Primary 6 PSLE Science Semestral Assessment 2 (End of Year) Paper 4

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Primary 6 PSLE Science

Version 4 of 5 - ANSWER KEY

SECTION A: Multiple-Choice Questions (20 marks)

SECTION B: Open-Ended Questions (60 marks)

Question 21 (3 marks)

(a) Organism most likely a fish: Organism A (1 mark)

Reason: It has scaly skin and its body temperature changes with the environment (it is cold-blooded/ecothermic) — these are characteristics of fish. (1 mark)

Alternative acceptable answer: It has a backbone AND scaly skin (fish have scales; reptiles also have scales but are less likely given the combination with changing body temperature and no mention of lungs).

Teaching note: Reptiles also have scales and changing body temperature, but they have legs typically (not specified here). The surest identification is the combination of aquatic adaptations typically associated with fish in P6 context. If students answer "Organism A because it has scaly skin and changes body temperature with environment," award full marks.

(b) Birds and mammals: Organism B and Organism D (0.5 mark)

How to tell them apart:

• Organism D has feathers → bird (0.5 mark)
• Organism B has no feathers but has constant body temperature → mammal (0.5 mark)

Teaching note: Birds are the only vertebrates with feathers. Mammals have fur/hair and maintain constant body temperature (warm-blooded/endothermic). Both B and D have constant body temperature, but only D has feathers.

Question 22 (4 marks)

(a) Why water plants are important for fish survival (2 marks):

• Water plants produce oxygen during photosynthesis in daylight (1 mark)
• Fish need this dissolved oxygen to breathe (respiration) underwater/through their gills (1 mark)

OR alternative valid point:

• Plants can provide food/shelter for small organisms that fish eat (1 mark)

Teaching note: The aquarium is a closed system. Without plants, oxygen would be depleted. Photosynthesis equation: carbon dioxide + water → glucose + oxygen (in presence of light and chlorophyll). Fish extract dissolved O₂ using gills.

(b) Effect of keeping aquarium in dark room (2 marks):

• The fish would die over time (or "population would decrease") (1 mark)
• Explanation: In darkness, plants cannot photosynthesize, so they stop producing oxygen (0.5 mark). Plants and fish both respire, using up oxygen (0.5 mark). Without oxygen production, oxygen levels fall until fish suffocate.

Teaching note: In darkness, plants do the opposite of photosynthesis — they only respire, consuming oxygen. This makes the oxygen depletion even faster. The key misconception to avoid: plants do NOT release oxygen in the dark.

Question 23 (4 marks)

(a) Two processes releasing CO₂ to atmosphere (2 marks):

Any two from:

1. Respiration (by plants, animals, decomposers) (1 mark)
2. Combustion/burning (of fossil fuels or wood) (1 mark)
3. Decomposition (by decomposers breaking down dead matter) (1 mark)

(b) Role of decomposers in carbon cycle (2 marks):

• Decomposers (bacteria/fungi) break down dead organisms and waste (1 mark)
• During respiration, they release carbon dioxide back into the atmosphere/soil (1 mark)

Teaching note: Decomposers are nature's recyclers. Without them, dead matter would pile up and nutrients/carbon would be locked away. They complete the cycle by returning carbon to atmospheric CO₂ pool.

Question 24 (4 marks)

(a) Aim and variables (2 marks):

Aim: To find out/how to investigate which material is the best insulator for keeping water hot/which material reduces heat loss from water the most. (0.5 mark)

Variables clearly stated:

• Changed variable (independent): Type of wrapping material / type of insulator (0.5 mark)
• Measured variable (dependent): Temperature of water / temperature drop over time / how hot the water stays (0.5 mark)
• Kept constant (control): Same amount/volume of water, same starting temperature, same size/type of beaker, same environment/room temperature (any one, 0.5 mark)

(b) Best insulator: Beaker Q (Wool) (1 mark)

