Biology · Introductory biology · Concept

Restriction digests and gel electrophoresis

A restriction enzyme cuts DNA wherever it finds its recognition sequence, so a digest turns one molecule into fragments of predictable lengths. Each fragment runs from one cut to the next, or to an end of a linear molecule, and the lengths add up to the whole. On an agarose gel shorter fragments travel farther, and fragments of the same length share one band.

A concrete case

Six pieces of a virus

The DNA of the bacteriophage lambda, a virus of E. coli, is one linear molecule $48, 502 bp$ long. The enzyme EcoRI cuts DNA wherever it finds the six bases GAATTC. Lambda DNA holds that sequence at five places.

Five cuts in a linear molecule leave six pieces. Reading along the molecule from one end, they are $21, 226$ , $4, 878$ , $5, 643$ , $7, 421$ , $5, 804$ and $3, 530 bp$ long. Together they make up the whole molecule, because cutting removes no DNA.

Run the digest on an agarose gel and six bands appear. The $21, 226 bp$ piece stays nearest the well and the $3, 530 bp$ piece travels farthest. The $5, 804$ and $5, 643 bp$ pieces differ by only $161 bp$ , so their bands sit close together. The order along the molecule is lost on the gel, which sorts the pieces by length alone.

21226 + 4878 + 5643 + 7421 + 5804 + 3530 = 48502

\begin{array}{l} 21226 + 4878 + 5643 \\ + 7421 + 5804 + 3530 \\ = 48502 \end{array}

Lambda DNA cut by EcoRI, in order along the molecule
Fragment	From $(bp)$	To $(bp)$	Length $(bp)$
$1$	$0$	$21, 226$	$21, 226$
$2$	$21, 226$	$26, 104$	$4, 878$
$3$	$26, 104$	$31, 747$	$5, 643$
$4$	$31, 747$	$39, 168$	$7, 421$
$5$	$39, 168$	$44, 972$	$5, 804$
$6$	$44, 972$	$48, 502$	$3, 530$

The rule

An enzyme cuts at its recognition site

A restriction enzyme reads the DNA sequence and cuts both strands at every copy of a short sequence, its recognition site. EcoRI recognizes GAATTC and cuts between the G and the first A, written G^AATTC. HindIII recognizes AAGCTT and cuts A^AGCTT.

Most recognition sites read the same on both strands, each read from its own 5′ end. The complement of GAATTC, read 5′ to 3′, is GAATTC again. Because the enzyme cuts each strand at the same place in the site, the two cuts are offset by four bases. Each end is left with a short single-stranded tail, AATT for EcoRI, called a sticky end.

A cut is named by its position, the number of base pairs from the left end of the molecule to the cut.

Linear or circular decides the count

Linear DNA has two ends. Each cut splits one piece into two, so $n$ cuts make $n + 1$ fragments. Lambda’s five EcoRI cuts make six.

Circular DNA, such as a bacterial plasmid, has no ends. The first cut only opens the circle into one linear piece, and each later cut adds one more. In a circle, $n$ cuts make $n$ fragments, for $n$ of at least $1$ . A circle with no cut is still one molecule.

Fragments lie between neighboring cuts

Put the cut positions in order and add the ends of a linear molecule as boundaries, $0$ and its full length $L$ . Each fragment is the distance from one boundary to the next. In a circle the last fragment wraps around, from the last cut, past the end of the numbering, to the first cut. Its length is $L - c_{last} + c_{first}$ , where $c_{last}$ is the position of the last cut and $c_{first}$ that of the first.

L_{wrap} = L - c_{last} + c_{first}

Linear DNA: keep both ends when finding fragment lengths

A gel sorts fragments by length

Agarose gel electrophoresis pulls DNA through a slab of gel with an electric field. DNA is negatively charged, so it moves away from the wells, toward the positive electrode. The gel is a mesh of agarose fibers, and longer molecules thread through it more slowly. Shorter fragments therefore travel farther.

