Week 10: Advanced Imaging & Measurement Technology Homework

Homework: Final Project

For my final project, Glowing Ice Cream, the main things I would measure are the identity, quantity, activity, and release behavior of the NanoLuc/substrate system.

1. Protein Identity and Purity

I need to confirm whether I successfully made the right protein, NanoLuc-GSGS-His₆.

I would measure:

• Protein molecular weight
  Use intact protein LC-MS or MALDI/ESI-MS to check whether the measured mass matches the expected NanoLuc-GSGS-His₆ molecular weight.

• Protein purity
  Use SDS-PAGE after Ni-NTA purification. A strong band around the expected NanoLuc size would support successful expression and purification.

• Sequence confirmation
  Use DNA sequencing for the plasmid and, if available, peptide mapping by LC-MS/MS to confirm that the expressed protein sequence matches the designed construct.

2. Protein Quantity

I need to know how much active or total protein I have before testing brightness.

I would measure:

• Total protein concentration using A280, Bradford, or BCA assay.
• Purified fraction concentration after Ni-NTA purification.
• Approximate NanoLuc concentration by combining A280 with SDS-PAGE purity estimation.

This matters because the visible glow depends heavily on protein concentration.

3. Bioluminescence Output

The most important functional measurement is whether the system glows visibly and for how long.

I would measure:

• Peak brightness
• Time to peak
• Glow duration
• Brightness decay curve
• Area under the brightness-time curve

Technologies:

• Plate reader or luminometer for quantitative luminescence.
• Dark-room phone imaging for the actual visual effect.
• ImageJ / Python image analysis to extract brightness from time-lapse videos.

For each protein:substrate condition, I would plot brightness over time and compare which ratio gives the best “lick-to-glow” effect.

4. Substrate Encapsulation and Temperature Release

Because the project uses cocoa butter / lipid encapsulation as a temperature trigger, I would measure:

• Whether the substrate stays separated from protein at cold temperature.
• Whether release happens after warming.
• Whether the glow starts only after the lipid melts.

Technologies:

• Cold-room vs warm-room luminescence test.
• Microscopy to observe lipid bead size and distribution.
• Time-lapse imaging to compare cold stability and warm-triggered release.
• If available, LC-MS or HPLC to measure how much substrate is retained or released.

5. Ice Cream Matrix Stability

I would also measure whether the system still works inside or on real ice cream.

I would test:

• Glow before freezing.
• Glow after frozen storage.
• Glow after short warming.
• Whether fat, sugar, water, and temperature affect signal.

The goal is not only to prove that NanoLuc works, but to measure whether it can work in the actual food-like prototype environment.

Waters Part I — Molecular Weight

Question 1

Based on the predicted amino acid sequence of eGFP and known modifications, what is the calculated molecular weight?

The sequence given in the assignment includes eGFP, the LE linker, and a His₆ tag.

MVSKGEELFTGVVPILVELDGDVNGHKFSVSGEGEGDATYGKLTLKFICTTGKLPVPWPTLVTTLTYGVQCFSRYPDHMKQHDFFKSAMPEGYVQERTIFFKDDGNYKTRAEVKFEGDTLVNRIELKGIDFKEDGNILGHKLEYNYNSHNVYIMADKQKNGIKVNFKIRHNIEDGSVQLADHYQQNTPIGDGPVLLPDNHYLSTQSALSKDPNEKRDHMVLLEFVTAAGITLGMDELYKLEHHHHHH

Calculated values:

Length = 247 amino acids
Unmodified average MW = 28,006.60 Da = 28.0066 kDa

Because this is eGFP, the mature chromophore formation causes a mass loss of approximately:

H2O + H2 ≈ 20.03 Da

So the mature eGFP theoretical molecular weight is:

28,006.60 Da - 20.03 Da = 27,986.57 Da

Answer:

Theoretical MW of mature eGFP-His₆ ≈ 27,986.57 Da
= 27.9866 kDa

If using only the raw Expasy sequence result without chromophore correction, the value is approximately:

28,006.60 Da
= 28.0066 kDa

For comparison with intact MS data, I would use the mature chromophore-corrected value.

