2024 AS Multiple Choice S2
Read each question and click on the ONE answer option you believe is correct. The explanation will appear after you click any option.
Score: 0
A prokaryotic cell which is 1 µm in diameter is magnified 50 000 times in an electron micrograph. What is the diameter of the cell in the electron micrograph in mm?
Click the CORRECT answer:
Reasoning for Correct Answer #3: The image size is calculated by multiplying the actual size by the magnification: Image Size = Actual Size × Magnification.
Image Size = 1 µm × 50 000 = 50 000 µm.
To convert micrometers (µm) to millimeters (mm), divide by 1000 (since 1 mm = 1000 µm).
Image Size in mm = 50 000 µm / 1000 µm/mm = 50 mm.
In standard form, 50 mm is written as 5 × 10¹ mm.
A diagram shows a plant cell with a nucleus (P), chloroplasts (Q), cell wall (R), and large central vacuole (S). Which labelled structures are bound by a double membrane?
Click the CORRECT answer:
Reasoning for Correct Answer #4: In eukaryotic plant cells:
– P (Nucleus) is enclosed by a double membrane (the nuclear envelope).
– Q (Chloroplasts) are also enclosed by a double membrane (outer and inner membranes).
– R (Cell wall) is an extracellular structure made primarily of cellulose, not a membrane.
– S (Large central vacuole) is enclosed by a single membrane called the tonoplast.
Therefore, both the nucleus (P) and chloroplasts (Q) are bound by a double membrane.
Which size of ribosome (based on sedimentation coefficient) is found in mitochondria and typical prokaryotic cells?
Click the CORRECT answer:
Reasoning for Correct Answer #3: Based on the endosymbiotic theory and molecular evidence, mitochondria (found in eukaryotic cells) share ancestry with prokaryotes. Both typical prokaryotic cells (like bacteria) and mitochondria possess ribosomes with a sedimentation coefficient of 70S. These are smaller than the 80S ribosomes found in the cytoplasm and on the rough endoplasmic reticulum of eukaryotic cells. The terms 50S and 30S refer to the large and small subunits of 70S ribosomes, while 60S and 40S refer to the subunits of 80S ribosomes.
Which row correctly compares typical prokaryotic cells and typical animal cells regarding the presence of lysosomes for organelle breakdown and whether ATP is produced by the cell? (✓ = correct for both, x = not correct for both but correct for one)
| Option | Lysosomes Present | ATP Produced |
|---|---|---|
| 1 | x | ✓ |
| 2 | ✓ | ✓ |
| 3 | x | x |
| 4 | ✓ | x |
| 5 | x | Not Applicable |
Click the CORRECT answer row number:
Reasoning for Correct Answer #1:
– **Lysosomes Present:** Lysosomes are membrane-bound organelles containing digestive enzymes, found typically in animal cells but absent in prokaryotic cells. Since they are present in one but not the other, the comparison is ‘x’.
– **ATP Produced:** All living cells require energy in the form of ATP. Both prokaryotic cells (using processes like respiration on the cell membrane or fermentation) and animal cells (using mainly mitochondrial respiration) produce ATP. Since both produce ATP, the comparison is ‘✓’.
Therefore, Row 1 (x, ✓) correctly represents the comparison.
Which row correctly identifies the presence or absence of specific structures in typical plant cells and typical animal cells?
| Option | Cell Structure | Plant Cell | Animal Cell |
|---|---|---|---|
| 1 | Golgi body | Present | Not Present |
| 2 | Plasmodesmata | Present | Present |
| 3 | Centriole | Not Present | Present |
| 4 | Cell Wall | Not Present | Present |
| 5 | Tonoplast | Not Present | Not Present |
Click the CORRECT answer row number:
Reasoning for Correct Answer #3:
– Row 1: Golgi bodies are present in both plant and animal cells. Incorrect.
– Row 2: Plasmodesmata (cytoplasmic connections through cell walls) are characteristic of plant cells, absent in animal cells. Incorrect.
– Row 3: Centrioles are present in animal cells (involved in cell division) but typically absent in higher plant cells. Correct.
– Row 4: Cell walls are present in plant cells but absent in animal cells. Incorrect.
– Row 5: Tonoplast (vacuolar membrane) surrounds the large central vacuole in plant cells, typically absent (or surrounding small vacuoles) in animal cells. Incorrect (shows absent in both).
Therefore, Row 3 provides the correct comparison for centrioles.
Which row correctly describes features of cellulose?
| Option | Rotation of alternate monomers by 180° | Shape of molecule | Hydrogen bonds between molecules |
|---|---|---|---|
| 1 | ✓ | Branched | x |
| 2 | ✓ | Unbranched | ✓ |
| 3 | x | Unbranched | x |
| 4 | x | Branched | ✓ |
| 5 | ✓ | Unbranched | x |
Click the CORRECT answer row number:
Reasoning for Correct Answer #2: Cellulose is a polymer of β-glucose units linked by β-1,4 glycosidic bonds.
– **Rotation:** The β-linkage causes alternate glucose monomers to be flipped 180° relative to their neighbours (✓).
