2024 IB DP Biology Practice Paper with Answers

Double circulation consists of two separate circuits: pulmonary and systemic. The pulmonary circuit operates at a lower pressure to protect the delicate capillaries of the lungs and prevent pulmonary edema, whereas the systemic circuit operates at a much higher pressure to distribute blood efficiently to all organs and tissues of the body.

Award 1 mark for selecting B. Option A is incorrect because double circulation prevents mixing. Option C is incorrect as blood does not flow slower in systemic compared to pulmonary. Option D is incorrect as it does not reduce blood volume.

An increase in cellular respiration during exercise leads to higher blood carbon dioxide concentration, which lowers blood pH. Chemoreceptors detect this change and stimulate the respiratory center in the medulla oblongata to increase the ventilation rate. Concurrently, the cardiovascular center stimulates the sinoatrial node via the sympathetic nervous system to increase heart rate, ensuring faster transport of oxygen to tissues and removal of carbon dioxide.

Award 1 mark for selecting C. Both ventilation rate and heart rate increase to meet metabolic demands and restore homeostatic blood gas levels.

The Lincoln Index is calculated using the formula: \(N = \frac{n_1 \times n_2}{m_2}\) where: \(n_1\) = number of individuals marked in the first sample (40), \(n_2\) = total number of individuals captured in the second sample (50), and \(m_2\) = number of marked individuals recaptured in the second sample (10). Therefore: \(N = \frac{40 \times 50}{10} = 200\).

Award 1 mark for the correct calculation yielding 200 (C).

The mitotic index is calculated as: \(\text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}}\). Cells in mitosis include prophase, metaphase, anaphase, and telophase: \(60 + 30 + 20 + 10 = 120\) cells. Total cells = 600. Therefore, \(\text{Mitotic Index} = \frac{120}{600} = 0.20\).

Award 1 mark for the correct mitotic index calculation of 0.20 (B).

Magnification is calculated as: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Convert the image size from millimeters to micrometers to align the units: \(60\text{ mm} = 60,000\ \mu\text{m}\). Using the formula: \(\text{Magnification} = \frac{60,000\ \mu\text{m}}{5\ \mu\text{m}} = 12,000\). Therefore, the magnification is \(\times 12,000\).

Award 1 mark for correct unit conversion and calculation of magnification (D).

Natural selection acts on pre-existing genetic variation within a population. When an environmental change or selection pressure is applied, individuals possessing advantageous traits (phenotypes) are more likely to survive and reproduce. They pass these beneficial alleles to their offspring, leading to an increase in the frequency of these alleles in the gene pool over generations.

Award 1 mark for B. A is incorrect because mutations are random and do not occur in response to an environmental need. C is out of logical order. D describes Lamarckian inheritance of acquired characteristics, which is incorrect.

Helicase is the enzyme that unwinds the double helix by breaking the hydrogen bonds between complementary base pairs. DNA polymerase can only synthesize the new strand in the 5' to 3' direction because it can only add new nucleotides to the free 3'-OH group of the preceding nucleotide.

Award 1 mark for the correct combination of Helicase and 5' to 3' synthesis (B).

During the repolarization phase of an action potential, voltage-gated sodium channels close (and become temporarily inactivated), preventing further influx of sodium ions. At the same time, voltage-gated potassium channels open, allowing potassium ions to diffuse out of the neuron down their concentration gradient, which makes the inside of the cell negative again.

Award 1 mark for the correct identification of channel states during repolarization (B).

During a stress response, the cardiovascular center in the medulla oblongata transmits excitatory signals along the sympathetic nerves to the sinoatrial (SA) node of the heart, stimulating it to increase heart rate. Simultaneously, the endocrine system is activated, causing the adrenal glands to secrete epinephrine (adrenaline) into the bloodstream, which also acts directly on the SA node to accelerate heart rate.

Award [1] for the correct answer (B). All other options represent incorrect combinations of nervous pathways, endocrine glands, or physiological responses.

As cellular respiration increases, more carbon dioxide is produced and dissolves in the blood plasma to form carbonic acid, which dissociates and lowers the blood pH. This change is detected by chemoreceptors in the medulla oblongata, carotid arteries, and aorta. In response, nerve impulses are sent to the diaphragm and intercostal muscles to increase the rate and depth of ventilation, thereby expelling carbon dioxide and returning blood pH to normal.

Award [1] for the correct answer (B). Award [0] for options involving an increase in pH (which would occur with hyperventilation/decreased CO2) or incorrect receptor types/muscles.

The hepatic portal vein carries blood directly from the capillaries of the small intestine villi to the liver. After a meal rich in carbohydrates, glucose is actively absorbed into this portal system, leading to a very high glucose concentration. In the liver, excess glucose is converted to glycogen under the influence of insulin. Consequently, the hepatic vein, which carries blood away from the liver toward the heart, has a lower and more regulated glucose concentration.

