Restriction Enzymes

Correct answer: A. Choosing it for starch digestion

Restriction enzymes are DNA-specific tools. Enzymes like amylase are needed for starch digestion.

Correct answer: A. neoschizomers

Neoschizomers can cut at different positions on the same recognition sequence. This is restriction enzyme terminology.

Correct answer: A. Bacterial enzyme that cuts a specific DNA sequence

The governing concept is the basic definition of a restriction enzyme. In school-level biotechnology, a restriction enzyme is generally described as a bacterial enzyme that recognizes a particular DNA sequence and cuts the DNA at or near that site. This concise description includes both its biological source and its essential function, so option A is the most accurate one-line answer. The wording does not claim that every bacterial enzyme is a restriction enzyme; it identifies the relevant enzyme by its sequence-specific DNA-cutting activity. The other options describe unrelated functions of pigments, vitamins, or hormones and do not involve sequence-specific DNA cleavage.

Correct answer: A. Vector and insert with compatible ends are mixed with ligase

Compatible ends and DNA ligase are needed to make a recombinant plasmid. Restriction enzymes can create compatible ends.

Correct answer: A. bacteria

Restriction enzymes were originally found in bacteria and are linked with defense. Remember this as a source-based fact.

Correct answer: A. Cutting at recognition site and recombinant DNA

The core idea of this topic is specific DNA cutting. It should be studied with recombinant DNA technology.

Correct answer: B. Specific nucleotide sequence

Restriction enzymes recognize specific nucleotide sequences in DNA. Keep sequence specificity as the key exam point.

Correct answer: C. Bacterial cell

Restriction enzymes are naturally found in bacteria. Bacteria use them to defend against foreign DNA.

Correct answer: A. Cutting DNA at specific sites

The direct answer is option A: cutting DNA at specific sites. Restriction enzymes are proteins that recognise particular short DNA sequences called recognition sites. When the matching sequence is present, the enzyme cuts the DNA backbone at a defined position, either within the site or close to it, depending on the enzyme. This makes controlled DNA manipulation possible in cloning and genetic analysis, so restriction enzymes are commonly called molecular scissors. Option A is correct because it states the essential general function. Option B, storing fat, is a storage function of cells and tissues, not a DNA-enzyme function. Option C, colouring starch, is not the biological role of restriction enzymes; starch tests involve other substances and indicators. Option D, making protein sweet, has no recognised molecular-biology meaning and does not describe enzyme action here. The word “specific” must not be missed: restriction enzymes do not cut randomly everywhere. Their recognition sequence controls where cutting occurs, and different enzymes recognise different sequences. The resulting fragments can then be separated, analysed, or joined to another DNA fragment. Memory cue: restriction enzymes are DNA scissors with an address—they cut only at their recognised address.

Correct answer: B. It is a recognized sequence of DNA

The direct answer is option B: a recognised sequence of DNA. A restriction site is a short, specific sequence of nucleotide bases that a particular restriction enzyme can identify. Recognition allows the enzyme to attach correctly and cut the DNA at a defined location or near it. Many well-known restriction sites are palindromic, meaning the sequence reads similarly in the relevant direction on the two complementary strands, although the key definition here is recognition by the enzyme. Option A is wrong because a restriction site is not a cell membrane or a membrane part. Option B is correct because it precisely describes a DNA sequence recognised by a restriction enzyme. Option C is wrong because the site is not only the end of a protein; restriction enzymes act on DNA, not on protein ends. Option D is wrong because a water molecule is a small chemical compound and cannot be a DNA recognition sequence. After recognising the site, the enzyme may create sticky ends or blunt ends, depending on its cutting pattern. Do not confuse “site” with a place in a cell; in this chapter it means a particular address in the DNA sequence. Memory cue: restriction site means the DNA address where the restriction enzyme finds its target.

Correct answer: D. They have no single stranded overhang

The governing concept is the structure of DNA ends produced by restriction enzymes. A blunt end is formed when both strands of double-stranded DNA are cut at the same position or directly opposite one another. Consequently, the resulting termini are flush and contain no protruding single-stranded DNA overhang. Therefore option D is correct. Sticky or cohesive ends are different because staggered cuts leave short single-stranded overhangs that can pair with complementary sequences. Blunt ends do not have a long protein projection, do not contain only RNA, and are not coloured; those descriptions are unrelated to DNA-end structure.

Correct answer: B. Restriction enzyme cuts and ligase joins

Restriction enzyme cuts DNA and DNA ligase joins cut fragments. This is a basic difference in recombinant DNA technology.

