Bio Zoology · Chapter 5

Samacheer Class 12 Bio Zoology - Molecular Genetics

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Complete Class 12 Bio Zoology book back solutions for Molecular Genetics with exam-ready answers.

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Q.1 Hershey and Chase experiment with bacteriophage showed that a) Protein gets into the bacterial cells b) DNA is the genetic material c) DNA contains radioactive sulphur d) Viruses undergo transformation
Answer: b

Hershey and Chase labelled phage protein with 35S and phage DNA with 32P. After infection only 32P entered bacteria and directed production of new phages, proving DNA is the genetic material. Key terms: bacteriophage, radioactive labelling, DNA.

Q.2 DNA and RNA are similar with respect to a) Thymine as a nitrogen base b) A single-stranded helix shape c) Nucleotide containing sugars, nitrogen bases and phosphates d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer: c

Both DNA and RNA are polymers of nucleotides; each nucleotide contains a sugar (deoxyribose in DNA, ribose in RNA), a nitrogenous base and a phosphate group. Key term: nucleotide.

Q.3 A mRNA molecule is produced by b) Transcription a) Replication c) Duplication d) Translation
Answer: b

mRNA is synthesized from a DNA template by RNA polymerase in the process called transcription. Key term: transcription.

Q.4 The total number of nitrogenous bases in human genome is estimated to be about a) 3.5 million b) 35000 d) 3.1 billion c) 35 million
Answer: d

The human haploid genome contains about 3.1 billion base pairs (≈3.1 billion nitrogenous bases). Key term: human genome size.

Q.5 E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment? (a) One high and one low density band. (b) One intermediate density band. (c) One high and one intermediate density band. (d) One low and one intermediate density band.
Answer: d

Meselson–Stahl prediction: after one generation DNA is hybrid (intermediate). After two generations half the DNA is light (14N–14N) and half remains hybrid (14N–15N), giving one light and one intermediate band. Key terms: Meselson–Stahl, semi-conservative.

Q.6 What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules? (a) Origin of replication occurs only at the 5' end of the molecules. (b) DNA ligase works only in the 3' → 5' direction. (c) DNA polymerase can join new nucleotides only to the 3' end of the growing stand. (d) Helicases and single-strand binding proteins that work at the 5' end.
Answer: c

DNA polymerase can add nucleotides only to the 3' hydroxyl end; because the two template strands are antiparallel this leads to continuous synthesis on the leading strand and discontinuous synthesis (Okazaki fragments) on the lagging strand. Key terms: DNA polymerase, 3' end, Okazaki fragments.

Q.7 Which of the following is the correct sequence of event with reference to the central dogma? (a) Transcription, Translation, Replication (b) Transcription, Replication, Translation (c) Duplication, Translation, Transcription (d) Replication, Transcription, Translation
Answer: d

Central dogma sequence: DNA replication (copying DNA), transcription (DNA → RNA), then translation (mRNA → protein). Key term: central dogma.

Q.8 Which of the following statements about DNA replication is not correct? (a) Unwinding of DNA molecule occurs as hydrogen bonds break. (b) Replication occurs as each base is paired with another exactly like it. (c) Process is known as semi conservative replication because one old strand is conserved in the new molecule. (d) Complementary base pairs are held together with hydrogen bonds.
Answer: b

Statement (b) is incorrect because replication pairs complementary bases (A–T and G–C), not 'exactly like' the same base. Other statements are correct: unwinding breaks hydrogen bonds, semi-conservative replication conserves one parental strand, complementary pairs held by hydrogen bonds. Key terms: complementary base pairing, semi-conservative.

Q.9 Which of the following statements is not true about DNA replication in eukaryotes? (a) Replication begins at a single origin of replication. (b) Replication is bidirectional from the origins. (c) Replication occurs at about 1 million base pairs per minute. (d) There are numerous different bacterial chromosomes, with replication ocurring in each at the same time.
Answer: a

Statement (a) is false for eukaryotes: eukaryotic chromosomes have multiple origins of replication (not a single origin), allowing replication of large genomes. Statement (d) refers to bacteria and is not about eukaryotes. Key term: origins of replication.

Q.10 The first codon to be deciphered was __________ which codes for ________. (a) AAA, proline (b) GGG, alanine (c) UUU, Phenylalanine (d)TTT, arginine
Answer: c

The first codon deciphered was UUU, which codes for the amino acid phenylalanine. Key term: codon.

Q.11 Meselson and Stahl's experiment proved (a)Transduction (b) Transformation (c) DNA is the genetic material (d) Semi-conservative nature of DNA replication
Answer: d

Meselson and Stahl used isotopic labelling of nitrogen to show each daughter DNA molecule contains one parental and one newly synthesized strand—demonstrating semi-conservative replication. Key term: semi-conservative replication.

