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AP Biology – Units 1–4 Practice Set

Chemistry of life, cells, energetics, communication and cell cycle.

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10 questions  ·  44 marks  ·  answers included  ·  AP Biology (College Board CED, Units 1–4)
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This worksheet covers:

  • Unit 1 (Chemistry of Life)
  • Unit 2 (Cell Structure and Function)
  • Unit 3 (Cellular Energetics)
  • Unit 4 (Cell Communication and Cell Cycle)

A calculator is permitted, in line with the AP Biology exam. Use $R = 0.0831\,\text{L bar mol}^{-1}\text{K}^{-1}$ for water potential calculations.

Questions

Section A – Chemistry of life and cell structure (Units 1–2)

1

(a) Describe two properties of water that result from hydrogen bonding between water molecules.(b) Explain how one of these properties is important for a biological process, such as transpiration in plants.

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(a) Any two of: cohesion (water molecules stick to each other via hydrogen bonds), adhesion (water molecules stick to other polar surfaces), high specific heat capacity (hydrogen bonds absorb energy as they break, resisting temperature change), high heat of vaporization, or being a good solvent for polar/ionic substances.(b) Cohesion allows water molecules to form a continuous column in the xylem, so that as water evaporates from leaf cells during transpiration, it pulls the whole column of water upward under tension; adhesion of water to the walls of the narrow xylem vessels also helps counteract the pull of gravity, supporting transport of water up the plant.
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2

A cube-shaped cell has a side length of $2\,\mu\text{m}$.(a) Calculate the surface area, volume, and surface area-to-volume ratio of this cell.(b) A second cube-shaped cell has a side length of $10\,\mu\text{m}$. Calculate its surface area-to-volume ratio.(c) Explain why a small surface area-to-volume ratio can limit the rate at which a cell can exchange materials with its environment, and explain how this constrains maximum cell size.

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(a) Surface area $= 6(2)^2 = 24\,\mu\text{m}^2$; volume $= 2^3 = 8\,\mu\text{m}^3$; ratio $= 24/8 = 3.0$.(b) Surface area $= 6(10)^2 = 600\,\mu\text{m}^2$; volume $= 10^3 = 1000\,\mu\text{m}^3$; ratio $= 600/1000 = 0.60$.(c) As a cell increases in size, its volume (and thus its metabolic demand for nutrients/oxygen and its waste production) increases faster than its surface area, so the surface area-to-volume ratio decreases. A low ratio means there is not enough membrane surface for diffusion to supply the cell’s interior with nutrients/oxygen and remove wastes fast enough, which limits how large a cell can grow before this exchange becomes rate-limiting.
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3

A red blood cell is placed in a solution with a solute potential ($\Psi_s$) more negative than the solute potential inside the cell, and the external pressure potential is zero.(a) Predict the direction of net water movement, and justify your answer in terms of water potential.(b) Predict what will happen to the cell, and explain why.

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(a) Water moves by osmosis down its water potential gradient, from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Since the external solution has a more negative solute potential (and hence more negative total water potential, as pressure potential is zero), water will move out of the cell into the solution.(b) The cell will shrink/crenate, because it is losing water to the hypertonic surrounding solution, reducing its internal volume.
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4

A solution of sucrose has a molar concentration of $0.30\,\text{mol L}^{-1}$ at $298\,\text{K}$. Assume the solution is ideal ($i=1$) and that pressure potential $\Psi_p = 0$.(a) Calculate the solute potential of this solution using $\Psi_s = -iCRT$.(b) State the total water potential of this solution and explain what this tells you about the direction water would move if this solution were separated from pure water by a selectively permeable membrane.

Show answer
(a) $\Psi_s = -(1)(0.30)(0.0831)(298) = -7.43\,\text{bar}$ (3 s.f.)(b) Total water potential $\Psi = \Psi_p + \Psi_s = 0 + (-7.43) = -7.43\,\text{bar}$. Since pure water has $\Psi = 0$, which is higher (less negative) than the sucrose solution’s water potential, water would move by osmosis from the pure water, across the membrane, into the sucrose solution.
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Section B – Cellular energetics (Unit 3)

5

(a) Name the two main stages of aerobic cellular respiration that occur after glycolysis, and state where each occurs in a eukaryotic cell.(b) Explain why the electron transport chain requires oxygen to keep functioning, and describe what happens if oxygen is not available.

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(a) The Krebs (citric acid) cycle occurs in the mitochondrial matrix; oxidative phosphorylation (the electron transport chain and chemiosmosis) occurs across/at the inner mitochondrial membrane.(b) Oxygen acts as the final electron acceptor at the end of the electron transport chain, combining with electrons and $\text{H}^+$ to form water; this keeps the chain running by continually removing electrons, allowing NADH and $\text{FADH}_2$ to keep being oxidised and pumping $\text{H}^+$ ions. Without oxygen, the electron transport chain backs up (electrons have nowhere to go), NADH cannot be recycled to $\text{NAD}^+$, and glycolysis/the Krebs cycle stall, so the cell must switch to fermentation (anaerobic respiration) to regenerate $\text{NAD}^+$.
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6

In an experiment, isolated mitochondria are supplied with pyruvate, ADP, and inorganic phosphate ($\text{P}_i$) in an oxygen-containing solution, and oxygen consumption is measured. When the drug oligomycin, which blocks ATP synthase, is added, oxygen consumption decreases sharply.(a) Explain why blocking ATP synthase reduces the rate of oxygen consumption by the electron transport chain, referring to the proton (H$^+$) gradient.(b) Predict what would happen to oxygen consumption if a chemical that makes the inner mitochondrial membrane freely permeable to $\text{H}^+$ (an uncoupler) were then added, and explain your reasoning.

