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Continuity and Change — Free Biology HL Practice Questions

1FoundationMCQOperon model (lac operon)1 markPaper 1~2 min
E. coli are grown in a medium containing only glucose as the sole carbon source. β-galactosidase activity is detected at very low levels. The bacteria are then transferred to a medium containing only lactose, and after 30 minutes β-galactosidase activity increases dramatically. Which statement best explains the low β-galactosidase activity in the glucose-only medium?
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2MasteryMCQRegulatory mechanisms of gene expression1 markPaper 1~2 min
A mutant strain of Escherichia coli carries a complete deletion of the promoter region of the lac operon, preventing RNA polymerase from binding. The mutant is grown in a medium containing only lactose as the sole carbon source. Which of the following best predicts the outcome for expression of lacZ, lacY, and lacA?
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3FoundationMCQMechanisms of temperature, pH, and water balance1 markPaper 1~2 min
A freshwater amoeba is transferred from its natural pond water into a beaker of distilled water. After 10 minutes, which statement correctly describes the contractile vacuole and explains its state?
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4MasteryMCQRegulation of blood glucose levels1 markPaper 1~2 min
A healthy volunteer drinks a solution containing 75 g of glucose after an overnight fast. Her blood glucose concentration peaks at 30 minutes and returns to the fasting level by 120 minutes. Which statement best explains the mechanism responsible for this return to the fasting blood glucose concentration?
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5MasteryMCQRegulation of blood glucose levels1 markPaper 1~2 min
A researcher investigates how insulin affects glucose uptake by liver cells in culture. Insulin is added to one dish of liver cells, and the glucose concentration in the culture medium decreases rapidly. The experiment is repeated with a chemical that blocks vesicle formation; the glucose concentration in the medium remains constant throughout. Which explanation most directly accounts for the result in the second experiment?
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6FoundationMCQAsexual vs sexual reproduction1 markPaper 1~2 min
A fungal disease spreads rapidly through a wild strawberry population. Some plants reproduce asexually via stolons, producing genetically identical offspring; others reproduce sexually, producing seeds dispersed by animals. Which statement best explains why sexually reproducing plants are more likely to have some survivors after the disease outbreak?
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7MasteryMCQSexual cycles in animals and plants1 markPaper 1~2 min
Pollen grains of Lilium each contain cells with 12 chromosomes. During double fertilization, one sperm nucleus fuses with the egg cell and a second sperm nucleus fuses with the two polar nuclei to form the endosperm. What is the chromosome number in an endosperm cell?
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8FoundationMCQAsexual vs sexual reproduction1 markPaper 1~2 min
A farmer grows a field of potato plants that are all clones of a single genotype. A new aphid species arrives and feeds on the leaves; nearly all plants die, but a few survive and produce small tubers. The farmer replants only tubers from the surviving plants the following season. What is the most likely outcome for the following season's crop?
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9FoundationMCQMitosis vs meiosis1 markPaper 1~2 min
A species of grasshopper has a diploid chromosome number of 24. A cell is observed in prophase I of meiosis. How many chromatids are present in this cell?
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10MasteryMCQMitosis vs meiosis1 markPaper 1~2 min
A drug prevents spindle fibre formation in human skin cells undergoing mitosis. The spindle assembly checkpoint monitors whether all chromosomes are correctly attached to spindle fibres before division proceeds. At which stage would these cells most likely become arrested?
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11FoundationMCQMitosis vs meiosis1 markPaper 1~2 min
A student examines a micrograph of a cell from the root tip of an onion. Individual chromosomes, each consisting of two sister chromatids, are aligned at the cell equator with no evidence of homologous pairing. Which stage does this cell most likely represent?
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12FoundationMCQRenewable resources and ecological conservation1 markPaper 1~2 min
A coastal mangrove ecosystem is assed for its ecosystem services. Which of the following services provided by mangroves is correctly classified as a regulating service?
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13MasteryMCQSustainable agricultural practices1 markPaper 1~2 min
A farmer uses a sustainable intercropping system combining legumes with fruit trees, permanent soil cover from crop residues, and contour ploughing. After three years, nitrogen fertilizer inputs have been reduced by 40% compared to conventional monoculture. Which of the following best explains this reduction in nitrogen fertilizer requirement?
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14FoundationMCQRenewable resources and ecological conservation1 markPaper 1~2 min
A conservation team is restoring a degraded grassland where a native grazer was removed 50 years ago. Monitoring shows that a single fast-growing grass species now dominates, suppressing most other plant species. The team predicts that reintroducing the grazer will increase plant species diversity. Which ecological principle best supports this prediction?
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15FoundationMCQMendelian inheritance1 markPaper 1~2 min
In Drosophila, red eye colour (allele XRX^R) is dominant over white eye colour (allele XrX^r). A female of genotype XRXrX^R X^r is crossed with a male of genotype XrYX^r Y. What is the probability that a randomly selected male offspring has white eyes?
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16MasteryMCQMendelian inheritance1 markPaper 1~2 min
An F1 plant from a cross between true-breeding yellow round (YYRR) and true-breeding green wrinkled (yyrr) pea plants is crossed with a homozygous recessive (yyrr) plant. Assuming independent assortment, what is the expected phenotypic ratio offspring from this test cross?
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17FoundationMCQMendelian inheritance1 markPaper 1~2 min