Question 2

Calculate the molecular weight using the adjacent charge state approach.

From Figure 1, I selected two adjacent charge state peaks:

m/z_n   = 1000.4302
m/z_n+1 = 965.9684

Using the adjacent charge state formula:

z = (m/z_n+1) / (m/z_n - m/z_n+1)

Substitute values:

z = 965.9684 / (1000.4302 - 965.9684)
z = 965.9684 / 34.4618
z = 28.03

So:

z ≈ 28

Using the proton-corrected version:

z = (965.9684 - 1.0073) / (1000.4302 - 965.9684)
z = 964.9611 / 34.4618
z = 28.00

So the higher m/z peak is:

1000.4302 = 28+

and the adjacent lower m/z peak is:

965.9684 = 29+

Now calculate molecular weight:

MW = z × (m/z - proton mass)

For the 28+ peak:

MW = 28 × (1000.4302 - 1.0073)
MW = 28 × 999.4229
MW = 27,983.84 Da

For the 29+ peak:

MW = 29 × (965.9684 - 1.0073)
MW = 29 × 964.9611
MW = 27,983.87 Da

Average experimental MW:

MW_exp = (27,983.84 + 27,983.87) / 2
MW_exp = 27,983.86 Da

Now calculate mass accuracy against the mature theoretical MW:

MW_theory = 27,986.57 Da
MW_exp    = 27,983.86 Da

Accuracy = |MW_exp - MW_theory| / MW_theory
Accuracy = |27,983.86 - 27,986.57| / 27,986.57
Accuracy = 2.71 / 27,986.57
Accuracy = 0.00009699

Convert to percent:

0.00009699 × 100 = 0.0097%

Convert to ppm:

0.00009699 × 1,000,000 = 97 ppm

Answer:

Experimental MW ≈ 27,983.86 Da
Theoretical mature MW ≈ 27,986.57 Da
Mass error ≈ 0.0097%
Mass error ≈ 97 ppm

Question 3

Can you observe the charge state for the zoomed-in peak in the intact eGFP mass spectrum?

Yes. In the zoomed-in peak around m/z ≈ 1473, the isotope peaks are separated by about:

Δm/z ≈ 0.053

Charge state can be estimated from isotope spacing:

z = 1 / isotope spacing
z = 1 / 0.053
z ≈ 18.9

So the charge state is approximately:

z = 19+

This also matches the molecular weight check:

MW ≈ 19 × (1473 - 1.007)
MW ≈ 27,900–28,000 Da

which is consistent with eGFP.

Answer:

Yes, the zoomed-in intact eGFP peak is approximately 19+.
The isotope spacing is about 1/19 ≈ 0.053 m/z.

Waters Part II — Secondary / Tertiary Structure

Question 1

Explain the difference between native and denatured protein conformations. What changes in the mass spectrum?

A native protein is folded. Its 3D structure is still mostly preserved, so many charged residues are buried inside the folded structure. Because fewer sites are exposed to solvent and protonation, the protein usually carries fewer charges.

A denatured protein is unfolded. When the protein unfolds, more basic and polar residues become exposed. This allows the protein to pick up more protons, so it appears in higher charge states.

In the mass spectrum:

Denatured eGFP:

• More charge states
• Higher charge numbers
• Peaks appear at lower m/z
• Broader charge-state distribution

Native eGFP:

• Fewer charge states
• Lower charge numbers
• Peaks appear at higher m/z
• More compact charge-state distribution

In Figure 2, the denatured spectrum has many peaks across the lower m/z range, while the native spectrum has fewer, stronger peaks at higher m/z values.

Answer:

The denatured protein unfolds, exposes more chargeable residues, and therefore produces higher charge states at lower m/z. The native protein remains folded, carries fewer charges, and appears at higher m/z with fewer charge states.

