Topic: 2.1
Scientists measured the concentration of sodium ions and potassium ions in the red blood cells and in the blood plasma of a group of people. The results are shown in Table 1.1.
(a) (i) Use the information in Table 1.1 to identify and describe the process by which potassium ions enter red blood cells from the blood plasma.
(ii) Sodium ions and oxygen molecules enter red blood cells.
State one similarity and one difference between the processes used by sodium ions and oxygen molecules to enter red blood cells.
(iii) Chloride ions move across the membrane of human red blood cells in a process called the chloride shift.
Explain why the chloride shift is important in the transport of carbon dioxide from respiring tissues.
(b) Scientists studied the uptake of a substance, F, by human red blood cells.
The red blood cells were immersed in a solution of substance F for 30 minutes. After this time the scientists recorded two observations:
- the cell surface membrane of the red blood cells showed infoldings (invaginations)
- an increase in the number of vesicles in the cytoplasm.
Identify the process by which substance F entered the red blood cells.
▶️ Answer/Explanation
(a)(i) Potassium ions enter red blood cells by active transport.
Explanation: The concentration of potassium ions is higher inside red blood cells (100 mmol dm-3) than in the plasma (4 mmol dm-3), indicating movement against the concentration gradient. This requires energy (ATP) and involves carrier proteins that undergo conformational changes to transport the ions. The process is specific to potassium ions due to the complementary shape of binding sites on the transport proteins.
(a)(ii) Similarity: Both processes involve movement down a concentration gradient (passive transport).
Difference: Sodium ions use facilitated diffusion (via protein channels/carriers) while oxygen uses simple diffusion (directly through the phospholipid bilayer).
Explanation: The similarity lies in both being passive processes that don’t require energy. The key difference is in their mechanisms – sodium ions, being charged, cannot cross the hydrophobic membrane core and require protein channels, while nonpolar oxygen molecules can diffuse directly through the lipid bilayer.
(a)(iii) The chloride shift maintains electrical neutrality as hydrogen carbonate ions leave the red blood cell, enabling more carbon dioxide transport.
Explanation: When CO2 enters red blood cells, it forms hydrogen carbonate ions (HCO3–) which diffuse out to plasma. To balance this negative charge loss, chloride ions (Cl–) move into the cell. This maintains electrochemical equilibrium, allowing continuous CO2 uptake from tissues and preventing charge buildup that would inhibit the carbonic anhydrase reaction.
(b) Endocytosis (specifically pinocytosis).
Explanation: The observations of membrane infoldings and increased vesicles indicate bulk transport of substance F via endocytosis. The membrane invaginates to form a vesicle around the substance, allowing its uptake without passing through the membrane. This is characteristic of large or polar molecules that cannot cross the membrane by simpler mechanisms.
Topic: 2.3
Fig. 2.1 is a transmission electron micrograph showing a section of a specialised epithelial cell found in the lining of the stomach. This cell produces extracellular proteins that are released into the bloodstream.
(a) Outline the role of structure X and structure Y, as shown in Fig. 2.1, in the production of extracellular proteins.
(b) The cell in Fig. 2.1 releases ghrelin, a small protein that acts as a cell signalling molecule.
Fig. 2.2 shows the sequence of amino acids in a ghrelin molecule.
The amino acid serine (Ser) in position 3 in Fig. 2.2 has been modified by the addition of a saturated fatty acid chain.
(i) State the level of protein structure shown in Fig. 2.2.
(ii) State one similarity between the structure of a saturated fatty acid molecule and an amino acid molecule.
(iii) The addition of the fatty acid chain allows ghrelin to function as a cell signalling molecule.
Suggest how the addition of this fatty acid chain allows a ghrelin molecule to act as a cell signalling molecule.
(c) Scientists have discovered that the gene coding for ghrelin contains 5000 base pairs. This is a much larger number of base pairs than is needed to code for the ghrelin molecule shown in Fig. 2.2.
