Topic: 1.1 (The microscope in cell studies)
A student used a light microscope to observe a blood smear on a microscope slide. An eyepiece graticule was used to measure the diameter of a white blood cell on the slide. The student recorded that the white blood cell was 5 eyepiece graticule units in diameter. Which additional information does the student need to determine the diameter of the white blood cell in micrometres?
▶️ Answer/Explanation
Ans: A
To convert eyepiece graticule units to micrometres, the student must calibrate the eyepiece graticule using a stage micrometer. The stage micrometer provides a known scale, allowing the student to determine the actual size (in µm) corresponding to each graticule unit. The magnification of the lenses is not required for this conversion, as calibration directly relates graticule units to real-world measurements.
Topic: 1.1 (The microscope in cell studies)
Which statement explains why it is necessary to use an electron microscope to see the cristae of a mitochondrion?
▶️ Answer/Explanation
Ans: C
Cristae are intricate internal folds of the mitochondrial inner membrane, with separations often less than 200 nm. The resolution of a light microscope is limited by the wavelength of visible light (~200 nm), making it insufficient to resolve such fine structures. An electron microscope uses a much shorter wavelength (electron beam), enabling higher resolution and visualization of cristae. Thus, option C correctly explains the necessity of an electron microscope.
Topic: 1.2 (cells as the basic units of living organisms)
Some stains can be used to identify cell structures in living cells.
A dilute solution of one stain causes the whole cell to appear blue.
The blue colour rapidly disappears from most cell structures. Those cell structures that release
energy stay blue.
Which type of cell structure is likely to stay blue?
▶️ Answer/Explanation
Ans: D
The stain remains blue in cell structures that release energy. The mitochondrion is the organelle responsible for energy production (ATP) through cellular respiration. Other options (Endoplasmic reticulum, Golgi body, lysosome) do not primarily function in energy release, making D the correct answer.
Topic: 9.1 (The gas exchange system)
When mucus is secreted from a goblet cell, these events take place.
1 addition of carbohydrate to protein
2 fusion of a vesicle with the cell surface membrane
3 extracellular release of a glycoprotein
4 separation of a vesicle from the Golgi body
What is the sequence in which these events take place?
▶️ Answer/Explanation
Ans: A
The correct sequence for mucus secretion in goblet cells is: 1 → 4 → 2 → 3. First, carbohydrate is added to protein (glycosylation) in the Golgi apparatus (1). Then, a vesicle containing the glycoprotein separates from the Golgi (4). The vesicle fuses with the cell membrane (2), releasing the glycoprotein extracellularly (3). This matches option A.
Topic: 8.1 (Classification)
The diagram shows four biological features.
Which biological features are present in typical prokaryotes?
▶️ Answer/Explanation
Ans: C
Prokaryotes (e.g., bacteria) have a cell wall (1), ribosomes (3), and genetic material (4) in the form of a circular DNA molecule. However, they lack membrane-bound organelles (2) like mitochondria or a nucleus. Thus, the correct combination is 1, 3, and 4 (Option C).
Topic: 7.1 (structure of transport tissues)
Which row shows features that occur in dicotyledonous plant cells and also in typical bacterial cells?
key
✓ = found in dicotyledonous plant cells and also in typical bacterial cells
✗ = not found in at least one of these two types of cell
▶️ Answer/Explanation
Ans: B
In dicotyledonous plant cells and typical bacterial cells, the common features are cell membrane and cytoplasm. However, features like cellulose cell wall (only in plants) and nucleus (absent in bacteria) are not shared. Row B correctly identifies these shared structures (✓) and differences (✗).
Topic: 2.1 (Testing for biological molecules)
Which flow chart outlining the test for non-reducing sugars is correct?
▶️ Answer/Explanation
Ans: C
The correct test for non-reducing sugars involves first hydrolyzing them into reducing sugars using hydrochloric acid (HCl) and then neutralizing with sodium hydrogen carbonate (NaHCO3). After that, Benedict’s solution is added and heated. Only option C follows this sequence, making it the correct choice.
