9.2Exam Guidance and Skills
Exam Technique — AQA A Level Biology 7402 Revision Notes & Topical Questions
40 min read 54 parts Lesson 6 of 13 Unit quiz
Command Words and Mark-scheme Reasoning — AQA A LEVEL BIOLOGY 7402 notes: Use command words correctly and give explanations, not just descriptions, in a data question.Command words are the instruction words in an exam question.
- Worked example 1
- Worked example 2
- Worked example 3
- Competitive inhibitor
- Describing when asked to explain
- Giving one-sided arguments in 'evaluate' questions
Section 1 of 54Why command words matter
Command words are the instruction words in an exam question. They show what the examiner wants you to do, such as describing data, explaining reasons or evaluating evidence, and they show the skill being assessed (recall, application or analysis), which guides how detailed and structured your answer should be.
| Command word | Meaning |
|---|---|
| Describe | Say what you see or what happens, using data where given |
| Explain | Give biological reasons, linking cause and effect |
| Draw | Produce, or add to, a diagram |
| Evaluate | Judge from the available evidence, considering arguments for and against |
| Give | Produce an answer from recall or from given information, without explanation or description |
| Justify | Support a case with evidence |
Section 2 of 54Worked example: inhibitors
Question
- Figure 1 shows the effect of increasing substrate concentration on enzyme activity with and without a competitive and a non-competitive inhibitor.
- Explain the difference between the effects of the inhibitors.
Mark scheme
- increasing substrate concentration reduces the effect of the competitive inhibitor, so competitive inhibition can be overcome by adding more substrate and the curve reaches the same Vmax as without inhibitor [1 mark].
- The non-competitive inhibitor changes the shape of the active site or reduces its function, so the curve plateaus at a lower maximum rate [1 mark].
A student answer that said only that the rate rises as substrate concentration increases and that substrate concentration can overcome one inhibitor but not the other, scored 0/2: it described how changing variables affects reaction rate and gave no biological explanation of the patterns. A correct description earns no credit when the command word is 'explain'.
Section 3 of 54Evaluate questions
Evaluate questions assess analytical skills and usually require arguments or evidence for and against a conclusion. A common error is for students to argue only in support of, or only against, a conclusion and fail to give a balance of both.
Section 4 of 54Worked examples
Distinguish between 'give' and 'explain'. [2]2 marks
'Give' needs only a recalled answer without reasons; 'explain' needs biological reasoning linking cause and effect.
Explain why a competitive inhibitor's effect is reduced by raising substrate concentration. [2]2 marks
Substrate competes with the inhibitor for the active site; more substrate increases the chance of binding, so the same maximum rate is reached.
Explain why a non-competitive inhibitor lowers the maximum rate. [2]2 marks
It binds away from the active site and alters its shape, so fewer enzymes work however much substrate is present.
Section 5 of 54The data question
Scientists planted plots with different numbers of native North American grassland species (for example 1, 2 ... 16 species) on former farmland and weeded all plots annually to remove additional species.
- and plots with 2 species 0.14 (± 0.06) and 0.31 (± 0.08).
Section 6 of 54What a high-scoring answer includes
- making the averages reliable.
- so the relationship is not consistently increasing; only carbon in the top 20 cm was measured,
- so it may not represent total storage at greater depths;
- the work was done on former farmland using native North American grassland plants,
- so results may not apply to other regions; the study needs replicating to back up the findings.
Section 7 of 54Why the weak answer scored 0/4
The weak answer described the pattern but did not link it to a supportive argument.
- and then contradicted itself.
'Suggest' and 'explain' command words require application of biological knowledge to unfamiliar situations. Common problems in suggest questions are not relating the answer directly to the context and relying on rote-learned facts rather than tailoring ideas to the scenario or data given.
Section 8 of 54Worked examples
State one reason the results may not apply to another region. [1]1 mark
The study used native North American plants on former farmland; other regions have different species and soils.
Explain why overlapping standard deviations matter. [2]2 marks
If the standard deviations overlap, the difference between the means may be due to chance; non-overlap suggests a significant difference.
Why is the phrase 'the sample size might be too small' not creditworthy here? [2]2 marks
It is a generic phrase; 168 plots is large, so the answer must use data from the study.
Section 9 of 54Efficient use of words
An answer that spends most of its first sentence repeating the question stem gains no credit for that sentence.
