4 th Review Project collection

4th Review Project collection

L.O. 1.9: The student is able to evaluate evidence provided by data from many scientific disciplines that support biological evolution. SP 5.3: The student can evaluate the evidence provided by data sets in relation to a particular scientific question. Explanation: Fossils can be dated by a variety of methods that provide evidence for evolution; these include the age of the rocks where a fossil is found, the rate of decay of isotopes including carbon-14, the relationships within phylogenetic trees, and the mathematical calculations that take into account information from chemical properties and/or geographical data. Morphological homologies represent features shared by common ancestry. Mathematical models and simulations such as sequence data sets and phylogenetic trees can be used to illustrate and support evolutionary concepts. Learning Log/FRQ-style Question: (a) DESCRIBE IN DETAIL the method by which separate species form from common ancestry. (b) Based on the data in the table below, DRAW a phylogenetic tree that reflects the evolutionary relationships of the organisms based on the differences in their cytochrome c amino-acid sequences. M.C. Question: The diagram below shows the bones in the forelimbs of three different organisms. The table to the left shows the Number of Amino Acid Differences in Cytochrome c Among Various Organisms. Which of the following statements is true based on the diagram above? A. These organisms evolved from common ancestry. B. The Human evolved directly from and is most closely related to the Bat. C. These organisms share the same genetic material and show no variation in amino acid sequences. D. Over the course of their lives, these organisms adapted to their environment by forming the forelimbs shown above.

Answer Key: L.O. 1.9 • M.C. Question: • Which of the following statements is true based on the diagram above? • These organisms evolved from common ancestry due to their similar homologous structures. • B. The Human’s forelimb is more similar to the Whale’s and it evolved from a common ancestor, not the Bat species itself • C. These organisms show similar homologous structures, but are NOT genetically identical • D. This theory is derived from a Lamarckian Concept, which we know to be false Learning Log/FRQ-style Question: (a) DESCRIBE IN DETAIL the method by which separate species form from common ancestry. (b) Based on the data in the table below, DRAW a phylogenetic tree that reflects the evolutionary relationships of the organisms based on the differences in their cytochrome c amino-acid sequences. (a) (b)

LO 4.13: The student is able to predict the effects of a change in the community’s populations on the community. SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. Explanation: In communities populations interact with each other through predator-prey, symbiotic (both species benefit), communalism (one species benefits, the other is unharmed), and parasitic (one species benefits, the other is harmed) relationships. Populations can also be effected by the introduction of species. Graphical representation of population growth patterns and interactions are common. Limits to growth outside of the species’ relationships include density-dependent and density-independent factors. These factors can be graphically shown in a logistics model which includes a carrying capacity (K) line. A community can be measure mathematically by calculating species composition and species diversity. In the before mentioned graphs and models can be used as grounds for claims and predictions of natural phenomena. M.C. Question: In an ecosystem frogs and flies have a predator-prey relationship. This ecosystem undergoes a massive cold wave in the middle of the fly mating season, killing most of their eggs. Based on the information, what will happen to the frog population in this community? • The frog population will decrease because the flies will migrate away. • The frog population will remain stable because the fly population will increase. • The frog population will decrease because the fly population decreased causing more competition among the frog population. • The frog population will move to another ecosystem because there is not enough food for all the frogs. Learning Log/ FRQ-style Question: A particular community has a problem with frogs. The frogs are too numerous and are eating most of the flies, cutting down of the food supply for spiders. • What will happen to the spider population if nothing changes? Explain why. • A new species is introduced into the community. This invasive species will eat the frogs and decrease their population within the community. What could happen to the community because of the addition of the foreign species? • The new species is not only eating the frogs, but also eating the keystone specie for that community. What will happen to the community if the invasive species continues to decrease the population of the keystone species?

ANSWER KEY-LO 4.13 In an ecosystem frogs and flies have a predator-prey relationship. This ecosystem undergoes a massive cold wave in the middle of the fly mating season, killing most of their eggs. Based on the information, what will happen to the frog population in this community? • The frog population will decrease because the flies will migrate away. • The frog population will remain stable because the fly population will increase. • The frog population will decrease because the fly population decreased causing more competition among the frog population. • The frog population will move to another ecosystem because there is not enough food for all the frogs. The frog population will decrease so choices B and D can be eliminated first. Based on the information given C is the correct answer because with a decrease in the fly population there will be an increase in competition among frogs causing the overall frog population to go down. If prey population decreases, so will the predator population. A particular community has a problem with frogs. The frogs are too numerous and are eating most of the flies, cutting down of the food supply for spiders. • What will happen to the spider population if nothing changes? Explain why. • A new species is introduced into the community. This invasive species will eat the frogs and decrease their population within the community. What could happen to the community because of the addition of the foreign species? • The new species is not only eating the frogs, but also eating the keystone specie for that community. What will happen to the community if the invasive species continues to decrease the population of the keystone species? a) The spider population will decrease in this community. This will happen because they are being outcompeted for their food resource by the excess amount of frogs. b) The foreign species could take over the community. Since the species is foreign, it will have no natural predators and any predators it may have in its new community won’t have adapted with the invasive species. The foreign species will have adapted to become more suitable for their environment, allowing them to get away from their new predators easily. The invasive species has essentially no predators and a practically unlimited food source. The new species will dominate the community and over run it with its massive population eventually. c) Keystone species hold the community together. If they die out so does the community. The keystone species is essential to the survival and growth of the community. If the invasive species continues to decrease the population of the keystone species the community will crash and no longer exist.

LO 4.26: The student is able to use theories and models to make scientific claims and/or predictions about the effects of variation within populations on survival and fitness. • SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. • Explanation: Genetic variation is introduced into a population mainly through mutations, which are a permanent changes in the chemical structure of a gene. Genetic variation affects a population’s ability to respond to changes in the environment. Species with variations have a higher chance of survival because they can adapt more easily to the changing environments (as opposed to species with little or no variation). Allelic variations can be modeled by the Hardy-Weinberg equations (p + q = 1; p2 + 2pq + q2 = 1; where p= frequency of the dominant allele in the population, q= frequency of the recessive allele in the population, p2= percentage of homozygous dominant individuals, q2= percentage of homozygous recessive individuals, and 2pa= percentage of heterozygous individuals). • M.C. Question: Based on your knowledge of the Hardy- Weinberg Theorem, which of the following best describes some requirements of a population that is in equilibrium? • Population must be large and natural selection must occur • Population cannot be isolated and natural selection must occur • No net mutations can occur and there must be random mating • Population must be small and isolated • A & c • Learning Log/ FRQ-style Question: • Explain and give 5 examples of how genetic variations affects the population’s ability to survive. • Mutations are the primary source of variation for all life forms. Describe at least 3 ways these mutations can come about.

M.C. Question: Based on your knowledge of the Hardy- Weinberg Theorem, which of the following best describes some requirements of a population that is in equilibrium? • Population must be large and natural selection must occur • Population cannot be isolated and natural selection must occur • No net mutations can occur and there must be random mating • Population must be small and isolated • A & c Learning Log/ FRQ-style Question: • Explain and give 5 examples of how genetic variations affects the population’s ability to survive. • Genetic variation ensures that no single event can cause the extinction of that species • Traits that are favored by natural selection is passed onto offspring • Reproduction of the fittest • Examples: • A sudden drop in temperature results in survival of more hairy animals, therefore furry trait is favored • A plague hits the U.S., those who are immune will reproduce and pass off the immunity to their children • Mutations are the primary source of variation for all life forms. Describe at least 3 ways these mutations can come about. • Duplications (removes a chromosomal segment) • Deletions (repeats a segment) • Inversions (reverses a segment within a chromosome) • Translocations (moves a segment from one chromosome to another non-homologous one)

