Biology · Biotechnology: Principles and Processes · NEET
It cuts BOTH strands. After binding the recognition site, the enzyme breaks the sugar-phosphate backbone at a specific point on each of the two strands. A common NEET trap statement says it 'cuts only one of the two strands' - that is WRONG. Because it cuts both strands, the double helix is fully separated into two pieces.
Cutting a little away from the centre is what makes the ends useful. Because the cut on the top strand and the cut on the bottom strand are at offset positions (not the middle), each fragment is left with a short single-stranded tail. These tails are called sticky ends. If the enzyme cut exactly at the centre, both strands would break at the same point and you would get flat blunt ends with no overhang.
The palindrome reads the same 5' to 3' on both strands, so the same two bases sit at offset positions on each strand. The enzyme cuts between those same two bases on each strand. Because the two cut points are staggered, one short stretch stays single-stranded on each fragment. These unpaired, complementary overhangs are the sticky ends. Sticky ends from the SAME enzyme are complementary, so they can base-pair back together.
EcoRI recognises 5'-GAATTC-3' (with 3'-CTTAAG-5' on the other strand). It cuts between G and A on each strand - after the G on the top strand and after the G reading the bottom strand the same way. This leaves a 5'-AATT overhang on each fragment. That AATT single-stranded tail is the sticky end.
Because the same restriction enzyme always cuts at the same sequence in the same staggered way, every fragment it makes has an IDENTICAL sticky end. A gene fragment and a plasmid vector cut by the same enzyme both get matching sticky ends, so their overhangs base-pair. DNA ligase then seals the joins to make recombinant DNA. This is why the same enzyme is used on both the foreign DNA and the vector.
Statement I: Restriction endonucleases recognise specific sequences to cut DNA known as palindromic nucleotide sequences. Statement II: Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site. In the light of the above statements, choose the most appropriate answer.
Following statements describe the characteristics of the enzyme Restriction Endonuclease. Identify the INCORRECT statement.
Which of the following statements are NOT true regarding restriction endonucleases? A. They are called molecular scissors. B. These are enzymes responsible for restricting the growth of bacteriophages in E. coli. C. They cut the DNA only at the centre of the palindromic sites. D. They remove nucleotides only from the ends of DNA fragments. E. They recognise specific palindromic base-pair sequences.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a short, specific palindromic base sequence (usually 4-8 base pairs) that the enzyme scans for and binds to. Each enzyme has its own site; for example EcoRI recognises 5'-GAATTC-3'. The enzyme cuts only at this sequence, which is why the same fragment ends are produced every time.
A sticky end is a short single-stranded overhang left when the enzyme cuts a little away from the centre of the palindrome (staggered cut). A blunt end is a flat, fully double-stranded end left when the enzyme cuts exactly at the centre so both strands break at the same point. Sticky ends can easily base-pair with matching ends; blunt ends cannot.
The bacterium protects its own DNA by adding methyl groups (methylation) at the recognition sites, so the enzyme cannot cut there. This is the basis of the host-controlled restriction system - the enzyme restricts foreign or viral DNA but spares the cell's own methylated DNA.
It is an endonuclease. It cuts DNA at specific internal sites within the molecule. Exonucleases remove nucleotides from the free ends of DNA. NEET often pairs this fact with the trap that restriction enzymes 'remove nucleotides from ends' - that describes exonucleases, not restriction endonucleases.
Because the same enzyme makes identical sticky ends on every fragment, the cut gene and the cut vector will have complementary overhangs that base-pair. DNA ligase then seals them into a recombinant molecule. If different enzymes were used, the ends would not match and would not join.
