IB Biology Quiz: Mathematics
16 questions · exam conditions
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MathematicsQuestion 1 of 16

The diameter of an E. coli bacterium is approximately 2.0 x 10⁻⁶ m. The diameter of a human red blood cell is approximately 8.0 x 10⁻⁶ m. How many E. coli bacteria could be placed side-by-side across the diameter of one red blood cell?

4
16
40
64
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IB Biology Quiz

IB Biology Quiz: Mathematics

Practice Mathematics in IB Biology 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 Mathematics, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Biology.

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

The diameter of an E. coli bacterium is approximately 2.0 x 10⁻⁶ m. The diameter of a human red blood cell is approximately 8.0 x 10⁻⁶ m. How many E. coli bacteria could be placed side-by-side across the diameter of one red blood cell?

  1. 4 (correct answer)
  2. 16
  3. 40
  4. 64
Explanation: To find how many bacteria fit across the diameter of the red blood cell, divide the diameter of the red blood cell by the diameter of the bacterium. Number = (Diameter of red blood cell) / (Diameter of bacterium) = (8.0 x 10⁻⁶ m) / (2.0 x 10⁻⁶ m). The 10⁻⁶ terms cancel out, leaving 8.0 / 2.0 = 4. Therefore, 4 bacteria can fit side-by-side across the diameter.

Question 2

An electron micrograph of a bacterial cell is taken. A scale bar labelled 0.5 µm measures 20 mm on the micrograph. The measured width of the bacterial cell on the micrograph is 35 mm. What is the actual width of the bacterial cell in nanometres (nm)?

  1. 700 nm
  2. 875 nm (correct answer)
  3. 1400 nm
  4. 1750 nm
Explanation: First, calculate the magnification. Ensure units are consistent. 20 mm = 20,000 µm. Magnification = Image size / Actual size = 20,000 µm / 0.5 µm = 40,000x. Next, calculate the actual size of the cell. Actual size = Image size / Magnification = 35 mm / 40,000. Convert 35 mm to µm: 35,000 µm. Actual size = 35,000 µm / 40,000 = 0.875 µm. Finally, convert µm to nm: 0.875 µm * 1000 nm/µm = 875 nm.

Question 3

In an osmosis experiment, a potato cylinder with an initial mass of 12.50 g is placed in a concentrated salt solution. After 30 minutes, its mass is 10.75 g. It is then transferred to distilled water for 30 minutes, and its final mass is 11.50 g. What is the net percentage change in mass of the potato cylinder from its initial mass to its final mass?

  1. -15.3%
  2. -14.0%
  3. -8.7%
  4. -8.0% (correct answer)
Explanation: The net percentage change is calculated based on the initial and final masses only. Initial mass = 12.50 g. Final mass = 11.50 g. Change in mass = Final mass - Initial mass = 11.50 g - 12.50 g = -1.00 g. Percentage change = (Change in mass / Initial mass) * 100 = (-1.00 g / 12.50 g) * 100 = -8.0%.

Question 4

In a large, randomly mating population of fruit flies (Drosophila melanogaster), 36% of the individuals exhibit a recessive phenotype for eye colour. Assuming the population is in Hardy-Weinberg equilibrium, what percentage of the population is expected to be heterozygous for this trait?

  1. 16%
  2. 24%
  3. 48% (correct answer)
  4. 64%
Explanation: The frequency of the recessive phenotype corresponds to q². So, q² = 0.36. The frequency of the recessive allele, q, is the square root of q², which is √0.36 = 0.6. The frequency of the dominant allele, p, is calculated as p = 1 - q, so p = 1 - 0.6 = 0.4. The frequency of heterozygous individuals is given by 2pq. So, 2 * 0.4 * 0.6 = 0.48. To express this as a percentage, multiply by 100, which gives 48%.

Question 5

A student uses the capture-mark-recapture method to estimate the population of woodlice in a forest. In the first sample, 60 woodlice are captured, marked, and released. One week later, a second sample of 90 woodlice is captured. In this second sample, 12 of the woodlice are marked. What is the estimated population size of the woodlice?

  1. 162
  2. 300
  3. 450 (correct answer)
  4. 540
Explanation: The Lincoln index formula is used for population estimation: N = (n1 * n2) / m, where N is the estimated population size, n1 is the number of individuals in the first capture, n2 is the number of individuals in the second capture, and m is the number of marked individuals in the second capture. Plugging in the values: N = (60 * 90) / 12 = 5400 / 12 = 450.

Question 6

The activity of an enzyme is measured by the rate of product formation. At time t=0 s, the product concentration is 0 mmol L⁻¹. At t=15 s, the concentration is 30 mmol L⁻¹, and at t=30 s, the concentration is 48 mmol L⁻¹. What is the average rate of reaction between 15 s and 30 s?

