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1 st Review Project collection

1 st Review Project collection. LO 2.7: The student is able to explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination. SP 6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices

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1 st Review Project collection

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  1. 1st Review Project collection

  2. LO 2.7: The student is able to explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination. SP 6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices Explanation:Cells typically are small, even microscopic, to maximize their ability to take in nutrients and eliminate waste. Aside from size, the shape and structure of a cell affects its functionality. The two types of cells, prokaryotic and eukaryotic, are structurally disparate. The eukaryotic cells are usually bigger than the prokaryotic cells and they are more sophisticated. Eukaryotic cells have internal membranes, which divide them into compartments. These membranes are composed of enzymes which contribute to the cell’s metabolism. The prokaryotic cells function at a very rudimentary level, expending just enough energy to sustain themselves and reproduce. The most efficient cells have a higher ratio of surface area to volume. The surface area of each cell type’s plasma membrane must be large enough to adequately exchange materials; smaller cells have a more favorable surface area to volume ratio for exchange of materials with the environment. In general, smaller eukaryotic cells have a higher rate of nutrient intake and waste elimination. Multiple Choice Question:You are performing a lab experiment where you are examining three cells. Cell A and Cell B each have a membrane bound nucleus and internal membranes. The surface area to volume ratio of Cell A is larger than that of Cell B. Cell C appears to have no true nucleus, and also lacks many of the membrane-enclosed organelles found in Cell A and Cell C. Which of these cells would have a higher rate of nutrient intake? Cell A Cell B Cell C Cell A and Cell B Free Response Question: Describe the differences between a prokaryotic and a eukaryotic cell in terms of their rate of metabolic potential. Why is one type of cell more efficient than the other? Give 2 examples of cells that have evolved to have increased surface area to volume ratios and as a result are highly efficient. Microvilli is an example of a cell that is critical in the exchange of materials. It has evolved to have increased surface area and high SA/V ratios and as a result is highly efficient.

  3. Answer Key LO 2.7 • Multiple Choice Question: You are performing a lab experiment where you are examining three cells. Cell A and Cell B each have a membrane bound nucleus and internal membranes. The surface area to volume ratio of Cell A is larger than that of Cell B. Cell C appears to have no true nucleus, and also lacks many of the membrane-enclosed organelles found in Cell A and Cell C. Which of these cells would have a higher rate of nutrient intake? • Cell A • Cell B • Cell C • Cell A and Cell B • Free Response Question: Describe the differences between a prokaryotic and a eukaryotic cell in terms of their rate of metabolic potential. Why is one type of cell more efficient than the other? Give 2 examples of cells that have evolved to have increased surface area to volume ratios and as a result are highly efficient. • A eukaryotic cell is much more efficient in its rate of metabolic potential than a prokaryotic cell. The eukaryotic cell’s various membranes which are filled with enzymes help to increase the metabolic rate of the cell. Eukaryotic cells maintain internal membranes that partition the cell into specialized regions. Internal membranes facilitate cellular processes by minimizing competing interactions and by increasing surface area where reactions can occur. A prokaryotic cells function a very basic level, expending just enough energy to sustain themselves and reproduce. They do not have as many membrane-enclosed organelles to help speed the process along. Two examples of cells that have evolved to have increased surface area to volume ratios and as a result are highly efficient are hair roots and cells of the Alveoli.

  4. 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: Phylogenetic trees and cladograms can represent traits that are either derived or lost due to evolution. Phylogenetic trees and cladograms illustrate speciation that has occurred, in that relatedness of any two groups on the tree is shown by how recently two groups had a common ancestor. Groups can be monophylectic, paraphylectic, or polyphylectic. Phylogenetic trees and cladograms can be constructed from morphological similarities of living or fossil species, and from DNA and protein sequence similarities. M.C. Question: Based on the character table, which organisms are most closely related, and why? The shark and the lamprey, because neither have lungs. The tiger and the shark, because both have jaws. The gorilla and the human, because both are bipedal. Both A and B. LO 1.15 The student is able to describe specific examples of conserved core biological processes and features shared by all domains or within one domain of life, and how these shared, conserved core processes and features support the concept of common ancestry for all organisms. FRQ: Define and explain the ways that phylogeny can be traced. Explain how these methods can help provide us with a clearer evolutionary history.

  5. Answer Key- LO 1.15 Multiple Choice Free Response Based on the character table, which organisms are most closely related, and why? • The shark and the lamprey, because neither has lungs. • The tiger and the shark, because both have jaws. • The gorilla and the human, because both are bipedal. • Both A and B. Define and explain the ways that phylogeny can be traced. Explain how these methods can help provide us with a clearer evolutionary history. Phylogeny can be traced through relative dating and absolute dating. Relative dating is based on the layer of soil that a fossil is found in. If the fossil is found in a higher soil level, then it is younger, and if it is found in a lower soil level, then it is older. Absolute dating is done through radiometric dating. It involves using an isotope such as carbon-14, and measuring the rate of decay to determine how many years ago a fossil was alive. These methods allow for a clearer evolutionary history, because they provide us with a timeline of evolution. We are able to track a fossil through the years and determine its features and see how they have changed several years later.

