Review Projects – 2013 Big Idea 2 - answers Mr. Bennett

1. Review Projects – 2013 Big Idea 2 - answers Mr. Bennett

2. LO 2.1 Answers • M.C. Question: Which of the following adaptations does NOT • illustrate energy conservation within an organism? • Large ears on the Fennec Fox • A thickened cuticle on plant leaves • American Black Bears hibernating through the winter • Male emperor penguins huddling in large rotating • masses during Antarctic winters • E) Changes in the E. Coli bacteria metabolism • dependent upon the outside environment • B is an incorrect answer because a plant’s cuticle has little to do • with energy conservation. The cuticle of a plant is adapted to • prevent water loss through transpiration, not to keep the plant • from having to expend more energy on a certain biological process. http://www.zooborns.com/.a/6a010535647bf3970b01116898bd5b970c-800wi Learning Log/FRQ-style Question: Over time, evolution has favored various methods of free energy capture in organisms. In plants, chloroplasts collect free energy from the sun and convert it to carbohydrates via the equation: 2 H2O + CO2 + light → carbohydrate (CH2O) + O2 + H2O. Suppose you are a scientist who wishes to conduct an experiment that will show how various environmental conditions can influence a plant’s ablility to capture free energy. If you wish to alter only the plant’s environment, what other variables need to be accounted for to ensure an accurate experiment and how might they be fulfilled? Answer: A variety of other factors that could change the outcome of the experiment. Leaf size, the number of leaves, the type and age of the plant, the density and location of the stomata, and the amount of chlorophyll (color) of leaves could change the observed rate of photosynthesis. These variables affect both photosynthesis itself (such as the amount of chlorophyll in the plant) and the rate of transpiration (such as the number and distribution of stomata), both of which could create inaccuracies. The use of cloned plants of the same age would eliminate many of these variables and greatly increase accuracy.

3. ANSWER KEY LO 2.2 MULTIPLE CHOICE: Which of the following events occur during the light-independent reactions of photosynthesis? I. Carbon gets reduced. II. ATP is produced. III. Oxidation of NADPH+ occurs. • I. only • II. only • I. and III. • II. and III. • I., II. and III. LEARNING LOG: Describe in detail the difference between substrate level phosphorylation and oxidative phosphorylation in producing ATP. Then classify glycolysis as one or the other and explain why. Finally, discuss how glycolysis directly relates to the citric acid cycle. Substrate level phosphorylation produces ATP when a kinase enzyme directly transfers a phosphate from a substrate to ADP. Less energy is produced this way in comparison to oxidative phosphorylation, which produces 90 percent of all ATP during cellular respiration. Oxidative phosphorylation occurs during chemiosmosis and involves the oxidation of the carrier molecules NADH and FADH2. Their protons are lost to the electron transport chain, which pumps the protons to the outer compartment of the mitochondria. As a result, a steep proton gradient is created between the outer compartment and inner matrix. Protons then flow down ATP synthase channels, generating energy to phosphorylate ADP into ATP. Glycolysis is classified as substrate level phosphorylation because a direct enzymatic transfer of a phosphate to ADP occurs. Glycolysis directly relates to the citric acid cycle because it produces 2 three-carbon molecules of pyruvate, and pyruvate is required for the citric acid cycle to begin.

4. Answer Key LO 2.3 M/C Question: If food is in short supply, which organism will feel the impact first? A) A cat B) A mouse C) An elephant D) A wolf A mouse would be the first organisms to be affected by the increase in energy supply because it metabolic rate is directly proportional to an organism’s size. Because the mouse is the smallest, it’s rate would be the fastest and thus energy would be used up more quickly. Without a constant food supply, there is no energy replacing the heat energy produced by the metabolic functions on the mouse. Learning Log/FRQ-style Question: A certain ecosystem experiences a volcanic eruption that prevents any light from penetrating its dust clouds for a period of time. What would be experienced by the autotrophs, primary consumers, secondary consumers. If no light is able to get to the photoautotrophs, there would be a halt in all photosynthesis. Photons would not be able to excite P680 and P700 electrons in photosystems I and II. Plants would not be able to produce NADPH or ATP, both of which are needed in the Calvin Cycle. If this environment persisted, most of the plant life would soon die off. As the number or producers becomes scare, their immediate consumers will also be feeling the pressure. In order to survive, they need to eat a lot since only 10% of the energy is transferred from trophic level to trophic level. That means less organisms can survive as the trophic level increase. So for secondary and tertiary consumers only the strongest organisms will be able to survive through a period of darkness, because they will have to compete with other organisms for the very limited number of resources and still have enough energy to reproduce. If the light does not return soon, the whole ecosystem could be wiped out.

5. Answer Key – LO 2.4 Which of the following reactions does not result in the release of free energy? A) glycolysis B) ATP hydrolysis C) lactic acid fermentation D) Oxidative phosphorylation The light reactions of photosynthesis can be represented by a “molecular mill,” to which it is sometimes referred. Describe the flow of free energy throughout the light reactions and be sure to include details regarding the production of ATP through ATP synthase. As photons enter the chloroplasts, they strike pigments molecules within Photosystem II, exciting electrons previously split from a water molecule and sending them down the electron transport chain, producing ATP. As this occurs, protons are translocated across the membrane to be diffused down their concentration gradient by ATP synthase, also producing ATP. Light energy continues to flow through the system as photons strike pigment molecules within Photosystem I, again sending electrons through the ETC via NADP+. The resulting products, including the ATP produced by ATP synthase, are then sent to the Calvin Cycle, where additional energy is produced. Pictures: www.bio.davidson.edu ; www.biologycorner.com

6. ANSWER KEY– LO 2.5 All of the following statements about glycolysis are true EXCEPT A) glycolysis has steps involving oxidation-reduction reactions. B) the enzymes of glycolysis are located in the cytosol of the cell.  C) glycolysis can operate in the complete absence of O2.  D) the end products of glycolysis are CO2 and H20. E) glycolysis makes ATP exclusively through substrate-level phosphorylations. The two stages of photosynthesis are known as the light reactions and the Calvin cycle. Identify and explain the steps within each of the stages and how light reactions and the Calvin cycle work together. Then describe the connection between photosynthesis and cellular respiration within plants. The light reactions are the steps of photosynthesis that convert solar energy to chemical energy. Light absorbed by chlorophyll drives a transfer of electrons and hydrogen from water to an acceptor called NADP+, which temporarily stores the energized electrons. The light reactions also generate ATP, using chemiosmosis to power the addition of a phosphate group to ADP, a process called photophosphorylation. Light reactions produce no sugar; that happens in the second stage of photosynthesis. The Calvin Cycle begins by incorporating CO2 from the air into organic molecules already present in the chloroplast. The initial incorporation of carbon into organic compounds is known as carbon fixation. Photorespiration is the connection between photosynthesis and cellular respiration. Photorespiration occurs in the light and consumes O2 while producing CO2. However, unlike normal cellular respiration, photorespiratoin generates no ATP. In certain plant species, alternate modes of carbon fixation have evolved that minimize photorespiration and optimize the Calvin Cycle.

