Practice Test: Biology (66)

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Question Number	Your Response	Correct Response (show answershide answers)	Related Objectives and Rationale (show objectives & rationaleshide objectives & rationales)
1	Answer to question 1	C	Objective 001 Response C is correct because the outermost shell of a carbon atom is incomplete, allowing carbon to form covalent bonds with up to four other types of atoms. These bonds may be single, double, or triple, and when bonded with other carbon atoms, carbon atoms can form chains, branched chains, and rings. Incorrect Responses A, B, and D, while describing some of carbon's characteristics, do not contribute primarily to carbon's ability to form large, complex molecules.
2	Answer to question 2	B	Objective 001 Response B is correct because sulfur is a major component of methionine and cysteine, amino acids that are important components of structural proteins in the human body. Carbon, hydrogen, oxygen, and nitrogen are the four most common atoms found in all molecules in the mammalian body (Incorrect Response A), while sodium and potassium are responsible for transmitting electrical impulses (Incorrect Response C). Finally, sulfur is an important part of the reaction that metabolizes ATP and releases free energy in some bacteria and archaea, not in humans (Incorrect Response D).
3	Answer to question 3	A	Objective 001 Response A is correct because enzymes are a type of protein, and their structure is dependent upon the hydrogen bonding that occurs between the amino acids that comprise the enzyme. A change in pH changes the concentration of the ions in the surrounding environment and affects the bonding between the amino acids, altering the structure and function of the enzyme. The kinetic energy of an enzyme is increased by an increase in temperature (Incorrect Response B). Altering the amino acid sequence of an enzyme is the result of changes in the DNA, such as those caused by frameshift mutations (Incorrect Response C). Finally, changes in pH can alter the active sites of enzymes but do not increase the numbers of active sites (Incorrect Response D).
4	Answer to question 4	A	Objective 001 Response A is correct because, in this experiment, the change in acidity is the independent variable, or the variable that will be manipulated. In line graphs, the independent variable values are on the x-axis. Catalase concentration (Incorrect Response B) and percent hydrogen peroxide (Incorrect Response D) are constants in this experiment; they do not change. As such, they do not get plotted along either axis on a line graph. The disk rise times (Incorrect Response C) are the dependent variable. Changes in the rise time depend on changes in the independent variable, such as acidity in this experiment. In a line graph, the dependent variable is plotted on the y-axis.
5	Answer to question 5	A	Objective 001 Response A is correct because water molecules ( H 2 O ) are polar and form hydrogen bonds between oxygen and hydrogen atoms on neighboring water molecules. This tendency of water molecules to "stick" to one another allows them to pull one another up through the roots and stem as water evaporates from the leaves of the plant. If the hydrogen bonds between water molecules are broken (Incorrect Response B), water loses its cohesion and thus its ability to draw water molecules up through the plant. Water transport in plants does not involve temperature-induced changes in water density (Incorrect Response C) or the high specific heat of liquid water (Incorrect Response D).
6	Answer to question 6	B	Objective 001 Response B is correct because the fact that water forms hydrogen bonds with other water molecules means that water molecules are attracted to one another (cohesive). Water can also absorb a large amount of heat while going through a phase change because an initial input of heat is required to break hydrogen bonds between water molecules before a phase change can occur. Ice is less dense than liquid water because the hydrogen bonds force water molecules into a more rigid, less dense structure as water cools. Water can form hydrogen bonds with other molecules as well, contributing to water's role as a solvent. Water does not tend to form double bonds (Incorrect Response A). The relatively low molecular weight of water (Incorrect Response C) and water's ability to dissociate (Incorrect Response D) do not contribute significantly to the collective set of properties listed.
7	Answer to question 7	A	Objective 001 Response A is correct because saturated fats have carbon chains that contain the maximum number of hydrogen atoms and only single bonds between adjacent carbon atoms. This results in carbon chains that are relatively linear and able to pack closely together. These properties tend to make saturated fats solid at room temperature. Although cholesterol (Incorrect Response B) is also a lipid, it has a molecular structure that is distinct from saturated and unsaturated fats. Saturated fats are more common in animals, not plants (Incorrect Response C). Unsaturated fats, rather than saturated fats, contain more double bonds (Incorrect Response D). These double bonds make unsaturated fats more rigid, less able to pack tightly together, and more likely to be liquid at room temperature.
8	Answer to question 8	D	Objective 001 Response D is correct because the flattening of the curve occurs when essentially all enzyme active sites are occupied at a given moment. At this point, the substrate concentration is so great that as soon as products are released from an enzyme molecule, new substrates bind to its active site. A competitive inhibitor (Incorrect Response A) would reduce the amount of product formed at low substrate concentrations, resulting in a curve with a more gradual increase or slope. If the enzyme had been denatured (Incorrect Response B), the amount of product would decrease. The graph does not depict a situation in which the substrate has been used up (Incorrect Response C), since the x-axis shows only a continuously increasing substrate concentration.
9	Answer to question 9	C	Objective 002 Response C is correct because the exchange of ions by way of a sodium-potassium pump is an example of active transport in which ions move against a concentration gradient. This transport mechanism uses ATP to move potassium ions across the cell membrane into a cell while sodium ions are moved out of the cell. The processes of simple diffusion (Incorrect Response A), movement of small ions through protein channels (Incorrect Response B), and movement of water into a cell in a hypotonic environment (Incorrect Response D) are processes that act to equalize concentrations, are passive transport processes, and do not require an input of energy.
10	Answer to question 10	A	Objective 002 Response A is correct because the breakdown of glucose is an exothermic reaction. Most of the energy produced is stored as ATP and is used for life processes in cells. A small amount is released in the form of thermal energy or heat. Evidence of this can be seen by measuring the increase in temperature of germinating seeds. Electromagnetic energy (Incorrect Response B) is used to produce energy via photosynthesis, radiant energy (Incorrect Response C) is the physical form of energy that results from electromagnetic energy, and mechanical energy (Incorrect Response D) is energy associated with the motion or position of an object.
11	Answer to question 11	C	Objective 002 Response C is correct because the process of cellular aerobic respiration is one in which glucose is broken down in the presence of oxygen to produce ATP, water, and carbon dioxide. There are several distinct stages of cellular respiration, and water is the byproduct of oxidative phosphorylation. Oxygen (Incorrect Response A) is a reactant in cellular respiration, as is glucose (Incorrect Response D). Diatomic hydrogen (Incorrect Response B) is neither a reactant nor a product.
12	Answer to question 12	B	Objective 002 Response B is correct because the process of photosynthesis uses light energy (a form of electromagnetic energy) from the sun in addition to water to produce chemical energy (in the form of glucose) and oxygen. The transformation of thermal into mechanical energy (Incorrect Response A) is associated with internal combustion engines and power plants. Roller coasters and water falling over a dam are examples of potential to kinetic energy transformations (Incorrect Response C), and radiant energy is the physical form of energy resulting from electromagnetic energy (Incorrect Response D).
13	Answer to question 13	D	Objective 002 Response D is correct because photosynthesis is used by green plants to synthesize glucose using electromagnetic energy from the sun along with water absorbed by plant roots and atmospheric carbon dioxide as reactants. As part of this process, N A D P H generated in the light-dependent reactions donates electrons in the form of H plus ions to produce three-carbon sugars in the Calvin cycle component of photosynthesis. These three-carbon sugars join up to form glucose. The function of chlorophyll molecules (Incorrect Response A) is to absorb electromagnetic energy. The stoma (Incorrect Responses B and C) are pores on the surface of leaves that allow for the exchange of carbon dioxide and oxygen, and are rarely involved in the uptake of water.
14	Answer to question 14	B	Objective 002 Response B is correct because glucose is one of the reactants in both aerobic and anaerobic respiration, both of which result in the production of ATP. Oxygen (Incorrect Response A) is a requirement for aerobic respiration that takes place in the mitochondria (Incorrect Response C) of the cell, while anaerobic respiration occurs in the absence of oxygen in the cell's cytoplasm and leads to the production of substances, such as lactic acid and ethanol (Incorrect Response D).
15	Answer to question 15	A	Objective 002 Response A is correct because microorganisms, such as bacteria and fungi, use the process of cellular respiration to break down the complex organic molecules in the compost into carbon dioxide and water. Cellular respiration is an exothermic process, but the heat produced does not convert all liquid water into water vapor (Incorrect Response B). Nitrogenous compounds are utilized by microorganisms, with excess nitrogen lost as ammonia gas (Incorrect Response C). Decomposers within the compost heap may undergo either aerobic or anaerobic respiration, so the use of oxygen contained within the pile is not likely to account for a noticeable change in the size of the compost heap (Incorrect Response D).
16	Answer to question 16	D	Objective 002 Response D is correct because fermentation is a process cells use to deal with the products of glycolysis in the absence of oxygen. During glycolysis, N A D plus is reduced to N A D H in order to produce ATP and pyruvate. The cell needs to regenerate N A D plus so that glycolysis can continue. Under anaerobic conditions, fermentation occurs. Pyruvate is reduced to lactic acid, and N A D H is oxidized to regenerate N A D plus . Hydrolysis (Incorrect Response A) is a general term that refers to reactions in which water is used to split a larger molecule. These reactions generally do not produce lactic acid. Glycolysis (Incorrect Response C) itself does not produce lactic acid. Rather, under aerobic conditions, the N A D H and pyruvate generated during glycolysis are shuttled into the mitochondria, which use them to produce more ATP via the process of cellular respiration (Incorrect Response B). Lactic acid is not a by-product of cellular respiration.
17	Answer to question 17	A	Objective 002 Response A is correct because during photosynthesis, the Calvin cycle uses ATP and N A D P H to incorporate carbon dioxide gas into carbohydrates. The production of N A D P H (Incorrect Response B) and ATP (Incorrect Response C), as well as the release of oxygen gas (Incorrect Response D), are all events that occur during the light reactions of photosynthesis, rather than during the Calvin cycle.
18	Answer to question 18	B	Objective 002 Response B is correct because during many metabolic reactions, enzymes facilitate the transfer of a phosphate group from adenosine triphosphate, or ATP, to one of the reactant molecules. Bonds with phosphate groups tend to contain a lot of chemical potential energy, which has now been transferred to the reactant in the metabolic process and can be used to drive the metabolic reaction forward. Although hydrolysis of a phosphate group will release heat (Incorrect Response A), this heat is dissipated, and its energy is unavailable to drive a metabolic reaction forward. Lowering the activation energy is often required for phosphorylation (Incorrect Response C), but this process is generally facilitated by enzymes. Absorption of light by chlorophyll does help provide the energy needed to generate ATP during photosynthesis. However, absorption of light by pigments (Incorrect Response D) is not a general mechanism used by the cell to drive metabolic reactions.
19	Answer to question 19	C	Objective 003 Response C is correct because growth, reproduction, and other life functions all require the production and breakdown of chemical compounds through metabolic processes. Many single-celled organisms rely primarily on asexual reproduction, and many plants and animals can also reproduce in this manner (Incorrect Response A). Although some organisms are capable of producing all of the amino acids needed to produce their own proteins, most organisms, including humans, must obtain essential amino acids from the food they consume (Incorrect Response B). Membrane-bound organelles (Incorrect Response D) are characteristic of eukaryotes, but prokaryotic organisms lack this structural feature.
20	Answer to question 20	B	Objective 003 Response B is correct because contractile vacuoles are found primarily in protists, such as Paramecia, and function to remove excess water from the cell. Vacuoles are responsible for storage of food molecules in these types of organisms (Incorrect Response A), while similar vesicles play a role in the capture and exocytosis of foreign matter and wastes by fusing with the cell membrane and expelling such material (Incorrect Response C). Blood sugar is not regulated in these types of organisms (Incorrect Response D), which lack a circulatory system. In more complex animals, blood sugar is regulated by the interactions between a number of different organs.
21	Answer to question 21	B	Objective 003 Response B is correct because, as a result of their molecular configuration, unsaturated fatty acids in the cell membranes do not solidify in cold temperatures. This allows the organism's cell membranes to remain fluid and functional. Integral membrane proteins (Incorrect Response A) transport materials across the cell membrane and act as receptors. Phosphate groups (Incorrect Response C) are part of the hydrophilic component of the phospholipids of the cell membrane. Carbohydrates (Incorrect Response D) in the cell membrane play a role in cell recognition and protection.
22	Answer to question 22	A	Objective 003 Response A is correct because, in an isotonic solution, the volume of an animal cell remains stable, as the movement of water into the cell is equal to the movement of water out of the cell. When the cell is moved to a saltwater solution, however, the concentration of solute outside the cell is higher than that inside the cell, and there would be a net movement of water out of the cell, via osmosis, through the cell membrane. Charged ions are unable to pass through the lipid bilayer due to the hydrophobic nature of the nonpolar tails (Incorrect Response B). Unlike sugar molecules and amino acids, water molecules are able to move directly through the cell membrane as a result of their small size and do not require carrier proteins for transport into and out of the cell (Incorrect Responses C and D).