Award [1] for the correct answer (B). All other options represent incorrect flow directions or incorrect relative concentrations of glucose post-meal.

To find the energy available to the tertiary consumers, we must track the energy transfer through the trophic levels starting from the primary producers:
1. Producers (Trophic Level 1): \(12,000 \text{ kJ m}^{-2}\text{ yr}^{-1}\)
2. Primary Consumers (Trophic Level 2): \(12,000 \times 0.10 = 1,200 \text{ kJ m}^{-2}\text{ yr}^{-1}\)
3. Secondary Consumers (Trophic Level 3): \(1,200 \times 0.10 = 120 \text{ kJ m}^{-2}\text{ yr}^{-1}\)
4. Tertiary Consumers (Trophic Level 4): \(120 \times 0.10 = 12 \text{ kJ m}^{-2}\text{ yr}^{-1}\).

Award [1] for the correct calculation and choice (C).

During the S phase of the cell cycle, DNA replication occurs, which doubles the mass of DNA from \(20\text{ pg}\) to \(40\text{ pg}\). However, the replicated sister chromatids remain joined at their centromeres, meaning they are still counted as individual chromosomes. Thus, the chromosome number remains 12 during metaphase. The chromosomes do not double in count until the sister chromatids separate during anaphase.

Award [1] for the correct selection of chromosome number and DNA mass (C). Incorrect options mistakenly double chromosome count before anaphase or fail to double the DNA content post-replication.

Prokaryotic bacterial cells are characterized by having 70S ribosomes, naked circular DNA (no histone proteins), and a cell wall made of peptidoglycan (murein). Eukaryotic plant cells have 80S ribosomes in the cytoplasm, linear DNA wrapped around histones, and cellulose cell walls. Fungal cell walls are composed of chitin, and animal cells lack cell walls entirely.

Award [1] for the correct identification of a prokaryotic bacterial cell (B).

Natural selection requires pre-existing genetic variation in the population, which arises from random mutations before exposure to the selective agent. When the population is exposed to an antibiotic (the selective pressure), non-resistant individuals are killed, whereas those with the pre-existing resistance mutation survive and reproduce, leading to an increase in the frequency of the resistance allele over generations.

Award [1] for the correct evolutionary order (B). Option A incorrectly implies Lamarckian-style directed mutation (mutating in response to the antibiotic).

DNA polymerase can only add free nucleotides to the 3' hydroxyl (-OH) group of the growing strand. This means that synthesis of the newly forming DNA strand always occurs in a 5' to 3' direction. Helicase and ligase have different functions (unwinding DNA and joining Okazaki fragments respectively) and do not assemble free nucleotides in this manner.

Award [1] for the correct direction of synthesis and function of DNA polymerase (B).

Chemoreceptors in the medulla oblongata, aorta, and carotid arteries detect the drop in pH caused by increased \(CO_2\) levels during exercise. They send nerve impulses to the cardiovascular centre in the medulla oblongata, which responds by increasing sympathetic nerve stimulation to the sinoatrial (SA) node, thereby increasing heart rate.

Award 1 mark for identifying that chemoreceptors detect pH changes and transmit signals to the cardiovascular centre to increase sympathetic stimulation to the SA node.

Dehydration leads to high blood osmolarity, detected by osmoreceptors in the hypothalamus. This stimulates the posterior pituitary gland to release Antidiuretic Hormone (ADH). ADH acts on the collecting ducts of the kidneys, increasing their permeability to water (via insertion of aquaporins), allowing more water reabsorption into the blood.

Award 1 mark for identifying that the posterior pituitary releases ADH, which increases the permeability of the collecting duct to water.

Bottom-up control occurs when the population of a species is limited by the availability of resources (such as food or space) from lower trophic levels. Since the insects are limited by their host plants (primary producers) rather than predators (wasps), this is an example of bottom-up regulation.

Award 1 mark for the correct identification and explanation of bottom-up control.

Cyclins regulate the cell cycle by binding to enzymes called cyclin-dependent kinases (CDKs). Once activated, these kinases phosphorylate specific target proteins that carry out tasks specific to that phase of the cell cycle.

Award 1 mark for stating that cyclins bind to and activate CDKs, which phosphorylate target proteins.

Compartmentalization in eukaryotes (via membrane-bound organelles) allows specific microenvironments to be created. This concentrates enzymes and substrates for particular reactions (e.g., inside lysosome or mitochondrion), improving metabolic efficiency.

Award 1 mark for identifying that compartmentalization concentrates enzymes and substrates for specific metabolic pathways.

Natural selection requires pre-existing genetic variation (e.g., through mutation). When an antibiotic is introduced, it acts as a selective pressure. The susceptible bacteria die, while those carrying the pre-existing resistance mutation survive and reproduce, passing the resistance gene to their offspring.

Award 1 mark for identifying that antibiotics act as selective agents, selecting for pre-existing resistant individuals.

