Assertion (A): Each restriction endonuclease functions by binding to the DNA and cutting each of the two strands of the double helix at specific points in their sugar-phosphate backbones. Reason (R): Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA by inspecting the length of the DNA sequence.
A is True: Restriction enzymes bind and cut the sugar-phosphate backbone. R is True: The cutting (A) is specific because the enzyme first inspects the DNA and finds its specific palindromic recognition sequence. R explains the prerequisite for A.
Question 502 of 535
📘 CLASS XII
Assertion (A): Restriction enzymes cut the DNA strands a little away from the center of the palindrome sites, leaving single-stranded portions called sticky ends. Reason (R): These sticky ends form strong covalent bonds with their complementary cut counterparts, which facilitates the action of DNA ligase.
A is True: Restriction enzymes cut away from the center, creating sticky ends. R is False: Sticky ends form *hydrogen bonds* with complementary counterparts. DNA ligase then seals the backbone.
Question 503 of 535
📘 CLASS XII
Assertion (A): Recombinant molecules of DNA are composed of DNA from different sources/genomes. Reason (R): Normally, the vector and the source DNA must be cut by the same restriction enzyme to ensure that the resultant fragments have the same kind of sticky ends.
A is True: Recombinant DNA combines DNA from different sources. R is True: Using the same enzyme is required because it generates matching sticky ends needed for ligation. R explains the necessary step to create A.
Question 504 of 535
📘 CLASS XII
Assertion (A): DNA fragments are separated in gel electrophoresis according to their size. Reason (R): The agarose matrix provides a sieving effect, causing smaller DNA fragments to move farther through the gel towards the anode.
A is True: Fragments separate based on size. R is True: The sieving effect of agarose causes smaller fragments to move farther. R explains the physical mechanism that achieves the separation described in A.
Question 505 of 535
📘 CLASS XII
Assertion (A): Pure DNA fragments are visible in the gel only after staining with ethidium bromide and exposure to UV radiation. Reason (R): Pure DNA fragments are not visible in visible light and without staining.
A is True: Visualization requires staining and UV exposure. R is True: This explains the necessity for the procedure described in A.
Question 506 of 535
📘 CLASS XII
Assertion (A): The process of cutting out the separated bands of DNA from the agarose gel and extracting them from the gel piece is known as elution. Reason (R): The DNA fragments purified through elution are then used in constructing recombinant DNA by joining them with cloning vectors.
A is True: Elution is the process of extracting DNA from the gel pieces. R is True: The purified fragments are used for ligation into vectors. R explains the importance of the DNA recovered by the process described in A.
Question 507 of 535
📘 CLASS XII
Assertion (A): The origin of replication ($ori$) sequence in a vector is responsible for controlling the copy number of the linked DNA. Reason (R): If a scientist wants to recover many copies of the target DNA, it should be cloned in a vector whose origin supports a high copy number.
A is True: The $ori$ controls the copy number. R is True: The desired copy number (R) dictates which vector (A) must be selected. R justifies the cloning strategy based on the function stated in A.
Question 508 of 535
📘 CLASS XII
Assertion (A): Vectors need to 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 complicate the gene cloning process by generating several fragments.
A is True: Vectors are engineered for single sites. R is True: The generation of multiple fragments (R) is the reason why A is essential. R explains A.
Question 509 of 535
📘 CLASS XII
Assertion (A): If a foreign DNA is ligated at the $Pst$ I site in the vector pBR322, the recombinant plasmid loses resistance to ampicillin. Reason (R): The $Pst$ I restriction site is present within the coding sequence of the $tet^R$ gene of pBR322.
A is True: $Pst$ I is located in the $amp^R$ gene, leading to ampicillin resistance loss [31, Figure 9.4]. R is False: $Pst$ I is in the $amp^R$ gene. The $BamH$ I and $Sal$ I sites are in the $tet^R$ gene [31, Figure 9.4].
Question 510 of 535
📘 CLASS XII
Assertion (A): Selection of recombinants using insertional inactivation of the $eta$-galactosidase gene is often preferred because it avoids the cumbersome procedure of simultaneous plating on two plates having different antibiotics. Reason (R): The general principle of insertional inactivation is that the foreign DNA is inserted within the coding sequence of a gene, making it non-functional.
A is True: Color-based selection avoids the cumbersome two-plate antibiotic method. R is True: R defines the general mechanism of insertional inactivation. R explains how the color system works, but not why it is *preferred* over the antibiotic method (A).
