The ESIMS spectrum in positive ionization mode for lysozyme

The ESI-MS spectrum in positive ionization mode for lysozyme is obtained.

Show all steps to the solutions of the following problems. Round your answers no more than three significant figures.

a. What is the Mr of the protein to 5 significant figures based on the two highlighted ion species?

b. What are the charges of the species having m/z = 2044.6, 1431.6 and 1301.4 respectively?

c. What is the approximate number of residues in the analyzed protein (using an average of 110 Da for residue)?

100 1431.6 1301.4 1193.1 1101.5 1000 1400 1200 1590.6 1600 1789.2 1800 2044.6 2000 2200 2400 m/z

1789.2 = (MW + nH+)/n

1590.6 = [MW + (n+1)H+] /(n+1)

n(1789.2) - nH+ = (n+1)1590.6 - (n+1)H+

n(1789.2) = n(1590.6) +1590.6- H+

n(1789.2 - 1590.6) = 1590.6 - H+

n = (1590.6 - H+) / (1789.2 - 1590.6)

n = 1589.6 / 198.6

n = 8

Substitute n= 8 in 1789.2 = (MW + nH+)/n

1789.2 = MW + (8 x 1)/8

1789.2 x 8 = MW + 8

MW = 14,305.6 Da

b) At 1789.2, the charge is 8

At 2044.6

2044.6 = (MW + nH+)/n

2044.6 x n = 14,305.6 + n

n = 7

At 1431.6, n= 10

At 1301.4, n = 11

c) MW = 14,305.6 Da, 110 Da per residue

Number of residues = 14, 305.6 Da / 110 Da = Approx. 130 residues