ACT Science Quiz: Interpreting Data From Tables
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Interpreting Data From TablesQuestion 1 of 20

PASSAGE VI

BIOLOGY / GENETICS: Research Summary

Introduction

Restriction enzymes are proteins that cut DNA molecules at specific, predictable sequences of base pairs (bp). Gel electrophoresis is a technique used to separate these resulting DNA fragments by size. When an electrical current is applied to the gel, the negatively charged DNA fragments migrate toward the positive electrode. Smaller DNA fragments move through the gel much faster and travel further than larger fragments.

A researcher isolated a circular bacterial plasmid (a ring of DNA) consisting of exactly 5,000 bp. To map the plasmid, the researcher treated identical samples of the plasmid with different restriction enzymes—Enzyme 1 (E1), Enzyme 2 (E2), and Enzyme 3 (E3)—both individually and in combinations.

After allowing the enzymes to cut the DNA, the researcher ran the samples on an electrophoresis gel. A dye was added to make the DNA bands visible. The size of the fragments in each band was recorded in Table 1.

Based on the passage, which of the following fragments from Table 1 would migrate the furthest distance from the starting well during electrophoresis?

Question graphic
The 5,000 bp fragment in Lane 2
The 3,000 bp fragment in Lane 3
The 2,000 bp fragment in Lane 5
The 500 bp fragment in Lane 5
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ACT Science Quiz

ACT Science Quiz: Interpreting Data From Tables

Practice Interpreting Data From Tables in ACT Science with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Interpreting Data From Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Science.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

PASSAGE VI

BIOLOGY / GENETICS: Research Summary

Introduction

Restriction enzymes are proteins that cut DNA molecules at specific, predictable sequences of base pairs (bp). Gel electrophoresis is a technique used to separate these resulting DNA fragments by size. When an electrical current is applied to the gel, the negatively charged DNA fragments migrate toward the positive electrode. Smaller DNA fragments move through the gel much faster and travel further than larger fragments.

A researcher isolated a circular bacterial plasmid (a ring of DNA) consisting of exactly 5,000 bp. To map the plasmid, the researcher treated identical samples of the plasmid with different restriction enzymes—Enzyme 1 (E1), Enzyme 2 (E2), and Enzyme 3 (E3)—both individually and in combinations.

After allowing the enzymes to cut the DNA, the researcher ran the samples on an electrophoresis gel. A dye was added to make the DNA bands visible. The size of the fragments in each band was recorded in Table 1.

Based on the passage, which of the following fragments from Table 1 would migrate the furthest distance from the starting well during electrophoresis?

  1. The 5,000 bp fragment in Lane 2
  2. The 3,000 bp fragment in Lane 3
  3. The 2,000 bp fragment in Lane 5
  4. The 500 bp fragment in Lane 5 (correct answer)
Explanation: The correct answer is D (the 500 bp fragment in Lane 5). The passage states that smaller fragments travel further during electrophoresis. Among all the fragments listed in Table 1 — 5,000 bp, 3,000 bp, 2,000 bp, 4,000 bp, 1,000 bp, 2,500 bp, and 500 bp — the smallest is 500 bp in Lane 5. The smallest fragment experiences the least resistance passing through the gel matrix, moves fastest under the electrical current, and therefore travels the greatest distance from the starting well. A (5,000 bp) is the largest fragment and would travel the least distance. B (3,000 bp) and C (2,000 bp) are intermediate sizes. On migration distance questions, identify the smallest fragment across all lanes in the table.

Question 2

PASSAGE VI

Fruit Fly Genetics

Introduction

In the fruit fly Drosophila melanogaster, eye color is a sex-linked trait determined by a gene on the X chromosome. The allele for the wild-type red eye color (XRX^R) is dominant, while the allele for the mutant white eye color (XrX^r) is recessive.

Females (XX): Inherit one X chromosome from each parent. A female will have white eyes only if she is homozygous recessive (XrXrX^r X^r).

