Notes · BiologyUK · A-Levels

Cells

Cells are the fundamental unit of life. This chapter examines the ultrastructure of eukaryotic and prokaryotic cells and the organelles that divide labour within them, the microscopy and fractionation methods used to study them, the cell cycle and its loss of control in cancer, transport across the cell-surface membrane, and how cell recognition drives the immune response.

6 sections~19 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 3

T·0222 / 8

Exam profile

AO1 · Describe the ultrastructure of cells and the stages of the cell cycle and immune responseAO2 · Apply magnification, water-potential and mitotic-index calculations and interpret electron micrographsAO3 · Analyse membrane-permeability and immunity data and evaluate the evidence and ethics of vaccination and monoclonal antibodies

Operators:describeexplaincalculatecompareanalyseevaluatesuggest

basic level

AS-Level covers cell ultrastructure, microscopy and magnification, the cell cycle, transport across membranes and the immune response.

higher level

The full A-Level adds quantitative work (mitotic index, water potential), the detail of antigenic variation and monoclonal antibodies, and synoptic links to later exchange and gene-expression topics.

Depth

Text

Contents · 6 sections

Cells

Eukaryotic and prokaryotic cell structure#

FoundationAQA 7402 3.2.1.1DfE GCE Biology - cell structure

Animal cell ultrastructure

Fig. 1A eukaryotic animal cell: membrane-bound organelles divide the cell into functional compartments.

A eukaryotic cell is defined by having membrane-bound organelles, including a true nucleus enclosed by a nuclear envelope. The division of the cell into compartments allows incompatible processes to occur simultaneously and efficiently, and each organelle has a structure suited to its role. Animal, plant, fungal and protoctist cells are all eukaryotic; plant cells additionally have a cellulose cell wall, a large permanent vacuole and, in photosynthetic tissue, chloroplasts.

The key organelles and their functions form the essential vocabulary of the topic. The nucleus contains the chromosomes (DNA) and controls the cell, with a nucleolus that makes ribosomes; mitochondria are the site of aerobic respiration and have a folded inner membrane (cristae) that increases surface area; ribosomes (on the rough endoplasmic reticulum or free) synthesise proteins; the rough endoplasmic reticulum processes and transports those proteins while the smooth endoplasmic reticulum makes lipids; the Golgi apparatus modifies, packages and labels proteins into vesicles; and lysosomes contain hydrolytic enzymes that digest worn-out organelles and pathogens.

Prokaryotic cells (bacteria) are much smaller and simpler: they have no membrane-bound organelles and no true nucleus. Their DNA is a single circular molecule free in the cytoplasm, often with additional small loops called plasmids; their ribosomes are smaller (70S versus the 80S of eukaryotes); and they have a cell wall made of murein (a glycoprotein), not cellulose. Some also have a protective capsule, flagella for movement and pili. These differences are examined directly and underpin how some antibiotics selectively attack bacteria.

Viruses are acellular and non-living: they are not cells at all. A virus consists of genetic material (DNA or RNA) enclosed in a protein coat (capsid), with attachment proteins on the surface that let it bind to specific host cells. Having no cytoplasm, ribosomes or organelles, a virus can only replicate inside a host cell whose machinery it hijacks - a point that becomes important when studying HIV later in this chapter.

Prokaryotic cell

Fig. 2A prokaryote has no membrane-bound organelles; its DNA is circular and free in the cytoplasm.

Comparing eukaryotic and prokaryotic cells

Fig. 3The presence of membrane-bound organelles is the defining eukaryotic feature.

Worked example

The secretory pathway

A cell secretes a protein hormone. Name, in order, the organelles involved and state the role of each.

Synthesis
Ribosomes on the rough endoplasmic reticulum synthesise the polypeptide.
Transport and processing
The rough ER folds and transports the protein in vesicles to the Golgi apparatus, which modifies and packages it.
Secretion
The Golgi releases the protein in secretory vesicles that fuse with the cell-surface membrane, releasing it by exocytosis (energy supplied by mitochondria as ATP).

Result: Ribosome / rough ER -> Golgi apparatus -> secretory vesicle -> cell-surface membrane (exocytosis).

