Molecular Basis of Inheritance – CBSE NCERT Study Resources

Previous Chapter Next Chapter

Study Materials

12th

12th - Biology

Molecular Basis of Inheritance

Jump to Question Sets

Overview

The chapter "Molecular Basis of Inheritance" explores the structure and function of DNA, RNA, and the processes involved in genetic information transfer. It covers key concepts such as DNA replication, transcription, translation, and regulation of gene expression, forming the foundation of molecular genetics in the CBSE Grade 12 Biology curriculum.

DNA: The Genetic Material

DNA (Deoxyribonucleic Acid) is the hereditary material in humans and almost all other organisms, carrying genetic instructions for development, functioning, growth, and reproduction.

The chapter begins with the discovery of DNA as the genetic material through experiments by Griffith, Avery, Hershey, and Chase. It explains the double-helix structure of DNA proposed by Watson and Crick, including the antiparallel strands, base pairing (A-T, G-C), and the sugar-phosphate backbone.

DNA Replication

DNA replication is the biological process of producing two identical replicas of DNA from one original DNA molecule, ensuring genetic continuity.

The semi-conservative mode of DNA replication, demonstrated by Meselson and Stahl, is discussed along with the roles of enzymes like DNA polymerase, helicase, and ligase. The chapter also explains the origin of replication, replication fork, and leading/lagging strands.

Transcription

Transcription is the process of copying a segment of DNA into RNA (mRNA, tRNA, rRNA) by the enzyme RNA polymerase.

The chapter covers transcription in prokaryotes and eukaryotes, including promoter regions, transcription factors, and post-transcriptional modifications like capping, tailing, and splicing in eukaryotes.

Genetic Code

The genetic code is the set of rules by which information encoded in genetic material (DNA or RNA sequences) is translated into proteins by living cells.

This section explains the triplet nature of the code, its universality, and characteristics like degeneracy and non-ambiguity. The chapter details how Marshall Nirenberg's experiments helped crack the genetic code.

Translation

Translation is the process of protein synthesis where the genetic code carried by mRNA is decoded to produce a specific polypeptide chain.

The chapter describes the machinery of translation including ribosomes, tRNA, and amino acids. It explains the three phases of translation: initiation, elongation, and termination, along with energy requirements.

Regulation of Gene Expression

Gene regulation refers to the mechanisms that cells use to turn genes on or off, controlling when and how much gene product is made.

The lac operon model in E. coli is discussed as a paradigm of gene regulation in prokaryotes. For eukaryotes, the chapter touches on differential gene expression and various levels of regulation.

Human Genome Project

The chapter concludes with an overview of the Human Genome Project, its goals, methodologies, and outcomes. It briefly discusses DNA fingerprinting and its applications in forensic science and paternity disputes.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the central dogma of molecular biology?
Answer:

DNA → RNA → Protein.

Question 2:
Name the enzyme that synthesizes RNA primers during DNA replication.
Answer:

Primase.

Question 3:
Which nucleotide is not present in RNA?
Answer:

Thymine (T).

Question 4:
What is the function of helicase in DNA replication?
Answer:

Unwinds the DNA double helix.

Question 5:
Define transcription.
Answer:

Synthesis of RNA from DNA template.

Question 6:
Which genetic material is found in tobacco mosaic virus?
Answer:

RNA.

Question 7:
What is the role of tRNA in protein synthesis?
Answer:

Transfers amino acids to ribosomes.

Question 8:
Name the stop codons in the genetic code.
Answer:

UAA, UAG, UGA.

Question 9:
What is semi-conservative DNA replication?
Answer:

Each new DNA has one parental and one new strand.

Question 10:
Which enzyme seals nicks in DNA strands?
Answer:

DNA ligase.

Question 11:
What is the function of the promoter in transcription?
Answer:

Initiation site for RNA polymerase.

Question 12:
Name the process by which mRNA is decoded into a protein.
Answer:

Translation.

Question 13:
What are exons in eukaryotic genes?
Answer:

Coding sequences retained in mRNA.

Question 14:
Which scientist discovered the transforming principle?
Answer:

Frederick Griffith.

Question 15:
Name the enzyme responsible for the polymerization of RNA during transcription.
Answer:

The enzyme responsible for the polymerization of RNA during transcription is RNA polymerase. It synthesizes RNA from a DNA template.

Question 16:
What is the function of promoter in transcription?
Answer:

The promoter is a DNA sequence that provides a binding site for RNA polymerase to initiate transcription. It determines the exact start point and direction of transcription.

Question 17:
Define codon and give an example.
Answer:

A codon is a sequence of three nucleotides in mRNA that codes for a specific amino acid. For example, AUG codes for Methionine and also acts as the start codon.

Question 18:
What is the significance of the lac operon in E. coli?
Answer:

The lac operon in E. coli regulates the metabolism of lactose. It consists of genes that produce enzymes for lactose digestion and is activated only when lactose is present and glucose is absent.

Question 19:
Differentiate between exons and introns.
Answer:

Exons are coding sequences in a gene that are expressed in proteins, while introns are non-coding sequences that are removed during RNA splicing.

Question 20:
Why is DNA replication called semi-conservative?
Answer:

DNA replication is semi-conservative because each new DNA molecule consists of one parental (old) strand and one newly synthesized strand, preserving half of the original DNA.

Question 21:
Name the scientist who proposed the transforming principle.
Answer:

Frederick Griffith proposed the transforming principle through his experiments on Streptococcus pneumoniae, showing genetic material could transfer traits.

Question 22:
Explain the term central dogma of molecular biology.
Answer:

The central dogma describes the flow of genetic information: DNA → RNA → Protein. It includes transcription (DNA to RNA) and translation (RNA to protein).

Question 23:
What is a mutation? Give one example.
Answer:

A mutation is a permanent alteration in the DNA sequence. Example: Sickle cell anemia is caused by a point mutation in the hemoglobin gene.

Question 24:
How does DNA fingerprinting work?
Answer:

DNA fingerprinting analyzes VNTRs (Variable Number Tandem Repeats) in DNA to create unique patterns for identification, used in forensics and paternity tests.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Name the enzyme responsible for the polymerization of RNA nucleotides during transcription.
Answer:

The enzyme responsible for the polymerization of RNA nucleotides during transcription is RNA polymerase. It synthesizes RNA in the 5' → 3' direction using a DNA template strand.