Males (XY): Inherit an X chromosome from the mother and a Y chromosome from the father. Because the Y chromosome does not carry the eye color gene, a male expresses whichever allele is present on his single X chromosome (X^R Y \= Red; X^r Y \= White).

Students conducted two studies to observe these inheritance patterns.

Study 1

The students crossed a homozygous red-eyed female (XRXRX^R X^R) with a white-eyed male (XrYX^r Y). To predict the genotypes of the offspring, they constructed a Punnett square (Figure 1).

They then collected 100 offspring (the F1 generation) and recorded the results in Table 1.

Study 2

The students performed the reciprocal cross. They crossed a white-eyed female (XrXrX^r X^r) with a red-eyed male (XRYX^R Y). A second Punnett square was constructed to predict the outcome (Figure 2).

They collected 100 offspring and recorded the results in Table 2.

Based on Table 1, what percent of the F1 offspring in Study 1 had white eyes?

  1. 0% (correct answer)
  2. 25%
  3. 50%
  4. 100%
Explanation: This is a straightforward data retrieval and percentage calculation question. You can identify this question type by the phrase "what percent" combined with "based on Table 1." To solve this, look at Table 1 and find all offspring with white eyes: Male White Eyes = 0, Female White Eyes = 0. Total white-eyed offspring = 0 out of 100 total. Therefore, 0/100 = 0%. Choice B (25%) represents a typical Mendelian ratio but doesn't match the actual data. Choice C (50%) is another common genetic ratio. Choice D (100%) is the opposite extreme. These wrong answers might tempt students who are trying to recall genetics ratios from memory instead of actually reading the table. Remember: Always use the actual data provided, not what you think "should" happen based on genetic theory—real experimental data may show all offspring with one trait, especially in sex-linked crosses!

Question 3

A chemistry lab measured the pH of a buffer solution after adding different volumes of acid. Use Table 1 to answer the question.

When 6.0 mL of acid was added, the pH was:

  1. 6.55 (correct answer)
  2. 6.35
  3. 6.10
  4. 6.80
Explanation: When 6.0 mL of acid was added, the pH was 6.55. In Table 1, locate the row for 6.0 mL acid volume and read across to the pH column to find the value of 6.55. This shows how the buffer solution's pH decreases as more acid is added, but the change is gradual due to the buffering capacity.

Question 4

Researchers tested how water temperature affects the time needed for a tablet to dissolve in 200 mL of water. Each trial used the same tablet mass and the same stirring rate. The results are shown in Table 1.

According to Table 1, what was the dissolve time when the water temperature was 40 °C?

  1. 120 s
  2. 70 s
  3. 95 s
  4. 55 s (correct answer)
Explanation: Table 1 shows dissolve times for tablets at different water temperatures. To find the dissolve time at 40 °C, locate the row for 40 °C in the temperature column and read across to the dissolve time column, where the value is 55 s. This is correct because the table organizes data by increasing temperature, ensuring accurate lookup for the specific condition. A key distractor might be misreading the row for a nearby temperature like 30 °C, which could show 70 s instead.

Question 5

An environmental lab measured nitrate concentration in river water at different distances downstream from a wastewater outlet. Samples were collected the same day and analyzed with the same instrument. Table 1 reports the results.

At a distance of 5.0 km downstream, the nitrate concentration was:

  1. 10.1 mg/L
  2. 8.5 mg/L
  3. 6.2 mg/L (correct answer)
  4. 3.8 mg/L
Explanation: Table 1 reports nitrate concentrations at various distances downstream from a wastewater outlet. To find the concentration at 5.0 km, locate the row for 5.0 km in the distance column and read the corresponding value in the nitrate concentration column, which is 6.2 mg/L. This approach is correct as it directly intersects the specific distance with the measurement column for precise data retrieval. Misreading might occur by selecting the value from an adjacent row, such as 3.0 km showing 8.5 mg/L.

Question 6

PASSAGE VI

PHYSICS: Research Summary

Introduction

Engineers test model wind turbines in wind tunnels to determine how different design parameters affect electrical power output. The power output of a turbine (measured in milliwatts, mW) depends on the wind speed and the pitch angle of the blades. The pitch angle is the angle at which the blades are twisted relative to the oncoming wind.