Fragments of the same length travel the same distance and make one band. A ladder, a mix of fragments of known lengths, runs in a lane beside the samples. Within the gel’s range, distance traveled falls in a straight line against the logarithm of length, which is how a band’s length is read from the ladder.

Why it works

Why cutting conserves length and a gel sorts by it

A restriction enzyme breaks bonds in the DNA backbone. It removes no base pairs and adds none. Every base pair of the original molecule ends up in exactly one fragment, so the fragment lengths add back to the whole.

Every nucleotide carries one negative charge on its phosphate, so the charge of a fragment grows in step with its length. Doubling the length doubles both the pull of the field and the amount of DNA it pulls. Charge alone would move every fragment at the same speed. The gel breaks the tie, because a longer molecule is held back more by the mesh, and length is what is left to sort by.

A site of six specified bases turns up by chance once in about $4^{6} = 4, 096 bp$ of random sequence. By that count lambda would hold about $11.8$ EcoRI sites. It holds $5$ , a reminder that real DNA is not a random string.

\begin{array}{l} 4^{6} = 4096 \\ \frac{48502}{4096} \approx 11.8 \end{array}

The method

Predict the fragments and the bands

The table follows the first worked example, a $6, 000 bp$ linear molecule cut at $1, 000$ and $3, 500 bp$ .

From cuts to bands
Step	What to do	For $6, 000 bp$ , linear, cut at $1, 000$ and $3, 500$
1	Decide whether the molecule is linear or circular.	Linear, so two cuts make three fragments.
2	List the boundaries in order, with $0$ and the full length for a linear molecule.	$0$ , $1, 000$ , $3, 500$ , $6, 000$
3	Subtract each boundary from the next.	$1, 000$ , $2, 500$ and $2, 500 bp$
4	Add the fragments and compare with the whole.	$1, 000 + 2, 500 + 2, 500$ $= 6, 000$
5	Group equal lengths into one band, and order the bands by length.	Two bands, $2, 500 bp$ nearer the well and $1, 000 bp$ farther

Read a band’s length from a ladder

Find the two ladder bands on either side of the unknown band, and note how far each has traveled. Lengths shrink in equal ratios over equal distances, so work in logarithms. A band halfway between ladder bands of $3, 000$ and $2, 000 bp$ has the logarithm halfway between theirs. Its length is the geometric mean, $\sqrt{3000 \cdot 2000}$ $\approx 2, 449 bp$ , not the plain average, $2, 500$ .

\sqrt{3000 \cdot 2000} \approx 2449

Where it breaks

The predicted fragments assume a complete digest, with every site cut in every molecule. Too little enzyme or time leaves some sites uncut, and a partial digest shows extra bands. Each extra band is the sum of neighboring fragments, such as $4, 878 + 5, 643$ $= 10, 521 bp$ from lambda.

An agarose gel separates linear DNA well from about $100 bp$ to about $25, 000 bp$ . Smaller fragments are better separated on a polyacrylamide gel, and larger ones need pulsed-field electrophoresis. Lambda’s $21, 226 bp$ piece sits near the top of the range, where bands crowd together.

The shape of a molecule changes how it runs. An uncut plasmid is usually supercoiled, and it travels faster than a linear molecule of the same length. A nicked, open circle travels slowest. The band of an uncut plasmid therefore does not give its length.

One band is not proof of one fragment. Fragments of the same length, or lengths too close to separate, share a band, and two fragments in one band need not have the same sequence.

4878 + 5643 = 10521

Check your answer

A digest prediction carries its own checks.

Add the fragments. They must sum to the length of the molecule. A sum that falls short has usually lost the piece after the last cut of a linear molecule.

Count the fragments. A linear molecule gives one more fragment than it has cuts, and a circle gives as many as it has cuts.

Compare with the gel. The bands must come in order of length, and the number of bands can be smaller than the number of fragments, never larger, in a complete digest.