Section 10 of 54Compare and contrast
These questions require identifying similarities and differences and describing them in comparative language. Example: compare prokaryotic chromosomes with eukaryotic chromosomes.
Use words such as 'both', 'while', 'whereas', 'is smaller than', 'is faster than' to make comparisons explicit.
Section 11 of 54Precision
Students often lose marks because answers are not precise enough.
- forming a phagosome in the cytoplasm [1]; the phagosome fuses with a lysosome [1]; lysozymes or hydrolytic enzymes are released and digest the pathogen [1]; the digested parts of the pathogen may be displayed on the cell surface membrane.
- (antigen presentation)
- and the phagosome step was missing.
Section 12 of 54Worked examples
Compare the chromosomes of prokaryotes and eukaryotes. [4]4 marks
Both are DNA; prokaryotic are circular whereas eukaryotic are linear; prokaryotic lack histones whereas eukaryotic have them; prokaryotic lack introns.
Describe the sequence of events in phagocytosis. [3]3 marks
Phagocyte engulfs the pathogen forming a phagosome; lysosome fuses with it; lysozymes digest the pathogen.
Explain why 'lysosomes break down the pathogen' loses a mark. [1]1 mark
Lysosomes are organelles that contain lysozymes (hydrolytic enzymes); the enzymes do the breaking down.
Section 13 of 54Cause and effect
Words such as 'so' and 'therefore' make cause and effect clear: for example, a membrane is folded so there is an increased surface area.
- so less lactate is produced, and fatigue is delayed [1].
Section 14 of 54A-level or GCSE?
If you could have written the same answer in a GCSE exam, it probably lacks the detail and terminology needed, so consider what else to add.
Section 15 of 54Avoiding ambiguity
Examiners will not infer meaning from vague statements; statements that can be interpreted in more than one way are not credited.
- 1/3 answer: used 'it' and 'them' ambiguously, so it was not clear where peptide bonds formed.
- 3/3 answer: the ribosome holds the tRNA molecules in place, each carrying a specific amino acid,
- and catalyses the formation of peptide bonds between two adjacent amino acids [1];
- it then moves to the next codon so more amino acids are added in sequence,
- and catalyses the formation of peptide bonds between two adjacent amino acids [1];
- forming a polypeptide chain.
Check each answer with four questions
- Have I chosen the exact term?
- Have I shown how one step leads to the next?
- Have I given enough A-level detail?
- Could any sentence be misunderstood?
- Assume that the examiner knows nothing and always say exactly what you mean;
- 'but that's what I meant' does not earn marks.
Section 16 of 54Worked examples
Suggest why more red blood cells benefit an endurance athlete. [4]4 marks
More haemoglobin carries more oxygen to muscles; higher rate of aerobic respiration; more ATP; less lactate so fatigue is delayed.
Describe the digestion of starch using A-level language. [4]4 marks
Amylase hydrolyses glycosidic bonds producing maltose; maltase hydrolyses maltose to glucose.
Describe the role of a ribosome in translation. [3]3 marks
Holds mRNA and tRNA in place; catalyses peptide bond formation between adjacent amino acids; moves along to the next codon.
Section 17 of 54Novel contexts
Questions with novel contexts let students move beyond basic description and make links between biological knowledge and the specific scenario or data given.
- and to apply your knowledge of the scientific method (for example to suggest how the validity of results could be improved,
- or to state what can be concluded).
Section 18 of 54Worked scenario: myoglobin in diving mammals
Myoglobin is a protein in muscle tissue that binds and stores oxygen.
- and what would support the hypothesis.
Section 19 of 54Worked scenario: heat shock proteins in whitefly
The whitefly Bemisia tabaci is an invasive agricultural pest that has spread to many regions; one factor in its success appears to be tolerance of heat, thought to involve heat shock proteins (HSPs) that help protect other cellular proteins from heat stress.
- or ensure proper formation of, hydrogen, ionic or disulfide bonds [1];
- assist protein folding so proteins fold correctly [1];
- stabilise or maintain the 3D structure,
- so preventing denaturation [1]; prevent agglutination or clumping together of proteins [1].
Section 20 of 54A routine for novel contexts
Read the stem and underline new terms.
Section 21 of 54Worked examples
Suggest how heat shock proteins protect enzymes in whiteflies. [3]3 marks
They help correct folding, maintain tertiary structure by preventing bonds breaking, and so prevent denaturation.
