Grade 8 · Science · curriculum collection
Grade 8 Science, NGSS domains
8 units for Grade 8 science, organised by NGSS disciplinary core idea domain: Physical Sciences, Life Sciences, Earth and Space Sciences. Each unit has a short lesson, a practice worksheet and a review worksheet, both with answer keys.
Answer-checked and approved on 29 October 2026
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Aligned by StudyBrix at domain level. Not an official publication of, or endorsed by, the standards' authors. Framework: Next Generation Science Standards.
Units in teaching order
PS Physical Sciences
Motion and energy
Calculations with velocity, acceleration, force, work, energy and power.
Short lesson: Energy efficiency
Study guide for Grades 7 to 11: 2 worked examples and 2 practice questions. Useful output cannot exceed total input in this simple energy accounting model.
efficiency % = useful output / input × 100
Useful output cannot exceed total input in this simple energy accounting model.
Worked example
A device receives 4100 J and delivers 615 J of useful energy. Find its efficiency as a percentage.
615/4100 × 100 = 15%
Answer: 15 %
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
Velocity changes from 4 m/s to 24 m/s in 4 s. Find average acceleration in m/s².
A device receives 1800 J and delivers 450 J of useful energy. Find its efficiency as a percentage.
A perpendicular force of 2992 N acts over 22 m². Find the pressure in Pa.
A constant 83 N force acts in the same direction as a 9 m displacement. Find work in J.
An object of mass 29 kg accelerates at 3 m/s². Find the net force in N.
An object moves at constant velocity 29 m/s for 38 s. Find displacement in m.
A device transfers 555 J in 15 s. Find average power in W.
Lift a 16 kg mass vertically by 4 m. Using g = 10 m/s², find the increase in gravitational potential energy in J.
Motion, forces and energy: which term fits this description? The rate at which distance is travelled.
- Impulse
- Weight
- Speed
- Elastic collision
Use g = 10 N/kg. A 2 kg object is raised vertically by 1.5 m. Find its increase in gravitational potential energy in J.
A student adds path lengths and calls the result displacement. Explain the distinction.
Velocity changes from 13 m/s to -35 m/s in 12 s. Find average acceleration in m/s².
Answer key
- 5 m/s² (24 − 4)/4 = 5
- 25 % 450/1800 × 100 = 25%
- 136 Pa 2992 ÷ 22 = 136
- 747 J 83 × 9 = 747
- 87 N 29 × 3 = 87 N
- 1102 m 29 × 38 = 1102
- 37 W 555 ÷ 15 = 37
- 640 J 16 × 10 × 4 = 640
- Speed Speed: The rate at which distance is travelled.
- 30 mgh = 2 × 10 × 1.5.
- Distance totals path length; displacement is the directed change from start to finish.
- -4 m/s² (-35 − 13)/12 = -4
Review worksheet: 12 questions
A device receives 6600 J and delivers 330 J of useful energy. Find its efficiency as a percentage.
An object of mass 13 kg accelerates at 8 m/s². Find the net force in N.
A constant 35 N force acts in the same direction as a 20 m displacement. Find work in J.
A perpendicular force of 558 N acts over 18 m². Find the pressure in Pa.
Velocity changes from 2 m/s to 68 m/s in 11 s. Find average acceleration in m/s².
An object moves at constant velocity -12 m/s for 23 s. Find displacement in m.
Lift a 12 kg mass vertically by 11 m. Using g = 10 m/s², find the increase in gravitational potential energy in J.
A device transfers 462 J in 7 s. Find average power in W.
Motion, forces and energy: which term fits this description? The directed change from initial position to final position.
- Acceleration
- Impulse
- Displacement
- Vector
In a simple model of a falling object, gravitational potential energy can become
- kinetic energy
- only chemical energy
- new mass created from nothing in this model
- zero energy everywhere
Explain why an object can move while the net force on it is zero.
A device receives 7400 J and delivers 2220 J of useful energy. Find its efficiency as a percentage.
Answer key
- 5 % 330/6600 × 100 = 5%
- 104 N 13 × 8 = 104 N
- 700 J 35 × 20 = 700
- 31 Pa 558 ÷ 18 = 31
- 6 m/s² (68 − 2)/11 = 6
- -276 m -12 × 23 = -276
- 1320 J 12 × 10 × 11 = 1320
- 66 W 462 ÷ 7 = 66
- Displacement Displacement: The directed change from initial position to final position.
- kinetic energy Motion increases as gravitational potential energy decreases.
- With zero acceleration it can continue at constant velocity; a nonzero net force is not needed to maintain that motion.
- 30 % 2220/7400 × 100 = 30%
PS Physical Sciences
Motion & graphs
Distance, displacement, speed, velocity and acceleration.
Short lesson: Acceleration from two readings
Study guide for Grades 8 to 11: 2 worked examples and 2 practice questions.
a = (v₂ − v₁)/Δt
Use consistent SI units and keep the stated direction convention.
Worked example
Signed velocity changes from 19 m/s to 5 m/s in 5 s. Find the average acceleration. Round to 3 decimal places if needed. Answer in m/s².
a = (v₂ − v₁)/Δt (5 − 19) / 5 = -2.8
Answer: -2.8 m/s²
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
On an east-positive axis, position changes from 32 m to -1 m in 27 s. Find the signed average velocity. Round to 3 decimal places if needed. Answer in m/s.
