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Mass spectrometry data analysis of compounds

The molecular mass of the compound is 298.16 Da because this m/z spectrum is calculated under positive ionisation conditions and the ions are singly charged and it is dominant in m/z 299.16. Since the charge is 1, the molecular mass is

299.16-1 = 298.16 Da

b) What is the peak at m/z 300.16?

The m/z at 300.16 shows a lower intensity of ions and the C12 atom has been replaced by the naturally occurring C13 isotope. This isotope is one Da higher than the C12 atom in given m/z graph as the charge is 1 (singly charged).

c) Theoretical, Monoisotopic mass of C18H22N2S

Monoisotopic Mass of C18H22N2S= (12*18)+(1.0078*22)+(14.0031*2)+(31.9721*1)

= 216+22.1716+28.0062+31.9721

= 298.1499 Da

Hence the Monoisotopic mass of the compound is 298.1499 Da.

a) What is the molecular mass of the sample?

The molecular mass of the given sample with positive electrospray ionisation spectrum is 452.3 Da.

b) Suggest the origin of the peak at m/z 475.2.

The peak at m/z 475.2 suggests that some molecules in the spectrum are being ionised by addition of Na+ ion because the peak at m/z 475.2 is about 23 Da higher than the expected molecular mass ( 475.2- 452.3=22.9 which is approximately 23 Da).

a) Explain the difference between the cluster of ions at ca. m/z 574-6 and those at ca. m/z 461-462.

The m/z 574.3,575.3,576.3 signifies that the ions at these peaks are singly charged because the difference is 1 Da (575.3-574.3= 1, 576.3-575.3=1) and the m/z 461.4 shows a difference of 0.5 Da (461.9-461.4=0.5, 462.4-461.9=0.5) which suggests that the ions are doubly charged.

b) What are the molecular masses of the two peptides?

The molecular mass at m/z 461.4 is 460.9 Da (461.4-0.5= 460.9) because it is doubly charged and the molecular mass at m/z 574.3 is 573.3 Da (574.3-1=573.3) since it is singly charged.

a) Calculate the average molecular mass of the protein.

The protein sample here give rise to multiple charged molecular ions. The molecular mass at each peak need to be calculated with the formula

m/z= (MW+nH+)/n

where MW is the molecular weight

n is the number of charge

H+= 1.0079

m/z is the mass to charge ratio of the peak

At the charge +3

1318.3= (MW + 3*1.0079)/3

MW = 3954.9-3.0237

MW = 3951.8763 Da

At the charge +4

MW= (988*4- 4*1.0079)

= 3950.7681 Da

At the charge +5

MW= (799*5-5* 1.0079)

= 3989.9605 Da

At the charge +6

MW= (666.1*6- 1.0079*6)

= 3990.5526 Da

Average Molecular mass = (3951.8763+ 3950.7681+3989.9605+3990.5526)/ 4

= 3970.789 Da

a) What do the ions at m/z 72.08 represent?

The ions at m/z 72.08 represent the presence of alanine in the peptide as the molecular weight of alanine is 71 Da which is approximately equal.

b) What do the ions at m/z 175.12 represent?

The ions at m/z 175.12 represent the presence of Cysteine in the peptide because the difference of 175.12 and 72.08 is 103.04 which is approximately equal to the molecular mass of Cysteine (103 Da ).

c) The Amino acid sequence is:

This sequence is obtained by subtracting the m/z ratios

For example: 246-175= 71- A

333-246= 87- S

480-333=147- F

627-480= 147 -F

684-627= 57- G

813-684= 129- E

942-813= 129-E

1056-942 = 114- N

1171-1056= 115- D

1285-1171= 114-N

1384-1285= 99- V

Therefore the amino acid sequence is ASFFGEENDNV

d) Give more possibilities with mass difference of 186.

The possibilities are

i) 813.40 and 627.34 and their difference is 186.06 which is approximately equal.

ii) 684.37 and 497.21.

e) Why is trypsin used?

Trypsin is commonly used in mass spectrometric analysis because every proteolytic element used in the analysis contain a basic arginine or lysine residue and this trypsin helps in cleaving the peptide bonds at the carbonyl side of lysine or arginine.

The amino acid sequence is:

The sequence found with differences between the alternative peaks like

148-205, 352-205, 449-352,548-449,651-548,764-651 gave the sequence

GFPVCI(L)

The next set say 288-175,391-288,491-390,587-490,734-587,791-734 gavethe sequence

I(L)CVPFG

These sequences are exactly the reverse of each other.

Question 7.

a) Yes the Mass spectrometric data confirms the changes because the Val which has the molecular mass 99.1 is replaced by Leu which is 113.1 and their difference is 13. The Ile which has the molecular mass 113.1 is relaced by Asn which is 114 and their difference is 1.

So the total difference in the mass is 14+1= 15 Da

Now when you see it is given that calculated mass of FBP aldolase is 37,964.3 Da and the mass found using Mass spectrometry is 37,979.3 Da and their difference is 15 Da which indicates that the mass spectrometry confirm the changes.

b) Overall mass difference of the derived product:

The derived equation is given as, RNH.CH.CH2OH.CH2.O.P=O.(OH)2

Lysine (NH.CH.[(CH2)4NH2].CO) + C3H8O5P = C6H14N2O2 + C3H8O5P = C9H22N2O7P

Molecular mass of C9H22N2O7P =

= (9*12.0110)+(22*1.0079)+(2*1)+(7*15.9994)+30.9738

=275.2424 Da

Therefore the overall mass difference = 275.2424 – 170 = 105.2424 Da.

c) Which Aldolase mutant reacts with DHAP?

K238A aldolase mutant reacts with DHAP.

d) Which Lysine is responsible for enzyme activity, K236 or K238?

The Lysine K238 is responsible for enzyme activity.

e) The other method for determining which lysine is the active site is peptide mass fingerprinting, tandem mass spectrometry and even DNA sequencing methods using restriction enzymes can also be used.

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