  6. 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. Explanation- Species diversity is a term used to describe the number of different species represented in various populations. It is directly proportional to an ecosystem’s stability, meaning that an ecosystem with low species diversity is more likely to have individual species go extinct than a species with great diversity. This is because populations with greater diversity are more resilient to changes in the ecosystem. Ecosystems with greater diversity are more resilient to change because they have more variation as a result of mutations which could be potentially beneficial to survival. Keystone species are vital in maintaining species diversity because they are species that make a large impact on the community. The loss of a keystone species will result in the lessening of species diversity, in turn decreasing the ecosystem’s stability. Other factors that contribute to the maintenance of species diversity, and the stability of the ecosystem as a whole, are producers, biotic factors, and abiotic factors. Multiple Choice Question: In 1950 there was a world wide drought that affected all of the ecosystems. Based on the chart to the right, keystone species played the most instrumental role in which of the ecosystems and why? A: Desert; it has the lowest number of species B: Desert; following the drought, it experienced the lowest drop C: Tropical Rainforest; it had the greatest number of species both before and after the drop, meaning it was stable and resilient to change D: Savanna; it experienced the greatest drop following the drought Free Response Question: The Winston-Salem City Council plans to build a road through a portion of land that is currently uninhabited, yet full of trees, bushes, grasses, and is home to a plethora of species. What would be an ecologist’s primary concern with the building of a road through the field? Why would this be an issue? Following the construction of the road, there is now two distinct habitats that evolve to be separate. How would the biodiversity of one compare to the biodiversity of the other? Would the ecosystem be more resilient to change before or after the construction of the road? Explain. Explain why the loss of habitat poses a threat to biodiversity and survival of species in the ecosystem.

  7. Multiple Choice Question: In 1950 there was a world wide drought that affected all of the ecosystems. Based on the chart to the right, keystone species played the most instrumental role in which of the ecosystems and why? A: Desert; it has the lowest number of species B: Desert; following the drought, it experienced the lowest drop C: Tropical Rainforest; it had the greatest number of species both before and after the drop, meaning it was stable and resilient to change D: Savanna; it experienced the greatest drop following the drought Based on the graph, the tropical rainforest had the greatest number of species, hovering around 190 different species. In that ecosystem, it is evident the the keystone species are doing their job to stabilize the biodiversity. Following the drought, the number of species stayed about constant because the high level of biodiversity made it resilient to change. Free Response Question: The Winston-Salem City Council plans to build a road through a portion of land that is currently uninhabited, yet full of trees, bushes, grasses, and is home to a plethora of species. What would be an ecologist’s primary concern with the building of a road through the field? Why would this be an issue? Following the construction of the road, there is now two distinct habitats that evolve to be separate. How would the biodiversity of one compare to the biodiversity of the other? Would the ecosystem be more resilient to change before or after the construction of the road? Explain. Explain why the loss of habitat poses a threat to biodiversity and survival of species in the ecosystem. A) An ecologist’s primary concern with the building of a road would be the habitat fragmentation. The road would cut the habitat in two, lessening the number of species in each, which would in turn make species in each habitat more susceptible to change, increasing their chance of extinction. B) Assuming the species were evenly distributed throughout the portion of uninhabited land, once the road was build each of the two new habitats should have roughly the same levels of biodiversity. Since species diversity is the number of species, rather than the number or organisms of a species, the biodiversity of the two distinct habitats should be fairly similar since the range of the species was cut in half. C) The ecosystem would be more resilient to change before the construction of the road. Since the ecosystem wouldn’t be divided, barriers would not be intact, preventing organisms from mating. This would promote species diversity which would make the the ecosystem, as a whole, less susceptible to change and extinction. D) When the road was built, the ecosystem became two distinct habitats. Since the ecosystem is no longer one, barriers such as temporal barriers and behavioral barriers would arise, eventually resulting in reduced hybrid variability and fertility. In the long run, the construction of the road would lessen species diversity in each of the two new habitats, eventually causing the demise of many of the species.

  8. LO 2.13: The student is able to explain how internal membranes and organelles contribute to cell functions. SP 6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices. Explanation: Every part of the cell is essential to the survival and productivity of the cell. For example, internal membranes are vital to cells’ homeostasis due to their selectively permeable properties. Each individual organelle, in eukaryotes that is, is responsible for carrying out a specific cell function. For example, the mitochondrion produce ATP through cellular respiration, and the chloroplasts are the sites of photosynthesis, where oxygen is produced in plants. Confirmation of these specialized organelle functions can be tested through various scientific experiments. For example, photosynthesis could be studied through the observation of percent light transmission in chloroplasts when placed in a spectrophotometer under a variety of conditions (different colors of light, different temperatures, boiling/not boiling chloroplasts, etc.) Through practices such as this one, experimenters are able to observe cell functions, in this case how and why photosynthesis takes place. M.C. Question:One of the most important concepts in many aspects of biology is the idea that structure determines function. This can be seen in proteins, for example in the “lock and key” binding of a specifically shaped substrate to an enzyme that fits that shape. If and only if these two fit can the enzyme catalyze a reaction. This theory is also seen in the functioning of various cell organelles. Which of the following is true concerning how a specialized structure of an organelle determines its function? A) Chloroplasts have thylakoid membranes which cause protons to accumulate, ultimately allowing the organelle to conduct photosynthesis: the production of oxygen in all eukaryotic cells. B) Mitochondrion have inner membranes that have an intermembrane space between them. This space causes protons to accumulate and move through ATP synthase, thus conducting cellular respiration and producing ATP. C) Ribosomes are able to conduct protein synthesis because of the binding of DNA to the large subunit, one of the ribosome’s two specific parts. D) The true nucleus of a prokaryotic cell has coiled chromatin that contains the majority of the genetic information, but some of the cell’s DNA is also found in ribosomes. FRQ: Photosynthesis is a complex cell process that only occurs in plant cells. In a laboratory experiment recorded in the graph below, light transmission was observed in different groups of chloroplasts, and one control group with no chloroplasts. Based on the diagram, which group had the most successful photosynthetic rate? Explain why this occurred. According to the graph, the “percentage of light transmission” was measured in different groups over time. Explain what technique could have been used to measure the percentage of light transmission. Describe in detail the process of photosynthesis, including the following: i) Where photosynthesis takes place ii) Products and reactants of the process iii) Why animal cells cannot do this process Graph source: http://www.grochbiology.org/ aplabwriteupexample_files/ image004.gif

