Restriction Enzymes

Correct answer: A. Its own DNA is protected from cutting

Methylation can protect self DNA from being cut. This is part of the restriction-modification system.

Correct answer: A. Cutting phage DNA

Restriction enzymes help bacterial defense by cutting foreign phage DNA. This is their natural role.

Correct answer: A. Enzyme cuts non-specifically under wrong conditions

In star activity an enzyme may cut outside its normal recognition specificity. It can be linked with improper reaction conditions.

Correct answer: A. Enzyme activity may decrease or its cutting pattern may change

A restriction enzyme needs an appropriate buffer containing the correct pH, salts, and cofactors such as Mg²⁺. With a wrong buffer, enzyme activity may fall, DNA cutting may fail, or non-specific cutting may occur. Enzyme activity cannot become infinite; it depends on pH and ionic conditions. Exam tip: in restriction digestion questions, check whether the recommended buffer and its concentration are used.

Correct answer: A. The number of restriction fragments will increase

EcoRI cuts DNA at its specific recognition sequence. Therefore, when the number of EcoRI sites increases, complete digestion produces more cuts and hence more restriction fragments. In linear DNA, n cuts usually give n+1 fragments, whereas in circular DNA, n cuts give n fragments. The option stating that fragment number decreases is incorrect. Exam tip: Count fragments by considering both the number of restriction sites and whether the DNA is linear or circular.

Correct answer: A. One expected cut will not occur and a larger fragment may appear

Loss of a restriction site can reduce cutting. Two previous fragments may appear as one larger fragment.

Correct answer: A. To cut DNA into specific fragments

In RFLP fragment length differences are observed after restriction cuts. Therefore restriction enzyme is the first important tool.

Correct answer: A. Cutting DNA into fragments that form a pattern

Restriction enzymes can cut DNA into fragments with a repeatable pattern. This helps in DNA comparison.

Correct answer: A. Which enzyme recognizes which sequence

The direct answer is option A: which enzyme recognises which sequence. Specificity means that an enzyme does not cut every DNA molecule randomly. It recognises a particular short DNA sequence, called its recognition sequence or restriction site, and cuts at a defined position related to that site. To test specificity, one should therefore match each enzyme with the sequence it can recognise. Option A is correct because enzyme-sequence matching directly tests the molecular rule that determines where cutting occurs. Option B, asking which water is sweet, concerns taste and has no relation to DNA or enzyme recognition. Option C, asking which leaf is bigger, compares physical size and does not test a molecular recognition sequence. Option D, asking which stone is heavier, compares mass and is also unrelated to enzyme action. The word “specificity” is important: it means selective recognition, not merely that an enzyme can cut something. For example, an enzyme that recognises one sequence will not normally cut a different sequence under the same conditions. The exact recognition sequence depends on the enzyme. Exam cue: when specificity appears, look for the link between a particular enzyme and its particular DNA sequence.

Correct answer: A. They will not pair easily

Direct answer: Option A, they will not pair easily. Sticky ends contain exposed single-stranded bases. Efficient pairing requires complementary base sequences: for example, A pairs with T and G pairs with C. If two fragments have non-complementary sticky ends, the exposed bases cannot form the correct hydrogen-bonded pairs, so the fragments will not align and remain together easily. DNA ligase also cannot efficiently seal a stable junction without proper alignment, although special laboratory methods may join non-compatible ends after further processing. Option A is correct. Option B, they will always join, is the opposite of the expected result. Option C, becoming chloroplasts, is biologically impossible; chloroplasts are cell organelles, not DNA fragments. Option D, becoming soil, is unrelated. Exam cue: compatible sticky ends pair; ligase seals the aligned backbone.

Correct answer: A. DNA ligase

DNA ligase is needed to permanently join compatible ends. Restriction enzymes act earlier in cutting.

