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Practice Materials

Sciences Resources

Biology resources written by Dr Krishan P. Chauhan — PhD in Metabolic Diseases & Medical Statistics. Chemistry and Physics resources aligned to GCSE specifications for all major exam boards.

GCSE Biology

AQA, Edexcel, OCR · Foundation & Higher · Click any question to see the solution

Cell Biology

2 marks
Give two differences between a plant cell and an animal cell.
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Worked Solution

1Plant cells have a cell wall (made of cellulose) — animal cells do not
2Plant cells may contain a permanent vacuole — animal cells do not (also accept: chloroplasts)
Examiner tip: Many students say ‘plant cells are bigger’ — this is not a reliable structural difference. Stick to cell wall, vacuole, and chloroplasts.
3 marks
Describe the process of osmosis.
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Worked Solution

1Osmosis is the movement of water molecules
2from a region of higher water potential (more dilute solution)
3to a region of lower water potential (more concentrated solution)
4through a partially permeable membrane
Examiner tip: Must include: water molecules, movement from high to low water potential, partially permeable membrane. Missing any of these loses marks.
4 marks
Describe and explain what happens when a plant cell is placed in a very concentrated salt solution.
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Worked Solution

1Water moves out of the cell by osmosis (from high to low water potential)
2The vacuole shrinks and the cytoplasm pulls away from the cell wall
3This is called plasmolysis — the cell becomes plasmolysed
4The cell becomes flaccid (not turgid) but the cell wall prevents it from bursting
Examiner tip: Distinguish between plasmolysis (plant cell) and crenation (animal cell). In plants, the cell wall prevents bursting even in extreme conditions.

Mitosis & Meiosis

3 marks
State three differences between mitosis and meiosis.
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Worked Solution

1Mitosis produces 2 daughter cells; meiosis produces 4 daughter cells
2Mitosis produces genetically identical cells; meiosis produces cells with genetic variation
3Mitosis is used for growth and repair; meiosis is used for production of gametes
Examiner tip: This is a very common question. Learn these three differences precisely. The examiner is looking for direct contrasts — say what happens in both processes.
5 marks
Explain how meiosis leads to genetic variation.
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Worked Solution

1During meiosis I, crossing over occurs between homologous chromosomes
2This exchanges segments of DNA between chromatids, creating new combinations of alleles
3Independent assortment means homologous chromosome pairs line up randomly at the cell equator
4This results in different combinations of maternal and paternal chromosomes in the gametes
5Combined with random fertilisation, this generates enormous genetic variation in offspring
Examiner tip: Mention both crossing over AND independent assortment for full marks. Many students know one but not both.

Digestion & Nutrition

4 marks
Describe how the structure of the small intestine is adapted for efficient absorption.
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Worked Solution

1Villi increase the surface area for absorption
2Microvilli on epithelial cells further increase surface area
3Rich blood supply (capillary network) maintains a concentration gradient for diffusion
4Thin walls (single layer of epithelium) reduce the diffusion distance
5Lacteals absorb fatty acids and glycerol (lipid absorption)
Examiner tip: This is a classic 4-5 mark ‘structure and function’ question. Each adaptation must be linked to the function it performs.

Genetics & Inheritance

5 marks
Cystic fibrosis is caused by a recessive allele (f). Two carrier parents have a child. Draw a genetic cross to find the probability of the child having cystic fibrosis.
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Worked Solution

1Both parents are carriers: genotype Ff × Ff
2Draw Punnett square: FF, Ff, Ff, ff
3Offspring: 1 FF (unaffected, not carrier), 2 Ff (carriers), 1 ff (affected)
4Probability of cystic fibrosis (ff) = 1/4 = 25%
5Probability of being a carrier (Ff) = 2/4 = 50%
Examiner tip: Draw the Punnett square in the exam even if you can do it in your head. The cross diagram earns a mark.

A-Level Biology

Year 1 & Year 2 · AQA, Edexcel, OCR · Click any question to see the solution

Dr Chauhan’s A-Level Biology advice: “A-Level Biology is overwhelmingly a subject of precise language. Examiners mark against specific words in the mark scheme. Vague answers that convey the right idea but use the wrong vocabulary will not earn marks. Learn the terminology — not as labels, but as concepts you genuinely understand.”

