ZLLM0421c – Medical Oral Microbiology I, practical sessions. Protocol to topic J08
Topic J08: Immunofluorescence, ELISA, immunoblotting
To study: Immunofluorescence, ELISA, radioimmunoassay, immunoblotting, immunochromatography
Task 1: Direct and indirect immunofluorescence
Evaluate a picture of direct immunofluorescence in diagnostics of syphilis and a picture of indirect
immunofluorescence in diagnostics of syphilis, which is called FTA-ABS, and answer following question:
Is there any difference in appearance between direct and indirect immunofluorescence? _______________
Add labels to schemes of direct and indirect diagnostics. Use red colour for components derived from the patient.
Task 2: ELISA – introduction
Look at an ELISA panel. As you can see, we use similar serological plates as for other serological reactions, but
each patient has just one well, eventually two or three (if we search for antibodies of various classes).
Write: positive wells have _____________ colour, negative wells are ___________
Task 3: ELISA in a proof of antigen (detection of galactomannan antigen in aspergilli)
In all ELISA reactions we use a reader (spectrophotometer) to evaluate absorbance (optical density) of
individual wells. So we measure absorbance of patient wells, but also of controls, eventually blank etc.
In process of assessment of positivity or negativity of results, we evaluate the reaction so that we compare
absorbance values of the patient wells with so named cut off value. Cut off value is either counted as average of
special wells, labelled usually “c. o.” (as „cut off“), or as average of negative controls + constant. The details
may differ for individual sets. Sometimes we consider positive result higher than e. g. 50 % of cut off.
Our reaction serves as a proof of an invasive infection caused by Aspergillus sp. (a filamentous fungus, more in
mycology practical session). Although it is an antigen detection, usually serum is the specimen. This is because
the fungus and so also its antigen are present in the bloodstream. Sometimes also bronchoalveolar lavage is used.
Read ELISA in a proof of galactomannan antigen in blood (serum) or bronchoalveolar lavage (BAL).
Check positivity / negativity of eight patients.
In this type of ELISA, first we have to check that cut off is OK: both c. o. absorbance values must be between
0.300 and 0.800. Then we count cut off value as the average of two “c. o.” wells (in wells B1 and C1, labelled c.
o. 1 and c. o. 2). We also check K+ and K– (see picture) and now we come to patient data (E1 to H1 and A2 to
D2). All absorbance values higher or equal than 50 % of cut off value are considered positive, all values lower
than 50 % of cut off are considered negative.
Add labels to schematic principle of the reaction. For substrate, write „substrate – changes colour“ or
„substrate – no change“. Colour red the component derived from the patient (missing in negative reaction).
Cut off = (B1 + C1)/2
K+ and K– are OK? yes – no
(delete as appropriate)
Cut off =
Cut off = (B1 + C1)/2
Cut off =
Positive control (K+) is*
 higher then 150 % of cut off (OK)
 lower than 150 % of cut off (bad)
Negative control (K–) is*
 lower then 40% of cut off (OK)
 higher than 40 % of cut off (bad)
*tick what is true
Conclusion:
Positive is patient No. ___
Name _____________________
Positive control (K+) is*
 higher then 150 % of cut off (OK)
 lower than 150 % of cut off (bad)
Negative control (K–) is*
 lower then cut off (OK)
 higher than cut off (not OK)
*tick what is true
Conclusion:
Positive is patient No. ___
Dental Medicine
Date 10. 4. 2014
Page 1/3
ZLLM0421c – Medical Oral Microbiology I, practical sessions. Protocol to topic J08
Task 4: ELISA in a proof of antibodies against Toxoplasma gondii in serum
4a) Proof of antibodies in class IgG, IgM and IgA
In proof of antibodies, we use to assess presence of antibodies against IgM (typical for fresh infection) and IgG
antibodies. In some cases, nevertheless, we assess IgA antibodies to check the mucosal immunity, too.