Signed velocity changes from 13 m/s to 14 m/s in 10 s. Find the average acceleration. Round to 3 decimal places if needed. Answer in m/s².
Two carts approach each other on a straight track at 24 m/s and 8 m/s. How fast is their separation decreasing? Round to 3 decimal places if needed. Answer in m/s.
A robot moves 146 m east and then 9 m west along a straight line. Find the total distance travelled. Round to 3 decimal places if needed. Answer in m.
A rover travels 390 m in 78 s, then 198 m in 41 s. Find its average speed for both stages. Round to 3 decimal places if needed. Answer in m/s.
Why can a round trip have zero average velocity but nonzero average speed?
A constant velocity of 4 m/s lasts 3 s. Find displacement.
A runner returns to the start after one complete lap. Which is zero?
- Elapsed time necessarily
- Displacement
- Distance travelled
On an east-positive axis, position changes from 51 m to 42 m in 17 s. Find the signed average velocity. Round to 3 decimal places if needed. Answer in m/s.
Signed velocity changes from 6 m/s to 3 m/s in 15 s. Find the average acceleration. Round to 3 decimal places if needed. Answer in m/s².
Two carts approach each other on a straight track at 16 m/s and 22 m/s. How fast is their separation decreasing? Round to 3 decimal places if needed. Answer in m/s.
A robot moves 96 m east and then 1 m west along a straight line. Find the total distance travelled. Round to 3 decimal places if needed. Answer in m.
Answer key
- -1.222 m/s v̄ = (x₂ − x₁)/Δt (-1 − (32)) / 27 = -1.222
- 0.1 m/s² a = (v₂ − v₁)/Δt (14 − 13) / 10 = 0.1
- 32 m/s closing speed = v₁ + v₂ 24 + 8 = 32
- 155 m distance = forward path + return path 146 + 9 = 155
- 4.941 m/s v̄ = (d₁ + d₂)/(t₁ + t₂) (390 + 198) / (78 + 41) = 4.941
- Displacement is zero while total distance is positive. Look for the different numerators rather than claiming the object never moved.
- 12 m Displacement equals velocity × time: 4 × 3 = 12 m.
- Displacement Final and initial positions coincide, so displacement is zero.
- -0.529 m/s v̄ = (x₂ − x₁)/Δt (42 − (51)) / 17 = -0.529
- -0.2 m/s² a = (v₂ − v₁)/Δt (3 − 6) / 15 = -0.2
- 38 m/s closing speed = v₁ + v₂ 16 + 22 = 38
- 97 m distance = forward path + return path 96 + 1 = 97
Review worksheet: 12 questions
On an east-positive axis, position changes from 63 m to -16 m in 10 s. Find the signed average velocity. Round to 3 decimal places if needed. Answer in m/s.
A robot moves 71 m east and then 65 m west along a straight line. Find the total distance travelled. Round to 3 decimal places if needed. Answer in m.
A rover travels 47 m in 16 s, then 263 m in 15 s. Find its average speed for both stages. Round to 3 decimal places if needed. Answer in m/s.
Signed velocity changes from 30 m/s to 29 m/s in 2 s. Find the average acceleration. Round to 3 decimal places if needed. Answer in m/s².
Two carts approach each other on a straight track at 7 m/s and 1 m/s. How fast is their separation decreasing? Round to 3 decimal places if needed. Answer in m/s.
Two travel stages take different amounts of time. Why is a simple mean of their speeds unreliable?
Position is 2 m at 1 s and 14 m at 5 s. Find average velocity.
An object begins at +12 m and ends at +3 m. Find signed displacement.
On an east-positive axis, position changes from -54 m to 53 m in 13 s. Find the signed average velocity. Round to 3 decimal places if needed. Answer in m/s.
A robot moves 128 m east and then 91 m west along a straight line. Find the total distance travelled. Round to 3 decimal places if needed. Answer in m.
A rover travels 207 m in 31 s, then 352 m in 27 s. Find its average speed for both stages. Round to 3 decimal places if needed. Answer in m/s.
Signed velocity changes from 13 m/s to 36 m/s in 13 s. Find the average acceleration. Round to 3 decimal places if needed. Answer in m/s².
Answer key
- -7.9 m/s v̄ = (x₂ − x₁)/Δt (-16 − (63)) / 10 = -7.9
- 136 m distance = forward path + return path 71 + 65 = 136
- 10 m/s v̄ = (d₁ + d₂)/(t₁ + t₂) (47 + 263) / (16 + 15) = 10
- -0.5 m/s² a = (v₂ − v₁)/Δt (29 − 30) / 2 = -0.5
- 8 m/s closing speed = v₁ + v₂ 7 + 1 = 8
- Longer stages have more time weight; use total distance divided by total time. Credit correct time weighting or a valid numerical counterexample.
- 3 m/s The slope is (14 − 2)/(5 − 1) = 3 m/s.
- -9 m 3 − 12 = −9 m on the chosen axis.
- 8.231 m/s v̄ = (x₂ − x₁)/Δt (53 − (-54)) / 13 = 8.231
- 219 m distance = forward path + return path 128 + 91 = 219
- 9.638 m/s v̄ = (d₁ + d₂)/(t₁ + t₂) (207 + 352) / (31 + 27) = 9.638
- 1.769 m/s² a = (v₂ − v₁)/Δt (36 − 13) / 13 = 1.769
PS Physical Sciences
Waves & sound
Wave speed, frequency, period and sound.