DNA, RNA & Protein Synthesis

4 marks
Describe the process of transcription.
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Worked Solution

1DNA double helix unwinds and unzips at the gene to be transcribed
2One strand (template strand) acts as the template
3RNA polymerase moves along the template strand (3′ to 5′ direction)
4Free RNA nucleotides bind to complementary bases on the template strand
5A strand of pre-mRNA is formed, which is processed to mRNA before leaving the nucleus
Examiner tip: The key enzyme is RNA polymerase. Always mention the direction of reading the template strand (3′ to 5′) for the highest-mark answers.
5 marks
Explain why a mutation in a gene does not always result in a change in the amino acid sequence of the protein.
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Worked Solution

1The genetic code is degenerate (redundant) — most amino acids are coded for by more than one codon
2A mutation may change the base sequence of the codon but produce a synonymous codon that codes for the same amino acid
3Some mutations occur in introns, which are removed during RNA processing and do not appear in mature mRNA
4Some mutations occur in non-coding regions of DNA that are not transcribed
5If the mutation is in a codon that is not translated (e.g., stop codon region may not be affected)
Examiner tip: The most important concept here is degeneracy of the genetic code. This is worth 2 marks on its own — make sure you explain it clearly.

Respiration

6 marks
Compare aerobic and anaerobic respiration in mammals.
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Worked Solution

1Aerobic requires oxygen; anaerobic does not
2Aerobic produces 38 ATP per glucose; anaerobic produces only 2 ATP per glucose
3Aerobic produces CO⊂2; and water; anaerobic in mammals produces lactate
4Aerobic occurs in the cytoplasm and mitochondria; anaerobic occurs only in the cytoplasm
5Aerobic involves the Krebs cycle and oxidative phosphorylation; anaerobic involves only glycolysis + lactate fermentation
6Aerobic uses NAD, FAD as electron carriers; anaerobic regenerates NAD without electron transport chain
Examiner tip: Six marks means six clear contrasting points. Structure your answer as a direct comparison for each point.

Ecology

4 marks
Explain the difference between a food chain and a food web, and explain why food webs are more useful models of ecosystems.
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Worked Solution

1A food chain shows a single linear sequence of feeding relationships
2A food web shows the interconnected network of all food chains in an ecosystem
3Food webs are more realistic because most organisms feed on more than one species
4They show how the loss or change in one species has knock-on effects through multiple pathways
5They demonstrate the complexity and interdependency of ecosystems more accurately
Examiner tip: Food webs show resilience and interdependency — these are the key ideas that the examiner wants to see addressed.

Degree & Postgraduate Biology

Undergraduate & Masters Level · Written by a PhD Researcher in Metabolic Diseases

Note on university biology: At degree level, biology becomes far more quantitative. Statistical analysis, experimental design, and data interpretation are as important as content knowledge. These resources focus on the areas where undergraduate students most commonly struggle.

Metabolic Diseases — Key Concepts

Type 2 Diabetes Mellitus — Pathophysiology

Type 2 DM is characterised by insulin resistance at peripheral tissues (skeletal muscle, adipose tissue, liver) combined with progressive β-cell dysfunction. In the early stages, the pancreatic β-cells compensate by producing more insulin (hyperinsulinaemia). Over time, β-cell mass and function decline, leading to relative insulin deficiency.

Key mechanisms of insulin resistance include: impaired insulin receptor signalling (reduced IRS-1 phosphorylation), ectopic lipid deposition in muscle and liver, chronic low-grade inflammation (elevated TNF-α, IL-6), and mitochondrial dysfunction in skeletal muscle.

Clinical significance: Understanding the distinction between insulin resistance (a tissue-level problem) and insulin deficiency (a pancreatic problem) is fundamental to understanding pharmacological treatments. Metformin primarily targets hepatic insulin resistance; GLP-1 agonists target β-cell function and appetite regulation.

Research Methods & Statistics

Concept check
What is the difference between a t-test and ANOVA? When would you use each?
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Worked Solution

1A t-test compares the means of two groups. It can be independent (between subjects) or paired (within subjects).
2ANOVA (Analysis of Variance) compares the means of three or more groups simultaneously.
3Using multiple t-tests instead of ANOVA inflates the Type I error rate (false positive rate) — this is known as the multiple comparisons problem.
4After a significant ANOVA result, post-hoc tests (Tukey, Bonferroni) identify which specific groups differ.
5Choose ANOVA when you have three or more groups; use a t-test for exactly two groups.
Examiner tip: This distinction comes up repeatedly in dissertation methods chapters. Examiners and supervisors will expect you to justify your choice of statistical test.
Concept check
Explain the difference between Type I and Type II errors in hypothesis testing.
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Worked Solution

