- A.Aerobic conditions
- B.Anaerobic conditions
- C.Aerobic conditions with the addition of a mitochondrial respiratory inhibitor
- D.Anaerobic conditions with the addition of a glycolytic inhibitor
HKDSE · Thinka-original Practice Paper
2024 HKDSE Biology Practice Paper | DSE Mock
Thinka 2024 DSE-Style Mock — Biology
Paper 1 Section A
- A.28%
- B.22%
- C.44%
- D.Cannot be determined from the given information
- A.The concentration of amino acids is higher in the hepatic portal vein than in the hepatic vein.
- B.The concentration of urea is higher in the hepatic portal vein than in the hepatic vein.
- C.The concentration of oxygen is higher in the hepatic vein than in the hepatic portal vein.
- D.The concentration of carbon dioxide is lower in the hepatic vein than in the hepatic portal vein.
- A.P > Q > R > S
- B.P > R > Q > S
- C.Q > P > S > R
- D.R > S > P > Q
- A.Turgid cells; water
- B.Turgid cells; 10% sucrose solution
- C.Plasmolyzed cells; 10% sucrose solution
- D.Plasmolyzed cells; air
- A.The rapid consumption of oxygen by the proliferating algae during daytime photosynthesis.
- B.The decrease in total oxygen production due to the death of submerged plants.
- C.The rapid multiplication of aerobic decomposers breaking down dead organic matter.
- D.The barrier formed by the dense algal bloom preventing oxygen from dissolving into the water.
- A.The student feels the pain but cannot withdraw the leg.
- B.The student cannot feel the pain and cannot withdraw the leg.
- C.The student cannot feel the pain but can still withdraw the leg automatically.
- D.The student feels the pain and withdraws the leg normally.
- A.The concentration of the toxic metal is highest in phytoplankton because they absorb it directly from water.
- B.The concentration of the toxic metal is highest in the osprey because it occupies the highest trophic level.
- C.The total amount of toxic metal in the osprey population is greater than that in the phytoplankton population.
- D.Organisms at higher trophic levels accumulate less toxic metal because energy is lost along the food chain.
- A.Lag phase: Primary is Shorter; Secondary is Longer
- B.Cells involved in initial recognition: Primary is Memory cells; Secondary is Naive B cells
- C.Peak antibody concentration: Primary is Lower; Secondary is Higher
- D.Rate of antibody production: Primary is Faster; Secondary is Slower
- A.12.5%
- B.25%
- C.50%
- D.0%
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
- A.The y-intercept (net \(\text{CO}_2\) uptake at zero light intensity) moves closer to zero.
- B.The light compensation point (X) shifts to a higher light intensity.
- C.The maximum rate of net \(\text{CO}_2\) uptake at high light intensity decreases significantly.
- D.The rate of photosynthesis at light intensity X becomes zero.
- A.The uptake of \(\text{K}^+\) ions is mainly an active process that relies on aerobic respiration.
- B.In the absence of oxygen, root cells completely lose their ability to absorb \(\text{K}^+\) ions.
- C.Nitrogen gas acts as a competitive inhibitor of the carrier proteins for \(\text{K}^+\) transport.
- D.Passive transport of \(\text{K}^+\) ions requires ATP produced from anaerobic respiration.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (4) only
- D.(3) and (4) only
- A.Muscle cells cannot be stimulated, leading to flaccid paralysis.
- B.Continuous stimulation of muscle cells, leading to repeated contractions or spasms.
- C.Failure of action potential propagation along the motor neurone.
- D.Rapid depletion of neurotransmitters in the synaptic vesicles of the motor neurone.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
- A.Valine - Leucine - Glutamine
- B.Glutamine - Aspartic acid - Valine
- C.Aspartic acid - Leucine - Glutamine
- D.Leucine - Valine - Aspartic acid
- A.Primary succession starts in an area where soil is already present, whereas secondary succession starts on bare rock.
