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Explanation: Signal transduction pathways lead to a cellular response such as a change in gene expression. There are four major types, g-protein linked receptors, hormonal receptors, tyrosine kinase receptors, and pathways usually using second messenger (cAMP, calcium). They are characterized by a signal that is then taken in by a receptor, which can either be intracellular or extracellular. For example, G-proteins and ligand gated ion channels are transmembrane proteins while cytoplasm and nuclear receptors would be intracellular. The cell response is usually gene activation, such as cellular growth, determined by stimulatory, inhibitory, and permissive factors. These pathways often activate protein kinase cascades. Major pathways usually have ligands binding to receptor that can affect second messengers and eventually result in cellular responses, include cAMP dependent pathways. Depending on the cell, the response alters the cell’s metabolism, shape, gene expression, or ability to divide, determined by cytokines. LO 3.22 The student is able to explain how signal transduction pathways mediate gene expression, including how this process can affect proteins. SP 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. Multiple Choice: Which of the following signal transduction pathways is not correctly paired to its respective response? cAMP—regulate metabolic gene expression Cytokines—allow for cell replication & division Ethylene—allows fruit to ripen HOX genes—lead to cancer Integrin—cell attachment Learning Log/FRQ style question: Explain the process of the signal transduction pathways in the development of the male sexual development pathway.
Answers Multiple Choice: Which of the following signal transduction receptors is not correctly paired to its respective response? cAMP—regulate metabolic gene expression Cytokines—allow for cell replication & division Ethylene—allows fruit to ripen HOX genes—lead to cancer Integrin—cell attachment D is incorrect because HOX genes play a major role in development, not cancer. Also it is not signal transduction receptor, but a hormone, which leads to the desired gene expression. Learning Log/FRQ style question: The hormone that triggers the development of the male development pathway is testosterone. Testosterone, coming from the extracellular tissue or from outside the cell, passes through the plasma membrane and attaches to an intracellular, more specifically cytoplasm, receptor. The hormone receptor then enters the nucleus, which houses DNA, and binds to a specific gene. The bounded protein then stimulated transcript of the gene, changing it from DNA to mRNA. The mRNA is then translated into a specific protein.
AP Biology Learning Objective Project • LO 1.22: The student is able to use data from a real or simulated population(s), based on graphs or models of types of selection, to predict what will happen to the population in the future. • SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. • Explanation: This LO involves the use of a variety of different graphs and measures to assess the growth and change of a population. Studies of logistic growth models allow students to estimate the carrying capacity of the selected population and to calculate other characteristics about the population as it changes over time. Knowledge of varying reproductive strategies (“K” or “r”) can be represented by graphs and, again, can indicate the future characteristics of a population in terms of size or survivorship. Both of these principles have direct practical applications in terms of predicting the behavior of a population over time, as a result of understood scientific principles. • MC Question: If a particular species exhibits the Type I survivorship curve shown in Figure 1, what kind of reproductive strategy does it follow, and what sort of behavior could be expected of it? • A) K; little parental involvement in rearing of offspring • B) r; high mortality rate early on • C) K; small numbers of offspring produced • D) r; low mortality rate early on • Free Response Question: • A. A population of deer in a forest currently contains 300 individuals. If the carrying capacity for the forest allows for no more than 500 deer and the rmax for the forest is .85, what will be the size of the population one year from now (round to the nearest whole deer)? • B. Draw the population curve for this deer population (with labeled axes and important values) • C. List 2 factors that could contribute to the limitation of the carrying capacity. Figure 1.
LO 1.22 Answers • MCQ: Correct Answer- C: The given curve is indicative of a “K” type reproductive strategy. K strategies are marked by several characteristics, including lots of parental care during maturation, low mortality rates early in life (since most organisms survive through most of their potential life spans), and small numbers of offspring. The other answers either had the wrong curve type (“r”) or the wrong associated behavior. • FRQ: • A: The population one year later will be 402 deer. This answer can be obtained using the logistic growth rate equation and the values given in the problem to find the growth rate for that year (102), and then by adding that value to the old population value (300) to get the final answer, 402. • B: As indicated in part A, the model here is logistic, so the graph should reflect that (and so bear a shape resembling that of the graph in Figure 2). It should also include the labeled carrying capacity (500). • C: Here, an adequate description of any 2 factors that limit carrying capacity can be included. Acceptable answers include: competition for food, competition for water, competition for space, predators, disease, natural disaster, and the introduction of new species. Figure 2.
LO 4.1 The student is able to explain the connection between the sequence and the subcomponents of a biological polymer and its properties. SP 7.1 The student can connect phenomena and models across spatial and temporal scales. Explanation Biological polymers contain 3 classes of molecules: carbohydrates, proteins, and nucleic acids, which are all composed of smaller molecules called monomers. Lipids are another large molecule often categorized with polymers, but instead of being made up of monomers, they are grouped together because of their hydrophobic qualities and long carbon chains. A good way to think of a polymer is a pearl necklace– the necklace itself is a polymer, while the individual pearls are all monomers. Polymers are similar in their formation and breakdown; via dehydration and hydrolysis reactions, where a water molecule is either removed or inserted between monomers to form or break a bond, respectively. Carbohydrates can be separated into sugars and polysaccharides, which serve as both biological fuel and structural material for cells. Proteins account for about half of the mass of each cell, as they can be enzymes and polypeptides, which are created from amino acids that form primary, secondary, tertiary, and sometimes quaternary structural levels. Proteins are also categorized based on the R-groups that cause different properties for each cell. Nucleic acids include DNA and RNA, which store and transmit genetic information to be passed on from one generation to the next. They have a sugar-phosphate backbone with a phosphate group and some kind of nitrogenous base (cytosine, thymine, adenine, guanine, or uracil in RNA.) Multiple Choice Question A student was given an unknown polymer and was asked to confirm whether it was a carbohydrate, protein, or nucleic acid. They were given molecular probes which attach to adenine and glow under fluorescent light, an iodine solution, which turns purple in the presence of starch, and a Biuret solution, which turns pink in the presence of peptides, and any additional lab materials they might need. Which of the following procedures is paired with the correct result? A. Probe the unknown polymer with the molecular tag, then mix with Biuret and iodine solutions. Shine the resulting purple/pink solution under fluorescent light. It will not appear to glow, indicating that the unknown polymer is a nucleotide. B. Mix the unknown polymer into the Biuret solution. When it has fully been incorporated, the solution appears light pink, indicating that the unknown polymer is a protein. C. Mix the unknown polymer with the iodine solution. When it has fully been incorporated, the solution appears dark pink, indicating that the unknown polymer is a carbohydrate. D. Obtain a brown paper bag, and place a sample of the unknown polymer on it. When it leaves traces of translucent oil behind, it indicates that the unknown polymer is a carbohydrate. Free Response Question Choose 2 of the 4 macromolecules and polymers. Describe the structure of each from monomer to polymer, and explain how the structure relates to its function. Include in your explanation a drawing with labels of at least 1 of the molecules you select. Lipids Nucleic acids Carbohydrates Proteins