In the myoglobin study, suggest why the domestic dog was included. [2]2 marks
As a non-diving comparison (control), to see whether diving species have higher oxygen binding.
Suggest how the validity of the myoglobin study could be improved. [2]2 marks
Match age, sex and diet of the animals, and use standardised extraction methods and muscle from the same tissue.
Section 22 of 54Evaluating a conclusion: monoclonal antibodies and tumours
Conclusion to evaluate'Monoclonal antibodies inhibit tumour growth by preventing cell division.' Figure 1 shows tumour size over time in treated and placebo mice.
For the conclusion
- mice treated with monoclonal antibodies (group A) had a reduction in average tumour size over time, whereas tumour size increased in the placebo group (group B) [1];
- the design includes a placebo control, so differences are more likely due to the active treatment and not another factor [1].
Against the conclusion
- the graph does not show the rate of mitosis (mitotic index), or the mechanism by which tumour size is affected [1];
- antibodies may reduce tumour size by stimulating the immune system to destroy tumour cells or inducing apoptosis;
- only the average is shown and no statistical test has been included, so we cannot judge how significant the difference is;
- evidence from a mouse experiment cannot be applied directly to other species, so more research is needed [1].
The 0/3 answer described the data but gave generic statements about correlation versus causation and statistical testing, without linking them to the context or experimental design. The 3/3 answer linked the description to both supporting and opposing arguments, applied correlation and causation and statistical testing to this study, and considered the design: a control makes causation more likely, but the data do not show whether cell division is prevented or more cells are killed, so the second half of the conclusion is not supported.
Section 23 of 54Investigative skills: describing procedures
Students often lose marks by giving unstructured or vague descriptions of practical procedures. Refer to the specification to check the required practicals for AQA A-level Biology, and describe steps in order with reasons.
Section 24 of 54Model answer: isolating nuclei from muscle tissue (6 marks)
Mark points
- homogenise the tissue to break open the cells and release the organelles [1];
- use a cold solution to prevent enzyme activity [1];
- the solution should be isotonic (equivalent water potential) to prevent organelles bursting or shrinking by osmosis [1];
- use a buffer to maintain a constant pH and stop enzymes and proteins denaturing [1];
- filter to remove large cell debris [1];
- centrifuge at low speed so that nuclei form a pellet at the bottom, then discard the supernatant [1].
A 2/6 answer put the tissue in a blender, mentioned a cold isotonic buffered solution, and then used an ultracentrifuge starting at a high speed; it missed the explanations and the filtration step, and the high speed would pellet other organelles with the nuclei.
Section 25 of 54Worked examples
Evaluate the conclusion that monoclonal antibodies prevent cell division. [4]4 marks
Supported: tumours shrank in treated mice and the placebo control suggests causation. Not supported: no data on mitosis, other mechanisms such as apoptosis or immune destruction possible, no statistics, mouse only.
Explain why the homogenate is placed in a cold, isotonic buffer. [3]3 marks
Cold reduces enzyme activity that would digest organelles; isotonic stops osmosis bursting or shrinking organelles; buffer prevents denaturation by pH changes.
State why low-speed centrifugation is used to isolate nuclei. [2]2 marks
Nuclei are the densest organelles and sediment at a low speed; lighter organelles remain in the supernatant.
Section 26 of 54Giving a clear, complete method
In questions about investigative procedures a clear, logical sequence of steps is essential. It is still possible to give an answer that is correct overall yet misses marks, because the specific details that the mark scheme looks for are not in place.
Your method should identify each variable clearly:
Section 27 of 54Naming the best piece of equipment
Always identify the equipment that is most appropriate for the measurement required, and name it precisely. Examples:
Vague answers such as 'measure the colour' or 'collect the gas' do not name the apparatus and usually lose the mark.
