A Level H2 Biology Practice Paper 2

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TuitionGoWhere Practice Paper - Biology H2 A-Level

TuitionGoWhere Practice Paper (AI) - Version 2

Subject: Biology H2
Level: A-Level
Paper: Structured Questions (Practice Set)
Duration: 2 hours
Total Marks: 75
Name: __________________________
Class: _________________________
Date: __________________________

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Use a black or blue pen.
4. Diagrams should be drawn with a sharp pencil.

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Section A: Cells and Biomolecules

Question 1 Fig 1.1 shows a simplified diagram of the $Na^+/K^+$ pump in a plasma membrane. (a) Describe the sequence of events that allows the pump to transport $Na^+$ and $K^+$ ions against their concentration gradients. [4]


(b) A drug, Ouabain, binds to the extracellular side of the pump and prevents the release of $K^+$ into the cell. Explain the likely effect of this drug on the membrane potential of the cell. [3]

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Question 2 The structure of proteins is critical to their function. (a) Explain how the primary structure of a protein determines its tertiary structure. [4]


(b) In certain neurodegenerative diseases, proteins misfold and aggregate. With reference to the properties of amino acid R-groups, explain why misfolded proteins tend to aggregate. [3]

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Question 3 Fig 3.1 shows the initial rate of reaction for an enzyme at varying substrate concentrations in the absence and presence of a competitive inhibitor. (a) Describe the effect of the inhibitor on the $V_{max}$ and the $K_m$ of the enzyme. [2]

(b) Explain the molecular basis for the changes observed in Fig 3.1. [3]

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Question 4 Mitochondria are the sites of aerobic respiration. (a) Describe the role of the electron transport chain in generating a proton gradient across the inner mitochondrial membrane. [4]


(b) A chemical inhibitor, sodium azide, blocks the final step of the electron transport chain. Explain how this affects the production of ATP via chemiosmosis. [3]

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Question 5 Gel electrophoresis is used to separate proteins or DNA fragments. (a) Explain how the principle of gel electrophoresis allows for the separation of two different alleles of a gene that produce fragments of different sizes. [4]


(b) An individual is found to be heterozygous for a specific genetic trait. Predict and explain the banding pattern that would be observed on a gel for this individual. [2]

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Section B: Genetics and Molecular Biology

Question 6 The lac operon in E. coli is an inducible system. (a) Explain why it is metabolically advantageous for a prokaryote to have an inducible operon rather than a constitutively expressed one. [3]


(b) Describe the role of the repressor protein and the inducer (allolactose) in the regulation of the lac operon. [4]

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Question 7 A specific trait in a species of plant is controlled by two genes, A and B. A cross between two dihybrids (AaBb $\times$ AaBb) produces offspring in a phenotypic ratio of 9 red : 3 pink : 4 white. (a) Determine the relationship between Gene A and Gene B. Justify your answer. [4]


(b) Suggest the possible biochemical pathway involving these two genes that leads to the observed phenotypes. [3]

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Question 8 The Philadelphia chromosome is associated with Chronic Myelogenous Leukemia (CML). (a) Describe the chromosomal event that leads to the formation of the Philadelphia chromosome. [2]


(b) Explain how the resulting fusion protein contributes to the uncontrolled division of white blood cells. [4]

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Question 9 Polymerase Chain Reaction (PCR) is a fundamental tool in biotechnology. (a) State the three main steps of a PCR cycle and the temperature required for each. [3]


(b) Explain the importance of using a thermostable DNA polymerase, such as Taq polymerase, in this process. [2]

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Question 10 (a) Distinguish between the processes of transcription and translation in terms of the template used and the final product. [3]


(b) Explain how a single base substitution in the DNA sequence could result in a protein that is completely non-functional. [3]

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Section C: Plant and Human Physiology

Question 11 C3 and C4 plants have different mechanisms for carbon fixation. (a) Explain the physiological advantage of the C4 pathway in environments with high temperatures and limited water availability. [4]


(b) Describe the role of bundle sheath cells in the C4 mechanism. [3]

