rona
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someoe help me and i said i should post the answer so any onw want to know how?
The enzymes they're using to produce cleavage tend to work only on certain sites.
- Trypsin cleaves at the carboxyl side of Lys and Arg except when they're followed by a Pro (no Pros here, so you don't have to worry about that).
- Chymotrypsin cleaves the carboxyl side of Tyr, Trp, and Phe.
- Edman degradation cleaves off the N-terminal only.
So then each line in your problem tells you something about what you're working with.
- You have the list of all the AAs involved (Arg, Glu, 2 Val, Gly, Lys, Tyr, Thr, Phe).
- You know the only Glu is at the N terminal of the whole.
- That's also in T-1, so T-1 must start with Glu. Trypsin cuts at Arg, so T-1's structure must be N-Glu-Tyr-Arg-C.
- Because of the Edman degradation, you know that T-3 starts with Phe, and you are told it ends in Thr. This tells you that T-3 is actually the third segment of the original protein because it doesn't end in something trypsin cuts.
- T-2 must be in the middle, and you know it starts with Val, so it must end with the only other AA you have the trypsin cuts, which is Lys. It's a dipeptide, so that's the whole thing.
- By eliminating the ones we've already used, the two last AAs in the middle of T-3 would be the Val and Gly, but we don't have a way of telling what order they're in yet.
So after the first test, have have: N-Glu-Tyr-Arg-Val-Lys-Phe- (Val, Gly) -Thr-C.
- Chymotrypisin cuts at Tyr, Trp (none), and Phe, so we should gets Gly-Tyr, Arg-Val-Lys-Phe, (something, something)-Thr. A dipeptide, a tetrapeptide, and a tripeptide (so it worked out so far).
- The N-terminal of the tripeptide was Gly, so that clears up the only uncertainty we had left. It's first.
Our structure is N-Glu-Tyr-Arg-Val-Lys-Phe-Gly-Val-Thr-C.
I don't know if that seems 'easy' or not. It's just a matter of being methodical and going through step by step. It helps sometimes if you cross off the clues you use as you use them. Hope that all made sense!
 Posted 39 day ago
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