Correct answer: C. Recognising and cutting a specific DNA sequence

EcoRI is a restriction endonuclease that recognises the specific DNA recognition sequence 5′-GAATTC-3′ and cuts it. This cleavage usually produces sticky ends, which are useful in recombinant DNA technology. Producing glucose or carrying out photosynthesis are not functions of EcoRI. Exam tip: Remember EcoRI by linking it with specific DNA-sequence recognition and cutting.

Correct answer: A. Escherichia

In EcoRI E is linked with the genus Escherichia of the source organism. Source organism is important in nomenclature.

Correct answer: B. Cutting foreign DNA

Bacteria can cut foreign DNA such as phage DNA using restriction enzymes. Remember this as the natural defense role.

Correct answer: A. Methylation

Bacteria protect their own DNA by methylation. Foreign DNA may be cut if it lacks this protection.

Correct answer: B. Cutting foreign DNA and protecting self DNA

In this system the restriction enzyme cuts foreign DNA and methylase protects self DNA. It is linked with bacterial defense.

Correct answer: A. They cut at or near recognition sites

Direct answer: Option A, they cut at or near recognition sites. Type II restriction enzymes recognize short, specific DNA sequences and usually cleave at a defined position within that sequence or very close to it. This predictable relationship between recognition and cutting makes the fragment sizes reproducible. Researchers can therefore use these enzymes in gene cloning, restriction mapping, and recombinant-DNA construction. Option A is correct. Option B, working only in leaves, is false because these enzymes are not limited to leaf cells and are commonly used in purified laboratory reactions. Option C, not recognizing DNA, contradicts their defining function. Option D, always joining proteins, is false because restriction enzymes cut DNA rather than join proteins; DNA ligase is the enzyme associated with sealing DNA fragments. Memory cue: Type II means a predictable recognition site and cut, making it convenient for laboratory work.

Correct answer: A. Cutting vector and gene

Vector and desired gene can be cut with a suitable restriction enzyme. This can produce compatible ends.

Correct answer: B. When both have compatible ends

The governing concept is compatibility of DNA ends during recombinant-DNA construction. When the foreign gene and vector are cut to produce complementary sticky ends, the exposed bases can hydrogen-bond with one another and temporarily hold the two DNA molecules in the correct alignment. This increases the efficiency of subsequent sealing by DNA ligase, which forms permanent phosphodiester bonds in the backbone. Therefore option B is correct. Incompatible ends cannot pair efficiently, burning destroys DNA, and absence of ligase prevents stable covalent joining even when temporary base pairing occurs. Compatible ends therefore make insertion easier, though ligase is still required for permanent joining.

Correct answer: C. Size

In gel DNA fragments separate mainly on the basis of size. Smaller fragments generally move farther.

Correct answer: B. To estimate fragment sizes

The direct answer is B: a DNA ladder is used to estimate fragment sizes. A DNA ladder contains several DNA fragments whose lengths are already known, usually measured in base pairs. During gel electrophoresis, the ladder is loaded in a separate lane beside the unknown sample. DNA fragments move through the gel according to size: smaller fragments generally travel farther than larger ones. By comparing a sample band with nearby ladder bands, we estimate its length. Option A is wrong because a ladder does not inactivate an enzyme; special chemicals or heat may do that. Option B is correct because known-size bands provide a measuring scale. Option C is wrong because the ladder does not burn DNA. Option D is wrong because it does not melt the gel; the gel remains the medium through which fragments move. The estimate is usually approximate, not an exact chemical measurement. Memory cue: ladder means a size ruler for DNA bands.

Correct answer: A. They have a single stranded overhang

Sticky ends have a single stranded overhang. This makes them able to join with complementary ends.

Correct answer: C. Three

The direct answer is option C: three fragments. A linear DNA molecule has two ends. Each complete cut made at an internal restriction site divides one continuous piece into two pieces. Therefore, for linear DNA, the number of fragments is the number of cuts plus one: \(F=C+1\). Here there are two restriction sites and complete digestion means both sites are cut, so \(F=2+1=3\). Visualise the order as end—site 1—middle section—site 2—end; the cuts separate the molecule into a left end fragment, a middle fragment, and a right end fragment. Option A, one fragment, would mean no effective cut. Option B, two fragments, would result from one cut in linear DNA, not two. Option C is correct because two cuts produce three separate pieces. Option D, four fragments, would require three cuts in linear DNA, assuming all cuts are distinct and complete. This rule is specifically for linear DNA. For a circular DNA molecule, two distinct cuts generally produce two fragments because there are no free ends, so do not apply the linear formula blindly. Memory cue: linear DNA fragments = cuts plus one; circular DNA fragments = cuts, under the usual complete-digestion assumptions.

Correct answer: A. Two

In circular DNA after complete digestion fragments equal the number of cuts. Two cuts give two fragments.