Question 2

What is the charge state of the native eGFP peak around m/z 2800?

For mature eGFP:

MW ≈ 27,986.57 Da

The native peak near:

m/z ≈ 2799.4

Charge state:

z ≈ MW / (m/z - proton mass)
z ≈ 27,986.57 / (2799.4 - 1.007)
z ≈ 27,986.57 / 2798.39
z ≈ 10.0

So the peak at around 2800 m/z is:

10+

This is also consistent with the adjacent native peaks around 2333, 2545, and 2799 m/z, which correspond approximately to 12+, 11+, and 10+.

Answer:

The native eGFP peak near 2800 m/z is 10+.

Small note: the actual zoomed inset in the provided figure appears to show the peak around 2545 m/z, which would correspond to 11+. But the peak specifically near 2800 m/z is 10+.

Waters Part III — Peptide Mapping / Primary Structure

Question 1

How many Lysines and Arginines are in eGFP? Highlight them.

Count from the full sequence:

Lysine (K) = 20
Arginine (R) = 6
Total K/R cleavage-related residues = 26

Highlighted sequence, with [K] and [R] marked:

001-050  MVS[K]GEELFTGVVPILVELDGDVNGH[K]FSVSGEGEGDATYG[K]LTL[K]FICT
051-100  TG[K]LPVPWPTLVTTLTYGVQCFS[R]YPDHM[K]QHDFF[K]SAMPEGYVQE[R]TIF
101-150  F[K]DDGNY[K]T[R]AEV[K]FEGDTLVN[R]IEL[K]GIDF[K]EDGNILGH[K]LEYNYNSHN
151-200  VYIMAD[K]Q[K]NGI[K]VNF[K]I[R]HNIEDGSVQLADHYQQNTPIGDGPVLLPDNH
201-247  YLSTQSALS[K]DPNE[K][R]DHMVLLEFVTAAGITLGMDELY[K]LEHHHHHH

Answer:

K = 20
R = 6
K + R = 26

Question 2

How many peptides are generated from tryptic digestion of eGFP?

Trypsin cleaves after:

K or R

With no missed cleavages, the full theoretical digest gives:

27 peptide fragments

However, the PeptideMass settings in Figure 4 only display peptides with mass larger than 500 Da.

Using that filter:

Displayed peptides > 500 Da = 19

The relevant peptides above 500 Da are:

Answer:

Full theoretical tryptic fragments = 27
PeptideMass displayed peptides above 500 Da = 19

For the homework, I would report:

19 peptides, using the Figure 4 PeptideMass output settings.

Question 3

How many chromatographic peaks are visible between 0.5 and 6 minutes, counting peaks >10% relative abundance?

From Figure 5a, visually counting peaks between 0.5 and 6.0 minutes that are above about 10% relative abundance gives approximately:

0.61
0.79
1.43
1.80
1.85
1.93
2.17
2.26
2.54
2.78
3.27
3.53
3.59
3.70
4.48
4.64
4.87

Total:

17 peaks

Answer:

Approximately 17 chromatographic peaks are visible between 0.5 and 6 minutes above 10% relative abundance.

Question 4

Does the number of peaks match the number of predicted peptides?

Predicted displayed tryptic peptides:

19 peptides above 500 Da

Observed chromatographic peaks:

~17 peaks

So the numbers are close, but they do not exactly match.

There are slightly fewer chromatographic peaks than predicted peptides. This makes sense because:

• Some peptides may co-elute at the same retention time.
• Some peptides may be too low in abundance to detect.
• Some peptides may ionize poorly.
• Some peaks may include modified peptides, missed-cleavage products, or non-peptide signals.
• The chromatogram peak count is not always equal to the theoretical peptide count.

Answer:

No, they do not exactly match. There are slightly fewer observed chromatographic peaks than predicted peptides.

Question 5

Identify the m/z of the peptide in Figure 5b, determine charge, and calculate singly charged mass.

