Cells

1. It has a nuclear envelope with pores.
2. It contains DNA associated with histones.
3. It contains a nucleolus.

1. It stores genetic material for the coding of polypeptides.
2. DNA replication occurs inside of it.
3. Transcription (production of mRNA and tRNA) occurs.
4. rRNA and ribosomes are produced.

1. They have circular DNA.
2. They have 70s ribosomes.
3. Their DNA has no introns.
4. Their DNA isn’t associated with histonesNote: Basically all the features of prokaryotic cells. These could also be applied to chloroplasts.

1. It contains ribosomes where protein synthesis occurs.
2. It transports, modifies and processes the proteins once they have been synthesised.

1. They contain chlorophyll in chloroplasts to absorb light.
2. The large permanent vacuole pushes the chloroplasts to the edge of the cell.
3. They have a thin/ permeable wall to absorb carbon dioxide.Note: Don’t confuse this with the adaptations of a leaf (e.g. gaps in the spongy mesophyll layer)

1. They have no nucleus or organelles to maximise the volume for haemoglobin.
2. They have a biconcave shape that allows them to bend and flow smoothly through the capillaries.

1. Cell membranes.
2. Ribosomes.
3. Cytoplasm.
4. DNA.

1. Prokaryotic has no membrane-bound organelles.
2. Prokaryotic has circular DNA (not linear) that lies free in the cytoplasm.
3. Prokaryotic’s DNA is not associated with histones/ proteins.
4. Prokaryotic’s cell wall is made of murein (instead of chitin for fungi or cellulose for plant cells)
5. Prokaryotes have smaller, 70S ribosomes (vs. 80S)
6. Prokaryotic DNA contains no introns.

1. Plasmids.
2. Flagellum.
3. Capsule.

1. Ribosomal RNA (rRNA)
2. Protein.

1. To create a single layer of cells.
2. This means light can pass through Bonus note: the specimen can be made thin by cutting a thin layer of tissue or by pressing down on it with a coverlip.

1. Electrons pass through the specimen.
2. The denser parts absorb more electrons.
3. The denser parts therefore appear darker.
4. Electrons have a short wavelength so produce a high resolution.Note: This should just be mentioned in every question about electron microscopes

Note: TEMs focus using magnets and optical microscopes focus using lenses.

1. Specimen cannot be living as it is viewed in a vacuum.
2. Specimen must be very thin.
3. Artifacts can be present.
4. The staining method is complex and the preparation time is therefore long.
5. Only 2D images can be produced (vs. 3D images with an SEM)
6. Only black and white images can be produced (vs. colour images with an optical)

1. Measure the specimen using an eyepiece graticule.
2. Calibrate the eyepiece graticule against a stage micrometer.
3. Take multiple measurements and calculate a mean.

1. They would measure the diameter of the field of view and calculate the area.
2. Using the stage micrometer and eyepiece graticule.
3. They would then count the number of capillaries in a large number of different fields of view and calculate a mean.
4. They would select the fields of view randomly to avoid bias.

1. Add a drop of water to the slide.
2. Get a section of the plant tissue and place on the slide, floating it on the water.
3. Stain with iodine in potassium iodide.
4. Lower cover slip at an angle using a mounted needle to avoid trapping air bubbles.

1. Homogenise the tissue to break open the cells and release the organelles.
2. Filter to remove intact tissue (connective tissue) and debris.
3. Place it in a cold solution to prevent enzyme activity.
4. Place it in an isotonic solution to prevent osmosis so the organelles don’t burst or shrink.
5. Place it in a buffered solution to prevent proteins/ enzymes denaturing.
6. Centrifuge/spin at a low speed so the nuclei form the pellet, then discard the supernatant.

1. It allows for growth.
2. It allows tissues and organs to be repaired as it allows cells to be replaced.
3. It produces genetically identical cells.Note: this is the only out of the 4 that wasn’t a function, suggesting it’s just a good thing to mention on any mitosis question.
4. It can be used in asexual reproduction.

1. The centromeres of the chromosomes attach to the spindle.
2. The chromosomes line up at the equator of the cell.

1. The centromeres divide.
2. The chromatids are pulled to opposite poles of the cell.
3. When the spindle fibres shorten and contract.

1. The cytoplasm dividing.
2. The cell membrane being pinched and a new cell membrane forming.
3. The chromosomes not being visible.

1. The chromosomes will be visible because they have condensed.
2. The chromosomes will be made of two chromatids because DNA has replicated.
3. The chromosomes are not arranged in homologous pairs.

