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Unity and Diversity — Free Biology HL Practice Questions

1FoundationMCQReplication of viruses1 markPaper 1~2 min
A virus has an outer lipid envelope with spike proteins, a protein capsid, and a single-stranded RNA genome. A student claims this virus is released from its host cell by lysis rather than by budding. Which feature of this virus most directly contradicts the student's claim?
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2MasteryMCQReplication of viruses1 markPaper 1~2 min
A virologist isolates a bacteriophage that completes a lytic cycle in 30 minutes. When a chemical that specifically inhibits the phage-encoded DNA polymerase is added at the start of infection, no new phage particles are produced and no bacterial lysis occurs. Which of the following conclusions is best supported by these results?
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3FoundationMCQRole of water in living organisms1 markPaper 1~2 min
A researcher adds molecule X to a liver cell extract. Molecule X is small, non-polar, and dissolves readily in the phospholipid bilayer. Which mechanism best describes how molecule X crosses the plasma membrane?
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4MasteryMCQRole of water in living organisms1 markPaper 1~2 min
Giant kelp (Macrocystis pyrifera) cells maintain turgor pressure in seawater by actively transporting ions into the cytoplasm, lowering intracellular water potential below that of the surrounding ocean. Water therefore moves into the cells down its own water potential gradient. Which property of water is most directly responsible for enabling this osmotic mechanism?
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5FoundationMCQDomain-based classification1 markPaper 1~2 min
A rooted phylogenetic tree of the three domains is constructed using substitutions in the SSU rRNA gene. The root is placed on the branch leading to Bacteria, which has the greatest number of substitutions from the root. Archaea and Eukarya diverge from a node that is closer to the root than the node separating any other pair of domains. Which statement is supported by this tree?
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6MasteryMCQDomain-based classification1 markPaper 1~2 min
A sediment sample from a deep-sea hydrothermal vent contains DNA from three previously unknown prokaryotic species. Sequencing of their small subunit ribosomal RNA (SSU rRNA) genes shows that Species X shares 95% nucleotide similarity with Archaeoglobus fulgidus (domain Archaea), Species Y shares 88% similarity with Thermotoga maritima (domain Bacteria), and Species Z shares 78% similarity with both reference sequences. Using the three-domain system and the principle that SSU rRNA similarity reflects evolutionary relatedness, which conclusion is most strongly supported?
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7FoundationMCQThe three domains of life1 markPaper 1~2 min
A microorganism isolated from a hydrothermal vent at 85 °C is found to be prokaryotic, with cell walls lacking peptidoglycan, membrane lipids containing isoprenoid chains linked to glycerol by ether bonds, and an RNA polymerase with subunit complexity comparable to that of eukaryotes. To which domain does this organism belong?
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8MasteryMCQThe three domains of life1 markPaper 1~2 min
Carl Woese's 1977 analysis of ribosomal RNA (rRNA) nucleotide sequences revealed that methanogenic prokaryotes from anaerobic swamp sediments were as genetically distant from typical bacteria such as Escherichia coli as they were from eukaryotes such as humans. This finding led Woese to propose a three-domain classification system. Which of the following best explains why rRNA sequence comparison was the most appropriate tool for identifying such deep evolutionary divergences?
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9FoundationMCQMiller-Urey experiment1 markPaper 1~2 min
The Miller-Urey experiment circulated a mixture of methane, ammonia, hydrogen, and water vapour through electrical discharges for one week. Analysis of the resulting liquid revealed the presence of amino acids and other small organic molecules. Which statement most accurately describes the significance of this finding?
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10MasteryMCQEndosymbiotic theory1 markPaper 1~2 min
A biologist studies a unicellular eukaryote that lacks mitochondria but posses other membrane-bound organelles. She hypothesises that this organism diverged from the eukaryotic lineage before the endosymbiotic event that gave rise to mitochondria. Which observation, if true, would most directly challenge her hypothesis?
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11FoundationMCQSpeciation: Allopatric and sympatric1 markPaper 1~2 min
A river forms through a forest habitat, physically separating two populations of a small rodent species. Over many generations, the isolated populations diverge in fur colour and mating calls. Which statement most accurately describes the mechanism of speciation occurring here?
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12MasteryMCQSpeciation: Allopatric and sympatric1 markPaper 1~2 min
Two populations of cichlid fish undergo allopatric speciation after a land bridge divides their lake for 10,000 years. When the lake reconnects and individuals from both populations are placed together, the females of each population only respond to the courtship displays of males from their own population, so fertilisation never occurs. Which type of reproductive isolating mechanism does this represent?