Multiple Choice Question A student was given an unknown polymer and was asked to confirm whether it was a carbohydrate, protein, or nucleic acid. They were given molecular probes which attach to adenine and glow under fluorescent light, an iodine solution, which turns purple in the presence of starch, and a Biuret solution, which turns pink in the presence of peptides, and any additional lab materials they might need. Which of the following procedures are paired with the correct results? A. Probe the unknown polymer with the molecular tag, then mix with Biuret and iodine solutions. Shine the resulting purple/pink solution under fluorescent light. It will not appear to glow, indicating that the unknown polymer is a nucleotide. This answer is incorrect. Assuming the probe attached to adenine, the probe would have glowed. This answer also does not demonstrate a solid lab procedure because it mixed all three tests into one. B. Mix the unknown polymer into the Biuret solution. When it has fully been incorporated, the solution appears light pink, indicating that the unknown polymer is a protein. This answer is correct. Because the solution turned pink, there are peptides in the unknown substance. Because there are peptides, we determine that the unknown polymer is a protein. C. Mix the unknown polymer with the iodine solution. When it has fully been incorporated, the solution appears dark pink, indicating that the unknown polymer is a carbohydrate. This answer is incorrect. The solution should have turned purple if it were indicating a starch, which is a carbohydrate– not a peptide like the color of the solution would point to. D. Obtain a brown paper bag, and place a sample of the unknown polymer on it. When it leaves traces of translucent oil behind, it indicates that the unknown polymer is a carbohydrate. This answer is incorrect. Brown paper bags test for the presence of lipids, not carbohydrates. Free Response Polymers essentially make up most of life’s organic compounds, and include four very diverse molecules– lipids, nucleic acids, carbohydrates, and proteins. Within those categories are more specific molecules which have vastly different structures, and ultimately, functions. Phospholipids (a subset to the lipid group) have a hydrophilic (polar) head, and two hydrophobic (nonpolar) tails. The head is primarily a phosphate group attached to a choline group, which allows for the polarity, whereas the tails are carbon skeletons of fatty acids. As a whole, this molecule forms bilayers as membranes to protect cells from the extracellular environment. Because the lipid’s outer surface (the heads) are polar, they only allow certain substances to pass through the membrane, thus shielding the nonpolar section of the cell from potentially harmful reactions to substances outside the cell. DNA (a subset of the nucleic acid polymer) is formed by the antiparallel connections of purines and pyrimidines through hydrogen bonding. These are held in place by the sugar-phosphate backbone. The DNA molecule in its entirety forms a spiral around an imaginary axis, becoming a double helix, while storing nucleotides that contain our genetic information. The structure of the double helix is especially important for when the genetic information is transcribed and translated to encode for other bodily proteins, making up the very genetic matter in all living things. Glucose is a monosaccharide that makes up polysaccharides, starch, and cellulose, all of which belong to the carbohydrate class of polymers. However, the monomer glucose itself plays a unique role because of its unique shape. It can be both linear or ring-shaped, and that allows it to exhibit flexibility as a molecule. The placement of the oxygen group in the glucose molecule (usually on the left/right side of the hexagonal-shaped molecule) allows it to hydrogen bond easily. This property then fosters the formation of larger carbohydrate polymer molecules such as cellulose found in the cell walls of plant cells. The organization of most cells is dependent upon the glucose and carbohydrate molecules that provide support. Proteins are all polymers constructed form the same set of 20 amino acids, which in a structure is called a polypeptide. The very basic level of protein structure is the primary structure, which is made up of the protein’s sequence of amino acids strung together in a strand that is determined through genetic inheritance. The secondary structure of proteins can be either an alpha helix, or beta pleated sheets, which are the primary structures of polypeptide chains coiled and folded in patterns. The protein’s tertiary structure is the interactions of the R-groups on the amino acids from the secondary structure. These interactions can be hydrogen bonds, ionic bonds, or disulfide bridges, that cause clusters in the proteins, making up the larger tertiary structure. Finally, the quaternary structure of a protein is only present in some of the protein polymers. If it does, it will be the overall protein structure that results from various tertiary structures forming a globular protein. Some of the most important proteins are enzymes, which, because of their specific amino acid makeup, and therefore shape, are able to accelerate chemical reactions within our bodies. This is done when the enzyme’s specific shape binds to a complimentary shape to induce a fit and catalyze biological reactions such as hydrolysis of polymers in food to make digestion easier.
This image is an sample of the phospholipid bilayer. There are thousands of these phospholipids that assemble themselves as a membrane. The polar heads, phosphate group, choline, and nonpolar tails are labeled. The purines are labeled as the A and G nucleosides, while the pyrimidines are the C and T nucleosides. The sugar-phosphate backbones are the ribbon-looking materials encompassing the nucleotides. The hydrogen bonds between the base pairs are present and labeled, as well as the double helix structure of this nucleic acid molecule of DNA. This is an example of an enzyme, which is an important protein for catalyzing reactions. The enzyme itself, active site, and complimentary-shaped substrates are labeled. The arrow indicates the formation of the induced substrate-enzyme relationship after binding has occurred. In this example, either drawing of the carbohydrate glucose (or both) is/are acceptable. The carbons, oxygens, and hydrogens are clearly labelled. Molecules identical to these are what power cellular respiration. They also make up cellulose in polymer form to act as a protecting element on cell walls.
Learning Objective Focus: • LO 2.27: The student is able to connect differences in the environment with the evolution of homeostatic mechanisms. • SP 7.1:The student can connect phenomena and models across spatial and temporal scales. • Explanation: Homeostatic mechanisms reflect both common ancestry and divergence due to adaptation in different environments. Homeostatic control systems in species of microbes, plants and animals support common ancestry. Homeostatic control system has three functional components: a receptor, a control center, and an effector. The receptor detects a change in some variable of the animal’s internal environment change. The control center processes information it receives from the receptor and directs an appropriate response by the effector. There is positive and negative feedback. Most homeostatic mechanisms operate on the principle of negative feedback (a change in the variable being monitored). Also dealing with how animals form and function together in regulating the internal environment having to do with the regulation of body temperature. • M.C Question: What is an example of negative feedback? • A) Blood clotting. • B) Low body temperature causes shivering. • C) Stools in rectum stimulate defecation. • D) Ingestion of salty food triggers salt excretion.
Response No heat produced Heater turned off Room temperature decreases Set point Too hot Set point Too cold Set point Control center: thermostat Room temperature increases Heater turned on Response Heat produced Learning Objective Focus: • Learning Log/ FRQ-style question: There are sea lions on a New Zealand beach in the winter time. • How do endotherms keep their internal temperature constant? • What is the difference between endotherms and ectotherms? • What type of animals are endotherms and what type of animals are ectotherms?
Answer Key: • M.C Question: What is an example of negative feedback? • A) Blood clotting. • B) Low body temperature causes shivering. • C) Stools in rectum arouse defecation. • D) Ingestion of salty food triggers salt excretion. • Answer A is wrong because it is an example of positive feedback. Once a vessel is damaged , the platelets start to cling to the injured spot causing blood clots. Positive feedback enhances the stimulus. • Answer B is the right answer because the hypothalamus of a human responds to temperature changes and responds accordingly. If the temperature drops, the body shivers to bring up the temperature and if it is too warm, the body will sweat to cool down due to evaporation. It also has to do with where the mechanism is according to the environment they are in. • Answer C is wrong because it is an example of positive feedback. It also has to do with positive feedback due to the contraction of muscles involved with defecation. • Answer D is wrong because the effect of the ingestion of salt is not decreasing. It is increasing due to the ingestion of the salty food. The feedback is enhancing it not inhibiting it.
Answer Key continued: • Learning Log/ FRQ-style question: There are sea lions on a New Zealand beach in the winter time. • How do endotherms keep their internal temperature constant? • What is the difference between endotherms and ectotherms? • What type of animals are endotherms and what type of animals are ectotherms? a) Endotherms keep their internal temperature constant by generating their own body heat. The heat comes from animals metabolism , chemical reactions in the body. Endotherms require less energy to maintain their body temperature as their surroundings increase. They still need to control heat loss. Without this the body temperature would fall too high or either too low. b) The difference between endotherms and ectotherms are that ectotherms gain most of their heat from the environment. They have such a low metabolic rate that the amount of heat it generates is too small to have much effect on the body temperature. Endotherms on the other hand can use metabolic heat to regulate their body temperature. In a cold environment, an endotherm’s high metabolic rate generates enough heat to keep it’s body substantially warmer than its surroundings. The main thing that distinguishes endotherms and ectotherm is the source of heat used to maintain body temperature. c) The type of animals that are ectotherms are most invertebrates, fishes, amphibians, lizards, snakes, and turtles since they are cold blooded. The type of animals that are endothermic are mammals and birds since they are warm blooded.