The most abundant peak in Figure 5b is:

m/z = 525.76712

The isotope spacing is approximately:

526.25918 - 525.76712 = 0.49206

This is close to:

0.5 m/z

Since isotope spacing is approximately:

1 / z

then:

z ≈ 1 / 0.5 = 2

So the peptide is:

2+

Now calculate the singly charged form, [M+H]+.

Formula:

[M+H]+ = z × (m/z) - (z - 1) × proton mass

Substitute values:

[M+H]+ = 2 × 525.76712 - 1 × 1.007276
[M+H]+ = 1051.53424 - 1.007276
[M+H]+ = 1050.52696 Da

This matches the singly charged peak shown around:

m/z = 1050.52438

Answer:

Most abundant m/z = 525.76712
Charge state = 2+
Calculated [M+H]+ = 1050.527 Da

Question 6

Identify the peptide and calculate ppm mass error.

From the theoretical tryptic peptide list, the peptide closest to:

[M+H]+ ≈ 1050.527 Da

is:

FEGDTLVNR

The theoretical [M+H]+ mass for this peptide is:

1050.52145 Da

Experimental mass from the 2+ peak:

1050.52696 Da

Mass difference:

Δ = 1050.52696 - 1050.52145
Δ = 0.00551 Da

PPM error:

ppm = (Δ / theoretical mass) × 1,000,000
ppm = (0.00551 / 1050.52145) × 1,000,000
ppm ≈ 5.25 ppm

Answer:

Peptide = FEGDTLVNR
Theoretical [M+H]+ = 1050.52145 Da
Experimental [M+H]+ = 1050.52696 Da
Mass error ≈ 5.25 ppm

If using the directly labeled singly charged peak at 1050.52438, the error is approximately:

2.8 ppm

Question 7

What percentage of the sequence is confirmed by peptide mapping?

Figure 6 reports:

Identified: 88%

Answer:

The peptide map confirms 88% of the eGFP sequence.

Bonus Question 8

What peptide sequence best matches the fragmentation spectrum in Figure 5c?

The peptide identified from the precursor mass is:

FEGDTLVNR

The fragmentation spectrum in Figure 5c is consistent with this peptide assignment.

Answer:

FEGDTLVNR

Bonus Question 9

Does the peptide map data indicate that the protein is eGFP?

Yes.

The data supports that the protein is eGFP because:

• The intact protein mass is close to the expected mature eGFP-His₆ mass.
• The tryptic peptide map contains peptides that match expected eGFP digestion products.
• The peptide at 2.78 min matches the expected peptide FEGDTLVNR.
• The peptide map confirms 88% sequence coverage.

Answer:

Yes. The intact mass, peptide masses, fragmentation data, and 88% sequence coverage are all consistent with the protein being the eGFP standard.

Waters Part IV — Oligomers

KLH subunit masses from the homework table:

7FU subunit = 340 kDa
8FU subunit = 400 kDa

Calculations

The figure shows strong peaks around 3.4, 4.013, 8.33, and 12.67 MDa. The 4.013 MDa peak likely corresponds to an 8FU decamer, although that specific species was not one of the four requested answers.

Answer:

7FU Decamer = 3.4 MDa
8FU Didecamer = 8.0 MDa, observed near 8.33 MDa
8FU 3-Decamer = 12.0 MDa, observed near 12.67 MDa
8FU 4-Decamer = 16.0 MDa, expected near 16 MDa but not as clearly resolved in the screenshot

Waters Part V — Did I Make GFP?

Using the intact LC-MS result from Part I:

Calculation:

Theoretical mature eGFP-His₆ MW = 27,986.57 Da
Observed intact LC-MS MW = 27,983.86 Da

ppm error = |27,983.86 - 27,986.57| / 27,986.57 × 1,000,000
ppm error ≈ 97 ppm

Answer:

Yes, the intact LC-MS data is consistent with eGFP. The measured molecular weight is close to the theoretical mature eGFP-His₆ molecular weight, and the peptide mapping data further supports the protein identity.