DNA polymerase I is responsible for removing the RNA primers from the lagging (and leading) strands and replacing them with DNA nucleotides. Helicase unwinds the helix, primase synthesizes RNA primers, and ligase joins the Okazaki fragments.

Award 1 mark for identifying that DNA polymerase I removes RNA primers and replaces them with DNA.

Cellular respiration increases the production of carbon dioxide, which dissolves in the blood to form carbonic acid, lowering blood pH. This change is detected by chemoreceptors in the medulla oblongata (respiratory centre) and the major arteries. In response, the medulla oblongata sends more frequent nerve impulses via the phrenic and intercostal nerves to the diaphragm and external intercostal muscles, increasing the rate and depth of ventilation to expel \(CO_2\).

Award 1 mark for identifying that increased blood \(CO_2\) lowers blood pH, which is detected by the respiratory centre in the medulla oblongata, leading to an increase in ventilation rate and depth.

During exercise, increased metabolic activity produces more carbon dioxide (
\(CO_2\)), which reacts with water to form carbonic acid, thereby decreasing blood pH. Chemoreceptors located in the medulla, carotid arteries, and aorta detect this drop in pH. They send sensory impulses to the cardiovascular control center in the medulla oblongata of the brain. The medulla oblongata then increases the frequency of impulses sent down the sympathetic nerve (accelerator nerve) to the sinoatrial (SA) node, increasing the heart rate to speed up the clearance of carbon dioxide and delivery of oxygen.

[1 mark] Award for selecting the correct option (A).
- Reject other options because baroreceptors detect pressure changes, not chemical/pH changes (ruling out B and D).
- The cerebellum does not act as the primary control center for heart rate regulation, and somatic motor nerves control voluntary skeletal muscle, not cardiac muscle (ruling out B and C).

Amino acids are hydrophilic molecules and are absorbed into the capillary network of the villus, from where they travel via the hepatic portal vein directly to the liver. Triglycerides are broken down into fatty acids and monoglycerides for absorption, reassembled into triglycerides (and packaged as chylomicrons) inside the epithelial cells, and then absorbed into the lacteals. The lacteals are part of the lymphatic system, which eventually empties into the systemic venous circulation.

[1 mark] Award for selecting the correct option (A).
- Options B, C, and D are incorrect because they misidentify the pathways: amino acids go to blood capillaries (and then the hepatic portal vein), whereas lipids go to the lacteals (and then the lymphatic system).

In a closed system with a finite amount of resources, the growth of a population eventually slows down due to limiting factors. As the yeast population increases, nutrients such as glucose are depleted and toxic waste products (like ethanol) accumulate in the medium. This leads to an increase in the mortality rate and a decrease in the natality rate, causing the growth curve to transition from exponential growth to a plateau (stationary phase).

[1 mark] Award for selecting the correct option (C).
- Option A is incorrect because natality decreases and mortality increases as resources deplete.
- Option B is incorrect because carrying capacity does not increase; rather, environmental resistance increases.
- Option D is incorrect because intra-specific competition increases, and in a closed yeast culture, there is no immigration.

During Prophase, DNA has already been replicated during the S phase of interphase. There are 8 chromosomes, but each chromosome consists of two sister chromatids joined at the centromere, meaning there are 16 DNA molecules in total. Following mitosis and cytokinesis, the sister chromatids have separated into individual chromosomes and are distributed into two daughter cells. Each daughter cell receives 8 chromosomes, each consisting of a single DNA molecule (double helix), so there are 8 chromosomes and 8 DNA molecules in one daughter cell.

[1 mark] Award for selecting the correct option (A).
- B is incorrect because during prophase, the individual sister chromatids are still joined at the centromere and counted as a single chromosome (8 chromosomes, not 16).
- C is incorrect because it implies DNA has not replicated or has halved inappropriately.
- D is incorrect because it overestimates both chromosomes and DNA molecules.

To resolve the internal ultrastructure of a mitochondrion (such as cristae), transmission electron microscopy (TEM) is required because of its high resolution (down to \(0.1\text{ nm}\)), which is much higher than the limit of resolution of a light microscope (around \(200\text{ nm}\)). However, because specimen preparation for electron microscopy requires a vacuum and kills the specimen, observing living organisms (like a moving bacterium) can only be done using light microscopy.

[1 mark] Award for selecting the correct option (A).
- Reject B because SEM is used for 3D surface topography, not internal ultrastructure, and TEM cannot be used on living specimens.
- Reject C and D because the light microscope lacks the resolution to clearly show the detailed internal membrane cristae of a mitochondrion.

Helicase functions by unwinding the double helix and breaking the hydrogen bonds holding complementary base pairs together. DNA polymerase synthesizes a new complementary DNA strand by bringing in free nucleotides and catalyzing the formation of covalent phosphodiester bonds between the 3' carbon of the pentose sugar and the phosphate group of the incoming nucleotide.

