BiologyCBSE Class 12 · Biotechnology: Principles and Processes

Restriction Enzymes: Molecular Scissors

Updated 15 October 2026 · Fact-checked

Restriction enzymes (restriction endonucleases) are 'molecular scissors' that recognise a specific palindromic DNA sequence and cut both strands at fixed points. EcoRI recognises GAATTC and cuts between G and A, leaving single-stranded sticky ends. Cutting vector and foreign DNA with the same enzyme gives matching sticky ends that DNA ligase joins 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 named a restriction endonuclease. Five years later the first restriction endonuclease whose action depended on a specific sequence, Hind II, was characterised: it always cuts DNA at a particular point by recognising a specific sequence of six base pairs (its recognition sequence). Today more than 900 restriction enzymes are known, from over 230 strains of bacteria, each recognising a different sequence.

Naming convention (EcoRI).

E = first letter of the genus (*Escherichia*);
co = first two letters of the species (*coli*);
R = the strain (RY 13);
I (Roman numeral) = the order in which enzymes were isolated from that strain.

So EcoRI is the first enzyme isolated from *Escherichia coli* RY 13.

Nucleases. Restriction enzymes belong to the class of nucleases:

Exonucleases remove nucleotides from the ends of DNA.
Endonucleases make cuts at specific positions within the DNA.

How a restriction enzyme cuts. It 'inspects' the length of the DNA; on finding its recognition sequence, it binds and cuts each of the two strands at specific points in their sugar-phosphate backbones.

Palindromes. The recognition sequences are palindromic: the sequence of base pairs reads the same on the two strands when the orientation of reading is kept the same (both 5'→3'). EcoRI's site:
5'-GAATTC-3'
3'-CTTAAG-5'
Read each strand 5'→3': GAATTC and GAATTC.

Sticky ends. Restriction enzymes cut a little away from the centre of the palindrome, but between the same two bases on both strands. This leaves single-stranded overhangs, the sticky ends, named so because they form hydrogen bonds with complementary cut counterparts. EcoRI cuts between G and A:
5'-G | AATTC-3' and 3'-CTTAA | G-5', giving 5'-AATT overhangs on each fragment.
The stickiness of the ends helps DNA ligase join them.

Making recombinant DNA.
1. Cut the vector and the source (foreign) DNA with the same restriction enzyme, so both get the same kind of sticky ends.
2. Mix them: the sticky ends pair by hydrogen bonding.
3. DNA ligase seals the sugar-phosphate backbone: recombinant DNA is formed.
Normally, unless vector and source DNA are cut with the same enzyme, a recombinant vector cannot be created.

History link. The first recombinant DNA was made by Stanley Cohen and Herbert Boyer in 1972: they cut out a piece of DNA carrying an antibiotic-resistance gene and linked it to a plasmid of *Salmonella typhimurium*, using restriction enzymes ('molecular scissors') and DNA ligase.

Flow chart: vector DNA + foreign DNA → same restriction enzyme → matching sticky ends → hydrogen bonding of overhangs → DNA ligase seals → recombinant DNA → into host.

Draw and label: action of EcoRI (three-step schematic). Checklist: (1) a double-stranded DNA with GAATTC/CTTAAG marked and zig-zag cut lines between G and A on both strands; (2) the two fragments with AATT overhangs shown; (3) vector and foreign DNA fragments with matching overhangs joined, with 'DNA ligase' written on the arrow. Labels: recognition site, EcoRI cut, sticky end, vector DNA, foreign DNA, DNA ligase, recombinant DNA, 5' and 3' ends.

Restriction Enzymes: key facts

Naming: $E co R I = E (genus) + co (species) + R (strain) + I (order)$; Escherichia coli RY 13, first enzyme isolated.
EcoRI site: $5^{'} -G ↓ AATTC- 3^{'} / 3^{'} -CTTAA ↑ G- 5^{'}$; Leaves 4-base AATT single-stranded overhangs.
Palindrome test: $top 5^{'} \to 3^{'} = bottom 5^{'} \to 3^{'}$; Equivalent: the sequence equals its reverse complement.
Fragments after cutting: $linear DNA: n sites \to n + 1 fragments; circular DNA: n sites \to n fragments$; Useful for plasmid questions.
Expected site frequency (random sequence): $one site per 4^{n} bp for an n -bp site$; 6-bp site: about 1 in 4096 bp; 4-bp site: about 1 in 256 bp.