Topic: 9.1 (Carbohydrates and Lipids)
Which diagrams show the release of a water molecule during the formation of a glycosidic bond?
▶️ Answer/Explanation
Ans: A
A glycosidic bond forms between two monosaccharides in a condensation reaction, releasing a water molecule. Diagrams 1, 2, and 3 all depict this process, where hydroxyl (–OH) groups from each sugar combine to form \( \text{H}_2\text{O} \). Thus, the correct answer is A (1, 2, and 3).
Topic: 2.2 (proteins)
Which molecules have a structural formula that contains C=O bonds?
1 amino acids
2 fatty acids
3 glycerol
4 protein
▶️ Answer/Explanation
Ans: B
The C=O (carbonyl) bond is present in:
- Amino acids (carboxyl group –COOH, which contains C=O).
- Fatty acids (carboxyl group –COOH).
- Proteins (due to peptide bonds, which include C=O).
Glycerol does not contain a C=O bond, as it has only hydroxyl (–OH) groups. Thus, the correct combination is 1, 2, and 4 (Option B).
Topic: 2.2 (carbohydrates and lipids)
Which statement about triglycerides is correct?
▶️ Answer/Explanation
Ans: D
Triglycerides consist of a glycerol molecule bonded to three fatty acids via ester (covalent) bonds. Option D is correct because glycerol forms covalent bonds with fatty acids during esterification. Option A is incorrect as triglycerides can have mixed fatty acids (saturated + unsaturated). Option B is false because triglycerides are individual molecules, not joined. Option C is incorrect as triglycerides are nonpolar and hydrophobic.
Topic: 2.1 (Testing for biological molecules)
Many flowers produce a sweet solution called nectar. Bees provided with nectar use enzyme Q to change the nectar into honey.
After testing a sample of nectar for the presence of reducing sugar using standard laboratory reagents, the sample was blue. After testing a sample of honey in the same way, the sample was orange.
Which conclusion about the reaction catalysed by enzyme Q is consistent with these results?
▶️ Answer/Explanation
Ans: C
The blue color in the nectar test indicates no reducing sugars (negative Benedict’s test), while the orange color in honey confirms their presence. This suggests enzyme Q converts non-reducing sugars (like sucrose in nectar) into reducing sugars (like glucose/fructose in honey). Thus, the correct conclusion is that enzyme Q produces reducing sugars from non-reducing sugars.
Topic: 1.2 (Cells as the basic units of living organisms)
The diagram shows the structure of part of a peptidoglycan molecule.
Which type of 1,4 linkage and how many peptide bonds are shown in this part of the molecule?
▶️ Answer/Explanation
Ans: D
Peptidoglycan consists of repeating units of N-acetylmuramic acid (NAM) and N-acetylglucosamine (NAG) linked by β 1,4-glycosidic bonds. In the given structure, two peptide bonds are visible: one between NAM and the first amino acid (L-alanine) and another between the last amino acids (D-alanine and another residue). Thus, the correct answer is β 1,4 linkage with 2 peptide bonds (Option D).
Topic: 3.1 (Mode of action of enzymes)
A scientist investigated the progress of two enzyme-catalysed reactions in separate test tubes, X and Y. Both reactions result in colour changes that can be detected using colorimetry.
0.5 cm samples were taken from each test tube at the start of the investigation and at regular intervals for the next 5 minutes. Copper ions were added to each sample as soon as the sample
was collected to inactivate the enzymes and stop the reactions from progressing further.
The absorbance of each sample was measured using a colorimeter.
The graph shows the results of this investigation.
Which statement is consistent with the results shown in the graph?
▶️ Answer/Explanation
Ans: B
From the graph:
- Test-tube Y shows a decrease in absorbance over time, indicating that the product has a lower absorbance than the substrate (as the reaction progresses, absorbance decreases).
- Test-tube X shows an increase in absorbance, but the question asks for a statement consistent with the results, and only option B correctly describes the relationship in test-tube Y.
- Options C and D incorrectly describe the reaction rates, as the graph shows linear trends (constant rates) for both reactions.