23	Answer to question 23	B	Objective 003 Response B is correct because cells cannot grow above a certain size due to the fact that the surface area of the cell decreases as the volume of the cell increases. Consequently, cells must remain a certain size so that nutrients and other materials can be transported efficiently into and out of the cell. Cell death is caused by trauma to the cell or may occur due to programmed cell death but should not occur primarily due to the lack of cell division (Incorrect Response A). An organism's growth is not limited by space between cells, especially since cells do not necessarily communicate with their immediate neighbors (Incorrect Response C). Finally, growth can occur via mitosis with no differentiation into different cell types, and growth that occurs during embryonic development is typically a small percent of what can occur in later stages of life (Incorrect Response D).
24	Answer to question 24	A	Objective 003 Response A is correct because the G1 phase of the cell cycle is the beginning of interphase and is the longest phase of the cell cycle. At this point, the cell is growing and getting ready for subsequent cell division. Therefore, it must synthesize all the components (e.g., building blocks of DNA, proteins) needed for this process. The remaining phases (Incorrect Responses B, C, and D) are all shorter in duration.
25	Answer to question 25	A	Objective 003 Response A is correct because the result of an asymmetric division of a stem cell is the production of one cell that is differentiated and a second cell that is, like the parent cell, a stem cell capable of additional proliferation. Symmetric division of a stem cell may result in the production of two differentiated daughter cells (Incorrect Response B) or two stem cells (Incorrect Response D). Having a finite capacity of cell division and another cell beginning apoptosis means that both cells are destined to stop growth altogether (Incorrect Response C).
26	Answer to question 26	C	Objective 003 Response C is correct because the preparation required to view a specimen under an electron microscope includes procedures such as thin sectioning and coating the specimen with a thin layer of metal. These steps, in addition to other associated procedures, result in the death of cells. Living cells and the process of cell division can be observed using a light microscope as cell function is not compromised. Specimens in electron microscopes are fixed and so do not move (Incorrect Response A). Electron microscopes do not create distortion in the image of the object being viewed (Incorrect Response B). An entire cell may be seen under an electron microscope, but the entire process of cell division would not be observable in a single cell (Incorrect Response D).
27	Answer to question 27	A	Objective 003 Response A is correct because the rough endoplasmic reticulum is the portion of the endoplasmic reticulum to which ribosomes are attached. Together, the endoplasmic reticulum and the ribosomes produce membrane proteins and proteins that are to be secreted from the cell. Additionally, carbohydrate chains are frequently attached to proteins in the rough endoplasmic reticulum to produce glycoproteins. For these reasons, the rough endoplasmic reticulum is well developed in cells that produce large quantities of secreted and/or glycoproteins. The detoxification of drugs or poisons (Incorrect Response B) occurs in the smooth endoplasmic reticulum. The movement of charged particles (ions) across the cell membrane is called chemiosmosis (Incorrect Response C), and the rough endoplasmic reticulum does not have a direct role in this. The hydrolysis of macromolecules (Incorrect Response D) frequently occurs in lysosomes and by proteasomes in the cytoplasm.
28	Answer to question 28	B	Objective 003 Response B is correct because the Golgi apparatus is an organelle that helps sort, modify, and transport membrane components and secreted proteins that were made in the endoplasmic reticulum. This function is required in all eukaryotic cells, including both plant and animal cells. Cell walls (Incorrect Response A) and central vacuoles (Incorrect Response D) are found in plant but not animal cells. Animal cells contain lysosomes, which perform hydrolytic functions for the cell (Incorrect Response C). These same functions are performed by vacuoles in plant cells.
29	Answer to question 29	B	Objective 004 Response B is correct because stomata, pores found on the epidermal surfaces of a green plant, allow carbon dioxide to enter the plant, while oxygen and water are able to exit the plant. The evaporation of water requires energy, and that energy is provided in the form of heat from the plant. Phloem (Incorrect Response A) transports carbohydrates in the plant, root hairs (Incorrect Response C) take up water and minerals from the soil, and sclerenchyma cells (Incorrect Response D) provide structural support.
30	Answer to question 30	D	Objective 004 Response D is correct because one of the main differences between the types of signaling is the distance that the signal needs to travel to reach the target cell. In synaptic paracrine signaling, neurotransmitters travel a short distance between nerve cells, resulting in an immediate response from the target cell. In endocrine signaling, hormones released from signaling cells in one part of the body move through the blood to transmit information to cells located in another region of the body. Responses A, B, and C incorrectly categorize these two types of cell signaling.
31	Answer to question 31	D	Objective 004 Response D is correct because xylem is responsible for transporting water and nutrients from the roots of plants upward to the stem and leaves. Placing the cut stem of a carnation in colored food dye will result in the movement of the colored material up the stem of the plant and into the flower part of the carnation. Exposing lettuce leaves to saline solution (Incorrect Response A) will result in wilting of the lettuce leaves as water moves from the leaves due to the process of osmosis. The plant pigments separated using paper chromatography (Incorrect Response B) are pigments found in the chloroplasts that play a role in photosynthesis. Channeling a seedling to grow in a downward direction (Incorrect Response C) is a process that tests gravitropism and the response of plant cells to hormones.
32	Answer to question 32	B	Objective 004 Response B is correct because spores are part of the haploid phase of the fern life cycle. When fern spores germinate, they undergo mitosis to produce a haploid gametophyte, the function of which is to produce gametes. When a male gamete fertilizes a female gamete, a fern can transition to the diploid phase of its life cycle. A rhizoid (Incorrect Response A) is a filamentous structure primarily produced on the underside of a moss thallus, and it functions in anchoring the plant or conducting water. The archegonium (Incorrect Response C) is the sex organ within the gametophyte that produces oocytes. The sporophyte (Incorrect Response D) is the multicellular diploid phase of the fern.
33	Answer to question 33	C	Objective 004 Response C is correct because stomata are openings found primarily on the underside of a plant's leaves and on its stem. When the stomata close, water vapor in the plant cannot escape, thus reducing the rate of water loss. Pressure in the root (Incorrect Response A) affects the amount of nutrients, including water, absorbed by the root from the environment, but root pressure is not affected immediately with the closure of the stomata. Transportation of nutrients (Incorrect Response B) throughout the plant uses special tissues; the closing of the stomata, however, does not affect the immediate movement of nutrients. A plant increases its frost resistance (Incorrect Response D) by changing the colloidal consistency of its protoplasm (or in the case of woody plants, its sap), but the closing of the stomata does not trigger or affect the frost resistance of the plant.
34	Answer to question 34	B	Objective 004 Response B is correct because the triple response occurs when the sprouting seedling encounters an obstacle, such as a rock, on its way to breaking through the soil's surface. The seedling then produces ethylene, which causes stem elongation to slow, the stem to thicken to be stronger, and the stem to have horizontal growth to find an easier path to the surface. Once the path is found, the seedling continues to move upward. Although ethylene does play a role in the triple response, phototropism and programmed cell deaths (Incorrect Response A) do not affect the triple response. Changes in turgor pressure, phytochromes, and blue-light receptors (Incorrect Response C) do not affect the triple responses of the seedling because the seedling has not yet broken through the soil. Auxin concentration, the production of growth inhibitors, and the abscission layers (Incorrect Response D) may occur in a plant, but they do not play a role in the triple response of a seedling that is still underground.
35	Answer to question 35	C	Objective 004 Response C is correct because the data in the graph indicate that the salt concentration in the cells of the spider crab is the same as that in its environment. As the salt concentration in the environment goes up, the salt concentration in the spider crab cells increases by the same amount. Because the data show that the salt concentration in the spider crab cells varies with the salt concentration of the environment, they demonstrate that the salt crab is unable to regulate its internal environment (Incorrect Response A). Nor is it able to maintain homeostasis, the ability to maintain a stable internal environment, by either a positive (Incorrect Response B) or negative (Incorrect Response D) feedback mechanism.
36	Answer to question 36	D	Objective 005 Response D is correct because red bone marrow is found in the flat bones of the body and contains the stem cells that grow into platelets, white blood cells, and red blood cells. These cells are important components of the circulatory system. Joints (Incorrect Response A) are the parts of the musculoskeletal system where two or more bones meet and allow movement. Collagen (Incorrect Response B) is found in many parts of the body (e.g., skin, bones, connective tissue) and provides structure and support, while cartilage (Incorrect Response C) is a type of smooth connective tissue.
37	Answer to question 37	C	Objective 005 Response C is correct because, among other functions, the liver helps control blood composition. It does this both by regulating the amount of sugar, fats, and proteins in the blood as well as by removing wastes and toxins, such as ammonia. Blood flow to the stomach (Incorrect Response A) is largely regulated by hormones. Generation of all new red blood cells and most white blood cells occurs in the bone marrow (Incorrect Response B). Control of peristalsis in the digestive tract is a function of the nervous system (Incorrect Response D).
38	Answer to question 38	B	Objective 005 Response B is correct because acquired immunity is a specific immune response that is generated after exposure to specific antigens. Each lymphocyte contains receptors that recognize a single antigen. When that specific antigen binds to a receptor on a lymphocyte, the lymphocyte divides to produce many cells capable of recognizing and attacking the pathogen source of the antigen. After the initial stimulation of a lymphocyte, the body can respond more rapidly upon exposure to the same pathogen. Immune responses as a result of increased body temperature (Incorrect Response A) and natural killer cells (Incorrect Response C) are part of the non-specific immune response rather than acquired immunity. Although certain phagocytes help trigger clonal selection by presenting antigens to lymphocytes, phagocytes (Incorrect Response D) are themselves part of the non-specific immune response in that they recognize and destroy pathogens indiscriminately.
39	Answer to question 39	D	Objective 005 Response D is correct because, when fighting an infection, the body's temperature increases, creating an environment that is hostile to pathogens. This increase in temperature is called a fever. The body begins to produce white blood cells—leukocytes—to detect and destroy the microorganisms that are causing the infection. Fevers increase, rather than reduce, the permeability of the plasma membrane (Incorrect Response A). A fever decreases the concentration of oxygen in the blood (Incorrect Response B) because it reduces the oxygen's ability to bind to hemoglobin. A fever increases blood pressure and heart rate, rather than lowering them (Incorrect Response C).
40	Answer to question 40	C	Objective 005 Response C is correct because cilia are hairlike extensions of the cell membrane of specialized cells that are found in the bronchi and bronchioles of the lungs. They have an upward sweeping motion that assists in moving dust and other particles that have entered the lungs back out into the environment. The lining of the lungs has specialized cells called goblet cells (Incorrect Response A) that produce secretions to trap dust and other particles that have entered the trachea. The lungs hold air that is brought in, and the body warms the air as it enters the lungs (Incorrect Response B). Oxygen moves from cells that line the lungs' alveoli into the blood by diffusion (Incorrect Response D). None of the processes in Incorrect Responses A, B, and D requires the use of cilia.
41	Answer to question 41	A	Objective 005 Response A is correct because the loading and unloading of oxygen from hemoglobin is sensitive to the pH of the blood. As blood becomes more acidic due to the release of carbon dioxide as a by-product of cellular respiration, the affinity of hemoglobin for oxygen decreases, and oxygen is released into the surrounding tissues. The affinity of hemoglobin for oxygen does increase after the initial oxygen molecule binds (Incorrect Response B), but this mechanism does not control the loading or unloading of the initial oxygen molecule onto hemoglobin. Blood arriving at the lungs does have a lower P O 2 than the air inside the lungs (Incorrect Response C), causing oxygen to diffuse into the blood from the lungs; however, this does not explain the mechanism by which hemoglobin loads and unloads oxygen. Although the heart rate does increase in response to increased levels of carbon dioxide in the blood (Incorrect Response D), this does not describe the mechanism by which hemoglobin loads and unloads oxygen.
42	Answer to question 42	D	Objective 005 Response D is correct because an increase in the concentration of dissolved chemicals present in the blood, or osmolarity, signals to the brain that the body needs more water to dilute the chemicals and return the body to homeostasis. A decrease in heart rate and blood pressure (Incorrect Response A) generally causes lightheadedness or dizziness rather than thirst. An increase in internal temperature (Incorrect Response B) can be an indication of infection and will signal the production of leukocytes to destroy pathogens. However, this does not usually cause the sensation of thirst. The first indication of less water in the body is a decrease in urinary output by the kidneys (Incorrect Response C), which are signaled to reabsorb the water instead of moving it to the urinary bladder. Only when reabsorption of water by the kidneys is unsuccessful at balancing the osmolarity of the blood is the sensation of thirst produced.