LO 4.11: The student is able to justify the selection of the kind of data needed to answer scientific questions about the interaction of populations within communities. SP 1.4: The student can use representations and models to analyze situations or solve problemsqualitatively and quantitatively. SP 4.1: The student can justify the selection of the kind of data needed to answer a particular scientific question. Explanation:Communities are made of populations that interact with the environment and with each other. A population is defined as a group of individuals of one species living in one area. Communities are characterized by how diverse they are and how dense they are. Species diversity has two components. One is species richness, the number of different species in the community, and the other is species evenness, referring to the relative abundance of the different species. The main interactions of populations within a community fall into the categories of: competition, predation, parasitism, mutualism, and commensalism. In nature, there are two related outcomes, besides extinction, if two species inhabit the same niche and therefore compete for resources; one of the species will evolve through natural selection to exploit different resources, called resource partitioning, or evolution through natural selection will cause a divergence in body structure in order to exploit different resources, e.g. the Galapagos finches studied by Darwin, called character displacement. Predation – one animal eating another animal or plants – has stimulated the evolution of defense mechanisms in both plants and animals; plants have evolved physical attributes, such as spines and thorns, and chemical poisons, and animals have evolved active defenses and passive defenses such as camouflages and mimicry. Mutualism, commensalism, and parasitism are symbiotic relationships in a community where both organisms benefit, one is benefited while one is neither harmed nor benefited, and one organism benefits while the other is harmed, respectively. In regards to assessing data to analyze interactions of populations within communities, the data provided, regarding to any aspect of interaction e.g. evolution of differing body structure resulting from competition, would have to fall under a experimentally acceptable time frame, usually an extensive time period, show relevant qualitative and quantitative data to the topic provided, and express measurements in a form that can be compared and analyzed effectively. A predator/prey spreadsheet model, graphical representation of field data, or illustrations of symbiotic relationships could be credible forms of data. Multiple Choice Question: Based on the data below, it could most definitively be assumed that: a) P. aureliaand P. caudatum have a mutualistic relationship b) P. caudatum has been transformed with antibiotic-resistant plasmids c) P. caudatumand P. aurelia inhabit the same environmental niche, and therefore are competing for resources; P. aureliais outcompeting P. caudatum d) P. aureliais exhibiting a form of mimicry FRQ-style question:The following graph illustrates the predator-prey dynamics between the snowshoe hare and the lynx in a certain community. If a disease decimated the lynx population, predict the effect this will have on the hare population over the next 20 years. Justify your prediction and justify the selection of the types of data that would need to be collected to document a change in the hare population.

Answer Key – LO 4.11 Based on the data below, it could most definitively be assumed that: a) P. aureliaand P. caudatum have a mutualistic relationship b)P. caudatum has been transformed with antibiotic-resistant plasmids c) P. caudatumand P. aurelia inhabit the same environmental niche, and therefore are competing for resources; P. aureliais outcompeting P. caudatum d) P. aureliais exhibiting a form of mimicry • Answer choice C is the correct answer. Choice A is incorrect because the data shown in graph “c” does not support the assumption that both species have a mutualistic relationship; P. caudatum actually has extremely low, declining growth when grown with P. aurelia, showing that both organisms are not being benefited by the presence of the other, the definition of mutualism. Choice B is incorrect because the graphs provide no evidence that would indicate that P. caudatum had been transformed with antibiotic-resistant plasmids, since there is no data given regarding bacteria with or without plasmids, or the presence of an antibiotic; additionally, if P. caudatum was antibiotic resistant, it would be outgrowing the other bacteria, not experiencing extremely low growth. Choice D is incorrect because this quantitative data provides no direct evidence that P. aurelia is exhibiting any form of mimicry; observations and qualitative data would be needed to prove this. Choice C is the correct answer because signs of competition are shown by the graphs; over the same period of time, both P. aurelia and P. caudatum showed approximately equally high growth, but when placed together, P. aurelia flourished while P. caudatumexperienced little growth that started to decline around day 7. This is evidence that both bacteria are competing in a shared habitat, and P. aurelia is outcompeting P. caudatum, possibly leading to its eventual extinction. FRQ-style question Answer Key: • Predict the effect this will have on the hare population over the next 20 years - justify your prediction • Hare population will increase - due to reduced predation from the lynx • Hare population will experience exponential growth and overshoot it’s carrying capacity, then crash – due to reduced predation from the lynx; since so few lynx now exist in the community, this virtually complete lack of predation would cause a surge in the hare population, but this extreme increase in the population of hares would lead them to grow beyond the limits of their resources (carrying capacity) and then crash, experiencing a loss in population due to a shortage of resources • Hare population will remain stable – even though most lynx have been eliminated as the main predators of the hare, another population of predators in the community may replace the lynx in the predator portion of the food chain of the hare and keep the hare population at the same level as it was with the presence of the lynx population • Justify the selection of the types of data that would need to be collected to document a change in the hare population • Predator/prey spreadsheet model – if a new predator has arisen or if the hare population is affected by the few remaining members of the lynx population • Quantitative data – documenting the increase/decrease in the number of hares in this community - in order to compare the numbers of hares from before the lynx epidemic and after to analyze a change in the number of hares • Qualitative data – less important, but notable if there are any physical characteristic that have evolved after the virtual removal of the hare as a predator • Graph showing the quantitative changes in the number of hares in the population over the 20-year period

Explanation: In all cell signaling there is a reception, this targets cell’s detection of signaling molecule coming from the outside cell then the signaling molecule binds to a receptor protein. Transduction then occurs which is when the binding of signaling molecules change receptor proteins, these signals can bring a certain cellular response. The response to a specific cellular response is then triggered. LO 3.31 The student is able to describe basic chemical processes for cell communication shared across evolutionary lines of descent. • M.C Question • What is the order in which cell communication occurs? • Transduction • Reception • Response • III,I,II • I,II,III • II,I,III • II,III,I SP 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. Learning Log/FRQ-Style Question: In Cell to cell communication across evolution there are bound to be some errors. In this case assume that the transduction function in cell communication is unresponsive. What would happen in this circumastance?

ANSWER KEY – LO 3.31 Learning Log/FRQ style question: If transduction was unable to function properly then the signaling molecule can’t change the receptor proteins. If this happened then all cellular responses wouldn’t be able to function. What is the order in which cell communication occurs? Transduction Reception Response III,I,II I,II,III II,I,III II,III,I

LO 2.11: The student is able to construct models that connect the movement of molecules across membranes with membrane structure and function. SP 1.1: The student can create representations and models of natural or man-made phenomena and systems in the domain. SP 7.1: The student can connect phenomena and models across spatial and temporal scales. SP 7.2: The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. Explanation: Cell membranes act as a divider between the external and internal environment of a cell. It is selectively permeable to allow certain molecules to pass through. Hydrophilic substances pass through the embedded channel and transport proteins, while water moves through aquaporins in the membrane. Along with embedded proteins and aquaporins, the cell membrane also consists of phospholipids, glycoproteins, cholesterol, and glycolipids. The hydrophilic phosphate head of the phospholipids faces the external environment or inside of the cell. The hydrophobic tails make up the middle of the membrane where cholesterol also resides. M.C. Question:If you were to freeze-fracture a cell membrane, what structures would you see on the inside? Glycolipids Carbohydrates Integral Proteins Peripheral Proteins Learning Log/FRQ-style Question: Explain the effect that saturated and unsaturated hydrocarbon tails have on the fluidity of a cell membrane. Also include the effect of cholesterol and temperature, as it relates to cholesterol, in the cell membrane.

Answer Key LO 2.11 M.C. Question: If you were to freeze-fracture a cell membrane, what structures would you see on the inside? C) Integral Proteins would be the only structure visible because the middle of the membrane would be presented and A, B, and D are all found on the exterior of the membrane. Learning Log/FRQ-style Question: Explain the effect that saturated and unsaturated hydrocarbon tails have on the fluidity of a cell membrane. Also include the effect of cholesterol and temperature, as it relates to cholesterol, in the cell membrane. If the membrane is made up of unsaturated hydrocarbons, it is more fluid because it’s not packed together tightly due to the kinks n the phospholipid’s tails allowing it more room to move. Saturated hydrocarbon tails cause the membrane to be more viscous because they are packed tightly together. At regular temperatures, cholesterol reduces fluidity. At low temperatures, cholesterol stop solidification of the phospholipids. At high temperatures, the cholesterol settles and the membrane solidifies.