[1 mark] Award for selecting the correct option (A).
- B is incorrect because Helicase breaks hydrogen bonds (not covalent bonds), and DNA polymerase forms phosphodiester (covalent) bonds (not peptide bonds).
- C is incorrect because primase (not DNA polymerase) synthesizes the RNA primer.
- D is incorrect because DNA ligase (not DNA polymerase) seals nicks in the sugar-phosphate backbone by joining Okazaki fragments.

(a) Percentage increase calculation:
\(\text{Percentage Increase} = \frac{\text{Value}_{\text{intense}} - \text{Value}_{\text{rest}}}{\text{Value}_{\text{rest}}} \times 100\)
\(\text{Percentage Increase} = \frac{24.2 - 5.0}{5.0} \times 100 = \frac{19.2}{5.0} \times 100 = 384\%\)

(b) The heart increases its output to meet the oxygen and nutrient demands of working muscles. Cardiac output is the product of heart rate and stroke volume. Increased carbon dioxide and hydrogen ions in the blood are detected by chemoreceptors. This triggers a sympathetic response from the medulla oblongata, releasing noradrenaline to increase heart rate. Adrenaline (epinephrine) released from the adrenal glands further boosts heart rate and cardiac muscle contractility.

(c) Anaerobic cell respiration yields only 2 ATP per glucose molecule compared to the ~36 ATP of aerobic respiration, but does so at a much faster rate. Pyruvate is converted to lactate to regenerate \(\text{NAD}^+\). If lactate production exceeds the rate of liver clearance (Cori cycle), it accumulates in the blood, serving as a marker for anaerobic threshold crossing.

Part (a) [1.5 marks]:
- Correct calculation of 384% [1]
- Show appropriate working (e.g., \((24.2 - 5.0) / 5.0\)) [0.5]

Part (b) [3 marks max]:
- Cardiac output increases due to increase in stroke volume and/or heart rate [1]
- Exercise increases carbon dioxide / decreases blood pH [1]
- Detected by chemoreceptors (in aorta/carotid/medulla) [1]
- Signals sent to the cardiovascular control center in the medulla [1]
- Sympathetic nerve impulses / epinephrine (adrenaline) stimulate the sinoatrial (SA) node [1]

Part (c) [4 marks max]:
- Intense exercise leads to an oxygen deficit / oxygen demand exceeds supply [1]
- Aerobic respiration cannot supply enough ATP [1]
- Muscle cells switch to anaerobic respiration / glycolysis [1]
- Pyruvate is converted to lactate [1]
- This reaction regenerates \(\text{NAD}^+\) [1]
- Allows glycolysis to continue producing ATP [1]
- Lactate diffuses out of muscles into the blood, causing accumulation [1]

(a) The competitive exclusion principle (or Gause's principle) states that two species competing for the exact same resources cannot stably coexist if other ecological factors remain constant.

(b) Fundamental niche refers to the potential niche space a species can occupy without interspecific competition. Since *T. castaneum* can live in both conditions, both are within its fundamental niche. Realized niche is the actual niche occupied under competitive pressure. In this case, *T. castaneum*'s realized niche does not include cool-dry conditions due to exclusion by *T. confusum*.

(c) Environmental fluctuations prevent competitive exclusion from running to completion. In warm-wet cycles, *T. castaneum* increases, while in cool-dry cycles, *T. confusum* increases. This temporal variation leads to a stable or dynamic coexistence.

(d) Acceptable answers include light intensity, space/volume of the habitat, pH of the medium, or presence of physical barriers.

Part (a) [1 mark]:
- Competitive exclusion principle / Gause's law [1]

Part (b) [3 marks max]:
- Fundamental niche is the theoretical range of conditions a species can occupy without competition [1]
- Realized niche is the actual range occupied in the presence of competition [1]
- Both environments are part of *T. castaneum*'s fundamental niche [1]
- Its realized niche is restricted to warm-wet environments when in competition with *T. confusum* [1]

Part (c) [3.5 marks max]:
- Prediction: Coexistence / both species persist [1]
- Explanation: Competitive advantage shifts back and forth as conditions change [1]
- Neither species has the advantage long enough to cause the extinction of the other [1]
- Temporal niche partitioning occurs [0.5]

Part (d) [1 mark]:
- Food availability / quality of medium / space / light levels / gas concentration (O2/CO2 levels) [1] (Do not accept temperature or humidity)

(a) DNA replication occurs during the S phase. Cyclin E peaks at the G1/S boundary, preparing the replication machinery.

(b) Cell cycle duration is 24 hours. Mitosis (M phase) is indicated by the peak of Cyclin B and its rapid degradation, occurring from hour 22 to 24 (2 hours). Interphase is the rest of the cycle: \(24 - 2 = 22\) hours (acceptable range: 21 to 23 hours based on reasonable estimation of mitosis duration).

