two channels, 

1. Phycoerythrin (PE) in FL1 detector (red)

2. Fluorescein isothiocynate (FITC) in FL2 detector (yellow)

From a purely theoretical physics perspective, FITC and PE show a consistent and reproducible  overlap in their emission spectrum (from Spectra analyser; https://www.biolegend.com/en-us/spectra-analyzer)

Looking at the two emission spectrum (X axis is wavelength in nm), one can calculate how much FITC emission overlaps with PE, and vice versa; it is evident that FITC has a significant component where PE peak (red) is located, but PE does not have much of a tail where FITC (yellow) peaks

Usually flow cytometers have detector which detect light within a band of wavelengths, and they are calibrated to detect certain peaks. For example, a flow cytometer might have a red detector (FL1) and a yellow detector(FL2).

By simple eyeballing, one can guess that 15% of the red peak is contributed by the yellow tail, which is very close to the correct figure

Looking at the graph, we can guess that the emission peak of PE is augmented by 15% because of 'spillover' from FITC. This percentage is quite invariant in the abstract, ideal world of theoretical physics. In such an ideal world, for each cell which is interrogated by the flow cytometer, we can deduce the 'true' PE emission by the equation:

PE_real = PE_observed - (0.15 x FITC_observed)

The graph also shows a 1% contribution of PE in FITC peak (the remnant of the red tail in the yellow band). Thus,

FITC_real = FITC_observed - 0.01 x PE_observed

The two equations are simultaneous, i.e. they have two unknown variables PE_real and FITC_real, and thus can be solved (the flow cytometer has already provided you with PE_observed and FITC_observed). In general, for n fluorochromes to be detected by n bands, you will have n such equations; with a little linear algebra, you can manually solve them all (or ask your favorite Flow Cytometry analyser software to do it for you!). All you need is a ready made compensation matrix, which is an assortment of the correction factors in the form of an n x n table.

	FITC	PE
FITC	1	0.15
PE	0.01	1

Is that it?

Unfortunately, no. Flow cytometers are real instruments who measure real cells with real lasers and detectors, which are prone to voltage fluctuations, dead cell artifacts, and errors which simply can not be traced to a particular cause. Plus, each flow cytometer can have a different detector and different band pass. Thus, using the blanket values 15% & 1% is not advised. It is now routine to determine these values every time.

Now how do you do that? Spectrometry is not everyone's cup of tea. Fortunately, we can make an assumption that will make the task simpler: fluorescence intensity in a channel is the sum of contributions from all fluorochromes, in the same ratio in which their emission spectra overlap. That is to say, if I get an intensity x in the FL1 channel and y in FL2 channel, I can be sure that some fixed fraction of x is always contributing to y.

Determination of this correction factor, now, becomes a matter of designing an experiment. We prepare 

1. a set of cells (C1) stained only with FITC combined with a diffusely positive marker (CD3)
2. another set of cells (C2) stained only with PE tagged to the same CD3

Now let's run C1 in our flow cytometer, and plot FITC vs PE. Ensure that

1. No part of the negative population does not go off the left of the chart, i.e. it must fully stay within the chart (adjust the voltage if necessary)
2. The positive population is at least as positive as you would expect cells to be in your eventual experiment
3. You don't use different cell lines (i.e. don't put lymphocytes in C1 and monocytes in C2; in such cases, better to use custom made beads which bind all IgG kappa antibodies for this purpose)

In this uncompensated graph, find the median fluorescence intensity (MFI) of both FITC and PE channel (although you haven't actually put PE in this tube, just FITC; whatever PE is showing is due to spill from FITC)

For example, let's say 

• MFI_FITC_Pos = 80
• MFI_FITC_Neg = 10
• MFI_PE_Pos = 30
• MFI_PE_Neg = 5

You can now make a few simple calculations

• Let x = (MFI_PE_Pos - MFI_PE_Neg) = (30-5) = 25, which represents the effective MFI in PE channel
• Let y = (MFI_FITC_Pos - MFI_FITC_Neg) = (80-10) = 70, which is the effective MFI in FITC channel

Now the factor x/y = 25/70 = 35% is the correction factor for spill of FITC into PE. You can now similarly calculate correction factor for spill of PE into FITC from the C2 tube.

Note that after compensation, the width of the FITC positive population will increase slightly; this is because the FITC information is now coming from two sources, both FITC and PE, thus increasing its error (degree of freedom); however, the median PE negative line (blue) must remain the same through both populations

With n fluorochromes, one can try n tubes with cells/ beads, each having a single fluorochrome, and acquire and plot them, and do the maths by hand, to come up with a compensation matrix. Or, you know, just let your favorite software do it magically.

