Cambridge IGCSE · thinka-original Practice Paper

2024 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Jun 2024 (V3) Cambridge IGCSE-Style Mock — Biology (0610)

160 marks180 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Section A

Answer all questions. Write your answers in the spaces provided on the question paper.

7 Question · 92 marks

Question 1 · structured

11 marks

Organisms can be classified as prokaryotes or eukaryotes based on their cellular structure.

(a) State **three** cell structures that are found in a eukaryotic plant cell, such as a palisade mesophyll cell, but are **not** present in a prokaryotic bacterial cell. [3]

(b) A student placed some cells of a freshwater plant into a concentrated sodium chloride solution.
Describe and explain the appearance of these plant cells after 30 minutes. [4]

(c) Red blood cells are eukaryotic cells that do not have a cell wall.
Explain why a red blood cell will burst when placed in pure water, whereas a plant cell will become turgid but not burst. [4]

Show answer & marking scheme

Worked solution

(a) Three structures present in eukaryotic plant cells but absent in prokaryotic bacterial cells include a membrane-bound nucleus (containing DNA), mitochondria, chloroplasts, and a large permanent vacuole.

(b) When placed in a concentrated sodium chloride solution (hypertonic solution), the plant cells undergo plasmolysis. Because the water potential of the solution is lower than that of the cell cytoplasm and vacuole, water moves out of the plant cells by osmosis, down a water potential gradient, through the partially permeable cell membrane. Consequently, the vacuole and cytoplasm shrink, causing the cell membrane to pull away from the rigid cell wall, leaving the cell flaccid/plasmolysed.

(c) Pure water has a higher water potential than the cytoplasm of both red blood cells and plant cells, so water moves into both cells by osmosis. Since a red blood cell lacks a cell wall, its delicate cell membrane cannot withstand the increased turgor pressure as it swells, causing it to burst (lysis). In contrast, the plant cell is surrounded by a strong, rigid cellulose cell wall. This wall resists expansion, prevents further water uptake, and keeps the cell from bursting, leaving it turgid.

Marking scheme

(a) [Max 3 marks]
- nucleus / nuclear envelope / nuclear membrane; [1]
- mitochondrion / mitochondria; [1]
- chloroplast(s); [1]
- permanent / large vacuole; [1]
(Note: Do not accept 'cell wall' as bacteria also possess a cell wall, though made of peptidoglycan, not cellulose.)

(b) [Max 4 marks]
- water moves out of the cell; [1]
- by osmosis; [1]
- from high water potential (inside cell) to low water potential (outside cell) / down a water potential gradient; [1]
- through/across a partially permeable membrane; [1]
- cytoplasm/vacuole shrinks OR cell membrane pulls away from the cell wall / cell becomes plasmolysed / cell becomes flaccid; [1]

(c) [Max 4 marks]
- water enters both cells by osmosis; [1]
- down a water potential gradient / from a region of higher water potential (pure water) to lower water potential (cytoplasm); [1]
- red blood cell has no cell wall so it swells and bursts / lyses; [1]
- plant cell has a rigid / strong / cellulose cell wall; [1]
- cell wall prevents the plant cell from bursting / exerts turgor pressure / cell becomes turgid; [1]

Question 2 · structured

13 marks

The flowers of the wild cherry tree, *Prunus avium*, are pollinated by insects.

(a) (i) Wild cherry trees are dicotyledons. State one visible feature of dicotyledonous leaves that distinguishes them from monocotyledons. [1]

(ii) Describe the advantages and disadvantages of sexual reproduction to a population of wild cherry trees in the wild. [4]

(b) Pollen grains from insect-pollinated flowers have different features compared to those from wind-pollinated flowers.

(i) Explain how the structure of a pollen grain from an insect-pollinated flower is adapted for its method of transfer. [1]

(ii) Describe how the petals of an insect-pollinated flower are adapted to attract pollinators. [1]

(c) Some plants can undergo self-pollination.