Question 2:
What is the role of promoter in transcription?
Answer:

The promoter is a DNA sequence that provides a binding site for RNA polymerase to initiate transcription. It determines the exact start point and direction of transcription.

Question 3:
What is the function of helicase during DNA replication?
Answer:

Helicase unwinds the double-stranded DNA by breaking hydrogen bonds between base pairs, creating two single strands that serve as templates for replication.

Question 4:
Define translation in molecular biology.
Answer:

Translation is the process where the sequence of nucleotides in mRNA is decoded by ribosomes to synthesize a specific polypeptide chain (protein) using tRNA molecules.

Question 5:
What is the significance of the start codon in protein synthesis?
Answer:

The start codon (AUG) initiates translation and codes for the amino acid methionine. It sets the reading frame for mRNA decoding.

Question 6:
Name the stop codons and state their function.
Answer:

The stop codons are UAA, UAG, and UGA. They signal the termination of translation by releasing the polypeptide chain from the ribosome.

Question 7:
How does lac operon regulate gene expression in E. coli?
Answer:

The lac operon regulates gene expression in response to lactose availability. Lactose acts as an inducer, inactivating the repressor and allowing transcription of genes for lactose metabolism.

Question 8:
What is the function of tRNA in protein synthesis?
Answer:

tRNA (transfer RNA) carries specific amino acids to the ribosome and matches its anticodon with the mRNA codon, ensuring accurate polypeptide assembly.

Question 9:
Explain the term genetic code and list its key features.
Answer:

The genetic code is the set of rules by which mRNA sequences are translated into proteins. Key features:

  • Triplet: 3 nucleotides code for 1 amino acid.
  • Universal: Same in most organisms.
  • Degenerate: Multiple codons can code for the same amino acid.
  • Non-overlapping: Codons are read sequentially.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Explain the significance of the Meselson and Stahl experiment in understanding DNA replication.
Answer:

The Meselson and Stahl experiment proved that DNA replication is semi-conservative, meaning each new DNA molecule consists of one parental strand and one newly synthesized strand.

They used heavy nitrogen (N15) and light nitrogen (N14) isotopes to label DNA. After replication in N14, the DNA formed a hybrid density, confirming semi-conservative replication.

This experiment provided crucial evidence against conservative and dispersive models of replication.

Question 2:
Differentiate between template strand and coding strand in transcription.
Answer:

Template Strand:

  • Acts as the guide for RNA synthesis.
  • Has a complementary sequence to the mRNA (except thymine replaced by uracil).
  • Also called the antisense strand.

Coding Strand:
  • Has the same sequence as mRNA (except thymine instead of uracil).
  • Does not participate directly in transcription.
  • Also called the sense strand.

Question 3:
Describe the role of RNA polymerase in transcription.
Answer:

RNA polymerase is the enzyme responsible for synthesizing RNA from a DNA template.

Functions:

  • Binds to the promoter region to initiate transcription.
  • Unwinds the DNA helix to expose the template strand.
  • Adds complementary RNA nucleotides (A, U, G, C) in the 5'→3' direction.
  • Terminates transcription upon reaching the terminator sequence.

Question 4:
What is the lac operon? Explain its regulation in E. coli.
Answer:

The lac operon is a group of genes in E. coli involved in lactose metabolism.

Regulation:

  • In the absence of lactose, the repressor protein binds to the operator, blocking transcription.
  • In the presence of lactose, lactose acts as an inducer, binding to the repressor and inactivating it, allowing RNA polymerase to transcribe the genes (lacZ, lacY, lacA).

This ensures energy-efficient enzyme production only when needed.

Question 5:
How does alternative splicing contribute to protein diversity in eukaryotes?
Answer:

Alternative splicing allows a single gene to produce multiple mRNA variants by selectively including or excluding exons.

Process:

  • Different combinations of exons are joined during mRNA processing.
  • This generates distinct protein isoforms from the same gene.

Significance: Enhances proteome diversity without increasing the number of genes, crucial for complex functions like immune response and neural development.

Question 6:
Explain the structure and function of tRNA in translation.
Answer:

tRNA (transfer RNA) is an adaptor molecule that delivers amino acids to the ribosome during translation.

Structure:

  • Cloverleaf-shaped with an anticodon loop (binds to mRNA codon) and an amino acid attachment site (3'-CCA end).
  • Contains modified bases like pseudouridine for stability.

Function: Ensures correct amino acid incorporation by matching its anticodon to the mRNA codon, maintaining the genetic code's accuracy.

Question 7:
Explain the significance of the Hershey-Chase experiment in confirming DNA as the genetic material.
Answer:

The Hershey-Chase experiment confirmed that DNA (not protein) is the genetic material. They used T2 bacteriophages labeled with radioactive phosphorus (DNA) and sulfur (protein) to infect E. coli. After blending and centrifugation, only the radioactive DNA entered bacterial cells, proving its role in heredity. This experiment laid the foundation for molecular genetics.

Question 8:
Differentiate between transcription and translation in eukaryotes.
Answer:

Transcription and translation are key steps in gene expression:

  • Transcription: Occurs in the nucleus, where DNA is copied into mRNA by RNA polymerase.
  • Translation: Occurs in the cytoplasm, where ribosomes decode mRNA into proteins using tRNA.

In eukaryotes, transcription precedes translation, and mRNA processing (splicing, capping, tailing) occurs before translation.

Question 9:
Describe the structure of a nucleosome with a labeled diagram.
Answer:

A nucleosome is the basic unit of DNA packaging in eukaryotes. It consists of:

  • Core histone octamer (2 molecules each of H2A, H2B, H3, H4) forming a bead-like structure.
  • DNA wrapped around the octamer (1.65 turns, ~146 bp).
  • Linker DNA (20-80 bp) connecting nucleosomes, bound by H1 histone.

This arrangement forms the "beads-on-a-string" chromatin structure, enabling efficient DNA compaction.

Question 10:
How does semi-conservative replication of DNA ensure genetic continuity?
Answer:

In semi-conservative replication, each new DNA molecule has one parental strand and one newly synthesized strand. This ensures:

  • Genetic continuity: The original sequence is preserved in one strand.
  • Error correction: The parental strand acts as a template for proofreading.
  • Stability: Reduces mutations over generations.

