Week 1 HW: Principles and Practices

pre-fire	post-fire

Week 1 Class Assignment

1. First, describe a biological engineering application or tool you want to develop and why.

I want to develop a biologically integrated microsensing tool that can detect heavy metal intoxication in living Lemna minor tissue at the cellular level. This idea grew out of a group project facilitated by Dr. Andrew Scarpelli through the ChiTownBio node, where we demonstrated Lemna minor’s ability to phytoremediate heavy metals using a ceramic kiln burnout process to visualize metal accumulation. While this approach confirmed effective metal uptake, its irreversible thermal processing of the plant tissue to generate a readout, making the method qualitative and not easily scalable. Developing a microsensing system that can read biochemical or protein-level responses in living plant cells would allow for more accurate, non-destructive, and low-cost detection of environmental metal contamination.

2. Next, describe one or more governance/policy goals related to ensuring that this application or tool contributes to an “ethical” future, like ensuring non-malfeasance (preventing harm). Break big goals down into two or more specific sub-goals.

A primary governance goal for this project is to ensure non-malfeasance and biological safety in the development of a microsensing tool that interfaces with living Lemna minor tissue to detect heavy metal intoxication. Key sub-goals include establishing risk assessment protocols for exposing living organisms to potentially toxic metal concentrations and ensuring that sensing mechanisms do not exacerbate cellular stress, impair plant viability, or introduce unintended environmental release of contaminants during testing or deployment. Clear system boundaries must also be defined to distinguish biological responses from engineered sensing components, particularly when integrating electronics or materials at the cellular or tissue scale.

A second governance goal is to promote transparency, traceability, and ethical biological use. Sub-goals include documenting the sourcing, cultivation, and post-use handling of Lemna minor, even though it is a non-engineered organism, and ensuring lifecycle considerations such as responsible disposal and environmental impact are addressed. Additionally, because the sensing output is empirical and interpretive, governance should include standards for public communication to prevent misinterpretation of results, especially in community or environmental decision-making contexts.

https://www.sciencedirect.com/science/article/pii/S0045653506013361?via%3Dihub

3. Next, describe at least three different potential governance “actions” by considering the four aspects below (Purpose, Design, Assumptions, Risks of Failure & “Success”).

Governance Action 1: Strengthened Safety and Containment Protocols for Heavy Metal Research

This governance action establishes dedicated safety and containment procedures for handling heavy metals in Lemna minor microsensing experiments. Because phytoremediation concentrates metals inside plant tissue, the resulting biomass becomes a hazardous waste stream that must be managed with the same rigor as chemical waste.

These protocols would define safe concentration limits, specify engineering controls and PPE, and outline approved disposal pathways for contaminated liquids, plant biomass, and sensing hardware. The action assumes that training and infrastructure vary widely, and that contaminated plant material is often underestimated as a hazard. Formalizing procedures and disposal routes can reduce risk while supporting responsible experimentation.

Governance Action 2: Open, Modular Standards for Microsensing Hardware

This governance action promotes transparency, accessibility, and ethical design by encouraging open standards for the microsensing hardware used with Lemna minor. Many microsensing systems—whether based on small PCBs, flexible circuits, microfabricated electrodes, or other substrates—are developed through proprietary processes that limit reproducibility and prevent external review of design choices that affect biological safety. By contrast, open, modular hardware architectures with publicly available schematics, fabrication files, and material specifications would allow researchers, community labs, and environmental groups to inspect, adapt, and audit the sensing system.

The action assumes that openness can improve safety and democratize participation, though it also introduces risks: users with varying levels of expertise may replicate the hardware, and widespread deployment of low‑cost sensors could lead to inconsistent calibration or uneven data quality. Still, by making the sensing system transparent and modular, this governance approach aims for reproducibility, ethical oversight, and broader community engagement.

Governance Action 3: Standards for Interpretation and Public Communication of Biological Sensing Data

This governance action establishes guidelines to prevent misinterpretation of biological sensing data generated from Lemna minor microsensing experiments. Plant‑based sensing outputs are inherently indirect: physiological changes in Lemna minor can be influenced not only by heavy metal exposure but also by light levels, nutrient availability, temperature, and general stress. Without proper context, these signals risk being treated as precise measurements of contamination when they are, in fact, biological indicators with meaningful uncertainty. To address this, the action proposes standardized communication practices that require contextual metadata, clear disclosure of uncertainty, and consistent visual or textual formats for reporting results. These standards help ensure that biological indicators are interpreted appropriately in environmental, regulatory, and community decision‑making contexts.

4. Next, score (from 1-3 with, 1 as the best, or n/a) each of your governance actions against your rubric of policy goals. The following is one framework but feel free to make your own:

5. Last, drawing upon this scoring, describe which governance option, or combination of options, you would prioritize, and why. Outline any trade-offs you considered as well as assumptions and uncertainties.

Based on the scoring, I would prioritize a combination of Option 1 and Option 3, since together they offer the strongest balance of safety, environmental protection, and constructive application without imposing excessive burdens on researchers. Option 1 directly reduces the most immediate risks—accidental exposure, improper disposal, and environmental contamination—making it essential for preventing harm at the source. Option 3 complements this by ensuring that any biological sensing data generated is communicated responsibly, reducing the likelihood of misinterpretation or misuse in public or regulatory contexts. While Option 2 has value, its lower scores in feasibility and safety‑related categories suggest it introduces more complexity than benefit at this stage. Overall, Options 1 and 3 provide the most robust and practical governance foundation, though this conclusion depends on the assumption that safety protocols will be consistently followed and that communication standards will be adopted across different settings.

Week 2 Lecture Prep

Homework Questions from Professor Jacobson:

1. Nature’s machinery for copying DNA is called polymerase. What is the error rate of polymerase? How does this compare to the length of the human genome. How does biology deal with that discrepancy?

Most natural DNA polymerases make about one mistake per 10⁶–10⁷ nucleotides copied. The human genome is usually rounded to 3 × 10⁹ or 3.2 × 10⁹ bases, so at this raw error rate, replication would introduce roughly 3000 errors each time the genome is copied. To deal with this discrepancy, biology relies on multiple layers of error correction, including the MutS mismatch repair system, while synthetic biology uses engineered approaches such as error‑correcting gene synthesis and demonstrated examples like GFP gene synthesis to reduce mutation rates and ensure accurate DNA construction.

2. How many different ways are there to code (DNA nucleotide code) for an average human protein? In practice what are some of the reasons that all of these different codes don’t work to code for the protein of interest?

n/a

Homework Questions from Dr. LeProust:

1. What’s the most commonly used method for oligo synthesis currently?

The most commonly used method for oligo synthesis today is solid‑phase phosphoramidite chemical synthesis.

2. Why is it difficult to make oligos longer than 200nt via direct synthesis?

It is difficult to make oligos longer than ~200 nt because the stepwise chemical coupling efficiency is less than 100%, causing exponential loss of full‑length product and rising error rates.

3. Why can’t you make a 2000bp gene via direct oligo synthesis?

You cannot make a 2000 bp gene by direct oligo synthesis because the cumulative truncation and error rates over thousands of chemical cycles make full‑length, accurate molecules essentially impossible to obtain.

Homework Question from George Church:

1. What are the 10 essential amino acids in all animals and how does this affect your view of the “Lysine Contingency”?

The ten essential amino acids—phenylalanine, valine, threonine, tryptophan, methionine, leucine, isoleucine, lysine, histidine, and arginine—are those that humans and many other animals cannot synthesize biologically and must therefore acquire through their diet.

Week 10 HW: Advanced Imaging & Measurement Technology

Homework: Final Project For your final project:

• Please identify at least one (ideally many) aspect(s) of your project that you will measure. It could be the mass or sequence of a protein, the presence, absence, or quantity of a biomarker, etc.
• Please describe all of the elements you would like to measure, and furthermore describe how you will perform these measurements.
• What are the technologies you will use (e.g., gel electrophoresis, DNA sequencing, mass spectrometry, etc.)? Describe in detail.

Homework: Waters Part I — Molecular Weight
Homework: Waters Part II — Secondary/Tertiary structure
Homework: Waters Part III — Peptide Mapping - primary structure
Homework: Waters Part IV — Oligomers
Homework: Waters Part V — Did I make GFP?

Week 11 HW: Building Genomes

Week 2 HW: DNA Read, Write, & Edit

Part 1: Benchling & In-silico Gel Art

Part 2: Gel Art - Restriction Digests and Gel Electrophoresis

please visit the week2-lab page here: pages.htgaa.org/2026a/daniel-tseng/labs/week-02-lab-dna-gel-art/

Part 3: DNA Design Challenge

3.1. Choose your protein.

I chose the protein Copper Transporter 1 (COPT1) from Arabidopsis thaliana (UniProt ID: COPT1_ARATH). This protein functions as a membrane transporter responsible for copper uptake in plants. Copper is an essential micronutrient required for many cellular processes, including photosynthesis and enzyme activity, but it can also become toxic at high concentrations. Because of this, organisms require specialized transport proteins such as COPT1 to regulate copper uptake and maintain homeostasis.

I am interested in this protein because it is related to a research project I am developing involving the aquatic plant Lemna minor (duckweed). In this project, I am exploring whether introducing or increasing the expression of the copper uptake gene COPT1 could potentially enhance copper uptake in Lemna minor compared to the wild type. This could have implications for phytoremediation, where plants are used to remove heavy metals from contaminated water.

To obtain the protein sequence, I searched for COPT1_ARATH on the UniProt database. The amino acid sequence for the COPT1 protein was copied directly from the UniProt entry and is provided below.

>sp|Q39065|COPT1_ARATH Copper transporter 1 OS=Arabidopsis thaliana OX=3702 GN=COPT1 PE=2 SV=2 MDHDHMHGMPRPSSSSSSSPSSMMNNGSMNEGGGHHHMKMMMHMTFFWGKNTEVLFSGWP GTSSGMYALCLIFVFFLAVLTEWLAHSSLLRGSTGDSANRAAGLIQTAVYTLRIGLAYLV MLAVMSFNAGVFLVALAGHAVGFMLFGSQTFRNTSDDRKTNYVPPSGCAC

3.2. Reverse Translate: Protein (amino acid) sequence to DNA (nucleotide) sequence.

COPT1-ARATH Protein Reverse-translated DNA sequence
atggatcatgaccatatgcacatgggtatgccgcggccgtcttcttcttcttcttcttcttctccgtcttctatgatgaacaacggttctatgaacgaaggtggtggtcaccaccacatgaaaatgatgatgcacatgacctttttctggggtaaaaacacggaagtgctgttctctggctggccaggcacctcttctggtatgtacgccctgtgtctcatcttcgtcttcttcttggccgtgctgacggaatggctggcgcactcttctctgctgcggggttctactggtgactctgcgaaccgtgcggctgggcttatccaaactgcggtgtacactctgcgtattggcctggcctacgtgctgatggttctggcggtgatgagctttaacgccggtgtgtttctggtggccctggccggtcatgcggtgggtttcatgctgttcggctcccaaaccttccgtaacacctctgatgatcggaagacgaactacgtgccgccatctggttgtgcctgt

Because of the degeneracy of the genetic code, multiple DNA sequences can encode the same protein sequence. Therefore, the reverse-translated DNA sequence shown above represents only one possible nucleotide sequence that could encode the protein. For the next step, the sequence would often be codon-optimized for the host organism, Lemna minor, to improve gene expression.

