4.1 Cell biology — coverage pack

Answer

• An animal cell's genetic material is enclosed in a nucleus, whereas a bacterium has no nucleus.
• A bacterium has one DNA loop and may have plasmids.

Method: Award one mark for the nucleus comparison and one for correctly describing the bacterial DNA loop or plasmids.

Answer

• It is a bacterial or prokaryotic cell.
• Its genetic material is a single DNA loop rather than being enclosed in a nucleus.
• The plasmids are small rings of DNA characteristic of bacterial cells.

Method: Use the genetic material as the decisive evidence: the absence of a nucleus and the presence of a DNA loop and plasmids identify a prokaryote.

Answer

• The animal cell is $10$ times longer.
• The lengths differ by one order of magnitude.

Method: Divide the lengths: $(3.0\times10^{-5})/(3.0\times10^{-6})=10^{1}=10$. A factor of $10$ is a difference of one order of magnitude.

Answer

• A ribosome is the site of protein synthesis.
• A mitochondrion is the site of aerobic respiration.

Method: Give one precise function for each named sub-cellular structure; do not describe mitochondria merely as producing energy.

Answer

• It is a plant cell.
• Chloroplasts absorb light for photosynthesis.
• The permanent vacuole contains cell sap.
• The cellulose cell wall strengthens the cell.

Method: Identify the cell from structures not normally found in animal cells, then link any two of chloroplast, vacuole and cell wall to their correct functions.

Answer

• Use clear, unbroken single lines and no shading.
• Draw the cells large and in proportion and add correct labels with ruled label lines.
• Include the magnification or a scale.
• Onion cells have a cellulose cell wall, but cheek cells do not.
• Onion cells have a large permanent vacuole, but cheek cells do not.
• Chloroplasts should not be claimed for onion epidermal cells because these cells are not exposed to light.

Method: Separate practical drawing conventions from biological comparison. The cell wall and large vacuole distinguish the onion cells; avoid the common assumption that every plant cell contains chloroplasts.

Answer

• Its tail allows it to swim towards the egg.
• Alternatively, many mitochondria provide energy from respiration for movement.

Method: Name one relevant feature and make the linked functional consequence explicit.

Answer

• The long projection gives a large surface area in contact with the soil solution.
• A larger surface area increases the uptake of mineral ions.
• The mitochondria carry out aerobic respiration.
• Respiration supplies energy for active transport of mineral ions.

Method: Build two feature-to-function chains: projection to larger exchange area, and mitochondria to respiration to the energy needed for active transport.

Answer

• It belongs to xylem tissue.
• Being hollow and joined end to end forms a continuous tube for water and mineral ions.
• The absence of cell contents reduces resistance to flow.
• The strengthening material supports the tube and helps prevent collapse.
• Its waterproof nature helps water remain in the transport pathway.

Method: Identify xylem from the hollow dead tube, then relate each supplied feature to efficient transport or structural support rather than listing features alone.

Answer

• It is the process by which a cell changes to become specialised for a particular function.

Method: Include both the change in the cell and the outcome: a specialised cell able to perform a particular function.

Answer

• Most animal cells differentiate at an early stage of development.
• Many plant cells retain the ability to differentiate throughout life.
• Differentiating cells acquire sub-cellular structures suited to their functions.

Method: Give the contrasting timing for animals and plants, then state the shared result of acquiring structures for a specialised function.

Answer

• Cell division produces new cells for repair or replacement in the animal.
• Division also produces new cells for growth of the plant root.
• The new plant cells differentiate into different specialised cell types.
• Differentiation involves acquiring sub-cellular structures suited to each function.
• Many plant cells can differentiate throughout life, whereas most animal cells differentiated earlier in development.

Method: Distinguish production of cells by division from development of specialised functions by differentiation, and apply the animal-plant timing difference to the contexts.

Answer

• It has higher resolving power.
• It also has higher magnification.

Method: Credit the two specification differences: greater resolution separates closer structures, and greater magnification produces a larger image.