How to answer Restriction Enzymes questions

Restriction enzyme questions test naming, palindromes, cutting and sticky ends. Draw the double strand every time.

1For naming, split the name into genus letter, species letters, strain letter and Roman numeral.
2To test a palindrome, write the complementary strand and read both 5'→3'; if they match, it is a palindrome.
3To cut, place the cut on the top strand as given, then place the matching cut on the bottom strand between the same two bases (reading that strand 5'→3').
4Write the fragments with polarity and identify the single-stranded overhang (sticky end); count its bases.
5For recombinant DNA, stress: same enzyme for vector and foreign DNA → complementary sticky ends → DNA ligase.
6For counting fragments, check whether the DNA is linear (n + 1) or circular (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.

Same enzyme, matching sticky ends

Suppose a student team in a college lab wants to insert a gene into a plasmid. They have two enzymes: Enzyme X recognises 5'-GGATCC-3' and cuts between the two G's; Enzyme Y recognises 5'-GAATTC-3' and cuts between G and A. The gene is flanked by GGATCC sites, and the plasmid has one GGATCC and one GAATTC site.

Which enzyme must they use on both the gene and the plasmid, what sticky ends will form, and how many fragments will the plasmid give?

Try it first, then show the solutionSolution

The gene can only be cut out with Enzyme X (its flanks are GGATCC), so the plasmid must also be cut with X to get matching ends.
X cuts G|GATCC on each strand, leaving 5'-GATC overhangs (4 bases) on each end.
The plasmid is circular with one X site, so cutting gives 1 fragment (it simply opens into a linear molecule).
Mixing the gene and opened plasmid lets the GATC ends pair; DNA ligase seals them into recombinant DNA.

Answer: Use Enzyme X on both; 5'-GATC sticky ends; the plasmid opens into 1 linear fragment; ligase joins gene and plasmid.

Memory hook: Same scissors, same edges, perfect fit.

Tricks to remember Restriction Enzymes

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

Eco-R-I = Every Country Reads In order
E genus, co species, R strain, I order of isolation.
Use it when: Naming convention.
MADAM in DNA
A DNA palindrome reads the same on both strands when each is read 5'→3', like MADAM.
Use it when: Recognising palindromes.
Exo = exit, Endo = inside
Exonucleases nibble from the ends; endonucleases cut inside.
Use it when: Types of nucleases.

Restriction Enzymes: solved examples

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

Example 1: Hind II's site · 1 mark · MCQBoard level

Hind II always cuts DNA at a particular point by recognising a specific sequence of how many base pairs?

Try it first, then show the solutionSolution

Hind II recognises a specific six-base-pair sequence.

Answer: Six base pairs.

Example 2: Naming EcoRI and its action · 3 marks · Short answerBoard level

Explain the convention for naming EcoRI. Show the action of EcoRI on its recognition site.

Try it first, then show the solutionSolution

E: genus Escherichia; co: species coli; R: strain RY 13; I: first enzyme isolated from this strain (2 marks).
5'-G↓AATTC-3' / 3'-CTTAA↑G-5' → fragments with 5'-AATT overhangs (sticky ends) (1 mark).

Answer: Genus, species, strain, order; cuts between G and A leaving AATT sticky ends.

Example 3: Sticky end length · 1 mark · MCQHarder board level

EcoRI cuts between G and A only where GAATTC is present. How many nucleotides are there in each single-stranded sticky end, and what are they?

Try it first, then show the solutionSolution

Top: G | AATTC; bottom (5'→3'): G | AATTC.
Each fragment has a 5' overhang of AATT: 4 nucleotides.

Answer: 4 nucleotides (AATT).