LO 3.45: The student is able to describe how nervous systems transmit information. SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: The nervous system in one’s body is able to transmit information throughout the body via the use of neurons. Neurons are cells with a soma that has dendrites on one side, with axons on the other. The dendrites are used to receive signals through the process while the axons are the part of the neurons that will send the signals. The axons will form a synapse with dendrites of another neuron, across which the connection will fire a signal from neuron to neuron. Neurotransmitters are the next big part of the process, and are used to send a specialized signal to the dendrites of the next neuron in the chain. They find the ‘active sites” of each and are very specific to where they will bind. No other neurotransmitter will be able to bind to this location and perform the function that is needed to progress the message down the chain of neurons. The neurotransmitters can then be used to excite or inhibit the reaction, and if excited, the reaction and signaling will continue down the chain. Once excited, it created the action potential, or the firing of the synapse across the axons. This is done once the cells can create a charge that is electric enough to force the signal to fire down the chain of neurons. The cell becomes more negative inside than on the outside to make the resting potential. Once the signal reaches the next axon terminal, positively charged ions are created which activate the neurotransmitters on the receiving neuron. The binding occurs and the new neurotransmitters are targeted and released outside the new neuron. The cycle will repeat for each receiving neuron all the way down the line from neuron to neuron, until the information has been transmitted all throughout the nervous system. M.C. Question: If the neurotransmitter becomes inhibitory rather than excitatory, which of the following will occur? The message will force it’s way past the receiving neuron regardless of it being inhibitory or excitatory. The receiving neuron will be unable to pass the message on down the chain, so the reaction will be stopped until the neurotransmitter can become excited again by the action potential with enough built up excited neurotransmitters. The inhibition of the neurotransmitter will cause the message to turn around and move in reverse back down the chain until it reaches the neuron the signal was created at. The neurotransmitter that has undergone inhibition will undergo apoptosis, or suicide of the cell that the message will never be passed on, and more neurons would be needed to replace it in the chain for any signals to be transmitted. Learning Log/FRQ-Style Question: In order for the message in a nervous system to be transmitted from neuron to neuron, an electric charge is created to force the action potential to activate the synapse and transmit the message. Why does the cell more become more negative on the inside? Are there positive charges anywhere on the cell? Why is it essential for the negative charge to activate the action potential prior to the end of the chain? Draw and label a diagram of the charged neuron.

Answer Key – LO 3.45 M.C. Question: If the neurotransmitter becomes inhibitory rather than excitatory, which of the following will occur? The message will force it’s way past the receiving neuron regardless of it being inhibitory or excitatory. The receiving neuron will be unable to pass the message on down the chain, so the reaction will be stopped until the neurotransmitter can become excited again by the action potential with enough built up excited neurotransmitters. The inhibition of the neurotransmitter will cause the message to turn around and move in reverse back down the chain until it reaches the neuron the signal was created at. The neurotransmitter that has undergone inhibition will undergo apoptosis, or suicide of the cell that the message will never be passed on, and more neurons would be needed to replace it in the chain for any signals to be transmitted. Learning Log/FRQ-Style Question: In order for the message in a nervous system to be transmitted from neuron to neuron, an electric charge is created to force the action potential to activate the synapse and transmit the message. Why does the cell more become more negative on the inside? Are there positive charges anywhere on the cell? Why is it essential for the negative charge to activate the action potential prior to the end of the chain? Draw and label a diagram of the electrical movement down the chain. The cell needs to change its charge during the reaction in order to progress the message further along down the chain of neurons. To do this, it must change from being positively charged to negatively charged in order to activate the electric potential to fire the action potential. The charged ions both in and out of the cell are switched and the electricity is created to speed the message and reaction further along the chain to the dendrites of the next, or receiving neuron. The positively charged ions do not disappear, but simply are placed on the outside of the cell as the negative ions create more electricity on the inside of the cell than the positively charged ions do. The positive ions will undergo another change later on towards the end of the reaction, completing the trip from one neuron to another. The electricity causes the action potential to be sent across the membrane due to the negatively charged cell. This allows for positive sodium ions to come both into and out of the cell, which keeps the cell negatively charged for the rest of the reaction. The negative charge is necessary in order for the final switch that occurs in the transmission because the positive sodium ions are what enter the next neuron and cause it to be positively charged at the start of the next reaction, which is a repeat of the cycle from the previous neurons. Labeled Diagram

LO 1.3 The student is able to apply mathematical methods to data from a real or simulated population to predict what will happen to the population in the future. SP 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. Explanation: Charles Darwin’s Theory of Natural Selection states that competition for limited resources leads to varied survival rates. Those that survive are more willing to pass their phenotype to their offspring, thus passing the trait on. This is also known as evolutionary fitness. Species are so different because of their diverse gene pool. This is a result of genetic variation and mutations. The environment and random events can also affect survival, especially in smaller populations. There is a scenario where allele frequency wont change and its called Hardy-Weinberg Equilibrium; It calls for 5 requirements to be met: (1) a large population size, (2) absence of migration, (3) no net mutations, (4) random mating and (5) absence of selection. Mathematics can be used to show changes in allele frequency by using the Hardy-Weinberg equations as well as using graphical analysis of frequencies in a population over time. M.C. Question: According to the graph to the right, what is the average rate of population growth (individuals/day) from day 3 to day 7? Also, at what population is the Population Size increasing the fastest? • 350 Individuals/day; 600 Individuals • 300 Individuals/day; 500 Individuals • 400 Individuals/day; 400 Individuals • 200 Individuals/day; 500 Individuals Learning Log/FRQ-style Question: A population of squirrels live in a newly created park. Initially, there are 15 squirrels. After 3 years, there are 40 squirrels. 3 years after that, there are a total of 42 squirrels. In two years, the park will begin to cut down trees as well as migrate 10 new squirrels.; What do you expect to happen to the population the next year? Why? What do you think will happen to allele frequency? Explain in terms of Hardy-Weinberg’s five requirements.

ANSWER KEY – LO 1.3 According to the graph to the right, what is the average rate of population growth (individuals/day) from day 3 to day 7? Also, at what population is the Population Size increasing the fastest? A) 350 Individuals/day; 600 Individuals B) 300 Individuals/day; 500 Individuals C) 400 Individuals/day; 400 Individuals D) 200 Individuals/day; 500 Individuals The Answer is D because average rate is found by Δy/ Δx so at x=3, y= 200, and at x=7, y=1000. (1000-200)/(7-3) = 800/4 = 200. The rate is fastest at 500 Individuals because it is half the carrying capacity (1000) A population of squirrels live in a newly created park. Initially, there are 15 squirrels. After 3 years, there are 40 squirrels. 3 years after that, there are a total of 42 squirrels. In two years, the park will begin to cut down trees as well as migrate 10 new squirrels.; What do you expect to happen to the population the next year? Why? What do you think will happen to allele frequency? Explain in terms of Hardy-Weinberg’s five requirements. You would expect the population to decrease even though 10 new squirrels are added because you can deduce that the population is reaching its carrying capacity under the original environment since the population barely increases from year 3 to 6. This would cause an increase in competition for resources and a lower survivability for some squirrels (more death). You would also expect the allele frequency to fluctuate a lot. This scenario would not be in Hardy-Weinberg equilibrium since: (1) the population is small, (2) there is migration, and (3) natural selection will be involved because of limited resources as stated above.

LO 3.23 The student can use representations to describe mechanisms of the regulation of gene expression.SP 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. Summary- Genes are expressed in numerous different ways in different types of organisms. Sometimes a signal determines how a gene is expressed, such as a morpheme. Other times, the signal dictates how a gene functions and how it is used within the cell or organism. Cytokines are among peptide and protein regulators and they play an important role in cell proliferation and differentiation. Cytokines are influential in the majority of growth factors, which have to be present in extracellular fluid in order to allow for usual growth, division, and develop. There is a pheromone in yeast that is crucial in haploid cell differentiation that prepare it for events such as mating. Cyclic AMP (also knows as cAMP) is a second messenger, which means that it carries a signal between cells. cAMP, specifically, usually activates protein kinase A, which then phosphorylates various other proteins. A lack of cyclic AMP levels in bacteria result in a failure to activate production of the enzymes necessary to metabolize lactose by lactase. (See Figure 2) In animals, the SRY gene determines gender by coding for the Y gene. Mutations in SRY can cause several different sexual issues. Plants produce ethylene when under stressful situations or as a response to high concentrations of externally applied auxin. The effects of ethylene include mechanical stress responses, apoptosis (programmed cell death), leaf abscission, and fruit ripening. Ethylene may also inhibit root elongation due to high auxin levels. A morphogen is essentially a signal that tells a cell its position and role within a body. The concentration of the morphogenis what actually determines the duty of a cell or, more commonly, a cluster of cells. There is a gene called p53 that, once activated(perhaps by damage to DNA), functions as an activator for several genes. When DNA is damaged, p53 can turn on DNA repair genes. When DNA cannot be repaired, p53 activates “suicide” genes that promote apoptosis. There is one severe issue that can occur if p53 is missing, damaged, or otherwise defective. Without the suicide genes, damaged or harmful cells can reproduce and form a tumor that may or may not be cancerous. P53 is certainly not the only factor in cancer formation, but it is certainly a major one. HOX genes are genes that are present in the embryo and determine that part of the developing organism will develop into the separate parts of the grown body. (See Figure 1 below for further explanation.) FIGURE 1 Learning Log/ Free Response Style Question- Choose two of the following, define the term, explain its function, and state whether it affects the function of a gene or the expression of a gene. a) cyclic AMP (cAMP) b) ethylene c) morphogen Figure 1 depicts how HOX genes work. The green section on the bottom picture (developing embryo) shows the placement in the fly’s genome, which is also the same anterior section on the adult Drosophila. Multiple Choice- Which of the following statements best explains how p53 normally functions and how it can be detrimental to a cell? a) p53 can code for many different genes such as suicidal genes, which can kill healthy cells unnecessarily. b) p53 acts as an activator for most genes that repair and sometimes kill defective cells. Without p53 working properly, existing damaged cells taking up room is a primary issue c) p53 enhances unnecessary cell production, replicating cells and resulting in cancer. d) p53 acts as an activator for numerous different genes such as suicidal genes. A problem arises when p53 isn’t functioning properly and there is no signal for harmful cells to undergo apoptosis, resulting in a tumor. FIGURE 2 Figure 2 shows the importance of cAMP. Observe that Adenylyl Cyclase could not send the signal to Kinase A without cAMP. It is a required step in cell communication.