43	Answer to question 43	A	Objective 005 Response A is correct because the central nervous system consists of the brain and spinal cord. The nerves of the peripheral nervous system originate in or near the central nervous system and extend to the different regions of the body. Because of this, the axons in the peripheral nervous system must be longer. Dendrites are the extensions of a nerve cell that receive signals from other nerve cells. The nerve cells within the peripheral nervous system do not have significantly more dendrites than those located in the central nervous system (Incorrect Response B). Synapses in the central and the peripheral nervous system both use neurotransmitters to transmit chemical signals (Incorrect Response C). Impulses in the peripheral nervous system flow both to and from the central nervous system (Incorrect Response D).
44	Answer to question 44	A	Objective 005 Response A is correct because oxytocin is a hormone made in the hypothalamus and stored in the posterior pituitary gland. This hormone is responsible for inducing contractions during labor. Follicle-stimulating hormone is responsible for stimulating the growth of follicles and the maturation of oocytes (Incorrect Response B). Estrogen and progesterone stimulate the development of the endometrium and maintain it in preparation for pregnancy (Incorrect Response D). If the oocyte is not fertilized, the levels of estrogen and progesterone drop off, which induces the disintegration of the uterine lining (Incorrect Response C).
45	Answer to question 45	B	Objective 005 Response B is correct because connective tissues join and/or support other tissues within the body and include such tissues as adipose tissue, bone, and cartilage. Cartilage is unique among connective tissue in that it does not contain blood vessels. Collagen (Incorrect Response A) is an important structural component of cartilage. Unlike bone, cartilage is not a significant reservoir of either phosphorus or calcium (Incorrect Response C). Chondrocytes, or cartilage cells, contain nuclei (Incorrect Response D).
46	Answer to question 46	C	Objective 006 Response C is correct because the pH meters may not be equally accurate. This can introduce errors in measurement associated with which meter is used by the students. It would be preferable for the students to use the same pH meter, or at least calibrate all four of the existing meters to some standard. The rounding errors (Incorrect Response A) are essentially random, not systematic, in nature. Students should use the same protocol when making the measurements (Incorrect Response B), which should reduce the associated error in measurement. Sampling at the same time every day should minimize daily fluctuations in the data (Incorrect Response D).
47	Answer to question 47	A	Objective 006 Response A is correct because the diagram shows that the population has a very large percentage of young people and a very small percentage of people who are past child-bearing age. These are characteristics of a population that will continue to grow rapidly. A population that would grow at a slow and steady rate (Incorrect Response B) would have a fairly even distribution of people among all age groups, with somewhat smaller numbers of older people. With such a large percentage of children and of people of child-bearing age, it is very unlikely that this population would grow for another generation and then stabilize (Incorrect Response C) or remain at its current size (Incorrect Response D).
48	Answer to question 48	B	Objective 006 Response B is correct because, in a mutualistic relationship, both organisms benefit from the relationship. Ants benefit by obtaining food from aphids, and aphids benefit from ants protecting them from predators. Hawks and owls (Incorrect Response A) have a predator/prey relationship with mice, and they compete with each other for mice as a food source. Both hawks and owls benefit from mice, but mice do not benefit from hawks and owls. The relationship between a tick and a moose (Incorrect Response C) is a parasitic relationship. The tick benefits from feeding on the moose's blood, but the moose does not benefit from the tick. In fact, the moose can become weak and debilitated from the loss of blood. Bees and siskins feed on different parts of the thistle plant, so they do not compete for thistle as a resource, but neither do they benefit from each other (Incorrect Response D).
49	Answer to question 49	D	Objective 006 Response D is correct because large bodies of water can be separated into layers according to temperature. A mixed layer of relatively warm water is near the surface, a cold deep layer is at the bottom, and a transition layer called a thermocline separates the mixed and deep layers. Ecotones (Incorrect Response A) are transition areas between ecological communities and often contain characteristics of both. Although an inverse relationship between temperature and elevation exists in the troposphere (Incorrect Response B), this is not a thermocline. Zonation (Incorrect Response C) is the distribution of organisms into specific geographic regions that occurs naturally based on the physical characteristics of the region. Zonation occurs in many habitats, including intertidal habitats. However, this phenomenon is distinct from thermoclines.
50	Answer to question 50	D	Objective 006 Response D is correct because estuaries are areas where freshwater rivers and streams meet the salty water of oceans; therefore, the salinity in an estuary varies widely. Organisms that live in these areas have adaptations that allow them to cope with varying salinity within the general range that occurs annually. A problem arises when salinity is not within the general range for the area. If the salinity is too low, excess water enters an organism's cells, and it cannot be removed quickly enough for the organism to survive. If the salinity is too high, excess water is removed from the organism's cells, and the organism's survival is in jeopardy. While availability of light (Incorrect Response A), leaching of nutrients (Incorrect Response B), and erosion of substrate (Incorrect Response C) are challenges for aquatic organisms, none of these are unique to estuaries.
51	Answer to question 51	A	Objective 006 Response A is correct because climate is largely responsible for determining the distribution of terrestrial biomes and is primarily determined by the mean temperature and precipitation in a region. For example, deserts and rain forests have similar mean temperatures but vary greatly in the amount of precipitation each receives. Although elevation is one factor that influences mean temperature (Incorrect Response B), it does not solely contribute as much to biome distribution as mean temperature and precipitation do. Biomes can have distinct soil types (Incorrect Response C). However, soil type is in part determined by the climate in a region, which is in turn determined by temperature and precipitation. Neither elevation nor soil type (Incorrect Response D) has as much influence on biome distribution as mean temperature and precipitation do.
52	Answer to question 52	D	Objective 006 Response D is correct because existing species can sometimes alter environmental conditions, such as by changing the mineral composition of the soil. When these changes inhibit the species' ability to reproduce relative to other species, this opens the door for other species that are better suited to the new environment to grow and thrive. These new, better-suited species begin to dominate the environment in a process called secondary succession. Density-independent selection (Incorrect Response A) refers to factors such as natural disasters that affect a species regardless of the species' population size. Habitat fragmentation (Incorrect Response B) affects a species by splitting its habitat into varying sizes, which might cause a change in the number of the species, but it is not caused by the self-inhibition of reproduction. Boom-and-bust population cycles (Incorrect Response C) occur as a result of the abiotic factors of space, nutrients, and water in an environment. When these factors are readily available, the population increases, but when they are depleted, the population decreases. This is not caused by a species inhibiting its own reproduction.
53	Answer to question 53	C	Objective 006 Response C is correct because the production of flowers and seeds requires relatively large amounts of energy, and after the shrub and tree leaf canopy has developed in the summer, the amount of sunlight on the forest floor is greatly reduced. Thus, there is a strong energetic advantage to reproduce before the leaf canopy develops. Temperatures are higher during the summer (Incorrect Response A), but these do not provide the energy required for reproduction. Fewer pollinators are active during the spring (Incorrect Response B), and although there are fewer insect herbivores in the spring (Incorrect Response D), this is not the primary reason for the frequency of early spring-blooming forest floor flowers.
54	Answer to question 54	A	Objective 006 Response A is correct because oceanic upwellings bring many nutrients that have fallen to the ocean floor back up to the surface, providing an area where ocean life thrives. The fact that fisheries over areas of regular upwellings are more productive suggests that the availability of nutrients may be a limiting factor in an open-ocean system. Light intensity (Incorrect Response B), salinity levels (Incorrect Response C), and water temperatures (Incorrect Response D) may vary throughout the ocean, including at the location of upwellings, so the productivity of fisheries over oceanic upwellings does not provide specific evidence that these factors are limiting.
55	Answer to question 55	C	Objective 007 Response C is correct because energy is passed up from the lower to higher trophic levels of a food web in the form of going from producers to consumers. However, only about 10 percent of the energy associated with any one trophic level is available to the subsequent trophic level. This is because energy within a given level must be used for metabolic processes or is released to the environment as heat. As a result, the energy available to be used to create biomass steadily decreases. Detritivores and decomposers play an important role at all trophic levels (Incorrect Response A). The diversity of species present does not directly affect the biomass of subsequent trophic levels (Incorrect Response B) and, in general, animals at lower trophic levels tend to be smaller than animals at higher trophic levels (Incorrect Response D).
56	Answer to question 56	B	Objective 007 Response B is correct because, since quagga mussels are filter feeders that feed on algae, the addition of quagga mussels to the food web shown would result in a decrease in algae. Algae are the primary food source for zooplankton and thus a reduction in algae would directly result in a decrease in zooplankton. Aquatic insects (Incorrect Response A) and tadpoles (Incorrect Response C) both depend on algae as their primary food source and thus a decrease in algae would cause a decrease, not increase, in the populations of these organisms. Lake trout (Incorrect Response D) feed primarily on small fish, which rely on a variety of food sources and thus would not be directly affected by the presence of quagga mussels.
57	Answer to question 57	D	Objective 007 Response D is correct because black bears occupy the level of primary consumer because they feed on blackberries and also are secondary consumers since they feed on insects. Thus, they occupy multiple trophic levels within this food web. Owls (Incorrect Response A), rabbits (Incorrect Response B), and insects (Incorrect Response C) utilize only one food source shown and occupy only a single trophic level.
58	Answer to question 58	D	Objective 007 Response D is correct because deep roots and growth from buds located below the surface of the soil are adaptations that allow grasses to survive wildfires and resprout, since the soil provides protection for these underground structures. Development of an extensive underground root network or a reduction in biomass would be an adaptation to limited nutrients (Incorrect Response A). Plant adaptations to winter blizzards would likely include plant dormancy and the dying back of all surface growth (Incorrect Response B). Plant adaptations to periodic floods include the development of specialized air spaces that allow plants to obtain oxygen when soils are water-logged (Incorrect Response C).
59	Answer to question 59	A	Objective 007 Net annual primary productivity refers to the net annual increase in biomass of producers in a given ecosystem. The relationship between net annual primary productivity and average annual precipitation is one in which the amount of precipitation is the most limiting factor. As the average annual precipitation increases, a high initial rate of productivity is observed (Correct Response A). However, beyond a certain amount, precipitation can have a negative effect on productivity, and productivity begins to level off. The graph shown for Incorrect Response B shows a lack of relationship between precipitation and primary productivity, and Incorrect Response D indicates a negative relationship, both patterns that do not reflect the importance of water as a reactant in photosynthesis. In Incorrect Response C, the graph indicates an exponential relationship between the two factors, which is also unrealistic as too much water results in the death of primary producers in most ecosystems.
60	Answer to question 60	C	Objective 007 Response C is correct because transpiration is the loss of water vapor from the surfaces of plants. The rate of transpiration decreases as humidity increases. This is because, as the surrounding air becomes more saturated with water molecules, there is a reduction in the driving force for transpiration. As temperature increases (Incorrect Response A), the rate of molecular movement also increases, resulting in an increased rate of transpiration. An increase in transpiration rate also accompanies an increased wind speed (Incorrect Response B), and water molecules are removed from plant surfaces at a higher rate. Finally, the rate of photosynthesis increases with increased light intensity (Incorrect Response D), since the stomata open and more water is produced as a result.
61	Answer to question 61	A	Objective 007 Response A is correct because a line graph can be used to show the changes or trends in a group of organisms over time, with different colored lines or dashed/dotted lines to compare the four different habitats at a given time. The comparison of the number of individuals of three different keystone species across four types of habitats (Incorrect Response B) would best be represented using a bar graph. A pie chart would provide the best representation of the contributions of seven different plant families to the total mean annual biomass of primary producers (Incorrect Response C). A histogram would best display the percentage distribution of plant species exposed to different soil conditions (Incorrect Response D).
