Biotechnology: Principles and Processes Previous Year Questions — NEET Biology
5 past-year Biotechnology: Principles and Processes questions from NEET (Biology).
Q98 (2024)
Which of the following statements are not true regarding restriction endonucleases?
A. They are called molecular scissors.
B. These are the enzymes responsible for restricting the growth of bacteriophages in E. coli.
C. They cut the DNA only at the centre of the palindromic sites.
D. They remove nucleotides only from the ends of DNA fragments.
E. They recognise specific palindromic base-pair sequences.
Choose the answer from the options given below:
- A and B only
- D and E only
- C and D only
- A and E only
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Biotechnology: Principles and Processes
Q110 (2024)
Match List I with List II:
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;"></th><th style="border: 1px solid #888; padding: 8px;">List-I</th><th style="border: 1px solid #888; padding: 8px;"></th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A.</td><td style="border: 1px solid #888; padding: 8px;">Genetically modified organism</td><td style="border: 1px solid #888; padding: 8px;">(I)</td><td style="border: 1px solid #888; padding: 8px;">Agrobacterium tumefaciens</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B.</td><td style="border: 1px solid #888; padding: 8px;">Thermostable DNA polymerase</td><td style="border: 1px solid #888; padding: 8px;">(II)</td><td style="border: 1px solid #888; padding: 8px;">Bt cotton</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C.</td><td style="border: 1px solid #888; padding: 8px;">Ti plasmid</td><td style="border: 1px solid #888; padding: 8px;">(III)</td><td style="border: 1px solid #888; padding: 8px;">Thermus aquaticus</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D.</td><td style="border: 1px solid #888; padding: 8px;">pBR322</td><td style="border: 1px solid #888; padding: 8px;">(IV)</td><td style="border: 1px solid #888; padding: 8px;">Escherichia coli</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- A-II, B-I, C-IV, D-III
- A-I, B-IV, C-III, D-II
- A-II, B-III, C-I, D-IV
- A-I, B-II, C-IV, D-III
### 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 $\rightarrow$ II
- Thermostable DNA polymerase: Isolated from *Thermus aquaticus* $\rightarrow$ III
- Ti plasmid: Natural vector from *Agrobacterium tumefaciens* $\rightarrow$ I
- pBR322: Plasmid vector of *Escherichia coli* $\rightarrow$ 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).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Biotechnology: Principles and Processes
Q119 (2024)
Identify the correct sequence of steps in each cycle of Polymerase Chain Reaction :
- Denaturation $\rightarrow$ Annealing $\rightarrow$ Extension
- Denaturation $\rightarrow$ Extension $\rightarrow$ Annealing
- Extension $\rightarrow$ Annealing $\rightarrow$ Denaturation
- Annealing $\rightarrow$ Denaturation $\rightarrow$ Extension
### Related Formula
PCR Cycle Steps:
1. Denaturation (High Temp $\sim 94^\circ\text{C}$)
2. Annealing (Primers attach $\sim 54^\circ\text{C}$)
3. Extension (Taq Polymerase $\sim 72^\circ\text{C}$)
### 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 $\rightarrow$ Annealing $\rightarrow$ Extension. Option (1) is correct.
### Pattern Recognition
PCR order = D $\rightarrow$ A $\rightarrow$ E (Denaturation, Annealing, Extension).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Biotechnology: Principles and Processes
Q120 (2024)
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 :
- A and D only
- B and D only
- B and C only
- A and B only
### Related Formula
DNA Visualization: Requires Ethidium Bromide ($EtBr$) staining AND exposure to UV radiation (gives bright orange bands).
### Core Logic
- Statement A: Restriction enzymes cut DNA and act as molecular scissors (True).
- Statement B: DNA fragments separate based on size through sieving effect of agarose gel (True).
- Statement C: DNA cannot be seen without staining (False).
- Statement D: Stained DNA requires UV light, not visible light, to be seen (False).
### Step 1: Selection
Statements A and B are correct. Correct option is (4).
### Pattern Recognition
EtBr stained DNA requires UV light, NOT visible light (D false). Staining IS required (C false). Correct statements = A and B.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Biotechnology: Principles and Processes
Q136 (2024)
Insertion of a foreign DNA at BamHI site in an E.coli cloning vector pBR322 results in the loss of antibiotic resistance towards:
- Gentamycin
- Ampicillin and tetracycline
- Tetracycline
- Ampicillin
### Related Formula
$\text{Vector pBR322} \rightarrow \text{BamHI site in } tet^R \text{ gene} \xrightarrow{\text{Insertion}} \text{Loss of Tetracycline Resistance}$
### Core Logic
In the cloning vector pBR322, the BamHI restriction site is located within the tetracycline resistance ($tet^R$) gene. When a foreign DNA fragment is inserted at the BamHI site, it disrupts the coding sequence of the $tet^R$ 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 $tet^R$ gene, leading specifically to the loss of tetracycline resistance while ampicillin resistance ($amp^R$) remains intact.
### Pattern Recognition
Sees: "pBR322" + "BamHI site insertion".
Shortcut: BamHI and SalI sites are in $tet^R$ gene $\rightarrow$ loss of tetracycline resistance. PstI and PvuI sites are in $amp^R$ gene $\rightarrow$ loss of ampicillin resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Biotechnology: Principles and Processes
