Answer: Here are 10 recombinant proteins and their therapeutic uses:

1. Human Insulin (Humulin): Used to treat Diabetes mellitus.
2. Human Growth Hormone (hGH): Treats dwarfism and growth hormone deficiency.
3. Interferons (\(\alpha, \beta, \gamma\)): Used in the treatment of viral infections and certain cancers.
4. Erythropoietin: Treats anemia by stimulating RBC production.
5. Tissue Plasminogen Activator (tPA): Dissolves blood clots in heart attack or stroke patients.
6. Hepatitis B Vaccine: Used for immunization against the Hepatitis B virus.
7. Blood Clotting Factor VIII: Treats Hemophilia A.
8. Interleukin-2: Used to stimulate the immune system in cancer therapy.
9. Platelet-derived Growth Factor (PDGF): Helps in wound healing.
10. Streptokinase: Used as a clot-buster in myocardial infarction.

Answer: Let's take the example of EcoRI.

• Restriction Enzyme: EcoRI
• Substrate DNA: Any double-stranded \(\text{DNA}\) containing the specific recognition sequence.
• Recognition Site: \(5^\prime\text{ - GAATTC - }3^\prime\) and \(3^\prime\text{ - CTTAAG - }5^\prime\)
• Cutting Site: Between \(\text{'G'}\) and \(\text{'A'}\) on both strands.
• Product: \(\text{DNA}\) fragments with sticky ends.

Exam Tip: You must draw the standard NCERT diagram showing the vector \(\text{DNA}\) and foreign \(\text{DNA}\) being cut by EcoRI, forming sticky ends, and then joining via \(\text{DNA}\) ligase to form recombinant \(\text{DNA}\).

Answer: \(\text{DNA}\) is significantly bigger in molecular size compared to enzymes.

Explanation: \(\text{DNA}\) is a massive polymer containing millions of base pairs carrying genetic information for the entire organism. An enzyme is a protein, which is synthesized based on a very small segment of \(\text{DNA}\) (a single gene). Since a whole chromosome (\(\text{DNA}\)) contains thousands of genes, the entire \(\text{DNA}\) molecule is structurally much larger than a single enzyme molecule.

Answer: To calculate the molar concentration, we need to know the total number of chromosomes and the volume of the nucleus.

• A diploid human cell has \(6.6 \times 10^9\) base pairs.
• \(1\) mole contains \(6.023 \times 10^{23}\) molecules (Avogadro's number).

$$\text{Molar concentration} = \frac{\text{Number of DNA molecules}}{\text{Avogadro's number} \times \text{Volume of the cell}} \quad \text{}$$

Since there are \(46\) \(\text{DNA}\) molecules (chromosomes) per diploid cell, the molar concentration of \(\text{DNA}\) molecules is extremely low, roughly in the range of \(10^{-11}\text{ M}\) depending on the exact nuclear volume.

Answer: No, eukaryotic cells do not have restriction endonucleases.

Justification: Restriction endonucleases are a defense mechanism found in bacteria (prokaryotes) to protect themselves from bacteriophages (viruses). They cut foreign viral \(\text{DNA}\). If eukaryotic cells had these enzymes, they would risk cutting and destroying their own genomic \(\text{DNA}\), leading to cell death. Eukaryotes have other mechanisms (like methylation and chromatin packaging) to protect their \(\text{DNA}\).

Answer: Stirred tank bioreactors offer several massive advantages over standard shake flasks for large-scale production:

1. Temperature Control: They have built-in cooling/heating jackets to maintain optimum temperature.
2. pH Control: They have sensors and automated systems to add acid/base to maintain the exact pH.
3. Foam Control: Built-in foam breakers prevent the culture from overflowing.
4. Sampling Ports: Small volumes of culture can be withdrawn periodically to check the quality and progression without contaminating the whole batch.
5. Large Volumes: They can process \(100\) to \(1000\text{ liters}\) of culture, which isn't possible in flasks.

Answer: A palindromic \(\text{DNA}\) sequence reads the same on both strands in the \(5^\prime \rightarrow 3^\prime\) direction.