- B.The pioneer community of primary succession usually consists of woody shrubs, whereas that of secondary succession consists of lichens.
- C.Secondary succession generally reaches a climax community faster than primary succession because soil and seeds are already present.
- D.Biodiversity decreases during primary succession but increases during secondary succession.
- A.The rapid decrease in DO is mainly due to the intensive respiration of aquatic plants.
- B.The high BOD is caused by a large amount of inorganic nutrients in the sewage.
- C.The population of aerobic decomposers increases rapidly, leading to the drop in DO.
- D.Fish populations increase because of the abundance of organic matter in the sewage.
- A.During the second injection, phagocytes recognize the antigen faster and produce more antibodies.
- B.Memory B cells produced after the first injection rapidly proliferate and differentiate into plasma cells upon the second exposure.
- C.T-killer cells directly transform into antibody-producing plasma cells during the secondary immune response.
- D.The antigen in the second injection stimulates helper T cells to secrete antibodies directly.
- A.Isolated chloroplasts suspended in an isotonic sucrose solution under blue light
- B.Isolated chloroplasts suspended in an isotonic sucrose solution under green light
- C.Boiled isolated chloroplasts suspended in an isotonic sucrose solution under red light
- D.Isolated chloroplasts suspended in distilled water in the dark
- A.I and II only
- B.I and III only
- C.II and III only
- D.I, II and III
- A.The water potential of the potato cells is lower than that of the 0.1 M sucrose solution.
- B.There is a net movement of water out of the potato cells.
- C.The potato cells will absorb water, expand, and eventually burst.
- D.The solute potential of the 0.1 M sucrose solution is lower than that of the potato cells.
- A.They can feel the pain but cannot withdraw their hand.
- B.They can withdraw their hand but cannot feel the pain.
- C.They can neither feel the pain nor withdraw their hand.
- D.They can feel the pain and withdraw their hand normally.
- A.Emulsification of lipids in the duodenum
- B.Hydrolysis of starch in the mouth
- C.Digestion of peptides into amino acids in the small intestine
- D.Conversion of proteins to polypeptides in the stomach
- A.Metaphase of mitosis
- B.Metaphase I of meiosis
- C.Metaphase II of meiosis
- D.Anaphase of mitosis
- A.0%
- B.25%
- C.50%
- D.75%
- A.Nitrification
- B.Nitrogen fixation
- C.Denitrification
- D.Decomposition of organic nitrogen to ammonium
- A.Rapid growth of algae -> Death of fish -> Decrease in dissolved oxygen -> Increase in aerobic decomposers
- B.Increase in dissolved oxygen -> Rapid growth of algae -> Increase in aerobic decomposers -> Death of fish
- C.Increase in organic matter -> Increase in aerobic decomposers -> Decrease in dissolved oxygen -> Death of fish
- D.Death of fish -> Increase in organic matter -> Decrease in dissolved oxygen -> Increase in aerobic decomposers
- A.20%
- B.30%
- C.40%
- D.70%
- A.Glucose
- B.Pyruvate
- C.Glycogen
- D.Starch
(1) The concentration of glucose in the hepatic portal vein is higher than that in the hepatic vein.
(2) The concentration of urea in the hepatic portal vein is lower than that in the hepatic vein.
(3) The concentration of carbon dioxide in the hepatic portal vein is higher than that in the hepatic vein.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
- A.Auxin accumulates on the upper side of the shoot, inhibiting cell elongation.
- B.Auxin accumulates on the upper side of the shoot, promoting cell elongation.
- C.Auxin accumulates on the lower side of the shoot, promoting cell elongation.
- D.Auxin accumulates on the lower side of the shoot, inhibiting cell elongation.
- A.20%
- B.30%
- C.40%
- D.Cannot be determined from the given information.
- A.12.5%
- B.25%
- C.50%
- D.100%
- A.Phytoplankton, because they directly absorb the toxin from the water over a large surface-area-to-volume ratio.
- B.Zooplankton, because they consume a massive quantity of phytoplankton.