Explanation using data: Wool kept the water hottest for longest / had slowest temperature drop. Temperature dropped from 80°C to only 59°C after 20 minutes (or dropped by only 21°C), compared to cotton (drop of 26°C), no wrapping (drop of 36°C), and aluminium foil (drop of 45°C). (1 mark)

(c) Why aluminium foil cooled fastest (1 mark):

• Aluminium is a metal / good conductor of heat (0.5 mark)
• It conducts heat away from the water to the surroundings instead of trapping it (0.5 mark)

Teaching note: Metals conduct heat well; insulators trap heat by reducing heat transfer through conduction, convection, and radiation. Wool and cotton trap air pockets. Aluminium being shiny might reflect some radiation, but as a thin sheet it conducts heat away effectively.

Question 25 (4 marks)

(a) Two structural adaptations (2 marks):

Any two:

1. Large, flat, round floating leaves / leaves on water surface (0.5 mark)
2. Waxy coating / cuticle on leaves (0.5 mark)
3. Long stems reaching to bottom / long petioles (0.5 mark)
4. Air spaces / aerenchyma in stems (0.5 mark)
5. Roots anchored in mud at bottom (0.5 mark)

(b) Explanation of one adaptation (2 marks):

Example 1 — Floating leaves:

• The large, flat leaves float on water surface to capture maximum sunlight for photosynthesis (1 mark)
• Being on surface prevents them from being submerged and unable to get light / CO₂ (1 mark)

Example 2 — Waxy coating:

• The waxy cuticle is waterproof / prevents water from sticking (1 mark)
• This allows water to run off and prevents leaves from rotting / being waterlogged / reduces water covering surface that would block light (1 mark)

Example 3 — Air spaces in stems:

• Air spaces make stems buoyant / less dense than water (1 mark)
• This helps leaves and flowers stay afloat at the surface rather than sinking (1 mark)

Teaching note: Water lilies face unique challenges: need light above water, need to stay afloat, need gas exchange. Their adaptations solve these specific problems.

Question 26 (4 marks)

(a) Part C / small intestine (1 mark)

(b) Why small intestine is well-adapted for absorption (2 marks):

• It is very long (about 6 metres) / has large surface area (0.5 mark)
• Has villi (and microvilli) / inner surface is folded — increases surface area for absorption (0.5 mark)
• Villi have thin walls / one cell thick — allows nutrients to pass through quickly (0.5 mark)
• Has blood capillaries close to surface — carries absorbed nutrients away (0.5 mark)

(c) Digestive enzyme in stomach: Pepsin (or protease) (0.5 mark)

Function: Breaks down proteins into smaller peptides / begins protein digestion (0.5 mark)

Teaching note: The stomach's acidic environment (HCl) activates pepsinogen → pepsin. This is the start of chemical digestion of proteins. Mechanical churning also occurs.

Question 27 (4 marks)

(a) Food chain (1 mark):

Algae → Mayfly nymph → Small fish → Large fish OR Algae → Mayfly nymph → Water spider → Small fish → Large fish OR Algae → Mayfly nymph → Water spider → (if spider is eaten by something else)

Any correct chain starting with algae as producer gets 1 mark.

(b) Effect of removing water spiders (2 marks):

• Mayfly nymph population would increase / go up (1 mark)
• Explanation: Water spiders eat / are predators of mayfly nymphs (0.5 mark). Without spiders, there is less predation / fewer nymphs are eaten (0.5 mark), so more nymphs survive and reproduce.

(c) Why few large fish compared to many mayfly nymphs (2 marks):

• Energy is lost at each trophic level (0.5 mark)
• From mayfly nymphs (primary consumers) to large fish (tertiary consumers), energy is lost three times / through multiple transfers (0.5 mark)
• Only about 10% of energy transfers between levels (0.5 mark)
• So large fish need many nymphs to support just a few individuals / fewer large fish can be supported (0.5 mark)

Teaching note: This is the pyramid of energy/biomass. With 45 mayfly nymphs at ~500 energy units each, next level gets ~2250, next ~225, enough for only 2 large fish needing ~100 each.

