BiologyISC Class 12 · Biotechnology - Principles and Processes

Restriction Enzymes (Molecular Scissors): ISC Class 12 Biology

Updated 15 October 2026 · Fact-checked

Restriction endonucleases are 'molecular scissors' that recognise specific palindromic DNA sequences and cut the sugar-phosphate backbone of both strands at fixed points. EcoRI cuts between G and A in GAATTC, leaving sticky ends; some enzymes leave blunt ends. Cutting vector and source DNA with the same enzyme lets DNA ligase join them into recombinant DNA.

Understand Restriction Enzymes

Discovery. In 1963 two enzymes that restrict the growth of bacteriophages in E. coli were isolated: one added methyl groups to DNA, the other cut DNA. The cutter was called a restriction endonuclease. Five years later the first one whose action depended on a specific sequence, Hind II, was characterised: it always cut DNA at one point by recognising a specific sequence of six base pairs, its recognition sequence. Today more than 900 restriction enzymes from over 230 strains of bacteria are known, each with its own recognition sequence. (In bacteria they cut invading phage DNA; the bacterium's own DNA is protected by methylation.)

Nomenclature (rule).

Part	Comes from	EcoRI	Hind II
1st letter (capital, italic)	genus	E (Escherichia)	H (Haemophilus)
next two letters (small, italic)	species	co (coli)	in (influenzae)
next letter	strain	R (RY13)	d (strain d)
Roman numeral	order of isolation from that strain	I (first)	II (second)

By this rule, a fifth enzyme isolated from E. coli strain RY13 would be named EcoRV (the ISC 2025 naming item).

Nucleases. Restriction enzymes belong to the nucleases: exonucleases remove nucleotides from the ends of DNA; endonucleases cut at specific positions within DNA. Restriction endonucleases inspect the length of DNA, bind at their recognition sequence and cut each strand in its sugar-phosphate backbone (phosphodiester bonds).

Palindromes in DNA. A sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same (5'→3' on both):
5'—G A A T T C—3'
3'—C T T A A G—5'
Check: the top strand 5'→3' is GAATTC; the bottom strand read 5'→3' (right to left) is also GAATTC.

Sticky ends. Many enzymes cut a little away from the centre of the palindrome but between the same two bases on opposite strands. EcoRI cuts between G and A on both strands (G^AATTC):
Fragment 1: 5'—G 3' over 3'—CTTAA 5'
Fragment 2: 5' AATTC—3' over 3' G—5'
Each end has a single-stranded overhang (AATT) called a sticky end, because it forms hydrogen bonds with a complementary cut end. This stickiness helps DNA ligase to seal the backbone.

Blunt ends. Some enzymes cut both strands at the centre of the site, leaving no overhang (blunt ends): Hind II, and EcoRV, which cuts GAT^ATC:
5'—GAT 3' + 5' ATC—3'
3'—CTA 5' + 3' TAG—5'
Blunt ends can be joined by ligase but less efficiently.

Role in rDNA technology.
1. Cut the source DNA to release the gene of interest.
2. Cut the vector (plasmid) at one site with the same enzyme, so both carry complementary sticky ends.
3. Mix: sticky ends pair by hydrogen bonds; DNA ligase joins the phosphodiester backbone → recombinant DNA.
Unless the vector and source DNA are cut with the same enzyme, a recombinant vector cannot be formed.

After cutting: gel electrophoresis. DNA fragments are negatively charged and move towards the anode through an agarose gel (from seaweed); smaller fragments move farther. They are stained with ethidium bromide and seen as orange bands under UV light; the wanted band is cut out (elution).

Fragment arithmetic.

n cuts in a linear DNA → n + 1 fragments; n cuts in a circular DNA (plasmid) → n fragments.
In random DNA a given 6-bp site is expected about once every 4⁶ = 4096 bp; a 4-bp site once every 4⁴ = 256 bp. Longer recognition sites cut less often.

Diagram checklist (EcoRI action). Draw double-stranded DNA with GAATTC/CTTAAG boxed; zig-zag cut line between G and A on both strands; the two products with 5' and 3' labels and the AATT overhangs labelled 'sticky end'; then a vector cut the same way, both sticky ends pairing, and 'DNA ligase' on the joining arrow.

