ANSWER KEY – LO 4.25/SP 6.1

1. LO 4.25:The student is able to use evidence to justify a claim that a variety of phenotypic responses to a single environmental factor can result from different genotypes within the population. SP 6.1:The student can justify claims with evidence. Explanation:In nature, environmental stimuli and changes in these stimuli provoke changes in phenotypes from all organisms in the environment—those species that do not adapt to these changes risk failure of surviving. Because of this one environmental change, many organisms make changes to their phenotype—such as a camouflage color change after pollutants distort environments color or larger wings that allow them to fly to higher regions where they can be safe from predators preying on those with the outdated camouflage—and these organisms all have different, respective genotypes. All of the organisms, with their unique genotypes, must adapt to this new environmental factor and produce a phenotypic response to better suit the environment’s demands. If not, they die. A real-life application of this scenario is the evolution of Peppered Moths in Industrial England, and provides solidified real-world evidence of this LO and SP. Moths in England were either light moths or dark moths, and their survival depended on the color of tree of which they were resting. When the industrial revolution occurred and copious amounts of soot was put in the air, trees turned dark from the pollution of the soot; this caused the light colored moths (evolved to be camouflaged while resting on the pre-industrial age trees) to be selected against by predators because the dark moths (previously selected against on the pre-industrial age trees) were now better camouflaged on the post-industrial age trees. The moths are now seeing more emergence of lighter colors, leading to their nomenclature. • M.C. Question:Which of the following is least likely to produce a new phenotypic response to a new environmental factor? • Transcriptional errors • Interbreeding between two identical genotypes • Disruptive selection of prey by local predators • Interbreeding between two different genotypes • Learning Log/FRQ-style question: • Identify and describe THREE possible environmental factors that could invoke a change in phenotypic responses in a ground squirrel • -and- • B) Discuss possible methods for the creation of two of the phenotypic responses in Earth’s oceans outlined in part A)

2. ANSWER KEY – LO 4.25/SP 6.1 • Multiple Choice solution: • Which of the following is least likely to produce a new phenotypic response to a new environmental factor? • Transcriptional errors • Interbreeding between two identical genotypes • Disruptive selection of prey by local predators • Interbreeding between two different genotypes • B is the best answer of the choices. This is correct because while interbreeding between two identical genotypes will certainly lead to a phenotypic response (through transcriptional errors [mutations]), it would take substantially longer than the other methods to create a phenotypic response due to the number of mutations needed from transcriptional errors after reproduction to create a response; also, transcriptional errors are already highlighted in choice A). • Sample FRQ response: • Identify and describe THREE possible environmental factors that could invoke a change in phenotypic responses in a ground squirrel • Three possible environmental factors that could invoke a change in phenotypic responses in ground squirrels are adaption to deforestation through new digestive capabilities to compensate for lack of acorns or other foods, bigger claws for dissuading ground-level predators, and darker fur to be better suiting for camouflaging among decomposing foliage on the forest floor. • -and- • B) Discuss possible methods for the creation of two of the phenotypic responses outlined in part A) • Two possible methods for creation of darker fur coloration and digestive adaptation are natural selection and coevolution. The squirrels who have the phenotype for darker fur color will have an advantage in this new environment, and will therefore more likely live long enough to reproduce and pass on the genes for darker fur. This will cause the fur of the squirrels to gradually become darker as long as the environmental change persists. The adaptation of the squirrels digestive system to tolerate a new diet will coevolve with the food it incorporates into it’s new diet—the food eaten at the squirrels height, for instance, will be eaten more than the taller plants. The squirrels will eat these low-hanging plants, and disperse their seeds through feces or their coats. The seeds from the droppings and coats will fall, germinate in the spring, and grow, and pass on the low-hanging genes to the next generation of plants. Eventually, the plants will evolve to be squirrel-height and the squirrel will adapt it’s diet to be most efficient off it’s diet of this new addition to it’s diet.

3. LO 3.13 The Student is able to pose questions about ethical, social or medical issues surrounding human genetic disorders. SP 3.1: The student is able to pose a question. Explanation: Privacy is one issue concerning new technology such as DNA probes and DNA chips that can scan a person for mutations in over 7000 genes. While many people may like to know if they are susceptible to genetic disorders the information may become public and cause controversy. One example is health insurance which might refuse to cover or raise insurance costs to people with dangerous genes. There is also the issue of a company refusing to higher someone who carries a genetic predisposition to a future illness in order to higher someone who is more likely to remain healthy. There is also the ethical issue of if someone either carries a gene for a genetic disease or one that raises their risk for cancer or some other illness would it be ethical for them to have children and risk passing the gene on to their child. The same standards apply to someone with a lethal genetic disorder such as Huntington’s disease. If an employer discovers that a person applying for a job suffers from this disease they may not hire them because of the imminent health issues they face in the future. To relate the LO to the SP the student should be able to recognize both sides of the argument and pose a question based on the issues they see problematic. M.C. Question: Which of these is true of Sickle cell anemia? A) The disorder is the result of a point mutation. B) This is most common is people of Asian descent. C) This trait had no genetically beneficial origins. D) This disease is usually related to Huntington's disease or some other serious genetic disorder. Learning Log/FRQ-style Question: Huntington’s disease is a dominant genetic disorder. What are two ethical issues that occur when health records of people with genetic diseases become public? Name a genetic disorder other then Huntington’s disease and explain two of the symptoms.

4. ANSWER KEY– LO 3.13 Which of these is true of Sickle cell anemia? A) The disorder is the result of a point mutation. B) This is most common is people of Asian descent. C) This trait had no genetically beneficial origins. D) This disease is usually related to Huntington's disease or some other serious genetic disorder Learning Log/FRQ-style Question: Huntington’s disease is a dominant genetic disorder. What are two ethical issues that occur when health records of people with genetic diseases become public? Name a genetic disorder other then Huntington’s disease and explain two of the symptoms. If a person has a genetic disorder such as Huntington’s and their health records become public the person may face persecution from employers because they are passed up on for job opportunities. They may find it more difficult to obtain life insurance either because they have to pay more for coverage or because companies refuse to give them a premium at all. Answers may very. Down syndrome or trisomy 21 is caused by a person having an extra copy of the 21st chromosome. A person with Down syndrome can exhibit, slanted eyes, short stature, weak muscles and a short wide neck. They may also suffer from some sort of cognitive disabilities.

5. LO 4.1: The student is able to make predictions about the effects of genetic drift, migration and artificial selection on the genetic makeup of a population. SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. Explanation: The genetic makeup of populations is controlled by factors such as: genetic drift, migration, and artificial selection. Genetic drift is when an unpredictable change occurs in a small population and as a result the genetic makeup is altered. Two situations can increase the likely hood of genetic drift having a large impact on a population: the bottleneck effect and founder effect. The bottleneck effect is when a disaster drastically changes an environment and its population size is reduced tremendously. This results in the few survivors not reflecting the original populations gene pool due to loss of many alleles. The founders effect is when a few individuals are isolated from the majority of their population and when the smaller group establishes a new population its gene pool doesn’t reflect it’s source population. Migration occurs in two ways, the exit of individuals from a population, called emigration, and the entrance of individuals into a population, called immigration. Both types of migration alter the genetic makeup of a population by either taking away genes or adding them. Artificial selection is when selective breeding of domesticated plants and animals encourage the occurrence of desired traits and therefore the undesired traits slowly occur less and less. M.C. Question: Many interesting birds inhabit islands, that are isolated from the main lands, off the cost of South America that scientist like to do research on. If a new species of birds emerged on one of the islands, after a hurricane struck. What would be a possible explanation? • Bottleneck Effect • Founders Effect • Migration • Artificial Selection • Immigration Free Response: Populations of the house mouse, or musmusculus, are found in in different areas and are isolated from one another within Mountain ranges. This isolation has caused genetic variation within the species. Scientist are trying to figure out what could have caused the separation of these mice in order to test why they are genetically different now. What are TWO possible reasons for the separation of this species and Why? Explain why each reason would lead to genetic variation?

6. Answer Key – L.O. 1.8 M.C Question: Many interesting birds inhabit islands, that are isolated from the main lands, off the cost of South America that scientist like to do research on. If a new species of birds emerged on one of the islands, after a hurricane struck. What would be a possible explanation? • Bottleneck Effect • Founders Effect • Migration • Artificial Selection • Immigration Free Response: Populations of the house mouse, or musmusculus, are found in in different areas and are isolated from one another within mountain ranges. This isolation has caused genetic variation within the species. Scientist are trying to figure out what could have caused the separation of these mice in order to test why they are genetically different now. What are TWO possible reasons for the separation of this species? And Why? Explain why each reason would lead to genetic variation? The separation of this species of mice could have been caused by either the founder effect or simply just emigration. The founders effect is an explanation because its when a few individuals of a population is isolated from the vast majority and start a new population. The mice could have been separated by the mountain ranges and therefore the smaller groups started their own colony. Another reason could just be emigration from the original population. Some mice could of left the majority just to leave. The founders effect would lead to the genetic variation of the groups because once the small group moved it now had a smaller gene pool to reproduce with. Since there’s a smaller number of mice there are less alleles, throughout time the offspring will gradually differ from that of the original population simply due to less options in mating. The emigration of the mice has the similar reason to why it would have genetic variation. When some left the majority they now limited themselves to a smaller gene pool with less options for mating.