Describe self-pollination. [2]

(d) Outline the events that occur inside the flower from the moment a pollen grain lands on a compatible stigma until fertilisation is complete. [4]

Show answer & marking scheme

Worked solution

(a) (i) Dicotyledonous leaves have branching/net-like veins (whereas monocotyledons have parallel veins) or a broad leaf blade/lamina.

(ii) Advantages of sexual reproduction include introducing genetic variation, allowing adaptation to changing environments, and reducing the speed of disease spread. Disadvantages include requiring more energy to produce flowers/pollen, relying on pollinators, and the risk that adapted parental gene combinations are disrupted.

(b) (i) The pollen grain is sticky, spiky, or has hooks so that it can easily adhere to the bodies of visiting insects.

(ii) Petals are large and brightly coloured to attract insects, and often feature nectar guides or scent.

(c) Self-pollination is the transfer of pollen grains from the anther to the stigma of the same flower, or to another flower on the same plant.

(d) Once on the stigma, the pollen grain germinates and a pollen tube grows down the style towards the ovary. It enters the ovule through the micropyle, allowing the male gamete/nucleus to travel down and fuse with the female gamete/nucleus to form a diploid zygote.

Marking scheme

**(a) (i)**
- leaf veins are net-like / branching / reticulate;
- broad leaf blade / lamina;
- presence of a petiole / leaf stalk;
[Max 1]

**(a) (ii)**
*Advantages (max 2):*
- introduces genetic variation / offspring are genetically different;
- population is more likely to survive environmental change / change in climate;
- population is more resistant to diseases / diseases do not spread as rapidly;
- allows for natural selection / adaptation to new environments;
- seeds can be dispersed further (reducing competition);
*Disadvantages (max 2):*
- requires more energy/resources (to produce flowers, nectar, pollen);
- relies on external factors / pollinators / successful fertilisation is not guaranteed;
- slower process / finding a mate;
- well-adapted parental gene combinations can be disrupted;
[Max 4]

**(b) (i)**
- (pollen grain is) sticky / spiky / has hooks to adhere/cling to the insect's body;
[Max 1]

**(b) (ii)**
- large / brightly coloured (to act as visual cues);
- have nectar guides / scent / produce nectar;
[Max 1]

**(c)**
- transfer of pollen grains from the anther to the stigma;
- of the same flower / of another flower on the same plant;
[Max 2]

**(d)**
- pollen grain germinates (on the stigma);
- pollen tube grows down the style (towards the ovary);
- enters the ovule through the micropyle;
- male nucleus / gamete travels down the pollen tube;
- male nucleus / gamete fuses with the female nucleus / egg cell;
- (to form a) diploid zygote;
- reference to digestive enzymes secreted by the pollen tube;
[Max 4]

Question 3 · structural

10 marks

Proteins are macromolecular polypeptides essential for all cellular processes. Enzymes are a class of proteins that act as biological catalysts.

**(a)** The synthesis of proteins involves several stages within a eukaryotic cell.

(i) Describe how a messenger RNA (mRNA) molecule is produced in the nucleus. [3]

(ii) State the role of ribosomes in the synthesis of a protein. [2]

**(b)** An enzyme-catalysed reaction can be regulated by molecules that inhibit enzyme activity.

Compound Z has a molecular structure very similar to the part of a polypeptide substrate that binds to the active site of a specific protease enzyme.

(i) Explain how compound Z decreases the rate of polypeptide digestion by the protease. [3]

(ii) State and explain how the rate of digestion of the polypeptide would change if the concentration of the polypeptide substrate is significantly increased while keeping the concentration of compound Z constant. [2]

Show answer & marking scheme

Worked solution

**(a)**
(i)
1. The DNA double helix unwinds or unzips to expose the specific gene sequence.
2. Free RNA nucleotides pair with the template DNA strand to form a complementary mRNA copy.
3. The completed mRNA molecule detaches from the DNA and exits the nucleus via a nuclear pore.

(ii)
1. The ribosome binds to the mRNA molecule and allows it to pass through.
2. It coordinates the joining of amino acids in a specific sequence to assemble the polypeptide chain.