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13FoundationMCQProkaryotic vs eukaryotic cells1 markPaper 1~2 min
A student observes a cell under a light microscope and notes the presence of a distinct nucleus. Which additional observation would provide the strongest evidence that this cell is eukaryotic?
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14MasteryMCQOrganelles and their functions1 markPaper 1~2 min
A mitochondrial gene mutation in a eukaryotic cell line causes a 60% reduction in ATP production. The mutation affects a multi-subunit protein complex embedded in the inner mitochondrial membrane that couples proton flow to nucleotide phosphorylation. Which process is most directly impaired by this mutation?
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15FoundationMCQHuman impact on ecosystems and biodiversity1 markPaper 1~2 min
In a coral reef ecosystem, an invasive predatory fish with no natural predators rapidly reduces the population of native herbivorous fish. Which of the following is the most likely long-term consequence for the reef?
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16MasteryMCQImportance of biodiversity1 markPaper 1~2 min
A fungal pathogen is introduced into a tropical rainforest, killing the dominant tree species. This tree provides fruit for 40 percent of bird species and nesting sites for 20 percent of mammal species in the ecosystem. Assuming this tree functions a keystone species, which outcome is most likely over the long term?
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17FoundationMCQHuman impact on ecosystems and biodiversity1 markPaper 1~2 min
A persistent pesticide is applied to a wheat field to control aphids. The pesticide is lipid-soluble and is not metabolised or excreted by organisms. The food chain the field is: wheataphidsladybirdssparrowhawks\text{wheat} \rightarrow \text{aphids} \rightarrow \text{ladybirds} \rightarrow \text{sparrowhawks} Which organism will have the highest concentration of pesticide per gram of body mass?
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18FoundationMCQStructure of DNA and RNA1 markPaper 1~2 min
A student analyses a segment of DNA. One strand has the sequence 5-ATGC-35'\text{-ATGC-}3'. Which of the following correctly identifies both the complementary strand the total number of hydrogen bonds formed between the two strands?
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19MasteryMCQStructure of DNA and RNA1 markPaper 1~2 min
In the Hershey–Chase experiment, two batches of bacteriophages were prepared: one labelled with 35^{35}S (incorporated into protein) and one labelled with 32^{32}P (incorporated into DNA). Each batch was used to infect E. coli, and after agitation and centrifugation, the radioactivity of the bacterial cell pellet was measured. The pellet showed high 32^{32}P activity but negligible 35^{35}S activity. Which conclusion is directly supported by this observation?
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20FoundationMCQStructure of DNA and RNA1 markPaper 1~2 min
The sugar in RNA differs from the sugar in DNA by the presence of a hydroxyl group on the 2' carbon. Which statement correctly describes this structural difference and its consequence?
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21MasterySAQ-SReplication of viruses2 marksPaper 2~3 min
Influenza A virus is an enveloped virus with a segmented single-stranded RNA genome. The virus binds to sialic acid receptors on host respiratory epithelial cells via its haemagglutinin (HA) glycoprotein. After endocytosis, the low pH of the endosome triggers fusion of the viral envelope with the endosomal membrane, releasing the viral RNA segments into the cytoplasm. The viral RNA-dependent RNA polymerase (RdRp) replicates the viral genome and transcribes mRNA. New viral particles assemble at the host cell plasma membrane, where neuraminidase (NA) cleaves sialic acid to release progeny virions. A researcher treats infected cell cultures with oseltamivir, a competitive inhibitor of neuraminidase. After 24 hours, the number of viral particles released into the culture medium and the number remaining attached to the host cell surface are measured.
(a)
State the role of the viral envelope during replication of Influenza A, with reference to events in the endosome. [1 mark]
(b)
In an oseltamivir-treated culture, 2.4×1062.4 \times 10^{6} virions are released into the medium and 1.6×1061.6 \times 10^{6} virions remain attached to cells. Calculate the percentage of total virions that remain attached to the host cell surface. [1 mark]
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22MasterySAQ-SReplication of viruses5 marksPaper 2~8 min
A laboratory culture of Escherichia coli is infected with bacteriophage T4. The phage injects its double-stranded DNA into the host cell, which then replicates the phage genome and synthesises phage proteins. New phage particles are assembled, and the host cell is lysed by phage-encoded lysozyme, releasing approximately 200 progeny phages per infected cell. This the lytic cycle. The culture initially contains 5.0×1085.0 \times 10^{8} bacterial cells and is infected at a multiplicity of infection (MOI) of 0.10.1 (1 phage per 10 bacterial cells).
