Cutting of DNA

Correct answer: A. DNA ends that can join with each other

Compatible ends are DNA termini whose shapes and exposed bases allow two fragments to align and associate. In many cloning experiments they are complementary sticky ends produced by the same restriction enzyme; their overhanging bases pair through hydrogen bonds. DNA ligase can then form covalent phosphodiester bonds across the aligned backbone, making a stable recombinant molecule. Some blunt ends can also be ligated, although they lack complementary overhangs and are generally less efficient. Therefore option A is correct. The other choices incorrectly describe ends as proteins, salts, or structures that necessarily break.

Correct answer: B. Joining can be difficult or impossible

Incompatible ends cannot pair properly. Therefore enzyme choice is important in cloning strategy.

Correct answer: A. DNA will not be specifically cut by that enzyme

A restriction enzyme cannot cut without its recognition site. Therefore sequence information is important.

Correct answer: C. Three fragments can form if DNA is linear

In linear DNA two cut sites can produce three fragments. Fragment number depends on site positions.

Correct answer: D. linear plasmid / Linear plasmid

A circular plasmid cut at one site can become a linear molecule. Supercoiled and linear forms can appear differently on gel.

Correct answer: A. Two DNA fragments

Two cuts on circular DNA generally give two fragments. This is useful for insert release or diagnostic digestion.

Correct answer: B. To estimate the sizes of DNA fragments

A DNA ladder contains DNA fragments of known base-pair lengths. During gel electrophoresis, the positions of sample bands are compared with ladder bands to estimate the sizes of DNA fragments produced by restriction digestion. The restriction enzyme cuts DNA; the ladder does not activate it. Exam tip: smaller DNA fragments generally migrate farther through an agarose gel.

Correct answer: A. DNA fragments separate according to size

The governing concept is agarose-gel electrophoresis. DNA molecules carry an overall negative charge because of their phosphate groups, so an electric field moves them through the gel toward the positive electrode. Smaller fragments pass through the gel pores more easily and migrate farther than larger fragments. After staining, many molecules of similar length accumulate in visible bands. Thus option A is correct: the band pattern represents the sizes of fragments produced by restriction digestion. The enzyme does not turn into protein, the buffer does not solidify to create bands, and water cannot become DNA.

Correct answer: B. More distance than large fragments

Direct answer: option B, more distance than large fragments. In agarose gel electrophoresis, DNA carries an overall negative charge because of its phosphate groups. An electric field pulls DNA toward the positive electrode. The gel acts like a tiny sieve: small DNA fragments pass through its pores more easily, so they move faster and travel farther during the same time. Large fragments face greater resistance and remain nearer the wells. Option A is opposite to the observed size-separation rule. Option B is correct. Option C is wrong because DNA does move when an electric field is applied. Option D is wrong in two ways: DNA generally moves toward positive, not from positive to negative, and the word “always” ignores the need for proper electrophoresis conditions. The distance also depends on voltage, gel concentration and time, but for fragments run together under the same conditions, smaller ones travel farther. Memory cue: small DNA slips through the gel sieve; negative DNA goes to positive.

Correct answer: A. Impurities can inhibit the enzyme

Contaminants such as phenol salt or ethanol can reduce enzyme activity. Clean DNA is better for digestion.

Correct answer: A. It can affect enzyme recognition or cutting

Some restriction enzymes do not cut methylated sites. Therefore methylation status can change digestion results.

Correct answer: B. By methylation

Bacteria can methylate their recognition sites. This protects their own DNA from restriction enzymes.

Correct answer: A. Cutting foreign DNA

In a restriction-modification system, a restriction enzyme cuts specific recognition sequences in foreign, usually unmethylated DNA. This protects bacteria from invading DNA such as bacteriophage DNA. Replication of host DNA is not the role of restriction; the modification enzyme methylates host DNA and helps protect it from cleavage. Exam tip: restriction means cutting, whereas modification protects the host DNA.