LO 3.50: The student is able to create a visual representation to describe how the vertebrate brain integrates information to produce a response. SP 1.1: The student can create representations and models of natural or man-made phenomena and systems in the domain. Background: The neuron is the basic structure of the nervous system that reflects function. A neuron has a cell body, axon and dendrites. Many axons have a myelin sheath that acts as an electrical insulator. The structure of the neuron allows for the detection, transmission and integration of signal information. Action potentials create impulses along neurons. In response to a stimulus, Na+ and K+ gated channels sequentially open and cause the membrane to become locally depolarized. Na+/K+ pumps, powered by ATP, work to maintain membrane potential. Transmission of information between neurons occurs across synapses. The transmission of information across the synapse triggers a response that can be either stimulatory or inhibitory. Different regions of the vertebrate brain have different functions including muscle movement, hearing, vision, neuro-hormone production and right and left cerebral hemispheres in us. M.C. Question: Which of the following statements is not true about how the vertebrate brain integrates information to produce a response? A.) In response to a stimulus, the NA+ gated channels will close B.) When NA+ and K+ gated channels open, the cause the membrane to become locally depolarized. C.) The transmission of information across the synapse does not trigger a response. D.) The structure of the neuron does not allow for the detection and transmission of signal information.
Learning Log/ FRQ-Style Question: Describe the three main parts of the neuron and explain how each part implies to how they integrate information to produce a response. Use Figure 1 and 2 to help explain the parts and their functions. Figure 2 Figure 1
Answer Key-LO3.50 M.C. Question: Answer: Which of the following statements is not true about how the vertebrate brain integrates information to produce a response? A.) In response to a stimulus, the NA+ gated channels will close B.) When NA+ and K+ gated channels open, the cause the membrane to become locally depolarized. C.) The transmission of information across the synapse does not trigger a response. D.) The structure of the neuron does not allow for the detection and transmission of signal information. Learning Log/ FRQ-Style Question: Answer: The three mains parts into which a neuron can be didactically divided are: dendrites, cell body and axon. Dendrites are projections of the plasma membrane that receive the neural impulse from other neurons. The cell body is where the nucleus and the main cellular organelles are located. The Axon is the long membrane projection that transmits the neural impulse at distance to other neurons, to muscle cells and to other effector cells. Once the information travels through the whole neuron and reaches its target, it produces a response.
LO 3.10: The student is able to represent the connection between meiosis and increased genetic diversity necessary for evolution. SP 7.1: The student can connect phenomena and models across spatial and temporal scales. Explanation: Meiosis is a reduction division, and, followed by fertilization, ensures genetic diversity in sexually reproducing organisms. The process of meiosis ensures that each gamete produced will have one complete haploid (1n) set of chromosomes. During meiotic division, homologous chromosomes are paired, with one homologue originating from the maternal parent and the other from the paternal parent. Location of the chromosome pairs is random with respect to the cell poles. Separation of the homologous chromosomes ensures that each gamete receives a haploid (1n) set of chromosomes composed of both maternal and paternal chromosomes. Homologous chromatids exchange genetic material through a process called “crossing over,” which increases genetic variation in the resultant gametes. Fertilization involves the fusion of two gametes, increases genetic variation in populations by providing for new combinations of genetic information in the zygote, and restores the diploid number of chromosomes. Genetic diversity is necessary for evolution because natural selection is based on differential success in reproduction: Individuals in a population exhibit variations in their heritable traits, and those with variations that are better suited to their environment tend to produce more offspring than those with variations that are less well suited. Sexual reproduction contributes to the genetic variation present in a population, which ultimately results from mutations. M.C. Question: Which of the following could not be an example of natural selection through genetic diversity derived from the process of meiosis? A) Independent assortment of chromosomes in an Emperor penguin population leads to most individuals being born with a body shape better suited for faster swimming and escape from predators. B) Frameshift mutation in a trout gene that codes for a digestive enzyme, enabling the trout to consume a wider variety of aquatic animals. C) A new species of rhinoceros beetle that feeds on the leaves of fruit trees, instead of the fruit, evolves from a current species through chance variation by crossing over; the populations of both species increase rapidly. D) Through the process of random fertilization, a woman received a allele combination that allowed her to bear children long after the average age for the onset of menopause; she goes on to birth 23 children, perhaps passing that allele combination on to them. Learning Log/FRQ-style Question: Identify five ways that sexual reproduction can expand genetic variability. For each, explain how it can expand genetic diversity among a population’s offspring. Explain at least one mechanism through the use of a diagram.
ANSWER KEY--LO 3.10 M.C. Question: Which of the following could not be an example of natural selection through genetic diversity derived from the process of meiosis? A) Independent assortment of chromosomes in an Emperor penguin population leads to most individuals being born with a body shape better suited for faster swimming and escape from predators. B) Frameshift mutation in a trout gene that codes for a digestive enzyme, enabling the trout to consume a wider variety of aquatic animals. C) A new species of rhinoceros beetle that feeds on the leaves of fruit trees, instead of the fruit, evolves from a current species through chance variation by crossing over; the populations of both species increase rapidly. D) Through the process of random fertilization, a woman received a allele combination that allowed her to bear children long after the average age for the onset of menopause; she goes on to birth 23 children, perhaps passing that allele combination on to them. Identify five ways that sexual reproduction can expand genetic variability. For each, explain how it can expand genetic diversity among a population’s offspring. Explain at least one mechanism through the use of a diagram. Crossing over, or recombination, during gamete formation generates new combinations of alleles, allowing for a greater number of physical differences between individuals of a population. Independent assortment, the random alignment of chromatids on the metaphase plate during meiosis, also contributes to greater genetic diversity because with 23 pairs of non-sister chromatids, the number of possible combinations is very high. Random fertilization plays a role in genetic variability because, for example, in humans each male and female gamete represents one of approximately 8 million possible chromosome combinations, meaning 64 trillion diploid combinations for a zygote. If each individual has 8 million unique gametes, and could possibly reproduce with any other individual in its population (random mating), then the number of possible combinations of zygotes is astronomical. Diploidy and polyploidy can expand genetic variability by preventing the expression of harmful mutations.
LO 3.32: The student is able to generate scientific questions involving cell communication as it relates to the process of evolutionSP 3.1: The student can pose scientific questions Explanation: An organisms interaction with its environment is essential to its survival, because it allows an individual to respond to a change in environmental conditions. For the organism to respond, cell communication must occur and through this fact, natural selection has selected for the trait of cell communication. Organisms whose cells could not communicate with each other could not respond to the change in environment that could lead to their survival, compared to those that could make changes to its body due to cell communication. The cells that were able to communicate with one another were capable of making changes within themselves to acclimate to change in environment, which will lead to their survival and their ability to pass on their traits to future generations. As these traits are passed down as time goes on, these traits become specialized in individual species and leads to the multiple types of cell signaling that can be found in organisms today. There are 3 types of cell-communication: cell-to-cell contact, local signaling, and long-distance signaling. Cell-to-cell contact is often a process of junctions and cell recognition. Junctions allow two cells to trade cytoplasmic properties without having to go through the selectively-permeable membrane. Recognition requires receptors which will relate to the other two types of signaling, because they require ligands and receptors. Local signaling is close range signaling, from one cell to another. Long-distance signaling requires hormones, produced by the endocrine system and travels via circulatory system to reach targeted cells throughout an organisms body. M.C. Questions: Mr. Bennett needs energy and has to rely on the sugar in his blood stream for energy. Which question is most relevant to the problem above? • How does the paracrine system signal nearby cells to take up sugar? • How does the endocrine system use insulin to fix Bennett’s problem? • How does the endocrine system use gulcagon to fix Bennett’s problem? • How do gap junctions pass necessary nutrients from one another? Learning Log/FRQ-style Question: Take into account how nerve cells communicate with one another. Why is it that natural selection favored paracrine signaling over endocrine signaling for nerve cell communication? Also describe how nerve cells communicate with one another.
