Molecular Basis of Inheritance — Class 12 Biology NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Biology chapter "Molecular Basis of Inheritance" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Biology chapter "Molecular Basis of Inheritance" — 7 important questions with detailed answers for CBSE…
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NCERT Solutions for Class 12 Biology — Molecular Basis of Inheritance. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Molecular Basis of Inheritance — All 7 Questions Solved
Q1. What is DNA? Describe the structure of DNA with a labeled diagram description.
DNA (Deoxyribonucleic acid): Macromolecule carrying genetic information for all living organisms. Double helix polymer of nucleotides.
Structure (Watson-Crick model):
1. Basic units (Nucleotides): Each contains three components:
- Nitrogenous base: Purines (Adenine-A, Guanine-G) or Pyrimidines (Cytosine-C, Thymine-T)
- Pentose sugar: Deoxyribose (5-carbon sugar, lacks hydroxyl at 2' carbon unlike RNA)
- Phosphate group: Links nucleotides
2. Primary structure (Backbone):
- Sugar-phosphate backbone: Alternating deoxyribose and phosphate groups
- Nitrogenous bases attached to 1' carbon of sugar
- 5' to 3' direction: Phosphodiester bonds between 3' carbon of one sugar and 5' phosphate of next
- Two strands run antiparallel: one 5' to 3', other 3' to 5'
3. Secondary structure (Double helix):
- Two complementary strands wound around common axis
- Right-handed double helix
- Major groove and minor groove between strands
- Base pairing (Chargaff's rules):
* Adenine-Thymine: 2 hydrogen bonds (A=T)
* Guanine-Cytosine: 3 hydrogen bonds (G≡C)
- Bases inside helix, sugar-phosphate backbone outside
- Pitch: 3.4 nm per complete turn; 10 base pairs per turn
- Diameter: 2 nm
4. Higher-order structure:
- Chromatin: DNA wrapped around histone proteins forming nucleosomes
- Nucleosome: 147 bp DNA around histone octamer
- Supercoiling: Further condensation into chromosomes
Function: Stores genetic information in sequence of bases; template for replication and protein synthesis.
Q2. Describe DNA replication and the role of DNA polymerase. What is semi-conservative replication?
DNA Replication: Process of exact duplication of DNA molecule, occurring during S phase of cell cycle. Produces two identical DNA molecules from one.
Semi-conservative replication (Meselson-Stahl experiment):
- Definition: Each new DNA molecule contains one original strand and one newly synthesized strand
- Mechanism: Original double helix separates; each strand serves as template for new complementary strand
- Result: Two daughter molecules, each 50% old (parental) DNA and 50% new DNA
Process:
1. Initiation:
- DNA helicase unwinds double helix, breaking hydrogen bonds between bases
- Opens replication fork (Y-shaped)
- Primase synthesizes short RNA primer (8-12 nucleotides)
2. Elongation:
- DNA polymerase III: Main replicative enzyme in prokaryotes; adds nucleotides to 3'-OH group
- Reads template strand in 3' to 5' direction
- Synthesizes new strand in 5' to 3' direction
- Leading strand: continuous synthesis 5' to 3'
- Lagging strand: discontinuous synthesis forming Okazaki fragments (1000-2000 bp in prokaryotes)
3. DNA polymerase I (Prokaryotes):
- Removes RNA primers
- Fills gaps with DNA
- 5' to 3' exonuclease activity removes primers
- Polymerase activity fills gaps
4. DNA ligase:
- Seals nicks between Okazaki fragments
- Forms phosphodiester bonds
5. Termination:
- Replication completes
- RNA primers removed
- Gaps filled and sealed
6. Proofreading:
- DNA polymerase 3' to 5' exonuclease activity checks nucleotides
- Removes incorrectly paired nucleotides
- Error rate: 1 per 10^9 to 10^10 nucleotides
Bidirectional replication:
- Replication proceeds in both directions from origin (ori)
- Forms two replication forks moving in opposite directions
- Prokaryotes: single circular chromosome, single origin
- Eukaryotes: linear chromosomes, multiple origins (every 50-300 kb)
Enzymes involved:
- Helicase: unwinds DNA
- Primase: synthesizes RNA primers
- DNA polymerase III (prokaryotes) or Pol α, δ, ε (eukaryotes): main synthesis
- DNA polymerase I (prokaryotes): primer removal
- Ligase: seals nicks
- Single-strand binding proteins: protect exposed strands
- Topoisomerase: relieves tension ahead of fork
Significance: Ensures accurate copying of genetic material; ensures each daughter cell receives complete genome.