1. Increased concentration of glucose - increases respiration.
2. Increased concentration of oxygen - increases respiration.
3. Increased temperature - increases enzyme activity.
4. Increased concentration of phosphate - increases ATP/DNA/RNA.

1. An abnormal mass of cells.
2. Where uncontrolled mitosis is occurringNote: Remember that uncontrolled mitosis occurs in both benign and malignant tumours, and that malignant tumours are cancerous because they invade other parts of the body.

1. Replication of (circular) DNA.
2. Replication of plasmids.
3. The cytoplasm divides and new membranes form to produce daughter cells.

Note: you could've have also just said 'cytokinesis'.

1. DNA replication will occur faster.
2. This causes a faster cell cycle and uncontrolled cell division.
3. This results in a mass of abnormal cells.

1. Squash it under the cover slip to spread the tissue and make a thin layer of cells, allowing light to pass through so the cells can be visible.Note: this also allows the stain to enter the cells.
2. Don’t push sideways to avoid rolling cells together and breaking chromosomes.

1. To break down its cell walls, allowing the stain to pass more easily into the cells and so the cells can be squashed more easily.
2. To stop mitosis from occurring.

1. The roots examined may be a different age or grown in different conditions.
2. The roots examined may be from different plants or even species.
3. The sample size used may not be large enough to be representative of a root tip.
4. They may have examined different parts of the root tip.
5. Different cells undergo mitosis at different times and different rates.

1. Wear eye protection.
2. Wear gloves.
3. Add water to spills immediately.
4. Don’t pour it down the sink.

1. Choose plants of the same age.
2. Grow the plants in the same conditions.
3. Examine cells the same distance from the root tip.
4. Put the tissue in the acid for the same time.
5. Use the same concentration of acid.
6. Squash them with the same pressure.
7. Use several fields of view for each species.
8. Conduct the experiment at the same time of day.

1. Only count whole cells to standardise counting.
2. Examine a large number of fields of view (and calculate a mean) to ensure a representative sample.
3. Choose fields of view randomly to remove bias.
4. Repeat the count to ensure the figures are correct.Note: I think this is the weakest one.

1. They form a bilayer, where water is present inside and outside of the cell.
2. The hydrophobic fatty acid tails point inwards as they are repelled by water.
3. The hydrophilic phosphate heads point outwards as they are attracted to water.

1. Enzyme activity decreases as the active sites denature.
2. Membrane fluidity increases due to the denaturing of carrier and channel proteins and the phospholipids gaining more kinetic energy.

Note: the water inside the cell gaining more kinetic energy and expanding also increases membrane fluidity, but I haven’t seen it mentioned on a mark scheme.

1. It isn’t lipid soluble, meaning it can’t diffuse through the phospholipid bilayer.
2. It is too large or the wrong shape, meaning it is unable to bind to or pass through any channel or carrier proteins.Note: This would be applied to a specific molecule that has been described in a way that shows its large/ has a specific shape (e.g. a polypeptide). If ‘molecules’ in general are not entering a cell via facilitated diffusion, it’s more likely because the proteins have denatured or they are all saturated.

1. The movement of polar/charged molecules.
2. Down a concentration gradient.
3. Via a carrier/channel protein.

1. Facilitated diffusion involves channel or carrier proteins whereas active transport only involves carrier proteins.Note: this one is often forgotten - never say active transport occurs via channel proteins.
2. Facilitated diffusion does not use ATP whereas active transport uses ATP;
3. Facilitated diffusion takes place down a concentration gradient whereas active transport occurs against a concentration gradient;

1. Sodium ions are actively transported from the ileum to blood via the sodium-potassium pump.
2. This forms a concentration gradient where there is a lower concentration of sodium ions inside the ileum than outside.
3. Sodium then enters the ileum via facilitated diffusion via co-transport proteins.
4. Glucose also enters via facilitated diffusion with the sodium ions.
5. Glucose then enters the blood via facilitated diffusion.

Note: 'Glucose' could of course be replaced with 'amino acids'.

1. They have folded membranes to increase the surface area for absorption.
2. They have a large number of carrier and channel proteins, allowing for a lot of facilitated diffusion and active transport.
3. They have a large number of mitochondria, producing the ATP that releases energy for active transport.

1. Plot calibration curve of concentration of solution against percentage change in mass.
2. Interpolate (using the line of best fit) to find what the concentration would be at a change in mass of 0.
3. Convert this concentration into water potential.

1. Water will affect the mass of the potatoes.
2. Drying them ensures there will be the same amount of water on the outside of each cylinderNote: I’d also mention each cylinder having a different mass of water on its outside before drying.