A pea plant heterozygous for both seed shape and seed colour (RrYyRrYy) is self-pollinated. Round (RR) is dominant over wrinkled (rr), and yellow (YY) is dominant over green (yy). What fraction of the offspring is expected to have round, green seeds?
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18FoundationMCQConcept of water potential in plants1 markPaper 1~2 min
A plant cell is placed in a sucrose solution and becomes flaccid but does not plasmolyse. Which statement best explains this observation?
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19MasteryMCQImportance in plant hydration and transport1 markPaper 1~2 min
A potometer is used to measure the rate of water uptake by a leafy shoot. An air bubble in the capillary tube moves 12 mm12 \text{ mm} in 55 minutes. The capillary tube has a cross-sectional area of 0.5 mm20.5 \text{ mm}^2. What is the rate of water uptake by the shoot?
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20FoundationMCQConcept of water potential in plants1 markPaper 1~2 min
A plant cell has a water potential of 0.6-0.6 MPa and is placed in a solution with a water potential of 0.3-0.3 MPa. Which of the following correctly describes the direction of net water movement and its effect on the cell's pressure potential?
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21FoundationMCQTranscription and translation processes1 markPaper 1~2 min
A scientist adds antibiotic to a bacterial culture. Transcription and mRNA production continue normally, but no functional proteins are produced. Which process is most likely being inhibited by the antibiotic?
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22MasteryMCQTranscription and translation processes1 markPaper 1~2 min
In a eukaryotic cell, a spliceosome fails to recognise the 5′ splice site of an intron in a pre-mRNA molecule. What is the most likely consequence for the mature mRNA produced?
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23FoundationMCQTranscription and translation processes1 markPaper 1~2 min
A DNA template strand has the sequence 3'-TACCCAUATG-5'. During translation, a ribosome reads the mRNA transcribed from this sequence. How many amino acids are incorporated into the polypeptide before translation terminates?
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24FoundationMCQDNA structure and replication mechanism1 markPaper 1~2 min
A segment of double-stranded DNA contains 20 percent of its nitrogenous bases as guanine. According to Chargaff's rules, what percentage of the bases in this segment is thymine?
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25MasteryMCQDNA structure and replication mechanism1 markPaper 1~2 min
In a double-stranded DNA molecule, 30 percent of the bases are adenine. What percentage of the bases is guanine?
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26FoundationMCQDNA structure and replication mechanism1 markPaper 1~2 min
In a DNA double helix, the two strands run antiparallel to each other. Each strand has a repeating structural framework to which the nitrogenous bases are attached on the interior. Which combination of molecules forms this repeating framework, known as the sugar-phosphate backbone, of each DNA strand?
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27FoundationMCQTypes of mutations: Point, frameshift, chromosomal1 markPaper 1~2 min
A gene in a bacterium has the base sequence 5'-ATG GCA TTT GGG-3'. A frameshift mutation inserts an extra adenine after the sixth nucleotide, giving 5'-ATG GCA ATT TGG G-3'. Which consequence of this mutation is most likely?
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28MasteryMCQTypes of mutations: Point, frameshift, chromosomal1 markPaper 1~2 min
A mutagen changes a single nucleotide in the coding strand of an E. coli gene, altering the sequence from 5'-GAA-3' to 5'-AAA-3'. GAA codes for glutamic acid and AAA codes for lysine. Which type of point mutation has occurred?
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29FoundationMCQTypes of mutations: Point, frameshift, chromosomal1 markPaper 1~2 min
A coding strand of DNA has the sequence 5'-TAC GCA TTG-3'. A single nucleotide deletion removes the second base (A), producing a mutated template. Which of the following best describes the mRNA transcribed from this mutated DNA?
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30FoundationMCQMitigation strategies and global agreements1 markPaper 1~2 min
Rice paddies are a significant source of atmospheric methane. Which statement best explains why modifying irrigation practices in rice paddies is an effective climate change mitigation strategy?
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31MasteryMCQMitigation strategies and global agreements1 markPaper 1~2 min
A research team measures net primary productivity (NPP) in three ecosystems over 20 years. Ecosystem A is a mature tropical rainforest with 400 tonnes of biomass per hectare, where growth is balanced by decay. Ecosystem B is a newly planted mangrove forest with rapid shoot and root growth but lower total biomass. Ecosystem C is a peat swamp where waterlogged, anaerobic conditions severely limit decomposition. Which ecosystem most likely has the highest NPP?
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32FoundationMCQCauses and effects of climate change1 markPaper 1~2 min
A student measures oxygen bubble production per minute in Elodea canadensis at 15°C, 25°C, and 35°C, keeping light intensity and CO₂ concentration constant. The bubble rate at 35°C is significantly lower than at 25°C. Which statement best explains this observation?
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33FoundationMCQMechanisms of natural selection1 markPaper 1~2 min
In a population of guppies (Poecilia reticulata), males with more orange spots gain greater mating success due to female preference, but also suffer higher predation rates. Which type of natural selection best describes the combined effect of these two opposing pressures on the number of orange spots in males?
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34MasteryMCQMechanisms of natural selection1 markPaper 1~2 min
In a population of 500 moths living in a forest with dark-coloured tree trunks, moth colour is controlled by a single gene with two alleles. The dark allele (DD) is dominant over the light allele (dd). Of the 500 moths, 180 are light-coloured. Assuming Hardy–Weinberg equilibrium, what is the frequency of the allele dd?
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35FoundationMCQAdaptations and evolutionary fitness1 markPaper 1~2 min
