Which of the following statements are correct with respect to DNA separation, isolation and visualization? A. The cutting of DNA is done by molecular scissors. B. The DNA fragments separate according to their size in an agarose gel, upon electrophoresis. C. The separated DNA fragments can be seen without staining when exposed to UV light. D. The separated DNA fragments, when stained with ethidium bromide, can be seen in visible light. Choose the correct answer from the options given below :

Related Formula

Restriction Endonucleases cut DNA within palindromic sites (slightly away from center). Exonucleases remove nucleotides from ends.

Core Logic

• Statement C is incorrect: Restriction endonucleases usually cut DNA slightly away from the center of palindromic sites, between same two bases on opposite strands.
• Statement D is incorrect: Removing nucleotides from the ends of DNA is the function of restriction exonucleases, not endonucleases.

Step 1: Selection

Statements C and D are NOT true regarding restriction endonucleases. Option (3) is correct.

Pattern Recognition

Note the question asks for 'NOT true'. Removing ends = Exonuclease (D false). Cut at center = slightly away from center (C false). Answer is C and D.

Chapter Mix

Class 12 Biology: Biotechnology: Principles and Processes

Related Formula

GMO: Bt cotton Taq Polymerase: Thermus aquaticus Ti Plasmid: Agrobacterium tumefaciens pBR322: Cloning vector constructed from E. coli

Core Logic

• Genetically modified organism: Bt cotton arrow$\rightarrow$ II
• Thermostable DNA polymerase: Isolated from Thermus aquaticus arrow$\rightarrow$ III
• Ti plasmid: Natural vector from Agrobacterium tumefaciens arrow$\rightarrow$ I
• pBR322: Plasmid vector of Escherichia coli arrow$\rightarrow$ IV

Step 1: Conclusion

Matches: A-II, B-III, C-I, D-IV. Correct option is (3).

Pattern Recognition

Thermostable DNA polymerase = Thermus aquaticus (B-III). Ti plasmid = Agrobacterium (C-I). Confirms option (3).

Chapter Mix

Class 12 Biology: Biotechnology: Principles and Processes

Related Formula

PCR Cycle Steps:

• Denaturation (High Temp 94^$\sim 94^\circ\text{C}$)
• Annealing (Primers attach 54^$\sim 54^\circ\text{C}$)
• Extension (Taq Polymerase 72^$\sim 72^\circ\text{C}$)

Core Logic

Each cycle of Polymerase Chain Reaction (PCR) consists of three basic steps in sequence: Denaturation of double-stranded DNA target, Annealing of oligonucleotide primers, and Extension using Taq DNA polymerase.

Step 1: Selection

The correct sequence is Denaturation arrow$\rightarrow$ Annealing arrow$\rightarrow$ Extension. Option (1) is correct.

Pattern Recognition

PCR order = D arrow$\rightarrow$ A arrow$\rightarrow$ E (Denaturation, Annealing, Extension).

Chapter Mix

Class 12 Biology: Biotechnology: Principles and Processes

Related Formula

Vector pBR322 arrow BamHI site in tet^R gene Insertion Loss of Tetracycline Resistance$\text{Vector pBR322} \rightarrow \text{BamHI site in } tet^R \text{ gene} \xrightarrow{\text{Insertion}} \text{Loss of Tetracycline Resistance}$

Core Logic

In the cloning vector pBR322, the BamHI restriction site is located within the tetracycline resistance (tet^R$tet^R$) gene. When a foreign DNA fragment is inserted at the BamHI site, it disrupts the coding sequence of the tet^R$tet^R$ gene, leading to insertional inactivation. Consequently, the recombinant plasmid loses its resistance to tetracycline.

Step 1: Evaluation of Options

Inserting foreign DNA at the BamHI site disrupts the tet^R$tet^R$ gene, leading specifically to the loss of tetracycline resistance while ampicillin resistance (amp^R$amp^R$) remains intact.

Pattern Recognition

Sees: "pBR322" + "BamHI site insertion". Shortcut: BamHI and SalI sites are in tet^R$tet^R$ gene arrow$\rightarrow$ loss of tetracycline resistance. PstI and PvuI sites are in amp^R$amp^R$ gene arrow$\rightarrow$ loss of ampicillin resistance.

Chapter Mix

Class 12 Biology: Biotechnology: Principles and Processes