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Lecture 34

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McGill University
Biology (Sci)
BIOL 300
Siegfried Hekimi

th BIOL 300 November 28 2012 Lecture 34 Dr. Shock Cytokine receptors control differentiation of many cell types: • Prolactin induces milk product in acinar cells • Erythropoietin (EPO) induces red blood cell proliferation and differentiation • This allows you to carry more oxygen and improve endurance, which is why it’s used in doping, but also to help people who have anemia (low red blood cell count) We can compare cytokine receptors (bottom right) to all other kinase receptors • The only important difference between RTKs and cytokine receptors is that the kinase in this pathway is a separate protein (whereas the kinase in RTKs is part of the receptor) • However, the kinase (JAK kinase) is constitutively bound to a cytokinase receptor, allowing cross phosphorylation of tyrosine residues to activate the pathway upon ligand binding 1 th BIOL 300 November 28 2012 Lecture 34 Dr. Shock Note that there is a single ligand which is able to bind the receptor, which is a dimer • This is unusual, because this means that the EPO in this case has to have two distinct surfaces with which to bind each monomer STAT, a transcription factor, is able to bind to phosphorylated residues on the EPO receptor, and STAT itself becomes phosphorylated by JAK kinase • Phosphorylated STAT will then dimerizes, which exposes an NLS allowing it to enter the nucleus • Therefore, in monomeric STAT, the NLS is masked, meaning it is cytoplasmic protein • STAT will then activate genes which activate proliferation STAT is also able to be phosphorylated by other kinases, but this is the most common way We can see that the EpoR mutant lacks all red blood cells (there is no red in the embryo); similarly, a JAK2 kinase null mutant would not be able to produce RBCs • These embryos are obviously non-viable There are multiple methods of termination, but the most important one involves the SHP-1 phosphates, which gets recruited to phosphorylated tyrosines (negative feedback to keep the signal short) • SHP-1 will dephosphorylate JAK2 which stops the pathway because the rest of the tyrosines on the receptor will eventually be dephosphorylated • Mice with SHP-1 null mutations die because of excess RBC production, because this pathway cannot be shut down • Therefore, you can die from having too little or too many blood cells, showing the importance of regulation of this pathway • An athlete that had a heterozygous mutation for one tyrosine on the receptor actually went on to win a gold medal in cross country skiing 2 th BIOL 300 November 28 2012 Lecture 34 Dr. Shock • They thought he was doping because he had a very high RBC count, but they sequenced his genome and found that he had a heterozygous mutation of the tyrosine on this receptor for an alanine, preventing binding of SHP-1 allowing more endurance and RBC production The next pathway is the TGF-beta (also called DPP or BMP) pathway involving a receptor serine/threonine kinase • This pathway is important for bone growth and early development, as well as the inhibition of proliferation • This pathway is unusual because 2 dimers actually come together to form a heterotetramer receptor when active Another unusual thing about this pathway is that the ligand also form dimers, using a disulfide bond to connect them together • Some TGF ligands prevent proliferation; all ligands are processed • You can achieve homodimeric ligands and heteromeric ligands, which allows a lot of variability to active many different receptors of this class This pathway is very similar to the cytokine pathway, involving only the ligand, the receptor, and one major transcription factor • The ligand in the extracellular space gets bound to a proteoglycan known as RIII which concentrates the ligand at the cell surface for more efficient receptor activation • The ligand always binds active RII receptors which then phosphorylates a kinase called R1, which in turn phosphorylates a transcription factor called R-SMAD, exposing its NLS • R-SMAD will then dimerize with a protein called co-SMAD, and this complex can then enter the nucleus to regulate gene expression, just like STAT Inhibitory SMADs (I-SMADs) cause inhibition of the pathway by binding active RI and preventing binding of R-SMADs to prevent continuation of the pathway • This is a negative feedback mechanism, because the I-SMAD gene is one of the genes activated by this pathway (i.e. the pathway produces its own inhibitory component) Usually, a cell has to respond to a combination of signals at the same time; every cell need at least a few pathways to just survive, and even more if it wants to grow or 3 th BIOL 300 November 28 2012 Lecture 34 Dr. Shock divide, and another set of pathways for differentiation into a terminal cell type (like a neuron) • If
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