Example 4: Recombinant DNA in three steps · 3 marks · Short answerHarder board level

With a schematic diagram, show in three steps how recombinant DNA is formed using EcoRI and DNA ligase. Why must the same enzyme be used on both DNAs?

Try it first, then show the solutionSolution

Step 1: EcoRI cuts the vector and the foreign DNA at GAATTC sites (1 mark).
Step 2: Both have AATT sticky ends; the complementary ends of the foreign DNA and vector pair by hydrogen bonds (1 mark).
Step 3: DNA ligase seals the backbones, forming recombinant DNA. Only the same enzyme gives the same (complementary) sticky ends on both, so they can join (1 mark).

Answer: Cut both with EcoRI → sticky ends pair → ligase seals; same enzyme gives matching ends.

Example 5: Is it a palindrome? · 2 marks · Very short answerHarder board level

Which of these double-stranded sequences (top strand given 5'→3') is a palindrome: (i) GAATTC (ii) GATCTA (iii) AAGCTT? Show the test.

Try it first, then show the solutionSolution

(i) Complement 3'-CTTAAG-5' read 5'→3' = GAATTC: same, palindrome. (iii) Complement 3'-TTCGAA-5' read 5'→3' = AAGCTT: same, palindrome (1 mark).
(ii) Complement 3'-CTAGAT-5' read 5'→3' = TAGATC ≠ GATCTA: not a palindrome (1 mark).

Answer: (i) and (iii) are palindromes; (ii) is not.

Example 6: Counting fragments · 3 marks · Short answerHOTS / competency

Suppose a linear 12 kb DNA has 4 BamHI sites and a circular plasmid has 3 BamHI sites. How many fragments does each give on complete digestion? Assuming a random sequence, how often would a 4-bp recognition site occur on average, compared with a 6-bp site such as BamHI's?

Try it first, then show the solutionSolution

Linear: 4 cuts → 4 + 1 = 5 fragments (1 mark).
Circular: 3 cuts → 3 fragments (a ring has no free ends, so n cuts give n pieces) (1 mark).
A specific base at one position has probability 1/4, so a 4-bp site occurs about once in 4⁴ = 256 bp, and a 6-bp site about once in 4⁶ = 4096 bp; enzymes with shorter sites cut far more often (1 mark).

Answer: Linear 5, circular 3; a 4-bp site about once per 256 bp versus once per 4096 bp for a 6-bp site.

Where students lose marks in Restriction Enzymes

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

Mistake: Wrong naming parts
Students say R = restriction.
Wrong working
R in EcoRI stands for restriction.
Right working
R stands for the strain RY 13.
Do this: Genus-species-strain-order.
Usually costs: 1 mark.
Mistake: Defining a palindrome as reading the same backwards on one strand
Word palindromes are used as the model.
Wrong working
GAATTC reads the same backwards.
Right working
It reads the same on both strands when each is read 5'→3'.
Do this: Always compare the two strands in the same orientation.
Usually costs: 1 mark.
Mistake: Cutting at the centre
Students cut GAA|TTC (blunt).
Wrong working
EcoRI cuts between A and T.
Right working
EcoRI cuts between G and A on both strands, off-centre, leaving AATT overhangs.
Do this: Off-centre cuts → sticky ends.
Usually costs: 1 mark.
Mistake: Forgetting ligase
Students stop at sticky-end pairing.
Wrong working
Sticky ends join and form recombinant DNA.
Right working
Sticky ends pair by hydrogen bonds; DNA ligase seals the backbone.
Do this: Always end with DNA ligase.
Usually costs: 1 mark.

Exam tips

Draw both strands with polarity and arrows marking the cut points.
Count overhang bases exactly; EcoRI gives 4 (AATT).
Learn the numbers: 1963 discovery, Hind II (6 bp), over 900 enzymes, over 230 strains, Cohen and Boyer 1972.
In recombinant DNA answers, write 'same restriction enzyme' explicitly.
Do not confuse restriction enzymes (cut) with DNA ligase (join) and DNA polymerase (copy).