7. ANSWER KEY-LO 2.6 Which would be more beneficial to a cell? Having a small surface area and a large volume. Having a small surface area and a small volume. Having a large surface area and a small volume. Having a large surface area and a large volume. How and why would evolution favor organisms that develop structures like the root hairs shown in the picture or like the cells of the villi that line humans intestines? What would be the consequences to the organisms if they did not develop these structures? Over a long period of time, natural selection would favor the organisms with these structures because it gives them a higher surface area. The larger the surface area, the more nutrients will be absorbed to carry out the cells functions. Natural selection would also favor these structures to have a smaller volume, because if the volume is too large, it will take more energy to sustain the cell. The organisms that had these structures that increased their surface area and maintained a lower volume would then have more energy that they could spend on reproduction and propogate their genes to later generations. If organisms never developed these structures, they would spend the majority of their energy on staying alive instead of reproducing and their genes would not be passed to later generations and evolution would be slowed.

8. LO 2.7 Answers • Multiple Choice Question: The cells of an ant and an elephant are, on average, the same size; an elephant just has more cells. What is the main advantage of small cell size? • Small cell has larger plasma membrane surface area than does a larger plasma membrane surface are than does a large cell, facilitating the exchange of sufficient materials with its environment. • B. A small cell has a larger plasma membrane surface area than does a large cell, facilitating the exchange of sufficient materials with its environment. • C. A small cell has a smaller cytoplasmic volume relative to its surface area, which helps to ensure the exchange of sufficient materials across its plasma membrane. • D. Small cells require less oxygen than do large cells. • FRQ/ Essay Question: Explain in detail why there would be both upper and lower limits to the size of a cell. • As an object of a particular shape increases in size, its volume grows proportionately more than its surface area. For each square micrometer of membrane, only so much of a particular substance can cross per second. Rates of chemical exchange with the extracellular environment will be inadequate to maintain a cell with a very large cytoplasm. The greater the surface area and smaller the volume, the faster the rate of diffusion; and vice versa. The need for a surface sufficiently large to accommodate the volume helps explain the microscopic size of most cells. Larger organisms do not generally have larger sized cells than smaller organisms but rather a greater number of cells.

9. LO 2.8 Answers • MC Question: Johnny is a young boy, but he loves to conduct experiments, because he wants to be a scientist when he grows up. For his latest experiment, he wants to test the impact of a light breeze on the growth of a species of water plant. He purchases three bowls to place the seeds inside, and two small fans to create the breeze; one plant will be his control and the other 2 will be his experimental plants. He places both plants in the same room under almost identical conditions, places the fans the same distance from each of the experimental plants, and fills the three bowls with the same amount of water. However, halfway through filling the bowl for his control plant, he runs out of distilled water and finishes filling it with tap water. After this, he turns the fans on low, and lets the plants grow for one week, before taking his first measurement and recording it. Is Johnny’s first measurement a justifiable selection of data? • Yes, all of the conditions are almost identical, and tap water in the control will not affect the outcome because the plant will take up the water regardless. • No, the tap water contains more solutes than distilled water, so the water potential of the water outside the plant in the control bowl will be higher than it should be, and the plant will take up water more quickly. • No, the tap water contains more solutes than distilled water, so the water potential of the water outside the plant in the control bowl will be lower than it should be, and the plant will take up water more slowly. • Yes, even though there are more solutes in the tap water, they will not make it past the selectively permeable Casparian strips. • No, the tap water will definitely be toxic to the plants. • FRQ Question: • The photosynthesis equation is 6CO2 + 6H2O + Energy --> C6H12O6 + 6O2, however that does not mean that the reactants work together immediately to produce sugar and oxygen. Photosyntheis actually has two very disticnt operations: the light reactions and the Calvin cycle. In general, the light reactions use H20 and light energy as reactants, and produce O2, ATP, and NADPH as products. The NADPH and ATP are then passed to the stroma to be used in the Calvin cycle, in conjunction with CO2, to produce the G3P sugar that the plant can use. The Calvin cycle also releases NADP+ and ADP + a phosphate, which will be cycled back into thylakoids for the light reactions. The creation of ATP and NADPH is based on the electron transport chains connecting the various photosystems within the thylakoids, so without light, no part of photosynthesis can occur because both the Calvin cycle and the light reactions are dependent of each other. • H20 is a crucial reactant in photosynthesis, as it is what provides the electrons for the electron transport chain between the photosystems. When H20 is broken down it gives off 2 Hydrogen ions, 2 electrons, and 1 oxygen. These electrons will travel between photosystems, and allow the production of ATP by an H+ gradient (H+ given off from H20 as well) which powers ATP synthase, and the electrons will then eventually reach the finall electron acceptor, NADP+, which will be reduced to NADH. • Oxygen is a waste product of photosynthesis. When H2O is broken down, the lone O molecules given off immediately combine with another one to become a stable O2 molecule. These will then exit as waste products in the reaction (oxygen being used in the Calvin cycle causes a process known as photorespiration, which is serves no productive purpose for the plant). • C02 is absolutely necessary in the Calvin cycle portion of photosynthesis, as it goes through carbon fixation, reduction (which outputs G3P), and then regenaration of RuBP, which accepts CO2. It is a crucial building block to the creation of sugar, and an integral part of photosynthesis. • Sugar is the desired product for the plant, and this is the accumulation of the work from all the building blocks in photosynthesis (H20, C02, ATP, electrons, NADH, etc.) After the Calvin cycle completes, then the plant recieves another molecule of G3P that it can use as energy.