1. EcoRI recognition site:
   \(5^\prime\text{-GAATTC-}3^\prime\)
   \(3^\prime\text{-CTTAAG-}5^\prime\)
2. BamHI recognition site:
   \(5^\prime\text{-GGATCC-}3^\prime\)
   \(3^\prime\text{-CCTAGG-}5^\prime\)
3. HindIII recognition site:
   \(5^\prime\text{-AAGCTT-}3^\prime\)
   \(3^\prime\text{-TTCGAA-}5^\prime\)
4. SalI recognition site:
   \(5^\prime\text{-GTCGAC-}3^\prime\)
   \(3^\prime\text{-CAGCTG-}5^\prime\)
5. SmaI recognition site:
   \(5^\prime\text{-CCCGGG-}3^\prime\)
   \(3^\prime\text{-GGGCCC-}5^\prime\)

Answer: In nature, recombinant \(\text{DNA}\) is formed during the Pachytene stage of Prophase I of meiosis. During this stage, homologous chromosomes pair up, and a process called crossing over occurs. Non-sister chromatids exchange genetic material with the help of the enzyme recombinase, naturally creating a "recombinant" \(\text{DNA}\) sequence.

Answer: A reporter enzyme acts as a visual indicator. Instead of using antibiotic resistance genes (which kill non-transformants), we can use a gene that produces a visible product.

Example: The \(lacZ\) gene produces the enzyme \(\beta\)-galactosidase. When grown on a chromogenic substrate (\(\text{X-gal}\)), non-recombinant colonies produce a blue color. If we insert our foreign \(\text{DNA}\) into the \(lacZ\) gene, the gene becomes inactivated (insertional inactivation). The enzyme is not produced, and these recombinant colonies appear white. This makes it incredibly easy to just look at the petri dish and identify recombinants.

Answer:

• (a) Origin of replication (ori): It is a specific sequence of \(\text{DNA}\) where replication initiates. Any piece of foreign \(\text{DNA}\) linked to this sequence can be made to replicate within the host cell. It also controls the copy number of the linked \(\text{DNA}\).
• (b) Bioreactors: These are large cylindrical vessels (\(100\text{–}1000\text{ liters}\)) in which raw materials are biologically converted into specific products, individual enzymes, etc., using microbial, plant, animal, or human cells. They provide optimal conditions for growth (temperature, pH, oxygen, etc.).
• (c) Downstream processing: The series of processes that a product is subjected to after the biosynthetic stage is complete, before it is ready for marketing. It primarily includes separation and purification of the desired product, followed by formulation with suitable preservatives.

Answer:

• (a) PCR (Polymerase Chain Reaction): A technique to amplify a specific gene segment. It involves three steps:
  1. Denaturation: Heating \(\text{DNA}\) to \(94^\circ\text{C}\) to separate strands.
  2. Annealing: Adding primers that bind to the \(\text{DNA}\) strands at \(\sim 54^\circ\text{C}\).
  3. Extension: Taq polymerase adds nucleotides to synthesize new strands at \(72^\circ\text{C}\). Millions of copies are made in a few hours.
• (b) Restriction enzymes and DNA: These are endonucleases that inspect the length of a \(\text{DNA}\) molecule, recognize a specific palindromic sequence, and cut the sugar-phosphate backbone of both strands. This is the foundational step for creating recombinant \(\text{DNA}\).
• (c) Chitinase: It is an enzyme used to break down the cell wall of fungi (which is made of chitin). This is done during \(\text{DNA}\) isolation to release the genetic material from fungal cells.

Answer:

Category	Feature A	Feature B
(a) Plasmid vs Chromosomal	Plasmid DNA: Extra-chromosomal, circular, smaller, contains non-essential but beneficial genes (like antibiotic resistance).	Chromosomal DNA: Main genomic \(\text{DNA}\), linear (in eukaryotes) or circular (in bacteria), very large, carries essential genes for survival.
(b) RNA vs DNA	RNA: Single-stranded, contains Ribose sugar, uses Uracil (\(\text{U}\)) instead of Thymine.	DNA: Double-stranded, contains Deoxyribose sugar, uses Thymine (\(\text{T}\)).
(c) Exonuclease vs Endonuclease	Exonuclease: Removes nucleotides from the ends of a \(\text{DNA}\) molecule.	Endonuclease: Makes cuts at specific positions within the \(\text{DNA}\) molecule.

