Integrated science — apply the concepts together

How to attempt these cases

The following problems combine ideas from more than one chapter. They are constructed-response exercises, separate from the 650 chapter MCQs. Cover the solution paragraph, write the physical quantity or biological process being tested, and attempt the calculation or explanation. Numerical values are invented teaching data unless explicitly identified as a standard constant. A correct number with the wrong unit is not a complete answer. For experimental cases, distinguish what was measured from the interpretation placed on that measurement.

Case 1 — A pump, an elevated tank and energy losses

A pump lifts 600 kg of water through a vertical height of 8 m in four minutes. Use g = 10 m/s² and ignore changes in speed. Its electrical input power is 300 W. Calculate the gravitational potential energy gained, useful output power and efficiency. Then explain why the useful mechanical output is lower than the electrical input even though energy is conserved.

The potential-energy increase is mgh = 600 × 10 × 8 = 48,000 J. Four minutes is 240 s, so useful power is 48,000/240 = 200 W. Efficiency is 200/300 × 100, approximately 66.7%. The difference corresponds to other energy transfers, including heating and mechanical losses. Conservation concerns all transfers, whereas efficiency compares the output defined as useful with the input. An answer of 12,000 W would have divided by four minutes as if four were seconds. Writing the time conversion before substitution prevents that error.

Case 2 — A lamp heats a photosynthesis experiment

Two identical aquatic-plant setups begin at the same temperature. A lamp is moved closer to one setup, and its observed bubble rate increases. A student concludes that doubling light intensity necessarily doubled photosynthesis. Identify two problems with the conclusion and propose a stronger comparison using the same general setup.

First, moving a lamp changes distance; it does not establish that the relevant light intensity at the leaf exactly doubled without measurement or a justified model. Second, the nearer lamp may heat the water, changing a second variable. Bubble number is also an approximate gas-release measure because bubble volumes vary. A stronger design measures or controls temperature, records an appropriate light measure, standardises plant material and observation time, and measures gas volume where practicable. Repeated observations test variability. The result may support an effect of light under those conditions, but its exact magnitude requires the measurements that the original conclusion omitted.

Case 3 — Electrical heating of water

An idealised 1,000 W heater supplies energy to 0.5 kg of water initially at 20°C. Take specific heat capacity as 4,200 J/(kg·°C). Calculate the minimum time to reach 60°C if all supplied energy heats the water and no phase change occurs. State how heat loss and heating the container would affect the required time in a real apparatus.

Temperature rise is 40°C, so Q = mcΔT = 0.5 × 4,200 × 40 = 84,000 J. Time is energy divided by power: 84,000/1,000 = 84 s. The real time would be longer if part of the energy warms the container or escapes to the surroundings under otherwise comparable conditions. The heater's power is energy per second, not its total energy use. This example links heat capacity with electrical power; it does not require the heater's resistance unless voltage or current is also being asked.

Case 4 — A changing circuit and brightness

Two identical ideal resistors, each 12 ohms, are connected first in series and then in parallel across the same ideal 12 V supply. Find the total current in each arrangement and the power in each resistor. Use the result to explain why “adding a branch always reduces current everywhere” is an unreliable statement.

In series, total resistance is 24 ohms and current is 12/24 = 0.5 A. Each resistor dissipates I²R = 0.25 × 12 = 3 W. In parallel, each branch receives 12 V, so each carries 1 A and dissipates V²/R = 144/12 = 12 W. Total supply current is 2 A. The parallel arrangement reduces total resistance and increases total current for this ideal constant-voltage supply. Real lamps may have temperature-dependent resistance, so the numerical result belongs to the ideal resistors stated. Identify what is held constant before extending a circuit rule.

Case 5 — Concentration is not the same as total load

Sample A contains 80 L of water with a dissolved substance at 5 mg/L. Sample B contains 500 L at 1 mg/L. Which has the higher concentration, and which contains more total dissolved substance? If both are mixed without reaction or volume change, find the resulting concentration.

Sample A has the higher concentration. Its total amount is 80 × 5 = 400 mg, while B contains 500 × 1 = 500 mg, so B has the larger load. Combined amount is 900 mg and volume is 580 L. Concentration is 900/580, approximately 1.55 mg/L. Averaging 5 and 1 to obtain 3 mg/L would be wrong because the volumes differ. The same distinction appears when comparing pollutant levels, nutrient solutions and laboratory mixtures. A measured concentration alone does not reveal the total amount discharged without a volume or flow measurement.

Case 6 — Evaporation, salt and water potential

A beaker contains 10 g of a nonvolatile salt dissolved in 190 g of water. Fifty grams of water evaporate without salt loss. Calculate the initial and final mass percentages of salt. Predict the direction in which this change would tend to alter water movement from an otherwise identical cell placed in the solution, assuming the salt does not cross its membrane freely.

Initial solution mass is 200 g and salt percentage is 10/200 × 100 = 5%. Final water mass is 140 g, so solution mass is 150 g and concentration is 10/150 × 100, approximately 6.67%. Evaporation concentrates the external solution. Other factors being equal, this favours water loss from a cell more strongly than the original solution did. It does not establish whether the cell actually loses water without knowing the internal condition and relevant pressure effects. Chemistry supplies the concentration change; Biology supplies the membrane comparison and limits of the inference.