7. LO 3.48 The student is able to create a visual representation to describe how nervous systems detect external and internal signals. SP 1.1 The student can create representations and models of natural or man-made and systems in the domain. Example: External: You are sitting in class and your friend calls your name. Your ears pick up the external stimulus, your friend calling your name, and sends a signal to your brain. Your brain then processes this signal and then sends a response to the rest of your body causing you to turn your head. In other words, a stimulus is picked up by your peripheral nervous system (PNS senses stimuli from outside) and is integrated by your brain and spinal cord (brain and spinal cord are the central nervous system) to produce a motor response , which is done by the PNS. Internal: You are sitting in class and you haven’t had any water all day. Receptor cells in your body sense a high osmolarity, or solute concentration, in your blood. This is the stimulus. This causes you to be thirsty so you know you need more water and the Hypothalamus signals the Pituitary gland to release the chemical hormone ADH. ADH moves to the kidneys and signals them to retain more water instead of releasing it as urine. This results in homeostasis of blood osmolarity. In other words, your body senses an internal imbalance and the hypothalamus sends signals to either the posterior or anterior pituitary causing them to release a hormone that tells the body how to act accordingly. • When you touch a hot stove your body reacts and moves your hand away from the stove in order to protect your hand from getting hurt. What is the correct pathway of neurological signals that move your hand away from the hot stove? • The CNS (central nervous system) senses pain in the hand and triggers a motor response, moving the hand to move away from the stove. • The PNS (peripheral nervous system) senses pain in the hand and triggers a motor response, moving the hand away from the stove. • The CNS (central nervous system) senses pain in the hand and sends a signal to the PNS (peripheral nervous system), the PNS then integrates this signal and tells the CNS to respond with a motor response, moving the hand from the stove. • The PNS (peripheral nervous system) senses pain in the hand and sends a signal to the CNS (central nervous system), the CNS then integrates this signal and tells the PNS to respond with a motor response, moving the hand from the stove. FRQ style question: Identify the role of tropic hormones and explain why they are necessary in the endocrine system. Give a specific of a neuroendocrine pathway.

8. Answer key LO 3.48 • When you touch a hot stove your body reacts and moves your hand away from the stove in order to protect your hand from getting hurt. What is the correct pathway of neurological signals that move your hand away from the hot stove? • The CNS (central nervous system) senses pain in the hand and triggers a motor response, moving the hand to move away from the stove. • The PNS (peripheral nervous system) senses pain in the hand and triggers a motor response, moving the hand away from the stove. • The CNS (central nervous system) senses pain in the hand and sends a signal to the PNS (peripheral nervous system), the PNS then integrates this signal and tells the CNS to respond with a motor response, moving the hand from the stove. • The PNS (peripheral nervous system) senses pain in the hand and sends a signal to the CNS (central nervous system), the CNS then integrates this signal and tells the PNS to respond with a motor response, moving the hand from the stove. FRQ style question: Identify the role of tropic hormones and explain why they are necessary in the endocrine system. Give a specific of a neuroendocrine pathway. Tropic Hormones are hormones that are released by the hypothalamus or anterior pituitary gland that regulate the function of endocrine organs. In other words they are signals from the brain that tell the rest of the body how to respond to a given stimulus. Tropic hormones themselves do not cause any changes in the body, but they signal other organs to release hormones that do. Tropic hormones are important because they allow for the endocrine system to be controlled at a centralized location. This allows the body to formulate a coordinated response against an environmental shift. An example of a neuroendocrine pathway is T3 and T4 secretion from the thyroid gland. The hypothalamus receives a stimulus that causes it to release TRH, a tropic hormone, that signals the Anterior pituitary to release TSH, another tropic hormone, which then travels to the thyroid and causes the thyroid to release T3 and T4. T3 and T4 are used in controlling metabolism in humans. T3 and T4 also exert negative feedback on the hypothalamus and anterior pituitary by inhibiting the release of TRH and TSH.

9. LO 4.15: The student is able to use visual representation to analyze situations or solve problems qualitatively to illustrate how interactions among living systems and with their environment result in the movement of matter and energy. SP 1.4: The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. Explanation: Primary producers harness light from the sun into energy through photosynthesis. However, approximately only 1% of sunlight is harnessed and remains after producers utilize the energy in respiration (known as the net primary production). Energy is transferred to the next trophic level of an ecosystem (the primary consumers who eat autotrophs)and approximately 10% of the energy present reaches the primary consumers. The autotrophs, like other trophic levels, lose energy through heat and feces, use some of the energy to generate new biomass, and some of the autotrophs are simply not eaten. The process continues to secondary consumers (eating herbivores), tertiary consumers (eating carnivores), and so on. Because of the increasingly smaller amount of energy available, the number of organisms also, normally, declines as trophic level increases along with weight of the organisms. This decline can be illustrated with a pyramid. The base, primary producers, is large while the top, e.g. tertiary consumers, is significantly smaller in both number and energy available. The pyramid additionally explains why large communities of lower trophic levels are necessary to support only a few higher level trophic consumers (with only 10% of the energy of a lower trophic level available, a higher trophic level must consume large quantities of the lower organisms). http://luptonbiology.weebly.com/iii-energy-transfer.html FRQ: Consider a terrestrial ecosystem containing four trophic levels. If the first trophic level has 10,000 J of energy available, how many J of energy is available to the fourth? Why does this number decrease, increase, or remain the same? Are there any cases in which energy transfers are similar to this terrestrial ecosystem but biomass comparisons are not? If the fourth trophic level of this ecosystem was removed would the third grow uncontrollably? What might “check” its growth? MCQuestion: A recent article published in the local newspaper claimed that if humans became herbivores world hunger could possibly end. Is this a BIOLOGICALLY sound claim?A) No. Humans must remain as omnivores to receive all necessary nutrients.. B) No. By moving humans to a lower trophic level more energy would need to be produced by producers. C) Yes. By moving humans to a lower trophic level more energy would be available and a greater number of organisms, people, could be supported from the same number of producers. D) Yes. Humans would be able to consume less energy per organism increasing the number of organisms that could survive.

10. Answer Key • MCQuestion: A recent article published in the local newspaper claimed that if humans became herbivores world hunger could possibly end. Is this a BIOLOGICALLY sound claim?A) No. Humans must remain as omnivores to receive all necessary nutrients.. • B) No. By moving humans to a lower trophic level more energy would need to be produced by producers. • C) Yes. By moving humans to a lower trophic level more energy would be available and a greater number of organisms, people, could be supported from the same number of producers. • D) Yes. Humans would be able to consume less energy per organism increasing the number of organisms that could survive. 10 J, only 10%, according to the 10% rule, of energy from one trophic level is available to the next. The number decreases because at each trophic level some energy is lost in feces and through heat. Other energy is used respiration and creating new biomass. While some other energy is never harnessed by the next trophic level as some organisms are never eaten. Yes, in some aquatic ecosystems primary producers still produce energy and primary consumers receive only about 10%. But in some exceptional ecosystems, such as the English Channel, a small producer population, phytoplankton measured by biomass, supports a larger consumer population, zooplankton measured by biomass. No, it would not grow uncontrollably. For sometime, without predators, the population would grow but this would end as energy supplies became low from the lower trophic level being exhausted by the growing population. Additionally, competitors at the third trophic level could “check” the growth of the population by using energy resources as well. • FRQ: Consider a terrestrial ecosystem containing four trophic levels. • If the first trophic level has 10,000 J of energy available, how many J of energy is available to the fourth? • Why does this number decrease, increase, or remain the same? • Are there any cases in which energy transfers are similar to this terrestrial ecosystem but biomass comparisons are not? • If the fourth trophic level of this ecosystem was removed would the third grow uncontrollably? What might “check” its growth?

11. LO 3.26: The student is able to explain the connection between genetic variations in organisms and phenotypic variations in populations.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: The main source of genetic variation is mutation, or changes in the nucleotide sequence of DNA. Mutations in reproductive cells are passed to the next generation. Mutations occur very often, but are usually caught before they become harmful. In a population, harmful mutations are selected against. Recombination is also a source of genetic variation. Recombination includes independent assortment and crossing over, which produce gametes with unique chromosomal sequences. Another cause of genetic variation within a population is gene flow. If an individual organism becomes a part of a new population, it brings its alleles to the new population. The genetic variation of this population will increase and phenotypic expression is likely to change. Genetic variation is shown through gene expression. DNA is transcribed into mRNA, which is then translated into proteins. Nitrogenous bases are the building blocks of DNA, and have an effect on the form and function of amino acid sequences. 3 base sequences of nitrogenous bases, called codons, code for specific amino acids. Genotype also affects phenotype. They are shown through dominant: AA; heterozygous: Aa; and recessive: aa. A punnet square can be used to determine the genotype. If a dominant allele is paired with a recessive allele, it takes on the phenotype of the dominant allele. An example is Mendel’s experiment with the flowers. MCQ: Which of the following is an example of a missense mutation (a type of point mutation), that is often common in African-Americans? This mutation is expressed through a phenotypically recessive characteristic and results in a defect in the hemoglobin gene. a) Down syndrome b) Sickle-cell anemia c) Klinefelter Syndrome d) Huntington’s Disease FRQ: Describe 3 ways genetic variation can occur and how each can affect phenotypic variation in a population.

12. LO 3.26- Answer Key MCQ: Which of the following is an example of a missense mutation (a type of point mutation), that is often common in African-Americans? This mutation is expressed through a phenotypically recessive characteristicand results in a defect in the hemoglobin gene. • Down Syndrome • Sickle-cell anemia • Klinefelter Syndrome • Huntington’s Disease FRQ: Describe 3 ways genetic variation can occur and how each can affect a phenotypic variation in a population. One way genetic variation can occur is through genetic recombination, or the process where 2 DNA molecules exchange genetic information, producing a new combination of alleles. Recombination results in an increase of genetic variation among a population. This also means physically, or phenotypically, the population will have more variation. Gene flow can also influence genetic variation. Also called genetic migration, this is when alleles or genes are transferred from one population to another. This may introduce new genes to a population that it has not been exposed to before. Mutations greatly influence genetic variation. A mutation is simply a change of an organism’s nucleotide sequence. Mutations occur sometimes because damage to DNA was not caught and repaired. Errors in the process of replication are common, but are usually caught before the cell growth becomes a problem. Mutations may or not result in a change in phenotype. It does however, affect the survival rate of the organism and can result in a less diverse gene pool.

13. Explanation: Natural and artificial ecosystems that lack species diversity have been proven to be often less resilient to environmental changes around them. This is because these changes could kill off a species important to the food chain, so one species begins to overpopulate and kill off its food supply because there are too many. This starts a chain that eventually reaches the plants themselves, which could cause erosion and soil degradation and cause the entire ecosystem to die out. Keystone organisms, producers (mostly if not always plants), and other important biotic and abiotic factors help maintain the diversity in ecosystems. However, their effect on the ecosystem is severely disproportionate to their relative population size, so when they are removed from the ecosystem, the ecosystem, in many cases, collapses. LO 4.27: The student is able to make scientific claims and predictions about how species diversity within an ecosystem influences ecosystem stability. SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models M.C. Question: If the keystone species in an ecosystem is removed or dies out, which of the following would be the most likely result? The primary consumers would overpopulate and kill off the main producer species. There would be too little prey for predators, so they would die out. There would be an increase in diversity because a predator would be eliminated. There would be no change, as keystone species aren’t essential to the ecosystem. FRQ: Suppose in an marine ecosystem, phytoplankton are the main producers for the ecosystem. What would happen to the population size of the phytoplankton if the primary consumer for that ecosystem was killed off? How would it effect the ecosystem as a whole? What if the primary predator for that ecosystem was killed off? How would it effect the ecosystem as a whole?