62	Answer to question 62	D	Objective 007 Response D is correct because biomagnification is the process by which toxins, such as mercury, accumulate within the organisms in a food web, with top predators affected to a greater degree than organisms such as producers and primary consumers. Secondary succession (Incorrect Response A) refers to the sequence of changes that take place in an ecosystem following an event such as a wildfire. Biodiversity (Incorrect Response B) is all the different life forms found in a given ecosystem. Carrying capacity (Incorrect Response C) is the average size of a given population in an ecosystem.
63	Answer to question 63	C	Objective 008 Response C is correct because bottom trawling, which is the practice of dragging fish nets along the ocean floor, crushes or smothers any organisms living on the sea floor. These changes may cause irreparable damage, both directly and indirectly, to these ecosystems. The sizes of male and female fish (Incorrect Response A) are not related to this fishing practice but have a genetic or evolutionary component. Eutrophication (Incorrect Response B) is the result of run-off of agricultural fertilizers or pollutants into coastal waters. The transmission of fish parasites occurs from fish to fish and is not directly correlated with disturbance of bottom sediments (Incorrect Response D).
64	Answer to question 64	A	Objective 008 Response A is correct because as the temperature of ocean water increases, there is an increase in the kinetic energy of molecules in sea water. This increase in kinetic energy is reflected by an increase in the volume of the ocean. Increased runoff from freshwater rivers directly impacts the salinity along coastal regions but not the overall volume of global sea levels (Incorrect Response B). Changing of ocean currents (Incorrect Response C) is related to global warming as increased temperatures may speed up ocean currents, but the changing of the ocean currents themselves is not responsible for the rise in global sea levels. Finally, increasing dissolved carbon dioxide (Incorrect Response D) results in ocean acidification, a process that is harmful to sea life that utilizes carbonate ions, like shellfish and corals, but it is not a factor in the rise of global sea levels.
65	Answer to question 65	D	Objective 008 An exotic species is any organism that is not native to a given ecosystem. These organisms lack natural predators, and native species may not have evolved defenses against them. Consequently, growth of populations of exotic species remains largely unchecked by typical limiting factors (Correct Response D). In general, exotic species do not alter the chemical balance (Incorrect Response A) or serve as a food source (Incorrect Response B) in a manner that is widespread and impactful on the entire ecosystem. Any interbreeding between the exotic species and native species should increase, rather than decrease, genetic diversity (Incorrect Response C).
66	Answer to question 66	B	Objective 008 Response B is correct because increased kinetic energy of the molecules comprising ocean waters results in an increase in volume and a rise in global sea levels. Acid rain is the result of atmospheric pollutants (Incorrect Response A). Ocean dead zones are areas of eutrophication caused by excess phosphorous and nitrogen (Incorrect Response C). Biomagnification (Incorrect Response D) is a process of accumulation of toxins within food webs.
67	Answer to question 67	C	Objective 008 Response C is correct because increased temperatures make regions previously uninhabitable during winter months more suitable for survival due to changes in both abiotic and biotic factors, causing ranges to shift northward toward generally cooler climates. Increasing fragmentation of wildlife habitat (Incorrect Response A) limits the mobility of wildlife and leads to decreased genetic diversity, while the increased use of wind turbines (Incorrect Response B) may result in an increased mortality of birds but not a shift northwards of wintering range. Rising numbers of invasive species (Incorrect Response D) would most likely result in a reduction in the size of an ecosystem population of hummingbirds but not necessarily a northern shift in range.
68	Answer to question 68	D	Objective 008 In contour plowing, the soil is tilled following the contours of hills rather than down the slope. This reduces the amount of soil that is washed away during rain and also increases the amount of water the soil can absorb (Correct Response D). Contour plowing by itself does not increase organic matter (Incorrect Response A); it just prevents it from being washed away. Contour plowing is a method of cultivation, or preparing the soil for planting, so it does not reduce the need to cultivate (Incorrect Response B). Following the existing contours of the land when plowing does not change the angle at which light would hit the crops (Incorrect Response C).
69	Answer to question 69	D	Objective 008 Response D is correct because aquatic dead zones are regions of water in which the dissolved oxygen content is so low that it cannot support life. Warmer water holds less oxygen, and discharge from nuclear power plants can contribute to the formation of dead zones by raising the temperature of the local water systems. Increased levels of acidity (Incorrect Response A) and decreased light penetration (Incorrect Response C) are conditions created by dead zones, rather than causes. Background radiation (Incorrect Response B) is the natural radiation that occurs in the environment, so any increased radiation in the region surrounding a nuclear power plant would not be considered background radiation.
70	Answer to question 70	A	Objective 008 Response A is correct because, when dissolved in water, especially rain, nitrogen oxides produce an acid that will cause the minerals in the soil to leach away as they dissolve in the acid rain. This can lead to reduced soil fertility, creating an environmental concern. The populations of nitrogen-fixing bacteria (Incorrect Response B), the rate of ozone depletion (Incorrect Response C), and the eutrophication of aquatic ecosystems (Incorrect Response D) are not affected by nitrogen oxides in the atmosphere.
71	Answer to question 71	B	Objective 008 Response B is correct because wetlands help absorb water that would otherwise directly enter downstream rivers and streams. Removal of wetlands along with their ability to store excess water has been shown to increase local flooding. Although human activity on the newly developed land may eventually lead to the excessive accumulation of nutrients in adjacent bodies of water (Incorrect Response A) and increased pressure on nearby landfills (Incorrect Response C), these potential environmental problems depend somewhat on how the land is used and are not the most direct consequence of draining wetlands. Removing the wetland habitat is likely to reduce biodiversity and is therefore unlikely to help establish exotic species in the area (Incorrect Response D).
72	Answer to question 72	C	Objective 008 Response C is correct because, when logging occurs in a mountainous area and all the trees are removed, the soil is no longer held in place, and erosion begins to occur. If some trees are left standing, there will be less erosion in the area, and environmental consequences will be reduced. Applying fertilizers (Incorrect Response A), closing the logged area (Incorrect Response B), and planting tree seedlings (Incorrect Response D) are not the best ways of reducing erosion in a clear-cut logged area.
73	Answer to question 73	D	Objective 009 Response D is correct because an mRNA sequence complementary to a given DNA sequence is created during transcription, a process that uses the rules of complementary base pairing in which cytosine is replaced with guanine (and vice versa), adenine is replaced by uracil, and thymine is replaced by adenine when the mRNA strand is created using the DNA sequence as a template. Incorrect complementary base pairing is exhibited by Responses A, B, and C.
74	Answer to question 74	D	Objective 009 Response D is correct because DNA replication is the process by which cellular DNA creates a copy of itself during cell division. The first step in this procedure is the opening (or unzipping) of the double-stranded DNA molecule by helicase enzyme, which breaks the bonds between complementary DNA base pairs. The operation of attaching nucleotides to the complementary strand (Incorrect Response A) and the deposit of primers along the DNA template (Incorrect Response C) occur after the DNA strands have separated. Reduction of the number of chromosomes (Incorrect Response B) by half occurs during the first division of meiosis.
75	Answer to question 75	C	Objective 009 Response C is correct because there are many species of fish and reptiles in which the ambient environmental temperature affects the sex of the developing embryo. Similar factors are thought to play a role in gender changes of adult species of some fish. Rapid vegetative growth in some types of food crops (Incorrect Response A) is triggered by high amounts of plant hormones and is not necessarily dependent on external factors. Tetraploidy in flowers (Incorrect Response B) is a heritable condition resulting from events, such as nondisjunction during meiosis. Gametes produced in this manner possess a complete set of duplicate chromosomes, so the fertilized zygote has multiple copies of chromosomes. The darkening of skin from Sun exposure (Incorrect Response D) is a temporary condition caused by the production of melanin in skin cells as a means of protecting the lower layers of skin cells from the Sun's ultraviolet radiation.
76	Answer to question 76	A	Objective 009 Response A is correct because interaction of ultraviolet light results in the oxidation of bases in the DNA such that subsequent base pairing that occurs during DNA replication is disrupted. If these mistakes are not repaired, cell death or diseases like cancer may occur. Blocking of DNA replication (Incorrect Response B), apoptosis of a large number of cells (Incorrect Response C), and the creation of small holes in the cell membranes (Incorrect Response D) are not the result of exposure to ultraviolet light.
77	Answer to question 77	C	Objective 009 Response C is correct because the fact that multiple codons all code for glycine is an example of the redundancy of the genetic code. Because multiple codons specify the same amino acid, there is no change in the protein coded for if a mutation occurs in the third codon of a triplet. A stop codon (Incorrect Response A) is coded for by U A G, U A A, and U G A , which would not occur by a mutation in the third codon of these triplet codes. A shift in the coding frame (Incorrect Response B) occurs when a base is inserted or deleted and causes the addition of the wrong amino acid, which is not the case for the given example. Inversion of the triplet code (Incorrect Response D) refers to the inversion and rejoining of a piece of DNA, and this would not be the result of the mutation changing the third codon.
78	Answer to question 78	B	Objective 009 Response B is correct because, during DNA replication, DNA polymerase binds to each of the separated strands of DNA and, using these strands as a template, begins to generate a new strand of DNA by adding nucleotides to the growing new strand. The DNA polymerase can only create a new strand of DNA by moving in the 5 prime to 3 prime direction, so the polymerase molecules must move in opposite directions as the original DNA is copied. Responses A and C incorrectly describe the number of polymerase enzyme molecules required for replication, and Response D incorrectly describes the action of these polymerase molecules.
79	Answer to question 79	D	Objective 009 Response D is correct because one advantage of generating transgenic plants with desirable traits is that the desirable traits will be passed from one generation to the next. For this to occur, the genes must be incorporated into the germline cells of the transgenic plants. Thus, the ability to be incorporated into new genomes is deliberately designed into the vectors used to transfer the genes to the target plants. Some people are concerned that under the right conditions, this vector design could allow the genes to be unintentionally incorporated into other organisms. There are structural differences between eukaryotic and prokaryotic gene promoters (Incorrect Response A). There is no evidence to date that transgenic crops are unusually susceptible to native plant diseases (Incorrect Response B). Some microbes can exchange DNA through conjugation (Incorrect Response C), but successful conjugation between microbes and eukaryotic cells is relatively infrequent under natural conditions.
80	Answer to question 80	A	Objective 009 Response A is correct because the primary role of tRNA molecules is to bring the appropriate amino acids to the ribosome during translation. A tRNA molecule accomplishes this by matching its anticodon (a sequence of three nucleotides that designates the specific amino acid attached to the tRNA molecule) to the codon sequences in the mRNA molecule being translated. The anticodon sequence will pair only with the complementary mRNA codon, thus ensuring that the correct amino acid is added to the polypeptide chain. The amino acids that are delivered to the ribosome by tRNA molecules are attached to the growing polypeptide chain by the large ribosomal subunit (Incorrect Response B). Introns (non-coding regions) are eliminated from the mRNA sequence (Incorrect Response C) by a structure called a spliceosome before the mRNA leaves the nucleus. Termination of the growing polypeptide chain (Incorrect Response D) occurs when the ribosomal complex reaches a termination codon, which signals a protein called a release factor to bind and release the polypeptide chain.
81	Answer to question 81	C	Objective 009 Response C is correct because the genetic code is redundant in the sense that there are 64 possible codons and only 20 different amino acids. This means that many amino acids are coded for by multiple codons. When a point mutation occurs that changes a codon but does not change the amino acid that it codes for, there will be no change in phenotype because the amino acid that is incorporated into the protein will not change. Somatic mutations occur in body cells and will not affect the germ line of the organisms (Incorrect Response A). The redundancy of the genetic code will not cause nucleotide base deletions to lead to frameshift mutations (Incorrect Response B). A missense mutation is a nucleotide substitution that results in a new amino acid being incorporated into the encoded protein. A translocation is when DNA is exchanged between chromosomes. Missense mutations do not result in translocations (Incorrect Response D), and neither are affected by the redundancy of the genetic code.
82	Answer to question 82	B	Objective 009 Response B is correct because to increase the quantity of DNA in a sample, a scientist would use the polymerase chain reaction technique, in which the polymerase enzyme replicates samples of DNA. DNA sequencing (Incorrect Response A) provides information about the genome but does not increase the quantity of DNA in a sample. Restriction fragment analysis (Incorrect Response C), which provides information about DNA that has been cut into smaller fragments, is not a technique that increases the quantity of DNA in a sample. Likewise, the quantity of a DNA sample is not increased using the technique of DNA hybridization (Incorrect Response D), which is a process that combines two single strands of DNA molecules into one single double-strand molecule.
