Basic Processes

Given / Concept: Gene expression is the process by which the information encoded in a gene is used to synthesize a functional gene product (usually a protein).

Importance of Gene Expression:
1. It converts the genetic information stored in DNA into functional proteins that carry out all biological activities of a cell.
2. It regulates growth, development, differentiation, and metabolism of an organism.
3. It allows cells to respond to environmental changes by switching genes on or off.
4. It determines the phenotype of an organism.
5. It is essential for maintaining cellular homeostasis.

Steps Involved in Gene Expression:

Step 1 — Transcription:
- The DNA double helix unwinds at the gene locus.
- RNA polymerase binds to the promoter region and reads the template (antisense) strand in the $3' \rightarrow 5'$ direction.
- A complementary mRNA strand is synthesized in the $5' \rightarrow 3'$ direction.
- In eukaryotes, the primary transcript (pre-mRNA) undergoes post-transcriptional modifications: addition of 5' cap, 3' poly-A tail, and splicing out of introns to produce mature mRNA.

Step 2 — Translation:
- The mature mRNA moves to the ribosome.
- The ribosome reads the mRNA codons (triplets of bases) in the $5' \rightarrow 3'$ direction.
- tRNA molecules carry specific amino acids; their anticodons pair with the mRNA codons.
- Peptide bonds form between successive amino acids, elongating the polypeptide chain.
- Translation ends at a stop codon (UAA, UAG, or UGA).
- The polypeptide is released and folds into a functional protein (sometimes requiring post-translational modifications).

Summary equation:
$$\text{DNA} \xrightarrow{\text{Transcription}} \text{mRNA} \xrightarrow{\text{Translation}} \text{Protein}$$

Final Answer: Gene expression is vital for converting genetic information into functional proteins. It involves two main steps: transcription (DNA → mRNA) and translation (mRNA → Protein).

Given / Concept: In prokaryotes, gene expression is regulated at the level of transcription. The lac operon (lactose operon) in *Escherichia coli* is the classic example, proposed by Jacob and Monod (1961).

Structure of the Lac Operon:
The lac operon consists of:
- Regulatory gene (i): Codes for the lac repressor protein.
- Promoter (P): Site where RNA polymerase binds.
- Operator (O): Site where the repressor binds to block transcription.
- Structural genes:
- lacZ — codes for $\beta$-galactosidase (breaks lactose into glucose + galactose)
- lacY — codes for permease (transports lactose into the cell)
- lacA — codes for transacetylase

Regulation — Two Situations:

Situation 1: Absence of Lactose (Operon OFF)
1. The regulatory gene i is continuously transcribed and translated to produce the lac repressor protein.
2. The repressor binds to the operator region.
3. RNA polymerase cannot move past the operator; transcription of structural genes is blocked.
4. No $\beta$-galactosidase, permease, or transacetylase is produced.

Situation 2: Presence of Lactose (Operon ON)
1. Lactose enters the cell and is converted to allolactose (the inducer).
2. Allolactose binds to the lac repressor, causing a conformational change.
3. The repressor–allolactose complex cannot bind to the operator.
4. RNA polymerase binds to the promoter and transcribes the structural genes.
5. A polycistronic mRNA is produced, which is translated into $\beta$-galactosidase, permease, and transacetylase.
6. These enzymes metabolize lactose.
7. When lactose is exhausted, allolactose levels fall, the repressor regains its active form, binds the operator, and transcription stops.

Conclusion: The lac operon is an example of negative inducible regulation — the repressor normally turns the operon off, and the inducer (allolactose) turns it on. This is an efficient, economical mechanism allowing the bacterium to produce enzymes only when needed.

Given: All proteins are lost from a cell.

Concept: DNA replication is a highly protein-dependent process. Multiple proteins/enzymes are required at every step.

