H2 Biology Comprehensive Review

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Comprehensive practice flashcards covering molecular techniques, viral and bacterial genetics, cancer biology, stem cells, cell signaling, gene expression, genetics and variation, evolution, enzymes, membrane transport, proteins, carbohydrates, climate change, photosynthesis, and cellular respiration.

44 Terms

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What are the three main steps and corresponding temperatures in one cycle of the Polymerase Chain Reaction (PCR)?

  1. Denaturation: heated to 94∘C−96∘C94^\circ\text{C}-96^\circ\text{C} to break hydrogen bonds and separate double-stranded DNA into single strands.
  2. Annealing: cooled to 54∘C−60∘C54^\circ\text{C}-60^\circ\text{C} to allow forward and reverse primers to base-pair to the 3′3' ends of the template strands flanking the target sequence.
  3. Extension: heated to 72∘C72^\circ\text{C} where Taq DNA polymerase synthesises the complementary strand by adding dNTPs to the free 3′-OH3'\text{-OH} end of primers.
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What is the biological source of Taq DNA polymerase, and what are its key kinetic and fidelity characteristics in PCR?

It is isolated from the thermophilic bacterium Thermus aquaticus (which thrives in hot springs up to 90∘C90^\circ\text{C}). It has an optimum temperature of approximately 70∘C70^\circ\text{C}, does not easily denature at high temperatures, but lacks a 3′→5′3' \rightarrow 5' exonuclease proofreading function, resulting in a relatively high error rate of about 1 in 90009000 bases.

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What are the key steps involved in the Southern blotting procedure for detecting specific DNA fragments?

  1. Digest genomic DNA with a restriction enzyme and separate fragments by agarose gel electrophoresis.
  2. Denature double-stranded DNA in the gel into single strands using an alkaline solution (NaOH\text{NaOH}).
  3. Transfer single-stranded DNA to a nitrocellulose or nylon membrane via capillary action.
  4. Bake the membrane (60∘C−100∘C60^\circ\text{C}-100^\circ\text{C}) to permanently cross-link DNA.
  5. Hybridise with a labeled single-stranded DNA or RNA probe complementary to the target sequence.
  6. Wash away unbound probes and detect bound probes via autoradiography, fluorescence, or chemiluminescence.
<ol>
<li>Digest genomic DNA with a restriction enzyme and separate fragments by agarose gel electrophoresis.</li>
<li>Denature double-stranded DNA in the gel into single strands using an alkaline solution ($$\text{NaOH}$$).</li>
<li>Transfer single-stranded DNA to a nitrocellulose or nylon membrane via capillary action.</li>
<li>Bake the membrane ($$60^\circ\text{C}-100^\circ\text{C}$$) to permanently cross-link DNA.</li>
<li>Hybridise with a labeled single-stranded DNA or RNA probe complementary to the target sequence.</li>
<li>Wash away unbound probes and detect bound probes via autoradiography, fluorescence, or chemiluminescence.</li>
</ol>
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<p>What are the main structural components of a typical T-even bacteriophage compared to animal naked and enveloped viruses?</p>

What are the main structural components of a typical T-even bacteriophage compared to animal naked and enveloped viruses?

A typical bacteriophage possesses an icosahedral capsid head containing double-stranded DNA, attached to a helical tail sheath, baseplate, baseplate pins, and tail fibres. In animal viruses, a naked virus consists solely of a nucleocapsid (capsomers surrounding nucleic acid), whereas an enveloped virus is enclosed by a host-derived phospholipid bilayer containing viral glycoproteins.

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How do the reproductive cycles of lambda (λ\lambda) bacteriophage differ between the lysogenic and lytic pathways?

In the lysogenic cycle, the phage DNA circularises, integrates into the bacterial chromosome via viral integrase to form a prophage, and replicates silently alongside host DNA under the control of lambda repressor protein. Environmental stress triggers spontaneous induction, cleaving the repressor, excising the prophage, and initiating the lytic cycle, where phage components are synthesised, assembled, and released by cell lysis.