This mechanism was proven by Meselson-Stahl using N15 and N14 isotopes.

Question 11:
What is the role of RNA polymerase in transcription?
Answer:

RNA polymerase is the enzyme responsible for synthesizing RNA from a DNA template. Its roles include:

  • Initiation: Binds to promoter regions with sigma factors (in prokaryotes).
  • Elongation: Adds ribonucleotides complementary to the DNA strand (3’→5’).
  • Termination: Stops at terminator sequences (e.g., rho-dependent in prokaryotes).

In eukaryotes, RNA polymerase II transcribes mRNA.

Question 12:
Explain the lac operon model of gene regulation in E. coli.
Answer:

The lac operon regulates lactose metabolism in E. coli:

  • Structure: Includes lacZ (β-galactosidase), lacY (permease), lacA (transacetylase), and regulatory lacI (repressor).
  • Repression: LacI binds to the operator, blocking transcription in absence of lactose.
  • Induction: Lactose (inducer) binds to repressor, freeing the operator for RNA polymerase.

This model illustrates negative inducible regulation.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the semi-conservative replication of DNA with experimental evidence. How does this mechanism ensure genetic stability?
Answer:
Theoretical Framework

We studied that DNA replication is semi-conservative, where each new DNA molecule has one parental and one newly synthesized strand. This was proven by Meselson and Stahl using E. coli and nitrogen isotopes.

Evidence Analysis
  • Meselson-Stahl experiment used 15N (heavy) and 14N (light) isotopes to label DNA.
  • After one generation, DNA had intermediate density, confirming semi-conservative replication.
Critical Evaluation

This mechanism prevents mutations by using the parental strand as a template, ensuring accuracy. Errors are minimized by proofreading by DNA polymerase.

Future Implications

Understanding this process aids in genetic engineering and cancer research, where DNA replication errors play a key role.

Question 2:
Describe the lac operon model of gene regulation in E. coli. How does it demonstrate inducible enzyme synthesis?
Answer:
Theoretical Framework

The lac operon, studied in our textbook, regulates lactose metabolism in E. coli. It consists of structural genes (lacZ, lacY, lacA) and regulatory sequences.

Evidence Analysis
  • In the absence of lactose, the lac repressor binds to the operator, blocking transcription.
  • When lactose is present, it acts as an inducer, inactivating the repressor and allowing enzyme synthesis.
Critical Evaluation

This model shows how prokaryotes efficiently regulate metabolism. It also explains how gene expression responds to environmental changes.

Future Implications

Studying operons helps in designing synthetic biology systems, like insulin production using recombinant DNA technology.

Question 3:
Compare transcription in prokaryotes and eukaryotes, highlighting key differences in initiation, elongation, and termination.
Answer:
Theoretical Framework

Transcription is the synthesis of RNA from DNA. Our textbook shows that prokaryotes and eukaryotes differ significantly in this process.

Evidence Analysis
  • Initiation: Prokaryotes use sigma factors, while eukaryotes require transcription factors (TFs) and RNA polymerase II.
  • Elongation: Eukaryotic mRNA undergoes splicing to remove introns, unlike prokaryotic mRNA.
  • Termination: Prokaryotes use rho-dependent or independent mechanisms, while eukaryotes rely on polyadenylation signals.
Critical Evaluation

These differences reflect the complexity of eukaryotic gene regulation, allowing for alternative splicing and post-transcriptional modifications.

Future Implications

Understanding these mechanisms aids in developing gene therapies targeting specific transcriptional errors.

Question 4:
Explain the central dogma of molecular biology. How do retroviruses challenge this concept?
Answer:
Theoretical Framework

The central dogma states that genetic information flows from DNA → RNA → protein. However, retroviruses like HIV reverse this flow.

Evidence Analysis
  • Retroviruses use reverse transcriptase to convert RNA into DNA, which integrates into the host genome.
  • This challenges the unidirectional flow proposed by Crick.
Critical Evaluation

Exceptions like retroviruses show that biological systems are more flexible than originally thought. This has implications for viral treatments.

Future Implications

Studying reverse transcription helps in designing antiretroviral drugs and understanding viral evolution.

Question 5:
Discuss the significance of Human Genome Project (HGP). How has it advanced our understanding of genetic disorders?
Answer:
Theoretical Framework

The HGP, completed in 2003, mapped the entire human genome. Our textbook highlights its role in identifying genes linked to diseases.

Evidence Analysis
  • HGP revealed ~20,500 genes, enabling the study of mutations causing disorders like cystic fibrosis.
  • It also discovered SNPs (single nucleotide polymorphisms) associated with diseases like diabetes.
Critical Evaluation

This project revolutionized personalized medicine by enabling gene therapy and early diagnosis of genetic conditions.

Future Implications

Ongoing research uses HGP data to develop CRISPR-based treatments and predict disease risks.

Question 6:
Describe the central dogma of molecular biology. How do exceptions like reverse transcription challenge this dogma?
Answer:
Theoretical Framework

The central dogma states genetic information flows from DNA → RNA → Protein. It was proposed by Francis Crick in 1958.

Evidence Analysis
  • In retroviruses like HIV, reverse transcriptase converts RNA to DNA, violating the dogma.
  • Prions replicate without nucleic acids, another exception.
Critical Evaluation

These exceptions show genetic information flow is more complex. However, the dogma remains foundational for most organisms.

Future Implications

Studying exceptions helps develop antiviral drugs targeting reverse transcriptase.

Question 7:
Compare transcription in prokaryotes and eukaryotes. Highlight three key differences with examples.
Answer:
Theoretical Framework

Transcription is the synthesis of RNA from DNA. Prokaryotes (e.g., E. coli) and eukaryotes (e.g., humans) differ significantly.

Evidence Analysis
  • Location: Prokaryotes lack nuclei; transcription occurs in cytoplasm. Eukaryotes transcribe in nuclei.
  • Processing: Eukaryotic pre-mRNA undergoes splicing; prokaryotic mRNA does not.
  • Initiation: Prokaryotes use sigma factors; eukaryotes use transcription factors (TFs).
Critical Evaluation

These differences reflect cellular complexity. Eukaryotic processing allows alternative splicing, increasing protein diversity.

Future Implications

Understanding transcription aids gene therapy development.