Study 1

Engineers built a model turbine and placed it in a wind tunnel. They locked the blade pitch angle at a constant 15°. They then turned on the wind tunnel, varying the wind speed from 2.0 meters per second (m/s) to 10.0 m/s. In Table 1, they recorded the power output of the turbine at each wind speed.

Study 2

The engineers wanted to find the optimal pitch angle for the turbine blades. They set the wind tunnel to blow at a constant wind speed of 8.0 m/s. They then adjusted the blade pitch angle from 0° to 40° and recorded the resulting power output. Findings are shown in Table 2.

Based on Table 2, what is the relationship between blade pitch angle and power output? As the pitch angle increases from 0° to 40°, the power output:

  1. increases only.
  2. decreases only.
  3. increases, reaches a maximum, and then decreases. (correct answer)
  4. decreases, reaches a minimum, and then increases.
Explanation: The correct answer is C. Table 2 shows power output rising from 20.0 mW (0°) through 75.0 (10°), 100.0 (15°), and peaking at 115.0 mW (20°) before declining to 80.0 (30°) and 35.0 mW (40°). The data traces a clear bell-curve pattern — increasing to a maximum, then decreasing. A is wrong — the data clearly shows a decrease after 20°. B is wrong — the data increases through the first four data points. D is wrong — the data starts by increasing, not decreasing. Pro tip: Bell-curve relationships (increase then decrease) are common in biology and physics. Scan the full column from top to bottom before deciding on the direction.

Question 7

PASSAGE VII

PHYSICS: Data Representation

Introduction

A student investigated the relationship between voltage (VV), current (II), and resistance (RR) in a simple direct current (DC) electrical circuit.

Voltage (VV) is the electrical potential difference provided by a power source, measured in volts (V).

Current (II) is the rate of flow of electrical charge, measured in amperes (A).

Resistance (RR) is the opposition to the flow of charge, measured in ohms (Ω\Omega).

The student set up a circuit containing a variable voltage power supply, a resistor, and an ammeter (a device used to measure current).

Experiment 1

In the first experiment, the student used a resistor with a constant resistance of 10.0 Ω10.0 \ \Omega. The student varied the voltage supplied to the circuit from 2.0 V to 10.0 V and recorded the resulting current measured by the ammeter. Results are shown in Table 1.

Experiment 2

In the second experiment, the student set the power supply to provide a constant voltage of 12.0 V12.0 \text{ V}. The student then swapped out the resistor, testing five different resistors with varying resistance values, and recorded the resulting current for each. Results are shown in Table 2.

Based on Table 2, as the resistance in the circuit increases from 2.0 Ω to 12.0 Ω, the current in the circuit:

  1. increases only.
  2. decreases only. (correct answer)
  3. increases, then decreases.
  4. remains constant.
Explanation: The correct answer is B (decreases only). Table 2 shows current values of 6.00, 3.00, 2.00, 1.50, and 1.00 A as resistance increases from 2.0 to 12.0 Ω. Every step shows a decrease — no increase, no plateau. This is consistent with the inverse relationship between current and resistance described by Ohm's Law: as resistance increases, it becomes harder for charge to flow, so the current drops. A (increases only) would require current to rise as resistance increases — the opposite of what the table shows. C (increases then decreases) would require at least one upward step in the current column before the decrease. D (remains constant) would require identical current values throughout. On trend questions, check each consecutive pair of values in the dependent variable column to confirm direction.

Question 8

PASSAGE II

BIOLOGY: Research Summary

Soil salinity (salt concentration) and pH can significantly affect seed germination. A botanist conducted two studies to determine how these factors influence the germination rate of Medicago sativa (alfalfa) seeds. •Note: Germination rate is the percentage of planted seeds that successfully sprout.

Study 1

The botanist prepared 5 identical planting trays. Each tray was filled with 1 kilogram (kg) of the same potting soil. The botanist adjusted the soil in each tray to have a different concentration of sodium chloride (NaCl), measured in millimoles per kilogram (mM/kg). The soil pH for all trays was kept constant at 6.5.

