APEX MOCK EXAM 4: Actual ADVANCED
SYNTHESIS REVIEW – Questions with Complete
Master Solutions
Question 1 (Multiple Select - Experimental Analysis):
Scenario: Researchers investigate cellular respiration in yeast. They set up four
respirometers with identical yeast suspensions in a glucose solution. Each is placed in
a water bath at a different temperature: 5°C, 20°C, 35°C, 50°C. They measure CO₂
production (mL/hr) as an indicator of respiration rate. Data shows increasing rates from
5°C to 35°C, then a sharp decline at 50°C.
Question: Which of the following statements are supported by this experimental data?
(Select TWO).
A. Yeast enzymes have an optimal temperature near 35°C.
B. Respiration is entirely inhibited at 5°C.
C. The decline at 50°C is likely due to enzyme denaturation.
D. Oxygen consumption would show an inverse pattern to CO₂ production.
E. The experiment effectively measures the rate of glycolysis only.
Solution:
Scientific Practice: Practice 4: Data Analysis; Practice 5: Statistical Tests & Data
Analysis.
Deconstruction:
● IV: Temperature (5°C, 20°C, 35°C, 50°C).
, ● DV: Rate of CO₂ production (proxy for metabolic rate/respirometer
displacement).
● Trend: Rate increases sigmoidally from 5°C to 35°C (Q₁₀ effect), then plummets
at 50°C.
● Controls: Implicit comparison across temperatures; respirometer without yeast
would control for abiotic pressure changes.
Conceptual Synthesis: Integrates enzyme kinetics (temperature effects on tertiary
structure and kinetic energy) with cellular respiration pathways (CO₂ production in
pyruvate oxidation/Krebs cycle).
Reasoning Pathway:
● A is SUPPORTED: The peak rate at 35°C defines the thermal optimum where
enzyme-substrate collision frequency is high without denaturation.
● B is NOT SUPPORTED: Low measurable rate at 5°C (cold denaturation is rare;
kinetic arrest is reversible) contradicts "entirely inhibited."
● C is SUPPORTED: Sharp decline beyond optimum indicates tertiary structure
disruption of respiratory enzymes (e.g., cytochrome oxidase, pyruvate
dehydrogenase).
● D is NOT SUPPORTED: In aerobic respiration, O₂ consumption and CO₂
production are coupled via the respiratory quotient (RQ ≈ 1.0 for glucose). They
correlate positively, not inversely.
● E is NOT SUPPORTED: CO₂ is produced in pyruvate oxidation (mitochondrial) and
the Krebs cycle, not glycolysis (cytosolic). The experiment measures total
respiration, not just glycolytic flux.
Final Answers: A, C
Higher-Order Insight: Demonstrates how abiotic factors act as density-independent
limiting factors on biological processes, connecting molecular biology (protein
structure) to ecosystem energetics (metabolic scaling).
Question 2 (Grid-In / Calculation - Genetics):
,Scenario: In a population of flowers, color is determined by a single gene with
incomplete dominance: RR=Red, Rr=Pink, rr=White. A survey of a population finds 36%
Red, 48% Pink, and 16% White flowers.
Question: Assuming the population is in Hardy-Weinberg equilibrium, what is the
frequency of the dominant allele (R)? Report your answer as a decimal to the nearest
hundredth.
Solution:
Scientific Practice: Practice 2: Mathematical Routines; Practice 3: Scientific
Questioning.
Deconstruction: Phenotype frequencies correspond directly to genotype frequencies
due to incomplete dominance: p² (Red/RR) = 0.36, 2pq (Pink/Rr) = 0.48, q² (White/rr) =
0.16.
Conceptual Synthesis: Applies the Hardy-Weinberg equilibrium model (p² + 2pq + q² = 1)
to a population exhibiting codominant/incompletely dominant inheritance, where
heterozygotes are phenotypically distinct.
Reasoning Pathway:
1. q² = frequency of rr (White) = 0.16.