MC: Mr. Bennett needs energy and has to rely on the sugar in his blood stream for energy. Which question is most relevant to the problem above?A) How does the paracrine system signal nearby cells to take up sugar?B) How does the endocrine system use insulin to fix Bennett’s problem?C) How does the endocrine system use gulcagon to fix Bennett’s problem?D) How do gap junctions pass necessary nutrients from one another? FRQ/Learning Log: Take into account how nerve cells communicate with one another. Why is it that natural selection favored paracrine signaling over endocrine signaling for nerve cell communication? Also describe how nerve cells communicate with one another. The nervous cells are found throughout the body and transfer information to one another by sending neurotransmitters from the axon into the synapse. The synapse is the small gap between nerve cells and the part of the nerve cell that takes up neurotransmitters is called dendrites. Neurotransmitters bind to a receptor on the nerve cell which opens up the ligand-gated ion channel. This allows the nerve cell to “fire,” because there is a change in charge in the cell body, which will push electrons from the cell body through the axon. For the neuron to fire, its charge must reach -55mV or it will not fire at all, its starting charge is -70mV. Since nerve cells are so close together, it is more beneficial to use paracrine signaling, because it is quicker and it already has a pathway to follow. Endocrine signaling is used between cells that are not connected and can only communicate with one another through hormones via bloodstream. This is an effective process, but a slow one as well. Neurons need to be able to pass the message to one another quickly, because the brain needs to receive this information quickly so we can react appropriately to the stimuli in our environment. Since neurons are practically connected and are only separated by the synapse, paracrine signaling would work, because paracrine signaling is defined as local signaling, communication between cell to cell.
LO 1.25: The student is able to describe a model that represents evolution within a population. SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: For quite some time, there were many views on how evolution occurred within populations of organisms. Lamarck had the model of use and disuse, where parts of the body become larger and more helpful, and the ones that are not used eventually disappear. This model, however, is incorrect. Using Darwin’s correct theory, we now know more about the many factors that cause evolution within populations, including environment, predation, or other factors. Changes in allele frequencies and mutations within a population can cause evolutionary changes in the organisms within a population over a long period of time. Emergent diseases within a population can also decrease the size of that population, making it lose important alleles that can help it evolve when the environmental conditions change. The Hardy-Weinberg theorem gives a basis of comparison for determining evolutionary change within a population. It gives unrealistic, ideal situations for a population, thus serving as a basis for comparison with any real population that actually does undergo some sort of evolutionary change.
M.C. Question: In a population of mongooses, with genetic variation present within this particular population, environmental pressures cause one of the extreme colors of the mongooses to be favored out of all others within the population. In other circumstance, with the same population, environmental changes cause the extremes of both kinds in the mongoose population to be favored. In a third circumstance, the changes in environmental pressures cause only the mongooses with intermediate fur coat color to be favored out of the population. Which of the following statements accurately describes the phenomenon of stabilizing selection? • Evolution favors only the extremely light colored rabbits out of the populaton. • Evolution favors only the intermediate colored rabbits out of the population. • Evolution favors only the extremely dark colored rabbits out of the population. • Evolution favors both the extremely light and dark rabbits out of the population. Learning Log/FRQ-style Question: There are many types of unforeseen factors that can alter the genetic composition of a population. Describe the phenomenon of the bottleneck effect and give an example. Why is it very important that humans have a knowledge of the bottleneck effect in terms of conservation?
M.C. Question and Answer: Which of the following statements accurately describes the phenomenon of stabilizing selection? • Evolution favors only the extremely light colored rabbits out of the populaton. • Evolution favors only the intermediate colored rabbits out of the population. • Evolution favors only the extremely dark colored rabbits out of the population. • Evolution favors both the extremely light and dark rabbits out of the population. It stabilizes the population, removing the extremes of the organism on both sides of the population over time. Learning Log/FRQ-style Question and Answer: There are many types of unforeseen factors that can alter the genetic composition of a population. Describe the phenomenon of the bottleneck effect and give an example. Why is it very important that humans have a knowledge of the bottleneck effect in terms of conservation? The bottleneck effect is caused by some sort of sudden change in the environment, usually by a disaster. This could greatly reduce the size of the population. With only a small number of survivors from the population, the new population may not have many of the qualities the larger original population once had. This would mean that certain, important alleles from the population may be lost forever. If a population of frogs was greatly reduced due to a massive hurricane, it would lose valuable genetic qualities that may be gone for good. This may cause them to eventually go extinct, mostly because there is very little room for evolutionary improvement since they have so few alleles left. From a perspective of conservation, humans need to be informed of the dangers of the bottleneck effect. Many of the activities humans do are very harmful to wildlife populations, which can cause disasters for them, making the bottleneck effect very common in these populations.
LO 1.20: The student is able to analyze data related to questions of speciation and extinction throughout the Earth’s history. SP 5.1: The student can analyze data to identify patterns or relationships. Explanation: Speciation is the origin of new species, and is the main point of the evolutionary theory because it causes biological diversity. In speciation gene flow is interrupted and changes the phenotype of an organism. Over time the organisms will reproduce and form a new species. There are two main types of speciation: allopatric speciation, and sympatric speciation. In allopatric speciation, a population forms a new species because it becomes isolated geographically from the other species. For example, if an earthquake caused the separation of a lake, then some fish in half the lake will evolve and change differently than the other fish. In sympatric speciation, a small population becomes a new species even though it is not geographically isolated. This occurs because of random chromosomal changes, and nonrandom mating that causes gene flow to reduce. Multiple Choice Question: Which of the following best explains sympatric speciation? A new population is formed while being geographically isolated. Species descended from a common ancestor gradually diverge more in morphology as they acquire more adaptations. A new species changes most as it buds from a parent species and then changes little for the rest of its existence. A small population becomes a new species without geographic separation. FRQ: A species of fish lives together in a pond. A tornado picks up some of the fish and moves them to a different pond. Over a period of time the same species of fish evolve into different species. What kind of speciation is this? Give two examples of how the gene pool can diverge.
Multiple Choice Answer: The correct answer to the multiple choice question is D. Answer A is an example of allopatric speciation. Answer B is an example of gradualism, and answer C is an example of punctuated equilibrium speciation. FRQ Response: This is an example of allopatric speciation. Allopatric speciation is when a new population is formed from a species while being geographically isolated from its parent population. When the population becomes isolated, they will evolve on their own. An example of how they will evolve is different mutations in their genotype. If the mutation causes a change that is beneficial it will be passed on to the next generation. Another example is through sexual selection. A trait may arise that the fish select for, such as color. This trait is then passed down due to selective pressure.
