CLASS 12 BIOLOGY · CHAPTER 5 · STUDY NOTES
Molecular Basis of Inheritance
A comprehensive study of the molecular mechanisms of heredity, covering DNA structure, replication, transcription, translation, gene regulation, and modern genomic technologies like HGP and DNA fingerprinting.
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The DNA and Polynucleotide Chain
Distinguish the parts before joining the strand.Build a strand from its chemical parts
- Nucleotide
- A nitrogenous base, pentose sugar and phosphate form a nucleotide; sugar plus base alone forms a nucleoside.
| Unit | What is counted |
|---|---|
| Nucleotides | Individual sugar–phosphate–base units |
| Base pairs | Paired units across a duplex |
DNA is a polymer of deoxyribonucleotides. A and G are purines; C, T and U are pyrimidines. DNA typically uses A, T, G, C; RNA typically uses A, U, G, C. The base attaches to sugar carbon 1′, while a phosphate shown at carbon 5′ helps define a nucleotide.
Adjacent nucleotides connect through 3′–5′ phosphodiester linkages. The backbone contains alternating sugar and phosphate; bases project from the sugar. A strand has different 5′ and 3′ ends. RNA’s ribose has a 2′ OH group, whereas deoxyribose has H there.
Let’s look a little closer.
Go deeper: Chemical identity and direction matter
Nucleoside names include adenosine, guanosine and cytidine. Phosphate changes a nucleoside into a nucleotide. Prime symbols distinguish sugar-carbon numbering from numbering within the base. The drawings use component symbols so the connections remain readable; they are not full chemical structures.
Lengths must specify nucleotides for a single strand or base pairs for a duplex. A base pair consists of one base from each of two strands; it is not one nucleotide. Viral genomes can be DNA or RNA and can be single- or double-stranded, so DNA is not a universal description of every genome.
Take a moment. What do you think?
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Question
An invented strand has 120 nucleotides. How many nucleotides are in its complete complementary duplex, and how many base pairs?
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The Double Helix Model and Central Dogma
Connect base pairing with copying and information flow.Complementary bases fit antiparallel strands
- Antiparallel complementarity
- Complementary DNA strands run in opposite 5′/3′ directions; A pairs with T and G with C.
Watson and Crick’s model drew on X-ray diffraction evidence, including Franklin’s and Wilkins’s work, and Chargaff’s observations. Two sugar–phosphate chains face opposite directions. A–T pairs have two hydrogen bonds and G–C three; purine–pyrimidine pairing supports an approximately uniform width.
The textbook B-DNA model has about10 base pairs per3.4nm turn, with 0.34nm spacing along the axis. Base stacking also contributes to stability. Complementarity makes copying possible: each parental strand can specify a new partner. DNA→RNA→polypeptide describes transcription and translation; RNA→DNA is reverse transcription.
Go deeper: Keep polarity on every sequence
Writing a complementary strand below a 5′→3′ strand gives a 3′→5′ partner. If asked to report that partner 5′→3′, reverse its written order after complementing. A sequence without direction can be ambiguous.
The central dogma concerns transfers of sequence information. DNA can copy DNA and RNA can template DNA in some systems; a protein does not provide a template for reconstructing its original nucleic-acid sequence. Functional RNA genes also matter, so not every gene’s product is a polypeptide.
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Question
Write the partner of5′-AGTCGA-3′ first antiparallel below it, then as a5′→3′ sequence.
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Packaging of DNA Helix
Explain how long DNA is organised without changing its sequence.Long DNA folds into an organised chromosome
- Nucleosome
- DNA wrapped around a histone octamer forms a nucleosome core; linker DNA connects neighbouring cores.
A histone octamer contains eight core histone proteins. About147bp wrap the core; the chapter’s approximately200bp repeat includes linker DNA, whose length varies. Positively charged histone residues interact with negatively charged DNA. H1 associates with linker organisation and is not one of the eight core subunits.
Chromatin forms higher organisation through fibres and loops, with additional proteins. It condenses strongly during cell division. Looser euchromatin is generally more accessible to transcription than compact heterochromatin. Prokaryotic DNA is organised in a nucleoid rather than a membrane-bounded nucleus.
Go deeper: Packaging is an organisation problem
Using the chapter’s 6.6×10⁹bp diploid estimate and 0.34×10⁻⁹m/bp gives 2.244 m of DNA before packaging. Folding lets long molecules fit a nucleus; the calculation is approximate and does not imply a literal straight DNA filament inside a cell.