  1. 1.2 mmol L⁻¹ s⁻¹ (correct answer)
  2. 1.6 mmol L⁻¹ s⁻¹
  3. 1.8 mmol L⁻¹ s⁻¹
  4. 2.0 mmol L⁻¹ s⁻¹
Explanation: The rate of reaction is the change in concentration divided by the change in time. For the interval between 15 s and 30 s: Change in concentration = 48 mmol L⁻¹ - 30 mmol L⁻¹ = 18 mmol L⁻¹. Change in time = 30 s - 15 s = 15 s. Rate = 18 mmol L⁻¹ / 15 s = 1.2 mmol L⁻¹ s⁻¹.

Question 7

An ecological survey of a quadrat reveals the following numbers of individuals for four plant species: Species P: 45, Species Q: 25, Species R: 20, Species S: 10. Calculate the Simpson reciprocal index of diversity for this community using the formula D = N(N-1) / Σn(n-1).

  1. 2.95
  2. 3.25 (correct answer)
  3. 3.88
  4. 4.15
Explanation: First, find the total number of individuals, N: 45 + 25 + 20 + 10 = 100. Next, calculate N(N-1): 100 × 99 = 9900. Then, calculate Σn(n-1) for each species: P: 45×44 = 1980, Q: 25×24 = 600, R: 20×19 = 380, S: 10×9 = 90. Sum = 1980 + 600 + 380 + 90 = 3050. Finally, calculate D: D = 9900 / 3050 = 3.25.

Question 8

Two populations of a species of grass were sampled. Population X, from a nutrient-rich field, had a mean height of 65 cm with a standard deviation of 4 cm. Population Y, from a nutrient-poor field, had a mean height of 40 cm with a standard deviation of 9 cm. What is the most accurate interpretation of these data?

  1. The grasses in Population X are taller on average and show less variation in height than Population Y. (correct answer)
  2. The grasses in Population X are shorter on average and show more variation in height than Population Y.
  3. The grasses in Population Y are taller on average and show less variation in height than Population X.
  4. The grasses in Population Y are shorter on average and show less variation in height than Population X.
Explanation: The mean is a measure of central tendency, and the standard deviation is a measure of the spread or variation in the data. Population X has a higher mean (65 cm vs 40 cm), so its grasses are taller on average. Population X has a smaller standard deviation (4 cm vs 9 cm), indicating that the heights of the individual plants in this population are more uniform and cluster more closely around the mean, meaning there is less variation compared to Population Y.

Question 9

A spherical prokaryotic cell has a diameter of 2 µm, while a spherical eukaryotic cell has a diameter of 20 µm. How many times greater is the surface area to volume ratio of the prokaryotic cell compared to the eukaryotic cell?

  1. 2 times
  2. 10 times (correct answer)
  3. 100 times
  4. 1000 times
Explanation: The surface area (SA) of a sphere is 4πr² and the volume (V) is (4/3)πr³. The SA:V ratio simplifies to 3/r. For the prokaryote, radius (r) = 1 µm, so its SA:V ratio is 3/1 = 3. For the eukaryote, radius (r) = 10 µm, so its SA:V ratio is 3/10 = 0.3. To find how many times greater the prokaryote's ratio is, divide the prokaryote's ratio by the eukaryote's ratio: 3 / 0.3 = 10. The prokaryotic cell has a SA:V ratio that is 10 times greater.

Question 10

In a microscopic field of view of a cancerous tissue sample, 450 cells are counted. Of these, 60 cells are in prophase, 25 are in metaphase, 10 are in anaphase, and 20 are in telophase. The remaining cells are in interphase. What is the mitotic index of this tissue?

  1. 0.13
  2. 0.26 (correct answer)
  3. 0.35
  4. 0.74
Explanation: The mitotic index is the ratio of the number of cells undergoing mitosis to the total number of cells. First, sum the number of cells in all stages of mitosis: 60 (prophase) + 25 (metaphase) + 10 (anaphase) + 20 (telophase) = 115 cells. The total number of cells counted is 450. The mitotic index = (Number of cells in mitosis) / (Total number of cells) = 115 / 450 ≈ 0.2556. This rounds to 0.26.

Question 11

In a forest ecosystem, the energy available from producers is 25,000 kJ m⁻² yr⁻¹. The energy transferred to primary consumers is 3,000 kJ m⁻² yr⁻¹, and the energy transferred to secondary consumers is 360 kJ m⁻² yr⁻¹. What is the efficiency of energy transfer between the primary and secondary consumers?

  1. 1.44%
  2. 10.8%
  3. 12.0% (correct answer)
  4. 88.0%
Explanation: The efficiency of energy transfer between trophic levels is calculated as (energy in the higher trophic level / energy in the lower trophic level) * 100. In this case, we are interested in the transfer from primary consumers (lower level) to secondary consumers (higher level). Efficiency = (360 kJ m⁻² yr⁻¹ / 3,000 kJ m⁻² yr⁻¹) * 100 = (0.12) * 100 = 12.0%.