Typical mistakes

Saying a prokaryote has 'no organelles' - it has ribosomes; the correct statement is no membrane-bound organelles.
Describing a virus as a very small cell - viruses are acellular and non-living.

Active revision

Using named organelles, describe the route taken by a protein from its synthesis to its secretion from the cell.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Studying cells: microscopy, magnification and fractionation#

StandardAQA 7402 3.2.1.2AQA 7402 3.2.1.3DfE GCE Biology - methods of studying cells

Cell fractionation

Fig. 4Increasing centrifuge speed sediments organelles in order of decreasing density.

Magnification is how many times larger the image is than the real object, whereas resolution is the smallest distance between two points that can still be distinguished as separate. These are different ideas: increasing magnification beyond the limit of resolution gives a bigger but blurred image. Resolution is limited by the wavelength of the radiation used, which is why electron microscopes, using electron beams of far shorter wavelength than light, resolve much finer detail.

The optical (light) microscope uses light, so its resolution is limited to about 200 nm; it is cheap, portable and can view living, coloured specimens. The transmission electron microscope (TEM) passes electrons through a thin section to reveal internal ultrastructure at very high resolution, while the scanning electron microscope (SEM) reflects electrons off the surface to give a three-dimensional image at somewhat lower resolution. Electron microscopes must operate in a vacuum, so they cannot view living material, and preparation can introduce artefacts.

Magnification is calculated with the equation

\text{magnification} = \dfrac{\text{size of image}}{\text{size of real object}}

. The commonest error is a unit mismatch, so always convert the image and actual sizes to the same unit first (remember

1\ \text{mm} = 1000\ \text{um}

and

1\ \text{um} = 1000\ \text{nm}

). The same equation rearranges to find the real size of a structure from a scale bar or a stated magnification.

Cell fractionation separates organelles so they can be studied. The tissue is first homogenised (blended) in a solution that is cold (to reduce enzyme activity), isotonic (the same water potential, so organelles do not burst or shrink by osmosis) and buffered (to keep pH constant). The homogenate is filtered and then spun in a centrifuge at increasing speeds - ultracentrifugation - so that the densest organelles (nuclei) sediment first, then mitochondria and chloroplasts, then the endoplasmic reticulum and finally ribosomes. This order reflects the relative density of the organelles.

\text{magnification} = \dfrac{\text{size of image}}{\text{size of real object}}

Magnification

Convert both sizes to the same unit first; rearrange to find the real size from a scale bar.

Worked example

A magnification calculation with unit conversion

A cell has an actual diameter of 6 um. In a micrograph it measures 30 mm across. Calculate the magnification.

Convert to the same unit
Image size = 30 mm = 30 000 um (since 1 mm = 1000 um).
Apply the equation
$\text{magnification} = \dfrac{30\,000}{6}$ .
$\text{magnification} = \frac{30\,000\ \text{um}}{6\ \text{um}} = 5000$
State the answer
The magnification is x5000 (no units, as it is a ratio).

Result: x5000.

Typical mistakes

Mixing units (e.g. dividing a millimetre image size by a micrometre actual size) so the magnification is out by a factor of 1000.
Saying the homogenising solution is isotonic 'to stop reactions' - it is isotonic to prevent osmotic damage; it is cold to reduce enzyme activity.

Active revision

A mitochondrion measures 8 um in length. In a photograph it appears 40 mm long. Calculate the magnification of the photograph.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

The cell cycle, mitosis and cancer#

StandardAQA 7402 3.2.2DfE GCE Biology - cell division

The cell cycle

Fig. 5DNA is replicated in the S phase; mitosis and cytokinesis then produce two identical cells.

The cell cycle is the sequence a cell passes through from one division to the next. It consists of a long interphase, during which the cell grows and copies its DNA by semi-conservative replication and doubles its organelles, followed by mitosis (nuclear division) and cytokinesis (division of the cytoplasm). Interphase is subdivided into G1 (growth), S (DNA synthesis) and G2 (further growth), and checkpoints regulate progress so that damaged or incompletely copied DNA is not passed on.