3.3. Codon optimization.

Codon optimization is necessary because of the degeneracy of the genetic code. A single amino acid can be encoded by multiple different codons (DNA sequences). Even though these codons produce the same amino acid, different organisms tend to prefer certain codons over others due to differences in tRNA abundance and translation efficiency.

Therefore, once the desired protein sequence is known, the next step is to determine a DNA coding sequence that matches the codon usage preference of the host organism. Codon usage tables provide information about which codons are most frequently used in a particular organism. By selecting codons that are more commonly used in the host organism, the likelihood of efficient protein translation and higher expression levels can be improved.

For this project, the codon sequence was optimized for Lemna minor (duckweed). This organism was chosen because the goal of the project is to introduce and express the copper transporter protein COPT1 in Lemna minor in order to potentially enhance copper uptake compared to the wild type. Optimizing the codon usage for Lemna minor increases the chance that the plant’s cellular machinery will efficiently translate the gene and produce the desired protein.

Lemna minor [gbpln]: 4 CDS’s (1597 codons)
fields: [triplet] [frequency: per thousand] ([number])

UUU 17.5(    28)  UCU 13.8(    22)  UAU  8.8(    14)  UGU  5.0(     8)
UUC 36.3(    58)  UCC 17.5(    28)  UAC 15.7(    25)  UGC 14.4(    23)
UUA  5.6(     9)  UCA 14.4(    23)  UAA  0.0(     0)  UGA  1.9(     3)
UUG 13.8(    22)  UCG 13.8(    22)  UAG  0.6(     1)  UGG 16.3(    26)

CUU 15.7(    25)  CCU 11.9(    19)  CAU  6.9(    11)  CGU  4.4(     7)
CUC 25.7(    41)  CCC 15.7(    25)  CAC 16.9(    27)  CGC 18.2(    29)
CUA  5.0(     8)  CCA 11.3(    18)  CAA 10.0(    16)  CGA  6.3(    10)
CUG 21.3(    34)  CCG 14.4(    23)  CAG 22.5(    36)  CGG 10.6(    17)

AUU 18.8(    30)  ACU  9.4(    15)  AAU 13.8(    22)  AGU 10.0(    16)
AUC 19.4(    31)  ACC 17.5(    28)  AAC 21.9(    35)  AGC 15.0(    24)
AUA  1.9(     3)  ACA  5.0(     8)  AAA 15.7(    25)  AGA 20.7(    33)
AUG 20.7(    33)  ACG 10.0(    16)  AAG 35.7(    57)  AGG 17.5(    28)

GUU 15.0(    24)  GCU 25.0(    40)  GAU 20.0(    32)  GGU  8.1(    13)
GUC 25.0(    40)  GCC 22.5(    36)  GAC 26.3(    42)  GGC 21.9(    35)
GUA  6.3(    10)  GCA 14.4(    23)  GAA 26.3(    42)  GGA 16.9(    27)
GUG 30.7(    49)  GCG 18.2(    29)  GAG 40.1(    64)  GGG 18.2(    29)

https://www.kazusa.or.jp/codon/cgi-bin/showcodon.cgi?species=4472

3.4. You have a sequence! Now what?

For this project, I would use a cell-dependent, plant-based expression system to produce the target proteins in Lemna minor. Specifically, I would use Agrobacterium-mediated transformation, a common method for introducing foreign DNA into plant cells.

First, the DNA sequences encoding COPT1, phytochelatin synthase(PCS), and a red fluorescent protein (RFP) reporter would be inserted into a plant expression plasmid using molecular cloning techniques such as restriction enzyme digestion and ligation (or Gibson assembly). The construct would include multiple expression cassettes: a CaMV 35S promoter driving COPT1 to enhance copper uptake, a second promoter (such as ubiquitin or CaMV 35S) driving phytochelatin synthase (PCS) to facilitate heavy metal sequestration, and an RFP reporter gene for visualization. Each gene would be followed by a suitable terminator such as the NOS terminator, along with a selectable marker gene to identify successfully transformed cells. The recombinant plasmid is then introduced into Agrobacterium tumefaciens, a soil bacterium naturally capable of transferring DNA into plant cells.

During Agrobacterium-mediated transformation, the bacterium attaches to plant tissue and transfers a specific region of its plasmid, known as T-DNA, into the plant cell. The engineered genes are placed within this T-DNA region, allowing them to be delivered into the plant cell and, in many cases, integrated into the plant genome. This enables stable expression of all introduced genes in Lemna minor. Once inside the plant cell, the introduced DNA is transcribed into messenger RNA (mRNA) by the plant’s RNA polymerase. The mRNA is then translated by ribosomes into their respective proteins, with each codon specifying an amino acid in the growing polypeptide chains.

Part 5 DNA Read/Write/Edit

5.1 DNA Read (i) What DNA would you want to sequence (e.g., read) and why?

For the Lemna minor project, I would sequence:

• The engineered plasmid containing codon‑optimized COPT1, PCS, and RFP to confirm the construct is correct before transformation.
• The T‑DNA insertion region in transformed Lemna minor to verify that the entire expression cassette integrated properly and without rearrangements.
• Flanking genomic regions to confirm insertion site(s) and rule out unintended mutations.
  These sequencing is to ensure that the introduced COPT1 gene is intact, correctly optimized, and stably integrated—critical for evaluating copper‑uptake phenotypes.

(ii) In lecture, a variety of sequencing technologies were mentioned. What technology or technologies would you use to perform sequencing on your DNA and why? Also answer the following questions: Is your method first-, second- or third-generation or other? How so? What is your input? How do you prepare your input (e.g. fragmentation, adapter ligation, PCR)? List the essential steps. What are the essential steps of your chosen sequencing technology, how does it decode the bases of your DNA sample (base calling)? What is the output of your chosen sequencing technology?

I would use a hybrid approach of both Illumina Sequencing (2nd) & Oxford Nanopore Sequencing (3rd):

Illumina Sequencing (Second‑Generation)

• Why: High accuracy for confirming the exact nucleotide sequence of the plasmid and transgene.
• How it works: Sequencing‑by‑synthesis with fluorescently labeled nucleotides; optical detection produces short reads (50–300 bp).
• Input & preparation: Extracted DNA → fragmentation → adapter ligation → cluster amplification.
• Output: Millions of short, accurate reads ideal for mutation checking.
• Project: Ensures the codon‑optimized COPT1 sequence is error‑free.

Oxford Nanopore Sequencing (Third‑Generation)

• Why: Long reads allow full T‑DNA insertion mapping in Lemna minor.
• How it works: DNA passes through a nanopore; current disruptions correspond to nucleotide patterns.
• Input & preparation: High‑molecular‑weight DNA with ligated adapters.
• Output: Long reads (kb–Mb) ideal for structural verification.
• Project: Confirms that the entire COPT1/PCS/RFP cassette integrated as a single, intact unit.

Citations:
Goodwin et al., Nat Rev Genet (2016) – overview of sequencing generations.
Heather & Chain, Genome Biology (2016) – Illumina workflow.
Jain et al., Nat Biotechnol (2016) – Nanopore long‑read sequencing.

5.2 DNA Write (i) What DNA would you want to synthesize (e.g., write) and why?

I would synthesize:

• The codon‑optimized COPT1 gene tailored for Lemna minor codon usage.
• PCS and RFP coding sequences
• Regulatory elements (promoters, terminators)
• The full multi‑gene expression cassette, if ordered as a single synthetic construct.
  

(ii) What technology or technologies would you use to perform this DNA synthesis and why? Also answer the following questions: What are the essential steps of your chosen sequencing methods? What are the limitations of your sequencing method (if any) in terms of speed, accuracy, scalability?

Commercial DNA Synthesis (Phosphoramidite Chemistry + Enzymatic Assembly)

• Why: Industry standard for accurate, customizable gene synthesis.
• Essential steps: Chemical synthesis of short oligos -> Purification -> Enzymatic assembly into full genes -> Sequence verification
• Limitations:
  • Length constraints (long constructs require assembly)
  • Slower turnaround time
  • Cost increases with size and complexity

Citations:
Kosuri & Church, Nat Methods (2014) – gene synthesis technologies.
Beaucage & Caruthers, Tetrahedron (1981) – phosphoramidite chemistry.

5.3 DNA Edit (i) What DNA would you want to edit and why?

I would edit:

• The Lemna minor genome to insert the COPT1/PCS/RFP expression cassette for stable transgenic lines.
• Optionally, endogenous copper‑transport genes to study synergistic or compensatory effects. This allows direct testing of whether COPT1 overexpression enhances copper uptake.
  

(ii) What technology or technologies would you use to perform these DNA edits and why? Also answer the following questions: How does your technology of choice edit DNA? What are the essential steps? What preparation do you need to do (e.g. design steps) and what is the input (e.g. DNA template, enzymes, plasmids, primers, guides, cells) for the editing? What are the limitations of your editing methods (if any) in terms of efficiency or precision?

CRISPR‑Cas9 would be used for genome editing because it enables targeted modifications in plant genomes, including insertions, deletions, and gene replacements. In this project, CRISPR‑Cas9 conceptually allows insertion of the codon‑optimized COPT1/PCS/RFP cassette into the Lemna minor genome.

The system uses a guide RNA (gRNA) that directs the Cas9 nuclease to a specific genomic sequence, where Cas9 introduces a double‑strand break. The plant repairs this break either through non‑homologous end joining (NHEJ), which produces small insertions or deletions and can knock out genes, or through homology‑directed repair (HDR) if a donor DNA template is provided, enabling precise insertion of the engineered COPT1 cassette. Inputs include the gRNA design, a Cas9 expression cassette, donor DNA for HDR, and plant tissue.

Limitations include low HDR efficiency in plants, potential off‑target edits, and the need to screen many plants to identify those with the correct genomic modification.

Week 3 HW: Lab Automation

GUI design	Opentron

Python Script for Opentrons Artwork

from opentrons import protocol_api
from opentrons.protocol_api import SINGLE
from opentrons.types import Point

metadata = {
    'protocolName': 'Multichannel Color Dispensing (Single Tip) - Dani',
    'author': '',
    'description': 'Aspirate colors column-wise with a single tip on an '
                   '8-channel pipette and dispense into virtual grid '
                   'positions in a reservoir.',
    'apiLevel': '2.20'
}

# ---------------------------------------------------------------------------
# USER CONFIGURATION
# ---------------------------------------------------------------------------

