Learn Extracted exam questions AP Biology 2019 Free Response
2019 Free Response
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[Figure 1. Model of two-step enzymatic plant pathway for synthesis of IAA from tryptophan: Gene Trp-T -> (arrow down) mRNA -> (arrow down) Enzyme Trp-T; Tryptophan --(Enzyme Trp-T)--> I3PA. Gene YUC -> (arrow down) mRNA -> (arrow down) Enzyme YUC; I3PA --(Enzyme YUC)--> IAA.]
Auxins are plant hormones that coordinate several aspects of root growth and development. Indole-3-acetic acid (IAA) is an auxin that is usually synthesized from the amino acid tryptophan (Figure 1). Gene Trp-T encodes an enzyme that converts tryptophan to indole-3-pyruvic acid (I3PA), which is then converted to IAA by an enzyme encoded by the gene YUC.
Circle ONE arrow that represents transcription on the template pathway. Identify the molecule that would be absent if enzyme YUC is nonfunctional.
Predict how the deletion of one base pair in the fourth codon of the coding region of gene Trp-T would most likely affect the production of IAA. Justify your prediction.
Explain one feedback mechanism by which a cell could prevent production of too much IAA without limiting I3PA production.
Rhizobacteria are a group of bacteria that live in nodules on plant roots. Rhizobacteria can produce IAA and convert atmospheric nitrogen into forms that can be used by plants. Plants release carbon-containing molecules into the nodules. Based on this information, identify the most likely ecological relationship between plants and rhizobacteria. Describe ONE advantage to the bacteria of producing IAA.
A researcher removed a plant nodule and identified several "cheater" rhizobacteria that do not produce IAA or fix nitrogen. Describe the evolutionary advantage of being a bacterial cheater in a population composed predominantly of noncheater bacteria. Plants can adjust the amount of carbon-containing molecules released into nodules in response to the amount of nitrogen fixed in the nodule. Predict the change in the bacterial population that would cause the plant to reduce the amount of carbon-containing molecules provided to the nodule.
A student studying two different aquatic, plant-eating, unicellular protist species (species A and B) designed an experiment to investigate the ecological relationship between the two species (Table 1).
TABLE 1. EXPERIMENTAL TREATMENT GROUPS
| Group | Description |
|---|---|
| Group I. | Species A and B are each grown in separate containers. |
| Group II. | Species A and B are grown together in the same container. |
In treatment group I, the student placed 10 individuals of species A into a container with liquid growth medium and 10 individuals of species B into a separate container with an equal amount of the same liquid growth medium. In treatment group II, the student placed 5 individuals of each species into a single container with the liquid growth medium. The student then maintained the containers under the same environmental conditions and recorded the number of individuals in each population at various time points. The results are shown in Table 2.
TABLE 2. NUMBER OF INDIVIDUALS IN EACH PROTIST POPULATION IN BOTH TREATMENT GROUPS
| Time (h) | Group I. Grown Separately — Species A | Group I. Grown Separately — Species B | Group II. Grown Together — Species A | Group II. Grown Together — Species B |
|---|---|---|---|---|
| 0 | 10 | 10 | 5 | 5 |
| 10 | 100 | 50 | 45 | 20 |
| 20 | 400 | 200 | 100 | 50 |
| 30 | 1100 | 500 | 250 | 25 |
| 40 | 1400 | 650 | 525 | 20 |
| 50 | 1500 | 700 | 900 | 10 |
| 60 | 1500 | 700 | 1250 | 0 |
| 70 | 1500 | 700 | 1400 | 0 |
[Graph template: a blank grid (graph paper) with two curves already plotted (shaded columns of Table 2): "Species A, Group II" - an S-shaped curve rising steeply then leveling off at the top right; "Species B, Group I" - an S-shaped curve rising then leveling off at a lower plateau, labeled partway up the right side. Axes are unlabeled on the template, to be completed by the student with the unshaded columns (Species A, Group I and Species B, Group II).]
The growth curves for species B in group I and for species A in group II (shaded columns) have been plotted on the template. Use the template to complete an appropriately labeled line graph to illustrate the growth of species A in treatment group I and species B in treatment group II (unshaded columns).
As shown in the table, the student established treatment group II with 5 individuals of each species. Provide reasoning for the reduced initial population sizes.
The student claims that species A and B compete for the same food source. Provide TWO pieces of evidence from the data that support the student's claim.
Predict TWO factors that will most likely limit the population growth of species A in treatment group I.
Many protists contain an organelle called a contractile vacuole that pumps water out of the cell. The student repeated the experiment using a growth medium with a lower solute concentration. Predict how the activity of the contractile vacuole will change under the new experimental conditions. Justify your prediction.
