AP Biology Midterm Answer Key

Multiple Choice

1. A2. A3. D4. A5. B6. B7. B8. B9. B10. D11. B12. B13. A

1999

14. B15. A16. A17. A18. D19. D20. B21. D22. A23. D

2012

24. C

2002

25. B26. D27. B28. B29. C30. C31. D

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)

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 axesData Plotted (1 point)•Correctly plotted points in proper orientation•Points may or may not be connected with a line•Bar graph acceptableOptimum 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 rateElaboration (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 baseControl – identification of a control treatmentMeasureable product per unit timeMultiple trialsPredication – 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)

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: 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.

Question 3 – 2012

a

b

c

d

(4 points maximum)

Type of mutation(not limited to the

following)

Description (1 point per box) Effect (1 point per box)

Silent Nucleotide change. No change in aminoacid/protein sequence

Missense/substitution Nucleotide change causes newcodon.

Different amino acid/proteinsequence.

Nonsense/substitution Nucleotide change causes stopcodon.

Protein not formed ORtruncated protein.

Frameshift(insertion/deletion) Nucleotide insertion/deletion altersreading frame after mutation.

Changes amino acid/proteinsequence OR nonfunctionalprotein OR no protein.

Regulatory region Nucleotideinsertion/deletion/substitution.

Alters gene expression ORalters splice site.

Translocation Chromosome segment moves todifferent site.

Alters gene expression.

Nondisjunction Chromosomes fail to separate. “”Duplication Chromosome segment doubles. “”Deletion Chromosome segment is

removed.“”

Inversion Chromosome segment is reversed.

“”

Transposition Chromosome segment moves to adifferent site.

“”

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