(a)
State the role of lysozyme in the T4 lytic cycle. [1 mark]
(b)
Calculate the total number of progeny phages produced after one complete lytic cycle, assuming all infected cells lyse successfully. [2 marks]
(c)
Some T4-related phages can enter a lysogenic cycle instead of a lytic cycle. Evaluate the consequences of lysogeny versus the lytic cycle for both phage survival and host cell fate under conditions of nutrient limitation. [2 marks]
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23ChallengeLAQReplication of viruses10 marksPaper 2~15 min
A culture of CD4+\text{CD4}^+ T-cells was infected with HIV at time =0h= 0\,\text{h}. The concentration of free virions in the culture medium was measured every 6h6\,\text{h} for 48h48\,\text{h}. At t=12ht = 12\,\text{h} the culture was divided into three treatment groups: - Group A: no drug (control) - Group B: treated with zidovudine (AZT), a nucleoside analogue that inhibits reverse transcriptase - Group C: treated with raltegravir, an inhibitor of viral integrase Viral concentrations (thousands of virions mL1\text{mL}^{-1}) are shown below. Time (h) — 0 — 6 — 12 — 18 — 24 — 36 — 48 Group A — 0.0 — 0.5 — 2.0 — 8.0 — 32.0 — 512.0 — 8192.0 Group B — 0.0 — 0.5 — 2.0 — 2.1 — 2.2 — 2.4 — 2.6 Group C — 0.0 — 0.5 — 2.0 — 1.9 — 1.7 — 1.4 — 1.1
(a)
Calculate the doubling time of the viral population in Group A between 6h6\,\text{h} and 12h12\,\text{h}, and between 12h12\,\text{h} and 18h18\,\text{h}[2 marks]
(b)
Explain the difference between the two doubling times calculated in (a) with reference to the HIV replication cycle. [2 marks]
(c)
Analyse the effect of AZT (Group B) and raltegravir (Group C) on viral concentration over the 48h48\,\text{h} period, identifying one similarity and one difference between the two treatments. [2 marks]
(d)
Evaluate the relative effectiveness of AZT and raltegravir inhibiting HIV replication, using data from the table to support your answer. [2] (e) A patient is treated with a combination of AZT and raltegravir from the time of infection. Predict the likely viral concentration trend over 48h48\,\text{h} and justify your prediction by discussing the mechanisms of both drugs and the concept of viral resistance. [2 marks]
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24MasterySAQ-SRole of water in living organisms5 marksPaper 2~8 min
The Namib Desert beetle (Stenocara gracilipes) survives in one of Earth's driest environments by harvesting water from coastal fog. The beetle's wing surface has alternating hydrophobic (waxy) regions and hydrophilic bumps rich in hydroxyl (OH-\text{OH}) groups. When fog rolls in, water droplets condense on the hydrophilic bumps. The beetle tilts its body forward and water rolls down the wing into its mouth.
(a)
Describe how the property of cohesion in water contributes to the movement of water droplets down the beetle's wing. [2 marks]
(b)
Explain how the hydroxyl (OH-\text{OH}) groups on the hydrophilic bumps promote condensation of water vapour onto the wing surface. [2 marks]
(c)
The beetle collects 15 spherical droplets, each with a radius of 1.2mm1.2\,\text{mm}. Show that the total volume of water collected is approximately 109mm3109\,\text{mm}^3. Vsphere=43πr3V_{\text{sphere}} = \frac{4}{3}\pi r^3 [1 mark]
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25MasterySAQ-SRole of water in living organisms5 marksPaper 2~8 min
The vascular system of a giant sequoia (Sequoiadendron giganteum) transports water from roots to leaves over 80m80\,\text{m} high. Xylem vessels are composed of dead cells with heavily lignified walls. Transpiration from leaf surfaces creates a negative pressure (tension) that pulls water upward through the xylem.