Mitosis produces two genetically identical diploid daughter nuclei and is described in four stages. In prophase the chromosomes condense and become visible (each as two sister chromatids joined at a centromere) and the nuclear envelope breaks down; in metaphase the chromosomes line up on the equator, attached to spindle fibres by their centromeres; in anaphase the centromeres divide and the spindle fibres pull the sister chromatids to opposite poles; in telophase the chromatids reach the poles, a nuclear envelope re-forms and the chromosomes decondense.

Mitosis is essential for growth, the repair and replacement of cells, and asexual reproduction, and because the daughter cells are genetically identical it preserves the chromosome number and the genome. The proportion of cells in mitosis in a sample is the mitotic index, calculated as the number of cells with visible chromosomes divided by the total number of cells; a high index indicates rapidly dividing tissue, which is used both in research and in assessing tumours.

Cancer results from a loss of control of the cell cycle. Mutations in the genes that regulate division (proto-oncogenes and tumour suppressor genes) can cause cells to divide uncontrollably, forming a tumour. Because cancer cells divide rapidly, many treatments target the cell cycle - for example some drugs disrupt spindle formation (preventing metaphase and anaphase) or interfere with DNA replication in the S phase - though these also affect healthy dividing cells, causing side effects.

\text{mitotic index} = \dfrac{\text{number of cells in mitosis}}{\text{total number of cells}}

Mitotic index

A proportion (0-1) or, times 100, a percentage; higher in rapidly dividing tissue.

The stages of mitosis

Fig. 6Remember the sequence PMAT: prophase, metaphase, anaphase, telophase.

Worked example

Calculating a mitotic index

A field of view contains 200 cells, of which 24 are undergoing mitosis. Calculate the mitotic index as a percentage.

Identify the numbers
Cells in mitosis = 24; total cells = 200.
Apply the equation
$\dfrac{24}{200} = 0.12$ .
Convert to a percentage
$0.12 \times 100 = 12\%$ .
$\text{mitotic index} = \frac{24}{200} \times 100 = 12\%$

Result: A mitotic index of 0.12, or 12%.

Typical mistakes

Confusing chromatids and chromosomes, or saying chromosomes (rather than centromeres) divide in anaphase.
Giving the mitotic index as a whole number of cells rather than a proportion or percentage.

Active revision

In a stained root-tip squash, 27 of 180 cells show condensed chromosomes. Calculate the mitotic index and suggest what it indicates about this tissue.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Transport across cell membranes#

AdvancedAQA 7402 3.2.3DfE GCE Biology - transport across membranes

The fluid-mosaic membrane

Fig. 7A fluid bilayer of phospholipids with proteins forming channels, carriers and receptors.

Cell membranes are described by the fluid-mosaic model: a bilayer of phospholipids (hydrophilic heads outward, hydrophobic tails inward) in which proteins are embedded like tiles in a mosaic, and which is fluid because the phospholipids move. The bilayer is the barrier - small, non-polar molecules (oxygen, carbon dioxide) pass straight through, but ions and large polar molecules cannot, so their transport depends on the proteins. Cholesterol between the phospholipids restricts movement and regulates fluidity and stability.

Diffusion is the net movement of molecules from a region of higher to lower concentration, down a concentration gradient, and is passive (no ATP). Small non-polar molecules cross by simple diffusion through the bilayer; charged or polar molecules cross by facilitated diffusion through specific channel or carrier proteins, still down their gradient. The rate depends on the steepness of the concentration gradient, the surface area, and the thickness of the membrane, and facilitated diffusion additionally plateaus when all the carrier proteins are saturated.

Osmosis is a special case: the net movement of water molecules from a region of higher water potential to a region of lower water potential across a partially permeable membrane. Water potential (

\Psi

, measured in kPa) is the tendency of water to leave a system; pure water has the highest value, defined as

0\ \text{kPa}

, and adding solutes makes it more negative. It is given by

\Psi = \Psi_s + \Psi_p

, the sum of the (negative) solute potential and the (usually positive) pressure potential, and water always moves towards the more negative value.