Fifty M. sativa seeds were planted in each tray. The trays were placed in a greenhouse with a constant temperature of 25C25^\circ\text{C} and watered equally every day for 14 days. On day 14, the germination rate was recorded. Results are shown in Table 1.

Study 2

The botanist prepared 5 new trays with the same potting soil. This time, the NaCl concentration in all trays was kept constant at 40 mM/kg. The botanist adjusted the soil pH in each tray to a different value, ranging from highly acidic to highly basic.

Fifty M. sativa seeds were planted in each tray. The greenhouse conditions, watering schedule, and duration were identical to those in Study 1. Results are shown in Table 2.

According to the results of Study 1, as the NaCl concentration increased from 0 mM/kg to 160 mM/kg, the germination rate of M. sativa seeds:

  1. increased only.
  2. decreased only. (correct answer)
  3. increased, then decreased.
  4. remained constant.
Explanation: The correct answer is B (decreased only). Study 1's table shows a consistent decline in germination rate as NaCl concentration increases: 96% → 82% → 54% → 28% → 6%. Every step shows a decrease with no reversal. A (increased only) directly contradicts the data. C (increased then decreased) would require the germination rate to rise at some point before falling — this does not occur. D (remained constant) would require identical values across all trays. On trend questions using tables, trace the dependent variable column from top to bottom and identify the direction of change at each step.

Question 9

PASSAGE VI

BIOLOGY / GENETICS: Research Summary

Introduction

Restriction enzymes are proteins that cut DNA molecules at specific, predictable sequences of base pairs (bp). Gel electrophoresis is a technique used to separate these resulting DNA fragments by size. When an electrical current is applied to the gel, the negatively charged DNA fragments migrate toward the positive electrode. Smaller DNA fragments move through the gel much faster and travel further than larger fragments.

A researcher isolated a circular bacterial plasmid (a ring of DNA) consisting of exactly 5,000 bp. To map the plasmid, the researcher treated identical samples of the plasmid with different restriction enzymes—Enzyme 1 (E1), Enzyme 2 (E2), and Enzyme 3 (E3)—both individually and in combinations.

After allowing the enzymes to cut the DNA, the researcher ran the samples on an electrophoresis gel. A dye was added to make the DNA bands visible. The size of the fragments in each band was recorded in Table 1.

According to Table 1, how many distinct DNA bands would be visible in Lane 3 of the electrophoresis gel?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: The correct answer is B (2 bands). Lane 3 was treated with E2 alone. Table 1 shows Lane 3 produced two fragment sizes: 3,000 bp and 2,000 bp. Each distinct size produces one visible band at a specific location in the gel — smaller fragments migrate further and appear lower, larger fragments appear higher. Two different sizes therefore produce exactly two distinct bands. A (1 band) would require only one fragment size, as in Lanes 1 or 2. C and D (3 or 4 bands) would require three or four different fragment sizes, which are not present in Lane 3. Note that the two fragments sum to 5,000 bp — consistent with the original plasmid size — confirming E2 cut the plasmid at exactly two locations.

Question 10

PASSAGE II

BIOLOGY: Research Summary

Introduction

Transpiration is the process by which moisture is carried through plants from roots to small pores on the underside of leaves, where it changes to vapor and is released to the atmosphere. A botanist conducted two studies to investigate how environmental factors affect the transpiration rate of Spathiphyllum (peace lily) plants.

Study 1

The botanist placed 5 identical Spathiphyllum plants into 5 identical environmentally controlled chambers. The relative humidity inside all chambers was kept constant at 40%, and the temperature was kept constant at 22°C. The botanist varied the light intensity—measured in micromoles of photons per square meter per second (μmol/m2/s\mu mol/m^2/s)—in each chamber. After 4 hours, the botanist measured the mass of water lost by each plant to calculate the transpiration rate in milligrams of water per square centimeter of leaf area per hour (mg/cm2/hrmg/cm^2/hr). Results are shown in Table 1.

