Principles and Applications of Gene Technology (Senior 6 Biology): free lessons | Ganzaa

Principles and Applications of Gene Technology

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Senior 6 · Biology

Principles and Applications of Gene Technology

5 lessons

Lesson 1 of 5

Recombinant DNA: restriction enzymes, ligase and plasmids

30 min

A bacterial defence turned into a tool

Bacteria are under constant attack by viruses called bacteriophages, which inject their DNA and take over the cell. Many bacteria fight back with enzymes that recognise a short base sequence in the invading DNA and cut it in two. The bacterium's own DNA carries the same sequences, but it is protected by methyl groups added to those bases. In the early 1970s biologists realised that these enzymes were a set of molecular scissors that cut DNA in exactly the same place every time. Together with an enzyme that joins DNA and a small circle of bacterial DNA to carry genes, they made it possible to combine DNA from different organisms. In Senior 3 you met this idea in outline. This lesson looks at the molecules in detail: which bases are cut, how the ends rejoin, and what makes a good vector.

Definition

Gene technology (genetic engineering): the techniques used to isolate, copy, change, transfer and analyse genes.

Recombinant DNA: DNA made by joining DNA from two different sources, often from different species.

Restriction enzyme (restriction endonuclease): a bacterial enzyme that cuts both strands of DNA at a specific base sequence, its recognition site.

Palindromic sequence: a double-stranded sequence that reads the same on both strands when each is read in the 5′ to 3′ direction, such as GAATTC.

Sticky ends: short single-stranded overhangs left by a staggered cut, which can base-pair with complementary overhangs. A straight cut leaves blunt ends.

DNA ligase: an enzyme that joins the sugar-phosphate backbones of two DNA fragments by forming phosphodiester bonds.

Vector: a DNA molecule, such as a plasmid or a virus, used to carry a gene into a host cell.

Plasmid: a small, circular, double-stranded DNA molecule found in bacteria, separate from the main chromosome and able to replicate independently.

Restriction enzymes

Each restriction enzyme is named after the bacterium it comes from. EcoRI was the first enzyme isolated from Escherichia coli strain RY13; BamHI comes from Bacillus amyloliquefaciens strain H. Their recognition sites are usually 4 to 8 base pairs long and palindromic:

Enzyme	Recognition site (5′ to 3′, ↓ = cut)	Ends produced
EcoRI	G↓AATTC	Sticky, with a 5′ overhang AATT
BamHI	G↓GATCC	Sticky, with a 5′ overhang GATC
HindIII	A↓AGCTT	Sticky, with a 5′ overhang AGCT
SmaI	CCC↓GGG	Blunt

Because the site is palindromic, the enzyme makes the same cut on both strands, but where the cut is off-centre the two cuts are staggered. That leaves each fragment with a short single-stranded tail. Any two fragments cut by the same enzyme have complementary tails, whatever organism they came from.

When the two kinds of fragment are mixed, their sticky ends pair up by hydrogen bonding between complementary bases. This is called annealing. The pairing is weak, and there is still a gap in each sugar-phosphate backbone. DNA ligase seals each gap by making a phosphodiester bond, so the joined molecule becomes one continuous, stable DNA molecule.

Worked example 1: finding the cuts

A short DNA fragment has the top strand 5′-TTGAATTCAGGCATGAATTCCA-3′. It is cut with EcoRI. (a) How many fragments are produced? (b) Write the base sequence of both strands of the middle fragment.

Search the top strand for GAATTC: it occurs twice, starting at base 3 and at base 15. Each site is cut once, so a linear molecule with 2 sites gives $2 + 1 = 3$ fragments.
The top strand is cut between G and A: TTG | AATTCAGGCATG | AATTCCA.
The bottom strand is complementary (3′-AACTTAAGTCCGTACTTAAGGT-5′) and is cut between A and G within each site, read from its own 5′ end.
Middle fragment, lined up:

5′-AATTCAGGCATG    -3′
3′-    GTCCGTACTTAA-5′

It has a four-base single-stranded tail at each end, AATT read 5′ to 3′, so it can anneal to any other DNA cut with EcoRI, including a plasmid.

Plasmids as vectors

Plasmids are ideal vectors because bacteria take them up, copy them and pass them on when they divide. A plasmid used as a cloning vector is engineered to carry:

an origin of replication (ori), so that it is copied inside the host cell;
one or more marker genes, often genes for resistance to antibiotics, so that cells that have taken up the plasmid can be identified;
unique restriction sites, each present only once, so that the plasmid is opened at a single point; some vectors have a cluster of them (a multiple cloning site);
for making a protein, a promoter that the host's RNA polymerase recognises, placed just before the inserted gene.

The steps of making recombinant DNA

Obtain the gene: cut it from the donor DNA with a restriction enzyme, make it from mRNA, synthesise it, or copy it by PCR (lesson 2).
Cut the vector with the same restriction enzyme, so that the ends match.
Mix and anneal the gene and the opened plasmid, then add DNA ligase to seal the backbones.
Transfer the recombinant plasmid into host cells (lesson 3).
Identify the cells that took up the recombinant plasmid, using the marker genes, and grow them to clone the gene or make its protein.