– **Shape:** This rotation results in long, straight, unbranched chains (✓).
– **H-bonds:** The straight chains lie parallel to each other, allowing extensive hydrogen bonding between hydroxyl groups on adjacent chains, forming strong microfibrils (✓).
Row 2 correctly identifies all three features.
Which statements about peptide bond formation during translation are correct?
- The bond forms between a carbon of one amino acid and a nitrogen of the next amino acid after they detach from tRNA.
- The bond formation occurs at the ribosome while amino acids are attached to tRNA, and it is a hydrolysis reaction.
- The bond formation is important for growth and involves the removal of a water molecule.
Click the CORRECT answer:
Reasoning for Correct Answer #3:
Statement 1: Incorrect. The bond forms while the growing polypeptide is attached to one tRNA and the incoming amino acid is attached to another tRNA; detachment from tRNA occurs after bond formation. Also, it’s between the carboxyl carbon of one and the amino nitrogen of the next.
Statement 2: Incorrect. Bond formation does occur at the ribosome while amino acids are attached to tRNA, but it is a condensation (dehydration) reaction, not hydrolysis.
Statement 3: Correct. Peptide bond formation links amino acids into polypeptide chains (proteins), which is essential for building tissues and thus for growth. The reaction involves the removal of a water molecule (condensation).
Therefore, only statement 3 is correct.
D-glucose and L-glucose are stereoisomers (mirror images). The enzyme glucose oxidase catalyses D-glucose oxidation but not L-glucose. Why?
Click the CORRECT answer:
Reasoning for Correct Answer #5: Enzymes exhibit high specificity due to the precise 3D shape of their active site, which is complementary to the shape of their substrate(s). Stereoisomers like D- and L-glucose, while having the same chemical formula and bonding, differ in their spatial arrangement (like left and right hands). This difference in 3D shape means that L-glucose, the mirror image of the natural substrate D-glucose (#1 is true but not the explanation), cannot bind effectively to the specifically shaped active site of glucose oxidase (#5). Therefore, the enzyme cannot catalyse its oxidation. Structural formulas (#2) are the same in terms of connectivity. Concentration (#3) and origin (#4) are irrelevant to the enzyme’s ability to bind the molecule.
Tests on three solutions (each containing only one biological molecule): Solution 1 gives positive Benedict’s test (blue to orange). Solution 2 gives positive Benedict’s test after acid hydrolysis (blue to red). Solution 3 gives positive biuret test (blue to purple). Which solutions would contain either sucrose or amylase?
Click the CORRECT answer:
Reasoning for Correct Answer #3:
– Solution 1: Positive Benedict’s test indicates a reducing sugar (e.g., glucose, fructose, maltose). This is neither sucrose nor amylase.
– Solution 2: Negative Benedict’s initially, but positive after acid hydrolysis. This indicates a non-reducing sugar that yields reducing sugars upon hydrolysis. Sucrose is a non-reducing disaccharide that hydrolyzes to glucose and fructose (both reducing sugars). This fits the description for sucrose.
– Solution 3: Positive biuret test indicates the presence of peptide bonds, meaning the molecule is a protein (or polypeptide). Amylase is an enzyme, which is a protein.
Therefore, Solution 2 contains sucrose, and Solution 3 contains amylase.
What is the expected effect on Vmax and Km when a competitive reversible inhibitor is added to an enzyme-catalysed reaction?
| Option | Effect on Vmax | Substrate concentration at Km |
|---|---|---|
| 1 | Decreases | Increases |
| 2 | No change | Increases |
| 3 | Decreases | No change |
| 4 | Increases | Decreases |
| 5 | No change | No change |
Click the CORRECT answer row number:
Reasoning for Correct Answer #2: A competitive inhibitor binds to the enzyme’s active site, competing with the substrate.
– **Effect on Vmax:** Since the inhibition is reversible and can be overcome by sufficiently high substrate concentrations (which outcompete the inhibitor for the active site), the maximum possible reaction rate (Vmax) remains unchanged.
– **Effect on Km:** Because the inhibitor competes for the active site, a higher concentration of substrate is required to achieve half of the maximum velocity (Vmax/2) compared to the uninhibited reaction. Therefore, the apparent Km (substrate concentration at Vmax/2) increases.
Row 2 correctly states Vmax has no change, and Km increases.
A graph shows rate of reaction vs substrate concentration for three enzymes (X, Y, Z). X reaches Vmax (~1500) quickly at low substrate conc (~200). Y increases almost linearly, not reaching Vmax (<2500 at substrate 1200). Z reaches a low Vmax (~200) quickly at low substrate conc (~200). Order the enzymes by affinity for their substrate (highest affinity first).
Click the CORRECT answer:
Reasoning for Correct Answer #4: Affinity is inversely related to Km (substrate concentration at Vmax/2). Lower Km means higher affinity.
– **Enzyme X:** Reaches its high Vmax quickly at low [S]. This means Vmax/2 is reached at a very low [S], indicating a very low Km and thus the highest affinity.
