Learn Extracted exam questions AP Biology 2017 Free Response
2017 Free Response
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TABLE 1. EFFECT OF 0.1 mM CAFFEINE ON MEMORY IN BEES
| Treatment | Memory (average probability of revisiting a nectar source $\pm 2SE_{\bar{X}}$) — 10 Minutes | Memory (average probability of revisiting a nectar source $\pm 2SE_{\bar{X}}$) — 24 Hours |
|---|---|---|
| Control | $0.72 \pm 0.09$ | $0.41 \pm 0.07$ |
| Caffeine | $0.83 \pm 0.07$ | $0.78 \pm 0.08$ |
In flowering plants, pollination is a process that leads to the fertilization of an egg and the production of seeds. Some flowers attract pollinators, such as bees, using visual and chemical cues. When a bee visits a flower, in addition to transferring pollen, the bee can take nectar from the flower and use it to make honey for the colony.
Nectar contains sugar, but certain plants also produce caffeine in the nectar. Caffeine is a bitter-tasting compound that can be toxic to insects at high concentrations. To investigate the role of caffeine in nectar, a group of researchers studied the effect of 0.1 mM caffeine on bee behavior. The results of an experiment to test the effect of caffeine on bees' memory of a nectar source are shown in Table 1.
On the axes provided, construct an appropriately labeled graph to illustrate the effect of caffeine on the probability of bees revisiting a nectar source (memory).
[A blank grid (graph paper) is provided below the question for the student to draw the graph.]
Based on the results, describe the effect of caffeine on each of the following:
- Short-term (10 minute) memory of a nectar source
- Long-term (24 hour) memory of a nectar source
Design an experiment using artificial flowers to investigate potential negative effects of increasing caffeine concentrations in nectar on the number of floral visits by bees. Identify the null hypothesis, an appropriate control treatment, and the predicted results that could be used to reject the null hypothesis.
Researchers found that nectar with caffeine tends to have a lower sugar content than nectar without caffeine. Plants use less energy to produce the caffeine in nectar than they do to produce the sugar in nectar. Propose ONE benefit to plants that produce nectar with caffeine and a lower sugar content. Propose ONE cost to bees that visit the flowers of plants that produce nectar with caffeine and a lower sugar content.
[Figure 1: Line graph titled "The effect of karrikins (KAR) and trimethylbutenolides (TMB) on seed germination in Lactuca plants. Error bars represent $\pm 2SE_{\bar{X}}$." Y-axis: "Seed Germination (%)", from 0 to 100 in increments of 10. X-axis: "Time (h)", with tick marks at 0, 12, 24, 36, 48. Four curves plotted:
- KAR (solid line, open squares): 0% until ~13 h, rises steeply to ~28% at 14 h, ~70% at 16 h, ~86% at 20 h, ~93% at 24 h, plateaus at ~96% from 36–48 h.
- Control (solid line, filled circles): 0% until ~16 h, rises to ~28% at 24 h, ~73% at 36 h, ~79% at 48 h.
- KAR + TMB (dashed line, open circles): 0% until ~16 h, rises slowly to ~11% at 20 h, ~19% at 22 h, ~28% at 36 h, ~28% at 48 h.
- TMB (solid line, filled squares): 0% until ~16 h, rises slowly to ~7% at 22 h, ~16% at 36 h, ~16% at 48 h. All curves start at 0% and are flat near 0 until roughly 12–16 h, then rise and plateau by 36–48 h, with KAR reaching the highest final germination percentage and TMB the lowest.]
Fires frequently occur in some ecosystems and can destroy all above-ground vegetation. Many species of plants in these ecosystems respond to compounds in smoke that regulate seed germination after a major fire. Karrikins (KAR) and trimethylbutenolides (TMB) are water-soluble compounds found in smoke that are deposited in the soil as a result of a fire. KAR and TMB bind to receptor proteins in a seed. In a study on the effects of smoke on seeds, researchers recorded the timing and percent of seed germination in the presence of various combinations of KAR and TMB. The results are shown in Figure 1.
In a second investigation into the effect of available water on seed germination after a fire, researchers treated seeds with KAR or TMB. The treated seeds were then divided into two treatment groups. One group received a water rinse and the other group received no water rinse. The seeds were then incubated along with a group of control seeds that were not treated. The results are shown in the table.
EFFECT OF CHEMICAL TREATMENT AND WATER RINSE ON GERMINATION
| Treatment Group | Chemical Treatment — KAR | Chemical Treatment — TMB | Water | Germination Result |
|---|---|---|---|---|
| 1 (control) | − | − | − | Control result |
| 2 | + | − | − | Different from control |
| 3 | − | + | − | Different from control |
| 4 (control) | − | − | + | Control result |
| 5 | + | − | + | Different from control |
| 6 | − | + | + | Same as control |
The researchers made the following claims about the effect of KAR and the effect of TMB on seed germination relative to the control treatment.