83	Answer to question 83	D	Objective 010 Response D is correct because a Punnett square is a graphical tool that uses a grid and letters to determine the range and percentage of genotypes of the potential offspring of two sexually reproducing organisms. It requires knowledge of the genetic makeup of the parents. Each box in the table represents the outcome of a possible fertilization event. A pedigree (Incorrect Response A) shows the generational relationships and phenotypes of family members. The gene sequence (Incorrect Response B) represents the order of DNA nucleotides in an organism and is obtained using a biological sample from that organism. A gene histogram (Incorrect Response C) displays the distribution of gene expression values for specific genes in an organism.
84	Answer to question 84	C	Objective 010 This cross involves two parents who are heterozygous for a single trait (fur color) and can be determined with the help of a Punnett square. The expected genotypic ratio resulting from this type of cross is 1 (homozygous dominant) to 2 (heterozygous) to 1 (homozygous recessive). In terms of fur color, these numbers occur in this response as 1 brown to 2 tan to 1 white (Correct Response C). The phenotypic ratios in Incorrect Responses A, B, and D do not exhibit the phenotypic ratios expected from this type of cross.
85	Answer to question 85	C	Objective 010 Response C is correct because asexual reproduction does not require the production of gametes and typically requires only one parent, both characteristics that reduce the amount of energy needed to produce offspring. Populations that reproduce asexually have higher survivorship (Incorrect Response A) but only in environments that are stable and unchanging where genetic variation and adaptation are not so critical. Having fewer mutations in the population (Incorrect Response B) is disadvantageous to the population, since there is less genetic diversity. All organisms compete for resources, regardless of their genetic make-up (Incorrect Response D).
86	Answer to question 86	A	Objective 010 Crossing over is the process by which material is exchanged between homologous chromosomes during meiosis. Alleles present at loci that are located near the ends of chromosomes are more likely to exhibit variation because the probability of crossing over increases with increasing distance between loci (Correct Response A). Because distance between loci is the primary factor affecting crossing over, the location of loci described in Incorrect Responses B, C, and D is not likely to increase variation in alleles.
87	Answer to question 87	C	Objective 010 Response C is correct because linked genes are located close together on the same chromosome and consequently are inherited together. A result of this is that there is usually a greater percentage of parental phenotypes in offspring than phenotypes that reflect recombination events between the linked genes. Incorrect Response A does not address variation in a population. Stabilizing selection (Incorrect Response B) refers to natural selection that favors the median trait in a population. This is not necessarily a trait important to physiological processes. Mutations are often changes at the molecular level in DNA, such as the insertion or deletion of a single base, and this occurs independently from the location of genes with respect to each other (Incorrect Response D).
88	Answer to question 88	A	Objective 010 Using a Punnett square, the expected genotypic ratio resulting from the cross of any two heterozygous individuals is 1 of the homozygous dominant genotype to 2 of the heterozygous genotype to 1 of the homozygous recessive genotype. Thus, the probability is 1 out of 4, or 25%, that a child is homozygous recessive for the trait of tongue rolling (Correct Response A). Half of the offspring (Incorrect Response B) would be heterozygous, and Incorrect Responses C and D would not likely result from the cross described.
89	Answer to question 89	B	Objective 010 Response B is correct because the phenotypic ratios produced in this cross are approximately 9 white and smooth to 3 white and wrinkled to 3 purple and smooth to 1 purple and wrinkled. This type of a ratio is associated with crossing of parental plants that are heterozygous for two traits (i.e., a dihybrid cross). Because white flowers and smooth pods are dominant traits, both parents can exhibit these traits as a phenotype while carrying the allele for the recessive trait (purple flowers and wrinkled seeds). If both parents were homozygous recessive for both traits (Incorrect Response A), none of the offspring would have white flowers and smooth pods. In the case of one parent being homozygous recessive for color and homozygous dominant for texture (Incorrect Response C), the resulting data would exhibit a different phenotypic ratio and would not include any offspring that are wrinkled (i.e., the recessive texture trait). Similarly, if one parent is heterozygous for texture but homozygous dominant for color (Incorrect Response D), the data would not include any offspring that are purple (i.e., the recessive color trait).
90	Answer to question 90	B	Objective 010 Since colorblindness is sex-linked and recessive, a female must receive two alleles for the trait to be expressed: one from the X-chromosome of the biological mother and one from the X-chromosome of the biological father (Correct Response B). A sex-linked trait is carried on the X-chromosome. Since a male has only one X-chromosome, and this father passed down the trait to his daughter, his X-chromosome must carry the trait; therefore, he is colorblind, so it is not true that he has normal vision (Incorrect Response C). A female has two X-chromosomes, of which only one is passed on to her child. Since the daughter received an X-chromosome that carried the trait from her mother, the mother could have been colorblind (Incorrect Response A) or a heterozygote carrier (Incorrect Response D), but it cannot be determined from the information given which genotype is true in this case.
91	Answer to question 91	D	Objective 010 Response D is correct because recombination of two genes occurs when a crossover event happens between homologous chromosomes in the region between the two genes. The probability of recombination increases as the distance between the genes increases, because there is a larger area of the chromosome in which a crossover event could occur. The diagram shows that genes A and D are farthest away from each other on this chromosome, so these two genes would have the highest recombination frequency. Gene pairs B and A (Incorrect Response A), A and C (Incorrect Response B), and B and C (Incorrect Response C) are all closer to each other on the chromosome than gene pair A and D. Therefore, each of these pairs would have a lower recombination frequency than gene pair A and D.
92	Answer to question 92	B	Objective 010 Each parent has a Big B little b Big C little c genome. This type of problem is often represented using a Punnett square. There are 16 possible combinations for the offspring: Big B Big C Big B Big C , Big B big C Big B little c , Big B big C little B big C , Big B big C little B little c , Big B little c big B big C , Big B little c big B little c , Big B little c little b big C , Big B little c little b little c , little b big C big B big C , little b big C big B little c , little b big C little b big C , little b big C little b little c , little b little c big B big C , little b little c big B little c , little b little c little b big C , and little b little c little b little c . In this dihybrid cross, only the genomes little b big C little b big C , little b big C little b little c , and little b little c little b big C , 3 sixteenths of the offspring, will inherit the two little b alleles and at least one big C allele needed to produce the brown coat phenotype (Correct Response B). Approximately 1 sixteenth (Incorrect Response A) of the offspring will have two little b alleles but still lack color because they also inherited two little c alleles. The probability that an offspring could have no color is 1 quarter (Incorrect Response C), and the probability that an offspring is black-coated is 9 sixteenths (Incorrect Response D).
93	Answer to question 93	B	Objective 011 Response B is correct because, to reduce competition, the two species of birds inhabiting the zone of sympatry have become specialized to fill distinct ecological roles (i.e., utilize different types of insects for food) as demonstrated by the lengths of their bills. Specifically, there is no longer an overlap in bill size. Hybridization between the two species (Incorrect Response A) would result in a similar length of bill for both species of birds. Since the abandoned agricultural land may support different insects than either bird's original range, it is possible that the birds are eating different insects (Incorrect Response C). However, their bill length should stay the same unless there is a selective pressure. Increased predator pressure on each of the two species (Incorrect Response D) would likely be reflected in changes other than length of bill.
94	Answer to question 94	A	Objective 011 Gene duplication is a type of mutation in which one or more genes or sections of DNA is copied. This can occur in all types of organisms and may be the result of events, such as nondisjunction or unequal crossing-over events. The resulting duplicate gene or pieces of DNA can accumulate mutations without affecting the function of the original gene, providing variation for selection to act on without affecting the organism carrying the duplication (Correct Response A). Gene duplication events are compatible, not incompatible, with crossing-over events (Incorrect Response B). Although gene duplication can result from the overproduction of a given protein (Incorrect Response C), it is unlikely to spread a decrease in adaptation throughout a species. A gene duplication event large enough to create reproductive isolation (Incorrect Response D) is exceedingly rare and is less likely than a given gene duplication event simply spreading through a species.
95	Answer to question 95	A	Objective 011 Response A is correct because genetic drift refers to change in the frequency of alleles, such as those responsible for skin coloration in frogs, that occurs by chance and does not affect the fitness of the individuals in the population in question but may result in an overall decrease in variation. This change in allelic frequency is more evident in small populations. Founder effect (Incorrect Response B) would mean that the frog population's coloration over time would look similar to the founding population of 50 to 50 yellow to green. If recombination was occurring in the frog population (Incorrect Response C), there would be an increase in the genetic variation of this frog population, which may result in a wider variety of skin coloration. Stabilizing selection (Incorrect Response D) would cause the heterozygous population to dominate, which would be a frog population with mostly yellow alleles.
96	Answer to question 96	D	Objective 011 Response D is correct because, when one ecosystem expands, there are more niches to occupy, which results in radiation, whereas when an ecosystem contracts, it decreases the number of available niches, which can lead to the extinction of some lineages. Although in some instances, one lineage can radiate into many as the older lineage goes extinct (Incorrect Response A), in this instance there are two separate lineages. Perissodactyls did not completely go extinct, and the ecological niche that they had occupied was gone, not emptied (Incorrect Response B). It is difficult for lineages to adapt their behaviors (Incorrect Response C) because behaviors are not necessarily inherited.
97	Answer to question 97	D	Objective 011 Certain types of viruses called retroviruses use RNA as their genetic material. Once inside the host cell, the RNA is converted to DNA and is incorporated into the DNA of the host cell, functions as a mutation or a gene-duplication event, and is incorporated (Correct Response D). Viruses that remain hidden in the body in small numbers (Incorrect Response A) cannot be passed down to subsequent generations as only the information contained in gametes is inherited by offspring. Viruses are not capable of sensing the presence of susceptible organisms in the surrounding environment and often cannot survive in the external environment for extended periods of time (Incorrect Response B). Rapid reproduction and lack of immune response (Incorrect Response C) are not characteristics related to acting like mutations in the genetic code of a host.
98	Answer to question 98	C	Objective 011 Response C is correct because the use of plants that have been propagated vegetatively from the same mother plant ensures that all the plants have the same initial genetic makeup, so any changes would likely be the result of exposure to the mutagens. Observation of the plants before they reach maturity (Incorrect Response A) may or may not be useful, depending on the effect of the mutagen. Limiting the number of mutagens to two (Incorrect Response B) does not impact the validity of the experiment if data are detailed and accurate. Finally, continuing the experiment for an additional generation does not add validity to the experiment (Incorrect Response D).
99	Answer to question 99	C	Objective 011 Response C is correct because evidence suggests that, at least for certain genes, the rate of mutation is fairly constant among different species, thus providing a sort of molecular clock. This means that the amount of time that has elapsed since two species diverged can be approximated by examining the number of differences in the DNA and protein sequences of the two species. These estimates can be made using tissue from living organisms, rather than relying on fossil evidence. In convergent evolution (Incorrect Response A), two species that are not closely related evolve similar traits as a result of living in similar environments. These similar structures may tend to obscure when the two species diverged in the absence of other types of evidence. The Hardy-Weinberg equilibrium (Incorrect Response B) assumes a stable population in which natural selection, and hence evolution, is not occurring. Thus, it provides no information about speciation. The biological species concept (Incorrect Response D) defines a species as organisms that are capable of interbreeding with each other. It provides criteria for whether two organisms belong to the same or different species but does not provide evidence about when speciation might have occurred.
100	Answer to question 100	B	Objective 011 Response B is correct because convergent evolution is the process followed by organisms that are not closely related but that independently evolved similar traits or structures as a result of having to adapt to similar environments or niches. Of these pairs, only the bird's wing and the bat's wing fit this description. Progressive evolution led to the amniotic turtle egg (Incorrect Response A), which evolved as organisms evolved into land organisms and which prevents the embryo from drying out when laid on land. The frog egg lacks a hard shell and needs to be laid in water to prevent the embryo from drying out. A housefly and a bee (Incorrect Response C) are both arthropods that have characteristic jointed legs. Through progressive evolution, arthropods developed the characteristic jointed appendages. The type of evolution for the human hand and the monkey hand (Incorrect Response D) is more divergent because humans and monkeys have a common ancestor and the hands of each have developed differently for different purposes.
Total Correct:	Total Correct	Review your results against the test objectives.