**(b)**
(i)
1. Compound Z has a shape complementary to the active site of the protease (similar to the substrate polypeptide).
2. Compound Z binds to and occupies the active site.
3. This prevents the polypeptide substrate from entering the active site, reducing the formation of enzyme-substrate complexes.

(ii)
1. **State:** The rate of digestion increases (and can reach its maximum rate).
2. **Explain:** There is a much higher probability of a substrate molecule colliding with and binding to the active site than compound Z, outcompeting the inhibitor.

Marking scheme

**(a)(i)** [Max 3 marks]
- DNA double helix unwinds / separates / unzips; [1]
- (Complementary) RNA nucleotides pair with DNA / complementary copy of DNA gene is made; [1]
- mRNA leaves / exits nucleus (through nuclear pore); [1]
- *Accept:* Role of RNA polymerase in transcribing the gene. [1]

**(a)(ii)** [Max 2 marks]
- mRNA passes through / binds to the ribosome; [1]
- Ribosome assembles / joins amino acids together (to form a protein); [1]
- The sequence of amino acids is determined by the sequence of bases in mRNA; [1]

**(b)(i)** [Max 3 marks]
- Compound Z has a complementary shape to the active site of the protease / similar shape to the substrate; [1]
- Compound Z binds to / fits into / blocks the active site; [1]
- Prevents the substrate / polypeptide from binding to the active site; [1]
- Fewer enzyme-substrate complexes are formed; [1]

**(b)(ii)** [Max 2 marks]
- (State) Rate of reaction / digestion increases; [1]
- (Explain) Substrate outcompetes the inhibitor / higher chance of substrate colliding with the active site (than compound Z); [1]

Question 4 · theory

12 marks

1 Barley (Hordeum vulgare) is a major cereal grain grown globally. A botanist studied a population of barley and recorded two features: seed coat colour (purple or yellow) and the mass of grain produced per plant (measured in grams).

(a) (i) Identify the type of variation shown by each of these two features.

Seed coat colour: ............................................................

Mass of grain: ...................................................................

[2]

(ii) State two different environmental conditions that could influence the mass of grain harvested from a barley plant.

1. .....................................................................................................................

2. .....................................................................................................................

[2]

(b) The botanist aims to develop a new commercial variety of barley that consistently yields a high mass of grain and possesses purple seed coats.

Explain the steps the botanist should take to breed this specific line of barley plants through selective breeding.

...........................................................................................................................................

[4]

(c) Compare natural selection and artificial selection, explaining how they differ.

...........................................................................................................................................

[4]

Show answer & marking scheme

Worked solution

(a) (i) Seed coat colour falls into distinct, non-overlapping categories (purple or yellow), which is discontinuous variation. Mass of grain shows a continuous range of quantitative values, which is continuous variation.

(ii) Environmental factors affecting plant growth include soil mineral/nutrient content, water availability, temperature, light intensity, or presence of pests/diseases.

(b) The process of selective breeding (artificial selection) involves:
1. Choosing/selecting parental plants that show purple seeds and produce a high mass of grain.
2. Cross-pollinating these selected parent plants.
3. Growing the resulting offspring to maturity.
4. Selecting only the offspring that exhibit both desired traits (purple seed coat + high yield) and breeding them together.
5. Repeating this process over many successive generations until the desired combination is stable and pure-breeding.

(c) Key differences between natural and artificial selection:
1. Natural selection is driven by environmental selection pressures (e.g., climate, predators, disease), whereas artificial selection is directed by human choice and needs.
2. Natural selection increases the evolutionary fitness of the organism in its wild habitat, whereas artificial selection develops traits beneficial to humans (which may decrease the organism's fitness/survival capability in the wild).
3. Natural selection is generally a slow process occurring over long geological timescales, whereas artificial selection is much faster.
4. Natural selection maintains or enhances genetic diversity/biodiversity, whereas artificial selection often reduces the gene pool/genetic variation within a crop line.

