How many amino acids span a 4.5 nm membrane in a 26 kDa protein?

[flunkie]

[SOLVED] BioChem Amino acid

Homework Statement

A protein has a MW of 26,000 Daltons. The protein consist of 7 alpha helix segments that cross a cell membrane of 4.5 nm.

Homework Equations

What they want to know is the minimum number of AA's for a single segment to cross the membrane completely and the fraction of the protein involved in the membrane spanning helices.

The Attempt at a Solution

I really do not know where to being with this question and have been looking for a beginning for a while. I know the answers are 30 and 89%, but I don't know how to calculate them.

My thought process has led me to try (26000/(110x7)) the weight of the protein by the average AA weight and number of chains, but that brings me to 33.8. I also know that the 4.5 nm factors in there somewhere, but where? I have also tried (26000/(45x7x3.6) 45 is 4.5 nm in angstroms, 7 is the number of helices and 3.6 is the amount of AA's in one helix turn, but that makes it too low at ~23. As far as the fraction thing goes I figured I would figure out the first part and go from there.

Thanks for any help.

 

[Moonbear]

You need to know something about the distance between amino acids in the alpha helix conformation described to figure out how many would span the membrane. Don't forget to count the number of amino acids required to make the turn at either end each time the protein passes through the cell membrane (I think that's what you forgot in your second attempt).

 

[flunkie]

Are you saying that I need to add on 2 turns of AA's to my final product?

If i did that for the second equation that would give me ~30 since there are 3.6 AA's per turn in a alpha helix, which is the right answer.

I figured out how to do the second part. That was just (30x110x7/26000) ~89%. AA's in helix, weight of AA, and 7 helices over the total weight of chain.