(c) Cyclins are regulatory subunits. CDKs are catalytic subunits that require cyclin binding to become active. Phosphorylation targets include transcription factors (e.g., to initiate S-phase genes) or structural proteins (e.g., to break down the nuclear envelope in mitosis).

(d) Cyclin B forms the Mitosis Promoting Factor (MPF) with CDK1. Without it, chromosome condensation, nuclear envelope breakdown, and spindle assembly cannot occur, causing G2/M arrest.

Part (a) [1 mark]:
- Cyclin E [1]

Part (b) [2.5 marks]:
- Estimate: 22 hours (allow 21 to 23 hours) [0.5]
- Reasoning: Mitosis (M phase) is represented by Cyclin B's peak and fall, which takes place in the final hours (approx. hours 22–24) [1]
- Interphase is the entire cell cycle minus mitosis (G1 + S + G2) [1]

Part (c) [3 marks max]:
- Cyclins bind to cyclin-dependent kinases (CDKs) [1]
- This binding activates the CDKs [1]
- Active CDKs phosphorylate target proteins [1]
- Phosphorylation activates/deactivates proteins that control cell cycle progress [1]
- Different cyclins peak at different times, ensuring sequential phase progression [1]

Part (d) [2 marks]:
- The cell cannot progress from G2 to mitosis / cannot enter M phase [1]
- Cell division is arrested / cell cycle stops at the G2/M checkpoint [1]

(a) The independent variable is manipulated (aphidicolin concentration), and the dependent variable is measured (rate of replication via CPM).

(b) Percentage inhibition calculation:
\(\text{Inhibition} = \frac{\text{Control} - \text{Treatment}}{\text{Control}} \times 100\)
\(\text{Inhibition} = \frac{15000 - 4500}{15000} \times 100 = \frac{10500}{15000} \times 100 = 70\%\)

(c) Helicase requires ATP to break the relatively weak hydrogen bonds between bases. DNA polymerase can only add nucleotides to a pre-existing 3'-OH end (requiring an RNA primer synthesized by primase) and works in opposite directions on the leading and lagging strands.

(d) Eukaryotes must unpack dense chromatin (nucleosomes) before replication can proceed. The absolute genome size of prokaryotes is also much smaller, enabling rapid duplication from a single origin of replication (oriC).

Part (a) [1 mark]:
- Independent: aphidicolin concentration AND Dependent: rate of DNA replication / thymidine uptake / CPM [1] (Both correct for 1 mark, 0.5 if only one is correct)

Part (b) [2 marks]:
- Correct working shown (e.g., \((15000 - 4500)/15000\)) [1]
- Correct final percentage: 70% [1]

Part (c) [3.5 marks max]:
- Helicase unwinds / unzips DNA double helix [1]
- Helicase breaks hydrogen bonds between nitrogenous bases [1]
- DNA polymerase adds complementary free nucleotides to the template strand [1]
- DNA polymerase synthesizes DNA in a 5' to 3' direction / forms phosphodiester bonds [1]
- DNA polymerase requires an RNA primer to initiate replication [0.5]

Part (d) [2 marks max]:
- Prokaryotic DNA is circular and not bound to histones / no nucleosomes to unwind [1]
- Prokaryotic genomes are much smaller than eukaryotic genomes [1]
- Lack of a nuclear envelope allows transcription, translation, and replication processes to occur in close proximity / simpler regulation [1]

### Part (a) Autonomic Nervous System and Heart Rate
* The cardiovascular control center is located in the medulla oblongata of the brain.
* At the start of exercise, the sympathetic nervous system is activated, while parasympathetic activity decreases.
* Sympathetic nerves (the cardiac nerve) release the neurotransmitter noradrenaline (norepinephrine) at the sinoatrial (SA) node.
* Noradrenaline binds to beta-receptors on the SA node, increasing the frequency of electrical impulses (pacemaker activity) and thereby raising the heart rate.
* Conversely, during rest, the parasympathetic nervous system (via the vagus nerve) releases acetylcholine to slow down the SA node; the withdrawal of this vagal tone contributes to the initial rise in heart rate during exercise.

### Part (b) Respiratory System Response to Carbon Dioxide
* Respiring muscle tissue produces carbon dioxide (
\(\text{CO}_2\)), which diffuses into the blood plasma.
* In the blood, \(\text{CO}_2\) reacts with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)), catalyzed by carbonic anhydrase, which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogen carbonate ions (\(\text{HCO}_3^-\)), lowering the pH of the blood.
* Central chemoreceptors in the medulla oblongata and peripheral chemoreceptors in the carotid and aortic bodies detect the increase in partial pressure of \(\text{CO}_2\) (\(p\text{CO}_2\)) and decrease in pH.
* These chemoreceptors send sensory nerve impulses to the respiratory control center in the medulla oblongata.
* The medulla oblongata sends increased frequency of motor impulses via the phrenic nerve (to the diaphragm) and intercostal nerves (to the external intercostal muscles).
* This stimulates an increase in both the rate and depth of ventilation (breathing), accelerating the excretion of \(\text{CO}_2\) across the alveoli and returning blood pH back to homeostatic limits (negative feedback).

