[qwiz dataset=”AP Bio FRQ Dataset 2022″]
[h] AP Bio FRQ Review Dataset, 2022, aligned to 2019 CED with topics
[i] This is the actual dataset
[q json=”true” xx=”1″ multiple_choice=”false” unit=”1.Chemistry of Life” dataset_id=”AP Bio FRQ Dataset 2022|2a51fe943407″ question_number=”1″ topic=”1.1.Structure_of_Water_and_Hydrogen_Bonding”]
List three phenomena that involve hydrogen bonding. Limit your response to processes that directly involve water molecules.
2) On a larger scale, hydrogen bonds allow large bodies of water to absorb lots of heat energy as the temperature of the environment increases. That helps to create moderate temperatures in any region near a large body of water (a large lake or an ocean). This also has a moderating effect on the climate of the Earth as a whole (since 3/4 of Earth’s surface is covered by oceans), keeping our planet’s temperature optimal for life.
3) Because the breaking of hydrogen bonds requires energy, water has a very high heat of vaporization. That means that when water is converted into water vapor (which happens when sweat evaporates from human skin or from a dog’s tongue), it carries away a lot of heat energy, lowering the temperature of the body that it evaporated from. This is how evaporative cooling works, and it’s a key thermoregulatory adaptation in humans and other animals.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”1.Chemistry of Life” dataset_id=”AP Bio FRQ Dataset 2022|2ac1c0cef407″ question_number=”2″ topic=”1.3.Monomers_and_Polymers”] Dehydration synthesis and hydrolysis are metabolic processes that happen in living things.
PART 1: What do these two processes have in common?
PART 2: Describe a biological use of each process.
[q json=”true” xx=”3″ multiple_choice=”false” unit=”1.Chemistry of Life” dataset_id=”AP Bio FRQ Dataset 2022|2b33d7159807″ question_number=”3″ topic=”1.4.Carbohydrates_and_Lipids”] Refer to the four molecules below to answer the following questions.
PART 1: Identify and state the name of the molecule that will organize itself into a bilayer when mixed with water.
PART 2: Sketch a bilayer.
PART 3: Explain bilayer formation.
PART 2: Bilayer sketch:
PART 3: Phospholipids have a polar head and a nonpolar tail. When mixed with water, they will spontaneously organize itself into several possible configurations, one of which is a phospholipid bilayer. The bilayer emerges because the polar heads are attracted to water, while the nonpolar tails are repelled by water. In the bilayer, the tails of the phospholipids form a hydrophobic, water free zone, while the heads face outward toward the cell exterior and the cytoplasm where they form hydrogen bonds with water molecules on the outside and inside of the cell.
If you want this same explanation in verse, here it is. This is from Mr. W’s sciencemusicvideos Membranes! Rap (which you can find on sciencemusicvideos.com).
There’s a head and a tail on every phospholipid,
The tail’s two long chain fatty acids
Bound to a glycerol, it’s made to order
The tail’s non-polar –hydrophobic– fears water
The head’s got a phosphate, it’s charged negatively,
makes the head hydrophilic–plays in water happily
So tail avoids water while the head’s attracted to it,
When phospholipids form the membrane that’s how they do it.
Cause when phospholipids into water get submerged,
A phospholipid bilayer structure will emerge
The tails hang together in a water free zone,
Hear their hydrophobic moan, “water leave me alone!”
While the heads are sticking out touching all those H2Os
Tails in, heads out, it’s how every membrane goes
Tails in, heads out, in a cellular sphere,
It’s the bilayered basis of membranes everywhere.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”1.Chemistry of Life” dataset_id=”AP Bio FRQ Dataset 2022|29d8ec29e407″ question_number=”4″ topic=”1.5.Proteins”]
Part 1: Identify the type of biological molecule shown below.
Part 2: Explain the molecular interactions that result in this molecule’s three dimensional structure (as shown in number 4).
UGFydCAy[Qq]: Protein structure results from at least 3 levels of interactions between the amino acids that make up the protein. In this case, however, there are four levels of interaction. The first level (shown at “1”) is the sequence of amino acids making up the protein. This is called primary structure. In secondary structure, the parts of the amino acid chain’s polypeptide backbone interact with one another to bend the chain into an alpha helix (shown at “2” above) or a beta pleated sheet. These interactions consist of hydrogen bonds between the carbonyl and amino groups within the polypeptide backbone. The third level, or tertiary structure (shown at “3”), involves interactions between amino acid side chains, or R-groups. These can include ionic bonds, hydrophobic clustering, hydrogen bonds, and covalent disulfide bridges between side chains that end in sulfhydryl groups. Finally, quaternary structure, shown in number “4” above, involves interactions between folded tertiary polypeptides.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”2.Cell Structure and Function” dataset_id=”AP Bio FRQ Dataset 2022|2620a52aa007″ question_number=”5″ topic=”2.1-2.2.Cell_Parts”] In terms of organelles, what specialized structures would you see in a cell whose function was either locomotion or transport?
[q json=”true” xx=”1″ multiple_choice=”false” unit=”2.Cell Structure and Function” dataset_id=”AP Bio FRQ Dataset 2022|1fe5c1598c07″ question_number=”6″ topic=”2.8.Tonicity_and_Osmoregulation”] The freshwater unicellular eukaryote Paramecium caudatum has a contractile vacuole that adapts to changes in osmolarity by adjusting pumping frequency. If Paramecia need to pump out more water from the cell, the contractile vacuole pumps at a faster rate (and the reverse is true as well).
A student does an experiment where she creates solutions of increasing osmolarity, adds Paramecia, and then counts the number of contractile vacuole pumps/minute. None of the solutions have a higher osmolarity than the osmolarity of the Paramecium’s cytosol. Predict what will happen.Justify your prediction using terms related to osmosis (hypotonic, hypertonic, isotonic) or water potential.
SlVTVElGSUNBVElPTiA6IEEg[Qq]Paramecium’s cytosol is hypertonic to the Paramecium’s freshwater environment. As a result, water constantly flows into the Paramecium. To maintain its osmotic balance (or “osmoregulate”) the Paramecium uses its contractile vacuole to pump water out.
As the experimenter increases the molarity of the surrounding solution, the osmolarity of that solution will become increasingly close to the osmolarity of the Paramecium. With a water concentration gradient that’s less steep, water will diffuse in at a slower rate, and the contractile vacuole won’t have to pump as often. Eventually, the environmental osmolarity will rise to a point where the environment is isotonic to the cytosol, and, at that point, the contractile vacuole wouldn’t have to contract at all.
ALTERNATIVE EXPLANATION (WATER POTENTIAL): In terms of water potential, water always flows from higher to lower water potential. A Paramecium in freshwater has a much higher solute concentration than its environment. That lowers the overall water potential in the paramecium, so water flows into the Paramecium. To counteract that flow, the contractile vacuole pumps water out. As the solute concentration of the surrounding solution increases, the water potential of the solution outside the paramecium decreases. As a result, less water flows into the paramecium, and the contractile vacuole responds by pumping at a slower rate. NOTE: there might be differences in the pressure potential related to the Paramecium’s pellicle: a kind of protein girdle that prevents over expansion (playing much the same role as the cell wall in a plant cell).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|294f8d6c5807″ question_number=”7″ topic=”3.1.Enzyme_Structure”] People with chronic high blood pressure have elevated blood levels of ACE (Angiotensin-converting enzyme). ACE acts on the polypeptide Angiotensin I to produce Angiotensin II, which has the effect of raising blood pressure.
Enzymes have complex three dimensional shapes. Additionally, the amino acids that make up enzymes can have charged side-chains. The diagram below represents the active site of ACE. The “+” sign represents positively charged regions in the active site.
A pharmaceutical company is trying to develop a drug to lower blood pressure. As part of this effort, a range of drugs was designed and manufactured. A sample of the molecular shape of each is shown below. Notice the “+” and “-” signs, which represent positively and negatively charged regions in the molecules making up the candidate drugs.
PART 1: IDENTIFY the drug that is most likely to be effective in preventing
excessive high blood pressure? JUSTIFY your choice.
PART 2: EXPLAIN the process by which this drug would contribute to lowering a person’s
UEFSVCAyOiA=[Qq]The drug would work through competitive inhibition, by blocking the active site of ACE and preventing it from acting on its substrate (Angiotensin I).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|23c7f12ed807″ question_number=”8″ topic=”3.3.Environmental_Impacts_on_Enzyme_Function”] You’re doing a lab with an enzyme called catalase, which you extracted from chicken liver. Catalase breaks down hydrogen peroxide into water and oxygen. Your lab group sets up the following experiment: you set up six test tubes, each with 5 mL of hydrogen peroxide. To each test tube, you add either an acid or a base to set to overall pH to 5, 6, 7, 8, and 9. You add 1 mL of catalase, and you measure the amount of oxygen produced.
PART 1: Knowing that the pH of liver tissue is about 7 (neutral) DRAW a graph predicting what the results will be.
PART 2: JUSTIFY your prediction. Make sure that your answer discusses the relationship between protein structure and factors in the environment.
PART 2: Catalase is an enzyme. Like almost all enzymes it’s a protein, with a very specific 3-dimensional shape. In enzymes, the shape of the enzyme’s active site is what enables it to bind with its substrate and catalyze the reaction changing substrate into product.
As with all proteins, the enzyme’s shape emerges as a result of a constellation of interactions between the amino acids that make up the enzyme. While some of these interactions are strong bonds (such as covalent bonds), others include hydrogen bonds and ionic attractions that are affected by the enzyme’s environment. Since each number on the pH scale represents a 10-fold shift in concentrations of protons or hydroxide ions, a change in pH represents an environmental change that can alter a the enzyme’s shape. In the case of this lab, the enzyme’s active site was altered to such a degree at pH 5 and 9 that the catalase couldn’t bind with the hydrogen peroxide, and, as a result, no oxygen was produced. On the other hand, at the enzyme’s pH optimum (7), the most substrate would be converted into product, and the most oxygen would be produced.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|1dab51f85807″ question_number=”9″ topic=”3.3.Environmental_Impacts_on_Enzyme_Function”] The diagram below shows an enzyme that plays a role in the metabolic pathways of glycolysis interacting with a metabolic poison. Name of the type of interaction shown, describe the interaction, and describe the impact of this interaction on cellular respiration.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|278bdc699007″ question_number=”10″ topic=”3.4.Cellular_Energy”] On the most fundamental level, what are the three ways that ATP can be made? Briefly describe each one.
In oxidative phosphorylation, electrons from food travel along an electron transport chain to oxygen, which acts as the final electron acceptor. Electron flow powers proton pumps that move protons from the mitochondrial matrix to the intermembrane space. This creates a concentration gradient. When protons flow down this gradient through ATP synthase, their kinetic energy powers the formation of ATP from ADP and phosphate.
Photophosphorylation is very similar to oxidative phosphorylation, except for that electron flow is powered by light, and protons are pumped from the stroma of the chloroplast into the thylakoid space. When these protons diffuse back into the stroma through ATP synthase, ATP is formed.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|23772d89abc806″ question_number=”106″ topic=”3.5.Photosynthesis”] The graph below shows the relationship between carbon dioxide uptake and light intensity in two plant species, A and B. One of these species is adapted to living in full sun, while the other is adapted to shade.
PART 1: Explain what processes can be measured by measuring carbon dioxide uptake.
PART 2: Using TWO pieces of information from this graph, determine which species is adapted to living in the shade. Explain your answer.
PART 3: Identify the species with the lower rate of respiration. Explain how this can be determined from the graph.
PART 4: Describe the significance that a low rate of respiration could have for this plant in its natural habitat.
[f]IFBBUlQgMTo=IENPMg==IGlzIGNvbnN1bWVkIGR1cmluZyBwaG90b3N5bnRoZXNpcyAod2hlcmUgaXQmIzgyMTc7cyBhbiBpbnB1dCBmb3IgZm9ybWF0aW9uIG9mIGNhcmJvaHlkcmF0ZSkuIENPMg==[Qq] is produced during cellular respiration. CO2 uptake indirectly measures the difference between the rates of photosynthesis and respiration, which also provides a measure of growth.
PART 2: Species B lives in the shade. This is indicated by 1) it reaches its maximum rate of CO2 uptake at a much lower light intensity than species A, 2) its CO2 uptake is lower in high light intensities, suggesting it is unable to make good use of increased light.
PART 3: Species B has a lower rate of respiration. At light intensity of zero, the plant is giving out 0.05 dm3/m-2 CO2 through respiration compared to 0.1 dm3/m-2 in species A.
PART 4:The amount by which total photosynthesis exceeds respiration determines the growth rate of a plant. Plant B is adapted to living in shady light conditions. Because of low light intensities in the shade, plant B would have a reduced rate of photosynthesis. A low rate of respiration would result in less glucose being broken down, thus more available for growth.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|28db2319d007″ question_number=”11″ topic=”3.5.Photosynthesis”] The action spectrum for photosynthesis (shown below) dips in the green part of the visible light spectrum (b), and rises in the red and blue parts (c and a) of the spectrum. How does this relate to what happens in photosynthesis?
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|286b60df1007″ question_number=”12″ topic=”3.5.Photosynthesis”] A student wants to set up a lab that compares overall photosynthetic activity in algae under varying light intensities. Describe at least two things that the student could measure, and predict the relationship between light intensity and these measured quantities.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|27fb9ea45007″ question_number=”13″ topic=”3.5.Photosynthesis”] What metabolic processes bring carbon into living systems? What processes return carbon to the atmosphere? For the purposes of reviewing for the AP exam, try to elaborate by describing some of the sub-processes involved.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|253c789d5807″ question_number=”14″ topic=”3.6.Respiration”] In one version of the cellular respiration lab, you measure oxygen consumption in organisms like peas in a device called a respirometer. If your experiment worked, you would have shown that respiring peas in a cold environment (often the temperature of ice water: about 4 C°) consume less oxygen than respiring peas in a warm environment (about 30 C°).
Imagine a version of this lab with huge respirometers into which you could compare the respiration rates of lizards and mice in cold and warm environments. You’re a good enough scientist to control for equal body mass of each type of organism. So you’ve got 1 kilogram of living mice stacked up against a kilogram of lizard. Compare and contrast rates of respiration in these two types of organisms in cold (about 4° C) and warm (about 30° C) and explain why one organism is metabolically more similar to peas than the other one is.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|228ff2f58007″ question_number=”15″ topic=”3.5.Photosynthesis”] Your teacher hands your lab group 20 radish seeds, two plastic bags, two paper towels, and a dropper bottle with water. She instructs you to divide the seedlings into two groups of 10 seeds, to mass each group, to water the seeds, and to keep one group in the dark and the second in the light. You continue to water the seeds for three weeks. After three weeks, you place the plants in a food dryer which removes all of their water, and you mass the plants. What happens to the mass in each group of plants, and why?