  9. LO 2.13 ANSWER KEY M.C. Question: Arguably one of the most important concepts in many aspects of biology is the idea that structure determines function. This can be seen in proteins, for example in the “lock and key” binding of a specifically shaped substrate to an enzyme that fits that shape. If and only if these two fit can the enzyme catalyze a reaction. This theory is also seen in the functioning of various cell organelles. Which of the following is true concerning how a specialized structure of an organelle determines its function? A) Chloroplasts have thylakoid membranes which cause protons to accumulate, ultimately allowing the organelle to conduct photosynthesis: the production of oxygen in all eukaryotic cells. - Incorrect – Only plant cells have chloroplasts, and thus only plant cells conduct photosynthesis. B) Mitochondrion have inner membranes that have an intermembrane space between them. This space causes protons to accumulate and move through ATP synthase, thus conducting cellular respiration and producing ATP. - Correct – The unique structure of the inner mitochondrial membrane allows for the intermembrane space as well as the mitochondrial matrix. Protons are forced through ATP synthase, which produces ATP in chemiosmosis, one of the steps of cellular respiration. C) Ribosomes are able to conduct protein synthesis because of the binding of DNA to the large subunit, one of the ribosome’s two specific parts. - Incorrect – Ribosomes conduct protein synthesis, but it is due to mRNA binding to the small subunit, not DNA binding to the large subunit. The function of ribosomes is to translate mRNA into chains of amino acids: proteins. D) The true nucleus of a prokaryotic cell has coiled chromatin that contains the majority of the genetic information, but some of the cell’s DNA is also found in ribosomes. - Incorrect – Prokaryotic cells do not have a true nucleus. They have one circular chromosome that is free-floating in the cytoplasm. FRQ: Photosynthesis is a complex cell process that only occurs in plant cells. In a laboratory experiment recorded in the graph below, light transmission was observed in different groups of chloroplasts, and one control group with no chloroplasts. Based on the diagram, which group had the most successful photosynthetic rate? Explain why this occurred. - The unboiled chloroplasts that were exposed to light had the most successful photosynthetic rate. This occurred because they absorbed the most light energy and were in their natural state: unboiled. Chloroplasts in the dark were not exposed to light, so they could not conduct photosynthesis because they could not take in any light energy. Being boiled also contributed to photosynthetic failure because they were disturbed and not in their natural, ideal conditions. According to the graph, the “percentage of light transmission” was measured in different groups over time. Explain what technique could have been used to measure the percentage of light transmission. - To measure percentage of light transmission, chloroplasts could be placed into cuvettes and into a spectrophotometer. This instrument, once blanked and set to a specific wavelength, can measure the percentage of light transmission through the chloroplasts. This data can then be used to calculate photosynthetic rates. c) Describe the process of photosynthesis, including the following: i) Where photosynthesis takes place – Photosynthesis takes place in the chloroplasts of plant cells, specifically in the thylakoid membranes which contain chlorophyll. ii) Products and reactants of the process – Photosynthesis is the process of plants taking in carbon dioxide, water, and light energy from the atmosphere and converting it into glucose, oxygen, water, and chemical energy in the form of ATP. iii) Why animal cells cannot do this process – Animal cells cannot conduct the process of photosynthesis because they do not have chloroplasts. Chloroplasts, and their thylakoid membranes they have inside, are unique to plant cells and are necessary to take in light energy and carry out photosynthesis properly.

  10. LO 2.18: The student can make predictions about how organisms use negative feedback mechanisms to maintain their internal environments. SP 6.4:The student can make claims and predictions about natural phenomena based on scientific theories and models. Explanation:Organisms use feedback mechanisms to maintain their internal environments and respond to external environmental changes. They do this by returning the changing condition back to its target set point to help maintain homeostasis. Negative feedback is a mechanism that either turns a process on or off, like a light switch.It occurs when the rate of the process decreases as the concentration of the product increases which controls the rate of a process to avoid accumulation of a product. For example, when humans get to hot, they sweat to help cool off. Another example is with operons in gene regulation. The repressor or protein that shuts off the operon is made when it binds to Trp. When there is enough Trp around, the Trp operon shuts off so that the cells can focus on other processes. M.C. Question: T3 and T4 are hormones produced by the Thyroid gland in the neck and are regulated by TSH from the pituitary gland in the brain. T3 acts by modifying gene transcription in virtually all tissues to alter cell metabolism. T4 is the precursor to T3. If a person has an overproduction of the thyroid hormones, T3 and T4, you would expect their TSH levels to do what? Refer to the diagram to the right. Rise Decline Stay the same Fluctuate Learning Log/FRQ-style Question: A person with Type II diabetes has insulin resistance forcing beta cells in the pancreas to produce large amounts of insulin to overcome that resistance. As the high levels of insulin in the diabetic bring the blood sugar down, what happens to the beta cells producing the insulin?Suppose a you have high blood glucose levels. Explain how negative feedback mechanisms will regulate your blood levels back to normal. How would negative feedback mechanisms regulate your blood back to normal if you have low blood glucose levels? "ReviseScience.co.uk." ReviseScience.co.uk. N.p., 1 Jan. 2005. Web. 27 Apr. 2014. <http://www.revisescience.co.uk/2011/schools/gaynes/21b210.asp>.

