Biology HL

1. 1.

   Compare a virus particle with a bacterial cell in terms of genetic material, protein synthesis and the production of new particles or cells. Explain why a virus cannot multiply on a sterile nutrient medium without suitable host cells.

   [4 marks] · no calculator

   Worked explanation

   1. Do not treat the capsid as a bacterial cell wall enclosing a normal cytoplasm.
   2. A host provides translation and metabolic resources for component production.
   3. Sterile nutrient medium supports some bacteria but cannot replace a suitable living host for a virus.

   Common mistake: Viruses do not divide by binary fission or make proteins using their own ribosomes.

   Marking points

   • A virus contains DNA or RNA in a protein capsid, whereas a bacterium is a cell with DNA and cytoplasm.
   • Bacteria have ribosomes; viruses rely on host translation machinery.
   • Bacteria grow and divide; viruses produce components in hosts that assemble into particles.
   • Nutrients alone cannot supply the cellular machinery and coordinated metabolism needed for viral replication.

   Examiner tip: Viruses do not divide by binary fission or make proteins using their own ribosomes.

2. 2.

   After lambda bacteriophage infects a bacterium, one pathway rapidly produces phages and lyses the host; another integrates viral DNA into the host chromosome. Contrast the pathways and explain how the integrated viral DNA can persist through many bacterial generations before phage production resumes.

   [4 marks] · no calculator

   Worked explanation

   1. Separate immediate production of infectious particles from inheritance of an integrated DNA sequence.
   2. Host chromosome replication can maintain the prophage without repeated release of phages.
   3. A later switch to lytic expression explains delayed phage production and lysis.

   Common mistake: A lysogenic bacterium need not be releasing new phage particles continuously while the prophage is inherited.

   Marking points

   • The lytic pathway involves viral component synthesis, assembly and host lysis releasing phages.
   • In lysogeny, phage DNA integrates as a prophage rather than immediately lysing the cell.
   • Prophage DNA is copied with the bacterial chromosome and inherited by daughter cells.
   • Induction can switch the prophage into a lytic pathway; integration is not irreversible destruction of viral information.

   Examiner tip: A lysogenic bacterium need not be releasing new phage particles continuously while the prophage is inherited.

3. 3.

   In a hypothetical influenza population, a pre-existing surface-protein variant is less strongly bound by antibodies than the common variant. Explain how its frequency could rise in an immune host population and why this does not mean antibodies deliberately induced a useful mutation.

   [4 marks] · no calculator

   Worked explanation

   1. Begin with variation already present, as explicitly stated in the hypothetical scenario.
   2. Compare relative reproduction under immune selection rather than claiming an individual virus adapts intentionally.
   3. Escape may be partial and environment-dependent; the question does not establish that all immunity is lost.

   Common mistake: Selection changes frequencies of heritable variants; it does not instruct a virus to make a needed mutation.

   Marking points

   • Mutation and/or reassortment can generate heritable viral variation before selection.
   • The stated variant is less effectively neutralized by the existing antibodies.
   • It can leave more viral descendants in that environment, increasing its frequency.
   • Antibodies select among variants; they do not cause targeted mutations because a virus needs them.

   Examiner tip: Selection changes frequencies of heritable variants; it does not instruct a virus to make a needed mutation.

4. 4.

   In a hypothetical bacterial culture infected with lambda phage, some cells divide for many generations while carrying integrated phage DNA. After a DNA-damaging treatment, many cells produce phage particles and lyse. Distinguish the two stages and explain why phage DNA can be inherited before cell lysis occurs.

   [4 marks] · no calculator

   Worked explanation

   1. Track the fate of the viral genome separately from the appearance of free viral particles.
   2. During lysogeny, chromosome replication can pass the integrated sequence to descendants without immediate destruction of the host.
   3. The treatment is a proposed trigger for induction in this scenario; the later production and lysis identify the lytic phase.

   Common mistake: Do not call every infected bacterium immediately lytic or describe inherited prophage DNA as a newly released virus particle.