14. Answer Key LO 4.27 If the keystone species in an ecosystem is removed or dies out, which of the following would be the most likely result? • The primary consumers would overpopulate and kill off the main producer species. • There would be too little prey for predators, so they would die out. • There would be an increase in diversity because a predator would be eliminated. • There would be no change, as keystone species aren’t essential to the ecosystem. Suppose in an marine ecosystem, phytoplankton are the main producers for the ecosystem. What would happen to the population size of the phytoplankton if the primary consumer for that ecosystem was killed off? Why? How would it effect the ecosystem as a whole? What if the primary predator for that ecosystem was killed off? Why? How would it effect the ecosystem as a whole? If the primary consumer in this ecosystem were to be killed off, the population size of the phytoplankton would increase drastically because they would have no predator. The population size is normally checked by the primary consumers consuming the phytoplankton, but if they are removed, they would have nothing checking their growth and they would grow drastically. This would eventually kill off the ecosystem because there would no longer be a food chain, so the predators would die along with everything but the phytoplankton. However, a new consumer may take the place of the removed one and the ecosystem would survive. Either way, ecosystem diversity would decrease. If the primary predator for that ecosystem was killed off, the phytoplankton population would essentially disappear because of overpopulation by the primary consumers. With no predator to check their population size, the consumers would grow and eventually consume most or all of the phytoplankton in the ecosystem. This would reduce ecosystem diversity because that would eliminate the producers and the predators above because they would have no food chain.

15. 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: The environment has played a large role in speciation and extinction throughout the Earth’s history. Today humans largely contribute to species extinction, by over-hunting and destroying species habitats (deforestation especially). Before, extreme climate change was the main cause of a species extinction due to a species inability to adapt quickly enough, and therefore they would die. Speciation is the formation of new species helping to create more biodiversity during evolution. The two main types of speciation are allopatric and sympatric speciation. For example, Charles Darwin researched finches, ground finches and tree finches, located in the Galapagos Islands, discovering that natural selection had taken place. Habitat isolation, or geographical isolation, caused the finches to interact rarely allowing for genetic mutations to occur. These mutations led to natural selection and two distinctive gene pools, ultimately resulting in two distinctive finch species. This resulted in the different beak shapes. This geographical/ habitat isolation speciation is called allopatric speciation. His data findings also indicated sympatric speciation in which several of the ground finches, or tree finches, which were not geographically separated, were still unable to reproduce with one another. This indicated they were two separate species. This was probably due to several types of isolation including: temporal isolation, meaning species do not breed at the same time; behavioral isolation, where a two species might have a different mating song/ritual; mechanical isolation, meaning reproductive parts do not match up…etc. Another example of sympatric speciation is by plants throughout history. Chromosomal changes can occur causing a plant to have extra sets of chromosomes, called being polyploidy. These extra sets of chromosomes do not allow for interbreeding because meiosis is unable to function normally. This creates reproductive isolation. Through analyzing data collected in experiments such as Darwin’s finches, we are able to gain further insight into how speciation occurs in populations. Multiple Choice Question: In a study conducted in the Grand Canyon, Harris found that two closely related antelope squirrels (the ammospermophilusharrisi and the ammospermophilusleucurus) living on different rims of a canyon were no longer able to reproduce. Which of the following statements is true? A) Geographical isolation caused for allopatric speciation to occur between the two squirrel relatives. B) Geographical isolation caused for sympatric speciation to occur between the two squirrel relatives. C) One of the squirrel populations will go extinct due to lack of reproductive ability. D) Due to behavioral isolation, sympatric speciation occurs causing the two squirrel species to distance themselves from one another. Learning Log/FRQ-style Question: Two species that once shared a common ancestor can become two separate species through two main types of speciation (allopatric and sympatric speciation). The eastern and western meadowlark are not separated by any geographical barriers. Although they look physically similar (feather color, body shape/size…etc), they are two distinct biological species. In detail, explain which of type of speciation would best explain this speciation that occurred, and why?

16. Answer Key-LO 1.20 In a study conducted in the Grand Canyon, Harris found that two closely related antelope squirrels (the ammospermophilusharrisi and the ammospermophilusleucurus) living on different rims of a canyon were no longer able to reproduce. Which of the following statements is true? A) Geographical isolation caused for allopatric speciation to occur between the two squirrel relatives. B) Geographical isolation caused for sympatric speciation to occur between the two squirrel relatives. C) One of the squirrel populations will go extinct due to lack of reproductive ability. D) Due to behavioral isolation, sympatric speciation occurs causing the two squirrel species to distance themselves from one another. Two species that once shared a common ancestor can become two separate species through two main types of speciation (allopatric and sympatric speciation). The eastern and western meadowlark are not separated by any geographical barriers. Although they look physically similar (feather color, body shape/size…etc), they are two distinct biological species. In detail, explain which of type of speciation would best explain this speciation that occurred, and why? The Western and Eastern Meadowlark Although the western and eastern meadowlark are physically similar their inability to reproduce is likely due to sympatric speciation. Sympatric speciation deals more with reproductive isolation than geographic isolation. Reproductive isolation is generated by extra chromosomes (polyploidy) that are unable to pair causing meiosis to be abnormal. This all happens without any type of geographic isolation. Behavioral isolation is a type of reproductive isolation that could have resulted in this sympatric speciation. This would mean that the two meadowlarks have different mating songs or other courtship rituals that prevent them from being attracted to one another. Another type of reproductive isolation that could have caused this sympatric speciation is temporal isolation, in which the two birds have different mating seasons/times and therefore would not mate. A third type of reproductive isolation that could cause this type of speciation is mechanical isolation. This is where the two birds reproductive parts do not fit together and therefore they are unable to reproduce, indicating two different species. For these reasons/ types of reproductive isolations the western and eastern meadowlark are different species due to sympatric speciation.

17. M.C. Question:A strand of E. coli bacteria is cultured for several generations in a medium containing a heavy isotope of nitrogen, 15N. The bacteria is then placed in a medium containing the lighter isotope of nitrogen, 14N, for one generation. How would you expect the DNA of the E. coli to have changed in construction due to their environment after this generation? • A. Every double helix of DNA will contain one old strand with nitrogenous bases composed entirely of 15N and one new strand composed with 14N nitrogen. • B. Each nitrogenous base will be made up of both 14N and 15N nitrogen. • C. Adenine and guanine nitrogenous bases will be composed of 14N while cytosine and thymine nitrogenous bases will be composed of 15N . • D. All of the DNA will be composed of the heavier, stronger nitrogen, 15N . • FRQ Question: In many labs, levels of photosynthesis or cellular respiration are collected and compared. In order to determine what conditions, such as what intensity of light, chloroplasts are most efficient at, describe the data you would collect, how you would collect it, and why this data would help in the overall determination of specific rates of photosynthesis. LO 2.8: The student is able to justify the selection of data regarding the types of molecules that an animal, plant or bacterium will take up as necessary building blocks and excrete as waste products. SP 4.1: The student can justify the selection of the kind of data needed to answer a particular scientific question. Explanation: In order to grow, reproduce, and maintain organization, animals, plants and bacteria must take up molecules including carbon, nitrogen, phosphorous, and water. Carbon is needed for forming carbohydrates, proteins, lipids, and nucleic acids which are the four classes of organic compounds. Carbohydrates are used for fuel and for building materials and composed of carbon, hydrogen and oxygen. Organisms take up oxygen through cellular respiration and take up both hydrogen and oxygen through water consumption. In plants, through the process of photosynthesis, carbon dioxide and water are catalyzed by light to convert to sugar, oxygen and water. Sugar is used to power the cell while oxygen is a waste product. The reactants and products are reversed in the process of cellular respiration, which is found in animals, plants and bacteria. In respiration, oxygen is taken up and carbon dioxide is released. The oxygen taken up is then used as the final electron receptor in the Electron Transport Chain in order to make ATP and power the cell. Nitrogen is also an essential building block for organisms as it is used both in building proteins and nucleic acids. Animals breathe in nitrogen in the air and plants absorb it from the soil with the assistance of nitrogen-fixing bacteria. Phosphorous is also taken up by animals, plants, and bacteria to form nucleic acids and some lipids. Organisms also depend on water for multiple reasons including its utilization in reactions. In hydrolysis, water molecules are used to break apart and break down other molecules in the process of digestion. In dehydration synthesis, on the other hand, water molecules are produced and released as other molecules join together. Different data sets can provide clues as to which processes are occurring and what organic compounds are being constructed from the necessary building blocks that are taken up. For photosynthesis and cellular respiration processes, rates of carbon dioxide or oxygen release signify the relative rates of the entire process. Rates in different conditions can then be compared.

18. ANSWER KEY-LO 2.8 A strand of E. coli bacteria is cultured for several generations in a medium containing a heavy isotope of nitrogen, 15N. The bacteria is then placed in a medium containing the lighter isotope of nitrogen, 14N, for one generation. How would you expect the DNA of the E. coli to have changed in construction due to their environment after this generation? A. Every double helix of DNA will contain one old strand with nitrogenous bases composed entirely of 15N and one new strand composed with 14N nitrogen. B. Each nitrogenous base will be made up of both 14N and 15N nitrogen. C. Adenine and guanine nitrogenous bases will be composed of 14N while cytosine and thymine nitrogenous bases will be composed of 15N . D. All of the DNA will be composed of the heavier, stronger nitrogen, 15N . In many labs, levels of photosynthesis or cellular respiration are collected and compared. In order to determine what conditions, such as what intensity of light, chloroplasts are most efficient at, describe the data you would collect, how you would collect it, and why this data would help in the overall determination of specific rates of photosynthesis. To determine the efficiency levels of chloroplasts I would utilize their production of oxygen to measure the overall rates of photosynthesis. The chemical reaction equation for photosynthesis is: 6 CO2 + 12 H2O + light energy → C6H12O6 + 6 O2 + 6 H2O, with sugars, oxygen and water as the products. Because of this, the quicker the rate of photosynthesis, the higher the levels of oxygen produced. In order to determine the relative rates of oxygen production I would use a respirometer to measure the levels of oxygen. In this way I could collect the amount of oxygen produced by each collection of chloroplasts and compare the rates of production. To test the effects of the intensity of light on overall photosythesis effectiveness I would set up several identical test tubes containing a chloroplast solution at varying intensities of light, including one facing direct light, one having semi-direct light, and one with indirect light. After measuring the amounts of oxygen produced I would have a more certifiable idea of how light intensity affects photosynthesis. Repeated trails and large sample sizes would allow my data to be even more reliable.