Thus, B is the correct answer.
Topic: 2.2 (carbohydrates and lipids)
A student investigated the hydrolysis of lipid in high-fat milk, using the enzyme lipase.
• 1 cm³ of enzyme solution was added to 10 cm³ of high-fat milk.
• The temperature was kept constant.
• The pH of the reaction mixture was recorded at time 0 minutes and every minute for 20 minutes.
Which statements correctly describe the expected results of this investigation?
1. The product forms more slowly as time proceeds because the concentration of the substrate is decreasing.
2. The pH of the reaction mixture increases rapidly in the first few minutes and then increases less rapidly.
3. The increase in the concentration of product eventually causes the lipase molecules to denature.
▶️ Answer/Explanation
Ans: B (1 and 2 only)
Statement 1 is correct because as lipids (substrate) are broken down, their concentration decreases, slowing the reaction rate.
Statement 2 is correct because lipase releases fatty acids, decreasing pH initially, but the rate slows as substrate depletes.
Statement 3 is incorrect because product accumulation does not denature lipase; only extreme pH or temperature does.
Topic: 4.1 (Fluid mosaic membrane)
What is the correct range of measurements for the width of the cell surface membrane?
▶️ Answer/Explanation
Ans: B
The width of the cell surface membrane (plasma membrane) is approximately 7–10 nm, as determined by electron microscopy. This matches option B (5–10 nm). The membrane consists of a phospholipid bilayer (~5 nm) with embedded proteins, giving a total thickness within this range. Options A and D are too small, while C is too large.
Topic: 4.1 (Fluid Mosaic Membrane)
Where in the cell surface membrane are the carbohydrate chains of glycoproteins and glycolipids mainly located?
▶️ Answer/Explanation
Ans: D
Carbohydrate chains of glycoproteins and glycolipids are attached to the outer surface (extracellular side) of the cell membrane (region 4 in the diagram). These carbohydrate moieties play key roles in cell recognition, signaling, and protection. The other regions (1, 2, 3) represent the hydrophobic core or cytosolic side, where carbohydrates are absent.
Topic: 4.2 (Movement into and out of cells)
Sodium ions can enter cells across the cell surface membrane.
Which methods could be used by sodium ions to cross a cell surface membrane and enter a cell?
▶️ Answer/Explanation
Ans: B
Sodium ions (Na+) are charged particles and cannot passively diffuse through the hydrophobic lipid bilayer. They enter cells via:
- Active transport (against concentration gradient, using ATP and carrier proteins).
- Facilitated diffusion (down concentration gradient, through channel or carrier proteins).
Simple diffusion is not possible due to their charge and polarity. Thus, B is correct.
Topic: 4.2 (Movement into and out of cells)
The diagram shows how an artificial partially permeable membrane was used to separate a 5% sodium chloride solution and a 10% sodium chloride solution in a beaker.
The two sides of the beaker were labelled R and S.
Which row correctly describes and explains what will happen in the half of the beaker labelled S?
▶️ Answer/Explanation
Ans: A
Since side S has a 10% NaCl solution (higher concentration) and side R has a 5% NaCl solution (lower concentration), water will move from R (lower solute concentration) to S (higher solute concentration) by osmosis. This will cause the volume in S to increase, and the NaCl concentration in S to decrease as it becomes diluted. Thus, option A is correct.
Topic: 4.2 (Movement into and out of cells)
Agar cubes can be used to demonstrate the effect on diffusion of changing the surface area to volume ratio. Three different agar cubes made using a coloured indicator solution were placed into a dilute acid that diffused into the cubes. As the acid diffused into the agar cubes, the colour of the indicator solution changed. The cubes had volumes of 1 cm, 2 cm, and 3 cm and were left in the dilute acid for 10 minutes. All other variables were kept the same. After 10 minutes, the agar cubes were removed from the dilute acid and cut in half. The cut surfaces were observed, and the results were recorded as diagrams. All diagrams were drawn to the same scale. The results for the 2 cm cube are shown.