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(a) Normally, $\text{H}^+$ ions flow back across the inner membrane through ATP synthase, down their concentration gradient, which drives ATP synthesis; this flow relieves the proton gradient built up by the electron transport chain. If ATP synthase is blocked, $\text{H}^+$ cannot flow back through it, so the proton gradient becomes very steep and it becomes increasingly difficult for the electron transport chain pumps to pump further $\text{H}^+$ against this gradient, so electron transport (and the coupled consumption of oxygen) slows down.(b) Oxygen consumption would increase again (even without ATP being made), because the uncoupler allows $\text{H}^+$ to leak back across the membrane independently of ATP synthase, dissipating the proton gradient. This removes the ‘back pressure’ on the electron transport chain, allowing electron transport (and oxygen consumption) to proceed rapidly, though the energy released is lost as heat rather than being captured as ATP.
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7

(a) State the overall balanced equation for photosynthesis.(b) Identify the products of the light-dependent reactions that are used in the Calvin cycle, and state where in the chloroplast each stage occurs.(c) Explain why, on a hot dry day, a plant may close its stomata, and describe one consequence this has for the rate of photosynthesis.

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(a) $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$ (in the presence of light)(b) The light-dependent reactions produce ATP and NADPH, which are used by the Calvin cycle to fix and reduce carbon dioxide into sugar. The light-dependent reactions occur in the thylakoid membranes; the Calvin cycle occurs in the stroma.(c) Closing stomata reduces water loss by transpiration, helping the plant avoid dehydration on a hot dry day. However, this also reduces $\text{CO}_2$ entry into the leaf, so as $\text{CO}_2$ is used up by the Calvin cycle and not replenished, $\text{CO}_2$ becomes limiting and the rate of photosynthesis decreases.
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Section C – Cell communication and the cell cycle (Unit 4)

Exam-style
8

A signaling molecule (ligand) binds to a G-protein-coupled receptor on the surface of a target cell, ultimately leading to a change in gene expression in the nucleus.(a) Name the three main stages of cell signaling, in order, and briefly describe what occurs at each stage in this example.(b) Explain one advantage to the cell of using a signal transduction pathway with multiple intermediate steps, rather than the ligand directly causing the cellular response.

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(a) Reception: the ligand binds to the specific G-protein-coupled receptor on the cell membrane, causing it to change shape/activate. Transduction: the activated receptor triggers a cascade of intracellular signaling molecules (e.g. via a second messenger and/or a phosphorylation cascade) that relays and often amplifies the signal toward the nucleus. Response: the pathway ultimately alters the activity of transcription factors, changing which genes are expressed.(b) A multi-step pathway allows the original signal to be amplified, so that a small number of ligand molecules binding at the membrane can produce a large intracellular response; it also provides multiple points at which the pathway can be regulated (e.g. feedback inhibition) or where a single signal can branch to produce several coordinated responses.
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9

The cell cycle includes checkpoints, such as the $G_1$ checkpoint, that regulate progression through the cycle.(a) Describe the role of the $G_1$ checkpoint in the cell cycle.(b) Explain why a mutation that causes a cell to no longer respond to signals at the $G_1$ checkpoint could lead to uncontrolled cell division (cancer).

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(a) The $G_1$ checkpoint checks internal and external conditions, such as cell size, availability of nutrients, and the presence of growth factors, and whether DNA is undamaged, before allowing the cell to commit to entering the S phase and replicating its DNA; if conditions are unfavorable, the cell is held at this checkpoint (or enters a resting $G_0$ state).(b) If a cell cannot respond correctly to the $G_1$ checkpoint (e.g. due to a mutation in a checkpoint protein such as p53 or a cyclin-dependent kinase regulator), the cell may proceed through the cycle and divide even when conditions are unfavorable or DNA is damaged, without the normal ‘stop’ signal being enforced. Repeated, unchecked division of cells with potentially damaged DNA can lead to the uncontrolled cell proliferation characteristic of cancer.
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10

In a population of $200$ cells sampled from actively dividing tissue, $12$ cells are observed to be in some stage of mitosis (as opposed to interphase). The total length of the cell cycle for this tissue is $24$ hours.(a) Calculate the proportion of cells observed in mitosis, and use this to estimate the length of time, in hours, that a cell spends in mitosis.(b) Explain the assumption you must make about the cell population for this calculation to be valid.

Show answer
(a) Proportion in mitosis $= \dfrac{12}{200} = 0.060$ (6.0 percent). Estimated time in mitosis $= 0.060 \times 24 = 1.44\,\text{hours}$ (about 86 minutes).(b) This calculation assumes that the sampled cells are representative of a population cycling continuously and asynchronously (not synchronized), so that the proportion of cells observed in a given stage at any snapshot in time is proportional to the fraction of the total cell cycle duration that stage normally occupies.
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