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What are the essential steps in the replication cycle of the Human Immunodeficiency Virus (HIV)?

  1. Gp120 binds to CD4 receptors (and CCR5/CXCR4 coreceptors) on host helper T cells or macrophages, and gp41 mediates membrane fusion.
  2. Capsid enters cytosol, uncoats, and reverse transcriptase transcribes (+)ssRNA into dsDNA.
  3. Integrase incorporates viral dsDNA into host chromosome to establish a provirus.
  4. Host RNA polymerase transcribes viral mRNA and genomic RNA; polyproteins are translated on ribosomes.
  5. Polyproteins, genomic RNA, and enzymes assemble at the membrane, bud out acquiring an envelope, and viral protease cleaves polyproteins into functional units.
<ol>
<li>Gp120 binds to CD4 receptors (and CCR5/CXCR4 coreceptors) on host helper T cells or macrophages, and gp41 mediates membrane fusion.</li>
<li>Capsid enters cytosol, uncoats, and reverse transcriptase transcribes (+)ssRNA into dsDNA.</li>
<li>Integrase incorporates viral dsDNA into host chromosome to establish a provirus.</li>
<li>Host RNA polymerase transcribes viral mRNA and genomic RNA; polyproteins are translated on ribosomes.</li>
<li>Polyproteins, genomic RNA, and enzymes assemble at the membrane, bud out acquiring an envelope, and viral protease cleaves polyproteins into functional units.</li>
</ol>
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<p>What are the main structural features of a typical bacterial cell?</p>

What are the main structural features of a typical bacterial cell?

A bacterial cell possesses a peptidoglycan cell wall, sometimes surrounded by a polysaccharide capsule. Internally, it lacks membrane-bound organelles and a true nucleus; its genetic material consists of a circular double-stranded DNA molecule concentrated in the nucleoid, alongside 70S70\text{S} ribosomes and autonomous circular plasmids in the cytoplasm. Surface appendages include flagella for motility and fimbriae or pili for attachment and conjugation.

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How does asexual reproduction occur in bacteria via binary fission?

  1. The circular bacterial chromosome attaches to the plasma membrane at the single origin of replication.
  2. Semiconservative bidirectional DNA replication proceeds until two copies of the chromosome are formed, each attached to the membrane.
  3. The cell elongates, moving the two origins and chromosomes toward opposite poles.
  4. The plasma membrane invaginates inward, new cell wall material is synthesised, and the parent cell cleaves into two genetically identical daughter cells.
<ol>
<li>The circular bacterial chromosome attaches to the plasma membrane at the single origin of replication.</li>
<li>Semiconservative bidirectional DNA replication proceeds until two copies of the chromosome are formed, each attached to the membrane.</li>
<li>The cell elongates, moving the two origins and chromosomes toward opposite poles.</li>
<li>The plasma membrane invaginates inward, new cell wall material is synthesised, and the parent cell cleaves into two genetically identical daughter cells.</li>
</ol>
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<p>What initiates bacterial conjugation between an $$\text{F}^+$$ donor and an $$\text{F}^-$$ recipient cell?</p>

What initiates bacterial conjugation between an F+\text{F}^+ donor and an F−\text{F}^- recipient cell?

The F+\text{F}^+ donor synthesises a sex pilus that makes direct physical contact with the F−\text{F}^- recipient. The pilus retracts to draw cells close together, establishing a temporary cytoplasmic conjugation bridge. One strand of the circular F plasmid is nicked at the origin of transfer and transferred unidirectionally into the recipient, followed by complementary strand synthesis in both cells.

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How does the antibiotic penicillin exert its bactericidal effect on bacteria?

Penicillin acts as a competitive inhibitor of transpeptidase enzymes, preventing peptide cross-linking between adjacent glycan chains during peptidoglycan synthesis in the bacterial cell wall. This weakens the mechanical strength of the cell wall, causing bacterial lysis under osmotic pressure.