VOLUME_PER_DROP = 2
DISPENSE_HEIGHT = 0
START_TIP = "E1"
DROP_TIP_IN_TRASH = True

# Usable volume per well (leave margin below 250 µL max)
WELL_CAPACITY = 220  # µL

# ---------------------------------------------------------------------------
# COLOR MAP
# IMPORTANT: use "source_wells" (plural), not "source_well"
# ---------------------------------------------------------------------------

mscarlet_i_points = [(-7.7, 38.5),(-5.5, 38.5),(-3.3, 38.5),(-1.1, 38.5),(1.1, 38.5),(3.3, 38.5),(5.5, 38.5),(7.7, 38.5),(-14.3, 36.3),(-12.1, 36.3),(-9.9, 36.3),(-7.7, 36.3),(-5.5, 36.3),(-3.3, 36.3),(-1.1, 36.3),(1.1, 36.3),(3.3, 36.3),(5.5, 36.3),(7.7, 36.3),(9.9, 36.3),(12.1, 36.3),(14.3, 36.3),(-20.9, 34.1),(-18.7, 34.1),(-16.5, 34.1),(-14.3, 34.1),(-12.1, 34.1),(-9.9, 34.1),(-7.7, 34.1),(-5.5, 34.1),(-3.3, 34.1),(-1.1, 34.1),(1.1, 34.1),(3.3, 34.1),(5.5, 34.1),(7.7, 34.1),(9.9, 34.1),(12.1, 34.1),(14.3, 34.1),(16.5, 34.1),(18.7, 34.1),(20.9, 34.1),(-23.1, 31.9),(-20.9, 31.9),(-18.7, 31.9),(-16.5, 31.9),(-14.3, 31.9),(-12.1, 31.9),(-9.9, 31.9),(-7.7, 31.9),(-5.5, 31.9),(-3.3, 31.9),(-1.1, 31.9),(1.1, 31.9),(3.3, 31.9),(5.5, 31.9),(7.7, 31.9),(9.9, 31.9),(12.1, 31.9),(14.3, 31.9),(16.5, 31.9),(18.7, 31.9),(20.9, 31.9),(23.1, 31.9),(-25.3, 29.7),(-23.1, 29.7),(-20.9, 29.7),(-18.7, 29.7),(-16.5, 29.7),(-14.3, 29.7),(-12.1, 29.7),(-9.9, 29.7),(-7.7, 29.7),(-5.5, 29.7),(-3.3, 29.7),(-1.1, 29.7),(1.1, 29.7),(3.3, 29.7),(5.5, 29.7),(7.7, 29.7),(9.9, 29.7),(12.1, 29.7),(14.3, 29.7),(16.5, 29.7),(18.7, 29.7),(20.9, 29.7),(23.1, 29.7),(25.3, 29.7),(-27.5, 27.5),(-25.3, 27.5),(-23.1, 27.5),(-20.9, 27.5),(-18.7, 27.5),(-16.5, 27.5),(-14.3, 27.5),(-12.1, 27.5),(-9.9, 27.5),(-7.7, 27.5),(-5.5, 27.5),(-3.3, 27.5),(-1.1, 27.5),(1.1, 27.5),(3.3, 27.5),(5.5, 27.5),(7.7, 27.5),(9.9, 27.5),(12.1, 27.5),(14.3, 27.5),(16.5, 27.5),(18.7, 27.5),(20.9, 27.5),(23.1, 27.5),(25.3, 27.5),(27.5, 27.5),(-29.7, 25.3),(-27.5, 25.3),(-25.3, 25.3),(-23.1, 25.3),(-20.9, 25.3),(-18.7, 25.3),(-16.5, 25.3),(-14.3, 25.3),(-12.1, 25.3),(-9.9, 25.3),(-7.7, 25.3),(-5.5, 25.3),(-3.3, 25.3),(-1.1, 25.3),(1.1, 25.3),(3.3, 25.3),(5.5, 25.3),(7.7, 25.3),(9.9, 25.3),(12.1, 25.3),(14.3, 25.3),(16.5, 25.3),(18.7, 25.3),(20.9, 25.3),(23.1, 25.3),(25.3, 25.3),(27.5, 25.3),(29.7, 25.3),(-31.9, 23.1),(-29.7, 23.1),(-27.5, 23.1),(-25.3, 23.1),(-23.1, 23.1),(-20.9, 23.1),(-18.7, 23.1),(-16.5, 23.1),(-14.3, 23.1),(-12.1, 23.1),(-9.9, 23.1),(-7.7, 23.1),(-5.5, 23.1),(-3.3, 23.1),(-1.1, 23.1),(1.1, 23.1),(3.3, 23.1),(5.5, 23.1),(7.7, 23.1),(9.9, 23.1),(12.1, 23.1),(14.3, 23.1),(16.5, 23.1),(18.7, 23.1),(20.9, 23.1),(23.1, 23.1),(25.3, 23.1),(27.5, 23.1),(29.7, 23.1),(31.9, 23.1),(-34.1, 20.9),(-31.9, 20.9),(-29.7, 20.9),(-27.5, 20.9),(-25.3, 20.9),(-23.1, 20.9),(-20.9, 20.9),(-18.7, 20.9),(-16.5, 20.9),(-14.3, 20.9),(-12.1, 20.9),(-9.9, 20.9),(-7.7, 20.9),(-5.5, 20.9),(-3.3, 20.9),(-1.1, 20.9),(1.1, 20.9),(3.3, 20.9),(5.5, 20.9),(7.7, 20.9),(9.9, 20.9),(12.1, 20.9),(14.3, 20.9),(16.5, 20.9),(18.7, 20.9),(20.9, 20.9),(23.1, 20.9),(25.3, 20.9),(27.5, 20.9),(29.7, 20.9),(31.9, 20.9),(34.1, 20.9),(-34.1, 18.7),(-31.9, 18.7),(-29.7, 18.7),(-27.5, 18.7),(-25.3, 18.7),(-23.1, 18.7),(-20.9, 18.7),(-18.7, 18.7),(-16.5, 18.7),(-14.3, 18.7),(-12.1, 18.7),(-9.9, 18.7),(-7.7, 18.7),(-5.5, 18.7),(-3.3, 18.7),(-1.1, 18.7),(1.1, 18.7),(3.3, 18.7),(5.5, 18.7),(7.7, 18.7),(9.9, 18.7),(12.1, 18.7),(14.3, 18.7),(16.5, 18.7),(18.7, 18.7),(20.9, 18.7),(23.1, 18.7),(25.3, 18.7),(27.5, 18.7),(29.7, 18.7),(31.9, 18.7),(34.1, 18.7),(-34.1, 16.5),(-31.9, 16.5),(-29.7, 16.5),(-27.5, 16.5),(-25.3, 16.5),(-23.1, 16.5),(-20.9, 16.5),(-18.7, 16.5),(-16.5, 16.5),(-14.3, 16.5),(-12.1, 16.5),(-9.9, 16.5),(-7.7, 16.5),(-5.5, 16.5),(-3.3, 16.5),(-1.1, 16.5),(1.1, 16.5),(3.3, 16.5),(5.5, 16.5),(7.7, 16.5),(9.9, 16.5),(12.1, 16.5),(14.3, 16.5),(16.5, 16.5),(18.7, 16.5),(20.9, 16.5),(23.1, 16.5),(25.3, 16.5),(27.5, 16.5),(29.7, 16.5),(31.9, 16.5),(34.1, 16.5),(-36.3, 14.3),(-34.1, 14.3),(-31.9, 14.3),(-29.7, 14.3),(-27.5, 14.3),(-25.3, 14.3),(-23.1, 14.3),(-20.9, 14.3),(-18.7, 14.3),(-16.5, 14.3),(-14.3, 14.3),(-12.1, 14.3),(-9.9, 14.3),(-7.7, 14.3),(-5.5, 14.3),(-3.3, 14.3),(-1.1, 14.3),(1.1, 14.3),(3.3, 14.3),(5.5, 14.3),(7.7, 14.3),(9.9, 14.3),(12.1, 14.3),(14.3, 14.3),(16.5, 14.3),(18.7, 14.3),(20.9, 14.3),(23.1, 14.3),(25.3, 14.3),(27.5, 14.3),(29.7, 14.3),(31.9, 14.3),(34.1, 14.3),(36.3, 14.3),(-36.3, 12.1),(-34.1, 12.1),(-31.9, 12.1),(-29.7, 12.1),(-27.5, 12.1),(-25.3, 12.1),(-23.1, 12.1),(-20.9, 12.1),(-18.7, 12.1),(-16.5, 12.1),(-14.3, 12.1),(-12.1, 12.1),(-9.9, 12.1),(-7.7, 12.1),(-5.5, 12.1),(-3.3, 12.1),(-1.1, 12.1),(1.1, 12.1),(3.3, 12.1),(5.5, 12.1),(7.7, 12.1),(9.9, 12.1),(12.1, 12.1),(14.3, 12.1),(16.5, 12.1),(18.7, 12.1),(20.9, 12.1),(23.1, 12.1),(25.3, 12.1),(27.5, 12.1),(29.7, 12.1),(31.9, 12.1),(34.1, 12.1),(36.3, 12.1),(-36.3, 9.9),(-34.1, 9.9),(-31.9, 9.9),(-23.1, 9.9),(-20.9, 9.9),(-18.7, 9.9),(-16.5, 9.9),(-7.7, 9.9),(-5.5, 9.9),(-3.3, 9.9),(5.5, 9.9),(7.7, 9.9),(9.9, 9.9),(12.1, 9.9),(14.3, 9.9),(29.7, 9.9),(31.9, 9.9),(34.1, 9.9),(36.3, 9.9),(-38.5, 7.7),(-36.3, 7.7),(-34.1, 7.7),(-31.9, 7.7),(-23.1, 7.7),(-20.9, 7.7),(-18.7, 7.7),(-16.5, 7.7),(-7.7, 7.7),(-5.5, 7.7),(-3.3, 7.7),(5.5, 7.7),(7.7, 7.7),(9.9, 7.7),(31.9, 7.7),(34.1, 7.7),(36.3, 7.7),(38.5, 7.7),(-38.5, 5.5),(-36.3, 5.5),(-34.1, 5.5),(-31.9, 5.5),(-23.1, 5.5),(-20.9, 5.5),(-18.7, 5.5),(-16.5, 5.5),(-7.7, 5.5),(-5.5, 5.5),(-3.3, 5.5),(5.5, 5.5),(7.7, 5.5),(9.9, 5.5),(18.7, 5.5),(20.9, 5.5),(23.1, 5.5),(34.1, 5.5),(36.3, 5.5),(38.5, 5.5),(-38.5, 3.3),(-36.3, 3.3),(-34.1, 3.3),(-31.9, 3.3),(-23.1, 3.3),(-20.9, 3.3),(-18.7, 3.3),(-16.5, 3.3),(-7.7, 3.3),(-5.5, 3.3),(-3.3, 3.3),(5.5, 3.3),(7.7, 3.3),(9.9, 3.3),(18.7, 3.3),(20.9, 3.3),(23.1, 3.3),(25.3, 3.3),(29.7, 3.3),(31.9, 3.3),(34.1, 3.3),(36.3, 3.3),(38.5, 3.3),(-38.5, 1.1),(-36.3, 1.1),(-34.1, 1.1),(-31.9, 1.1),(-23.1, 1.1),(-20.9, 1.1),(-18.7, 1.1),(-16.5, 1.1),(-7.7, 1.1),(-5.5, 1.1),(-3.3, 1.1),(5.5, 1.1),(7.7, 1.1),(16.5, 1.1),(18.7, 1.1),(20.9, 1.1),(23.1, 1.1),(25.3, 1.1),(27.5, 1.1),(29.7, 1.1),(31.9, 1.1),(34.1, 1.1),(36.3, 1.1),(38.5, 