19. Signal Transduction and Responses • LO: 3.38 – The student is able to describe a model that expresses key elements to show how a change in signal transduction can alter cellular responses. • SP: 1.5 – The student can re-express key elements of natural phenomena across multiple representations of the domain • There are three main steps in signal transduction. Reception occurs when a ligand binds to a receptor protein embedded within the cell membrane. Transduction is the cascade of chemical signals that takes place within the cell after reception has occurred that triggers the third step of this process, the response. • A couple examples of a cellular response: • a liver cell releasing glucose into the bloodstream • the activation of transcription enzymes in the nucleus. • In this example, we will look at how the hormone epinephrine triggers the release of glucose from liver cells. • Epinephrine binds to a G-protein linked receptor embedded in the cell membrane. The activation of this protein causes the release of a section of the G-protein called the alpha subunit. This subunit travels through the cytosol and binds to a receptor protein called adenylyl cyclase. This protein is also found embedded within the cell membrane. When the alpha subunit attaches to adenylyl cyclase, that protein becomes active. The job of adenylyl cyclase is to convert ATP molecules to cAMP (cyclic adenosine-monophosphate). ATP is chemically attracted to adenylyl cyclase, and when it reaches it, two phosphate groups are cleaved off, leaving one phosphate group behind. The open bonds that are left form a ring structure that gives cAMP its name. Adenylyl cyclase is the enzyme that is most responsible for amplifying the G-protein linked response simply because as long as the alpha subunit is bound to it, ATP will be continuously converted to cAMP. The cAMP that is converted will bind to protein kinases (action proteins). Specifically, they will bind to the regulatory subunits of the protein kinase, allowing the catalytic subunits to break off and acquire phosphate groups from ATP. The catalytic subunits are now active and can bind to other proteins that directly trigger responses. The catalytic subunits bind to a protein called phosphorylase which moves along chains of glucose (glycogen), separating individual glucose molecules to be released into the bloodstream. *This cellular response elevates blood sugar levels, increases the heart rate, and dilates skeletal blood vessels for easier transport of blood and hormones through the body.* • M.C. Question: What protein or chemical directly influences the activation of a protein kinase in signal transduction? • A) Epinepherine • B) Adenylyl Cyclase • C) cAMP • D) the G-protein • Learning Log/FRQ: Explain why it is important for ATP to be converted to cAMP in a signal transduction pathway. Which protein is responsible for facilitating this conversion? You may draw a picture as an aid to your explanation, but be sure to adequately address the question in writing.

20. Answer Key – LO 3.38 M.C. Question: What protein or chemical directly influences the activation of a protein kinase in signal transduction? A) Epinepherine B) Adenylyl Cyclase C) cAMP D) the G-protein Learning Log/FRQ: Explain why it is important for ATP to be converted to cAMP in a signal transduction pathway. Which protein is responsible for facilitating this conversion? You may draw a picture as an aid to your explanation, but be sure to adequately address the question in writing. Answer: ATP must be converted to cAMP so that it can bind to a protein kinase’s regulatory subunits, freeing its catalytic subunits. These catalytic subunits become active by binding to ATP so that they can directly trigger a response. Adenylyl cyclase is the protein that is responsible for converting ATP to cAMP.

21. 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 6.4The student can make claims and predictions about natural phenomena based on scientific theories and models. • Explanation: Fitness is the contribution that an individual makes to the gene pool of the next generation, relative to the contributions of other individuals. It involves the contribution of a genotype to the next generation compared to other genotypes of the same locus. The distribution of these traits can be altered in three ways. Directional selection, which is the most common, is when the overall makeup of the population is shifted toward one extreme of the phenotype. Disruptive selection is when the selection is favored at both extreme ends of the phenotypic distribution. Stabilizing selection removes the extreme variants and preserves the intermediate phenotypes. Hardy-Weinberg equations allow us to predict genotypic frequencies of future generations using the allelic frequencies that we have in the current population. p2 + 2pq + q2 is the equation that you use, where p2 is the percentage of homozygous dominant individuals, q2 is the percentage of homozygous recessive individuals, and 2pq is the percentage of heterozygous individuals. Polymorphism, when two or more distinct alleles are present at a specific locus, is a way that genetic variation is displayed within a population. Average heterozygosity is a quantitative way that shows the proportion of individuals in a population that are heterozygous at a certain locus. Why is genetic polymorphism important to evolution? a) Variability between individuals provides the basis for which natural selection acts upon b) Genes can’t mutate unless polymorphic c) Only heterozygous individuals are selected in natural populations d) Hardy-Weinberg equilibrium is less likely to be disturbed in polymorphic populations e) None of the above; polymorphism isn’t important to evolution Successfully doing the tango is a homozygous recessive trait (tt). In all Americans, 16% of people have this genotype. What percentage of the population would be heterozygous? What would be an advantage of being heterozygous?

22. Why is genetic polymorphism important to evolution? a) Variability between individuals provides the basis for which natural selection acts upon b) Genes can’t mutate unless polymorphic c) Only heterozygous individuals are selected in natural populations d) Hardy-Weinberg equilibrium is less likely to be disturbed in polymorphic populations e) None of the above; polymorphism isn’t important to evolution Successfully doing the tango is a homozygous recessive trait (tt). In all Americans, 16% of people have this genotype. What percentage of the population would be heterozygous? What would be an advantage of being heterozygous? q2 = 0.16, so q = 0.4 If q = 0.4, then p = 0.6 p2 + 2pq + q2 = (0.4)2 + 2(0.4)(0.6) + (0.6)2 2pq = percentage of heterozygous individuals 2pq = 2(0.4)(0.6) = 0.48 48% of individuals will be heterozygous for the tango success trait Sometimes, individuals who are heterozygous for a particular trait have greater fitness that homozygotes. Heterozygotes can usually be protected against effects that may affect the homozygotes. In this case, you will never be the worst tango dancer on the dance floor, but you will also not be the best. Not being the best can, however, be an advantage because you don’t want everyone to watch you. Being heterozygous for tango dancing can definitely have its advantages. I love FITNESS!

23. LO 1.30 The student is able to evaluate scientific hypotheses about the origin of life on Earth. SP 6.5 The student can evaluate alternative scientific explanations. • Explanation: Scientific evidence has shown that processes on early Earth could produce simple cells through four main steps: 1.the abiotic synthesis of organic molecules;2.the joining of these small molecules into polymers3.the packing of molecules into protobionts4.the origin of self-replicating molecules that eventually made inheritance possible. The 1rst atmosphere was had water vapor, N, CO2, CH4, NH3, H in it. A. I. Oparin & J. B. S. Haldane postulated that Earth’s early atmosphere had been a reducing (electron-adding) environment, in which organic compounds could have formed simple molecules using energy from lightning and intense UV radiation. Haldan suggested that the early oceans were a solution of organic molecules from which life arouse. Stanley Miller & Harold Urey tested the Oparin-Haldane hypothesis by creating laboratory conditions close to those that were on early earth which yielded many amino acids found in organisms today. The joining of these monomers produces polymers with the ability to replicate, store and transfer information. Some of the organic compounds may have come from space. Protobionts are aggregates of abiotically produced molecules surrounded by a membrane with some of the properties associated with life, including simple reproduction and metabolism and the maintaining of an internal chemical environment different from that of their surroundings. The RNA World hypothesis proposes that RNA could have been the earliest genetic material. • MC Q: What are protobionts? • aggregates of biotically produced molecules surrounded by a membrane with some of the properties associated with life. • aggregates of abiotically produced molecules surrounded by a membrane with none of the properties associated with life. • aggregates of biotically produced molecules surrounded by a membrane with none of the properties associated with life. • aggregates of abiotically produced molecules surrounded by a membrane with some of the properties associated with life. • FRQ: Explain the environment & process on early Earth that created molecules such as protobionts and experiments that proved it.

24. Answer Key L.O 1.30 • MC Q: What are protobionts? • aggregates of biotically produced molecules surrounded by a membrane with some of the properties associated with life • aggregates of abiotically produced molecules surrounded by a membrane with none of the properties associated with life • aggregates of biotically produced molecules surrounded by a membrane with none of the properties associated with life • aggregates of abiotically produced molecules surrounded by a membrane with some of the properties associated with life FRQ: Explain the environment and process on early Earth that created molecules such as protobionts and experiments that proved it. Earth and the other planets formed about 4.6 billion years ago and it is unlikely that life could have survived those first hundred million years because chunks of rock and ice were still crashing into the planet. Once that slowed down, the conditions on Earth changed. This first atmosphere probably had water vapor, nitrogen and its oxides, methane, ammonia, hydrogen , and hydrogen sulfide. The water vapor turned into oceans as the Earth cooled. Chemist Oparin and Haldane independently postulated that the early atmosphere had been a reducing-electron adding- environment which would have allowed organic compounds to form from simple molecules using energy from lightening or intense UV radiation and it is from this in which life could have arouse. Scientists Miller and Urey test their hypothesis by creating conditions in a closed system close to those thought to have existed on early Earth. A warm flask of water represented the ocean, the atmosphere was filled with H2, CH4, NH3 and water vapor. Sparks were discharged in the representative atmosphere to represent lightening. A condenser cooled atmosphere, raining water and any dissolved compounds into the “sea.” As the material circulated through the apparatus they collected sample for analysis. They found many different organic molecules including amino acids found in organisms today. The conclusion was that organic molecules, a first step in the origin of life, can form in a strongly reducing atmosphere. The chemical and physical processes on early Earth produced very simple cells through sequence of four main stages: 1. the abiotic synthesis of small organic molecules, such as amino acids and nucleotides; 2. the joining of these monomers into polymers including proteins and nucleic acids. 3. the packaging of these molecules into protobionts, droplets with membranes that maintained an internal chemist y different from that of their surrounding s 4. the origin of self-replicating molecules that eventually made inheritance possible.