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What is the genetic distinction between a proto-oncogene gain-of-function mutation and a tumour suppressor gene loss-of-function mutation?

A gain-of-function mutation in a proto-oncogene (converting it to an oncogene) is genetically dominant; mutating only one of two alleles produces hyperactive or overexpressed oncoproteins that stimulate excessive cell division. In contrast, a loss-of-function mutation in a tumour suppressor gene is genetically recessive; both alleles must be inactivated or deleted to abolish cell-cycle inhibitory controls or apoptotic signaling.

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How does colorectal cancer illustrate the multi-step model of carcinogenesis?

Carcinogenesis requires an accumulation of multiple mutations over time:

  1. Loss of the tumour suppressor gene APC initiates benign polyp formation.
  2. Activation of the ras proto-oncogene promotes adenoma growth.
  3. Loss of the DCC tumour suppressor gene triggers larger adenoma formation.
  4. Loss of the p53 tumour suppressor gene and additional mutations result in a malignant carcinoma capable of angiogenesis and metastasis.
<p>Carcinogenesis requires an accumulation of multiple mutations over time: </p>
<ol>
<li>Loss of the tumour suppressor gene APC initiates benign polyp formation.</li>
<li>Activation of the ras proto-oncogene promotes adenoma growth.</li>
<li>Loss of the DCC tumour suppressor gene triggers larger adenoma formation.</li>
<li>Loss of the p53 tumour suppressor gene and additional mutations result in a malignant carcinoma capable of angiogenesis and metastasis.</li>
</ol>
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What are the three unique hallmark properties of stem cells?

  1. They are unspecialised cells without tissue-specific structures.
  2. They are capable of prolonged self-renewal by dividing indefinitely via mitosis in an undifferentiated state.
  3. Under appropriate chemical and physical cues, their daughter cells can differentiate into one or more specialised cell lineages.
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What therapeutic and research promises are held by cultured pluripotent stem cells?

Pluripotent stem cells can differentiate into virtually all cell types of the three embryonic germ layers (endoderm, mesoderm, ectoderm). They can be directed to generate bone marrow for leukemia therapy, nerve cells for Parkinson's and Alzheimer's disease, heart muscle cells for heart disease, and pancreatic islet cells for diabetes, as well as serve in drug target identification, toxicity screening, and birth defect studies.

<p>Pluripotent stem cells can differentiate into virtually all cell types of the three embryonic germ layers (endoderm, mesoderm, ectoderm). They can be directed to generate bone marrow for leukemia therapy, nerve cells for Parkinson's and Alzheimer's disease, heart muscle cells for heart disease, and pancreatic islet cells for diabetes, as well as serve in drug target identification, toxicity screening, and birth defect studies.</p>
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<p>What are the three fundamental stages of a cell signaling pathway?</p>

What are the three fundamental stages of a cell signaling pathway?

  1. Reception: An extracellular signaling molecule (ligand) binds specifically to a cell-surface or intracellular receptor.
  2. Transduction: The ligand-bound receptor changes conformation, triggering a multi-step relay involving protein kinases in phosphorylation cascades and/or second messengers.
  3. Response: Transduced signals activate specific cellular outcomes, such as changes in gene transcription, enzyme activity, or cytoskeleton rearrangement.
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How does insulin binding to its Receptor Tyrosine Kinase (RTK) stimulate glucose uptake in target cells?

Insulin binds to the extracellular domains of RTK monomers, inducing dimerisation and cross-phosphorylation of tyrosine residues on their intracellular tails. The activated receptor recruits and phosphorylates cytoplasmic relay proteins, triggering a signaling cascade that causes intracellular vesicles containing glucose carrier proteins (GLUTs) to translocate and fuse with the plasma membrane, increasing membrane permeability to glucose.

