Assertion (A): Restriction enzymes cut the strand of DNA a little away from the centre of the palindromic site, creating sticky ends. Reason (R): These sticky ends are necessary because they facilitate the action of the enzyme DNA ligase by forming hydrogen bonds with their complementary cut counterparts.
A is True: Restriction enzymes cut away from the center of the palindrome, leaving single-stranded portions called sticky ends. R is True: The stickiness facilitates DNA ligase action as they bond with complementary ends. R explains why the structure created in A is useful.
Question 442 of 535
📘 CLASS XII
Assertion (A): The linking of the cut antibiotic resistance gene with the plasmid vector to form recombinant DNA became possible with the enzyme DNA ligase. Reason (R): If the vector DNA and the source DNA are cut by the same restriction enzyme, the recombinant molecule cannot be created.
A is True: DNA ligase acts on cut DNA molecules and joins their ends, creating recombinant DNA. R is False: Normally, the vector and source DNA must be cut by the *same* restriction enzyme to generate the necessary matching sticky ends for ligation.
Question 443 of 535
📘 CLASS XII
Assertion (A): Exonucleases remove nucleotides from the ends of the DNA, whereas endonucleases make cuts at specific positions within the DNA. Reason (R): Restriction enzymes belong to a larger class of enzymes called nucleases.
A is True: This correctly defines the functional difference between exonucleases and endonucleases. R is True: Restriction enzymes are a type of nuclease. R provides the class name, but does not explain the functional differences detailed in A.
Question 444 of 535
📘 CLASS XII
Assertion (A): The palindromic sequence in DNA reads the same on the two strands when the orientation of reading (e.g., 5' to 3') is kept the same. Reason (R): The first restriction endonuclease, Hind II, always cut DNA molecules by recognising a specific sequence of six base pairs.
A is True: This is the definition of a palindromic sequence in DNA. R is True: Hind II recognizes a specific sequence of six base pairs. Both statements are facts, but R does not explain or justify the definition of the DNA palindrome in A.
Question 445 of 535
📘 CLASS XII
Assertion (A): When an alien piece of DNA is transferred into an alien organism, it needs to be linked with the origin of replication (`ori`). Reason (R): The origin of replication sequence is responsible for initiating replication and also controls the copy number of the linked DNA.
A is True: For an alien DNA piece to multiply in the host, it must be linked to the `ori`. R is True: The `ori` sequence is specifically responsible for initiating replication and controlling the copy number. R explains why the linkage described in A is essential for multiplication.
Question 446 of 535
📘 CLASS XII
Assertion (A): Cloning vectors must have very few, preferably single, recognition sites for commonly used restriction enzymes. Reason (R): Presence of more than one recognition site within the vector will generate several fragments, which will complicate the gene cloning process.
A is True: Vectors are engineered to have very few restriction sites. R is True: The complication (R) resulting from multiple cuts is the specific reason for the necessity stated in A.
Question 447 of 535
📘 CLASS XII
Assertion (A): Genes conferring resistance to antibiotics are considered useful selectable markers in cloning experiments. Reason (R): Selectable markers help in identifying and eliminating non-transformants by selectively permitting the growth of only the transformants.
A is True: Antibiotic resistance genes (like ampicillin or tetracycline) are commonly used selectable markers. R is True: R defines the general function of a selectable marker. Although related, R is the general goal, not the specific mechanism or justification for *using* antibiotic resistance genes (which rely on the fact that normal E. coli lacks resistance).
Question 448 of 535
📘 CLASS XII
Assertion (A): If a foreign DNA is ligated at the $BamH$ I site in the vector pBR322, the transformant cells will lose resistance to ampicillin. Reason (R): The $BamH$ I restriction site is present within the coding sequence of the tetracycline resistance gene ($tet^R$) of pBR322.
A is False: Ligation at the $BamH$ I site inactivates the *tetracycline* resistance gene ($tet^R$), not the ampicillin resistance gene ($amp^R$). R is True: The $BamH$ I site is located within the $tet^R$ gene of pBR322.
Question 449 of 535
📘 CLASS XII
Assertion (A): Selection of recombinants using insertional inactivation of the $eta$-galactosidase gene is often preferred over selection using two antibiotic resistance genes. Reason (R): Non-recombinant colonies carrying the intact $eta$-galactosidase gene appear blue in the presence of a chromogenic substrate.
A is True: Using color production is an alternative developed because selection involving two antibiotic plates is cumbersome. R is True: Non-recombinants, which have the active $eta$-galactosidase gene, produce blue colonies in the presence of a chromogenic substrate. R describes the visible outcome but does not explain *why* the method is preferred (which is because it avoids simultaneous plating).
Question 450 of 535
📘 CLASS XII
Assertion (A): Bacterial cells are made 'competent' to take up DNA by treating them with divalent cations such as calcium. Reason (R): DNA is a hydrophilic molecule, so it cannot pass easily through the hydrophobic cell membrane without chemical manipulation.
A is True: Competence is achieved by treatment with divalent cations like calcium. R is True: DNA is hydrophilic, which is the chemical reason why it needs the competency treatment (A) to cross the hydrophobic membrane. R explains A.