Q.12Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for _________ and the larger subunit has two binding sites for two __________. (mRNA, tRNA)v
Solution

The small ribosomal subunit binds mRNA. The large subunit contains the A (aminoacyl) and P (peptidyl) sites for two tRNA molecules during translation. Key terms: mRNA, tRNA, A site, P site.

Answer:

Smaller subunit: mRNA; Larger subunit: two tRNA binding sites (A and P sites).

Q.13 An operon is a: (a) Protein that suppresses gene expression (b) Protein that accelerates gene expression (c) Cluster of structural genes with related function (d) Gene that switched other genes on or off
Answer: c

An operon is a cluster of functionally related structural genes plus their regulatory sequences (promoter, operator) that are transcribed together (e.g., lac operon). Key term: operon.

Q.14 When lactose is present in the culture medium: (a) Transcription of lac y, lac z, lac a genes occurs. (b) Repressor is unable to bind to the operator. (c) Repressor is able to bind to the operator. (d) Both (a) and (b) are correct.
Answer: d

When lactose (actually allolactose, an inducer) is present it binds the lac repressor causing a conformational change so the repressor cannot bind the operator; RNA polymerase transcribes lacZ, lacY and lacA, so both (a) and (b) are correct. Key terms: lac operon, inducer, repressor, operator.

Q.15Give reasons: Genetic code is 'universal'.v
Solution

The genetic code is called 'universal' since the same triplet codons largely specify the same amino acids across different species (e.g., AUG → methionine as start codon). Minor exceptions exist in mitochondria and some protozoa, but the universality supports common ancestry and enables genetic engineering. Key terms: genetic code, codon, AUG, universal.

Answer:

Because nearly all organisms use the same codon–amino acid assignments; a given codon specifies the same amino acid in bacteria, plants and animals.

Q.16Name the parts marked 'A' and 'B' in the given transcription unit: 5' 3' A 3' 5' Bv
Solution

In a transcription unit the coding (sense) strand runs 5' → 3' and has the same sequence as the mRNA (T in DNA = U in RNA); the template (antisense) strand runs 3' → 5' and is used by RNA polymerase to synthesize complementary mRNA. Key terms: coding strand, template strand, mRNA.

Answer:

A: Coding (sense or non-template) strand (5' → 3'); B: Template (antisense) strand (3' → 5').

Q.17Differentiate - Leading stand and lagging strandv
Solution

Leading strand: continuous synthesis, DNA polymerase adds nucleotides 5'→3' in same direction as fork movement, requires a single RNA primer. Lagging strand: discontinuous synthesis in short Okazaki fragments, each fragment begins with an RNA primer and fragments are joined by DNA ligase; synthesis still 5'→3' on each fragment. Key terms: leading strand, lagging strand, Okazaki fragments, DNA ligase, RNA primer.

Answer:

Leading strand: synthesized continuously toward the replication fork. Lagging strand: synthesized discontinuously away from the replication fork as Okazaki fragments.

Q.18Differentiate - Template strand and coding strand.v
Solution

Template strand (antisense): serves as the template for transcription; the RNA produced is complementary to it. Coding strand (sense): not used as template; its sequence matches the mRNA (except T→U). Key terms: template strand, coding strand, antisense, sense, mRNA.

Answer:

Template (antisense) strand: used by RNA polymerase, is complementary to mRNA and oriented 3'→5'. Coding (sense, non-template) strand: has same sequence as mRNA (T instead of U), oriented 5'→3'.

Q.19Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.v
Solution

1) Pharmacogenomics: SNPs affecting drug-metabolizing enzymes or drug targets allow tailoring drug type and dose to individuals, reducing adverse reactions and improving efficacy. 2) Genome-wide association studies (GWAS) use SNPs to locate loci associated with diseases (e.g., diabetes, cancer), enabling genetic risk profiling, predictive diagnostics and new therapeutic targets. Key terms: single nucleotide polymorphism (SNP), pharmacogenomics, biomarkers, GWAS.

Answer:

1) Pharmacogenomics — SNPs predict individual drug responses enabling personalized medicine. 2) Disease risk and diagnosis — SNPs identified by GWAS serve as biomarkers for susceptibility and early detection of complex diseases.

Q.20State any three goals of the human genome project.v
Solution

Primary aims of the Human Genome Project included sequencing the entire human genome, locating and cataloguing all human genes (gene mapping and annotation), creating publicly accessible databases and improving sequencing technologies, and investigating ethical, legal and social implications of genomic information (ELSI). Key terms: sequencing, gene mapping, ELSI, databases.