Case 7 — A balanced equation and a limiting reactant

Use the equation 2H₂ + O₂ → 2H₂O. A sealed teaching model begins with six hydrogen molecules and two oxygen molecules. Assume the reaction proceeds until one reactant is exhausted. How many water molecules form, and what remains? Explain why the total number of molecules changes without violating conservation of atoms.

Two oxygen molecules require four hydrogen molecules and produce four water molecules. Oxygen is exhausted and two hydrogen molecules remain. Initially there are twelve hydrogen atoms and four oxygen atoms; finally the four water molecules contain eight hydrogen atoms and four oxygen atoms, while the remaining hydrogen molecules contain four more hydrogen atoms. Atom counts are conserved. Molecules have been rearranged, so their count need not remain fixed. The coefficients describe particle ratios, not equal masses of all substances. This is a molecular-counting model, not an instruction to carry out the reaction.

Case 8 — A falling ruler and experimental uncertainty

Two ruler-drop distances are 0.15 m and 0.25 m. Use g = 10 m/s² and the free-fall relation d = gt²/2 to estimate the associated response times. Does the larger distance mean a faster response? Can one pair of trials establish a permanent difference between two people?

The first time is the square root of 0.03, approximately 0.173 s. The second is the square root of 0.05, approximately 0.224 s. A larger fall distance corresponds to a longer time, hence a slower response in this model. One pair of trials cannot establish a permanent difference because anticipation, attention, technique and measurement vary. The square-root relation also means that doubling distance does not double time. This example connects mechanics with nervous-system response while keeping the classroom measurement distinct from direct measurement of nerve-conduction speed.

Case 9 — Microscope images and comparable scale

A plant cell appears 24 mm long in an image labelled 600×. An animal cell appears 18 mm long in another image labelled 300×. Which represented cell is actually longer? Why would comparison of the printed lengths alone give the wrong conclusion?

The plant cell's represented length is 24/600 = 0.04 mm, or 40 micrometres. The animal cell's represented length is 18/300 = 0.06 mm, or 60 micrometres. The animal cell is longer in this example despite having the smaller printed image. The images use different magnifications. This does not establish a universal plant–animal size rule, since cells vary greatly by type. When an image is resized, a scale bar included in the same proportional transformation is especially useful. Both image size and magnification must be known before recovering the represented size.

Case 10 — Ideal DNA amplification and contamination

A PCR teaching model begins with four copies of a target and assumes perfect doubling for six cycles. Calculate the final number. A negative control from the same run also produces the target signal. Explain why the numerical amplification model does not resolve the control problem.

Ideal final count is 4 × 2⁶ = 256 copies. This describes amplification under the model, not the origin of the starting material. An unexpected target signal in a negative control raises concern about contamination or another process problem. Copying an unwanted target efficiently does not make it valid case evidence. The run needs investigation before its sample results are relied upon. This distinction links exponential arithmetic with laboratory quality: mathematical amplification and evidential reliability answer different questions. Real PCR also departs from ideal doubling, so 256 is a model answer rather than a guaranteed laboratory yield.

Case 11 — A genetic ratio and small samples

Two heterozygous parents Aa are crossed under complete dominance and ordinary independent offspring outcomes. Calculate the probability that two offspring both show the dominant phenotype, and the probability that exactly one shows the recessive phenotype. Does observing two dominant offspring prove either parent was AA?

Each offspring has dominant-phenotype probability 3/4. Both dominant therefore has probability 3/4 × 3/4 = 9/16. Exactly one recessive has two orders, giving 2 × 1/4 × 3/4 = 6/16, or 3/8. Two dominant offspring do not prove either parent was AA; the Aa × Aa model itself gives that observation substantial probability. Genotype, phenotype and sampling variation must be kept separate. A ratio such as 3:1 describes expected proportions over repeated outcomes, not a compulsory sequence of four offspring. The assumptions are biological as well as mathematical.

Case 12 — Oxygen exchange at leaf and pond scale

A leaf has gross photosynthetic oxygen production of 18 units per hour and respiratory use of 6 units per hour in a simplified model. Calculate net release. A nearby pond has dense algal growth but low dissolved oxygen before sunrise. Explain why the leaf calculation does not justify assuming the pond must always be oxygen-rich.

Net leaf release is 18 − 6 = 12 units per hour under the specified conditions. The pond contains many organisms and decomposing material, all contributing to oxygen use. In darkness, light-dependent photosynthetic production does not continue at the daytime rate, while respiration continues. Gas exchange, temperature, mixing and decomposition also affect dissolved oxygen. One illuminated leaf's net rate cannot represent a whole pond at every time. The numerical example is local and conditional; the ecosystem explanation requires the complete set of gains and losses across the chosen boundary.

Case 13 — Cardiac output and circuit direction

A fictional heart-rate measurement is 80 beats per minute and stroke volume is 65 mL per beat. Find cardiac output in litres per minute. Then trace the blood leaving the right ventricle to the left atrium and explain why multiplying output by two for two ventricles would misrepresent the usual circulation.