Question 28 (4 marks)

(a) Why green parts turn blue-black with iodine (2 marks):

• The green parts contain chlorophyll and can carry out photosynthesis (0.5 mark)
• Photosynthesis produces glucose / starch as a food store (0.5 mark)
• Starch turns blue-black when tested with iodine solution (1 mark)

(b) Role of chlorophyll in photosynthesis (2 marks):

• Chlorophyll is the green pigment in plant cells that traps/absorbs light energy from the Sun (1 mark)
• Without chlorophyll (white parts), no light is absorbed, so no photosynthesis occurs and no starch is produced (1 mark)
• This shows chlorophyll is essential / necessary for photosynthesis to take place

Teaching note: Classic experiment using variegated leaves (e.g., Coleus or geranium) where white parts lack chlorophyll. The comparison proves chlorophyll's requirement for photosynthesis and starch production.

Question 29 (4 marks)

(a) Main difference (1 mark):

Butterfly has complete metamorphosis with four distinct stages (egg, caterpillar/larva, pupa/chrysalis, adult), while grasshopper has incomplete metamorphosis with three stages where the young (nymph) resembles the adult.

(b) Advantage of complete metamorphism (1 mark):

Any one:

• The caterpillar and adult eat different foods — reduces competition for food within the species (0.5 mark), allowing more individuals to survive (0.5 mark)
• The pupa is less active / hidden — protected from predators during vulnerable transformation (1 mark)
• Different life stages can exploit different niches/environments (1 mark)

(c) What happens during pupa stage (2 marks):

• The caterpillar's body breaks down / reorganizes during pupation (1 mark)
• It transforms/metamorphoses into the adult butterfly with wings, legs, antennae, and reproductive organs (1 mark)
• This is a period of major change / restructuring with no feeding

Teaching note: Inside the chrysalis, histolysis (breakdown of larval tissues) and histogenesis (formation of adult tissues) occur. The dramatic transformation is why it's called "complete" metamorphosis.

Question 30 (5 marks)

(a) Aim (1 mark):

To find out/investigate how the direction/position of light affects the direction of seedling growth / whether seedlings grow towards light.

(b) Variables (3 marks):

(i) Changed variable (independent): Direction/position/source of light / whether light comes from above or the side (1 mark)

(ii) Measured variable (dependent): Direction of seedling growth / how much seedlings bend / angle of bending (1 mark)

(iii) Two variables kept the same:

• Type/amount of soil (0.5 mark)
• Number of seedlings / type of seeds (0.5 mark)
• Amount of water given
• Size/type of pot
• Temperature
• Duration of experiment

Any two correct — 0.5 mark each, total 1 mark.

(c) Why seedlings in Pot B grew towards light (2 marks):

• Seedlings need light for photosynthesis to make food / energy for growth (0.5 mark)
• The tip of the shoot produces auxin (plant hormone) (0.5 mark)
• Auxin moves away from light / accumulates on shaded side (0.5 mark)
• More auxin on shaded side causes faster cell elongation / cells grow longer on dark side, bending shoot towards light — phototropism (0.5 mark)

Teaching note: This is positive phototropism. The hormone auxin (indole-3-acetic acid) promotes cell elongation. Unequal distribution causes differential growth. At P6 level, "plant hormone causes shaded side to grow faster" is sufficient detail.