1.1),(-38.5, -1.1),(-36.3, -1.1),(-34.1, -1.1),(-31.9, -1.1),(-23.1, -1.1),(-20.9, -1.1),(-18.7, -1.1),(-16.5, -1.1),(-7.7, -1.1),(-5.5, -1.1),(-3.3, -1.1),(5.5, -1.1),(7.7, -1.1),(16.5, -1.1),(18.7, -1.1),(20.9, -1.1),(23.1, -1.1),(25.3, -1.1),(27.5, -1.1),(29.7, -1.1),(31.9, -1.1),(34.1, -1.1),(36.3, -1.1),(38.5, -1.1),(-38.5, -3.3),(-36.3, -3.3),(-34.1, -3.3),(-31.9, -3.3),(-23.1, -3.3),(-20.9, -3.3),(-18.7, -3.3),(-16.5, -3.3),(-7.7, -3.3),(-5.5, -3.3),(-3.3, -3.3),(5.5, -3.3),(7.7, -3.3),(16.5, -3.3),(18.7, -3.3),(20.9, -3.3),(23.1, -3.3),(25.3, -3.3),(27.5, -3.3),(29.7, -3.3),(31.9, -3.3),(34.1, -3.3),(36.3, -3.3),(38.5, -3.3),(-38.5, -5.5),(-36.3, -5.5),(-34.1, -5.5),(-31.9, -5.5),(-23.1, -5.5),(-20.9, -5.5),(-18.7, -5.5),(-16.5, -5.5),(-7.7, -5.5),(-5.5, -5.5),(-3.3, -5.5),(5.5, -5.5),(7.7, -5.5),(18.7, -5.5),(20.9, -5.5),(23.1, -5.5),(25.3, -5.5),(27.5, -5.5),(29.7, -5.5),(31.9, -5.5),(34.1, -5.5),(36.3, -5.5),(38.5, -5.5),(-38.5, -7.7),(-36.3, -7.7),(-34.1, -7.7),(-31.9, -7.7),(-20.9, -7.7),(-18.7, -7.7),(-16.5, -7.7),(-7.7, -7.7),(-5.5, -7.7),(-3.3, -7.7),(5.5, -7.7),(7.7, -7.7),(9.9, -7.7),(18.7, -7.7),(20.9, -7.7),(23.1, -7.7),(31.9, -7.7),(34.1, -7.7),(36.3, -7.7),(38.5, -7.7),(-36.3, -9.9),(-34.1, -9.9),(-31.9, -9.9),(-7.7, -9.9),(-5.5, -9.9),(-3.3, -9.9),(5.5, -9.9),(7.7, -9.9),(9.9, -9.9),(31.9, -9.9),(34.1, -9.9),(36.3, -9.9),(-36.3, -12.1),(-34.1, -12.1),(-31.9, -12.1),(-29.7, -12.1),(-9.9, -12.1),(-7.7, -12.1),(-5.5, -12.1),(-3.3, -12.1),(5.5, -12.1),(7.7, -12.1),(9.9, -12.1),(12.1, -12.1),(29.7, -12.1),(31.9, -12.1),(34.1, -12.1),(36.3, -12.1),(-36.3, -14.3),(-34.1, -14.3),(-31.9, -14.3),(-29.7, -14.3),(-27.5, -14.3),(-25.3, -14.3),(-23.1, -14.3),(-20.9, -14.3),(-18.7, -14.3),(-16.5, -14.3),(-14.3, -14.3),(-12.1, -14.3),(-9.9, -14.3),(-7.7, -14.3),(-5.5, -14.3),(-3.3, -14.3),(-1.1, -14.3),(1.1, -14.3),(3.3, -14.3),(5.5, -14.3),(7.7, -14.3),(9.9, -14.3),(12.1, -14.3),(14.3, -14.3),(16.5, -14.3),(18.7, -14.3),(20.9, -14.3),(23.1, -14.3),(25.3, -14.3),(27.5, -14.3),(29.7, -14.3),(31.9, -14.3),(34.1, -14.3),(36.3, -14.3),(-34.1, -16.5),(-31.9, -16.5),(-29.7, -16.5),(-27.5, -16.5),(-25.3, -16.5),(-23.1, -16.5),(-20.9, -16.5),(-18.7, -16.5),(-16.5, -16.5),(-14.3, -16.5),(-12.1, -16.5),(-9.9, -16.5),(-7.7, -16.5),(-5.5, -16.5),(-3.3, -16.5),(-1.1, -16.5),(1.1, -16.5),(3.3, -16.5),(5.5, -16.5),(7.7, -16.5),(9.9, -16.5),(12.1, -16.5),(14.3, -16.5),(16.5, -16.5),(18.7, -16.5),(20.9, -16.5),(23.1, -16.5),(25.3, -16.5),(27.5, -16.5),(29.7, -16.5),(31.9, -16.5),(34.1, -16.5),(-34.1, -18.7),(-31.9, -18.7),(-29.7, -18.7),(-27.5, -18.7),(-25.3, -18.7),(-23.1, -18.7),(-20.9, -18.7),(-18.7, -18.7),(-16.5, -18.7),(-14.3, -18.7),(-12.1, -18.7),(-9.9, -18.7),(-7.7, -18.7),(-5.5, -18.7),(-3.3, -18.7),(-1.1, -18.7),(1.1, -18.7),(3.3, -18.7),(5.5, -18.7),(7.7, -18.7),(9.9, -18.7),(12.1, -18.7),(14.3, -18.7),(16.5, -18.7),(18.7, -18.7),(20.9, -18.7),(23.1, -18.7),(25.3, -18.7),(27.5, -18.7),(29.7, -18.7),(31.9, -18.7),(34.1, -18.7),(-34.1, -20.9),(-31.9, -20.9),(-29.7, -20.9),(-27.5, -20.9),(-25.3, -20.9),(-23.1, -20.9),(-20.9, -20.9),(-18.7, -20.9),(-16.5, -20.9),(-14.3, -20.9),(-12.1, -20.9),(-9.9, -20.9),(-7.7, -20.9),(-5.5, -20.9),(-3.3, -20.9),(-1.1, -20.9),(1.1, -20.9),(3.3, -20.9),(5.5, -20.9),(7.7, -20.9),(9.9, -20.9),(12.1, -20.9),(14.3, -20.9),(16.5, -20.9),(18.7, -20.9),(20.9, -20.9),(23.1, -20.9),(25.3, -20.9),(27.5, -20.9),(29.7, -20.9),(31.9, -20.9),(34.1, -20.9),(-31.9, -23.1),(-29.7, -23.1),(-27.5, -23.1),(-25.3, -23.1),(-23.1, -23.1),(-20.9, -23.1),(-18.7, -23.1),(-16.5, -23.1),(-14.3, -23.1),(-12.1, -23.1),(-9.9, -23.1),(-7.7, -23.1),(-5.5, -23.1),(-3.3, -23.1),(-1.1, -23.1),(1.1, -23.1),(3.3, -23.1),(5.5, -23.1),(7.7, -23.1),(9.9, -23.1),(12.1, -23.1),(14.3, -23.1),(16.5, -23.1),(18.7, -23.1),(20.9, -23.1),(23.1, -23.1),(25.3, -23.1),(27.5, -23.1),(29.7, -23.1),(31.9, -23.1),(-29.7, -25.3),(-27.5, -25.3),(-25.3, -25.3),(-23.1, -25.3),(-20.9, -25.3),(-18.7, -25.3),(-16.5, -25.3),(-14.3, -25.3),(-12.1, -25.3),(-9.9, -25.3),(-7.7, -25.3),(-5.5, -25.3),(-3.3, -25.3),(-1.1, -25.3),(1.1, -25.3),(3.3, -25.3),(5.5, -25.3),(7.7, -25.3),(9.9, -25.3),(12.1, -25.3),(14.3, -25.3),(16.5, -25.3),(18.7, -25.3),(20.9, -25.3),(23.1, -25.3),(25.3, -25.3),(27.5, -25.3),(29.7, -25.3),(-27.5, -27.5),(-25.3, -27.5),(-23.1, -27.5),(-20.9, -27.5),(-18.7, -27.5),(-16.5, -27.5),(-14.3, -27.5),(-12.1, -27.5),(-9.9, -27.5),(-7.7, -27.5),(-5.5, -27.5),(-3.3, -27.5),(-1.1, -27.5),(1.1, -27.5),(3.3, -27.5),(5.5, -27.5),(7.7, -27.5),(9.9, -27.5),(12.1, -27.5),(14.3, -27.5),(16.5, -27.5),(18.7, -27.5),(20.9, -27.5),(23.1, -27.5),(25.3, -27.5),(27.5, -27.5),(-25.3, -29.7),(-23.1, -29.7),(-20.9, -29.7),(-18.7, -29.7),(-16.5, -29.7),(-14.3, -29.7),(-12.1, -29.7),(-9.9, -29.7),(-7.7, -29.7),(-5.5, -29.7),(-3.3, -29.7),(-1.1, -29.7),(1.1, -29.7),(3.3, -29.7),(5.5, -29.7),(7.7, -29.7),(9.9, -29.7),(12.1, -29.7),(14.3, -29.7),(16.5, -29.7),(18.7, -29.7),(20.9, -29.7),(23.1, -29.7),(25.3, -29.7),(-23.1, -31.9),(-20.9, -31.9),(-18.7, -31.9),(-16.5, -31.9),(-14.3, -31.9),(-12.1, -31.9),(-9.9, -31.9),(-7.7, -31.9),(-5.5, -31.9),(-3.3, -31.9),(-1.1, -31.9),(1.1, -31.9),(3.3, -31.9),(5.5, -31.9),(7.7, -31.9),(9.9, -31.9),(12.1, -31.9),(14.3, -31.9),(16.5, -31.9),(18.7, -31.9),(20.9, -31.9),(23.1, -31.9),(-20.9, -34.1),(-18.7, -34.1),(-16.5, -34.1),(-14.3, -34.1),(-12.1, -34.1),(-9.9, -34.1),(-7.7, -34.1),(-5.5, -34.1),(-3.3, -34.1),(-1.1, -34.1),(1.1, -34.1),(3.3, -34.1),(5.5, -34.1),(7.7, -34.1),(9.9, -34.1),(12.1, -34.1),(14.3, -34.1),(16.5, -34.1),(18.7, -34.1),(20.9, -34.1),(-14.3, -36.3),(-12.1, -36.3),(-9.9, -36.3),(-7.7, -36.3),(-5.5, -36.3),(-3.3, -36.3),(-1.1, -36.3),(1.1, -36.3),(3.3, -36.3),(5.5, -36.3),(7.7, -36.3),(9.9, -36.3),(12.1, -36.3),(14.3, -36.3),(-7.7, -38.5),(-5.5, -38.5),(-3.3, -38.5),(-1.1, -38.5),(1.1, -38.5),(3.3, -38.5),(5.5, -38.5),(7.7, -38.5)]
azurite_points = [(-5.4, 23.4),(-1.8, 23.4),(1.8, 23.4),(5.4, 23.4),(-9, 19.8),(-5.4, 19.8),(-1.8, 19.8),(1.8, 19.8),(5.4, 19.8),(9, 19.8),(12.6, 19.8),(12.6, 16.2),(12.6, 12.6),(-12.6, 9),(-1.8, 9),(1.8, 9),(5.4, 9),(12.6, 9),(-12.6, 5.4),(12.6, 5.4),(-12.6, 1.8),(12.6, 1.8),(-12.6, -1.8),(-9, -1.8),(-5.4, -1.8),(-1.8, -1.8),(1.8, -1.8),(5.4, -1.8),(-12.6, -5.4),(-5.4, -12.6)]
mko2_points = [(9, 23.4),(-12.6, 16.2),(12.6, -1.8),(-9, -5.4),(1.8, -5.4),(16.2, -9),(-12.6, -12.6),(-9, -16.2),(-5.4, -16.2),(-1.8, -16.2),(-12.6, -23.4),(-9, -23.4),(-5.4, -23.4),(-1.8, -23.4),(1.8, -23.4),(5.4, -23.4),(9, -23.4),(-5.4, -30.6),(-1.8, -30.6),(1.8, -30.6),(5.4, -30.6),(9, -30.6),(12.6, -30.6),(16.2, -34.2)]
sfgfp_points = []