Q3. What is the central dogma of molecular biology? Explain the process of transcription.
Central Dogma: Fundamental principle stating genetic information flows from DNA to RNA to Protein. Information in DNA transferred to RNA, which directs protein synthesis. Unidirectional for most organisms (exception: reverse transcriptase in retroviruses).
Flow: DNA → RNA (transcription) → Protein (translation)
Transcription: Process synthesizing RNA from DNA template. Occurs in nucleus (eukaryotes) or cytoplasm (prokaryotes).
Types of RNA produced:
- mRNA (messenger RNA): carries genetic information for protein synthesis
- tRNA (transfer RNA): brings specific amino acids to ribosome
- rRNA (ribosomal RNA): structural and catalytic component of ribosome
Transcription process (in prokaryotes):
1. Initiation:
- RNA polymerase binds to promoter (specific DNA sequence upstream of gene)
- Promoter recognition: -10 box (Pribnow box: TATAAT) and -35 box in prokaryotes
- Sigma factor helps RNA polymerase recognize promoter
- DNA unwinds at promoter
2. Elongation:
- RNA polymerase moves along template strand (3' to 5' direction)
- Synthesizes mRNA in 5' to 3' direction
- Complementary base pairing: A-U, T-A, G-C, C-G
- Catalyzes phosphodiester bond formation between ribonucleotides
- Forms RNA-DNA hybrid helix
- Nucleotide incorporation rate: ~50 nucleotides/second in prokaryotes
3. Termination:
- Two mechanisms:
* Rho-independent: GC-rich inverted repeat in RNA forms hairpin; poly-U tract; destabilizes RNA-DNA hybrid; transcript releases
* Rho-dependent: Rho protein recognizes C-rich region in RNA; catches up to polymerase; dissociates complex
- RNA polymerase releases
Differences in eukaryotes:
- Three RNA polymerases: Pol I (rRNA), Pol II (mRNA), Pol III (tRNA)
- Promoter elements: TATA box (-25 to -30), CAAT box, GC box
- Transcription factors (TFIID) recognize promoter
- Transcription occurs in nucleus
- mRNA processing: 5' capping, 3' polyadenylation, splicing
Post-transcriptional modifications (eukaryotes):
- 5' cap: 7-methylguanosine added for stability and translation
- 3' poly-A tail: 200-250 adenines for stability and translation
- Splicing: removal of introns and joining of exons
Significance: Transfers genetic information from DNA to mRNA; mRNA carries instructions for protein synthesis; regulates gene expression.
Q4. What is the genetic code? Explain its characteristics and significance.
Genetic code: Sequence of bases in DNA/mRNA specifying amino acid sequence in proteins. Language translating genetic information into protein structure.
Characteristics of genetic code:
1. Triplet code:
- Three consecutive nucleotides (codon) code for one amino acid
- 64 possible codons (4^3 combinations of 4 bases)
- 61 codons code for amino acids; 3 stop codons
2. Universal code:
- Same codons represent same amino acids across all organisms
- Evidence: bacteria and humans share many codon assignments
- Minor variations in mitochondria and some organisms
- Supports common evolutionary origin of life
3. Non-overlapping:
- Codons read sequentially without overlap
- Genetic code organized in triplets from start to stop
- Example: AUGCCGUAA read as AUG-CCG-UAA, not AUG-UGC-GCC-CUG-CUA-UAA
4. Degenerate (redundant):
- Multiple codons code for same amino acid (synonymous codons)
- Example: Leucine coded by UUA, UUG, CUU, CUC, CUA, CUG
- Protects against mutations: wobble position (3rd base) changes often without altering amino acid
- Wobble pairing: tRNA anticodon third position pairs flexibly with mRNA codon third position
5. Start and stop codons:
- Start codon: AUG (methionine); initiates translation
- Stop codons: UAA, UAG, UGA; terminate translation; no tRNA for these
6. Specific:
- Each codon specifies particular amino acid
- No comma punctuation between codons
- Reading frame: particular grouping of codons determines meaning
- Frameshift mutations change reading frame, produce different protein
7. Comma-less:
- No spacing or punctuation between codons
- Continuous sequence from start to stop
Codon table organization:
- First position: defines amino acid family
- Second position: further defines amino acid
- Third position (wobble): often variable without changing amino acid
Significance:
- Explains how DNA sequence determines protein sequence
- Basis of genetic engineering: ability to predict protein from gene
- Universal code allows cross-species protein expression
- Wobble explains genetic variation tolerance
- Enables understanding of mutations and their effects
Q5. Describe the process of translation. Explain the role of tRNA and ribosome.
