ap biology assessment 2 - oea.nyc · q1 d big idea 2, enduring understanding 2.a, essential...

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AP Biology Assessment 2

DATE OF ADMINISTRATION: JANUARY 8 – 12, 2017

TOPICS COVERED: ALL TOPICS THROUGH BASIC GENETICS

In the second assessment, the focus will be on all topics through basic Mendelian Genetics. There

will be 25 multiple choice questions and 3 grid-in responses (45 minutes) which will account for

50% of the exam score. There will also be 4 free-response questions (45 minutes) accounting for

the remaining 50%.

Multiple Choice Key and crosswalk

Question Key Description

Q1 D Big Idea 2, Enduring Understanding 2.A, Essential Knowledge 2.A.3

Q2 C Big Idea 4, EU 4.A, EK 4.A.1 a1,2 SP 7.1 LO 4.1

Q3 A Big Idea 2, EU 2.A, EK 2.A.2g

Q4 D Big Idea 4, EU 4.A, EK 4.A.1 a

Q5 D Big Idea 4, EU 4.B, EK 4.B.1d

Q6 A Big Idea 4, EU 4.B, EK 4.B.1d

Q7 B Big Idea 3, EU 3.A, EK 3.A.2b

Q8 C Big Idea 4, EU 4.A, EK 4.A.1

Q9 D Big Idea 4, EU 4.A, EK 4.A.2d

Q10 A Big Idea 1, EU 1.B, EK 1.B1

Q11 A Big Idea 3, EU 3.A, EK 3.A.2a

Q12 D Big Idea 2, EU 2.A, EK 2.A.2f

Q13 D Big Idea 2, EU 2.A, EK 2.B.1ab

Q14 C Big Idea 2, EU 2.B, EK 2.B.1ab

Q15 C Big Idea 1, EU 1.B, EK 1.B.1

Q16 B Big Idea 2, EU 2.A, EK 2.A.2c

Q17 D Big Idea 2, EU 2.A, EK 2.A.2c

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Question Key Description

Q18 A Big Idea 2, EU 2.A, EK 2.A.2c

Q19 C Big Idea 3, EU 3.A, EK 3.A.2a2

Q20 A Big Idea 3, EU 3.A, EK 3.A.2c

Q21 A Big Idea 1, EU 1.A, EK 1.B.1

Q22 A Big Idea 2, EU 2.A, EK 2.A.3

Q23 C Big Idea 3, EU 3.A, EK 3.A.2c

Q24 B Big Idea 2, EU 2.A, EK 2.A.1

Q25 A Big Idea 2, EU 2.A, EK 2.A.1d3

Q26 See Rubric Big Idea 2, EU 2.A, EK 2.A.3b

Q27 See Rubric Big Idea 2, EU 2.E, EK 2.E.1b3

Q28 See Rubric Big Idea 3, EU 3.A, EK 3.A.3a

Q29 See Rubric Big Idea 2, EU 2.A, EK 2.A.2dg

Q30 See Rubric Big Idea 1, EU 1.A, EK 1.A.1, EK 1.A.2

Q31 See Rubric Big Idea 4, EU 4.B, EK 4.B.1d

Q32 See Rubric Big Idea 3, EK 3.A.3, SP 2.2, LO 3.14

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GRID-IN ANSWERS (26-28)

Question 26

A hypothetical organism is composed of 27 cells, and each cell is in the approximate

shape of a cube. The side of each cube is 1 unit long. Determine the surface area-to-

volume ratio of the cells contained in this hypothetical organism. When gridding the

answer, use a fraction format. For example, a surface area-to-volume ratio may be 5:1

but the fraction format is 5/1

Each cube (cell) has six faces, thus each has a surface area of 6 and a volume of 1 the

surface to area ratio is 6:1 or 6

1

Accepted answers:

6/1

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Question 27

Respirometer readings from an experiment measuring oxygen consumption in germinating

peas are shown in the table below. An initial reading was recorded at time zero followed by

a final reading recorded after the experiment ran for 10 minutes.