Restriction Enzymes: key facts

Naming (EcoRI): E: genus Escherichia capital, italic
co: species coli italic
R: strain RY13
I: first enzyme isolated from that strain Roman numeral; EcoRI, EcoRV, HindII.
EcoRI cut: $5^{'} -G ↓ AATTC- 3^{'} / 3^{'} -CTTAA ↑ G- 5^{'}$; Leaves 5'-AATT sticky ends.
Number of fragments: $linear: n + 1; circular: n$; n = number of cut sites.
Expected site frequency: $one k -bp site per 4^{k} bp$; Assumes equal, random base composition.

E: genus Escherichia	capital, italic
co: species coli	italic
	R: strain RY13
I: first enzyme isolated from that strain	Roman numeral

How to answer Restriction Enzymes questions

ISC asks the naming rule, the role of restriction endonucleases, drawing fragments with 5' and 3' ends, sticky versus blunt ends, palindromes, and how ligase forms rDNA.

1For naming, take genus initial (capital), two species letters, strain letter and Roman numeral; never Arabic numbers.
2To check a palindrome, read the top strand 5'→3' and the bottom strand 5'→3'; they must match.
3To draw fragments, write both strands with 5' and 3' labels, put the cut between the stated bases on each strand, and show the overhangs.
4Name the ends: overhang = sticky; flush = blunt.
5For the role, say: recognise a specific palindromic site, cut phosphodiester bonds, produce sticky ends that allow joining by ligase.
6In fragment sums, decide linear or circular first, then use n + 1 or n.

Restriction Enzymes in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Why the same enzyme must cut vector and gene

Suppose a student cuts a human gene out with EcoRI but cuts the plasmid with BamHI (which leaves GATC overhangs), mixes them with ligase, and finds no recombinant plasmids.

Explain the failure and how to fix it.

Try it first, then show the solutionSolution

EcoRI leaves AATT overhangs; BamHI leaves GATC overhangs.
The overhangs are not complementary, so they cannot pair by hydrogen bonds, and ligase has no aligned ends to seal.
Cut both the gene and the plasmid with the same enzyme (EcoRI), so the sticky ends match.
The plasmid should have a single EcoRI site, so it opens once to accept the insert.

Answer: Non-matching sticky ends cannot pair; use the same restriction enzyme for both.

Memory hook: Same scissors, matching edges.

Tricks to remember Restriction Enzymes

Short memory aids. Say each one aloud once, then use it on a question the same day.

E-co-R-I = family, first name, house, birth order
Genus, species, strain, order of isolation.
Use it when: Naming any enzyme.
Gee, All Are There, Come!
G | AATTC: EcoRI cuts after the G.
Use it when: Drawing EcoRI fragments.
Sticky = Stuck out
Sticky ends stick out as single strands; blunt ends are flush.
Use it when: Naming ends.

Restriction Enzymes: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Naming rule · 2 marks · Short answerBoard level

Bacillus amyloliquefaciens strain H yielded its first restriction enzyme. Derive the enzyme's name, then explain what each part of the name EcoRI stands for.

Try it first, then show the solutionSolution

B + am + H + I: BamHI (1 mark).
EcoRI: E = genus Escherichia, co = species coli, R = strain RY13, I = first enzyme isolated from this strain (1 mark).

Answer: BamHI; E-co-R-I = genus, species, strain, order.

Example 2: Palindrome and sticky ends · 3 marks · Short answerBoard level

What is a palindromic sequence in DNA? Explain, with the EcoRI site, how sticky ends are formed.

Try it first, then show the solutionSolution

A sequence of base pairs that reads the same on both strands when read in the same orientation (5'→3'): 5'-GAATTC-3' / 3'-CTTAAG-5' (1 mark).
EcoRI cuts each strand between G and A, a little away from the centre but between the same two bases on both strands (1 mark).
This leaves single-stranded 5'-AATT overhangs (sticky ends) that can form hydrogen bonds with complementary ends (1 mark).

Answer: Same sequence on both strands 5'→3'; off-centre cuts leave AATT overhangs.

Example 3: Draw the fragments · 3 marks · Short answerHarder board level

Two enzymes recognise these sites: RE-1 5'-GTTAAC-3'/3'-CAATTG-5', cutting at the centre (GTT^AAC); RE-2 5'-GAATTC-3'/3'-CTTAAG-5', cutting between G and A. Draw the fragments produced at each site and name the type of ends.

Try it first, then show the solutionSolution

RE-1: 5'-GTT 3' and 3'-CAA 5' on one fragment; 5' AAC-3' and 3' TTG-5' on the other: blunt ends (1½ marks).
RE-2: 5'-G 3' over 3'-CTTAA 5' on one fragment; 5' AATTC-3' over 3' G-5' on the other: sticky ends with AATT overhangs (1½ marks).