Translation: Process synthesizing proteins from mRNA template. Converts genetic information in nucleic acids to amino acid sequence in proteins.
Stages of translation (in prokaryotes):
1. Initiation:
- mRNA binds to small ribosomal subunit (30S in prokaryotes, 40S in eukaryotes)
- Shine-Dalgarno sequence in prokaryotes positions mRNA correctly
- Initiator tRNA (fMet-tRNA in prokaryotes, Met-tRNA in eukaryotes) binds to start codon (AUG)
- Large ribosomal subunit (50S prokaryotes, 60S eukaryotes) joins small subunit
- Forms 70S (prokaryotes) or 80S (eukaryotes) ribosomal complex
- Three tRNA binding sites on ribosome: A (aminoacyl), P (peptidyl), E (exit)
- Initiator tRNA occupies P site
2. Elongation:
- Elongation factor Tu brings aminoacyl-tRNA (aa-tRNA) to A site
- Codon-anticodon pairing (mRNA codon pairs with tRNA anticodon)
- Proofreading: ribosome checks base pairing accuracy
- Peptidyl transferase center: catalyzes peptide bond between amino acids
- Peptide bond forms between carboxyl group (P site aa) and amino group (A site aa)
- tRNA in P site moves to E site (exit)
- aa-tRNA from A site moves to P site
- Empty tRNA in E site exits
- Ribosome translocates (moves) one codon forward (3 nucleotides)
- Cycle repeats: A site ready for next aa-tRNA
- Rate: 20 amino acids per second in prokaryotes
3. Termination:
- Stop codon (UAA, UAG, UGA) enters A site
- Release factors (RF1, RF2) recognize stop codon
- Release factors catalyze hydrolysis of bond between polypeptide and tRNA
- Deacylated tRNA released
- Ribosomal subunits dissociate from mRNA
Role of tRNA (transfer RNA):
Structure:
- Cloverleaf secondary structure (2D)
- L-shaped tertiary structure (3D)
- Contains 76-90 nucleotides
- Anticodon arm: contains anticodon (3 bases complementary to mRNA codon)
- Amino acid arm: 3' CCA sequence where amino acid attaches
- Acceptor stem: where aminoacyl transferase attaches amino acid
- D arm, TψC arm: provide stability
Functions:
- Recognition: specific tRNA recognizes specific mRNA codon via anticodon pairing
- Amino acid activation: aminoacyl-tRNA synthetase charges tRNA with specific amino acid
- Wobble pairing: anticodon third position pairs flexibly with codon first position
- Adapter role: brings correct amino acid to ribosome based on codon
- Quality control: mismatched aa-tRNA rejected by ribosome
Aminoacy-tRNA synthetase:
- Catalyzes aa + ATP + tRNA → aa-tRNA + AMP + PPi
- Highly specific: ensures correct amino acid attached to correct tRNA
- High fidelity: error rate 1 in 10,000
- 20 enzymes for 20 amino acids
Role of ribosome:
Structure:
- Prokaryotic: 70S (50S + 30S subunits); ~3 million daltons
- Eukaryotic: 80S (60S + 40S subunits); ~4.5 million daltons
- RNA (rRNA) + proteins (ribosomal proteins)
- rRNA: catalytic (peptidyl transferase activity)
- Binding sites: A (aminoacyl), P (peptidyl), E (exit)
Functions:
- Template reading: positions mRNA correctly
- tRNA binding: holds tRNA in correct positions
- Catalysis: peptidyl transferase center catalyzes peptide bond formation
- Proofreading: checks codon-anticodon pairing accuracy
- Translocation: moves along mRNA
- Protein synthesis: catalyzes formation of 3000-8000 amino acid chains
Polyribosomes:
- Multiple ribosomes translate single mRNA simultaneously
- Increases efficiency: same protein made by many ribosomes
- Ribosomes spaced ~80 nucleotides apart
- Prokaryotes: translation coupled with transcription
- Eukaryotes: transcription in nucleus, translation in cytoplasm (spatially separated)
Significance: Converts mRNA sequence into functional protein; ensures accurate amino acid incorporation; highly regulated process determining gene expression.