Question 8:
Explain the lac operon model with a labeled diagram. How does it regulate gene expression in E. coli?
Answer:
Theoretical Framework

The lac operon in E. coli controls lactose metabolism. It consists of promoter, operator, and structural genes (lacZ, lacY, lacA).

[Diagram: Lac operon with labeled components]Evidence Analysis
  • Without lactose, the repressor binds the operator, blocking transcription.
  • Lactose acts as an inducer, inactivating the repressor and allowing transcription.
Critical Evaluation

This model demonstrates negative regulation. It ensures enzymes are only produced when needed, conserving energy.

Future Implications

Operon studies help design synthetic biology systems.

Question 9:
What is genetic code? Discuss its salient features with examples.
Answer:
Theoretical Framework

The genetic code is the set of rules by which DNA/RNA sequences are translated into proteins. It is nearly universal.

Evidence Analysis
  • Triplet: Codons are 3-nucleotide sequences (e.g., AUG for methionine).
  • Degenerate: Multiple codons code for one amino acid (e.g., UCU, UCC both code for serine).
  • Non-overlapping: Each nucleotide is part of only one codon.
Critical Evaluation

These features ensure accurate protein synthesis. Mutations may cause diseases like sickle-cell anemia.

Future Implications

Deciphering the code aids in genetic engineering.

Question 10:
Describe the Hershey-Chase experiment. How did it confirm DNA as the genetic material?
Answer:
Theoretical Framework

Hershey and Chase (1952) used radioactive isotopes to prove DNA, not protein, carries genetic information.

Evidence Analysis
  • They labeled phage DNA with P32 and protein with S35.
  • Only P32 entered bacterial cells during infection, confirming DNA’s role.
Critical Evaluation

This experiment resolved the debate about genetic material. It paved the way for molecular genetics.

Future Implications

Their work underpins modern biotechnology, like CRISPR.

Question 11:
Differentiate between exons and introns. How does alternative splicing increase proteome diversity?
Answer:
Theoretical Framework

Exons are coding sequences, while introns are non-coding and spliced out during mRNA processing.

Evidence Analysis
  • In humans, the dystrophin gene has 79 exons; mutations cause muscular dystrophy.
  • Alternative splicing combines exons differently, producing multiple proteins from one gene (e.g., tropomyosin).
Critical Evaluation

This mechanism allows ~20,000 genes to encode ~100,000 proteins, enhancing complexity.

Future Implications

Studying splicing aids in treating genetic disorders.

Question 12:
Explain DNA fingerprinting. How is it used in forensic science and paternity testing?
Answer:
Theoretical Framework

DNA fingerprinting identifies individuals by analyzing VNTRs (Variable Number Tandem Repeats), which are unique to each person.

Evidence Analysis
  • In forensics, it matches crime scene DNA to suspects (e.g., the 1986 Colin Pitchfork case).
  • In paternity tests, child DNA is compared to parents’ to confirm biological relationships.
Critical Evaluation

This technique is highly accurate (1 in a billion chance of error).

Future Implications

Advances like PCR have made fingerprinting faster and cheaper.

Question 13:
What are transposons? Discuss their role in evolution and disease.
Answer:
Theoretical Framework

Transposons (jumping genes) are mobile DNA sequences that can change position within a genome.

Evidence Analysis
  • They cause mutations by inserting into genes (e.g., Alu elements in humans).
  • In maize, Barbara McClintock showed transposons alter kernel color.
Critical Evaluation

Transposons drive evolution by creating genetic diversity but can also cause diseases like hemophilia.

Future Implications

Gene therapy may target transposon-related disorders.

Question 14:
Describe the human genome project (HGP). What were its major findings and applications?
Answer:
Theoretical Framework

The HGP (1990-2003) mapped all human genes to understand genetic diseases and evolution.

Evidence Analysis
  • Found only ~20,000 genes, fewer than expected.
  • Identified genes linked to diseases like BRCA1 (breast cancer).
Critical Evaluation

HGP revolutionized medicine but raised ethical issues like genetic privacy.

Future Implications

It enables personalized medicine and CRISPR-based therapies.

Question 15:
Explain the process of DNA replication in prokaryotes with emphasis on the role of enzymes involved. How does it differ from eukaryotic DNA replication?
Answer:

The process of DNA replication in prokaryotes is semi-conservative and involves several key enzymes. Here's a step-by-step explanation:

  • Initiation: Replication begins at the origin of replication (ori), where the enzyme helicase unwinds the DNA double helix, forming a replication fork. Single-stranded binding proteins (SSBs) stabilize the unwound strands.
  • Elongation: The enzyme DNA polymerase III adds nucleotides to the growing DNA strand in the 5'→3' direction, using the parental strand as a template. RNA primase synthesizes a short RNA primer to initiate DNA synthesis.
  • Termination: Replication ends when the replication forks meet, and DNA polymerase I replaces RNA primers with DNA. DNA ligase seals the gaps between Okazaki fragments on the lagging strand.

Differences from eukaryotic replication:
1. Prokaryotes have a single origin of replication, while eukaryotes have multiple.
2. Prokaryotic DNA polymerase III is faster than eukaryotic DNA polymerases.
3. Eukaryotic DNA is linear and requires telomerase to prevent shortening, unlike circular prokaryotic DNA.

Question 16:
Describe the central dogma of molecular biology with a detailed explanation of transcription and translation. Include the significance of this process.
Answer:

The central dogma explains the flow of genetic information: DNA → RNA → Protein. Here's a breakdown:

  • Transcription: In the nucleus, RNA polymerase synthesizes mRNA from a DNA template strand.
    1. Initiation: RNA polymerase binds to the promoter region of DNA.
    2. Elongation: Nucleotides are added complementary to the DNA strand (A-U, T-A, G-C).
    3. Termination: mRNA detaches after reaching the terminator sequence.
  • Translation: Occurs in the cytoplasm at ribosomes.
    1. Initiation: mRNA binds to the ribosome; the start codon (AUG) is recognized.
    2. Elongation: tRNA brings amino acids, forming a polypeptide chain.
    3. Termination: A stop codon signals the release of the protein.

Significance: This process ensures accurate protein synthesis, which is vital for cellular functions, growth, and heredity. Errors can lead to mutations or diseases.