1. Each cylinder will have a different starting mass.
2. So using percentage change in mass allows for fair comparison, as it makes the change in mass proportional to the starting mass of each cylinder.

1. Because sodium/ sucrose concentration stops being the limiting factor, and another factor (e.g. temperature) becomes limiting.
2. The reason concentration stops being a limiting factor is because water is being taken up through the pores at the maximum possible rate.

1. Take readings of the temperature at (stated e.g. 5 minute) intervals with a thermometer to make sure the temperature remains constant.
2. Use an electric water bath.

1. Surface area of the potato cylinders.
2. Mass of the potato cylinders
3. Time spent in solution.
4. Volume of solution.
5. Surface area in contact with the solution (make sure they’re all fully submerged)

1. Ensure all cut surfaces of the cubes are exposed to the sucrose solution. This could be done by hanging them by a piece of thread in the solution so they are fully submerged but not touching the bottom.
2. Control the temperature using an electric water bath set at 30 degrees C or taking multiple readings of the temperature to ensure it remains at the same room temperature (or 30 degrees C)
3. Dry the cubes before measuring them by patting them dry with paper towels.
4. Measure the mass of cubes every minute.

1. Temperature.
2. pH.
3. Type/ age of beetroot.
4. Amount of stirringNote: cholesterol content in the cell membranes as well as saturated vs unsaturated fatty acid composition of the cell membranes also affect the results.

1. It causes the membrane proteins to become denatured.
2. It causes the fluidity of the phospholipid bilayer to increase.

1. It ensure all surfaces of the discs are submerged and exposed to water.
2. Which maintains a diffusion gradient for the pigment out of the discs.

1. A foreign protein.
2. That stimulates an immune response and causes antibody production.

1. Pathogens OR cells from an organism of another species.Note: notice that these fall under the same mark point, so giving both would only gain you 1 mark. I’m not sure why they weren’t credited separately.
2. Cells from organisms of the same species.
3. Abnormal body cells.
4. Antigen-presenting cells.

1. The phagocyte is attracted to the pathogen by chemo attractants released by the pathogen. It moves towards the pathogen along a concentration gradient via diffusion.
2. The phagocyte has several receptors on its cell-surface membrane that attach to the chemicals on the surface of the pathogen. The phagocyte then engulfs the bacterium, forming a phagosome.
3. Lysosomes within the phagocyte migrate towards and fuse with the phagosome (vesicle), forming a phagolysosome.
4. The lysosomes release their lysozymes into the phagosome which then hydrolyse and break down the bacterium.
5. The hydrolysis products of the bacterium are absorbed by the phagocyte.
6. The pathogen's antigens are displayed on the phagocyte's cell surface membrane, making it become an antigen presenting cell.

1. Lysosomes fuse with the vesicle.
2. Then release hydrolytic enzymes.
3. These breakdown and digest the pathogen.

1. More unsaturated fatty acids increases the fluidity of the cell membrane.
2. Making it easier for the phagocytes to engulf pathogens.

1. The thicker capsule means phagocytes are less likely to bind to and get stimulated by the murein.
2. This causes reduced phagocytosis so more bacterial growth via binary fission.Note: When explaining why something causes more disease, always link it to more replication.

1. After a macrophage/ phagocyte engulfs a pathogen, it presents its antigens on its cell surface and becomes an antigen presenting cell.
2. The T cell’s specific receptor then binds to the antigen, activating the T cell.
3. This stimulates the T cell to divide by mitosis.
4. They then produce stimulate B cells, activate phagocytosis and activate cytotoxic cells.

1. Macrophages present antigens to B lymphocytes.
2. The antigen is complementary to receptors on the lymphocyte.
3. They therefore bind to a specific lymphocyte.
4. This causes the lymphocyte to become sensitised.
5. This allows T helper cells to activate them to divide by mitosis.

1. A complementary antigen on the surface of the pathogen binds to the surface protein/ receptor on a specific B Cell.
2. This B Cell is stimulated to divide by a helper T Cell (previously activated by the same antigen) binding to it. (clonal selection)
3. The activated B Cell then divides by mitosis to form a clone of plasma cells (clonal expansion)
4. These plasma cells then release antibodies.
5. Some B Cells become memory cells.
6. Memory cells produce antibodies faster.

1. The antibody has a specific tertiary structure.
2. Its binding site is therefore complementary to only one specific antigen.
3. This specific antigen is only found on one type of cell, causing the antibody to only bind to this cell, and only form antigen-antibody complexes with antigens on this cell.

1. They bind to their complementary antigens on the pathogen, forming an antibody-antigen complex.
2. This causes agglutination and attracts phagocytes.