Open Responses, Sample Responses, and Analyses

Question Number

Your Response
Read about how your responses are scored and how to evaluate your practice responses

101

Answer to 101

Open Response Item Assignment #1

For each assignment, you may type your written response on the assigned topic in the box provided.

Note: The actual test allows you to handwrite your responses on separate response sheets to be scanned for upload to the test. For this practice test, you may handwrite each response on 1–2 sheets of paper.

Essay Response Field

First Sample Weak Response

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First Sample Weak Response to Open-Response Item Assignment #1

The beads will be used to make a necklace that translates into a chain. Because of the three-digit code, the chain of beads will all be in an order that comes from the key, as stated in the question. This chain is an amino acid chain and comes from DNA in a step that is called translation. The key is what tells the cell how the DNA should turn into RNA, which is a messenger to make protein. This process is called transcription.

DNA is transcribed into RNA is translated into PROTEIN

The teacher will be sure to have the students explain what they have completed by labeling the colored beads to explain why they put them in the order they did.

First Weak Response Analysis

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Analysis of First Weak Response to Open-Response Item Assignment #1

This is an example of a weak response because it is characterized by the following:

Purpose: The purpose is partially achieved in this response. There is no detailed explanation of either translation or transcription. This should have been done when addressing the first bullet that asks for the key concepts. Limited scientific language is used. The response demonstrates partial understanding of transcription and translation.

Subject Matter Knowledge: Only limited and inaccurate knowledge of the subject is evident (e.g., explanations of the mRNA sequence, polypeptide chain, the function of codons). There is too much repetition of the prompt without revealing a deeper understanding. The candidate does not adequately explain the two-step process of transcription, in which the sequence of one gene is replicated in an mRNA molecule, and translation, in which the RNA molecule serves as a code for the formation of an amino acid chain. The location of these two processes is not addressed.

Support: The response is limited in providing relevant examples such as nitrogenous base pairs, genetic connection, transcription occurring in the nucleus, transcribing as in copying, and translation, occurring in the cytoplasm.

Rationale: The response employs only a limited rationale. The degree of understanding of the subject is vague: the operation of the triplet code and tRNA are not included at all. The formula is limited in detail. There is no explanation as to why the beads are colored and how that applies to the scientific concepts being examined.

Second Sample Weak Response

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Second Sample Weak Response to Open-Response Item Assignment #1

Transcription and translation are the way that a cell makes proteins. This involves first making a copy of the DNA in the nucleus and sending that copy out into the cytoplasm where RNA is used to make the protein. Proteins are made of amino acids. The beads in the scenario are probably those. (See scanned diagram.) Explaining this to students will help them see the connection between the bead necklace and a protein molecule.

A drawing of a protein is shown as a series of seven linked amino acids A A.

Second Weak Response Analysis

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Analysis of Second Weak Response to Open-Response Item Assignment #1

This is an example of a weak response because it is characterized by the following:

Purpose: This response addresses most elements of the prompt but has very little detail. The diagram especially does not adequately explain the process of transcription and translation. There is limited knowledge and understanding of the subject matter. The final bullet is not addressed and lacks a discussion of how a biology teacher could use the science and engineering practice of “engaging in argument from evidence.”

Subject Matter Knowledge: There is very little scientific language utilized. The lack of detail fails to explain the terms transcription and translation and only partially describes the phenomenon of transcription and translation of DNA and RNA.

Support: The response is most lacking in this area. High quality, relevant examples are missing from the brief cursory description of the process of protein synthesis. There is no mention of the base pairing in nucleic acids and the triplet code, and the response fails to adequately explain the intermediate steps arising with the creation of mRNA and tRNA. The diagram provides little support.

Rationale: The response demonstrates a poorly reasoned understanding of the topic. The diagram shows only the final product and does not connect with the prior steps in the process.

First Sample Strong Response

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First Sample Strong Response to Open-Response Item Assignment #1

Key concepts -- The instructions that direct the chemistry of the cell, and specifically the making of proteins, are stored in the DNA molecules of the nucleus. The "words" are formed by the order of the nitrogenous bases that make up the "rungs" of the DNA ladder, or double helix. There are four of these bases: adenine, thymine, guanine, and cytosine. They occur in pairs across the rungs. Adenine is always paired with thymine and guanine with cytosine. One pair makes up a single rung.

When protein synthesis happens, the DNA unwinds, the two sides split apart, and one side forms the template for carrying the message of what proteins to build. This message is transported to the cytoplasm by a single-stranded molecule of nucleic acid (the family of macromolecules to which DNA belongs) called RNA. The RNA is formed when an enzyme molecule arranges complementary bases along the template side of the DNA. Adenine is again paired with thymine, but a molecule of uracil takes the place of the thymine that would normally be paired with the template's adenine. Cytosine is paired with guanine and vice versa. This assembled macromolecule is specifically referred to as messenger RNA. This is the step in the process that is called transcription. Much like transcribing a spoken sentence into a written one, the language remains the same.

When the messenger RNA enters the cytoplasm, it attaches to a ribosome and at this point another form of RNA called transfer RNA starts to bring amino acid molecules to the site. Each amino acid is joined to another as a protein is assembled. These would represent the necklace beads in the scenario. The necklace itself would represent the protein being constructed. This is the step in the process called translation -- much like translating one language (written in nucleic acids) into another (written in amino acids). A particular sequence of three nucleic acids on the mRNA chain, called a codon, will give the instructions for a single amino acid's location in the growing protein. Each amino acid is carried to its location there by another RNA molecule called transfer RNA. The tRNA is directed to insert the amino acid by the arrangement of the codons on the mRNA. A sequence on the tRNA, which complements a codon, is called an anticodon. The scenario's number of 21 beads is significant since there are only 21 amino acids from which all proteins are constructed.

