📐 SAT
📝 ACT
🎓 AP Exams
About Contact Blog Privacy Policy

AP Biology — Unit 2 — Cell Size & Surface Area — Drill 7

Drill 7 ·

0 / 5
0/5 correct

Nice work!

Review your answers above to learn from any mistakes.

← More Drills
←
Previous drill
Drill 6
Next drill
Drill 8
→
More AP Bio drills
Drill 1 5 questions → Drill 2 5 questions → Drill 3 5 questions → Drill 4 5 questions → Drill 5 5 questions → Drill 6 5 questions →
Drill 7 — current you are here
Drill 8 5 questions → Drill 9 5 questions → Drill 10 5 questions → Drill 11 5 questions → Drill 12 5 questions → Drill 13 5 questions → Drill 14 5 questions → Drill 15 5 questions → Drill 16 5 questions → Drill 17 5 questions → Drill 18 5 questions → Drill 19 5 questions → Drill 20 5 questions → Drill 21 5 questions → Drill 22 5 questions → Drill 23 5 questions → Drill 24 5 questions → Drill 25 5 questions → Drill 26 5 questions → Drill 27 5 questions → Drill 28 5 questions → Drill 29 5 questions → Drill 30 5 questions → Drill 31 5 questions → Drill 32 5 questions → Drill 33 5 questions → Drill 34 5 questions → Drill 35 5 questions → Drill 36 5 questions → Drill 37 5 questions →

About This Drill

AP Biology — Unit 2 — Cell Size & Surface Area — Drill 7 is a practice drill. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.

Practice applying surface area-to-volume relationships to cell size constraints and membrane adaptations with this AP Biology drill.

Passage

As a cell increases in size, its volume grows faster than its surface area. This relationship creates a fundamental constraint on cell size: a cell must have sufficient surface area to support the metabolic demands of its volume. The surface area-to-volume ratio (SA:V ratio) determines how efficiently a cell can exchange materials -- nutrients, gases, and waste -- with its environment across the plasma membrane. For a spherical cell, surface area = 4(pi)r^2 and volume = (4/3)(pi)r^3. As radius increases, SA:V ratio decreases proportionally. For a sphere, SA:V ratio = 3/r. Cells that are highly metabolically active or that must exchange materials rapidly tend to be small, or have structural adaptations (such as microvilli or folded membranes) that increase surface area without proportionally increasing volume. A student measures three spherical cells of different radii and calculates their surface areas, volumes, and SA:V ratios. Cell Radius (um) Surface Area (um^2) Volume (um^3) SA:V Ratio A 1 12.6 4.2 3.0 B 2 50.3 33.5 1.5 C 4 201.1 268.1 0.75

Questions in This Drill

  1. Based on the data in the table, which of the following correctly describes the relationship between cell radius and SA:V ratio?
  2. A student calculates the SA:V ratio for a spherical cell with a radius of 3 um. Using the formulas provided, which of the following is closest to the correct SA:V ratio?
  3. A researcher observes that intestinal epithelial cells have densely packed microvilli on their apical surface. Based on the passage, which of the following best explains the functional significance of this adaptation?
  4. Two cells are identical in all respects except that Cell X has a radius of 1 um and Cell Y has a radius of 4 um. Both cells have the same metabolic rate per unit volume. Which of the following best predicts the consequence for Cell Y?
  5. A student claims that a cell could indefinitely increase in size as long as it continues to synthesize new membrane to expand its surface area. Which of the following best identifies the flaw in this reasoning?