(a)
Describe how adhesion contributes to water transport in xylem vessels. [2 marks]
(b)
Describe how the cohesion-tension theory explains the upward movement of water from roots to leaves. [2 marks]
(c)
The tension measured in xylem vessels near the top of a giant sequoia is approximately 1.8MPa-1.8\,\text{MPa}, compared with 0.3MPa-0.3\,\text{MPa} near the base. Evaluate whether the cohesion-tension mechanism alone is sufficient to account for water transport to a height of 80m80\,\text{m}, given that a water column of 10m10\,\text{m} requires approximately 0.1MPa0.1\,\text{MPa} to lift against gravity. [1 mark]
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26ChallengeLAQRole of water in living organisms10 marksPaper 2~15 min

Data

- Cytoplasm is approximately 70%70\% water by volume, with a viscosity 55 times greater than pure water - Diffusion coefficient of glucose in pure water at 37°C37°\text{C}: Dglucose=6.7×1010m2s1D_{\text{glucose}} = 6.7 \times 10^{-10}\,\text{m}^2\,\text{s}^{-1} - Diffusion coefficient of oxygen in pure water at 37°C37°\text{C}: DO2=2.1×109m2s1D_{\text{O}_2} = 2.1 \times 10^{-9}\,\text{m}^2\,\text{s}^{-1} - Glucose must diffuse from the cell membrane to an enzyme complex 2.0μm2.0\,\mu\text{m} away - Diffusion time may be estimated using t=x22Dt = \dfrac{x^2}{2D}, where xx is distance and DD is diffusion coefficient - Stokes–Einstein relation: D1ηD \propto \dfrac{1}{\eta}, where η\eta is viscosity
Researchers investigate the role of water as a solvent in the cytoplasm of Escherichia coli using a theoretical diffusion model.
(a)
Calculate the time for glucose to diffuse 2.0μm2.0\,\mu\text{m} (i) in pure water and
(ii) in the cytoplasm. [3 marks]
(b)
Explain, with reference to molecular size and polarity, why oxygen has a higher diffusion coefficient than glucose in pure water. [2 marks]
(c)
A student claims: "Water's polarity makes it a universal solvent for all biological molecules." Using the data provided and your knowledge of water chemistry, evaluate this claim. In your answer, consider both polar and non-polar solutes and the biological consequences for E. coli of any limitation you identify. [5 marks]
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27MasterySAQ-SDomain-based classification6 marksPaper 2~9 min
A phylogenetic tree was constructed using small subunit ribosomal RNA (SSU rRNA) sequences from three domains of life. The tree is rooted at a trifurcation point, with three branches of approximately equal length diverging simultaneously to Domain Bacteria (E. coli, 1542 nucleotides), Domain Archaea (Methanococcus, 1430 nucleotides), and Domain Eukarya (Homo sapiens, 1869 nucleotides). Within Domain Eukarya, the branch to Saccharomyces cerevisiae is longer than the branch to Homo sapiens.
(a)
State the domain that is the sister group to Domain Eukarya, based on this tree. [1 mark]
(b)
The SSU rRNA gene of E. coli contains 1542 nucleotides and that of Methanococcus contains 1430 nucleotides. Both genes are double-stranded. Calculate the total number of phosphodiester bonds present in both genes combined. [2 marks]
(c)
The branch lengths from the trifurcation point to each domain are approximately equal. Evaluate whether equal branch lengths alone are sufficient evidence to conclude that all three domains have evolved at the same rate since their divergence. [3 marks]
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28MasterySAQ-SEvolutionary relationships and cladistics5 marksPaper 2~8 min
A cladogram for five mammal species (platypus, kangaroo, bat, whale, and human) is constructed using three shared derived characters:
(1) mammary glands,
(2) placenta,
(3) echolocation. Presence is indicated by 11, absence by 00. Species — Mammary glands — Placenta — Echolocation Platypus — 1 — 0 Kangaroo — 1 — 0 Bat — 1 — 1 Whale — 1 — 0 Human — 1 — 0
(a)
State why the character 'mammary glands' cannot be used to determine branching order among these five species. [1 mark]
(b)
A cladogram groups bat, whale, and human into a monophyletic clade, with platypus and kangaroo as successive outgroups. Assuming no character reversals, determine the minimum number of evolutionary changes required for each of the two characters — placenta and echolocation — and state the total number of changes. [3 marks]
(c)
Explain why echolocation being an autapomorphy of bat means it cannot be used to unite bat with any other species in this cladogram into a clade. [1 mark]
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29ChallengeLAQDomain-based classification10 marksPaper 2~15 min

Data

- DNA sequence comparisons of the small subunit ribosomal RNA (SSU rRNA) gene from the deep-sea hydrothermal vent archaeon Thermococcus celer show 85%85\% sequence similarity with Archaea and only 65%65\% similarity with Bacteria. - A new, uncultured microbial lineage discovered in a subglacial Antarctic lake has membrane lipids composed of ester-linked fatty acids (characteristic of Bacteria and Eukarya), but its SSU rRNA gene sequences cluster phylogenetically with Archaea. - The genome of Methanopyrus kandleri, an archaeon, contains a complete set of genes encoding histones, previously considered exclusive to Eukarya.
The three-domain system (Bacteria, Archaea, Eukarya) is based on fundamental molecular and cellular differences. However, discoveries of organisms with mixed characteristics challenge the assumption of monophyly — a group containing a common ancestor and all its descendants — for each domain.