Q6. What is protein synthesis? Explain the synthesis of a specific protein starting from DNA to final protein.
Protein synthesis: Overall process of producing functional proteins from genetic information stored in DNA. Two-stage process: transcription and translation.
Complete pathway from DNA to functional protein:
1. Gene selection:
- Specific gene on DNA selected for expression
- Gene contains:
* Regulatory sequences: promoter, enhancer, silencer
* Exons: coding sequences (in eukaryotes)
* Introns: non-coding sequences (in eukaryotes)
2. Transcription:
- RNA polymerase binds to promoter
- DNA unwinds; template strand serves as template
- mRNA synthesized 5' to 3' complementary to template strand
- Primary transcript (pre-mRNA in eukaryotes)
- Rate: 50 nucleotides/second (prokaryotes)
3. RNA processing (eukaryotes only):
- 5' capping: 7-methylguanosine cap added to 5' end
* Protects mRNA from degradation
* Enhances translation efficiency
- 3' polyadenylation: poly-A tail (200-250 adenines) added to 3' end
* Protects mRNA from degradation
* Enhances translation
- Splicing: removes introns, joins exons
* snRNPs (small nuclear ribonucleoproteins) catalyze splicing
* Produces mature mRNA with only coding sequence
* Can produce different proteins from same gene (alternative splicing)
4. mRNA export (eukaryotes):
- Mature mRNA exported from nucleus through nuclear pores
- Transported to ribosomes in cytoplasm
- Prokaryotes: mRNA used immediately by ribosomes
5. Translation initiation:
- mRNA binds to small ribosomal subunit
- Initiator tRNA binds to start codon (AUG)
- Large ribosomal subunit joins
- Translation complex ready
6. Elongation:
- Aminoacyl-tRNAs sequentially bring amino acids to ribosome
- Codon-anticodon recognition ensures correct amino acid
- Peptide bonds form between consecutive amino acids
- Ribosome advances along mRNA
- Growing polypeptide chain emerges from ribosome
7. Termination:
- Stop codon reached
- Release factors cause polypeptide release
- Ribosomal subunits dissociate
- mRNA released
8. Post-translational modifications:
- Signal peptides removed
- Disulfide bonds form
- Phosphorylation, glycosylation, ubiquitination occur
- Proteolytic cleavage removes propeptides
- Proper folding assisted by chaperone proteins
9. Protein folding:
- Heat shock proteins (chaperones) assist proper folding
- Prevent aggregation
- Ensure functional 3D structure
- Misfolded proteins targeted for degradation
10. Transport and localization:
- Signal sequences direct protein to specific cellular compartment
- Mitochondria, endoplasmic reticulum, Golgi, plasma membrane
- Secretory proteins: ER → Golgi → secretory vesicles → plasma membrane
11. Functional protein:
- Mature, fully processed protein
- Reaches target location
- Performs specific cellular function
- Example: enzyme catalyzes reaction, transcription factor regulates genes, structural protein provides support
Example: Insulin synthesis
- Gene: INS gene on chromosome 11
- Transcription: pre-mRNA (1500 bp) synthesized
- Processing: capping, polyadenylation, splicing
- mRNA: 443 nucleotides
- Translation: produces preproinsulin (110 amino acids)
- Processing: signal peptide removed → proinsulin (86 amino acids)
- Cleavage: C-peptide removed → mature insulin (51 amino acids in two chains)
- Secretion: insulin stored in secretory granules
- Function: regulates blood glucose
Regulation points:
- Transcriptional: which genes expressed
- Post-transcriptional: mRNA processing, alternative splicing
- Translational: which mRNAs translated
- Post-translational: protein modifications, degradation
Significance: Explains how genetic information becomes functional cellular machinery; basis of life itself.