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Question 12 The rate of photosynthesis is affected by various environmental factors. (a) Explain why an increase in oxygen concentration in the atmosphere can lead to a decrease in the net rate of photosynthesis in C3 plants. [4]


(b) Define the "compensation point" and explain its significance in relation to the survival of a plant. [3]

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Question 13 The human body maintains a constant internal environment through homeostasis. (a) Describe the negative feedback mechanism that regulates blood glucose levels when they rise above the set point. [5]


(b) Explain the role of the hypothalamus and the pituitary gland in the regulation of water potential in the blood. [4]

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Question 14 Neural transmission involves both electrical and chemical signals. (a) Describe the changes in membrane potential that occur during the generation of an action potential. [4]


(b) Explain how the neurotransmitter acetylcholine facilitates the transmission of an impulse across a synapse. [4]

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Question 15 (a) Explain the process of allopatric speciation, using a hypothetical example of a population divided by a geographic barrier. [5]


(b) Discuss the extent to which DNA sequence analysis can be used to determine the evolutionary relationship between two species. [4]

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Question 16 A population of organisms shows a logistic growth curve. (a) Explain why the growth rate decreases as the population size approaches the carrying capacity ($K$). [3]


(b) Predict and explain the effect on the carrying capacity if a new, more efficient predator is introduced into the ecosystem. [3]

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Question 17 (a) Describe the structure and function of the thylakoid membrane in the chloroplast. [3]


(b) Explain the role of electrons as they move from Photosystem II to Photosystem I during the light-dependent reactions. [4]

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Question 18 (a) Compare and contrast the roles of DNA and RNA in the expression of genetic information. [4]


(b) Explain how the use of restriction endonucleases and DNA ligase allows for the creation of recombinant DNA. [4]

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Question 19 (a) Describe the mechanism of facilitated diffusion and explain how it differs from active transport. [4]


(b) Explain why the rate of facilitated diffusion reaches a maximum (saturation) at high solute concentrations. [2]

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Question 20 (a) Explain the importance of the $S$ phase of the cell cycle. [3]


(b) Describe the role of spindle fibres during the process of mitosis to ensure that each daughter cell receives an identical set of chromosomes. [4]

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TuitionGoWhere Practice Paper - Biology H2 A-Level

Answer Key (Version 2)

Q1 (Cells/Membrane) (a) 1. Binding of 3 $Na^+$ ions to the pump from the intracellular side. [1] 2. ATP hydrolysis occurs, phosphorylating the pump. [1] 3. Conformational change occurs, releasing $Na^+$ to the extracellular space. [1] 4. 2 $K^+$ ions bind from the extracellular side, triggering dephosphorylation and a return to original shape, releasing $K^+$ inside. [1] (b) Ouabain prevents $K^+$ influx. [1] This reduces the positive charge returning to the interior. [1] The membrane potential becomes more depolarized (less negative) or the resting potential cannot be maintained. [1]

Q2 (Biomolecules) (a) Primary structure is the linear sequence of amino acids. [1] This sequence determines the specific R-groups present. [1] Interactions between R-groups (e.g., hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions) [1] cause the polypeptide to fold into a specific 3D shape. [1] (b) Misfolding exposes hydrophobic R-groups that are normally buried in the protein core. [1] These exposed hydrophobic regions seek to avoid the aqueous environment. [1] They interact with similar hydrophobic regions of other misfolded proteins, leading to aggregation. [1]

Q3 (Enzymes) (a) $V_{max}$ remains unchanged. [1] $K_m$ increases. [1] (b) Competitive inhibitors bind to the active site, competing with the substrate. [1] This reduces the affinity of the enzyme for the substrate, increasing $K_m$. [1] However, at very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the enzyme to reach its maximum velocity ($V_{max}$). [1]

Q4 (Mitochondria) (a) Electrons from NADH/FADH2 pass through a series of protein complexes. [1] As electrons move, energy is released. [1] This energy is used to pump $H^+$ ions from the matrix into the intermembrane space. [1] This creates a high concentration of $H^+$ (proton gradient/electrochemical gradient). [1] (b) Sodium azide blocks the final electron acceptor (oxygen). [1] Electron flow stops, and the proton gradient is no longer maintained. [1] Without the proton motive force, ATP synthase cannot phosphorylate ADP to ATP. [1]