Question 17:
What is the lac operon? Explain its regulation in E. coli in the presence and absence of lactose. Include a labeled diagram of the operon.
Answer:

The lac operon is a gene regulatory system in E. coli that controls lactose metabolism. It consists of:

  • Structural genes: lacZ (β-galactosidase), lacY (permease), and lacA (transacetylase).
  • Regulatory elements: Promoter, operator, and repressor gene (lacI).

Regulation:
1. Absence of lactose: The repressor protein binds to the operator, blocking RNA polymerase → no transcription.
2. Presence of lactose: Lactose acts as an inducer, binding to the repressor and inactivating it. RNA polymerase transcribes the genes, producing enzymes to metabolize lactose.

Diagram (description):
1. Label the lacI gene, promoter, operator, and structural genes (lacZYA).
2. Show the repressor bound to the operator (no lactose) and detached (with lactose).
3. Indicate mRNA synthesis during active transcription.

Significance: This model demonstrates gene regulation in prokaryotes, conserving energy by producing enzymes only when needed.

Question 18:
Explain the process of transcription in prokaryotes with a detailed diagram. Highlight the role of RNA polymerase and promoter regions.
Answer:

Transcription is the process by which the genetic information from DNA is copied into RNA. In prokaryotes, it occurs in three main stages: initiation, elongation, and termination.


1. Initiation: The RNA polymerase enzyme binds to the promoter region on the DNA, which is a specific sequence marking the start of a gene. The promoter helps in the correct positioning of RNA polymerase. The enzyme then unwinds the DNA helix, exposing the template strand.


2. Elongation: RNA polymerase synthesizes the RNA strand in the 5' to 3' direction, using the DNA template strand. Nucleotides are added complementary to the DNA template (A-U, T-A, G-C, C-G). The DNA helix rewinds behind the enzyme.


3. Termination: Transcription stops when RNA polymerase encounters a terminator sequence. The newly formed RNA strand is released, and the RNA polymerase detaches from the DNA.


Role of RNA polymerase: It catalyzes the formation of phosphodiester bonds between nucleotides and ensures accurate copying of the genetic code.


Role of promoter: It signals where transcription begins and ensures RNA polymerase binds correctly.


Diagram: (A labeled diagram showing DNA unwinding, RNA polymerase, promoter, and the growing RNA strand should be drawn for full marks.)

Question 19:
Describe the Meselson and Stahl experiment and explain how it provided evidence for the semi-conservative mode of DNA replication.
Answer:

The Meselson and Stahl experiment (1958) demonstrated that DNA replication follows a semi-conservative mechanism, where each new DNA molecule consists of one parental strand and one newly synthesized strand.

The experiment involved:

  • Growing E. coli in a medium containing heavy nitrogen (¹⁵N) to label parental DNA.
  • Transferring the bacteria to a light nitrogen (¹⁴N) medium and allowing replication.
  • Analyzing DNA density using cesium chloride (CsCl) gradient centrifugation after each generation.

Results:

  • Generation 0 (¹⁵N-DNA): Only heavy DNA (one band).
  • Generation 1 (hybrid DNA): Intermediate-density DNA (one band), proving semi-conservative replication.
  • Generation 2: Two bands (light and hybrid), confirming the mechanism.

This experiment conclusively disproved conservative and dispersive models, establishing the semi-conservative model as the universal mode of DNA replication.

Question 20:
Describe the Meselson and Stahl experiment and explain how it proved the semi-conservative mode of DNA replication.
Answer:

The Meselson and Stahl experiment (1958) provided evidence for the semi-conservative replication of DNA. Here’s how it was conducted and interpreted:


1. Experimental Setup:
- E. coli bacteria were grown in a medium containing heavy nitrogen (¹⁵N) for several generations, ensuring all DNA was labeled with ¹⁵N.
- The bacteria were then transferred to a medium with light nitrogen (¹⁴N) and allowed to replicate.


2. Observations:
- After one generation, the DNA formed a single band of intermediate density (hybrid ¹⁵N-¹⁴N DNA) in a cesium chloride density gradient.
- After two generations, two bands appeared: one intermediate and one light (¹⁴N-¹⁴N DNA).


3. Conclusion:
- The results matched the semi-conservative model, where each new DNA molecule consists of one parental strand (old) and one newly synthesized strand.
- This disproved the conservative (parental DNA remains intact) and dispersive (mixed fragments) models.


Significance: This experiment confirmed Watson and Crick’s hypothesis about DNA replication, a cornerstone of molecular biology.

Question 21:
Explain the process of transcription in prokaryotes with a detailed emphasis on the role of RNA polymerase and the significance of promoter and terminator regions.
Answer:

Transcription is the process by which the genetic information from DNA is copied into RNA. In prokaryotes, this process involves the enzyme RNA polymerase, which synthesizes RNA from a DNA template.

The key steps are:

  • Initiation: RNA polymerase binds to the promoter region, a specific DNA sequence upstream of the gene. The sigma factor helps the enzyme recognize the promoter.
  • Elongation: RNA polymerase unwinds the DNA and synthesizes RNA in the 5' to 3' direction, using one strand (template strand) as a guide.
  • Termination: Transcription stops when RNA polymerase reaches the terminator region, causing the RNA transcript and enzyme to detach.

The promoter ensures accurate initiation, while the terminator signals the end of transcription. Prokaryotic transcription is efficient due to the absence of a nucleus, allowing simultaneous transcription and translation.

Question 22:
Explain the process of transcription in prokaryotes with a focus on the role of RNA polymerase and promoter. How does it differ from transcription in eukaryotes?
Answer:

In prokaryotes, transcription is the process of synthesizing RNA from a DNA template. The key enzyme involved is RNA polymerase, which binds to a specific region on the DNA called the promoter. The promoter signals the start of transcription and helps RNA polymerase recognize where to begin.


Steps in prokaryotic transcription:

  • Initiation: RNA polymerase binds to the promoter with the help of a sigma factor, forming a holoenzyme complex.
  • Elongation: RNA polymerase unwinds the DNA and synthesizes RNA in the 5' to 3' direction, using one strand (template strand) as a guide.
  • Termination: Transcription stops when RNA polymerase encounters a terminator sequence, leading to the release of the RNA transcript.