10. ANSWER KEY--LO 2.10 • M.C. Question: A plant cell starts off in an isotonic solution with 20 water molecules, 15 sodium molecules, 15 glucose molecules, and additional molecules necessary for photosynthesis. After completing photosynthesis, the plant cell still has the 20 water molecules and 15 sodium molecules, but now it has 30 glucose molecules. Do not account for the change in concentrations of the molecules needed for photosynthesis. What happens to the movement of water and other materials into and out of the cell? • A) 15 glucose molecules leave the cell to create an isotonic environment again. • B) 15 sodium molecules leave the cell to create an isotonic environment again. • C) With help from aquaporins, 15 water molecules enter the cell to balance out the increased glucose concentration with the cell. • D) The cell is a plant cell and has no plasma membrane for water and other materials to move in and out of. • Learning Log/FRQ-style Question: Suppose you place a cell in a colder environment. a). What does this environment do to the membrane’s fluidity? Why? b). Draw what the phospholipids in the membrane would look like. c). Explain the ways that membranes remain fluid in spite of temperature changes. d). Explain the importance of membrane fluidity with regards to selective permeability. e). Hypothesize ways that cells might evolve in warmer and colder environments if their membranes no longer had cholesterol. • a. The colder environment decreases the membrane’s fluidity. The colder temperatures decrease the movement among the phospholipids, which causes them to get closer together, and become more solid. • b. (Sample drawing) • c. If the membrane’s phospholipids have unsaturated hyrdocarbon tails, the tails, with their kinks, keep the phospholipids more separated, which helps the membrane remain fluid in colder temperatures. Cholesterol helps keep membranes fluid in colder temperatures by keeping phospholipids separated. Cholesterol also makes the membrane more rigid in warmer temperatures because it restricts the movement of the phospholipids. • d. Membranes must be fluid to work properly. If a membrane solidifies, its permeability will change. • e. The cells might evolve to having more phospholipids with unsaturated hydrocarbon tails, or they may evolve to changing their lipid compositions to the unsaturated lipids during the colder times of year. Another molecule, probably a steroid, could act like cholesterol as well, which would help the membrane stay fluid in the cold and not too fluid in the heat.

11. LO 2.11 Answers: Multiple Choice Question: In a U-tube such as the one in the diagram to the right, the concentration of solutes in the left side X is 2M Sucrose and 1M Glucose while the concentration on the right side Y is 1M Sucrose and 2M Glucose. Note: The membrane pores are only permeable to Glucose. After the system reaches equilibrium, what changes are observed?a. The water level is higher in side X than in side Y.b. The water level is higher in side Y than in side X.c. The water level is unchanged. d. The molarity of sucrose and glucose are equal on both sides.e. The molarity of glucose is higher in side X than in side Y. b) FRQ: Cell membranes separate the internal environment of a cell from the outside environment. a) Explain why small plants or animals often die after a winter freeze. Be sure to include the effects of water crystallization on plasma membranes. b) Create a diagram and explain how the Casparian strip blocks the passive flow of dissolved materials into the interior of a plant. The Casparian strip is composed of hydrophobic suberin. Answers: a) The expansion of water as it freezes often causes cell membranes to lyse, or split open. When the structural integrity of a phospholipid bilayer is compromised, its function as a barrier fails. In the case of a cell with an irreparably damaged membrane, substances immediately begin to flow with their respective concentration gradients. When this is coupled with the inability of the cell to keep foreign substances out, the no longer favorable conditions within the cell cause it to die. The destruction of cells caused by freezing is often too damaging to be repaired, and organism is unable to survive. b) The Casparian strip is highly hydrophobic because it is composed mainly of waxy suberin. This hydrophobicity means that water is unable to pass through the strip, despite its relative thinness. When water absorbed through the root hairs takes the apoplastic route, it eventually encounters the Casparian strip as it slides between cell walls. Upon this encounter, the water and any dissolved substances it contains must pass through at least one cell membrane before continuing into the plant. This barrier exists in order to ensure that the water is filtered by the structure of the semipermeable before it can enter the xylem and proliferate throughout the plant.

12. LO 2.12 Answers The compounds in biological membranes that form a barrier to the movement of hydrophilic materials across the membrane are ?A. LipidsB. Carbohydratesc. Nucleic AcidsD. Integral membrane proteins Using a diagram describe the fluid-mosaic model of the cell membrane. Indicate the following; phospholipid molecules, hydrophobic and hydrophilic ends, types of membrane proteins and glycoproteins. List substances to which the membrane is relatively permeable and those substances to which it is relatively impermeable. The fluid-mosaic model describes the plasma membrane of animal cells. The plasma membrane that surrounds these cells has two layers of phospholipids. Substances to which the membrane is relatively impermeable include: (Hardly any) ex: water-soluble molecules such as amino acids, sugars, proteins, nucleic acids, and various ion Those Permeable include: ( a variety of substances) ex: Ions and molecules

13. LO 2.13 Answer Key • MC Question: Which of the following statements is/are true? A) Smooth ER is abundant in cells that synthesize large amounts of lipids B) Chloroplasts have their own DNA, while mitochondria do not C) Molecules travel from the rough ER to the Golgi apparatus via vesicles that break off the ER and fuse to the Golgi D) Both A and C are true E) A, B, and C are true • FRQ Question: It’s a grim prognosis. Without functional mitochondria, your cousin’s cells will not have enough ATP to carry out normal cell processes, thus leading to cell death. It won’t happen immediately, as the disease will take a few years to destroy the mitochondria completely; the disease will be most evident in the brain and muscles because these cells have a high energy requirement. Your cousin will likely be weak as the mitochondria produce less and less ATP for use throughout the body, until eventual cell death when no ATP can be produced. Cell processes that use passive transport, however, will not be affected because they don’t require energy from ATP.

