Short Essay Practice ap exam Review Fall 2014 short essay #1

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Membranes are vital to the transport of substances into and out of cells. Three important forms of cellular transport include:

Active Transport


Facilitated Diffusion
For each of the forms listed above, explain how the organization of the cell membrane functions in the movement of specific molecules across membranes.

Explain the significance of each type of transport to a specific cell.

Active Transport (1.5 Points Max)

  • (+ ½ ) Describing active transport as the movement of molecules against a gradient or in bulk across membranes, which require energy.

  • (+ ½ ) Mentioning that it allows the cell to concentrate substances within the cell membrane.

  • (+ ½ ) Mentioning that it is performed by protein pumps embedded within the membrane.

  • (+ ½ ) Possible Example: sodium-potassium pump moves potassium into the cell and sodium out of the cell against steep gradients. Other relevant biological examples are acceptable.

Endocytosis/Exocytosis (1.5 Points Max)

  • (+ ½ ) Mentioning that endocytosis brings a substance into a cell by enclosing it within a membrane-created vesicle.

  • (+ ½ ) Mentioning that the vesicle then fuses with lysosome containing hydrolytic enzymes.

  • (+ ½ ) Possible Example: phagocytic white blood cells of the immune system engulf foreign invaders. Other relevant biological examples are acceptable.

  • (+ ½ ) Mentioning that exocytosis expels waste substances for export by enclosing these substances in a vesicle that fuses with the membrane.

  • (+ ½ ) Possible Examples: cells expelling waste or a pancreatic cell exporting insulin protein into the bloodstream. Other relevant biological examples are acceptable.

Facilitated Diffusion (1.5 Points Max)

  • (+ ½ ) Describing facilitated diffusion as the diffusion of particles with the assistance of membrane transport proteins.

  • (+ ½ ) Mentioning that transport proteins are specific and have a binding site for the specific molecules of interest.

  • (+ ½ ) Mentioning that it does not require energy.

  • (+ ½ ) Possible Examples: osmosis occurs from hypotonic solution to a hypertonic solution across aquaporins. Other relevant biological examples are acceptable.


The complete oxidation of a mole of glucose produces 686 kcal of free energy. The oxidation of a mole of glucose in a cell generates a maximum of 38 moles of ATP. Each mole of ATP stores about 7.3 kcal of energy. The efficiency of the ATP energy yield from the complete aerobic respiration of glucose is about 40 percent.
Using the laws of thermodynamics, explain what happens to the rest of the energy.

Describe how humans benefit from this energy loss.

Discuss why hibernating animals possess an adaptation to reduce efficiency of cellular respiration even further.

Loss of Energy (2 Points Max)

  • (+1) Mentioning that energy cannot be created or destroyed but only transformed from one form to another (first law of thermodynamics).

  • (+1) Mentioning that every energy transformation results in some energy loss as heat (second law of thermodynamics).

Benefits from Loss of Energy (1 Point Max)

  • (+1) Mentioning that humans use some of the heat to maintain body temperature.

Adaptation of Hibernating Animals (1 Point Max)

  • (+ ½ ) Mentioning that hibernating animals don’t need a lot of ATP because they are inactive.

  • (+ ½ ) Mentioning that hibernating animals must still maintain internal body heat.

  • (+ ½ ) Mentioning that lower efficiency means heat generation without much ATP being produced.


Surface area is an important concept in biology.
Explain how surface area plays a critical role in the digestive system of animals.

SA and the Small Intestine (3 Points Max)

  • (+1) Mentioning that the majority of absorption occurs in small intestine.

  • (+1) Mentioning that numerous folds and ridges increase surface area.

  • (+1) Mentioning that villi and microvilli increase surface area.

  • (+1) Mentioning that large surface area leads to greater absorption of nutrients.


The following table includes data from scan samples conducted on a fictional mammal called a googabear every 10 minutes over the course of 42 hours. At each scan, it was noted whether the googabear was active or inactive. The percentage of active (feeding, moving, engaging in social behavior) and inactive (resting or sleeping) scans recorded for each time period are shown in the table below.
Time 600-1200 1200-1800 1800-2400 0000-600 600-1200 1200-1800 1800-2400

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