### Part (c) Cardiovascular and Muscular System Coordination
* **Cardiovascular Adjustments:**
1. Cardiac output is significantly increased by raising both heart rate and stroke volume.
2. Vasodilation occurs in the arterioles supplying active skeletal muscles, mediated by local metabolic factors (e.g., high \(\text{CO}_2\), low pH, lactic acid, and adenosine).
3. Vasoconstriction occurs in arterioles supplying non-essential organs (such as the digestive tract and kidneys), redirecting blood flow toward active muscle beds.
4. Precapillary sphincters in skeletal muscle capillaries relax, increasing capillary perfusion and maximizing surface area for gas exchange.
* **Muscular System Adaptations and Coordination:**
5. Rhythmic contraction of skeletal muscles compresses nearby veins, acting as a "skeletal muscle pump" that enhances venous return of blood to the heart.
6. The local environment of working muscles (elevated temperature, high \(p\text{CO}_2\), and lower pH) causes a rightward shift in the oxygen-hemoglobin dissociation curve (the Bohr effect), reducing hemoglobin's affinity for oxygen and promoting its release to active tissues.
7. Active muscle fibers contain myoglobin, which has a higher affinity for oxygen than hemoglobin, facilitating rapid extraction and storage of oxygen within the muscle sarcoplasm.
8. Mitochondria within muscle fibers consume oxygen rapidly during aerobic respiration (electron transport chain), maintaining a steep concentration gradient that drives the continuous diffusion of oxygen from blood plasma into muscle cells.

### Part (a) [Max 4 marks]
* **MP1:** Cardiorespiratory control center is located in the medulla oblongata.
* **MP2:** Exercise triggers activation of the sympathetic nervous system (and withdrawal of parasympathetic activity).
* **MP3:** Sympathetic cardiac nerves release noradrenaline.
* **MP4:** Noradrenaline binds to receptors on the sinoatrial (SA) node to increase heart rate.
* **MP5:** Parasympathetic vagus nerve releases acetylcholine, which slows heart rate at rest (withdrawal allows rapid heart rate increase).
*(Accept any four of the above points)*

### Part (b) [Max 5 marks]
* **MP1:** Carbon dioxide (\(\text{CO}_2\)) reacts with water in blood to form carbonic acid (\(\text{H}_2\text{CO}_3\)) which dissociates to release hydrogen ions (\(\text{H}^+\)) / decreases blood pH.
* **MP2:** Chemoreceptors in the medulla (central) or aorta/carotid arteries (peripheral) detect the low pH / high \(p\text{CO}_2\).
* **MP3:** Nerve impulses are transmitted to the respiratory control center in the medulla oblongata.
* **MP4:** Medulla oblongata sends motor signals via the phrenic and/or intercostal nerves.
* **MP5:** Diaphragm and external intercostal muscles contract more rapidly/strongly, increasing rate and depth of ventilation.
* **MP6:** Increased ventilation enhances gas exchange at the alveoli, removing excess \(\text{CO}_2\) and restoring pH (negative feedback).
*(Accept any five of the above points)*

### Part (c) [Max 7 marks]
* **MP1:** Cardiac output is increased (by raising heart rate and stroke volume) to meet systemic oxygen demand.
* **MP2:** Vasodilation of arterioles in active muscle tissues increases local blood flow.
* **MP3:** Vasoconstriction of arterioles in non-essential systems (e.g. gut/kidneys) redistributes blood to working skeletal muscles.
* **MP4:** Precapillary sphincters open in muscle beds, increasing surface area for oxygen diffusion.
* **MP5:** Skeletal muscle pump (contraction of muscles) compresses veins, increasing venous return to the heart.
* **MP6:** Bohr effect occurs: high temperature / low pH / high \(\text{CO}_2\) in active muscles decreases hemoglobin's affinity for oxygen, promoting oxygen unloading.
* **MP7:** Myoglobin in muscle cells acts as an oxygen store/facilitator due to its high oxygen affinity.
* **MP8:** Mitochondrial respiration maintains a steep concentration gradient of oxygen from capillaries into muscle fibers.
*(Accept any seven of the above points)*

a) The independent variable is the factor systematically changed by the experimenter, which is the sucrose concentration (measured in \( \text{mol~dm}^{-3} \)).
b) Percentage change accounts for differences in the initial starting mass of the individual potato cylinders, allowing for a fair comparison between trials.
c) The isotonic point is the solute concentration at which there is no net movement of water into or out of the cells, resulting in a 0% change in mass. By interpolating between 0.2 \( \text{mol~dm}^{-3} \) (+5.2%) and 0.4 \( \text{mol~dm}^{-3} \) (-1.8%), the 0% change lies around 0.35 \( \text{mol~dm}^{-3} \).
d) Temperature must be controlled because it affects the rate of diffusion/osmosis. Alternatively, the source/type of potato must be kept the same because different varieties or ages of potatoes have different initial solute concentrations.