Output is 80 × 65 = 5,200 mL per minute, or 5.2 L per minute for one ventricle. Blood leaving the right ventricle travels through pulmonary outflow to the lungs and returns by pulmonary veins to the left atrium. The circuits are connected in sequence, with approximately matched average output in a stable circulation. Adding the outputs as though they supplied two separate whole-body streams would double-count the flow around the circuit. The example tests units and anatomy; it does not define a clinical target for an individual.

Case 14 — Filtration and mass balance

Over a stated interval, a model nephron filters 150 units of substance X, reabsorbs 120 and secretes 10 into the tubule. Calculate excretion. Another substance Y is filtered at 50 units and entirely reabsorbed, with no secretion. Does zero excretion of Y prove it never crossed the initial filter?

For X, excretion = 150 − 120 + 10 = 40 units. For Y, excretion is 50 − 50 = 0. Zero final excretion does not imply zero filtration because a later process can return the substance to blood. This is a biological example of a general accounting principle: a final output may conceal opposing intermediate flows. Identify the direction associated with each term. Reabsorption subtracts from the tubular amount, while secretion adds to it. All values must cover the same interval and use the same quantity units.

Case 15 — Food-chain arithmetic with unequal efficiencies

A hypothetical producer level makes 40,000 energy units available in the stated model. Transfer to primary consumers is 15%, and transfer from primary to secondary consumers is 8%. Calculate the energy reaching each consumer level. Compare the final value with a calculation that incorrectly assumes 10% at every step.

Primary consumers receive 40,000 × 0.15 = 6,000 units. Secondary consumers receive 6,000 × 0.08 = 480 units. Assuming 10% at both steps would instead produce 400 units. The supplied efficiencies must take precedence over a memorised teaching approximation. Across both transfers, the combined fraction is 0.15 × 0.08 = 0.012, or 1.2%, not 23%. Successive fractions multiply; they do not add. This is the same mathematical structure as successive percentage changes, applied to ecological energy transfer under explicitly stated assumptions.

Case 16 — Choosing a denominator for an outbreak comparison

Town A records 24 new cases among 1,200 initially disease-free people during a specified month. Town B records 30 among 3,000 over the same period, using comparable case definitions and follow-up. Which has more new cases, and which has the higher cumulative incidence in this simplified comparison?

Town B has more cases in absolute number. Town A's cumulative incidence is 24/1,200 = 2%, while Town B's is 30/3,000 = 1%. Town A therefore has twice the proportion developing the condition in this example. This does not prove a particular environmental cause. Population structure, exposure and reporting still matter for causal interpretation. The calculation separates burden as a count from risk as a proportion under the stated assumptions. The two comparisons can point in different directions without either numerical answer being inconsistent.

Case 17 — A test that is accurate in two different senses

A fictional test is applied to 2,000 people. A reference method identifies 200 with the target condition. Sensitivity is 95% and specificity is 90%. Find true positives, false negatives, true negatives and false positives. Then calculate the fraction of positive results that are true positives.

Among the 200 with the condition, 190 test positive and 10 negative. Of the 1,800 without the condition, 1,620 test negative and 180 positive. There are 370 positive results, of which 190 are true positives, so positive predictive value is 190/370, approximately 51.4%. Neither 95% sensitivity nor 90% specificity directly answers that final question. Drawing a two-by-two table prevents the denominator error. All figures are synthetic; they do not state the performance of a named medical test or provide a diagnosis.

Case 18 — Mineral nutrients, dry mass and conservation

A seedling gains dry mass while growing in light with water and an adequate mineral supply. A student says all the added dry material must have come from soil minerals. Explain why that conclusion is incomplete. Identify a major source of carbon and distinguish the source of matter from the source of energy.

Carbon dioxide from the surroundings supplies carbon incorporated into organic material during photosynthesis. Water supplies matter too, while mineral nutrients provide essential elements in appropriate forms. Light supplies energy; it is not a bag of carbon atoms added to the plant. Conservation of matter requires tracing the chemical inputs, including gases that may be overlooked because they are invisible. Drying removes much water before weighing, but does not mean the remaining organic matter came entirely from soil. This case joins chemical accounting with plant physiology and explains why “food from soil” is an inadequate model.

Final error diagnosis

If a calculation is wrong, classify the mistake before repeating it. A unit error involves quantities such as minutes and seconds or millilitres and litres. A denominator error involves choosing the wrong population, mass, volume or total number of outcomes. A model error applies a formula outside its assumptions, such as treating every trophic transfer as exactly 10% or every genetic trait as completely dominant. A route error attaches the right substance to the wrong organ or vessel. A causal error treats a single association or uncontrolled observation as proof of one explanation.

Rewrite the failed step in one sentence and attempt a changed example. For an incorrect concentration average, use unequal volumes again and calculate total solute before dividing. For a genetic-ratio error, list possible gametes and distinguish phenotype from genotype. For a circulation error, trace the loop from the right ventricle and check direction before oxygen content. For an experiment error, name the variable changed, the outcome measured and one plausible confounding factor. This practice turns an answer key into a method that transfers to unfamiliar questions.