In a meadow, grasshoppers vary in colour from bright green to brownish-green. During a prolonged drought, dry brown patches replace most of the green grass. A biologist records that the proportion of brownish-green grasshoppers increases significantly over several generations. Which of the following best explains this change in allele frequency?
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36ChallengeSAQ-LRegulatory mechanisms of gene expression7 marksPaper 2~11 min
Alternative splicing is a key post-transcriptional regulatory mechanism in eukaryotes. The Drosophila gene dsx (doublesex) is alternatively spliced to produce two different transcription factors that control sex determination. In females, the splicing pattern includes exon 4, producing the DsxF\text{Dsx}^\text{F} protein, which represses male-specific genes. In males, exon 4 is skipped, producing DsxM\text{Dsx}^\text{M} protein, which represses female-specific genes. The splicing decision is regulated by the female-specific protein Tra (transformer), which binds to a splicing enhancer sequence in the pre-mRNA. Researchers created a transgenic Drosophila line in which the splicing enhancer sequence in exon 4 was mutated so that Tra protein cannot bind. The table below summarises exon 4 inclusion data. Genotype — Exon 4 inclusion (%) Wild-type females — 95 Wild-type males — 5 Mutant females — 15 Mutant males — 4
(a)
Calculate the percentage of dsx transcripts in mutant females that exclude exon 4. [1 mark]
(b)
Explain the molecular mechanism by which the mutation in the splicing enhancer leads to the observed change in splicing pattern in females. [3 marks]
(c)
Evaluate whether mutant females would be expected to develop as phenotypic males, phenotypic females, or an intersex condition, justifying your answer with reference to the proportions of DsxF\text{Dsx}^\text{F} and DsxM\text{Dsx}^\text{M} produced. [3 marks]
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37MasterySAQ-SOperon model (lac operon)5 marksPaper 2~8 min
A strain of E. coli is cultured in a medium containing only lactose as a carbon source, in the absence of glucose. After 3030 minutes, glucose is added to the medium. The graph below shows the relative concentration of β\beta-galactosidase over time.
(a)
State the role of the lac repressor protein when E. coli is grown in the presence of lactose and absence of glucose. [1 mark]
(b)
Calculate the percentage decrease in β\beta-galactosidase concentration between t=30t = 30 minutes and t=35t = 35 minutes. [2 marks]
(c)
Explain the molecular mechanism by which glucose addition at t=30t = 30 minutes causes the observed decrease in β\beta-galactosidase concentration. [2 marks]
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38ChallengeLAQRegulatory mechanisms of gene expression10 marksPaper 2~15 min
The lac operon in Escherichia coli is a classic model for transcriptional regulation. The operon contains three structural genes (lacZ, lacY, lacA) controlled by a promoter and operator region. A regulatory gene, lacI, encodes the Lac repressor. When lactose is present, allolactose (an isomer of lactose) binds to the Lac repressor, causing a conformational change that prevents operator binding, allowing transcription. Additionally, when glucose is absent, elevated intracellular cAMP binds to CAP (catabolite activator protein); the cAMP–CAP complex binds to the CAP site upstream of the promoter, enhancing RNA polymerase binding and increasing transcription rate. A research team studied a mutant strain of E. coli (strain M1) carrying a point mutation in lacI that alters the allolactose-binding site of the repressor. The mutant repressor retains full operator-binding ability but cannot bind allolactose. β\beta-galactosidase activity (encoded by lacZ) was measured under three conditions: - Condition A: 2%2\% glucose, no lactose - Condition B: 2%2\% lactose, no glucose - Condition C: 1%1\% glucose and 1%1\% lactose Wild-type β\beta-galactosidase activity was high only in Condition B. Strain M1 showed negligible β\beta-galactosidase activity in all three conditions.
(a)
Explain the molecular mechanism by which the M1 mutation leads to negligible β\beta-galactosidase activity in Condition B, contrasting each regulatory step with the equivalent events in wild-type cells under the same condition. [5 marks]
(b)
The lac operon is described as subject to both negative and positive gene regulation. (i) Identify the molecular component responsible for each mode of regulation and state the specific interaction through which each component acts. [2 marks]
(ii) Using the M1 mutant data from all three conditions, evaluate whether the experimental results provide evidence for both negative and positive regulation, or for only one of these modes. Justify your answer with reference to specific observations. [3 marks]
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39MasterySAQ-SRegulation of blood glucose levels5 marksPaper 2~8 min
A marathon runner consumes a high-carbohydrate meal 2 hours before a race. During the race, her blood glucose concentration is monitored. Time (min) — Blood glucose (mmolL1)(\text{mmol}\,\text{L}^{-1}) 0 — 5.2 30 — 4.8 60 — 4.5 90 — 4.3 120 — 4.0
(a)
State the hormone primarily responsible for raising blood glucose concentration during prolonged exercise. [1 mark]
(b)
Calculate the percentage decrease in blood glucose concentration from t=0t = 0 to t=120mint = 120\,\text{min}. Show your working. [2 marks]
(c)
Using the data and your knowledge of homeostasis, evaluate whether the negative feedback mechanism controlling blood glucose is fully effective during this race. [2 marks]
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40MasterySAQ-SRegulation of blood glucose levels5 marksPaper 2~8 min
A patient with Type 2 diabetes mellitus has a fasting blood glucose concentration of 9.8mmoldm39.8\,\text{mmol}\,\text{dm}^{-3} (normal range: 3.95.6mmoldm33.9\text{–}5.6\,\text{mmol}\,\text{dm}^{-3}). The patient's pancreatic beta cells still produce insulin, but body cells show reduced responsiveness to the hormone. A doctor prescribes Metformin, a drug that increases the sensitivity of liver and muscle cells to insulin.
(a)
Describe the mechanism by which insulin normally causes cells to absorb glucose from the blood. [2 marks]
(b)
Explain why the patient's blood glucose concentration remains elevated despite normal or raised insulin levels. [2 marks]
(c)
Deduce how Metformin could restore normal blood glucose concentration in this patient. [1 mark]
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41MasterySAQ-SSexual cycles in animals and plants5 marksPaper 2~8 min