Marking scheme

(a) (i) [Total: 2 marks]

- Seed coat colour: discontinuous (variation) ; [1]

- Mass of grain: continuous (variation) ; [1]

(ii) [Total: 2 marks]

Any two from:

- water availability / soil moisture ;

- mineral ions / nutrients / fertilizer (levels) ;

- light intensity / sunlight ;

- temperature ;

- carbon dioxide concentration ;

- spacing / competition (with other plants) ;

- presence of pathogens / pests / weeds ; [2]

(b) [Total: 4 marks]

Any four from:

- select / choose parent plants with purple seeds AND high grain yield ;

- cross-pollinate / breed / mate these selected plants together ;

- grow the resulting seeds / offspring ;

- select the offspring that show both desired features (purple seeds and high yield) ;

- repeat the process of selection and breeding over many generations ;

- until the crop is pure-breeding / all offspring consistently show both traits ; [4]

(c) [Total: 4 marks]

Any four from:

- (presence of humans) natural selection occurs without human intervention / artificial selection is controlled/directed by humans ;

- (selection pressure) selection pressure in natural selection is the environment / in artificial selection, it is human choice/needs ;

- (fitness/adaptation) natural selection leads to adaptation to the environment / survival of the fittest ;

- (benefit) artificial selection leads to traits that benefit humans (even if they decrease the organism's fitness in the wild) ;

- (speed) natural selection is generally a slower process / artificial selection is faster ;

- (diversity) natural selection maintains/increases genetic diversity / artificial selection can reduce genetic diversity/gene pool ; [4]

Question 5 · theory

12 marks

This is a placeholder for standard array returns.

Show answer & marking scheme

Worked solution

N/A

Marking scheme

N/A

Question 6 · structured

19 marks

1 (a) State the balanced chemical equation for aerobic respiration.

_____________________________________________________________________________________[2]

(b) A scientist made notes about a temperate grassland food web.

Fig. 1.1 shows the notes she made.

- In the grassland, clover plants photosynthesise.
- Clover plants are eaten by grasshoppers.
- Field mice and prairie dogs eat clover plants. Field mice also eat grasshoppers.
- Weasels eat other animals such as field mice.
- Garter snakes eat grasshoppers.

**Fig. 1.1**

(i) Using the information in Fig. 1.1, complete the food web in Fig. 1.2 by writing the names of the four missing organisms in the empty boxes.

```
[ weasels ]
▲
│
[ ]
▲ ▲
/ \
[ garter snakes ] │
▲ │
│ /
[ ] ◄───┘
▲ ▲
\ /
[ ]
```
*(Note: [ prairie dogs ] is also present in the food web connected directly from clover plants)*

_____________________________________________________________________________________[2]

(ii) State the principal source of energy for this food web.

_____________________________________________________________________________________[1]

(iii) Table 1.1 describes features of the food web in Fig. 1.2.

Complete Table 1.1 by writing the maximum number for each feature.

**Table 1.1**

| Feature of the food web | Maximum number in the food web in Fig. 1.2 |
| :--- | :--- |
| Trophic levels | |
| Primary consumers | |
| Tertiary consumers | |

_____________________________________________________________________________________[3]

(iv) Using the information in Fig. 1.2, predict and explain the most likely effect of a large decrease in the weasel population on the population size of:

**field mice**

prediction: ___________________________________________________________________________

explanation: __________________________________________________________________________

**clover plants**

prediction: ___________________________________________________________________________

explanation: __________________________________________________________________________[2]

(c) Grasshoppers and field mice can be harvested by some predators to obtain lipids (fats). Many animals store lipids as a major energy reserve.

(i) List the chemical elements found in all fats.

_____________________________________________________________________________________[1]

(ii) Using the information in Fig. 1.2, explain why it is more energy efficient for a predator to eat grasshoppers rather than weasels.

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________[4]

(d) Describe how fats ingested by humans are digested and absorbed.

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________

_____________________________________________________________________________________[4]

Show answer & marking scheme

Worked solution

(a) Aerobic respiration equation:
$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$

(b)(i) From bottom to top, the missing boxes in Fig. 1.2 are:
- Bottom box: clover plants
- Lower middle box: grasshoppers
- Upper right box (preceding weasels): field mice
- Left box: garter snakes (pre-filled in diagram)

(b)(ii) The principal source of energy is the Sun / sunlight.

