Abstract
Digestion of the rat liver glucocorticoid receptor with chymotrypsin results in the generation of a 42-kDa fragment which contains the steroid-binding and DNA-binding domains and the antigenic site for the BuGR anti-glucocorticoid receptor monoclonal antibody, while digestion with trypsin generates a 15-kDa receptor fragment containing only the DNA-binding function and the BuGR epitope (Eisen, L.P., Reichman, M.E., Thompson, E.B., Gametchu, B., Harrison, R. W., and Eisen, H.J. (1985) J. Biol. Chem. 260, 11805-11810). In this paper, glucocorticoid receptor of mouse L cells that were grown in the presence of [32P]orthophosphate was digested with trypsin or chymotrypsin (either before or after immune purification with BuGR antibody) and analyzed by sodium dodecyl sulfate-polyacrylamide gel electrophoresis, autoradiography, and Western blotting. The receptor is endogenously phosphorylated only on serine residues. Chymotrypsin digestion results in a 32P-labeled 42-kDa receptor fragment which contains steroid-binding, DNA-binding, and BuGR-reactive sites. Trypsin digestion generates a 27-kDa steroid-bound fragment (meroreceptor) which is not labeled with 32P and a 32P-labeled 15-kDa fragment which contains both the DNA-binding domain and the BuGR epitope. We have calculated that there are 4 times as many phosphate residues in the intact receptor than in the 42-kDa chymotrypsin fragment. From examination of 32P-labeled receptor fragments, we have deduced that one phosphate is located between amino acids 398 and 447, a region containing the BuGR epitope and about one-third of the DNA-binding domain, and the remaining three phosphates appear to be clustered just to the amino-terminal side of the BuGR epitope in a region defined by amino acids 313 to 369. Treatment of intact 32P-labeled receptor in cytosol with alkaline phosphatase removes these three phosphates, but it does not remove the phosphate from the DNA-binding-BuGR-reactive fragment and it does not affect the ability of the transformed receptor to bind to DNA-cellulose.
Highlights
DNA-binding domains and the antigenic site for the occurs on serine, butthe protein that was analyzed was
L cells that were grown in the presencoef [32Plorthophosphate was digested with trypsin or chymotrypsinand analyzed by sodium dodecyl sulfatecinoma cells revealed 89% phosphoserine and 11%phosphotyrosine [13].Miller-Diener et al [14] have reported that cellfree phosphorylation of purified rat liver glucocorticoid receptor occurs on threonine residues, but theamino acids that are phosphorylated in intact rat liver cells have not been demonstrated
From examination of ”P-labeled receptor fragments, we have deduced that one phosphate is located between amino acids 398 and 447,a region containing the BuGR epitope and about one-third of the DNAbinding domain, and the remaining three phosphates appear to be clustered just to the amino-terminal side of the BuGR epitope in a region defined by amino acids review)
Summary
Ner, but does not react with BuGR antibody [3]. it is Identification of Proteolytic Fragments of the Mouse L Cell not a highmolecularweightform of the receptor, and its Glucocorticoid Receptor-In the experiment of Fig. 1,the [3H] function is unknown [3]. Digestion of the demonstrate that themeroreceptor species are not phosphoreceptor with chymotrypsin (lane 3)yields a 42-kDa steroid- rylated In this experiment, 32P-labeledand [3H]DM-labeled bound fragment and a minor band at 30 kDa. The presence receptors were first immunoadsorbed with BuGR2 antibody of 30-kDa meroreceptor following chymotrypsin treatment is and thendigested with trypsin. 5. Proteolysis of the s2P-labeled glucocorticoid receptor yields phosphorylated 15-kDa tryptic and 42-kDa chymotryptic fragments.Aliquots (1 ml) of 32P-labeledL cell cytosol were incubated for 1 h at 0 “Cwith the concentrations of trypsinor chymotrypsin indicated belowT.he intact receptor and receptor fragments wereimmunoadsorbed to protein A-Sepharose with BuGR2 monoclonal antibodyor nonimmune mouse IgG,and the immunoad-. After normalizing the amount of 32Pto the amount of intact receptor protein or42-kDa fragment by dividing the 32Pcpm by the lZ6Ior 3H cpm, the ratio of 32Pin intact 100-kDa receptor to the32Pin the 42-kDa receptor fragment was calculated
Sign-in/Register to view highlights & summary Sign-in/Register to access full text options 
Free) 
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Schedule a call