Correct answer: A. Enzyme like SmaI

Direct answer: option A, an enzyme such as SmaI. Restriction enzymes cut DNA at specific recognition sequences. The cut pattern determines whether the DNA ends are sticky or blunt. SmaI cuts straight across the two DNA strands at its recognition site, producing blunt ends, in which no single-stranded overhang remains. Option A is therefore correct. Option B, EcoRI, commonly produces sticky ends with a characteristic overhang, so it is not a blunt-end example here. Option C, HindIII, also produces sticky ends rather than blunt ends. Option D, BamHI, produces sticky ends as well. The word “always” in the other options is especially unsuitable because the question asks for an enzyme known for blunt-end production, not merely any restriction enzyme. Blunt ends can be joined by ligase, although ligation is often less efficient than joining compatible sticky ends. Memory cue: SmaI is a standard blunt cutter; EcoRI, HindIII and BamHI are familiar sticky cutters.

Correct answer: B. Restriction digestion of DNA

HindIII is a restriction endonuclease that recognizes a specific DNA recognition sequence and cuts the DNA at or near that site. Hence, it is associated with restriction digestion of DNA. Protein translation occurs on ribosomes using RNA, not through HindIII. Exam tip: Enzymes named HindIII and EcoRI are commonly restriction enzymes used to cut DNA.

Correct answer: A. Complete digestion may not occur

Restriction digestion depends on an adequate amount of active enzyme relative to the DNA quantity, recognition-site number, reaction time, and buffer conditions. If too little enzyme is added, some recognition sites may remain uncut, so a fraction of the starting DNA stays intact or only partially digested. This produces an incomplete or mixed band pattern during gel analysis. Therefore option A is correct. A low enzyme amount does not make DNA overcut; overdigestion is not an inevitable result and would require different conditions. Protein formation and gel dissolution are unrelated to this enzyme-to-substrate limitation.

Correct answer: B. Incomplete digestion can occur

With too little time the enzyme may not cut all sites. Proper incubation helps complete digestion.

Correct answer: C. Stopping enzyme activity

Some enzymes can be inactivated by heat treatment. This stops unwanted cutting in the next step.

Correct answer: C. To keep enzyme stable

Restriction enzymes are temperature-sensitive proteins. Cold storage helps preserve enzyme activity.

Correct answer: A. To interpret results correctly

A control shows whether DNA and enzyme reaction are working properly. It helps troubleshooting.

Correct answer: A. To compare cut pattern with original DNA form

An uncut control shows how DNA looked before digestion. This helps understand digestion success.

Correct answer: A. The uncut DNA pattern in the absence of restriction enzyme

The governing concept is the use of a negative control in restriction digestion. A no-enzyme control contains the DNA and the reaction mixture but omits the restriction enzyme. Therefore, any band pattern observed in this tube represents the original, undigested DNA, including its initial form and possible pre-existing degradation. The enzyme-treated sample is compared with this control to identify bands produced by cleavage. Option A is correct because it shows the baseline uncut pattern. Option B is not measured by this control, option C confuses DNA with protein, and option D is incorrect because the buffer is normally still present.

Correct answer: A. To use it in cloning

Gel extraction separates the desired DNA fragment from other fragments. It is useful in insert preparation.

Correct answer: A. To prepare DNA fragments and compatible vector ends

The governing concept is restriction-based DNA cutting during recombinant DNA construction. A restriction enzyme recognizes a particular sequence and cleaves the DNA, producing a desired insert fragment and an opened, or linearized, vector. If the same or compatible enzymes are used, the ends of the insert and vector can pair by complementary base pairing, allowing DNA ligase to join them in the next step. Option A is correct because cutting creates the physical DNA pieces and compatible ends required for assembly. Option B concerns protein digestion, not DNA cloning. Options C and D describe biologically impossible conversions and are unrelated to recombinant DNA construction.

Correct answer: A. Remember restriction enzyme recognition site sticky end and gel check

In Cutting of DNA a specific enzyme cuts DNA into predictable fragments. Fragments are checked by gel.