▶️ Answer/Explanation
Ans: C
The 1 cm cube has a higher surface area-to-volume ratio, allowing the acid to diffuse completely, resulting in a full colour change. The 3 cm cube has a lower ratio, so diffusion is slower and only partial colour change occurs in the same time. The 2 cm cube (shown) exhibits an intermediate change, supporting option C as correct.
Topic: 5.1 (Replication and division of nuclei and cells)
During sperm formation in mammals, part of the structure of each chromosome is replaced with proteins called protamines.
This replacement allows the DNA to be packaged much more densely than would otherwise be possible.
Which part of the chromosome is replaced by protamines?
▶️ Answer/Explanation
Ans: C
In sperm formation (spermatogenesis), the DNA is tightly packed to fit inside the sperm head. Normally, DNA is wrapped around histones to form nucleosomes. However, in sperm cells, histones are replaced by smaller, positively charged proteins called protamines, which allow for extreme DNA condensation.
Centromeres (A), chromatids (B), and telomeres (D) are structural parts of chromosomes but are not replaced by protamines. Thus, the correct answer is C (histones).
Topic: 5.1 (The mitotic cell cycle)
The diagram shows the mitotic cell cycle.
During which phase do chromosomes condense and become visible?
▶️ Answer/Explanation
Ans: D
Chromosomes condense and become visible during prophase of mitosis. This is when chromatin fibers coil tightly, making chromosomes distinct under a microscope. The other phases (G1, S, G2) are part of interphase, where chromosomes remain uncondensed as chromatin. Thus, the correct answer is D (Prophase).
Topic: 5.1 (The mitotic cell cycle)
Which statements correctly describe features of stem cells that are essential for their role in cell replacement and tissue repair?
1. After mitosis of stem cells, the daughter cells can either remain as stem cells or follow a developmental pathway that leads to the formation of specialised cells.
2. Stem cells are different to all other body cells because they retain all of the genetic information in their DNA throughout the life of the organism.
3. A small population of stem cells is retained in the body of adults throughout their life time.
4. Stem cells have more telomeres than other body cells and this allows them to undergo an unlimited number of mitotic divisions.
A. 1, 2, 3 and 4
B. 1, 2 and 3 only
C. 1 and 3 only
D. 2, 3 and 4 only
▶️ Answer/Explanation
Ans: C
Statement 1 is correct: Stem cells can self-renew (remain undifferentiated) or differentiate into specialized cells.
Statement 3 is correct: Adult stem cells persist in tissues (e.g., bone marrow) for repair.
Statement 2 is incorrect: All body cells retain the same genetic information; stem cells differ in their ability to express it variably.
Statement 4 is incorrect: Stem cells have active telomerase (not more telomeres), enabling prolonged division, but not unlimited.
Topic: 6.1 (Structure of nucleic acids and Replication of DNA)
Which statements are correct for all nucleotides?
- The nitrogen-containing base is always attached to carbon atom 1 of the pentose.
- The phosphate group is always attached to carbon atom 5 of the pentose.
- A condensation reaction occurs to join the nitrogen-containing base to the pentose.
- Nucleotides are linked together by condensation reactions between phosphate groups.
▶️ Answer/Explanation
Ans: A
Statements 1, 2, and 3 are correct for all nucleotides:
- The nitrogenous base always attaches to the 1′ carbon of the pentose sugar.
- The phosphate group always binds to the 5′ carbon of the pentose.
- The base-sugar bond forms via a condensation reaction (removing water).
Statement 4 is incorrect because nucleotides are linked by phosphodiester bonds (condensation between a phosphate group and the 3′ hydroxyl of another nucleotide), not just phosphate-phosphate bonds. Thus, Option A (1, 2, 3) is correct.
Topic: 6.1 (Structure of nucleic acids and Replication of DNA)
How many phosphodiester bonds are present in a circular DNA molecule of 2700 base pairs?
▶️ Answer/Explanation
Ans: D
In a double-stranded circular DNA molecule:
- Each strand has 2700 phosphodiester bonds (one bond per nucleotide linkage).
- Since there are two strands, the total number of phosphodiester bonds is \( 2700 \times 2 = 5400 \).