2. q = √0.16 = 0.40 (frequency of recessive allele r).
3. p + q = 1 → p = 1 - 0.40 = 0.60 (frequency of dominant allele R).
4. Verification: p² = (0.6)² = 0.36 (Red), 2pq = 2(0.6)(0.4) = 0.48 (Pink). Data
matches.
Final Answer: 0.60
Higher-Order Insight: Illustrates that H-W equations model allele frequencies regardless
of dominance pattern; incomplete dominance simply allows direct observation of
, genotype frequencies, bypassing the need for additional statistical inference (unlike
complete dominance where p² = dominant phenotype).
Question 3 (Free-Response Stem with MCQ - Systems & Ecology):
Stem: A simplified nitrogen cycle diagram is shown (described: N₂ in atmosphere ->
[Nitrogen Fixation] -> NH₃ -> [Nitrification] -> NO₂⁻ & NO₃⁻ -> [Assimilation] -> Organic N
-> [Ammonification] -> NH₃ -> [Denitrification] -> N₂). An industrial farm applies
excessive ammonium (NH₄⁺) fertilizer.
Q3a: Which process, if accelerated by the excess ammonium, could most directly
contribute to the eutrophication of a nearby lake?
A. Nitrification
B. Nitrogen Fixation
C. Denitrification
D. Assimilation
Q3b: The farm's practice also increases microbial activity that converts the ammonium
into nitrous oxide (N₂O), a potent greenhouse gas. This is an example of:
A. Assimilation
B. Denitrification (incomplete)
C. Eutrophication
D. Nitrogen Fixation
Solution for Q3a:
Scientific Practice: Practice 1: Models & Representations; Practice 7: Relating
Knowledge Across Scales.
Deconstruction: Eutrophication requires bioavailable nitrogen (nitrate, NO₃⁻) that
leaches into aquatic systems, causing algal blooms and hypoxia.
Conceptual Synthesis: Links biogeochemical cycling (nitrogen transformation) to
ecosystem dynamics (nutrient limitation and productivity).
SYNTHESIS REVIEW – Questions with Complete
Master Solutions
Question 1 (Multiple Select - Experimental Analysis):
Scenario: Researchers investigate cellular respiration in yeast. They set up four
respirometers with identical yeast suspensions in a glucose solution. Each is placed in
a water bath at a different temperature: 5°C, 20°C, 35°C, 50°C. They measure CO₂
production (mL/hr) as an indicator of respiration rate. Data shows increasing rates from
5°C to 35°C, then a sharp decline at 50°C.
Question: Which of the following statements are supported by this experimental data?
(Select TWO).
A. Yeast enzymes have an optimal temperature near 35°C.
B. Respiration is entirely inhibited at 5°C.
C. The decline at 50°C is likely due to enzyme denaturation.
D. Oxygen consumption would show an inverse pattern to CO₂ production.
E. The experiment effectively measures the rate of glycolysis only.
Solution:
Scientific Practice: Practice 4: Data Analysis; Practice 5: Statistical Tests & Data
Analysis.
Deconstruction:
● IV: Temperature (5°C, 20°C, 35°C, 50°C).
, ● DV: Rate of CO₂ production (proxy for metabolic rate/respirometer
displacement).
● Trend: Rate increases sigmoidally from 5°C to 35°C (Q₁₀ effect), then plummets
at 50°C.
● Controls: Implicit comparison across temperatures; respirometer without yeast
would control for abiotic pressure changes.
Conceptual Synthesis: Integrates enzyme kinetics (temperature effects on tertiary
structure and kinetic energy) with cellular respiration pathways (CO₂ production in
pyruvate oxidation/Krebs cycle).
Reasoning Pathway:
● A is SUPPORTED: The peak rate at 35°C defines the thermal optimum where
enzyme-substrate collision frequency is high without denaturation.
● B is NOT SUPPORTED: Low measurable rate at 5°C (cold denaturation is rare;
kinetic arrest is reversible) contradicts "entirely inhibited."