Answer:

Three goals: 1) Determine the complete sequence of human DNA (all base pairs). 2) Identify and map all human genes (locations and functions). 3) Develop databases and tools for data analysis and address ethical, legal and social issues (ELSI).

Q.21In E. coli, three enzymes β-galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.v
Solution

The lac operon is an inducible operon under negative control. The lacI gene produces a repressor that binds to the operator sequence and physically prevents RNA polymerase from transcribing the structural genes (lacZ, lacY, lacA). When lactose (actually allolactose) is absent there is no inducer to bind and inactivate the repressor, so transcription and thereby synthesis of β-galactosidase, permease and transacetylase is repressed. (Note: repression is not absolute—there is low basal or "leaky" expression.)

Answer:

In the absence of lactose a repressor protein (product of lacI regulatory gene) binds the operator of the lac operon and blocks RNA polymerase binding, preventing transcription of the structural genes lacZ, lacY and lacA.

Q.22Distinguish between structural gene, regulatory gene and operator gene.v
Solution

Definitions and features: - Structural gene: encodes enzymes or structural proteins; its coding sequence is transcribed and translated into functional polypeptides. Example: lacZ encodes β-galactosidase. - Regulatory gene: usually trans-acting; encodes regulatory molecules (repressors or activators) that modulate transcription of structural genes; may be located away from the operon. Example: lacI produces the lac repressor. - Operator (operator gene): a short cis-acting DNA segment adjacent to the promoter that serves as the binding site for repressors/activators; it does not code for protein but controls access of RNA polymerase to structural genes.

Answer:

Structural gene: codes for a polypeptide or RNA (e.g., lacZ). Regulatory gene: encodes a product (usually a protein) that controls expression of other genes (e.g., lacI repressor). Operator gene: a cis-acting DNA sequence where regulatory proteins bind to control transcription (e.g., lac operator).

Q.23A low level of expression of lac operon occurs at all the time. Justify the statement.v
Solution

Repressor binding to the operator is reversible and in equilibrium with free repressor. Sometimes the repressor transiently dissociates, permitting RNA polymerase to initiate transcription at a low frequency. This basal expression (leakiness) ensures small amounts of β-galactosidase and permease are present so the cell can respond rapidly when lactose becomes available.

Answer:

The lac repressor–operator interaction is dynamic and not absolute; occasional dissociation of repressor or incomplete repression allows basal (leaky) transcription of the lac operon.

Q.24HGP is the window for treatment of various genetic disorders. Justify the statement.v
Solution

Justification points: - Gene identification: HGP located and annotated thousands of human genes including many disease-associated genes, enabling molecular diagnosis. - Mutations and markers: catalogs of mutations/SNPs facilitate genetic screening, carrier testing and prenatal diagnosis. - Target discovery: knowledge of causal genes/proteins allows development of targeted drugs and biologics. - Gene therapy and genome editing: knowing exact gene sequences enables design of vectors, CRISPR targets and corrective strategies. - Pharmacogenomics: HGP data permit tailoring drug choice/dose based on genotype, reducing adverse effects. - Bioinformatics and databases: public data resources hasten research and clinical translation. Together these advances make HGP a foundation (a "window") for treating genetic disorders.

Answer:

The Human Genome Project (HGP) provided a reference sequence and gene maps that enabled identification of disease genes, diagnostic markers, targets for gene therapy and personalized medicine, thereby opening pathways for diagnosis, prevention and targeted treatment of genetic disorders.

Q.25Why the Human Genome Project is called a mega project?v
Solution

Reasons: huge scale of sequencing (entire human genome ~3 × 10^9 bp); involvement of many countries and institutions; large budgets and specialized infrastructure; requirement for advanced technologies, automation and bioinformatics; generation and management of enormous datasets. These features qualify HGP as a "mega" scientific project.

Answer:

Because it was an international, large-scale, long-term undertaking involving sequencing ~3 billion base pairs, vast funding, multidisciplinary teams, high-throughput technology and massive data generation and analysis.

Q.26From their examination of the structure of DNA, what did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?v
Solution

Inferences from the double helix: - Replication: Complementary base pairing (A–T, G–C) implies each strand can serve as a template for synthesis of a new complementary strand, leading to semi-conservative replication (each daughter molecule has one parental strand). - Coding capability: The sequence of four bases along a strand can specify amino acid sequence (information storage)—the linear arrangement carries genetic code. - Mutation: Any change in base sequence will change the information; base substitutions, insertions or deletions can produce heritable variation. These conceptual insights explained fidelity of replication, mechanism for copying information and how genetic variation (mutation) arises.