Differences in eukaryotes:

  • Eukaryotes have three types of RNA polymerases (I, II, III) for different RNAs, while prokaryotes have only one.
  • Eukaryotic transcription involves transcription factors and enhancers for regulation, unlike prokaryotes.
  • Post-transcriptional modifications (e.g., splicing, capping, tailing) occur in eukaryotes but not in prokaryotes.
Question 23:
Describe the Meselson and Stahl experiment and explain how it provided evidence for the semi-conservative replication of DNA. Include a labeled diagram to support your answer.
Answer:

The Meselson and Stahl experiment (1958) demonstrated that DNA replication is semi-conservative, meaning each new DNA molecule consists of one parental strand and one newly synthesized strand.


Experimental Procedure:

  • E. coli bacteria were grown in a medium containing heavy nitrogen (¹⁵N), making their DNA denser.
  • These bacteria were then transferred to a medium with light nitrogen (¹⁴N) and allowed to replicate.
  • DNA samples were extracted after each generation and analyzed using density gradient centrifugation.

Observations & Conclusions:

  • First generation: DNA formed a single band of intermediate density (hybrid of ¹⁵N and ¹⁴N), ruling out conservative replication.
  • Second generation: Two bands appeared—one intermediate and one light (only ¹⁴N), confirming semi-conservative replication.

Diagram: (Illustrate the centrifugation tubes showing bands for ¹⁵N, hybrid, and ¹⁴N DNA after generations.)


This experiment conclusively proved that DNA replication is semi-conservative, as predicted by Watson and Crick.

Question 24:
Explain the process of DNA replication in eukaryotes, highlighting the role of enzymes involved. How does this process ensure high fidelity in the transmission of genetic information?
Answer:

The process of DNA replication in eukaryotes is a semi-conservative mechanism where each parental DNA strand serves as a template for the synthesis of a new complementary strand. This ensures the accurate transmission of genetic information during cell division. The key steps and enzymes involved are:

  • Initiation: Replication begins at specific sites called origins of replication, where the helicase enzyme unwinds the double helix, creating replication forks. Single-stranded binding proteins (SSBs) stabilize the unwound strands.
  • Primer Synthesis: The enzyme primase synthesizes short RNA primers complementary to the DNA template, providing a starting point for DNA synthesis.
  • Elongation: DNA polymerase III adds nucleotides to the 3' end of the RNA primer, synthesizing the new strand in the 5'→3' direction. The leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously as Okazaki fragments.
  • Proofreading and Repair: DNA polymerase I replaces RNA primers with DNA, and DNA ligase seals the gaps between Okazaki fragments. DNA polymerase also proofreads the newly synthesized strand, correcting mismatches to maintain fidelity.

The high fidelity of DNA replication is ensured by:
1. The complementary base pairing rule (A-T, G-C), which reduces errors.
2. The proofreading activity of DNA polymerase, which removes incorrect nucleotides.
3. Mismatch repair mechanisms post-replication, which correct errors missed during synthesis.

This meticulous process ensures that genetic information is passed accurately to daughter cells, maintaining genetic stability across generations.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A research team identified a point mutation in the β-globin gene causing sickle-cell anemia. Explain how this mutation affects hemoglobin and discuss its inheritance pattern.
Answer:
Case Deconstruction

A single nucleotide change (GAG to GTG) replaces glutamic acid with valine in the β-globin chain, altering hemoglobin's structure.

Theoretical Application
  • Inheritance follows autosomal recessive pattern: carriers (heterozygotes) show mild symptoms, while homozygotes exhibit severe anemia.
Critical Evaluation

Our textbook shows this mutation confers malaria resistance in carriers, demonstrating balanced polymorphism. Example: Higher prevalence in malaria-endemic regions.

Question 2:
Compare semi-conservative DNA replication with dispersive replication using Meselson-Stahl's experiment. Which model did their results support?
Answer:
Case Deconstruction

Meselson-Stahl used N15 isotope labeling to track DNA strands across generations.

Theoretical Application
  • Semi-conservative: Each new DNA has one parental strand.
  • Dispersive: Parental DNA fragments scatter randomly.
Critical Evaluation

Their density gradient results matched only the semi-conservative model. Example: First-generation hybrids showed intermediate density.

Question 3:
Analyze how lac operon regulation in E. coli differs under high lactose vs. high glucose conditions using a diagram of the control mechanism.
Answer:
Case Deconstruction

The operon includes structural genes (lacZ, lacY) regulated by a repressor and CAP-cAMP.

Theoretical Application
  • High lactose: Allolactose inactivates repressor, enabling transcription.
  • High glucose: Low cAMP prevents CAP activation.
Critical Evaluation

[Diagram: Operon switching] Example: Mutant studies proved repressor-operator binding.

Question 4:
A patient's DNA fingerprint shows 12 STR repeats at one locus and 15 at another. Explain how this profile is unique and its forensic application.
Answer:
Case Deconstruction

STRs are non-coding tandem repeats with high polymorphism.

Theoretical Application
  • Combined probabilities of multiple loci create uniqueness (1 in billion).
  • Used in paternity tests and crime investigations.
Critical Evaluation

Example: 1987 Colin Pitchfork case first used DNA fingerprinting. Limitations include sample degradation risks.

Question 5:
A research team discovered a point mutation in the β-globin gene causing sickle-cell anemia. Analyze how this mutation affects protein structure and function.
Answer:
Case Deconstruction

The point mutation replaces glutamic acid with valine at the 6th position of the β-globin chain. Our textbook shows this alters hemoglobin's solubility.

Theoretical Application
  • Mutated hemoglobin polymerizes under low oxygen, distorting RBCs into sickle shapes.
  • This reduces oxygen transport efficiency, causing anemia.
Critical Evaluation

While harmful, heterozygous carriers gain malaria resistance, demonstrating evolutionary trade-offs.

Question 6:
Compare semi-conservative and dispersive DNA replication models using Meselson-Stahl's experimental evidence.
Answer:
Case Deconstruction

Meselson-Stahl used N15 isotope labeling to trace DNA strands across generations.

Theoretical Application
  • Semi-conservative model produced hybrid (N15-N14) DNA after 1st generation, matching observed results.
  • Dispersive model would yield uniformly mixed strands, which wasn't detected.
Critical Evaluation

This experiment conclusively supported Watson-Crick's prediction of semi-conservative replication.

Question 7:
Explain how lac operon regulation in E. coli demonstrates gene expression control with lactose metabolism.
Answer:
Case Deconstruction

The lac operon contains genes for lactose digestion (lacZ, lacY) controlled by a repressor protein.