25. 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 Explanation: Gene expression is regulated at the DNA, RNA, and the protein level. All somatic cells in one organism contain the same DNA, however, there are switches/activators that turn these genes on or off. This process generates different functions in cells. Before DNA undergoes transcription, it can be regulated a few ways: DNA methylation is a process that turns off DNA and involves CH3 caps (blocks RNA polymerase); histone acetylation turns on a gene by unwinding DNA; and epigenetic inheritance, which is the inheritance of genes other than the nucleotide sequence. DNA is also regulated at the RNA level. First, RNA polymerase must recognize and bind to DNA at the promoter. Then, transcription is triggered at the TATA box, which is where transcription factors bind. There are also enhancers (activators) and silencers (repressors) involved in transcription. RNA processing involves spliceosomes, which remove introns and allow exons, expressed sequences, to leave the nucleus. Once DNA is transcribed into a complementary RNA sequence and translated into codons of mRNA, the amino acids eventually form a protein. Once the protein is in the cytoplasm it can tagged with ubiquitin and degraded. P53 protein assists in apoptosis (programmed cell death). If this pathway is blocked it can lead to cancer. M.C. Question: What effect doeshistone acelylation have on transcription and gene expression? A)By unwinding the chromatin, the DNA is accessible for transcription, allowing gene expression to occur B)It adds methyl groups to DNA and turns off gene expression C)It forces the chromatin into a compact structure, inhibiting gene expression to occur D)It activates transcription of a gene by binding to the enhancer Learning Log/FRQ-style Question: Regulation of gene expression involves certain mechanisms that activate and inhibit. Discuss how these mechanisms operate and their effect on the gene. -dna methylation vs histone acetylation -activators vs repressors -introns vs exons -mRNA degradation Alternative gene splicing

26. ANSWER KEY- LO 3.23 Regulation of gene expression involves certain mechanisms that activate and inhibit. Discuss how these mechanisms operate and their effect on the gene. -dna methylation vs histone acetylation -activators vs repressors -introns vs exons -mRNA degradation DNA methylation represses transcription in the nucleus by capping of the DNA strand with CH3 caps. In contrast, histone acetylation unwinds compact chromatin structures, allowing DNA access to undergo transcription. Enhancers are involved at the RNA level by activating transcription. There are different activators present in different cells that activate different genes. These operate by binding to other transcription factors and mediator proteins which trigger transcription on the promoter. Repressors operate differently in that they repress gene expression by blocking the binding of activators. Introns, noncoding segments of mRNA, only exist in eukaryotes, and are removed by spliceosomes. This process allows for exons to be expressed. MicroRNA (miRNA) blocks expression of mRNA molecules. This involves a Dicer that that cuts RNA into shorter segments. Similarily, RNA interference (RNAi) pathway can cause the destruction of RNA, and ultimately contributes to regulating gene expression in the cell. What effect doeshistone acelylation have on transcription and gene expression? A)By unwinding the chromatin, the DNA is accessible for transcription, allowing gene expression to occur B)It adds methyl groups to DNA and turns off gene expression C)It forces the chromatin into a compact structure, inhibiting gene expression to occur D)It activates transcription of a gene by binding to the enhancer Activators & repressors mRNA degradation

27. A) LO 3.27 The student is able to compare and contrast processes by which genetic variation is produced and maintained in organisms from multiple domains. B) 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. C) In this topic there are 3 main things we need to focus on: Mutation, crossing over, and independent assortment. Mutations are caused by mistakes in DNA of an organism and are passed on to the next generations if it makes the organism more likely to survive. Crossing over happens in sexual reproduction when chromosome segments “cross over” between chromatids during prophase I of meiosis, which creates a completely new combination of genes. If an organism uses mitosis to asexually reproduce, then crossing over will NOT occur. Independent assortment is when chromosomes are randomly chosen between parents for the offspring. This will NOT happen in asexual bacteria. When asexual bacteria reproduce, their offspring have exactly the same set of chromosomes that the parent did. D) Which of the following statements are TRUE? a.) Bacteria use meiosis to grow and to increase genetic variation. b.) Mitosis increases genetic variation by randomly selecting chromosomes between parents for the offspring. c.) Crossing over involves chromatids of homologous chromosomes pairing up to decrease genetic variation. d.) Mutations originate because a trait helped the organism in survival and reproduction. e.) None of the above. E) Discuss 3 mechanisms that help increase genetic variation.

28. G) The correct answer to the multiple choice question is E, None of the above. a) Bacteria use meiosis to grow and toincrease genetic variation. FALSE. Bacteria use mitosis to grow and these cells are identical, there is no genetic variation here. b) Mitosis increases genetic variation by randomly selecting chromosomes between parents for the offspring. FALSE. Mitosis is the division of cells into identical cells, there is no genetic variation here. Randomly selecting chromosomes between parents for the offspring doesn’t even exist in mitosis. c) Crossing over involves chromatids of homologous chromosomes pairing up to decrease genetic variation. FALSE. Crossing over involve chromatids of homologous chromosomes pairing up, but to increase genetic variation, not decrease. d) Mutations originate because a trait helped the organism in survival and reproduction. FALSE. This obviously leaves e) None of the above to be the answer. This answer choice is false because mutations occur randomly in the genetic code, they are not chosen to occur through evolution. FRQ: Discuss 3 mechanisms that help increase genetic variation. Three mechanisms which help increase genetic variation are crossing over, mutations, and independent assortment. Crossing over is what happens in prophase 1 in meiosis, and this is where two chromatids literally cross over and mix up their genes, which obviously results in more variation. Mutations occur when there are mistakes in DNA replication, so the genetic code is changed. The organism with the mutation will either die, be unaffected, or benefit. An example would be if a giraffe had a mutation where its neck was twice the length it normally is so it can reach higher branches. Natural selection would take effect and that trait will be chosen for the next generation, because it helped this giraffe survive. Independent assortment is when chromosomes are randomly selected from the parents. The result of it are genetically unique gametes.

29. LO 1.4: The student is able to evaluate data-based evidence that describes evolutionary changes in the genetic makeup of a population over time. SP 5.3: The student can evaluate the evidence provided by data sets in relation to a particular scientific question. Explanation: Evolution is the gradual change over time in which you can tell that it occurs by fossils and the similarities in homologous structures. To know if a certain population is evolving, you would do it though the Hardy-Weinberg Theorem. (Micro)Evolution occurs through gene flow, mutations, genetic drift, bottle neck population, founder effect, non random sampling, and natural selection. M.C Question: Darwin's evolutionary theory included which of the theory...? A. Natural selection: occurs through interactions between the environment and the variability among individual organisms in a population B. Evolution: a gradual change over time C. Sexual Selection: marked differences between the sexes in secondary sexual characteristics D. A & B Learning Log/ FRQ- style Question: Discuss the differences between Darwin’s and Lamarck’s theory of evolution. Identify and explain two of Darwin’s theory in how it could alter a population’s genetic composition. Include the evidence of evolution.

30. ANSWER KEY LO 1.4 M.C Question: Darwin's evolutionary theory included which of the theory...? A. Natural selection: occurs through interactions between the environment and the variability among individual organisms in a population B. Evolution: a gradual change over time C. Sexual Selection: marked differences between the sexes in secondary sexual characteristics D. Evolutionary Adaptation: an accumulation of inherited characteristics that enhance organisms' ability to survive and reproduce in specific environment E. A & B & D Learning Log/ FRQ- style Question: The difference between Darwin’s and Lamarck’s theory of evolution was that Darwin believed that the main mechanism for evolution was through a process called natural selection where the favorable traits are passed on causing it to be more common while the unfavorable traits are not passed down causing it to be less common which also cause the favorable traits to be more fitted to the environment. On the other hand, Lamarck believed that organisms that acquired certain traits during their lifetime will be passed down to their offspring, in other words, Darwin believed that traits were already present and gained and changed during their lifetime while Lamarck believed that traits were only acquired through lifetime and then passed to the offspring. Two types that could alter a population’s genetic composition are genetic drift in which is the change in the frequency of a gene allele in one’s population due to random sampling and the bottleneck effect in which occurs when one’s population is dramatically decreased due to environmental factors. This causes one’s population to be at different places (genetic drift being a factor) and cause it to have many substantial effects on the gene pool. This then causes the evidence of evolution due to comparing the similar homologous structures with different functions and show signs of evolution of a common ancestor which may include vestigial organs. Other evidence may include comparative embryo and biogeography.

31. LO 1.10 The student is able to refine evidence based on data from many scientific disciplines that support biological evolution. SP 5.2 The student can refine observations and measurements based on data analysis. By reviewing evidence based on data from scientific disciplines such as molecular biology, biogeography, and comparative embryology, one can make a strong support for biological evolution. Using tools such as BLAST one can prove that organisms with similar DNA could have been related. If a student were to use the BLAST program to compare two organisms’ nucleotide DNA sequences for a gene, the more nucleotide similarities they had together, the greater the chances are that they are related, or evolved from one another. Biogeography is the study of the geographical distribution of species over time. One could prove two organisms evolved from one another by proving that they have similar structures and occupy similar niches in similar environments. Comparative embryology helps to prove evolution because the embryo’s of most animals look very similar. As seen in the picture below, the embryo of a chicken and human early in embryonic development look similar. M.C. Question: Which of the following is NOT a supporting factor for the theory of biological evolution? Homologous structures Comparative Embryology Molecular Biology Similar Adaptations Fossils Learning Log/FRQ-style Question: Explain how using such tools as BLAST could prove that a newly discovered fossil (with a soft tissue sample) could be related to a previously related organism.