14. LO 2.14 Answers Multiple Choice Answer: B The Mitochondria in figure 1.1 is used to assist in cellular respiration in the creation of ATP molecules. The DNA that is being pointed out in figure 1.2 is used to store genetic information that is used in protein synthesis. Learning Log Response Eukaryotic cells, figure 1.1, have several differences to that of prokaryotic cells, figure 1.2. A Eukaryotic cell has a nucleus that is semi-permeable and is used to protect the DNA from changes beyond those of natural variation through crossing over and random fertilization. Prokaryotic cells incorporate new DNA much more easily than Eukaryotic cells do, thus making them more susceptible. A second difference is the presence of organelles in Eukaryotic cells. These organelles wrapped in selectively permeable membranes perform specialized functions for the cell such as cellular respiration in the mitochondria, photosynthesis in chloroplasts. Eukaryotic cells do not have these specialized organelles. They primarily grow and divide. A third difference is the presence of a rough and smooth endoplasmic reticulum in Eukaryotic cells as opposed to prokaryotic cells. In Eukaryotic cells, this is the site of protein synthesis. In prokaryotic cells, protein synthesis is done very shortly after the RNA polymerase transcribes the genes from the DNA. A fourth difference is that prokaryotic cells reproduce by means of binary fission. This process is where the bacteria copies its DNA and divides into two cells. Mitosis function much the same that binary fission does. Mitosis is only see in eukaryotes and involves the condensing of DNA into chromosomes that then line up in the center of the cell and divide into two cells. Meiosis, also only found in eukaryotes, takes a diploid cell and, by dividing twice, makes 4 gametes with half of a regular cells DNA. Sources:http://www.scienceprofonline.com/publishImages/cell-biology~prokaryotic-and-eukaryotic-two-types-of-biological-cells~~element156.jpg http://www.yellowtang.org/images/structure_animal_ce_cu_la_784.jpg

15. LO 2.15 Answers What physiological change might occur in an individual if there is an inhibition of ADH? The individual would experience an increase in urination because of lack of reabsorption in the kidney tubules. The individual would not feel the need to rehydrate because inhibition of ADH results in water reabsorption. Inhibiting ADH would cause an individual to feel dehydrated as there would be an increase in water loss through urination resulting in “hangover” like symptoms A and C A, B, and C Osmoreceptors in hypothalamus Thirst Hypothalamus Drinking reduces blood osmolarity to set point Alcohol inhibits ADH Increased permeability ADH Frq style question: Using specific evidence justify the following claim. Why is that individual who consumes alcohol experiences water loss in urine instead of reabsorbing water back into the blood stream? Pituitary gland H2O reab- sorption helps prevent further osmolarity increase Normally, an increase in osmolarity in the blood is detected by receptors in the hypothalamus, the hypothalamus then releases ADH which increases the permeability of the distal tubules in the kidneys to water. The increase in permeability promotes water reabsorption in order to maintain homeostasis in blood osmolarity. Once blood osmolarity is lowered ADH is no longer released to prevent a decrease in osmolarity. However, when alcohol is consumed blood osmolarity is increased but, alcohol inhibits the release of ADH into the blood stream. Thus the distal tubules in the kidneys do not receive the signal that signals them to become more permeable to water. The lack of permeability means that water just passes through kidneys and exits the body in urine. STIMULUS: The release of ADH is triggered when osmo- receptor cells in the hypothalamus detect an increase in the osmolarity of the blood Collecting duct Homeostasis: Blood osmolarity

16. ANSWER KEY – LO 2.16 All of the following internal variables are moderated by negative feedback circuits EXCEPT • Blood pressure • Respiration rate • Uterine contractions • Blood glucose level Antidiuretic hormone (ADH) enhances fluid retention by making the kidneys reclaim more water. Discuss how negative feedback circuits regulate water balance in kidneys. Be sure to reference specific endocrine glands in your response. ADH is produced in the hypothalamus of the brain and is stored in and released from the posterior pituitary gland. Osmoreceptor cells in the hypothalamus monitor the osmolarity of blood. When it rises above a set point , more ADH is released into the bloodstream and reaches the kidney. The main targets of ADH are the distal tubules and collecting ducts of the kidney, where the hormone increases the permeability of the epithelium of water. This amplifies water reabsorption, which reduces urine volume and helps prevent further increase of blood osmolarity above the set point. By negative feedback, subsiding the osmolarity of the blood reduces the activity of osmoreceptor cells in the hypothalamus, and less ADH is then secreted.

17. LO 2.17 Answer M.C. Answer: D) D FRQ-Style Answer: a) I. When the body consumes carbohydrates, or sugars, the body breaks down the sugars into glucose to be used as fuel for cellular respiration. After eating a carbohydrate-rich meal, the concentration of glucose in the blood will increase, stimulating the release of insulin. II. When there is an increase in blood glucose, the beta cells of the pancreas releases the hormone insulin into the blood. Insulin targets all body cells, except brain cells, and causes them to take up glucose from the blood. Insulin also targets the liver, causing it to store glucose as glycogen and slowing the breakdown of existing glycogen. As a result, blood glucose levels decrease to the homeostatic point of about 90 mg/100 mL and the stimulus for insulin release diminishes. b) I. When a person skips a meal, his or her body does not get replenished with carbohydrates to be broken down to glucose. Thus, there is less glucose in the blood after skipping a meal and blood-glucose concentration drops, stimulating the release of glucagon. II. When there is a decrease in blood glucose concentration, the alpha cells of the pancreas will release the hormone glucagon into the blood. Glucagon targets the liver and stimulates the breakdown of stored glycogen and the release of glucose into the blood. As a result, the blood glucose concentration rises to the homeostatic level of about 90 mg/100 mL and the stimulus for glucagon release diminishes.