Theoretical Application
  • Without lactose, repressor binds operator, blocking transcription.
  • Lactose acts as inducer by inactivating repressor, allowing enzyme production.
Critical Evaluation

This dual regulation (repression/induction) optimizes energy use, showing prokaryotic adaptability.

Question 8:
Analyze how alternative splicing in eukaryotes enables single genes to produce multiple proteins, using tropomyosin as example.
Answer:
Case Deconstruction

Alternative splicing rearranges exons to create variant mRNAs from one gene.

Theoretical Application
  • The tropomyosin gene produces 5+ isoforms through exon skipping in different tissues.
  • Smooth muscle isoforms differ from skeletal muscle versions in function.
Critical Evaluation

This process exponentially increases proteome diversity without increasing gene count.

Question 9:
A research team discovered a point mutation in the β-globin gene causing sickle-cell anemia. Explain how this mutation affects hemoglobin and discuss its inheritance pattern.
Answer:
Case Deconstruction

A single nucleotide change (GAG to GTG) substitutes glutamic acid with valine in the β-globin chain, altering hemoglobin's structure.

Theoretical Application
  • Inheritance is autosomal recessive; heterozygous individuals are carriers.
  • Mutated hemoglobin polymerizes under low oxygen, distorting RBCs into sickle shapes.
Critical Evaluation

Our textbook shows this mutation confers malaria resistance in carriers, demonstrating balanced polymorphism. Example: Higher allele frequency in malaria-endemic regions.

Question 10:
Analyze how lac operon regulation in E. coli exemplifies gene expression control. Include roles of lacI and cAMP-CRP complex.
Answer:
Case Deconstruction

The lac operon switches on only when lactose is present and glucose is absent.

Theoretical Application
  • lacI encodes a repressor blocking transcription without lactose.
  • cAMP-CRP enhances RNA polymerase binding when glucose is low.
Critical Evaluation

We studied this as a classic example of dual regulation. Example: Mutations in lacI can cause constitutive expression.

Question 11:
A patient has fragile X syndrome due to trinucleotide repeat expansion (CGG) in the FMR1 gene. Explain molecular consequences and why males are more severely affected.
Answer:
Case Deconstruction

CGG repeats exceeding 200 cause FMR1 silencing via DNA methylation.

Theoretical Application
  • Absence of FMR protein disrupts neural synapse development.
  • X-linked inheritance: Males (XY) lack a compensatory X chromosome.
Critical Evaluation

Our textbook shows this as the most common inherited intellectual disability. Example: Females may be mildly affected due to X-inactivation.

Question 12:

In a research lab, scientists isolated a segment of DNA from a bacterial cell and observed that it contained 20% adenine bases. Based on this information:

  • Calculate the percentage of cytosine bases in the DNA segment.
  • Explain the principle used to arrive at this conclusion.
Answer:

According to Chargaff's rule, in a DNA molecule, the amount of adenine (A) is equal to thymine (T), and the amount of guanine (G) is equal to cytosine (C). This is due to complementary base pairing.


Given that adenine (A) is 20%, thymine (T) will also be 20%.


Total percentage of A + T = 20% + 20% = 40%.


Therefore, the remaining 60% must be G + C.


Since G = C, cytosine (C) will be 30%.


Principle: Chargaff's rule states that in any DNA molecule, the ratio of purines (A + G) to pyrimidines (T + C) is always 1:1, ensuring complementary base pairing.

Question 13:

A student was studying the process of transcription in eukaryotes and noted the following observations:

  • The RNA polymerase binds to the promoter region.
  • The primary transcript undergoes splicing before becoming functional mRNA.

Explain the significance of these observations in the context of gene expression.

Answer:

1. RNA polymerase binding to the promoter:

The promoter region is a specific DNA sequence where RNA polymerase binds to initiate transcription. This ensures that only specific genes are transcribed at the right time and place, regulating gene expression precisely.


2. Splicing of the primary transcript:

In eukaryotes, the primary transcript (pre-mRNA) contains both exons (coding regions) and introns (non-coding regions). Splicing removes introns and joins exons to form mature mRNA. This allows:

  • Generation of multiple proteins from a single gene (alternative splicing).
  • Increased diversity in protein synthesis.
  • Efficient translation as only coding sequences are retained.

Thus, these steps ensure accurate and regulated gene expression in eukaryotes.

Question 14:
A research team isolated a segment of DNA from a bacterial cell and observed that it contained 20% adenine bases. Using this information, answer the following:

a) Calculate the percentage of cytosine bases in this DNA segment.
b) Explain the significance of Chargaff's rules in determining the base composition of DNA.
Answer:

a) According to Chargaff's rules, in a DNA molecule, the amount of adenine (A) is equal to thymine (T), and the amount of guanine (G) is equal to cytosine (C).

Given that adenine (A) = 20%, thymine (T) will also be 20%.

Total percentage of A + T = 20% + 20% = 40%.

Remaining percentage for G + C = 100% - 40% = 60%.

Since G = C, cytosine (C) = 60% / 2 = 30%.

b) Chargaff's rules are significant because they provided the foundational evidence for the double-helix structure of DNA. The rules state that:

  • A pairs with T, and G pairs with C, ensuring complementary base pairing.
  • The rules helped Watson and Crick deduce that DNA strands are antiparallel and held together by hydrogen bonds.
  • They also explain the uniformity in base ratios across species, supporting the universality of DNA structure.

Question 15:
A student was studying the process of transcription in eukaryotes and noted the following observations:

1. The RNA polymerase binds to a specific region of DNA.
2. The synthesized RNA has non-coding sequences removed before translation.

Based on these observations, answer:

a) Identify the specific DNA region where RNA polymerase binds and its significance.
b) Explain the process of removal of non-coding sequences from the RNA transcript.
Answer:

a) RNA polymerase binds to the promoter region of DNA, which is located upstream of the gene to be transcribed.

Significance:

  • The promoter provides a binding site for RNA polymerase and transcription factors.
  • It determines the direction and efficiency of transcription.
  • Mutations in the promoter can disrupt gene expression.

b) The removal of non-coding sequences (introns) from the RNA transcript occurs during RNA splicing, a part of post-transcriptional modification.