In this type of ELISA, again, we count cut off value as the average of negative controls (in wells C1and D1 for
IgG, in wells C3 and D3 for IgG and in wells C5 and D5 for IgA). All absorbance values higher than 110 % of
cut off value (so called positivity index > 1.1) are considered positive, all values lower than 90 % of cut off
(positivity index < 0.9) are considered negative. Values between 90 % and 110 % of cut of value are considered
borderline.
Read ELISA in a proof of IgG, IgM and IgA antibodies to Toxoplasma gondii from serum. Check
positivity / negativity of all 8 patients. Compare with the work sheet presented to you by the teacher.
Remark also CFT results (CFT was performed as a parallel reaction).
Add labels to schematic principle of the reaction. For substrate, write „substrate – changes colour“ or
„substrate – no change“. Also use red colour for component derived from the patient (logically, that
component is supposed to be missing in negative reaction).
IgG
Cut off = (C1 + D1) / 2
Cut off = ______________
Positive patients:
IgM
Cut off = (C3 + D3) / 2
Cut off = ______________
Positive patients:
IgA
Cut off = (C5 + D5) / 2
Cut off = ______________
Positive patients:
4b) Checking of avidity for the positive patient
Avidity is the strength of binding between the antigen and antibody. Antibodies in fresh infection use to have low
avidity, antibodies from older infections tend to have higher avidity value. Avidity assessment is often more
precise than just considering IgM predictive for an acute infection, especially in some infections like
toxoplasmosis. Avidity is usually checked for IgG antibodies. We use a special “avidity solution” that releases
part of the antibodies from their binding to their antigens. So, “avidity well” absorbance is always lower then
“normal well”, where the avidity solution was not used.
Dental students do not perform this part
4c) Conclusion
Add suitable numbers of patients to following conclusion.
Patient(s) No. Conclusion
The infection of this patient is fresh or at least active as IgM antibodies were found
The patient met the infection before, now the infection is not present or not active
These patients probably did not have contact with the disease
Task 5: Immunoblotting
5a) Classic Western blotting for diagnostics of Lyme borreliosis
Classic Western blotting is very similar to enzyme immunoassays (ELISA), but the antigen is divided by
electrophoresis to individual antigenic determinants, ant then blotted to nitrocellulose membrane, where
“proper ELISA” is performed.
Name _____________________
Dental Medicine
Date 10. 4. 2014
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ZLLM0421c – Medical Oral Microbiology I, practical sessions. Protocol to topic J08
Read strips according to teacher’s instructions in order to assess antibodies to Borrelia afzelii in serum. Draw a
scheme of positions for all important antigens according to a given pattern; chose one negative and one positive
patient and draw their findings into a bottom strip. Evaluate the result (patient that is considered positive must
have at least two bands corresponding to the pattern, or it should have positive the “important band” – in this
case, one band is sufficient) and write it into the protocol.
Find at least three positive patients and write their numbers here: ______________________________
5b) Modern type immunoblots for diagnostics of Lyme borreliosis and tick-borne
anaplasmosis
In modern type blots there exist recombinant antigens that are placed on the surface of strips by special
technologies (instead of electrophoresis). The immunoblot is then placed to a scanner that scans it and evaluates
the degree of “darkness” using a special software. Nevertheless, it is also possible to read it visually.
Check the positive control (supposed to be positive for borreliosis, not necessarily for anaplasmosis) and
negative control, and evaluate specimens No. 1 to 7.
Positive control is – is not OK (delete as appropriate)
Lyme disease: positive are specimens No. _________________ borderline specimens No. _________________
Anaplasmosis: positive are specimens No. _________________ borderline specimens No. _________________
Task 6: Immunochromatography
Immunochromatography is a method that also uses labelled components, but there is no washing and it is very
simple for manipulation. Therefore it is used not only in laboratory, but also at the patients. Some uses outside
microbiology (e. g. pregnancy test) are even available by laïcs themselves.
Draw and write the result for two stool specimens, tested for presence of Clostridium difficile toxins. Evaluate
not only the positivity (band TEST), but also validity of the test (band CTRL)
Name _____________________
Dental Medicine
Date 10. 4. 2014
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