Short lesson: Sound, echoes and travel time
Study guide for Grades 8 to 11: 4 key ideas, including sound as a wave, echo and frequency; 2 worked examples and 3 practice questions. Use round-trip reasoning and distinguish medium effects from source pitch.
- Sound as a wave. Sound in air is a mechanical longitudinal wave in the basic model; compressions and rarefactions move through the medium.
- Echo. An echo's travel time includes the outward and return path, so a single-reflector distance is vt/2.
- Frequency. Pitch is associated with frequency. Medium and conditions influence sound speed.
- Model limits. Use the sound speed supplied in the problem. Do not assume one value applies to every material and temperature.
Worked example
An echo returns in 0.6 s with sound speed 340 m/s. Find distance to the reflector.
The round-trip length is 204 m; divide by two.
Answer: 102 m
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
A periodic wave has frequency 17 Hz and wavelength 4 m. Find its speed in m/s.
A wave has frequency 32 Hz and wavelength 3.1 m. Find its speed. Round to 3 decimal places if needed. Answer in m/s.
An echo returns after 1420 ms. The sound speed in this medium is 330 m/s. Find the distance to the stationary reflector. Round to 3 decimal places if needed. Answer in m.
21 complete oscillations take 50 s. Find the period. Round to 3 decimal places if needed. Answer in s.
With speed 330 m/s and frequency 110 Hz, find wavelength.
Why is saying the water itself travels across the entire ocean with a wave misleading?
A periodic wave has frequency 67 Hz and wavelength 26 m. Find its speed in m/s.
A wave has frequency 112 Hz and wavelength 4.3 m. Find its speed. Round to 3 decimal places if needed. Answer in m/s.
An echo returns after 1630 ms. The sound speed in this medium is 340 m/s. Find the distance to the stationary reflector. Round to 3 decimal places if needed. Answer in m.
21 complete oscillations take 53 s. Find the period. Round to 3 decimal places if needed. Answer in s.
Sound travels 680 m at 340 m/s. Find travel time.
A wave's amplitude doubles. Does that alone tell you that its frequency doubles?
Answer key
- 68 m/s 17 × 4 = 68
- 99.2 m/s v = fλ 32 × 3.1 = 99.2
- 234.3 m d = vt/2 330 × (1420/1000) / 2 = 234.3
- 2.381 s T = total time / number of cycles 50 / 21 = 2.381
- 3 m λ = v/f = 330/110 = 3 m.
- In a simple wave model, local particles oscillate while the disturbance transfers energy. Credit local motion versus wave propagation; complex currents are separate.
- 1742 m/s 67 × 26 = 1742
- 481.6 m/s v = fλ 112 × 4.3 = 481.6
- 277.1 m d = vt/2 340 × (1630/1000) / 2 = 277.1
- 2.524 s T = total time / number of cycles 53 / 21 = 2.524
- 2 s t = d/v = 680/340 = 2 s.
- No. Amplitude and frequency describe different properties. Credit independence in the simple model rather than inventing a fixed link.
Review worksheet: 12 questions
An echo returns after 870 ms. The sound speed in this medium is 1500 m/s. Find the distance to the stationary reflector. Round to 3 decimal places if needed. Answer in m.
A periodic wave has frequency 13 Hz and wavelength 27 m. Find its speed in m/s.
36 complete oscillations take 33 s. Find the period. Round to 3 decimal places if needed. Answer in s.
A wave has frequency 38 Hz and wavelength 0.9 m. Find its speed. Round to 3 decimal places if needed. Answer in m/s.
Crest-to-trough height is 10 cm for a symmetric wave. Find amplitude.
Why is an echo distance often divided by two?
- All waves are half their stated speed.
- Two ears hear it.
- The sound travels to the reflector and back.
An echo returns after 1720 ms. The sound speed in this medium is 1500 m/s. Find the distance to the stationary reflector. Round to 3 decimal places if needed. Answer in m.
A periodic wave has frequency 79 Hz and wavelength 25 m. Find its speed in m/s.
9 complete oscillations take 15 s. Find the period. Round to 3 decimal places if needed. Answer in s.
A wave has frequency 25 Hz and wavelength 4.3 m. Find its speed. Round to 3 decimal places if needed. Answer in m/s.
A source makes 20 complete cycles in 5 s. Find frequency.
An echo returns in 0.6 s with sound speed 340 m/s. Find distance to the reflector.
Answer key
- 652.5 m d = vt/2 1500 × (870/1000) / 2 = 652.5
- 351 m/s 13 × 27 = 351
- 0.917 s T = total time / number of cycles 33 / 36 = 0.917
- 34.2 m/s v = fλ 38 × 0.9 = 34.2
- 5 cm The amplitude is half the full crest-to-trough height.
- The sound travels to the reflector and back. The measured duration is for a round trip.
- 1290 m d = vt/2 1500 × (1720/1000) / 2 = 1290
- 1975 m/s 79 × 25 = 1975
- 1.667 s T = total time / number of cycles 15 / 9 = 1.667
- 107.5 m/s v = fλ 25 × 4.3 = 107.5
- 4 Hz 20/5 = 4 cycles per second.
- 102 m The round-trip length is 204 m; divide by two.
PS Physical Sciences
Electricity & circuits
Charge, current, voltage, resistance and electrical power.
Short lesson: Charge transferred
Study guide for Grades 8 to 12: 2 worked examples and 2 practice questions. This relation uses constant current in amperes and time in seconds.