18. Free Response Explanation • (A) When blood Ca+2 level falls below this set point, parathyroid hormone (PTH) is released. PTH is produced by four small structures, the parathyroid glands, that are embedded in the surface of the thyroid. PTH raises the level of blood Ca+2 by indirect and direct effects. PTH induces specialized cells called osteoclasts to decompose the mineral matrix of bone to release Ca+2 into the blood. The kidneys directly stimulate absorption of the Ca+2. PTH has an indirect effect on kidneys, which promotes Vitamin D conversion to its hormonal form. The active form of Vitamin D acts directly on the intestines, stimulating the uptake of Ca+2 and augmenting the effect of PTH. A rise in blood Ca+2 level above the set point promotes release of calcitonin from the thyroid gland. Calcitonin exerts effects on bone and kidney opposite to those of PTH and thus lower the blood Ca+2 level. Each hormone functions in a simple endocrine pathway in which the hormone-secreting cells themselves monitor the variable being regulated. • (B) Vitamin D is an integral part of the PTH response to the deviance in the Ca+2 levels. This steroid driven molecule is obtained from food or synthesized in the skin. Activation of vitamin D begins in the liver and is completed in the kidneys, a process stimulated by PTH. Vitamin D acts directly on the intestines, which stimulates the uptake of Ca+2 from food, and thus increasing the effect of PTH. A decrease or deficiency in vitamin D would result in a dampened effect of PTH, as the response would still occur from the mineralized portions of the bone. However, there would be no release of Ca+2 from the liver. In simpler terms, there would be a decrease of the amount of Ca+2 released, which would then slow the homeostasis reactions. The optimum level would be reached, but at a slower rate. Completed Image from the FRQ Question: • LO 2.18 Multiple Choice Explanation • (D) is the correct answer • (A) is incorrect as negative feedback occurs with the blood calcium level, with the parathyroid hormones and calcitonin. PTH, produced by the parathyroid glands of the thyroid, works to raise the calcium level back to homeostasis, when levels become lower than the optimum level. Calcitonin, another hormone from the thyroid, works to lower the calcium ion levels when it is too high. • (B) is incorrect as negative feedback occurs with blood glucose levels, with insulin and glucagon hormones. Insulin, secreted by the pancreas, works to lower the blood glucose level back to homeostasis, when levels become higher than the optimum level. Glucagon, another hormone released by the pancreas, works to raise the glucose levels when it is too low. • (C) is incorrect as negative feedback occurs with thermoregulation, with fluctuating temperatures. When it becomes too cold and deviates from the set point, the heater is turned on to produce heat, which then raises the room temperature. By the same token, when it becomes too hot, the heater is turned off and no heat is produced, which decreases the room temperature. In both scenarios the control center detects a deviation in the temperature change and activates mechanisms that reverse that change. • (E) is incorrect as negative feedback does occur with these control pathways. Negative feedback occurs with the thyroid glands as calcitonin is produced here, which lowers the blood calcium levels. Also, negative feedback does occur in the adrenal glands: the adrenal medulla and adrenal cortex helps with the blood glucose homeostasis. • (D) is the correct answer. During childbirth, the pressure of the baby’s head against receptors near the opening of the uterus stimulates uterine contraction, which cause greater pressure against the uterine opening, heightening the contractions, which causes sill greater pressure. This is an example of positive feedback because a stimulus activates mechanisms that escalate the response towards the same direction as the stimulus, until completion. This is opposite to negative feedback. • SUMMARY: In Choices A,B,C, and E, a deviation from optimum levels was detected, which activated mechanisms that reversedthat change. This is negative feedback.

19. ANSWER KEY- LO 2.19 Which of the following processes has a positive feedback mechanism? • Heating and cooling a house. • Placing a banana in a brown paper bag. • A Pancreas releasing insulin. • Shivering. Timmy was racing his friend down the road when he fell and a rock created a gash in his knee. What inflammatory response would follow this incident which exemplifies a positive feedback mechanism? How is this process have a positive feedback mechanism? What would happen if this response did not occur? (example of blood clotting) As soon as the skin is punctured blood clotting begins to form. It is a positive feedback mechanism in the way that once clotting starts and the platelets adhere to the site, with the help of fibrin, they release chemicals that attract more platelets. This rapid cascade of platelets adhering to each other, with fibrin, forms the clot, blocking passage of more blood out of the wound. If organisms did not have the ability to clot then they would bleed to death.

20. Answer slide for LO 2.20 • MC Answer: The correct answer would be A. As your body senses the change in the environment (your stomach) it reacts by converting pepsinogen to the enzyme pepsin which helps your body to break down the food. The creation of pepsin stimulates a the continuing change of pepsinogen molecules into pepsin. This cascade effect continues until the body has enough pepsin so all the food can be digested. • Free Response Question: After your friend Sara accidentally cuts her finger you inform her not to worry because her blood will start to clot and she then informs you she is a hemophiliac. Explain the normal positive feedback response that would occur in a non hemophiliac individual. Identify the receptor, control center and effector in this response and explain their role in feedback mechanisms. • FRQ Answer: In a non hemophiliac individual the normal response would a positive feedback loop. The broken skin would be the receptor, which detects a change in some variable in the organisms internal environment, so in this case the breaking of the blood vessels and decrease in blood pressure would relay a message to the control center. The control center, or the brain, reads the message the receptor sends and initiates an appropriate response via the effector. The effector is what essentially carries out the response. In this case the tear in a blood vessel would stimulate the body to respond through secreting coagulation proteins which clott the blodd and stop the bleeding.

21. LO 2.22 Answers MC Answer: D: A and B – the grasshopper and mouse gain 10% of the 100% of energy originally retained by grass through photosynthesis. FRQ Answer: There are 3 different symbiotic relationships. The first, mutualism, is a unique relationship in which 2 separate organisms benefit from each other. For example, a bee’s relationship with a flower. As the bee collects nectar from flower to flower, it helps the plant by spreading pollen. Another symbiotic relationship, is commensalism. Commensalism is a relationship in which one organism benefits, while the other is unaffected. An example of this would be a bird that lives in a tree. Finally, the last symbiotic relationship is parasitism. Parasitism occurs when one organism benefits from the other, while harming it. This occurs when ticks latch onto dogs for nutritents in their blood while harming the dog in the process.

22. Answer Key-LO 2.23 • MC Question- Which of the following reproductive action would you expect from a fish in shallow water that is hypersensitive to light and predators, when there is a scarcity of food and weather is turbulent? • A) The fish would move out to deeper water, in order to find food, and reproductive success would increase • B) The fish would remain in its habitat and abstain from reproduction because of the food scarcity • C) The fish’s reproductive success would plummet and the population size would decrease constantly • D) No change in behavior or population size would occur • E) Both A and C • Explain the difference between 3 different types of marine biomes and their interactions with species that inhabit those regions. • -Wetlands, estuaries and lakes all are similar in that they are marine biomes, but they vary in biotic and abiotic diversity, and that results in there different climates. Wetlands are covered with water to support aquatic plants, have low dissolved oxygen, and most photosynthetic organisms live here, some birds. Estuaries differ in that they transition between river and sea, salty bottom layer, salinity varies, and grasses along with algae grow, crabs and fish are more abundant here, and it is a good breeding ground. Finally Lakes are either oligotrphic and are nutrient poor and oxygen rich, or eutrophic lakes are nutrient rich and oxygen poor. Littoral zone is the shallow, well lighted zone, home to more plants and animals. One can see that depending on climate, biological systems are affected by complex biotic and abiotic interactions.