Study 2

The botanist obtained 5 new, identical Spathiphyllum plants and placed them in the chambers. This time, the light intensity in all chambers was kept constant at 400 μmol/m2/s\mu mol/m^2/s and the temperature at 22°C. The botanist varied the relative humidity in each chamber. The transpiration rates were calculated after 4 hours. Results are shown in Table 2.

Based on Study 2, what is the relationship between relative humidity and transpiration rate?

  1. As relative humidity increases, the transpiration rate decreases. (correct answer)
  2. As relative humidity increases, the transpiration rate increases.
  3. As relative humidity increases, the transpiration rate remains constant.
  4. There is no clear relationship between relative humidity and transpiration rate.
Explanation: The correct answer is A. Table 2 shows a consistent inverse relationship: as relative humidity increases from 20% to 100%, the transpiration rate drops from 6.5 mg/cm²/hr to 0.2 mg/cm²/hr without any reversals. This makes biological sense — higher humidity means the air is already saturated with water vapor, reducing the concentration gradient that drives water loss from the leaf. B is wrong — this describes the opposite relationship from what the data show. C is wrong — the rate clearly changes across humidity levels. D is wrong — the trend is very clear and consistent across all five data points. Pro tip: Relationship questions on the ACT always have a correct answer supported directly by the data. Scan the table for whether values increase, decrease, or hold steady before reading the choices.

Question 11

A research vessel lowered a probe at a single ocean station and recorded water temperature and salinity at five depths on the same cast. Table 1 shows the measurements.

Based on Table 1, as depth increased, temperature and salinity changed in which of the following ways?

  1. Both temperature and salinity increased.
  2. Temperature decreased and salinity increased. (correct answer)
  3. Both temperature and salinity decreased.
  4. Temperature increased and salinity decreased.
Explanation: Reading down Table 1 from 0 m to 200 m, temperature falls steadily from 24.8 °C to 6.5 °C while salinity rises steadily from 34.2 to 35.6 practical salinity units. The two quantities change in opposite directions.

Question 12

A radioactive sample was placed next to a detector and its count rate was recorded every 10 minutes. The half-life is the time required for the count rate to fall to half its earlier value. Table 1 shows the measurements.

Based on Table 1, the half-life of this sample is closest to which time?

  1. 10 min
  2. 20 min (correct answer)
  3. 30 min
  4. 40 min
Explanation: The count rate falls from 800 counts per minute at 0 min to 400 at 20 min, which is half. It falls again from 400 at 20 min to 200 at 40 min, another halving over the same 20 minutes. The half-life is therefore about 20 minutes.

Question 13

A researcher cut a core from one tree and measured the width of the growth ring formed in each of five years. Rainfall totals for the same years were taken from a nearby weather station. Table 1 shows both sets of records.

Based on Table 1, what was the ring width in the year with the least rainfall?

  1. 1.1 mm (correct answer)
  2. 1.8 mm
  3. 2.6 mm
  4. 3.2 mm
Explanation: The lowest rainfall total in Table 1 is 34 cm, in 2021. Reading across that row gives a ring width of 1.1 mm. The other options are ring widths from years with more rainfall: 1.8 mm in 2019, 2.6 mm in 2020, and 3.2 mm in 2018.

Question 14

Table 1 lists the surface temperature and luminosity of four stars. Luminosity is given relative to the Sun, so a value of 25 means the star emits 25 times as much energy per second as the Sun.

Based on Table 1, which star has a lower surface temperature but a greater luminosity than Star W?

  1. Star X
  2. Star Y
  3. Star Z (correct answer)
  4. No star listed
Explanation: Star W is at 9,900 K with a luminosity of 25. Star Z is cooler at 3,500 K yet far more luminous at 100,000, so it satisfies both conditions. Star X is more luminous but hotter, and Star Y is cooler but much less luminous.

Question 15

A student was given an unlabeled metal sample. She measured its melting point as 963 °C and its density as 10.5 g/cm³, using the same methods that produced the accepted values in Table 1.

