Making Human Insulin in E. coli: Why the First Clone Failed
A state research laboratory is asked to help make human insulin affordable. The team clones the human insulin gene from a genomic DNA library into a pBR322-derived plasmid carrying an ampicillin-resistance marker and a single EcoRI site, then transforms E. coli K-12. Induced cultures give a band of the expected size on Western blot, but the protein is immunologically inactive and is not processed. A second attempt uses cDNA prepared from mRNA of pancreatic islet beta cells. The plasmid marker is used to select transformants, but the production medium is being designed for a product intended for human injection, and a reviewer objects to the use of ampicillin throughout. The team also proposes injecting the bacterial lysate directly, arguing that purification is expensive, and wants to know why the C-peptide cannot simply be left in.
1. Explain why the genomic clone produced an inactive protein and why the cDNA strategy is better. Note any remaining problems with expressing cDNA in E. coli.
Read the worked answer
Why the genomic clone failed.
- The human insulin gene contains introns. E. coli has no spliceosome and cannot remove them, so the transcript is not correctly processed.
- The protein made would be preproinsulin, carrying the signal peptide and C-peptide; in a bacterium it folds without the correct disulphide bonds, so it is the right size on a gel but biologically inactive.
- The human promoter and other eukaryotic regulatory signals are not recognised by bacterial RNA polymerase.
Why cDNA is better. cDNA is made by reverse transcriptase from mature beta-cell mRNA, so it contains only exons - no introns and no need for splicing.
What still must be added or solved:
- A bacterial promoter and ribosome-binding site must be fused upstream, since the eukaryotic promoter is useless in E. coli.
- The cDNA still encodes preproinsulin; the signal peptide and C-peptide must be dealt with.
- E. coli cannot form the disulphide bonds reliably; correct folding requires the A and B chains to be produced separately and joined chemically, or the use of a secretory host such as yeast.
- Codon usage, plasmid stability and proteolysis all affect yield.
So cDNA solves the intron problem but not the processing problem - the two must be answered separately.
2. Explain the normal processing of insulin and judge whether the C-peptide can be left in the product.
Read the worked answer
Normal pathway.
- Ribosomes on the rough ER make preproinsulin (signal peptide + A chain + C-peptide + B chain).
- The signal peptide is cleaved in the ER → proinsulin, which folds and forms disulphide bonds.
- In the Golgi and secretory granules, endopeptidases remove the C-peptide, yielding mature insulin (A chain + B chain joined by disulphide bridges) plus free C-peptide, both released in equimolar amounts.
Judgement on C-peptide. It cannot be left in a therapeutic product:
- Mature insulin's three-dimensional structure and receptor binding depend on the A and B chains being correctly linked; proinsulin has only a few percent of insulin's biological activity.
- Proinsulin and residual bacterial proteins are immunogenic; they provoke antibody formation, allergy and, at worst, neutralising antibodies that make dosing unpredictable.
- Regulatory approval requires a product identical or highly similar to human insulin; animal insulins differ by a few residues and were themselves a reason for immune reactions.
The correct design is therefore to express A and B chains separately (often as fusion proteins), purify them, and join them by controlled disulphide bond formation, or to use a host that secretes correctly processed insulin.
On injecting the lysate directly: this must be refused. Bacterial endotoxin, host proteins and plasmid products would cause pyrogenic and allergic reactions; purification and quality control are not optional cost items but part of the safety of the product.
3. Assess the objection to ampicillin in the production medium and describe how transformants should be screened.
Read the worked answer
The objection is valid. The plasmid carries bla, encoding β-lactamase, so ampicillin is destroyed as the culture grows, meaning selection weakens; more importantly, a product for human injection should not be manufactured in a medium containing an antibiotic that could leave residues or select for resistance genes. A marker-free or non-β-lactam system is preferred: for example, a kanamycin marker with strict removal during purification, an auxotrophic complementation marker (a host lacking an essential gene, complemented by the plasmid), or a selection system that can be cured before the production run.
Screening and confirmation of transformants, in order:
- Plate on selective medium - only cells carrying the plasmid grow.
- Distinguish recombinants from empty plasmids by insertional inactivation of a marker (for example lacZ blue/white screening, or loss of a second resistance gene) followed by replica plating onto the two media.
- Confirm the insert by colony PCR, restriction mapping of the isolated plasmid, and sequencing of the cloned cDNA - size and orientation both matter for expression.
- Confirm expression after induction by SDS-PAGE and Western blot, then by bioassay of the purified product.
- Finally, quality control on the purified protein: identity, potency, sterility, endotoxin and host-protein contamination.