23. Answer Key LO 2.24 Which of the following is an example of an abiotic factor at work? A) A deadly virus that mutates every 30 seconds and destroys vital body organs. B) A storm generating an enormous tidal surge at 6 meters in height, inundating thousands of acres of coastal marshland. C) A decomposing body being broken down by bacteria. D) A swarm of locus in the gregarious phase devastate crops and cause major agricultural damage. Vast amounts of crocodiles once populated the city of New Orleans. Over time however, the population has declined significantly. Using the graph provided, list TWO abiotic factors and ONE biotic factor that could have contributed to the fluctuation in the New Orleans crocodile population. Include the specific year(s) the population experienced the most dramatic change. There are a variety of factors, both biotic and abiotic that may have contributed to the dramatic decrease in the alligator population, namely from 2005 to 2006. One biotic factor could be a decrease in the availability of prey such as small fish, sharks, turtles, snakes, and mammals. Other biotic factors contributing to the population decrease could be predators of crocodile eggs and young crocodiles such as winged birds like herons, egrets or eagles, cats , and even carnivorous mammals, such as wild. An abiotic factor that could effect alligator population is agricultural pollution. When swamps alligators live in are polluted with pesticides and chemicals, it can take as long as 10,000 years to repair the damage done. Hormone changes called anomalies, could affect alligator reproductive organs. Ultimately, these hormonal changes can alter the future of the species because the alligators are not able to reproduce effectively, which may have contributed to the population decrease from 2005-2006. Another abiotic factor that may have contributed to the population decline could have been a major storm such as a hurricane or other natural disaster. It can be observed that in 2005, New Orleans experienced an incredibly destructive hurricane; Hurricane Katrina. Ironically the same time the storm occurred was when the alligator population declined. The storm may have generated a massive tidal surge that drowned or destroyed many alligators on contact. On the contrary, the opposite abiotic scenario could have occurred. Instead of there being a storm generating thousands of gallons of water, a drought may have occurred. Alligators reside in mainly freshwater rivers, lakes, swamps, and marshes but a drought could radically change the salinity of these bodies of water. As a result, an alligator can experience changes in it’s blood diffusion. For example, blood samples could show a marked elevation in plasma – having a high concentration of solute in the blood in contrast to a low concentration of water in the blood. Since the marsh water may have became more salty, the marsh is hypertonic compared to the alligators’ blood cells. The marsh contains a higher concentration of solute and a lower concentration of water than the alligators’ blood cells. Therefore fluid would flow from the area of low solute concentration (the blood cells) to an area of high solute concentration (the marsh). As a result the alligators’ blood cells would shrink, and most likely, many would die. Another possible abiotic factor could be dense fog and cloud cover blocking the sun. Like all reptiles, alligators must bask in the sun to regulate their body temperatures. If alligators cannot bask in the sun, they cannot regulate their temperatures, either becoming too hot or too cool which in any case could lead to death and thus proving to be yet another possible scenario of how the alligator population decreased from 2005-2006. In 2005, the alligator population in New Orleans decreased significantly.

24. Answer Key – LO 2.25 The lungs that amphibians and mammals developed in order to breathe on land are an example of what in relation to a fish’s gills? A.) a vestigial organ B.) a homologous structure C.) an analogous structure D.) convergent evolution How does osmoregulation occur in most marine animals? Freshwater? Land? The problem marine animals face in saltwater is dehydration because the ocean is much saltier than their internal fluids, so water is lost from their bodies because osmosis. To balance the loss of water, marine animals drink a great amount of saltwater to gain water and salt ions. They use their gills and kidneys to excrete salt and water. Freshwater animals have the opposite problem marine animals have. Their high osmolarity causes the freshwater animal to gain water through osmosis. To maintain water balance freshwater animals uptake water and salt through their mouth and gills and excrete water through very large amounts of very dilute urine. Land animals must retain water in order to survive, they have body coverings that prevent water loss like exoskeletons or shells. However these animals lose a decent amount of water through urine and feces so they must make up for this water loss by eating and drinking moist foods.

25. LO 2.27 Answer key MC: The habitat of a species of rabbit has a particularly cold winter six years in a row. Which is most likely to happen to the rabbits? • Stabilizing selection because the rabbits of small and large size will be more commonly preyed upon. • Divergent selection because rabbits with medium color fur will not blend into their environment well. C. Directional selection because rabbits with thicker fur will have higher survivorship and reproductive success. • The rabbits will likely die because of a lack of food. FRQ: Explain in detail the effect of a steady increase in temperature of a desert environment on a species of lizard. Why would this occur? What effects would it have on the species in the future? If there were a steady increase in temperature in a desert environment, lizards in that environment would benefit from mutations that help them stay cool and conserve water. Therefore, in directional selection, the lizards with these beneficial mutations would reproduce more frequently and with greater success than those without. The environment would force this change through natural selection, and would cause the species to evolve into a different, more adapted species over generations.

26. ANSWER KEY – LO 2.29 M.C. Question: Why can normal immune responses be described as polyclonal? A) Blood contains many different antibodies to many different antigens. B) Construction of a hybridoma requires multiple types of cells. C) Multiple immunoglobulins are produced from descendants of a single B cell. D) Diverse antibodies are produced for different epitopes of a specific antigen. E) Macrophages, T cells, and B cells all are involved in normal immune response. Describe the main role of each of the following cell types, once it is activated by antigens and cytokines: helper T cells, cytotoxic T cells, and B cells. Answer: An activated helper T cell secretes cytokines that promote activation of both cytotoxic T cells and B cells. An activated cytotoxic T cell kills infected cells and tumor cells by apoptosis. An activated B cell differentiates into plasma cells that secrete antibodies.

