imagine you treated one dish with lysophosphatidic acid (LPA); you treated anoth
ID: 58878 • Letter: I
Question
imagine you treated one dish with lysophosphatidic acid (LPA); you treated another dish with epidermal growth factor (EGF); and you left a third dish untreated. When you returned after three days, the untreated cells looked the same, while the EGF- and LPA-treated cells had definitely proliferated and were more crowded in the dishes.
1. If you measured the protein levels of cyclin D in these three populations of cells, how would they compare to each other?
2. What if you measured the phosphorylation state of pRb in these three?
Explanation / Answer
Cyclin D is one major protein who functional properties are quite conspicuous during cell division. It is the protein which interacts with Cdks 2,4 and 6 to form holoenzymes which have a key role in the proliferation of the cell.
Cyclin-Cdks complexes phosphorylate retinoblastoma tumour suppressor protein Rb and thus inhibits its interaction with the E2F which is a transcriptional factor and thus foster E2F mediated cell division and proliferation of course.
Coming back to the Q., if the Cyclin D level in these three populations of cells would be compared then In the case of EGF and LPA-treated cells then obviously the expression level of Cyclin in these cells would be investigated to high because both EGF and LPA are involved in cell signalling which ultimately involved in cell proliferation. However, in untreated cells Cyclin D level should be normal.
If the phosphorylation state of RB is measured then EGF and LPA-treated cells RB would be found in phosphorylated state as an effect of the treatment already discussed above. However, In untreated cells, it may is found in the phosphorylated state i.e in the form of pRB-E2F.
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