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Any phd experts there??? Question 34. In the course material you have been introduced to multiple charging for electrospray spectra. Each peak in the spectrum below can be used to provide a measure of the mass of this protein, you can work out the charge associated with each peak, and measures can be averaged to find the overall mass. Use simultaneous equations to find the charge associated with each peak in the spectra below, which will allow you to calculate the average mass of each species. Include an estimate of the error associated with this measure. Graph for Intrpt. Below is samples see calculation etc

Any phd experts there??? Question 34. In the course material you have been introduced to multiple charging for electrospray spectra. Each peak in the spectrum below can be used to provide a measure of the mass of this protein, you can work out the charge associated with each peak, and measures can be averaged to find the overall mass. Use simultaneous equations to find the charge associated with each peak in the spectra below, which will allow you to calculate the average mass of each species. Include an estimate of the error associated with this measure. Graph for Intrpt. Below is samples see calculation etc

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Any phd experts there??? Question 34. In the course material you have been introduced to multiple charging for electrospray spectra. Each peak in the spectrum below can be used to provide a measure of the mass of this protein, you can work out the charge associated with each peak, and measures can be averaged to find the overall mass. Use simultaneous equations to find the charge associated with each peak in the spectra below, which will allow you to calculate the average mass of each species. Include an estimate of the error associated with this measure. Graph for Intrpt. Below is samples see calculation etc
x305 10- 02 00 600 659.8 714.7 700 779.5 800 857.4 900 952.6 1000 1071.5 1100 1200 1300 1400 +M5, 0.1-1.3min 1(7-29)
Transcribed Image Text:x305 10- 02 00 600 659.8 714.7 700 779.5 800 857.4 900 952.6 1000 1071.5 1100 1200 1300 1400 +M5, 0.1-1.3min 1(7-29)
4. B/S [1 509 ← 2023 MSC CHROMATOGRAPHY AN... Q L : 687.56 20-651,34 600 700 1:39 800 883 68 900 951.56 958.91 1030.74 1124.43 1000 1158 23 1281 67 1373.99 1463.06 1570.06 1100 1200 1300 1400 1500 1600 Each of the peaks in this spectrum correspond to the intact protein (no fragmentation!) but with different levels of charge. This is an electrospray mass spectrum, hence you get multiple charging of the analyte. So To find molecular weight of this protein you need to multiply the m/z of each peak by the number of charges (and then remember to subtract the number of charges, cos in electrospray the charges are protons and have a mass of 1 each, and you don't want to include this in your sum. You can do this process for quite a few peaks in the spectrum and average the calculated masses to give you a more accurate answer, and this will also give an error associated with the measurement. Firs, however, you need to work out what the charge is for each peak. To do this you use simultaneous equations. 773.362-727.942=727.94 Choose two adjacent peaks in the series (e.g. 773.36 & 727.94). Use the following equations: 773.36 = M/Z (where M is the mass of the protein and Z is the number of charges that gives rise to the peak at 773.36 727.94 = M/(2+1) (M is still the same, but to give the next peak along in the series the charge has gone up by 1! Rearrange these equations and solve for Z: 773.362 = M and 727.94Z+727.94 M therefore 773.362 = 727.94Z+727.94 O 1729.74 1840 28 1932 07 1274 m 1700 1800 1900 2000 Z=16.03. Since the charge is a proton, it needs to be an integer, hence Z=16. Z(773.36-727.94) = 727.94 therefore Z= 727.94 / (773.36-727.94) Lets work out the mass: M = 773.36*16= 12,373.76 take off 16 (the proton charge) = 12,357.76 And 727.94 * 17 12,374.98 take off 17 12,357.98 And 824.85 15 = 12,372.75 take off 15 = 12,357.75 And 687.56* 18 = 12,376.08 take off 18 = 12,358.08 etc PAGE 6 OF 6 O
Transcribed Image Text:4. B/S [1 509 ← 2023 MSC CHROMATOGRAPHY AN... Q L : 687.56 20-651,34 600 700 1:39 800 883 68 900 951.56 958.91 1030.74 1124.43 1000 1158 23 1281 67 1373.99 1463.06 1570.06 1100 1200 1300 1400 1500 1600 Each of the peaks in this spectrum correspond to the intact protein (no fragmentation!) but with different levels of charge. This is an electrospray mass spectrum, hence you get multiple charging of the analyte. So To find molecular weight of this protein you need to multiply the m/z of each peak by the number of charges (and then remember to subtract the number of charges, cos in electrospray the charges are protons and have a mass of 1 each, and you don't want to include this in your sum. You can do this process for quite a few peaks in the spectrum and average the calculated masses to give you a more accurate answer, and this will also give an error associated with the measurement. Firs, however, you need to work out what the charge is for each peak. To do this you use simultaneous equations. 773.362-727.942=727.94 Choose two adjacent peaks in the series (e.g. 773.36 & 727.94). Use the following equations: 773.36 = M/Z (where M is the mass of the protein and Z is the number of charges that gives rise to the peak at 773.36 727.94 = M/(2+1) (M is still the same, but to give the next peak along in the series the charge has gone up by 1! Rearrange these equations and solve for Z: 773.362 = M and 727.94Z+727.94 M therefore 773.362 = 727.94Z+727.94 O 1729.74 1840 28 1932 07 1274 m 1700 1800 1900 2000 Z=16.03. Since the charge is a proton, it needs to be an integer, hence Z=16. Z(773.36-727.94) = 727.94 therefore Z= 727.94 / (773.36-727.94) Lets work out the mass: M = 773.36*16= 12,373.76 take off 16 (the proton charge) = 12,357.76 And 727.94 * 17 12,374.98 take off 17 12,357.98 And 824.85 15 = 12,372.75 take off 15 = 12,357.75 And 687.56* 18 = 12,376.08 take off 18 = 12,358.08 etc PAGE 6 OF 6 O
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