a) [1 max]
- Sucrose concentration / concentration of sucrose solution (do not accept 'amount of sucrose' or 'water potential').

b) [1 max]
- Initial/starting masses of potato cylinders were not identical / allows for valid comparison between cylinders with different starting masses.

c) [2 max]
- 0.35 \( \text{mol~dm}^{-3} \) (accept range 0.33 to 0.37 \( \text{mol~dm}^{-3} \)); [1]
- The isotonic concentration is where there is no net water movement / 0% mass change / where the line of best fit crosses the x-axis / between 0.2 and 0.4 \( \text{mol~dm}^{-3} \); [1]

d) [1 max]
- Temperature (of the solutions);
- Source/variety/age of potato tissue;
- Volume of sucrose solution in each tube;
- Duration of immersion/time left in solution.

a) Mitotic Index is calculated as: \( \text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} \times 100 \).
- Control (0%): \( \text{Total} = 450 + 50 = 500 \). \( \text{Mitotic Index} = \frac{50}{500} \times 100 = 10.0\% \).
- 10% Extract: \( \text{Total} = 490 + 10 = 500 \). \( \text{Mitotic Index} = \frac{10}{500} \times 100 = 2.0\% \).
b) Increasing the concentration of garlic extract causes a decrease in the mitotic index / inhibits cell division.
c) Root tips contain meristematic tissue (apical meristems) which actively undergoes mitosis for growth, making it easy to find cells in different stages of cell division.
d) Staining root tips typically involves concentrated acids (e.g., hydrochloric acid) for maceration and toxic/staining dyes (e.g., aceto-orcein). Safety measures include wearing safety goggles, protective gloves, or performing the heating steps in a well-ventilated area/fume hood.

a) [2 max]
- Control = 10% (or 0.10) AND 10% extract = 2% (or 0.02); [1]
- Correct working shown: \( \frac{\text{cells in mitosis}}{\text{total cells}} \times 100 \) (e.g., \( \frac{50}{500} \times 100 \) and \( \frac{10}{500} \times 100 \)); [1]

b) [1 max]
- Garlic extract inhibits cell division / decreases the mitotic index;
- Higher concentrations of garlic extract lead to lower mitotic indices / fewer cells entering mitosis;

c) [1 max]
- Root tip contains the (apical) meristem;
- Active site of growth / high rate of cell division / many dividing cells;

d) [1 max]
- Wear safety goggles / protective gloves to avoid contact with corrosive acids / toxic stains;
- Heat slide gently or use a fume hood to avoid inhaling harmful chemical fumes.

a) Substitute the values into the formula: \(n_1 = 80\), \(n_2 = 75\), \(m_2 = 15\). Calculation: \(N = \frac{80 \times 75}{15} = \frac{6000}{15} = 400\) beetles.
b) Key assumptions of the Lincoln Index include: the population is closed (no births, deaths, immigration, or emigration during the study period); the marks do not wear off, fade, or get lost; the marks do not affect the survival or behavior of the individuals; marked individuals mix completely and randomly back into the population before the second sample is taken.
c) The paint must be non-toxic so it does not harm/kill the beetles, and not highly visible to prevent making them easier targets for predators, which would violate the assumption that marking does not affect survival rates.

a) [2 max]
- 400 (beetles); [1]
- Correct substitution shown: \( \frac{80 \times 75}{15} \); [1]

b) [2 max]
- No immigration or emigration / closed population; [1]
- No births or deaths / population size remains constant; [1]
- The mark does not affect survival / does not increase predation risk; [1]
- The mark does not rub off / remains clearly visible; [1]
- Marked individuals mix randomly with the rest of the population; [1]
(Award [1] for each correct assumption up to [2])

c) [1 max]
- To prevent increasing the death rate / mortality of the marked beetles (which would skew the population estimate / violate the assumption of equal survival probability).

Kupffer cells are specialized macrophages in the liver that perform phagocytosis to engulf aged or damaged red blood cells. Once inside the cell, hemoglobin is broken down. The protein component, globin, is hydrolyzed into its constituent amino acids, which are released back into the bloodstream for protein synthesis. The non-protein heme group is split into iron and an organic residue. Iron is stored within the liver as ferritin or transported back to the bone marrow via transferrin to synthesize new hemoglobin. The remaining organic portion of the heme group is converted into bilirubin (bile pigment), which is secreted into the bile canaliculi by hepatocytes.

Award 1 mark for each of the following points, up to a maximum of 4 marks: 1. Correctly naming phagocytosis as the mechanism used by Kupffer cells. 2. Stating that globin is broken down/hydrolyzed into amino acids (which are recycled). 3. Explaining that iron is removed from heme and stored (as ferritin) or sent to bone marrow. 4. Stating that the remaining heme group is converted into bilirubin / bile pigment.