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|221934971407″ question_number=”16″ topic=”3.6.Respiration”] Sometimes, as you respire, your cells completely oxidize food to carbon dioxide and water. At other times, you partially oxidize food to lactic acid. When does each process happen, and why? As you answer, use this as an opportunity to review by describing how each process works.
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IChhIHN1YnN0cmF0ZSBmb3Igb25lIG9mIHRoZSBlbnp5bWVzIGluIHRoZSBnbHljb2x5c2lzIHBhdGh3YXkpLiBUbyByZWdlbmVyYXRlIHRoZSBOQUQ=[Qq]+, your cells chemically reduce pyruvate to lactic acid, while simultaneously oxidizing the NADH produced in glycolysis back to NAD+. As opposed to aerobic respiration, which can be sustained for very long periods of time, completely anaerobic respiration can only be sustained for a few minutes (after which you need to stop, rest, and switch back to aerobic metabolism).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|21a4ca448c07″ question_number=”17″ topic=”3.6.Respiration”] NADH and NADPH play similar roles in different processes. Explain, using this as an opportunity to explain each process in some detail.
NADPH carries electrons from Photosystem 1, at the end of the light reactions, to the Calvin cycle. In the Calvin cycle, NADPH’s job is to drop off these energetic electrons (along with hydrogen) in order to chemically reduce carbon dioxide to carbohydrate. As NADPH drops off its electrons and hydrogen, it becomes oxidized to NADP+. In this oxidized form, NADP+ can act as the final electron acceptor in the light reactions, accepting electrons (originally from water). This chemically reduces the NADP+ into NADPH, which can then bring this reducing power over to the Calvin cycle (which is where we started).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|1d0856b80007″ question_number=”18″ topic=”3.6.Respiration”] The diagram below represents a portion of the electron transport system in the inner mitochondrial membrane. As is shown, this pathway transfers electrons from NADH (and FADH2) to O2. Energy from this reaction is coupled to pumping of H+.
Normally, cells carrying out respiration consume O2, produce CO2, and produce 36 ATP for each glucose consumed.
As is shown in the diagram, two drugs, drug X and drug Y, can pick up electrons from particular intermediates in this pathway.
PART 1: Cells carrying out respiration are treated with a saturating dose of drug X, so that all the electrons which would normally continue along the pathway are captured by drug X. Under these conditions, PREDICT whether:
* The cells will continue to consume O2.
* The cells will continue to produce CO2.
* The rate of ATP synthesis will increase, decrease or stay the same.
For each prediction, EXPLAIN your reasoning.
PART 2: Cells carrying out respiration are treated with a saturating dose of drug Y, so that all the electrons which would normally continue along the pathway are captured by drug Y. Under these conditions, PREDICT whether:
* The cells will continue to consume O2.
* The cells will continue to produce CO2.
* The rate of ATP synthesis will increase, decrease or stay the same.
For each prediction, EXPLAIN your reasoning.
[f]IFBBUlQgMTo=IFdpdGggZHJ1ZyBYLCB0aGUgY2VsbHMgd29uJiM4MjE3O3QgY29uc3VtZSBPMg==LiBUaGV5IHdpbGwgcHJvZHVjZSBDTw==Mg==[Qq]. ATP production will decrease. That’s because the effect of drug X is to intercept energetic electrons from NADH and FADH2. That keeps these electrons from powering the electron flow that powers the proton pumps that create the proton gradient that powers ATP synthesis through ATP synthase. With no electron flow, there will be no consumption of oxygen, because oxygen is the final electron acceptor of the mitochondrial electron transport chain. However, CO2 will continue to be produced by the link reaction and the Krebs cycle, and a small amount of ATP will continue to be produced by Glycolysis and the Krebs cycle.
PART 2: With drug Y, the cells won’t consume O2. They will produce CO2. ATP production will stay the same because protons are still being pumped. The difference between drug X and drug Y is that Y is intercepting electrons at the very end of the electron transport chain, and essentially substituting for oxygen as the final electron acceptor. While this stops oxygen consumption, it still allows for ATP production through chemiosmosis, and it still (as with drug X) allows for CO2 production during the link reaction and the Krebs cycle.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|1b9d1f791007″ question_number=”19″ topic=”3.6.Respiration”] Aquaporins (one of which is shown below) are membrane channels that allow water molecules to enter into cells. While aquaporins increase the permeability of cell membranes to water, they are impermeable to protons.
PART 1: As specifically as possible, IDENTIFY the structure at T.
PART 2: With reference to the figure, EXPLAIN why aquaporins are not permeable to protons.
PART 3: DESCRIBE a biological process that depends on a cell’s preventing the free passage of protons through the cell membrane, and briefly explain how that process works.
[Qq]PART 2: The + signs indicate areas of positive charge, associated with amino acids with positively charged R groups at that position in the structure. Because like charges repel, positively charged protons will be repelled by these positively charged regions, and therefore won’t be able to pass through the aquaporin.
PART 3: Oxidative phosphorylation in mitochondria and photophosphorylation in chloroplasts is based on protons becoming sequestered in a membrane-enclosed compartment (the intermembrane space in mitochondria, and the thylakoid space in chloroplasts). This sequestration of protons forces them to diffuse through the ATP synthase channel, providing the kinetic energy to drive ATP synthase to be able to create ATP from ADP and inorganic phosphate. (Note from Mr. W: there might be other processes: send me an email or ask your teacher if you used one).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|25ae8ee3fc07″ question_number=”20″ topic=”4.2.Intro_to_Signal_Transduction”] Compare and contrast signal transduction pathways in protein hormones (like epinephrine) and steroid hormones (like estrogen).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|198eecf9c807″ question_number=”38″ topic=”4.2.Intro_to_Signal_Transduction”] The diagram below represents signal transduction.
PART 1: Explain how the initiation of signal transduction ensures that a cell’s response to a signal will be controlled and specific.
PART 2: List the type of molecules that generally act as intermediate or relay molecules.
PART 3: In a multicellular organism, signals can come in various forms. Explain how you know that in the system shown above, the ligand is NOT a steroid hormone.
PART 4: Describe a specific example of a cellular response that could occur in the kind of system depicted above.
UEFSVCAyOg==[Qq] A common second messenger is cAMP; others include IP3 (inositol trisphosphate), and calcium ions.
PART 3:In this system, the ligand is binding at the membrane, making it most likely a protein hormone. Steroid hormones would diffuse through the membrane and bind with a cytoplasmic receptor. A hormone-receptor complex would then diffuse into the nucleus.
PART 4: One example of possible cellular response is activation of enzymes, such as when liver cells respond to epinephrine by activating enzymes that break down glycogen into glucose as part of the fight or flight response. (note: MANY other examples are possible)
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|1e1d683efc07″ question_number=”21″ topic=”4.5.Feedback”] Anti-diuretic hormone (ADH) is synthesized in the hypothalamus and released into the blood when body fluids become too hypertonic. When ADH reaches the kidney, it increases the amount of water reabsorbed from the kidney’s filtrate back into the blood, decreasing urine output.
Different drugs can affect the production of ADH. MDMA (methylenedioxymethamphetamine, also known as Ecstasy) can increase the amount of ADH produced. Alcohol, by contrast, decreases ADH production.
The graph below shows data from a clinical trial for a new medicine. Three groups of patients (A, B, and C) had their total urine output measured over a four day period. The data have been adjusted to account for differences in body size and fluid intake in the patients’ diets.
Patient Group B had normal results. IDENTIFY which group of patients was likely to include people consuming alcohol, and which group of patients was likely to include people taking the drug Ecstasy. Explain your response.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|1c1d2e070c07″ question_number=”22″ topic=”4.5.Feedback”] During the fight or flight response, a molecule of epinephrine, released from the adrenal glands, can induce liver cells to convert glycogen to glucose, providing essential fuel for survival. Because epinephrine (also known as adrenaline) is not lipid-soluble, it cannot diffuse through the cell membrane.
PART 1: Describe the events that occur after epinephrine reaches its target cells that result in changes in target cell activities.
PART 2: Epinephrine relies on a second messenger system that involves multiple steps. Explain how having a number of steps in a signaling pathway can promote survival for an organism as a whole.
[Qq]PART 2: The many steps involved in the phosphorylation cascade that result from binding of epinephrine to its receptor allow for amplification of the signal. One molecule of epinephrine can result in the release of many molecules of cAMP, each one of which can initiate a phosphorylation cascade. Thus one molecule of epinephrine can ultimately wind up inducing the activation of millions of enzyme molecules, and the release of tens of millions of glucose molecules as glycogen is broken down. This can have strong survival benefits, as it can quickly make fuel available during the fight or flight response.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|1b081c801007″ question_number=”35″ topic=”4.5.Feedback”] Type 2 diabetes and certain heart conditions can lead to fatigue, which can be described as extreme tiredness, due to a lack of energy. Fatigue is also associated with chronic fatigue syndrome, which has been linked to viral infections, immune system problems, and hormonal imbalances; and emphysema, in which impaired lung function creates difficulty breathing.
PART 1: In Type 2 diabetes, the target cells do not respond correctly to the insulin produced when there is an increase in blood glucose concentration. Explain why fatigue may occur in a person with Type 2 diabetes who is not taking medication.
PART 2: Certain heart conditions result in a weak and irregular heartbeat. Explain how a weak and irregular heartbeat could result in fatigue.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|1a880df21407″ question_number=”36″ topic=”4.2.Intro_to_Signal_Transduction”] Epinephrine, also known as adrenaline, is produced by the adrenal glands. The structural formula for epinephrine is shown below.
These next two diagrams are general models for signal reception and transduction.
PART 1: Identify which model, S or T, best corresponds to the mechanism of epinephrine. Justify your response.
PART 2: While epinephrine has effects throughout the body, some tissues are not affected. Explain.
PART 3: Identify the model, S or T, that would bring about longer lasting effects. Justify your response.
UEFSVCAyOg==[Qq] The cells in tissues that don’t respond to epinephrine lack the appropriate receptors for this hormone.
PART 3: Model S, in which hormones bind with cytoplasmic receptors, which then diffuse into the nucleus to turn on genes, would result in much longer lasting changes. That’s because activating genes and the resulting production of proteins tends to persist for long periods of time. Model T, by contrast, is connected to cellular responses that are quickly mobilized and just as quickly shut down (an example being epinephrine mobilizing enzymes in the liver to convert glycogen to glucose as part of the fight or flight response).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”4.Cell Communication, Cell Cycle, Feedback” dataset_id=”AP Bio FRQ Dataset 2022|23739cd776a806″ question_number=”107″ topic=”4.5.Feedback”] The table below summarizes the results of a study of two mammal species living in their natural habitat. The study’s focus was upon the ways in which the mammals gained and lost water.
Identify the type of environment in which each of these species normally lives. Support your answer with evidence from the table.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|24514fec6407″ question_number=”23″ topic=”5.2.Meiosis_and_Genetic_Diversity”] Your imaginary fourteen year old younger brother (Kiddo) and sister (Kidda), who are fraternal twins, confront you at dinner with the question, “If we’re twins, with the same mother and father, how can we be so different?” They know you’re taking AP biology, and you don’t have to hold back. Explain it to them.
Kiddo and Kidda are impressed. You continue. “But that’s not all. Even before independent assortment happens, there’s another process that made you different. We’ll think of Dad this time. Before his chromosome pairs can be separated during independent assortment, they have to line up. It’s weird to think of this, but when Dad made sperm, his parents’ chromosomes had to line up inside the cell that was about to divide. In other words, chromosome 1 from his dad had to find chromosome 1 from his mom. These matching chromosomes are called homologues, and the lining up is called synapsis. And when these homologous chromosomes found each other, they exchanged pieces of their DNA, making new DNA that’s never been seen before in this universe. It’s called crossing over. And, that, my brother and sister, is why you’re different from each other, and from me.”
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|118707f65c07″ question_number=”44″ topic=”5.2.Meiosis_and_Genetic_Diversity”] The diagrams below represent three reproductive strategies.
Here’s the approximate scale for these images:
- A is a multicellular organism about 1 centimeter in length
- B is much smaller. The second image in row B shows individual cells
- C is about 10 centimeters in length.
Identify which of the strategies above will produce offspring with the greatest genetic variation. Justify your selection.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|ec58de38007″ question_number=”50″ topic=”5.2.Meiosis_and_Genetic_Diversity”] The diagram below shows two cells, P and Q, that during sexual reproduction combine to form cell R. Cell R and its daughter cells (not shown) divide and to produce an embryo.
PART 1:Label each of the following cells as haploid or diploid: Cell P, Cell Q, Cell R, Cells of the embryo
PART 2: Each cell of the embryo in the diagram has the potential to generate any human cell. Explain.
PART 3: For a zygote to be formed and for that zygote to develop into an embryo, two different types of cell division are involved. Define and describe each type of cell division, and explain its significance.
[f]IFBBUlQgMTogQ2VsbCBQOiA=SGFwbG9pZA==LCBDZWxsIFE6IA==SGFwbG9pZA==[Qq], Cell R: Diploid, Cells of the embryo: Diploid
PART 2: Each cell in the embryo is a diploid cell that possesses the entire genome for the whole organism. Because the cells are early stage embryonic cells, epigenetic changes (such as methylation) will not yet have occurred, allowing each cell to be the source of an entirely new organism.
PART 3: Zygote formation and embryonic development require meiosis and mitosis. Meiosis occurs during the formation of gametes, or sex cells. Sex cells are shown as P & Q in the diagram. Meiosis produces daughter cells that are haploid, with half the number of chromosomes and, because of the way that meiosis occurs, half of the genome. Meiosis sets the stage for the full diploid number to be restored at fertilization (at R), when two haploid gametes fuse together. During mitosis, daughter cells are produced that are genomic clones of the parent cell, containing the entire genome. At the same time, these daughter cells can differentiate into the specialized cells and tissues of a multicellular organism.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|e5a73c08807″ question_number=”51″ topic=”5.2.Meiosis_and_Genetic_Diversity”] Identical twins are genetic clones, formed when the cells in a dividing zygote (a fertilized egg) separate into two physically distinct individuals, instead of remaining connected within the same body.