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 organism whose plasmid was used by Cohen and Boyer for the first recombinant DNA (1972) is:
(a) Escherichia coli (b) Salmonella typhimurium (c) Agrobacterium tumefaciens (d) Bacillus thuringiensis
Hint
Native plasmid with an antibiotic-resistance gene linked.
Answer
(b) Salmonella typhimurium.
Q2 · 2 marks · Very short answerBoard level
Differentiate between exonucleases and endonucleases.
Hint
Where they cut.
Answer
Exonucleases remove nucleotides from the ends of DNA; endonucleases make cuts at specific positions within the DNA. Restriction enzymes are endonucleases.
Q3 · 2 marks · Very short answerHarder board level
The enzyme PstI has the site 5'-CTGCA↓G-3' / 3'-G↑ACGTC-5'. Write the two fragments formed and state which end (5' or 3') the overhang is on.
Hint
Cut after A on the top; before G on the bottom.
Answer
Fragments: 5'-CTGCA-3' / 3'-G-5' and 5'-G-3' / 3'-ACGTC-5'. Each fragment has a 4-base single-stranded overhang, TGCA, ending at the 3' end: a 3' overhang (unlike EcoRI's 5' AATT overhang).
Q4 · 3 marks · Short answerHOTS / competency
Suppose a vector is cut with EcoRI and the foreign DNA with an enzyme that recognises GGATCC and cuts between the two G's. Will ligase join them easily? Explain and suggest a correction.
Hint
Compare the overhangs.
Answer
No. EcoRI leaves AATT overhangs; the other enzyme leaves GATC overhangs. These are not complementary, so the ends cannot pair by hydrogen bonds and ligase cannot join them efficiently. Correction: cut both vector and foreign DNA with the same enzyme so their sticky ends match.
Q5 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a DNA segment reads 5'-ATGAATTCGGTACGAATTCAT-3' (top strand) and is treated with EcoRI.
(a) How many EcoRI sites are there? (1)
(b) How many fragments are formed if the DNA is linear? (1)
(c) Write the top-strand sequence of the middle fragment and explain why its ends are 'sticky'. (2)
Hint
Look for GAATTC; EcoRI cuts G|AATTC.
Answer
(a) Two (GAATTC at bases 3-8 and 14-19). (b) Linear DNA with 2 sites → 3 fragments. (c) Top strand of the middle fragment: 5'-AATTCGGTACG-3' (from the first cut after G to the next G before AATTC). Its ends carry single-stranded AATT overhangs that can form hydrogen bonds with complementary AATT overhangs from any EcoRI-cut DNA, so they are sticky.
Q6 · 2 marks · Very short answerHarder board level
Why do restriction enzymes not cut the bacterium's own DNA? (Use the 1963 discovery of two enzymes.)
Hint
One enzyme methylated DNA.
Answer
Along with the cutting enzyme, the bacterium has an enzyme that adds methyl groups to DNA; its own DNA is methylated at the recognition sites and is protected, while unmethylated phage DNA is cut, restricting phage growth.

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 'I' in EcoRI indicate?Show answer
The order in which the enzyme was isolated from that strain (first).
2.EcoRI recognition sequence?Show answer
5'-GAATTC-3'.
3.What are sticky ends?Show answer
Single-stranded overhangs left by off-centre cuts, which pair with complementary ends.
4.Which enzyme joins DNA fragments?Show answer
DNA ligase.
5.How many restriction enzymes are known?Show answer
More than 900, from over 230 bacterial strains.
6.Fragments from a circular plasmid with 3 sites?Show answer
3.

Restriction Enzymes: frequently asked questions

What are restriction enzymes?

Endonucleases that recognise specific palindromic sequences in DNA and cut both strands at specific points; they are called molecular scissors.

How is EcoRI named?

E from the genus Escherichia, co from the species coli, R from the strain RY 13, and I because it was the first enzyme isolated from that strain.

What is a palindromic sequence in DNA?

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

Why are sticky ends important?

They form hydrogen bonds with complementary sticky ends, so DNA cut with the same enzyme can be joined by DNA ligase to make recombinant DNA.

Why must the vector and foreign DNA be cut with the same enzyme?

Only the same enzyme produces matching sticky ends, which can pair and be sealed by ligase.