Q5 (Gel Electrophoresis) (a) An electric field is applied across the gel. [1] DNA/proteins are negatively charged and migrate toward the anode. [1] The gel matrix acts as a sieve. [1] Smaller fragments move faster/further than larger fragments, separating them by size. [1] (b) Two distinct bands will be observed. [1] Because the individual has two different alleles, each allele produces a fragment of a different size/length. [1]

Q6 (Operons) (a) Prevents the waste of energy and resources. [1] Enzymes for lactose metabolism are only produced when lactose is actually present. [1] This allows the cell to adapt to changing nutrient availability. [1] (b) The repressor protein binds to the operator, blocking RNA polymerase from transcribing the genes. [1] When allolactose (inducer) is present, it binds to the repressor. [1] This changes the repressor's shape, preventing it from binding to the operator. [1] RNA polymerase can then transcribe the operon. [1]

Q7 (Epistasis) (a) Recessive Epistasis. [1] The 9:3:4 ratio is characteristic of recessive epistasis. [1] Gene A is epistatic to Gene B; if the genotype is 'aa', the phenotype is white regardless of the B allele. [1] If A is present, B determines red or pink. [1] (b) Precursor $\xrightarrow{Gene A}$ Intermediate $\xrightarrow{Gene B}$ Final Pigment. [1] Gene A converts white precursor to pink intermediate. [1] Gene B converts pink intermediate to red pigment. [1] (Accept similar logical pathways).

Q8 (Cancer/Genetics) (a) Translocation between chromosomes 9 and 22. [1] A piece of chromosome 9 swaps with a piece of chromosome 22. [1] (b) The translocation creates a BCR-ABL fusion gene. [1] This gene produces a fusion protein with constitutive (always active) tyrosine kinase activity. [1] This leads to continuous phosphorylation of signaling proteins. [1] This signals the cell to divide uncontrollably, leading to leukemia. [1]

Q9 (PCR) (a) Denaturation ($\sim 95^\circ\text{C}$), Annealing ($\sim 55^\circ\text{C}$), Extension ($\sim 72^\circ\text{C}$). [3] (b) The denaturation step occurs at very high temperatures which would denature normal human DNA polymerase. [1] Taq polymerase is heat-stable and remains functional across multiple cycles. [1]

Q10 (Molecular Bio) (a) Transcription: Template = DNA; Product = mRNA. [1.5] Translation: Template = mRNA; Product = Polypeptide/Protein. [1.5] (b) A substitution could create a premature stop codon (nonsense mutation). [1] This truncates the protein. [1] The resulting protein is too short to fold correctly or lack essential active sites, rendering it non-functional. [1]

Q11 (Plant Bio) (a) C4 plants use PEP carboxylase to fix $\text{CO}_2$ into a 4-carbon compound in mesophyll cells. [1] This allows $\text{CO}_2$ to be concentrated in bundle sheath cells. [1] This reduces photorespiration by ensuring RuBisCO is saturated with $\text{CO}_2$ even when stomata are partially closed to save water. [1] Higher efficiency in hot/dry conditions. [1] (b) Bundle sheath cells are the site of the Calvin cycle. [1] They are arranged in a ring around the vascular bundle. [1] They maintain a high $\text{CO}_2$ concentration to minimize oxygenation of RuBP. [1]

Q12 (Plant Bio) (a) RuBisCO can act as both a carboxylase and an oxygenase. [1] High $\text{O}_2$ concentrations favor the oxygenation of RuBP (photorespiration). [1] This process consumes ATP and releases $\text{CO}_2$ without producing sugar. [1] Consequently, the net rate of carbon fixation/photosynthesis decreases. [1] (b) The light intensity at which the rate of photosynthesis equals the rate of respiration. [1] At this point, net gas exchange is zero. [1] If a plant stays below this point for too long, it will consume its energy reserves and die. [1]

