Restriction Enzymes

For linear DNA after complete digestion fragments are one more than cuts. Apply this in small numerical MCQs.

A restriction enzyme can cut DNA only when its particular recognition sequence is present. If a linear DNA molecule contains no such site, the enzyme has no suitable location to bind and cleave. Consequently, the DNA molecule remains intact under the stated treatment. Because the molecule is linear and is not cut, it is counted as one continuous DNA fragment. This conclusion assumes the enzyme is active and the conditions are suitable, as the question intends.

Therefore option A, one uncut fragment, is correct. Two fragments would require one cut in a linear molecule, while three or four fragments would require still more cuts. Since the question states that there is no recognition site at all, none of those cuts can occur. The original DNA does not disappear; it remains as a single fragment of its original length.

Differences in DNA sequences can change number and positions of restriction sites. This can change the fragment pattern.

Direct answer: Option A, cutting may be blocked or reduced. A methyl group added to DNA bases can change the chemical environment around a restriction recognition site. For many restriction enzymes, methylation prevents the enzyme from binding correctly or cutting efficiently. Bacteria use restriction-modification systems: methylation can protect their own DNA, while foreign unmethylated DNA may be cut. This is not universal for every enzyme, so “may” is the accurate wording. Option A is correct. Option B, cutting always increases, is wrong because methylation commonly inhibits or reduces cutting. Option C, DNA becoming protein, is impossible by methylation; DNA and protein are different biological molecules. Option D, an enzyme becoming a vitamin, is unrelated and chemically incorrect. Memory cue: methyl marks can act like a “do not cut” sign, but check enzyme-specific exceptions.

Without methylation foreign DNA is not protected like self DNA. Therefore restriction enzymes may cut it.

The substrate of a restriction enzyme is DNA. It recognises a specific recognition sequence in DNA and cuts it at or near that site. RNA and proteins are not the usual substrates of restriction endonucleases. Exam tip: remember that restriction endonucleases act on DNA recognition sites.

Restriction enzymes recognise specific DNA recognition sequences and cut phosphodiester bonds. This produces cut DNA ends, which may be sticky ends or blunt ends. Newly synthesised mRNA is produced during transcription, not by restriction enzyme action. Exam tip: remember that the key function of a restriction enzyme is to cut DNA at specific sites.

The governing concept is the extent of restriction-enzyme digestion. Partial digestion means that a restriction enzyme has cut only some available recognition sites or only some DNA molecules, rather than cleaving every susceptible site. Consequently, an experiment may contain completely cut fragments, larger partially cut fragments, and sometimes uncut DNA. This mixture can produce extra or unexpected bands during gel electrophoresis. Option A is correct because it directly describes DNA digestion that has not gone to completion. Complete digestion would mean that all accessible target sites were cleaved under the reaction conditions. Photosynthesis, translation, and osmosis are unrelated biological processes and cannot name the incomplete cleavage of DNA by a restriction enzyme.

Complete restriction digestion means that the selected restriction enzyme has cleaved every accessible copy of its recognition sequence in the DNA sample under the stated reaction conditions. Consequently, the DNA produces the expected set of fragments for that restriction map, without a substantial population of molecules remaining uncut or only partly cut. Option A expresses this governing principle. Complete digestion does not mean that DNA is converted into protein, that the electrophoresis gel disappears, or that a cell changes colour. Those alternatives confuse DNA cleavage with translation, laboratory material behaviour or cell physiology. Proper enzyme amount, buffer, temperature and incubation time help achieve complete digestion.

For restriction enzymes name source sequence and cut end are all important. Study them like a table for exams.

The direct answer is A: both can get compatible sticky ends. A restriction enzyme cuts DNA at a particular recognition sequence. EcoRI cuts the vector and the gene at its recognition sites and usually leaves short single-stranded overhangs called sticky ends. Because the same enzyme makes matching bases exposed on both DNA pieces, the gene can temporarily pair with the vector by complementary base pairing. DNA ligase can then seal the sugar-phosphate backbones and form recombinant DNA. Option A is correct because compatible ends make joining easier and more specific. Option B is wrong: cutting DNA does not convert either DNA molecule into a protein; proteins are made from amino acids. Option C is wrong: cell walls are cellular structures, not products formed by restriction digestion. Option D is wrong: DNA is not changed into water by an enzyme cut. The complete process is cut, allow matching ends to pair, and seal with ligase. Memory cue: same cutter, matching sticky ends, easier joining.