<p>Insulin binds to the extracellular domains of RTK monomers, inducing dimerisation and cross-phosphorylation of tyrosine residues on their intracellular tails. The activated receptor recruits and phosphorylates cytoplasmic relay proteins, triggering a signaling cascade that causes intracellular vesicles containing glucose carrier proteins (GLUTs) to translocate and fuse with the plasma membrane, increasing membrane permeability to glucose.</p>
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How do histone acetylation and DNA methylation reciprocally regulate chromatin structure and eukaryotic gene transcription?

Histone acetylation by histone acetyltransferases (HATs) attaches acetyl groups to lysine residues on histone tails, neutralising positive charges, weakening ionic attraction to negatively charged DNA, decondensing chromatin into euchromatin, and allowing transcription machinery access. Conversely, DNA methylation by DNA methyltransferases attaches methyl groups to cytosine bases, recruiting histone deacetylases (HDACs) to condense chromatin into transcriptionally silent heterochromatin.

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How does iron concentration regulate ferritin protein synthesis at the translational level via the Iron Response Element (IRE)?

Under low iron concentrations, an active IRE-binding protein (IRP) binds to the stem-loop IRE located in the 5′-UTR5'\text{-UTR} of ferritin mRNA, physically blocking the ribosome from binding and inhibiting translation. When iron concentration is high, iron binds to the IRP, inactivating it and causing it to dissociate from the IRE; the ribosome can then scan the mRNA and translate ferritin to store excess iron.

<p>Under low iron concentrations, an active IRE-binding protein (IRP) binds to the stem-loop IRE located in the $$5'\text{-UTR}$$ of ferritin mRNA, physically blocking the ribosome from binding and inhibiting translation. When iron concentration is high, iron binds to the IRP, inactivating it and causing it to dissociate from the IRE; the ribosome can then scan the mRNA and translate ferritin to store excess iron.</p>
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What is the molecular basis of sickle-cell anaemia at the DNA, mRNA, and polypeptide levels?

A single base-pair substitution occurs in the β\beta-globin gene on chromosome 11, where thymine replaces adenine on the coding strand (template strand changes from 3′-CTT-5′3'\text{-CTT-}5' to 3′-CAT-5′3'\text{-CAT-}5'). This transcribes a mutant mRNA codon 5′-GUA-3′5'\text{-GUA-}3' instead of 5′-GAA-3′5'\text{-GAA-}3', substituting a hydrophilic glutamic acid with a hydrophobic valine at position 6 of the β\beta-globin polypeptide chain.

<p>A single base-pair substitution occurs in the $$\beta$$-globin gene on chromosome 11, where thymine replaces adenine on the coding strand (template strand changes from $$3'\text{-CTT-}5'$$ to $$3'\text{-CAT-}5'$$). This transcribes a mutant mRNA codon $$5'\text{-GUA-}3'$$ instead of $$5'\text{-GAA-}3'$$, substituting a hydrophilic glutamic acid with a hydrophobic valine at position 6 of the $$\beta$$-globin polypeptide chain.</p>
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How did the Meselson-Stahl experiment provide definitive evidence for the semi-conservative model of DNA replication?

E. coli cultured in heavy 15N^{15}\text{N} medium were transferred to light 14N^{14}\text{N} medium. After one generation of replication, density-gradient centrifugation in CsCl\text{CsCl} yielded a single hybrid 14N/15N^{14}\text{N}/^{15}\text{N} DNA band of intermediate density, eliminating the conservative model. After two generations, two distinct bands appeared—one light (14N/14N^{14}\text{N}/^{14}\text{N}) and one hybrid (14N/15N^{14}\text{N}/^{15}\text{N})—ruling out the dispersive model.

<p>E. coli cultured in heavy $$^{15}\text{N}$$ medium were transferred to light $$^{14}\text{N}$$ medium. After one generation of replication, density-gradient centrifugation in $$\text{CsCl}$$ yielded a single hybrid $$^{14}\text{N}/^{15}\text{N}$$ DNA band of intermediate density, eliminating the conservative model. After two generations, two distinct bands appeared—one light ($$^{14}\text{N}/^{14}\text{N}$$) and one hybrid ($$^{14}\text{N}/^{15}\text{N}$$)—ruling out the dispersive model.</p>
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Why does the end-replication problem occur exclusively on linear eukaryotic chromosomes?