Answer: B) Electrophoresis
Explanation: Gel electrophoresis separates \(\text{DNA}\) fragments based on their size under an electric field.
Difficulty Level: Easy

Answer: B) pBR322
Explanation: pBR322 is one of the most widely used *E. coli* cloning vectors. The others are enzymes.
Difficulty Level: Easy

Answer: C) Is thermostable and remains active at high temperatures
Explanation: Derived from Thermus aquaticus, it doesn't denature during the high heat (\(94^\circ\text{C}\)) step of PCR.
Difficulty Level: Medium

Answer: \(\text{DNA}\) is hydrophilic because of the negatively charged phosphate groups in its backbone. To introduce it into a bacterial cell, the host cells are made "competent" by treating them with a specific concentration of a divalent cation, such as calcium (\(\text{Ca}^{2+}\)). The cells are then incubated with recombinant \(\text{DNA}\) on ice, followed by a brief heat shock at \(42^\circ\text{C}\), and then put back on ice.

Answer: \(\text{DNA}\) fragments are negatively charged molecules. In gel electrophoresis, they are forced to move towards the anode (positive electrode) under an electric field through a medium (agarose gel). The smaller fragments move faster and further than the larger ones.

Answer: It is a technique where a recombinant \(\text{DNA}\) is inserted within the coding sequence of an enzyme (like \(\beta\)-galactosidase). This insertion inactivates the gene, preventing the production of the enzyme. This phenomenon is used to differentiate recombinant colonies from non-recombinant ones.

Answer: A selectable marker helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants. Common examples are genes coding for resistance to antibiotics like ampicillin or tetracycline.

Answer: (Note for students: Practice the pBR322 diagram extensively!)

• ori (Origin of replication): Sequence where replication starts.
• Selectable markers: \(amp^\text{R}\) (ampicillin resistance) and \(tet^\text{R}\) (tetracycline resistance) genes.
• Cloning sites: Unique recognition sites for restriction enzymes. E.g., BamHI and SalI within the \(tet^\text{R}\) gene, and PvuI and PstI within the \(amp^\text{R}\) gene.
• rop: Codes for the proteins involved in the replication of the plasmid.

Answer: PCR amplifies \(\text{DNA}\) *in vitro*. The three main steps are:

1. 1. Denaturation: The double-stranded \(\text{DNA}\) template is heated to around \(94^\circ\text{C}\). The hydrogen bonds break, resulting in single-stranded \(\text{DNA}\).
2. 2. Annealing: The temperature is lowered to around \(54^\circ\text{C}\). Two sets of oligonucleotide primers bind (anneal) to the complementary regions on the single-stranded \(\text{DNA}\) templates.
3. 3. Extension: The temperature is raised to \(72^\circ\text{C}\). The thermostable enzyme Taq DNA polymerase extends the primers by adding nucleotides complementary to the template.

This cycle is repeated \(30\) times to get \(\sim 1\text{ billion}\) copies.

Answer:

1. 1. Cell Wall Breakdown: Treat plant tissues with enzymes like Cellulase and Pectinase to break the cell wall.
2. 2. Cell Membrane Lysis: Use detergents to dissolve the lipid membrane and release cellular contents.
3. 3. Removal of RNA and Proteins: Treat the extract with Ribonuclease (to digest \(\text{RNA}\)) and Proteases (to digest proteins).
4. 4. Precipitation: Add chilled ethanol to the purified mixture. This causes the \(\text{DNA}\) to precipitate out as fine threads in the suspension.
5. 5. Spooling: The precipitated \(\text{DNA}\) threads can be removed by spooling using a glass rod.

(a) Identify the critical mistake the scientist made. (1 Mark)
Answer: She used a "normal DNA polymerase" instead of a thermostable polymerase like Taq polymerase.

(b) Why did her mistake lead to no amplification? (2 Marks)
Answer: The first step of PCR involves heating the mixture to \(94^\circ\text{C}\) for denaturation. At this high temperature, normal \(\text{DNA}\) polymerase denatures (gets destroyed) and loses its catalytic activity. Hence, no extension occurred.

(c) Name the source organism of the correct enzyme required for this process. (1 Mark)
Answer: The bacterium Thermus aquaticus.

Answer: (A) Both A and R are true and R is the correct explanation of A.
Difficulty Level: Medium

Answer: (D) A is false but R is true. (\(\text{DNA}\) moves towards the positive electrode, which is the anode, not cathode).
Difficulty Level: Easy

Answer: (A) Both A and R are true and R is the correct explanation of A.
Difficulty Level: Medium

Answer: (D) Both A and R are false. Downstream processing (purification, clinical trials, formulation) is absolutely strictly required for human consumption products.
Difficulty Level: Hard