   Marking points

   • Integrated phage DNA persisting with host division is the lysogenic stage, with a prophage.
   • The prophage is copied with the bacterial chromosome and passed to daughter cells.
   • Induction can switch the system to a lytic cycle involving phage genome replication and particle assembly.
   • Lysis releases newly assembled phages; host division during lysogeny is not itself release by lysis.

   Examiner tip: Do not call every infected bacterium immediately lytic or describe inherited prophage DNA as a newly released virus particle.

5. 5.

   Two influenza strains infect the same host cell. Some offspring viruses contain genome segments from both strains. Explain how this can generate variation, distinguish it from a point mutation, and explain why it does not guarantee a successful new epidemic strain.

   [4 marks] · no calculator

   Worked explanation

   1. The supplied co-infection creates an opportunity to package segments from different parents.
   2. Reassortment recombines existing segments; it is not the same process as introducing a base substitution.
   3. Generation of variation and selection of a transmissible lineage are separate steps, so a new combination is not automatically an epidemic.

   Common mistake: Reassortment is not possible for every virus in the same way and does not mean every new combination is more dangerous.

   Marking points

   • Influenza has a segmented RNA genome.
   • Co-infection can allow reassortment of segments from the two parental strains.
   • A point mutation changes nucleotides within a sequence rather than exchanging whole genome segments.
   • The resulting combination must still replicate, transmit and compete in its environment; many combinations are not successful.

   Examiner tip: Reassortment is not possible for every virus in the same way and does not mean every new combination is more dangerous.

6. 6.

   Two hypothetical viruses contain RNA in protein capsids. E also buds through a host-cell membrane and retains a lipid envelope with proteins needed for entry; N has no envelope. A supplied lipid-disrupting treatment is applied to particles, then removed before they contact healthy host cells. E's RNA remains detectable, but its infectivity falls sharply. Explain the source of E's envelope and why retaining RNA need not preserve infectivity. Explain why this result alone cannot show that N is unaffected, and identify a control addressing host-cell damage.

   [5 marks] · no calculator

   Syllabus placement evidence: Biology guide, first assessment 2025, updated March 2024: A2.3.1-A2.3.2 AHL, printed pp.41-42 (PDF pp.47-48); Inquiry 1, Controlling variables, p.32. A2.3.2 explicitly distinguishes viruses with host-derived envelopes from non-enveloped viruses. The supplied treatment/entry scenario applies that prescribed AHL structural distinction and control design; disinfectant effectiveness is not unsupplied recall. RNA detection is not taken as proof of infectivity, and no practical work is performed.

   Worked explanation

   1. Separate the protein capsid from the additional host-derived lipid membrane. Removing or damaging the latter need not destroy all detectable genome fragments.
   2. Infectivity requires successful delivery and replication, not just a positive RNA signal. A processed treatment blank with an appropriate host-cell baseline tests residual host toxicity. Matched untreated-virus controls are also recommended for a broader infectivity comparison, but they are not an additional requirement for the requested host-damage-control mark.

   Common mistake: A viral envelope is not a bacterial cell wall, and a positive RNA assay is not proof of an infectious intact virus.

   Marking points

   • E acquires its lipid envelope from a host-cell membrane during budding.
   • Disrupting an envelope needed for entry can prevent effective host-cell infection even while RNA remains detectable.
   • Detection of RNA establishes the presence of the target sequence, not intact entry machinery or an infectious particle.
   • N lacks that envelope target, but other treatment effects on its capsid/genome or entry remain untested; compare treated and untreated N rather than declaring it unaffected.
   • Proposes uninfected host cells exposed to a matched processed treatment/removal blank, with viability assessed against an appropriate untreated or vehicle host-cell baseline, to check residual treatment toxicity. An untreated-virus infectivity control is not required for this host-damage-control mark.

   Examiner tip: A viral envelope is not a bacterial cell wall, and a positive RNA assay is not proof of an infectious intact virus.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.