DNA polymerases can only add deoxyribonucleotides to the free 3′-OH3'\text{-OH} group of a pre-existing strand and require an RNA primer to initiate synthesis. When the terminal RNA primer at the 5′5' end of a newly synthesised daughter lagging strand is removed, DNA polymerase cannot fill the resulting gap because there is no upstream 3′-OH3'\text{-OH} group, resulting in progressive chromosome shortening with each round of replication.

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What are the structural features and catalytic mechanism of telomerase?

Telomerase is a ribonucleoprotein complex composed of a reverse transcriptase protein and a non-coding RNA molecule containing the sequence 3′-CAAUCCCAA-5′3'\text{-CAAUCCCAA-}5' (in humans). The RNA component base-pairs with the 3′3' telomeric overhang and acts as a template, while the reverse transcriptase catalyses the addition of telomeric DNA repeats (5′-TTAGGG-3′5'\text{-TTAGGG-}3') to the 3′3' end of the parental strand in the 5′→3′5' \rightarrow 3' direction.

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How do mitosis and meiosis differ in terms of chromosomal behaviour and final cellular yield?

In mitosis, chromosomes replicate once during S phase and undergo one nuclear division without synapsis or crossing over; sister chromatids segregate at anaphase to produce 2 genetically identical diploid (2n2n) daughter cells. In meiosis, homologous chromosomes undergo synapsis and crossing over in prophase I, homologous pairs separate in anaphase I, and sister chromatids separate in anaphase II, producing 4 genetically non-identical haploid (nn) daughter gametes.

<p>In mitosis, chromosomes replicate once during S phase and undergo one nuclear division without synapsis or crossing over; sister chromatids segregate at anaphase to produce 2 genetically identical diploid ($$2n$$) daughter cells. In meiosis, homologous chromosomes undergo synapsis and crossing over in prophase I, homologous pairs separate in anaphase I, and sister chromatids separate in anaphase II, producing 4 genetically non-identical haploid ($$n$$) daughter gametes.</p>
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<p>How does an enzyme catalyze a biochemical reaction according to the energy profile of a reaction?</p>

How does an enzyme catalyze a biochemical reaction according to the energy profile of a reaction?

An enzyme acts as a biological catalyst by lowering the free energy of activation (EAE_A) required to reach the unstable transition state where reactant bonds can break. It does not alter the overall standard free energy change (ΔG\Delta G) of the reaction or shift the equilibrium point, but dramatically increases the rate at which equilibrium is attained.

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<p>How do allosteric activators and inhibitors regulate a multi-subunit allosteric enzyme?</p>

How do allosteric activators and inhibitors regulate a multi-subunit allosteric enzyme?

Allosteric enzymes oscillate between active and inactive conformations. Binding of an allosteric activator to a regulatory site stabilises the active conformation across all subunits, maintaining functional active sites with high affinity for the substrate. Conversely, binding of an allosteric inhibitor to the regulatory site stabilises the inactive conformation, distorting active sites into a non-functional state.

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<p>What is the transport mechanism and stoichiometry of the sodium-potassium pump ($$\text{Na}^+/\text{K}^+$$ ATPase)?</p>

What is the transport mechanism and stoichiometry of the sodium-potassium pump (Na+/K+\text{Na}^+/\text{K}^+ ATPase)?

The pump binds 3 Na+3\,\text{Na}^+ ions from the cytoplasm, stimulating ATP hydrolysis and phosphorylation of the transport protein. This triggers a conformational change that expels 3 Na+3\,\text{Na}^+ ions into the extracellular fluid. Two extracellular K+\text{K}^+ ions then bind to the protein, triggering dephosphorylation, which restores the original conformation and releases 2 K+2\,\text{K}^+ ions into the cytoplasm against their electrochemical gradients.