Two identical twin brothers (David and John) married two identical twin sisters (Sarah and Anne). Each couple had a single child, as shown in the diagram below.
Using your knowledge of mitosis and meiosis, estimate the percentage of alleles shared by the individuals listed in the table below, and write a short justification of your answer.
|Individuals||% of alleles shared||Justification|
|David and John|
|Anne and Lisa|
|Sarah and Lisa|
|[Qq]Individuals||% of alleles shared||Justification|
|David and John||100%||David and John arose from the same fertilized egg cell. All of their alleles will be the same (except for ones that change through mutation)|
|Anne and Lisa||50%||Through meiosis, Anne produced the unfertilized egg that John’s sperm fertilized, producing the zygote that became Lisa. Meiosis involves passing on 50% of one’s alleles to one’s offspring (thus a 50% match)|
|Sarah and Lisa||50%||Because Sarah is Anne’s identical twin, they both possess the same genes. During meiosis, they’ll randomly pass on half of their alleles to their egg cells. In the same way that Lisa has 50% of Anne’s alleles, she’ll also have 50% of Sarah’s alleles.|
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|8c03723e807″ question_number=”62″ topic=”5.2.Meiosis_and_Genetic_Diversity”] Figures A, B, and C below represent stages of cell division from the same organism. One shows the first division of meiosis, one the second division of meiosis and one shows division by mitosis.
PART 1: Identify which figure shows an event from mitosis, which figure shows an event from meiosis 1, and which figure shows an event from meiosis 2. Justify your choices.
PART 2: Explain two ways in which meiosis leads to genetic variation in gametes.
[Qq]In figure B we see two doubled chromosomes lined up on the cell equator. Because the diploid number is four, the division of this haploid cell has to be meiosis 2. The result will be haploid gametes with half the number of chromosomes than the cell which started this process.
PART 2: Two processes that occur during meiosis create variation. The first is independent assortment of paternal and maternal chromosomes. The result of independent assortment is that each gamete contains a unique combination of maternal and paternal chromosomes which gets passed on to the offspring. The second is crossing over. During crossing over, homologous chromosomes exchange segments of DNA, creating new, never before seen recombinant DNA sequences. During the fertilization that follows, DNA from the mother and the father are combined, creating yet another source of variation (though you should give yourself full credit and press that “Got it!” button if you just wrote about the first two).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|7e55ac63007″ question_number=”64″ topic=”5.2.Meiosis_and_Genetic_Diversity”] In a honey bee colony, the only reproductive female is the queen. The queen mates with a male that is not related to her. When the queen reproduces, she fertilizes her eggs with sperm that she has stored in her body from that one mating.
In bees, sex is determined by the number of chromosomes, and the sex of her offspring is under the queen’s physiological control. If, during ovulation, the queen’s body prevents sperm from reaching an egg passing down her oviduct, the egg will develop into a male drone. If a sperm is allowed to enter the oviduct and fertilization occurs, a female worker develops.
PART 1: If female worker bees have 32 chromosomes in their body cells, how many do male drones have? Explain.
PART 2: In the same colony, which is more closely related to the queen: a female worker or a male drone? Explain.
PART 3: What kind of cell division takes place during the production of gametes by the male drone? Explain.
PART 4: Are all drones born from the same queen genetically identical? Explain.
UEFSVCAyOg==[Qq] The drone is more closely related to the queen, because he gets 100% of his genes/chromosomes from her. By contrast, the female worker is like a human daughter in terms of her relationship to her mother: she gets only half of her chromosomes from the queen, with the other half coming from the drone who fertilized the queen.
PART 3: Drones, who are haploid, produce their sperm by mitosis (not meiosis). Because they’re already haploid, there’s no way they could undergo a meiotic reduction of their chromosome number.
PART 4: No, the drones are not identical, because each drone is produced by meiosis from a germ cell in his mother, the queen. The same meiotic processes that ensure that all haploid gametes are genetically unique also ensure that each drone is genetically unique.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|769f44ffc07″ question_number=”65″ topic=”5.2.Meiosis_and_Genetic_Diversity”] In Tasmania, grapevines are usually propagated from cuttings that have been removed from existing vines. A recent study determined that in the young cells in the plant’s root tips, about 5% of the total cell volume consists of mitochondria. In a mature cell in the root of the same plant, however, the mitochondrial volume is only about 1%.
PART 1: Briefly explain this difference in mitochondrial volume.
PART 2: What kind of cell division occurs when these grapevine cuttings start growing, and how is this significant to the grower?
UEFSVCAyOg==[Qq] The grapevine cuttings grow through mitosis. This is significant because this type of propagation maintains an identical genetic makeup among all the newly propagated plants.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|6fc86212007″ question_number=”66″ topic=”5.2.Meiosis_and_Genetic_Diversity”]
A pair of homologous chromosomes involved in normal meiosis in an ovary carries the alleles shown below.
Now, look at the chromosomes below.
Identify the only viable chromosome that could result from meiosis. Justify your choice.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|6757b6f7807″ question_number=”67″ topic=”5.2.Meiosis_and_Genetic_Diversity”] In the diagram below, maternal chromosomes are colored white, and paternal chromosomes are colored black.
A cell with a diploid number of six is undergoing a type of cell division. Identify which one of the diagrams below could represent the arrangement of chromosomes in this cell at some stage of the cell division process. Justify your answer.
Q2hvaWNlICYjODIyMDtCJiM4MjIxOyBpcyB0aGUgb25seSBwb3NzaWJsZSByZXByZXNlbnRhdGlvbiBvZiB3aGF0IHRoaXMgY2VsbCB3b3VsZCBsb29rIGxpa2UgaWYgaXQgd2VyZSBkaXZpZGluZy4gSWYgdGhlIGNlbGwgd2VyZSBnb2luZyB0aHJvdWdoIA==bWVpb3Npcw==LCB0aGVuIHlvdSBtaWdodCBzZWUgdGhyZWUgZG91YmxlZCBjaHJvbW9zb21lcyA=[Qq]after meiosis 1 (which is what you see here). Because of independent assortment, it’s completely possible that two of the chromosomes would be maternal, and one would be paternal.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|235386dc5007″ question_number=”24″ topic=”5.3.Mendelian_Genetics”] PART 1: Draw a pedigree chart that shows inheritance of an autosomal dominant allele.
PART 2: Describe the principles you used to draw this pedigree. As you do, describe how autosomal dominant inheritance works.
PART 2: Any pedigree showing autosomal dominant inheritance will have the following features. First (unless there’s a mutation), two unaffected mates will never pass the condition on to their offspring. This is shown in the mating between II-1 and II-2. That’s because the nature of dominance is to show up in the phenotype. So, unaffected individuals don’t have the dominant allele, and can’t pass it on to their offspring. By contrast, as is shown by couples I-1 / 1-2 and II-3 / II-4, an individual with a dominant phenotype (represented by a filled-in circle or square) who mates with an unaffected partner has a either a 50 percent chance of passing on the dominant allele (if he or she is heterozygous) or a 100% chance (if he or she is homozygous). That means that in the pedigree above, I-1 and II-4 must be heterozygotes. Finally, because the allele is autosomal (meaning that it’s found on any chromosome but the X), it will show up equally in males and females.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|20c09db74407″ question_number=”25″ topic=”5.3.Mendelian_Genetics”] PART 1: Draw a pedigree chart that shows inheritance of an autosomal recessive allele. PART 2: Describe the principles you’ll use to draw this pedigree. As you do, describe how autosomal recessive inheritance works.
PART 2: In a pedigree chart, autosomal recessive traits will show the following patterns. First, because the condition is recessive, you’ll see two parents who are unaffected produce one or more offspring with the condition. This is shown in the mating between individuals 1 and 2 above. The genotype of both I-1 and I-2 is “Aa,” and they have a 1/4 chance of any offspring (such as II-2) inheriting genotype “aa.” The second pattern (only indirectly shown above) is that you’ll often see the condition skip a generation. That’s because the affected individual (aa), if they mate with a homozygous dominant individual (AA) will pass on their allele to their offspring, creating heterozygous dominant carrier (Aa). If this second generation carrier mates with another carrier, as described above, then the grandchildren (or F2s) can wind up inheriting the condition. Finally, because the condition is autosomal (not on the X chromosome), you’ll see the condition show up equally in males and females.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|dda65328c07″ question_number=”52″ topic=”5.3.Mendelian_Genetics”] In a certain breed of dogs, eyes can be blue (B) or grey (b), the tail can be rigid (R) or stumpy (r) and hair can be long (Q) or short (q). B, R and Q are the dominant alleles and the three genes are on different chromosomes.
- Betty is a dog that shows the three dominant traits. Indicate the genotype of the dog you should cross Betty with in order to determine her genotype. Explain your reasoning.
- All of the progeny of the cross described above have blue eyes and long hair. However half have a stumpy tail and half have a rigid tail. What is Betty’s genotype?
- What fraction of the offspring from the cross BBRrQq X BbrrQq would be blue-eyed with a stumpy tail and short hair?
- The cross above produced 64 dogs from several matings. How many dogs would be predicted to have the genotype BBrrqq?
UEFSVCAxOiA=[Qq]bbrrqq. Explanation: This kind of cross is what is known as a test cross, and here’s how it works. To determine the genotype of an organism whose genotype is unknown, you cross it with an individual who is recessive (and therefore homozygous) for all the traits. If many offspring are produced and all of the offspring have the dominant trait, then the test crossed parent has to be homozygous dominant. If there’s a mix of the dominant and recessive trait in the offspring, then you know that the test crossed parent was a heterozygote for that trait.
PART 2. BBRrQQ. Explanation: If all of the progeny of the cross described above have blue eyes and long hair, then Betty must have been homozygous dominant for those two traits, and her genotype must be BB and QQ. If you’re not clear on this, just draw a punnett square that crosses BB x bb. All of the offspring will be Bb (with blue eyes). The same works for QQ x qq. On the other hand, if half of Betty’s offspring have a stumpy tail and half have a rigid tail, then Betty must have been heterozygous for tail type, or Rr. Again, imagine (or draw) a Punnett square that’s Rr x rr. Half the offspring will be Rr, and half will be rr.
PART 3: 1/8. Explanation: Use the rule of multiplication to solve this problem. In a cross between BB x Bb, all of the offspring will have blue eyes (BB or BB). Crossing Rr with rr, half of the offspring will have stumpy tails (rr). Crossing Qq with Qq, 1/4 will have short hair (qq). 1 x 1/2 x 1/4 = 1/8.
PART 4: 4. Explanation. Again, use the rule of multiplication to solve this problem. In a cross between BB x Bb, 1/2 of the genotype will have genotype BB. Crossing Rr with rr, 1/2 will have genotype rr. Crossing Qq with Qq, 1/4 will have genotype qq. 1/2 x 1/2 x 1/4 = 1/16.
64 times 1/16= 4
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|bfac3a11407″ question_number=”56″ topic=”5.3.Mendelian_Genetics”] A plant with genotype AaBbCc was crossed with a plant with genotype aaBBCc.
PART 1: Determine the probability of obtaining AaBbCc progeny.
PART 2: Explain how you can solve the problem above without creating a huge Punnett Square.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|cd59ffecc07″ question_number=”54″ topic=”5.4.Non-Mendelian_Genetics”] In a newly discovered species of fruit fly, the allele for black bristles (B) is dominant to green bristles (b) and red eyes (R) is dominant to white eyes (r).
Several BbRr flies are testcrossed with bbrr flies. The observed distribution of offspring is black-red, 1070; black-white,177; green-red, 180; green-white 1072.
PART 1: Use a Punnett square to prove that the alleles for bristle color and eye color do not independently assort.
PART 2: Calculate the recombination frequency between the bristle color and eye color alleles.
PART 2: Instead of 1:1:1:1, 2142 offspring have parental phenotypes (black-red or green-white), while 357 have recombinant phenotypes (black-white or green-red). This can be explained by interpreting the results as follows: the bristle color and eye color alleles are linked on the same chromosome. As a result, they usually stay together (producing parental phenotypes), but because of crossing over, they’ll also recombine (producing recombinant phenotypes) with a frequency of 357/2499, or 14.2%. Another way to say this is that the bristle color and eye color alleles are on the same chromosome, 14.2 cM apart.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|c65ddc40c07″ question_number=”55″ topic=”5.4.Non-Mendelian_Genetics”] You are studying the inheritance of flower color in poppies. Suppose that flower color is determined by a single gene with 3 alleles:
H = purple allele
I = orange allele
J = red allele
The phenotypes produced by the different genotypes are:
HH = purple flowers
HI = purple flowers
II = orange flowers
HJ = white flowers
JJ = red flowers
IJ = lethal (seed does not germinate and no flowers are produced)
Suppose you cross a plant producing purple flowers (genotype HI) with a
plant producing white flowers (genotype HJ).
Predict the frequencies of the genotypes and the phenotypes in the offspring. Justify your prediction.
[Qq]JUSTIFICATION: We’re told that the parents are HI and HJ.
Here’s the punnett square.
To determine the phenotypes look up the phenotype associated with each genotype. HH is purple, as is HI. HJ is white. IJ is lethal.
So the phenotype ratio is 2/3 purple (HH or HI) and 1/3 white (HJ).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|b88ad5a7007″ question_number=”57″ topic=”5.4.Non-Mendelian_Genetics”] Cynthia and Neil have just had twins, a boy named Jadin and a girl named Kacee. Both of the babies’ grandfathers are red-green color blind, while their grandmothers have normal vision. Neil, who is an only child, has normal vision; Cynthia and her brother also have normal vision.
PART A: Construct a family pedigree to show the inheritance of this recessive sex-linked genetic disorder. (Use circles for females, squares for males; shade those affected by color blindness.)
PART B: Explain what can be deduced about the genotypes of both Cynthia and Neil with respect to red-green color blindness from the given information.
PART C: Show how you can use your answers to part (b) to predict the probabilities that Jadin and Kacee will be color blind.
[Qq]PART B: XC denotes gene for normal vision. Xc denotes gene for normal vision. Y denotes Y chromosome. Neil must be XCY because he has normal vision (his only X chromosome must carry the gene for normal vision). Cynthia (because her dad was colorblind and therefore had to pass on his Xc allele to his daughters), must be a carrier, making her genotype XCXc.
PART C: Jadin’s genotype, as shown below, could beXCY or XcY). Therefore, he has a 50% chance of being colorblind. Kacee’s genotype could be XCXC or XCXc , which means that she has 0% chance of being colorblind, and a 50% chance of being a carrier .
