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biology question and answer, Exercises of Biology

Biology

biology question and answers full

Typology: Exercises

2025/2026

Uploaded on 04/15/2026

prasanna-lakshmi-3 🇮🇳

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2025 Biology Revision

Q1.1 — The digestion and absorption of lipids (4 marks)

The importance of emulsification in the digestion of lipids:

• Bile salts break up large lipid droplets into smaller droplets

(emulsification).

• This increases the surface area of lipids.

• Lipase enzymes can act more efficiently on a larger surface area.

• Therefore, lipids are hydrolysed faster into monoglycerides and fatty

acids.

The role of micelles in the absorption of lipids:

• Micelles transport monoglycerides and fatty acids to the epithelial

surface of the ileum.

• Monoglycerides and fatty acids diffuse from micelles into epithelial cells.

• Inside the cells, they are re-formed into triglycerides.

• Triglycerides are packaged into chylomicrons for transport in the lymph.

Q1.2 — Arrangement of phospholipids in the cell-surface

membrane (2 marks)

• Phospholipids form a bilayer with hydrophilic heads facing outwards

towards the water (cytoplasm and extracellular fluid).

• Hydrophobic fatty-acid tails face inwards, away from water.

• This arrangement creates a partially permeable barrier allowing

movement of small, non-polar molecules while preventing entry of polar

or charged molecules.

Q2.1 — Describe how mRNA is obtained from pre-mRNA (1 mark)

• Introns (non-coding sequences) are removed, and exons are joined

together by a process called splicing.

Q2.2 — Describe how RNA polymerase joins RNA nucleotides (2

marks)

• RNA polymerase joins RNA nucleotides together by condensation

reactions.

• Phosphodiester bonds form between the phosphate group of one

nucleotide and the ribose sugar of the next.

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2025 Biology Revision Q1.1 — The digestion and absorption of lipids (4 marks) The importance of emulsification in the digestion of lipids:

Bile salts break up large lipid droplets into smaller droplets (emulsification).
This increases the surface area of lipids.
Lipase enzymes can act more efficiently on a larger surface area.
Therefore, lipids are hydrolysed faster into monoglycerides and fatty acids. The role of micelles in the absorption of lipids:
Micelles transport monoglycerides and fatty acids to the epithelial surface of the ileum.
Monoglycerides and fatty acids diffuse from micelles into epithelial cells.
Inside the cells, they are re-formed into triglycerides.
Triglycerides are packaged into chylomicrons for transport in the lymph. Q1.2 — Arrangement of phospholipids in the cell-surface membrane (2 marks)
Phospholipids form a bilayer with hydrophilic heads facing outwards towards the water (cytoplasm and extracellular fluid).
Hydrophobic fatty-acid tails face inwards, away from water.
This arrangement creates a partially permeable barrier allowing movement of small, non-polar molecules while preventing entry of polar or charged molecules. Q2.1 — Describe how mRNA is obtained from pre-mRNA (1 mark)
Introns (non-coding sequences) are removed, and exons are joined together by a process called splicing. Q2.2 — Describe how RNA polymerase joins RNA nucleotides ( marks)
RNA polymerase joins RNA nucleotides together by condensation reactions.
Phosphodiester bonds form between the phosphate group of one nucleotide and the ribose sugar of the next.

Q2.3 — Figure shows the energy in molecules during an enzyme- controlled reaction. Which expression represents the activation energy (Eₐ) of this reaction? (1 mark)

A – C Q2.4 — The 1914 model suggested that enzymes speed up reactions without combining with substrates. The induced-fit model was later proposed. Explain how the induced-fit model differs from the 1914 model. (4 marks) Similarities:
Both models state that enzymes act as biological catalysts and speed up reactions.
Enzymes are not used up or permanently changed in the reaction. Differences:
In the induced-fit model, the enzyme’s active site changes shape when the substrate binds.
The active site becomes complementary to the substrate only after binding, forming an enzyme–substrate complex.
This stresses the bonds in the substrate, lowering the activation energy and allowing the reaction to proceed. Q3.1 — The table shows information about glycogen, sucrose and DNA. Complete the table. (3 marks) Feature Glycog en Sucros e

DN

A

Contains glucose ✓ ✓ ✗ Contains galactose ✗ ✗ ✗ Contains fructose ✗ ✓ ✗ Contains phosphate ✗ ✗ ✓ Formed by condensation reaction(s)

Is a polymer ✓ ✗ ✓

Q4.1 — Some, but not all, prokaryotic cells have certain structures. Name three such structures. (2 marks) Any three from:

Plasmids
Capsule / slime layer
Flagellum / flagella
Pili / fimbriae
Photosynthetic membranes
Storage granules / lipid droplets Q4.2 — Two investigations used microscopes to observe cells. For each, name the most suitable microscope and give a reason. ( marks) Investigation 1 – Living yeast cells stained to show dead cells:
Light (optical) microscope.
Can be used to observe living cells.
Has sufficient resolution (≈0.25 μm) to view whole yeast cells (≈3 μm).
Stain allows dead cells to be identified. Investigation 2 – 3D structure of mitochondria:
Scanning electron microscope (SEM).
Produces a 3D image of the cell surface.
Has a higher resolution (≈0.02 μm) than a light microscope.
Mitochondria too small to be seen in detail with a light microscope. Q5.1 — Pear cells increase in vacuole size as the fruit ripens. Explain the roles of ATP hydrolase and carrier proteins X and Y in this process. (4 marks) Role of ATP hydrolase:
Hydrolyses ATP to release energy for active transport.
Uses this energy to pump H⁺ ions into the vacuole via carrier protein X.
This creates a proton gradient across the vacuole membrane and lowers the pH inside the vacuole. Role of carrier proteins X and Y:
Carrier X actively transports H⁺ ions into the vacuole using ATP energy.

Carrier Y uses the H⁺ concentration gradient to co-transport fructose into the vacuole.
The build-up of solutes (H⁺ and fructose) lowers the water potential inside the vacuole.
Water enters by osmosis, causing the vacuole to expand and increasing cell size. Q5.2 — The effect of a respiratory inhibitor on vacuole pH. ( marks) Mark-scheme answer:
The respiratory inhibitor reduces ATP production by preventing oxidative phosphorylation.
Less ATP is available for the ATP hydrolase pump to actively transport H⁺ ions into the vacuole.
As fewer H⁺ ions are pumped in, the proton gradient collapses, so the vacuolar pH increases (becomes less acidic). Q5.3 — The relationship between expression of carrier proteins and the rate of ripening. (2 marks) Mark-scheme answer:
Increased gene expression leads to the production of more carrier proteins X and Y in the vacuole membrane.
This increases H⁺ pumping and fructose co-transport, causing faster solute accumulation and water uptake.
As a result, vacuoles expand more quickly, leading to a faster rate of fruit ripening. Q6.1 — Figure shows oxygen dissociation curves for haemocyanin at 4°C and 15°C. Explain how temperature affects the loading and unloading of oxygen. (2 marks) Mark-scheme answer:
At a higher temperature (15°C), haemocyanin has a lower affinity for oxygen.
Therefore, it loads less oxygen at the gills (high pO₂).
But it unloads more oxygen to the tissues (low pO₂).
This ensures greater oxygen delivery to respiring tissues when temperature is higher.

Q7.1 — Calculate the minimum number of A₁ epithelial cells needed to produce 1.5 × 10⁷ sperm cells. (2 marks)

Each A₁ cell → 6 mitoses → (^26) = 64 primary spermatocytes (PS).
Each PS → meiosis → 4 sperm ⇒ 256 sperm per A₁.
Required A₁ =

1.5 × 10

7 256

=5.86 × 10

4 . ✅ Answer: 5.86 × 10⁴ cells (accept 5.9 × 10⁴ or 58,600). Q7.2 — State the DNA mass per cell at each stage shown. ( marks)

Start of interphase: 8 (a.u.)
End of interphase (after replication): 16 (a.u.)
End of first meiotic division: 8 (a.u.)
End of second meiotic division (sperm): 4 (a.u.)

(Award 3 marks for the correct pattern across stages.)

Q7.3 — Draw/describe the chromosomes at the end of meiosis I. (1 mark)

Homologous chromosomes separated; each chromosome still duplicated (two sister chromatids, X-shaped) in one cell. Q7.4 — Name and explain the mutation that produces a gamete with one extra chromosome. (3 marks)
Non-disjunction.
Failure of homologous chromosomes (meiosis I) or sister chromatids (meiosis II) to separate.
Results in a gamete with n + 1 (and its pair with n − 1). Q8.1 — Explain why antibiotic-resistant bacterial strains are common in hospitals. (3 marks)
High use of antibiotics in hospitals creates a strong selection pressure.
Non-resistant bacteria are killed, but resistant bacteria survive and reproduce.
The resistance allele (often on a plasmid) is passed on to offspring and may spread between species by horizontal gene transfer.