  11. Answer Key- LO 2.18 T3 and T4 are hormones produced by the Thyroid gland in the neck and are regulated by TSH from the pituitary gland in the brain. T3 acts by modifying gene transcription in virtually all tissues to alter cell metabolism. T4 is the precursor to T3. If a person has an overproduction of the thyroid hormones, T3 and T4, you would expect their TSH levels to do what? Refer to the diagram to the right. • Rise- wrong because the body would think there was already enough • Decline- correct because as the thyroid hormones rise, they inhibit TSH (thyroid stimulating hormone) slowing the production of hormones from the thyroid gland • Stay the same- wrong because the body is constantly trying to achieve homeostasis • Fluctuate – wrong because there is a negative correlation between the thryoid hormones and the TSH A person with Type II diabetes has insulin resistance forcing beta cells in the pancreas to produce large amounts of insulin to overcome that resistance. As the high levels of insulin in the diabetic bring the blood sugar down, what happens to the beta cells producing the insulin? Suppose a you have high blood glucose levels. Explain how negative feedback mechanisms will regulate your blood levels back to normal. How would negative feedback mechanisms regulate your blood back to normal if you have low blood glucose levels? As the high levels of insulin in the diabetic bring the blood sugar down, the beta cells are turned off to maintain a minimum level of glucose in the blood stream. When blood glucose levels are high, the production of insulin from the pancreas increases. Insulin transports the glucose into the cells to be used for energy. As the glucose levels fall it slows the production of insulin to maintain homeostasis. If the blood glucose levels fall to far, the production of insulin decreases while the production of adrenalin increases to stimulate the muscle to breakdown glycogen into glucose. This also stimulates gluconeogenesis in the liver.

  12. LO 3.6: The student can predict how a change in a specific DNA or RNA sequence can result in changes in a gene expression. SP 6.4: Students can make claims and predictions about natural phenomena based on scientific theories and models. Explanation: Random changes in DNA nucleotide sequences can lead to heritable mutations if they are not repaired. These mutations can consist of translocation (the chromosome moves to a different site), nondisjunction (chromosome fails to separate), duplication (chromosome segment doubles),and deletion (chromosomal segment is removed). Mutations can potentially alter levels of gene expression, pr remain silent and unnoticeable. These changes in a nucleotide sequence, if present in a protein-coding region, can change the amino acid sequence of the polypeptide. The product of the cells are controlled by the expression of the genetic material inside of it, and the product determines the nature of the cell. So if it is alter in any way, the nature of the cell could potentially be changed. Gene expression is regulation by environmental factors, like cell communication and cell peer pressure. But it is also regulated by developmental factors, which could be the sequence of DNA that is given . All of these are factors that can alter DNA or RNA sequences, causing them to change. The result of these alterations could be mutations, diseases, which can lead to the gene expression to physically change. Just because the genotype changes doesn’t mean the phenotype does, but the gene expression will. MC: Which of these figures best describes genetic deletion? A) B) C) D) FRQ: Mutations in a DNA sequence can alter gene expression. Define the following examples and explain how each example can alter a gene expression. a) translocation b) nondisjunction c) duplication d) deletion

  13. FQR pointers and starting points: a) Translocation is the movement of a chromosome segment to a different site. Translocation can be balanced, which means there is an even exchange in genetic material. But it can also be unbalanced, causing extra or missing genes. You can test for this with prenatal diagnosis. • Nondisjunction is the separation failure of a chromosome. This means sister chromatids did not separate properly during cell division. It can result in an abnormal amount of chromosomes, like aneuploidy. • Duplication is the doubling of chromosomal segments. Duplication can result in recombination, aneuploidy and polyploidy. It can cause many errors in the DNA sequence and genetic material, • Deletion is the removal of a chromosomal segment. It is the direct loss of genetic material.. Deletion can cause many problems with crossing over during meiosis, and also scan lead to serious genetic diseases. Deletion also can cause a frameshift in the DNA sequencing meaning that a whole fragment of the DNA strand moves over completely jumbling up the DNA code. It causes losses from translocation, unequal crossing over, breaking without rejoining the segment, and chromosomal crossover within a chromosomal inversion.

  14. LO 1.8: 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: Evolutionary mechanisms such as genetic drift, migration, and artificial selection, disrupt Hardy-Weinberg Equilibrium by changing genotype frequencies in populations. Genetic drift (changes in relative allele frequency) is a nonselective process that usually occurs in small populations where infrequently-occurring alleles face a greater chance of being lost. Migration is the movement of organisms from one place to another and it can increase or decrease allele frequencies in populations. After mating is established between the migrating and the original individuals, the migrating individuals will contribute gametes carrying alleles that can alter the existing proportion of alleles in the original population. Selective breeding of organisms, also known as artificial selection, produces organisms with more desirable traits. Many researchers can use these organisms to test genetic variation in a population. Hardy-Weinberg Equilibrium states that allele and gene frequencies in a population will stay the same unless acted on by other evolutionary influence. All three of these disrupt the Equilibrium and decrease genetic variations which can then lead to an increase in the differences between two populations of the same species. M.C. Question: According to the Hardy-Weinberg Equilibrium there are five factors that play a role in evolutionary changes in populations. Which of the following is NOT one of these factors. A. Natural selection B. Non-random mating C. Mutations D. Habitat fragmentation E. Genetic drift Learning Log/FRQ-style Question: You are a researcher that has gone into the field and sampled a population of birds in which you find that the percentage of the bird’s that display the recessive phenotype of black wings is 49%, calculate the following allele frequencies: A. The frequency of the “aa” genotype. B. The frequency of the “a” allele. C. The frequency of the “A” allele. D. The frequencies of the genotypes “AA” and “Aa.”