– **Enzyme Z:** Reaches its low Vmax quickly at low [S]. Vmax/2 is reached at a low [S], indicating a low Km and high affinity, but likely slightly higher Km than X since it saturates around the same [S] but achieves a much lower Vmax.
– **Enzyme Y:** Requires very high [S] to even approach Vmax. This means Vmax/2 occurs at a very high [S], indicating a high Km and the lowest affinity.
Therefore, the order from highest affinity to lowest affinity is X > Z > Y.
Which row correctly identifies weak and strong bonds contributing to the tertiary and quaternary structure of a typical protein?
| Option | Disulfide | Hydrogen | Hydrophobic Interactions | Ionic |
|---|---|---|---|---|
| 1 | Weak | Weak | Strong | Strong |
| 2 | Strong | Weak | Weak | Weak |
| 3 | Weak | Weak | Weak | Strong |
| 4 | Strong | Strong | Weak | Weak |
| 5 | Strong | Weak | Strong | Strong |
Click the CORRECT answer row number:
Reasoning for Correct Answer #2: Comparing bond strengths involved in protein tertiary and quaternary structure:
– **Disulfide bonds:** These are covalent bonds, formed between cysteine residues. Covalent bonds are significantly stronger than non-covalent interactions. So, classified as Strong.
– **Hydrogen bonds:** These are non-covalent interactions formed between polar groups. Individually, they are relatively weak compared to covalent bonds. So, classified as Weak.
– **Hydrophobic interactions:** These arise from the tendency of nonpolar groups to cluster together away from water. While collectively important, the individual interactions are driven by entropy and are considered non-covalent and relatively weak. So, classified as Weak.
– **Ionic bonds (salt bridges):** These are electrostatic attractions between oppositely charged R-groups. They are non-covalent and considered relatively weak in the aqueous cellular environment (easily disrupted by water and ions). So, classified as Weak.
Row 2 correctly identifies disulfide bonds as strong and the others (hydrogen, hydrophobic, ionic) as weak in this comparative context.
Which row correctly describes features of haemoglobin?
| Option | Structure 1 | Structure 2 | Structure 3 | Function |
|---|---|---|---|---|
| 1 | Polypeptides form helical shape | Hydrophobic R-groups near iron | Hydrophobic R-groups surround iron | Transports 8 O atoms total |
| 2 | Polypeptides interact, spherical shape | Iron ion in each haem group | Quaternary structure has 2α and 2β chains | Transports 4 O atoms total |
| 3 | Four polypeptides, each with haem group | Polypeptides form globular chain | Hydrophobic R-groups point inwards | 2 O₂ molecules at 50% saturation |
| 4 | Polypeptides interact, globular chain | Each chain has haem group | Two identical α, two identical β chains | Each chain transports O₂ |
| 5 | Two large polypeptides, two small lipids | Iron ion binds O₂ | Hydrophobic groups create binding pocket for lipids | Transports O₂ and CO₂ equally efficiently |
Click the CORRECT answer row number:
Reasoning for Correct Answer #3: Analyzing the statements in Row 3:
– Structure 1: “Four polypeptides, each with haem group” – Correct. Haemoglobin is a tetramer, typically 2 alpha and 2 beta chains, and each chain binds one haem group.
– Structure 2: “Polypeptides form globular chain” – Correct. Each globin chain folds into a complex globular tertiary structure.
– Structure 3: “Hydrophobic R-groups point inwards” – Correct. In globular proteins like haemoglobin, hydrophobic residues tend to be buried in the interior, away from the aqueous environment.
– Function: “2 O₂ molecules at 50% saturation” – Correct. Haemoglobin has 4 oxygen binding sites. 50% saturation means half the sites are occupied, which corresponds to 2 O₂ molecules bound per haemoglobin molecule on average.
Other rows have errors: Row 1 says 8 O atoms (should be 4 O₂ molecules). Row 2 misses that each chain has haem. Row 4 suggests each chain transports O₂ independently (binding is cooperative). Row 5 mentions lipids and equal O₂/CO₂ efficiency (incorrect).
Which process always takes place without the direct involvement of energy from ATP hydrolysis?
Click the CORRECT answer:
Reasoning for Correct Answer #2: Facilitated diffusion is a form of passive transport. It involves the movement of molecules across a cell membrane down their concentration gradient, aided by membrane proteins (channels or carriers). Because the movement follows the concentration gradient, it does not require the cell to expend metabolic energy directly (i.e., no ATP hydrolysis is directly coupled to the transport protein). Active transport (#3) moves substances against their gradient and requires ATP. Endocytosis (#5, including pinocytosis #4 – cell drinking) and exocytosis (#1) are bulk transport mechanisms involving membrane vesicle formation and movement, which are energy-dependent processes requiring ATP.