Process:

1. The primary transcript (pre-mRNA) contains both exons (coding sequences) and introns (non-coding sequences).

2. A spliceosome (a complex of proteins and snRNAs) recognizes specific sequences at the exon-intron boundaries.

3. The introns are excised, and exons are joined together to form mature mRNA.

4. This ensures only functional coding regions are translated into proteins.

Question 16:
A research team isolated a segment of DNA from a bacterial cell and observed that it contained 20% adenine bases. Based on this information, answer the following:
a) Calculate the percentage of cytosine bases in the DNA segment.
b) Explain the principle used to arrive at this calculation.
Answer:

a) According to Chargaff's rule, in a DNA molecule, the amount of adenine (A) is equal to thymine (T), and the amount of guanine (G) is equal to cytosine (C). Since the DNA segment has 20% adenine, thymine will also be 20%.

Total percentage of A + T = 20% + 20% = 40%.

Remaining percentage for G + C = 100% - 40% = 60%.

Since G = C, cytosine will be 60% / 2 = 30%.

b) The principle used is Chargaff's rule, which states that in any DNA molecule, the ratio of purines (A + G) to pyrimidines (T + C) is always 1:1. This forms the basis for understanding base pairing in the double-helix structure of DNA, where A pairs with T and G pairs with C via hydrogen bonds.

Question 17:
A student was studying the process of transcription in eukaryotes and noted the following observations:
1. The RNA polymerase binds to a specific region upstream of the gene.
2. The primary transcript undergoes modifications before becoming functional mRNA.
Based on these observations, answer:
a) Identify the specific region where RNA polymerase binds.
b) Describe two major modifications the primary transcript undergoes.
Answer:

a) The RNA polymerase binds to the promoter region, which is a specific DNA sequence located upstream of the gene. In eukaryotes, this region includes sequences like the TATA box, which helps in the accurate initiation of transcription.

b) The two major modifications are:

  • Capping: A 7-methylguanosine cap is added to the 5' end of the primary transcript, which protects it from degradation and aids in ribosome binding during translation.
  • Tailing: A poly-A tail (a string of adenine nucleotides) is added to the 3' end, which increases stability and facilitates export of mRNA from the nucleus to the cytoplasm.
Additionally, splicing removes non-coding introns and joins coding exons to form mature mRNA.

Question 18:
A research team isolated a segment of DNA from a bacterial cell and observed that it contained 20% adenine bases. Based on this information, answer the following:
(a) Calculate the percentage of cytosine bases in the DNA segment.
(b) Explain the significance of Chargaff's rule in determining the base composition of DNA.
Answer:

(a) According to Chargaff's rule, in a DNA molecule, the amount of adenine (A) is equal to thymine (T), and the amount of guanine (G) is equal to cytosine (C). Since the DNA segment contains 20% adenine, it will also contain 20% thymine.
Total percentage of A + T = 20% + 20% = 40%.
The remaining 60% must be G + C. Since G = C, the percentage of cytosine is 30%.

(b) Chargaff's rule is fundamental in understanding the double-helical structure of DNA. It states that the ratio of purines (A + G) to pyrimidines (T + C) is always 1:1, ensuring complementary base pairing. This rule helped Watson and Crick propose the base-pairing model, where A pairs with T (via 2 hydrogen bonds) and G pairs with C (via 3 hydrogen bonds). This uniformity maintains the structural stability and genetic consistency of DNA across generations.

Question 19:
A student performed an experiment to study the process of transcription in a prokaryotic cell. The following observations were made:
(1) RNA polymerase binds to a specific region on the DNA template.
(2) Only one strand of DNA is transcribed into RNA.
Based on these observations, answer:
(a) Identify the specific DNA region where RNA polymerase binds and its significance.
(b) Why is only one DNA strand transcribed, and how does the cell determine which strand to transcribe?
Answer:

(a) RNA polymerase binds to the promoter region of the DNA, which is a specific sequence located upstream of the gene to be transcribed. The promoter contains consensus sequences like the Pribnow box (TATAAT) in prokaryotes, which help in the accurate initiation of transcription. The significance of the promoter includes:

  • Providing a binding site for RNA polymerase.
  • Determining the start point and direction of transcription.
  • Regulating gene expression through transcription factors.

(b) Only one DNA strand (the template strand) is transcribed because RNA synthesis occurs in the 5' → 3' direction, and the template strand runs in the 3' → 5' direction, allowing complementary base pairing. The cell determines which strand to transcribe based on the promoter's orientation and the presence of specific regulatory sequences. The coding strand (non-template strand) has the same sequence as the RNA (except thymine replaced by uracil) and is not transcribed.

Question 20:
A research team isolated a segment of DNA from a bacterial cell and observed that it contained 20% adenine bases. Based on this information, answer the following:
(a) Calculate the percentage of cytosine bases in the DNA segment.
(b) Explain the principle used to arrive at this calculation.
Answer:

(a) According to Chargaff's rule, in a DNA molecule, the amount of adenine (A) is equal to thymine (T), and the amount of guanine (G) is equal to cytosine (C). Since adenine is 20%, thymine will also be 20%.
The remaining 60% is shared equally between guanine and cytosine.
Thus, cytosine = 30%.

(b) The principle used is Chargaff's rule, which states that in any DNA molecule:

  • A = T
  • G = C
  • Purines (A + G) = Pyrimidines (T + C)
This rule is fundamental to understanding the double-helix structure of DNA, where base pairing occurs between complementary strands.

Question 21:
A student was studying the process of transcription in eukaryotes and noted the following observations:
(1) The RNA polymerase binds to the promoter region.
(2) The primary transcript undergoes splicing before becoming functional.
Based on these observations, answer:
(a) Why is the promoter region essential for transcription?
(b) What is the significance of splicing in eukaryotic transcription?
Answer:

(a) The promoter region is essential because:

  • It provides a binding site for RNA polymerase and transcription factors.
  • It determines the start point and direction of transcription.
  • It regulates the efficiency of transcription initiation.
Without the promoter, transcription cannot begin as the machinery cannot assemble properly.

(b) Splicing is significant because:

  • It removes introns (non-coding sequences) and joins exons (coding sequences) to form mature mRNA.
  • It enables alternative splicing, allowing a single gene to produce multiple proteins.
  • It ensures the production of functional proteins by eliminating unnecessary sequences.
This process is crucial for gene expression regulation in eukaryotes.

Chat on WhatsApp