(a)
(i) Analyse how the SSU rRNA sequence data for Thermococcus celer support the monophyly of a combined Archaea–Eukarya clade within the three-domain system. [2]
(ii) Analyse how the membrane lipid composition of the Antarctic microbe challenges the monophyly of the domain Archaea. [3 marks]
(b)
Evaluate the view that 'mix-and-match' organisms, such as the Antarctic microbe and Methanopyrus kandleri, require a fundamental revision of the domain-based classification system, referring to all three data sets provided. [5 marks]
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30MasterySAQ-SThe three domains of life5 marksPaper 2~8 min
A simplified phylogenetic tree was constructed using rRNA gene sequences from three organisms: a methanogen (Methanococcus), a cyanobacterium (Anabaena), and a human (Homo sapiens). The tree indicates that the methanogen and human share a more recent common ancestor than either shares with the cyanobacterium. The methanogen's rRNA gene sequence is 85%85\% similar to the human sequence; the cyanobacterium's rRNA gene sequence is 60%60\% similar to the human sequence.
(a)
State the domain to which Methanococcus belongs. [1 mark]
(b)
Describe two differences between the cell membrane structure of Methanococcus and that of Anabaena. [2 marks]
(c)
Using the rRNA similarity data and the phylogenetic tree, evaluate whether rRNA sequence analysis a reliable method for determining evolutionary relationships between these three organisms. [2 marks]
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31MasterySAQ-SThe three domains of life5 marksPaper 2~8 min
A phylogenetic tree of the three domains of life is constructed using rRNA sequence comparisons. Branch lengths are proportional to genetic distance, measured in substitutions per nucleotide site. The root is placed between Bacteria and the Archaea–Eukarya clade. - Branch length, Bacteria: 0.200.20 substitutions per nucleotide site - Branch length, Archaea: 0.120.12 substitutions per nucleotide site - Branch length, Eukarya: 0.050.05 substitutions per nucleotide site
(a)
State which two domains share the most recent common ancestor, as shown by the tree. [1 mark]
(b)
Calculate how many times greater the substitution rate in Bacteria is compared to Eukarya, assuming a constant molecular clock and that all branches diverged at the same time. [2 marks]
(c)
Evaluate one limitation of using rRNA sequence comparisons alone to determine the evolutionary relationships among the three domains. [2 marks]
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32ChallengeLAQThe three domains of life10 marksPaper 2~15 min
The three-domain classification system proposes that all life is divided into Bacteria, Archaea, and Eukarya.
(a)
State two structural differences between the cell walls of Archaea and Bacteria. [2 marks]
(b)
The table below shows the percentage similarity of small subunit ribosomal RNA (SSU rRNA) gene sequences between selected organisms. Explain how these data support the three-domain classification system. Organism pair — SSU rRNA sequence similarity Escherichia coli (Bacteria) vs. Methanococcus jannaschii (Archaea) — 68%68\% Escherichia coli (Bacteria) vs. Saccharomyces cerevisiae (Eukarya) — 65%65\% Methanococcus jannaschii (Archaea) vs. Saccharomyces cerevisiae (Eukarya) — 78%78\% Homo sapiens (Eukarya) vs. Saccharomyces cerevisiae (Eukarya) — 85%85\% Homo sapiens (Eukarya) vs. Methanococcus jannaschii (Archaea) — 74%74\% [3 marks]
(c)
The eocyte hypothesis proposes that Eukarya evolved from within the Archaea through a symbiotic merger with a bacterium, rather than as a separate domain diverging from a common ancestor of Archaea and Bacteria. Using the data in (b) and your knowledge of eukaryotic cellular structures, evaluate the extent to which the available molecular and structural evidence supports the eocyte hypothesis over a strict three-domain model. [5 marks]
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33MasterySAQ-SEndosymbiotic theory9 marksPaper 2~14 min
The endosymbiotic theory proposes that mitochondria originated from free-living prokaryotes that were engulfed by a host cell.
(a)
State the type of prokaryote that is believed to have given rise to mitochondria. [1 mark]
(b)
A mitochondrion appears an image of length 15mm15\,\text{mm} at a magnification of ×5000\times 5000. Calculate the actual length of the mitochondrion in micrometres (μm\mu\text{m}). [2 marks]
(c)
Explain how the double membrane of mitochondria provides evidence for the endosymbiotic theory. [2 marks]
(d)
Evaluate the strength of the evidence for the endosymbiotic theory by discussing two additional features of mitochondria, other than the double membrane, that support this theory. [4 marks]
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34MasterySAQ-SEndosymbiotic theory6 marksPaper 2~9 min
Endosymbiotic theory proposes that mitochondria and chloroplasts evolved from free-living prokaryotes that were engulfed by a host eukaryotic cell.