A diagram shows molecule X moving down its concentration gradient into a cell through a transmembrane protein channel. Which row shows a likely property of molecule X and the effect of cytoplasmic ATP concentration on its entry rate?
| Option | Property of X | Effect of ATP concentration |
|---|---|---|
| 1 | Polar | Affects rate |
| 2 | Non-polar | Affects rate |
| 3 | Lipid-soluble | Has no effect on rate |
| 4 | Non-polar | Has no effect on rate |
| 5 | Polar | Has no effect on rate |
Click the CORRECT answer row number:
Reasoning for Correct Answer #5: The diagram shows movement down a concentration gradient through a protein channel. This describes facilitated diffusion.
– **Property of X:** Protein channels are typically required for molecules that cannot easily cross the lipid bilayer directly, such as polar molecules or ions. Non-polar, lipid-soluble molecules usually diffuse directly through the membrane. Thus, X is likely polar.
– **Effect of ATP:** Facilitated diffusion is passive transport; it does not directly consume ATP. Therefore, the cytoplasmic concentration of ATP should have no direct effect on the rate of transport via this channel.
Row 5 correctly pairs “Polar” property with “Has no effect on rate” for ATP concentration.
An electron micrograph shows red blood cells in a solution. Cell X appears shrunken and crenated (spiky surface). What was the net movement of water by osmosis, and how did the water potential (Ψ) of cell X’s cytoplasm compare with the solution?
| Option | Net water movement | Ψ of cytoplasm vs Solution |
|---|---|---|
| 1 | Out of the cell | Lower |
| 2 | Into the cell | Lower |
| 3 | No net movement | Equal |
| 4 | Into the cell | Higher |
| 5 | Out of the cell | Higher |
Click the CORRECT answer row number:
Reasoning for Correct Answer #5: Crenation (shrinking) of a red blood cell indicates that it has lost water to the surrounding solution.
– **Net water movement:** For the cell to shrink, the net movement of water must have been out of the cell.
– **Water Potential Comparison:** Water moves by osmosis from a region of higher water potential (Ψ) to a region of lower water potential. For water to move out of the cell, the water potential inside the cytoplasm must have been higher (less negative) than the water potential of the external solution.
Row 5 correctly states that the net water movement was out of the cell and the cytoplasm’s Ψ was higher than the solution’s Ψ.
Agar blocks containing red indicator were placed in acid (turns indicator yellow). Block 1 (SA:V 1.33:1) turned yellow in 4 mins. Block 2 (SA:V 0.67:1) took 5 mins. Block 3 (SA:V 0.5:1) took 11 mins. Which row correctly matches SA:V ratios and times?
| Option | Block 1 (SA:V=1.33) | Block 2 (SA:V=0.67) | Block 3 (SA:V=0.5) |
|---|---|---|---|
| 1 | Time = 11 min | Time = 5 min | Time = 4 min |
| 2 | Time = 4 min | Time = 5 min | Time = 11 min |
| 3 | Time = 4 min | Time = 11 min | Time = 5 min |
| 4 | Time = 5 min | Time = 4 min | Time = 11 min |
| 5 | Time = 11 min | Time = 4 min | Time = 5 min |
Click the CORRECT answer row number:
Reasoning for Correct Answer #2: Diffusion time is related to the distance the substance needs to travel and the surface area available for entry relative to the volume it needs to fill. A higher Surface Area to Volume (SA:V) ratio generally means faster diffusion into the entire volume. The data shows:
– Highest SA:V (Block 1, 1.33) corresponds to the shortest time (4 min).
– Lowest SA:V (Block 3, 0.5) corresponds to the longest time (11 min).
– Intermediate SA:V (Block 2, 0.67) corresponds to the intermediate time (5 min).
Row 2 correctly matches these SA:V ratios with their corresponding times.
Which metabolic processes will be very active in a cell during the G1 phase (immediately after cytokinesis)?
- ATP formation
- DNA replication
- Protein synthesis
Click the CORRECT combination:
Reasoning for Correct Answer #3: G1 phase is the primary growth phase of the cell cycle.
– 1 ATP formation: Cell growth and protein synthesis are highly energy-demanding processes, so cellular respiration producing ATP will be very active. Correct.
– 2 DNA replication: This process occurs specifically during the S phase, which follows G1. Incorrect for G1.
– 3 Protein synthesis: The cell needs to synthesize numerous proteins for growth, increasing organelle numbers, and carrying out its specific functions. Correct.
Therefore, ATP formation and protein synthesis (1 and 3) are very active during G1.
Consider the eukaryotic cell cycle (G1 → S → G2 → M → Cytokinesis). Points V, W, X, Y mark positions: V=mid-S, W=mid-G2, X=start-M (Prophase), Y=mid-Cytokinesis. Which points represent: 1) Cell where DNA replication is complete but cell not max size? 2) Cell where preparation for microtubule formation is underway & chromosomes condense?
| Option | Point for Description 1 | Point for Description 2 |
|---|---|---|
| 1 | W | X |
| 2 | V | W |
| 3 | W | Y |
| 4 | V | X |
| 5 | Y | X |
Click the CORRECT answer row number:
Reasoning for Correct Answer #1:
– Description 1: DNA replication is completed during S phase. The cell continues to grow during G2 phase before reaching its maximum size just prior to division. Therefore, a cell in mid-G2 (Point W) fits this description.