Q = It
This relation uses constant current in amperes and time in seconds.
Worked example
A constant current of 6 A flows for 88 s. How much charge passes in C?
6 × 88 = 528
Answer: 528 C
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
A constant current of 12 A flows for 57 s. How much charge passes in C?
An ideal ohmic resistor of 9 Ω carries 8 A. Find the voltage across it in V.
A constant current of 1 A flows for 121 s. Find transferred charge. Round to 3 decimal places if needed. Answer in C.
A model device uses constant power 925 W for 9 h. Find the energy in kilowatt-hours. Round to 3 decimal places if needed. Answer in kWh.
An ohmic resistor of 41 Ω has potential difference 58 V. Find current. This is a paper calculation, not a construction instruction. Round to 3 decimal places if needed. Answer in A.
Waves and electric circuits: which term fits this description? A circuit arrangement with components connected along one current path.
- Reflection
- Open circuit
- Longitudinal wave
- Series circuit
What is the unit of charge?
- Watt
- Coulomb
- Ampere
Ideal parallel branches connected across the same nodes share the same
- wire length
- current in every case
- resistance in every case
- potential difference
A constant current of 15 A flows for 71 s. How much charge passes in C?
An ideal ohmic resistor of 42 Ω carries 3 A. Find the voltage across it in V.
A constant current of 1.4 A flows for 141 s. Find transferred charge. Round to 3 decimal places if needed. Answer in C.
A model device uses constant power 575 W for 11 h. Find the energy in kilowatt-hours. Round to 3 decimal places if needed. Answer in kWh.
Answer key
- 684 C 12 × 57 = 684
- 72 V 8 × 9 = 72
- 121 C Q = It 1 × 121 = 121
- 8.325 kWh E(kWh) = P(W) × t(h)/1000 925 × 9 / 1000 = 8.325
- 1.415 A I = V/R 58 / 41 = 1.415
- Series circuit Series circuit: A circuit arrangement with components connected along one current path.
- Coulomb An ampere is coulombs per second, not charge itself.
- potential difference
- 1065 C 15 × 71 = 1065
- 126 V 3 × 42 = 126
- 197.4 C Q = It 1.4 × 141 = 197.4
- 6.325 kWh E(kWh) = P(W) × t(h)/1000 575 × 11 / 1000 = 6.325
Review worksheet: 12 questions
An ideal ohmic resistor of 72 Ω carries 7 A. Find the voltage across it in V.
A constant current of 1 A flows for 78 s. How much charge passes in C?
An ohmic resistor of 32 Ω has potential difference 7 V. Find current. This is a paper calculation, not a construction instruction. Round to 3 decimal places if needed. Answer in A.
A constant current of 2.8 A flows for 41 s. Find transferred charge. Round to 3 decimal places if needed. Answer in C.
Waves and electric circuits: which term fits this description? Energy transferred per unit charge between two points.
- Open circuit
- Potential difference
- Transverse wave
- Reflection
A model device uses constant power 750 W for 7 h. Find the energy in kilowatt-hours. Round to 3 decimal places if needed. Answer in kWh.
Why must a simple steady current model have a complete conducting path?
Adding a positive-resistance parallel branch to an existing positive resistance makes equivalent resistance
- always larger
- smaller
- unchanged in all cases
- negative
An ideal ohmic resistor of 27 Ω carries 9 A. Find the voltage across it in V.
A constant current of 14 A flows for 27 s. How much charge passes in C?
An ohmic resistor of 70 Ω has potential difference 44 V. Find current. This is a paper calculation, not a construction instruction. Round to 3 decimal places if needed. Answer in A.
A constant current of 0.5 A flows for 103 s. Find transferred charge. Round to 3 decimal places if needed. Answer in C.
Answer key
- 504 V 7 × 72 = 504
- 78 C 1 × 78 = 78
- 0.219 A I = V/R 7 / 32 = 0.219
- 114.8 C Q = It 2.8 × 41 = 114.8
- Potential difference Potential difference: Energy transferred per unit charge between two points.
- 5.25 kWh E(kWh) = P(W) × t(h)/1000 750 × 7 / 1000 = 5.25
- A sustained current needs a circuit path and source conditions, not just one isolated connection. Accept elementary closed-loop reasoning without unsafe construction steps.
- smaller
- 243 V 9 × 27 = 243
- 378 C 14 × 27 = 378
- 0.629 A I = V/R 44 / 70 = 0.629
- 51.5 C Q = It 0.5 × 103 = 51.5
LS Life Sciences
Plant biology
Plant parts, what plants need, photosynthesis and plant reproduction.
Short lesson: Photosynthesis and matter sources
Study guide for Grades 7 to 10: 4 key ideas, including process, carbon and chloroplasts; 2 worked examples and 3 practice questions. Trace carbon as well as energy.
- Process. Photosynthesis uses light energy to build energy-rich organic compounds from simpler starting materials.
- Carbon. In the familiar plant model, carbon dioxide supplies carbon for sugar production.
- Chloroplasts. Photosynthetic machinery is located in chloroplasts in many plant cells.
- Whole plant. Plants also respire; photosynthesis is not their only energy-related process.
Worked example
Where does much of the carbon in newly made plant sugar come from?
CO₂ supplies carbon in the photosynthesis model.
Answer: Carbon dioxide.
