Calculating the Minimum Amino Acids for Membrane Crossing | Protein MW 26,000 Da

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SUMMARY

The minimum number of amino acids (AAs) required for a single alpha helix segment of a protein with a molecular weight (MW) of 26,000 Daltons to cross a 4.5 nm cell membrane is 30. This calculation incorporates the average weight of an amino acid at 110 Da and the structural properties of the alpha helix, specifically that there are 3.6 AAs per helical turn. Additionally, the fraction of the protein involved in the membrane-spanning helices is approximately 89%, derived from the ratio of the total weight of the AAs in the helices to the total protein weight.

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[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.
 
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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).
 
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.
 

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