Not every plasmid in the mixture ends up carrying the gene. Some simply close up again with their own sticky ends, and some gene fragments join to each other. That is why step 5 is needed, and why the marker genes in Figure 2 are placed where they are.

Worked example 2: predicting fragment sizes

(a) A circular plasmid of 5 000 base pairs (bp) has EcoRI sites at positions 400 and 3 100. (b) A linear DNA molecule of 10 000 bp has EcoRI sites at 1 500, 4 000 and 8 200. Predict the fragments produced by EcoRI in each case.

(a) In a circle, $n$ cuts give $n$ fragments. Between the sites: $3\,100 - 400 = 2\,700$ bp. The rest of the circle: $5\,000 - 2\,700 = 2\,300$ bp. Check: $2\,700 + 2\,300 = 5\,000$ .
(b) A linear molecule with $n$ cuts gives $n + 1$ fragments: $1\,500$ ; $4\,000 - 1\,500 = 2\,500$ ; $8\,200 - 4\,000 = 4\,200$ ; $10\,000 - 8\,200 = 1\,800$ bp. Check: the four add up to 10 000 bp.

Use. Comparing the predicted sizes with the bands on a gel (lesson 4) shows whether a plasmid really contains the inserted gene.

Warning

Common mistakes

"Ligase makes the hydrogen bonds between sticky ends." The sticky ends pair by themselves; ligase forms phosphodiester bonds in the backbone.
"Any two restriction enzymes give matching ends." Only ends with complementary overhangs can anneal; normally the gene and the vector are cut with the same enzyme.
"A palindrome reads the same backwards on one strand." In DNA, a palindromic site reads the same on the two strands, each read 5′ to 3′.
Counting fragments. A linear molecule cut at $n$ sites gives $n + 1$ pieces; a circle gives $n$ .

DNA Learning Center (Cold Spring Harbor Laboratory): "Mechanism of Recombination". A 3D animation of EcoRI cutting a plasmid, sticky ends annealing and ligase sealing the joins.

✦

Practice: recombinant DNA

Explain why bacteria produce restriction enzymes, and why their own DNA is not cut.
Write the complementary strand of the BamHI site 5′-GGATCC-3′, and explain why it is described as palindromic.
Distinguish between sticky ends and blunt ends.
Explain why the gene and the plasmid are normally cut with the same restriction enzyme.
State the role of DNA ligase and name the bond it forms.
List four features of a good plasmid vector and give the reason for each.
A circular plasmid of 6 400 bp is cut by an enzyme with sites at positions 900, 2 500 and 5 100. Calculate the sizes of the fragments.
A student cuts a gene with EcoRI and a plasmid with BamHI, then adds ligase. Predict the result and explain it.

Show answersHide answers

To cut up the DNA of invading viruses (bacteriophages). The bacterium's own recognition sites are protected by methylation, so its enzymes do not cut them.
3′-CCTAGG-5′. Read 5′ to 3′, the bottom strand is GGATCC, the same as the top strand.
Sticky ends have short single-stranded overhangs, from a staggered cut, which can base-pair with complementary overhangs. Blunt ends have no overhang, from a cut straight across both strands.
So that both have complementary sticky ends that can anneal by base pairing.
Joins the sugar-phosphate backbones of adjacent DNA fragments; it forms phosphodiester bonds.
Origin of replication (copied in the host); marker gene (identify transformed cells); unique restriction site (opened at one point only); promoter (so the inserted gene is transcribed). Small size also helps it enter cells.
Three sites in a circle give three fragments: $2\,500 - 900 = 1\,600$ ; $5\,100 - 2\,500 = 2\,600$ ; $6\,400 - 5\,100 + 900 = 2\,200$ bp. Check: total 6 400 bp.
The overhangs are different (AATT and GATC) and not complementary, so the gene's ends cannot anneal to the plasmid's ends; recombinant plasmids will not form (or only very rarely).

Key points

Restriction enzymes cut DNA at specific, usually palindromic, recognition sites; staggered cuts leave sticky ends, straight cuts leave blunt ends.
DNA cut with the same enzyme has complementary sticky ends that anneal; DNA ligase then forms phosphodiester bonds to seal the backbone.
A plasmid vector needs an origin of replication, marker genes, unique restriction sites and, for expression, a promoter.
Steps: obtain gene → cut vector with the same enzyme → anneal and ligate → transfer into host → identify transformed cells.
A linear molecule with $n$ cut sites gives $n + 1$ fragments; a circular one gives $n$ .

Sources and credits

17.1 Biotechnology, from Biology 2e by Mary Ann Clark, Matthew Douglas and Jung Choi, OpenStax (Rice University), licensed under CC BY 4.0. Used for restriction enzymes, ligase and molecular cloning. Adapted by Ganzaa.
Video: "Mechanism of Recombination", DNA Learning Center, Cold Spring Harbor Laboratory, embedded from YouTube.

Figures 1 and 2, the table, the worked examples and the practice questions were written by Ganzaa.

1A bacterial defence turned into a tool

2Restriction enzymes

3Plasmids as vectors

4The steps of making recombinant DNA

A bacterial defence turned into a tool

Restriction enzymes

Plasmids as vectors

The steps of making recombinant DNA