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<p>What are the common quaternary structural arrangements in proteins, and what are classical examples of each?</p>

What are the common quaternary structural arrangements in proteins, and what are classical examples of each?

Quaternary structure involves the association of two or more folded polypeptide subunits:

  • Dimer: two subunits, either identical (homodimer) or different (heterodimer).
  • Trimer: three subunits, such as the triple helix of collagen (tropocollagen).
  • Tetramer: four subunits, such as the adult hemoglobin molecule consisting of an α2β2\alpha_2\beta_2 tetramer.
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<p>How do the straight-chain Fischer projections of glucose, galactose, and fructose differ structurally?</p>

How do the straight-chain Fischer projections of glucose, galactose, and fructose differ structurally?

Glucose and galactose are aldohexoses with a terminal carbonyl group (C=O\text{C}=\text{O} at C1); they are optical isomers differing only at C4, where the -OH\text{-OH} group is on the right in glucose and on the left in galactose. Fructose is a ketohexose containing an internal carbonyl group (C=O\text{C}=\text{O} at C2).

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<p>Why is maltose classified as a reducing sugar while sucrose is a non-reducing sugar?</p>

Why is maltose classified as a reducing sugar while sucrose is a non-reducing sugar?

Maltose possesses an unbonded anomeric carbon (C1 on the second glucose residue) that can open into a straight-chain form to expose a free reactive aldehyde group, which reduces alkaline copper(II) in Benedict's reagent. In sucrose, the glycosidic bond joins the anomeric carbon C1 of α\alpha-glucose directly to the anomeric carbon C2 of β\beta-fructose, locking both anomeric carbons and leaving no free carbonyl group.

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What structural features distinguish amylose, amylopectin, glycogen, and cellulose?

  • Amylose: unbranched helical polymer of α\alpha-glucose linked by α-(1,4)\alpha\text{-}(1,4) glycosidic bonds.
  • Amylopectin: branched polymer of α\alpha-glucose with α-(1,4)\alpha\text{-}(1,4) chains and α-(1,6)\alpha\text{-}(1,6) branch points every 24–30 residues.
  • Glycogen: extensively branched polymer of α\alpha-glucose with α-(1,4)\alpha\text{-}(1,4) chains and frequent α-(1,6)\alpha\text{-}(1,6) branch points (every 8–12 residues).
  • Cellulose: unbranched straight polymer of β\beta-glucose joined by β-(1,4)\beta\text{-}(1,4) glycosidic bonds, with alternate glucose residues rotated by 180∘180^\circ, bundled into rigid microfibrils via extensive intermolecular hydrogen bonds.
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What are the structural features and functions of an Immunoglobulin G (IgG) antibody molecule?

IgG is a Y-shaped globular glycoprotein consisting of four polypeptide chains: two identical heavy (H\text{H}) chains and two identical light (L\text{L}) chains, linked by covalent disulfide bridges and non-covalent interactions. It has two identical Fab regions formed by the variable domains (VH\text{V}_\text{H} and VL\text{V}_\text{L}) that provide specific antigen-binding sites, connected by a flexible hinge region to a constant Fc region (CH2\text{C}_\text{H}2, CH3\text{C}_\text{H}3) that mediates effector functions like opsonisation and macrophage binding.

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How do somatic recombination, somatic hypermutation, and class switching generate antibody diversity and affinity maturation?

  • Somatic Recombination: in developing naïve B cells, random combinatorial joining of V and J segments (light chain) and V, D, and J segments (heavy chain) creates millions of unique variable domain sequences.
  • Somatic Hypermutation: in activated B cells, point mutations are introduced at a high rate into rearranged V(D)J exons, generating variants with altered antigen affinities; cells with higher affinity are selected (affinity maturation).
  • Class Switching: activated B cells excise constant heavy-chain segments (CH\text{C}_\text{H}) to switch antibody isotype (e.g. from IgM\text{IgM} to IgG\text{IgG}) without changing antigen specificity.
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How does Mycobacterium tuberculosis evade the host's innate immune defences inside alveolar macrophages?