- The circular nature means there are no free ends, so all bonds are intact.
Thus, the correct answer is D (5400).
Topic: 6.1 (Structure of nucleic acids and replication of DNA)
Which statements about complementary base pairing are correct?
1. Purines and pyrimidines are different sizes.
2. Complementary base pairing occurs during translation.
3. The base pairs are of different lengths.
4. Uracil forms two hydrogen bonds with adenine.
▶️ Answer/Explanation
Ans: B (1, 2, and 4)
Statement 1 is correct: Purines (A, G) are double-ring structures, while pyrimidines (C, T, U) are single-ring, making them different sizes.
Statement 2 is correct: Complementary base pairing occurs during translation (e.g., tRNA anticodon pairs with mRNA codon).
Statement 3 is incorrect: All base pairs (A-T/U, C-G) have uniform lengths due to consistent hydrogen bonding.
Statement 4 is correct: Uracil pairs with adenine via two hydrogen bonds, similar to thymine in DNA.
Topic: 6.1 (Structure of nucleic acids and replication of DNA)
During the mitotic cell cycle, the chromosomal DNA is replicated. The specific points in DNA molecules where replication is occurring are known as replication forks.
A typical human chromosome has about 150 million base pairs of DNA. It takes about 1 hour to replicate the DNA of a typical human chromosome.
The rate of replication using a single replication fork is approximately 50 base pairs per second.
Approximately how many replication forks must occur in a typical human chromosome during DNA replication?
▶️ Answer/Explanation
Ans: A
Step 1: Calculate the total base pairs replicated per fork in 1 hour.
Rate = 50 base pairs/second × 3600 seconds = 180,000 base pairs/fork.
Step 2: Divide the total DNA (150 million base pairs) by the output per fork.
Number of forks = 150,000,000 ÷ 180,000 ≈ 833.33.
Since forks work bidirectionally, the closest option is A (835).
Topic: 3.1 (Mode of Action of Enzymes)
Which molecule has its synthesis directly controlled by DNA?
▶️ Answer/Explanation
Ans: A
Amylase is an enzyme (a protein), and proteins are directly synthesized based on the genetic code in DNA through transcription and translation. Cholesterol (B), glycogen (C), and phospholipids (D) are lipids or carbohydrates, which are synthesized via metabolic pathways controlled by enzymes (indirectly by DNA), not directly by DNA itself.
Topic: 7.1 (Structure of transport in plants)
Which statement correctly describes the association between a companion cell and its sieve tube cell?
▶️ Answer/Explanation
Ans: D
Sieve tube cells lack nuclei and rely on their adjacent companion cells for metabolic support. The key roles of companion cells include:
- Providing a nucleus to regulate cellular activities in both cells (since sieve tubes are enucleated).
- Supplying ATP and proteins via plasmodesmata, but not all ATP for both cells (A is incorrect).
Options B and C describe incorrect or partial functions. Thus, D is the most accurate statement.
Topic: 7.1 (Structure of transport tissues)
A maize seedling was grown in soil that contained lanthanum ions labelled with a chemical that fluoresces under ultraviolet light. The diagram represents what was observed when a section of root was examined using a light microscope with ultraviolet illumination.
What is a correct conclusion about the transport of lanthanum ions in maize roots?
A. The ions are not transported through the apoplast pathway or the symplast pathway.
B. The ions are transported through the apoplast pathway only.
C. The ions are transported through the apoplast pathway and symplast pathway.
D. The ions are transported through the symplast pathway only.
▶️ Answer/Explanation
Ans: B
The diagram shows fluorescence only in the cell walls (apoplast pathway) and not inside the cytoplasm or plasmodesmata (symplast pathway). This indicates that lanthanum ions move through the apoplast pathway only, as they cannot cross the plasma membrane into the symplast. Thus, Option B is correct.
Topic: 7.2 (Transport Mechanisms)
Which description of adhesion and cohesion is correct?