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
In supplied data, 91 of 94 seeds germinated under the stated test conditions. Find the sample germination percentage. Round to 3 decimal places if needed. Answer in %.
Which comparison can test a light-related effect more fairly?
- Different species with different water, soil and age at once.
- One leaf drawn twice.
- Comparable plants differing in light exposure while other conditions are controlled.
What are stomata?
- Small chromosomes.
- Adjustable pores involved in leaf gas exchange.
- Only roots.
Which event is pollination?
- Every seed germinating.
- Pollen reaching a receptive stigma.
- Roots absorbing water.
In supplied data, 6 of 126 seeds germinated under the stated test conditions. Find the sample germination percentage. Round to 3 decimal places if needed. Answer in %.
Why does an oxygen-bubble count give only an indirect measure of photosynthesis?
Which tissue is associated with water and mineral transport?
- Only phloem in every case.
- Only pollen.
- Xylem.
Why is a seed not the same thing as pollen?
In supplied data, 26 of 194 seeds germinated under the stated test conditions. Find the sample germination percentage. Round to 3 decimal places if needed. Answer in %.
A simplified equation uses 6 CO₂ to represent formation of one glucose molecule. How many carbon atoms are supplied?
Why is a potometer's water-uptake reading not exactly the same as water lost by transpiration?
What does dispersal describe?
- Movement of seeds or other reproductive units away from their source.
- Only DNA copying.
- Always gamete fusion.
Answer key
- 96.809 % percentage = germinated/tested × 100 91/94 × 100 = 96.809
- Comparable plants differing in light exposure while other conditions are controlled. Control other influential variables.
- Adjustable pores involved in leaf gas exchange. Stomata connect internal leaf spaces with the atmosphere.
- Pollen reaching a receptive stigma. Pollination is pollen transfer, not fertilization itself.
- 4.762 % percentage = germinated/tested × 100 6/126 × 100 = 4.762
- Bubble size, dissolved gas and other processes can affect the observation. Treat a proxy as limited evidence, not an exact molecule count.
- Xylem. Xylem provides a water-transport pathway.
- A seed contains an embryo after reproductive development; pollen has a different role in reproduction. Keep the biological stages distinct.
- 13.402 % percentage = germinated/tested × 100 26/194 × 100 = 13.402
- 6 Each CO₂ contains one carbon atom.
- Some absorbed water is retained or used by the plant, and the apparatus has measurement limits. Treat uptake as an estimate rather than an identity.
- Movement of seeds or other reproductive units away from their source. Dispersal concerns location changes.
Review worksheet: 12 questions
In supplied data, 162 of 164 seeds germinated under the stated test conditions. Find the sample germination percentage. Round to 3 decimal places if needed. Answer in %.
Can the direction of phloem transport be described as always downward in every plant?
- Yes, without exceptions.
- No.
- Only if all leaves are green.
Must pollination always be carried out by insects?
- No; wind and other agents can be involved.
- Only if a flower is yellow.
- Yes, all plants use bees.
Why is saying plants get all their food from soil misleading?
In supplied data, 152 of 193 seeds germinated under the stated test conditions. Find the sample germination percentage. Round to 3 decimal places if needed. Answer in %.
Why might greater humidity reduce transpiration under otherwise similar conditions?
A sample has 40 seeds and 30 germinate under the stated test conditions. Find the germination percentage.
What should be included when reporting a photosynthesis investigation?
- Only the expected conclusion.
- Conditions, method, measurements and uncertainty.
- Only a picture with no labels.
In supplied data, 20 of 176 seeds germinated under the stated test conditions. Find the sample germination percentage. Round to 3 decimal places if needed. Answer in %.
A model shoot loses 8 g of water in 4 hours. Find the average mass-loss rate.
Which event is fertilization?
- Fusion of gametes.
- Only petals opening.
- Only wind moving a seed.
Where does much of the carbon in newly made plant sugar come from?
- Only mineral particles from soil.
- Sunlight atoms.
- Carbon dioxide.
Answer key
- 98.78 % percentage = germinated/tested × 100 162/164 × 100 = 98.78
- No. Movement is from sources to sinks, whose locations vary.
- No; wind and other agents can be involved. Pollination mechanisms vary.
- Plants generally make sugars using carbon dioxide and water, while soil supplies water and mineral nutrients. Distinguish organic food molecules from mineral resources.
- 78.756 % percentage = germinated/tested × 100 152/193 × 100 = 78.756
- It reduces the water-vapor gradient from leaf air spaces to the surrounding air. State the gradient explanation and that other conditions matter.
- 75 % 30/40 × 100 = 75%.
- Conditions, method, measurements and uncertainty. Interpretation depends on the measurement context.
- 11.364 % percentage = germinated/tested × 100 20/176 × 100 = 11.364
- 2 g/hour 8/4 = 2 g per hour.
- Fusion of gametes. Fertilization combines genetic material through gamete fusion.
- Carbon dioxide. CO₂ supplies carbon in the photosynthesis model.
LS Life Sciences
Cells & microscopy
Cells, organelles, microscopes and cell division.
Short lesson: Cells and their components
Study guide for Grades 5 to 8: 4 key ideas, including cell, membrane and nucleus; 2 worked examples and 3 practice questions. Match structures to functions without implying every cell is identical.
- Cell. Cells are basic structural and functional units of living organisms.
- Membrane. A cell membrane helps regulate exchange with the surroundings.