Fragment sizes are subtractions between cut sites, and Chalk can check them. Write each one, and the sum that must give back the whole molecule, as plain numbers on its own row of a Math box, such as $3500 - 1000 = 2500$ and $1000 + 2500 + 2500$ $= 6000$ . A wrong fragment reads ✗. A row with the unit written in, $6000 bp$ , gets no mark, so keep units in your words. Restriction digest & gel, in Biology, cuts any molecule you enter and draws the gel it predicts.

Sticky ends, brightness and maps

Every EcoRI cut leaves the same four-base tail, AATT. Any two EcoRI ends can therefore pair, even if they came from different organisms, and the enzyme DNA ligase can seal them. Joining a gene into a plasmid this way is the start of DNA cloning.

Band brightness reports the mass of DNA in a band, because the stain slips between base pairs. In a complete digest every fragment comes from each molecule once, so a longer fragment carries more stain. Lambda’s $21, 226 bp$ band holds about $6$ times the DNA of its $3, 530 bp$ band.

Comparing single and double digests locates the sites. If enzyme A alone opens a plasmid and enzymes A and B together cut out a short piece, B lies that far from A, on one side or the other. Another enzyme or a sequence settles which side.

\frac{21226}{3530} \approx 6.0

Common mistakes

Forgetting the fragment between the final cut and the end of linear DNA. Count from the last cut to the full length, or the fragments will not add up to the molecule.
Using the same fragment-count rule for linear and circular DNA. A linear molecule gives one more fragment than it has cuts, but a circle gives exactly as many fragments as cuts.
Equating the number of bands with the number of fragments. Fragments of the same length run together, so $1, 000$ , $2, 500$ and $2, 500 bp$ make only two bands.
Subtracting each cut from the total length. The fragments lie between neighboring boundaries, so subtract each boundary from the next one in order.
Leaving out the wraparound in a circle. The last fragment runs from the last cut past the end of the numbering to the first cut, $L - c_{last} + c_{first}$ .
Reading an uncut plasmid’s band as its length. A supercoiled circle travels faster than linear DNA of the same length, so only linear fragments compare fairly with the ladder.
Averaging lengths to read a band between two ladder bands. Distance tracks the logarithm of length, so halfway between $3, 000$ and $2, 000 bp$ is about $2, 449 bp$ , not $2, 500$ .

Key terms

Restriction enzyme: An enzyme that cuts DNA at one short, specific sequence, its recognition site. Where it cuts decides the sizes of the fragments a digest produces.
Restriction site: A short DNA sequence that a restriction enzyme recognizes and cuts at. The cut may fall inside the site or a set distance away, so the cut position comes from the enzyme’s own definition.
Complete digest: A digest in which the enzyme cuts at every one of its recognition sites, so every fragment ends at a site. A partial digest leaves some sites uncut and also gives longer fragments that span them.
Circular DNA: A DNA molecule joined into a closed loop with no free ends, as in bacterial chromosomes and plasmids. A circle cut at $n$ sites gives $n$ fragments, where a linear molecule gives $n + 1$ , because the two end pieces of a line are one piece in a circle.
Plasmid: A small ring of DNA that is copied separately from a cell’s chromosome, common in bacteria. Plasmids often carry genes such as antibiotic resistance, and biotechnology uses them to carry genes into cells.
Gel electrophoresis: Separating charged molecules such as DNA fragments by pulling them through a gel with an electric field. DNA is negatively charged, so it runs toward the positive electrode, and smaller fragments travel farther, so the bands show fragment sizes.
DNA ladder: A mixture of fragments with known lengths used as a size reference in electrophoresis. A shared band position does not establish sequence identity.
Base pair: Two complementary bases, one on each strand, joined by hydrogen bonds: A with T and G with C (A with U where RNA is involved). The length unit $bp$ counts these pairs, not the nucleotides on both strands added together.

Work through an example

A linear $6000 bp$ molecule has cuts at $1000$ and $3500 bp$ . Find its fragments and predict the visible band positions.

Predict restriction fragments and gel bands →

Cut a plasmid with two enzymes →

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