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i)The MH1 domain of Smad 3 is missing.
The MH1 domain contains a NLS sequence. Thus Smad 3 will not localize to the nucleus
even in the presence of TGFb57 and will not activate transcription of SloGro.
ii)R1 has a truncated cytoplasmic tail and cannot be phosphorylated.
The R1 kinase is activated by phosphorylation of its cytoplasmic tail by RII. A truncated
version of R1 that cannot be phosphorylated will be unable to phosphorylate Smad3, and
will not activate the signaling cascade. 7.06 Spring 2004 PS 2 KEY 4 of 7 iii) Smad 7 is constitutively active.
Smad 7 blocks the ability of R1 to phosphorylate Smad 3, which prevents it from entering
the nucleus and activating transcription.
B) You also isolate a version of Smad3 that cannot bind DNA, and find that this mutant version
exerts a dominant negative effect on the TGFß57 pathway. Give one possible mechanistic
explanation for how mutant Smad3 could act dominant negatively on the pathway.
One possible scenario is as follows: Smad 3 can’t bind DNA, and therefore the complex
between mutant Smad3 and Smad4 can’t activate transcription of target genes. Because
the complex that binds DNA and activates transcription involves two Smad 3 proteins, the
mutant version gets incorporated into complexes with wild type Smad 3, effectively
inactivating complexes with wild type protein as well.
3) Epineprhine increases the concentration of glucose 1-phosphate in wild type liver cells.
Explain how the following mutations would affect the glucose 1-phosphate concentration within
an epinephrine stimulated liver cell.
i) The cell contains a PKA with a nulceotide binding site that is unable...
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This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore.
- Spring '11