The pyruvate dehydrogenase complex (PDC) catalyzes the conversion of pyruvate to acetyl-CoA, a substrate for the Krebs (citric acid) cycle. The rate of pyruvate conversion is greatly reduced in individuals with PDC deficiency, a rare disorder.
Identify the cellular location where PDC is most active.
Make a claim about how PDC deficiency affects the amount of NADH produced by glycolysis AND the amount of NADH produced by the Krebs (citric acid) cycle in a cell. Provide reasoning to support your claims based on the position of the PDC-catalyzed reaction in the sequence of the cellular respiration pathway.
PDC deficiency is caused by mutations in the PDHA1 gene, which is located on the X chromosome. A male with PDC deficiency and a homozygous female with no family history of PDC deficiency have a male offspring. Calculate the probability that the male offspring will have PDC deficiency.
[Figure 1. Release of neurotransmitters into the synapse in response to an action potential: a diagram of a presynaptic neuron terminal releasing "Neurotransmitter" molecules (packaged in vesicles) into the "Synapse", which are received by receptor channels on the "Postsynaptic Neuron" membrane.]
[Figure 2. Model of a typical action potential in a neuron: a graph of Membrane Potential (mV) on the y-axis (labeled -70, -55, 0, 30) versus Time (milliseconds) on the x-axis (0 to 5); a dashed horizontal line at -55 mV labeled "Threshold"; the curve starts at resting potential -70 mV, rises sharply to a peak of about +30 mV around t=1.5 ms, then falls rapidly, undershoots to about -75 mV around t=3 ms, and returns gradually toward -70 mV by t=5 ms.]
Acetylcholine is a neurotransmitter that can activate an action potential in a postsynaptic neuron (Figures 1 and 2). A researcher is investigating the effect of a particular neurotoxin that causes the amount of acetylcholine released from presynaptic neurons to increase.
Describe the immediate effect of the neurotoxin on the number of action potentials in a postsynaptic neuron. Predict whether the maximum membrane potential of the postsynaptic neuron will increase, decrease, or stay the same.
The researcher proposes two models, A and B, for using acetylcholinesterase (AChE), an enzyme that degrades acetylcholine, to prevent the effect of the neurotoxin. In model A, AChE is added to the synapse. In model B, AChE is added to the cytoplasm of the postsynaptic cell. Predict the effectiveness of EACH proposed model. Provide reasoning to support your predictions.
TABLE 1. DIVERGENCE (IN PERCENT) OF MITOCHONDRIAL DNA SEQUENCES AMONG FIVE PRIMATE SPECIES
| Human | Gorilla | Orangutan | Gibbon | Chimpanzee | |
|---|---|---|---|---|---|
| Human | - | 10.3 | 16.1 | 18.1 | 8.8 |
| Gorilla | - | 16.7 | 18.9 | 10.6 | |
| Orangutan | - | 18.9 | 17.2 | ||
| Gibbon | - | 18.9 | |||
| Chimpanzee | - |
A researcher studying the evolutionary relationship among five primate species obtained data from a sequence of mitochondrial DNA (mtDNA) from a representative individual of each species. The researcher then calculated the percent divergence in the sequences between each pair of primate species (Table 1).
Based on fossil data, the researcher estimates that humans and their most closely related species in the data set diverged approximately seven million years ago. Using these data, calculate the rate of mtDNA percent divergence per million years between humans and their most closely related species in the data set. Round your answer to two decimal places.
Using the data in the table, construct a cladogram on the template provided. Provide reasoning for the placement of gibbons as the outgroup on the cladogram.
On the cladogram, draw a circle around all of the species that are descended from the species indicated by the node within the square.
[Cladogram template: a branching diagram with a single trunk splitting into a ladder of five branch tips (unlabeled horizontal lines) at increasing distances from the root; a small square marks one internal node partway along the tree; the bottom-most (most divergent) branch is labeled "Gibbon".]
TABLE 1 (data table referenced by the question, reproduced below as Table 1 with the growth results for the three haploid strains):
| MEDIUM | STRAINS — Wild Type | STRAINS — Mutant 1 | STRAINS — Mutant 2 | |
|---|---|---|---|---|
| Treatment I | All amino acids present | + | + | + |
| Treatment II | No amino acids present | + | − | − |
| Treatment III | All amino acids present EXCEPT methionine | + | − | + |
| Treatment IV | All amino acids present EXCEPT leucine | + | + | − |
Table 1. The data show the growth of haploid Saccharomyces cerevisiae yeast strains on media that differ in amino acid content. A plus sign (+) indicates that the yeast strains grow, and a minus sign (−) indicates that the strains do not grow.