Multiple Choice Answer-Which of the following statements best explains how p53 normally functions and how it can be detrimental to a cell? a) p53 can code for many different genes such as suicidal genes, which can kill healthy cells unnecessarily. b) p53 acts as an activator for most genes that repair and sometimes kill defective cells. Without p53 working properly, existing damaged cells taking up room is a primary issue c) p53 enhances unnecessary cell production, replicating cells and resulting in cancer. d) p53 acts as an activator for numerous different genes such as suicidal genes. A problem arises when p53 isn’t functioning properly and there is no signal for harmful cells to undergo apoptosis, resulting in a tumor. Learning Log/ Free Response Style Question Answer- Choose two of the following, define the term, explain its function, and state whether it affects the function of a gene or the expression of a gene. a) Cyclic AMP (cAMP) Cyclic AMP is a signal molecule (ligand) that binds to an existing inactive CAP, making it active. A CAP is a transcription activator that binds to a specific site at the upstream end of the lac promoter. An active CAP molecule allows for RNA polymerase to bind and transcribe large amounts of RNA. Therefore, the attachment of CAP to the promoter directly stimulates gene expression. b) Ethylene Ethylene is a plant hormone that comes in the form of a gas. It ripens fruit, programs cell death (apoptosis), and responds to mechanical stress and leaf abscission (loss of leaves). Added ethylene makes a plant stem shorter, thicker, and have more curvature. This is called the triple response. Apoptosis calls for new gene expression. Newly formed enzymes break down chemical components of the leaf, and a burst of ethylene is almost always associated with this process. Leaf abscission is when leaves fall of the trees. When this happens, there is a change in ethylene and auxin levels. The aging leaf produces less and less auxin, which makes it more susceptible to ethylene which digests cellulose and other components of the cell walls. Lastly, ethylene ripens fruit. It breaks down the cell walls of the fruit which makes it softer and the starches and acids are converted to sugars, sweetening the fruit. For each of these products of ethylene, it is apparent that ethylene causes changes in the expression of a gene. c) Morphogen Morphogens are substances that establish the axes of an embryo, along with various other features. One such morphogen is bicoid. It determines which end is the anterior and which is the posterior of the embryo. This is obviously a substance that determines the function of a gene. If a morphogen is distorted in any way, the function of the cell it is a part of will change drastically, usually for the worse.

LO: 3.19- The student is able to describe the connection between the regulation of gene expression and observed differences between individuals in a population.SP: 7.1- The student can connect phenomena and models across spatial and temporal scales Explanation: Gene expression can be regulated by sequences of DNA, such as promoters, terminators, and enhancers, which, when interacting with regulatory proteins, cause the transcription, end transcription, and increase transcription of certain genes respectively. A gene itself can help regulate; a regulatory gene codes for a regulatory section of RNA or a regulatory protein, which then can interact with the promoters, terminators, and enhancers. In viruses and bacteria, regulation is simple. Inducers and repressors interact with these proteins and sequences to either have positive (transcription) control or negative (no transcription) control. In both cases, a protein binds to the DNA to control transcription positively or negatively. In eukaryotes, regulation is more complex. Transcription factors bind to the DNA sequences or regulatory proteins. They are either activators (increasing transcriptions) or repressors (decreasing transcription.) The amount and combination of transcription factors present causes variation in the degree to which the gene is expressed. Within organisms with similar genetic codes, regulation causes some phenotypic variance. Free Response: Phenotype can be altered by gene regulation. Describe the events in Steps 1-4. What will happen next? If two members of a species both have the same levels of tryptophan, but one of the organisms has a nonsense mutation in the trpR gene, how would the two organisms, phenotype differ? Multiple Choice: Which of the following is false regarding the regulation of gene expression in organisms with the same genotype? A. Organisms with the same gene and same exons may undergo alternative splicing. If the gene has multiple exons, some may be spliced out, and the fmRNA of the two organisms will be different. They will then produce different proteins from the same gene. B. Gene expression can be altered pre-transcription (by regulatory proteins) and pre-translation (by RNA interference) but once the gene has been translated, the protein products cannot be altered and two organisms will have the same phenotype. C. Even though the organisms have the same genotype, they may have inherited different patterns of methylation. Access to the gene may be different in each, and therefore they will have different phenotypes. D. Even if the two organisms have the same genotype, they may not be exposed to the same environmental factors. Therefore, a repressor may bind to operator of one organism but not the other. The gene will not be expressed in that organism, but will in the other, and so the two will have a different phenotype.

LO 3.19 Free Response: a. • Step 1- the trpR gene is transcribed and translated to produce the regulatory protein for the trpE gene. • Step 2- tryptophan is currently being produced by the organism. The cycle of producing tryptophan is a negative feedback loop. • Step 3- when a molecule of tryptophan binds to the protein, the protein now has quaternary structure and changes shape. • Step 4- with its new shape, the regulatory protein binds to the promoter of the trpE gene, blocking RNA polymerase from transcribing the gene. Because the gene is not being transcribed, no more tryptophan is produced. • Next- When levels fall below a certain amount, the protein will unbind from the promoter and the gene will resume being expressed. • In the organism that does not have the mutation, the regulatory process will be the same as in part A. In the other organism, the nonsense mutation will change the expression of the trpR gene. The protein produced by the mutated gene will have different primary structure and therefore different function. It will no longer function as a regulatory protein. Even as tryptophan levels rise, the regulatory protein will be unable to bind upstream to the trpE gene. RNA polymerase will continue to transcribe the trpE gene. Tryptophan will continue to be synthesized and levels will rise. The two organisms, despite having identical copies of the trpE gene will have vastly different levels of tryptophan. Multiple Choice: Answer B is false. (Gene expression can be altered pre-transcription (by regulatory proteins) and pre-translation (by RNA interference) but once the gene has been translated, the protein products cannot be altered and two organisms will have the same phenotype.) Even after a gene has been translated, the regulatory protein ubiquitin can alter expression. It can alter protein interactions, signal a proteasome to destroy the protein, or otherwise alter protein function. If two organisms with the same genotype make equal amounts of the same protein after translation, their phenotype could still be different if one organism signals ubiquitin to alter protein expression. The other responses are correct. Alternative splicing occurs after transcription to remove exons from the RNA in order to derive different proteins from the same gene. Aside from the actual code of the DNA, the way DNA is coiled can alter access to certain genes and may be heritable. An organism with more tryptophan for example, will have a repressor protein bind to the DNA and less tryptophan will be produced.