Question 31 (4 marks)

(a) Processes A and B (2 marks):

• A = Evaporation — process where water changes from liquid to water vapour/gas, usually due to heat from Sun (1 mark)
• B = Condensation — process where water vapour cools and changes back to liquid droplets, forming clouds (1 mark)

(Accept: Transpiration for A if referring to plant water loss, but evaporation is primary correct answer for ocean)

(b) Importance of forests in water cycle (2 marks):

• Trees absorb water through roots and release it through transpiration — adds water vapour to atmosphere (1 mark)
• Forests increase evaporation from soil and plant surfaces → more cloud formation → more rainfall (0.5 mark)
• Tree roots and leaf litter help water infiltrate/soak into ground rather than run off → replenishes groundwater / prevents flooding (0.5 mark)
• Forests protect watersheds and maintain steady water supply

Teaching note: Deforestation reduces transpiration, leading to less rainfall and more surface runoff (erosion, flooding). Singapore's forest reserves help maintain local water cycle despite urbanization.

Question 32 (5 marks)

(a) Sun highest at 12 noon (1 mark)

How to tell: Shadow is shortest at 12 noon (only 20 cm) / shortest shadow when Sun is directly overhead (1 mark)

Physics principle: When Sun is highest/zenith, light rays hit most vertically, creating shortest shadow.

(b) Why shadow length changes (2 marks):

• The Sun appears to move across the sky due to Earth's rotation (1 mark)
• In morning/evening, Sun is lower on horizon → light hits at lower angle/ more slanted → shadow is longer (0.5 mark)
• At noon, Sun is highest/above head → light hits more vertically/straighter down → shadow is shorter (0.5 mark)

(c) Results on cloudy day (1 mark):

• Shadows would be faint or not visible / hard to measure accurately (0.5 mark)
• Because clouds block/scatter sunlight — diffuse light comes from many directions instead of single point source (0.5 mark)

Teaching note: On overcast days, shadows are "soft" with blurred edges. Position determination becomes unreliable. Ancient sundials don't work without direct sunlight.

Question 33 (5 marks)

(a) Bulb will light up / glow (1 mark)

Explanation: When switch is closed, circuit is complete/closed loop (0.5 mark). Electric current can flow from battery through bulb and back, so bulb has energy to produce light (0.5 mark).

(b) Expected ammeter reading: 0.2 A (or 0.1–0.5 A range) (1 mark)

Explanation: A small torch bulb with 2-cell battery (3V total) typically draws small current. 5A and 50A are far too large for this simple circuit — would require much higher voltage/power source. (1 mark)

Accept reasonable estimates with correct reasoning about scale.

(c) Drawing of circuit with two bulbs in series (1 mark):

Description for correct drawing: Battery → Switch → Bulb 1 → Bulb 2 → back to battery, all in single loop. Ammeter still in series.

[Battery +|−]──[Switch]──[Ammeter]──[Bulb]──[Bulb]──┐
                                                   │
                    └──────────────────────────────┘

Or text description: Two bulbs connected one after another in the same loop with the battery and switch.

Question 34 (5 marks)

(a) Conclusion about water and growth (2 marks):

• Bean plants need some water to grow — seeds in Pot W with no water did not grow at all (0.5 mark)
• Optimal amount is around 30 mL — plants in Pot Y grew tallest (18 cm) (0.5 mark)
• Too much water (100 mL in Pot Z) is harmful — plants grew poorly (only 5 cm), worse than moderate water (0.5 mark)
• Water is needed for germination and growth, but excess water causes problems (0.5 mark)

(b) Why seeds in Pot W did not grow (1 mark):

• Water is needed for germination — activates enzymes, allows seed coat to soften, enables metabolic processes to begin (1 mark)

(c) Why Pot Z grew less than Pot Y (2 marks):

Any two points:

• Too much water fills air spaces in soil — roots cannot get oxygen for respiration (1 mark)
• Roots may rot due to waterlogged conditions / lack of gas exchange (1 mark)
• Excess water may wash away nutrients / cause root diseases (1 mark)
• Seeds may have drowned / suffocated before germinating (1 mark)

Teaching note: This demonstrates the concept of limiting factors and optimal conditions. More is not always better — excess becomes detrimental. Real agricultural problem: overwatering causes crop failure.