Answer: RE-1 gives blunt ends; RE-2 gives sticky ends (5'-AATT overhangs), drawn with 5' and 3' labels.

Example 4: Counting fragments · 3 marks · Short answerHarder board level

A linear DNA of 10 000 bp has EcoRI sites at positions 2000, 5500 and 9000. (a) How many fragments are formed and of what sizes? (b) If the same sites were in a circular plasmid of 10 000 bp, how many fragments would form, and of what sizes?

Try it first, then show the solutionSolution

(a) 3 cuts in linear DNA → 4 fragments: 2000, 3500, 3500 and 1000 bp (sum 10 000) (1½ marks).
(b) 3 cuts in circular DNA → 3 fragments: 3500, 3500 and 3000 bp (the last joins 9000→10 000 and 0→2000) (1½ marks).

Answer: (a) 4 fragments: 2000, 3500, 3500, 1000 bp. (b) 3 fragments: 3500, 3500, 3000 bp.

Example 5: Role of restriction enzymes and ligase · 3 marks · Short answerHarder board level

Explain the roles of restriction endonuclease and DNA ligase in making recombinant DNA. Why must the vector have only one site for the enzyme used?

Try it first, then show the solutionSolution

Restriction endonuclease recognises a specific palindromic site and cuts phosphodiester bonds in both strands of the source DNA and the vector, producing complementary sticky ends (1 mark).
Ligase joins the sugar-phosphate backbones of the paired ends, sealing the gene into the vector to give recombinant DNA (1 mark).
A single site opens the vector once; several sites would cut it into pieces and destroy essential parts such as ori or marker genes (1 mark).

Answer: Cutter makes matching sticky ends; ligase seals; one site keeps the vector intact.

Example 6: Choosing a cutter · 4 marks · Short answerHOTS / competency

Suppose a researcher wants to cut a bacterial genome of about 4 million bp into a few large pieces for mapping. Enzyme P recognises a 4-bp site and enzyme Q an 8-bp site. (a) Estimate how often each site occurs in random DNA. (b) Roughly how many fragments would each give? (c) Which enzyme should she choose?

Try it first, then show the solutionSolution

(a) 4-bp site: once every 4⁴ = 256 bp; 8-bp site: once every 4⁸ = 65 536 bp (1½ marks).
(b) P: about 4 000 000/256 ≈ 15 600 fragments; Q: about 4 000 000/65 536 ≈ 61 fragments (the genome is circular, so fragments = cuts) (1½ marks).
(c) Q, the 8-bp cutter, because it gives a few large pieces (1 mark).

Answer: (a) 1 per 256 bp and 1 per 65 536 bp. (b) about 15 600 versus about 61. (c) Enzyme Q.

Where students lose marks in Restriction Enzymes

The slips examiners see most often, and what to write instead.

Mistake: Wrong naming format
In 2025 examiners reported EcoRy13V, EcoRyV or EcoR5 for the fifth enzyme from E. coli RY13.
Do this: Use only the strain letter and a Roman numeral: EcoRV.
Usually costs: 1 mark.
Mistake: Drawing fragments without bases or ends
Examiners (2025) found only 'blunt end' and 'sticky end' written, box diagrams without base sequences, and one of two sites ignored.
Do this: Write the bases on both strands with 5' and 3' labels for every site asked.
Usually costs: Up to 2 marks.
Mistake: Not mentioning recognition or palindromic sites
Examiners (2025) noted that the role of restriction endonucleases was given without restriction sites or palindromes.
Do this: Say: recognise a specific palindromic sequence and cut phosphodiester bonds there.
Usually costs: 1 mark.
Mistake: Mixing up exonuclease and endonuclease
Examiners (2025) suggested teaching the difference between endo-, exo- and restriction endonucleases because answers confused them.
Do this: Exo: from ends; endo: within; restriction endo: within, at a specific site.
Usually costs: 1 mark.

Exam tips

Italicise or underline the first three letters of an enzyme name (Eco, Hin); the strain letter and numeral are not italic.
Always label 5' and 3' on both strands of every fragment.
Say 'same restriction enzyme' when explaining rDNA formation.
For fragment counts, state linear or circular before answering.
Use ethidium bromide (a chemical) and UV light (radiation) as two separate words.