M. tuberculosis is coated with a thick layer of mycolic acid that resists digestion by lysozymes. Following phagocytosis, it secretes proteins that prevent the fusion of the phagosome with lysosomes, preventing phagolysosome formation. Alternatively, it acidifies the phagolysosome to inhibit hydrolytic enzymes, allowing the bacteria to survive and replicate by binary fission intracellularly within macrophages, leading to chronic granuloma (tubercle) formation.

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What is the distinction between antigenic drift and antigenic shift in the evolution of the influenza virus?

  • Antigenic Drift: gradual accumulation of point mutations in the RNA genes encoding haemagglutinin (HA) and neuraminidase (NA) due to error-prone viral RNA-dependent RNA polymerase, leading to seasonal epidemics.
  • Antigenic Shift: sudden, drastic genetic reassortment occurring when two or more distinct influenza strains infect a single host cell simultaneously, swapping RNA segments to produce a novel HA/NA subtype that can jump species and trigger global pandemics.
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What is the definition and epidemiological significance of the basic reproduction number (R0\text{R}_0)?

R0\text{R}_0 is the average number of secondary cases an infected individual generates in a completely susceptible population. If R0>1\text{R}_0 > 1, the infection will spread exponentially through the population; if R0=1\text{R}_0 = 1, the disease remains endemic at stable rates; if R0<1\text{R}_0 < 1, the number of cases declines and the outbreak will die out.

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What are the five essential conditions required for a population to remain in Hardy-Weinberg equilibrium?

  1. No mutations (no new alleles formed).
  2. Random mating (no mate choice or sexual selection).
  3. No natural selection (equal reproductive success for all genotypes).
  4. Extremely large population size (negligible genetic drift).
  5. No gene flow (no migration of individuals into or out of the population).
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How does allopatric speciation differ from sympatric speciation?

Allopatric speciation occurs when a physical geographic barrier divides an ancestral population, preventing gene flow; the isolated gene pools diverge through independent mutations, genetic drift, and natural selection under differing selective pressures until reproductive isolation evolves. Sympatric speciation occurs without geographic separation within the same area, driven by mechanisms such as polyploidy (autopolyploidy/allopolyploidy), habitat differentiation, or sexual selection.

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What are the expected phenotypic ratios in dihybrid crosses involving unlinked vs autosomal linked genes?

  • Unlinked genes: A dihybrid cross between double heterozygotes (AaBb×AaBb\text{AaBb} \times \text{AaBb}) yields a 9:3:3:19:3:3:1 ratio; a testcross (AaBb×aabb\text{AaBb} \times \text{aabb}) yields a 1:1:1:11:1:1:1 ratio.
  • Linked genes without crossing over: Heterozygote selfing yields a 3:13:1 ratio; a testcross yields a 1:11:1 ratio.
  • Linked genes with crossing over: Produces parental phenotypes in significantly higher proportions and recombinant phenotypes in smaller proportions, deviating from 1:1:1:11:1:1:1.
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How do recessive epistasis and dominant epistasis modify the classic 9:3:3:19:3:3:1 dihybrid phenotypic ratio?

  • Recessive Epistasis: Homozygous recessive alleles at one gene locus mask the phenotypic expression of alleles at a second locus, modifying the 9:3:3:19:3:3:1 ratio to 9:3:49:3:4 (e.g. mouse coat colour) or 9:79:7 when both homozygous recessive conditions produce the same phenotype.
  • Dominant Epistasis: The presence of at least one dominant allele at an epistatic locus suppresses the expression of the hypostatic gene, producing modified ratios such as 12:3:112:3:1 (e.g. summer squash) or 15:115:1 (duplicate dominant epistasis).
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How does global warming accelerate the transmission and geographic spread of mosquito-borne diseases like dengue?