Based on Table 1, the sample is most likely which metal?

  1. Silver (correct answer)
  2. Copper
  3. Zinc
  4. Iron
Explanation: Silver is listed in Table 1 with a melting point of 962 °C and a density of 10.49 g/cm³, matching both of the student's measurements. Copper has a similar density (8.96 g/cm³) but melts about 120 °C higher, and iron and zinc differ in both properties.

Question 16

An astronomer catalogued the mass and volume of four planets. Average density is mass divided by volume. Table 1 shows the catalogue entries.

Based on Table 1, which planet has the greatest average density?

  1. Planet P
  2. Planet Q
  3. Planet R
  4. Planet S (correct answer)
Explanation: Dividing mass by volume for each planet gives 12 ÷ 3 = 4 for Planet P, 45 ÷ 15 = 3 for Planet Q, 8 ÷ 4 = 2 for Planet R, and 30 ÷ 6 = 5 for Planet S. Planet S is the densest at about 5 g/cm³. Planet Q is the most massive of the four, so reading the mass column alone gives the wrong planet.

Question 17

The same reaction was run six times with different masses of catalyst and no other changes. Each run was stopped after exactly 30 minutes, and the percentage of the starting material converted to product was recorded. Table 1 shows the results.

Based on Table 1, above which catalyst mass does adding more catalyst produce almost no further gain in yield?

  1. 0.5 g
  2. 1.0 g
  3. 1.5 g (correct answer)
  4. 2.0 g
Explanation: Yield climbs by 19, 13, and 8 percentage points across the first three increases, reaching 82% at 1.5 g. It then stays at 82% at both 2.0 g and 2.5 g, so adding catalyst beyond 1.5 g changes nothing.

Question 18

A chemist measured the initial rate at which hydrogen peroxide decomposed in five solutions of different concentration. Each solution was tested once at 25 °C and once at 35 °C. Table 1 shows the results. The chemist then prepared a 0.60 M solution and planned to repeat the 25 °C trial with it.

Based on Table 1, the initial rate for the 0.60 M solution at 25 °C would most likely be closest to which rate?

  1. 0.012 mol/L·s
  2. 0.036 mol/L·s
  3. 0.060 mol/L·s
  4. 0.072 mol/L·s (correct answer)
Explanation: In the 25 °C column of Table 1, each 0.10 M increase in concentration raises the rate by 0.012 mol/L·s (0.012, 0.024, 0.036, 0.048, 0.060). Extending that pattern to 0.60 M gives 0.060 + 0.012 = 0.072 mol/L·s. The other options are rates the table already lists for lower concentrations.

Question 19

A fixed amount of gas was sealed in a syringe held at constant temperature. The plunger was moved to five volumes and the pressure was recorded at each. Table 1 shows the results.

Which of the following statements about the measurements in Table 1 is correct?

  1. The sum of pressure and volume is the same for every row.
  2. The product of pressure and volume is the same for every row. (correct answer)
  3. Pressure increases by the same amount for each 10 mL increase in volume.
  4. Pressure and volume both increase together.
Explanation: Multiplying the two columns in Table 1 gives 10 × 240 = 2,400 for the first row, and 20 × 120, 30 × 80, 40 × 60, and 50 × 48 all give 2,400 as well. The sums are not constant, pressure falls rather than rises as volume increases, and the drops between rows shrink from 120 kPa to 12 kPa.

Question 20

A student hung masses from a spring and measured how far the spring stretched below its resting length. The spring returned to its resting length after each mass was removed. Table 1 shows the results.

Based on Table 1, a 700 g mass would most likely stretch the spring by about how far?

  1. 16.8 cm (correct answer)
  2. 19.2 cm
  3. 21.6 cm
  4. 24.0 cm
Explanation: Each additional 100 g in Table 1 stretches the spring another 2.4 cm. Continuing the pattern past 500 g gives 14.4 cm at 600 g and 16.8 cm at 700 g. The larger options correspond to 800 g, 900 g, and 1,000 g.