The shape of the lesson
By the end of the lesson, learners can
- Three ways to get the gene: Pick each source in turn and press Run.
- Building and checking a recombinant plasmid: Press Next step to work through all five stages.
- The toolkit in one table: Every gene transfer uses the same small set of tools.
- Source: donor DNA (restriction enzyme), mRNA (reverse transcriptase + DNA polymerase) or chemical synthesis.
- Vector: plasmid cut with the same enzyme; ligase seals it.
- Switch: a promoter the host recognises, so the gene is transcribed.
- Signal: a fluorescent marker shows which cells express the DNA.
One picture every idea returns to
Recombinant DNA is DNA made by joining together DNA from two different sources, often from different species. Genetic engineering is the deliberate manipulation of genetic material to modify specific characteristics of an organism; often a gene is transferred into another organism so that it is expressed there. The gene can be extracted from donor DNA with restriction endonucleases, made from mRNA with reverse transcriptase and DNA polymerase, or synthesised chemically from nucleotides. A plasmid cut with the same restriction endonuclease has complementary sticky ends; DNA ligase seals the gene into it. A promoter that the host's RNA polymerase recognises is transferred with the gene so that it is transcribed, and a marker gene coding for a fluorescent protein such as GFP shows which cells are expressing the new DNA.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Cut, paste and switch on.. A human gene can be made to work inside a bacterium. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Three ways to get the gene | Pick each source in turn and press Run. Project tab 1 of the lab (Genetic technology lab, getting the gene) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Building and checking a recombinant plasmid | Press Next step to work through all five stages. Project tab 2 of the lab (Genetic technology lab, recombinant DNA and marker genes) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: The toolkit in one table | Every gene transfer uses the same small set of tools. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 18 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The toolkit in one table, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Reverse transcriptase makes DNA from an RNA template (the reverse of transcription). | It does not make mRNA from DNA, and it is not the enzyme that cuts DNA. |
| Seeing the gene in the cell is not enough: the marker shows that the inserted DNA is being expressed (transcribed and translated), because the green colour only appears if a fluorescent protein has been made. | Return to Building and checking a recombinant plasmid: ask a pair to state the correct version and show it on the lab before moving on. |
Recombinant DNA: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Define recombinant DNA and explain what is meant by genetic engineering.
Explain how a gene for human insulin can be obtained from the mRNA of pancreatic β cells, and why this method is preferred to cutting the gene from human DNA when the gene is to be expressed in bacteria.
Describe the roles of restriction endonucleases and DNA ligase in inserting a gene into a plasmid.
Explain why a promoter may have to be transferred into an organism as well as the desired gene.
Bacteria were treated with a recombinant plasmid carrying a human gene and a gene for green fluorescent protein (GFP). They were spread on an agar plate and 400 colonies grew. Under ultraviolet light, 12 colonies glowed green. Calculate the percentage of colonies expressing the recombinant plasmid and explain why the marker gene was needed.
Outline how a gene can be synthesised chemically, and give two advantages of this method.
Answer key and marking notes
- Recombinant DNA is DNA made by joining DNA from two different sources (often different species).
- Genetic engineering is the deliberate manipulation of genetic material to modify specific characteristics of an organism, for example by transferring a gene so that it is expressed in another organism.
- β cells make insulin, so they contain many copies of insulin mRNA, which is extracted.
- Reverse transcriptase uses the mRNA as a template to make a single strand of complementary DNA (cDNA).
- DNA polymerase makes the second strand, giving double-stranded DNA of the gene.
- mRNA has already had its introns removed, so the cDNA has no introns; bacteria cannot remove introns, so a gene cut from human DNA would not give a functional protein.
- A restriction endonuclease cuts DNA at a specific base sequence (restriction site), cutting the gene out of the donor DNA.
- The same restriction endonuclease cuts the plasmid open, so both have staggered cuts with complementary sticky ends.
- The sticky ends of the gene and the plasmid pair by complementary base pairing (hydrogen bonds).
- DNA ligase forms phosphodiester bonds, joining the sugar-phosphate backbones to make a recombinant plasmid.
- RNA polymerase must bind to a promoter next to the gene to start transcription.
- The promoter from the donor (eukaryotic) gene may not be recognised by the host's RNA polymerase (or may be absent from the inserted DNA), so without a suitable promoter the gene would not be expressed.
- 12 / 400 × 100 = 3 %.
- Only a small proportion of bacteria take up the plasmid, so most colonies do not contain the gene.
- Colonies that fluoresce are making GFP, so they contain and are expressing the recombinant DNA; these can be selected and the rest discarded.
- The base sequence of the gene is known, or is worked out from the amino acid sequence of the protein (using the genetic code).
- Nucleotides are joined one at a time in the correct order by a machine; short sections are joined to make the whole gene.
- Advantages (any one, plus the above or a second): there are no introns; codons can be chosen to suit the host organism; no donor tissue or mRNA is needed; sequences such as restriction sites can be added at the ends.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which description of a plasmid is correct?
- Aa small circle of double-stranded DNA in a bacterium, separate from its main chromosome
- Ba short single-stranded piece of DNA that starts copying of a template strand
- Ca single-stranded RNA molecule that carries a gene from the nucleus to a ribosome
- Da protein coat that surrounds a gene and injects it into a host cell
A restriction endonuclease recognises the sequence 5' GAATTC 3' and cuts each strand between G and A. The complementary strand also reads GAATTC in the 5' to 3' direction. What is the result of the cut?