  15. ANSWER KEY- LO 1.8 M.C. Question: According to the Hardy-Weinberg Equilibrium there are five factors that play a role in evolutionary changes in populations. Which of the following is NOT one of these factors. A. Natural selection B. Non-random mating C. Mutations D. Habitat fragmentation E. Genetic drift Habitat fragmentation is the right answer because although it would cause evolutionary changes in a population, it is specifically not one of the five factors that is listed in the Hardy-Weinberg Theory. Learning Log/FRQ-style Question: You are a researcher that has gone into the field and sampled a population of birds in which you find that the percentage of the bird’s that display the recessive phenotype of black wings is 49%, calculate the following allele frequencies: A. The frequency of the “aa” genotype. B. The frequency of the “a” allele. C. The frequency of the “A” allele. D. The frequencies of the genotypes “AA” and “Aa.” • aa= 49%=q²=.49 • q²= .49 • q= .7 • C. p+q=1 • p+.7=1 • p= .3 • AA= p²= (.3)²= .09 • Aa= 2pq= 2(.3)(.7)= .42

  16. Explanation: Meiosis allows offspring to inherit traits from its parents without being a genetic copy of the previous generation. Any cell with two chromosome sets is a diploid cell, abbreviated 2n. Haploid cells contain a single chromosome set, abbreviated n. In humans, n=23, meaning that in each gamete the haploid number of chromosomes is 23, and our diploid number would be 46 (23x2). In meiosis, chromosomea from each parent are donated. The homologous pairs of chromosomes in a diploid parent cell each replicate, forming 2 sister chromatids in the diploid. In Meiosis I, these homologous chromosomes separate, forming two haploid cells with replicated chromosomes from parents. Then in meiosis II, these two separate sister chromatids split again, forming four haploid cells, all with unreplicated chromosomes. In prophase 1 of the meiosis 1 stage, these dna molecules in the nonsisterchromatids break at a specific place, then rejoin at the other chromatid. This is called crossing over, which is one of the main causes for genetic variation in meiosis and reproduction. LO 3.12: The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring.

  17. LO 3.12 Practice • MC Question: Genetic rearrangements between two non sister chromatids, commonly known as crossing over, occurs in which of the following stages of meiosis? • A. Prophase II B. Metaphase I C. Prophase I D. Metaphase II • FRQ Question: Sexual reproduction needs chromosomes donated from two seperate parent organisms to ignite meiosis and the growth of an offspring, with each parent giving half of the total chromosomes to the mix. Name on species that reproduces sexually, and describe the mode in which genetic information is transferred to keep variation between the parent and offspring organisms.

  18. Practice Answers • Multiple Choice: Answer is C, Crossing over occurs in prophase I of meiosis I. • FRQ: Humans are a species that use sexual reproduction to generate new, genetically differing offspring. It is necessary to have two parents, one male and one female, to produce one egg and one sperm cell to begin meiosis. These two gametes (1n each) join in fertilization, creating a diploid cell (2n). Once the diploid cell is created, chromosomes replicate to form sister chromatids. These chromatids meet in the center of the parent cell, and exchange parts of the chromatids to one another in prophase I, prior to the division of these chromatids. This step is known as switching over, which takes chromatin from one chromatid and exchanges it with another, ensuring that the offspring is not a genetic copy of the two parents.

  19. LO 4.16: The student is able to predict the effects of a change of matter or energy availability on communities SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. Competition and cooperation are vital in ecosystems. As is seen in food webs and other interactions, organisms in a community are heavily connected. Competition and competition go all the way down to molecular levels. For example, inhibitors may compete for the same active site of enzymes. Cells also compete with each other; for example, spores have to compete with each to survive. Cooperation is also vital within an organism. The organism’s parts have to cooperate all the way from a molecular level to the organ systems cooperating. Finally, organisms interact with each other and their environment. The exchange of oxygen and carbon dioxide in plants and animals is vital to both groups. A change in matter or energy can disrupt this entire web of interactions and cooperation and force more competition among an ecosystem. Multiple Choice Question: Use the food web to the left to answer the following question. If hunting became an issue in the ecosystem and the biomass of deer was greatly diminished, which of the following populations would be the most directly and immediately affected? The shrew population would increase as a result of more insects The cougar would begin to eat mice as an alternative food source The snake would face new competition from the cougar The rabbit population would decrease as a result of increased predation A new herbivore would replace the deer in the web, which would initially decrease the grass population. Consider the following: Consider a small ecosystem with producers, primary consumers, and secondary consumers. If sunlight was somehow blocked and the ecosystem received less light energy, describe the effect it would have on the producers, primary consumers, and secondary consumers. Explain these predictions. Draw a graph showing the population size of each in relation to available light energy.

  20. Use the food web to the left to answer the following question. If hunting became an issue in the ecosystem and the biomass of deer was greatly diminished, which of the following populations would be the most directly and immediately affected? The shrew population would increase as a result of more insects- while this could potentially happen as a result of more grass, it is not the most direct and immediate result. The cougar would begin to eat mice as an alternative food source- while this could again in theory happen, it is unlikely that the dietary limitations of the cougar could change like that and even if they could it would not be immediate and direct The snake would face new competition from the cougar-this is very similar to b. Just as in b, it is unlikely that the cougar would begin to prey on the snake’s food source and even if it did it is not the most direct and immediate result. The rabbit population would decrease as a result of increased predation- if the deer population is greatly diminished, the cougars will have to consume more rabbits in order to sustain themselves. A new herbivore would replace the deer in the web, which would initially decrease the grass population.-this is incredibly unlikely and if it replaced the deer, the grass population would remain fairly constant, not decrease. Consider the following: Consider a small ecosystem with producers, primary consumers, and secondary consumers. If sunlight was somehow blocked and the ecosystem received less light energy, describe the effect it would have on the producers, primary consumers, and secondary consumers. Explain these predictions. Draw a graph showing the population size of each in relation to available light energy. If sunlight was blocked, there would be less light energy available to the population. Light energy is used in photosynthesis, meaning producers would struggle. Producers are photosynthetic organisms that convert light energy into usable energy. Because of decreased available light energy, the producer population would likely decrease. Also, without the photosynthesis, there will be more carbon dioxide and less oxygen in the air, which could be harmful to organisms that require oxygen. The primary consumers will suffer mostly from the decrease in the producer population. Producers are their food source, so as the producer population decreases, the primary consumer population will decrease. In the same manner, the secondary consumer population will decrease as the population of their food source, the primary producer, decreases. Key Producers Primary consumers Secondary consumers Population size As the available light energy increases, the populations of all three groups increase. Producers have to have the highest population because only about ten percent of energy is passed on to each trophic level. Available sunlight energy

