00:02
The problem contains an octa peptide.
00:10
So the octa peptide we have 1, 2, 3, 4, 5, 6, 7, 8.
00:32
In which the n terminal is l .eu.
00:43
So the edmund reagent, that man reels l -p -u as the n terminal so we put l -p -u here carboxy bacteries gives the the c terminal so c terminal and carboxy bacteria releases s -e -r serene so the c terminal remember this is the n side and this is the n -terror terminal always peptides are written with the end terminal on the left side and c terminal on the right end side so c terminal.
01:49
Now let us look at what cyanogen bromide do.
01:55
Cyanogen bromide hydrolysis the c -1cionide cyanbronide c -n -br hydrolyse hydro -nice c -side.
02:16
Now in the given data in the given data, methionine is contained in a polypeptide having one, two, three, four, five amino acids, in which l u is present.
02:41
That means from here fifth amino acid is methyonine.
02:49
Now in between we have thyrin, lycine and arson, thyrusin, lycine and arson.
03:05
These are there in between we are to find out the order on the other side we have s -e -r -a -r -g -p -h -e so s -er -g is here now we have a -r -g -g and p -h -e here okay now next tripsin catalytic hydrolysis forms the two amino acids and two peptides so tripsin catalyze what does the trypsin do that is a lot that is no doubt that tripsin will hydrolyse the sea side of arginine and lysine tripsin will hydrolyte hydrolyte sea side of arginine sea side of arginine and lysine so that means here there is licein now licein containing peptide let us look at the lysine containing peptide so where is lysinease.
04:45
So the lysine containing peptide is lys l -e -u -t -y -r...