(i) Calculate the percentage of base pairs found in the gene that codes for the ghrelin molecule shown in Fig. 2.2.
Show your working and give your answer to one decimal place.
(ii) The primary transcript produced from the ghrelin gene is a longer molecule than the mRNA found in the cytoplasm.
Explain how the primary transcript is modified before translation.
▶️ Answer/Explanation
(a)
Structure X (nucleolus): The nucleolus is responsible for the synthesis of ribosomal subunits or ribosomes. These ribosomes are essential for protein synthesis as they are the sites where translation occurs.
Structure Y (chromatin): Chromatin contains the genes that code for the polypeptides or proteins. It serves as the site for transcription and the production of mRNA, which carries the genetic information from DNA to the ribosomes for protein synthesis.
(b)(i) Primary structure.
Explanation: The figure shows the linear sequence of amino acids in the ghrelin molecule, which represents its primary structure – the most basic level of protein organization.
(b)(ii) Both contain a carboxylic acid/carboxyl group.
Explanation: Both saturated fatty acids and amino acids contain a carboxyl group (-COOH), which is a key functional group in these molecules. Additionally, both types of molecules contain carbon, hydrogen, and oxygen atoms.
(b)(iii)
The fatty acid chain plays several important roles in allowing ghrelin to function as a cell signalling molecule:
- It contributes to the overall 3D shape (tertiary structure) of the ghrelin molecule, helping to form a hydrophobic region that is important for binding.
- It allows ghrelin to bind specifically to its receptor by creating a complementary shape that fits the receptor site.
- The hydrophobic nature of the fatty acid chain may help the molecule interact with or embed in the hydrophobic portion of cell membranes, bringing the signaling portion of the molecule closer to its receptor.
- The modification may help the molecule be transported in the bloodstream by binding to transport proteins.
(c)(i) 1.8%
Working: The ghrelin molecule shown has 28 amino acids (positions 1-28) plus a stop codon, totaling 29 codons × 3 base pairs each = 87 base pairs needed. Percentage = (87/5000) × 100 = 1.74%, which rounds to 1.8% to one decimal place.
(c)(ii)
The primary transcript undergoes several modifications before becoming mature mRNA:
- Splicing: Introns (non-coding regions) are removed and exons (coding regions) are joined together through a process called RNA splicing.
- 5′ Capping: A modified guanine nucleotide is added to the 5′ end of the mRNA to protect it from degradation and help with ribosome binding.
- Polyadenylation: A poly-A tail (many adenine nucleotides) is added to the 3′ end to increase stability and facilitate export from the nucleus.
These modifications result in a much shorter, mature mRNA molecule that can be translated into protein.
Topic: 3.1
Plants have specialised cells for the efficient transport of assimilates.
(a) Table 3.1 shows some of the features of two different types of cell found in plant tissue, which are adapted for the efficient transport of assimilates.
Many plasmodesmata connect type A cells with type B cells.
Identify cell type A and explain why the plasmodesmata are important.
(b) The polysaccharide callose is found in the cell walls of cells close to plasmodesmata.
Fig. 3.1 is a simplified diagram showing the structure of callose. Not all hydrogen atoms are shown.
Complete Table 3.2 to compare the structure of a callose molecule with a cellulose molecule.
(c) Hydrogen bonding is important for movement of water through xylem vessels in a plant.
Describe the roles of hydrogen bonding in the movement of water through xylem vessels.
(d) Water that has travelled through xylem vessels reaches the leaves. Cooling of the leaf occurs as a result of the evaporation of water during transpiration.
Water has a high latent heat of vaporisation because water molecules form hydrogen bonds.
With reference to hydrogen bonding, suggest why cooling of the leaf occurs as a result of evaporation of water during transpiration.
▶️ Answer/Explanation
(a)
Cell type A: Phloem sieve tube element
Explanation of plasmodesmata importance:
- Plasmodesmata connect the cytoplasm of adjacent cells, allowing for the symplast pathway of transport.