A metaphase chromosome drawn as an X contains two sister chromatids after replication. It is not a pair of maternal and paternal homologues. Open chromatin is not a guarantee that every associated gene is expressed; transcription also depends on regulatory factors.
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Question
Using an invented1.0×10⁸bp duplex and 0.34nm/bp, calculate its uncoiled length in metres.
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The Search for Genetic Material
Follow the evidence from bacteria to molecular tracers.Distinguish an observation from its molecular explanation
- Transformation
- A heritable change caused by uptake of genetic material; Griffith observed it before its chemical basis was identified.
Griffith compared live smooth S, live rough R, heat-killed S, and mixed live R plus heat-killed S bacteria. The mixture caused disease in the historical mouse experiment and live S bacteria were recovered. A transferable factor had changed R properties, but Griffith’s result alone did not identify DNA.
Avery, MacLeod and McCarty showed that DNase inhibited transformation whereas protease and RNase did not. Hershey and Chase traced phage DNA with³²P and protein coat with³⁵S: after blending and centrifugation, DNA-associated tracer mainly accompanied bacteria while coat tracer mainly remained outside. Together these experiments supported DNA as hereditary material in the systems studied.
Go deeper: What each experiment can establish
DNase means DNA-digesting enzyme. Comparing enzyme treatments helped distinguish molecular candidates; DNA and DNase are different objects. Hershey–Chase separated entry of phage material from the surrounding coat, rather than deciding based on where phosphorus occurs in every biological molecule.
Genetic material must be copied, maintained sufficiently stably, expressed and able to vary. DNA’s structure supports long-term information storage, while RNA can also be a genome in some viruses and has messenger, adapter, structural and catalytic roles. Generalising beyond the experimental organism requires additional evidence.
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Question
A hypothetical transforming extract still works after protease but stops after DNase. What inference is supported, and what does it not prove?
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The RNA World Hypothesis
Separate modern RNA capabilities from proposed early history.RNA can carry information and catalyse reactions
- Ribozyme
- An RNA molecule that catalyses a reaction through its folded structure.
RNA can store information in sequence and catalyse reactions when appropriately folded. The ribosome’s RNA-centred peptide-bond-forming machinery and other ribozymes support the plausibility of an RNA-rich early biological stage.
The RNA-world hypothesis proposes early systems with RNA carrying information and providing catalysis before DNA and proteins took major specialised roles. The chapter presents this idea strongly; it remains a hypothesis about a past origin pathway, not a directly observed complete history.
Go deeper: Stability and evidence
RNA’s2′ hydroxyl makes it more chemically susceptible to cleavage under some conditions. DNA’s deoxyribose and complementary duplex support stability and repair. Catalysis does not automatically mean a molecule destroys itself, and not all RNAs have identical stability.
A folded RNA that cleaves a substrate demonstrates catalysis. It does not by itself demonstrate self-replication, an entire metabolism or the historical origin of life. Separate the capabilities shown by modern experiments from the additional steps a complete origin model would need.
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Question
Does finding one catalytic RNA demonstrate that all steps of an RNA-world origin pathway occurred? Explain.
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DNA Replication
Track old strands and synthesis direction separately.Preserve one old strand in each new double helix
- Semiconservative replication
- Each daughter DNA duplex retains one parental strand and gains one newly synthesised strand.
Complementarity guides copying of each separated parental strand. Meselson and Stahl grew bacteria with stable¹⁵N, then transferred them to¹⁴N medium. After one generation the extracted duplex DNA was hybrid; after two, half was hybrid and half light. Taylor’s chromosome experiments provided related evidence in plants.
DNA polymerase extends an existing3′ end, synthesising5′→3′. Primers are required. At a fork, one strand is made continuously toward fork progression; the other is made in fragments away from the fork. Primer removal and replacement precede ligase sealing backbone nicks. Replication begins at origins; eukaryotic nuclear DNA replication occurs in S phase.
Go deeper: Interpret bands and enzyme roles
In the ideal semiconservative model, after n≥1 generations in light medium, there are2ⁿ duplexes from the starting molecule, two of which retain a heavy strand. Hybrid fraction=2/2ⁿ. At generation1 this is100%; generation2 is50%; generation4 is12.5%, with the remainder light.¹⁵N is not radioactive.