Question 12

An individual has a resting heart rate of 60 beats per minute and a stroke volume of 80 mL per beat. During vigorous exercise, their cardiac output increases to 19.2 L per minute, and their heart rate rises to 160 beats per minute. What is the stroke volume in mL per beat during this exercise?

  1. 100 mL
  2. 120 mL (correct answer)
  3. 140 mL
  4. 160 mL
Explanation: Cardiac Output (CO) = Heart Rate (HR) × Stroke Volume (SV). We need to find the stroke volume during exercise. First, ensure the units are consistent. The cardiac output is given in L/min, so convert it to mL/min: 19.2 L/min * 1000 mL/L = 19,200 mL/min. Now, rearrange the formula to solve for SV: SV = CO / HR. SV = 19,200 mL/min / 160 beats/min = 120 mL/beat.

Question 13

In a population that is in Hardy-Weinberg equilibrium, the frequency of heterozygous individuals is 0.42. The frequency of the recessive allele (q) is less than the frequency of the dominant allele (p). What is the frequency of individuals with the homozygous dominant genotype?

  1. 0.09
  2. 0.30
  3. 0.49 (correct answer)
  4. 0.70
Explanation: We are given 2pq = 0.42 and p + q = 1. We can substitute p = 1 - q into the first equation: 2(1 - q)q = 0.42, which simplifies to 2q - 2q² = 0.42, or 2q² - 2q + 0.42 = 0. Solving this quadratic equation (q² - q + 0.21 = 0) gives two possible values for q: q = 0.3 and q = 0.7. The corresponding values for p are p = 0.7 and p = 0.3. The problem states that q < p, so we must choose the solution where q = 0.3 and p = 0.7. The frequency of the homozygous dominant genotype is p², which is (0.7)² = 0.49.

Question 14

A respirometer measures the gas exchange of germinating seeds. Over 20 minutes, the seeds consumed 210 mm³ of oxygen and produced 150 mm³ of carbon dioxide. What is the respiratory quotient (RQ) for these seeds, and what substrate are they primarily metabolizing?

  1. RQ = 0.71, primarily lipids (correct answer)
  2. RQ = 0.71, primarily carbohydrates
  3. RQ = 1.40, primarily lipids
  4. RQ = 1.40, anaerobic respiration
Explanation: The respiratory quotient (RQ) is calculated as the volume of CO₂ produced divided by the volume of O₂ consumed. RQ = 150 mm³ / 210 mm³ ≈ 0.71. An RQ value of approximately 0.7 is characteristic of the aerobic respiration of lipids. An RQ of 1.0 indicates carbohydrate metabolism, and an RQ of around 0.8 indicates protein metabolism.

Question 15

A turgid plant cell has a solute potential (ψs) of –1.2 MPa and a pressure potential (ψp) of +0.7 MPa. The cell is placed in a sucrose solution and begins to lose water until it becomes flaccid (loses all turgor pressure), at which point there is no net movement of water. What is the solute potential of the sucrose solution?

  1. +0.7 MPa
  2. –0.5 MPa
  3. –1.9 MPa
  4. –1.2 MPa (correct answer)
Explanation: Water potential (ψw) = solute potential (ψs) + pressure potential (ψp). When there is no net movement of water, the water potential inside the cell equals the water potential of the external solution. When the cell is flaccid, its pressure potential (ψp) is zero. Therefore, the cell's final water potential is ψw = ψs + 0 = –1.2 MPa. Since the cell is in equilibrium with the solution, the water potential of the solution must also be –1.2 MPa. As the solution is open to the atmosphere, its pressure potential is zero, so its water potential is equal to its solute potential. Thus, the solute potential of the solution is –1.2 MPa.

Question 16

In a genetic cross between two heterozygous pea plants for flower colour (Pp x Pp), the observed offspring are 730 purple-flowered and 270 white-flowered. The expected Mendelian ratio is 3:1. A chi-squared test is performed, and the calculated value is χ² = 2.13. Given a critical value of 3.84 at a significance level of p = 0.05 for one degree of freedom, what conclusion should be drawn?

  1. The null hypothesis is accepted; the observed ratio is significantly different from the expected ratio.
  2. The null hypothesis is rejected; the observed ratio is not significantly different from the expected ratio.
  3. The null hypothesis is rejected; any deviation from the expected ratio is not due to chance.
  4. The null hypothesis is accepted; any deviation from the expected ratio is likely due to chance. (correct answer)
Explanation: The null hypothesis (H₀) states that there is no significant difference between the observed and expected results. The calculated chi-squared value (2.13) is compared to the critical value (3.84). Since 2.13 < 3.84, the calculated value does not fall into the rejection region. Therefore, we accept (or fail to reject) the null hypothesis. This means there is no statistically significant difference between the observed and expected ratios, and any deviation is likely due to random chance.