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Re: Yeast Enolase_2 Help# Biology - 生物学
m*h
1
If you want to try top-down. You need definitely two things:
1) A proper fragment method for break down large molecule. low energy
CID can handle some small proteins, but I think Enolase is too big
for low energy CID. High energy CID, ECD may help. But the fragmentation
mechanism are quite different from low engergy CID.
2) you need a high accuracy in mass determination. some time, 100ppm
error will give you so many possible matches.
It is not necessary that the weakest bond be broken first
avatar
h*n
2
You are right that the CID of proteins usually won't
give rise to anther two charge state distribution.
However, this may be one of the rare cases that you
did observe this phenomenon. In fact, I will predict
that this phenomenon is observed if a protein is
dissociated into two big pieces. As you know, the
charges do not distribute evenly in the protein. When
you fragment it into two big pieces, there ought to
be different charge distributions on these two big
pieces. If you break the protein in
avatar
h*n
3

I guess you didn't pick the right protein. :)
My boss did quite a lot of such study before.
For transferrin, a ~75kD glycoprotein, he was able to get
a lot of fragmentation at the C-terminus and got a good
ID using that stretch of sequence.
However, for most other protein, it is really hard to get
enough sequence information for a good protein ID.
ECD, UV laser photodissociation or other high energy
fragmentation methods may producce more fragments.
Low enerygy ways, such as IRMPD, CID and BIRD

【在 m*******h 的大作中提到】
: If you want to try top-down. You need definitely two things:
: 1) A proper fragment method for break down large molecule. low energy
: CID can handle some small proteins, but I think Enolase is too big
: for low energy CID. High energy CID, ECD may help. But the fragmentation
: mechanism are quite different from low engergy CID.
: 2) you need a high accuracy in mass determination. some time, 100ppm
: error will give you so many possible matches.
: It is not necessary that the weakest bond be broken first

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