32. M.C. Question: Which of the following is NOT a supporting factor for the theory of biological evolution? Homologous structures Comparative Embryology Molecular Biology Similar Adaptations Fossils Learning Log/FRQ-style Question: Explain how using such tools as BLAST could prove that a newly discovered fossil (with a soft tissue sample) could be related to a previously related organism. Since the fossil has a soft tissue sample, once could collect a DNA sample and determine the sequence of nucleotides by doing a process such as automated chain-termination DNA sequencing. Once the nucleotide is known, one could use the BLAST website to cross-reference the DNA in question against all preexisting DNA sequences. If there is an organism with a high percentage match, then the two organisms are/were related.

33. LO 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: Evidence of biological evolution uses information from geographical, geological, physical, chemical , as well as mathematical applications. These applications can help us look back at data and the genetic information of living and extinct animals to help give us a greater understanding of evolution. Data such as graphs or charts can be used to show the growth or decline in a specific trait, gene, or physical feature in a species. Biochemical and genetic similarities in certain DNA nucleotides and protein sequences provide evidence of evolution and natural selection. Geographical evolution can be seen through the sugar glider and flying squirrel, both live on different continents but have similar appearances which suggest that through natural selection and similar environments they have adapted the same characteristics as each other. Although the sugar glider is closely related to other Australian marsupials rather then North American eutherian’s. Physical evolution characteristics through data can be seen below in the figure displaying the bone structures. Humans along with other mammals have similar bone structure, although the homologous structures have different functions the have the same skeletal elements. M.C. Question: Of the following graph which is the correct statement about the finches beaks during the drought? A) As the drought goes on the beaks get smaller. B) The smaller beaked birds dyed out during the drought only leaving the larger beaked birds to survive. C) The large beaked bids ate the smaller beaked birds which caused them to become the dominate species. D) Not enough data to conclude. Learning Log/FRQ-style Question: Compare and contrast the Anatomical similarities and differences in the chicken and human embryos, and what can be determined about there ancestry? Be sure to label the similar parts between the chicken and human. Also include one other structures that humans have while they are still embryos?

34. Answer Key LO 1.9 M.C. Question: 1)Of the following graph which is the correct statement about the finches beaks during the drought? A) As the drought goes on the beaks get smaller. B) The smaller beaked birds dyed out during the drought only leaving the larger beaked birds to survive. C) The large beaked bids ate the smaller beaked birds which caused them to become the dominate species. D) Not enough data to conclude. Learning Log/FRQ-style Question: Compare and contrast the Anatomical similarities and differences in the chicken and human embryos, and what can be determined about there ancestry? Be sure to label the similar parts between the chicken and human. Also include one other structures that humans have while they are still embryos? -Some of the anatomical similarities of the chicken and human embryos are that they both have post-anal tail’s and pharyngeal pouches or gills, this can be determined that at one point in time ourselves and chickens had a common ancestor due to similarities. Another structures along with gills and a tail that humans have are webbed feet and hands, which suggest that through evolution we have adapted to land.

35. LO 2.32 – The student is able to use a graph or diagram to analyze situations or solve problems (quantitatively or qualitatively) that involve timing and coordination of events necessary for normal development in an organism. Interdigital tissue SP 1.4 – The student can use representations and models to analyze situations or solve problems quantitatively or qualitatively. Explanation – Many mechanisms regulate timing and coordination of molecular, physiology and behavioral events that are needed for an organism’s development and its survival. In order for a cell to develop, cell differentiation, cell division or morphogenesis must occur. This comes from the expression of genes, and the induction of transcription factors during development result in different gene expression. Once those genes have been expressed and structures have been established, apoptosis plays a major role in normal development by eliminating certain regions in order to form digits in organisms. Timing and Coordination come into play for normal development because individuals can act on information and communicate it to others and those responses to information become important to natural selection. This is evident in example behaviors like courtship rituals, hibernation and even migration. This communication contributes to survival of populations and individuals and to normal development within individuals and their ecosystems. Multiple Choice Question: The ultimate basis for the differences between cells is transcription regulation, but it’s induction that brings about differentiation in the genes. Which of the following model organisms is a more useful model for investigating the roles of cell signaling, induction, and programmed cell death? Drosophila Arabidopsis C. elegans MyoD Zygote (fertilized egg) Adult animal (sea star) Blastula (Cross Section) Gastrula (cross section) Eight Cells Cell Division Animal and Plant Development Morphogenesis Observable Cell Differentiation • Free Response Question: A newly cloned species of goats were successfully created and were released into the a controlled community over a year ago. Researchers saw a multitude of abnormalities and deaths in goats each month and a few conclusions were made about the as to why the abnormalities and deaths were occurring: • Nothing was changed in apoptotic cells, and programmed cell death did not occur • There were more methyl groups than the DNA equivalent cells from uncloned embryos of the goat species • Explain why each could be a cause of the prevention of normal development, use a Diagram to further explain or prove one of the two possibilities. Zygote (fertilized egg) Two Cells Embryo inside seed Plant Apoptosis

36. Answer Key LO 2.32 Ced-9 protein (active) inhibits Ced-4 activity Mitochondrion Ced-4 Ced-3 Death signal receptor Inactive proteins Cell forms blebs (a) No death signal Ced-9 (inactive) Death signal Active Ced-3 Active Ced-4 Other proteases Nucleases Activation cascade (b) Death signal Multiple Choice Answer: The ultimate basis for the differences between cells is transcription regulation, but it’s induction that brings about differentiation in the genes. Which of the following model organisms is a more useful model for investigating the roles of cell signaling, induction, and programmed cell death? Drosophilia Arabidopsis C. elegans MyoD • Free Response Answer: • Nothing was changed in apoptotic cells, and programmed cell death did not occur • There were more methyl groups than the DNA equivalent cells from uncloned embryos of the goat species • Explain why each could be a cause of the abnormalities or deaths in the population, use a Diagram to further explain or prove one of the three possibilities. • On the diagram to the right, we see that for a change to be seen and for cell death to occur, a cell would have to receive a death signal, the Ced-9 would be inactivated and it would relieve its inhibition of Ced-3 and Ced-4. That Ced-3 would trigger a lot of actions leading up to the activation of nucleases and proteases. There could have also been no death signal and the Ced-9 may have not been located on the outer membrane, and apoptosis would’ve been inhibited • Due to the fact that DNA methylation helps regulate gene expression, the misplaced methyl groups in the DNA of donor nuclei may interfere with the pattern of gene expression necessary for normal embryonic development

37. Learning Objective 2.24: The student is able to analyze data to identify possible patterns and relationships between a biotic or abiotic factor and a biological system (cells, organism, populations, communities, or ecosystems. Science Practice 5.1: The student can analyze data to identify patterns or relationships. Explanation: Abiotic factors in the environment are non living factors such as temperature, water availability, and sunlight. Biotic factors are all of the living things in the environment. The cell can be affected by abiotic factors, and water can cause a cell to expand and contract. Also, in plants sunlight is required for the cells to undergo photosynthesis in order for the plant to grow. Organisms can also be effected by these factors through predator-prey relationships. These are biotic factors that can help to grow or shrink a population. If there is an abundance of a predator such as a snake, then there will be a smaller mouse population; this also can happen in reverse. The predator-prey relationship keeps the population to a size that is conducive to its environment. Other factors can effect the success of a population in regards to these relationships. This is seen in red kangaroos. The distribution of red kangaroos may be able to be determined by both abiotic and biotic factors or patterns. These kangaroos only live in stable environments that do not have a variable climate that changes from wet to dry. These kangaroos mostly enjoy dry climates(which are abiotic), the higher temperatures allow them to thrive. In many other organisms symbiosis is a huge relationship. Biotic factors feed off of each other in order to survive. One way this is seen is through mutualism where both organisms involved are benefitted. An example of mutualism is the relationship between a human and microorganisms. The “good” bacteria in our digestive tract breaks down food into small useable compounds, while they benefit from the nutrients. Scientists can use data such as soil pH levels and nutrient levels to determine the optimal soil conditions for plants to grow. A healthy, thriving ecosystem that flows well with the perfect mixture of factors will live in harmony and keep the biological systems running. • Multiple Choice Question: Which of the following interactions between biotic and abiotic factors would benefit the entire ecosystem? • The abundance of light and warm temperatures allowing for more plants to grow a plentiful amount of food for primary and secondary consumers. • Mosses growing on the side Oak trees taking in the water for themselves, but still giving water to the trees. • Foxes overpopulating an area, and killing all of the rabbits that eat the wildflowers. This leads to rapid growth of wildflowers. • Temperature fluctuates in an area in the Tropic of Cancer, causing different seasons of extreme hot and extreme cold. This leads to the decrease in parasites and disease causing fungi leading to exponential growth of producers. FRQ Style Question: Discuss how abiotic and biotic factors work together to create a positive environment in the food chain. How do these factors benefit the individuals in the ecosystem, and name two ways that biotic factors work to control populations, and give two examples.

38. Multiple Choice Question: Which of the following interactions between biotic and abiotic factors would benefit the entire ecosystem? a. The abundance of light and warm temperatures allowing for more plants to grow a plentiful amount of food for primary and secondary consumers. b. Mosses growing on the side Oak trees taking in the water for themselves, but still giving water to the trees. c. Foxes overpopulating an area, and killing all of the rabbits that eat the wildflowers. This leads to rapid growth of wildflowers.d. Temperature fluctuates in an area in the Tropic of Cancer, causing different seasons of extreme hot and extreme cold. This leads to the decrease in parasites and disease causing fungi leading to exponential growth of producers. FRQ Style Question: Discuss how abiotic and biotic factors work together to create a positive environment in the food chain. How do these factors benefit the individuals in the ecosystem, and name two ways that biotic factors work to control populations, and give two examples. Answer: Without abiotic and biotic factors working in harmony the food chain would be disrupted. There must be a balance of light, water, temperature, and a perfect pH level in the soil in order for first level producers to grow. These producers can then grow and use light to provide oxygen to the environment as well as be a source of nutrients for herbivores. Without the first levels of the food chain, nutrients could not be passed through the ecosystem to the larger predators who get their energy and nutrients from smaller herbivores and carnivores. Individuals in an ecosystem are benefited in different ways. Plants are benefitted greatly from abiotic factors. Without light, and a neutral pH level they would not be able to survive. Abiotic factors must be perfect, this includes the amount of water availability that is accessible to plants and animals for survival. Predator-prey relationships control populations by keeping that amount of animals at each trophic level conducive to the resources in the area. Foxes keep the rabbit population down by hunting the proper amount and staying in their territories. Parasites also help to keep populations under control because they infect populations, allowing for the strongest individuals to survives. This helps keep the food chain healthy, and helps keep the area proportionate to the resources that can be abiotic or biotic. *Examples can be anything that represents relationships between living and/or nonliving factors. *