- They enable rapid diffusion of sucrose and other molecules between sieve tube elements and companion cells.
- This connection increases the efficiency of loading assimilates into and unloading from the phloem sieve tubes.
- The presence of many plasmodesmata creates a continuous pathway for assimilate transport throughout the plant.
(b)
Completed Table 3.2
Explanation: Callose and cellulose are both composed of β-glucose monomers but differ in their glycosidic bonds (1,3 vs 1,4) which affects their molecular shape and properties. The 1,3 bonds in callose create a helical structure, while the 1,4 bonds in cellulose create straight chains that can form strong microfibrils.
(c)
Roles of hydrogen bonding in water movement:
- Hydrogen bonding between water molecules creates cohesion, forming a continuous column of water through the xylem vessels.
- Hydrogen bonding between water molecules and cellulose/lignin in xylem walls (adhesion) helps prevent the water column from collapsing.
- The cohesive forces maintain water tension as it’s pulled upward by transpiration.
- These hydrogen bonds allow water to be pulled up against gravity in the transpiration stream.
- The combination of cohesion and adhesion creates capillary action that helps water rise in narrow xylem vessels.
(d)
Cooling effect explanation:
- Water molecules are held together by hydrogen bonds which require significant energy to break.
- During evaporation, heat energy from the leaf is used to break these hydrogen bonds between water molecules.
- This energy (latent heat of vaporization) is taken from the leaf surface, resulting in cooling.
- The high number of hydrogen bonds in water means more energy is required for evaporation, leading to greater cooling.
- This cooling mechanism helps prevent overheating of leaves during photosynthesis in sunlight.
Topic: 11.1
Some people who are infected with HIV develop HIV/AIDS.
(a) Fig. 4.1 shows the number of people that have been newly infected with HIV (new infections) in 2018 across the world and the percentage changes in the number of new infections since 2010.
(i) State the full name of the pathogen HIV.
(ii) Using only the data in Fig. 4.1, state what can be concluded about the change in number of new HIV infections across the world between 2010 and 2018. You may use the letters in Fig. 4.1 to identify the regions of the world.
(b) People who develop HIV/AIDS have fewer T-helper cells as the pathogen destroys these cells. This makes them more susceptible to tuberculosis (TB). Explain why people with HIV/AIDS are more likely to develop TB.
(c) A person infected with HIV may develop a heart condition called cardiac tamponade. The pericardium is a thin sac surrounding the heart, as shown in the healthy heart in Fig. 4.2. When a person develops cardiac tamponade, the pericardium starts to fill with blood. This prevents the ventricles functioning efficiently.
People with cardiac tamponade experience a variety of symptoms including low blood pressure and an increased breathing rate. Suggest why a person with cardiac tamponade would experience symptoms of low blood pressure and an increased breathing rate.
▶️ Answer/Explanation
(a)(i) Human immunodeficiency virus.
Explanation: HIV stands for Human Immunodeficiency Virus, which is the pathogen that attacks the body’s immune system, specifically the CD4 cells (T-helper cells), weakening the immune system over time and leading to AIDS if not treated.
(a)(ii)
From the data in Fig. 4.1, we can conclude that:
- There is a general decreasing trend in new HIV infections across most regions of the world between 2010-2018.
- Region E (East and Southern Africa) shows the most significant decrease at 28%.
- However, two regions show increases: Region B (Eastern Europe and Central Asia) with a 29% increase and Region D (Middle East and North Africa) with a 10% increase.
- Region G (Latin America) shows the smallest decrease at just 7%.
- The data suggests that while progress is being made in most areas, certain regions are experiencing worsening HIV infection rates.
(b)
People with HIV/AIDS are more susceptible to TB because:
- HIV specifically targets and destroys T-helper cells (CD4 cells), which are crucial for coordinating the immune response.
- With fewer T-helper cells, the immune system becomes weakened and less able to fight off infections like TB.