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|6080d409c07″ question_number=”68″ topic=”5.4.Non-Mendelian_Genetics”] The shapes of tabby stripes in a cat’s fur is controlled by an autosomal gene with two alleles:
T: vertical colored stripes (called mackerel tabby)
t: swirly colored stripes (called blotched tabby.}
The patterns are illustrated as follows.
A second autosomal gene (not linked to the tabby-striped gene)is the Agouti gene. This gene has the alleles
A: tabby fur pattern
a: no tabby pattern.
A cat that is aa for the Agouti locus fails to show any stripes, regardless of its genotype at the tabby stripes locus. These cats are solid black in color.
A cross was carried out between cats of the following genotypes.
TTAa X ttAa
The cats had two kittens. Identify which of the images below shows a possible phenotype for these kittens. Justify your answer.
SGVyZSYjODIxNztzIGhvdyB0byBmaWd1cmUgdGhpcyBvdXQuIFRoZSBjcm9zcyBpcyBiZXR3ZWVuIA==VFRBYQ==IGFuZCA=[Qq]ttAa. Here’s the Punnett square
There are three possible genotypes: 1/4 TtAA, 1/2 TtAa, and 1/4 Ttaa. The Tt genotype produces a vertical striped phenotype, unless it’s paired with aa. Ttaa produces a black phenotype. No cats with blotched phenotype are possible, which eliminates choice b, c, and d. There would be a 1/16 chance of two black cats (unlikely, but possible, unlike the other three choices).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|238227cdbbc406″ question_number=”89″ topic=”5.4.Non-Mendelian_Genetics”] In a newly developed variety of apple, the skin of the apple has three phenotypes: red, yellow, and green. The expression of skin pigment is under genetic control. Pigment production is controlled by two genes, each gene has two alleles, as shown in the table below.
Apples expressing the N allele are green.
PART 1: Genes 1 and 2 are unlinked. Briefly explain what this means.
PART 2: If an apple plant has genotype NnRr , determine the color of its fruit. Justify your answer.
PART 3: Two apple plants with genotypes NnRr and nnRr were crossed. Use a Punnett square and the information above to calculate the proportion of skin color phenotypes in the offspring.
UEFSVCAyOg==[Qq] Apples with genotype NnRr will be green, because the allele N prevents pigment production. Without production of pigment, the apple will be green.
PART 3: Here’s a modified Punnett square for the cross between NnRr and nnRr . The phenotype for each genotype is also included (which is why it’s a modified Punnett square)
4/8 (or 1/2) of the apples will be green; 3/8 will be red; and 1/8 will be yellow.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|238192cac2c406″ question_number=”90″ topic=”5.4.Non-Mendelian_Genetics”] In fruit flies, allele W codes for normal wings, while allele w codes for vestigial wings. Allele B codes for normal body color, while allele b codes for ebony. A heterozygous normal winged, ebony male fly is crossed with a vestigial winged, heterozygous normal body color female fly.
PART 1: List the genotypes of the two fruit flies.
PART 2: Determine the expected ratio of the phenotypes in the offspring of this cross.
PART 2: Here’s the Punnett square, and the phenotypic results
¼: normal wing, normal body
¼ : vestigial wing, normal body
¼ : normal wing, ebony
¼ : vestigial wing, ebony body.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|238120b47c2006″ question_number=”91″ topic=”5.4.Non-Mendelian_Genetics”] Hemophilia A is an X-linked recessive disorder in which the ability of the blood to clot is severely reduced. Hemophilia is caused by a mutation in the gene for Factor VIII, a clotting component.
Rachel’s brother has hemophilia A. Neither Rachel or her parents show symptoms of hemophilia A.
PART 1: If Rachel has a son with a man who is not a hemophiliac, what is the probability that the son will have hemophilia A?
PART 2: If Rachel’s husband has hemophilia, what is the probability their son will have hemophilia A.
RXhwbGFuYXRpb246IElmIFJhY2hlbOKAmXMgYnJvdGhlciBpcyBhIGhlbW9waGlsaWFjLCB0aGF0IG1lYW5zIHRoYXQgUmFjaGVs4oCZcyBtb3RoZXIgd2FzIGEgY2FycmllciwgYW5kIHdlIGNhbiBkZXNpZ25hdGUgUmFjaGVs4oCZcyBtb3RoZXLigJlzIGdlbm90eXBlIGFzIFg=SA==[Qq]Xh. Rachel’s father was not a hemophiliac, so his genotype is XHY. The cross between Rachel’s parents would have been the following.
Notice that the difference is completely in the status of the daughters. Compared to the Punnett square above, one of the daughters is a carrier, and the second is a hemophiliac. But the chance of a son being a hemophiliac remains one in two.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|2380a9f61db406″ question_number=”92″ topic=”5.4.Non-Mendelian_Genetics”] In a newly discovered species of parakeets, having green feathers (G) is dominant to having blue feathers (g) and a yellow beak (Y) is dominant to a black beak (y). A parakeet with genotype GgYy is mated with a parakeet with genotype ggyy.
PART 1: Determine the expected phenotypic ratio of the offspring. Support your prediction with brief explanation and a Punnett square.
PART 2: Scientists observed 2500 parakeet offspring. The phenotype distribution is shown in the table below. Explain.
1 green feathers, yellow beak: 1 green feathers, black beak: 1 blue feathers, yellow beak: 1 blue feathers, black beak
PART 2: The alleles for feather color and beak color are linked (on the same chromosome). They have a recombination frequency of 14.28%, and are located 14.3 centimorgans away from one another.
EXPLANATION: Instead of seeing a 1:1:1:1 ratio, the results show that most of the offspring have are the parental phenotypes (green feathers with a yellow beak or blue feathers with a black beak), while a much smaller proportion have recombinant phenotypes (green feathers with a black beak or blue feathers with a yellow beak). Whenever you see this pattern in a test cross (a cross between a heterozygote and a recessive), you know that this is occurring because of linkage (which is why most of the offspring have the parental phenotypes, with recombination accounting for the recombinant phenotypes).
Of the 2500 offspring, 181 + 176 are recombinant. That means that the recombination frequency is 14.28%, and that the alleles are 14.3 Centimorgans away from one another.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|2382bcd0b4c406″ question_number=”88″ topic=”5.6.Chromosomal_Inheritance”] The following question is based on the diagram below.
The diagram shows a germ cell (top center) dividing to become egg cells, each of which is fertilized by sperm cells (the flagellated cells) to become zygotes (bottom row).
PART 1: Identify the process indicated by the blue arrow.
PART 2: Assuming that the chromosomes shown in the germ cell represent one of the 22 autosomes, predict which zygote on the bottom row could develop into an individual with Down syndrome. Justify your prediction.
PART 3: If the chromosomes shown in the germ cell represent X chromosomes, predict which zygote on the bottom row could develop into an individual with Turner syndrome. Justify your prediction.
UEFSVCAyOg==[Qq] Zygote C has an extra chromosome, a condition known as a trisomy. Down syndrome is caused by a trisomy of chromosome 21.
PART 3: Zygote D has a missing chromosome. Turner syndrome is caused by a monosomy of the X chromosomes: a female with Turner syndrome has one X chromosome instead of two.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|11094d744407″ question_number=”45″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below is a representation of a two fragments of DNA, each with the same nucleotide sequence, from the same species of bacterium. The fragment of DNA on the left was grown in a medium containing 14N, the most common isotope of nitrogen. The fragment on the right was grown in a medium containing 15N, a less common, radioactive isotope of nitrogen that contains an additional neutron.
Bacteria with DNA consisting of 14N were transferred to a medium consisting of 15N, and allowed to replicate once. Identify which of the diagrams below correctly shows the daughter DNA molecules that would result. Justify your answer.
SlVTVElGSUNBVElPTg==[Qq]: Model D represents the results of semi-conservative DNA replication, which is the model that Meselson and Stahl established in 1958 (Note: this historical information is not required). During DNA replication, a DNA molecule will “unzip” and free nucleotides will move in and pair up according to the base pairing rules (A bonds with T; and C bonds with G). In the scenario described above, the DNA molecule that’s unzipping will consist of 14N, but the free nucleotides coming in will contain the 15N nitrogen in their nucleotide bases. As a result, in each of the duplicated DNA molecules, one strand will be consist of 14N and the paired complementary strand will consist of 15N.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|b03f6b4ac07″ question_number=”58″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below shows a short section of a DNA molecule.
Explain the semiconservative model of DNA replication. Add a simple sketch to accompany your explanation.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|a7a97f72007″ question_number=”59″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below depicts semi-conservative replication of DNA.
Explain how the Meselson-Stahl experiment proved the semiconservative model to be correct.
[Qq]Meselson and Stahl were able to prove the semiconservative model to be correct by using isotopes of nitrogen to label original DNA and newly synthesized DNA in replicating E. coli cells.
They started by growing a culture of E. coli on N15, a heavy isotope of nitrogen so that all of the nitrogen in this population of E. coli’s DNA was composed of this heavier isotope. Then they transferred this E. coli to a medium containing N14, a lighter isotope of nitrogen (and the kind that is much more commonly found in nature). Knowing that E. coli cells can replicate themselves in 20 minutes, they allowed the E. coli to replicate itself once, and then used centrifugation to determine the weight of the nitrogen in the E. coli’s DNA. They found that the DNA’s weight was halfway between N14 and N15. This established that the new DNA was composed of 50% N14 and 50%, N15, which proved an alternative theory (the conservative model of DNA replication) to be incorrect. That’s because if the conservative model were correct, then there would still be some DNA that was all composed of N15 (DNA that was conserved), with other DNA being composed exclusively of N14 (the newly synthesized DNA). The conservative model is represented by image A below, which shows two daughter double helices.
Meselson and Stahl then let the bacteria replicated for a second generation, and again sampled the DNA. They found that in the second generation, there continued to be DNA whose weight was half way between N14 and N15, and new DNA that was completely N14. This proved another model, the dispersive model to be incorrect), and corroborated what one would predict if DNA were semi-conservatively replicated.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|96a2e87ec07″ question_number=”61″ topic=”6.1.DNA_Structure_and_Replication”] During their work in determining the structure of DNA, Watson and Crick examined published results from three different laboratories. Each laboratory was reporting on the proportion of nucleotides in skin cells from the same organism.
Which laboratory’s results did Watson and Crick disregard? Explain why they had to disregard the results, and speculate about what might have occured in the lab with the flawed data that caused them to have incorrect results.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|8551d00f007″ question_number=”63″ topic=”6.1.DNA_Structure_and_Replication”] The diagram below shows the amount of DNA per cell in a dividing cell.
PART 1: Identify what kind of cell division must be occurring?
PART 2: Justify your answer.
PART 3: Explain what’s happening in each numbered stage.
UEFSVCAyOg==[Qq] It has to be meiosis because the cell starts with a two units of DNA, and ends with 1 unit. The type of cell division that reduces the amount of DNA/cell is meiosis.
PART 3: At I we have a germ cell. II is interphase 1, which doubles the amount of DNA. In III, prophase 1 through telophase 1 are occurring, leading to the first cell division, cytokinesis 1, between III and IV. Meiosis 1 pulls homologous chromosomes apart, Now, in IV, all the steps of meiosis 2 would occur, separating the sister chromatids. In V we have cytokinesis 2, resulting in the haploid gametes in step VI.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|212e0be62007″ question_number=”26″ topic=”6.3.Transcription_and_RNA_Processing”] You’re interviewing for a summer internship at a local research lab. During the interview, your graduate student mentor tells you that the lab has just discovered a new eukaryotic protein that’s 197 amino acids long, and that the two of you will be working on finding the gene for this protein. To check your knowledge of biology, the grad student asks, “About how big is this gene going to be?” Provide an answer, and a justification.
[Qq]First, you think “The protein is 197 amino acids long. It takes 3 nucleotides to code for one amino acid, and 197 x 3 = 591.” But then you pause, and think some more. You know that during protein synthesis, the first amino acid is methionine, which is coded for by the start codon AUG. However, this first methionine is often snipped off. So you’ll want to add three nucleotides to your total.
But don’t just blurt out “594!” Think about the end of protein synthesis, when the stop codon signals for a release factor. The stop codon is another three nucleotides, so you have to add three more nucleotides, which brings you to 597. But you don’t say that, either. You remember that this is a eukaryotic gene, and might be consist of exons and introns that are excised before translation. So, your final answer is this:
“The minimum number of nucleotides that could code for a 197 amino acid polypeptide would be 597. That’s 591 nucleotides to code for the 197 amino acids, plus three for the start codon and three for the stop codon. But because this is a eukaryotic gene, it might contain introns, and we can’t tell, just from the number of amino acids in the protein, how big it is.”
With a smile, you add, “I’m really looking forward, however, to working with you to figure this out.”
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|26a7afdc4807″ question_number=”27″ topic=”6.4.Translation”] Starting from its synthesis at a ribosome, describe the pathway of a protein (such as a protein hormone or an antibody) that is going to be secreted from the cell.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|fb0b6947407″ question_number=”48″ topic=”6.4.Translation”] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.
List the remaining amino acids in the sequence. Here’s a genetic code chart to use for reference.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|237ff6ae8a2006″ question_number=”93″ topic=”6.4.Translation”] The following questions are based on the figure below.
PART 1: Identify whether the figure depicts a eukaryotic cell or a prokaryotic cell. Support your response with evidence from the diagram.
PART 2: Assume that letter B represents a protein, such as insulin, which is exported from cells in the pancreas. Beginning with information in a gene, describe the flow of information and matter that results in the secreted protein at B.
UEFSVCAy[Qq]: The information for a protein like insulin would be encoded in the cell’s DNA. The DNA would be transcribed to RNA, which would be translated into protein. Because insulin is a protein destined for export, it would contain a signal peptide that would bind with a signal recognition particle, which would diffuse toward and bind with the ER membrane. The insulin polypeptide would be synthesized inside the rough ER, and then move to the Golgi for modification. A vesicle would move the insulin to the membrane, and exocytosis would release insulin out of the cell.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1e967aa94c07″ question_number=”28″ topic=”6.5.Regulation_of_Gene_Expression”] Jacques Monod, who co-developed the operon concept (and – I can’t resisting adding – was also leader in the French Resistance to the Nazis during World War II), noticed that when E. coli bacteria were fed lactose, they were able to activate a regulatory pathway that produced lactose-digesting enzymes. When the lactose was digested, this regulatory pathway was turned off, and the enzymes for digesting lactose were no longer produced. Explain the biology behind what Monod discovered.
