Fab, Fc and F(ab')2 in antibodies (immunoglobulins) (FL-Immuno/36)

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[@ali-wc5sz]

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[@sciencelover526]

Amazing video! But do the Fab regions cleaved off by papain still act as opsonins and bind complement?

[@FrankLectures]

Thanks. Please watch video lecture on OPSONIZATION for your query.

[@sciencelover526]

@@FrankLectures I watched the video on opsonization but it still did not answer my question :-( I'm wondering if the fragments themselves are opsonins and if they can bind complement since the Fc region is no longer attached.

[@FrankLectures]

Answer is NO. This is because opsonins act as bridge for eg. Between pathogen and phagocyte. The antigen binding sites bind to antigens on pathogen. And Fc region bind the Fc receptors on phagocytes. When cleaved by papain, the Fab fragments can still bind to antigen BUT they cannot act as opsonins because Fc region is not present. I hope it is clear to you now :).

[@sciencelover526]

@@FrankLectures Thank you!!! That makes sense :-)

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[@Viridian88]

Very nice video. I've got a question - how did they actually discover that they were composed by multiple units? For instance, reduction by mercaptoethanol performed by Edelman yields to 4 peptides, but since they're identical 2 by 2, you're supposed to see just 2 bands on an electrophoretic separation.

[@BioMedUSA]

They would compare the staining intensity of the 2 different bands referenced to a set of standards (various concentrations) and also referenced to the starting amount (untreated antibody). So untreated Antibody (150kDa mol. wt.) say for convenience 15 ug on a non-denaturing gel would have staining intensity X; the lane containing the mercaptoethanol reduced antibody fragments would show two bands at 25kDa and 50kDa and at twice the staining intensity as what would be expected if there were just one of each fragment, i.e total staining intensity would be conserved (not linearly) compared to starting material and the molecular weights would also have to be approximately the same, so if only 1 molecule, 1 X staining intensity of 25kDa + 1 X staining intensity of 50kDa would be ~ half 1 X staining intensity of 150kDa starting material - this does not add up, so could deduce there must be 2X so molecular weights add up correctly.

[@Viridian88]

@@BioMedUSA Thank you so much.💚

[@Viridian88]

@@BioMedUSA Nowadays, rather than comparing standards, you could probably just go for a densitometer, is it correct?

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