Deoxynucleoside triphosphates provide building blocks and energy. Polymerase does not seal every nick by itself, and ligase does not replace missing bases or bond opposite DNA strands together. Illustrative fork arrows must show both new strands synthesised5′→3′ despite their different directions relative to the fork.
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Question
In an ideal experiment, one heavy duplex is moved to light medium and completes three replication generations. How many duplexes are hybrid and light?
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Transcription
Keep template identity, polarity and RNA processing visible.Read the template; grow a complementary RNA
- Transcription
- RNA synthesis from a DNA template; the template is read3′→5′ while RNA grows5′→3′.
A promoter and termination signals define a transcription unit with its transcribed region. RNA polymerase reads one template strand for that transcript and makes complementary RNA using U rather than T. RNA matches the coding strand when both are reported5′→3′, apart from U/T.
Bacterial RNA polymerase works with initiation factors such as sigma; termination has several mechanisms, including rho-dependent termination. In eukaryotes, nuclear polymeraseII produces an mRNA precursor, I major rRNAs, and III tRNA/5S rRNA among other products. Pre-mRNA processing includes splicing,5′ capping and3′ polyadenylation in the illustrated pathway; mature mRNA can then exit the nucleus.
Go deeper: Genes are more than protein instructions
Exons are retained in processed RNA; introns are removed by splicing. Exons can include untranslated regions, so “exon” is not synonymous with “protein-coding bases”. A gene can produce functional tRNA or rRNA rather than a polypeptide. Regulatory sequences help determine expression.
Bacterial transcription and translation can overlap in one compartment. Nuclear eukaryotic transcription and processing are separated from cytoplasmic translation. Opposite DNA strands can act as templates for different genes; one selected transcription unit need not copy both simultaneously.
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Question
For coding5′-ATGCCATTA-3′, give the template below it and the RNA written5′→3′.
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The Genetic Code and Mutations
Compare triplet meaning with reading-frame changes.Read triplets in a defined frame
- Codon
- A three-nucleotide mRNA unit interpreted in a defined reading frame by a genetic code.
The standard code has 64 triplets: 61 specify amino acids and UAA, UAG, UGA are stops. AUG specifies methionine and commonly serves as a start codon. Several codons may specify one amino acid, which is degeneracy; the standard code is nearly universal with known exceptions.
Insertions or deletions not divisible by three can shift the downstream reading frame in a coding region. Three-base changes retain the downstream frame but can add or remove an amino acid. Substitutions can be synonymous, missense or introduce a stop depending on sequence. The HbS example changes coding DNA GAG→GTG and mRNA GAG→GUG, altering glutamate to valine.
Go deeper: Decoder orientation matters
A tRNA anticodon pairs antiparallel with an mRNA codon. An amino acid attaches at the3′ acceptor end; an aminoacyl-tRNA synthetase supplies charging specificity. Stop codons recruit release factors rather than a tRNA carrying a stop amino acid.
An inserted base affects triplet grouping from its position onward, not necessarily the earlier AUG. An incomplete terminal group is not a full codon. The effect of a mutation cannot be inferred from “one base changed” alone; its location and the resulting message matter.
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Question
Under the standard code, compare AUG GAA UUC UAA with AUG GAG UUC UAA. Must the peptide sequence differ?
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Translation
Show charging, decoding, peptide transfer and release as different steps.Watch the ribosome turn codons into a chain
- Translation
- Ribosomal assembly of a polypeptide according to an mRNA sequence.
Before decoding, an aminoacyl-tRNA synthetase uses ATP to attach the correct amino acid to its tRNA. At initiation, the small ribosomal subunit and initiator tRNA recognise a start site and assemble the translation machinery.
During elongation, charged tRNAs pair with successive codons. The growing chain transfers from P-site tRNA to the amino acid on A-site tRNA, then translocation advances the reading position. Ribosomal RNA contributes catalysis; bacterial23S rRNA participates in peptide-bond formation. A release factor recognises a stop and releases the chain.
Go deeper: Distinguish message length from peptide length
The ribosome reads mRNA5′→3′ and builds a peptide from amino to carboxyl end. The invented AUG GAA UUC UAA message has four codons but only three encoded amino acids; the stop contributes no amino acid.
UTRs occur before the translated start and after the stop in the illustrated message. They can regulate translation and stability. The chapter’s two binding sites simplify a ribosome that also has an exit site; schematic motion is a teaching sequence rather than a measured molecular animation.