Data

- Concentration of egg jelly glycoprotein: 0.5μgmL10.5\,\mu\text{g}\,\text{mL}^{-1} - Percentage of sperm undergoing acrosomal reaction at this concentration: 78%78\% - Percentage of sperm undergoing acrosomal reaction in control (no glycoprotein): 2%2\%
In a species of sea urchin (Arbacia punctulata), external fertilization occurs in seawater. Researchers observed that the acrosomal reaction in sperm is triggered by a specific glycoprotein on the surface of the egg's jelly coat.
(a)
Describe the events of the acrosomal reaction in sea urchin sperm upon contact with the egg jelly coat. [2 marks]
(b)
Determine the difference in percentage points between sperm undergoing the acrosomal reaction in the experimental group and the control. [1 mark]
(c)
Predict how the percentage of sperm undergoing the acrosomal reaction would change if the glycoprotein concentration were increased to 2.0μgmL12.0\,\mu\text{g}\,\text{mL}^{-1}. Explain your reasoning with reference to receptor behaviour and the consequence for fertilization. [2 marks]
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42MasterySAQ-SSexual cycles in animals and plants5 marksPaper 2~8 min
A student investigated sporophyte production in Funaria hygrometrica over two years. The table shows the number of sporophytes counted in sample plots. Year — Rainfall classification — Sporophytes counted Year 1 — Wet — 8484 Year 2 — Dry — 1212 The life cycle of Funaria hygrometrica involves a dominant haploid gametophyte generation. The gametophyte produces gametes by mitosis, and fertilization requires water for sperm to swim to the egg.
(a)
Describe how the gametophyte of Funaria hygrometrica produces both male and female gametes. [2 marks]
(b)
Calculate the ratio of sporophyte production in the wet year to the dry year. Express your answer as a whole number ratio (x:1)(x:1)[1 mark]
(c)
Explain why dependence on water for fertilization limits moss reproduction in terrestrial environments. [2 marks]
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43ChallengeSAQ-LMitosis vs meiosis7 marksPaper 2~11 min
In Drosophila melanogaster the diploid chromosome number is 2n=82n = 8. A geneticist studies a temperature-sensitive mutation that causes complete failure of spindle fibre formation at the restrictive temperature (30C30^\circ\text{C}). Cells are assumed to begin G1 of the cell cycle before entering division.
(a)
Calculate the number of chromosomes and the number of DNA molecules present in a single cell at each stage below, at the restrictive temperature. - (i) At the end of mitosis, after cytokinesis has failed (spindle fibres did not form; chromatids did not separate). -
(ii) At the end of meiosis I, after cytokinesis has occurred (spindle fibres did not form in meiosis I). -
(iii) In a cell that has completed S phase (DNA replication) and entered G2, but has not yet undergone any division. [3 marks]
(b)
Explain why failure of spindle fibre formation produces different chromosomal outcomes in mitosis compared to meiosis I, even though the same molecular mechanism is disrupted. [2 marks]
(c)
Evaluate the usefulness of this temperature-sensitive mutant strain for investigating the role of meiosis in generating genetic variation compared with mitosis. [2 marks]
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44MasterySAQ-SPhases of mitosis and meiosis5 marksPaper 2~8 min
In a laboratory, a culture of human skin cells is treated with a chemical that prevents the formation of spindle fibres during mitosis.
(a)
Describe what would happen to the chromosomes during anaphase in these treated cells. [2 marks]
(b)
The cell cycle of untreated skin cells lasts 24 hours. After treatment, 40% of cells fail to complete mitosis and undergo apoptosis; the remaining cells divide normally. A sample initially contains 800 cells. Calculate the number of cells expected after one full cell cycle. Show your working. [3 marks]
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45ChallengeSAQ-LRenewable resources and ecological conservation7 marksPaper 2~11 min
A research team investigated the use of microalgae (Chlorella vulgaris) as a renewable biofuel feedstock. The algae were grown in open raceway ponds (500m2500\,\text{m}^2 each) under natural sunlight over a 30-day growth period. Data collected: - Initial algal biomass: 0.25gL10.25\,\text{g\,L}^{-1} (dry mass) - Final algal biomass: 1.80gL11.80\,\text{g\,L}^{-1} (dry mass) - Pond volume: 100,000L100{,}000\,\text{L} - Lipid content of dried algae: 32%32\% by mass - Energy content of algal lipids: 37.8MJkg137.8\,\text{MJ\,kg}^{-1} - Energy required for harvesting and lipid extraction: 8.2MJ8.2\,\text{MJ} per kg of dry algae processed - CO2\text{CO}_2 absorbed during growth: 1.83kg CO21.83\,\text{kg CO}_2 per kg dry algae - CO2\text{CO}_2 emitted during biodiesel combustion: 2.68kg CO22.68\,\text{kg CO}_2 per kg biodiesel - CO2\text{CO}_2 emitted from processing energy: 0.94kg CO20.94\,\text{kg CO}_2 per kg dry algae Useful relationships: Net energy=Energy outputEnergy input\text{Net energy} = \text{Energy output} - \text{Energy input} Mass=concentration×volume\text{Mass} = \text{concentration} \times \text{volume}
(a)
Calculate the net energy yield (in MJ) from the algae grown in one pond over 30 days. [3 marks]
(b)
Explain how using algal biodiesel contributes to renewable energy sustainability, with reference to the carbon cycle. [2 marks]
(c)
Using the CO2\text{CO}_2 data provided, calculate the net CO2\text{CO}_2 balance (in kg) for one pond over 30 days and determine whether this system is carbon-neutral. [2 marks]
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46ChallengeSAQ-LRenewable resources and ecological conservation7 marksPaper 2~11 min