Section 28 of 54Aseptic technique: vague versus precise answers
A typical 3-mark question asks how to carry out a procedure safely and avoid contamination when transferring bacteria. The mark scheme credits any three of these points:
A student who wrote only that they would wash their hands, briefly flame the bottle neck, pour the culture, keep the lid off the plate for the shortest time and work near the heat 'so that the air is sterilised' scored 0/3. Each statement was too vague or was explicitly rejected:
| Student wording | Why it fails |
|---|---|
| 'wash my hands' | No reference to soap (MP1) |
| 'briefly flame the bottle neck' | The required step is remove the bottle lid and flame the neck (MP2) |
| 'pour the culture' | No pipette or syringe mentioned (MP3) |
| 'lid off the plate for the shortest time' | Does not describe lifting the lid slightly or at an angle (MP4) |
| 'heat sterilises the surrounding air' | Rejected - the flame creates upward air movement; it does not sterilise the air (MP5) |
The model answer scored 3/3: wash hands with soap and disinfect the bench, remove the lid and flame the neck of the bottle, use a sterile pipette, and lift the Petri dish lid only slightly. Because the maximum is 3, any three valid techniques are credited.
Section 29 of 54Validity and experimental controls
Students often lose marks through a lack of understanding of experimental validity and experimental controls. Validity is how well a scientific test actually measures what it sets out to measure. Features that support validity include:
Section 30 of 54Worked examples
Describe how you would transfer a bacterial culture to an agar plate while avoiding contamination. [3]3 marks
Any three of: wash hands with soap and disinfect the bench [1]; remove the lid of the culture bottle and flame the neck, replacing the lid quickly [1]; use a sterile pipette to transfer the bacteria [1]; lift the lid of the Petri dish only slightly while transferring [1].
Name the most appropriate apparatus to collect and measure the oxygen released by pondweed in a photosynthesis investigation. [1]1 mark
A gas (measuring) syringe, or a capillary tube, connected to the apparatus to collect the oxygen and measure its volume [1].
A student plans to judge the types of photosynthetic pigment in a leaf by looking at leaf colour. Explain why this lacks validity. [2]2 marks
Leaf colour is not a suitable measure of the dependent variable [1] because it is subjective and cannot separate or identify individual pigments; chromatography (with Rf values) would give a valid measurement [1].
Section 31 of 54Worked case: validity of a weight-loss study
Scientists investigated whether a weight-loss programme would result in type II diabetes remission (remission being achieved without taking medication). They:
The questionsuggest and explain three features that justify the statement that the design supports valid conclusions.
Section 32 of 54How to earn the marks
Each feature must be linked to how it contributes to validity. The credited points were:
A weak answer just repeats details from the stem (age range, splitting into two groups, measuring mass and remission) without explaining how these features give validity. Always write 'feature + because it ...'.
Section 33 of 54Required practicals: what examiners test
There are 12 required practicals, which focus on a range of different skills, and AQA exam questions assess whether you can apply them. Typical lost marks in the water potential practical (practical 3) are:
- The intercept is where there is no net change in mass, meaning the solution has the same water potential as the tissue.
Where a practical involves tissue, remember also that breaking down connections between cells may be needed so that cells or tissue behave consistently.
Section 34 of 54Worked examples
Explain how selecting volunteers using a computer-generated list improves validity. [2]2 marks
It makes selection random [1], which reduces bias, so differences between groups are due to the treatment rather than the selection [1].
In a water potential practical, explain why potato cylinders are blotted dry before they are reweighed. [2]2 marks
Surface liquid would add extra mass [1], so the mass change would not truly reflect water gained or lost by osmosis [1].
A graph of percentage change in mass against sucrose concentration crosses the x-axis at 0.30 mol dm-3. What does this value tell you? [2]2 marks
At this concentration there is no net movement of water [1], so the water potential of the sucrose solution equals the water potential of the tissue [1].
Section 35 of 54Microbiology and aseptic practicals
Students commonly lose marks for:
Section 36 of 54Photosynthetic pigments by chromatography
Chromatography separates biological compounds. Marks are lost for:
Section 37 of 54Taxis, kinesis and glucose in urine
Animal behaviour practicals
- students confuse taxis (directional movement towards or away from a stimulus) with kinesis (a change in the rate of movement or turning that is not directional), and do not explain how movement is measured.
- For ethical handling of organisms, avoid generic statements;
- refer to abiotic factors that can be controlled in the lab, such as pH and temperature, and give context.
Determining glucose concentration in urine uses a colorimeter and a calibration curve. Typical errors:
Section 38 of 54Dealing with data: read the question precisely
Many students struggle with data-heavy questions through misreading the question, mishandling graphs or calculations, and weak description. Data questions contain precise instructions, and marks are lost when students answer the question they expect rather than the one set.