- T-helper cells normally stimulate macrophages to destroy TB bacteria and help B-cells produce antibodies against TB.
- The reduced immune response allows latent TB infections to become active.
- There’s less cytokine production to activate other immune cells against TB.
- The weakened immune system may not respond effectively to TB vaccines.
This makes TB an “opportunistic infection” that takes advantage of the weakened immune system in HIV/AIDS patients.
(c)
A person with cardiac tamponade experiences:
Low blood pressure because:
- The blood filling the pericardial sac compresses the ventricles externally.
- This compression reduces the volume of blood the ventricles can hold during filling (diastole).
- The ventricles cannot expand fully, so less blood enters them between beats.
- With less blood entering, less blood is pumped out during contraction (systole).
- This reduced stroke volume leads to decreased cardiac output and consequently lower blood pressure.
Increased breathing rate because:
- The reduced cardiac output means less oxygen is delivered to tissues.
- The body compensates by increasing breathing rate to try to get more oxygen into the blood.
- Also, the reduced blood flow leads to carbon dioxide buildup, which stimulates faster breathing to remove it.
- The respiratory center in the brain detects these changes and increases the breathing rate as a compensatory mechanism.
These symptoms represent the body’s attempt to maintain adequate oxygenation despite the impaired heart function caused by the tamponade.
Topic: 5.1
Fig. 5.1 is a photomicrograph of a transverse section of a bronchus in the lungs.
(a) (i) Identify the tissue labelled T in Fig. 5.1.
(ii) Describe the function of smooth muscle in the bronchus.
(b) The walls of alveoli contain some specialised epithelial cells called type II epithelial cells. These cells secrete surfactant. Surfactant helps to prevent the alveoli collapsing during breathing.
Surfactant contains phospholipid, cholesterol and protein.
The components of surfactant are synthesised in the rough endoplasmic reticulum and smooth endoplasmic reticulum and then passed to the Golgi body.
The surfactant that is produced is stored in secretory organelles called lamellar bodies.
The surfactant in the lamellar bodies is released onto the surface of the alveolar epithelium by exocytosis, as shown in Fig. 5.2.
(i) Each lamellar body is surrounded by a single membrane.
Draw a diagram to show the arrangement of phospholipid molecules in the membrane surrounding the lamellar body.
(ii) Scientists studying the production and secretion of lung surfactant have discovered that a reduction in cholesterol in the cell surface membrane of type II epithelial cells reduces the secretion of surfactant.
Suggest why secretion of surfactant is affected by a reduction in cholesterol in the cell surface membranes of type II epithelial cells.
(iii) Lung surfactant is engulfed by macrophages that are in close contact with the type II epithelial cells.
Suggest why macrophages engulf surfactant.
▶️ Answer/Explanation
(a)(i) Cartilage.
Explanation: The tissue labelled T is cartilage, which provides structural support to the bronchus and helps maintain an open airway. Cartilage is a firm but flexible connective tissue that prevents the bronchus from collapsing during inhalation and exhalation.
(a)(ii)
Explanation: The smooth muscle in the bronchus has two main functions:
- It contracts and relaxes to change the diameter of the airway, controlling the size of the lumen. This helps regulate airflow into and out of the lungs.
- During conditions like asthma, the smooth muscle can constrict excessively (bronchoconstriction), narrowing the airways and reducing airflow. Conversely, relaxation of the smooth muscle increases airflow.
(b)(i)
Explanation: The diagram should show a phospholipid bilayer with:
- Phospholipid molecules arranged with their hydrophilic (water-loving) heads facing outward toward the aqueous environment and hydrophobic (water-fearing) tails facing inward
- Each phospholipid molecule should be represented with a circular head and two parallel tails
- The membrane should show two layers of these molecules (bilayer)
(b)(ii)
Explanation: Cholesterol plays several important roles in membrane function that affect surfactant secretion:
- Cholesterol helps maintain optimal membrane fluidity – not too rigid in cold temperatures and not too fluid in warm temperatures.