During the cephalic phase, the sight, smell, or thought of food sends sensory inputs to the brain, which transmits parasympathetic impulses via the vagus nerve directly to the gastric glands, stimulating the secretion of pepsinogen, acid, and gastrin. During the gastric phase, the physical distension of the stomach walls by food activates stretch receptors, and the presence of peptides activates chemoreceptors. This triggers local nerve reflexes and further release of the hormone gastrin from endocrine G-cells in the stomach mucosa. Gastrin travels via the blood to stimulate parietal cells to produce more hydrochloric acid (HCl) and chief cells to secrete pepsinogen, ensuring a highly acidic and enzymatic environment for protein digestion.

Award 1 mark for each of the following points, up to a maximum of 4 marks: 1. Describing the cephalic phase as nervous stimulation via the vagus nerve in response to sensory anticipation of food. 2. Stating that stomach distension / chemical detection of peptides triggers the gastric phase. 3. Explaining that the hormone gastrin is secreted by G-cells in response to stomach stretching or chemical stimuli. 4. Detailing that gastrin travels via the bloodstream to stimulate parietal cells (for HCl) and/or chief cells (for pepsinogen).

The PR interval measures the time from the beginning of atrial depolarization (the P wave) to the beginning of ventricular depolarization (the QRS complex). This interval represents the passage of the cardiac impulse from the sinoatrial (SA) node, through the atrial myocardium, and down to the atrioventricular (AV) node. The conduction of the action potential slows down significantly at the AV node. This delay is biologically essential because it gives the atria enough time to complete their contraction (atrial systole), ensuring that the ventricles are fully filled with blood before they begin to contract (ventricular systole). Without this delay, the atria and ventricles would contract almost simultaneously, leading to incomplete filling of the ventricles and severely reduced cardiac output.

Award 1 mark for each of the following points, up to a maximum of 4 marks: 1. Identifying that the PR interval represents the conduction of the electrical signal from the SA node through the atria and the AV node. 2. Explaining that the AV node causes a delay in the transmission of the action potential to the ventricles. 3. Stating that this delay allows the atria to contract fully and empty their blood into the ventricles. 4. Explaining that this maximizes ventricular filling and prevents the ventricles from contracting before they are filled, thus maintaining efficient stroke volume/cardiac output.

As skeletal muscles respire intensely, they release carbon dioxide, which reacts with water to form carbonic acid, and lactic acid from anaerobic pathways. Both metabolites increase the concentration of hydrogen ions, thereby lowering the local blood pH. The increased acidity and high partial pressure of carbon dioxide promote the binding of hydrogen ions and carbon dioxide to hemoglobin, causing a conformational change. This change reduces hemoglobin's affinity for oxygen, shifting the oxygen-hemoglobin dissociation curve to the right (known as the Bohr shift). The physiological benefit is that hemoglobin releases (unloads) oxygen much more readily at the tissue level, ensuring that actively respiring muscle cells receive a greater supply of oxygen to maintain aerobic respiration and minimize fatigue.

Award 1 mark for each of the following points, up to a maximum of 4 marks: 1. Stating that carbon dioxide and lactic acid accumulation lowers the local blood pH / increases acidity. 2. Explaining that lower pH/higher carbon dioxide decreases hemoglobin's affinity for oxygen. 3. Describing that this causes the oxygen dissociation curve to shift to the right (Bohr shift). 4. Identifying the physiological benefit as enhanced unloading/delivery of oxygen to the active skeletal muscles to support aerobic respiration.

Helicobacter pylori is adapted to the highly acidic environment of the stomach by secreting the enzyme urease, which converts urea into ammonia and carbon dioxide, neutralizing local gastric acid. Once established, the bacteria burrow into the protective mucus layer lining the stomach wall, producing toxins and initiating an inflammatory response that degrades the mucin gel. This localized destruction leaves the underlying gastric epithelial cells exposed to the highly corrosive hydrochloric acid and pepsin of the stomach, leading to tissue damage and the formation of a gastric ulcer. To treat this, proton pump inhibitors (PPIs) are administered. PPIs target and irreversibly inhibit the H+/K+-ATPase pump proteins in the membrane of parietal cells. By shutting down these pumps, PPIs dramatically reduce gastric acid secretion, raising the pH of the stomach and allowing the damaged mucosal lining to heal.

Award 1 mark for each of the following points, up to a maximum of 4 marks: 1. Explaining that H. pylori secretes urease to neutralize gastric acid, allowing survival/colonization. 2. Describing how the bacteria damage the protective mucus layer of the stomach lining. 3. Explaining that this exposure of the gastric epithelium to stomach acid/pepsin results in inflammation and ulceration. 4. Stating that proton pump inhibitors (PPIs) inhibit the H+/K+-ATPase pump in parietal cells, reducing acid secretion to facilitate healing.