- Aa staggered cut leaving four unpaired bases, AATT, at each end (sticky ends)
- Ba straight cut across both strands, leaving blunt ends with no unpaired bases
- Ca cut through one strand only, so the DNA molecule stays in one piece
- Dremoval of the whole GAATTC sequence from both strands of the DNA
A scientist wants to make cDNA for the human insulin gene. Why is mRNA taken from pancreatic β cells rather than from skin cells?
- Aβ cells are transcribing the insulin gene, so they contain many insulin mRNA molecules
- Bonly β cells contain the insulin gene in their DNA
- Cskin cell mRNA still contains introns, but β cell mRNA does not
- Dβ cell mRNA is double-stranded, so it is easier for reverse transcriptase to copy
Which information is needed before a gene can be synthesised chemically from nucleotides?
- Athe base sequence of the gene, or the amino acid sequence of its protein
- Bthe positions of restriction sites around the gene in the donor chromosome
- Ca sample of mRNA from cells that express the gene
- Dthe base sequence of the gene's promoter in the donor organism
Bacteria take up a recombinant plasmid carrying the desired gene and a GFP marker gene, but the GFP gene has no functioning promoter. What would be seen under ultraviolet light?
- Ano green fluorescence, even though the bacteria contain the plasmid
- Bgreen fluorescence in every bacterium on the plate
- Cgreen fluorescence only in bacteria that did not take up the plasmid
- Dgreen fluorescence that appears only after the bacteria have died
The figure shows a 5.0 kbp plasmid with one site for restriction endonuclease R, and a 1.2 kbp gene fragment cut out with R. The gene is ligated into the plasmid. The recombinant plasmid is then cut again with R. Which fragments are produced?
- Atwo fragments: 5.0 kbp and 1.2 kbp
- Bone fragment of 6.2 kbp
- Ctwo fragments of 3.1 kbp each
- Done fragment of 1.2 kbp only
A human gene is inserted into bacteria together with its own human promoter. The gene is present in the bacteria, but none of the protein is produced. What is the most likely explanation?
- Abacterial RNA polymerase does not recognise the human promoter, so the gene is not transcribed
- Bbacterial ribosomes cannot translate any mRNA made from a human gene
- Cbacteria do not have tRNA molecules for the codons used in human genes
- DDNA ligase cut the gene into pieces while joining it into the plasmid
A human gene cut from a chromosome is 15 000 base pairs long. cDNA made from its mature mRNA is 2400 base pairs long. Assuming the whole cDNA corresponds to exons, what percentage of the gene consists of exons?
- A16 %
- B84 %
- C6.25 %
- D32 %
Quiz key
A plasmid is a small, circular, double-stranded DNA molecule found in many bacteria in addition to the main chromosome; it replicates independently, which makes it a useful vector. A short single-stranded start point for copying is a primer. An RNA molecule carrying a gene's code to a ribosome is mRNA. A protein coat that injects DNA describes a virus (which can be a vector, but is not a plasmid).
Because the site is palindromic and each strand is cut between G and A, the cuts are four bases apart on the two strands. Each fragment is left with a single-stranded AATT overhang: a sticky end that can pair with any other end cut by the same enzyme. Blunt ends would need both cuts at the same point. Both strands are cut, so the molecule is split. Restriction enzymes cut within or near the site; they do not remove it.
All nucleated body cells contain the insulin gene, but only β cells express it, so their cytoplasm is rich in insulin mRNA, the template for reverse transcriptase. The gene is not missing from skin cells. Mature mRNA has had its introns removed in every cell type, not only in β cells. mRNA is single-stranded in all cells.
The machine needs to know the order of bases. This can come from a known gene sequence or be worked out from the amino acid sequence using the genetic code. Restriction sites matter when cutting a gene out of donor DNA. An mRNA sample is needed for making cDNA with reverse transcriptase. A promoter suited to the host is added separately; the donor's own promoter is not needed.
Without a promoter, RNA polymerase cannot bind to start transcription, so no GFP is made and no cell fluoresces, even though the DNA is present. This is why the marker shows expression, not just presence. Every bacterium glowing would need all to have the plasmid and a working promoter. Bacteria without the plasmid have no GFP gene at all. Fluorescence needs the protein, which dead cells do not suddenly produce.
Ligating the gene into the single R site recreates an R site at each junction, so the 6.2 kbp recombinant plasmid has two R sites. Cutting both releases the 1.2 kbp insert and the 5.0 kbp plasmid backbone. A single 6.2 kbp fragment would come from a circle with only one site (cutting a circle once opens it into one linear piece). Equal 3.1 kbp halves would need sites exactly opposite each other, which this map does not show. The plasmid DNA does not disappear when the insert is released.
Promoters are recognised by RNA polymerase (with transcription factors in eukaryotes). Eukaryotic promoters differ from prokaryotic ones, so the bacterial polymerase does not bind and no mRNA is made; a bacterial promoter must be inserted instead. The genetic code is universal, so bacterial ribosomes and tRNAs can translate human codons once mRNA exists. Ligase joins DNA; it does not cut it.
Exons = 2400 / 15 000 × 100 = 16 %; the other 84 % is introns, which are removed from pre-mRNA during splicing and so are absent from cDNA. 84 % is the intron fraction, not the exon fraction. 6.25 comes from dividing 15 000 by 2400 (an inverted ratio, and not a percentage). 32 % doubles the cDNA length, wrongly counting both strands, but base pairs already count both strands of double-stranded DNA.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.