LO 3.4:The student is able to describe representations and models illustrating how genetic information is translated into polypeptides. • TSP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. • Explanation:The process in which genetic information is converted into polypeptides is translation. Genetic information in DNA is transcribed onto mRNA which then travels out of the nucleus to a ribosome; The ribosome acts as a catalyst by breaking and forming bonds between molecules. The first step of Translation is the scanning of the mRNA’s nucleotide codons, which are groupings of three nucleotides that correspond to a specific amino acid. The translation from the nucleotide “language” involves tRNA which binds to the amino acids, specific to the codons that are present in the sequence. The amino acid, carried by the tRNA molecule, binds to the ribosome’s A-site through the expenditure of GTP(an energy releasing molecule like ATP). To convert the sequence into the Amino acid “language” that proteins are derived from, the amino acids are bonded through a dehydration synthesis reaction. The reaction occurs between the carboxyl group of one molecule and the amino group of the other, releasing a water molecule to form covalent peptide bonds that join single amino acids into chains called polypeptides. These polypeptide proteins are used in cells as enzymes, structural supports, transporters, etc. Proteins are vital to cellular functions because they can manipulate organelles, ion concentrations, and membranes to ensure that all the proper conditions are met for the processes to work properly. • M.C. Question:The correct sequence of events in protein translation is… • Peptide bonds chain amino acids together a.)I,II,III,IV • The anticodon in tRNA matches a codon in mRNA b.) II,IV,I,III • A protein product is released from the ribosome c.)IV,III,I,II • tRNA binds to an amino acid d.)III,I,IV,II • FRQ:Using the figure provided • describe the process of translation. • Make sure to address the roles of RNA , • peptide bonds, amino acids, codons, • and ribosomes. If a mutation were to occur in • The DNA segment that codes for elongation factors, • What effect would it have on protein synthesis?

Answer to M.C. question: The correct sequence of events in protein synthesis is… • I,II,III,IV • II,IV,I,III • IV,III,I,II • III,I,IV,II • The first step of protein synthesis is tRNA recognizing a codon on mRNA. There are amino acids throughout the cell that are resting until their codon is found in a strand of mRNA. Transfer RNA has a segment in its nucleotide composition that corresponds to messenger RNA called an anticodon, this structure is a sequence of three nucleotides that forms the mirrored version of a codon on mRNA. Once the anticodon detects its match the tRNA can then find the proper amino acid for the protein product. The tRNA binds to the amino acid with the help of the enzyme aminacyl tRNA synthetase. The enzyme binds the tRNA to the amino acid allowing it to be transported to the ribosome. There, other proteins called elongation factors • bind separate amino acids together. This process repeats until the polypeptide • is completed and the protein product of the gene • is ready for use. Answer to FRQ: Using the figure provided describe the process of translation. Make sure to address the roles of RNA , peptide bonds, amino acids, codons, anticodons, and ribosomes. If a mutation were to occur in The DNA segment that codes for elongation factors, What effect would it have on protein synthesis? Translation occurs after Transcription when mRNA reaches a ribosome. The mRNA carries a series of nucleotides copied from DNA; every three nucleotides compose a codon that corresponds to an anticodon segment of tRNA that signals for a specific amino acid. Amino acids are comprised of an amino functional group and a carboxyl group along with a side compoud that differentiates each amino acid. There are 64 possible codons with only approximately 20 amino acids including the “start/stop” codons, so multiple codons can call for the same amino acid, and each codon has a different tRNA that holds its anticodon. The Amino acids are bonded to tRNA so that they can be transported to the ribosome, where a peptide bond is formed between the acids to form a long chain of them called a polypeptide. The chain is stable because of strong covalent bonds formed through dehydration reactions. Finally, The polypeptides are transported to the areas within the cell in which they are used. A mutation that interrupted the production or function of elongation factors would debilitate the process of translation. Amino acids would reach the ribosome bound to tRNA, but once they arrived there a polypeptide couldn’t form. Elongation factors perform the dehydration reactions that bind peptides through a carboxyl/amine interaction. Without them the amino acids would bunch up in the ribosome and never bind, thus never achieve tertiary structure that would give a protein its function.

LO 2.16: The student is able to connect how organisms use negative feedback to maintain their internal environments. SP 7.2: The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. Explanation: Organisms are able to maintain internal consistency through negative feedback loops. Sensors in the organisms monitor internal conditions and extracellular conditions which are relayed by nerves to the control center. The control center, usually the brain, contains the set point for the organism, which allows it to react to changes in the external environment to maintain the set point, or homeostasis, in the internal environment. When stimuli are sent to the brain, the control center responds by either increasing or decreasing activities of various effectors (muscles or glands) which then work to return the internal environment back to the set point. M.C. question: Which of the following is NOT an example of a negative feedback loop? a) Body temperature b) Blood clotting c) Blood pressure Predator/Prey relationship FRQ-style question: Knowing that organisms rely on negative feedback loops to maintain their internal consistencies, explain how the production of insulin and glucagon in the pancreas utilizes the negative feedback loop to monitor glucose levels in the body. Why is this particular function an important function in the body? What would be the result of this system failing?

Answer Key—LO 2.16 a) Body temperature--The hypothalamus of a human responds to temperature fluctuations 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. b) Blood clotting--Once a vessel is damaged, platelets start to cling to the injured site and release chemicals that attract more platelets. The platelets continue to pile up and release chemicals until a clot is formed. c) Blood pressure--When blood pressure increases, signals are sent to the brain from the blood vessels. Signals are sent to the heart from the brain and heart rate slows down, thus helping blood pressure to return to normal. d) Predator/Prey relationship--If the numbers of prey decreases, then some predators will starve, and their numbers will decrease, causing the numbers of the prey to increase followed by the increase in the numbers of the predators. Knowing that organisms rely on negative feedback loops to maintain their internal consistencies, explain how the production of insulin and glucagon in the pancreas utilizes the negative feedback loop to monitor glucose levels in the body. Why is this particular function an important function in the body? What would be the result of this system failing? When blood glucose levels rise, insulin is secreted by the pancreas, lowering blood glucose by increasing its uptake in cells and stimulating the liver to convert glucose to glycogen, in which form it can be stored. When blood glucose levels fall, glucagon is secreted by the pancreas, which increases blood glucose levels by stimulating the breakdown of glycogen into glucose and the creation of glucose from amino acids. Both high and low levels of blood glucose stimulate the pancreas to respond in such a way that the blood glucose levels will return to normal. This makes the process of maintaining blood glucose levels in the body a work of the negative feedback loop. This is an important function in the body because it allows us to properly digest foods and it allows for the homeostasis of blood glucose. If insulin secretion is impaired, it can result in diabetes mellitus. Diabetes mellitus is a disease in which blood glucose levels remain high, leading to excess glucose in the urine, increased urine output, and dehydration.

LO 3.44: The student is able to describe how nervous systems detect external and internal signals. SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: The nervous system has three basic functions: sensation (gathering of the information), integration (processing the information) and response (coordinated action appropriate to environment). Sensation and response are carried out by the peripheral nervous system, while integration is carried out by the central nervous system. Externally, receptors monitor things such as temperature, light, and sound. Internally, receptors detect variations in pressure, pH, carbon dioxide concentration, and the levels of various electrolytes. Sensory neurons, which are a type of nerve cell within the nervous system, are responsible for converting external stimuli from the organisms environment into internal electrical impulses. There are different types of sensory neurons in different areas of the body, responsible for detecting specific stimuli. Examples are chemoreceptors that sense chemicals, thermoreceptors that sense temperature, auditory receptors that detect sound, and olfactory receptors that detect smells. Next, sensory transduction occurs, which is the conversion of stimulus energy into a change in the membrane potential of a sensory receptor. In response to a stimulus/electrical pulse received by the dendrites of a nerve cell, Na+ and K+ gated channels sequentially open and cause the membrane to become locally depolarized. If the interior potential reaches -55 mV from -70 mV, an action potential is initiated and transmission of information between neurons occurs. This is known as an excitatory postsynaptic potential (EPSP), and is a temporary depolarization of postsynaptic membrane potential caused by the flow of prohibitively charged ions into the postsynaptic cell. A post synaptic potential is defined as excitatory if it makes it easier for the neuron to fire an action potential. In most animals, transmission across synapses involved chemical messengers called neurotransmitters. The transmission of information along neurons and synapses results in a response. The opposite of an EPSP is an inhibitory postsynaptic potential (IPSP), which usually result form the flow of negative ions into the cell. An IPSP is the change in membrane voltage of a postsynaptic neuron which results from synaptic activation of inhibitory neurotransmitter receptors. A postsynaptic potential is considered inhibitory when the resulting change in membrane voltage makes it more difficult for the cell to fire an action potential, lowering the firing rate of the neuron. The neurotransmitter released form the cell will bind to the postsynaptic cell and elicit an action potential there as well. Eventually the signal reaches the brain where it is integrated and is sent back out which results in a response. • Learning Log/FRQ-style Question: • Describe the process of the signal detection, integration, and subsequent response by the nervous system. Be sure to include the effect of polarization or depolarization on Na+ and K+ gated ion channels, and the creation of an action potential. • Name two types of sensory neurons, and where they would be found in the body. • Would a charge of -34 mV initiate an action potential? Explain M.C. Question: Which of the following is incorrect involving the detection of a signal and the subsequent response? If the neuron depolarizes enough and reaches the threshold of -55 mV, then the neuron will fire, initiating an action potential. B) Sensory receptors are specialized (respond to specific energy stimuli), and are part of the CNS (central nervous system). C) No energy is required for the creation of the action potential by the Na+ and K+ gated ion channels. D) Sensory transduction is the conversion of stimulus energy into a change in the membrane potential of a sensory receptor.