1. They contain antigens.
2. Usually from dead or weakened pathogens.
3. Memory cells are produced when they are injected.
4. On second exposure these memory cells are activated as they recognise the antigens on the pathogen.
5. These memory cells then rapidly produce more antibodies.
6. The antibodies destroy the pathogens.
7. The herd immunity effect occurs, as there are few people to pass on the disease.Note: this is probably worth mentioning on any vaccine question.Note: This is just the general description of how vaccines cause immunity, there are some questions that require more detailed explanations of the humoural response and how it leads to the production of memory cells, which will be covered later in the doc.

1. It causes the production of more memory cells.
2. This means there is a more rapid production of antibodies and the concentration of antibodies in the blood rises more on further infection.

1. HIV has a high rate of mutation.
2. HIV has high genetic diversity.
3. There may be a lack of funding for research.
4. HIV causes there to be fewer T cells, so the immune response to the vaccine doesn’t happen.

1. Active produces memory cells, passive does not.
2. Active involves the production of antibodies by plasma cells.
3. Passive involves antibodies introduced from an outside source.
4. Active is a long-term response as an antibody is produced in response to an antigen.
5. Passive is short-term as the antibody is eventually broken down.
6. Active can take time to work, passive is fast acting.

1. They have weakened immune systems so are more susceptible to the disease.
2. They are more likely to get infected as they come into contact with a large number of people.
3. There will be more people with the disease in the hospital so they are more likely to come in contact with it.

1. Production involves the use of animals.
2. There is the potential for dangerous side effects that people may not be aware of.
3. People may get ill or die during clinical trials of the treatments.

1. Genetic material - codes for viral proteins.
2. Capsid - protects the genetic material.
3. Attachment proteins - binds to receptors on cells.

It contains:

1. RNA.
2. Reverse transcriptase.
3. A protein capsid.
4. A lipid envelope.
5. Attachment proteins.

1. Attachment proteins attach to receptors on helper T Cell.
2. Viral RNA enters the cell.
3. Reverse transcriptase converts RNA to DNA.
4. DNA inserted into T-helper cell’s nucleus and modifies the cell’s DNA.
5. DNA transcribed into HIV mRNA.
6. HIV mRNA translated on the ribosomes into viral proteins.
7. Viral proteins/ capsids (name capsids) are produced.
8. Virus is assembled and released from the cell via budding off.Note: When adding the sources, I noticed that a few of these points didn’t seem to actually appear in any of the mark schemes. However, they’re correct and must’ve come from somewhere so I’d still include them.

1. Viruses don’t have a cell wall so don’t have a site for antibiotics to work.
2. They don’t have any metabolic processes or cell structures that antibiotics can disrupt.
3. Antibiotics can’t reach viruses inside cells.

1. Wash hands with soap and disinfect the surfaces to kill microbes.
2. Sterilise equipment by boiling them or using an autoclave.
3. Work near to a bunsen burner to create upwards movement of air (convection currents).
4. Sterilise the inoculating loop (or pipette) by flaming it with a bunsen burner.
5. Lift the lid of the agar plate at an angle and for as little time as possible to prevent the entry of microbes.
6. Flame the neck of any bottles with a bunsen burner to sterilise them.
7. Place all equipment into disinfectant after use.
8. Tape the plate lightly to prevent the lid from opening but do not seal it so air exchange can still happen so anaerobic bacteria do not grow.
9. Incubate at 25 degrees C to reduce rate of pathogen growth and cause them to not grow at and be adapted to human body temperature.

1. To create upwards air movement via convection currents,.
2. This draws microbes away from the culture.Note: its purpose is not to sterilise the working environment (never say this).

1. They should have the same area.
2. The should be made of the same material so they have the same absorbency.
3. They should be soaked in the solution for the same duration.

1. They target specific antigens on specific cells.Note: you could mention cancer cells as an example of a specific cell.
2. They carry drugs to specific cells.
3. They block antigens on cells.

Note: The idea of specific targeting is obviously important.

1. The first antibody (attached to the well) binds to the complementary antigen in the sample.
2. The second antibody, with the enzyme attached, is added.
3. The second antibody attaches to the antibody-antigen complex.
4. You should then wash to remove any unattached antibodies to prevent a false positive.
5. The substrate solution is added and the colour changes when enzyme-substrate complexes form.Note: You could have either described direct or indirect, this is the answer for indirect. Either way, you had to get the idea of the antibody being complementary to the antigen, an antibody with an enzyme attached being added, and the colour changing when a substrate is added. Personally, I’m going to describe indirect if a similar question comes up in the exam.