TLS Online TPP Program

#Id: 3992


Because nucleosomal DNA wraps around the histone protein 1.65 times, the formation of a single nucleosome using ccc plasmid would create a writhe of –1.65 and thus change the linking number by an equivalent amount. As described above, when the change in linking number associated with each nucleosome was measured, the number was lower than this, approximately –1.2 for each nucleosome added.
This discrepancy is referred to as the “nucleosome linking  number paradox,”

#Unit 2. Cellular Organization #CELL CYCLE #Part B Pointers
More Pointers
TLS Online TPP Program

#Id: 4663

#Unit 13. Methods in Biology

TUNEL method uses TdT to add BrdU to fixed and permeabilized cells which can only incorporated into fragmented DNA of apoptotic cells.

TLS Online TPP Program

#Id: 4664

#Unit 13. Methods in Biology

BrdU is incorporated into the newly synthesized DNA, and is then detected with fluorescently labeled anti-BrdU antibodies


TLS Online TPP Program

#Id: 4665

#Unit 13. Methods in Biology

Caspases western blotting assay is also used to detect apoptosis



TLS Online TPP Program

#Id: 4666

#Unit 13. Methods in Biology

The total concentration of antibody in the equilibrium dialysis chamber is known, the equilibrium equation can be rewritten as:


TLS Online TPP Program

#Id: 4667

#Unit 13. Methods in Biology

Scatchard plots are based on repeated equilibrium dialyses with a constant concentration of antibody and varying concentration of ligand. 

If all antibodies have the same affinity, then a Scatchard plot yields a straight line with a slope of Ka. The x intercept is n, the valency of the antibody, which is 2 for IgG and other divalent Igs. For IgM, which is pentameric, n  10, and for dimeric IgA, n  4. In this graph, antibody #1 has a higher affinity than antibody #2. .  







TLS Online TPP Program

#Id: 4668

#Unit 13. Methods in Biology

If the antibody preparation is polyclonal and has a range of affinities, Scatchard plot yields a curved line whose slope is constantly changing. The average affinity constant K0 can be calculated by determining the value of Ka when half of the binding sites are occupied (i.e., when r  1 in this example). In this graph, antiserum #3 has a higher affinity (K0  2.4  108) than antiserum #4 (K0  1.25  108)