ANSWER KEY– LO 3.44. M.C. Question: Which of the following is incorrect involving the detection of a signal and the subsequent response? A) If the neuron depolarizes enough and reaches the threshold of -55 mV, then the neuron will fire, initiating an action potential. B) Sensory receptors are specialized (respond to specific energy stimuli), and are part of the CNS (central nervous system). C) No energy is required for the creation of the action potential by the Na+ and K+ gated ion channels. D) Sensory transduction is the conversion of stimulus energy into a change in the membrane potential of a sensory receptor. Answer: Answer choice B is incorrect because the sensory receptors are part of the PNS (peripheral nervous system). FRQ answer: a) The nervous system has three basic functions: sensation (gathering of the information), integration (processing the information) and response (coordinated action appropriate to environment). Sensation and Response are carried out by the peripheral nervous system, while integration is carried out by the central nervous system. Externally, receptors monitor things such as temperature, light, and sound. Internally, receptors detect variations in pressure, pH, carbon dioxide concentration, and the levels of various electrolytes. Sensory neurons, which are a type of nerve cell within the nervous system, are responsible for converting external stimuli from the organisms environment into internal electrical impulses. Next, sensory transduction occurs, which is the conversion of stimulus energy into a change in the membrane potential of a sensory receptor. In response to a stimulus/electrical pulse received by the dendrites of a nerve cell, Na+ and K+ gated channels sequentially open and cause the membrane to become locally depolarized. If the interior potential reaches -55 mV from -70 mV, an action potential is initiated and transmission of information between neurons occurs. This is known as an excitatory postsynaptic potential (EPSP), and is a temporary depolarization of postsynaptic membrane potential caused by the flow of prohibitively charged ions into the postsynaptic cell. A post synaptic potential is defined as excitatory if it makes it easier for the neuron to fire an action potential. In most animals, transmission across synapses involved chemical messengers called neurotransmitters. The transmission of information along neurons and synapses results in a response. The opposite of an EPSP is an inhibitory postsynaptic potential (IPSP), which usually result form the flow of negative ions into the cell. An IPSP is the change in membrane voltage of a postsynaptic neuron which results from synaptic activation of inhibitory neurotransmitter receptors. A postsynaptic potential is considered inhibitory when the resulting change in membrane voltage makes it more difficult for the cell to fire an action potential, lowering the firing rate of the neuron. The neurotransmitter released form the cell will bind to the postsynaptic cell and elicit an action potential there as well. Eventually the signal reaches the brain where it is integrated and is sent back out which results in a response. b) There are different types of sensory neurons in different areas of the body, responsible for detecting specific stimuli. Examples are chemoreceptors that sense chemicals, thermoreceptors that sense temperature, auditory receptors that detect sound, and olfactory receptors that detect smells. c) In response to a stimulus/electrical pulse received by the dendrites of a nerve cell, Na+ and K+ gated channels sequentially open and cause the membrane to become locally depolarized. If the interior potential reaches -55 mV from -70 mV, an action potential is initiated and transmission of information between neurons occurs. Since -34 mV has reached -55mV and continues past it, this will occur.

INTERPHASE S(DNA synthesis) G1 CytokinesisMitosis G2 MITOTIC(M) PHASE LO 3.8: The student can describe the events that occur in the cell cycle. SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: The cell cycle is important for reproduction, growth and development of fertilized eggs, and repair/ replacement of damaged cells. The cell cycle contains TWO major phases: interphase (90%) and mitotic phase (10%). There are SIX steps in the cell cycle and an easy way to remember them is IPMAT (Interphase, Prophase, Metaphase, Anaphase, and Telephase). The cell cycle is regulated by certain “check points” throughout the cell cycle. These “check points” look to make sure the cell has enough ingredients to pass on to the next stage of the cycle. Multiple Choice: What is the main difference between the way that animal cells and plant cells divide during cytokinesis? • Two diploid daughter cells split into four Haploid daughter cells. • Cleavage furrows start from the middle in animal cells, and starts at the top and works itself down in plant cells. • Cleavage furrows in animal cells and Cell plates in plants. • Cell Plates in animal cells and Cleavage in plants. Learning Log/ FRQ- style Question: Using the diagram, describe in detail the parts of the cell cycle and what phases they are in. Also explain the significance of the check points.

Cleavage furrow Cell plate Daughter cells (a) Cleavage of an animal cell (SEM) (b) Cell plate formation in a plant cell (SEM) Answer Key- LO 3.8 What is the main difference between the way that animal cells and plant cells divide during cytokinesis? Two diploid daughter cells split into four Haploid daughter cells. Cleavage furrows start from the middle in animal cells, and starts at the top and works itself down in plant cells. Cleavage furrows in animal cells and Cell plates in plants. Cell Plates in animal cells and Cleavage in plants. Cleavage Furrows: The contraction of the dividing cell’s ring of microfilaments is like the pulling of drawstrings. The cleavage furrow deepens until the parent cell is pinched in two, producing two completely separated cells, each with its own nucleus and share of cytosol and organelles. Cell Plates: During telophase vesicles derived from the Golgi apparatus move along microtubules to the middle of the cell, where they coalesce, producing a cell plate. Cell wall materials carried in the vesicles collect in the cell plate as it grows. The cell plate enlarges until its surrounding membrane fuses with the plasma membrane along the perimeter of the cell. Multiple Choice: Learning Log/ Free Response: Below are suggestions of what the student needs to write about in order to hit all of the main points in the question. Using the diagram, describe in detail the parts of the cell cycle and what phases they are in. Also explain the significance of the check points. • Two phases: Interphase (90%), contains the G1, S, and G2 sub phases, and Mitotic M Phase (10%), contains mitosis and cytokinesis • Interphase: the cell grows by producing proteins and cytoplasmic organelles such as mitochondria and endoplasmic reticulum. Chromosomes are only duplicated in S phase. • Mitosis: the division of one nucleus into two genetically identical nuclei • Prophase and Prometaphase: Chromatin fibers in the cell become more dense so that two sister chromatids join together while the nuclear envelope begins to fragment. • Metaphase: The centresomes move to opposite ends and the chromosomes convey the metaphase plate in the middle of the cell. • Anaphase: Two sister chromosomes part to become individual chromatids. The chromatids start to move towards the ends of the cell. At the end of anaphase the two ends have equivalent collections of chromosomes. • Telophase: Two daughter nuclei begin to form and the nuclear envelope arises. After this is done mitosis is complete. • Cytokinesis: The division of the cytoplasm is usually well underway by late telophase, so the two daughter cells appear shortly after the end of mitosis. • Check Points: There to make sure that the cell has developed in the necessary way to continue on in the cell division process. For example a cell might not have separated their chromosomes right, so they may not be able to pass the check point.