1A Type I error (false positive) occurs when you reject a null hypothesis that is actually true. The probability of this is α (significance level, usually 0.05).
2A Type II error (false negative) occurs when you fail to reject a null hypothesis that is actually false. The probability is β.
3Power = 1 − β = the probability of detecting an effect when one truly exists.
4Reducing α (being more strict) decreases Type I errors but increases Type II errors.
5Increasing sample size reduces both error types by increasing statistical power.
Examiner tip: Exam questions at degree level often ask you to discuss what factors influence the reliability of a conclusion. Always link Type I/II errors to the practical implications for the study.

GCSE Chemistry

AQA, Edexcel, OCR · Foundation & Higher · Click any question to see the solution

3 marks
A student reacts 4.0 g of calcium carbonate (CaCO⊂3;) with excess hydrochloric acid. Calculate the mass of carbon dioxide produced. (Relative atomic masses: Ca=40, C=12, O=16)
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Worked Solution

1Write the equation: CaCO⊂3; + 2HCl → CaCl⊂2; + H⊂2;O + CO⊂2;
2Molar mass of CaCO⊂3; = 40 + 12 + 48 = 100 g/mol
3Moles of CaCO⊂3; = 4.0/100 = 0.04 mol
41:1 ratio, so moles of CO⊂2; = 0.04 mol
5Molar mass of CO⊂2; = 12 + 32 = 44 g/mol
6Mass of CO⊂2; = 0.04 × 44 = 1.76 g
Examiner tip: Show every step: equation, molar masses, moles, ratio, answer. An error in one step does not cost you all the marks if method is shown.
4 marks
Explain why ionic compounds have high melting points.
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Worked Solution

1Ionic compounds have a giant ionic lattice structure
2The lattice consists of oppositely charged ions held together by strong electrostatic forces of attraction
3A large amount of energy is required to overcome these forces
4Therefore the melting point is high
Examiner tip: Must mention: giant lattice, electrostatic forces, and that energy is needed to overcome them. ‘Strong bonds’ alone is insufficient — specify electrostatic forces.
3 marks
Describe the structure of an atom.
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Worked Solution

1The atom has a central nucleus containing protons (positive charge) and neutrons (no charge)
2Electrons (negative charge) orbit the nucleus in energy levels (shells)
3The atom is mostly empty space — the nucleus is tiny compared to the overall size of the atom
Examiner tip: The relative mass and charge of each particle is a required fact: proton (1, +1), neutron (1, 0), electron (negligible, −1).

GCSE Physics

AQA, Edexcel, OCR · Foundation & Higher · Click any question to see the solution

3 marks
A car of mass 1200 kg accelerates from 0 to 20 m/s in 8 seconds. Calculate the force required.
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Worked Solution

1Find acceleration: a = (v−u)/t = (20−0)/8 = 2.5 m/s²
2Use F = ma: F = 1200 × 2.5
3Force = 3000 N
Examiner tip: Always state the formula, substitute the values, then calculate. This is the structure the mark scheme rewards.
4 marks
A 2 kg object is dropped from a height of 10 m. Calculate its velocity just before it hits the ground. (g = 10 m/s²)
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Worked Solution

1The object falls from rest (u = 0), so all GPE converts to KE
2Method 1 (energy): mgh = ½mv² ⇒ v² = 2gh = 2×10×10 = 200 ⇒ v = √200 ≈ 14.1 m/s
3Method 2 (SUVAT): v² = u² + 2as = 0 + 2(10)(10) = 200 ⇒ v = √200 ≈ 14.1 m/s
4Note: the mass cancels — all objects fall at the same rate in the absence of air resistance
Examiner tip: Both methods give the same answer. In GCSE, either approach earns full marks. In A-Level Physics, the energy method is often more elegant.
5 marks
Explain how a transformer works, including why they only work with alternating current.
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Worked Solution

1A transformer consists of two coils of wire (primary and secondary) wound around an iron core
2An alternating current in the primary coil creates a changing magnetic field
3The iron core concentrates and transfers this field to the secondary coil
4The changing magnetic field in the secondary coil induces an alternating EMF (by electromagnetic induction)
5A direct current would produce a constant magnetic field, which would NOT induce any EMF in the secondary coil — hence transformers only work with AC
Examiner tip: The phrase ‘changing magnetic field’ is essential for the induced EMF marks. A constant field induces nothing.

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