Engraftment monitoring: part I

Following transplant of hematopoeitic stem cells, one must carefully follow up the patient for months. This is because the aim of a bone marrow transplant is to reconstitute the immune system, or - to be more precise - to borrow some immunity from a donor. Which means, eventually, that at some point of time, most of immune system (i.e. the white blood cells) of the recipient has to be replaced entirely by the donor's. In essence, the recipient will become a genetic chimera, carrying a distinct cell population with the donor's genome.

Now, how do you know when that happens? White blood cells look all alike under the microscope, and there is no telling whether it belongs to the patient or donor just by looking at them. Which brings us to they key question of identity, i.e. how am I any different from you?

If the donor and recipient are of opposite gender, a simple karyotype from a sample of blood cells would easily identify which one is from whom. The Y chromosome will stand out. However:

1. If you have ever done cytogenetics, you know the mind-numbing labor that goes into karyotyoing

2. If multiple donors of different genders have donated stem cells to a single recipient (as is often the case), the method falls on it face.

So let's move on to genetic differences between two individuals. As members of the same species, we are more or less similar, except for a few things. Short tandem repeats (STR) are repetitions of a short nucleotide core (usually less than 9 nucleotides), which vary in repeat numbers between people. For example the donor might have:

GTCGTCGTCGTC

is a 4-repeat of a 3-sequence. Again, the recipient could have:

GTCGTCGTCGTCGTCGTCGTC

is a 7-repeat of the same. Which is to say that if you perform a polymerase chain reaction for this particular STR, the band for the donor and the recipient would show up at different places.

For small differences like a few nucleotides, gel electrophoresis is too low res. On capillary electrophoresis, the same would show up as two peaks.

Capillary electrophoresis of STR PCR; the exact height and area of the peaks are subject to PCR conditions, the loci being chosen and the relative difference in repeats between the two

Remember: PCR is competitive; shorter segments will eat up reagent nucleotides faster than longer ones (simply because the longer repeats need more time to copy). In this case, the difference 7-4=3 is not much; but if you pick a VNTR (variable number of tandem repeats) which are much more variable in repeats than STR, the difference can show up and skew the analysis in favor of the shorter allele.

Like every other gene, STRs are also inherited in two alleles: one maternal and one paternal. A person might have

• homozygous at an STR locus, i.e. both alleles might have the same number of repeats
• heterozygous, i.e. one allele has a 7-repeat, another a 3-repeat (i.e. the aforementioned scenario)

In an ideal scenario, the donor and the recipient have completely different STR alleles: maybe the donor has a 7-repeat and a 4-repeat, and the recipient has a 10-repeat and a 15-repeat. Which makes things really simple to analyse:

The proportion of donor cells in this case is simply the proportion of donor alleles (a & b) in the post transplant mix (green), i.e. 

The height/ area of the peaks reflect the relative proportion of the alleles in the mix (remember: all are competing for the same primer!)

(a + b)/ (a + b + c + d)

This simplest scenario is of course, quite idealistic. One won't often find such an STR between two individuals; and even if one does, it is not wise to assess engraftment by only one STR marker. Consider this particular situation where the exact same allele pair is carried by both donor and recipient.

This is a non-informative STR marker - because the donor and recipient carry the same allele pair

In other situations, there might be one common allele between the two.

The allele at extreme right is ahred between the recipient; the one at extreme left belongs to donor only. Thus, the proportion of the left allele in the final sample is the amount of chimerism

This is akin to the schema:

Here, the amount of chimerism is b / (b + d), i.e. pretend that 'a' is not there.

Antigen titration for T cell stimulation

T lymphocytes don't 'naturally' respond to an antigen; for an antigen to elicit a T cell response

1. A T cell clone with that particular receptor must be present

2. There must be antigen presenting cells (dendritic cells/ macrophages/ B cells)

In vitro, in isolated mononuclear cell preparations from whole blood, the second point is not usually a problem. However, the first condition, presence of antigen specific T cell clone, varies. 

For example, for a disease like tuberculosis which is very prevalent in India, one might expect that everyone has at least a few T cell clones for tubercular antigens. However, if the person has never got actual tuberculosis disease, this clone will not be very large. For rare diseases (like Scrub typhus), one would expect an even lower proportion of antigen specific T cells.

Empirically, it is prudent to do a titration of a specific antigen - if only to see which dose does the T cell respond the most. It is not a very tightly controlled experiment, because

1. The total number of lymphocytes vary between people

2. The size of antigen specific T cell clones vary between person to person

3. The T cell response is not dose dependent

However, like all biological phenomenon, T cell stimulation is not an 'all or none' phenomenon, and some correspondence with dose of antigen is always observed. In this case, a whole cell lysate of Mycobacterium tuberculosis (the antigen) was used in increasing doses to stimulate T cells from a healthy person. Production of interferon gamma (IFNG) was used as a marker for T cell response.