Worked case 'Explain the rate of transpiration between 5 am and midday.' The marking points were:
References to tide are ignored or rejected - tide does not affect transpiration or water potential.
A weak answer was purely descriptive (rate increases then decreases), spent time on what happens after midday (outside the time window), discussed tide height and made no attempt to explain. A full-mark answer stayed in the 5 am - midday window and linked temperature, kinetic energy, diffusion gradients, light and stomatal opening to the rate.
Section 39 of 54Worked examples
Explain why the origin line in chromatography is drawn in pencil. [1]1 mark
Pencil is insoluble in the solvent and so does not move up the paper or interfere with the pigments, unlike ink [1].
Explain why a calibration curve is used in the glucose-in-urine practical. [2]2 marks
Absorbance of known glucose concentrations (made by a dilution series) is plotted [1]; the absorbance of the urine sample is read from the curve to find its concentration [1].
Explain the rate of transpiration between 5 am and midday. [4]4 marks
Rate increases due to higher temperature or light intensity, lower humidity or more air movement [1]; greater water loss because kinetic energy and the diffusion gradient increase [1]; stomata open after 5 am allowing gas exchange [1]; some stomata close around 11 am, reducing transpiration [1].
Section 40 of 54Technical slips in data questions
Data questions often test your ability to read scales, handle numbers and interpret statistical markers.
Section 41 of 54Worked case: bacteriophage treatment in mice
Mice in group A were not treated with bacteriophage.
(i) With some samples a series of dilutions was needed before counting.
- so they had to be diluted to give a number that could be counted,
- while samples with low numbers did not need diluting.
(ii) State what can be concluded about the effectiveness of the bacteriophage:
- so it does not work equally well in all mice; or the bacteria are not reduced to zero in any mice [1].
Ignore references to significance; reject references to standard deviation or standard error.
Section 42 of 54Why the weak answer scored 1/3
The weak answer said bacteriophage reduces the number of live bacteria (credited) but then claimed 'the results are significant because the standard deviation bars do not overlap'. This is wrong for several reasons:
The 3/3 answer stated the reduction, noted that the ranges do not overlap so treated mice always had fewer bacteria than any untreated mouse, and commented that the large range for the treated group shows effectiveness varies greatly between mice.
Section 43 of 54Describing data well
Even when a question asks only for a description, students lose marks through vague or unsupported summaries. A good description should:
Section 44 of 54Worked examples
A graph shows bacteria on a log10 scale. A bar reaches 8. How many bacteria does this represent? [1]1 mark
108 (100 000 000), not 8 [1].
Use Figure 2 to explain why dilution was needed for some samples but not others. [2]2 marks
Untreated samples contained very large numbers of bacteria (quote the standard-form value) which could not be counted directly, so were diluted [1]; treated samples had low numbers (quote the value) which could be counted without dilution [1].
State what can be concluded about the effectiveness of the bacteriophage. Do not consider statistical analyses. [3]3 marks
Bacteriophage reduces the number of bacteria [1]; the ranges do not overlap, so treatment worked in all mice [1]; wide variation in response, and bacteria were not reduced to zero in any mouse [1].
Section 45 of 54Describing a scatter graph
Some strains of E. coli have developed resistance to the antibiotic Colomycin, which binds to a surface protein called RCP1. If enough Colomycin binds to RCP1 on a cell, the cell is killed. Researchers studied the relationship between the amount of RCP1 on the cell surface and the percentage of cells killed (Figure 3). 'Describe the results' is worth 3 marks:
Approximate values read from the graph are accepted. The 1/3 answer only said that as RCP1 increases more cells are killed and there is a positive correlation: no numbers and no mention that some points do not fit. The 3/3 answer gave the weak positive correlation, quoted two values, then identified the anomaly and quoted values for it.
Section 46 of 54The list rule
Understanding how the examiner marks helps you avoid losing marks. The list rule applies to any question requiring a set number of responses, e.g. 'give three reasons why...'. Apply it like this:
Resist adding more answers 'just in case' - you risk being penalised.
- and DNA has hydrogen bonds and tRNA does not.