Imagine the promoter (5) as being upstream of the structural genes (7): in other words, RNA polymerase (4) will bind with the promoter, and then proceed on to the structural genes. However, between the promoter and the structural genes is an operator (6): a DNA region that allows for regulatory control. In the case of the lac operon, there’s a regulatory gene (1) that produces a regulatory protein (3). In the absence of lactose, the regulatory protein can bind with the operator in a way that blocks RNA polymerase, preventing it from transcribing the structural genes (that code for lactose-digesting enzymes). In other words, when lactose is not present, the cell won’t transcribe the structural genes or translate the resulting mRNA into protein (both of which would be a waste of energy in lactose’s absence).
However, the regulatory protein, in addition to having a site for binding with the operator, also has a site for binding with lactose (8, below).
When lactose (8) binds to this regulatory protein, it changes the regulator’s shape so that it can no longer bind with the operator. This allows RNA polymerase to move along from the promoter to transcribe the structural genes. In other words, the presence of lactose induces changes that allow for production of lactose-digesting enzymes.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1a184bb75407″ question_number=”37″ topic=”6.5.Regulation_of_Gene_Expression”] Bacteria require amino acids to produce proteins. For example, bacteria in a human intestine may absorb amino acids from digested food, but at times there may be a deficiency of a particular amino acid. If this is the case, the bacteria will produce the necessary amino acid themselves.
The diagram below depicts a system in bacterial cells that regulates the production of the amino acid tryptophan. This system involves two pathways (X and Y). In both pathways, tryptophan acts as a repressor.
PART 1: Describe the immediate outcome when tryptophan activates pathway X.
PART 2: In the 1950s, biologists Francois Jacob and Jacques Monod developed the concept of the operon to explain how bacterial cells were able to regulate certain biochemical pathways. Explain the operation of the trp (tryptophan) operon.
Cg==UEFSVCAyOg==IFRoZSA=[Qq]trp operon is a repressible operon, and it works as shown below. (Note from Mr. W: I’ve included diagrams to assist your learning, but they weren’t required in your answer.)
The trp regulatory gene creates a repressor protein (3, below). The repressor protein has two binding sites: one for tryptophan (8), and one for the operator region of the trp operon (6).
When tryptophan is present in the cell’s environment, it binds to the repressor protein, which changes shape so that it can bind to the operator region of the operon. Because the operator is just downstream of the promoter region of the operon (5), the binding of the repressor to the operator blocks RNA polymerase (4), keeping it from transcribing the structural genes (7) that synthesize tryptophan.
However, when tryptophan is absent from the cell’s environment, the repressor protein (3, below) can’t bind with the operator, allowing RNA polymerase to bind with the promoter (5) and transcribe the structural genes (which results in production of the enzymes that synthesize tryptophan (A-E).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1c8f444db007″ question_number=”29″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] The image below is a schematic diagram of a signal transduction pathway within a single cell. Arrows represent activation, while a “T” (which may be upside down: see the one between JNI and Bcl2 below) represents inhibition. The overall result of this pathway is apoptosis (programmed cell death).
Some of the key events that occur within this pathway are as follows: 1) Tumor necrosis factor (TNF) binds to and activates its specific receptor (tumor cell necrosis factor receptor, or TNFR). 2) The activated TNFR activates JNK. 3) The activated JNK activates caspases by causing mitochondria to release cytochrome C, and also by inhibiting BCl-2. BCl-2, when active, promotes cell survival by inhibiting caspases. 4) The active caspases result in DNA damage and cell death.
PART 1: Several elements of the system above have been implicated in the development of cancer. Some of these are loss of function mutations, where a previously functioning component of a cell stops working. Based on the signaling pathway shown, describe the effect of a loss of function mutation of each of the following upon apoptosis (compared with wild type cells lacking this mutation).
* TNFR * BCl2 * Caspases
PART 2: Gain of function mutations are mutations where a cell component gains a capability that it previously lacked. Based on the schematic above, identify the component(s) of this signaling pathway that might promote cancer cell growth if they had a gain-of-function mutation that increased that component’s activity.
PART 3: Tumor suppressor genes inhibit cell division. List the component(s) of the system above that may be the product of tumor suppressor genes and that would promote cancer cell growth if they had a loss-of-function mutation.
PART 1b: If BCl-2 lost its activity, cell death would INCREASE because the inhibitory effect of Bcl2 would be removed, allowing caspases to cause nuclear damage, leading to cell death.
PART 1c: If caspases lost their function then cell death would DECREASE, because caspases cause cell death.
PART 2: If a mutation caused BCl-2 to increase its activity (gain of function mutation), it would block caspases, which would decrease cell death/apoptosis. Less cell death means higher survival rates for cancer cells, increasing cancer cell growth.
PART 3: TNFR, JNK, Caspases, and Cytochrome C released from mitochondria are all associated with apoptosis/cell death, which would decrease the spread of cancer cells. Loss of function of any of these could be associated with promoting cancer cell growth.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|1219b6e37807″ question_number=”43″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] Proto-oncogenes, when they mutate to become oncogenes, can cause cells to become cancerous.
A lysogenic virus infects a particular cell type and integrates its genome into a site that contains a proto-oncogene. This transforms the cell and increases the level of a protein X. Protein X, in turn, increases cellular proliferation (division and spread).
Tumor suppressor genes code for proteins that prevent cells from becoming cancerous. A compound P is known to increase the level of tumor suppressor proteins in that cell type. A second compound, Q helps in stimulating protein Z. Protein Z has been shown to be capable of binding to X rendering it inactive.
In the graphs below, a minus sign indicates that substance P or Q was not applied to experimental cell cultures, while a plus sign indicates that substance P or Q was applied. Identify which of the following graphs correctly represents the mode of action of P and Q. Justify your response.
SlVTVElGSUNBVElPTjogUCBpbmNyZWFzZXMgdHVtb3Igc3VwcHJlc3NvciBwcm90ZWlucywgd2hpY2ggd291bGQgZGVjcmVhc2UgY2VsbCBwcm9saWZlcmF0aW9uLiA=UQ==IGluZGlyZWN0bHkgaW5hY3RpdmF0ZXMg[Qq]X. Because X increases cell proliferation, inactivating X (which is what substance Q does) should decrease cellular proliferation. Therefore a combination of P and Q should lead to the lowest amount of cellular proliferation, which is shown in the last bar of graph C. P and Q used separately should also decrease cell proliferation, which is what is shown in the second and third bars of graph C. Absence of both substances would lead to an increase in cellular proliferation, which is what is shown in the first bar of graph C.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|14d688de8c07″ question_number=”40″ topic=”6.6.Gene_Expression_and_Cell_Specialization”] The bos/seven receptor, shown below, is a receptor tyrosine kinase. This receptor is required for the differentiation of a particular cell, called R7. Like all receptor tyrosine kinases, the protein is inactive as a monomer. Binding of ligand causes the monomeric form of the receptor to form a dimer (shown on the right). Dimerization causes phosphorylation of the receptor’s intracellular domain, activating the protein.
During protein synthesis and processing of the receptor, the extracellular domain is cleaved and a disulfide bridge forms between two cysteine amino acid residues. The disulfide bridge tethers the ligand-binding domain to the rest of the protein.
PART 1: Predict the effect of a mutation in the DNA for the receptor that resulted in changing one of the cysteine amino acid residues into an alanine. Justify your prediction.
PART 2: How would this mutation affect differentiation in cell type R7?
UEFSVCAyOg==[Qq] This would prevent the differentiation of the R7 cell type.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|271c1a2ed007″ question_number=”30″ topic=”6.7.Mutation”] List four types of mutations and describe their effect upon the protein that the mutated DNA codes for.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|205583944c07″ question_number=”31″ topic=”6.7.Mutation”] Bacteria reproduce asexually (through binary fission). Yet bacterial species can show significant genetic diversity. List and describe 5 processes that generate genetic diversity within bacteria.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|12a0c1952007″ question_number=”42″ topic=”6.7.Mutation”] In Tay Sachs disease, a genetic mutation in an enzyme called beta-hexosaminidase A results in the build-up of a molecule called GM2 ganglioside. GM2 ganglioside buildup is toxic and ultimately fatal. Victims usually die before the age of five, and there is no treatment.
The table below shows a portion of a DNA sequence for a person with a normal version of beta-hexosaminidase A and for a person with the Tay Sachs mutation.
|DNA nucleotide Sequence|
|Person with Tay Sachs||…GAGGGGTTT|
PART 1: Base on the DNA sequence above, list the amino acid sequence for an individual with Tay Sachs. Use the genetic code chart below
PART 2: With reference to protein structure and amino acid chemistry, explain how the Tay Sachs mutation affects the hexosaminidase A enzyme. Use the same chart listed above for reference.
PART 3: Part of the metabolic pathway connected with Tay Sachs is shown below. Using this diagram, describe some of the effects of the Tay Sachs mutation.
PART 2: The mutation changes only one amino acid: Serine to Proline. However, Serine (ser) has an polar, while proline has a nonpolar side chain. Because a protein’s 3-D conformation (shape) is based on interactions between side chains, this single shift could change the shape of the enzyme. If this changes the enzyme’s active site, then the enzyme’s ability to break down its substrate could be effected. Since Tay Sachs results in the buildup of a toxic substance that is normally broken down by the healthy version of the enzyme, then that is probably the molecular cause of Tay Sachs disease.
PART 3: If the Tay Sachs mutation has occurred and the Hex A enzyme has been affected, then it will not be able to break down the glycolipid into compound Q. Compound Q, in turn, will not be able to be converted into compounds S and T. Instead, compound Q will build up to toxic levels.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|9de98da7407″ question_number=”60″ topic=”6.7.Mutation”] Fungi have metabolic pathways that enable them to synthesize their own amino acids. This enables them to grow on a substance called minimal media (a gel that supports fungal growth, but which lacks any amino acids).
A mutant strain of fungus that could not grow on minimal media was discovered. This fungus was able to grow on minimal media to which all of the twenty amino acids were added. These observations are summarized in the diagram below.
In work that was done in the 1940s, George Beadle hypothesized that the abnormal fungus had a fault in a gene coding for the production of one amino acid. To determine which amino acid was involved, Beadle set up the following experiment involving 22 tubes.
Tube 1 had minimal media.
Tube 2 had minimal media, plus all 20 amino acids.
Tubes 3 to 22 had minimal media, plus one of the 20 amino acids, with each tube containing a different amino acid.
Spores from the mutant fungal strain were inoculated onto the media in each test tube
The tubes were incubated under the same conditions for several days and then examined.
Part A: Why were tubes 1 and 2 included in this experiment?
Part B: The amino acid histidine had been added to tube 12. What specific conclusion can be drawn about the mutant fungus based on this result.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|237f5d03795806″ question_number=”94″ topic=”6.7.Mutation”] The diagram below illustrates the effect of a single point mutation in protein P. Protein P is normally an individual protein molecule found in the cytosol of eukaryotic cells. As a result of this mutation, protein P molecules will aggregate together forming chains. The binding that underlies this chain is represented by letters M and S below. The black half-circle S represents a small hydrophobic cavity on one side of protein P. The black ball at M represents a cluster of hydrophobic R groups on the other side of protein P,
Provide a real-life example of this type of mutation and its effects. Identify the human disease with which it is associated.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|24c80e4ad007″ question_number=”32″ topic=”6.8.Biotechnology”] In your new job at the Department of Interspecific Genetic Comparison, you’re told to use biochemical methods to compare two newly discovered bacterial species. What features might you compare?
[q json=”true” xx=”1″ multiple_choice=”false” unit=”6.Gene Expression and Regulation” dataset_id=”AP Bio FRQ Dataset 2022|f40f459b407″ question_number=”49″ topic=”6.8.Biotechnology”] The image below represents the pGLO plasmid, which is used both in research and in education.
As indicated, the plasmid contains a restriction site and three genes:
- ampR: confers resistance to ampicillin, and antibiotic.
- gfp: encodes the green fluorescent protein (GFP), which glows green (fluoresces) under ultraviolet (UV) light
- araC: encodes a protein required to promote the expression of GFP when arabinose, a disaccharide, is present
The results of a bacterial transformation experiment using the pGLO plasmid are shown in the table below.
PART 1: Explain the purpose of plates W and X.
PART 2: Explain why there is a lawn of bacteria on plate W, as opposed to only discrete colonies on plates Y and Z.
PART 3: Explain why only the bacteria on plate Z glow under ultraviolet light.
PART 4: Explain why there is no growth on plate X.
[Qq]PART 2: Any bacterium will grow on plate W. Because the nutrient agar has everything that E. coli needs, it grows everywhere it’s deposited (which is all over the plate). By contrast, only transformed bacteria which took up the plasmid (which contains an antibiotic resistance gene)will grow on plates with ampicillin. Because only a small proportion of the cells exposed to the plasmid will be transformed, the resulting growth will show up as discrete colonies (with every colony representing a successfully transformed cell).
PART 3: GFP is only expressed when the araC gene produces a regulatory protein that permits transcription of the GFP genes. araC, in turn, will only produce this regulatory protein in the presence of arabinose. While the details are somewhat more complicated, you can think of GFP as equivalent to the structural genes in an inducible operon, with arabinose as the inducer). Only plate Y has arabinose on the plate
PART 4: The bacteria on plate X are untransformed, which means they haven’t taken up the plasmid, which means that they lack the ampicillin resistance gene, and won’t grow in the presence of ampicillin.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”7.Evolution” dataset_id=”AP Bio FRQ Dataset 2022|10893ee64807″ question_number=”46″ topic=”7.1-7.3_Natural_Selection”] In selective breeding, humans look for variation between members of the same species and use this variation to bring about a desired trait.
PART 1: Describe two different causes of variation.
PART 2: Describe two negative consequences of selective breeding.
PART 3: The wild ancestor of the domestic chicken is the red jungle fowl found in the rainforests of Southeast Asia. Explain why it is important to preserve the population of the red jungle fowl.
UEFSVCAyOg==[Qq] Selective breeding can lead to increased susceptibility to genetic disorders by increasing the frequency of harmful alleles within a population’s gene pool. Also, by decreasing diversity, selective breeding can increase susceptibility to infectious disease by creating a single, uniform target for parasites.