Q7. Explain gene expression and its regulation. What is the lac operon and how does it regulate genes?
Gene expression: Process of converting genetic information into functional gene products (proteins). Determining which genes are expressed in which cells at which times.
Levels of regulation:
1. Transcriptional regulation: controlling whether gene is transcribed
2. Post-transcriptional regulation: processing and degradation of mRNA
3. Translational regulation: controlling whether mRNA is translated
4. Post-translational regulation: protein modification and degradation
Lac operon (classic prokaryotic gene regulation model):
Structure:
- Three structural genes:
* lacZ: encodes β-galactosidase (hydrolyzes lactose to glucose and galactose)
* lacY: encodes permease (transports lactose into cell)
* lacA: encodes transacetylase (modifies lactose metabolites)
- Regulatory elements:
* Promoter: where RNA polymerase binds
* Operator: where repressor protein binds
* CAP binding site: where catabolite activator protein binds
Regulatory genes:
- lacI: encodes lac repressor protein
Function: Regulates genes required for lactose metabolism
Negative regulation (repression):
In absence of lactose (lactose OFF):
- Lac repressor protein binds to operator
- Blocks RNA polymerase from transcribing structural genes
- Prevents synthesis of lactose-metabolizing enzymes
- Energy conservation: no enzymes made if lactose absent
In presence of lactose (lactose ON):
- Lactose (allolactose) acts as inducer
- Binds to repressor protein, changing its shape
- Repressor releases from operator
- RNA polymerase can transcribe structural genes
- Enzymes synthesized for lactose metabolism
- When lactose depleted, inducer removed, repressor rebinds
Positive regulation (CAP-cAMP):
In presence of glucose (glucose OFF):
- Glucose represses cAMP synthesis
- No CAP-cAMP complex formed
- CAP cannot bind to promoter
- Low transcription of lac operon
- Even with lactose present
In absence of glucose (glucose ON):
- Low glucose increases cAMP levels
- CAP binds cAMP forming complex
- CAP-cAMP binds to CAP binding site
- Increases affinity of RNA polymerase for promoter
- Increases transcription when lactose also present
- Lactose preferred fuel source if available
Mechanism summary:
- Glucose present: cAMP low, CAP-cAMP cannot form, operon weakly transcribed (even with lactose)
- Lactose absent: repressor binds operator, operon off (even with glucose absent)
- Both glucose absent and lactose present (ideal conditions): operon fully ON
- Glucose present, lactose absent: operon OFF
Significance:
- Demonstrates negative and positive regulation
- Shows inducible gene regulation: turned on by substrate
- Shows catabolite repression: glucose preferred over lactose
- Efficient use of cellular resources
- Model for understanding eukaryotic gene regulation
Differences in eukaryotic gene regulation:
- Chromatin remodeling: DNA packaged with histones; must be accessible
- Multiple transcription factors: enhancers, silencers act at distance
- Mediator complex: bridges transcription factors and RNA polymerase
- More complex post-transcriptional regulation: alternative splicing, mRNA stability
- Nuclear organization: transcription and translation separated
- Hormonal signals: growth factors, steroid hormones regulate transcription
Molecular Basis of Inheritance — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Who experimentally proved that DNA is the genetic material?
- A. Alfred Hershey and Martha Chase
- B. Frederick Griffith
- C. Oswald Avery, Colin MacLeod, and Maclyn McCarty
- D. James Watson and Francis Crick
Answer: A. Alfred Hershey and Martha Chase — The Hershey-Chase experiment (1952) using bacteriophages conclusively demonstrated that DNA, not protein, is the genetic material that enters the bacterial cell to direct synthesis of new virus particles.
MCQ 2. Which of the following statements about the double helical structure of DNA is INCORRECT?
- A. The two chains have antiparallel polarity.