Effect — Step-by-step Analysis:

| Protein Lost | Role in Replication | Effect of Loss |
|---|---|---|
| Helicase | Unwinds the double helix at the origin of replication | Replication fork cannot form; replication cannot begin |
| Primase | Synthesizes the RNA primer on the template strand | DNA polymerase has no starting point; replication cannot proceed |
| DNA Polymerase III | Synthesizes new DNA strand in $5'\rightarrow 3'$ direction | No new DNA strand can be synthesized |
| DNA Polymerase I | Removes RNA primers and fills the gaps | Gaps remain in the newly synthesized strand |
| DNA Ligase | Joins Okazaki fragments on the lagging strand | Lagging strand remains fragmented |
| Single-Strand Binding Proteins (SSBPs) | Stabilize the unwound single strands | Template strands re-anneal; replication fork collapses |
| Topoisomerase | Relieves torsional stress ahead of the replication fork | Supercoiling prevents further unwinding |
| Sliding clamp (β-clamp) | Holds DNA polymerase on the template | Polymerase falls off; replication is incomplete |

Overall Effect:
If all proteins are lost, DNA replication would be completely impossible. Not a single step — from unwinding of the double helix to the joining of Okazaki fragments — can occur without proteins. The cell would be unable to divide, would lose its genetic material upon cell division, and would ultimately die.

Final Answer: Loss of all proteins would completely halt DNA replication because every step (unwinding, priming, synthesis, gap-filling, ligation) is catalyzed by specific proteins/enzymes.

Given: Effect of UV radiation on DNA structure.

Part 1 — Effect of UV Rays on DNA Structure:
- UV radiation (particularly at wavelength ~260 nm) is absorbed by the nitrogenous bases of DNA.
- The most common damage is the formation of thymine dimers (pyrimidine dimers).
- Two adjacent thymine bases on the same strand become covalently linked to each other, forming a cyclobutane ring between them:
$$\text{T} - \text{T} \xrightarrow{\text{UV}} \text{T} \Leftrightarrow \text{T (thymine dimer)}$$
- This distorts the DNA double helix, preventing normal base pairing and blocking replication and transcription.
- Cytosine dimers (C–C) and cytosine–thymine dimers (C–T) can also form, but thymine dimers are most common.

Part 2 — Molecular Basis of Mutation:
- When DNA polymerase encounters a thymine dimer during replication, it may insert incorrect bases (often adenine opposite both thymines — the "A rule"), or it may stall.
- If the dimer is not repaired before replication, it leads to a substitution mutation in the daughter strand.
- Accumulation of such mutations can lead to skin cancer (e.g., xeroderma pigmentosum in humans with defective repair).

Part 3 — Repair Mechanisms:

(a) Photoreactivation (Light Repair):
- An enzyme called photolyase (DNA photolyase) binds to the thymine dimer.
- In the presence of visible light (300–600 nm), the enzyme uses light energy to break the covalent bonds of the cyclobutane ring.
- The two thymines are restored to their original monomeric form.
- This is an error-free, direct reversal mechanism.

(b) Nucleotide Excision Repair (NER) — Dark Repair:
- This mechanism operates without light.
- Step 1: Specific endonucleases recognize the distortion caused by the thymine dimer.
- Step 2: The damaged strand is nicked on both sides of the dimer (excision).
- Step 3: A short oligonucleotide segment (~12–13 nucleotides in prokaryotes; ~27–29 in eukaryotes) containing the dimer is removed by a helicase.
- Step 4: DNA polymerase fills in the gap using the complementary strand as a template.
- Step 5: DNA ligase seals the nick.
- This is also an error-free repair mechanism.

Final Answer: UV rays cause thymine dimer formation, distorting the DNA helix and causing substitution mutations. Cells correct this by photoreactivation (using photolyase + light) or nucleotide excision repair (dark repair involving excision of the damaged segment and resynthesis).