– Description 2: Preparation for microtubule (spindle) formation and the condensation of chromosomes are key events that begin in prophase, the start of M phase (Point X).
Therefore, Row 1 correctly matches Description 1 with W and Description 2 with X.
A graph shows the mean length of spindle fibres increasing (A), peaking/plateauing (B), then decreasing (C), finally becoming zero (D) during mitosis. When do centromeres detach from the spindle fibres?
Click the CORRECT answer:
Reasoning for Correct Answer #2: Spindle fibres attach to centromeres (kinetochores) during prometaphase/metaphase (Region B shows fully formed spindle). They shorten during anaphase (Region C) to pull sister chromatids apart. The chromosomes reach the poles and the spindle begins to break down during telophase, continuing into cytokinesis (leading into Region D where length goes to zero). The centromeres effectively remain associated with the remnants of the spindle fibres until the spindle structure is completely disassembled as the new nuclear envelopes form and the cell divides. Therefore, the final detachment occurs as the spindle disappears in Region D.
The mRNA codons ACU, ACC, ACA, and ACG all code for threonine. Which tRNA anticodons (reading 3′ to 5′) could potentially pair with an mRNA codon specifying an amino acid other than threonine? Consider anticodons: 1 UCA, 2 ACC, 3 UGU, 4 UGC.
Click the CORRECT answer:
Reasoning for Correct Answer #2: Let’s determine the mRNA codons (5′-3′) that each anticodon (3′-5′) would pair with, considering potential wobble:
– Anticodon 1 (3′-ACU-5′): Pairs with 5′-UGA-3′ (Stop codon). Could potentially wobble pair with 5′-UGG-3′ (Tryptophan). Neither is Threonine. **Pairs with non-Thr.**
– Anticodon 2 (3′-ACC-5′): Pairs with 5′-UGG-3′ (Tryptophan). Not Threonine. **Pairs with non-Thr.**
– Anticodon 3 (3′-UGU-5′): Pairs perfectly with 5′-ACA-3′ (Threonine). Could potentially wobble pair with 5′-ACG-3′ (Threonine). Codes for Threonine.
– Anticodon 4 (3′-CGU-5′): Pairs with 5′-GCA-3′ (Alanine). Could potentially wobble pair with 5′-GCG-3′ (Alanine). Not Threonine. **Pairs with non-Thr.**
Based on standard pairing and common wobble rules, anticodons 1, 2, and 4 can pair with codons specifying non-Threonine outcomes. *However, the provided answer key selected ‘B’ (1 and 2 only) in the original context. This suggests either specific wobble rules are assumed that exclude #4, or there’s an error in the original key/question. Adhering to the provided key leads to selecting option 2 (1 and 2 only).*
Which bond formation does DNA polymerase catalyse during DNA replication?
Click the CORRECT answer:
Reasoning for Correct Answer #1: DNA polymerase synthesizes new DNA strands by adding nucleotides complementary to a template strand. The core catalytic activity is the formation of phosphodiester bonds, which link the 5′ phosphate group of an incoming deoxynucleotide triphosphate to the 3′ hydroxyl group of the last nucleotide on the growing strand, forming the sugar-phosphate backbone. Peptide bonds (#2) link amino acids. Hydrogen bonds (#3, #5) form spontaneously between complementary base pairs (A-T, G-C) holding the two DNA strands together, but are not catalysed by DNA polymerase. Glycosidic bonds (#4) link the sugar to the base within a nucleotide.
In eukaryotes, primary RNA transcripts are modified by removing non-coding sequences and joining coding sequences together to form mature mRNA. What are the coding sequences called?
Click the CORRECT answer:
Reasoning for Correct Answer #5: The process described is RNA splicing. The coding sequences within a eukaryotic gene and its primary transcript, which are retained and joined together in the mature mRNA, are called exons (Expressed regiONs). The non-coding intervening sequences that are removed are called introns (#1). Codons (#2) are three-nucleotide sequences within the exons that specify amino acids. Primary transcripts (#3) are the initial RNA molecules before splicing. Promoters (#4) are DNA sequences where transcription starts.
Which row correctly identifies typical sinks for sucrose transported by mass flow in plants?
| Option | Root storage organ | Growing leaf bud | Growing shoot tip | Mature Leaf |
|---|---|---|---|---|
| 1 | ✓ | ✓ | ✓ | ✓ |
| 2 | ✓ | x | ✓ | x |
| 3 | x | ✓ | x | x |
| 4 | ✓ | ✓ | ✓ | x |
| 5 | x | x | x | ✓ |
Click the CORRECT answer row number:
Reasoning for Correct Answer #4: Plant sinks are tissues or organs that consume or store sugars (primarily sucrose) transported via the phloem. They represent areas of net sugar import. Typical sinks include:
– Root storage organs (like tubers, taproots) (✓)
– Growing, developing tissues like leaf buds (✓) and shoot tips (apical meristems) (✓)
– Developing fruits and seeds
Mature, photosynthetically active leaves, on the other hand, are typically sources, producing excess sugar and exporting it. Therefore, a mature leaf is generally not a sink (x). Row 4 correctly identifies root storage organs, growing buds, and growing tips as sinks, and mature leaves as non-sinks.