  21. SP 5.1 The student can analyze data to identify patterns or relationships. LO 2.38 The student is able to analyze data to support the claim that responses to information and communication of information affect natural selection [SP 5.1] Multiple choice: In the graph below shows the number of finches in the area during a drought. The larger seeds come from tall trees that have longer, wider roots that allow for more water to be absorbed. While the small seeds come from ground plants with smaller roots that cannot soak up the little water as easily. From the data given, which answer choice best explains why small beak size is NOT favored during the period of drought? a) The smaller beak size allows the birds to penetrate the larger nut. Natural selection has favored the smaller, more brittle beaks. b) The smaller beak size does not have the power to consume the larger seeds. Natural selection has favored the larger beak size and these birds have passed on their favorable trait. c) The smaller beak size does not weigh the birds down. Therefore they are more likely to survive when running from predators. d) The smaller beak size does not have the ability to absorb nutrient from the large nuts. This is why they moved away. Connection: • The responses to information and communication of information affects natural selection come from the way organisms interact with the environment and with other organisms. A graph showing the period of dormancy during hibernation can be evaluated to determine how the individuals body functions are slowed down to the point where normal energy consumption is temporarily slowed. Other graphs that can be drawn from include courtship periods of the year or even migratory patterns due to the environment. These graphs that are analyzed can show patterns or relationships between different species. The student can then claim that the survival and reproduction that presents natural selection is valid in the way organisms interact to the environment as well as other organisms. Sometimes populations interact with each other to better themselves and the community. This role it plays is called its niche. The way organisms interact can be described as mutualism, commensalism, or parasitism. These responses help a species survive and reproduce to pass on their favorable traits. This leads to biodiversity and the strongest will survive according to natural selection. Free response: According to the data of both the graph and the table, answer the following question. There was a drought in the years 2000-2002. Why did the drought affect the size of the birds beaks? Explain the food sources as well as the size of the beaks. Use the graph and chart to justify your answer. Page 1/2

  22. Answers: Page 2/2 Free response: According to the data of both the graph and the table, answer the following question. There was a drought in the years 2000-2002. Why did the drought affect the size of the birds beaks? Explain the food sources as well as the size of the beaks. Use the graph and chart to justify your answer. Answer: The drought has an effect on the birds because the drought causes the larger trees to soak up the little water that is present in the ground. Large trees have larger roots and cover a larger surface area. These large trees have caused the smaller shrubs and smaller trees to not absorb enough water needed to survive. Thus a shortage of smaller seeds is present and larger seeds are the remaining food source. The birds with the small beaks are not able to eat the large seeds therefore they have run out of their food supply. The large beaked birds survive and are able to reproduce. This is natural selection playing its role in the environment. When an environment lacks the right food supply for a species, it will die and another species can take its place. As the graph shows with the drought comes a shift and an increase in the number of large beak birds and a decline in small and medium beaked birds. The table has the same information in number form. The table shows that during the drought the small beaked bird numbers decline along with the medium beak birds until the small beak birds reaches zero. On the other hand the large beak birds are able to find food and survive to reproduce and increase their population to 1400 birds. Once the drought is over, the numbers begin to return to the original state and the small beak to large beak ratio almost equalizes again while the medium beaks rise. Multiple choice: In the graph below shows the number of finches in the area during a drought. The larger seeds come from tall trees that have longer, wider roots that allow for more water to be absorbed. While the small seeds come from ground plants with smaller roots that cannot soak up the little water as easily. From the data given, which answer choice best explains why small beak size is NOT favored during the period of drought? • a) The smaller beak size allows the birds to penetrate the larger nut. Natural selection has favored the smaller, more brittle beaks. • b) The smaller beak size does not have the power to consume the larger seeds. Natural selection has favored the larger beak size and these birds have passed on their favorable trait. • c) The smaller beak size does not weigh the birds down. Therefore they are more likely to survive when running from predators. • d) The smaller beak size does not have the ability to absorb nutrient from the large nuts. This is why they moved away.

  23. LO 3.9: The student is able to construct an explanation, using visual representations or narratives, as to How DNA in chromosomes is transmitted to the next generation via mitosis, or meiosis followed by Fertilizations SP 6.2:The student can construct explanations of phenomena based on evidence produced through scientific practices. Explanation: During meiosis, DNA begins on the chromosomes that replicate during S-phase and proceed to prophase I where crossing occurs that helps create genetic variability by also exchanging DNA. The cell then undergoes Telephase I to divide it (this process was studied by Mendel who theorized the idea of homologous chromosomes separating). The resulting two cells under go meiosis II and produce the 4 haploid gametes that contain single chromosomes containing the unique genes/DNA. This is where recessive and dominant alleles get put on the locus on the final gametes that are produced. The gametes from each parent come into contact with each other during reproduction and fertilization (combination of the haploid sperm cell with the haploid ovum) occurs. The resulting zygote is diploid and represents the combined haploid sets of chromosomes/DNA from the parents. M.C. Question: A homozygous dominant red flower (RR) is crossed with a heterozygous white flower (rr). What is the ratio of the colors in the F2 generation. A) 1:3 B) 1:1:1 C) 1:2:1 D) 2:1:1 Learning Log/FRQ-style Question: Attached is a pedigree of Bailey’s family. Her great grandmother, and both uncles have both been afflicted with Red-Green color blindness. One of her brothers is also afflicted with the disease. However neither her nor her sister have the disease. When she marries, does her son have chance of getting the disease? Also, explain why one brother and not the other has the disease.