- KAR alone affects the timing of seed germination.
- KAR alone affects the percentage of seeds that germinate.
- TMB alone affects the timing of seed germination.
- TMB alone affects the percentage of seeds that germinate.
Provide support using data from Figure 1 for each of the researchers' claims.
Make a claim about the effect of rinsing on the binding of KAR to the receptor in the seed and about the effect of rinsing on the binding of TMB to the receptor in the seed. Identify the appropriate treatment groups and results from the table that, when compared with the controls, provide support for each claim.
There is intense competition by plants to successfully colonize areas that have been recently cleared by a fire. Describe ONE advantage of KAR regulation and ONE advantage of TMB regulation to plants that live in an ecosystem with regular fires.
Gibberellin is the primary plant hormone that promotes stem elongation. GA 3-beta-hydroxylase (GA3H) is the enzyme that catalyzes the reaction that converts a precursor of gibberellin to the active form of gibberellin. A mutation in the GA3H gene results in a short plant phenotype. When a pure-breeding tall plant is crossed with a pure-breeding short plant, all offspring in the $F_1$ generation are tall. When the $F_1$ plants are crossed with each other, 75 percent of the plants in the $F_2$ generation are tall and 25 percent of the plants are short.
Second Base in Codon
| First \ Second | U | C | A | G | Third |
|---|---|---|---|---|---|
| U | UUU, UUC = Phe; UUA, UUG = Leu | UCU, UCC, UCA, UCG = Ser | UAU, UAC = Tyr; UAA = Stop; UAG = Stop | UGU, UGC = Cys; UGA = Stop; UGG = Trp | U, C, A, G |
| C | CUU, CUC, CUA, CUG = Leu | CCU, CCC, CCA, CCG = Pro | CAU, CAC = His; CAA, CAG = Gln | CGU, CGC, CGA, CGG = Arg | U, C, A, G |
| A | AUU, AUC, AUA = Ile; AUG = Met or Start | ACU, ACC, ACA, ACG = Thr | AAU, AAC = Asn; AAA, AAG = Lys | AGU, AGC = Ser; AGA, AGG = Arg | U, C, A, G |
| G | GUU, GUC, GUA, GUG = Val | GCU, GCC, GCA, GCG = Ala | GAU, GAC = Asp; GAA, GAG = Glu | GGU, GGC, GGA, GGG = Gly | U, C, A, G |
Figure 1. The universal genetic code
The wild-type allele encodes a GA3H enzyme with alanine (Ala), a nonpolar amino acid, at position 229. The mutant allele encodes a GA3H enzyme with threonine (Thr), a polar amino acid, at position 229. Describe the effect of the mutation on the enzyme and provide reasoning to support how this mutation results in a short plant phenotype in homozygous recessive plants.
Using the codon chart provided, predict the change in the codon sequence that resulted in the substitution of alanine for threonine at amino acid position 229.
Describe how individuals with one (heterozygous) or two (homozygous) copies of the wild-type GA3H allele can have the same phenotype.
DIETARY COMPOSITION OF ORGANISMS IN AN AQUATIC ECOSYSTEM
| Organism | Food Source (% of diet) — Algae | Food Source (% of diet) — Stoneflies | Food Source (% of diet) — Midges | Food Source (% of diet) — Hellgrammites | Food Source (% of diet) — Caddisflies |
|---|---|---|---|---|---|
| Algae | |||||
| Stoneflies | 90 | 10 | |||
| Midges | 100 | ||||
| Hellgrammites | 20 | 10 | 70 | ||
| Caddisflies | 70 | 30 |
The table above shows how much each organism in an aquatic ecosystem relies on various food sources. The rows represent the organisms in the ecosystem, and the columns represent the food source. The percentages indicate the proportional dietary composition of each organism. High percentages indicate strong dependence of an organism on a food source.
[Figure: A food-web template with blank horizontal lines to be labeled with organism names by the student, arrows already drawn between the (student-labeled) boxes toward a box labeled "Midges," and a vertical double-headed arrow labeled "Increasing Trophic Level" alongside the template. Three arrows point into "Midges" from three blank labeled lines above/below it, indicating energy flow directions the student must complete by naming the organisms at each blank line.]
Based on the food sources indicated in the data table, construct a food web in the template below. Write the organism names on the appropriate lines AND draw the arrows necessary to indicate the energy flow between organisms in the ecosystem.
In an effort to control the number of midges, an area within the ecosystem was sprayed with the fungus Metarhizium anisopliae, which significantly decreased the midge population. Based on the data in the table, predict whether the spraying of the fungus will have the greatest short-term impact on the population of the stoneflies, the caddisflies, or the hellgrammites. Justify your prediction.