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Question
A short invented translated region contains AUG, three further sense codons and then UGA. How many amino acids are specified, and how many peptide bonds join them?
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Regulation of Gene Expression: The Lac Operon
Separate inducer binding from making lactose-processing proteins.An operator responds to the repressor’s state
- Lac operon
- A bacterial regulatory arrangement in which a common promoter controls lac structural genes; repressor binding at the operator affects transcription.
lacI encodes a repressor from its own transcription unit. Without inducer, the repressor can bind the lac operator and reduce transcription. Allolactose binding changes the repressor so operator binding decreases. Lactose availability thus connects to production of proteins used in its metabolism.
The structural genes z, y, a encode β-galactosidase, permease and transacetylase. They are transcribed together as a polycistronic message. β-galactosidase hydrolyses lactose; permease facilitates entry. Very low basal expression helps avoid the false idea that no lactose could ever enter before induction.
Go deeper: Control depends on more than one switch
The visual isolates negative regulation by the repressor. Glucose-dependent positive regulation also affects expression, so “lactose present means maximum transcription” is too simple. The repressor is not physically destroyed when inducer binds.
Eukaryotic expression can be regulated at transcription, RNA processing, transport and translation. These control points act on different objects and can operate together. Expression depends on cell type and conditions; the same genome does not imply identical protein output in every cell.
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Question
Why does a diagram showing lacI on the same DNA not mean that lacI is part of the z, y, a message?
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Human Genome Project (HGP)
Connect genome assembly, interpretation and responsible use.Assemble a reference, then ask what it means
- Sequence annotation
- Assigning supported biological features and functions to a sequence; distinct from determining base order.
The Human Genome Project began in 1990 and concluded in 2003, creating a reference resource and improved sequencing, bioinformatics and data-sharing methods. Whole-genome sequencing includes coding and noncoding regions; EST approaches identify sequences expressed as RNA. Historical workflows used BAC/YAC cloning, sequencing and computational alignment.
The HGP reference covered about 92% of the genome, with hard-to-sequence gaps addressed by later work, including the 2022 T2T milestone. The textbook includes historical gene-count estimates; NHGRI education now describes roughly 20,000 protein-coding genes. Reference genomes continue to improve and do not capture every person’s complete variation.
Go deeper: A sequence is a starting point
Overlapping reads support assembly, but repeats and sequencing errors can make alignments ambiguous. Annotation needs evidence: a region may be a protein-coding exon, a retained untranslated exon, a functional RNA gene or another regulatory/noncoding sequence. Noncoding does not mean useless.
Genome information can support biology and health research, while consent, privacy and fair treatment remain essential. A sequence or one variant alone usually cannot predict a person’s abilities, appearance or health. The HGP’s historical cost illustration should not be presented as a current sequencing price.
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Question
Align invented reads CGATTA, TTAGCA and GCACCT using their three-base overlaps. Write the assembled sequence and name one reason real assembly is harder.
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DNA Fingerprinting
Explain fragment patterns without overstating identity evidence.Repeat counts change fragment length
- DNA polymorphism
- Variation in DNA sequence among individuals; repeat-count differences can provide informative profiling markers.
VNTR loci contain tandem repeat units whose copy number can vary between alleles. Shared flanking sites with different repeat counts can yield fragments of different lengths. Such variation can be inherited and need not alter a protein.
Historical DNA fingerprinting involved isolation, restriction digestion, gel separation, Southern transfer, labelled-probe hybridisation and detection. Smaller DNA fragments generally travel farther through a gel toward the positive electrode. Modern methods often amplify STR loci. Multiple loci and statistical context are needed to interpret consistency.
Go deeper: A classroom match is not a verdict
At an ordinary autosomal locus, a child receives one allele from each biological parent. Hypothetical band patterns can illustrate inheritance, but mixtures, measurement error, mutation and laboratory quality affect real interpretation. Identical twins generally share conventional DNA profiles.
A pattern matching at a few markers does not establish guilt or unique identity by itself. Evidence must be considered with population frequencies, contamination, chain of custody and the question being asked. A whole genome is not necessary for routine profiling, and DNA from different sampled tissues generally represents the same person despite important biological exceptions.
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Question
At a hypothetical repeat locus with a 6 bp motif, two alleles have4 and7 repeats plus identical 40 bp flanking DNA. Calculate both fragment lengths and which usually travels farther in a gel.
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