A scientific exercise starts by equating the pop-it bead necklace to the assembled protein, with the beads representing the individual amino acids in that protein. The teacher gives each student group an mRNA sequence of the same number of amino acids as they have beads. With designated colors representing amino acids, student groups assemble their proteins by consulting a codon wheel. Student groups exchange their necklaces. By decoding the other group's protein, students reinforce by argument from evidence how the mRNA sequence leads to a particular combination.

A diagram of the process of transcription and translation is shown. First, D N A unzips and unwinds with the M R N A formed as a copy of the D N A template strand in the process of transcription. This all takes place inside the nucleus. Then, the M R N A passes through the nuclear envelope and enters the cytoplasm. There, the M R N A binds with a ribosome, where it is presented with the anticodon end of a T R N A that recognizes the exposed M R N A. The other end of the T R N A has an amino acid, which is linked together to form a protein chain in the process of translation.

First Strong Response Analysis

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Analysis of First Strong Response to Open-Response Item Assignment #1

This is an example of a strong response because it is characterized by the following:

Purpose: The response strongly addresses all aspects of the prompt including the key scientific concepts, a representative diagram, and a classroom activity to further help students understand the phenomenon. The purpose is thoroughly achieved.

Subject Matter Knowledge: Thorough understanding is evident in the candidate's accurate use of key scientific terms such as messenger RNA, transfer RNA, and codon. The application of the scenario is substantial, appropriate, and accurate. All elements of the process are fully explained.

Support: High-quality, relevant examples of the steps in transcription and translation—such as transcription occurring in the nucleus and translation occurring in the cytoplasm—support the subject matter knowledge. The accompanying diagram is sound and fully supports the description with graphics and labels.

Rationale: Ably reasoned and comprehensive, this response demonstrates soundness of argument and a deep understanding of the topic by providing a logically connected and appropriate follow-up exercise that engages and challenges students to apply the key concepts of the lesson. The response provides an accurate explanation as well as a sound application.

Second Sample Strong Response

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Second Sample Strong Response to Open-Response Item Assignment #1

Transcription is the process of transferring information from DNA to RNA; messenger RNA (mRNA) is formed when a segment of DNA known as a gene becomes active and serves as a sort of template for the formation of a piece of RNA. That piece of RNA carries a sequence of bases that is the mirror image of the strand of DNA and exactly like the other strand of DNA. A sequence of three bases on the mRNA is called a codon.

The four bases that make up the code are adenine (A), thymine (T), guanine (G), and cytosine (C). Bases pair off together in a double helix structure at A and T, and C and G. However, RNA does not contain thymine bases, replacing them with uracil (U) bases, which pair to adenine. The four bases of RNA can be combined into 64 different three-base combinations.

Messenger RNA carries the genetic message from the nucleus of the cell into the cytoplasm to the ribosomes. In the ribosomes, mRNA is ready to direct the synthesis of a protein.

Translation is the process of bonding amino acids to form proteins. Translation refers to the fact that the order of the bases in a portion of DNA is translated into the order of amino acids of a particular protein. Transfer RNA (tRNA) is also involved in protein synthesis. These are shaped with three projections--one end is for the attachment site of an amino acid; the other end has an exposed base triplet pair called an anticodon, which has bases that are complementary to the bases of its corresponding codon. (See table.)

To help students gain an understanding of the concept of transcription, I would provide them with a chart listing amino acids and their mRNA codons. For example, they could see that the amino acid glycine has the codons CGU CGC and GGG. I would point out the U which would have to pair with A. We could also examine how the third position in transfer RNA anticodon is not specific so that the tRNA can match with multiple codons for any specific amino acid. For example, for phenylalanine the mRNA codon is UUU UUC and the anticodon is AAG. Redundancy is one way to limit the effect of mutations as changes in the third position will not affect the amino acids. Students can examine the concepts of mutations, protein synthesis, and further discuss the developing practice of gene splicing and its implications for disease intervention and ethics. Students could also be given an mRNA codon list and write corresponding tRNA sequences. By giving students an example of a mutated gene such as the BRCA gene, students will discover how a single or very small protein substitution can lead to serious health consequences. Comparing mutated genes with their normal sequences will show students how disease develops through the practice of engaging argument from evidence.

DNA	RNA
Bases: adenine, thymine, guanine, cytosine Double strand Deoxyribose sugar	adenine, uracil, guanine, cytosine single strand ribose sugar

Second Strong Response Analysis

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Analysis of Second Strong Response to Open-Response Item Assignment #1

This is an example of a strong response because it is characterized by the following:

Purpose: The response addresses all aspects of the prompt with clear explanations of the differences between DNA and RNA and their chemical compositions, a representative graph, and several classroom activities that are directly related to the topic and challenge students to further their investigations such as being given an mRNA codon list with the instructions to write corresponding tRNA sequences.

Subject Matter Knowledge: An in-depth knowledge of the subject matter is provided through the candidate's use of scientific terms such as gene, sequence of bases, codon, genetic message, order of amino acids, etc. The transcription and translation process of DNA and RNA are clearly explained, demonstrating substantial, accurate, and appropriate application. A chart graphically and accurately shows the relationships between the different nucleic acids.

Support: High-quality examples of the types of bases, possible arrangements in forming the triplet code, which operates in the steps of transcription and translation, support the subject matter knowledge. Support and explanations are clear and thorough, demonstrating a deep understanding of the topic.

Rationale: The response is ably reasoned. Transcription and translation are not only fully understood but the follow-up activities are well chosen to clearly demonstrate how a biology teacher could use this science to further engage in argument and evidence-based investigation. This response shows a comprehensive understanding of the subject matter.

102

Answer to 102

Open-Response Item Assignment #2

For each assignment, you may type your written response on the assigned topic in the box provided.

Essay Response Field

First Sample Weak Response

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First Sample Weak Response to Open-Response Item Assignment #2

Energy is transferred from one trophic level to another. A food web can demonstrate how this works while showing how inefficient the energy gets as it moves up the web. Students work with another person to create their own web of an ecosystem with organisms to show how it works.

Each pair of students will also be given a 6 inches by 14 inches unlined paper and, in cm., will draw what they think a food pyramid would look like if energy efficiency changed at different levels of the pyramid and name each level.

Their web should show where there is more energy and where it is the least efficient. They would then present it to the class and explain how it works.

First Weak Response Analysis

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Analysis of First Weak Response to Open-Response Item Assignment #2

This is an example of a weak response because it is characterized by the following:

Subject Matter Knowledge: There is little scientific vocabulary utilized in the response. It is unclear how the energy is being transferred. Although there is a direction stated, the details of the inefficiency of the energy transfer are vague. Because there is no clear testable claim, the investigation itself lacks specificity and demonstrates little understanding of the topic.

Support: There is a lack of high-quality, relevant examples to support the subject matter knowledge such as how much energy is lost at each succeeding trophic level. There is no data indicated and, therefore, no evidence that supports or refutes an imprecise claim.

Rationale: The use of a simulation or demonstration might adequately serve the same purpose as an investigation, but more support is needed to explain the reasons for that substitution in this particular case, such as the need to overcome difficulties inherent in recreating complete trophic levels under experimental conditions. Drawing a food pyramid does not constitute an investigation. The response demonstrates a limited, poorly reasoned understanding of the topic.

Second Sample Weak Response

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Second Sample Weak Response to Open-Response Item Assignment #2

The class will be shown a YouTube video about energy loss at each trophic level. There also will be a large pyramid graphic displayed showing the levels, with examples of the organisms in each.

The class will be divided into groups of 4 students and each group will be given a beaker with 75 ml. of water, which is one of the variables.

Groups of students will also be supplied with additional empty beakers and sponges.

Each beaker is a different trophic level and should be identified. They will pour water onto the sponge inside the first beaker, then pour the water into the next beaker and so on.

The students will conclude their observations by agreeing or not about energy transfer.

Second Weak Response Analysis

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Analysis of Second Weak Response to Open-Response Item Assignment #2

This is an example of a weak response because it is characterized by the following:

Purpose: The response is limited and only partially achieved. It is unclear where a testable claim has been stated. The procedure does not mention elements such as safety considerations and controls. The explanation of the water variable is vague and not explained how the activity is connected to the topic.

Subject Matter Knowledge: The simulation could serve the purpose of demonstrating energy loss at succeeding trophic levels but needs more explanation as to how that is to be accomplished. There is very little scientific vocabulary utilized in the response, and no data collected. The demonstration reflects an inaccurate application of the subject matter as its relevance is questionable.

Support: There is a lack of high-quality relevant examples to support the subject matter knowledge, such as how much energy is lost at each succeeding trophic level and how the sponges can be analogs for this process. The lack of a testable claim and little evidence make support difficult to demonstrate.

Rationale: The description of the final steps of the project are not sufficiently grounded to explain how the students are to make sense of the activity.

First Sample Strong Response

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First Sample Strong Response to Open-Response Item Assignment #2

The claim to be tested is the expected transfer of energy will be only 10% between trophic levels. Students will be involved in setting up an experiment involving euglena and their predators. If there is an ideal ratio between predators and prey, the students could possibly discover that ratio experimentally. A typical student experiment could be conducted in the following fashion:

Four colonies of euglena, all with similar populations, are started. By adding an appropriate number of euglena predators, such as larval salamanders or small minnows, one euglena colony receives a predator population less than the expected maximum of approximately one tenth of its mass, another receives the expected maximum, and a third is overburdened with a number that is greater than the expected maximum. A fourth colony is maintained without predators as a control.

Euglena in all colonies receive adequate food, as well as light (being both a heterotroph and autotroph). Students will discover if the claim is supported when the numbers of euglena per colony show trend lines for their populations such that the under-predated colony's numbers show an increase, the maximum-predated colony's numbers remain level, and the over-predated colony's numbers decrease. The control shows an undisturbed increase. The independent variable in this case is the number of predators per colony and the dependent variable the number of euglenas per colony. All other variables remain constant between all the colonies.

Students will be instructed in the proper procedures for using microscopes and associated glassware as well as the steps for correct hygiene when working with colonies of microorganisms. While using this activity, students will learn the difficulties involved in the science and engineering practice of planning and carrying out scientific investigations.

First Strong Response Analysis

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Analysis of First Strong Response to Open-Response Item Assignment #2

This is an example of a strong response because it is characterized by the following:

Purpose: The response addresses all aspects of the prompt such as forming a testable scientific claim, designing an investigation, explaining how to test the claim, and suggesting how the evidence supports or refutes the claim. This response thoroughly achieves the purpose of the assignment.

Subject Matter Knowledge: Thorough understanding of the components of a scientific investigation is evident in the candidate's attention to the use of a control, proper identification of the independent and dependent variables, and the need to keep conditions constant between experimental groups and the control group. The investigation is clearly explained using scientific vocabulary that demonstrates substantial, accurate, and appropriate application of the topic.