- Nucleus. A typical eukaryotic nucleus contains most of the cell’s DNA; prokaryotes lack a membrane-bound nucleus.
- Organelles. Mitochondria, chloroplasts and vacuoles have different roles and are not found in the same form in every cell.
Worked example
Which structure forms a selective boundary around a cell?
The membrane regulates exchange with the surroundings.
Answer: Cell membrane.
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
Supplied plant-tissue data show initial mass 6.4 g and final mass 8.7 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.
A cell is modeled as a cube of side 18 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.
A printed image is 59 mm long at 400× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.
Which statement is most accurate?
- Every cell has the same shape and number of organelles.
- Different cell types can have different structures and functions.
- Cells contain no chemical reactions.
Which term means distinguishing two nearby points?
- Only magnification.
- Specimen color.
- Resolution.
One side of a simplified model has 30 solute units in 10 volume units. Find its concentration.
Supplied plant-tissue data show initial mass 12.8 g and final mass 13.8 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.
A cell is modeled as a cube of side 19 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.
A printed image is 67 mm long at 400× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.
Do all cells have a membrane-bound nucleus?
- Only all bacteria do.
- No; prokaryotic cells do not.
- Yes, without exceptions.
An object is 20 µm long and its image is 2000 µm long. Find magnification.
Why cannot active transport be identified from movement direction alone without knowing the relevant gradient?
Answer key
- 35.937 % (final − initial)/initial × 100 (8.7 − 6.4)/6.4 × 100 = 35.937
- 0.333 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/18 = 0.333
- 147.5 µm actual size (µm) = image size (mm) × 1000 / magnification 59 × 1000/400 = 147.5
- Different cell types can have different structures and functions. Specialization produces structural and functional differences.
- Resolution. Resolution concerns detail separation.
- 3 30/10 = 3 solute units per volume unit.
- 7.813 % (final − initial)/initial × 100 (13.8 − 12.8)/12.8 × 100 = 7.813
- 0.316 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/19 = 0.316
- 167.5 µm actual size (µm) = image size (mm) × 1000 / magnification 67 × 1000/400 = 167.5
- No; prokaryotic cells do not. A membrane-bound nucleus is a eukaryotic feature.
- 100 2000/20 = 100 times.
- The concentrations, charges and membrane conditions determine whether movement is uphill. For ions, the electrical contribution may matter as well.
Review worksheet: 12 questions
Supplied plant-tissue data show initial mass 6.3 g and final mass 5.6 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.
A printed image is 47 mm long at 400× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.
A cell is modeled as a cube of side 34 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.
A model tissue has 5 equal rows of 6 cells. How many cells are modeled?
Why can digitally enlarging an image fail to show new cellular detail?
Which process directly requires energy coupling when moving a solute uphill?
- Simple passive diffusion.
- Active transport.
- Unrestricted random mixing at equilibrium.
Supplied plant-tissue data show initial mass 9.8 g and final mass 11.2 g. Find signed percentage mass change. No experiment is required. Round to 3 decimal places if needed. Answer in %.
A printed image is 57 mm long at 40× magnification. Find the actual length in micrometers. Round to 3 decimal places if needed. Answer in µm.
A cell is modeled as a cube of side 38 µm. Find surface area divided by volume. This geometry is a simplified model. Round to 3 decimal places if needed. Answer in µm⁻¹.
Which structure makes proteins from amino acids using genetic instructions?
- Starch grain.
- Ribosome.
- Cell wall only.
Why must image size and actual size use matching units in a magnification ratio?
- Otherwise the numerical ratio is wrong.
- Only for colored pictures.
- Because units have no meaning.
Osmosis specifically describes net movement of what in these examples?
- Water across a selectively permeable membrane.
- Any solid object.
- Only DNA through a nucleus.
Answer key
- -11.111 % (final − initial)/initial × 100 (5.6 − 6.3)/6.3 × 100 = -11.111
- 117.5 µm actual size (µm) = image size (mm) × 1000 / magnification 47 × 1000/400 = 117.5
- 0.176 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/34 = 0.176
- 30 5 × 6 = 30.
- It may enlarge existing pixels without improving the captured resolution. Bigger does not necessarily mean more information.
- Active transport. The unfavorable movement is coupled to an energy source.
- 14.286 % (final − initial)/initial × 100 (11.2 − 9.8)/9.8 × 100 = 14.286
- 1425 µm actual size (µm) = image size (mm) × 1000 / magnification 57 × 1000/40 = 1425
- 0.158 µm⁻¹ SA/V = 6s²/s³ = 6/s for a cube 6/38 = 0.158
- Ribosome. Ribosomes carry out translation.
- Otherwise the numerical ratio is wrong. Magnification is a dimensionless ratio of comparable lengths.
- Water across a selectively permeable membrane. The moving substance is water, and membrane selectivity matters.
LS Life Sciences
Ecology & populations
Ecosystems, food webs, populations and biodiversity.
Short lesson: Food-web transfer model
Study guide for Grades 7 to 11: 2 worked examples and 2 practice questions. Convert the supplied transfer efficiency to a fraction. Do not assume that all food webs have a fixed 10% efficiency.
transferred energy = available energy × stated efficiency/100
Convert the supplied transfer efficiency to a fraction. Do not assume that all food webs have a fixed 10% efficiency.