Q8.2 — Calculate the Spearman rank correlation coefficient (rₛ) for the data in Table 5. (2 marks) Step 1 – Data extraction: From Table 5 → squared differences (d²): 9, 4, 0, 9, 16

Σ d

2 = 9 + 4 + 0 + 9 + 16 = 38 Step 2 – Substitution into formula:

r s = 1 −

6 Σ d

2

n ( n

2 − 1 ) r (^) s = 1 −

6 × 38

2 − 1 )

Step 3 – Final answer: ✅ rₛ = – 0. Step 4 – Interpretation (if required):

There is a strong negative correlation between plasmid length and mean plasmid copy number — as plasmid length increases, mean copy number decreases. Q8.3 — The rs value between plasmid length and mean plasmid copy number is –0.94. What can you conclude from this? (1 mark) Mark-scheme answer:
There is a strong negative correlation between plasmid length and mean plasmid copy number.
This means that as plasmid length increases, the mean copy number decreases.

✅ Only 1 mark available → must state both direction (negative)

and strength (strong).

Q8.4 — Calculate the total number of kbp in all plasmid molecules in a B. thuringiensis cell. (2 marks) Given: Length of single circular DNA = 5400 kbp Total plasmid DNA = 65% greater than this Working:

65% of 5400 = 5400 × 0.65= 3510

Total plasmid DNA = 5400 + 3510 = 8910 ✅ Answer: 8910 kbp

Mark-scheme answer:

The mutant chloroplasts have fewer thylakoids / grana → smaller surface area for chlorophyll and light-dependent reactions.
Therefore, they will absorb less light and produce less ATP and reduced NADP for the light-independent stage.
As a result, the overall rate of photosynthesis will be lower in mutant plants. Q9.4 — Use Table 6 to describe and explain the effect of the mutation on photosynthesis. (3 marks) Mark-scheme answer:
The mutant plants have a much lower total chlorophyll concentration (9 μg cm⁻³ compared to 45 μg cm⁻³).
The chlorophyll a : chlorophyll b ratio is higher (6 : 1) in mutants, meaning less chlorophyll b is present to absorb additional wavelengths of light.
Despite less chlorophyll, the rate of photosynthesis is higher (5×), suggesting the mutation increases photosynthetic efficiency — possibly by altering chlorophyll structure or enhancing light absorption/energy transfer by chlorophyll a.
Therefore, the mutation affects chlorophyll composition rather than simply reducing photosynthetic capacity. Q10.1 — Describe the roles of ribosomes, tRNA and ATP in the process of translation. (6 marks) Mark-scheme answer: Role of ribosomes:
Ribosomes bind to mRNA at the start codon (AUG).
They move along the mRNA, reading the codons in sequence.
The ribosome provides the site for tRNA anticodon–codon pairing.
It catalyses the formation of peptide bonds between adjacent amino acids. Role of tRNA:
Each tRNA molecule carries a specific amino acid corresponding to its anticodon.
The anticodon on tRNA binds to a complementary codon on the mRNA via base pairing.
tRNA brings amino acids together in the correct order to form a

polypeptide chain.

The tRNA is released after the amino acid has been added to the growing polypeptide. Role of ATP:
ATP provides energy to attach amino acids to their specific tRNA molecules (aminoacyl-tRNA formation).
ATP hydrolysis also supplies energy for the formation of peptide bonds during translation. Q10.2 — Describe and explain the adaptations of leaves that maximise gas exchange. (4 marks) Mark-scheme answer:
Thin leaf — short diffusion distance for gases (CO₂ and O₂).
Many stomata — large surface area for gas exchange; gases can enter and leave rapidly.
Air spaces in the spongy mesophyll — allow rapid diffusion of gases between stomata and photosynthesising cells.
Large surface area of mesophyll cells — maximises the area for gas exchange with air spaces.
Flattened shape of the leaf — large surface area to volume ratio; ensures all cells are close to the surface for efficient diffusion. Q10.3 — Describe the pathway taken by blood between a kidney and the coronary arteries. (5 marks) Mark-scheme answer:
Blood leaves the kidney through the renal vein.
The renal vein joins the vena cava, which carries blood to the right atrium of the heart.
Blood passes through the right atrioventricular (tricuspid) valve into the right ventricle.
It leaves the right ventricle through the pulmonary valve into the pulmonary artery and travels to the lungs.
In the lungs, blood becomes oxygenated.
Oxygenated blood returns to the left atrium via the pulmonary vein.
It passes through the left atrioventricular (bicuspid / mitral) valve into the left ventricle.