(a)
State two features of mitochondrial DNA that are similar to prokaryotic DNA. [2 marks]
(b)
A chloroplast genome is found to be 120kb120\,\text{kb} in length and replicates independently of the nuclear genome. Deduce how independent replication of the chloroplast genome supports the endosymbiotic theory. [2 marks]
(c)
Explain how the presence of 70S70\text{S} ribosomes in chloroplasts provides evidence for the endosymbiotic theory. [2 marks]
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35ChallengeLAQOrigin of eukaryotic cells10 marksPaper 2~15 min
The endosymbiotic theory proposes that mitochondria and chloroplasts originated from free-living prokaryotes engulfed by a host cell. The following data were obtained from comparative genomic studies. Parameter — Value Genome size of free-living α\alpha-proteobacterium (ancestral to mitochondria) — 4.0×106bp\approx 4.0 \times 10^{6}\,\text{bp} Genome size of modern human mitochondrial DNA (mtDNA) — 1.66×104bp\approx 1.66 \times 10^{4}\,\text{bp} Genome size of free-living cyanobacterium (ancestral to chloroplasts) — 3.5×106bp\approx 3.5 \times 10^{6}\,\text{bp} Genome size of modern chloroplast DNA (cpDNA) in Arabidopsis thaliana — 1.54×105bp\approx 1.54 \times 10^{5}\,\text{bp} Protein-coding genes in free-living α\alpha-proteobacterium — 3,000\approx 3{,}000 Protein-coding genes in modern human mtDNA — 3737 Mitochondrial proteins in humans (nuclear-encoded) — 1,100\approx 1{,}100
(a)
Using only the data in the table, analyse the evidence that supports the endosymbiotic theory. [4 marks]
(b)
Explain the discrepancy between the number of protein-coding genes retained in modern human mtDNA and the total number of mitochondrial proteins found in human cells. [2 marks]
(c)
Evaluate the strengths and limitations of the endosymbiotic theory in explaining the origin of eukaryotic cells, with reference to both mitochondria and chloroplasts. [4 marks]
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36MasterySAQ-SSpeciation: Allopatric and sympatric6 marksPaper 2~9 min
The three-spined stickleback (Gasterosteus aculeatus) is a small fish found in both marine and freshwater environments in the Northern Hemisphere. After the last Ice Age (approximately 10,000 years ago), many marine stickleback populations became trapped in newly formed freshwater lakes and streams. These isolated populations have since diverged into distinct freshwater forms. Data from a study of two stickleback populations in British Columbia, Canada, are shown below: - Population A: Found in a large, deep lake (Loon Lake). Males build nests in open, sandy areas. Breeding occurs in June. Average number of lateral bony plates =32= 32. - Population B: Found in a small, shallow stream (Beaver Creek) that flows into Loon Lake. Males build nests under rocks in fast-flowing water. Breeding occurs in May. Average number of lateral bony plates =8= 8.
(a)
State the process that initially isolated these stickleback populations from the ancestral marine population. [1 mark]
(b)
Explain how differences in nesting site and breeding time could each contribute to reproductive isolation between Population A and Population B. [2 marks]
(c)
Evaluate the claim that the difference in lateral plate number between the two populations is entirely due to genetic drift, using evidence from the data and your knowledge of evolutionary mechanisms. [3 marks]
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37MasterySAQ-SSpeciation: Allopatric and sympatric6 marksPaper 2~9 min
The apple maggot fly (Rhagoletis pomonella) originally infested only hawthorn (Crataegus) fruits in North America. Around 1850, a population began to infest introduced apple (Malus domestica) fruits. Today, the two host races are genetically distinct and show partial reproductive isolation. Mating occurs on the host fruit on which the female lays her eggs. In a laboratory experiment, female flies from the apple race were given a free choice of fruits on which to lay eggs. Results: - Apple fruits: 7878 eggs laid - Hawthorn fruits: 2222 eggs laid
(a)
State the type of speciation occurring in Rhagoletis pomonella. [1 mark]
(b)
Calculate the percentage of eggs laid on apple fruits by the apple-race females. [1 mark]
(c)
Using your answer to (b) and the information provided, explain how host-fruit preference contributes to reproductive isolation between the two races. [2 marks]
(d)
Evaluate the extent to which natural selection alone could drive complete speciation in Rhagoletis pomonella, given that the two races remain the same geographic area and gene flow has not been entirely eliminated. [2 marks]
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38ChallengeLAQMechanisms of evolution: Natural selection, genetic drift, gene flow10 marksPaper 2~15 min

Data

- Before drought (1976): mean beak depth =9.2mm= 9.2\,\text{mm}, standard deviation =1.1mm= 1.1\,\text{mm}, population size N1200N \approx 1200 - After drought (1978): mean beak depth =10.4mm= 10.4\,\text{mm}, standard deviation =0.9mm= 0.9\,\text{mm}, population size N180N \approx 180 - Heritability (h2h^2) of beak depth =0.65= 0.65 - Effective population size during drought: Ne=200300N_e = 200\text{–}300 - A second island population (Isla Santa Cruz) maintained mean beak depth =9.3mm= 9.3\,\text{mm} over the same period with no drought
The Grant family's long-term study of Darwin's finches (Geospiza spp.) on Daphne Major island in the Galápagos has provided direct evidence for natural selection in the wild. In 1977, a severe drought reduced seed availability. Small, soft seeds were rapidly depleted, leaving primarily large, hard seeds. Researchers measured beak depth in the medium ground finch (Geospiza fortis) population before and after the drought.