Answer

• $\times 2000$

Method: Convert the image length to micrometres: $24\,\mathrm{mm}=24\,000\,\mathrm{\mu m}$. Then $\text{magnification}=24\,000/12=2000$, written as $\times 2000$.

Answer

• $20\,\mathrm{\mu m}$
• The second microscope may have higher resolving power, allowing close structures to be distinguished and finer detail to be seen.

Method: Rearrange to $\text{real size}=\text{image size}/\text{magnification}$. This gives $30/1500=0.020\,\mathrm{mm}=20\,\mathrm{\mu m}$. Magnification is unchanged, so improved usefulness must come from greater resolution.

Answer

• $800$ bacteria

Method: There are $60/20=3$ divisions. Each division doubles the population, so $100\times2^3=800$.

Answer

• Sterilise the Petri dish and agar before use.
• Sterilise the inoculating loop by passing it through a flame.
• Open the lid as little as possible, then secure it with adhesive tape.
• Store the plate upside down.
• Incubating at $25\,{}^\circ\mathrm{C}$ reduces the chance that harmful pathogens will grow.

Method: Each aseptic step reduces entry or spread of unwanted microorganisms. The lower incubation temperature is a safety measure, not the temperature that maximises bacterial growth.

Answer

• $3.2\times10^4$ bacteria

Method: Convert $2.0\,\mathrm{hours}$ to $120\,\mathrm{minutes}$, giving $120/20=6$ divisions. The population is $500\times2^6=500\times64=32\,000=3.2\times10^4$ bacteria.

Answer

• Chromosomes are found in the nucleus.
• They are made of DNA molecules.

Method: Give the cell location and molecular material as two separate facts.

Answer

• The nucleus contains chromosomes.
• Chromosomes are made of DNA molecules.
• Each chromosome carries many genes.
• Chromosomes in body cells are normally found in pairs.

Method: Build the hierarchy from cell structure to genetic unit: nucleus contains chromosomes, chromosomes consist of DNA, and genes occur on the chromosomes.

Answer

• The statement is incorrect.
• A chromosome is made of a DNA molecule.
• A chromosome carries a large number of genes.
• The chromosomes are contained in the nucleus and are normally paired in body cells.

Method: Reverse the student's incorrect nesting: DNA makes up the chromosome, while genes are sections carried on that chromosome; then place chromosomes in the nucleus.

Answer

• The cell grows and increases its number of sub-cellular structures.
• The DNA replicates to form two copies of each chromosome.

Method: Choose the growth or organelle-increase stage and the DNA-replication stage; both occur before the chromosomes separate in mitosis.

Answer

• The cell grows and increases the number of sub-cellular structures.
• Its DNA replicates to make two copies of each chromosome.
• In mitosis, one set of chromosomes moves to each end and the nucleus divides.
• The cytoplasm and cell membrane divide to form two identical cells.

Method: Present the three overall stages in order: preparation and DNA copying, chromosome and nuclear division, then division of the rest of the cell.

Answer

• Cells with increased ribosome and mitochondrion numbers are growing and preparing to divide.
• Copied chromosomes show that DNA replication has occurred.
• Dividing nuclei show that mitosis is occurring.
• One set of chromosomes is moved to each end of a dividing cell.
• Division of the cytoplasm and membranes produces two identical cells.
• Repeated mitosis increases cell number for growth of the root.

Method: Match each observation to its cell-cycle stage, then connect the identical daughter cells to an increase in cell number and therefore growth.

Answer

• It is an undifferentiated cell that can divide to make more cells of the same type and can differentiate into other cell types.

Method: Include both defining abilities: self-renewal by division and production of specialised cells by differentiation.

Answer

• Embryonic stem cells can differentiate into most types of human cell.
• Adult bone-marrow stem cells can form several types of cell, including blood cells.
• Meristem cells can differentiate into any type of plant cell throughout the plant's life.
• All are undifferentiated cells able to divide and then differentiate.

Method: State the different ranges of cell types each source can form and finish with the shared stem-cell properties.