Support: High-quality, relevant examples of the organisms being used, the method of their use, the expected outcomes, and the areas of safety concern demonstrate strong subject matter knowledge. The description of the investigation and the conclusions are sound and well-reasoned.

Rationale: The response employs a sound rationale by providing a logical outcome and an appropriate follow-up exercise to help students make sense of the phenomena related to energy transfer between trophic levels.

Second Sample Strong Response

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Second Sample Strong Response to Open-Response Item Assignment #2

Trophic levels are a series of levels in the ecosystem or the position of an organism in the food chain. Organisms within each level share the same function in the food chain and use the same nutritional sources, for example herbivores feed primarily on vegetation. The amount of energy at each trophic level decreases as it moves through an ecosystem. As little as ten percent of the energy at any trophic level is transferred to the next with the rest being lost through metabolic processes as heat. To understand this concept of energy transfer, students would form a claim and since we are near a river and a forest environment, a claim could be that the patterns of energy transfer are similar in both of these environments.

Students use data from the Department of Fish and Wildlife to assess the biomass of trophic levels of these two environments. Students will test the claim that as little as 10 percent of the energy is transferred, by comparing information regarding numbers of species at each level in both environments. The conclusion would be reached that there is a loss of energy if the biomass at a higher level was 90% less than the one below it.

Other possible results could be noticed by the students if producer biomass consumed by herbivores is greater in the river environment than in the forest ecosystem because in a terrestrial setting, all biomass is not fully available to herbivores such as tree roots, large tree trunks, etc. With more available biomass in the aquatic setting of the river, that food ecosystem could consist of more trophic levels than can occur in the forest/terrestrial ecosystem.

Students also use computer simulation models of both environments to develop questions regarding the commonalities and differences at the trophic levels. Students are put into two different groups, one to look at the river environment and one to look at the forest environment, and results are then analyzed by the class. For these assessments no safety precautions are needed other than normal classroom hygiene.

Second Strong Response Analysis

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Analysis of Second Strong Response to Open-Response Item Assignment #2

This is an example of a strong response because it is characterized by the following:

Purpose: The response fully addresses all aspects of the prompt. It describes the process of energy transfer and has a claim that a terrestrial and an aquatic environment would have a similar loss of energy through the trophic levels. The response includes an investigation that students could carry out and suggests possible results of data collection. Further investigative activities are also included.

Subject Matter Knowledge: This response demonstrates sound knowledge of the subject matter. Trophic levels are accurately defined. There is a substantial amount of subject area information provided using scientific vocabulary such as herbivore, biomass, and terrestrial. An explanation is provided using the collected data.

Support: This response includes many high-quality supporting details such as herbivores feeding upon vegetation. Use of existing research data is suggested. A potential result of the investigation, that patterns of energy transfer are not the same in both environments, is included. Suggestions are made for student projects. All areas of the prompt are addressed and supported with examples.

Rationale: The response employs a sound rationale by employing a logical approach. The concept of trophic levels is defined and explained clearly. More than one suggestion for student projects is provided. All suggested investigations are reasonable and could be carried out by students. The response demonstrates an ably reasoned, comprehensive understanding of the topic.

Review the Performance Characteristics and Score Scale for Written Performance Assignments.

Multiple Choice Question
Practice Test Evaluation Chart

In the evaluation chart that follows, the multiple-choice questions are arranged in numerical order and by test objective. Check your responses against the correct responses provided to determine how many questions within each objective you answered correctly.

Subarea 1 : Molecules: Structures and Processes

Objective 0001: Apply knowledge of the chemical components of living systems and basic principles of biochemistry.

Question Number	Your Response	Correct Response
1		C
2		B
3		A
4		A
5		A
6		B
7		A
8		D

Total Correct for Objective 1 out of 8

Objective 0002: Apply knowledge of the processes that generate cellular energy.

Question Number	Your Response	Correct Response
9		C
10		A
11		C
12		B
13		D
14		B
15		A
16		D
17		A
18		B

Total Correct for Objective 2 out of 10

Subarea 1 (Objectives 0001–0002) Total Total Correct for Subarea 1 out of 18

Subarea 2 : Organisms: Structures and Processes

Objective 0003: Apply knowledge of cell structure and function and of the cell cycle.

Question Number	Your Response	Correct Response
19		C
20		B
21		B
22		A
23		B
24		A
25		A
26		C
27		A
28		B

Total Correct for Objective 3 out of 10

Objective 0004: Apply knowledge of the structures, structural organization, and life processes of unicellular and multicellular organisms (i.e., archaea, bacteria, protists, fungi, plants, and animals).

Question Number	Your Response	Correct Response
29		B
30		D
31		D
32		B
33		C
34		B
35		C

Total Correct for Objective 4 out of 7

Objective 0005: Apply knowledge of human anatomy and physiology.

Question Number	Your Response	Correct Response
36		D
37		C
38		B
39		D
40		C
41		A
42		D
43		A
44		A
45		B

Total Correct for Objective 5 out of 10

Subarea 2 (Objectives 0003–0005) Total Total Correct for Subarea 2 out of 27

Subarea 3 : Ecosystems: Interactions, Energy, and Dynamics

Objective 0006: Analyze interactions and dynamics of populations, communities, ecosystems, and biomes.

Question Number	Your Response	Correct Response
46		C
47		A
48		B
49		D
50		D
51		A
52		D
53		C
54		A

Total Correct for Objective 6 out of 9

Objective 0007: Apply knowledge of the cycling of materials and the transfer of energy through an ecosystem.

Question Number	Your Response	Correct Response
55		C
56		B
57		D
58		D
59		A
60		C
61		A
62		D

Total Correct for Objective 7 out of 8

Objective 0008: Analyze the effects of human activities on ecosystems and the environment.

Question Number	Your Response	Correct Response
63		C
64		A
65		D
66		B
67		C
68		D
69		D
70		A
71		B
72		C

Total Correct for Objective 8 out of 10

Subarea 3 (Objectives 0006–0008) Total Total Correct for Subarea 3 out of 27

Subarea 4 : Heredity and Biological Evolution

Objective 0009: Apply knowledge of the molecular basis of genetics.

Question Number	Your Response	Correct Response
73		D
74		D
75		C
76		A
77		C
78		B
79		D
80		A
81		C
82		B

Total Correct for Objective 9 out of 10

Objective 0010: Apply knowledge of the principles of genetics to understand the inheritance and variation of traits.

Question Number	Your Response	Correct Response
83		D
84		C
85		C
86		A
87		C
88		A
89		B
90		B
91		D
92		B

Total Correct for Objective 10 out of 10

Objective 0011: Apply knowledge of the theories and mechanisms of evolution to understand biological change and the diversity of life.

Question Number	Your Response	Correct Response
93		B
94		A
95		A
96		D
97		D
98		C
99		C
100		B

Total Correct for Objective 11 out of 8

Subarea 4 (Objectives 0009–0011) Total Total Correct for Subarea 4 out of 28

Practice Test Score Calculation

The practice test score calculation is provided so that you may better gauge your performance and degree of readiness to take an MTEL test at an operational administration. Although the results of this practice test may be used as one indicator of potential strengths and weaknesses in your knowledge of the content on the official test, it is not possible to predict precisely how you might score on an official MTEL test.

The Sample Responses and Analyses for the open-response items may help you determine whether your responses are more similar to the strong or weak samples. The Scoring Rubric can also assist in estimating a score for your open responses. You may also wish to ask a mentor or teacher to help evaluate your responses to the open-response questions prior to calculating your total estimated score.

How to Calculate Your Practice Test Score

Review the directions in the sample below and then use the blank practice test score calculation worksheet to calculate your estimated score.

Multiple-Choice Section

Enter the total number of multiple-choice questions you answered correctly:

Use Table 1 below to convert that number to the score and write your score in Box A:

190

Open-Response Section

Enter the number of points (1 to 4) for your first open-response question:

Enter the number of points (1 to 4) for your second open-response question:

Add those two numbers (Number of open-response question points):

Use Table 2 below to convert that number to the score and write your score in Box B:

Total Practice Test Score (Estimated MTEL Score)

Add the numbers in Boxes A and B for an estimate of your MTEL score:

A + B =

242

Practice Test Score Calculation Worksheet: Biology (66)

Table 1:

Number of Multiple-Choice Questions Correct	Estimated MTEL Score
0 to 25	121
26 to 30	129
31 to 35	136
36 to 40	144
41 to 45	152
46 to 50	159
51 to 55	167
56 to 60	175
61 to 65	183
66 to 70	190
71 to 75	198
76 to 80	206
81 to 85	214
86 to 90	221
91 to 95	229
96 to 100	237

Table 2:

Number of Open-Response Question Points	Estimated MTEL Score
2	36
3	40
4	44
5	48
6	52
7	56
8	60

Use the form below to calculate your estimated practice test score.

Multiple-Choice Section

Enter the total number of multiple-choice questions you answered correctly:

total correct

Use Table 1 above to convert that number to the score and write your score in Box A:

estimated multiple-choice score

Open-Response Section

Enter the number of points (1 to 4) for your first open-response question:

open-response points for first assignment

Enter the number of points (1 to 4) for your second open-response question:

open-response points for second assignment

Add those two numbers (Number of open-response question points):

total open-response score

Use Table 2 above to convert that number to the score and write your score in Box B:

estimated open response score

Total Practice Test Score (Estimated MTEL Score)

Add the numbers in Boxes A and B for an estimate of your MTEL score:

A + B =

total estimated score

Practice Test: Biology (66)

Answer Key, Sample Responses, Evaluation Chart, and Score Calculation Tool

Answer Key

Open Responses, Sample Responses, and Analyses

Multiple Choice QuestionPractice Test Evaluation Chart

Subarea I 1 : Molecules: Structures and Processes

Objective 0001: Apply knowledge of the chemical components of living systems and basic principles of biochemistry.

Objective 0002: Apply knowledge of the processes that generate cellular energy.

Subarea II 2 : Organisms: Structures and Processes

Objective 0003: Apply knowledge of cell structure and function and of the cell cycle.

Objective 0004: Apply knowledge of the structures, structural organization, and life processes of unicellular and multicellular organisms (i.e., archaea, bacteria, protists, fungi, plants, and animals).

Objective 0005: Apply knowledge of human anatomy and physiology.

Subarea III 3 : Ecosystems: Interactions, Energy, and Dynamics

Objective 0006: Analyze interactions and dynamics of populations, communities, ecosystems, and biomes.

Objective 0007: Apply knowledge of the cycling of materials and the transfer of energy through an ecosystem.

Objective 0008: Analyze the effects of human activities on ecosystems and the environment.

Subarea IV 4 : Heredity and Biological Evolution

Objective 0009: Apply knowledge of the molecular basis of genetics.

Objective 0010: Apply knowledge of the principles of genetics to understand the inheritance and variation of traits.

Objective 0011: Apply knowledge of the theories and mechanisms of evolution to understand biological change and the diversity of life.

Practice Test Score Calculation

How to Calculate Your Practice Test Score

Practice Test Score Calculation Worksheet: Biology (66)

Multiple Choice Question
Practice Test Evaluation Chart

Subarea 1 : Molecules: Structures and Processes

Subarea 2 : Organisms: Structures and Processes

Subarea 3 : Ecosystems: Interactions, Energy, and Dynamics

Subarea 4 : Heredity and Biological Evolution