A diagram shows a cross-section of phloem. Cell 1 is small with dense cytoplasm and a nucleus. Cell 2 is larger, lacks a nucleus, has sparse cytoplasm, and is adjacent to cell 1. Identify cells 1 and 2.
| Option | Cell 1 | Cell 2 |
|---|---|---|
| 1 | Phloem sieve tube element | Xylem vessel element |
| 2 | Companion cell | Phloem sieve tube element |
| 3 | Parenchyma cell | Companion cell |
| 4 | Companion cell | Xylem vessel element |
| 5 | Phloem sieve tube element | Phloem sieve tube element |
Click the CORRECT answer row number:
Reasoning for Correct Answer #2: The phloem conductive pathway consists primarily of sieve tube elements and companion cells.
– **Sieve Tube Elements (Cell 2):** These are the main transport cells. They are large, elongated cells that lose their nucleus, vacuole, and most other organelles at maturity, resulting in sparse cytoplasm to facilitate bulk flow. They are connected end-to-end via sieve plates.
– **Companion Cells (Cell 1):** These are closely associated with sieve tube elements. They are smaller, metabolically active cells that retain their nucleus and dense cytoplasm with many organelles. They provide metabolic support and are involved in loading/unloading sugars into/from the sieve tube elements.
The description matches Cell 1 as the companion cell and Cell 2 as the phloem sieve tube element.
Which statement correctly describes the movement of solutes in the symplast pathway in a plant root?
Click the CORRECT answer:
Reasoning for Correct Answer #2: The symplast pathway refers to the movement of water and solutes through the continuous network of cytoplasm within plant cells, connected by plasmodesmata. To enter the symplast pathway from the outside environment (soil solution), solutes must first cross the selectively permeable cell surface membrane of an epidermal or cortex cell. This uptake is regulated by transport proteins embedded in the membrane, allowing the cell to selectively control which solutes enter the symplast. Option 1 describes the apoplast pathway. Option 3 is incorrect; the Casparian strip blocks the apoplast pathway, forcing substances into the symplast. Option 4 is incorrect; movement within the symplast involves diffusion/cytoplasmic streaming, not bulk flow between vacuoles. Option 5 is incorrect; plasmodesmata facilitate movement within the symplast.
Which statement helps to explain why water moves up through xylem vessel elements as a continuous column during transpiration?
Click the CORRECT answer:
Reasoning for Correct Answer #3: The upward movement of water in xylem is primarily explained by the cohesion-tension theory. Transpiration creates tension (negative pressure) in the leaves, pulling water up. Cohesion refers to the strong attraction between water molecules due to hydrogen bonding. This cohesive force allows the water molecules to stick together, forming a continuous, unbroken column that can be pulled up through the xylem vessels under tension without breaking. Adhesion (#5), the attraction of water to the xylem walls, also plays a role in counteracting gravity and maintaining the column, but cohesion is essential for the continuity of the column under tension.
Where is the atrioventricular (AV) node located in the mammalian heart?
Click the CORRECT answer:
Reasoning for Correct Answer #4: The atrioventricular (AV) node is located in the lower part of the interatrial septum (the wall separating the right and left atria), specifically near the junction of the atria and ventricles, close to the opening of the coronary sinus and the septal leaflet of the tricuspid valve (one of the atrioventricular valves). It receives the impulse from the SA node (located as described in #2) and introduces a slight delay before relaying it to the ventricles via the Bundle of His.
A graph shows pressure changes in the left atrium, left ventricle, and aorta. At point X (~0.15s), ventricular pressure rises sharply and exceeds atrial pressure. What event happens at X?
Click the CORRECT answer:
Reasoning for Correct Answer #4: Point X marks the beginning of isovolumetric ventricular contraction. As the ventricle starts to contract, its internal pressure rapidly increases. The moment the left ventricular pressure exceeds the left atrial pressure, the atrioventricular (mitral) valve is pushed shut to prevent backflow into the atrium. This valve closure produces the first heart sound (“lub”). The semilunar (aortic) valve (#5) opens only later when ventricular pressure surpasses aortic pressure. Semilunar valves close (#1) at the end of ventricular ejection when ventricular pressure falls below aortic pressure.
Which components of blood are normally present in tissue fluid?
- Phagocytes
- Some (smaller) proteins
- Sodium ions
Click the CORRECT combination:
Reasoning for Correct Answer #2: Tissue fluid (interstitial fluid) is essentially blood plasma minus most of the large proteins and blood cells.
– 1 Phagocytes: Certain white blood cells, including phagocytes like neutrophils and macrophages, can actively squeeze out of capillaries (diapedesis) into the tissue fluid to perform immune surveillance. Present.
– 2 Some (smaller) proteins: While large proteins like albumin are mostly retained in the plasma, smaller plasma proteins can leak through capillary walls into the tissue fluid, contributing to its osmotic potential. Present.
– 3 Sodium ions: Small ions like sodium pass freely from plasma into tissue fluid, maintaining similar concentrations. Present.