Answer:

They inferred: replication is semi-conservative via complementary base pairing; genetic information is encoded in the linear sequence of bases (providing coding capability); and changes in base sequence (mutations) can alter genetic information and be inherited.

Q.27Why tRNA is called an adapter molecule?v
Solution

tRNA has two functional sites: an anticodon triplet that base-pairs with an mRNA codon, and a 3' terminal acceptor site (CCA) to which a specific amino acid is attached by an aminoacyl‑tRNA synthetase. This bridging role—matching codons to their amino acids—makes tRNA an adapter molecule in protein synthesis.

Answer:

tRNA acts as an adaptor by carrying a specific amino acid at its 3' CCA end and recognizing the corresponding mRNA codon via its anticodon loop, thus linking codons to amino acids during translation.

Q.28What are the three structural differences between RNA and DNA?v
Solution

Key structural differences: (i) Ribose vs deoxyribose: the 2' hydroxyl in RNA makes it chemically more reactive. (ii) Base composition: RNA uses uracil (U) in place of thymine (T) found in DNA. (iii) Secondary structure: most RNAs are single-stranded with local folding; DNA forms a long stable double helix. (Additional differences: RNA molecules are generally shorter and less stable; cellular localization differs: RNA in nucleus and cytoplasm, DNA mainly nuclear in eukaryotes.)

Answer:

1) Sugar: RNA has ribose (2'‑OH) while DNA has deoxyribose (2'‑H). 2) Bases: RNA contains uracil (U) instead of thymine (T). 3) Strandness: RNA is usually single‑stranded; DNA is usually double‑stranded (double helix).

Q.29Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.v
Solution

Anticodons are complementary and antiparallel to codons. Writing anticodons in 3'→5' orientation to pair with 5'→3' codons: - Codon 5' AAU 3' pairs with anticodon 3' UUA 5'. - Codon 5' CGA 3' pairs with anticodon 3' GCU 5'. - Codon 5' UAU 3' pairs with anticodon 3' AUA 5'. - Codon 5' GCA 3' pairs with anticodon 3' CGU 5'. (Note: the same anticodons may be written 5'→3' as AUU, UCG, AUA, UGC respectively.)

Answer:

Provide anticodons in 3'→5' orientation: AAU → 3' UUA 5'; CGA → 3' GCU 5'; UAU → 3' AUA 5'; GCA → 3' CGU 5'.

Q.30a) Identify the figure given below. b) Redraw the structure as a replicating fork and label the parts (3' 5' 3' 5'). c) Write the source of energy for this replication and name the enzyme involved in this process. d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.v
Solution

a) Identification: the diagram shows antiparallel DNA strands poised for replication — the replication fork.

b) Replication fork features to include when redrawing (text labels): - Parental DNA strands (antiparallel): one strand 3'→5' and the other 5'→3'. - Replication fork (Y-shaped region). - Leading strand (new strand synthesized continuously in 5'→3' direction toward the fork). - Lagging strand (new strand synthesized away from the fork as Okazaki fragments). - RNA primers attached to Okazaki fragments. - DNA polymerase (extends primers), primase (makes RNA primers), helicase (unwinds helix), ligase (joins Okazaki fragments), single-strand binding proteins.

c) Source of energy: polymerization is driven by cleavage of the high-energy phosphate bonds of incoming deoxynucleoside triphosphates (dNTPs) — each nucleotide brings its own energy. Enzyme: DNA polymerase (DNA‑dependent DNA polymerase; e.g., DNA pol III in bacteria).

d) Effect of strand polarity on protein synthesis: - DNA strands are antiparallel; RNA polymerase synthesizes mRNA 5'→3' using the template strand which runs 3'→5'. - For genes on the two opposite strands transcription occurs in opposite directions; the non‑template (coding) strand has the same sequence as mRNA (T→U). - Thus, which strand serves as template determines the orientation of transcription and the reading frame for translation; different genes may be encoded on either strand and are transcribed accordingly. (Translation itself always proceeds 5'→3' on mRNA and polypeptide synthesis from N-terminus to C-terminus.)

Answer:

a) The figure represents antiparallel DNA strands at a replication site (replication fork). b) Key labels for a replication fork: replication fork, leading strand (synthesized continuously 5'→3'), lagging strand (synthesized as Okazaki fragments 5'→3'), RNA primer, DNA polymerase, helicase, ligase, primase, single‑stranded binding proteins, parental (template) strands with 3' and 5' ends. c) Energy: incoming deoxynucleoside triphosphates (dNTPs); enzyme: DNA polymerase (DNA‑dependent DNA polymerase). d) Because DNA strands are antiparallel, transcription/translation proceed with mRNA synthesized 5'→3' using the 3'→5' template; genes on opposite strands are transcribed in opposite directions, so protein synthesis initiation sites and reading frames differ depending on which strand is the template.

