Question 1
A single-celled organism like Amoeba can rely on simple diffusion across its cell membrane for gas exchange. Which feature of large multicellular animals makes this strategy insufficient?No clue? Show me the answer
Correct answer
Correct!
IncorrectStep-by-step walkthrough
Choose a solution method
Method #1Approach 1Step 1: Identify the key concept
The question asks why large multicellular animals cannot rely solely on simple diffusion across their body surface for gas exchange.
Step 2: Apply Fick's Law to body size
Fick's Law states that rate of diffusion is inversely proportional to diffusion distance. As organisms grow larger, the distance from the outer surface to internal cells increases dramatically.
Step 3: Consider the path length for diffusion
In a large animal, cells deep within organs may be centimetres from the body surface. Diffusion over such distances is far too slow to deliver sufficient oxygen to meet metabolic demands.
Step 4: Select the correct answer
The limiting factor is the diffusion distance — internal cells are simply too far from the external surface. This is why specialised gas exchange organs and circulatory systems are necessary.
Method #2Approach 2Step 1: Identify what is being asked
We need to identify which feature of large multicellular animals makes simple diffusion across the body surface insufficient for gas exchange.
Step 2: Eliminate: 'Large animals produce less CO₂ per kilogram'
This is incorrect — large, metabolically active animals actually produce more CO₂ overall, making diffusion even less adequate, not more.
Step 3: Eliminate: 'Outer surface is proportionally smaller relative to internal volume'
While it is true that surface area to volume ratio decreases with increasing size, this describes a contributing factor but does not directly address why diffusion fails — the core issue is the path length to internal cells.
Step 4: Eliminate: 'Thicker cell membranes slow diffusion'
Cell membrane thickness does not significantly differ between small and large animals, and this is not a reason why diffusion across the body surface becomes insufficient.
Step 5: Select: 'Cells deeper inside the body are too far from the external surface'
This correctly identifies the diffusion distance problem — as body size increases, internal cells are too far from the surface for oxygen to reach them fast enough by simple diffusion alone.
Question 2
A student counts stomata in five fields of view on a nail varnish cast of a leaf. The counts are: 28, 32, 26, 30, and 34. Each field of view has an area of 0.20 mm². What is the stomatal density of this leaf?No clue? Show me the answer
Correct answer
Correct!
IncorrectStep-by-step walkthrough
Choose a solution method
Method #1Approach 1Step 1: Identify the formula
Stomatal density is calculated as:
Step 2: Calculate the mean number of stomata
Add all counts and divide by the number of fields of view:
Step 3: Divide by field of view area
Step 4: State the answer with units
The stomatal density is 150 mm⁻². Note that using multiple fields of view and calculating the mean first is essential to account for uneven distribution of stomata across the leaf surface.
Method #2Approach 2Step 1: Identify what is being asked
We need to calculate stomatal density using the mean count divided by the field of view area.
Step 2: Eliminate: 30 mm⁻²
A value of 30 mm⁻² would result from using the mean count (30) as the final answer without dividing by the area (0.20 mm²) — a common error of omitting the final step.
Step 3: Eliminate: 750 mm⁻²
750 mm⁻² could arise from using the total count (150) divided by 0.20, without first calculating the mean — incorrectly summing all five counts instead of averaging them.
Step 4: Eliminate: 300 mm⁻²
300 mm⁻² might arise from dividing the mean (30) by 0.10 instead of 0.20, using the wrong area value.
Step 5: Select: 150 mm⁻²
Mean = 30 stomata; area = 0.20 mm²; density = 150 mm⁻². This is the correct calculation.