27. ANSWER KEY -- LO 2.30 M.C. Question: Which of the following serve as examples of internal nonspecific immune responses? A) Phagocytosis B) Mucous membranes C) Natural killer cells D) A & B E) A & C The response would begin when chemical signals released by activated macrophages and mast cells to the injury site cause nearby capillaries to widen and become more permeable. Next, fluid, antimicrobial proteins and clotting elements move from the blood to the site. Clotting begins. Chemokines released by various cells attract more phagocytic cells from the blood to the injury site. Lastly, neutrophils and macrophages phagocytose pathogens and cell debris at the site and the tissue heals. You were slicing an apple, and accidentally cut your finger. Discuss the inflammatory response that would occur in response to the cut on your finger (use the following: macrophage, blood clotting, pathogens, chemokines, etc.).

28. ANSWER KEY– LO 2.31 • Exposure to certain types of ultraviolet radiation can degrade the spindle fibers in meiotic cells. If an animal cell in prophase I was irradiated with this type of radiation but went on to fertilize a gamete, which of the following would most likely afflict the resultant offspring? • Red-green color blindness • Huntington’s disease • Klinefelter syndrome • Multiple sclerosis • Human immunodeficiency virus Biologists studying thigmomorphogenesis, a type of physical change in plants induced by mechanical stress, notice that the plants raised inside a greenhouse grow taller and skinnier than plants of the same species outside, even when both groups receive identical sunlight, water, weather exposure, and nutrients. What environmental conditions might have induced thigmomorphogenesis outside of the greenhouse? Why might natural selection continue to favor thigmomorphogenetic plants? Physical contact with any condition not kept constant—such as mechanical input from animals or people in the area—could have induced thigmomorphogenesis in the outdoor plants. The shorter, stockier builds of plants affected by stimulated growth reduction could benefit individual organisms by allowing them to adapt to the stresses presented in each’s environment. By growing lower to the ground and thicker, plants could develop resistance to extreme wind or storm condition while still able to produce offspring that could grow taller and skinnier if doing so conferred an advantage (e.g., growing above shorter plants to deal with competition for resources). A comparison of two plants of the same species affected and unaffected, respectively, by thigmomorphogenesis

29. Ced-9 protein (active) inhibits Ced-4 activity Mitochondrion Ced-4 Ced-3 Death signal receptor Inactive proteins Cell forms blebs (a) No death signal Ced-9 (inactive) Death signal Active Ced-3 Active Ced-4 Other proteases Nucleases Activation cascade (b) Death signal Learning Log Answer: Messenger RNA, proteins, other substances, and organelles are spread out unevenly in the unfertilized egg; this has a significant effect on the development of the future embryo. The maternal substances, the cytoplasmic determinants, influence the organism’s development. The set of determinants a cell receives determines its fate through regulating the expression of the cell’s genes during cell differentiation. Through induction, the molecules conveying signals cause changes in nearby cells. The molecules which carry signals are proteins expressed by the embryo’s own genes. Signal molecules cause a change in the cell’s gene expression resulting in genotypical changes. In conclusion, interaction between these embryonic cells induce differentiation resulting in a new organism. LO 2.33 Answers Multiple Choice Answer: D

30. ANSWER KEY– LO 2.34 Which of the following statement about the apoptotic process it the development of paws in the mouse is TRUE? a. Bcl7 regulates the activation of procaspases b. Protein Bax activate procaspases directly, by inducing the release of cytochrome c from mitochondria into the cytosol signaling the beginning of cell death. c. Cytochrome C is released from lysosome into cytosol to kill the cell and thus eliminates interdigital webs. d. During apoptosis, the cell shrinks and breaks up into membrane-enclosed fragments called apoptotic bodies with subsequent elimination by a process similar to phagocystosis. e. All of the above Discuss how the immune system responds to an initial pathogenic exposure, and how this initial exposure can lead to a quicker response following a second exposure to the same pathogen. Draw and label the processes that occur. Pathogens such as macrophages, dendritic cells, and B cells, are presented as an antigen. Due to the fact that this is an antigen, B cells/plasma cells secrete antibodies. Then helper T cells active B cells, cytotoxic T cells, and/or macrophages. Moreover, cytotoxic T cells can cause cell death, also known as apoptosis. Apoptosis plays an important role in the immune system of vertebrates. It is used to regulate the function of immune system cells as well as to eliminate infected cells. Apoptosis plays a major role in the regulation of T lympocytes, therefore, it’s essential for programmed cell death of these harmful autoreactive T cells to prevent them for attacking normal cells of the body. Specifically, antibodies bind to these antigens that may possibly be viruses, and these antibodies secrete phagocytes that destroy these antigens. In the end, memory cells are produced in order for the secondary immune response, thus producing a faster response.

31. Answer Key- L.O. 2.35 M.C. Question: Quorum sensing would most likely occur when: A bacterium senses the production of cAMP Tobacco plants are exposed to sunlight Bacteria reach a certain critical concentration Predator insects eat herbivorous insects Insulin binds to target cells FRQ: Describe the difference between long-day and short-day plants. How can a long-day plant be induced to bloom in the middle of winter? Photoperiodism is the response of an organism to seasonal changes in day length. In plants, photoperiodism usually influences the physiological changes required to initiate changes in leafing or blooming of flowers in angiosperms. A photoreceptor protein, called a phytochrome, is usually found in plants that respond to photoperiods. The phytochrome is responsible for absorbing light and often initiating a signal transduction pathway that induces a response. The flowering of long-day plants was initially thought to be stimulated by the length of daylight being longer than the critical photoperiod for that plant, although recent research suggests that a critical “night length” is responsible for the control of blooming. (Thus, ‘long-day’ plants may be more accurately referred to as ‘short-night’ plants.) In the same way, ‘short-day’ plants may be more accurately referred to as ‘long-night’ plants. A long-day plant may be induced to bloom in the middle of winter by interrupting the long nights of winter with a regular “flash” of light.

32. ANSWER KEY: LO 2.36 Which of the following is not an example of regulation in animals? • Release and reaction to pheromones • Visual displays in the reproductive cycle • Sensory cues in the nervous system • Circadian rhythms in eukaryotes • Jet lag in humans What are examples of physiological events involved with regulation in plants, animals and fungi? Be sure to include one example for each category with elaboration on the process. In plants, phototropism is the growth of a plant shoot toward or away from light. If the plant grew toward the light it would be an example of positive phototropism, while negative phototropism is growth away from light. The regulation is seen in how in an ecosystem that is crowded, phototropism will direct seedlings to grow toward the sun in order to power photosynthesis. In animals, hibernation is a regulatory event. It is a long term physiological state of low activity and decreased metabolism. It is an adaptation to winter cold and scarcity of food. When certain vertebrate endotherms hibernate, their body temperatures are lowered so that less energy is used. Metabolic rate is also lowered so that these animals can survive on a limited supply of energy. In fungi, a fruiting body is formed. It is a special spore-producing structure. The fruiting bodies of fungi are commonly seen as mushrooms, which are edible.

