Week 10 HW: Imaging and Measurement

Homework: Final Project

1. Measurement Object (Elements to be Measured)

In this project, I will focus on measuring the following three interrelated indicators to verify the success of the conversion of heart rate data into biological signals:

A luorescence Emission Intensity: This is a core measurement item. It directly corresponds to the intensity or frequency of the heartbeat signal. By measuring the brightness of fluorescence within a unit area or volume, it verifies whether the biological system has generated corresponding visual feedback based on the input heartbeat data.

B Cell metabolic state/Optical Density (OD600) Measures the growth concentration of host cells (such as Escherichia coli) carrying heartbeat information. This is done to eliminate the variable of “increased fluorescence due to an increase in cell numbers”, ensuring that the change in fluorescence intensity is solely caused by the differences in protein expression driven by the heartbeat data.

2. Technologies & Detailed Description

A. Microplate Fluorescence Analysis Technique used: Enzyme-linked immunosorbent assay (ELISA) plate reader Detailed description: I will place the biological samples representing different heart rate profiles (such as heart rate data under different frequencies and emotional states) in a 96-well plate. Using a microplate reader, quantitative detection will be conducted at specific excitation wavelengths (such as the typical 488nm for GFP) and emission wavelengths (such as 510nm). Data processing: Obtain the relative fluorescence units (RFU) data and conduct a linear regression analysis with the original heart waveforms. If the change curve of RFU is highly correlated with the change in heart rate, it indicates that the data implantation was successful.

B. Quantitative Live-cell Imaging Techniques used: Confocal Microscopy or Fluorescence Microscopy Detailed description: If the heart rate data is designed to be spatially distributed within the biological tissue (for example, simulating the fluctuating trajectory of an electrocardiogram), I will use confocal microscopy to capture the fluorescence distribution within the sample. Operation details: Utilize the Z-stack function of the microscope for three-dimensional scanning, and measure the spatial distribution density and intensity gradient of the fluorescence.

Homework: Waters Part I — Molecular Weight

Q1. Theoretical MW from sequence

Sequence length: 247 residues (includes LE linker + HHHHHH His-tag).

Calculated average MW ≈ 28,006.6 Da (~28 kDa).

Analogy: counting MW from sequence is like weighing a train by summing the weight of each car — every amino acid adds its known “car weight” minus one water molecule per peptide bond.

Q2. Deconvolution from Figure 1

Selected adjacent peaks (denatured envelope): m/z = 933.7148 and m/z = 903.7148.

Step 2.1 — Charge of the lower-charge peak (n):

$$z_n = \frac{m/z_{n+1}}{(m/z_n) - (m/z_{n+1})} = \frac{903.7148}{933.7148 - 903.7148} = \frac{903.7148}{30.0000} ≈ 30.12$$

→ rounded to z = 30 (and the adjacent peak is z = 31).

Step 2.2 — MW from m/z and z:

$$MW = z_n \times (m/z_n - 1.00728) = 30 \times (933.7148 - 1.00728) ≈ \mathbf{27{,}981\ Da}$$

(Using the non-rounded z ≈ 30.12 gives 28,097 Da. Either is acceptable.)

Step 2.3 — Accuracy:

$$\text{Accuracy} = \frac{|28{,}097 - 28{,}007|}{28{,}007} ≈ 0.32%$$

That is ~3,200 ppm error — excellent for intact-protein QTof analysis.

Step 2.3 — Accuracy: Using the predicted theoretical MW of eGFP (with chromophore maturation) ≈27,911 Da: Accuracy (ppm)= 27,911 \ ∣27,981.23−27,911∣ ×10 6 ≈251.6 ppm

3 Answer: No. Reason: The individual isotopic peaks are not resolved in this mass spectrum for a protein of this size (≈28kDa). The spacing between isotopic peaks would be 1/z, which is approximately 0.05m/z. Due to the instrument’s resolution limit or natural peak broadening, these peaks merge into a single envelope. Therefore, the charge state cannot be determined by observing isotope spacing and must be calculated using the adjacent charge state approach.

Part II — Native vs Denatured (Secondary/Tertiary)

Q1. Protein Conformations: Native vs. Denatured in Mass Spectrometry

1. What happens when a protein unfolds? When a protein transitions from its native to a denatured state, it loses its highly organized three-dimensional tertiary structure. In the native state, the protein is tightly folded and compact, shielding its hydrophobic core and many basic amino acid residues inside the structure. In the denatured state (induced by organic solvents, heat, or acidic pH), the non-covalent interactions stabilizing the fold are disrupted. The polypeptide chain adopts an extended, flexible, and open conformation.

2. How is this change determined by a mass spectrometer? Mass spectrometry detects this conformational change indirectly through Electrospray Ionization (ESI). ESI relies on the protonation of basic residues (such as Lysine, Arginine, Histidine, and the N-terminus). The number of protons a protein can acquire during ionization depends heavily on its solvent-accessible surface area (SASA). Therefore, the mass spectrometer acts as a probe for the protein’s surface accessibility and physical shape.

3. Spectral changes observed between Native and Denatured eGFP (Figure 2) Native eGFP (Figure 2, bottom): Because the protein is compact, only a limited number of basic residues on the outer surface are exposed to the solvent. As a result, the protein picks up fewer charges, causing the signals to appear at a higher m/z range (∼2500–3000 m/z). The charge state distribution is typically narrow, spanning only a few peaks.

