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!
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
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
Well, it seems like you've got the right idea, but your calculations might have taken a wrong turn somewhere. Let's break it down.
A) To calculate the minimum molecular weight of BSA, we need to use the percentage of tryptophan by weight. You correctly used the formula weight of tryptophan (204 g/mol) divided by the percentage (0.58%). However, your math seems to have gone astray.
If we correct your calculation, it should be:
Minimum molecular weight of BSA = 204 / 0.0058 = 35,172 g/mol.
So, the correct answer for part A is not 32,000 but actually 35,172.
B) To determine the number of tryptophan residues, we can use the molecular weight estimate obtained from gel filtration (70,000 g/mol) and the percentage of tryptophan (0.58%). Your calculation this time appears to be multiplying the molecular weight estimate (70,000) with the percentage (0.58) and then dividing by the molecular weight of tryptophan (204 g/mol).
However, the correct calculation should be:
Number of tryptophan residues = (70,000 * 0.0058) / 204 = 1.978.
Since we can't have fractional amino acid residues, we round down to the nearest whole number. So, the correct answer for part B is 2, not 199.
I hope this helps clarify things for you. Keep up the good work, and don't let the math monsters scare you away!
Your calculations for part A seem to be incorrect. Let me walk you through the correct process:
A) To calculate the minimum molecular weight of BSA, we need to determine the weight of the tryptophan portion. Given that tryptophan represents 0.58% of the total weight of BSA, we can set up the following equation:
Weight of tryptophan = (0.58% / 100) * Weight of BSA
Since the molecular weight of tryptophan (Mr 204) is known, we can rearrange the equation to solve for the weight of BSA:
Weight of BSA = Weight of tryptophan / (0.58% / 100)
Now, let's substitute the values into the equation:
Weight of BSA = 204 / (0.58% / 100)
Weight of BSA = 204 / (0.0058)
Weight of BSA ≈ 35,172 daltons
Thus, the minimum molecular weight of BSA is approximately 35,172 daltons (not 32,000 daltons as mentioned in the book).
Now, let's move on to part B:
B) Gel filtration estimates the molecular weight of a protein by comparing its elution volume with the elution volumes of standard proteins. In this case, the molecular weight estimate of BSA is given as 70,000.
However, we need to determine the number of tryptophan residues present in the BSA molecule. To do so, we can use a similar calculation to what you have done:
Number of tryptophan residues = (Weight of BSA * Percentage of tryptophan) / Molecular weight of tryptophan
Let's substitute the values into the equation:
Number of tryptophan residues = (70,000 * 0.58) / 204
Number of tryptophan residues ≈ 197
Therefore, based on the given values, there are approximately 197 tryptophan residues present in a molecule of bovine serum albumin (BSA), not 2 as mentioned in the book.
It seems that there might be an error in the answers provided in the book.