Practice questions on Restriction Enzymes

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The enzyme that removes nucleotides from the ends of DNA is a/an:
(a) endonuclease (b) exonuclease (c) ligase (d) polymerase
Hint
Exo = outside.
Answer
(b) exonuclease.
Q2 · 2 marks · Short answerBoard level
Why are restriction enzymes called 'molecular scissors'? Name the first sequence-specific restriction enzyme characterised.
Hint
Six base pairs.
Answer
They cut DNA at specific recognition sequences, like scissors cutting at fixed points. Hind II, which recognises a specific six-base-pair sequence.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): DNA fragments with sticky ends join more easily than fragments with blunt ends.
Reason (R): Sticky ends have single-stranded overhangs that form hydrogen bonds with complementary ends.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Pairing holds the ends for ligase.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Which of these are palindromes: 5'-GGATCC-3', 5'-GAATTG-3', 5'-AAGCTT-3'? Show your check for one of them.
Hint
Write the complementary strand and read it 5'→3'.
Answer
GGATCC and AAGCTT are palindromes; GAATTG is not. Check for GAATTG: complementary strand 3'-CTTAAC-5', read 5'→3' gives CAATTC, which differs from GAATTG.
Q5 · 2 marks · Short answerHarder board level
A circular plasmid of 6000 bp has EcoRI sites at 1000 and 4500 bp. How many fragments does EcoRI give, and of what sizes?
Hint
Circular: fragments = cuts.
Answer
Two fragments: 4500 − 1000 = 3500 bp, and (6000 − 4500) + 1000 = 2500 bp.
Q6 · 2 marks · Short answerHarder board level
In gel electrophoresis of digested DNA, why do fragments move towards the anode, and why does a 500 bp fragment travel farther than a 5000 bp fragment?
Hint
Charge; sieving.
Answer
DNA is negatively charged (phosphate groups), so it moves towards the positive electrode. Agarose acts as a sieve; smaller fragments pass through the pores more easily and move farther.
Q7 · 3 marks · Short answerHOTS / competency
Suppose bacteria make a restriction enzyme that cuts GAATTC, and their own chromosome contains many GAATTC sequences. Suggest why the bacterium does not destroy its own DNA, and why the enzyme is useful to it.
Hint
The other enzyme isolated in 1963.
Answer
The bacterium's own recognition sites are methylated by a companion methylase (the methyl-adding enzyme discovered with restriction endonuclease), so the restriction enzyme does not cut them. Invading bacteriophage DNA is not methylated and is cut, which restricts the growth of the phage.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a plasmid has one BamHI site inside its tetracycline-resistance gene and one EcoRI site outside any gene. A researcher has a gene cut out with BamHI.
(a) Which enzyme should be used to open the plasmid, and why? (1)
(b) Name the enzyme that seals the insert. (½)
(c) What will happen to tetracycline resistance after insertion, and how is this useful? (1½)
Hint
Matching ends; insertional inactivation.
Answer
(a) BamHI, so that the plasmid and the insert have complementary sticky ends. (b) DNA ligase. (c) The insert interrupts the tetracycline-resistance gene, so recombinant cells lose tetracycline resistance (insertional inactivation); cells that grow on ampicillin (if the plasmid also has amp^R) but not on tetracycline are identified as recombinants.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does the R in EcoRI stand for?Show answer
The strain RY13 of E. coli.
2.Where does EcoRI cut its site?Show answer
Between G and A in GAATTC on both strands.
3.What are sticky ends?Show answer
Single-stranded overhangs left after staggered cuts, which pair with complementary ends.
4.Which enzyme joins DNA fragments?Show answer
DNA ligase.
5.A linear DNA has 4 sites for an enzyme. How many fragments?Show answer
Five.
6.How many restriction enzymes are known?Show answer
More than 900, from over 230 bacterial strains.

Restriction Enzymes: frequently asked questions

What is a restriction enzyme?

A restriction endonuclease that recognises a specific (usually palindromic) DNA sequence and cuts both strands at fixed points; it is used to cut DNA in genetic engineering.

How are restriction enzymes named?

Genus initial, two letters of the species, a strain letter and a Roman numeral for the order of isolation, for example EcoRI.

What is the difference between sticky and blunt ends?

Sticky ends have single-stranded overhangs from staggered cuts; blunt ends come from cuts at the same point on both strands and have no overhang.

Why is the same restriction enzyme used for vector and gene?

So that both have complementary sticky ends that pair and can be joined by DNA ligase.

What is a palindromic sequence?

A DNA sequence that reads the same on both strands when read in the same 5'→3' direction, such as GAATTC.