▶️ Answer/Explanation
Ans: D
Adhesion is the attraction between water molecules and the xylem vessel walls (due to polar interactions), while cohesion is the attraction between water molecules themselves (due to hydrogen bonding). Option D is the only choice that accurately describes both terms, making it correct.
Topic: 7.2 (Transport mechanisms)
Which conditions are needed to allow the mass flow of sucrose in phloem sieve tubes?
▶️ Answer/Explanation
Ans: B
The mass flow of sucrose in phloem sieve tubes occurs due to the pressure flow hypothesis, which requires:
- Active loading of sucrose into sieve tubes at the source (e.g., leaves), creating a high solute concentration.
- Water uptake by osmosis, increasing hydrostatic pressure.
- Unloading of sucrose at the sink (e.g., roots or fruits), reducing pressure and maintaining flow.
The image likely depicts these conditions, with option B correctly representing the necessary steps for mass flow.
Topic: 8.3 (The heart)
One type of congenital heart defect is where the left and right atria are not completely separated. This is called an atrial septal defect (ASD).
ASD usually results in blood moving from the left atrium into the right atrium. This causes increased blood pressure in the right atrium and decreased blood pressure in the left atrium.
Which row describes other effects caused by ASD?
▶️ Answer/Explanation
Ans: D
In atrial septal defect (ASD), oxygenated blood from the left atrium leaks into the right atrium, causing:
- Increased pulmonary blood flow (right atrium pumps extra blood to the lungs).
- Reduced systemic circulation efficiency (less oxygenated blood reaches the body).
- Right ventricular hypertrophy (due to increased workload).
These effects match row D, where pulmonary circulation increases while systemic circulation decreases.
Topic: 2.4 (Water)
Which property of water, related to its role in blood and tissue fluid, is correctly described?
A. Water is a solvent for all biological molecules.
B. Water is a solvent for most non-polar molecules.
C. Water requires little energy to increase its temperature because it has a high specific heat capacity.
D. Water cools down slowly because it has a high specific heat capacity.
▶️ Answer/Explanation
Ans: D
Option D is correct: Water’s high specific heat capacity (4.18 J/g°C) means it absorbs/releases large amounts of heat with minimal temperature change, enabling stable body temperatures and slow cooling in blood/tissue fluid.
Option A is incorrect: Water dissolves polar/ionic molecules (e.g., glucose, salts) but not all (e.g., lipids).
Option B is incorrect: Non-polar molecules (e.g., oils) are insoluble in water.
Option C is incorrect: High specific heat capacity implies water resists temperature changes, requiring more energy to heat.
Topic: 8.1 (The circulatory system)
Which statement about tissue fluid formation is correct?
▶️ Answer/Explanation
Ans: A
Statement A is correct because hydrostatic pressure is highest at the arteriole end of the capillary (due to heart pressure) and decreases toward the venule end as energy is lost to friction. This pressure gradient drives tissue fluid formation (ultrafiltration) at the arteriole end and reabsorption at the venule end.
Why other options are incorrect:
- B: Water potential is lower in blood at the arteriole end (due to high solute concentration from plasma proteins) but becomes higher at the venule end after fluid loss.
- C: White blood cells (e.g., phagocytes) can squeeze through capillary walls into tissues.
- D: Most tissue fluid forms at the arteriole end where hydrostatic pressure exceeds osmotic pressure.
Topic: 9.1 (The gas exchange system)
Oxyhaemoglobin, carbaminohaemoglobin, hemoglobinic acid, and carbonic anhydrase are found inside red blood cells.
How many of these substances will show an overall decrease in concentration as a red blood cell passes through capillaries in the lungs?
▶️ Answer/Explanation
Ans: B
In the lungs, red blood cells undergo changes to facilitate oxygen uptake and CO2 release:
- Decrease in concentration:
- Carbaminohaemoglobin (CO2 bound to hemoglobin) decreases as CO2 diffuses out.
- Hemoglobinic acid (H+ bound to hemoglobin) decreases as H+ combines with HCO3− to form CO2 and H2O.
- No change/increase:
- Oxyhaemoglobin increases (O2 binds to hemoglobin).