Red blood cells are normally confined to blood vessels. Therefore, phagocytes, some proteins, and sodium ions are typically found in tissue fluid.
What is the minimum number of times a carbon dioxide molecule, initially inside a red blood cell in an alveolar capillary, must cross a cell surface membrane to reach the alveolar air space?
Click the CORRECT answer:
Reasoning for Correct Answer #1: For a CO₂ molecule to travel from the cytoplasm of a red blood cell (RBC) to the air inside an alveolus, it must cross the following barriers, each involving at least one cell surface membrane (phospholipid bilayer):
1. The cell surface membrane of the red blood cell (to exit into the plasma).
2. The cell surface membrane of the capillary endothelial cell (to exit the capillary). This involves crossing the membrane twice (apical and basal) if passing through the cell, or potentially between cells, but minimum one layer.
3. The cell surface membrane of the alveolar epithelial cell (Type I pneumocyte) lining the alveolus (to enter the air space). Again, this involves crossing the membrane twice if passing through the cell.
However, the question asks for the number of *cell surface membranes* crossed. Path: RBC cytoplasm -> RBC membrane -> Plasma -> Capillary endothelial membrane(s) -> Basement membrane -> Alveolar epithelial membrane(s) -> Alveolar air.
Minimum path: Cross RBC membrane (1), cross capillary endothelial cell membrane (2), cross alveolar epithelial cell membrane (3). Total = 3 membranes.
What maintains the steep concentration gradients needed for successful gas exchange between alveoli and blood in the lungs?
- Air flow in alveoli is counter-current to blood flow.
- Blood arriving has lower O₂ / higher CO₂ concentration than alveolar air.
- Blood constantly flows through, removing oxygenated blood / bringing deoxygenated blood.
Click the CORRECT combination:
Reasoning for Correct Answer #2: Efficient diffusion requires steep concentration (partial pressure) gradients. These are maintained by:
– **Ventilation:** Continuously replacing alveolar air keeps alveolar PO₂ high and PCO₂ low. This establishes the initial difference relative to the blood arriving (#2 is correct – blood arriving has lower O₂/higher CO₂).
– **Perfusion:** Continuous blood flow through the pulmonary capillaries ensures that as blood picks up O₂ and releases CO₂, it is quickly moved away and replaced by deoxygenated blood, preventing equilibrium from being reached (#3 is correct).
– **Gas Properties:** Rapid diffusion of O₂ and CO₂ across the thin respiratory membrane.
Counter-current flow (#1) is a highly efficient mechanism used in fish gills but not in mammalian lungs. Active transport is not involved in respiratory gas exchange. Therefore, statements 2 and 3 correctly identify key factors maintaining the gradients.
Where is cartilage tissue always found in the human gas exchange system?
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Reasoning for Correct Answer #5: Cartilage provides structural support to keep the larger airways open. In the human respiratory system:
– **Trachea:** Supported by C-shaped rings of hyaline cartilage. Always present.
– **Bronchi (primary, secondary, tertiary):** Supported by irregular plates of cartilage, which decrease in amount as the bronchi get smaller. Always present in larger bronchi.
– **Bronchioles:** Lack cartilage; their walls contain smooth muscle and elastic fibers.
– **Alveoli:** Thin-walled sacs for gas exchange, composed mainly of epithelial cells and capillaries, lacking cartilage.
Therefore, cartilage is always found in both the trachea and the (larger) bronchi.
Data compares non-smokers (Goblet 19 cells/mm², Mucus 6 units), smokers without lung disease (Goblet 54, Mucus 26), and smokers with lung disease (Goblet 37, Mucus 15). Which conclusions are indicated?
- Positive correlation between goblet cell density and mucus density across the groups shown.
- Lung disease in smokers results in a further increase in goblet cell density compared to healthy smokers.
- There is an association between smoking (comparing smokers vs non-smokers) and increased mucus density.
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Reasoning for Correct Answer #3:
– Statement 1: Comparing non-smokers (19 goblets, 6 mucus) to healthy smokers (54 goblets, 26 mucus), both values increase. Comparing healthy smokers to smokers with disease (37 goblets, 15 mucus), both values decrease. This suggests a positive correlation overall – higher goblet density tends to occur with higher mucus density in this dataset. Correct.
– Statement 2: Comparing smokers with lung disease (37 goblets) to smokers without lung disease (54 goblets), the goblet cell density actually *decreases* in the group with lung disease according to this data. Incorrect.
– Statement 3: Comparing both smoker groups (healthy: 26 mucus; diseased: 15 mucus) to non-smokers (6 mucus), both smoker groups show significantly increased mucus density. Correct.
Therefore, only statements 1 and 3 are supported by the data.
Which disease does Mycobacterium bovis cause?
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Reasoning for Correct Answer #4: Mycobacterium bovis is a slow-growing bacterium that belongs to the Mycobacterium tuberculosis complex. It is the causative agent of tuberculosis in cattle (bovine TB) and can also infect humans and other mammals, causing a form of tuberculosis that is clinically similar to that caused by Mycobacterium tuberculosis. Transmission to humans often occurred historically through consumption of unpasteurized milk from infected cows.