Q.31If the coding sequence in a transcription unit is written as follows: 5' TGCATGCATGCATGCATGCATGCATGC 3' Write down the sequence of mRNA.v
Solution

The coding (non-template) DNA strand has the same sequence as the mRNA except T is replaced by U. Replace every T with U to get the mRNA: T→U, so 5' TGCATG... 3' becomes 5' UGCAUG... 3'.

Answer:

mRNA sequence (5'→3'): 5' UGCAUGCAUGCAUGCAUGCAUGCAUGC 3'.

Q.32How is the two stage process of protein synthesis advantageous?v
Solution

Advantages: - Amplification: many protein molecules can be made from each mRNA, and many mRNAs from one gene. - Regulation: gene expression can be regulated at transcriptional and post‑transcriptional levels. - Compartmentalization (in eukaryotes): transcription in nucleus and translation in cytoplasm allows processing (splicing, capping, polyadenylation). - Flexibility: mRNA serves as a portable template that can be translated multiple times and in different locations. - Error control: separation allows quality control and modification (e.g., proofreading during translation is distinct from DNA replication).

Answer:

Two stages (transcription and translation) allow amplification of message, regulation at multiple steps, use of mRNA as a mobile template, simultaneous translation by many ribosomes (polysomes), and separation of information storage (DNA) from protein synthesis machinery.

Q.33Why did Hershey and Chase use radioactively labelled phosphorus and sulphur only? Would they have got the same result if they used radiolabelled carbon and nitrogen?v
Solution

Rationale: phosphorus is a constituent of DNA (phosphate groups) but not of proteins, while sulfur is present in some amino acids of proteins but not in DNA. Thus 32P and 35S uniquely labeled DNA and protein respectively, allowing clear determination of which entered bacterial cells. Radiolabelled carbon or nitrogen would have labeled both macromolecules and so could not differentiate DNA from protein in the experiment.

Answer:

They used 32P to label DNA (phosphate backbone contains P, protein does not) and 35S to label protein (sulfur in methionine/cysteine, DNA lacks S). Using labelled C or N would not distinguish DNA from protein because both contain carbon and nitrogen.

Q.34Explain the formation of a nucleosome.v
Solution

Formation steps and components: - Histone octamer: two copies each of core histones H2A, H2B, H3 and H4 assemble into an octamer. - DNA wrapping: ~146 bp of linker-free DNA wraps around the octamer ~1.65 times, forming the nucleosome core particle ("bead"). - Linker DNA: stretches of linker DNA (≈20–80 bp) connect adjacent nucleosome cores and are bound by histone H1 which stabilizes the entry/exit of DNA and promotes folding into higher-order chromatin. - Result: the "beads-on-a-string" 10 nm fiber (nucleosomes + linker DNA) which is further compacted into 30 nm fiber and higher-order structures to package chromatin within the nucleus. Nucleosomes both condense DNA and regulate access to transcription machinery.

Answer:

A nucleosome is formed when ~146 base pairs of DNA wrap ~1.65 turns around a histone octamer (2 each of H2A, H2B, H3 and H4) forming the core particle; linker DNA connects nucleosomes and histone H1 binds linker DNA to stabilize higher-order chromatin structure.

Q.35It is established that RNA is the first genetic material. Justify giving reasons.v
Solution

Key reasons supporting RNA-first (RNA world) hypothesis: - Dual function: RNA can carry genetic information (sequence) and catalyze chemical reactions (ribozymes), fulfilling roles of both genes and enzymes in early life. - Experimental support: catalytic RNAs (e.g., self‑splicing introns, ribozymes) show RNA can catalyze polymerization and cleavage reactions. - Centrality in modern biology: RNA is central to translation (rRNA, tRNA, mRNA) and ribosome catalytic activity is RNA-based, suggesting ancient origin. - Prebiotic chemistry: nucleotides and short RNAs can be synthesized in prebiotic simulations and may self‑assemble/replicate more readily than proteins or DNA. - Later evolution: DNA (more stable) likely evolved as a superior long-term repository for genetic information and proteins became primary catalysts, but RNA preceded them as the primordial genetic material.

Answer:

RNA is considered the first genetic material because it can both store information and act as a catalyst (ribozyme), can self-replicate under plausible prebiotic conditions, and modern evidence (ribozymes, central role of RNA in translation) supports an RNA world preceding DNA/protein biology.