Section 47 of 54Calculation questions
Calculation questions are often worth several marks, so always show your working. Scientists investigated how axon diameter and myelination affect conduction speed in neurones. At 20 micrometres the mean speeds were 10.5 (myelinated) and 8.5 (unmyelinated) in the figure's units. The mark scheme for the percentage calculation was:
Error carried forward (ECF) marks are awarded for answers that come from slightly different readings of the graph, and the correct answer without working gains full marks. Note that strictly the myelinated speed is 123.5% of the unmyelinated speed, a rise of 23.5%; read the exact wording of the question and the scheme.
Section 48 of 54Worked examples
Describe the results shown in a scatter graph of RCP1 level against percentage of cells killed. [3]3 marks
Weak positive correlation: higher RCP1 means a higher percentage killed [1]; quote values, e.g. 9% killed at RCP1 1.8 and 60% at 15.2 [1]; some points do not fit, e.g. 30% killed at 4.2 but 25% at 6.6 [1].
Give three differences between DNA and tRNA. [3]3 marks
Write exactly three correct differences: DNA is double-stranded, tRNA single-stranded [1]; DNA contains deoxyribose, tRNA ribose [1]; tRNA folds into a three-lobed shape, DNA is a double helix [1]. Do not add a fourth, because an incorrect extra answer cancels a mark.
Calculate the conduction speed of the myelinated axon as a percentage of the unmyelinated axon, given values of 10.5 and 8.5. [2]2 marks
(10.5 ÷ 8.5) × 100 [1] = 123.5% [1]. A speed increase of 23.5%.
Section 49 of 54Follow the exact instruction
Sometimes students lose marks for failing to follow the exact instructions in a question. Examples:
Extra content that falls outside the scope is not creditworthy and wastes time.
Section 50 of 54Worked case: protein synthesis scope
- A 3/3 answer focused entirely on translation: the mRNA leaves the nucleus and binds to a ribosome [1];
- tRNA molecules with anticodons bind to the mRNA codons [1];
- the ribosome catalyses the formation of a peptide bond between the amino acids held by the tRNA [1];
- finally the ribosome moves along to the next codon to continue the chain.
- It was long, so the student thought it was enough, even though most of the content was not creditworthy.
Section 51 of 54Small errors that cost marks
Even in simple questions, students lose marks through small errors. Check this table and avoid them:
| Mistake | Explanation | Correction |
|---|---|---|
| Microvilli are described as 'small folds on the villi' or confused with villi | Microvilli are small folds of the cell-surface membrane | Microvilli are present on the epithelial cells, which are on the villi |
| 'The alveolus has a thin cell wall' | Only plants, algae, fungi and prokaryotes have cell walls; animal cells may group together to form tissue walls, such as those of the alveoli or blood vessels, but they do not have cell walls | The alveolar epithelium is a single layer of thin cells |
| 'Hydrogen bonds between bases are hydrolysed' | Hydrolysis is breaking a covalent bond using water, e.g. in ATP or polypeptides | Hydrogen bonds between bases are broken |
Section 52 of 54Worked examples
Describe the role of the ribosome in translation. Do not describe transcription. [3]3 marks
mRNA binds to the ribosome [1]; tRNA anticodons bind to mRNA codons [1]; the ribosome catalyses a peptide bond between amino acids [1]; then moves to the next codon.
A student writes 'the alveolus has a thin cell wall'. Correct this. [1]1 mark
Animal cells have no cell walls; the alveolar epithelium is a single layer of thin cells [1].
A student states 'hydrogen bonds between bases are hydrolysed'. Explain the error. [2]2 marks
Hydrolysis breaks covalent bonds using water [1]; hydrogen bonds are simply broken (not hydrolysed) [1].