Data

Maximum sustainable yield (MSY) is the largest catch that can be taken from a population without reducing its size over time.
The island nation of Palau has committed to protecting 80%80\% of its marine territory as a 'no-take' marine reserve. A team of ecologists is evaluating the sustainability of this conservation strategy by comparing fish communities inside the reserve with those in adjacent fishing zones. Data from a 2023 survey: - Simpson's diversity index (DD) inside the reserve: 0.920.92 - Simpson's diversity index (DD) in the fishing zone: 0.680.68 - Total fish biomass inside the reserve: 4500kg km24500\,\text{kg km}^{-2} - Total fish biomass in the fishing zone: 2100kg km22100\,\text{kg km}^{-2} - Annual fish catch in the fishing zone: 380kg km2year1380\,\text{kg km}^{-2}\,\text{year}^{-1} - Estimated annual fish recruitment in the fishing zone: 420kg km2year1420\,\text{kg km}^{-2}\,\text{year}^{-1} - Reserve area: 250km2250\,\text{km}^2; fishing zone area: 800km2800\,\text{km}^2
(a)
State the Simpson's diversity index value for the reserve and identify which zone has the higher value. [1 mark]
(b)
Explain what the difference in Simpson's diversity index between the two zones suggests about the ecological stability of the reserve compared with the fishing zone. [2 marks]
(c)
Using the biomass and recruitment data, determine whether the current fishing rate in the fishing zone is sustainable with reference to maximum sustainable yield. [2 marks]
(d)
Evaluate the effectiveness of the marine reserve as a conservation tool. In your answer, discuss one reason why the data collected may not be sufficient to conclude that the reserve alone guarantees long-term sustainability of the fishing zone. [2 marks]
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47ChallengeSAQ-LMendelian inheritance9 marksPaper 2~14 min
A genetic counsellor analyses a family pedigree for a rare autosomal recessive disorder controlled by a single gene (alleles DD and dd). The disorder causes severe metabolic dysfunction and is often fatal in early childhood. An affected individual (genotype dddd) has two unaffected parents. The affected individual marries an unaffected woman with no known family history of the disorder.
(a)
State the genotypes of the affected individual's parents. [1 mark]
(b)
The recessive allele frequency in the general population is q=0.02q = 0.02 and the population is in Hardy–Weinberg equilibrium, where p2+2pq+q2=1p^2 + 2pq + q^2 = 1. Calculate the probability that the unaffected woman is a carrier (heterozygous DdDd). [3 marks]
(c)
Explain how the Law of Segregation accounts for the production of the dddd child from two heterozygous parents. [2 marks]
(d)
Evaluate the use of population-wide genetic screening to identify carriers of this disorder prior to reproduction. [3 marks]
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48MasterySAQ-SPunnett squares and genetic ratios6 marksPaper 2~9 min
In a species of fruit fly (Drosophila melanogaster), grey body colour (GG) is dominant over black body colour (gg). A female fly heterozygous for body colour is crossed with a male fly that has a black body. In a large experiment, 800 offspring are produced: 560 have grey bodies and 240 have black bodies.
(a)
State the expected phenotypic ratio and the expected percentage of black-bodied offspring from this cross. [2 marks]
(b)
Explain how meiotic drive in the female parent could account for a deviation from the expected ratio. [2 marks]
(c)
Using the experimental data, evaluate whether the results are consistent with meiotic drive acting on the GG allele in the female parent. [2 marks]
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49MasterySAQ-SOsmosis and water movement in cells6 marksPaper 2~9 min
A student investigates osmosis using cylinders of potato (Solanum tuberosum) tissue. She cuts five cylinders of equal size and mass, then places each in a different concentration of sucrose solution. After 30 minutes, she removes the cylinders, blots them dry, and measures their mass again. Her results are shown below. Sucrose concentration (mol dm3)(\text{mol dm}^{-3})0.00.00.20.20.40.40.60.60.80.8 Percentage change in mass (%)(\%)+12+12+5+52-28-815-15
(a)
Describe the relationship between sucrose concentration and the percentage change in mass of the potato cylinders. [2 marks]
(b)
Using the data, estimate the sucrose concentration at which the potato tissue would show no net change in mass. [1 mark]
(c)
State what the concentration estimated in (b) represents in terms of the water potential of the potato cells. [1 mark]
(d)
The student repeats the experiment but does not blot the cylinders dry before re-weighing. Evaluate how this procedural change would affect the reliability of her conclusion about the isotonic concentration. [2 marks]
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50MasterySAQ-SOsmosis and water movement in cells5 marksPaper 2~8 min
A researcher studies water uptake in the freshwater plant Elodea canadensis. She places a leaf in a 10%10\% sucrose solution and observes the cells under a microscope. Initially, the chloroplasts are distributed throughout each cell. After 10 minutes, the chloroplasts are clumped in the centre of each cell, and the cell membrane has pulled away from the cell wall.
(a)
State the process by which water leaves the cell, causing the cell membrane to pull away from the cell wall. [1 mark]
(b)
The initial volume of the cell sap (vacuole) is 2000μm32000\,\mu\text{m}^3 and the final volume after the membrane pulls away from the cell wall is 1400μm31400\,\mu\text{m}^3. Calculate the percentage decrease in the volume of the cell sap. [2 marks]
(c)
The researcher replaces the 10%10\% sucrose solution with distilled water and waits 10 minutes. Explain whether the cells will return to their original turgid state, with reference to water potential and pressure potential. [2 marks]
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51ChallengeSAQ-LPost-translational modifications10 marksPaper 2~15 min