- Carbonic anhydrase (enzyme) concentration remains constant.
Thus, 2 substances decrease, making B the correct answer.
Topic: 9.1 (The gas exchange system)
Which of these structures typically contain cartilage and cilia?
1. bronchi
2. bronchioles
3. trachea
▶️ Answer/Explanation
Ans: C (1 and 3 only)
Trachea (3) and bronchi (1) contain both cartilage rings (for structural support) and cilia (to move mucus and trapped particles).
Bronchioles (2) lack cartilage and have only smooth muscle, though some may have cilia in larger bronchioles. Since the question specifies “typically,” bronchioles are excluded.
Thus, only 1 (bronchi) and 3 (trachea) definitively meet both criteria.
Topic: 9.1 (The gas exchange system)
Which statements about the function of tissues found in the human gas exchange system are correct?
1. Collagen in the bronchi prevents them collapsing.
2. Smooth muscle in the bronchioles can contract to increase the flow of air into the alveoli.
3. Elastic fibres in the alveoli stretch and recoil during breathing.
▶️ Answer/Explanation
Ans: D
Analysis of Statements:
1. False: While collagen provides structural support, it is cartilage (not collagen alone) that prevents bronchial collapse.
2. False: Smooth muscle contraction in bronchioles reduces airflow (during bronchoconstriction), not increases it.
3. True: Elastic fibres in alveoli stretch during inhalation and recoil during exhalation, aiding breathing efficiency.
Thus, only statement 3 is correct, making D the right choice.
Topic: 10.2 (Antibiotics)
Antibiotic-resistant strains of Mycobacterium tuberculosis are a major problem when treating TB.
A new antibiotic, teixobactin, could be very effective at killing M. tuberculosis with only a small risk that the bacteria will evolve teixobactin resistance.
Penicillin and similar antibiotics bind to a single protein, but teixobactin binds to two lipids that are needed for the formation of the bacterial cell wall. Teixobactin binds to regions of the two lipids that do not vary across many different species of bacteria.
Which statements help to explain why the use of teixobactin is thought to be less likely to lead to the evolution of antibiotic resistance than the use of many other antibiotics, such as penicillin?
▶️ Answer/Explanation
Ans: C
Statement 1 is correct because teixobactin targets two lipids, requiring multiple mutations for resistance, unlike penicillin (single protein target). Statement 3 is correct as conserved lipid structures across species indicate evolutionary constraints, making resistance less likely. Statement 2 is incorrect because while lipids are not directly encoded by DNA, their synthesis depends on enzymes (proteins) that can mutate, indirectly affecting lipid structure.
Topic: 11.2 (Antibodies and vaccination)
Monoclonal antibodies are now being used to treat some human diseases.
What explains why monoclonal antibodies are suitable for this purpose?
- They can divide by mitosis to produce the large numbers of antibodies required for treatment.
- They are specific to a particular antigen.
- They can be modified so that they do not act as antigens themselves.
▶️ Answer/Explanation
Ans: C
Monoclonal antibodies are suitable for disease treatment because:
- Statement 1 is incorrect: Antibodies cannot divide by mitosis; they are produced by B-cells or hybridoma cells in culture.
- Statement 2 is correct: They bind specifically to target antigens (e.g., pathogens or cancer cells).
- Statement 3 is correct: They can be “humanized” to avoid immune rejection.
Thus, 2 and 3 are valid reasons, making C the correct choice.
Topic: 11.2 (Antibodies and vaccination)
Measles is an infectious disease caused by a virus.
The graph shows the number of cases of measles each year in a country before and after a vaccine was introduced.
What could have caused the decrease in the number of cases of measles after vaccination was introduced?
▶️ Answer/Explanation
Ans: A
The graph shows a sharp decline in measles cases after vaccination was introduced. This is because vaccines stimulate the production of memory cells, which provide long-term immunity. When a vaccinated person encounters the measles virus, these memory cells quickly produce antibodies, preventing infection. This mechanism of herd immunity (option A) explains the decrease, as widespread vaccination reduces the virus’s ability to spread in the population.