Section 53 of 54Key terms
| Term | Meaning |
|---|---|
| Command word | The word in a question that tells you what to do. |
| Competitive inhibitor | Molecule that competes with substrate for the active site. |
| Non-competitive inhibitor | Molecule binding elsewhere on the enzyme and changing the active site. |
| Vmax | Maximum rate of an enzyme-catalysed reaction. |
| Term | Meaning |
|---|---|
| Species richness | Number of different species in an area. |
| Standard deviation | Measure of spread of data around the mean. |
| Replication | Repeating an investigation to check results. |
| Sample size | Number of samples used. |
| Term | Meaning |
|---|---|
| Phagocytosis | Engulfing and digesting a particle or pathogen by a cell. |
| Phagosome | Vesicle containing an engulfed pathogen. |
| Lysosome | Organelle containing hydrolytic enzymes. |
| Lysozyme | Hydrolytic enzyme that digests bacterial cell walls. |
| Histone | Protein around which eukaryotic DNA is wound. |
| Term | Meaning |
|---|---|
| Hydrolysis | Breaking a bond by adding water. |
| Glycosidic bond | Bond between monosaccharides in carbohydrates. |
| Ambiguity | A statement that can be interpreted in more than one way. |
| Peptide bond | Bond between amino acids in a polypeptide. |
| Term | Meaning |
|---|---|
| Novel context | An unfamiliar situation used to test application of knowledge. |
| Myoglobin | Oxygen-binding protein that stores oxygen in muscle. |
| Heat shock protein | Protein that helps other proteins fold correctly and resist heat damage. |
| Validity | How well an investigation tests the intended hypothesis. |
| Term | Meaning |
|---|---|
| Placebo | An inactive treatment given to a control group. |
| Homogenise | Break up tissue to release the organelles. |
| Isotonic | Having the same water potential as the cells. |
| Supernatant | The liquid above the pellet after centrifugation. |
| Monoclonal antibody | Identical antibodies from one clone of cells. |
| Term | Meaning |
|---|---|
| Independent variable | The factor deliberately changed by the investigator. |
| Dependent variable | The factor that is measured as the response to the change. |
| Control variable | A factor kept constant so it cannot affect the results. |
| Colorimeter | An instrument that measures how much light a solution absorbs, giving a quantitative measure of colour. |
| Aseptic technique | Procedures that prevent contamination of cultures and the environment by unwanted microorganisms. |
| Validity | How well a test measures what it sets out to measure. |
| Term | Meaning |
|---|---|
| Random selection | Choosing participants by chance so that every individual has the same likelihood of being chosen. |
| Bias | A systematic distortion of results caused by how a study is designed or carried out. |
| Control group | A group that does not receive the treatment, used as a comparison. |
| Representative sample | A sample whose characteristics reflect those of the whole population. |
| Water potential | The tendency of water to move from one place to another; pure water has the highest value. |
| Dilution series | A set of solutions made by repeatedly diluting a stock solution by a known factor. |
| Term | Meaning |
|---|---|
| Taxis | A directional movement of an organism towards or away from a stimulus. |
| Kinesis | A change in rate of movement or turning in response to a stimulus, without direction. |
| Rf value | Distance moved by a compound divided by the distance moved by the solvent front. |
| Calibration curve | A graph of a measured property (e.g. absorbance) against known concentrations, used to find unknown values. |
| Serial dilution | Repeated dilution of a sample by the same factor. |
| Abiotic factor | A non-living factor of the environment, such as temperature or pH. |
| Term | Meaning |
|---|---|
| Log10 scale | A scale where each unit increase represents a ten-fold increase in the value. |
| Range | The difference between, or the span from, the lowest to the highest value in a data set. |
| Mean | The sum of all values divided by the number of values. |
| Standard form | A way of writing numbers as a value between 1 and 10 multiplied by a power of ten. |
| Anomaly | A data point that does not fit the general pattern. |
| Bacteriophage | A virus that infects and destroys bacteria. |
| Term | Meaning |
|---|---|
| Positive correlation | A relationship in which one variable tends to increase as the other increases. |
| Anomaly | A result that does not fit the pattern of the rest of the data. |
| List rule | Marking rule: only the first set number of answers are credited, and incorrect extra answers can cancel marks. |
| ECF (error carried forward) | Marks awarded for correct method applied to an incorrect earlier value. |
| tRNA | Transfer RNA, a single-stranded folded molecule carrying an amino acid and an anticodon. |
| Myelin sheath | A fatty layer around some axons that speeds up conduction. |
| Term | Meaning |
|---|---|
| Hydrolysis | Breaking of a covalent bond by the addition of water. |
| Microvilli | Small folds of the cell-surface membrane that increase surface area. |
| Villi | Finger-like folds of the intestinal lining, covered by epithelial cells. |
| Translation | The stage of protein synthesis where a ribosome assembles amino acids according to the mRNA codons. |
| Epithelial cell | A cell forming a covering or lining layer of tissue. |
| Cell wall | A rigid layer outside the membrane of plants, algae, fungi and prokaryotes (not animal cells). |
Section 54 of 54Exam technique
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