Answer

• Embryonic stem cells can differentiate into most human cell types, so they may replace damaged cells.
• The treatment could improve movement or quality of life.
• The cloned embryo has the same genes as the patient.
• The cells are therefore less likely to be rejected by the patient's body.
• There is a risk of transferring a viral infection.
• Some people have ethical or religious objections to producing and using an embryo, so benefits and risks must be weighed before a conclusion.

Method: For an evaluation, give explained benefits, the genetic-match advantage, practical risk and ethical issue. A justified conclusion may support or oppose use if it weighs this evidence.

Answer

• Diffusion is the net movement of particles from an area of higher concentration to an area of lower concentration.

Method: Both 'net movement' and the direction down the concentration gradient are required for the complete definition.

Answer

• Many air sacs provide a large surface area for diffusion.
• A thin wall gives a short diffusion distance.
• A good blood supply carries oxygen away and maintains a steep concentration gradient.
• Ventilation replaces the air and maintains a high oxygen concentration in the air sac.

Method: Link each adaptation to surface area, diffusion distance or maintenance of the concentration gradient; any three complete links can score.

Answer

• Cube A has surface area $6\,\mathrm{mm^2}$, volume $1\,\mathrm{mm^3}$ and ratio $6:1$.
• Cube B has surface area $96\,\mathrm{mm^2}$, volume $64\,\mathrm{mm^3}$ and ratio $1.5:1$.
• The larger cube has the smaller surface area to volume ratio.
• A large organism therefore has too little external surface for diffusion alone to supply all cells quickly enough.
• Specialised exchange surfaces increase exchange area and shorten the diffusion path.
• A transport system moves substances between the exchange surface and internal cells.

Method: For a cube, surface area is $6s^2$ and volume is $s^3$. Thus A gives $6/1=6:1$, while B gives $96/64=1.5:1$. Apply the falling ratio to the need for rapid exchange and internal transport in a large organism.

Answer

• Water moves from the surrounding dilute solution into the more concentrated cell contents.
• The process is osmosis through the partially permeable cell membrane.

Method: Apply the osmosis definition: water moves from dilute to concentrated solution through a partially permeable membrane.

Answer

• $+15\%$
• The potato gained water.

Method: The mass change is $4.60-4.00=0.60\,\mathrm{g}$. Percentage change is $(0.60/4.00)\times100=15\%$. The positive change shows a gain of water by osmosis.

Answer

• $-15\%$
• Water moved from the more dilute solution inside the tissue to the more concentrated external solution.
• Water crossed partially permeable cell membranes by osmosis.
• Control variables include the initial dimensions or surface area of the tissue, time, temperature, tissue source and volume of solution.

Method: The change is $4.25-5.00=-0.75\,\mathrm{g}$, so percentage change is $(-0.75/5.00)\times100=-15\%$. Link the mass loss to water moving from dilute to concentrated solution through partially permeable membranes, then name any two valid controls.

Answer

• It is movement of a substance from a more dilute solution to a more concentrated solution, against the concentration gradient, using energy from respiration.

Method: State both the direction against the concentration gradient and the energy requirement.

Answer

• Nitrate ions move from the more dilute soil solution into the more concentrated cell contents.
• This movement is against the concentration gradient.
• Active transport uses energy transferred by respiration.
• The ions are needed for healthy plant growth.

Method: Use the supplied concentrations to rule out diffusion, identify movement against the gradient, give the respiration-energy requirement and state the biological benefit.

Answer

• Diffusion is the net movement of particles from higher to lower concentration and does not require energy from respiration.
• Osmosis is the diffusion of water from a dilute to a concentrated solution through a partially permeable membrane and does not require energy from respiration.
• Active transport moves substances from a more dilute to a more concentrated solution, against the concentration gradient.
• Active transport requires energy transferred by respiration.
• Diffusion and active transport can move solutes across cell membranes, whereas osmosis refers only to water.
• Only osmosis requires the membrane to be described as partially permeable in its definition.

Method: Organise the comparison by particle, direction, membrane and energy. Keep osmosis restricted to water, and distinguish passive movement down a gradient from energy-requiring movement against it.