39. LO 3.30: The student is able to use representations and appropriate models to describe how viral replication introduces genetic variation in the viral population. SP 1.4: The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. Explanation: Viruses cannot reproduce on their own so they depend on host cells to allow them to produce copies of themselves and transfer genetic material. When found outside of host cells, viruses exist as a protein coat or capsid, sometimes enclosed within a membrane. The capsid encloses either DNA or RNA which codes for viral elements. Viral replication allows for mutations to occur through usual host pathways. Mutation is the major source of genetic variation in viruses. Mutations occur in retroviruses and RNA viruses due to the error prone nature of RNA polymerases. They have a higher rate of mutation than DNA viruses because DNA polymerases have proofreading activity whereas RNA viruses do not. Another major source of genetic variation comes from transduction. Transduction happens through either the lytic cycle or the lysogenic cycle in bacteria. In the lytic cycle, the viral DNA will infect and break up the host cell’s chromosome so that only the viral DNA remains. Thus, it can replicate and create many phages which will then cause the cell to lyse (burst into pieces) and release all the phages. These released phages can go on to infect other host cells and can initiate the lysogenic cycle which is the cycle where most of the genetic incorporation and variation can occur. In the lysogenic cycle, a phage DNA is inserted into the host cell by a bacteriophage (bacterial virus), which then integrates itself into the host cell’s bacterial chromosome. Thus, when the host cell replicates, the viral DNA will also be reproduced along with the bacteria’s own original DNA. These mechanisms provide the much needed genetic diversity to the viral population. M.C Question: What is one reason why it is difficult to make effective vaccines to prevent diseases caused by RNA viruses? A) The glycoproteins on the viral envelope of the RNA virus fight off the vaccines B) Antibodies only respond to the vaccines that prevent diseases caused by DNA viruses C) DNA polymerase used by RNA viruses is error prone and lacks efficient RNA proofreading capabilities, causing frequent mutations D) RNA polymerase used by RNA viruses is error prone and lacks efficient DNA proofreading capabilities, causing frequent mutations Learning Log/FRQ-style Question:Viruses cannot replicate on their own so they must seek out host cells. Describe in detail the two methods in which the genome of bacteriophages can become incorporated into a host cell through transduction. Which method is the quickest? Why?

40. ANSWER KEY- LO 3.30 M.C Question: What is one reason why it is difficult to make effective vaccines to prevent diseases caused by RNA viruses? A) The glycoproteins on the viral envelope of the RNA virus fight off the vaccines B) Antibodies only respond to the vaccines that prevent diseases caused by DNA viruses C) DNA polymerase used by RNA viruses is error prone and lacks efficient RNA proofreading capabilities, causing frequent mutations D) RNA polymerase used by RNA viruses is error prone and lacks efficient DNA proofreading capabilities, causing frequent mutations Viruses cannot replicate on their own so they must seek out host cells. Describe in detail the two methods in which the genome of bacteriophages can become incorporated into a host cell through transduction. Which method is the quickest? Why? Bacteriophages first uses its tail fibers to bind to specific receptor sites on the outer surface of a bacterial cell. Then, the sheath of the tail contracts, injecting the phage DNA into the cell and leaving an empty capsid outside. The phage DNA will then circularize and after this, certain factors will determine whether the lytic cycle is induced or the lysogenic cycle. If the lytic cycle is induced, the new phage DNA “hijacks” the bacterial host cell and degrades its DNA. It then synthesizes new phage DNA and proteins using components within the cell. Three separate sets of proteins self-assemble to form phage heads, tails, and tail fibers. The phage DNA is packaged inside the capsid as the head forms. The accumulation of phages directs production of an enzyme that breaks the cell wall, allowing fluid to enter. The cell swells and finally bursts, releasing 100 to 200 phages. Now if the lysogenic cycle is induced, the phage DNA integrates into the bacterial chromosome, becoming a prophage. The bacterium reproduces normally, copying the prophage and transmitting it to daughter cells. Many cell divisions produce a large population of bacteria infected with the prophage. This cycle repeats until an environmental signal causes a prophage to switchover from the lysogenic cycle to the lytic cycle. The lytic cycle is the quickest cycle because it almost immediately degrades the host cells DNA and starts the synthesizing of new phages. It can take many generations and generations of division through the lysogenic cycle to create a pathogenic enough prophage to match the virulence of a lytic cycle phage.

41. LO 4.4 The student is able to make a prediction about the interactions of subcellular organelles. SP 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. EXPLANATION: Eukaryotic cells contain subcellular organelles that help maintain life within the cell. The brain center of the cell, or the nucleus, is the home of the DNA that controls the activities within the cell. Within the nucleus is the nucleolus, which is a non-membrane bound tangle of chromatin which synthesizes ribosomes. Ribosomes are made on RNA that produces proteins within the cell and are sometimes attached to the membrane of the rough endoplasmic reticulum (ER). The smooth ER helps with the synthesis of steroid hormones/lipids, detoxifies the cell, and connects the rough ER with the Golgi bodies. Golgi bodies receive the vesicles that were sent by the ER and repackage and send off the substances to their designated location, often the lysosomes. Lysosomes digest the dead or useless cell parts with digestive enzymes and then recycle them. Similar to the lysosome, peroxisomes contain catalase which converts H2O2 into H20 and 02. H202 is a waster product of cell respiration. The site of cellular respiration is the mitochondria. Mitochondria have 2 membranes: a smooth outer membrane and the convoluted cristae. Mitochondria produce ATP for the cell. Mitochondria contain their own DNA and are able to reproduce without the help of the nucleus. The cytoskeleton, made up microtubules and microfilaments, help support the structure of the cell. The centrioles help organize the spindle fibers during cell division. The plasma membrane encases the entire cells and helps transport necessary molecules in and out of the cell. M.C. Question: Smooth endoplasmic reticulum does all of the following for the cell EXCEPT: A) Detoxification B) Lipid production C) Connection to the Golgi body D) Steroid hormone production E) Protein synthesis FRQ: What is the importance of the mitochondria? Why is it better for the cell that the mitochondria can reproduce on its own?

42. ANSWER KEY – LO 4.4 M.C. Question: Smooth endoplasmic reticulum does all of the following for the cell EXCEPT: A) Detoxification B) Lipid production C) Connection to the Golgi body D) Steroid hormone production E) Protein synthesis FRQ: What is the importance of the mitochondria? Why is it better for the cell that the mitochondria can reproduce on its own? Without a mitochondria, the cell would not have enough energy, or ATP, to be able to sustain function. Cell respiration requires ATP to be able to break down glucose to form energy to power all kinds of cellular work. Also active transport of vital molecules across the plasma membrane requires ATP. Cell division (mitosis) requires ATP to occur successfully. Since the mitochondria have their own DNA, they have the capability to reproduce on their own. This benefits the cell because the more mitochondria they have the more energy they will have to maintain their function. Also it helps to quickly replace the old/dead mitochondria that get digested by the lysosomes.

43. L.O 3.29 The student is able to construct an explanation of how viruses introduce genetic variation in host organisms. S.P Example of phenomena based on scientific practices. Explanation: Viruses inject themselves into host cells in order to be reproduced and introduce genetic variation. A membranous envelope (sometimes) surrounds the capsid of most viruses. The viral envelope (capsid) is derived from the membrane of the host cell allowing it to target that particular cell. Due to the virus having the same phospholipids and membrane proteins (capsomeres) it can enter the host cell. Once it enters the host cell, host enzymes replicate the viral DNA. Meanwhile host enzymes also transcribe the viral genome (capsid proteins) into mRNA that can be used to make more viral proteins. Newly formed viral genomes and capsid proteins self-assemble into new virus particles which exit the cell, thus creating variation. Another way this can happen is through the lysogenic cycle, with the lytic cycle being described earlier. In the lysogenic cycle the once the virus enters the cell it can integrate into the genome of the host cell. This is now a prophage that will continue to copy it’s DNA, transmitting daughter cells. Many cell divisions result in a large population of bacterial chromosomes, yielding genetic variation. M.C. Question: What allows a virus to enter a host cell? Certain cells can only be affected by certain viruses, for example respiratory cells can only be infected by the influenza virus. The capsomeres of the virus allow it to diffuse through the membrane of the host cell. The capsid of the virus attaches to the host cell and breaks down the membrane of the host cell allowing it to enter the host cell. Viruses have designated cells to target upon entering the body. A virus changes it’s structure upon deciding which cell to infect. Learning Log/FRQ-style Question: What do you expect to happen to the surrounding cells of a cell infected with a bacteriophage virus and how does this contribute to genetic variation?

44. ANSWER KEY – LO 3.29 What allows a virus to enter a host cell? A) Certain cells can only be affected by certain viruses, for example respiratory cells can only be infected by the influenza virus. B) The capsomeres of the virus allow it to diffuse through the membrane of the host cell. C) The capsid of the virus attaches to the host cell and breaks down the membrane of the host cell allowing it to enter the host cell. D) Viruses have designated cells to target upon entering the body. E) A virus changes it’s structure upon deciding which cell to infect. What do you expect to happen to the surrounding cells of a cell infected with a bacteriophage virus and how does this contribute to genetic variation? Phages make a lysozyme, which degrades the bacterial cell wall. Once the cell wall is broken, lysisoccurs. This allows the virus to be released into the cytoplasm and the newly made copies of the virus from the host cell can now attach to new cells and continue to enter the cell and weave their DNA into newly entered cells, further contributing to genetic variation.