PART 3: Preserving the ancestral population is a way of maintaining a source of biodiversity from a population that has not been subjected to artificial selection.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”7.Evolution” dataset_id=”AP Bio FRQ Dataset 2022|101bd0b76c07″ question_number=”47″ topic=”7.1-7.3_Natural_Selection”] In 1959, Russian scientist Dmitri K. Belyaev began a long-term experiment to investigate the genetic basis of the tame behavior that can be observed in dogs and other domesticated animals. Belyaev’s experimental subject was the silver fox. Over the course of the experiment, which ran for decades, foxes were bred together and the resulting pups were assessed each month between the ages of 1 and 8 months to see how tame they were.
The following system was used to categorize the foxes based on their tameness
- Class 3: Not tame – These foxes flee from or bite their handlers.
- Class 2: Neutral – These foxes allow handling, but show no friendly response.
- Class 1: Tame – These foxes are friendly toward humans. They whine for attention and wag their tails.
- Elite/Very tame – These foxes are eager for human contact. Their behavior includes everything listed above for tame foxes, plus sniffing and licking their handlers’ hands, and whimpering to attract attention
To breed the next generation, Belyaev selected the tamest 5% of the males foxes and the tamest 20% of the female foxes. His institute repeated this processes for over forty generations. The results are shown in the table below.
|Number of generations||Foxes in elite class (%)|
PART 1: List the name of the process used by Belyaev, and discuss the biological basis for the changes he was able to produce.
PART 2: Explain why only 5% of the male foxes were allowed to breed, while 20% of the female foxes were chosen.
[Qq]Note from Mr. W: Give yourself extra credit if you mentioned the possible role of dopamine receptors (with lower levels of dopamine reception correlated with lowered anxiety).
PART 2: Fewer males were used because males can father many offspring, while females produce relatively few offspring. Using a small number of males allows for increased selection pressure. At the same time, there could be genetic problems if the breeding population is too small, so using more females than males allows for maintenance of adequate genetic diversity.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”7.Evolution” dataset_id=”AP Bio FRQ Dataset 2022|23861c9df03006″ question_number=”81″ topic=”7.1-7.3_Natural_Selection”] The following facts can be used in interpreting the graph shown below.
* Concentrations of sulfur dioxide (SO2) can be used as a general indicator of air pollution.
* Air pollution can cause tree trunks in forests to darken through two mechanisms: by depositing dark soot on tree trunks, and by killing lichens (light colored growths) on tree trunks.
* In certain moth species, a single allele controls color, which can range from light (peppered) to melanistic (dark).
* Certain moth species are mostly active at night, and spend their days resting on tree trunks.
* Birds prey on moths.
The graph above shows sulfur dioxide concentrations in New England between about 1960 and 1980, after which they remained at 1980 levels. The graph also shows the percentage of dark colored (melanistic) moths found in New England forests from 1960 to 2010.
Explain the relationship (if any) between the SO2 levels shown in the graph, and the changing percentage of melanistic moths.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”7.Evolution” dataset_id=”AP Bio FRQ Dataset 2022|23839c552a4406″ question_number=”86″ topic=”7.1-7.3_Natural_Selection”] The graph below displays the results of an experiment in artificial selection that used fruit flies, Drosophila melanogaster. During the first 25 generations, the smallest flies were selected to produce the next generation. The process was reversed for generations 25 through 35, the largest flies were selected to produce the next generation.
Explain why no further increase in body size was observed after generation 25, despite selection for larger size.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”7.Evolution” dataset_id=”AP Bio FRQ Dataset 2022|23832a3ee3a006″ question_number=”87″ topic=”7.10.Speciation”] For the past 40 years, the London Zoo has been working to prevent British insects from extinction through a program that involves captive breeding, followed by re-release into the environment. An example of this program involves the field cricket (Gryllus campestris). The males of this species dig burrows in the ground, making distinctive platforms at the entrance where they stand and “sing” to attract females for mating. Their song has a distinctive pitch and rhythm. By 1991, this cricket population was reduced to only 30 individuals.
In 1992, biologists captured twelve field crickets, six male and six female. Their population was increased to over 4000 through captive breeding under controlled laboratory conditions. The 4000 crickets were released at carefully chosen spots in southern England. Each spot was within the cricket’s original range. As of 2010, there were six self-sustaining colonies, shown in the map below. Note that although field crickets have long wings, they can only use these wings to assist their hopping, and can’t fly long distances.
PART 1: The biologists made an effort to keep conditions constant in the laboratories where the crickets were being bred. Explain how this decision might have affected diversity within the population of the field crickets.
PART 2: The biologists predicted that the newly established colonies will be geographically isolated from one another. Define “geographic isolation” and give two reasons to support the prediction.
PART 3: Carry out this thought experiment. Suppose that the newly established colonies are left to breed for hundreds of years. After that time period, six males from colony 1 and six females from colony 2 are collected. They’re placed together in a laboratory environment that is conducive to cricket mating. However, the biologists observe that no mating takes place. Propose an explanation for the lack of mating between these crickets.
PART 4: Predict how the genetic diversity in the re-established populations of this species could affect its long term survival.
UEFSVCAyOg==[Qq] Geographic isolation occurs when sub-populations of a species become physically isolated from one another, preventing alleles from flowing between these sub-populations through interbreeding. The cricket colonies are 10 – 15 km apart and they do not fly. Secondly the habitat in between is not the acidic grassland that these crickets require, and so there would be no chance of them moving across the land.
PART 3: The two populations have diverged and have become reproductively isolated from one another. That is, they’ve become separate species. It’s impossible to tell from the description whether this isolation is pre or post zygotic.
PART 4: Genetic diversity in the populations is low because breeding program had so few animals to work with. This is not desirable as any change in the environment could leave the crickets without the necessary genetic diversity and phenotypic variation to adapt to the change.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”7.Evolution” dataset_id=”AP Bio FRQ Dataset 2022|1320d0231c07″ question_number=”41″ topic=”7.13_Origin_of_Life”] The “RNA World Hypothesis” posits that RNA is the most likely candidate for the being the first molecule with some of the essential properties of living things. Explain why this is thought to be so.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2386af4cdd4c06″ question_number=”80″ topic=”8.1.Responses_to_the_Environment”] In bee colonies, worker bees take care of bee larvae, a behavior known as “nursing.” Scientists studied the nursing behavior using glass-walled hives. They observed the percentage of nursing that occurred over 24 hours. The data they collected is shown on the graph below. A value of 100% indicates only day-time nursing activity, and 0% indicates only night-time nursing activity.
As part of their data analysis, the scientists compared the actual amount of day-time and night-time nursing activity to an ideal normal curve of activity.
State a null hypothesis. Then, using the data in the graph from 30% to 70%, decide whether to accept or reject your null hypothesis. Justify your response.
The formula for chi square is
The sum of all of the (O-E)2/E is 7.47 Because there are five categories (30, 40, 50, 60, and 70% daytime nursing), there are four degrees of freedom. I looked up the critical value at 0.05 (the commonly accepted value in the sciences: ask a statistics teacher if you want more information about that!) for four degrees of freedom in the critical values table (see above), and the value is 9.49. Because 7.47 is less than 9.49, I can assume that the difference between observed and expected values is insignificant.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2377b240518c06″ question_number=”105″ topic=”8.1.Responses_to_the_Environment”] The little penguin, Eudyptula minor, lives on the southern Australian coast. It feeds on fish, cephalopods (octopi and squid), and crustaceans. Its breeding cycle starts in October, when eggs are laid. The eggs hatch during the end of November. Small chicks can be observed during December, and older chicks, about 4-8 weeks old, can be observed during January and February. The chicks are fed by their parents and are ready to leave the nest by the end of February.
Over the course of 15 month study, penguins were captured and, using a method that causes no harm to the penguins, the contents of their stomachs were sampled. In the graph below, “relative mass occurrence” refers to the comparative dry weight of various food sources in the penguins’ diet. For example, in December on 1995, 80% of the penguins’ diet was from fish, 15% from cephalopods, and 5% from crustaceans.
During the course of the study, the breeding success rate of the penguins was above average, based on studies conducted during previous years.
PART 1: Explain how the diet of the penguins varied through the time period measured. Connect these variations to the penguins’ breeding
PART 2: In 1995, there was a high mortality of small fish known as the pilchard, Sardinops sagaux. Pilchards are a major prey species of little penguins. Based on the data given, suggest why the mortality of pilchards had no apparent effect on the ability of little penguins to rear chicks.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|66d4805007″ question_number=”79″ topic=”8.2.Energy_Flow_through_Ecosystems”] The diagram below shows the food web for an estuary. Estuaries are transition zones that are open to the sea, but which are also fed by one or more rivers or streams. The result is that fresh water mixes with salt water and creates a variety of aquatic and terrestrial habitats.
PART 1: Heterotrophic bacteria (not shown in this food web) play an important role in estuarine waters and sediments. Describe their role and its importance in this food web.
PART 2: In a recent study, ecologists measured the concentration of the heavy metal cadmium in the water and various organisms in the estuary. The concentration of cadmium in the estuary water is 0.005µg/g. Explain the levels of cadmium in the organisms below.
PART 3: The concentration of cadmium considered acceptable for human consumption is less than 3µg/g. Explain whether any of the organisms in the table above would be safe for humans to consume. Justify your answer.
UEFSVCAyOiA=[Qq]The table shows biomagnification. Cadmium, like most other substances that biomagnify, isn’t excreted or exhaled from the organisms that ingest it or absorb it. As a result, it stays within the body of the organism, and gets passed from one trophic level to the next, accumulating in concentration as it goes up from producers, to primary consumers, to secondary consumers, and so on.
PART 3: Despite the fact the the mollusk’s cadmium level is 2.4 µ g/g, human consumption of mollusks would concentrate the cadmium to levels that might be toxic. A safe recommendation would be for humans to avoid eating any organisms from this estuary.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2385acdbb57006″ question_number=”82″ topic=”8.2.Energy_Flow_through_Ecosystems”] One way that ecologists characterize an ecosystem is through a transect: a path along which occurrences of a species, or abiotic features of an ecosystem, are measured. The diagram below shows a transect across a sand-dune ecosystem.
In this transect, soil was analyzed at locations A, B and C. The table below shows the results
PART 1: Describe the difference in the percentage of humus at A and C. Explain the reasons for this difference.
PART 2: Describe the difference in nitrate concentration between location C and location A. Use information in the table and your own knowledge to explain this difference.
PART 3: A student predicted that at sites A and B legumes would grow better than non-legumes. Explain
UEFSVCAyOg==[Qq] At A the nitrate composition is 27.6 ppm, while at C the concentration is 165.6 ppm. Soil nitrates are produced by nitrifying bacteria which act on organic matter in the soil (humus). With less organic matter at A (because of the harsher, more abiotic conditions) there is less plant growth, less decomposition, and less release of nitrates into the soil.
PART 3: Sites A and B are relatively low in soil nitrates, which might restrict the growth of most plants. Legumes, however, have symbiotic nitrogen fixing bacterial nodules in their roots, which would enable them to grow in the low nitrate conditions at sites A and B.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|23851a2cc85406″ question_number=”83″ topic=”8.2.Energy_Flow_through_Ecosystems”] The questions that follow are based on the following food web, which shows some of the organisms living in Antarctic waters.
PART 1: List an autotroph in the food web above.
PART 2: Using information in the food web, explain the difference between the zooplankton that are labeled carnivorous, and those that are labeled herbivorous.
PART 3: The leopard seal can be placed into two trophic levels. Identify which ones, and justify your response.
PART 4: Blue whales, the largest animals ever known to exist, can be as large as 200 tons. They feed almost exclusively on krill. Krill are crustaceans weighing 1-2 grams. They are present in vast numbers in Antarctic waters. The largest Orcas, or killer whales, are about 10 tons. They feed on a wide range of foods as shown in the food web. Identify the ecological principle that enables the blue whale to grow to a much larger size than the Orca. Provide reasoning to support your response.
UEFSVCAyOg==[Qq] Herbivorous zooplankton feed on the phytoplankton, which are autotrophs or ecological producers. Carnivorous phytoplankton feed on the herbivorous zooplankton and krill, which are primary consumers.
PART 3: Leopard seals are secondary consumers when they eat krill (which is a primary consumer). Leopard seals are tertiary consumers when they eat fish, other birds or penguins (each of which can be classified as a secondary consumer, since they all consume krill).
PART 4: The key principle is the 10% rule. Only about 10% of the energy available at a particular trophic level is transferred to the trophic level above it, due to inefficiencies in harvest and assimilation, and to energy being lost as wasted heat during cellular respiration. By feeding on krill, which are primary consumers, the blue whale is a secondary consumer and is able to consume a significant amount of the original energy captured by the phytoplankton, allowing it to grow to a much larger size. Killer whales feed on higher order consumers such as penguins and leopard seals (making orcas 4th, 5th, and 6th level consumers). Orcas, as a result, have less energy available for them to consume and grow, as much of the original energy present in lower trophic levels has been lost.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237e7d7f03d806″ question_number=”96″ topic=”8.2.Energy_Flow_through_Ecosystems”] An ecologist studied the behavior of four mammals in a suburban environment. The observations are summarized in the following table.
PART 1: Draw a food web that shows how the species in this system Include the seeds.
PART 2: Based on the ecologist’s observations, identify each species’ trophic level.
PART 3: Homeowners have begun to poison the coyotes because of concerns about coyotes preying on pets. This has caused a decline in the coyote population. Make a short term and long term prediction about what will happen to the population of each of the other three mammals. Justify your reasoning.
PART 2: Rat – primary consumer; Mouse – primary consumer; Raccoons – secondary consumer; Coyotes – tertiary consumer; Seeds – producer.
PART 3: In the short term, raccoons should increase in population (since they’re no longer subject to predation by coyotes). The increase in raccoons could cause a decline in rats, as they become subject to increased predation by raccoons. With the rats in decline, the mice should expand.
Longer term, it’s possible that the expansion of the mice could lead to over harvesting of seeds, and that could lead to a decline in the rat population, which in turn would lead to a decline in the raccoon population.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237ddcd7cf6406″ question_number=”97″ topic=”8.2.Energy_Flow_through_Ecosystems”]
PART 1: The time it takes for a carbon atom that is leaving the atmosphere to be recycled back into the atmosphere can vary from days to millions of years. Describe a sequence of events, involving living organisms, that would result in carbon being absent from the atmosphere for each of the following periods of time, and then being returned to the atmosphere.
- Two days
- 100 years
- Millions of years
PART 2: Nitrogen makes up approximately 78% of the molecules in the atmosphere. Carbon dioxide, by contrast, makes up less than 0.04% of the air. Explain why plants can experience a deficiency of nitrogen, but never experience a deficiency of carbon.