[Figure 1: Line graph titled "Characteristics of a pond community over time." Y-axis: "Relative Concentration" (unscaled). X-axis: "Time", with four labeled time points I, II, III, IV. Legend shows three lines: solid = Oxygen, dashed = Cyanobacteria, dotted = Decomposers. The Cyanobacteria (dashed) curve starts at a low flat baseline, rises steeply to peak at time II, then declines back toward baseline by time III–IV. The Oxygen (solid) curve starts at a slightly higher flat baseline, stays flat until just before time II, rises to peak just after time II (slightly later/higher than cyanobacteria's peak), then declines steeply, dropping below the starting baseline by time IV. The Decomposers (dotted) curve starts at the lowest flat baseline, stays low through time II, then rises to peak between times II and III (later than both other curves), and declines gradually toward IV, remaining above its original baseline.]
Microcystis aeruginosis is a freshwater photosynthetic cyanobacterium. When temperatures increase and nutrients are readily available in its pond habitat, M. aeruginosis undergoes rapid cell division and forms an extremely large, visible mass of cells called an algal bloom. M. aeruginosis has a short life span and is decomposed by aerobic bacteria and fungi. Identify the metabolic pathway and the organism that is primarily responsible for the change in oxygen level in the pond between times I and II AND between times III and IV.
[Figure 1: Diagram titled "Comet assay to detect double-stranded breaks in DNA," showing three sequential panels connected by labeled steps. Panel 1: "Cell with DNA Damage" — a large irregular circle (the cell) containing a smaller circle (the nucleus) with a tangled DNA symbol inside, with two lines extending from the nucleus to a small circle in Panel 2. Panel 2: "Individual Nucleus Placed on Slide" — a rectangle (representing the slide/gel) containing a single small filled circle (the nucleus). An arrow labeled "Electrophoresis" points from Panel 2 to Panel 3. Panel 3: "Intact DNA Remains in the Head and Damaged DNA Is in the Tail" — a rectangle with a "−" (minus) label on the top-left corner and a "+" (plus) label on the top-right corner, containing a shape labeled "Head" (a larger dark circular/oval region on the left, representing intact DNA) connected to a smaller lighter oval labeled "Tail" (representing damaged DNA fragments) extending to the right, toward the "+" electrode.]
A comet assay is a technique used to determine the amount of double-strand breaks in DNA (DNA damage) in cells. The nucleus of an individual cell is placed on a microscope slide coated with an agarose gel. An electric current is applied to the gel that causes DNA to move (electrophoresis), and the DNA is stained with a fluorescent dye. When viewed using a microscope, undamaged DNA from the nucleus appears as a round shape (the head), and the fragments of damaged DNA extend out from the head (the tail). The length of the tail corresponds to the amount of the damage in the DNA (see Figure 1).
To explain the movement of DNA fragments in the comet assay, identify one property of DNA and provide reasoning to support how the property contributes to the movement during the comet assay technique.
In a different experiment, cells are treated with a chemical mutagen that causes only nucleotide substitutions in DNA. Predict the likely results of a comet assay for this treatment.
Many species of bacteria grow in the mouths of animals and can form biofilms on teeth (plaque). Within plaque, the outer layers contain high levels of oxygen and the layers closest to the tooth contain low levels of oxygen. The surface of the tooth is covered in a hard layer of enamel, which can be dissolved under acidic conditions. When the enamel breaks down, the bacteria in plaque can extract nutrients from the tooth and cause cavities.
Certain types of bacteria (e.g., Streptococcus mutans) thrive in the innermost anaerobic layers of the plaque and are associated with cavities. Other types of bacteria (Streptococcus sanguinis) compete with S. mutans but are unable to thrive in acidic environments.
Identify the biochemical pathway S. mutans uses for metabolizing sugar and describe how the pathway contributes to the low pH in the inner layers of plaque.
Normal tooth brushing effectively removes much of the plaque from the flat surfaces of teeth but cannot reach the surfaces between teeth. Many commercial toothpastes contain alkaline components, which raise the pH of the mouth. Predict how the population sizes of S. mutans AND S. sanguinis in the bacterial community in the plaque between the teeth are likely to change when these toothpastes are used.
Estrogens are small hydrophobic lipid hormones that promote cell division and the development of reproductive structures in mammals. Estrogens passively diffuse across the plasma membrane and bind to their receptor proteins in the cytoplasm of target cells.
Describe ONE characteristic of the plasma membrane that allows estrogens to passively cross the membrane.
In a laboratory experiment, a researcher generates antibodies that bind to purified estrogen receptors extracted from cells. The researcher uses the antibodies in an attempt to treat estrogen-dependent cancers but finds that the treatment is ineffective. Explain the ineffectiveness of the antibodies for treating estrogen-dependent cancers.