After purification of a protein, protein contained 0.51% methionine by weight(me

Question

After purification of a protein, protein contained 0.51% methionine by weight(methionine Mr=149) (account for loss of molecule of H2O.

Calculate the min molecular weight of protein (one methionine residue per protein molecule)

Size-exclusion chromatography of protein gave molecular weight estimate of 53900. How many methionine residues are present in a molecules of new protein?

How do you solve this? Full solution please!

:or. methionine by weight(meth ioni n e Mrii ya) b) 92e exclusiun chromotography of psutei e , g ht-e st i mas e of S 39 00. How many nmethionine esidues arepresen+ nolecule arthenew piotei

Explanation / Answer

1. The molecular weight of methionine has to adjusted for removal of water during peptide bond formation. Peptide bond is formed when water is removed from the two amino acids. Molecular weight of water is 18.

Molecular weight of methionine in the protein= 149-18= 131

Molecular weight of the protein is calculated as follows:

Weight of methionine/ weight of the protein= n (molecular weight of methionine)/Molecular weight of the protein

Where n= number of methionine residues in the protein

n=1 (as there is one methionine residue pr protein molecule). Hence, the molecular weight becomes minimum molecular weight.

Substituting all the values in the formula,

0.51 g/100g= 1 (131)/ Minimum Molecular weight of protein

Minimum molecular weight of protein= 100 X 1(131)/0.51= 25686.28 =25686

2) Molecular weight of the protein= 53900

Protein has 53900/25686=2.098=2.1

Hence, there are 2 (or 2.1) methionine residues of methionine in the new protein.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.