33. Germinating pea seedlings were placed in the dark and exposed to varying ethylene concentrations. Their growthwas compared with a control seedling not treated with ethylene. EXPERIMENT All the treated seedlings exhibited the tripleresponse. Response was greater with increased concentration. RESULTS 0.10 0.40 0.00 0.20 0.80 Ethylene concentration (parts per million) Ethylene induces the triple response in pea seedlings,with increased ethylene concentration causing increased response. CONCLUSION LO 2.37 Multiple Choice Answer: C The answer is C because ethylene’s major function opposes auxin growth. While the other answer choices like thicker plant stem, curvature and environmental stress are results of ethylene. When plants become stress like during drought, flooding or injury ethylene is produced. In figure 39.13 in the textbook the plant shows increase curvature and stem thickness in a higher ethylene concentration. Therefore C is the answer. Free Response Answer: Yes, the plant will grow towards the light because the plant has the ability to do phototropism. Photo meaning light and tropism meaning to turn in a particular direction because of a external stimulus. Once the plan breaks the soil the coleoptile will ben towards the light because its dark side had a higher concentration of a growth promoting chemical –auxin. Auxin is in the apical meristem of the plan and it affects the apical dominance, meaning growing stronger on ones side than the other. The side of the stem that’s closer to the light will grow more in that direction. Gibberellin is a hormone in the fertilizer that makes a plant increase its productive size . Promoting apical meristem elongation and flowering makes Gibberellin a plant hormone perfect for fertilizers.

34. ANSWER KEY– LO 2.38 The prayer plant folds up its leaves each night in accordance with a circadian rhythm. If the plant is shipped halfway around the world to a location where it is daytime there when it is night here, the plant will A) slowly adjust to synchronize w/the new day-night cycle. B) immediately switch to a new cycle & begin to open leaves in day and close them at night. C) detect the change but remain on its original cycle, therefore still fold leaves in day and open them at night. D) not detect the change, therefore remain on its original cycle. • Behaviors of organisms may be influenced by environmental factors. For each of the following behaviors explain: i. how the environment affects the behavior, and ii) why this behavior increases the survivorship of individuals of a species. • A) Taxis/Kinesis • B) Migration • Algae/plant moves towards the directional stimulus of light in positive phototaxis as it needs light for photosynthesis. Another example would be an animal shivering in colder temperatures in order to stay warm. • Many animals migrate during the winter to warmer climates. This is because the animal is not equipped for colder climates, or food sources are scarce.

35. LO 2.39 M.C. Question: What would be an example of a behavioral inhibition to the mating process of a bowerbird? A) The sperm fails to fertilize the egg. B) Two birds live on opposite sides of a mountain range. C) A male bird builds an insufficient nest and is rejected. D) Two birds have different mating seasons. FRQ Response: (a) The symbiotic relationship between angler fish and bacteria is mutualistic because both organisms benefit – while the bacteria facilitates the hunting process for angler fish, the angler fish provides the bacteria with nutrients. This is in contrast with commensalism (where one organism benefits and the other is unaffected) and parasitism (where one organism benefits from the harm of another). An example of commensalism (though these are rare) is the relationship between cattle egrets and cattle. As the cattle graze, they stir up insects which are then more accessible to the egrets, which follow the cattle. Thus, the egrets benefit from the cattle’s daily business, which is unaffected. An example of parasitism is the relationship between ticks and dogs – while the tick benefits from the blood of the dog, the dog loses blood and is irritated by itchiness. (b) The angler fish demonstrates a rare example of polyandry, in which a female mates with multiple males. The sexual dimorphism reinforces this behavior, since polygynous species are generally known for large, ornamental males – in this case, the female is large and adorned with the bioluminescent lure, while the males are so small that many diffuse their sperm into the bloodstream of the female all at once. This sort of reproductive behavior is in contrast with polygyny, displayed by most bowerbirds, in which the males are generally larger and more ornamented and often mate with multiple females. Monogamous relationships are characterized by less sexual dimorphism and one mate per organism – an example of monogamy occurs in swans, which often choose a mate for life to maximize reproductive time.

36. LO 2.40 Answer key • Multiple choice answer : A European robins learning to open milk bottles, robbing cream from the top of milk bottles and adapting to the use of aluminum foil seals on the bottles, learning to tear them to access the cream Free response answer : Phototropism is the growth of organisms toward light . This is often observed in plants and fungi. The cells on the plant that are farthest from the light have a chemical called auxin that elongates the farther side of the plant. The lack of light or concentrated light in one area is the environment interacting with the organism causing it to grow toward the light. Hibernation is a state of inactivity and metabolic depression in endotherms. Hibernation refers to a season of heterothermy that is characterized by low body temperature, slow breathing and heart rate, and low metabolic rate. Organisms go into this because their environments are lacking what they need or nutrients are scarce or the climate isn’t favorable. An example of an organism hibernating would be a squirrel. Slide 1 of 2

37. LO 2.40 Answer key continued… • Estivation is another form of torpor dormancy ,Animals that estivate are trying to escape things happening in their environment.  This happens in hot, desert climates where heat and water are so important to the animals that live there.  Estivation protects these animals from high temperatures and drought. Just as animals hibernate in order to stay alive in cold places, animals estivate in hot, dry places.  The bodies of estivators will slow down.  Breathing and heartbeat get very slow.  An example of this would be Madagascan fat-tailed dwarf lemur . Migration is the relatively long-distance movement of individuals, usually on a seasonal basis. It is found in all major animal groups, including birds, mammals, fish, reptiles, and amphibians. The trigger for the migration may be local climate, local availability of food, the season of the year or for mating reasons.To be counted as a true migration, and not just a local dispersal or irruption, the movement of the animals should be an annual or seasonal occurrence, such as birds migrating south for the winter, or a major habitat change as part of their life. • Slide 2 of 2