(a) Leading Strand vs. Lagging Strand

| Feature | Leading Strand | Lagging Strand |
|---|---|---|
| Direction of synthesis | Synthesized continuously in the $5' \rightarrow 3'$ direction | Synthesized discontinuously in the $5' \rightarrow 3'$ direction |
| Template strand direction | Runs $3' \rightarrow 5'$ (same direction as fork movement) | Runs $5' \rightarrow 3'$ (opposite to fork movement) |
| Primers required | Only one primer needed | Multiple primers needed (one per Okazaki fragment) |
| Okazaki fragments | Not formed | Formed (short fragments later joined by ligase) |
| Enzymes | DNA Pol III synthesizes continuously | DNA Pol III, DNA Pol I, and Ligase all required |

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(b) Transcription vs. Translation

| Feature | Transcription | Translation |
|---|---|---|
| Definition | Synthesis of RNA from a DNA template | Synthesis of protein from an mRNA template |
| Template | DNA (antisense/template strand) | mRNA |
| Product | mRNA (also tRNA, rRNA) | Polypeptide (protein) |
| Enzyme | RNA polymerase | Ribosomes (rRNA + proteins) |
| Location (eukaryotes) | Nucleus | Cytoplasm (ribosomes) |
| Monomers used | Ribonucleotides (ATP, GTP, CTP, UTP) | Amino acids |
| Direction | $5' \rightarrow 3'$ (RNA strand) | $5' \rightarrow 3'$ (mRNA read) |

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(c) Transition vs. Transversion Mutation

| Feature | Transition | Transversion |
|---|---|---|
| Definition | Substitution of a purine by another purine, or a pyrimidine by another pyrimidine | Substitution of a purine by a pyrimidine or vice versa |
| Type of bases exchanged | Purine ↔ Purine (A ↔ G) or Pyrimidine ↔ Pyrimidine (C ↔ T) | Purine ↔ Pyrimidine (A or G ↔ C or T) |
| Frequency | More common | Less common |
| Example | A → G or C → T | A → C or G → T |
| Structural change | Less drastic (same class of base) | More drastic (different class of base) |

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(d) Codon vs. Anticodon

| Feature | Codon | Anticodon |
|---|---|---|
| Definition | A triplet of nucleotides on mRNA that codes for a specific amino acid | A triplet of nucleotides on tRNA complementary to the codon |
| Location | mRNA | tRNA (anticodon loop) |
| Function | Specifies which amino acid is to be incorporated | Recognizes and base-pairs with the codon on mRNA |
| Total number | 64 codons (61 sense + 3 stop codons) | Corresponds to the number of tRNA species |
| Example | 5'-AUG-3' (codes for Methionine / start codon) | 3'-UAC-5' (anticodon on tRNA$^{Met}$) |
| Direction of reading | $5' \rightarrow 3'$ | $3' \rightarrow 5'$ (antiparallel to codon) |

Correct Answer: (b) Ultraviolet rays

Justification:
- Gamma rays, X-rays, and alpha rays are ionizing radiations — they carry sufficient energy to ionize atoms, break chemical bonds, and cause double-strand breaks in DNA, leading to severe and often irreparable damage.
- Ultraviolet (UV) rays are non-ionizing radiation. They do not have enough energy to ionize atoms; instead, they cause specific, repairable damage such as thymine dimer formation.
- Cells possess efficient repair mechanisms (photoreactivation and nucleotide excision repair) to correct UV-induced damage.
- Therefore, among the given options, UV rays are least likely to be harmful compared to the highly penetrating and ionizing gamma rays, X-rays, and alpha rays.

$$\boxed{(b) \text{ Ultraviolet rays}}$$

Correct Answer: (a) Excision repair

Justification:
- An apurinic (AP) site is a location in DNA where a purine base (adenine or guanine) has been removed, and an apyrimidinic (AP) site is where a pyrimidine base (cytosine or thymine) has been removed.
- In Base Excision Repair (BER) — a type of excision repair — a specific enzyme called DNA glycosylase recognizes and cleaves the N-glycosidic bond between the damaged/incorrect base and the deoxyribose sugar, releasing the base and leaving behind an AP site (also called an abasic site).
- Subsequently, AP endonuclease nicks the DNA backbone at the AP site, and the gap is filled by DNA polymerase and sealed by ligase.
- Mismatch repair does not involve the creation of AP sites; it recognizes mismatched base pairs and excises a stretch of the newly synthesized strand containing the mismatch.

$$\boxed{(a) \text{ Excision repair}}$$