The yeast Saccharomyces cerevisiae is a single-celled organism. Amino acid synthesis in yeast cells occurs through metabolic pathways, and enzymes in the synthesis pathways are encoded by different genes. The synthesis of a particular amino acid can be prevented by mutation of a gene encoding an enzyme in the required pathway.
A researcher conducted an experiment to determine the ability of yeast to grow on media that differ in amino acid content. Yeast can grow as both haploid and diploid cells. The researcher tested two different haploid yeast strains (Mutant 1 and Mutant 2), each of which has a single recessive mutation, and a haploid wild-type strain. The resulting data are shown in Table 1.
Identify the role of treatment I in the experiment.
Provide reasoning to explain how Mutant 1 can grow on treatment I medium but cannot grow on treatment III medium.
Yeast mate by fusing two haploid cells to make a diploid cell. In a second experiment, the researcher mates the Mutant 1 and Mutant 2 haploid strains to produce diploid cells. Using the table provided, predict whether the diploid cells will grow on each of the four media. Use a plus sign (+) to indicate growth and a minus sign (−) to indicate no growth.
Table for recording predictions:
| MEDIUM | Wild Type (haploid) | Mutant 1 (haploid) | Mutant 2 (haploid) | Diploid Cells Produced by Mating Mutant 1 and Mutant 2 | |
|---|---|---|---|---|---|
| Treatment I | All amino acids present | + | + | + | |
| Treatment II | No amino acids present | + | − | − | |
| Treatment III | All amino acids present EXCEPT methionine | + | − | + | |
| Treatment IV | All amino acids present EXCEPT leucine | + | + | − |
mRNA EXPRESSION LEVELS
[Figure 1. mRNA expression levels of six genes — a table with rows Liver, Heart, Brain, Kidney, Pancreas, Skeletal Muscle and columns Gene E, Gene F, Gene G, Gene H, Gene I, Gene J; each cell is shaded white (No mRNA), gray (Moderate amount of mRNA), or black (High amount of mRNA), as follows:
| Tissue | Gene E | Gene F | Gene G | Gene H | Gene I | Gene J |
|---|---|---|---|---|---|---|
| Liver | High | High | High | No | No | No |
| Heart | High | Moderate | Moderate | No | No | No |
| Brain | High | High | High | Moderate | Moderate | No |
| Kidney | No | Moderate | Moderate | High | High | No |
| Pancreas | No | No | Moderate | High | Moderate | No |
| Skeletal Muscle | No | No | High | Moderate | High | No |
| ] |
A researcher is studying patterns of gene expression in mice. The researcher collected samples from six different tissues in a healthy mouse and measured the amount of mRNA from six genes. The data are shown in Figure 1.
Based on the data provided, identify the gene that is most likely to encode a protein that is an essential component of glycolysis. Provide reasoning to support your identification.
The researcher observed that tissues with a high level of gene H mRNA did not always have gene H protein. Provide reasoning to explain how tissues with high gene H mRNA levels can have no gene H protein.
TABLE 1. CHANGES IN MORNING GLORY PETAL CELLS DURING FLOWER OPENING
[Diagram, BUD: a petal cell containing a central vacuole. On the plasma membrane: a "K+ Channel" (dark circle) allowing K+ to cross; a "Proton Pump" (light gray circle) pumping H+ out of the cell (labelled H+, H+ outside). On the vacuole membrane: a "K+/H+ Transport Protein (inactive)" (black oval, inactive) with H+ and K+ labelled nearby but not being actively transported.]
[Diagram, OPEN FLOWER: a larger petal cell/vacuole. On the plasma membrane: "K+ Channel" (dark circle) with K+ moving in; "Proton Pump" (light gray circle) pumping K+ in and H+ out (labelled K+, H+). On the vacuole membrane: "K+/H+ Transport Protein (active)" (black oval, active) transporting H+ out of the vacuole and K+ into the vacuole (labelled H+, K+, with multiple K+ ions shown accumulating inside the vacuole); H2O arrow entering the cell, labelled H2O.]
| BUD | OPEN FLOWER | |
|---|---|---|
| Vacuole pH | 6.6 | 7.7 |
| Flower Color | Red | Blue |
| Cell Volume | Small | Large |
The petal color of the Mexican morning glory (Ipomoea tricolor) changes from red to blue, and the petal cells swell during flower opening. The pigment heavenly blue anthocyanin is found in the vacuole of petal cells. Petal color is determined by the pH of the vacuole. A model of a morning glory petal cell before and after flower opening is shown in Table 1.
Identify the cellular component in the model that is responsible for the increase in the pH of the vacuole during flower opening AND describe the component's role in changing the pH of the vacuole.
A researcher claims that the activation of the K$^+$/H$^+$ transport protein causes the vacuole to swell with water. Provide reasoning to support the researcher's claim.