A quantitative amino acid analysis reveals that bovine serum albumin (BSA) contains 0.58% tryptophan (Mr 204) by weight.
A) Calculate the minimum molecular weight of BSA (i.e. assume there is only one Trp residue per protein molecule)
B) Gel filtration of BSA gives a molecular weight estimate of 70,000. How many Trp residues are present in a molecule of serum albumin?
In the back of the book, the answers are:
A) 32000
B) 2
I got
A) 204/.58 = 352 daltons.
B) 70000*.58/204 = 199 Trp residues
My calculations seems correct, and I can't see what I did wrong. Can anyone help? Thanks!
3 answers
It is .58% so you should be using .0058 as a decimal.
when the trp is bound inside the protein chain its effective molecular weight in that peptide chain will be 204-18=186, as to form a peptide bond one water(molecular weight=18) molecule must be displaced.
A) (186/0.58)*100= 32068.9 ~ 32000
B) ((70000*0.58)/100)186=2.18 ~ 2
A) (186/0.58)*100= 32068.9 ~ 32000
B) ((70000*0.58)/100)186=2.18 ~ 2
see, as the question says that W contributes to 0.58% of the proteins weight, so assuming the protein weighs 100g than W contributes to 0.58g of protein weight. Now we know that peptide bond formation releases water so subtraction 18 from W mol. weight. We get 186!
0.58g = 100g
186 = ?
and when you calculate you get approximately 32000
0.58g = 100g
186 = ?
and when you calculate you get approximately 32000
0 answers