45. LO 3.45 - The student is able to describe how nervous system transmit information SP 1.2 - The student can describe representations of natural or man-made phenomena and systems in the domain. Explanation: Inside a neuron cell is at -70 mV due to the negatively charged substances (DNA, RNA, and proteins). The neuron has potassium ions (K+) inside, and sodium ions (Na+) surround the outside of the cell. When the potassium channels open, the cell becomes more negative (hyperpolarization). When the sodium channels open, the cell becomes less negative (depolarization). If the cell depolarizes to the threshold of -55 mV, the signal fires, generating action potential, which causes more Na+ channels to open and the response to become stronger. The nerve impulse starts at the axon hillock and runs down the axon following the negative charge (because the more positive signal is attracted to the negative part of the cell). Depolarization and re-polarization occurs at the Nodes of Ranvier (gaps in the Myelin sheath) and so the impulse is not continuous, but almost ‘jumps’ from node to node. Active transport is used to re-polarize the cell, pumping the entered Na+ ions back out of the cell, and exited K+ ions into the cell. The chemical signal reaches the axon terminals, where it is converted to a chemical signal (a neurotransmitter) and sent across the synapse with the help of Ca ions, where it becomes a ligand to ligand-gated ion channels, which can allow K+ to leave (hyperpolarize) or Na+ to enter (depolarize – the signal continues). FRQ: Explain why nerve signaling is not as efficient in patients with Multiple Sclerosis. Multiple Choice: A nerve cell needs to re-polarize after firing because… K+ ions need to enter the cell again, and Na+ ions need to exit in order to restore the cell to its original state using active transport. Na+ ions need to enter the cell again, while K+ ions need to exit in order to restore the cell to its original state using active transport. K+ ions need to enter the cell again and Na+ ions need to exit in order to restore the cell to its original state using facilitated diffusion. Na+ ions need to enter the cell again, while K+ ions need to exit in order to restore the cell to its original state using passive transport through the outer membrane. Na+ ions and K+ ions need to exit the cell in order to make the cell less positive and to restore the charge of -70 mV.

46. Answer Key – LO 3.45 FRQ: Explain why nerve signaling is not as efficient in patients with Multiple Sclerosis. Multiple Sclerosis (MS) is an autoimmune disease, where the immune system attacks self cells, specifically the insulating Myelin Sheath that surrounds the axon of a nerve cell. The Myelin Sheath is made of individual Schwann Cells, allowing the signal to “jump” from the small gaps between the insulation (an area called a Node of Ranvier). When the immune system targets the Myelin, the nodes become wider, making the signal not be able to travel down the axon as efficiently. In a healthy individual, the nodes of Ranvier are where depolarization and re-polarization occur, causing the action potential to travel quickly. When the nodes are more spaced out due to Myelin deterioration, the signal cannot move through the axon as quickly. A nerve cell needs to re-polarize after firing because… • K+ ions need to enter the cell again while Na+ ions need to exit in order to restore the cell to its original state using active transport. • Na+ ions need to enter the cell again while K+ ions need to exit in order to restore the cell to its original state using active transport. • K+ ions need to enter the cell again while Na+ ions need to exit in order to restore the cell to its original state using facilitated diffusion. • Na+ ions need to enter the cell again, while K+ ions need to exit in order to restore the cell to its original state using passive transport through the outer membrane. • Na+ ions and K+ ions both need to exit the cell in order to make the cell less positive and to restore the charge of -70 mV. Nerve Cell that has yet to fire

47. 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. After T.H. Morgan discovered that genes were located on chromosomes, the scientific community became obsessed with the search for the “genetic material.” By the 1940’s, they had narrowed the candidates down to two: DNA and proteins. Through many famous experiments, scientists finally discovered the role of DNA. Frederick Griffith discovered by experimenting with mice in 1928 that the genetic trait of pathogenicity could be transferred between bacteria through what he called a transformation. This indicated the presence of an unknown, heritable substance. Oswald Avery built off of Griffith’s experiment, purifying various types of molecules from the heat-killed pathogenic bacteria, then tried to transform live nonpathogenic bacteria with each type. Only DNA worked. He announced along with Maclyn McCarty and Colin MacLeod in 1944 that DNA was the transforming agent, but was greeted by much skepticism by the scientific community. In 1952, Alfred Hershey and Martha Chase used radioactive sulfur and phosphorus to trace the fates of protein and DNA of T2 phages that infected bacterial cells. In the experiment, phage proteins remained outside of the bacterial cells during infection while phage DNA entered the cell. When cultured, bacterial cells with radioactive phage DNA released new phages with some radioactive phosphorus. Hershey and Chase concluded that DNA, not protein, functions as the T2 phage’s genetic material. Multiple Choice Question In the Hershey-Chase experiment, _______ entered the cells during infection. A) Sulfur-tagged protein B) Phage heads C) Sulfur-tagged DNA D) Phosphorus-tagged DNA E) Phosphorus-tagged protein Free Response Question Explain in detail TWO of the following historical experiments and describe their significance in proving that DNA was the source of heritable information. I. Griffith’s Transformation experiment II. Avery-McCarty-MacLeod Experiment III. Hershey-Chase experiment The Hershey-Chase Experiment

48. Answer Key Multiple Choice Question In the Hershey-Chase experiment, _______ entered the cells during infection. A) Sulfur-tagged protein B) Phage heads C) Sulfur-tagged DNA D) Phosphorus-tagged DNA E) Phosphorus-tagged protein Free Response Question Explain in detail TWO of the following historical experiments and describe their significance in proving that DNA was the source of heritable information. I. Griffith’s Transformation experiment II. Avery-McCarty-MacLeod Experiment III. Hershey-Chase experiment I. Griffith experimented with mice and two strains of bacteria: pathogenic S-cells and nonpathogenic R-cells. He killed the S-cells with heat and then mixed them with the R-cells. Though the R-cells alone did not harm the mice, and the dead S-cells alone did not harm the mice, when mixed, the mice were killed. Griffith hypothesized that the R cells had been “transformed” into S-cells by some unknown “transforming principle.” Though the identity of the transforming principle remained unknown, Griffith’s experiment set the stage for the many experiments yet to come that would prove DNA to be the source of heritable information. II. The Avery-McCarty-MacLeod ultimately proved that the transforming principle Griffith had discovered was indeed DNA. By purifying various types of molecules from the heat-killed pathogenic bacteria, then attempting to transform live nonpathogenic bacteria with each type, they discovered that DNA alone was successful. This provided concrete evidence that DNA was the explanation behind bacterial transformation and therefore was the source of heritable information. III. In the Hershey-Chase experiment, E-coli bacteria and bacteriophages were studied to test what entered the bacteria when it was infected by the phage. First, phage proteins were tagged with radioactive sulfur. After the phages infected the bacteria, a blender was used to separate phages outside the bacteria from the bacterial cells . They then centrifuged the mixture so that bacteria formed a pellet at the bottom of the test tube. The radioactivity in the pellet and the liquid were measured. They found radioactivity in the liquid, indicating that the proteins had not entered the bacteria. They then repeated the process with radioactive-phosphorus tagged DNA and found radioactivity in the pellet, indicating that DNA had entered the bacteria at the time of infection. By proving that DNA was the material that was transferred during infection, the Hershey-Chase experiment provided very convincing evidence that DNA was the source of heritable information.

49. LO 2.5: The student is able to construct explanations of the mechanisms and features of cells that allow organisms to capture, store or use free energy. [See SP 6.2] • SP 6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices. • Explanation: Biological processes (e.g., active transport across the plasma membrane) require energy. Organisms use two main processes (chemosynthesis and photosynthesis) to take free energy in from their environments and convert it into a more usable form of energy, but photosynthesis is the most notable. In photosynthesis, environmental free energy in the form of light energy is used to excite electrons in the photosystems of the chloroplasts, the energy from which is ultimately used to build up complex carbohydrates, such as glucose, from atmospheric carbon dioxide. These carbohydrates represent a mechanism for the storage and transfer of free energy; the energy is stored in the structure of the carbohydrate, and transferring of the carbohydrate between organisms (e.g., by predation) also transfers the free energy. Cellular respiration, which occurs in the mitochondria, is the reverse process wherein this stored free energy is released into ATP as a result of the mitochondrion’s breakdown of the food molecule to reduce electron carriers and then using the resultant electrons to create an H+ concentration gradient to power ATP synthase. This ATP is de-phosphorylated back into ADP by sub-cellular structures, such as transport pumps in the plasma membrane which need the energy stored in the bond between the phosphate molecules in order to operate. The science practice requires that students understand the mechanism of these metabolic processes and how a change in environmental conditions (e.g., oxygen availability) could affect them. • MC Question: In which of the following scenarios does the cell serve as a sink for environmental free energy: • I. A photosynthesizing leaf cell on a tree II. A muscle cell in a rhinoceros III. A lung cell in a human • I only • III Only • I and III only • I, II, and III • FRQ • The electron transport chain plays a key role in many cellular processes involved in the fixing or consumption of environmental free energy. Discuss the role and importance of the electron transport chain in both photosynthesis and cellular respiration. Image Credit: http://media-1.web.britannica.com/eb-media/89/22489-004-EE9DC4F5.jpg Mitochondria convert free energy stored in food molecules into more easily used free energy stored in the bond between the last three phosphates in ATP.

50. Answer to MC Question: A, as only a photosynthesing cell absorbs and stores environmental free energy, the others consume it by breaking down glucose and then using the resulting ATP. Answer to FRQ: In both cellular respiration and photosynthesis, the electron transport chain accepts excess electrons from other sources , such as electron carriers NADH and NAPDH, and uses the energy released by those electrons moving down the chain to pump hydrogen ions against their concentration gradient, thus providing ATP synthase an H+ gradient with which to power its spinning that allows it to create ATP. In photosynthesis, the electron transport chain is used to transfer energy between Photosystem II and Photosystem I, and it also produces ATP by ATP synthase that is used to provide the energy to build up complex carbohydrates during the Calvin Cycle. The added input from ATP is necessary to drive the Calvin cycle, as building molecules up is necessarily an endergonic reaction, thus necessitating the use of two electron transport chains in photosynthesis (one between the photosystems and a shorter, non-ATP producing one for NAPDH). The electron transport chain is more fundamental to the successfully completion of cellular respiration than it is for photosynthesis, as the electron transport chain is directly responsible for most of the product of cellular respiration. However, the mechanics of the chain in both processes are essentially identical, with electrons moving “down” series of proteins in a membrane, with the change in the electrons’ energy being used to power a proton pump. The electron transport chain is the final step in cellular respiration. It produces far more ATP than glycolysis, and its consumption of the electron acceptors NADH and FADH2 from the Krebs cycle rather than food molecules allows it t be much more specialized and efficient.