When both DNA strands of a circular plasmid are cut at one restriction site, the circle opens to form one linear DNA molecule. A nicked circular plasmid results when only one strand is broken, whereas a restriction enzyme cut normally cleaves both strands. Exam tip: one complete cut in circular DNA gives one linear fragment; two cuts give two linear fragments.

In linear DNA, the restriction enzyme makes one cut at each restriction site. Therefore, the number of fragments equals the number of cuts plus one. Three cuts produce four fragments. The option three is incorrect because it equals the number of cuts, not the number of fragments. Exam tip: For linear DNA, remember: fragments = restriction sites + 1.

In circular DNA after complete digestion fragments equal the number of cuts. Three cuts give three fragments.

BamHI is a restriction endonuclease that cuts DNA only at its specific recognition sequence, 5′-GGATCC-3′. If this site is absent from linear DNA, there is no position for BamHI to cleave, so the DNA remains uncut. Linear DNA would form two fragments only if it had one BamHI site. Exam tip: for linear DNA, the number of fragments is usually one more than the number of cuts.

The governing concept is that restriction enzymes cut DNA at specific recognition sequences, producing fragments of different lengths. During agarose-gel electrophoresis, DNA fragments move toward the positive electrode because DNA carries a negative charge. Smaller fragments pass more easily through the gel network and travel farther, whereas larger fragments remain closer to the wells. Each visible band usually represents many DNA molecules of approximately the same length, not one unrelated substance. Therefore option A is correct. Hormones, fat droplets, and soil particles do not explain the characteristic DNA separation pattern produced after restriction digestion.

Two different ends help control insert orientation. Treat this as an advantage of directional cloning.

Sticky ends have single-stranded DNA overhangs. Complementary nitrogenous bases on these overhangs join through hydrogen bonds; this is complementary base pairing. It holds the DNA fragments together in the correct alignment temporarily, after which DNA ligase forms permanent phosphodiester bonds. Protein coagulation is unrelated to joining DNA ends. Exam tip: for questions on sticky ends and ligation, first identify complementary base pairing.

The governing concept is restriction-enzyme specificity. EcoRI is a bacterial restriction endonuclease that recognizes the six-base-pair palindromic sequence 5′-GAATTC-3′ and cuts within this site, producing characteristic sticky ends. A palindrome reads correspondingly on the two complementary DNA strands, which is important for recognition by many restriction enzymes. Therefore option A, GAATTC, is correct. GGATCC is the familiar recognition sequence for BamHI, AAGCTT is associated with HindIII, and GGCC is associated with other restriction enzymes such as HaeIII. The alternatives are therefore plausible enzyme sequences but not EcoRI’s sequence.

The restriction enzyme BamHI recognizes the palindromic sequence 5′-GGATCC-3′ and usually cuts between G and GATCC. GAATTC is the recognition sequence of EcoRI, while AAGCTT is recognized by HindIII. Exam tip: learn common restriction enzymes together with their specific recognition sequences.

HindIII is a restriction enzyme that recognises the palindromic sequence 5′-AAGCTT-3′ and cuts between the two A bases. Therefore, option A is correct. GAATTC is the recognition sequence of EcoRI, whereas CTGCAG is associated with PstI. Exam tip: Memorise HindIII with AAGCTT and EcoRI with GAATTC as a pair.

HaeIII recognizes the sequence 5′-GGCC-3′ and cuts at the same position on both DNA strands. Therefore, it produces blunt ends with no single-stranded overhang. A 5′ or 3′ overhang, also called a sticky end, results from staggered cutting and is not produced by HaeIII. Exam tip: a straight cut gives blunt ends, whereas a staggered cut gives sticky ends.

A DNA ladder is a mixture of DNA fragments of known sizes. During gel electrophoresis, the positions of sample bands are compared with ladder bands to estimate their fragment sizes. A restriction enzyme cuts DNA; the ladder serves as a molecular size standard. Exam tip: remember that a DNA ladder is also called a molecular size marker.

Incomplete digestion or low DNA can make a band weak or absent. Remember such reasons for practical questions.

Restriction enzymes cut DNA at particular recognition sites. If digestion is complete, the molecules are cut wherever the relevant sites are present, producing the expected smaller fragments. If some DNA molecules are not cut at all sites, partly cut or uncut molecules remain larger than the expected fragments.

On a gel, these incompletely digested molecules can appear as an additional band of larger size, often alongside the normal digestion bands. Thus an extra larger band may indicate incomplete digestion, making option A the best answer. It does not indicate complete DNA loss, chlorophyll synthesis, or translation alone, because those processes do not explain the presence of an unexpectedly large DNA fragment in a restriction-digestion sample.