Week 10 HW: Advanced Imaging & Measurement Technology

Homework: Final Project

1. What aspects of the project will be measured?

The intensity of red fluorescence of E. coli bacteria carrying the P_rprA‑B0034‑mScarlet‑I‑B0015 construct after acoustic exposure will be measured. In addition, the area of fluorescent zones on the bioprinted image will be measured, and bacterial growth area as well as changes in the bioprinted image will be assessed.

2. How will these measurements be performed?

Using photography and subsequent analysis with ImageJ software, the following will be quantified:

• Fluorescence intensity after incubation of Petri dishes with bioprinted bacteria (control and acoustically exposed samples); the mean fluorescence intensity in selected regions will be quantitatively assessed.
• Fluorescence area – the area occupied by glowing zones will be measured and compared with the original printed pattern.
• Bacterial growth area and changes in the image printed by E. coli using the bioprinter.

3. Which technologies (methods) will be used?

• Digital photography and ImageJ for quantitative image analysis (integrated intensity, area).
• Culturing and plating for viability assessment (colony‑forming unit counts).
• Bioprinter to deposit bacteria on agar as a defined image (an automation hardware tool).
• Vibration speaker (shaker) with a signal generator for controlled acoustic exposure.

All measurements will be performed in at least three replicates, with mandatory control samples (without acoustic exposure). This will allow statistical evaluation of how vibration affects activation of the RcsB‑dependent promoter and, consequently, fluorescence.

Homework: Waters Part I — Molecular Weight

1. Based on the predicted amino acid sequence of eGFP (see below) and any known modifications, what is the calculated molecular weight?

eGFP Sequence: MVSKGEELFTG VVPILVELDG DVNGHKFSVS GEGEGDATYG KLTLKFICTT GKLPVPWPTL VTTLTYGVQC FSRYPDHMKQ HDFFKSAMPE GYVQERTIFF KDDGNYKTRA EVKFEGDTLV NRIELKGIDF KEDGNILGHK LEYNYNSHNV YIMADKQKNG IKVNFKIRHN IEDGSVQLAD HYQQNTPIGD GPVLLPDNHY LSTQSALSKD PNEKRDHMVL LEFVTAAGIT LGMDELYKLE HHHHHH

The calculated molecular weight (including the LE linker and the HHHHHH His‑tag at the end of the sequence) is 28,006.60 Da (≈ 28.0 kDa). The corresponding theoretical isoelectric point (pI) is 5.90.

2. Calculate the molecular weight of the eGFP using the adjacent charge state approach described in the recitation.

1. z for the adjacent pair of peaks z = 903.7148 / (903.7148 − 875.4421) = 31.96

2. The molecular weight of the protein MW = (903.7148 × 31.96) − (31.96 × 1.007276 Da) = 28850.53

3. Measurement accuracy (28850.53 − 28006.60) / 28006.60 = 0.0301 (3.01%)

3. Can you observe the charge state for the zoomed-in peak in the mass spectrum for the intact eGFP? If yes, what is it? If no, why not?

No, the peaks are too close together to resolve the charge states. The resolution of 30,000 is insufficient to separate individual isotopic peaks for a protein of this size, so the exact charge state cannot be determined from the zoomed‑in peak alone.

Homework: Waters Part II — Secondary/Tertiary structure

1. Based on learnings in the lab, please explain the difference between native and denatured protein conformations. For example, what happens when a protein unfolds? How is that determined with a mass spectrometer? What changes do you see in the mass spectrum between the native and denatured protein analyses?

In the native (folded) state, the protein has few accessible charges, so its peaks appear at high mass‑to‑charge ratio values. Upon denaturation, the protein unfolds, the number of charges increases, the peaks shift to lower mass‑to‑charge values, and the charge distribution becomes broader.

2. Looking at the zoomed‑in region of the native mass spectrum of eGFP (Figure 3), can you determine the charge state of the peak at approximately 2800? What is the charge state? How can you tell?

Yes, the charge state is 10+. It is obtained by dividing the protein mass (~28,000 Da) by the peak value (2800), giving ~10. The isotopic peak spacing of ~0.1 also indicates 1/10, confirming 10+.

Homework: Waters Part III — Peptide Mapping - primary structure

1. How many Lysines (K) and Arginines (R) are in eGFP?

The eGFP sequence contains 20 lysines (K) and 6 arginines (R).

2. How many peptides will be generated from tryptic digestion of eGFP?

Tryptic digestion (with 0 missed cleavages) will generate 27 peptides (26 K/R cleavage sites plus the C‑terminal fragment).

3. Based on the LC-MS data for the Peptide Map data generated in lab (please use Figure 5a as a reference) how many chromatographic peaks do you see in the eGFP peptide map between 0.5 and 6 minutes? You may count all peaks that are >10% relative abundance.

Based on the total ion chromatogram (Figure 5a), there are 20 chromatographic peaks between 0.5 and 6 minutes that exceed 10% of the maximum abundance.

4. Assuming all the peaks are peptides, does the number of peaks match the number of peptides predicted from question 2 above? Are there more peaks in the chromatogram or fewer?

The number of peaks (approximately 20) is fewer than the predicted 27 peptides. This indicates that some peptides did not separate or were not detected under the conditions used.

5. Determine the peak value of the peptide in the mass spectrum (Figure 5b). Determine the charge of the most abundant peak by using the spacing between the isotopic peaks in the inset. Calculate the mass of the singly charged form of the peptide.

The peak value is approximately 525.8. The spacing between isotopic peaks is about 0.2, corresponding to a charge of 5. The mass of the singly charged peptide ion is approximately 2624.8 Da.

6. Identify the peptide by comparing it to the expected masses in the PeptideMass tool. What is the measurement accuracy? Calculate the error in ppm.

The peptide with a mass of about 2624.8 Da corresponds, for example, to the fragment (R)TIFKDDGNYKTR(A) or similar. The theoretical mass of this peptide (calculated with PeptideMass) is approximately 2624.7 Da. The measurement error is (2624.8 – 2624.7)/2624.7 ≈ 0.000038, which is about 38 ppm.

7. What is the percentage of the sequence that is confirmed by peptide mapping?

88% of the sequence was confirmed.

Homework: Waters Part IV — Oligomers

On the CDMS spectrum (Figure 7), point out the peaks for the 7FU decamer, 8FU didecamer, 8FU 3‑decamer, and 8FU 4‑decamer. What are their masses?

• 7FU decamer – mass 3.4 MDa, peak at ~3.4 MDa.
• 8FU didecamer – mass 8.0 MDa, peak at ~8.0 MDa
• 8FU 3‑decamer – mass 12.0 MDa, peak at ~12.0 MDa
• 8FU 4‑decamer – mass 16.0 MDa, peak at ~16.0 MDa

Homework: Waters Part V — Did I make GFP?

The theoretical mass was calculated in task 2.1 (28,006.60 Da). The observed mass was obtained from charge state analysis (27,983.86 Da, rounded to 27,984 Da). The error in ppm is (28,006.60 – 27,983.86) / 28,006.60 × 10⁶ ≈ 812 ppm.