1. AP Biology Midterm Answer Key
Multiple Choice
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
A
A
D
A
B
B
B
B
B
D
B
B
A
1999
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
B
A
A
A
D
D
B
D
A
D
2012
24. C
2002
25.
26.
27.
28.
29.
30.
31.
B
D
B
B
C
C
D

2. Grid-In/Calculations
2012
32. Correct answer is 340-360. The graph depicts a logistic growth curve for a population. The
formula for calculating the per capita rate increase between days 3 and 5 is ΔN/ΔT. Where
ΔN=change in population size and ΔT=time interval. In other wards, ΔN/ΔT = 900 individuals-200
individuals/2 days = 700 individuals/ 2 days. However, the mean rate of population growt is for one
day, or 350 individuals/day
33.
No Test
34. Surrounding solution: Ψ = ΨP + ΨS = 0 bars + (1)(0.5 mol/L)(0.0831 L*bars/mol*K)(293 K) = 12.2 bars
Cell at equilibrium: -12.2 bars = ΨP + (1)(0.6 mol/L)(0.0831 L*bars/mol*K)(293 K) = ΨP + (-14.6
bars)
ΨP = -12.2 bars – (-14.6 bars) = 2.4 bars
FRQ – Long
Question 1 - 2005
Part A: Graph and Optimum Temperature (3 points maximum)

3. Graph Setup (1 point)
Must contain:
• Title/Legend and Y-axis [Bubbles of
gas/Min]
• X-axis [ Temperature (°C)]
• Correct measurement units and scaling of axes
Data Plotted (1 point)
•Correctly plotted points in proper orientation
•Points may or may not be connected with a line
•Bar graph acceptable
Optimum Temperature (1 point)
• 30° C. or between 20o C and 40° C either
clearly indicated on the graph or in a sentence
Part B: Analyze and Explain the Results
(4 points maximum)
Analysis (1 point)
• Provide range of the change in respiration activity (increase and decrease) to
temperature change
(increase and decrease)
Explanations (1 point each)
• Below optimum—Increase in molecular movement leads to increase in reaction rate
• Above optimum—Denaturing of enzymes leads to decrease in reaction rate
Elaboration (2 points maximum, 1 point each)
• Relating enzyme function (effect on reaction rates) to allosteric site, active site, H bond,
B
groups
• Gas production due to respiration (can use either aerobic respiration or fermentation)
• Induced fit
• Lowering energy of activation
• Enzyme specificity
Part C: Experimental Design (4 points maximum)
NOTE: Experiment must be feasible. Must include sugar solutions of varying pH and an organism. If
experiment is not reasonable, no points are awarded in the design structure section below.
Design Structures (3 points maximum, 1 point each)
• Two experimental constants—constant amounts of yeast or sugar, or temperature held constant
• Independent variable tested—reasonable pH range must be stated, including acid through base
Control – identification of a control treatment
Measureable product per unit time
Multiple trials
Predication – designated pH at which enzyme will function optimally
FRQ – Short
Question 2 - 2008
Explain the data presented by the graph, including a description of the relative rates of metabolic
processes occurring at different depths of the pond. (1 point for each bullet; 4 points maximum)

4. Explanation of data:
• As depth is increased, the net primary productivity decreases because light decreases/lower
rates of photosynthesis.
Description of relative rates of metabolic process occurring at specific depths according to the graph
(letters added to graph to simplify rubric):
a
b
c
d
• A: The upper area of the graph is equally productive because light availability is not a limiting
factor at the surface/ photosynthesis is not limited.
• B: The rapidly decreasing productivity region is a result of decreasing light available for
photosynthesis/photosynthesis is decreasing rapidly.
• C: At 0 (the compensation point) the photosynthetic product is equal to the cell respiration
requirements due to light availability/photosynthesis equals cell respiration.
• D: Below 0 the photosynthetic product does not meet the cell respiration requirements due
to insufficient light. Photosynthesis less than respiration.
Describe how the relationship between net primary productivity and depth would be expected to
differ if new data were collected in mid-summer from the same pond. Explain your prediction.
(1 point for each bullet; 2 points maximum)
• Description of a plausible prediction of a change in graph or a change in the relationship
between productivity and depth from spring graph to mid-summer graph.
• Explanation of a plausible prediction of a shift in the graph must be tied to a valid or plausible
reason.

5. Question 3 – 2012
(4 points maximum)
Type of mutation
(not limited to the
following)
Description (1 point per box)
Silent
Nucleotide change.
Missense/substitution
Nucleotide change causes new
codon.
Nucleotide change causes stop
codon.
Nucleotide insertion/deletion
alters
reading frame after mutation.
region Nucleotide
insertion/deletion/substitution.
Chromosome segment moves to
different site.
Chromosomes fail to separate.
Chromosome segment doubles.
Chromosome segment is
removed.
Chromosome segment is
reversed.
Chromosome segment moves to
a
different site.
Nonsense/substitution
Frameshift(insertion/deletion)
Regulatory
Translocation
Nondisjunction
Duplication
Deletion
Inversion
Transposition
Effect (1 point per box)
No change in amino
acid/protein sequence
Different amino acid/protein
sequence.
Protein not formed OR
truncated protein.
Changes amino acid/protein
sequence OR nonfunctional
protein OR no protein.
Alters gene expression OR
alters splice site.
Alters gene expression.
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Question 4 – 2006b
(6 points maximum) One for each of the following:
Correct description of meiosis (simply rephrasing the question earns no point)
Sister chromosomes pair in prophase I
Spindles move chromosome pairs to poles in anaphase I
Two cycles/rounds od division in meiosis
Sister chromatids separate in anaphase II
1 germ cell = 4 gametes
DNA replicates in interphase
No additional replication in Meiosis II