MS4gREFZUzog[Qq]Carbon dioxide → photosynthesis → glucose in plant →respiration in plant → carbon dioxide
2. 100 Years: Carbon dioxide → photosynthesis→ wood of tree→decomposition or combustion → carbon dioxide
3. Millions of Years: Carbon dioxide → photosynthesis → marine plants → consumption by a marine animal with shell → rock formation → eventual weathering → carbon dioxide
PART 2: All plants readily access CO2 by diffusion from the air into their leaves via stomata, eventually producing glucose during photosynthesis (a process also referred to as carbon fixation). Although the concentration of CO2 is just over 0.04% in the atmosphere, CO2 levels are maintained by combustion (natural forest fires, and human-caused burning of wood, biomass, and fossil fuels) and continuous cellular respiration in plants and animals. As a result, plants are never in short supply of carbon dioxide.
On the other hand, although about 78% of the atmosphere is nitrogen gas, plants cannot utilize this nitrogen directly in its gaseous form, N2. Plants can only absorb soluble nitrate ions (NO3) from the soil, through active transport into their roots. Nitrate ions are produced from atmospheric nitrogen either by lightning or volcanoes, or by the action of nitrogen fixing bacteria, which are found in soil or in nodules on the roots of legumes. Other bacteria convert ammonia from animal remains and wastes into nitrites and then nitrates, which can be assimilated by plants.
Nitrate ions are required by plants for the synthesis of proteins and nucleic acids. So plant growth relies upon the uptake of adequate soil nitrogen. However, many soils are deficient in nitrate ions or nitrogen fixing bacteria, so plants growing in them have poor growth and are significantly “nitrogen deficient.” Furthermore, in conditions of oxygen shortage (compacted or waterlogged soils), denitrifying bacteria compound this problem by reducing nitrates to either nitrites, ammonia or nitrogen gas, and so deplete much of the usable nitrogen from the soil.
If this description made you want to review the nitrogen cycle, study the diagram below.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237981958fc806″ question_number=”101″ topic=”8.2.Energy_Flow_through_Ecosystems”] The diagram below shows the nitrogen cycle for a field of wheat. The figures are in kilograms of nitrogen per hectare per year.
PART 1: Assuming that pathway A involves leaching on nitrogen from the soil by runoff, describe what’s happening in pathway D.
PART 2: A farmer is considering growing peas instead of wheat. Predict the effect that growing peas would have on the nitrogen levels in the soil.
PART 2: Next to the farm is a native grassland. Explain how the cycling of nitrogen in this area would be different from that in the area growing wheat above.
PART 2: Peas are legumes which have nodules of nitrogen-fixing bacteria on their roots. Nitrogen fixed by a legume crop becomes available, which increases nitrogen levels in the soil. This is represented by the downward arrow on the far left above.
PART 3: There would be no equivalent to a crop harvest, so to a much greater degree, nitrogen would remain within the ecosystem (as shown in the nitrogen cycle diagram above).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2379142760ec06″ question_number=”102″ topic=”8.2.Energy_Flow_through_Ecosystems”] The diagram below represents the components of a freshwater aquarium. A fluorescent light bulb (not shown), provides the aquarium with the equivalent of natural light.
PART 1: Using the diagram below, show how carbon moves through this ecosystem. Draw at least four arrows, and label each one with the name of the process that is occurring.
PART 2: Describe how nitrogen would be made available to the freshwater plants in the aquarium. In your response, assume that the plants are not capable of fixing their own nitrogen.
PART 2: If the plants cannot fix their own nitrogen, then nitrogen would become available to the plants by excretion of waste by snails, along with death of snails and plants. This detritus would be broken down by bacteria to provide nitrates that could be taken up by the plants in the aquarium.
For reference, here’s a terrestrial version of this same process. Follow the “death” arrows to decomposers that change proteins and nucleic acids into ammonia (ammonification), followed by nitrification with creates nitrates that plants can assimilate.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2378a2111a4806″ question_number=”103″ topic=”8.2.Energy_Flow_through_Ecosystems”] The movement of carbon between the atmosphere, land systems, and ocean systems is shown in the figure below. The values are in gigatons of carbon per year. Note that 1 gigaton = 109 tons.
PART 1: Identify and describe all of the processes that result in
the movement of carbon indicated in the diagram by:
- the arrow marked Y
- the arrow marked Z.
PART 2: Atmospheric carbon dioxide has increased in concentration from 317 ppm (parts per million) in 1958 to 409 ppm in 2018. Explain how TWO components of the global movement of carbon have contributed to this change.
PART 3: Describe how the movement of energy through ecosystems differs from the movement of matter.
UEFSVCAyOg==[Qq] increased combustion of fossil fuels converts fossilized carbon that was sequestered as petroleum, coal, and natural gas into carbon dioxide in the atmosphere. This has the effect of increasing atmospheric carbon dioxide (because it adds carbon that had been removed from the carbon cycle back into the cycle)
PART 3: Energy flows through ecosystems. By contrast matter is recycled. Another way of saying this is that the Earth is an open system with regards to energy, but a closed system with regards to matter.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|1f063ce40c07″ question_number=”33″ topic=”8.3.Population_Ecology”] In a computer simulation of population growth, a small number of dandelion seeds are introduced into a field. The number of dandelions grows slowly at first, then accelerates, then levels off to stabilize at a constant number.
Part 1: Draw a representation of the growth of this dandelion population.
Part 2: Define and explain the rules of this simulation, and, as you do, explain the relevant principles of population growth. Make sure your explanation includes the concepts of carrying capacity and limiting factors.
Part 2: As logistic growth begins, it looks like exponential growth. That’s because logistic growth is an extension of the exponential growth model that includes the concept of carrying capacity: the maximum number of individuals that an environment can support.
The exponential growth model involves a population growing at a fixed rate. As the population increases, the number of individuals added over time increases. For example, if a population consists of 100 individuals, and it grows by 1% each year, then in the first year it will add only 1 individual. But when the population reaches 1,000, it will add 10 individuals a year. When the population reaches 10,000 it adds 100 individuals/year, and so on.
But here’s where the logistic growth model comes into play. The logistic model takes the population’s growth rate, and multiplies it by (K-N)/K, where K is carrying capacity and N is the size of the population. When a population is well below the carrying capacity, this expression has a low value. But as the population approaches carrying capacity, this expression can halt population growth.
Imagine for example, that the carrying capacity for a population in a specific area is 1000 individuals. If the population is 10, then (K-N)/K is 990/1000. You’ll get 99% of the exponential growth rate. But when the population is 900, then (K-N)/K is 100/1000. You’ll get only 10% of the exponential growth rate.
As a result, logistic growth looks like what’s shown for the dandelions in this computer model: population growth begins slowly, and then accelerates. But as the population approaches its carrying capacity, growth over time slows and then stops. In the real world this is where environmental resistance emerges. This resistance includes density-dependent limiting factors, which can be intrinsic (stress-related changes that limit birth rate or increase death rate) or extrinsic (increased predation, parasitism, or competition), any or all of which act to increase the death rate and/or lower the population’s birth rate.
[q json=”true” xx=”1″ multiple_choice=”false” dataset_id=”AP Bio FRQ Dataset 2022|4fc4be93007″ question_number=”70″ unit=”8.Ecology” topic=”8.3.Population_Ecology”] In 1975, a small flock of five English house sparrows was blown by a storm to an oceanic island. At the end of 10 years the island supported a stable population of about 10,000 individuals. In a subsequent census of the house sparrows in 2000, the population was still about 10,000.
When compared to the house sparrows found on the mainland, the male island house sparrows have, on average, darker plumage.
PART 1: Sketch a simple graph that shows the probable change in population size (N) from the time of colonization to 15 years post colonization.
PART 2: Describe three evolutionary mechanisms by which the island males could have evolved their darker plumage.
[Qq]PART 2: One possible mechanism would be sexual selection. If among the small number of female founders there was a heritable preference for darker males, then these females would preferentially mate with darker males. Assuming that the darker plumage was also heritable, the male offspring would inherit alleles for darker plumage, and the female offspring would inherit alleles for the preference for darker males. This creates a kind of positive feedback loop that can, with relative rapidity, shift allele frequencies in a population (see https://en.wikipedia.org/wiki/Sexual_selection)
A second mechanism would be natural selection. If the environment on the island had a darker background, and some other environmental feature made camouflage advantageous for males more so than females, then male birds with darker plumage would be selected.
Finally, genetic drift and the founder effect could explain the darker coloration (with an allele for dark coloration that is mostly expressed in males being found in higher frequency among the founding population).
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|39f0cf19807″ question_number=”73″ topic=”8.4.Effect_of_Density_of_Populations”] Parus major is a songbird species in which males and females form monogamous pairs. The male of each pair then establishes a territory that he defends against other males by singing and threat displays. The male and female cooperate in building a nest within the territory where they lay eggs and rear their chicks.
Weasels, Mustela nivalis, prey upon Parus major eggs and chicks. The graph below illustrates how the territory size of Parus major affects the risk of nest predation by weasels.
Describe and explain the relationship shown in the graph.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|1f78532ab007″ question_number=”34″ topic=”8.5.Community_Ecology”] At one time, parts of the San Francisco bay were used as landfills. But once the dumping stopped, life began to invade. Lichens formed on concrete remnants of house foundations. Then mosses entered, then grasses, then shrubs. Insects arrived. Birds built nests. Rodents invaded from nearby areas. Now, in certain areas, a forest has started to grow. What is this process called, what are the underlying dynamics that are at work, and what are some of the key trends?
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|315ae340c07″ question_number=”74″ topic=”8.5.Community_Ecology”] In their study of the Galapagos finches, Peter and Rosemary Grant discovered that when G. fulginosa, and G. fortis lived on separate islands, their beak sizes overlapped. However, on the two islands where both species lived together (Floreano and San Cristobal) their beak sizes diverged. Explain, citing both the evolutionary and ecological principles that are at work.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2a143e18407″ question_number=”75″ topic=”8.5.Community_Ecology”] In 1958, Biologist Carl Huffaker published a study on predator-prey relationships involving two species of mites, Eotetranychus sexmaculatus and Typhlodromus occidentals. By increasing the size and complexity of an experimental system (including hiding places and dispersal barriers), he was able to bring about relatively long-term coexistence for the two species in his study. The results are shown in the graph below.
PART 1: Based on the diagram above, identify which species was the predator, and which was the prey. Justify your answer.
PART 2: In another experiment, Huffaker set up a simpler ecosystem (no hiding places, no barriers) with the same two species. The results are shown below.
Describe how these results of the two experiments are different, and explain why.
[f]IFBBUlQgMTo=IFRoZSBwcmVkYXRvciBpcyA=VC4gb2NjaWRlbnRhbHM=LCBhbmQgdGhlIHByZXkgaXMgRS4gc2V4bWFjdWxhdHVz[Qq]. This can be determined by the fact that the population of the predator is almost always lower than that of the prey (which conforms with how energy passes from lower to higher trophic levels, with less energy available at higher trophic levels, so predators are almost always less common than prey). In addition, there’s a clear time lag, where prey population rises, then predator population follows, and vice versa. This makes sense: more food available in terms of more prey will (at least temporarily) support higher numbers of predators. When prey population falls, so will predator population. This also known from other predator-prey relationships, such as lynx and hare.
PART 2: The simpler model provided for fewer hiding places and refuges for the prey. As a result, the prey were quickly consumed by the predator, which overshot the carrying capacity, leading the entire system to collapse in just a few weeks.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|1eba8420c07″ question_number=”76″ topic=”8.5.Community_Ecology”] Mauritius is an island nation in the Indian ocean, located about 2000 kilometers from Africa.
From a biological perspective, the island is famous for being the home of the dodo bird, which was made extinct by human activities shortly after the island was settled in the 1500s.
Mauritius’ calvaria tree, which produces fruit that fall to the ground near the parent tree, is critically endangered. Seeds contained within the fruit germinate at a frequency of less than 1%. Any seedlings appearing under the canopy of the parent tree suffer very high rates of disease.
As part of a program to prevent the calvaria trees’ extinction, domestic turkeys and giant tortoises have been released on Mauritius. Both turkeys and tortoises are known to consume fruit from the calvaria tree.
PART 1: In areas where only giant tortoises were released, young calvaria trees grew in new locations and these new trees showed no sign of disease. Explain these observations.
PART 2: In areas where only domestic turkeys were released, no new calvaria trees were found. Explain this observation.
UEFSVCAy[Qq]: It’s possible that the turkeys (unlike the tortoises) do not pass out viable seeds. Possibly, the seeds may be damaged in the turkey’s gizzard.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|eddf31f807″ question_number=”78″ topic=”8.5.Community_Ecology”] The diagram below depicts the life cycle of a mistletoe plant. Mistletoe grows on tree branches, and never roots itself to the ground.
For each pair of organisms listed below, identify the type of ecological relationship. Justify your response.
- Bird and mistletoe
- Mistletoe and tree
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|23847bd99fc406″ question_number=”84″ topic=”8.5.Community_Ecology”] The table below summarizes the results of a study of coyotes (Canis
latrans) in western Arkansas. In the ecosystems that were studied, the following relationships were observed:
* Rodents and jackrabbits consume plants.
* Both coyotes and mustelids (weasels, skunks, etc,) are rodent predators.
* Coyotes also consume some plant material, jackrabbits, and very rarely, mustelids.
Four 5000 acre sites were studied. In two of these sites, coyotes were removed. However, coyote removal is only partially effective, with many coyotes migrating back to the removal sites. Ultimately, coyote density is about half of that in the areas where the coyotes were not removed.
PART 1: One of hypotheses underlying this study is that reduction in numbers of coyotes causes an increase in the total number of rodents. Explain how the study’s findings either support or reject this hypothesis.
PART 2: A second hypothesis underlying the study is that reduction in numbers of coyotes causes (directly or indirectly) a decrease in mammal species diversity. Using evidence from the study, accept or reject this hypothesis.
PART 3: Explain why the areas with fewer coyotes have more total individual rodents but have fewer species of rodents.
UEFSVCAyOiA=[Qq] Coyote removal resulted in a decrease in mammal species diversity. In both areas where coyotes remain, 17 species of mammals can be found (11 rodent species, 5 mustelids, and jackrabbits). In both areas were coyotes were removed, species diversity fell to 6 (1 rodent, 5 mustelids, and jackrabbits).