Explanation: A membrane’s ability to permit certain molecules to pass through it and prevent the passage of others is its selective permeability. In cell membranes, this property is accomplished by the phospholipid bilayer. A model of the membrane on a molecular scale would show two layers of amphipathic phospholipids with polar (thus hydrophobic) heads and nonpolar (thus hydrophilic) tails. With the polar heads on the outside and the nonpolar tails on the inside of the membrane, and since polar objects oppose each other, polar molecules (like water) cannot • pass through the polar heads of the membrane, while nonpolar molecules (like steroids) pass through which cannot come through the membrane because of size of charge. Selective permeability can also be affected by proteins in the membrane, which can transport molecules across the membrane . This transport can be facilitated diffusion (not require ATP, because molecule moves down concentration gradient) , a type of passive transport, or it can be active transport (requires ATP to move molecule against concentration gradient). For a facilitated diffusion example, a ligand-gated ion channel involves an integral protein that, when a ligand (such as a neurotransmitter) binds to it, changes shape to allow ions to pass through the membrane, down the concentration gradient. Regarding selective permeability and other properties of membranes (like the fluid mosaic model), questions can be posed and predictions made about how certain molecules would interact with a membrane based on the membrane’s molecular structure. LO 2.10: The student is able to use representations and models to pose scientific questions about the properties of cell membranes and selective permeability based on molecular structure. SP 1.4: The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. SP 3.1: The student can pose scientific questions. • Multiple Choice Question: • A sealed dialysis bag containing Solution A (an aqueous solution of 0.2M) is placed in a beaker holding Solution B (an aqueous solution of 0.4M). The membrane pores of the dialysis bag are selectively permeable to small molecules such as water, but are too small for the solute to pass through. Which of the following statements accurately describes the ensuing movement of substance(s) across the membrane and the resulting change in solution concentrations? • A) Because Solution A and Solution B are isotonic, the volume of water moving from Solution A to Solution B equals that moving from Solution B to Solution A, and their concentrations remain in homeostasis. • B) Because the solute is larger than the membrane pores, there is no movement across the membrane and the solutions remain the same concentration. • C) Water from Solution A travels by osmosis across the dialysis bag to Solution B, resulting in a higher solute concentration in Solution B and a lower solute concentration in Solution A. • D) Water from Solution A travels by osmosis across the dialysis bag to Solution B, resulting in a lower solute concentration in Solution B and a higher solute concentration in Solution A. • Free Response Question: • a) From the model of a cell membrane cross section, identify all 5 of the labeled parts and describe how each functions within/around the cell membrane. • b) Name three factors that influence the fluidity of a cell membrane, and describe and explain the specific effect each factor has on fluidity, connecting your explanation to the molecular structure of the membrane.
Multiple Choice Question: • A sealed dialysis bag containing Solution A (an aqueous solution of 0.2M) is placed in a beaker holding Solution B (an aqueous solution of 0.4M). The membrane pores of the dialysis bag are selectively permeable to small molecules such as water, but are too small for the solute to pass through. Which of the following statements accurately describes the ensuing movement of substance(s) across the membrane and the resulting change in solution concentrations? • A) Because Solution A and Solution B are isotonic, the volume of water moving from Solution A to Solution B equals that moving from Solution B to Solution A, and their concentrations remain in homeostasis. • B) Because the solute is larger than the membrane pores, there is no movement across the membrane and the solutions remain the same concentration. • C) Water from Solution A travels by osmosis across the dialysis bag to Solution B, resulting in a higher solute concentration in Solution B and a lower solute concentration in Solution A. • D) Water from Solution A travels by osmosis across the dialysis bag to Solution B, resulting in a lower solute concentration in Solution B and a higher solute concentration in Solution A. • Free Response Question: • a) From the model of a cell membrane cross section, identify all 5 of the labeled parts and describe the structure and function of each within/around the cell membrane. • b) Name three factors that influence the fluidity of a cell membrane. Describe and explain the specific effect each factor has on fluidity, referring in your explanation to molecular structure. a) “A” is a carbohydrate, which attach to the external surface of the membrane and therefore can act to hold adjoining cells together or serve as an attachment site for chemical messengers like hormones or viruses. “B” is a phospholipid, two rows of which make up the phospholipid bilayer of the cell membrane. Phospholipids have polar heads and nonpolar tails, and since the heads are on the outside of the membrane (in contact with the extracellular matrix and the cytoplasm), the polarity of the heads prevents polar molecules from diffusing passively through the membrane. This gives the membrane its selective permeability. “C” is an integral protein, which can serve many purposes, including structural support for the membrane and, significantly, active transport or facilitated diffusion of molecules across the membrane. These functions are possible since the integral protein lies embedded within the membrane. “D” is cholesterol, which is found only in animal cells and is has a strong fused-ring structure that, because it lies within/between the phospholipid bilayer, gives structural support and regulates fluidity in the membrane. “E” is a peripheral protein, which attach to the side of the membrane at the site of an integral protein and play an essential role in the signal reception process, often initiating a reaction in a chain of proteins by its “activation” at the time of signal reception. b) Temperature, cholesterol, and the presence of fatty acids affect fluidity in a cell membrane. An increase in temperature increases fluidity because there is more kinetic energy that forces individual phospholipids away from each other. Cholesterol embedded in the phospholipid bilayer maintains fluidity and prevents extreme fluidity fluctuations since it has a strong fused ring structure. In high temperatures, phospholipids cannot be pushed as far away as normal because of cholesterol holding them together (so fluidity is reduced), but in low temperatures, phospholipids cannot get as close together as normal, for the same reason (so fluidity is increased). Thirdly, the presence of unsaturated fatty acids within the membrane increases fluidity because the hydrocarbon tails of unsaturated fats are kinked, and therefore prevent the components of the membrane from being as tightly packed.
Introduction LO 3.2: The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of heritable information. SP 4.1: The student can justify the selection of the kind of data needed to answer a particular scientific question. Explanation: As it was being discovered that DNA is the genetic information of a cell, many scientists performed experiments to provide proof for this hypothesis. It needed to be proven that DNA is what enters a cell and becomes part of the genome, and that other molecules (such as proteins) don’t have the ability to be assimilated into a cell in the same fashion. Alfred Hershey and Martha Chase illustrated this concept in their experiment dealing with bacteriophages containing either a protein or DNA. By demonstrating that only the DNA phage became part of the cell’s genome, they further proved the identity of DNA as the genetic information. This experiment reinforced the concept previously shown by scientist Frederick Griffith, who used dead pathogenic cell DNA to prove that living cells could assimilate new genetic information into their genome successfully. Bacteriologist Oswald Avery demonstrated this same concept with his later experiment in which he proved that DNA was the only substance able to be successfully transformed to bacteria. Erwin Chargaff furthered the world’s understanding of DNA when he demonstrated that within a specific species, there is a regular ratio of base pairs (meaning cytosine-guanine and adenine-thymine) found in DNA. This discovery became part of what was soon known as “Chargaff’s rules.” Each of these experiments provided undisputable proof that DNA is the best candidate for genetic material, as it was consistently the only substance able to change the genome of a cell, and there is diversity in DNA throughout different species due to base pair differences, which accounts for the diversity of life.
Questions • M.C. Question: In order to determine that DNA was the source of heritable information, scientists needed to show that DNA A) Replicates according to the dispersive model B) Serves as the main site for protein synthesis C) Contains the base uracil D) Enters a host cell and provides genetic information after being assimilated into the genome E) Is the only substance in a cell subject to mutations
Questions *These cells were placed together in an environment and left to interact themselves. Look at the information presented in the graph above and answer the following questions. • Why would all of the cells be pathogenic by the end of the experiment? • Which researcher’s scientific experiment supported these results? Explain the results obtained from the experiment. • Why did these results support the hypothesis that DNA is genetic material?