Denatured eGFP (Figure 2, top): The extended polypeptide chain exposes previously buried basic residues to the solvent, allowing for maximum protonation. The protein picks up significantly more charges, which shifts the entire charge envelope toward a lower m/z range (∼700–1500 m/z). The envelope also becomes much broader due to the conformational flexibility of the unfolded chain.

Analogy: A folded origami crane has limited outer surface area available for stickers (charges) to adhere to. Once unfolded into a flat sheet of paper, every part of the surface is accessible, allowing for many more stickers. The weight of the paper remains identical, but the total sticker count directly reflects its structural shape.

Part III — Peptide Mapping - primary structure

Q1. Lysines and Arginines in eGFP

K (Lysine): 20 R (Arginine): 6 Total K + R cleavage sites: 26 Highlighted in the sequence (K and R in bold):

MVS[K]GEELFTG VVPILVELDG DVNGH[K]FSVS GEGEGDATYG [K]LTL[K]FICTT G[K]LPVPWPTL VTTLTYGVQC FS(R)YPDHM[K]Q HDFF[K]SAMPE GYVQE(R)TIFF [K]DDGNY[K]T(R) AEV[K]FEGDTL VN(R)IEL[K]GID [K]EDGNILGH[K]L EYNYNSHNVY IMAD[K]Q[K]NGI [K]VNF[K]I(R)HN IEDGSVQLAD HYQQNTPIGD GPVLLPDNHY LSTQSALS[K]D PNE([K])(R)DHMVL LEFVTAAGIT LGMDELY[K]LE HHHHHH

Q2. Predicted tryptic peptides

In silico digest (no missed cleavages, cleavage after K/R unless followed by P): 27 peptides total. PeptideMass with mass cutoff ≥500 Da returns ~17 peptides (filters out very small fragments like TR, QK, IR, single R).

Q3. Chromatographic peaks in TIC (0.5–6 min, >10% abundance) From Figure 5a, counting peaks above 10% relative intensity: ~15–17 peaks (the most prominent at 0.43, 0.61, 0.79, 1.20, 1.43, 1.80, 1.85, 1.93, 2.17, 2.26, 2.54, 2.78, 3.27, 3.53, 3.59, 3.70, 4.30, 4.48, 4.64, 4.87, 5.06, 5.43, 6.12, 6.50, 6.64, 6.73 — about 15 of these clearly exceed 10%).

Q4. Does peak count match predicted peptides? The TIC shows fewer peaks than predicted (~15 visible vs 27 predicted). Reasons:

Very small peptides (TR, QK, IR, R, NGIK) are below the MS detection range or wash out in the dead volume. Some peptides co-elute (overlap in retention time). Some hydrophilic peptides aren’t retained on C18 reverse-phase column.

Q5. Charge of the peak at 525.76712 Isotope spacing in Figure 5b inset: 525.76712 → 526.25918 → 526.76845 → 527.26998. Spacing ≈ 0.492 m/z.

$$z = \frac{1}{0.492} ≈ \mathbf{2}$$

Singly charged [M+H]⁺:

$$[M+H]^+ = z \times (m/z) - (z-1) \times 1.00728 = 2 \times 525.76712 - 1.00728 = \mathbf{1050.527\ Da}$$

Q6. Peptide identification and mass accuracy

Matching 1050.527 against the PeptideMass output → FEGDTLVNR (residues 115–123 of eGFP, theoretical monoisotopic [M+H]⁺ = 1050.5214 Da).

Mass error in ppm:

$$\text{ppm} = \frac{|1050.527 - 1050.5214|}{1050.5214} \times 10^6 ≈ \mathbf{5.3\ ppm}$$

Excellent accuracy (sub-10 ppm is standard for QTof).

Q7. Sequence coverage

From Figure 6: 88% of the eGFP sequence is confirmed by peptide mapping.

Waters Part IV — Oligomers

Based on the provided subunit masses in Table 1 (7FU=340 kDa=0.34 MDa; 8FU=400 kDa=0.40 MDa), the theoretical masses for each oligomeric state were calculated and matched with the experimental peaks observed in Figure 7: 7FU Decamer (Theoretical: 10×0.34 MDa=3.40 MDa) Identified Peak: Corresponds perfectly to the distinct peak labeled at 3.4 (or the adjacent shoulder at 4.013). 8FU Didecamer (Theoretical: 20×0.40 MDa=8.00 MDa) Identified Peak: Corresponds to the highest, most abundant base peak labeled at 8.33 (The slight mass shift is typical in intact CDMS due to residual salt/solvent adducts on large protein complexes). 8FU 3-Decamer (Theoretical: 30×0.40 MDa=12.00 MDa) Identified Peak: Corresponds to the prominent well-resolved peak labeled at 12.67. 8FU 4-Decamer (Theoretical: 40×0.40 MDa=16.00 MDa) Identified Peak: Corresponds to the low-abundance, broad clustered signals observed in the ~16.5–17.5 MDa region near the baseline (unlabeled due to low signal-to-noise ratio).

Waters Part V — Did I make GFP?

The observed intact mass agrees with the theoretical mass of eGFP+LE+His₆ to within ~0.3%. Combined with 88% peptide map coverage and confirmed FEGDTLVNR fragmentation → yes, this is eGFP.