Question 35 (5 marks)

(a) Materials attracted to magnet (1 mark):

Iron nail and steel paperclip (0.5 mark each)

(b) Difference between temporary and permanent magnets (2 marks):

• Temporary magnet: Becomes magnetized when near a magnet but loses magnetism quickly when removed / can be easily demagnetized (0.5 mark)
• Permanent magnet: Retains magnetism for a long time / permanently after being magnetized (0.5 mark)

Example: The iron nail is a temporary magnet — it is attracted and may become slightly magnetic near the bar magnet, but loses this quickly (1 mark) OR The steel paperclip can become permanent magnet if stroked repeatedly — retains magnetism longer.

(c) Why aluminum, plastic, and copper not attracted (2 marks):

• They are not magnetic materials / they do not contain iron, nickel, or cobalt (1 mark)
• Only ferromagnetic materials (iron, nickel, cobalt, and some steels) can be attracted by magnets / become magnetized (1 mark)
• Aluminum, plastic, and copper do not have magnetic domains that can align to produce attraction

Teaching note: Common misconception that all metals are magnetic. Only 3 elements (Fe, Ni, Co) and their alloys are ferromagnetic. Copper is diamagnetic (very weakly repelled). Plastic is non-metallic.

Question 36 (revised Diversity question) (5 marks)

(a) Compare fish and amphibian reproduction (2 marks):

Similar: Both reproduce by laying eggs in water / external fertilization often occurs in water (1 mark)

Different:

• Fish: Eggs usually have no protective shell / jelly-like, no metamorphosis, young are miniature adults (0.5 mark)
• Amphibians: Eggs have soft jelly covering, undergo metamorphosis (tadpole with gills → adult with lungs), can live on land as adults (0.5 mark)

(b) Why birds/mammals have lungs, fish have gills (2 marks):

• Fish live entirely in water — gills are adapted to extract dissolved oxygen from water / have filaments and lamellae for gas exchange in aquatic environment (1 mark)
• Birds and mammals live on land / breathe air — lungs are adapted to extract oxygen from air / air has higher oxygen concentration than water, lungs provide larger surface area for gas exchange in air (1 mark)

Extension: Gills would collapse in air; lungs would not work underwater without special adaptations.

(c) Is the claim correct? (1 mark):

No / Not entirely correct (0.5 mark)

Explanation: Most mammals give live birth, but some mammals lay eggs — e.g., platypus and echidna (monotremes) are mammals that lay eggs (0.5 mark). The table shows "Live birth (mostly)" which indicates exceptions exist.

Teaching note: Monotremes are the exception that proves students need to be careful with absolute statements. The "~mostly" in the table is a deliberately placed hint.

Question 37 (5 marks)

(a) Why pyramid base is wider than top (2 marks):

• Energy decreases/is lost at each trophic level (1 mark)
• At each level, ~90% energy is lost as heat (respiration), undigested waste, and movement — only ~10% transfers to next level (0.5 mark)
• So less energy supports fewer organisms at higher levels / more producers needed to support top predators (0.5 mark)

(b) Energy transfer percentage calculation (2 marks):

From primary consumers (500) to secondary consumers (50):

$\text{Percentage} = \frac{50}{500} \times 100\% = 10\%$ (1 mark for correct working)

Answer: 10% (1 mark)

Accept 10% or any working showing understanding of ratio.

(c) Effect of disease killing grass (1 mark):

• All levels above would decrease/collapse — no food/energy source for primary consumers, so whole web suffers (0.5 mark)
• Or: Pyramid becomes very narrow/unstable at all levels, ecosystem collapses (0.5 mark)

MARKING SUMMARY

Common marking principles across all questions:

• Accept synonyms and equivalent scientific terminology unless specifically prohibited
• Spelling errors: accept if meaning is clear
• For "explain" questions: must provide mechanism/reason, not just restatement
• For "suggest" questions: accept reasonable inferences based on scientific principles
• Consequential marks (follow-through errors): award method marks if working is correct from wrong value

END OF ANSWER KEY