IB Biology A2.3: Viruses worked solutions

The capsid is the protein packaging associated with a viral genome. RNA can be the hereditary material, so DNA is not a universal capsid constituent. A lipid envelope is optional, and cellulose is not viral packaging. Award 1 for C.

Viral genes specify the amino acid sequence, but host ribosomes assemble the polypeptide by translating viral mRNA. Award 1 for ribosomes and 1 for translation. Merely stating that the host helps replication does not identify either requirement.

Each HIV genome copy is single-stranded RNA (1). Having two copies counts molecules; double-stranded describes complementary pairing within a duplex (1). The number two by itself supplies no evidence of such pairing.

The acquired outer bilayer is a viral envelope (1). In budding, a patch of host membrane surrounds the viral components and pinches off with them (1). The internal protein capsid is a different layer.

Assembly puts genome copies and coat proteins into complete progeny phages (1). Host lysis then allows those particles to leave (1). Copying DNA is not enough to create a complete particle.

Integrated phage DNA copied in a surviving bacterial lineage is a prophage in the lysogenic phase. Lysis would destroy the cell; assembly is particle construction and translation makes proteins. Award 1 for B.

Award 1 for a valid non-error-rate factor and 1 for its causal link. Short generations permit repeated replication and differential reproduction over a short interval. Alternatively, explain how strong selection changes variant frequencies, or how compatible coinfection permits genetic exchange. Large numbers alone do not establish a frequency change.

Award one mark for each paired criterion: lambda dsDNA versus coronavirus ssRNA (1); absent versus present envelope (1); head-and-tail form versus roughly spherical enveloped form with spikes (1). A list of names without matched features earns no comparison marks.

Award 1 for HIV RNA versus lambda DNA, 1 for single-stranded copies versus double-stranded DNA, 1 for envelope present versus absent, and 1 for HIV conical internal capsid versus lambda head-and-tail organisation. Do not infer genome chemistry from shape alone.

(a) Ribosomes perform translation (1), using viral mRNA to assemble a viral polypeptide (1). (b) Amino acids supply monomers for the coat proteins (1). ATP supplies energy for biosynthetic processes (1). Ribosome inhibition prevents translation even with these supplies present. The question gives a selective effect; do not infer that DNA replication is also inhibited.

(a) Assembled progeny and rupture identify lytic development (1); rupture releases particles (1). (b) Joined bacterial and viral sequences identify an integrated prophage (1); host chromosome replication and binary fission pass it to daughter bacteria without immediate virion production (1). Both routes involve viral DNA copying, so copying alone is not a sufficient distinction.

(a) Four doublings give 6\times2^4=96 bacteria (1). One prophage per bacterium gives 96 integrated copies (1). (b) The prophage is replicated with the bacterial chromosome (1). Binary fission gives each daughter a chromosome containing the integrated sequence (1). (c) Induction can switch a prophage into lytic development (1), after which genome copying, assembly and lysis can produce free particles.

(a) Escape proposes cellular genetic material gaining independent transmission between hosts (1). A homologous cellular gene is compatible with a cellular connection (1). (b) Gene similarity can also follow later transfer or common ancestry; one gene cannot establish the history of all viruses (1). (c) Reduction proposes loss of cellular functions from parasites (1); pre-cellular replicator models propose earlier genetic entities later associated with cells (1). Present host dependence does not by itself resolve which route occurred.

(a) Reassortment packages RNA segments from different parental strains into progeny (1). Mutation changes the nucleotide sequence within genetic material rather than requiring a mixed parental segment set (1). (b) A changed surface protein may have altered antigenic sites (1), so antibodies generated against earlier variants may bind less effectively, complicating vaccine matching (1). This does not prove complete immune escape or pandemic potential.

(a) M directly provides a viral surface protein antigen (1). N provides mRNA instructions that host ribosomes translate into that protein (1). (b) The purified spike lacks the genome and particle needed for viral replication (1). The stated mRNA encodes only one protein, not the complete genome and structures for a replicating virion (1). Detailed immune-cell pathways are not required by the supplied model.