Active transport moves substances against their concentration gradient, from low to high concentration, and therefore requires ATP and carrier proteins that act as pumps. Co-transport is a linked form used, for example, to absorb glucose in the ileum: sodium ions are actively pumped out of the epithelial cell, creating a gradient down which sodium then diffuses back in through a co-transporter protein, dragging glucose in with it against the glucose gradient. This mechanism reappears in the exchange chapter, so it is worth mastering here.

\Psi = \Psi_s + \Psi_p

Water potential

$\Psi_s$ (solute potential) is negative; $\Psi_p$ (pressure potential) is usually positive; pure water has $\Psi = 0$ . Water moves to the more negative $\Psi$ .

Rate of uptake against oxygen concentration

Fig. 8Active transport depends on ATP from respiration, so its rate rises with oxygen availability; facilitated diffusion does not.

Worked example

Predicting water movement

A plant cell with a water potential of -500 kPa is placed in a solution of water potential -200 kPa. Determine the direction of net water movement and describe what happens to the cell.

Compare the water potentials
The solution (-200 kPa) is higher (less negative) than the cell (-500 kPa).
Apply the rule
Water moves by osmosis from higher to lower water potential, i.e. from the solution into the cell.
Describe the effect
The cell gains water, its volume and pressure potential rise, and it becomes turgid (the wall prevents bursting).

Result: Water enters the cell (solution -> cell); the cell becomes turgid.

Typical mistakes

Describing osmosis in terms of solute concentration rather than water potential, or forgetting that water moves to the more negative (lower) water potential.
Saying facilitated diffusion needs ATP - it is passive; only active transport (against the gradient) needs ATP.

Active revision

Two adjacent cells have water potentials of -600 kPa and -450 kPa. State the direction of net water movement and explain your answer.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Cell recognition and the immune response#

AdvancedAQA 7402 3.2.4DfE GCE Biology - the immune system

The stages of the immune response

Fig. 9Phagocytosis leads to antigen presentation, T-cell activation, and the humoral B-cell response.

The immune system distinguishes self from non-self using antigens - molecules (usually proteins or glycoproteins) on cell surfaces that trigger an immune response. Each cell of an organism carries its own antigens, so the immune system can recognise pathogens, cells from other organisms (relevant to transplants), toxins and abnormal (cancer) cells as foreign. This specificity of recognition is the foundation of the whole response.

The first specific response is often phagocytosis, a non-specific engulfing: a phagocyte recognises and engulfs a pathogen into a vesicle (phagosome), which fuses with a lysosome so that hydrolytic enzymes (lysozymes) digest it. The phagocyte then displays the pathogen's antigens on its surface, becoming an antigen-presenting cell that activates the specific lymphocytes - the link between the non-specific and specific responses.

The cellular response is driven by T lymphocytes. A specific helper T cell with a complementary receptor binds to the presented antigen and is activated; it then stimulates phagocytes, activates cytotoxic T cells (which kill infected cells) and, importantly, stimulates the B lymphocytes. This selection of the one lymphocyte with the matching receptor from the enormous existing variety is called clonal selection.

The humoral response is driven by B lymphocytes and produces antibodies. Once selected and activated by a helper T cell, the specific B cell divides rapidly by mitosis (clonal expansion) into plasma cells, which secrete large quantities of a specific antibody, and memory cells, which remain in the blood. An antibody is a Y-shaped protein (quaternary structure) with variable regions that form two antigen-binding sites complementary to a specific antigen; binding causes agglutination (clumping) of pathogens, marking them for destruction by phagocytes.

The distinction between the primary and secondary responses explains immunity. On first exposure the primary response is slow because the specific lymphocyte must be selected and expanded, so symptoms appear. Memory cells then persist, so on a second exposure to the same antigen the secondary response is faster and stronger - memory cells divide rapidly into plasma cells and destroy the pathogen before symptoms develop. This is the basis of long-term immunity and of vaccination.

Worked example

Why the secondary response is faster

Using the idea of memory cells, explain why the secondary immune response destroys a pathogen before symptoms appear.

Memory cells persist
After the primary response, memory B and T cells specific to the antigen remain in the blood for years.
Rapid recognition
On re-exposure the correct lymphocyte is already present, so no time is lost in clonal selection.
Faster, larger response
Memory cells divide rapidly into plasma cells that secrete antibody sooner and in greater quantity, clearing the pathogen before it can multiply enough to cause symptoms.