(a)
Calculate the response to selection (RR) for beak depth, using the breeder's equation R=h2×SR = h^2 \times S, where the selection differential SS is the difference between the mean beak depth of surviving parents and the mean of the original population. Assume the mean beak depth of surviving parents equals the 1978 mean. [2 marks]
(b)
Explain how phenotypic variation, heritability, and differential survival each contributed to the observed shift in mean beak depth between 1976 and 1978. [3 marks]
(c)
Using the data provided, evaluate whether natural selection or genetic drift was the dominant mechanism driving the change in beak depth in the G. fortis population Daphne Major during the drought. [3 marks]
(d)
A researcher proposes using FSTF_{ST} to determine whether the beak-depth allele frequency difference between Daphne Major and Isla Santa Cruz after the drought reflects local adaptation or genetic drift. The total heterozygosity across both populations is HT=0.48H_T = 0.48 and the average within-population heterozygosity is HS=0.31H_S = 0.31. Calculate FSTF_{ST} and deduce what this value indicates about the relative roles of selection and drift in differentiating the two populations. [2 marks]
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39MasterySAQ-SRole of ribosomes, mitochondria, and chloroplasts7 marksPaper 2~11 min
A researcher compares the efficiency of mitochondria from different tissues. Liver mitochondria consume 1.8×1016mol1.8 \times 10^{-16}\,\text{mol} of O2\text{O}_2 per second when respiring with excess substrate. Each liver mitochondrion contains 5000 copies of cytochrome c oxidase, the terminal enzyme of the electron transport chain.
(a)
State the final electron acceptor in the electron transport chain. [1 mark]
(b)
Explain how the electron transport chain establishes a proton gradient across the inner mitochondrial membrane. [2 marks]
(c)
Calculate the number of O2\text{O}_2 molecules consumed per cytochrome c oxidase molecule per second. (Avogadro's constant =6.02×1023mol1= 6.02 \times 10^{23}\,\text{mol}^{-1}.) Give your answer to 2 significant figures. [2 marks]
(d)
The researcher finds that muscle mitochondria consume O2\text{O}_2 at three times the rate of liver mitochondria but contain the same number of cytochrome c oxidase molecules. Deduce what this difference indicates about the rate-limiting step of oxidative phosphorylation in muscle mitochondria compared with liver mitochondria. [2 marks]
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Solutions

40MasterySAQ-SRole of ribosomes, mitochondria, and chloroplasts6 marksPaper 2~9 min
A chloroplast in a leaf cell fixes 1.2×1016mol1.2 \times 10^{-16}\,\text{mol} of CO2\text{CO}_2 per second during peak sunlight. The Calvin cycle requires 3 molecules of ATP and 2 molecules of NADPH to fix one molecule of CO2\text{CO}_2.
(a)
State the location within the chloroplast where the Calvin cycle occurs. [1 mark]
(b)
Explain why the Calvin cycle is described as a cyclic process, referring to RuBP. [2 marks]
(c)
Calculate the rate of ATP consumption in the chloroplast in mols1\text{mol}\,\text{s}^{-1}. Give your answer in standard form to 2 significant figures. [1 mark]
(d)
Determine the ratio of ATP to NADPH consumed per molecule of CO2\text{CO}_2 fixed, and explain why a higher proportion of ATP than NADPH is required in the Calvin cycle. [2 marks]
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41ChallengeLAQOrganelles and their functions10 marksPaper 2~15 min

Data

- Healthy cell: 3232 ATP molecules produced per glucose molecule during complete aerobic respiration - Patient cell: 88 ATP molecules produced per glucose molecule - Oxygen consumption rate in patient cells: 15%15\% of the rate in healthy cells - Electron transport chain (ETC) complexes show normal activity when assayed in isolated inner mitochondrial membrane preparations from patient cells
A research team investigates a rare mitochondrial disorder in human fibroblasts. Cells from a patient show normal rates of glycolysis but significantly reduced overall ATP production. Electron microscopy reveals sparse, disorganised cristae and a less dense matrix compared with healthy cells.