● C is SUPPORTED: Sharp decline beyond optimum indicates tertiary structure
disruption of respiratory enzymes (e.g., cytochrome oxidase, pyruvate
dehydrogenase).
● D is NOT SUPPORTED: In aerobic respiration, O₂ consumption and CO₂
production are coupled via the respiratory quotient (RQ ≈ 1.0 for glucose). They
correlate positively, not inversely.
● E is NOT SUPPORTED: CO₂ is produced in pyruvate oxidation (mitochondrial) and
the Krebs cycle, not glycolysis (cytosolic). The experiment measures total
respiration, not just glycolytic flux.
Final Answers: A, C
Higher-Order Insight: Demonstrates how abiotic factors act as density-independent
limiting factors on biological processes, connecting molecular biology (protein
structure) to ecosystem energetics (metabolic scaling).
Question 2 (Grid-In / Calculation - Genetics):
,Scenario: In a population of flowers, color is determined by a single gene with
incomplete dominance: RR=Red, Rr=Pink, rr=White. A survey of a population finds 36%
Red, 48% Pink, and 16% White flowers.
Question: Assuming the population is in Hardy-Weinberg equilibrium, what is the
frequency of the dominant allele (R)? Report your answer as a decimal to the nearest
hundredth.
Solution:
Scientific Practice: Practice 2: Mathematical Routines; Practice 3: Scientific
Questioning.
Deconstruction: Phenotype frequencies correspond directly to genotype frequencies
due to incomplete dominance: p² (Red/RR) = 0.36, 2pq (Pink/Rr) = 0.48, q² (White/rr) =
0.16.
Conceptual Synthesis: Applies the Hardy-Weinberg equilibrium model (p² + 2pq + q² = 1)
to a population exhibiting codominant/incompletely dominant inheritance, where
heterozygotes are phenotypically distinct.
Reasoning Pathway:
1. q² = frequency of rr (White) = 0.16.
2. q = √0.16 = 0.40 (frequency of recessive allele r).
3. p + q = 1 → p = 1 - 0.40 = 0.60 (frequency of dominant allele R).
4. Verification: p² = (0.6)² = 0.36 (Red), 2pq = 2(0.6)(0.4) = 0.48 (Pink). Data
matches.
Final Answer: 0.60
Higher-Order Insight: Illustrates that H-W equations model allele frequencies regardless
of dominance pattern; incomplete dominance simply allows direct observation of
, genotype frequencies, bypassing the need for additional statistical inference (unlike
complete dominance where p² = dominant phenotype).
Question 3 (Free-Response Stem with MCQ - Systems & Ecology):
Stem: A simplified nitrogen cycle diagram is shown (described: N₂ in atmosphere ->
[Nitrogen Fixation] -> NH₃ -> [Nitrification] -> NO₂⁻ & NO₃⁻ -> [Assimilation] -> Organic N
-> [Ammonification] -> NH₃ -> [Denitrification] -> N₂). An industrial farm applies
excessive ammonium (NH₄⁺) fertilizer.
Q3a: Which process, if accelerated by the excess ammonium, could most directly
contribute to the eutrophication of a nearby lake?
A. Nitrification
B. Nitrogen Fixation
C. Denitrification
D. Assimilation
Q3b: The farm's practice also increases microbial activity that converts the ammonium
into nitrous oxide (N₂O), a potent greenhouse gas. This is an example of:
A. Assimilation
B. Denitrification (incomplete)
C. Eutrophication
D. Nitrogen Fixation
Solution for Q3a:
Scientific Practice: Practice 1: Models & Representations; Practice 7: Relating
Knowledge Across Scales.
Deconstruction: Eutrophication requires bioavailable nitrogen (nitrate, NO₃⁻) that
leaches into aquatic systems, causing algal blooms and hypoxia.
Conceptual Synthesis: Links biogeochemical cycling (nitrogen transformation) to
ecosystem dynamics (nutrient limitation and productivity).