Answers • M.C. Answer: Choice D. • Essay Answer a) By the end of the experiment, all of the previously living nonpathogenic cells had become pathogenic. This is because the living cells probably took up the DNA of the dead pathogenic cells, and this dangerous DNA became part of the cell’s genome. Once this happened, the previously harmless cells became pathogenic, as this new genetic information had been taken up and integrated into the cell’s genes. Because of this, no nonpathogenic cells remained by the end of the experiment. Additionally, the number of pathogenic cells by the end of the experiment is 20+ because now that the pathogen is part of the cells’ DNA, as these cells continue to divide they will pass this trait on, resulting in pathogenic offspring. (Continues on next slide)
Answers b) Frederick Griffith performed an experiment in which he found results consistent with those in this experiment. When Griffith placed heat killed pathogenic cells with nonpathogenic cells, the nonpathogenic cells soon became pathogenic. They also passed on the pathogenic trait to their offspring. Griffith called this process transformation, and it is now understood to be the process of a change in genotype or phenotype due to the assimilation of external DNA by a cell. c) This experiment is consistent with the hypothesis that DNA is genetic material because when transformation is attempted with other various molecules, it is not effective. Only DNA can be assimilated by a cell and made part of the genome, and this demonstrates that DNA is the genetic material. In Griffith’s experiment, it was DNA making the cells pathogenic, as it is the only substance able to be transformed into a cell.
LO 3.42 The student is able to describe how organisms exchange information in response to internal changes or environmental cues. S.P. 7.1 The student can connect phenomena and models across spatial and temporal scales. Explanation: Organisms within populations have a lot of social interaction that requires signals as a main form of communication. Animals can communicate by visual, auditory, chemical, or electrical signals based on their response to their environmental niches. For example, Terrestrial mammals are nocturnal so they depend more on olfactory and auditory signals while in contrast bird species often use visual and auditory signals. These signals make exchanging information in response to external or internal changes possible. Honey bees and wasps are both organisms that are known for their painful stings. Only female bees and wasps have the ability to sting. When a female wasps stings a person the nest of wasps will be alarmed and also come to attack. Which statement correctly describes why the other wasps responded? A- They generally respond to auditory signals given off by other wasps. B- They generally respond to the presence of a threat instantly. C- They generally use chemicals they can produce from their body to emit signals that cause a response. D- They always use chemical emissions at low concentrations to cause a response. FRQ: A.) Provide two ways that organisms exchange information in response to an environmental change. B.) Provide 1 specific example for each way provided above. C.) Explain the significance of animals being able to exchange information.
Honey bees and wasps are both organisms that are known for their painful stings. Only female bees and wasps have the ability to sting. When a female wasps stings a person the nest of wasps will be alarmed and also come to attack. Which statement correctly describes why the other wasps responded? A- They generally respond to auditory signals given off by other wasps. B- They generally respond to the presence of a threat instantly. C- They generally use chemicals they can produce from their body to emit signals that cause a response. D- They always use chemical emissions at low concentrations to cause a response. FRQ: A.) Provide two ways that organisms exchange information in response to an environmental change. B.) Provide 1 specific example for each way provided above. C.) Explain the significance of animals being able to exchange information. Organisms are able to exchange information in response to environmental changes by the ability to communicate by chemical signals or auditory signals. One way that animals use chemical signals to exchange information is when they’re trying to mate in their natural environments. Certain pheromones that are released will help attract males to females that wouldn’t be able to locate each other otherwise. Organisms also use auditory signals to exchange information by composing songs to communicate those signals so that a response can be initiated. Exchanging information is significant among organisms because it determines how organisms communicate either dangers or signs of mating season. By them being able to exchange information based on environmental changes it helps them maintain their populations and aids their ability to respond to external or internal forces.
LO 3.46: The Student is able to describe how the vertebrate brain integrates information to produce a response SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: The Brain is able to integrate information due to the nervous system, which utilizes neurons and their structures to be able to process information from sight, scent, taste, scent or sound. The neuron will then send an action potential down its axon, which will connect to the dendrites of another neuron and allow for the continuation of the signal. The receiving neuron will either receive enough positively charged neurotransmitters to continue the response or receive more negative neurotransmitters and will cease the response. M.C. Question: Neurotransmitters are released in response to any event to cause us to react to it. Neurons will only activate if their charge increases to a certain amount, -55mv. Using the following graph, determine at what point the charge will stop. • Point 1 • Point 2 • Point 3 • Point 4
Learning Log/FRQ Style: A neuron suddenly is unable to send or receive any signals. Using the figure, determine what may have cause This. FRQ Answer: According to the figure, it is possible that the calcium has been prevented from leaving the neuron, causing a synaptic response to be impossible. Another probable cause is that the neurotransmitter may be unable to bind with the receptors of other neurons.
Answer Key- LO 3.46 Neurotransmitters are released in response to any event to cause us to react to it. Neurons will only activate if their charge increases to a certain amount, -55mv. Using the following graph, determine at what point the charge will stop. • Point 1 • Point 2 • Point 3 • Point 4
LO 4.18 The student is able to use representations and models to analyze how cooperative interactions within organisms promote efficiency in the use of energy and matter. SP 1.4 The student can use representations and models to analyze or solve problems qualitatively and quantitatively. Explanation: Many organisms are able to promote energy by having emergent properties that help the organism be most efficient. These emergent properties can be found in many different levels of life. A large-scale example can be found in the genetic make-up of a herbivore population. This biodiversity allows the population to be more flexible to problems, such as a disease that only infects certain genetically inferior deer. Also, this diversity helps increase capacity and saves resources. If different parts of the population eat different plants at different locations in their ecosystem, then the plant population is able survive and provide for other organisms as well. Overall, if the herbivores listed in the example above continue to feed and live in different areas, they are able to avoid competition. This will increase their quality of life and keep their population at a healthy level near the carrying capacity. It also helps keep the quality and number of plans they consume at a natural level. M.C. Question: How do plants have interactions within organelles that improve efficincy? Plants have a cell wall that keep too much water out, preventing it from bursting. Plants have a nucleus that stores DNA Chloroplasts convert sunlight to ATP, which is then used to feed the cell and help other organelles function. Plants have a cell membrane that it selectively permeable. L.L. Question: Explain one molecular way an organism has interactions that make it more efficient. Please be as specific as possible.
M.C. Question: How do plants have interactions within organelles that improve efficincy? Plants have a cell wall that keep too much water out, preventing it from bursting. Plants have a nucleus that stores DNA Chloroplasts convert sunlight to ATP, which is then used to feed the cell and help other organelles function. Plants have a cell membrane that it selectively permeable. L.L. Question: Explain one molecular way an organism has interactions that make it more efficient. Please be as specific as possible. An example of molecules in a cell working together to promote efficiency can be found in DNA. Each DNA strand is made up of nucleotides that are in a certain sequence based on the order of the hereditary material that is needed to make proteins for the cell. This polymer created is important because it controls which genes are expressed and which proteins are made. The proper order and binding of nucleotides in DNA helps make the cell more efficient, because the genes can only be properly expressed if the nucleotides are in the right order. So, DNA undergoes actions such as transcription and translation to help benefit the rest of the cell by expressing its desired genes.