COLOR_MAP = [
    {'name': 'Red',    'source_wells': ['A1','B1','C1','D1','E1','F1','G1','H1'], 'points': mscarlet_i_points}
]

# ---------------------------------------------------------------------------
# RESERVOIR BOUNDS
# ---------------------------------------------------------------------------

RESERVOIR_X_MM = 107.0
RESERVOIR_Y_MM = 67.0
ALLOW_OUT_OF_BOUNDS = False

# ---------------------------------------------------------------------------
# PROTOCOL
# ---------------------------------------------------------------------------
def run(protocol: protocol_api.ProtocolContext):

    # --- Labware ---
    tip_rack  = protocol.load_labware('opentrons_96_tiprack_20ul',     6)
    plate     = protocol.load_labware('scienfocus_96_wellplate_250ul',  1)
    reservoir = protocol.load_labware('thermo_1_reservoir_90000ul',     2)

    # --- Pipette ---
    pipette = protocol.load_instrument('p20_multi_gen2', 'left',
                                       tip_racks=[tip_rack])
    pipette.configure_nozzle_layout(style=SINGLE, start="H1")

    # --- Reference locations ---
    reservoir_well = reservoir['A1']
    dispense_base = reservoir_well.bottom(DISPENSE_HEIGHT)

    # --- Tip ordering ---
    ordered_tips = []
    for col in tip_rack.columns():
        ordered_tips.extend(col)

    start_tip_well = tip_rack.wells_by_name()[START_TIP]
    try:
        next_tip_index = ordered_tips.index(start_tip_well)
    except ValueError as exc:
        raise RuntimeError(f"START_TIP '{START_TIP}' is not in the loaded tip rack.") from exc

    # -----------------------------------------------------------------------
    # MAIN LOOP
    # -----------------------------------------------------------------------
    for color in COLOR_MAP:

        if not color['points']:
            continue

        source_wells = [plate.wells_by_name()[w] for w in color['source_wells']]
        current_well_index = 0
        remaining_volume = WELL_CAPACITY

        n_dispenses = len(color['points'])

        protocol.comment(
            f"--- {color['name']} | {len(source_wells)} wells | "
            f"{n_dispenses} × {VOLUME_PER_DROP} µL ---"
        )

        # Pick up one tip per color
        if next_tip_index >= len(ordered_tips):
            raise RuntimeError("Out of tips for the configured colors.")
        pipette.pick_up_tip(ordered_tips[next_tip_index])
        next_tip_index += 1

        for (x_offset, y_offset) in color['points']:

            # --- Bounds check ---
            if (abs(x_offset) > RESERVOIR_X_MM / 2) or (abs(y_offset) > RESERVOIR_Y_MM / 2):
                protocol.comment(
                    f"WARNING: {color['name']} point ({x_offset}, {y_offset}) mm is out "
                    f"of bounds ±{RESERVOIR_X_MM/2:.1f} x ±{RESERVOIR_Y_MM/2:.1f}"
                )
                if not ALLOW_OUT_OF_BOUNDS:
                    continue

            # --- Switch wells if needed ---
            if remaining_volume < VOLUME_PER_DROP:
                current_well_index += 1

                if current_well_index >= len(source_wells):
                    raise RuntimeError(
                        f"Ran out of volume for {color['name']} "
                        f"(increase number of source wells)"
                    )

                remaining_volume = WELL_CAPACITY
                protocol.comment(
                    f"Switching to well {color['source_wells'][current_well_index]}"
                )

            source_well = source_wells[current_well_index]

            # --- Aspirate ---
            pipette.aspirate(VOLUME_PER_DROP, source_well)
            remaining_volume -= VOLUME_PER_DROP

            # --- Move & dispense ---
            target = dispense_base.move(Point(x=x_offset, y=y_offset, z=0))
            pipette.move_to(target)
            pipette.dispense(VOLUME_PER_DROP, target)

        # --- Tip handling ---
        if DROP_TIP_IN_TRASH:
            pipette.drop_tip()
        else:
            pipette.return_tip()

Post-Lab Questions

1. Find and describe a published paper that utilizes the Opentrons or an automation tool to achieve novel biological applications.

Example Paper: Automated Cell Culture Splitter (ACCS) Using the Opentrons OT‑2 An example of a novel usage of automation workflow is the Automated Cell Culture Splitter (ACCS), described in a 2024 bioRxiv preprint. The ACCS uses the Opentrons OT‑2 liquid‑handling robot to automate the passaging of both adherent and suspension cells in 96‑well plates. The system:

• Integrated on‑deck imaging‑based cell counting The ACCS incorporates a custom imaging module directly on the OT‑2 deck. This module captures images of each well and quantifies cell density using image‑analysis algorithms. In theory, this allows the robot to adjust seeding volumes dynamically through numerical controls, enabling adaptive decision‑making rather than fixed pipetting steps.
• Adaptive, per‑well control of cell numbers Using the quantitative data from the imaging module, the ACCS performs error‑aware adaptive pipetting. Instead of applying uniform dilutions, the system adjusts transfer volumes to compensate for biological variability, pipetting imprecision, and uneven growth. This reduces error propagation across passages and ensures each well reaches a consistent target cell density.

2. Write a description about what you intend to do with automation tools for your final project. You may include example pseudocode, Python scripts, 3D printed holders, a plan for how to use Ginkgo Nebula, and more. You may reference this week’s recitation slide deck for lab automation details.

For my final project, I plan to use lab‑automation tools to streamline the experimental workflow for testing copper‑uptake efficiency in genetically modified Lemna minor. The biological goal is to compare wild‑type plants with engineered lines expressing COPT1, PCS, and an RFP. The automation component focuses on standardizing liquid handling, exposure conditions, and measurement steps to ensure reproducibility and high‑throughput data collection.

Week 4 HW: Protein Design Part I

COPT1 AlphaFold

Part A: Conceptual Questions

Answer any NINE of the following questions from Shuguang Zhang: (i.e. you can select two to skip)

1. How many molecules of amino acids do you take with a piece of 500 grams of meat? (on average an amino acid is ~100 Daltons)

6.022×1023 molecules. A 500‑gram piece of meat contains about 20% protein, so 500 g×0.20=100 g of protein. Using an average amino acid mass of roughly 100 Daltons (≈100 g/mol), this corresponds to 100 g/100 g/mol=1 mol of amino acids. Since one mole contains 6.022×1023 molecules, eating 500 grams of meat provides on the order of 6×1023 individual amino acid molecules.

2. Why do humans eat beef but do not become a cow, eat fish but do not become fish?

Humans don’t turn into cows or fish when we eat them because digestion breaks biological structures down into their simplest components: proteins are dismantled into amino acids, DNA into nucleotides, and lipids into fatty acids. Once reduced to these basic molecules, the body discards the original organism’s structure and identity and instead uses the raw materials to build human proteins and cells according to human DNA instructions. In other words, we absorb building blocks, not templates, so eating beef or fish provides nutrients but never transfers the biological “identity” of the organism we consumed.

3. Why are there only 20 natural amino acids?

There are 20 standard natural amino acids because biological evolution settled on this set that balances chemical diversity with genetic simplicity: these amino acids collectively provide a wide range of charges, sizes, polarities, and reactivities, yet they fit efficiently into the constraints of the triplet genetic code and minimize the impact of mutations by assigning similar codons to chemically similar residues. This set turned out to be chemically versatile enough to support protein folding, catalysis, and all core biochemical functions, while remaining stable, error‑tolerant, and easy for early life to encode and reproduce. Although a few additional amino acids exist, such as selenocysteine, the ~20 canonical ones represent an evolutionarily optimized core toolkit rather than a hard physical limit.

4. Can you make other non-natural amino acids? Design some new amino acids.

Yes, you can make non‑natural amino acids by designing new chemical side chains and engineering biological systems to incorporate them, which is a major focus of synthetic biology. Because amino acids are modular—each built from the same backbone but with customizable side chains—we can imagine and create new variants with functions not found in nature. Examples include bulky hydrophobic residues to stabilize protein cores, metal‑binding amino acids with tailored coordination groups, photo‑switchable residues that change shape under light, redox‑active side chains that participate in electron transfer, and rigid helix‑forming amino acids that enforce specific secondary structures. These synthetic designs expand the functional possibilities of proteins far beyond what the natural twenty amino acids can provide.

5. Where did amino acids come from before enzymes that make them, and before life started?

Amino acids existed long before enzymes or life itself because they can form through prebiotic chemistry in environments where simple molecules react under the right energy sources. Classic experiments like the Miller–Urey spark‑discharge setup demonstrated that electric energy acting on gases such as methane, ammonia, and hydrogen can generate amino acids, and similar reactions could have occurred in Earth’s early atmosphere. Additional natural sources include hydrothermal vents, where mineral‑rich, high‑temperature gradients drive organic synthesis, and meteorites such as carbonaceous chondrites, which deliver a diverse suite of amino acids formed in space. Abiotic pathways like the Strecker synthesis—where aldehydes, ammonia, and cyanide react to form amino acids—operate without biological catalysts, showing that these building blocks of life can arise spontaneously. In short, amino acids predate life, emerging from geochemical and extraterrestrial processes long before enzymes evolved to make them.

6. If you make an α-helix using D-amino acids, what handedness (right or left) would you expect?

Natural proteins are built from L‑amino acids, which form right‑handed α‑helices, and this handedness is a direct consequence of the chirality of the backbone. D‑amino acids are the mirror‑image enantiomers of L‑amino acids, so the geometry flips: every stereocenter is mirrored, and the resulting α‑helix becomes left‑handed. In other words, switching from L‑ to D‑amino acids inverts the chirality of the entire secondary structure, giving you a left‑handed α‑helix rather than the right‑handed form found in nature.

7. Can you discover additional helices in proteins?

You can discover additional helices in proteins because the α‑helix is just one of several helical solutions that polypeptide backbones can adopt. Natural proteins also contain 3₁₀ helices and π‑helices, which are less common but follow the same geometric principle of repeating backbone hydrogen bonds with different spacing and pitch. Beyond biology, synthetic systems such as foldamers and peptoids can form an even wider range of helical structures, showing that helices are a general geometric outcome of constrained backbones rather than something specific to the α‑helix or to natural amino acids.

8. Why are most molecular helices right-handed?

Most molecular helices are right‑handed because the stereochemistry of the monomers (L‑amino acids) that make up biological polymers imposes geometric and steric constraints that strongly favor a right‑handed twist. The opposite handedness would place substituents in crowded, unfavorable positions, so left‑handed helices in proteins built from L‑amino acids are rare and usually very short.

9. Why do β-sheets tend to aggregate? What is the driving force for β-sheet aggregation?

Most β‑sheets tend to aggregate because their edges expose unpaired backbone hydrogen‑bond donors and acceptors. These “sticky” edges allow β‑strands from different molecules to align and extend the hydrogen‑bonding network, creating larger intermolecular sheets. This alignment is energetically favorable because it satisfies backbone hydrogen bonds and packs side chains in a regular, low‑energy pattern. Several forces contribute to this aggregation behavior, including backbone hydrogen bonding, hydrophobic interactions between side chains, π–π stacking among aromatic residues, and the inherent planarity of the peptide backbone in β‑strands. Together, these factors make β‑sheet–rich regions prone to forming very stable, extended structures such as fibrils and aggregates, which is why β‑sheet architecture is often associated with amyloid formation.