(a)
Explain how sparse cristae could account for the reduced ATP yield observed in the patient cells. [4 marks]
(b)
The following two hypotheses have been proposed to explain the disorder. - Hypothesis 1: A mutation in the gene encoding ATP synthase reduces its catalytic activity. - Hypothesis 2: A defect in the mitochondrial inner membrane transporter reduces import of pyruvate into the matrix. Using all of the data provided, evaluate which hypothesis better supported. [6 marks]
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Solutions

42MasterySAQ-SHuman impact on ecosystems and biodiversity5 marksPaper 2~8 min
A population of sea turtles nests on a beach undergoing tourism development. The bar chart shows the number of successful nests (eggs hatched) from 2015 to 2025.
(a)
Describe the trend shown in the bar chart from 2015 to 2025. [1 mark]
(b)
Explain one human activity that could account for the decline in nesting success. [2 marks]
(c)
Calculate the average rate of decrease in successful nests per year over the period 2015 to 2025. [2 marks]
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Solutions

43MasterySAQ-SHuman impact on ecosystems and biodiversity9 marksPaper 2~14 min
A factory releases sulfur dioxide (SO2\text{SO}_2) into the atmosphere, contributing to acid rain. The table shows the pH of rainwater collected at four sites at increasing distances from the factory. Site — Distance from factory / km — Rainwater pH A — 2 — 4.2 B — 5 — 4.5 C — 10 — 5.0 D — 20 — 5.6 Normal, unpolluted rainwater has a pH of approximately 5.6.
(a)
State the relationship between distance from the factory and rainwater pH. [1 mark]
(b)
Explain one consequence of acid rain for biodiversity in a lake ecosystem. [2 marks]
(c)
The pH scale is logarithmic; each whole-number change represents a tenfold change in hydrogen ion concentration [H+][\text{H}^+]. Calculate how many times more acidic the rainwater at Site A is compared to Site D. [2 marks]
(d)
Evaluate the use of rainwater pH alone as an indicator of the impact of SO2\text{SO}_2 pollution a lake ecosystem. [4 marks]
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Solutions

44MasterySAQ-SGene expression regulation7 marksPaper 2~11 min
The bacterium E. coli can use either glucose or lactose as a carbon source. When both sugars are present, the lac operon is repressed until glucose is depleted. The graph below shows the relative intracellular lactose concentration over time after both sugars are added simultaneously at t=0t = 0.
(a)
State the role of the lac repressor protein regulating the lac operon when lactose is absent. [1 mark]
(b)
Explain why intracellular lactose concentration remains low during the first 20 minutes, even though lactose is present in the growth medium. [2 marks]
(c)
Calculate the average rate of increase of intracellular lactose concentration between t=20mint = 20\,\text{min} and t=40mint = 40\,\text{min}. Give your answer in arbitrary units per minute. [2 marks]
(d)
The plateau at 9.09.0 arbitrary units is reached at t=40mint = 40\,\text{min} and maintained thereafter. Deduce what this plateau indicates about the balance between lactose uptake and lactose metabolism in E. coli after t=40mint = 40\,\text{min}[2 marks]
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Solutions

45MasterySAQ-SDNA replication and transcription8 marksPaper 2~12 min
A researcher studies transcription of a gene in a eukaryotic cell. The gene is 12001200 base pairs long. RNA polymerase II moves along the template strand at a rate of 5050 nucleotides per second. - The gene contains 44 introns, each 150150 base pairs long - Average mass of one ribonucleotide in mRNA: 5.0×1022g5.0 \times 10^{-22}\,\text{g}
(a)
State the role of RNA polymerase during transcription. [1 mark]
(b)
Calculate the time, in seconds, taken for RNA polymerase II to transcribe the entire gene. [2 marks]
(c)
Calculate the mass, in grams, of the mature mRNA molecule after splicing. [2 marks]
(d)
Explain how RNA splicing produces a functional mRNA molecule from the pre-mRNA transcript, and suggest one advantage this process provides to the cell. [3 marks]
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Solutions