  24. Learning Log question Bailey

  25. ANSWER KEY– LO 3.9 M.C. Question: A homozygous dominant red flower (RR) is crossed with a heterozygous white flower (rr). What is the ratio of the colors in the F2 generation. A) 1:3 B) 1:1:1 C) 1:2:1/ 1: homozygous dominant red, 2: heterozygous pink, and one homozygous recessive white D) 2:1:1 Learning log question: Attached is a pedigree of Bailey’s family. Her great grandmother, and both uncles have both been afflicted with Red-Green color blindness. One of her brothers is also afflicted with the disease. However neither her nor her sister have the disease. When she marries, does her son have chance of getting the disease? Also, explain why one brother and not the other has the disease. Answer: Her son does have a chance of getting the disease, because Red-Green color blindness is an X-linked disease. That is why it makes sense for her, her sister, and mother to not show signs of the trait but her brother does have it. However, this does mean that she could very well be a carrier. With her mother having two daughters, the chances of one of them at least being a carrier is close to, if not 100%. The reason that one of her brothers has the disease and the other one doesn’t, is a result of the mother being a carrier. She has two X chromosomes with one of them having the disease on it. This results in a 50% chance of having a son with the disease. In this case, the family is a perfect statistical representation. Each of the sons has a 50% of catching the disease; and one did while the other did not. While one of the daughters is most likely a carrier.

  26. Works cited http://classconnection.s3.amazonaws.com/65/flashcards/816065/png/x-linked_r1318385614753.png http://mrsmarsigliano.weebly.com/uploads/2/4/8/0/24807467/2299956_orig.jpg

  27. Learning Objective 1.30The student is able to evaluate scientific hypotheses about the origin of life on Earth. Science Practice 6.5The student can evaluate alternative scientific explanations. The Miller-Urey Experiment MCQ:Which of the following hypothesis the origin of life would & most closely relate to the belief that the interaction of terrestrial elements and electrostatic discharge contributed to the formation of Earth’s first biotic substances? A. Ocean Vent B. Primordial Soup C. Extraterrestrial Theory D. both A and C Explanation: The “Organic Soup” theory hypothesized that electrical energy could catalyze the gases in the atmosphere of the early Earth to create biomolecules that. Experiments (such as the Miller-Urey Experiment) were able to mimic electrical discharges through which water and gasses (hydrogen, methane, and ammonia gas) passed in order to mimic the atmospheric makeup of early Earth and its interaction with lightning, radioactivity, and UV light. The energy from the charge caused various amino acids and some nucleic acids to form and over time, these molecules became more complex and acted together to run metabolic processes. Some models suggest that primitive life developed on biogenic surfaces, such as clay, that served as templates and catalysts for assembly of macromolecules. Under laboratory conditions, complex polymers and self-replicating molecules can spontaneously assemble. It remains an open question whether the first genetic and self-replicating material was DNA or RNA. The “Ocean Vents Theory” hot, pressurized water mixed with dissolved gases (including hydrogen sulfide, carbon monoxide, carbon dioxide, hydrogen cyanide and ammonia) passed out of prehistoric vents and over various minerals containing iron, nickel and other metals within the rocks around the vents. These metals served as catalysts for a chain reaction that synthesized organic compounds and coupled some of them with other metals to form new compounds with greater ability to yield new chemical reactions. This coupling between the catalyst and the product of an organic reaction is the key first step of Wächtershäuser’s theory. Starting with these metals and gases reacting together as life emerged, Wächtershäuser says evolution starts with the beginning of a primitive metabolism that created increasingly complex chemical reactions (chemosynthesis), eventually leading over time to the formation of DNA — life’s blueprints for making more living cells today. Chemosynthesis is the biological conversion of carbon molecules and nutrients into organic matter — the stuff of life. Whereas photosynthesis uses energy from sunlight to convert carbon dioxide into that organic matter, giving off oxygen as a byproduct, chemosynthesis uses inorganic molecules (such as hydrogen sulfide) or methane and combines them with an oxygen source (in this case seawater) to create simple sugars. The Extraterrestrial theory suggests that comets and meteorites striking the Earth attributed to the creation of organic matter. In support of this theory, we know that organic compounds are relatively common in space and that Uracil, a component of RNA, likely formed extraterrestrially. We also know that comets contain complex organic material such as coal. Though there are differing hypotheses about the origin of life, the following scientific evidence supports the various models: 1. Primitive Earth provided inorganic precursors from which organic molecules could have been synthesized due to the presence of available free energy and the absence of a significant quantity of oxygen. 2. In turn, these molecules served as monomers or building blocks for the formation of more complex molecules, including amino acids and nucleotides. 3. The joining of these monomers produced polymers with the ability to replicate, store and transfer information. 4. These complex reaction sets could have occurred in solution (organic soup model) or as reactions on solid reactive surfaces. 5. The RNA World hypothesis proposes that RNA could have been the earliest genetic material. FRQ:Identify and explain one of the theories on the origin of life and briefly explain how, in two ways, it specifically differs from another hypothesis of biological merit.