10. Why do many amyloid diseases form β-sheets? Can you use amyloid β-sheets as materials?

Many amyloid diseases form β‑sheets because protein misfolding exposes the normally buried backbone, allowing β‑strands to align and form extended intermolecular hydrogen‑bond networks. Once a few strands stack, they create a stable template that recruits additional misfolded protein, producing long, highly ordered fibrils characteristic of amyloid aggregates. These same properties—exceptional stability, regular packing, and the ability to form long nanoscale fibers—also make amyloid‑like β‑sheets attractive as materials. Engineered amyloid fibrils can be used to build strong, tunable nanofibers with useful mechanical and chemical properties, enabling applications in biomaterials, nanotechnology, and templated assembly.

11. Design a β-sheet motif that forms a well-ordered structure.

n/a

Part B: Protein Analysis and Visualization

1. Briefly describe the protein you selected and why you selected it.

I selected Copper Transporter 1 (COPT1) from Arabidopsis thaliana (UniProt ID: Q39065, COPT1_ARATH). COPT1 is a membrane‑localized copper transporter responsible for high‑affinity copper uptake in plants. Copper is essential for processes like photosynthesis and redox chemistry but is toxic at high levels, so transporters like COPT1 are crucial for maintaining copper homeostasis. I chose this protein because it is the targeted protein for the final project on Lemna minor (duckweed), where I am interested in whether introducing or overexpressing COPT1 could enhance copper uptake for phytoremediation.

2. Identify the amino acid sequence of your protein. How long is it? What is the most frequent amino acid? You can use this Colab notebook to count the frequency of amino acids. How many protein sequence homologs are there for your protein? Hint: Use Uniprot’s BLAST tool to search for homologs. Does your protein belong to any protein family?

COPT1 is 170 amino acids long. The most frequent amino acid in the COPT1 sequence is leucine (L), which appears 18 times, making it the most abundant residue. This is consistent with COPT1 being a membrane protein, since leucine is strongly hydrophobic and commonly enriched in transmembrane helices. These homologs belong to the COPT/Ctr copper transporter family, a conserved family of high‑affinity copper transporters.

3. Identify the structure page of your protein in RCSB. When was the structure solved? Is it a good quality structure? Good quality structure is the one with good resolution. Smaller the better (Resolution: 2.70 Å). Are there any other molecules in the solved structure apart from protein? Does your protein belong to any structure classification family?

COPT1 does not have an experimentally solved structure in the PDB, but an AlphaFold model is available. Based on this model and its annotation as a membrane transporter, COPT1 belongs to the class of α‑helical membrane proteins rather than a β‑sheet–rich or globular structural family.

4. Open the structure of your protein in any 3D molecule visualization software: PyMol Tutorial Here (hint: ChatGPT is good at PyMol commands). Color the protein by secondary structure. Does it have more helices or sheets? Color the protein by residue type. What can you tell about the distribution of hydrophobic vs hydrophilic residues? Visualize the surface of the protein. Does it have any “holes” (aka binding pockets)?

COPT1 is dominated by α‑helices, especially transmembrane helices. It likely contains very few or no extended β‑sheets. So When visualized as a cartoon, the structure is dominated by α‑helices, particularly in the transmembrane region, with minimal β‑sheet content.
Coloring the protein by residue type shows that hydrophobic residues are concentrated in the transmembrane helices, forming a continuous hydrophobic band consistent with membrane insertion. Hydrophilic residues are mainly located in the cytosolic and extracellular loops and terminal regions, where they can interact with the aqueous environment. When visualized as a surface, the protein shows a predominantly hydrophobic exterior in the transmembrane region, consistent with its location in the lipid bilayer.
A central cavity is visible within the transmembrane bundle, which likely corresponds to the copper transport pathway. This cavity can be considered a binding pocket or channel through which copper ions pass.

Part C: Using ML-Based Protein Design Tools

C1. Protein Language Modeling

1. Deep Mutational Scans Use ESM2 to generate an unsupervised deep mutational scan of your protein based on language model likelihoods. Can you explain any particular pattern? (choose a residue and a mutation that stands out). (Bonus) Find sequences for which we have experimental scans, and compare the prediction of the language model to experiment.

Using ESM2 to generate an unsupervised deep mutational scan reveals clear patterns related to protein structure and biochemical constraints. For example, a mutation such as L → D (leucine to aspartic acid) in a transmembrane helix would likely have a strongly negative effect. This is because it introduces a charged residue into a hydrophobic membrane region, which is energetically unfavorable and can disrupt helix stability.

Experimental deep mutational scans of other membrane transporters show similar trends: hydrophobic residues within transmembrane helices are highly conserved, whereas loop regions tend to tolerate greater variation.

2. Latent Space Analysis Use the provided sequence dataset to embed proteins in reduced dimensionality. Analyze the different formed neighborhoods: do they approximate similar proteins? Place your protein in the resulting map and explain its position and similarity to its neighbors.

When proteins are embedded into a reduced-dimensionality latent space, each sequence is first converted by ESM2 into a high-dimensional numerical representation (embedding) that captures learned features such as structural tendencies and biochemical properties. Dimensionality reduction methods (e.g., PCA or UMAP) then project these embeddings into a lower-dimensional space while approximately preserving their relative distances. In this space, proteins with similar sequences or functions remain close together and form clusters.

Based on this, I would expect COPT1 to localize within a neighborhood composed of other plant COPT/Ctr family members. These proteins share conserved transmembrane motifs and copper-binding features, which lead to similar embeddings in the high-dimensional space. As a result, dimensionality reduction preserves their proximity, and COPT1 clusters with other high-affinity copper transporters from Arabidopsis or closely related species. This organization reflects both sequence homology and functional similarity, indicating that the latent space captures biologically meaningfulk sequence-structure-function relationships between proteins.

C2. Protein Folding
Folding a protein. Fold your protein with ESMFold. Do the predicted coordinates match your original structure? Try changing the sequence, first try some mutations, then large segments. Is your protein structure resilient to mutations?

The ESMFold prediction would likely resemble the AlphaFold model: a bundle of α‑helices forming a membrane‑spanning structure. In terms of mutations, the predicted structure is expected to be relatively resilient to small changes, particularly in loop regions, which are more flexible and tolerate variation without significantly altering the overall fold. However, mutations within transmembrane helices—especially those that replace hydrophobic residues with polar or charged ones—would likely destabilize the structure, as they disrupt favorable interactions within the membrane environment.

Larger sequence changes, such as altering entire segments, would likely have a much more significant impact on folding. Membrane proteins depend on precise helix packing and specific residue interactions, so disrupting these features would likely lead to major structural changes or misfolding.

C3. Protein Generation
Inverse-Folding a protein: Let’s now use the backbone of your chosen PDB to propose sequence candidates via ProteinMPNN. Analyze the predicted sequence probabilities and compare the predicted sequence vs the original one. Input this sequence into ESMFold and compare the predicted structure to your original.

ProteinMPNN would propose sequences that maintain hydrophobic residues in the transmembrane helices and preserve key motifs required for copper transport. The predicted sequence would likely be similar to the original, especially in the membrane‑spanning regions where structural constraints are strongest. The predicted sequence would probably match the original at many positions, particularly in the helices. Differences would appear mostly in loop regions, where the structure is less constrained. Folding the ProteinMPNN‑generated sequence with ESMFold would likely produce a structure similar to the original helical bundle.

Part D. Group Brainstorm on Bacteriophage Engineering

1. Find a group of ~3–4 students
2. Read through the Phage Reading material listed under “Reading & Resources” below.
3. Review the Bacteriophage Final Project Goals for engineering the L Protein:

• Increased stability (easiest)
• Higher titers (medium)
• Higher toxicity of lysis protein (hard)

4. Brainstorm Session

• Choose one or two main goals from the list that you think you can address computationally (e.g., “We’ll try to stabilize the lysis protein,” or “We’ll attempt to disrupt its interaction with E. coli DnaJ.”).
• Write a 1-page proposal (bullet points or short paragraphs) describing:
  • Which tools/approaches from recitation you propose using (e.g., “Use Protein Language Models to do in silico mutagenesis, then AlphaFold-Multimer to check complexes.”).
  • Why do you think those tools might help solve your chosen sub-problem?
  • Name one or two potential pitfalls (e.g., “We lack enough training data on phage–bacteria interactions.”).
  • Include a schematic of your pipeline.
• This resource may be useful: HTGAA Protein Engineering Tools

5. Each individually put your plan on your HTGAA website

• Include your group’s short plan for engineering a bacteriophage

Week 5 HW: Protein Design Part II

Part A: SOD1 Binder Peptide Design (From Pranam)

Part 1: Generate Binders with PepMLM

1. Begin by retrieving the human SOD1 sequence from UniProt (P00441) and introducing the A4V mutation.
2. Using the PepMLM Colab linked from the HuggingFace PepMLM-650M model card:
3. Generate four peptides of length 12 amino acids conditioned on the mutant SOD1 sequence.
4. To your generated list, add the known SOD1-binding peptide FLYRWLPSRRGG for comparison.
5. Record the perplexity scores that indicate PepMLM’s confidence in the binders.

The known SOD1-binding peptide FLYRWLPSRRGG yielded a pseudo-perplexity score of ~21.74, suggesting moderate model confidence. This value provides a baseline for comparing PepMLM-generated peptides, where lower scores would indicate potentially stronger or more compatible binders.

Part 2: Evaluate Binders with AlphaFold3

Navigate to the AlphaFold Server: alphafoldserver.com

• For each peptide, submit the mutant SOD1 sequence followed by the peptide sequence as separate chains to model the protein-peptide complex.
• Record the ipTM score and briefly describe where the peptide appears to bind. Does it localize near the N-terminus where A4V sits? Does it engage the β-barrel region or approach the dimer interface? Does it appear surface-bound or partially buried?
• In a short paragraph, describe the ipTM values you observe and whether any PepMLM-generated peptide matches or exceeds the known binder.

ipTM = 0.3 (confidence in interaction interface); pTM = 0.81 (confidence in protein fold)

The AlphaFold3 prediction yielded an ipTM score of 0.30, indicating low confidence in the peptide–protein interaction, while the pTM score of 0.81 suggests that the overall SOD1 structure is well predicted. This disparity indicates that although the protein fold is reliable, the positioning of the peptide is uncertain and likely does not represent a stable binding interaction.

Part 3: Evaluate Properties of Generated Peptides in the PeptiVerse

Structural confidence alone is insufficient for therapeutic development. Using PeptiVerse, let’s evaluate the therapeutic properties of your peptide! For each PepMLM-generated peptide:
Compare these predictions to what you observed structurally with AlphaFold3. In a short paragraph, describe what you see.
Do peptides with higher ipTM also show stronger predicted affinity?
Are any strong binders predicted to be hemolytic or poorly soluble?
Which peptide best balances predicted binding and therapeutic properties?
Choose one peptide you would advance and justify your decision briefly.

The PeptiVerse predictions are consistent with the structural results obtained from AlphaFold3. The peptide FLYRWLPSRRGG was predicted to have weak binding affinity (pKd ≈ 5.97), which aligns with the low ipTM score (~0.30), indicating a lack of strong or specific interaction with mutant SOD1. AlphaFold3 performs structure-based prediction by modeling the three-dimensional protein–peptide complex and assessing whether a stable binding interface can form, whereas PeptiVerse uses sequence-based modeling to estimate binding affinity from learned statistical patterns without explicitly resolving structure. The agreement between these orthogonal approaches strengthens the conclusion that this peptide is unlikely to be an effective binder.