Insects are poikilotherms with narrow thermal tolerances. Rising ambient temperatures increase the metabolic rate of Aedes aegypti, accelerating larval development, shortening maturation time, increasing adult emergence, and elevating feeding/biting frequency. Furthermore, elevated temperatures shorten the extrinsic incubation period of the dengue virus within the mosquito salivary glands, while expanding the vector's geographic range poleward into previously temperate zones.

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How are ATP and NADPH synthesised via non-cyclic photophosphorylation during the light-dependent stage of photosynthesis?

Light harvested by PS II excites P680, ejecting electrons replaced by the photolysis of water (2H2O→4H++4e−+O22\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4e^- + \text{O}_2). Electrons pass down an electron transport chain to PS I, releasing energy to actively pump protons from the stroma into the thylakoid space; the resulting proton motive force drives ATP synthesis via ATP synthase (chemiosmosis). Light exciting P700 in PS I boosts electrons down a second ETC to ferredoxin-NADP+ reductase, which reduces NADP+\text{NADP}^+ to NADPH.

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What are the three phases of the Calvin cycle in C3 photosynthesis, and what are their biochemical requirements per glucose molecule synthesised?

  1. Carbon fixation: Rubisco catalyses the carboxylation of 6 RuBP (5C5\text{C}) with 6 CO26\,\text{CO}_2 to form unstable 6C6\text{C} intermediates that split into 12 PGA (3C3\text{C}).
  2. Reduction: 12 PGA are phosphorylated by 12 ATP12\,\text{ATP} and reduced by 12 NADPH12\,\text{NADPH} to form 12 G3P (3C3\text{C}); 2 G3P exit the cycle to form 1 glucose (6C6\text{C}).
  3. Regeneration: The remaining 10 G3P are rearranged using 6 ATP6\,\text{ATP} to regenerate 6 RuBP. Total cost per glucose: 6 CO26\,\text{CO}_2, 18 ATP18\,\text{ATP}, and 12 NADPH12\,\text{NADPH}.
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What are the four successive stages of aerobic cellular respiration, and where do they take place within a eukaryotic cell?

  1. Glycolysis: Cytosol; splits glucose (6C6\text{C}) into 2 pyruvate (3C3\text{C}), yielding net 2 ATP2\,\text{ATP} and 2 NADH2\,\text{NADH}.
  2. Link Reaction: Mitochondrial matrix; oxidises 2 pyruvate into 2 acetyl-CoA (2C2\text{C}), yielding 2 CO22\,\text{CO}_2 and 2 NADH2\,\text{NADH}.
  3. Krebs Cycle: Mitochondrial matrix; completely oxidises 2 acetyl-CoA to 4 CO24\,\text{CO}_2, yielding 2 ATP2\,\text{ATP}, 6 NADH6\,\text{NADH}, and 2 FADH22\,\text{FADH}_2.
  4. Oxidative Phosphorylation: Inner mitochondrial membrane (cristae); transfers electrons through the ETC to O2\text{O}_2, generating a proton gradient across the inner membrane to drive ATP synthesis via chemiosmosis (≈34 ATP\approx 34\,\text{ATP}).
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How do alcoholic and lactic acid fermentation regenerate NAD+\text{NAD}^+ under anaerobic conditions?

  • Alcoholic fermentation (yeast): Pyruvate (3C3\text{C}) is decarboxylated by pyruvate decarboxylase to form ethanal (2C2\text{C}) and CO2\text{CO}_2. Alcohol dehydrogenase then reduces ethanal to ethanol (2C2\text{C}), oxidising NADH to regenerate \text{NAD}^+$.\n- Lactic acid fermentation (mammalian muscle): Pyruvate (3\text{C})isdirectlyreducedbylactatedehydrogenasetolactate() is directly reduced by lactate dehydrogenase to lactate (3\text{C})withoutreleasing) without releasing\text{CO}_2,oxidisingNADHtoregenerate, oxidising NADH to regenerate\text{NAD}^+,whichallowsglycolysistocontinueproducingnet, which allows glycolysis to continue producing net2\,\text{ATP}$$.