PART 3: Coyotes might act as keystone predators in the study areas. Specifically, the role of the coyote might be to keep the rodent species under control. When the coyote is removed, one rat species (the Ord’s kangaroo rat) increases in number, and probably out-competes the other rodent species (reducing the number of rodent species from 11 to 1). Thus, coyote predation increases overall mammal diversity.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237d2be447b406″ question_number=”98″ topic=”8.5.Community_Ecology”] Explain how the re-introduction of a predator can increase ecosystem diversity.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|238409c3592006″ question_number=”85″ topic=”8.7.Disruptions_to_Ecosystems”] The table below shows concentrations of nitrates and phosphates in three lakes in Minnesota (all units are in mg/L). All three of the lakes have little input of nutrients from streams, and no outlets.
A developer proposed a housing development around Lake Balance. Because the area is out of the reach of sewage lines, the developer has proposed to provide each house with a septic system for human waste disposal. Septic systems are known to leach nitrogen and phosphate-rich compounds (derived from human wastes) into the groundwater.
PART 1: Predict whether, over time, Lake Balance will come to be more like Trout Lake or Big Lake. Justify your response.
PART 2: Describe the biological changes that can be expected in Lake Balance in the coming years.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237c8b3d134006″ question_number=”99″ topic=”8.7.Disruptions_to_Ecosystems”] A team of student researchers experimented with building closed, self-sustaining aquatic ecosystems that consisted of goldfish and aquatic plants.
In one experiment, the team assembled two equivalent systems. They filled the first system (A) with water from an oligotrophic lake. They filled the second system (B) with water from a eutrophic lake. The next morning, the fish in system A were alive, but the fish in system B died, Explain these results.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237826aaa41406″ question_number=”104″ topic=”8.7.Disruptions_to_Ecosystems”] Ecologists are studying the relationships among the species surrounding an island off of the southern coast of Australia. The community living on the island and in its surrounding waters includes phytoplankton (microscopic plants); zooplankton (microscopic animals); krill (small shrimp-like organisms); various species of fish; squid; and birds such as the Australian gannet and the shy albatross.
PART 1:The ecologists have identified the following relationships:
- gannets eat fish that feed on krill,
- the shy albatross eats fish and squid,
- squid feed on fish,
- krill eats phytoplankton and zooplankton and
- zooplankton eats phytoplankton.
Draw and label a food web that represents these relationships.
PART 2: Within the past few years, a commercial fishery has developed that harvests large amounts of krill. Predict what effect this could have on the population of the shy albatross.
[Qq] PART 2:Krill are the source of food for the fish. Albatross feed on fish or squid, which have krill as their source of food. A reduction in the number of krill through harvesting by commercial fisheries will in turn reduce the fish, squid and shy albatross populations.
Commercial fisheries will reduce the krill population and there will be a reduction in energy/biomass flow in the food web through the fish and squid. The shy albatross population will decrease in numbers.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|2379ef03bea406″ question_number=”100″ topic=”8.Ecology”] The fish tank below represents an artificial aquatic ecosystem. The system has the following features.
- Light is the only energy input.
- The whole tank is aerobic.
- Nothing eats the bacteria.
- The fish only consume the phytoplankton.
- Humans eat the fish.
PART 1: Construct an energy flow diagram for this ecosystem. Label all the energy flows, compartments, and products. (Note from Mr. W. Creating this kind of diagram goes somewhat beyond the requirements of AP Biology. However, give it a try, look at my sample answer, and make sure that you understand the flows, compartments, energy losses, and energy transfers)
PART 2:In an experiment, a worm was added to this ecosystem. The worm only consumes the fish waste products. The waste and detritus produced by the worm are not consumed by the detritivores. Predict the effect of the worm on the level of respiration by the heterotrophic bacteria. Explain your prediction.
PART 3:A second experiment with the same setup as part (b) was conducted. In this experiment, the waste and detritus produced by the worm are consumed by the detritivores. Predict the effect of the worm on the level of respiration by the heterotrophic bacteria. Explain your prediction.
[c*] Show the answer
[f] PART 1:
PART 2: Ecologically, the worm is intercepting fish waste products (7 above) that would power metabolic processes (such as cellular respiration) in the heterotrophic bacteria. The worm is going to use some of the energy in those waste products to power its life processes, so there will be energy lost to the system (and the level below) as cellular respiration. In addition, because the waste and detritus from the worm are not passed on the the bacteria below, the effect of this would be to decrease the amount of cellular respiration performed by these bacteria.
PART 3: Similar to what’s above, the worm is still intercepting the fish waste products (7 above) that would power metabolic processes (such as cellular respiration) in the heterotrophic bacteria. The worm will continue to power its own life processes through the energy consumed in these waste products (as above). However, because the worm’s detritus and wastes are passed on to the heterotrophic bacteria, the rate of cellular respiration in these bacteria will be greater than in the scenario above, though less than in an ecosystem without the worm.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”3.Cellular Energetics” dataset_id=”AP Bio FRQ Dataset 2022|18285dd2a007″ question_number=”39″] In all eukaryotes, DNA is one of the main products that results from the cell cycle.
PART 1: If, at a particular moment, the main energy source for cell division is glucose, which metabolic pathways will be needed to create the ATP required for production of DNA?
PART 2: To investigate how energy is provided for DNA synthesis during cell division, yeast are fed with 14C-labeled glucose under anaerobic conditions. Assume that the only biosynthetic product during yeast cell division is DNA. List the end products of metabolic pathways where the 14C would be found.
[c*] Show the answer
[f] PART 1: Glycolysis, TCA/Krebs cycle, electron transport chain.
PART 2: In yeast in anaerobic conditions, the end products of anaerobic metabolic pathways for DNA production would include CO2, ethanol, and DNA.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”AP Bio FRQ Dataset 2022|d580298ac07″ question_number=”53″] Assume that a DNA-based marker for hemophilia-A (an X-linked recessive blood clotting disorder) has been identified on the X chromosome. The marker is 3 cM (centimorgans) away from the Factor VIII gene that is associated with the disease. The marker has two alleles, designated xA and xB. Each marker can be identified by analysis of the DNA in a saliva sample.
A woman named Alyssa has a brother who is a hemophiliac, but neither Alyssa nor anyone else in her family has this condition. DNA testing has uncovered the following markers among Alyssa’s family members.
Alyssa: xA/xB. Note that this means that Alyssa has the “xA” marker on one of her X chromosomes, and the xB marker on her other X chromosome.
Alyssa’s mother: xA/xB.
Alyssa’s father: xB
Alyssa’s brother: unknown.
PART 1: Illustrate the possible arrangements of the given alleles for both of Alyssa’s X chromosomes. Use xA or xB to indicate the markers; “+” to indicate the normal (wild type) allele, and “m” to indicate the mutant allele that causes hemophilia A.
PART 2: Write a short explanation of your diagram.
PART 3: State the probability of each of the above arrangements occurring.
PART 4: Assuming that Alyssa is a carrier, explain which of the arrangements of alleles on the chromosomes must be correct.
[c*] Show the answer
[f] Note from Mr. W. This is a really hard problem. If you didn’t get it, just study the explanation below, click “need more practice,” and see if you can recreate it the next time you see it.
Part 2 (Explanation)
Alyssa’s father was not a hemophiliac. His marker is xB. Therefore, one of Alyssa’s chromosomes (the one she inherits from her father) has to be xB…+ (with “+” indicating the normal allele for blood clotting). It’s stated in the data table for this question that Alyssa is a heterozygote for the marker (xA/xB). We know, therefore, that Alyssa inherited her xA marker from her mother. Because Alyssa’s mother was a carrier for hemophilia, she could have passed on either of her X chromosomes to Alyssa. The two possibilities of this chromosome are xA…+ (with the normal allele) or xA…m (with the mutant allele)
There’s a 50% probability of either arrangement (because we don’t know if Alyssa is a carrier or not)
If Alyssa is a carrier, then she must have arrangement “b.”
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|581028ef407″ question_number=”69″] The graph below shows the results of a seven year study of a grassland community. Based on the data, what would you conclude about the relationship between species richness and ecosystem resilience (as measured by drought resistance)?
[c*] Show the answer
[f] The study shows a correlation between increasing levels of species richness and increased resilience (as measured by drought resistance) in these grassland communities.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|49131c63807″ question_number=”71″] The diagram below represents the transect of a small river which flows through a woodland.
PART 1: Briefly describe two negative effects of heavy boat traffic on the ecosystem of a small river.
PART 2: The diagram below represents how development along some rivers has changed the landscape.
Describe two ways in which development could change river ecosystems.
[c*] Show the answer
[f] PART 1: Heavy boat traffic could result in erosion of river banks. Erosion could cause an increase in turbidity, resulting in decreased light penetration and photosynthesis. In addition, boat traffic could result in the release of pollutants, including spilled fuels and exhaust gases, which could have a detrimental effect on organisms living in the river.
PART 2: Development of the land along the riverbanks could result in runoff of fertilizers and/or pesticides causing (respectively) algal blooms, or killing insect or other river life. Also, development could cause a loss of habitat or food sources from the removal of the vegetation (aquatic plants, bulrushes, sedges, etc.) along the river’s edges.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|40c7b207407″ question_number=”72″] The graph below shows the results of a study of the effects of biological disturbance on three locations in the North Sea. The North Sea is surrounded by densely populated and highly industrialized European countries. Over a three year period, scientists examined the abundance of a number of fauna (animal species) in three locations. The locations were similar in terms of the depth the water, but their distance to the coast varied. The species were divided into four broad groups:
- disturbance sensitive
- disturbance indifferent
- disturbance tolerant
The results are shown in the following graphs. A month without a column indicates no measurements were taken.
PART 1: Identify the location that corresponds with each of the following descriptions. Justify your choices.
- Highest level of disturbance
- Lowest level of disturbance
PART 2: Describe the characteristics of an opportunistic species.
[c*] Show the answer
[f] PART 1: Based on the three graphs, it appears that Location A is subject to the highest level of disturbance. In a disturbed environment, one would expect few disturbance sensitive species, and many disturbance tolerant species. This is what is seen in A, which has the lowest proportion of disturbance sensitive species, the highest proportion of disturbance tolerant species, and the highest proportion of disturbance indifferent species. In an undisturbed region, one would expect to see few disturbance sensitive species, and many disturbance tolerant or indifferent species. That is what is seen in area C, which has the highest proportion of disturbance sensitive species, with the rest of the fauna consisting of a large proportion of disturbance indifferent species, and a small proportion of disturbance tolerant species.
PART 2: Weeds (like the well known dandelion) are opportunistic species. Opportunistic species are characterized by high rates of reproduction, rapid development to reproductive maturity, and low adult survival rates. Opportunistic species are adapted for colonizing newly available habitats and/or exploiting newly available resources. Their offspring are able to quickly establish themselves and rapidly grow to maturity as they carry out the process of rapid colonization.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|17be6074c07″ question_number=”77″] The diagram below represents five butterfly species. The species vary in both coloration and in toxicity.
The grey background shown in the box represents the natural background on which these butterfly species occur.
Three different evolutionary strategies are listed below. DEFINE each one, IDENTIFY which butterfly species fits each of the following strategies, and describe some of the evolutionary dynamics involved in each strategy. Note that a species may fit with more than one category.
PART 1: Mullerian mimicry
PART 2: Batesian mimicry
PART 3: Cryptic coloration
[c*] Show the answer
[f] PART 1: Mullerian mimicry occurs when two toxic species converge on the same appearance. This sends a signal to potential predators that consuming animals with this pattern can be harmful. Species A and E are Mullerian mimics of one another. A useful way to think of this is to imagine each species as a company that sells an electrical fence. They want to protect their customers (the property owners), but they also want to avoid liability (from people accidentally touching their fence). If they both paint their fence the same color, then people will learn to avoid that color (protecting the property inside and avoiding accidents).
PART 2: Batesian mimicry is when a non-toxic species evolves the coloration and form of a toxic species. Species D is a Batesian mimic of species E and A. To use the electric fence advertising analogy above, species D is a dishonest advertiser, an evolutionary parasite. It’s using the same color and form as species E and A, but that increases the chance that predators might learn that the form is not as toxic as feared. However, Batesian mimicry does provide a selective advantage to species D.
PART 3: Cryptic coloration is an adaptation in which organisms blend into their environment, and it’s in use by species C.
[q json=”true” xx=”1″ multiple_choice=”false” unit=”8.Ecology” dataset_id=”AP Bio FRQ Dataset 2022|237eed413e9806″ question_number=”95″ topic=”8.Ecology”] An ecologist has been hired by an aquaculture company which is interested in studying the commercial viability of raising squid. The ecologist constructs a large self-supporting tank with the food web shown below.
The ecologist’s first goal is to bring the tank to a steady state, and then establish a sustainable harvest of 10 kg of squid carbon per week. Assume that all the other biomasses (phytoplankton, fish, squid, and detritivores) were constant.
The following values are known.
- Plant production efficiency is 60%.
- Fish exploitation efficiency is 50%.
- Fish assimilation efficiency is 20%.
- Fish net production efficiency is 5%
- Squid exploitation efficiency is 50%.
- Squid assimilation efficiency is 80%.
- Squid net production efficiency is 40%.
Given the goal of 10 kg of squid carbon per week, what is the required net primary production, in kilograms of carbon? Show your work.
[c*] Show the answer
[f] You’ll need 12,500 kg in net primary production.
Here’s the work solving this problem:
NPP * 0.5 * 0.2 * 0.05 * 0.5 * 0.8 * 0.4 = 10 Kg
0.0008 NPP = 10Kg
NPP = 10 Kg/0.0008
NPP = 12,500 Kg
EXPLANATION: Two concepts are being addressed in this question. The first is NPP: that’s the amount of energy available in plants after they’ve carried out cellular respiration. The 60% listed in # 1 above is what the plants consume themselves between photosynthesis (which is gross primary production) and respiration. In other words, what’s left after 1 is NPP. That means that to determine the amount of plant biomass that we’ll need to sustainably harvest 10 kg of squid, we can ignore # 1.
The second concept is ecological efficiency: the amount of energy and matter that gets transferred between trophic levels. That includes the exploitation efficiency (the percentage of plant matter eaten by the fish, for example), the assimilation efficiency (the percentage of ingested matter that can be used) and the production efficiency (what’s left in the organism after it’s used the assimilated biomass to take care of its own energy needs).
Basically, that means that NPP required = 10 kg divided by the product of all the efficiency losses above (2 – 7 above).
IMAGES THAT NEED REDRAWING
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