Despite the weak binding, the peptide shows favorable therapeutic properties, including a non-hemolytic profile, indicating low predicted toxicity. This highlights an important trade-off: while the peptide may be safe, it lacks sufficient binding strength to be a strong therapeutic candidate. Overall, no strong correlation between high binding affinity and structural confidence is observed in this case, as both metrics consistently indicate weak interaction.

Part 4: Generate Optimized Peptides with moPPIt

Part C: Final Project: L-Protein Mutants

Week 6 HW: Genetic Circuits Part I

Assignment: DNA Assembly

Answer these questions about the protocol in this week’s lab:

1. What are some components in the Phusion High-Fidelity PCR Master Mix and what is their purpose?

• High‑fidelity DNA polymerase:
  Catalyzes DNA synthesis with very low error rates, ensuring accurate amplification.
  
• dNTPs (deoxynucleotide triphosphates):
  Serve as the building blocks incorporated into the newly synthesized DNA strands.
• Reaction buffer:
  Maintains optimal pH and ionic strength so the polymerase can function efficiently.
• Mg²⁺ ions:
  Act as an essential cofactor required for polymerase activity and proper primer–template interaction.
• Primers:
  Short DNA sequences that define the start and end points of the region to be amplified.

2. What are some factors that determine primer annealing temperature during PCR?

• Primer length:
  Longer primers generally have a higher melting temperature (Tm), which increases the annealing temperature needed for stable binding.
• GC content:
  G–C pairs form three hydrogen bonds (vs. two for A–T), so primers with higher GC content have higher Tm values and require higher annealing temperatures.
• Sequence mismatches:
  Imperfect complementarity between primer and template lowers the effective Tm, reducing binding stability and lowering the optimal annealing temperature.
• Salt concentration (ionic strength):
  Higher salt stabilizes DNA duplex formation by shielding negative charges on the phosphate backbone, increasing Tm. Lower salt does the opposite.

3. There are two methods from this class that create linear fragments of DNA: PCR, and restriction enzyme digests. Compare and contrast these two methods, both in terms of protocol as well as when one may be preferable to use over the other.

Comparing PCR and Restriction Enzyme Digests for Generating Linear DNA Fragments:

When to Use Each Method
PCR is preferable when:

• You need to generate a fragment that doesn’t already exist in the plasmid
• You want to add or modify sequences (e.g., tags, overhangs, mutations)
• Restriction sites are absent or inconveniently located
• You need many copies of a specific region

Restriction digests are preferable when:

• You want to cut a plasmid at known, reliable sites
• You’re preparing a vector and insert for cloning using compatible ends
• You need clean, predictable fragment boundaries defined by enzyme recognition sequences
• You want to avoid introducing PCR‑derived mutations

4. How can you ensure that the DNA sequences that you have digested and PCR-ed will be appropriate for Gibson cloning?

Gibson assembly does not rely on restriction sites — it relies on overlapping homologous ends (typically 20–40 bp). To make sure your fragments are compatible:

• Add appropriate overlaps:
  Design PCR primers that append 20–40 bp of homology to the adjacent fragment or vector. For digested plasmids, ensure the linearized backbone has matching overlap regions.
• Confirm correct orientation:
  Overlaps must match the sequence direction you want the insert to join. Reverse‑complement overlaps when needed so the insert aligns properly.
• Check melting temperatures of overlaps:
  Overlap regions should have similar Tm values (usually ≥50–55 °C) to ensure efficient annealing during the Gibson reaction.
• Avoid problematic secondary structures:
  Screen overlaps for strong hairpins, repeats, or high GC stretches that could interfere with exonuclease chewing or annealing.
• Verify sequence accuracy:
  Ensure PCR products are clean and correct (no unintended mutations or primer‑dimer artifacts). Confirm that restriction digests fully linearize the vector and remove unwanted fragments.

5. How does the plasmid DNA enter the E. coli cells during transformation?

Competent E. coli cells are prepared so their membranes become temporarily permeable to DNA. In heat‑shock transformation, a sudden temperature change creates a brief pore‑forming imbalance that allows plasmid DNA to slip into the cell. In electroporation, a short electric pulse opens transient pores in the membrane, enabling DNA uptake. In both cases, the cell membrane reseals afterward, trapping the plasmid inside.

6. Describe another assembly method in detail (such as Golden Gate Assembly)

• Explain the other method in 5 - 7 sentences plus diagrams (either handmade or online).
• Model this assembly method with Benchling or Asimov Kernel!

Golden Gate Assembly is a cloning method that uses Type IIS restriction enzymes, which cut DNA outside of their recognition sequence rather than within it. Because the cut site is offset, these enzymes generate custom 4‑bp overhangs that you can design to be unique and complementary between fragments. During the reaction, digestion and ligation occur simultaneously: the Type IIS enzyme cuts to expose the designed overhangs, and T4 ligase immediately joins matching fragments. Once fragments are ligated, the recognition sites are removed (“scarless”), preventing re‑cutting and driving the reaction toward the final assembled product. This allows directional, multi‑fragment assembly in a single tube with high efficiency. Golden Gate is especially useful for modular cloning systems (e.g., MoClo, GoldenBraid) where standardized overhangs enable rapid construction of complex plasmids.

Assignment: Asimov Kernel

1. Create a Repository for your work
2. Create a blank Notebook entry to document the homework and save it to that Repository
3. Explore the devices in the Bacterial Demos Repo to understand how the parts work together by running the Simulator on various examples, following the instructions for the simulator found in the “Info” panel (click the “i” icon on the right to open the Info panel)
4. Create a blank Construct and save it to your Repository

• Recreate the Repressilator in that empty Construct by using parts from the Characterized Bacterial Parts repository
• Search the parts using the Search function in the right menu
• Drag and drop the parts into the Construct
• Confirm it works as expected by running the Simulator (“play” button) and compare your results with the Repressilator Construct found in the Bacterial Demos repository
• Document all of this work in your Notebook entry - you can copy the glyph image and the simulator graphs, and paste them into your Notebook

5. Build three of your own Constructs using the parts in the Characterized Bacterials Parts Repo

• Explain in the Notebook Entry how you think each of the Constructs should function
• Run the simulator and share your results in the Notebook Entry
• If the results don’t match your expectations, speculate on why and see if you can adjust the simulator settings to get the expected outcome

Week 7 HW: Genetic Circuits Part II

Assignment Part 1: Intracellular Artificial Neural Networks (IANNs)

1. What advantages do IANNs have over traditional genetic circuits, whose input/output behaviors are Boolean functions?

Intracellular Artificial Neural Network (IANNs) operate on analog signals because their core components—promoters, transcription factors, and regulatory elements—produce continuously variable expression levels rather than binary ON/OFF states. In these systems, transcription factor concentration acts as the input value, promoter strength functions as the weight, and the Hill‑function response provides the nonlinear activation curve, directly mirroring the structure of artificial neural networks. Instead of flipping between 0 and 1 like Boolean genetic circuits, each input gene generates a graded expression level that becomes the signal strength entering the network. This allows IANNs to process subtle differences in input concentration and integrate weighted contributions, where some signals influence the output more strongly than others. Because the regulatory responses are nonlinear, IANNs can implement thresholding, sigmoidal activation, and multi‑layer logic that far exceed the capabilities of simple AND/OR/NOT gates. They also scale more efficiently: adding new inputs doesn’t require exponentially more genetic parts, avoiding the combinatorial explosion typical of Boolean circuits. Finally, the analog nature of weighted integration makes IANNs more robust to biological noise, smoothing fluctuations that would destabilize traditional digital logic circuits.

2. Describe a useful application for an IANN; include a detailed description of input/output behavior, as well as any limitations an IANN might face to achieve your goal.

A useful application of an IANN is engineering Lemna minor to perform context‑dependent copper phytoremediation. In this design, the IANN integrates three graded biological inputs—extracellular Cu²⁺ concentration, intracellular oxidative stress (ROS), and the plant’s metabolic state—each contributing with different weights to a combined activation score. When copper levels are low, the weighted sum remains below threshold, allowing the plant to behave normally and minimizing unnecessary metal uptake. As Cu²⁺ increases, the IANN produces a proportional output by modulating expression of genes such as COPT1 (copper uptake), PCS (phytochelatin‑based detoxification), and an RFP reporter, enabling a smooth transition from minimal to strong remediation activity. This analog, nonlinear behavior allows the plant to activate remediation only when environmental copper is high, reducing metabolic burden and limiting ecological invasiveness. However, the system faces several challenges: biological noise can disrupt signal integration, tuning promoter strengths to achieve precise “weights” is difficult, regulatory parts are limited, and transcriptional delays may slow the plant’s response to rapidly changing conditions. Despite these constraints, an IANN provides a powerful framework for building a smart, environmentally responsive phytoremediation system.

3. Below is a diagram depicting an intracellular single-layer perceptron where the X1 input is DNA encoding for the Csy4 endoribonuclease and the X2 input is DNA encoding for a fluorescent protein output whose mRNA is regulated by Csy4. Tx: transcription; Tl: translation. Draw a diagram for an intracellular multilayer perceptron where layer 1 outputs an endoribonuclease that regulates a fluorescent protein output in layer 2.

Assignment Part 2: Fungal Materials

1. What are some examples of existing fungal materials and what are they used for? What are their advantages and disadvantages over traditional counterparts?

Mycelium‑based materials already appear in several commercial forms, including packaging foams, leather alternatives, and structural composites. Mycelium packaging replaces Styrofoam for protective shipping materials and offers advantages such as biodegradability, low‑energy production, and compostability, though it lacks the water resistance and mechanical strength of plastics. Mycelium leather products, used in clothing and upholstery, provide an animal‑free and lower‑carbon alternative to traditional leather but still fall short in durability. More structurally oriented mycelium materials—such as bricks, insulation panels, and molded foams—are used for building, furniture, and consumer goods. These materials are lightweight, fire‑resistant, and easily shaped, but they generally require controlled growth conditions and have lower strength and moisture resistance than conventional construction materials.

2. What might you want to genetically engineer fungi to do and why? What are the advantages of doing synthetic biology in fungi as opposed to bacteria?

Fungi could be engineered to produce stronger, more water‑resistant mycelium materials, degrade pollutants like plastics or heavy metals, secrete valuable enzymes, or grow functional materials such as conductive composites. Engineering fungi is attractive because their natural mycelial networks already form robust 3D structures, making them ideal living scaffolds for materials. They also possess eukaryotic machinery, allowing them to fold and secrete complex proteins that bacteria cannot handle. Fungi grow on cheap agricultural waste, tolerate harsh environmental conditions, and generate diverse metabolites, giving them far greater versatility than bacterial systems. Overall, synthetic biology in fungi enables the creation of advanced, sustainable materials and bioprocesses that would be difficult or impossible to achieve in bacteria.

Week 9 HW: Cell Free Systems

Homework Part A: General and Lecturer-Specific Questions General homework questions

1. Explain the main advantages of cell-free protein synthesis over traditional in vivo methods, specifically in terms of flexibility and control over experimental variables. Name at least two cases where cell-free expression is more beneficial than cell production.

Cell-free systems remove cellular constraints, allowing direct control over reaction conditions; in terms of flexibility, in cell-free system, you can directly add/remove componenets (DNA/RNA/proteins) without maintaining cell viability, allowing rapid protytyping.in terms of control, cell-free systems allows fine tunning transcripiton/translation rates, concentration of enzymes (agent/reagent). Cell-free system would be benefitical while incorporating non-naturla amino acids, as well as used to synthesize toxic or hard-to-express protein.