1. CD4+ T cells were selected (first panel): the anti-CD4 antibody is tagged with APC

2. The IFNG producing CD4+ T cells are shown in sequence, with increasing doses of the antigen producing a higher proportion of IFNG+ T cells

The unstimulated population is used as a control to create the box ('gate') for IFNG+ T cells

The proportion of IFNG+ T cells plateaus at 7.5 microL of antigen dose, and thus this dose was selected.

Thanks Uddeep & Dr Abhinav 

Oxidative burst

Neutrophils (& monocytes) produce hydrogen peroxide (through superoxide production) to kill phagocytosed bacteria. 

2O₂ + NADPH —> 2O₂^•– + NADP⁺ + H⁺

This reaction is catalaysed by NADPH oxidase. A series of reactions then generates hydrogen peroxide from this superoxide radical, and finally hypochlorous acid.

In vitro, this reaction can be visualised by Dihydrorhodamine (DHR), which is oxidised by H2O2 from normal, stimulated neurophils to give fluorescence (in the FITC channel). EDTA/ heparin blood must be used within 2 hours of collection, in room temparature all the time.

Gating neutrophils, lymphocytes and monocytes from whole blood

Before and after stimulation; note presence of DHR peak ('oxidative burst') in neutrophils and monocytes, and not in lymphocytes

The lack of superoxide produces a (weirdly named) disease.

Chronic granulomatous disease

Well, this has got (almost) nothing to do with granulomas (maybe a little, indrectly). This is a defect in several components of NADPH oxidase, resulting in inability to produce hydrogen peroxide. An X linked form (75%) & an autosomal recessive form (25%) occur.

Agent

XR: Mutation in phagocyte NAPH-O complex due to defect in gene for gp91phox.

AR: Defects in genes p47,67,22 phox or RAC2

It manifests as recurrent pyogenic infection with catalase +ve organisms.

Pathology

Three scenarios:

1. Normal neutrophils accumulate H2O2 in the phagosome containing ingested bacterium → MPO (myeloperoxidase) is delivered to the phagosome by degranulation, → H2O2 acts as a substrate for MPO to oxidize halide to hypochlorous acid and chloramines, which then kill the microbes. The quantity of hydrogen peroxide produced by normal neutrophils is sufficient to exceed the capacity of catalase, a hydrogen peroxide-catabolizing enzyme of many aerobic microorganisms.

2. When catalase +ve organisms such as E. coli gain entry into the CGD neutrophils, they are not exposed to hydrogen peroxide because the neutrophils do not produce it, and the hydrogen peroxide generated by microbes themselves is destroyed by their own catalase. Thus catalase-positive microbes, such as E. coli, can survive within the phagosome of the CGD neutrophil.

3. When CGD neutrophils ingest catalase negative organisms such streptococci or pneumococci, the organisms generate enough hydrogen peroxide to result in a microbicidal effect; i.e. they are killed by their own hydrogen peroxide.

(Ref: Williams, Hematology)

Thus there is a preponderane by recurrent infections by Catalse positive organisms.

CGD; note the lack of oxidative burst in stimulated neutrophils

Foxp3

Titration of FoxP3 (tagged with PE). Note the gradual and slow population shift over increasing doses of antibody.  (UU - unstimulated unstained, US - unstimulated stained, SS - stimulated & stained)

As the dose of antibody is inreased, the entire helper T cell population is taking up the stain (which is unwanted behavior)

There is no drop in median PE expression with doses upto 10 microL. This means there might be yet more antibody targets left to bind

In this case, 2.5 miroL dose which gives good positive population (2.86%) without staining the negative population. So we select this dose.

Antibody clone: 236A/E7, FoxP3 - PE

Transforming growth factor beta 1

Titration of anti TGFB1 antibody

Without stimulation

Titration plots for increasing doses of anti-TGFB1. Note the vertical line drawn at unstained (UU) and successive population shifts (non specific binding?). The ideal dose in this case must be close to 0.625 microL.

Doses ranging from 0.625 to 10 microL; note that entire cell population is moving

Plotting median FITC versus dose shows a continuously increasing trend. so no help from there!

With antigenic (MTb) stimulation

Does not make much of a difference

Thus, still undecided on the dose.

(Antibody clone TW4-9E7 bound to Alexa fluor 488, detected on FITC channel; cells - healthy peripheral blood mononuclear cells).