(a) Divergence is difference from the original reference; diversity is variation within the current population (1). They compare different pairs of sequences, so one may rise while the other falls (1). (b) Relative change is (0.4-1.6)/1.6\times100=-75\%: a 75% decrease (1). (c) Selection favouring a subset, or a bottleneck, could reduce within-population variation while retaining divergent descendants; these means cannot distinguish causes (1).

(a) X can make coat proteins but cannot supply newly copied genomes for packaging, so no complete progeny form (1). Y can make genome copies but not the new protein coats needed for complete particles (1). (b) Z allows both components and assembly (1); complete particles can therefore accumulate inside cells although none are released by lysis (1). (c) Release after removing Z is consistent with liberation of particles assembled earlier (1). Absence of extracellular particles alone cannot distinguish an assembly block from a release block; intracellular particle measurement supplies that distinction (1). Treat each stipulated block as selective, without inventing secondary effects.

(a) Three independent binary choices give 2^3=8 distinguishable combinations (1). Subtract the two all-one-parent sets to obtain 6 mixed combinations (1). (b) With the stated equal-choice assumption, 6/8\times100=75\% are mixed (1). Without equal packaging probabilities, the combination count alone would not determine that percentage (1). (c) Reassortment changes the combination of whole segments without requiring a new mutation within a segment (1). Efficient infection, host-to-host spread and immune effects require separate evidence; a mixed genome alone proves none of these (1).

(a) An insertion before the host split is at least 84 million years old (1). This assumes one ancestral insertion inherited by both lineages rather than independent insertion or later transfer (1). (b) Independent A-to-G changes produce convergence at this site (1); the shared endpoint alone cannot identify a recent common ancestor or a whole-genome relationship (1). (c) Inherited viral DNA documents a historical association (1). It does not identify the origin of every viral lineage; shared code, diverse genomes and extreme parasitism are compatible with more than one history (1).

(a) Treated descendants: susceptible 80\times1=80; resistant 20\times6=120 (1). Resistant fraction 120/200=60\%, up from 20% (1). (b) Drug-free descendants: susceptible 80\times4=320; resistant 120\times2=240 (1). Resistant fraction 240/560\times100=42.9\% to three significant figures (1). (c) The drug changes relative reproductive success, favouring an already-present resistant variant (1). Without drug, resistance carries the supplied reproductive disadvantage, so its frequency falls; this is selection under a changed environment rather than directed mutation (1).

(a) GGU-to-GGC leaves glycine unchanged because different codons can encode the same amino acid (1). GGU-to-GUU substitutes valine for glycine, changing primary structure (1). (b) Different side-chain interactions may alter folding or the surface antigen shape (1). Binding or immune-response measurements are needed because one substitution does not guarantee an altered epitope or loss of protection (1). (c) Host ribosomes translate the RNA into protein (1). A single-protein RNA lacks the complete viral genome and particle required to represent a replicating virion (1).

(a) With antibodies, Q is 80/(20+80)=80\% (1). Without them, Q is 60/(90+60)=40\% (1). (b) Altered antigen shape may reduce binding by antibodies elicited against P (1), so more Q particles remain able to infect and produce descendants (1). (c) Q has higher relative reproductive success in the antibody condition (1) but lower success in its absence, so the favoured variant depends on the environment (1). (d) These data do not show that antibody exposure generated the variant, nor do they demonstrate escape from every immune response; state either inference limit (1).

(a) Six doublings give 1\times2^6=64 bacteria (1); one prophage per chromosome gives 64 integrated copies at the stated sampling time (1). (b) 64/4=16 cells are induced, giving 16\times45=720 released phages (1). 64-16=48 intact lysogens remain (1). (c) Host DNA replication copies the integrated prophage (1), and binary fission distributes chromosomes to daughter bacteria (1). Integrated genome copies, cell numbers and released particles refer to different entities and times.