- C.Sea eagle, because the toxin is non-biodegradable and accumulates progressively along the trophic levels.
- D.Small fish, because they have a high metabolic rate which accelerates toxin absorption.
Paper 1 Section B
Tube A: 5 mL fresh milk + 1 mL lipase solution + 1 mL distilled water
Tube B: 5 mL fresh milk + 1 mL lipase solution + 1 mL bile salt solution
The pH of both mixtures was monitored at 37 °C for 20 minutes.
(a) Explain why the pH in both tubes decreased during the reaction. (2 marks)
(b) Explain why the rate of pH decrease was faster in Tube B than in Tube A. (3 marks)
(c) Why was the temperature maintained at 37 °C? (1.3 marks)
(d) Suggest a control setup to show that the pH change was due to the action of lipase. (1 mark)
(a) Explain why the gas exchange surfaces of both alveoli and gills must be kept moist. (2 marks)
(b) Describe how the ventilation mechanism in bony fish maintains a continuous concentration gradient of oxygen across the gill lamellae. (3.3 marks)
(c) Emphysema is a lung disease where the alveolar walls break down, merging small alveoli into larger air sacs. Explain how this structural change affects the efficiency of gas exchange. (2 marks)
(a) When the left ventricle begins to contract, the pressure inside it quickly exceeds the pressure in the left atrium. State which valve closes at this point, and explain how its closure prevents backflow. (2 marks)
(b) When the pressure in the left ventricle exceeds the pressure in the aorta, another valve opens. Identify this valve and state the destination of the blood pumped out. (2 marks)
(c) Compare the thickness of the muscular walls of the left ventricle and the right ventricle. Explain the physiological importance of this difference. (3.3 marks)
- Tube 1: Chloroplast suspension + DCPIP, kept in the dark.
- Tube 2: Chloroplast suspension + DCPIP, exposed to bright light.
- Tube 3: Boiled chloroplast suspension + DCPIP, exposed to bright light.
After 10 minutes, only Tube 2 turned from blue to colorless.
(a) Explain why the blue dye in Tube 2 was decolorized under light. (3 marks)
(b) Explain the scientific purpose of including Tube 1 and Tube 3 in this experiment. (2 marks)
(c) What is the immediate source of energy for active transport of hydrogen ions into the thylakoid space during the light stage? (1 mark)
(d) Briefly describe how the chemical products of the light-dependent stage are used in the carbon-fixing (light-independent) stage. (1.3 marks)
(a) State the end products of anaerobic respiration in:
(i) yeast cells. (1 mark)
(ii) human muscle cells. (1 mark)
(b) Explain why anaerobic respiration releases much less energy (ATP) per glucose molecule than aerobic respiration. (2 marks)
(c) During heavy exercise, lactic acid accumulates in muscle cells. Describe how lactic acid is cleared from the body after exercise ceases. (3.3 marks)
(a) Describe the sequence of events that occurs from the arrival of an action potential at the synaptic knob to the generation of a new nerve impulse in the postsynaptic neurone under normal conditions. (4 marks)
(b) Predict and explain the effect of Drug X on nerve impulse transmission across the synapse. (2 marks)
(c) Acetylcholine is broken down by acetylcholinesterase in the synaptic cleft. Explain the importance of this enzymatic breakdown. (1.3 marks)
(a) Describe how the organic matter in the sewage causes a rapid decline in the level of dissolved oxygen (DO) in the lake. (3 marks)
(b) Sewage also contains high concentrations of inorganic minerals such as nitrates and phosphates. Explain how these minerals lead to 'algal blooms' and how this process further depletes dissolved oxygen. (3.3 marks)
(c) State one consequence of this pollution event on the animal community in the lake. (1 mark)
- Parents I-1 and I-2 are both phenotypically normal.
- They have two children: a daughter II-3 who has Condition Y, and a son II-4 who is phenotypically normal.