Result: Memory cells give a faster, stronger secondary response that eliminates the pathogen before symptoms develop.

Typical mistakes

Confusing the roles of T and B lymphocytes, or of plasma cells (secrete antibody now) and memory cells (persist for future).
Saying antibodies destroy pathogens directly - they agglutinate or mark them, and phagocytes then destroy them.

Active revision

Explain why a person who has recovered from a particular infection usually shows no symptoms if exposed to the same pathogen again.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

Vaccination, HIV and monoclonal antibodies#

AdvancedAQA 7402 3.2.4DfE GCE Biology - vaccination and antibodies

Primary and secondary antibody responses

Fig. 10The secondary response is faster and produces far more antibody, thanks to memory cells.

Vaccination exploits the secondary response by exposing the immune system to a safe form of an antigen (a dead or attenuated pathogen, or an isolated antigen). This triggers a primary response and, crucially, the production of memory cells, so if the real pathogen is later encountered a rapid secondary response prevents disease. Herd immunity arises when a large enough proportion of a population is vaccinated that the pathogen cannot spread easily, protecting even the unvaccinated.

Vaccines do not always eradicate a disease. Some pathogens show antigenic variation - they change their surface antigens (as the influenza virus does frequently), so memory cells produced against the old antigen no longer recognise the new one and a fresh primary response is needed. This is why new flu vaccines are produced each year and why a single vaccine cannot give lifelong immunity to some diseases.

HIV (human immunodeficiency virus) is a retrovirus that infects and destroys helper T cells. Its structure - RNA and the enzyme reverse transcriptase inside a capsid, surrounded by a lipid envelope studded with attachment proteins - lets it enter a helper T cell, make a DNA copy of its RNA that is inserted into the host DNA, and replicate. As helper T cells are destroyed, the whole immune response is weakened; AIDS is the condition in which the immune system can no longer fight off other infections. Because a virus has no metabolism of its own, antibiotics (which target bacterial processes) are useless against HIV.

Monoclonal antibodies are identical antibodies produced from a single clone of plasma cells, all specific to one antigen. Their specificity makes them powerful tools: in targeted medical treatment (delivering a drug directly to cancer cells that bear a particular antigen, reducing side effects) and in diagnosis. The ELISA test uses a monoclonal antibody attached to an enzyme; if the target antigen (or antibody) is present it binds, and adding the substrate produces a colour change whose intensity indicates the amount present - the principle behind many pregnancy and disease tests. Their use raises ethical considerations around animal use in production and the testing of new treatments.

Worked example

Interpreting an ELISA result

In an ELISA test for a viral antigen, a patient's sample produces a strong colour change while a negative control stays colourless. Explain what this shows and how the colour is produced.

Binding step
A monoclonal antibody specific to the viral antigen is immobilised; if the antigen is present in the sample it binds to this antibody.
Enzyme and substrate
A second antibody carrying an enzyme then binds; after washing, adding the substrate lets the enzyme catalyse a colour change.
Interpretation
A strong colour shows the antigen is present (the patient has been exposed/infected); the colourless control confirms the colour is not an artefact.

Result: The colour change shows the viral antigen is present; its intensity reflects the amount.

Typical mistakes

Saying a vaccine 'gives antibodies' - it stimulates the body to make its own antibodies and, crucially, memory cells.
Suggesting antibiotics could treat HIV - antibiotics target bacterial structures/processes that viruses lack.

Active revision

Explain why the influenza vaccine has to be changed and re-administered each year, whereas a single childhood vaccine can protect against measles for life.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Biology 7402 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Biology 7402 specification

Cells

6 sections~19 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 3

Cells

Key points

The secretory pathway

Key points

A magnification calculation with unit conversion

Key points

Calculating a mitotic index

Key points

Predicting water movement

Key points

Why the secondary response is faster

Key points

Interpreting an ELISA result

Sources

Cells

Key points

The secretory pathway

Key points

A magnification calculation with unit conversion

Key points

Calculating a mitotic index

Key points

Predicting water movement

Key points

Why the secondary response is faster

Key points

Interpreting an ELISA result

Sources