Calculate the rate of respiration per minute for the peas in Respirometer Number 2. Give your answer to the nearest one thousandth mL.

𝑟𝑎𝑡𝑒 = Vol O2 final − Vol O2 initial

final time − initial time

𝑟𝑎𝑡𝑒 = 0.14 mL − 0.01mL

10 min − 0 min= 0.013

ml

min

Accepted answers: 0.013

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Question 28

In sheep, eye color is controlled by a single gene with two alleles. When a homozygous

brown-eyed sheep is crossed with a homozygous green-eyed sheep, blue-eyed offspring

are produced. If the blue-eyed sheep are mated with each other, what percent of their

offspring will most likely have brown eyes?

25% are Brown; 50% Blue; 25% Green

Accepted answers:

• 25 • 1/4

Brown

Eyes X Green Eyes

BB X B’B’

Blue Eyed Offspring = BB’

Blue Eyes X Blue Eyes

BB’ X BB’

B B’

B BB -

Brown

BB’ –

Blue

B’ BB’ –

Blue

B’B’ -

Green

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FREE RESPONSE SCORING GUIDELINES (29-32)

QUESTION 29

10 POINTS

Question 29 is a long free-response question. Questions 30 through 32 are short free-response questions. Read each question carefully and write your response. Answers must be written out. Outline form is not acceptable. It is important that you read each question completely before you begin to write.

An absorption spectrum indicates the relative amount of light absorbed across a range of wavelengths. The graphs above represent the absorption spectra of individual pigments isolated from two different organisms. One of the pigments is chlorophyll a, commonly found in green plants. The other pigment is bacteriorhodopsin, commonly found in purple photosynthetic bacteria. The table above shows the approximate ranges of wavelengths of different colors in the visible light spectrum.

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a. Identify the pigment (chlorophyll a or bacteriorhodopsin) used to generate the

absorption spectrum in each of the graphs above. Explain and justify your answer.

(3 points maximum)

1 point for each

Identify BOTH pigments: Graph 1 = bacteriorhodopsin AND graph 2 = chlorophyll a

Explain that an organism containing bacteriorhodopsin appears purple because the pigment absorbs light in the green range of the light spectrum and/or reflects violet or red and blue light. The reflected red and blue light appears purple.

Explain that an organism containing chlorophyll a appears green because the pigment absorbs light in the red and blue ranges of the light spectrum and/or reflects green light.

b. In an experiment, identical organisms containing the pigment from Graph II as the

predominant light- capturing pigment are separated into three groups. The organisms

in each group are illuminated with light of a single wavelength (650 nm for the first

group, 550 nm for the second group, and 430 nm for the third group). The three light

sources are of equal intensity, and all organisms are illuminated for equal lengths of

time. Predict the relative rate of photosynthesis in each of the three groups. Justify

your predictions.

(5 points maximum)

Wavelength (Group)

Prediction (1 point each)

Justification (1 point each)

650 nm

(1st Group)

Intermediate rate

An intermediate level of absorption occurs at 650 nm (compared to 430 nm and 550 nm); therefore, an intermediate amount of energy is available to drive photosynthesis.

550 nm

(2nd Group)

Lowest rate

The lowest level of absorption occurs at 550 nm; therefore, the least amount of energy is available to drive photosynthesis.

430 nm

(3rd Group)

Highest rate

The highest level of absorption occurs at 430 nm; therefore, the greatest amount of energy is available to drive photosynthesis.

NOTE: A student who combines two groups (e.g., “the 650 nm and 430 nm groups have

higher rates of photosynthesis compared to the 550 nm group”) can earn a maximum of 4

points: up to 2 points for the prediction and up to 2 points for the justification.

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c. Bacteriorhodopsin has been found in aquatic organisms whose ancestors existed before

the ancestors of plants evolved in the same environment. Propose a possible

evolutionary history of plants that could have resulted in a predominant photosynthetic

system that uses only some of the colors of the visible light spectrum.