Q13 (Physiology) (a) High blood glucose detected by $\beta$-cells of pancreas. [1] Insulin is secreted. [1] Insulin increases glucose uptake by muscle/fat cells and stimulates glycogenesis in the liver. [1] Blood glucose levels fall. [1] This removes the stimulus for insulin secretion (negative feedback). [1] (b) Hypothalamus detects high blood osmotic pressure (low water potential). [1] It stimulates the pituitary gland to release Antidiuretic Hormone (ADH). [1] ADH increases the permeability of collecting ducts in the kidney to water. [1] More water is reabsorbed into the blood, increasing water potential. [1]

Q14 (Physiology) (a) Resting potential is maintained. [1] Stimulus reaches threshold, $\text{Na}^+$ channels open. [1] $\text{Na}^+$ influx causes depolarization (potential becomes positive). [1] $\text{Na}^+$ channels close and $\text{K}^+$ channels open, $\text{K}^+$ efflux causes repolarization (potential returns to negative). [1] (b) Action potential triggers $\text{Ca}^{2+}$ influx into the presynaptic terminal. [1] This causes exocytosis of vesicles containing ACh into the synaptic cleft. [1] ACh binds to receptors on the postsynaptic membrane. [1] This opens $\text{Na}^+$ channels, triggering a new action potential in the postsynaptic neuron. [1]

Q15 (Evolution) (a) A population is split by a physical barrier (e.g., mountain range/river). [1] The two groups are reproductively isolated. [1] Each group faces different selective pressures/environmental conditions. [1] Mutations accumulate independently in each group. [1] Over time, they diverge so much that they can no longer interbreed even if the barrier is removed. [1] (b) Comparing base sequences of a conserved gene (e.g., Cytochrome c). [1] Fewer differences indicate a more recent common ancestor. [1] More differences indicate a more distant relationship. [1] It provides quantitative evidence for phylogeny. [1]

Q16 (Ecology) (a) Resources (food, space, light) become limiting. [1] Competition increases. [1] The birth rate decreases and/or death rate increases until they are equal. [1] (b) Carrying capacity ($K$) will decrease. [1] The predator increases the death rate of the prey population. [1] The environment can support fewer individuals of the prey species due to higher predation pressure. [1]

Q17 (Plant Bio) (a) A network of membranes (thylakoids) stacked into grana. [1] It provides a large surface area for photosystems and electron carriers. [1] It allows for the establishment of a proton gradient. [1] (b) Light excites electrons in PSII. [1] Electrons move through the ETC (plastoquinone, cytochrome complex). [1] Energy released is used to pump $H^+$ into the thylakoid lumen. [1] Electrons are then re-excited at PSI to eventually reduce $\text{NADP}^+$ to $\text{NADPH}$. [1]

Q18 (Molecular Bio) (a) DNA: Double-stranded, deoxyribose, thymine, stores genetic info long-term. [2] RNA: Single-stranded, ribose, uracil, acts as a messenger/catalyst for protein synthesis. [2] (b) Restriction endonucleases cut DNA at specific recognition sites (often leaving sticky ends). [2] DNA ligase joins the complementary sticky ends of the target DNA and the plasmid together by forming phosphodiester bonds. [2]

Q19 (Cells/Membrane) (a) Movement of molecules via a transmembrane protein (channel or carrier) down a concentration gradient. [2] Facilitated diffusion is passive (no ATP); active transport requires ATP to move molecules against a gradient. [2] (b) There are a limited number of carrier proteins in the membrane. [1] Once all binding sites are occupied, the rate cannot increase further (saturation). [1]

Q20 (Cell Cycle) (a) DNA replication occurs. [1] Each chromosome is duplicated to form two sister chromatids. [1] This ensures that each daughter cell receives a complete set of genetic information. [1] (b) Spindle fibres attach to the centromeres of sister chromatids. [1] They pull the chromatids apart toward opposite poles of the cell. [1] This ensures an equal distribution of chromosomes. [1] The result is two genetically identical nuclei. [1]