(a) Deduce whether the allele responsible for Condition Y is dominant or recessive. Explain your answer. (3 marks)
(b) Deduce whether the gene is located on an autosome or the X-chromosome. Explain your answer. (3.3 marks)
(c) If son II-4 marries a woman who is a carrier (heterozygous) for Condition Y, what is the probability that their first child will have Condition Y? (1 mark)
3'- T A C G G G C T A A T T A C T -5'
(a) Write down the complementary sequence of the mRNA strand transcribed from this DNA template strand, indicating the 5' and 3' ends. (2 marks)
(b) Describe the role of transfer RNA (tRNA) during the process of translation. (2 marks)
(c) Suppose a mutation occurs, substituting the fifth base (G) on this DNA template strand with T.
(i) State the resulting change in the corresponding mRNA codon. (1 mark)
(ii) Explain why some single-base substitution mutations in DNA do not result in any changes in the amino acid sequence of the polypeptide chain produced. (2.3 marks)
- Grass (Producers): 2,000,000 kJ m^-2 y^-1
- Rabbits (Primary consumers): 180,000 kJ m^-2 y^-1
- Foxes (Secondary consumers): 15,000 kJ m^-2 y^-1
(a) Calculate the efficiency of energy transfer from the producers to the primary consumers. (2 marks)
(b) Explain why a large amount of energy is lost as it flows from one trophic level to the next. State three pathways of energy loss. (3.3 marks)
(c) Explain why a pyramid of energy is always upright (pyramid-shaped), unlike a pyramid of numbers which can sometimes be inverted. (2 marks)
Paper 2 Elective Modules (Choose 2 out of 4)
(a) Describe how the design of the aeration tank in secondary sewage treatment promotes the decomposition of organic pollutants. (6 marks)
(b) An ecological survey measured the Biochemical Oxygen Demand (BOD) and Dissolved Oxygen (DO) levels upstream and downstream of a discharge point where partially treated sewage was accidentally released into a river.
(i) Describe and explain the change in DO level immediately downstream of the discharge point. (3 marks)
(ii) Explain how the river system eventually recovers its DO level further downstream. (3 marks)
(c) To prevent such pollution, a city plans to construct an artificial wetland as a tertiary treatment step.
Describe how the interactions between wetland plants and microorganisms in the soil help remove nitrogen-containing pollutants (such as ammonium ions, \(NH_4^+\)) from the wastewater. (5 marks)
(d) Suggest two limitations of using constructed wetlands for wastewater treatment compared to conventional chemical treatment plants. (3 marks)
(a) To detect whether a corn product contains a transgenic Bt gene (which confers insect resistance), polymerase chain reaction (PCR) is used to amplify the target gene.
(i) Explain why two different primers (forward and reverse) are required in a PCR mixture. (2 marks)
(ii) Explain the importance of primer design in ensuring only the transgenic Bt sequence is amplified, rather than host corn genomic DNA. (2 marks)
(b) The PCR reaction involves repeated thermal cycling. State the biological/chemical event occurring at each of the following temperatures in a typical PCR cycle, and explain its significance:
(i) \(94^\circ\text{C}\) to \(98^\circ\text{C}\) (2 marks)
(ii) \(50^\circ\text{C}\) to \(65^\circ\text{C}\) (2 marks)
(iii) \(72^\circ\text{C}\) (2 marks)
(c) After PCR, agarose gel electrophoresis was performed. Lane 1 is a DNA ladder. Lane 2 is a positive control (known Bt corn). Lane 3 is a negative control (non-GM corn). Lane 4 is the test sample of corn starch. The gel shows a distinct band of \(350\text{ bp}\) in Lane 2 and Lane 4, but no band in Lane 3.
(i) Interpret these results and state whether the corn starch contains GM ingredients. Explain your reasoning. (3 marks)
(ii) Why is it essential to include a negative control (Lane 3) in this analysis? (3 marks)
(d) Discuss one potential ecological risk and one potential human health concern of cultivating this transgenic insect-resistant corn. (4 marks)