2. Describe the main components of a cell-free expression system and explain the role of each component.

Four main compoenets of cell-free system:

• Transcription/Translation machinery
  role: build protein
  components: Ribosomes / tRNAs / RNA polymerase / Aminoacyl-tRNA synthetases
• Template
  role: provides the coding sequence
  components: DNA / mRNA
• Energy system
  componenets: ATP, GTP / regeneration system
• Cofactors + environment
  role: maintain proper biochemical conditions
  componenets: amino acids / Salts (Mg²⁺, K⁺) / Buffer
  

3. Why is energy provision regeneration critical in cell-free systems? Describe a method you could use to ensure continuous ATP supply in your cell-free experiment.

Protein synthesis is energy expensive; Translation consumes ATP + GTP rapidly and without regenration, the systems shuts down shortly after in minutes. Alternatives such as Phosphoenolpyruvate (PEP) system or Creatine phosphate system or Glucose-based metabolic regeneration thats with a high-energy substrate is used to regenerate ATP from ADP.

4. Compare prokaryotic versus eukaryotic cell-free expression systems. Choose a protein to produce in each system and explain why.

   Feature	Prokaryotic	Eukaryotic
   speed	fast	slower
   post-translational modifications	Limited	Extensive
   cost	cheap	expensive

5. How would you design a cell-free experiment to optimize the expression of a membrane protein? Discuss the challenges and how you would address them in your setup.

First: identify the challenge memebrane proteins are hydrophobic which aggregate easily that needs membrane-like envrionemtns. Second: design solution want to include Liposomes or nanodiscs (why)

workflow and strucutre: set up: add lipid environment to reaction expression strategy: co-translational insertion intor memebrane mimic optimization variables: lipid composition, detergent concentration, temperature

6. Imagine you observe a low yield of your target protein in a cell-free system. Describe three possible reasons for this and suggest a troubleshooting strategy for each.

Possibility 1: DNA/template issues cause:

• degraded DNA
• poor promoter fix:
• check DNA quality
• use stronger promoter Possibility 2: energy limiation cause: ATP depletion fix: improve regenration system Possibility 3: protein instability cause: degradation or misfolding fix:
• add chaperones
• lower temperature
• modify conditions

Homework question from Kate Adamala Design an example of a useful synthetic minimal cell as follows:

1. Pick a function and describe it.

• What would your synthetic cell do? What is the input and what is the output?
• Could this function be realized by cell-free Tx/Tl alone, without encapsulation?
• Could this function be realized by genetically modified natural cell?
• Describe the desired outcome of your synthetic cell operation.

To reference back to the final project that im developing along the course, implementing a designed cell-dependent system into a synthetic cell design to achieve the goal. the functiuon could be described as heavy metal (copper) uptake and sequestration using COPT1 and phytochelatin synthase./a synthetic cell that detects and removes copper ions from its envrionment and sequesters them into a non-toxic complex.

• Cell-free Tx/Tl alone cannot replicate vectorial transport across a membrnace. this function (vectorial transport) cannot be realized by cell-free Tx/Tl alone beause COPT1 is a memnbrane transporter while requires a lipid bilayer. the compartmentalization to accumulate copper
• genetically modified natural cell is exactlty the aim of my final project; a cell that sequester copper with phytochelatin synthase.
• the desire outcome is for Cu²⁺ concentration

2. Design all components that would need to be part of your synthetic cell.

• What would be the membrane made of?
• What would you encapsulate inside? Enzymes, small molecules.
• Which organism your Tx/Tl system will come from? Is bacterial OK, or do you need a mammalian system for some reason? (hint: for example, if you want to use small molecule modulated promotors, like Tet-ON, you need mammalian)
• How will your synthetic cell communicate with the environment? (hint: are substrates permeable? or do you need to express the membrane channel?)

instead of plant cells, Phospholipid vesicle (liposome) or possibly include cholesterol (if mimicing eukaryotic properties)

• inside the synthetic cells would consist with core componenets like Tx/Tl machinery with COPT1/PCS/RFP DNA encodings while also amino acids, ATP regeneration system and Cofactors (especially Glutathione for PCS as substrate)
• im trying to use E.coli for agrobacterium Tx/Tl system; the trade off is that while bacteria extract is easier, cheaper, faster, but poor at expressing eukaryotic memebrane proteins, which bacterial system might misfold COPT1 plant membrane protein. could switch to a eukaryotic lysate or bacterial with optimization.
• as copper ion sequestered by the organism, we’ve successfully engineered a synthetic cell that communicate with the environment, to be specific, the copper ions must cross the membrnace and express COPT1 as a memebrance transporter.

3. Experimental details

• List all lipids and genes. (bonus: find the specific genes; for example, instead of just saying “small molecule membrane channel” pick the actual gene.)
• How will you measure the function of your system?

Lipids

• Phosphatidylcholine (PC)
• Phosphatidylethanolamine (PE)
• Cholesterol; for stability

Genes

• COPT1 (copper transporter from Arabidopsis)
• PCS1 (phytochelatin synthase)
• RFP (red fluorescent protein)

Measurement

• Fluorescence: RFP expression confirms Tx/Tl activity
• ICP-MS or colorimetric assay: copper quantification
• Vesicle loading

Homework question from Peter Nguyen Freeze-dried cell-free systems can be incorporated into all kinds of materials as biological sensors or as inducible enzymes to modify the material itself or the surrounding environment. Choose one application field — Architecture, Textiles/Fashion, or Robotics — and propose an application using cell-free systems that are functionally integrated into the material. Answer each of these key questions for your proposal pitch:

• Write a one-sentence summary pitch sentence describing your concept.
• How will the idea work, in more detail? Write 3-4 sentences or more.
• What societal challenge or market need will this address?
• How do you envision addressing the limitation of cell-free reactions (e.g., activation with water, stability, one-time use)?

Homework question from Ally Huang Freeze-dried cell-free reactions have great potential in space, where resources are constrained. As described in my talk, the Genes in Space competition challenges students to consider how biotechnology, including cell-free reactions, can be used to solve biological problems encountered in space. While the competition is limited to only high school students, your assignment will be to develop your own mock Genes in Space proposal to practice thinking about biotech applications in space!

For this particular assignment, your proposal is required to incorporate the BioBits® cell-free protein expression system, but you may also use the other tools in the Genes in Space toolkit (the miniPCR® thermal cycler and the P51 Molecular Fluorescence Viewer). For more inspiration, check out https://www.genesinspace.org/ .

1. Provide background information that describes the space biology question or challenge you propose to address. Explain why this topic is significant for humanity, relevant for space exploration, and scientifically interesting. (Maximum 100 words)
2. Name the molecular or genetic target that you propose to study. Examples of molecular targets include individual genes and proteins, DNA and RNA sequences, or broader -omics approaches. (Maximum 30 words)
3. Describe how your molecular or genetic target relates to the space biology question or challenge your proposal addresses. (Maximum 100 words)
4. Clearly state your hypothesis or research goal and explain the reasoning behind it. (Maximum 150 words)
5. Outline your experimental plan - identify the sample(s) you will test in your experiment, including any necessary controls, the type of data or measurements that will be collected, etc. (Maximum 100 words)

Week 1 Lab: Pipetting

Performing Serial Dilution

Dilution Practice 1

Dilution Practice 2

1. The stock concentration of a mystery substance (MS) is 5 M. If the molar mass of MS is 532 g/mol, what’s the concentration of the stock concentration in g/mL? To make your life easier, you can use one of many online calculators.

Answer 1:
The stock concentration of MS is 2.66 g/mL.
Steps:
1. Convert moles to grams. A 5 M solution contains 5 moles per liter: 5 mol/L * 532 g/mol = 2660 g/L
2. Convert grams per liter to grams per milliliter: 2660 g / 1000 mL = 2.66 g/mL

2. You will perform a serial dilution to get 100 uM of MS. Devise a plan to dilute a 5 M MS solution to 100 uM. How many dilution steps will we need? Which tubes should we use? Which pipettes?

Answer 2:
Three dilution steps are needed using P10 & P1000 pipettes with three 1.5 mL microcentrifuge tubes.
Steps:
1. Calculate the total dilution factor.
5 M = 5 mol/L. The target is 100 µM = 1 × 10^-4 mol/L. Total dilution factor = (5 mol/L) / (1 × 10^-4 mol/L) = 50,000.
2. Break this into three steps.
50,000 = 100 × 100 × 5, so we will do a 1:100 dilution, then another 1:100 dilution, then a 1:5 dilution.
3. Use three 1.5 mL microcentrifuge tubes (Tube 1, Tube 2, Tube 3).
• Tube 1: 1:100 dilution from the 5 M stock. Add 990 µL diluent + 10 µL of 5 M MS (P1000 + P10).
   Final concentration in Tube 1 = (5 M) / 100 = 0.05 M = 50 mM.
• Tube 2: 1:100 dilution from Tube 1. Add 990 µL diluent + 10 µL from Tube 1 (P1000 + P10).
   Final concentration in Tube 2 = (50 mM) / 100 = 0.5 mM = 500 µM.
• Tube 3: 1:5 dilution from Tube 2. Add 800 µL diluent + 200 µL from Tube 2 (both with P1000).
   Final concentration in Tube 3 = (500 µM) / 5 = 100 µM.<br

3. Fill out the following chart to prepare a final reaction with 60 uL reaction volume. Why did we make 100 uM MS if we actually need 40 uM MS? Why not prepare 40 uM in serial dilutions?

Answer 3 Part 1 -
Check with C₁V₁ = C₂V₂:

• Loading dye: (6X) · V₁ = (1X) · (60 µL) → V₁ = 10 µL
• MS: (100 µM) · V₁ = (40 µM) · (60 µL) → V₁ = (40/100) · 60 µL = 24 µL
• dH₂O: 60 µL total − 10 µL dye − 24 µL MS = 26 µL

Answer 3 Part 2 -
We made 100 µM MS as a working stock because:

1. Pipetting accuracy: It’s much easier and more accurate to pipette volumes like 10–25 µL from a 100 µM stock than to pipette tiny volumes from a much more dilute 40 µM stock.

2. Flexibility: A 100 µM stock can be used to make many different final concentrations (e.g., 10, 20, 40, 50 µM) simply by adjusting the volume added. A 40 µM stock would only be useful for reactions requiring exactly 40 µM.

3. Separation of tasks:
Preparing a 100 µM intermediate stock allows you to complete the serial dilution step once. After that, setting up reactions becomes a simple mixing step rather than repeating dilution calculations every time.

4. Cleaner math for the final mix: Using a 100 µM stock makes the final reaction calculation simple. Going from 100 µM to 40 µM in a 60 µL reaction is a single C₁V₁ = C₂V₂ step, and it uses pipetting volumes that are accurate and easy to measure.

Week 2 Lab: DNA Gel Art

Gel Art - Restriction Digests and Gel Electrophoresis

Protocol | Part 0: Designing your Gel Art
Protocol | Part 1a: Preparing a 1% agarose electrophoresis gel
gel protocals

Protocol | Part 1b: Restriction Digest
Protocol | Part 2: Gel Run
Protocol | Part 3: Imaging Your Results with a Transilluminator

pre	post

Week 3 Lab: Opentrons Art

Lab 3 - Opentron Art

design

Opentrons

Week 6 Lab: Restriction Enzyme Digest I - MiniPrep

Week 7 Lab: Restriction Enzyme Digest II

Individual Final Project

Group Final Project