An antibiotic inhibits the formation of cross-links between peptidoglycan molecules in bacterial cell walls. Which statements explain why bacteria are killed?
- The bacterial cell is destroyed by osmotic lysis.
- Cellulose molecules cannot form hydrogen bonds.
- The cell wall is no longer partially permeable.
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Reasoning for Correct Answer #2: The bacterial cell wall, made primarily of peptidoglycan cross-linked into a strong mesh, protects the bacterium from osmotic stress, particularly in hypotonic environments. If the formation of cross-links is inhibited, the cell wall becomes weak. Water entering the cell by osmosis generates internal turgor pressure. A weakened cell wall cannot withstand this pressure, leading to cell rupture or osmotic lysis (Statement 1). Statement 2 is irrelevant (cellulose is in plants). Statement 3 is incorrect; the cell wall’s primary role isn’t permeability regulation (that’s the cell membrane), and weakening it doesn’t make it impermeable. Therefore, only statement 1 correctly explains the mechanism of killing for this type of antibiotic (like penicillin).
Graphs show bacterial growth (total and resistant populations) at different antibiotic concentrations (0, 90, 215, 600 mg/dm³). Higher concentrations suppress total growth more. The resistant population grows at 90 and 215 mg/dm³ but is suppressed at 600 mg/dm³. Which conclusions are correct?
- Increasing antibiotic conc decreases non-resistant bacteria at 24h.
- Proportion of resistant bacteria increases with increasing antibiotic conc (up to 215).
- Increasing antibiotic conc always increases the number of resistant bacteria.
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Reasoning for Correct Answer #1:
– Statement 1: Correct. The description states higher concentrations suppress total growth more. Since the total includes resistant and non-resistant, and the resistant population *grows* at lower concentrations, the suppression of the total must be due to the decrease in non-resistant bacteria.
– Statement 2: Correct. As the antibiotic concentration increases from 0 to 90 to 215, the non-resistant population is increasingly suppressed, while the resistant population still grows. This means the resistant bacteria make up an increasing proportion of the total population.
– Statement 3: Incorrect. The resistant population grows at 90 and 215, but is suppressed (decreases) at 600 mg/dm³. Therefore, increasing antibiotic concentration does not *always* increase the number of resistant bacteria.
Thus, only statements 1 and 2 are correct based on the description.
What is the correct sequence of key events in a primary humoral immune response leading to antibody production?
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Reasoning for Correct Answer #3: A typical sequence for T-dependent humoral immunity is:
1. **Antigen Presentation:** An antigen-presenting cell (APC, e.g., macrophage or dendritic cell) or a B cell presents processed antigen peptides on MHC class II molecules to a specific T-helper (Th) cell.
2. **T-helper Cell Activation & Cytokine Release:** The Th cell becomes activated and releases cytokines (signalling molecules).
3. **B-cell Activation & Differentiation:** A B cell that has bound the same antigen via its B cell receptor receives signals from the activated Th cell (via direct contact and cytokines). This triggers the B cell to proliferate (clonal expansion) and differentiate into antibody-producing plasma cells and memory B cells.
Option 3 correctly places antigen presentation first, followed by Th cytokine release leading to B cell differentiation. Other options have incorrect orders or involve memory cells which are key in secondary, not primary, responses.
Which statement about the properties of the antigen-binding sites in different antibody molecules is correct?
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Reasoning for Correct Answer #1: The ability of the immune system to recognize a vast array of antigens relies on the diversity of antibody antigen-binding sites. These sites are formed by the variable regions of the heavy and light polypeptide chains. The amino acid sequences within these variable regions, particularly in the hypervariable or complementarity-determining regions (CDRs), differ significantly between antibodies produced by different B cell clones, creating unique binding pockets specific for different antigens. Option 2 is incorrect; phagocyte binding sites are on the constant (Fc) region. Option 3 is incorrect; binding sites involve both heavy and light chain variable regions. Option 4 is incorrect; an individual produces many different antibodies. Option 5 describes the hinge region, not the binding site itself.
In monoclonal antibody production, what cell type (X) is fused with cancer cells (myeloma cells) to form hybridoma cells?
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Reasoning for Correct Answer #5: Hybridoma technology, used to produce monoclonal antibodies, involves fusing two types of cells:
1. **Myeloma cells:** These are cancerous plasma cells that are immortal (divide indefinitely in culture) but have typically been selected to no longer produce their own antibodies.
2. **B-lymphocytes (specifically plasma cells or activated B cells):** These are isolated from an animal (like a mouse) that has been immunized with the target antigen. These cells produce the desired specific antibody but have a limited lifespan in culture.
The fusion creates hybridoma cells that inherit immortality from the myeloma parent and the ability to produce the specific antibody from the B-lymphocyte parent. Antibodies (#1) and antigens (#3) are molecules, not cells. T-lymphocytes (#2) and macrophages (#4) are other immune cells not directly used in this fusion process.