(b)(iii) Maximum numbers:
- Trophic levels: 4 (clover plants → grasshoppers → field mice → weasels)
- Primary consumers: 3 (grasshoppers, field mice, prairie dogs)
- Tertiary consumers: 1 (weasels)

(b)(iv) Effect of decrease in weasel population:
- **field mice**: population increases, because there is less predation by weasels.
- **clover plants**: population decreases, because there are more field mice feeding on them.

(c)(i) Chemical elements in fats: carbon, hydrogen, and oxygen.

(c)(ii) Energy transfer efficiency:
- Energy is lost at each trophic level as it is transferred along a food chain.
- Energy is lost through respiration, heat loss, movement, excretion, or undigested waste (faeces).
- Grasshoppers are at a lower trophic level (second trophic level / primary consumers) compared to weasels (third or fourth trophic level / secondary or tertiary consumers).
- Therefore, less energy has been lost before reaching the grasshoppers, making them a more energy-efficient food source.

(d) Fat digestion and absorption:
- Emulsification of fats by bile in the small intestine (duodenum).
- Bile increases the surface area of fat droplets for the action of lipase.
- Chemical breakdown of fats by lipase (secreted by the pancreas into the small intestine).
- Lipase digests fats into fatty acids and glycerol.
- Absorption occurs in the small intestine via the villi / microvilli.
- Fatty acids and glycerol are absorbed into the lacteals.

Marking scheme

**(a)**
- $C_6H_{12}O_6 + 6O_2$ on reactant side [1]
- $6CO_2 + 6H_2O$ on product side [1]
*(Accept word equation for 1 mark if chemical equation is completely incorrect or missing)*

**(b)(i)**
- clover plants and grasshoppers in the correct positions [1]
- field mice in the correct position [1]

**(b)(ii)**
- the Sun / sunlight [1]

**(b)(iii)**
- Trophic levels: 4 [1]
- Primary consumers: 3 [1]
- Tertiary consumers: 1 [1]

**(b)(iv)**
- field mice: increases AND because there is less predation [1]
- clover plants: decreases AND because there are more herbivores / field mice eating them [1]

**(c)(i)**
- carbon, hydrogen, (and) oxygen (all three required) [1]

**(c)(ii)**
- Any four from:
1. Energy is lost at / between each trophic level [1]
2. Energy is lost through respiration / movement / heat / excretion / faeces / undigested parts [1]
3. Grasshoppers are at a lower trophic level / weasels are at a higher trophic level [1]
4. More energy is available at the grasshopper level / less energy has been lost before reaching grasshoppers [1]

**(d)**
- Any four from:
1. (Physical digestion) fats are emulsified by bile [1]
2. Emulsification increases surface area (for lipase action) [1]
3. (Chemical digestion) fats are broken down by lipase [1]
4. Lipase breaks down fats into fatty acids and glycerol [1]
5. (Absorption) occurs in the small intestine / duodenum / ileum / villi [1]
6. Fatty acids and glycerol enter the lacteals / lymphatic system [1]

Question 7 · either_or

15 marks

The Emerald Valley skink, *Oligosoma smaragdinum*, is a rare reptile species endemic to fragmented temperate rainforests.

**(a)** State the genus of the Emerald Valley skink. [1]

**(b)** Table 1.1 shows the estimated population sizes of the Emerald Valley skink in four separate forest patches in 2010 and 2022.

| Forest Patch | Population in 2010 | Population in 2022 |
| :--- | :---: | :---: |
| Patch 1 | 160 | 42 |
| Patch 2 | 75 | 82 |
| Patch 3 | 210 | 195 |
| Patch 4 | 95 | 30 |

Table 1.1

**(i)** Calculate the percentage decrease in the skink population in **Patch 1** from 2010 to 2022. Show your working and give your answer to **two significant figures**. [3]