- Reduced cholesterol makes the membrane less stable, potentially affecting the fusion process between lamellar bodies and the cell membrane during exocytosis.
- Cholesterol is important for the formation of membrane microdomains (lipid rafts) that may be involved in the secretory process.
(b)(iii)
Explanation: Macrophages engulf surfactant for several important reasons:
- To prevent excessive accumulation of surfactant in the alveoli, which could impair gas exchange.
- To recycle and reuse the components of surfactant (phospholipids, proteins, cholesterol) in the production of new surfactant.
- To remove any pathogens or foreign particles that might have become trapped in the surfactant layer.
Topic: 6.1
(a) (i) State the name of the phase of the mitotic cell cycle during which DNA replication occurs.
(ii) During research, scientists can use modified nucleotides to prevent elongation of a polynucleotide chain during DNA replication.
Fig. 6.1 shows the structure of a DNA nucleotide found in the nucleus of a cell.
Fig. 6.2 shows the structure of a modified nucleotide.
Suggest how the structure of the modified nucleotide prevents DNA polymerase joining it to another nucleotide.
(b) RNA aptamers are short, single-stranded RNA molecules that can be used to study some infectious diseases.
Scientists studying an infectious disease in animals investigated the effect of RNA aptamers on the activity of RNA polymerase that is produced by the pathogen.
The aptamers bind to a specific region of the RNA polymerase.
Table 6.1 shows the effect of two aptamers, F47 and F52, on the activity of RNA polymerase produced by the pathogen.
(i) With reference to Table 6.1, compare the effect of the aptamers on the affinity of RNA polymerase for its substrate.
(ii) With reference to Table 6.1, suggest explanations for the effect of the presence of an aptamer on the rate of transcription catalysed by the RNA polymerase.
▶️ Answer/Explanation
(a)(i) S phase / synthesis phase / interphase
Explanation: The S phase (Synthesis phase) of the cell cycle is when DNA replication occurs. This happens during interphase, specifically in the S phase, where the cell’s DNA is duplicated in preparation for cell division. The cell ensures that each daughter cell will receive an identical copy of the genetic material.
(a)(ii)
The modified nucleotide prevents DNA polymerase from joining it to another nucleotide because:
- It lacks a hydroxyl group (OH) on the 3′ carbon of the deoxyribose sugar, which is essential for forming the phosphodiester bond with the next nucleotide.
- Without this hydroxyl group, the condensation reaction between the phosphate group of one nucleotide and the hydroxyl group of another cannot occur.
- The modified structure means the nucleotide cannot properly fit into the active site of DNA polymerase, preventing the formation of enzyme-substrate complexes.
- The hydrogen atom replacing the hydroxyl group changes the shape of the nucleotide, making it incompatible with the polymerase’s active site.
(b)(i)
From Table 6.1, we can observe that:
- Both aptamers F47 and F52 increase the Km value of RNA polymerase (from 346 to 508 and 523 respectively), indicating they both reduce the enzyme’s affinity for its substrate.
- F52 reduces the affinity more than F47, as shown by its higher Km value (523 vs 508).
- The greater the Km value, the lower the enzyme’s affinity for the substrate, meaning more substrate is needed to achieve half of Vmax.
(b)(ii)
The presence of aptamers affects transcription rate because:
- Both aptamers decrease the maximum reaction rate (Vmax), with F47 having a more significant effect (reducing Vmax from 1664 to 1072) than F52 (reducing to 1467).
- The aptamers likely bind to RNA polymerase, changing its tertiary structure and altering the shape of the active site, making it less complementary to the substrate.
- With a distorted active site, fewer enzyme-substrate complexes form, and the enzyme cannot catalyze the reaction as efficiently.
- The aptamers might physically block the active site or prevent RNA polymerase from binding properly to the DNA template.
- The binding could interfere with the polymerase’s ability to form phosphodiester bonds between RNA nucleotides, slowing down the overall transcription process.