[qbkr21]

Q: You wish to determine the sequence of a protein from the plant Brassica nigra.  Cleavage of the protein with chymotrypsin yields three fragments:  (1) LHKQANQSGGGPS; (2) QQAQHLRACQQW; (3) RIPKCRKF.  Cleavage with trypsin yields seven fragments: (1) R; (2) ACQQWLHK; (3) CR; (4) QANQSGGGPS; (5) FQQAQHLR; (6) IPK; (7) K.  What is the primary sequence of the protein? 

so i am able to get half of the problem, but cannot finish it off and that is where i need your help.  i know that chymotrypsin cleaves on the carboxy side at phe(f), trp(w), and tyr(y).  so in the first breakage piece (1) must be on the far right carboxy side because there are no residues in the sequence.  now i have

n'-----------------------------------------------------c'
?--------------------?------------------LHKQANQSGGGPS

i also realize after searching google that trypsin cleaves on the carboxy terminus side at arg(r) and lys(k), but this is where i am stuck and i can find which order pieces (2) and (3) should go in.

please help.

thank you.

[Dan]

I agree that fragment 1 from the chymotrypsin treatment (1A) must be at the C-terminus. This shows that fragment 4 from the trypsin treatment (4B) is also at the C-terminus. As 1A is longer than 4B you know that the next fragment in the set B must end in LHK, find that fragment (XB), stick it onto 4B to give (XB4B). Now fragment XB4B is longer than set A, and you know the next fragment in set A must end in.., find that fragment YA, stick it onto 1A to give YA1A. Now there is only one fragment left in set A, ZA, and by process of elimination the sequence is ZAYA1A.