- B. The backbone is formed by sugar-phosphate, and the bases project inward.
- C. A purine always pairs with a pyrimidine.
- D. The two strands are joined by covalent bonds between the bases.
Answer: D. The two strands are joined by covalent bonds between the bases. — The two strands of DNA are held together by hydrogen bonds between complementary base pairs (A-T and G-C), not by covalent bonds. Covalent bonds form the sugar-phosphate backbone.
MCQ 3. If a double-stranded DNA molecule has 20% guanine, what will be the percentage of adenine?
- A. 20%
- B. 30%
- C. 40%
- D. 50%
Answer: B. 30% — According to Chargaff's rules, the amount of guanine (G) equals cytosine (C), and adenine (A) equals thymine (T). If G=20%, then C=20%, making G+C = 40%. The remaining 60% must be A+T, so A=T=30%.
MCQ 4. The enzyme DNA-dependent RNA polymerase catalyses the transcription of:
- A. tRNA only
- B. mRNA only
- C. rRNA only
- D. All types of RNA
Answer: D. All types of RNA — In prokaryotes, a single DNA-dependent RNA polymerase transcribes all three types of RNA (mRNA, tRNA, and rRNA). In eukaryotes, there are different RNA polymerases for different RNA types.
MCQ 5. Which of the following is NOT a feature of the genetic code?
- A. Degenerate
- B. Universal
- C. Overlapping
- D. Non-ambiguous
Answer: C. Overlapping — The genetic code is non-overlapping, meaning that each nucleotide in the mRNA sequence is part of only one codon and is read sequentially without overlap.
MCQ 6. In Meselson and Stahl's experiment, E. coli was grown in a medium containing 15NH4Cl for several generations. When transferred to 14NH4Cl medium, the first generation of progeny showed DNA with:
- A. Heavy density DNA
- B. Light density DNA
- C. Hybrid density DNA
- D. No DNA
Answer: C. Hybrid density DNA — The Meselson and Stahl experiment demonstrated semi-conservative DNA replication. After one generation in 14N medium, each daughter DNA molecule contained one old (15N) and one new (14N) strand, resulting in hybrid density DNA.
MCQ 7. Which of these is directly involved in the initiation of translation?
- A. Stop codons
- B. Release factors
- C. Initiator tRNA
- D. Sigma factor
Answer: C. Initiator tRNA — Translation initiation involves the binding of the mRNA, the small ribosomal subunit, and the initiator tRNA (carrying methionine) to the start codon (AUG).
MCQ 8. In eukaryotes, the primary transcript (hnRNA) undergoes processing. Which of the following steps is NOT part of this processing?
- A. Splicing
- B. Capping
- C. Tailing
- D. Degradation by nucleases
Answer: D. Degradation by nucleases — Eukaryotic hnRNA processing involves splicing (removal of introns), capping (addition of methyl guanosine triphosphate at 5' end), and tailing (addition of poly-A tail at 3' end). Degradation by nucleases is generally prevented by these modifications.
MCQ 9. The Lac operon is an example of:
- A. Only positive regulation
- B. Only negative regulation
- C. Both positive and negative regulation
- D. Feedback inhibition
Answer: C. Both positive and negative regulation — The Lac operon is primarily regulated negatively by the repressor protein (inhibited in presence of lactose) and also positively by catabolite activator protein (CAP) in the absence of glucose.
MCQ 10. Which enzyme is responsible for joining the Okazaki fragments during DNA replication?
- A. DNA Polymerase I
- B. DNA Polymerase III
- C. DNA Ligase
- D. Helicase
Answer: C. DNA Ligase — DNA ligase forms phosphodiester bonds to connect the discontinuous Okazaki fragments synthesized on the lagging strand into a continuous strand.
MCQ 11. What is the function of the 'i' gene in the Lac operon?
- A. Codes for beta-galactosidase
- B. Codes for permease
- C. Codes for transacetylase
- D. Codes for the repressor protein
Answer: D. Codes for the repressor protein — The 'i' gene (inhibitor gene) in the Lac operon codes for the repressor protein, which regulates the transcription of the structural genes.