Explanation: Prokaryotic cells have DNA that is concentrated in the nucleoid. Unlike prokaryotes, eukaryotes have DNA concentrated within a membrane-bound nucleus. Both cell types have a cytoplasm, or the area between the nucleus (for eukaryotes) or the interior cell between the plasma membrane. Within the cytosol of a eukaryotic cell, organelles perform specialized functions to help the cell efficiently perform metabolic processes. Internal membranes minimize competing interactions in the cell and increase surface area, creating more space for reactions to take place. Prokaryotes do not have organelles. Organelles include the endoplasmic reticulum, the nucleus, ribosomes, and golgi apparatus, although these are not organelles. Eukaryotes are also usually bigger than prokaryotic sizes. When looking at diagrams and models, the difference between prokaryotes and eukaryotes is plainly visible due to the presence or lack of organelles within the cell and the concentration of DNA. LO 2.14: The student is able to use representations and models to describe differences in prokaryotic and eukaryotic cells. MC Question: Biology majors at a local University are practicing using cell fractionation by centrifugation. The students centrifuged two types of cells. After an hour of centrifugation, one tube contained a very large pellet rich in microsomes as well as pieces of nuclei and ribosomes (cell type 1) and the other formed a noticeably smaller, lighter pellet rich in plasma membrane pieces and ribosomes (cell type 2). Which classification of each cell best supports the findings of the students? A. Cell type one is a prokaryote because it contains pieces of internal membranes. Cell type two is a eukaryote because ribosomes are present. B. The pellets do not provide enough information to classify each cell type. C. Both are prokaryotic because both pellets contain ribosomes and pieces of plasma membrane. D. Cell type one is a eukaryotic cell; the microsomesin the pellet are pieces of internal membranes, proving organelles are present. Cell type two is prokaryotic because organelles can not be detected. SP 1.4: The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. Free Response: Look at the models on the next slide. Classify each cell as prokaryotic or eukaryotic. For each cell, describe what makes each cell type different.Then, list an example of an organism that contains each cell type.
MC Answer: D is the correct answer because the pellet rich in inner membrane components proves that the cell is eukaryotic since eukaryotes have internal membrane-bound organelles. Eukaryotes also have nucleus, which explains the nuclear components found in the pellet. Prokaryotes only have a plasma membrane and ribosomes, and have no inner membrane-bound components. Since pellet 2 is much lighter and has only ribosome and plasma membrane pieces, it must be cell type 2. FRQ Answer: The far left model is a eukaryotic cell, and the right model is a prokaryotic cell. This is evident because the left model contains DNA within a bound membrane, the nucleus and has many apparent components bound by membranes within the cytoplasm. The right model shows DNA located, somewhat freely, throughout the cytoplasm as well as scattered ribosomes. Both cells are surrounded by a plasma membrane, but prokaryotes have no other membrane-bound components within the cell. Prokaryotic DNA is found in the cytoplasm , and the cell also facilitates ribosomes. Eukaryotic cells contain membrane-bound organelles which compartmentalize specific metabolic processes within the cell to reduce competition and increase efficiency. DNA is stored within the nucleus enclosed by the nuclear membrane. Examples of organelles include the Golgi apparatus, mitochondria, nuclear membrane, and endoplasmic reticulum. An example of a prokaryotic organism is bacterial cell and an example of a eukaryotic cell is a plant cell.
Learning objective 2.17 The student is able to evaluate data that show the effect(s) of changes in concentrations of key molecules on negative feedback mechanisms. S.p. 5.3 The student can evaluate the evidence provided by data sets in relation to particular scientific question. Explanation: Negative feedback in homeostasis describes a process of how bodily systems maintain their normal environments or states. Homeostasis describes the body's overall regulation of its internal systems. When changes in a state such as body temperature occur, negative feedback responses are triggered to bring the temperature back to its normal point. For example, when the body has to much of something specific, it can release a hormone to counteract it and bring the body back to its homeostatic state. Multiple choice question: Failure of negative-feedback mechanisms to maintain homeostasis A) may produce disease. B) occurs when blood pressure increases during exercise. C) can be corrected by stimulating positive-feedback mechanisms. D) cannot be corrected by medical therapy. E) all of these Free Response: The following chart contains different components found in the body the body. In the first column are the normal levels found in the body and in 2-4 are the levels at that time of the test. Triglycerides is one of the two chemicals that is not within its normal range. Please indicate the other that is not within its range and explain what negative feedback mechanism will be used to bring it back to the homeostatic rate.
ANSWER KEY LO 2.17 Multiple choice question: Failure of negative-feedback mechanisms to maintain homeostasis A) may produce disease B) occurs when blood pressure increases during exercise. C) can be corrected by stimulating positive-feedback mechanisms. D) cannot be corrected by medical therapy. E) all of these The following chart contains different components found in the body the body. In the first column are the normal levels found in the body and in 2-4 are the levels at that time of the test. Triglycerides is one of the two chemicals that is not within its normal range. Please indicate the other that is not within its range and explain what negative feedback mechanism will be used to bring it back to the homeostatic rate. The other molecule that is not within its range is glucose. It is way above the normal range. The negative feedback mechanism when blood glucose is high, the beta cells of the pancreas release insulin, which causes the glucose to be converted into glycogen.
LO 1.27: The student is able to describe a scientific hypothesis about the origin of life on Earth. SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: The Oparin and Haldane hypothesis of the 1920s postulated that that early earth conditions favored reactions that formed organic compounds from inorganic compounds. They suggested that the oxygen free atmosphere of the time, combined with an appropriate supply of energy, would be favorable conditions for numerous different organic compounds to synthesize. Later, in 1953, the Miller and Urey experiment took place to to test Oparin and Haldane’s hypothesis. The experiment simulated the conditions of early earth by sealing water, methane, ammonia, and hydrogen in a sterile array of interconnected flasks and tubes. One flask was filled halfway with liquid water, which was heated to induce evaporation. Sparks were fired between electrodes to simulate lightning, which traveled through the simulated atmosphere and the water vapor. The simulated atmosphere was then cooled so water would liquify and run back down to its original flask to repeat the process. Within two weeks in between 10% and 15% of the carbon was in the form of organic compounds, and thus life had been created within the early earth simulation. Multiple Choice Question: There is abundant evidence now of large Volcanic eruptions in early earth, which would have released carbon dioxide, nitrogen, hydrogen sulfide, and sulfur dioxide into the atmosphere. If these gasses had been included in the Miller Urey experiment, how would their results have differed? Organic molecules still would have been produced, only they would not have been racemic. B) A more diverse array of molecules would have been produced. Interactions between the volcanically introduced molecules and the originally used molecules would have prevented any organic molecules from synthesizing. Lightning would not have been needed to induce the synthesis. Learning Log/FRQ-style Question: Describe the Miller-Urey experiment and explain how it supports Abiogenesis (the process by which life arose from non life). Another theory about the origins of life postulate that life may have began first in the hydrothermal vents of the deep sea. Describe what this hypothesis might entail.
Answer Key- LO 1.27 Organic molecules still would have been produced, only they would not have been racemic. – Molecules are naturally synthesized to have Left and Right enantiomers, the extra gasses would not affect this. B) A more diverse array of molecules would have been produced. – the extra gasses would have provided more resources for a more diverse array of molecules to be synthesized. Interactions between the volcanically introduced molecules and the originally used molecules would have prevented any organic molecules from synthesizing.- Gasses cannot prevent other gasses from reacting with each other. Lightning would not have been needed to induce the synthesis.- Reactions require energy, lightning was the form in which this energy was provided in early earth. Describe the Miller-Urey experiment and explain how it supports Abiogenesis (the process by which life arose from non life). Another theory about the origins of life postulate that life may have began first in the hydrothermal vents of the deep sea. Describe what this hypothesis might entail. The Miller and Urey experiment simulated the conditions of early earth by sealing water, methane, ammonia, and hydrogen in a sterile array of Interconnected flasks and tubes. One flask was filled halfway with liquid water, which was heated to induce evaporation. Sparks were fired between electrodes to simulate lightning, which traveled through the simulated atmosphere and the water vapor. The simulated atmosphere was then cooled so water would liquify and run back down to its original flask to repeat the process. After some time, the mixture contained numerous organic compounds. Because the connected flasks and tubes simulated the conditions of early earth, and organic compounds were synthesized over the course of time, then the experiment was able to demonstrate and prove the theory that abiogenesis took place in early earth. The hydrothermal vent hypothesis postulates that life may have formed after deep sea vents released hot gasses from the center of the earth into cold water, which would have promoted the creation of peptides.