Data

- Molecular mass of wild-type pre-DNase I (including signal peptide): 32.3kDa32.3\,\text{kDa} - Molecular mass of active, mature DNase I (after signal peptide removal): 30.5kDa30.5\,\text{kDa} - Average molecular mass of one amino acid: 110Da110\,\text{Da} - Rate of intracellular degradation of engineered DNase I in E. coli: 5%5\% of total protein per minute - Human wild-type DNase I is stabilized by two disulfide bonds between cysteine residues; the engineered E. coli protein does not form these bonds
A research team investigates the production of a therapeutic enzyme, human deoxyribonuclease I (DNase I), in E. coli bacteria. The wild-type human DNASE1 gene is inserted into a bacterial expression plasmid. The protein produced is enzymatically inactive despite being successfully translated. Analysis reveals that the recombinant DNase I lacks removal of an N-terminal signal peptide that targets the protein for secretion in human cells. The team engineers a modified DNASE1 gene coding directly for the mature form of the protein (without the signal peptide). This modified gene is expressed in E. coli. The resulting protein is active but is degraded by intracellular proteases within 30 minutes, whereas human wild-type DNase I is stable for over 24 hours in the bloodstream.
(a)
Calculate the number of amino acids in the signal peptide. [2 marks]
(b)
Explain why the absence of disulfide bonds leads to rapid degradation of the engineered DNase I in E. coli. [3 marks]
(c)
Evaluate the effectiveness of the team's genetic engineering strategy for producing a stable, active therapeutic enzyme, and propose one further modification that could improve protein stability, justifying your answer. [3] (formerly ) [2 marks]
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52ChallengeSAQ-LPost-translational modifications7 marksPaper 2~11 min
A group of students studies the effects of a novel drug, Modifin, on post-translational modification of the tumour suppressor protein p53 in human cell cultures. p53 is activated by phosphorylation at serine residues in response to DNA damage. Two sets of cells are exposed to UV radiation. Set A receives no Modifin. Set B receives Modifin immediately after UV exposure. Phosphorylated p53 (detected at serine 15) and total p53 are measured over 6 hours. - Set A (control): phosphorylated p53 peaks at 2 hours (100units100\,\text{units}), declines to baseline by 6 hours; total p53 remains constant at 100units100\,\text{units} throughout. - Set B (Modifin-treated): phosphorylated p53 peaks at 2 hours (100units100\,\text{units}), remains at 80units80\,\text{units} at 6 hours; total p53 increases by 50%50\% over 6 hours. - Modifin specifically inhibits phosphatase PP2A, which removes phosphate groups from proteins. - Rate of p53 translation is constant and identical in both sets.
(a)
Calculate the percentage of total p53 that is phosphorylated at serine 15 at the 6-hour time point in Set B. Assume the initial total p53 level is 100units100\,\text{units} and the increase in total p53 is due entirely to reduced degradation. [2 marks]
(b)
Explain how inhibition of PP2A by Modifin leads to the observed changes in phosphorylated p53 levels in Set B. [3 marks]
(c)
Using the data and your knowledge of p53's role in the cell cycle, evaluate whether the sustained phosphorylation of p53 observed in Set B is likely to beneficial or harmful to the cell. Justify your conclusion. [2 marks]
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Solutions

