https://pages.htgaa.org/2026a/flavoris-belue/homework/week-09-cell-free-systems/index.htmlGeneral and Lecturer-Specific Questions General homework questions 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 protein synthesis has a big advantage over in vivo methods because it gives you direct control over the reaction environment without needing to keep cells alive. You can precisely tune things like DNA concentration, energy sources, cofactors, salts, and even add or remove specific components in real time, which is much harder to do inside living cells where metabolism and regulation get in the way. It’s also faster since you skip cloning, transformation, and cell growth steps. This makes it especially useful for expressing toxic proteins that would kill or stress cells, and for rapid prototyping or screening large libraries of genetic constructs where you want quick, iterative testing without waiting on cultures to grow. Describe the main components of a cell-free expression system and explain the role of each component. A cell-free expression system is mainly made up of a cell extract, a DNA template, and a reaction mix that supports transcription and translation. The cell extract provides the core molecular machinery, like ribosomes, tRNAs, aminoacyl-tRNA synthetases, transcription and translation factors, which are all needed to actually make protein. The DNA template contains the gene of interest along with the regulatory sequences needed for expression. The reaction mix supplies the raw materials and energy needed to drive the system, including amino acids, nucleotides, salts, cofactors, ATP regeneration components, and buffering agents to keep conditions stable. Together, these components recreate the basic protein production machinery of a cell, but in a much more controllable format. 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. Energy provision and regeneration are critical in cell-free systems because transcription and translation burn through ATP and GTP fast, so without a way to replenish that energy, protein synthesis stalls. Since there are no living cells to continuously regenerate energy through metabolism, the reaction depends entirely on whatever energy system you build into it. Basically, if the reaction runs out of usable energy, the whole system stalls, so energy regeneration is what keeps protein production going for longer and improves overall yield. One common way to maintain ATP supply is to include an energy regeneration substrate such as phosphoenolpyruvate (PEP), which can be used to help regenerate ATP during the reaction. In the reaction, PEP transfers a phosphate group to ADP through the enzyme pyruvate kinase, which regenerates ATP that can then be used to keep transcription and translation going. Compare prokaryotic versus eukaryotic cell-free expression systems. Choose a protein to produce in each system and explain why. Prokaryotic and eukaryotic cell-free systems each have different strengths depending on the type of protein being produced. Prokaryotic systems, like E. coli extracts, are usually faster, cheaper, and great for making simple proteins that do not need complex folding or post-translational modifications. In contrast, eukaryotic cell-free systems are better for proteins that require more advanced folding, disulfide bond formation, or modifications that bacteria cannot do well. For a prokaryotic system, a strong candidate would be Luz (luciferase) from the fungal bioluminescence pathway, since it is a relatively compact enzyme that folds well in bacterial extracts and does not require eukaryotic post-translational modifications; producing it cell-free would allow rapid screening of variants and direct assay of luminescence activity by simply adding the 3-hydroxyhispidin substrate to the reaction. For a eukaryotic system, a suitable target would be H3H (hispidin-3-hydroxylase) or another upstream enzyme in the caffeic acid–to–luciferin pathway, since these fungal oxidative enzymes often depend on proper folding, cofactor incorporation, and a eukaryotic redox environment to remain active. Expressing the pathway enzymes in their appropriate systems enables modular prototyping of the bioluminescence circuit before committing to stable plant transformation. 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. To optimize expression of a membrane protein in a cell-free system, I would design the reaction so it not only makes the protein but also gives it a membrane-like environment to fold into correctly. One of the main challenges with membrane proteins is that they tend to misfold, aggregate, or precipitate because their hydrophobic regions do not stay stable in plain aqueous solution. To deal with that, I would test conditions that include detergents, liposomes, or nanodiscs so the protein has somewhere to insert during or right after translation. I would also optimize variables like magnesium concentration, temperature, reaction time, and DNA concentration, since these can strongly affect yield and folding quality. On top of that, I would check expression using something like SDS-PAGE or a tagged reporter, then compare solubility and activity across conditions. 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. Low protein yield in a cell-free reaction can arise from numerous sources, but three common causes are the following. First, degradation of the DNA template or mRNA transcript by nucleases present in the extract can sharply reduce output. This can be addressed by switching from linear PCR products to circular plasmid DNA, adding RNase inhibitors, and verifying template integrity by gel electrophoresis before use. Second, depletion of energy substrates or accumulation of inhibitory byproducts such as inorganic phosphate can stall translation mid-reaction. This is best addressed by switching to a more robust energy regeneration system (e.g., PEP/pyruvate kinase), adjusting the starting concentrations of NTPs and amino acids, and running time course sampling to identify when the reaction plateaus. Third, poor translation efficiency caused by suboptimal codon usage, weak ribosome binding site strength, or mRNA secondary structure near the start codon can limit ribosome loading. This can be addressed by codon optimizing the gene for the extract source, redesigning the 5’ UTR and RBS using established calculators, and introducing silent mutations to disrupt inhibitory secondary structures near the translation initiation site. Homework questions from Kate Adamala Design an example of a useful synthetic minimal cell as follows:Hugoen