L.O. 4.20: The student is able to explain how the distribution of ecosystems changes over time by identifying large-scale events that have resulted in these changes in the past. S.P. 6.3: The student can articulate the reasons that scientific explanations and theories are refined or replaced. Explanation: The distribution of ecosystems has altered due to large-scale events that significantly impact the environments in which populations live. The theory of continental drift states that all continents were once joined in a subcontinent called Pangea and over a vast period of time drifted apart to their current locations. Current scientists credit this movement to Earth’s tectonic activity. With the movement of landmasses, ecosystems separated over time and organisms within the ecosystems adapted to changing environmental conditions. Other events include: El Niño, the abnormal warming of the ocean’s waters in the eastern tropical Pacific; the mass extinction of populations (i.e. dinosaurs) due to a massive asteroid/comet impact; volcanic eruptions; climate change; and sea level change. Scientists continue to refine theories of the effects of large-scale events as time reveals further consequences and provides more advanced technological mediums that can be used for effective research. M.C. Question: The Galápagos Islands provide habitats for numerous distinctive organisms. The Marine Iguana is one organism that can only be found in that area due to the fact that: a) sympatric speciation led the organisms to consciously adapt to changing environmental conditions b) temporal isolation resulting from the differences in time zones on the islands c) evolution of these organisms occurred at a considerably lower rate d) allopatric speciation has occurred since the islands have been isolated from other landmasses for a vast period of time Learning Log/FRQ-style Question: Large-scale phenomenon alter the distributions of ecosystems in significant ways. • Explain why continental drift has resulted in the allopatric speciation of the populations of various organisms in terms of habitat isolation and behavioral isolation. b) How could El Niño alter ecosystems?
The Galápagos Islands provide habitats for numerous distinctive organisms. The Marine Iguana is one organism that can only be found in that area due to the fact that: • Sympatric speciation occurs as chromosomal changes and non-random mating alter gene flow. Rather, a major change in the genome of the subpopulation reproductively isolates the subpopulation from the parent population. The organisms, in this case the Marine Iguana, do not “consciously” speciate and evolve. • Temporal isolation refers to a reproductive barrier in which populations breed at different times. While the islands are contained in one time zone, temporal isolation refers to the internal reproductive time of the Marine Iguana. • The rate of evolutionary change does not explain why the Marine Iguana is distinct to the Galápagos Islands; it suggests only why the Marine Iguana is what it is. • Allopatric speciation occurs when a geographic barrier isolates a population and blocks gene flow. The geographic barrier for the Galápagos Islands is a vast amount of water that was created by continental shift and the development of the islands from tectonic activity. Large-scale phenomenon alter the distributions of ecosystems in significant ways. • Explain why continental drift has resulted in the allopatric speciation of the populations of various organisms in terms of habitat isolation and behavioral isolation. Continental drift results in the separation of landmasses by massive bodies of water serving as a geographic barrier for species. Tectonic activity associated with continental drift has led to the development of mountains and other landforms that isolate populations through habitat isolation, thus altering the distribution of ecosystems. Behavioral isolation occurs as natural selection favors certain behaviors in different ecosystems. The ecosystems are determined from the landmasses and landforms that have resulted from continental drift. For example, a population of birds on one side of a mountain close to a body of water attracts mates by splashing in the water. A similar population of birds on the side of the mountain closest to the forest attracts mates by flying into the trees and showing off their wings. • How could El Niño alter ecosystems? Because El Niño is a warming of the waters in certain areas of the Pacific Ocean, the affected ecosystems will be altered as environmental conditions change. Many fish populations do not survive in warmer waters. There is a reduction in the supply of chemical nutrients to the euphotic zone, thus causing a decline in phytoplankton production and a shift in the functioning of ecosystems.
LO 1.2: The student is able to evaluate evidence provided by data to qualitatively and quantitatively investigate the role of natural selection in evolution. SP 2.2 and 5.3, which state that the student can properly use mathematics (2.2) and can perfrom data analysis and evaluation of evidence (5.3). Students can look at data that explains evolution, such as fossil records, biogeography, and comparative anatomy, biochemistry and embryology, and evaluate this evidence to understand how evolution occurs, and the roles these types of data play. Homologous structures provide solid evidence to support evolution because a) They have the same function, so they show how species in similar environments have similar adaptations. b) They have no function, because they evolved from ancestors eating a different diet. c) They have the similar form but different function, showing a common ancestor. d) They are new body parts that have evolved due to a change in climate, diet, or other factors. Explain how ages of fossils are measured, and how this data can help us understand evolution. This image shows the homologous structures of 5 different organisms. They have very similar structure, but different functions.
Multiple choice answer: C, because they are the same in structure but different in function, which defines homologous structures. • Answer choices A and B are analogous and vestigial structures, respectively, and although they do provide evidence for evolution, they are different from homologous structures. D is not a correct response because it is not something that is recognized to aid in proving evolution occurs. • FRQ answer: Radiometric and relative dating are used to measure fossil age. Through radiometric dating, you compare the amount of an element in the fossil like carbon and its decay rates. In relative dating, you look at factors like the fossil’s position in layers of rock and then compare this to how old fossils in other layers are. This can help us understand evolution by seeing how long it took for a species to evolve, and what types of structures ancestors of certain species had along with how long ago they lived.
LO 3.29: the student is able to construct an explanation of how viruses introduce genetic variation in host organisms. SP 6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices. Explanation: Viruses contain either DNA or RNA and incorporate their genetic material into host cells in several different ways. Some viruses, called phages, attach to a host cell and inject their DNA which is then integrated into a bacterial chromosome. Another type of virus, called a retrovirus, fuses with the plasma membrane of a host cell and viral proteins and RNA are released. Reverse transcriptase catalyzes the viral RNA to DNA and the DNA enters the nucleus and is incorporated into the host cell’s DNA. Once viruses integrate their DNA or RNA into the host cell’s genome, they can produce genetic variations in the host organisms. For example, bacteria require either mutations or the incorporation of foreign DNA for there to be genetic variation in the population. Phages help provide this foreign DNA by incorporating pieces of a host chromosome from one bacterial cell to another recipient bacterial cell. The new recombinant cells are genetically varied from the other bacterial cells in a population. Which of the following does not relate to the incorporation of genetic material from viruses to host cells? Transduction Lysogenic Cycle Conjugation Reverse Transcriptase Explain how viruses transfer genetic information from one bacterial cell to another by the process of transduction.
Which of the following does not relate to the incorporation of genetic material from viruses to host cells? • Transduction • Lysogenic Cycle • Conjugation • Reverse Transcriptase In the lytic cycle, when a phage attaches to a bacterial cell, new phage DNA and proteins are synthesized in order to assemble new phages. Sometimes, bits of the degraded host cell’s DNA is incorporated into the assembly of new phages. When the cell lyses and releases phages, they carry bits of the host cell’s DNA. In the process of transduction, phages release their DNA, which also contains genes from the original bacteria cell, into the new recipient bacteria cell. The transferred DNA can then recombine with the genome of the recipient, thus creating a recombinant cell. Transduction is an important process that causes genetic variation in bacteria.