(2 points maximum; 1 point for proposal including a selective pressure; 1 point for reasoning to support the proposal)

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QUESTION 30

4 POINTS

Many aquatic animals can hold their breath and dive underwater for long periods of time. During long dives, lactic acid builds up in tissues of the organism. The graph above shows blood lactate levels following dives of different duration for three species. Lactate is produced when lactic acid is dissolved in the blood and becomes ionized.

a. Based on the data, propose a hypothesis to explain the change in blood lactate

levels in Weddell seals for dives lasting longer than 20 minutes. (1 point)

• Proposal: Increase in lactate levels are due to

fermentation/anaerobic metabolism

b. Describe the most likely shape of the curve if blood oxygen levels of Weddell

seals were plotted rather than blood lactate levels. Include in your description

the likely shape of the curve between 0 and 20 minutes and the shape of the

curve after 20 minutes. (2 points maximum; points may be earned from only

one row)

0-20 min (1 point) After 20 min (1 point)

Oxygen levels start high and decline steadily

Oxygen levels decline more slowly or remain flat

Oxygen levels start and remain high Oxygen levels decline

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c. The data suggest that Baikal seals can sustain much longer dives than Emperor

penguins. Propose a hypothesis that could explain the evolution of different dive

responses in Emperor penguins and Baikal seals. (1 point)

• The (genetic/heritable) capacity to sustain dives for longer periods of

time provides selective/reproductive advantages (access to

food/avoidance of predators) for seals but not for penguins.

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QUESTION 31

3 POINTS.

The graph above shows the initial rate of an enzyme-catalyzed reaction at

different substrate concentrations in the presence of a constant concentration

of enzyme.

a. Connect the primary structure of the enzyme to its overall shape. (1 point)

• The amino acid sequence determines the overall shape (of the

polypeptide/protein/enzyme)

• R-groups interact and stabilize the structure

b. Predict the effect of adding a noncompetitive inhibitor to the reaction

mixture on the rate of reaction at a high substrate concentration. Support

your prediction by describing how a noncompetitive inhibitor affects the

structure and function of an enzyme. (2 points maximum; 1 point for

prediction, 1 point for support)

• Noncompetitive/allosteric inhibitor will decrease the (initial) rate of

reaction (at any substrate concentration)

• Inhibitor binding to site other than active site (allosteric site) changes

shape of enzyme, which alters the interaction of substrate with active

site

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QUESTION 32

3 POINTS

Fruit flies (Drosophila melanogaster) with a wild-type phenotype have gray bodies and red eyes. Certain mutations can cause changes to these traits. Mutant flies may have a black body and/or cinnabar eyes. To study the genetics of these traits, a researcher crossed a true-breeding wild-type male fly (with gray body and red eyes) with a true-breeding female fly with a black body and cinnabar eyes. All of the F1 progeny displayed a wild-type phenotype.

Female flies from the F1 generation were crossed with true-breeding male flies with black bodies and cinnabar eyes. The table below represents the predicted outcome and the data obtained from the cross. Explain three (3) differences between the expected data and the actual numbers observed.

(3 points maximum; 1 point for each difference, only the first 3 differences are accepted.) Student explanations could include the following:

• Prediction of a 1:1:1:1 ratio with correct phenotypes based on

independent assortment.

• Support for prediction with a diagram of the cross of BbEe x bbee.

• Correct application of chi-square analysis to show that observed

results do not conform to expected Mendelian frequencies.

• Identification of body color and eye color as linked genes/loci.

• Use of ratios to show linkage and independent assortment of wing type

versus linked traits.

• Identification of the bottom two phenotypes as products of crossing

over (recombinant chromosome).

• Mentioning that crossover rate is approximately 9–10 percent.

F2 Generation Phenotypes

Body Color

Eye Color

Number

Predicted

Number

Observed

Gray Red 244 455

Black Cinnabar 244 432

Gray Cinnabar 244 42

Black Red 244 47

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ap biology assessment 2 - oea.nyc · q1 d big idea 2, enduring understanding 2.a, essential...

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