MCQ 12. The technique of DNA fingerprinting involves identifying differences in specific regions of DNA called:
- A. Exons
- B. Introns
- C. Variable Number Tandem Repeats (VNTRs)
- D. Single Nucleotide Polymorphisms (SNPs)
Answer: C. Variable Number Tandem Repeats (VNTRs) — DNA fingerprinting primarily relies on the analysis of Variable Number Tandem Repeats (VNTRs), which are highly polymorphic, repetitive DNA sequences unique to individuals (except identical twins).
MCQ 13. Which type of RNA brings amino acids from the cytoplasm to the ribosome during protein synthesis?
- A. mRNA
- B. tRNA
- C. rRNA
- D. snRNA
Answer: B. tRNA — Transfer RNA (tRNA) acts as an adaptor molecule, picking up specific amino acids from the cytoplasm and carrying them to the ribosome for incorporation into the polypeptide chain, guided by its anticodon.
MCQ 14. The process of unwinding of DNA double helix during replication is carried out by:
- A. DNA Polymerase
- B. Primase
- C. Helicase
- D. Topoisomerase
Answer: C. Helicase — Helicase is the enzyme responsible for unwinding the double helix of DNA by breaking the hydrogen bonds between complementary base pairs, creating a replication fork.
MCQ 15. Consider a segment of DNA with the sequence 5'-ATGGTAC-3' on one strand. If this is the coding strand, what would be the sequence of the mRNA transcribed from it?
- A. 5'-AUGGUAC-3'
- B. 3'-TAC CATG-5'
- C. 5'-UACCAUG-3'
- D. 3'-AUGGUAC-5'
Answer: A. 5'-AUGGUAC-3' — If 5'-ATGGTAC-3' is the coding strand, then the mRNA sequence will be identical to it, except that thymine (T) will be replaced by uracil (U).
MCQ 16. Which of the following is true regarding the histone proteins?
- A. They are negatively charged and acidic.
- B. They are found only in prokaryotes.
- C. They are rich in basic amino acid residues like lysine and arginine.
- D. They bind to DNA non-specifically without specific sequences.
Answer: C. They are rich in basic amino acid residues like lysine and arginine. — Histone proteins are positively charged due to the abundance of basic amino acids (lysine and arginine), which allows them to bind tightly to the negatively charged DNA, facilitating DNA packaging in eukaryotes.
MCQ 17. A gene 'A' is transcribed to mRNA, which is then translated into a protein. If a mutation occurs that changes a single nucleotide in gene 'A', resulting in a premature stop codon, this type of mutation is called:
- A. Missense mutation
- B. Silent mutation
- C. Nonsense mutation
- D. Frameshift mutation
Answer: C. Nonsense mutation — A nonsense mutation occurs when a point mutation results in a premature stop codon, leading to the production of a truncated, non-functional protein.
MCQ 18. The human genome has approximately how many base pairs?
- A. 3 x 10^6 bp
- B. 3 x 10^7 bp
- C. 3 x 10^8 bp
- D. 3 x 10^9 bp
Answer: D. 3 x 10^9 bp — The human genome (haploid content) is estimated to contain approximately 3.1647 billion base pairs (3 x 10^9 bp).
MCQ 19. In the process of translation, what is the role of release factors?
- A. Initiate protein synthesis
- B. Bind to the start codon to recruit tRNA
- C. Bind to the stop codon, leading to the termination of translation
- D. Catalyze the formation of peptide bonds
Answer: C. Bind to the stop codon, leading to the termination of translation — Release factors recognize the stop codons (UAA, UAG, UGA) and bind to the ribosome, leading to the hydrolysis of the ester bond between the polypeptide and the tRNA, thus terminating translation.
MCQ 20. Which of the following describes the 'wobble hypothesis' correctly?
- A. The first nucleotide of the anticodon dictates amino acid specificity.
- B. The first two bases of the codon are most critical, and the third base can vary.
- C. An amino acid can be specified by only one codon.
- D. It explains why there are more tRNAs than codons.
Answer: B. The first two bases of the codon are most critical, and the third base can vary. — The wobble hypothesis states that the first two base pairs between a codon and an anticodon are always strong, but the third base pair can 'wobble', allowing a single tRNA to recognize multiple synonymous codons.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.