53ChallengeSAQ-LDNA structure and replication mechanism7 marksPaper 2~11 min

Data

- Genome size: 1.9×106bp1.9 \times 10^6\,\text{bp} - Replication rate: 500500 nucleotides s1\text{s}^{-1} per fork - Error rate of polymerase: 1×1071 \times 10^{-7} errors per nucleotide incorporated - GC content: 42%42\%
A research team studies DNA replication in thermophilic archaeon Pyrococcus furiosus, which lives at 100C100^\circ\text{C}. The DNA polymerase incorporates nucleotides at 500500 nucleotides per second at 100C100^\circ\text{C}. Replication is bidirectional from a single origin of replication.
(a)
Calculate the minimum time required to replicate the P. furiosus genome, assuming both replication forks operate simultaneously at maximum rate. [2 marks]
(b)
Explain two reasons why the actual replication time is longer than the minimum calculated in (a), linking each reason to a named molecular mechanism of DNA replication. [3 marks]
(c)
Evaluate the likely impact of the 42%42\% GC content on the stability of the DNA double helix during replication at 100C100^\circ\text{C}, with reference to the molecular structure of DNA. [2 marks]
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Solutions

54ChallengeSAQ-LDNA structure and replication mechanism9 marksPaper 2~14 min

Data

- Density of 15^{15}N/15^{15}N DNA (both strands heavy): 1.724g cm31.724\,\text{g cm}^{-3} - Density of 14^{14}N/14^{14}N DNA (both strands light): 1.710g cm31.710\,\text{g cm}^{-3} - Density of hybrid 15^{15}N/14^{14}N DNA: 1.717g cm31.717\,\text{g cm}^{-3} - After one round of replication in 14^{14}N: one band at 1.717g cm31.717\,\text{g cm}^{-3} - After two rounds of replication in 14^{14}N: two bands at 1.717g cm31.717\,\text{g cm}^{-3} and 1.710g cm31.710\,\text{g cm}^{-3}, of equal intensity
In 1958, Meselson and Stahl investigated the mechanism of DNA replication using E. coli. Bacteria were grown for many generations in a medium containing heavy nitrogen (15^{15}N), ensuring all DNA was fully labelled. They were then transferred to a medium containing light nitrogen (14^{14}N) and allowed to replicate. After each round of replication, DNA was extracted and its density analysed by caesium chloride density gradient centrifugation.
(a)
State the model of DNA replication supported by these data. [1 mark]
(b)
Calculate the expected number of bands and their densities after three rounds of replication in 14^{14}N, assuming the supported model is correct. [2 marks]
(c)
Explain how the observed band pattern after two rounds of replication eliminates both the conservative model and the dispersive model. [3 marks]
(d)
Evaluate the strength of the Meselson–Stahl experiment as evidence for semi-conservative replication. In your answer, identify one limitation of the experiment and explain how the full dataset (across both rounds) addresses it. [3 marks]
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Solutions

55ChallengeSAQ-LTypes of mutations: Point, frameshift, chromosomal8 marksPaper 2~12 min
The CFTR gene encodes the CFTR chloride ion channel protein. Mutations in this gene cause cystic fibrosis. The wild-type CFTR protein is 14801480 amino acids long. A patient with severe cystic fibrosis has a single cytosine (C) nucleotide deleted from exon 10 of the CFTR gene. The wild-type mRNA sequence around this region is: 5-AAG AAC AUA UCC UUC GAU GAG-35'\text{-} \ldots \text{AAG AAC AUA UCC UUC GAU GAG}\ldots\text{-}3' corresponding to the amino acid sequence: Lys – Asn – Ile – Ser – Phe – Asp – Glu. The deletion removes the first nucleotide (U) of the UCC codon (Ser) from the mRNA.
(a)
State the type of mutation caused by this single-nucleotide deletion. [1 mark]
(b)
Determine the new mRNA sequence from the point of deletion onwards, and identify the first four amino acids produced from that point. Use the genetic code table from the IB Data Booklet. [3 marks]
(c)
Explain why this deletion is classified as a frameshift mutation and describe the likely effect on the length and function of the CFTR protein. [2 marks]
(d)
A second patient carries a point mutation that changes the UCC codon (Ser) to UCA (also Ser). Evaluate whether this point mutation would have a less severe effect on CFTR function than the frameshift mutation described above, justifying your answer with reference to protein structure and function. [2 marks]
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Solutions