LO 2.31: The student can connect concepts in and across domains to show that timing and coordination of specific events are necessary for normal development in an organism and that these events are regulated by multiple mechanisms. SP 7.2: The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. Explanation: Many biological processes involved in growth, reproduction and dynamic homeostasis include temporal regulation and coordination. Multiple mechanisms regulate the timing and coordination of molecular, physiological and behavioral events that are necessary for an organism’s development. For example, cell differentiation results from the expression of genes for tissue-specific proteins, and the induction of transcription factors during development results in sequential gene expression. Homeotic genes are involved in developmental patterns and sequences and embryonic induction in development results in the correct timing of events. MC Question: Which of the following statements concerning morphogenesis is untrue? A.It mostly occurs in the embryonic stages of animals B.It occurs continuously in plants C.It requires reverse transcriptase to convert RNA back to DNA D.It consists of molecular cues that direct formation of specific genes LL: Explain each of the following and describe its effect on the normal development of an organism: A. Homeotic Genes B. cell differentiation/morphogenesis C. mutations Homeotic Genes

MC Question: Which of the following statements concerning morphogenesis is untrue? A. It mostly occurs in the embryonic stages of animals B. It occurs continuously in plants C. It requires reverse transcriptase to convert RNA back to DNA D. It consists of molecular cues that direct formation of specific genes LL: A. Homeotic genes: -genes that regulate anatomical identify of segments -specify types of structures that each segment will form -gene activators and repressors -affects development by turning undifferentiated cells into specialized structures, such as appendages B. Morphogenesis: -involved in the transformation of a zygote into an organism -cells become specialized in function and are organized into tissues and organs -Must occur very early in embryonic development in order to determine the body axes of the organism -Requires the appropriately timed death of certain cells C. Mutations: -changes in the genetic material of a cell -can cause death at the embryonic stage -may alter DNA, affecting the normal development of an embryo -can impact the phenotype of an organism

LO 3.40 The student is able to analyze data that indicate how organisms exchange information in response to internal changes and external cues, and which can change behavior. SP 5.1 The student can analyze data to identify patterns or relationships. Explanation: Responses to information and communication of information are vital to natural selection and evolution. Natural selection favors innate and learned behaviors that increase survival and reproductive fitness. In natural selection there are specific interactions and climates that help favor animals. Darwin’s Finches are a great example of this because there were originally one species of finch, but then over the course of time and migration due to favorable traits the species subdivided into twenty different species of finches with a shared common ancestor. Cooperative behavior tends to increase the fitness of the individual and the survival of the population. With that being said, this is commonly seen in pack behavior animals such as herds or flocks of birds, sheep, or other schooling behavior animals. The increased behavior of fitness allows them to be more susceptive to predator warnings and swarming behaviors in insects. This allows them to take in their surroundings and be able to make the best decisions based off this for survival. Animals such as bees use migratory flight patterns to signal how close or far away their food sources are. They fly in patterns of an “8” or a similar shape and depending on how long or short the distances between the loops of the “8” is, leads them to be able to determine how near or far their food source is. MC: Cooperative Behavior is the interaction of two or more organisms directed toward a common goal which is mutually beneficial. Which of the following does not meet this criteria? A) Compliant Behavior B) Territorial choruses C) The inter- and intra-relationships between various microorganisms. D) Ideas going against The Evolutionary Theory FRQ: Read and analyze the graphs below. I. Looking at Chart 16 and using prior knowledge what is the difference between r selected and K selected species? How does the difference play a role in their survival rating and presence on Earth? II. Describe what types of phenotypic traits might favored by natural selection for both the r and the K species previously discussed species. Give a hypothesis for why these might be favored? III. Using prior knowledge about natural selection explain why it is so important that animals change and adapt instead of remaining constant.

MC: Cooperative Behavior is the interaction of two or more organisms directed toward a common goal which is mutually beneficial. Which of the following does not meet this criteria? A) Compliant Behavior B) Territorial choruses C) The inter- and intra-relationships between various microorganisms. D) Ideas going against The Evolutionary Theory FRQ: Read and analyze the graph below. I. Looking at Chart 16 and using prior knowledge what is the difference between r selected and K selected species? How does the difference play a role in their survival rating and presence on Earth? • The production of numerous small offspring followed by exponential population growth is the defining characteristic of r-selected species. They require short gestation periods, mature quickly (and thus require little or no parental care), and possess short life spans. • selection occurring when a population is at or near the carrying capacity of the environment, which is usually stable: tends to favor individuals that successfully compete for resources and produce few, slowly developing young, and results in a stable population of long-lived individuals as the k species. • The r selected will be fewer and live the same lifespan as the k selected species. The r selected are equally as numerous when born but quickly die off, however the surviving few will be just as prosperous as the k selected. The k selected are numerous and last for many years before their population declines. • Overall this plays a role on earth by having a proportionate volume of each species. It is this way because the environment can only support so much and the predator/pray relationships as well. II. Describe what types of phenotypic traits might be favored by natural selection for both the r and the K species previously discussed. Give a hypothesis for why these might be favored? • Phenotypic traits could include: coloration, camouflage, taste, smell, size, physical adaptations such as specific body parts size and function, disappearance of a physical appendages that once were present, habitat change, etc. • Hypothesis should make sense with topics and is up for interpretation of the reader. It should include either the k or r selected species or both. The hypothesis should also include why they believe what they said was true and provide supporting details and viable statements as to why they believe their hypothesis to be correct III. Using prior knowledge about natural selection explain why it is so important that animals change and adapt instead of remaining constant. • Reproduction of the fittest • the process by which forms of life having traits that better enable them to adapt to specific environmental pressures, as predators, changes in climate, or competition for food or mates, will tend to survive and reproduce in greater numbers than others of their kind, thus ensuring the perpetuation of those favorable traits in succeeding generations.

LO 3.33 The student is able to use representation(s) and appropriate models to describe features of a cell signaling pathway. SP 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. Explanation: The first step in cell signaling pathways is reception which is where the ligand, or signal molecule, binds to the receptor protein. The second step is transduction which changes the signal so that the cell can comprehend it. The third and final step is the response which is the receiving cell’s reaction to the signal. There are three types of cell signaling pathways. The first is direct signaling which includes cell- to- cell contact, presumably physical. Second is short range cell signaling which includes paracrine signaling that occurs between cells that are in close proximity. Another example of short range cell signaling is the pathway between neurotransmitters. The ligand binds to the dendrites of the neuron which causes a signal to be sent down the axon and across the synapse to be received by the dendrites of the next cell; the process continues and cycles among neurons. The third type of cell signal pathway is long range which involves endocrine signaling and the releasing of pheromones. An example of a long range cell signaling pathway is that of insulin which begins as carbohydrates are digested; the pancreas is signaled that the blood glucose levels are too high, so it releases insulin to incite the uptake of glucose. The insulin must bind to its receptor in order to transmit the signal. MC: Scientists are testing the effect of the inhibition of insulin by a new drug. Which of the following statements is true? A. Glucose uptake will increase as insulin does not bind to the insulin receptor. B. Glucose uptake will be unaffected as the insulin signal will be received by another insulin receptor that is not inhibited.C. Glucose levels will be unaffected because insulin is only in short range signaling thus making the new drug ineffective. D. Glucose uptake will decrease as insulin does not bind to the insulin receptor. FR/ LL: Suppose a new drug is being tested to cure a disease that inhibits dendrites. How is the cell signaling pathway affected if this disease inhibits dendritic function? How would this new drug cause the dendrites to respond in cell signaling if it cures this disease? Use a diagram to help explain your answer.

MC: Scientists are testing the effect of the inhibition of insulin by a new drug. Which of the following statements is true? • A. Glucose uptake will increase as insulin does not bind to the insulin receptor. • B. Glucose uptake will be unaffected as the insulin signal will be received by another insulin receptor that is not inhibited.C. Glucose levels will be unaffected because insulin is only in short range signaling thus making the new drug ineffective. • D. Glucose uptake will decrease as insulin does not bind to the insulin receptor. FR/LL: Suppose a new drug is being tested to cure a disease that inhibits dendrites. How is the cell signaling pathway affected if this disease inhibits dendritic function? How would this new drug cause the dendrites to respond in cell signaling if it cures this disease? Use a diagram to help explain your answer. If this disease inhibits dendritic function, then cell signals will not be transmitted between neurons in cell- to – cell communication. This is because dendrites act as the receptor site for various cell signals. If the drug cures this disease, then the dendrites will no longer be inhibited and the cell signal will be received by the dendrite, transmitted down the axon, and express across the synapse between two neurons.