Worked example
A supplied food-web model transfers 10% of 524 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
transferred energy = available energy × stated efficiency/100 524 × 10/100 = 52.4
Answer: 52.4 kJ
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
A fictional population starts at 221. During one interval: 77 births, 38 arrivals, 30 deaths and 38 departures. Find the final count. Round to 3 decimal places if needed.
A supplied food-web model transfers 13% of 92 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
A supplied survey counts 37 plants in 11 non-overlapping quadrats, each 2 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².
Why should season be recorded in a field survey?
Why can removing one species affect more than one part of a food web?
A fictional population starts at 812. During one interval: 66 births, 9 arrivals, 34 deaths and 32 departures. Find the final count. Round to 3 decimal places if needed.
A supplied food-web model transfers 5% of 1123 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
A supplied survey counts 192 plants in 11 non-overlapping quadrats, each 1 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².
Using that mean of 5 plants/m² for a comparable 20 m² area, estimate the plant count.
Which is a primary producer in this simple land food chain?
- Grass.
- Rabbit.
- Fox.
A fictional population starts at 109. During one interval: 20 births, 25 arrivals, 23 deaths and 26 departures. Find the final count. Round to 3 decimal places if needed.
A supplied food-web model transfers 5% of 262 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
Answer key
- 268 N₂ = N₁ + births + immigration − deaths − emigration 221 + 77 + 38 − 30 − 38 = 268
- 11.96 kJ transferred energy = available energy × stated efficiency/100 92 × 13/100 = 11.96
- 1.682 plants/m² density = total count / total sampled area 37/(11 × 2) = 1.682
- Presence, activity and abundance can change with season. A time-specific survey may not represent all periods.
- Multiple organisms are linked through feeding and other interactions. Trace at least one connected consequence.
- 821 N₂ = N₁ + births + immigration − deaths − emigration 812 + 66 + 9 − 34 − 32 = 821
- 56.15 kJ transferred energy = available energy × stated efficiency/100 1123 × 5/100 = 56.15
- 17.455 plants/m² density = total count / total sampled area 192/(11 × 1) = 17.455
- 100 5 × 20 = 100 under the representativeness assumption.
- Grass. Grass produces organic matter using photosynthesis.
- 105 N₂ = N₁ + births + immigration − deaths − emigration 109 + 20 + 25 − 23 − 26 = 105
- 13.1 kJ transferred energy = available energy × stated efficiency/100 262 × 5/100 = 13.1
Review worksheet: 12 questions
A fictional population starts at 771. During one interval: 22 births, 21 arrivals, 45 deaths and 5 departures. Find the final count. Round to 3 decimal places if needed.
A supplied food-web model transfers 22% of 1095 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
A supplied survey counts 23 plants in 10 non-overlapping quadrats, each 0.25 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².
Why is a food web often more realistic than one chain?
- Organisms can have multiple feeding relationships.
- Every predator eats exactly one species.
- Only one organism exists in nature.
Four equal 1 m² quadrats contain 2, 4, 6 and 8 plants. Find the mean per m².
A fictional population starts at 838. During one interval: 52 births, 34 arrivals, 8 deaths and 29 departures. Find the final count. Round to 3 decimal places if needed.
A supplied food-web model transfers 15% of 1188 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
A supplied survey counts 125 plants in 3 non-overlapping quadrats, each 1 m². Find the mean density in the sampled area. Round to 3 decimal places if needed. Answer in plants/m².
What is a decomposer's role in a simple nutrient model?
- Breaking down organic material and helping recycle nutrients.
- Stopping every chemical reaction.
- Creating new atoms from nothing.
A sample estimate is 120 while a later complete count is 100. What is the absolute error?
A fictional population starts at 159. During one interval: 69 births, 17 arrivals, 13 deaths and 3 departures. Find the final count. Round to 3 decimal places if needed.
A supplied food-web model transfers 7% of 640 kJ to the next level. Find transferred energy. The stated efficiency is not a universal ecological constant. Round to 3 decimal places if needed. Answer in kJ.
Answer key
- 764 N₂ = N₁ + births + immigration − deaths − emigration 771 + 22 + 21 − 45 − 5 = 764
- 240.9 kJ transferred energy = available energy × stated efficiency/100 1095 × 22/100 = 240.9
- 9.2 plants/m² density = total count / total sampled area 23/(10 × 0.25) = 9.2
- Organisms can have multiple feeding relationships. A web allows connected pathways.
- 5 The total 20 divided by four is 5 plants/m².
- 887 N₂ = N₁ + births + immigration − deaths − emigration 838 + 52 + 34 − 8 − 29 = 887
- 178.2 kJ transferred energy = available energy × stated efficiency/100 1188 × 15/100 = 178.2
- 41.667 plants/m² density = total count / total sampled area 125/(3 × 1) = 41.667
- Breaking down organic material and helping recycle nutrients. Decomposition transfers matter into reusable forms.
- 20 The absolute difference is 20.
- 229 N₂ = N₁ + births + immigration − deaths − emigration 159 + 69 + 17 − 13 − 3 = 229
- 44.8 kJ transferred energy = available energy × stated efficiency/100 640 × 7/100 = 44.8
ESS Earth and Space Sciences
Earth & space
The Sun, Moon and planets, day and night, the water cycle and Earth's layers.
Short lesson: Earth systems
Study guide for Grades 4 to 9: 20 key ideas, including crust, mantle and outer core; 3 practice questions.
Weathering breaks material down; erosion transports it. These processes can happen together, but they are not the same.