56ChallengeSAQ-LTypes of mutations: Point, frameshift, chromosomal8 marksPaper 2~12 min
A geneticist studies a population of Drosophila melanogaster examining mutations in a gene responsible for eye colour. The wild-type allele produces red eyes. Three mutant strains are isolated: - Strain 1: A single base substitution in the coding region changes codon GAG (Glu) to GUG (Val). - Strain 2: A deletion of a single adenine nucleotide in the coding region. - Strain 3: A chromosomal translocation moves a segment of chromosome 2 containing the eye colour gene to chromosome 3. All three strains show white eyes instead of red.
(a)
State which strain carries a frameshift mutation. [1 mark]
(b)
Describe the effect of each mutation the DNA sequence and the resulting protein for all three strains. [3 marks]
(c)
Explain why the chromosomal translocation in Strain 3 might have a different impact on offspring compared to the point mutation in Strain 1, even though both result in white eyes. [2 marks]
(d)
Evaluate the usefulness of frameshift mutations compared to point mutations for studying gene function in Drosophila, using evidence from all three strains. [2 marks]
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Solutions

57ChallengeSAQ-LMitigation strategies and global agreements11 marksPaper 2~17 min

Data

- Conventional tillage: soil organic carbon =1.2%= 1.2\% of soil mass (0030cm30\,\text{cm} depth) - No-till agriculture after 10 years: soil organic carbon =1.8%= 1.8\% of soil mass - Soil bulk density =1.3gcm3= 1.3\,\text{g}\,\text{cm}^{-3} - Farm area =100hectares= 100\,\text{hectares}; 1hectare=10000m21\,\text{hectare} = 10\,000\,\text{m}^2 - Soil depth considered =30cm= 30\,\text{cm}
International climate agreements such as the Paris Agreement rely on Nationally Determined Contributions (NDCs), which are each country's self-defined targets for reducing greenhouse gas emissions. A key biological mitigation strategy promoted in many NDCs is the adoption of agricultural practices that increase soil carbon sequestration, including no-till farming, cover cropping, and agroforestry.
(a)
Calculate the additional mass of carbon, in tonnes, sequestered in the soil after converting a 100100-hectare farm from conventional tillage to no-till agriculture over 10 years. [3 marks]
(b)
Explain one biological mechanism by which no-till agriculture reduces the rate of decomposition of soil organic matter compared to conventional tillage. [2 marks]
(c)
Explain one biological mechanism by which no-till agriculture increases the input of organic matter into the soil compared to conventional tillage. [2 marks]
(d)
Evaluate the extent to which agricultural soil carbon sequestration can realistically contribute to meeting NDC targets under the Paris Agreement. In your answer, refer to at least one scientific limitation, one economic factor, and one political factor. [4 marks]
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Solutions

58MasterySAQ-SCauses and effects of climate change5 marksPaper 2~8 min
Coral reefs are highly sensitive to ocean warming. In 2016, a marine heatwave caused mass coral bleaching on the Great Barrier Reef. Scientists monitored two coral species: the branching coral Acropora and the massive coral Porites. Before the heatwave, Acropora covered 40%40\% of the reef area and Porites covered 30%30\%. After the heatwave, Acropora cover fell to 8%8\% and Porites cover fell to 21%21\%.
(a)
State the name of the photosynthetic symbiont expelled from coral tissues during bleaching. [1 mark]
(b)
Calculate the percentage decline in coral cover for each species. [2 marks]
(c)
Explain, using named biological processes, why the loss of the symbiont causes coral death if bleaching is prolonged. [2 marks]
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Solutions

59MasterySAQ-SMechanisms of natural selection6 marksPaper 2~9 min
In a population of marine iguanas (Amblyrhynchus cristatus) on the Galápagos island of San Cristóbal, researchers measured the body length of 500 adult iguanas. Iguanas with a medium body length (606070cm70\,\text{cm}) were best able to dive and feed on algae in the intertidal zone. Very small iguanas (under 50cm50\,\text{cm}) were easily swept away by strong currents, while very large iguanas (over 80cm80\,\text{cm}) had difficulty manoeuvring in rocky crevices where algae grow. Over a 10-year period, researchers recorded lifetime reproductive output: Size class — Body length — Mean offspring per lifetime Small — <50cm<50\,\text{cm}22 Medium — 606070cm70\,\text{cm}88 Large — >80cm>80\,\text{cm}33
(a)
State the type of natural selection acting on body length in this population. [1 mark]
(b)
Explain why medium-sized iguanas have greater reproductive fitness than both small and large iguanas in this environment. [2 marks]
(c)
Explain the long-term effect of this selection pressure on the distribution of body lengths in the population. [3 marks]
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Solutions

60MasterySAQ-SMechanisms of natural selection5 marksPaper 2~8 min
A population of Darwin's finches (Geospiza fortis) on the island of Daphne Major in the Galápagos experiences a severe drought in 1977. The drought reduces the availability of small, soft seeds, forcing finches to eat large, hard seeds that require a deep beak to crack open. Researchers measured beak depth (a heritable trait) before and after the drought. Before drought (1976) — 9.2mm9.2\,\text{mm}0.8mm0.8\,\text{mm} After drought (1978) — 10.1mm10.1\,\text{mm}0.6mm0.6\,\text{mm}
(a)
State the type of natural selection that occurred during the drought. [1 mark]
(b)
Calculate the change in mean beak depth between 1976 and 1978. [1 mark]
(c)
Calculate the change in standard deviation of beak depth between 1976 and 1978. [1 mark]
(d)
Explain, using the mechanism of natural selection, why the standard deviation of beak depth decreased after the drought, and evaluate what this change in variation suggests about the future adaptive potential of the population. [2 marks]
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Solutions