LO 3.17: The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. • SP 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. • Explanation: Not all traits that are inherited follow the Mendelian Model. For example: any type of mutation in the DNA cant be explained by the Mendelian Model. Nonmendelian Models prove there is much more to genetics than just doing a Punnet Square of Pp X Pp. Another example of Nonmendelian genetics is sex-linked genes. Sex linked genes are genes located on either sex chromosome (Men XY and Women XX). In humans the term is typically referring to the X chromosome. The Y chromosome is typically small and carries few genes. In this case, men can only pass a sex-linked gene to their daughter. Women can pass it to either son or daughter. As shown in the picture, the father is unable to pass the disease (hemophilia) to his son, but he can to his daughter. The mother can pass it down to either child. Another example that doesn’t follow the Mendelian Model is how mitochondrial DNA is transmitted to an offspring. It is transmitted by the egg and not the sperm. This means that the mitochondrial determined traits are maternally inherited. • Multiple Choice: A couple has different types of blood. The male has Type A blood while the female has Type B blood. Which of the following Non Mendelian terms describes why their offspring may have Type AB blood? • Incomplete Dominance • Codominance • Polygenic Inheritance • Genetic Recombination • Law of Independent Assortment • Free Response: Hemophilia affects 1 in 5,000 male babies each year. Hemophilia occurs due to a defect in the X chromosome. Using the figure on the right to help answer the following questions. • Can daughters become infected? Why? • Is this disease caused by a dominant allele? Explain.

Multiple Choice: A couple has different types of blood. The male has Type A blood while the female has Type B blood. Which of the following Non Mendelian terms describes why their offspring may have Type AB blood? • Incomplete Dominance- This is not the answer because incomplete dominance is more of a blending of two alleles. • Codominance- This is the answer because both alleles are expressed. • Polygenic Inheritance- This is not the answer because polygenic inheritance would call for multiple genes to control a single trait. • Genetic Recombination- This is not the answer because there is no separation or replacement of different alleles (such as in crossing over) • Law of Independent Assortment- Not only is this not an example of a Non Mendelian term, but it does not explain why Type AB blood can be a result from the two parents. • Free Response: Hemophilia affects 1 in 5,000 male babies each year. Hemophilia occurs due to a defect in the X chromosome. Using the figure on the right, • Can daughters have the genetic mutation? Why? • Is this disease caused by a dominant allele? Explain. • Daughters can have the genetic mutation. The daughter may become infected only if the father is infected and the mother is a carrier or infected. In the case where both parents are infected, there is a 25% chance of having a daughter with the genetic mutation. In a case where only one parent is infected, the daughters may only be carriers. B) This mutation is caused by a recessive allele. As seen in the picture, the daughter is a carrier despite having the hemophilia gene. Since she is only a carrier and not fully infected, the disease is caused by being homozygous recessive.

Learning Objective 2.6 F. • Objective 2.6: The student is able to use calculated surface area-to-volume ratios to predict which cell(s) might eliminate wastes or procure nutrients faster by diffusion. • Science Practice 2.2: The student can apply mathematical routines to quantities that describe natural phenomena. • Explanation: The rate at which chemical reactions occur within a cell is dependent upon the volume of that cell. Larger cells are host to greater volumes of substances, and carry out more chemical reactions. As such, larger cells require a higher rate of diffusion over the plasma membrane to transport needed materials like nutrients and water into the cell and to excrete wastes from the cell. A cell’s rate of diffusion is directly proportional to the ratio of its surface area to volume. When a cell’s volume increases, its surface area does not increase by a similar scale. Since the volume of that cell is the denominator, and the surface area of that cell is the numerator, and the growth of the volume outpaces the growth of the surface area, the surface area-to-volume ratio decreases as volume grows. When an inadequate amount of surface area exists to carry out diffusion over the plasma membrane, a cell cannot utilize needed nutrients and water and excrete wastes at a fast enough rate. Since a cell can only support itself to a minimum surface area-to- volume ratio, the cell remains small enough to ensure necessary rates of diffusion. Cells with smaller radii (smaller volumes and larger surface area-to-volume ratios) can carry out diffusion more efficiently than can cells with larger radii. • Multiple Choice Question: Two cells are found in a culture. Cell A, which carries out basic respiratory functions, possesses a radius of 3 micrometers. Cell B, which carries out basic digestive functions, possesses a radius of 5 micrometers. Given this information, which of the following statements is true? Use the formulas at right to answer this question. a. Cell A carries out diffusion of nutrients at a slower rate than Cell B. b. Cell B carries out diffusion of nutrients at a slower rate than Cell A. c. Each cell carries out diffusion of nutrients at a rate equivalent to that of the other cell. d. Cell B carries out its metabolic processes more efficiently that Cell A. • Learning Log/FRQ-style Question: Genetic engineers working in a laboratory are trying to determine which of three liver cells would be more effective as a template for the creation of artificial liver tissue. Cell A is considered the ideal candidate for the template cell. Cell B is 17 square micrometers in surface area and 68 cubic micrometers in volume. Cell C is 10 square micrometers in surface area and 100 cubic micrometers in volume. a. Based on the above information, which cell would be second best choice for the template, assuming that maximum diffusion rates are most preferable? Explain your reasoning in terms of the surface area-to-volume ratio. b. Why do limitations exist on the potential size of a cell?

Answer Key- Learning Objective 2.6 Two cells are found in a culture. Cell A, which carries out basic respiratory functions, possesses a surface area of 24 square micrometers and a volume of 72 cubic micrometers. Cell B, which carries out basic digestive functions, possesses a surface area of 28 square micrometers and a volume of 56 cubic micrometers. Given this information, which of the following statements is true? a. Cell A carries out diffusion of nutrients at a slower rate than Cell B. b. Cell B carries out diffusion of nutrients at a slower rate than Cell A. c. Each cell carries out diffusion of nutrients at a rate equivalent to that of the other cell. d. Cell B carries out its metabolic processes more efficiently that Cell A. • By plugging in 3 for the radius in the formulas for r of Cell A, the student derives V=(4/3)(84.823) and SA=(4)(28.274). SA/V= ((4)(28.274))/((4/3)(84.823))=113.096/113.096=1. By plugging in 5 for r of Cell B, the student derives V=(4/3)(392.699) and SA=(4)(78.540). SA/V= ((4)(78.540)/(4/3)(392.699))=314.16/523.599=0.60. This means that Cell A possesses a larger surface area-to-volume ratio and that it can transport nutrients and water by diffusion more quickly and efficiently than Cell B. This makes choice B the correct answer. Choice A is incorrect, because it holds that Cell A carries out diffusion more slowly than Cell B. Since Cell A possesses more membrane proportionate to volume with which to transport nutrients to each region of the inner cell, it can carry out diffusion more rapidly than Cell B. The differences of SA/V calculations for each of the two cells (1 for Cell A and 0.6 for Cell B) automatically disqualify choice C. It has already been established that Cell A carries out diffusion more efficiently than Cell B. It follows that Cell A can provide nutrients and water to drive metabolic processes and remove excretory wastes more rapidly and efficiently than Cell B as well. Therefore, choice D is incorrect. Genetic engineers working in a laboratory are trying to determine which of three liver cells would be more effective as a template for the creation of artificial liver tissue. Cell A is considered the ideal candidate for the template cell. Cell B is 17 square micrometers in surface area and 68 cubic micrometers in volume. Cell C is 10 square micrometers in surface area and 100 cubic micrometers in volume. a. Based on the above information, which cell would be second best choice for the template, assuming that maximum diffusion rates are most preferable? Explain your reasoning in terms of the surface area-to-volume ratio. b. Why do limitations exist on the potential size of a cell? a. Cell C would be the second best choice, because it possesses a surface area-to-volume ratio of 0.1. This is opposed to Cell B’s surface area-to-volume ratio of 0.25. Because surface area and volume are more equitably sized in Cell C, the cell can absorb nutrients and water for all parts of its interior volume more rapidly than can Cell B. b. When a cell grows in volume, its surface area does not grow at an equivalent rate. It grows at a slower rate. Because the surface area grows at slower rate, it cannot carry out diffusion rapidly enough to service metabolic chemical reactions taking place in each region of the cell’s interior volume. When chemical reactions within the cell’s interior are not serviced, the cell will begin to shut down from an inability to maintain homeostatic operations. Therefore, cells can only grow to a certain size before they can no longer carry out sustainable and efficient diffusion.

4 th Review Project collection

4 th Review Project collection

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1 st Review Project collection

2 nd Review Project Collection

4 th 6 Weeks Test Review

4 th Grade Social Studies Review

8 th Grade 4 th Quarter Project

4 th Six Weeks Test Review

2 nd /4 th Quarter Review

4 th -8 th Social Studies Generalist Review

4 th -8 th Social Studies Generalist Review

4 th Grade Review

4 th Grade Review

CRCT Review: 4 th Grade ELA

4 th Quarter Review

1 st /2 nd /4 th Review Project Collection #2

7 th /8 th Review Project Collection

4 th Grade research project

COSMOS 4 th Project Meeting

Review 4 th Quarter

4 th Grade Science Review