Ligase Enzymes

Correct answer: A. NAD⁺

E. coli DNA ligase uses NAD⁺ as a cofactor to form phosphodiester bonds between broken DNA ends. In contrast, many viral DNA ligases, such as T4 DNA ligase, are ATP-dependent. FAD and biotin are not cofactors for this ligation reaction. Exam tip: remember bacterial DNA ligase as generally NAD⁺-dependent and T4 DNA ligase as ATP-dependent.

Correct answer: B. T4 DNA ligase

The direct answer is option B: T4 DNA ligase. T4 is a bacteriophage, meaning a virus that infects bacteria. The enzyme named T4 DNA ligase is encoded by and associated with bacteriophage T4. It joins DNA ends by catalysing phosphodiester-bond formation and is widely used in molecular cloning, especially for joining DNA fragments. Option B is correct because its name directly identifies the T4 bacteriophage source or association. Option A, E. coli ligase, is associated with the bacterium Escherichia coli, not T4 phage. Option C, DNA polymerase I, is mainly involved in DNA synthesis and replacement of primers or gaps, not identified as the T4-linked ligase in this question. Option D, amylase, digests starch into smaller carbohydrates and is unrelated to DNA joining. Exam cue: the name “T4 DNA ligase” itself points to T4 bacteriophage.

Correct answer: C. phosphodiester bond in DNA

DNA ligase repairs or creates phosphodiester bonds in the DNA backbone. It is needed for DNA continuity.

Correct answer: D. Ligase joins Okazaki fragments

Ligase joins Okazaki fragments and makes the lagging strand continuous. This is an important point in replication.

Correct answer: A. Okazaki fragments will not remain joined

If ligase is absent nicks between Okazaki fragments will not be sealed. The lagging strand will remain incomplete.

Correct answer: A. 3 prime OH and 5 prime phosphate

DNA ligase forms a covalent phosphodiester bond between 3 prime OH and 5 prime phosphate. This is the chemical basis of ligation.

Correct answer: A. Self-ligation can decrease

Removing 5 prime phosphate from vector can reduce joining of its own ends. This helps control self-ligation.

Correct answer: A. alkaline phosphatase treatment

Alkaline phosphatase can remove phosphate from vector ends. This can reduce vector self-ligation.

Correct answer: A. Insert has proper phosphate ends

Dephosphorylated vector reduces self-ligation but proper insert ends help ligation. Chemical ends must be considered.

Correct answer: A. compatible designed ends of insert and vector

In directional cloning different ends are designed. Ligase joins compatible insert and vector ends.

Correct answer: A. Insert orientation can be controlled

Directional cloning helps join the insert in the desired orientation. Ligase then permanently joins the correct ends.

Correct answer: A. desired insert DNA

A self-ligated vector may close with its own ends but lacks the desired insert. Therefore it can be unwanted in cloning.

Correct answer: A. To increase chance of desired recombinant molecule formation

A proper vector and insert ratio can improve desired ligation. Too little insert can form fewer recombinant plasmids.

Correct answer: A. Desired recombinants may be fewer

Low insert can reduce the chance of desired joining with vector. Ratio is a practical factor in cloning experiments.

Correct answer: A. DNA end compatibility

If DNA ends are compatible ligation can be easier. Incompatible ends can make joining difficult.

Correct answer: A. To preserve enzyme activity

Ligase is a protein enzyme and can become inactive by heat. Cold condition helps preserve activity.

Correct answer: A. The ligase enzyme is inactive

Successful ligation requires an active ligase, suitable DNA ends, an appropriate reaction buffer, and the chemical energy needed to activate the DNA ends. If the enzyme has lost its activity because of improper storage, excessive heat, unsuitable pH, or another damaging condition, it cannot catalyse phosphodiester-bond formation even when the DNA fragments are compatible. Thus option A is the correct reason for reaction failure. Compatible ends in B generally promote joining, while a proper buffer in C supports enzyme function. The availability of ATP or another required cofactor in D supports ligation rather than causing failure.

Correct answer: A. Ligation may be greatly reduced or fail

T4 DNA ligase requires ATP to form phosphodiester bonds between DNA ends. Without ATP, enzyme activation and the ligation reaction can stop, so ligation may be greatly reduced or fail. Cutting DNA is the role of a restriction endonuclease, not a ligase. Exam tip: Remember that T4 DNA ligase is ATP-dependent.

Correct answer: A. Enzyme activity may decrease

Proper buffer gives ligase correct pH and salt conditions. Wrong conditions can reduce activity.

Correct answer: A. Mg ion can help enzyme activity as a cofactor

Many ligation buffers contain magnesium ions for enzyme activity. Cofactors can be important in practical conditions.

Correct answer: A. Ligase can denature and become inactive

The direct answer is option A: ligase can denature and become inactive. Ligase is a protein enzyme that joins DNA ends by forming a phosphodiester bond. Every enzyme works best within a suitable temperature range. First, moderate warmth may increase molecular movement and help the reaction. However, excessive heat disrupts the weak bonds that maintain the enzyme’s three-dimensional shape. The active site then changes, so ligase cannot bind and join DNA efficiently. Therefore, ligation usually decreases or stops; the exact effect depends on the enzyme and conditions. Option A is correct because it describes this heat-induced loss of activity. Option B is wrong because ligation will not always increase; too much heat can destroy enzyme activity. Option C is wrong because DNA does not turn into chlorophyll, which is a pigment made through biological pathways. Option D is wrong because ATP is not converted into sugar by heating a ligase reaction. In an experiment, use the recommended temperature and time, and avoid unnecessary heating. Memory cue: mild heat may help a reaction, but excess heat can unfold a protein enzyme.

Correct answer: A. Okazaki fragments and vector-insert joining

In replication ligase joins Okazaki fragments and in cloning it joins vector-insert. These are two main contexts.

Correct answer: A. Ligase can seal a break or nick in a DNA strand

DNA repair often leaves a nick or discontinuity in the sugar-phosphate backbone after damaged portions have been removed and the correct nucleotides have been inserted. DNA ligase seals this remaining break by catalysing formation of the missing phosphodiester bond, thereby restoring continuity of the strand. Hence option A is correct. Ligase does not generally destroy DNA, so B reverses its actual role. It is not a protein-staining reagent, making C irrelevant, and although water is present in biological reactions, D is scientifically meaningless because ligase requires appropriate substrates, ions, pH, and energy conditions rather than water alone.

Correct answer: A. By sealing a nick in the DNA backbone

DNA ligase joins a nick in the sugar-phosphate backbone of DNA by restoring the phosphodiester bond, making the strand continuous again. Cutting DNA is the role of nucleases, not ligase. Exam tip: remember ligase as the enzyme that “joins” DNA fragments or seals nicks.

Correct answer: A. Ligase seals a nick by joining the sugar-phosphate backbone of DNA

A nick is a small break in the sugar-phosphate backbone of one DNA strand. DNA ligase seals this break by forming a phosphodiester bond, so option A is correct. DNA polymerase adds nucleotides, whereas restriction enzymes cut DNA at specific sequences. Exam tip: remember ligase as the enzyme that ‘joins’ broken DNA ends or seals nicks.