Restriction Enzymes

Correct answer: A. CTGCAG

Restriction enzymes recognize particular short DNA sequences and cut DNA at or near those sites. PstI is a restriction endonuclease whose recognition sequence is 5′-CTGCAG-3′. It produces compatible sticky ends when it cleaves the DNA at its specific site. Therefore, option A, CTGCAG, is correct. AAGCTT is recognized by HindIII, GAATTC is associated with EcoRI, and GGCC is the central recognition sequence associated with HaeIII in standard examples. The key principle is that each restriction enzyme has a characteristic recognition sequence, so the enzyme name must be matched with the correct DNA base order.

Correct answer: A. More often than a six-base sequence

A shorter recognition sequence can statistically occur more often. In exams consider four-base sites more frequent.

Correct answer: A. Less frequent

A longer recognition sequence is less likely to occur frequently. This can affect the number of restriction fragments.

Correct answer: A. They cut at defined positions

Defined cut positions keep experiments controlled. Therefore Type II enzymes are convenient for cloning.

Correct answer: A. Magnesium ion

Many Type II restriction enzymes require Mg ions. In laboratory conditions buffer and ions are important.

Correct answer: A. Non-specific cutting under improper reaction conditions

Improper buffer or excess enzyme may cause star activity. Remember it as non-specific cutting in exams.

Correct answer: A. Providing proper pH and ionic conditions for the enzyme

Buffer provides the right environment for enzyme activity. In exams treat buffer as an essential reaction condition.

Correct answer: A. To preserve enzyme activity

Protein enzymes can become inactive by heat. Therefore they are stored in cold conditions in laboratories.

Correct answer: A. About 37 degrees Celsius

Many bacterial enzymes are active around 37 degrees Celsius. However each enzyme may have its own optimal condition.

Correct answer: A. Cutting the same DNA with two restriction enzymes

In double digestion two enzymes act on the same sample. This allows use of two different cut sites.

Correct answer: A. Directional cloning can become easier

Two different ends make correct orientation of an insert easier. In exams link directional cloning with two enzymes.

Correct answer: A. Different enzymes recognizing the same sequence

Isoschizomers may come from different sources but recognize the same sequence. This term is useful in enzyme selection.

Correct answer: A. Positions of restriction enzyme cut sites on DNA

A restriction map shows the relative positions of recognition or cutting sites for restriction enzymes on a DNA molecule. It can be used to predict the order and sizes of DNA fragments formed after digestion. It does not show the total number of genes. Exam tip: focus on the cut sites and the distances between them on a map.

Correct answer: A. Differences in lengths of cut DNA fragments

In RFLP variation in fragment length is observed after restriction cutting. It is useful in DNA identification.

Correct answer: A. Cutting DNA into specific fragments

Specific cutting helps analyze individual fragment patterns. In exams link it with DNA fragment pattern.

Correct answer: A. Both are restriction enzymes

EcoRI and BamHI are both restriction endonucleases used in molecular biology to cut double-stranded DNA at defined recognition sequences. Their exact recognition sequences and cleavage patterns are different, so they should not be treated as identical enzymes. Nevertheless, the common feature asked here is that both belong to the restriction-enzyme class and act as sequence-specific DNA-cutting tools. Therefore option A is correct. Options B, C and D refer to hormones, vitamins and lipids, which are different biological categories and do not describe either EcoRI or BamHI. The question asks for a shared classification, not for their individual recognition sequences.

Correct answer: A. EcoRI

The direct answer is A: EcoRI. Sticky ends are short single-stranded overhangs produced when the two DNA strands are cut at staggered positions. These complementary overhangs can pair with matching ends, which makes joining DNA fragments easier. EcoRI is a standard example: it recognizes the sequence GAATTC and cuts in a staggered way, producing compatible sticky ends. Step by step: EcoRI finds its recognition site; it cuts the two strands at offset positions; an overhanging end remains; a matching DNA fragment can base-pair with it; DNA ligase can then seal the backbone. Option A is correct. Option B, HaeIII, is commonly taught as producing blunt ends because it cuts across both strands at the same position. Option C, SmaI, is also a standard blunt-end cutter. Option D, AluI, is generally classified as a blunt-end cutter as well. Thus the other choices are not the usual sticky-end example in this syllabus. Memory cue: EcoRI is the classic sticky-end enzyme; staggered cuts leave helpful overhangs.

Correct answer: A. DNA ligase and proper reaction conditions

Blunt ends lack complementary overhangs so joining can be harder. Proper ligase conditions are needed.

Correct answer: A. Sticky ends

Sticky ends have single-stranded overhangs. If the overhangs are complementary, they can base-pair first, holding the DNA fragments together before DNA ligase forms permanent phosphodiester bonds. Blunt ends have no overhangs, so their ligation is generally less efficient. Exam tip: an overhang that undergoes complementary base pairing indicates sticky ends.

Correct answer: A. It may be difficult to see DNA fragment bands on the gel

When the DNA amount is very low, the amount of restriction fragments produced is also low. Therefore, their bands may be very faint or may not be visible on the gel. A low DNA amount does not change the sequence specificity of the enzyme, and DNA normally moves towards the positive electrode because of its negatively charged phosphate groups. Exam tip: insufficient DNA sample is a common reason for faint or absent bands.

Correct answer: A. Cutting will decrease or stop

An inactive enzyme will not cut properly even if recognition sites are present. Treat enzyme activity as an essential condition.

Correct answer: A. Two

In linear DNA, the enzyme makes one cut at its single restriction site. One cut divides one linear DNA molecule into two separate fragments, so the correct answer is two. The option ‘one’ would apply if no cut were made. Exam tip: for linear DNA, number of fragments = number of cuts + 1.

Correct answer: A. One linear DNA molecule

When a circular plasmid has only one restriction site, the restriction enzyme makes one cut at that site. The closed circle opens to form one linear DNA molecule. Two linear fragments would require at least two cuts in the plasmid. Exam tip: for circular DNA, the number of linear fragments equals the number of cuts.

Correct answer: A. Three

Two cuts in linear DNA produce three fragments. For exams remember fragments equal cuts plus one for linear DNA.

Correct answer: A. Fragments equal number of cuts

For circular DNA after complete digestion fragments equal the number of cuts. This is useful in mapping.