Correct answer: A. EcoRI-sticky ends

The direct answer is A: EcoRI produces sticky ends. Restriction enzymes are proteins that recognize particular DNA sequences and cut the DNA backbone. EcoRI is a well-known restriction enzyme that cuts a specific site and can leave compatible sticky ends. Therefore the enzyme-product pair in A is scientifically appropriate. Option B is wrong because chlorophyll is a photosynthetic pigment made by cells, not a product made by EcoRI. Option C is wrong because glucose is a sugar and is not produced when BamHI cuts DNA. Option D is wrong because a cell wall is a structural layer, not the product of HindIII digestion. The word “product” here means the immediate result of enzyme action on DNA, not something made by metabolism. A DNA cut may generate fragments and sticky or blunt ends, depending on the enzyme. Memory cue: restriction enzymes cut DNA; they do not make food pigments, sugars, or cell structures.

Correct answer: A. Use cold storage and proper buffer

Restriction enzymes are proteins and can be condition-sensitive. Proper storage and buffer preserve activity.

Correct answer: A. Methylation may have blocked enzyme cutting

Direct answer: Option A, methylation may have blocked enzyme cutting. The expected result assumes that the restriction enzyme can recognize and cut its target site. If a methyl group is attached at or near that recognition sequence, the enzyme may be unable to bind properly or may fail to cleave the DNA. Consequently, the expected smaller restriction fragment or gel band pattern may not appear. Option A is correct and explains the missing cut. Option B, DNA had no bases, is impossible because DNA is made of nucleotides containing bases; without bases there would be no DNA recognition site. Option C, the enzyme made protein, is not what restriction enzymes do; they catalyse DNA cleavage. Option D, no current is needed in gel, is false and irrelevant: electrophoresis normally uses an electric field, but that does not explain methylation-dependent failure of cutting. Memory cue: methylated site can hide the recognition site from the cutter.

Correct answer: A. Incorrect because different enzymes can recognize different sequences

Restriction enzymes are sequence-specific and the sequence can differ by enzyme. This is a fundamental fact.

Correct answer: A. specific cutter

Restriction enzymes cut at specific recognition sequences. Therefore they are considered specific cutters.

Correct answer: A. When complementary overhangs are absent

Blunt ends lack overhangs so base pairing support is absent. This can make ligation comparatively harder.

Correct answer: A. First cut DNA then separate fragments

First the restriction enzyme cuts DNA then the fragments are separated in a gel. This is the logical order of the experiment.

Correct answer: A. Positions of restriction sites

The governing concept is the restriction map of a plasmid. A plasmid map is a schematic representation of important sequence features, including the locations of restriction-enzyme recognition sites. If EcoRI and HindIII sites are marked, the map tells the researcher where those enzymes can cut the circular DNA and helps predict the fragments produced by single or double digestion. It can therefore guide vector selection, insert placement, and cloning strategy. Option A is correct. A restriction map does not describe the taste of a protein, the colour of a cell, or the amount of water; those properties are unrelated to enzyme-site positions on DNA.

Correct answer: A. It contains many restriction sites

A multiple cloning site provides options for inserting DNA through many restriction sites. Link it with cloning flexibility.

Correct answer: A. Insert may ligate in either direction

With same compatible ends orientation control may be reduced. Directional cloning reduces this issue.

Correct answer: A. Because it cuts internally

Restriction enzymes are endonucleases because they cut within DNA. Exonucleases remove nucleotides from ends.

Correct answer: A. Defense in bacteria and DNA cloning in lab

The natural role is defense against foreign DNA and the lab role is DNA manipulation. This gives a complete view of the topic.

Correct answer: A. It can cut a specific sequence and make sticky ends

The usefulness of EcoRI is linked with specific cutting and sticky ends. Connect it with recombinant DNA.

Correct answer: A. Recognition sequence of that enzyme

A restriction enzyme recognizes and cuts DNA only at its specific recognition sequence. Therefore, to predict a cut site, the DNA must be checked for that enzyme’s recognition sequence. The GC percentage describes a general property of DNA but does not identify the cut site of a particular restriction enzyme. Exam tip: Learn each enzyme together with its recognition sequence.