**(ii)** State **two** reasons, other than disease, why skink populations can decline in fragmented forest patches. [2]

**(c)** To conserve the Emerald Valley skink, a captive breeding programme has been established.

**(i)** In captive breeding programmes, scientists use a 'studbook' to record the genetic pedigree of all individuals. Explain why it is important to avoid inbreeding when breeding endangered species in captivity. [3]

**(ii)** Explain the advantages of using artificial insemination rather than natural mating in captive breeding programmes. [3]

**(d)** Describe the term *biodiversity* and explain why the conservation of forest ecosystems is important. [3]

Show answer & marking scheme

Worked solution

**(a)**
The genus is the first part of the scientific name, capitalized: *Oligosoma*.

**(b)(i)**
1. Calculate the difference in population size for Patch 1:
$$\text{Decrease} = 160 - 42 = 118$$

2. Divide the decrease by the original population in 2010 and multiply by 100 to get the percentage:
$$\text{Percentage Decrease} = \left(\frac{118}{160}\right) \times 100 = 73.75\%$$

3. Round to two significant figures:
$$74\%$$

**(b)(ii)**
Acceptable reasons include: habitat destruction/deforestation, increased predation by invasive species, lack of food/prey, inability to find mates due to isolation, and illegal collection for the pet trade.

**(c)(i)**
Avoiding inbreeding is crucial because it reduces the risk of inbreeding depression, prevents harmful recessive genetic defects from being expressed in homozygous form, and maintains a wider genetic diversity to help the population adapt when reintroduced into the wild.

**(c)(ii)**
Advantages of artificial insemination include: eliminates the need to transport fragile or dangerous animals between zoos, allows genetic material (semen) to be frozen and stored long-term, prevents physical injuries during mating, and avoids transmission of reproductive diseases.

**(d)**
Biodiversity refers to the number of different species living in a specific area. Conserving forest ecosystems is vital because it preserves food webs, maintains ecosystem services (like carbon absorption and oxygen production), and provides valuable resources such as medicines and food for humans.

Marking scheme

**(a)** [1 mark total]
* *Oligosoma* ; (Ignore case, reject full binomial name)

**(b)(i)** [3 marks total]
* MP1: Correct calculation of the population decrease: $160 - 42 = 118$ ;
* MP2: Correct calculation of percentage to any decimal place: $\frac{118}{160} \times 100 = 73.75(\%)$ ;
* MP3: Correct rounding to two significant figures: $74(\%)$ ;
*(Award 3 marks for a correct final answer of 74 without working. Accept ecf from MP1 to MP2/MP3)*

**(b)(ii)** [2 marks total]
Any two from:
* Habitat destruction / deforestation / logging ;
* Increased predation (by introduced predators) ;
* Loss of food sources / lack of prey ;
* Difficulty finding mates due to physical isolation ;
* Hunting / poaching / pet trade ;
* Climate change / extreme weather ;

**(c)(i)** [3 marks total]
Any three from:
* Reduces the risk of inbreeding depression ;
* Prevents inheritance of harmful homozygous recessive alleles / genetic defects ;
* Preserves genetic variation / keeps a large gene pool ;
* Enables future adaptation of populations to changing environments when reintroduced ;
* Increases survival rates / reproductive success of offspring ;

**(c)(ii)** [3 marks total]
Any three from:
* No need to transport large/sensitive/stressed animals over long distances (only transport semen) ;
* Can easily breed individuals residing in separate geographical locations/zoos ;
* Semen can be frozen/stored indefinitely for long-term genetic preservation ;
* Prevents physical trauma/fighting during natural courtship/mating attempts ;
* Overcomes behavioral issues where partners refuse to mate naturally ;
* Lowers the transmission rate of sexually transmitted infections ;

**(d)** [3 marks total]
Any three from:
* Biodiversity is the number of different species in a given area ;
* Forests act as global carbon sinks / produce oxygen / regulate climate ;
* Forest roots bind the soil to prevent erosion / flooding ;
* Preserving ecosystems maintains complex food webs / prevents secondary extinctions ;
* Source of potential medicines / food / resources for humans ;
* Moral / ethical duty / aesthetic value / ecotourism ;

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