- Crust. Earth's thin outermost solid rocky layer.
- Mantle. The thick rocky region between Earth's crust and core.
- Outer core. The mostly liquid metallic layer surrounding Earth's inner core.
- Inner core. Earth's solid central metallic region.
Worksheets
A practice worksheet and a review worksheet, 12 and 12 questions.
Practice worksheet: 12 questions
Space and the solar system: which term fits this description? The fourth planet from the Sun.
- Uranus
- Mars
- Lunar eclipse
- Comet
When the Northern Hemisphere tilts toward the Sun, the Southern Hemisphere generally has
- winter
- no gravity
- the same summer at the same time
- no daylight anywhere
Earth systems: which term fits this description? The breakdown of rock in place at or near Earth's surface.
- Transpiration
- Weathering
- Water cycle
- Outer core
How could paving a previously grassy area change the route taken by rainwater?
Must every rock pass through the same cycle steps in the same order?
- Yes, exactly once each.
- Only if it contains quartz.
- No.
What additional information is needed before attributing a change in a stream’s level to rainfall alone?
Why should a solar-angle diagram state the location and time of year?
Space and the solar system: which term fits this description? The second planet from the Sun.
- Venus
- Saturn
- Milky Way
- Asteroid
Waxing describes a visible illuminated portion that is
- increasing
- always absent
- permanently frozen
- decreasing
Earth systems: which term fits this description? Rock changed by heat, pressure or fluids without fully melting.
- Crust
- Metamorphic rock
- Magma
- Weather
A puddle disappears on a dry day. Does that mean its water no longer exists? Explain.
What can transform existing rock without fully melting it?
- Removing every atom.
- Heat and pressure during metamorphism.
- Only writing a new label.
Answer key
- Mars Mars: The fourth planet from the Sun.
- winter The hemispheres receive different seasonal sunlight conditions.
- Weathering Weathering: The breakdown of rock in place at or near Earth's surface.
- It may reduce infiltration and increase or redirect surface runoff, depending on drainage. Avoid claiming that all rain follows one route.
- No. The cycle shows alternative pathways.
- Other relevant factors can include upstream flow, storage, evaporation, withdrawals, snowmelt and measurement timing.
- The geometry depends on latitude and orbital position. A single drawing does not describe every location simultaneously.
- Venus Venus: The second planet from the Sun.
- increasing Waxing refers to the growing lit fraction seen from Earth.
- Metamorphic rock Metamorphic rock: Rock changed by heat, pressure or fluids without fully melting.
- No. Much may have evaporated or infiltrated, moving to another place or state. Credit conservation and a plausible pathway.
- Heat and pressure during metamorphism. Metamorphic change occurs under changed conditions without complete melting in the basic distinction.
Review worksheet: 12 questions
Earth systems: which term fits this description? Earth's solid central metallic region.
- Inner core
- Transpiration
- Water cycle
- Precipitation
Space and the solar system: which term fits this description? A large gravitationally bound system of stars, gas, dust and dark matter.
- Mars
- Galaxy
- Revolution
- Jupiter
What is weathering?
- All melting deep in Earth.
- Breakdown or alteration of material at or near the surface.
- Only transport by a river.
At full Moon, which description is approximately correct?
- The Moon's near side appears largely sunlit.
- Earth's shadow covers the visible Moon.
- The Moon has no far side.
- The Moon creates its own sunlight.
Water flowing over the land after rain is
- a magnetic field
- a phase of the Moon
- runoff
- a chemical element
Should a model use a light beam as a safe instruction to look directly at the Sun?
- Only at midday.
- Yes, for better detail.
- No.
Liquid water becoming water vapour is
- condensation
- sedimentation
- evaporation
- melting
Earth systems: which term fits this description? Rock formed as magma or lava cools and solidifies.
- Transpiration
- Plate tectonics
- Igneous rock
- Runoff
Space and the solar system: which term fits this description? An event in which the Moon blocks our view of part or all of the Sun.
- Moon
- Uranus
- Solar eclipse
- Galaxy
A fictional sediment layer is 2 cm thicker each year for 5 years under a stated model. What added thickness is predicted?
Which is a sound safety rule for a classroom Sun discussion?
- Never look directly at the Sun without appropriate certified protection and expert guidance.
- Use binoculars with no filter.
- Look longer when it is bright.
- Stare at it through ordinary sunglasses.
Why is the water cycle better shown as connected pathways than one single circle?
Answer key
- Inner core Inner core: Earth's solid central metallic region.
- Galaxy Galaxy: A large gravitationally bound system of stars, gas, dust and dark matter.
- Breakdown or alteration of material at or near the surface. Weathering differs from the transport commonly called erosion.
- The Moon's near side appears largely sunlit. We see most of the illuminated near side.
- runoff Runoff travels across the surface.
- No. Never look directly at the Sun for this activity; use diagrams or safe supplied data.
- evaporation
- Igneous rock Igneous rock: Rock formed as magma or lava cools and solidifies.
- Solar eclipse Solar eclipse: An event in which the Moon blocks our view of part or all of the Sun.
- 10 cm 2 × 5 = 10 cm; real deposition need not be constant.
- Never look directly at the Sun without appropriate certified protection and expert guidance. Unprotected solar viewing can damage eyes; use safe indirect classroom models instead.
- Water can be stored in many places and move through different routes, including groundwater and ice. Credit multiple possible pathways or different storage times.