  28. MCQ:Which of the following hypothesis the origin of life would & most closely relate to the belief that the interaction of terrestrial elements and electrostatic discharge contributed to the formation of Earth’s first biotic substances? A. Ocean Vent B. Primordial Soup C. Extraterrestrial Theory D. both A and C Explanation: The correct answer is B, as the Miller-Urey experiment (supporting this theory) explains and demonstrates the relationship between lightning and other photo-interference and the gases present on early Earth, and how this interaction produces life from non-life. FRQ:Identify and explain one of the theories on the origin of life and briefly explain how, in two ways, it specifically differs from other hypotheses. Response: An often-told origin-of-life story is that complex biological compounds assembled by chance out of an organic broth on the early Earth's surface, many times referred to the Primordial Soup Theory. This pre-biotic synthesis culminated in one of these bio-molecules being able to make copies of itself. The first support for this idea of life arising out of the primordial soup came from the famous 1953 experiment by Stanley Miller and Harold Urey, in which they made amino acids—the building blocks of proteins—by applying sparks to a test tube of hydrogen, methane, ammonia, and water. If amino acids could come together out of raw ingredients, then bigger, more complex molecules could presumably form given enough time. Biologists have devised various scenarios in which this assemblage takes place in tidal pools, near underwater volcanic vents, on the surface of clay sediments, or even in outer space. Many researchers, therefore, think that RNA—a cousin of DNA—may have been the first complex molecule on which life was based. RNA carries genetic information like DNA, but it can also direct chemical reactions as proteins do. However in the deep sea vent theory poses that hot, pressurized water mixed with dissolved gases (including hydrogen sulfide, carbon monoxide, carbon dioxide, hydrogen cyanide and ammonia) passed out of prehistoric vents and over various minerals containing iron, nickel and other metals within the rocks around the vents. These metals served as catalysts for a chain reaction that synthesized organic compounds and coupled some of them with other metals to form new compounds with greater ability to yield new chemical reactions. The Primordial Soup theory does not take into account the potential of hydrogen sulfide, and it also focuses its attention on the interaction between electrical stimulus rather than heat stimulus.

  29. LO 3.3: The student is able to describe representations and models that illustrate how genetic information is copied for transmission between generations. Sp 1.2: The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation: Heritable information that is needed for organizational purposes of all living organisms, is stored and passed to generations by DNA. Chromosomes contain the genetic information necessary for protein synthesis and transmission into daughter cells. Prokaryotes contain circular chromosome, while Eukaryotes have multiple linear chromosome. Before the process of mitosis and meiosis, DNA is replicated during Interphase. Interphase contains three parts, G1, S, and G2. DNA synthesis occurs in the S phase, while growth of cell occurs in G1 and G2. DNA synthesis, replication, ensures continuity of heritable information. This is because daughter cells receive an identical and complete set of chromosomes, which are copied in during DNA synthesis and distributed during mitosis/meiosis. The enzymes, DNA polymerase, adds on corresponding nucleotides to the new strand by reading the template strand. DNA polymerase works in a 5’ to 3’ manner by adding each new base to the 3’ end. Because DNA is antiparallel, there will be a leading and lagging strand. MC Question: What is the role of molecule 5 in the diagram of DNA synthesis, Adds new nucleotides to the new strand Proofreads the new DNA strand Joins to together fragments with the use of Hydrogen bonds Unwinds double helix, breaking the hydrogen bonds Free Response/Learning Log Question a) Using figure 1, describe the role of the molecules (by name) in the process of DNA replication. b) DNA replication is necessary for transmission of genetic material between generations, explain Meselson-Stahl’s experiment in determining how DNA replicates in a semiconservative manner and provide a diagram of this replication

  30. What is the role of molecule 5 in the diagram of DNA synthesis, Adds new nucleotides to the new strand Proofreads the new DNA strand Joins to together fragments with the use of Hydrogen bonds Unwinds double helix, breaking the hydrogen bonds C is the correct answer because in the diagram, molecule 5 represents DNA ligase. The role of DNA ligase is to bind okazaki fragments together forming a complete strand. a) Using figure 1, describe the role of the molecules (by name) in the process of DNA replication. b) DNA replication is necessary for transmission of genetic material between generations, explain Meselson-Stahl’s experiment in determining how DNA replicates in a semiconservative manner and provide a diagram of this replication The template DNA strand is first unzipped at the origins of replication by the enzyme, helicase. On the leading strand, DNA polymerase reads the template strand and continuously adds on nucleotides to the 3’ end of the new strand. On the lagging strand, primasesynthesizes RNA primers at the beginning of each Okazaki fragment. DNA polymerase can then add nucleotides to each primer until it reaches the next one. DNA ligase comes in to form bonds between each fragment, thus completing the strand. Meselson and Stahl performed an experiment using E. coli bacteria. They first cultured the bacteria in a heavy isotope of nitrogen, N15. Then, they transferred the bacteria into a lighter isotope, N14, in order for the bacteria to synthesize. The DNA was centrifuged for 40 minutes to show two replications. The first replication displayed a band of DNA that was a hybrid of the two and the second replication displayed band of both a hybrid and light DNA. This represented the semiconservative model. Continue to slide 3 for the diagram.

  31. First replication Second Replication

  32. (LO 4.4) The student is able to make a prediction about the interactions of subcellular organelles.(SP 1.5) The student can make claims and predictions about natural phenomena based on scientific theories and models. MC Question: The endomembrane system is a complicated system involving multiple organelles. Which of the following correctly describes the movement through this system of a protein that functions best in the ER? A. The mRNA is synthesized in the ER, translated on a free ribosome to the Golgi, and then a transport vesicle carries the mRNA back to the ER for modification. The mRNA is then sent to the nucleus. B. The mRNA is synthesized in the Golgi, is translated on a free ribosome to the ER for modification, and then a transport vesicle carries the mRNA back to the Golgi. C. The mRNA is synthesized in the nucleus, is translated on a free ribosome, then is sent to the Golgi and modified. A transport vesicle carries the protein to the ER for further modification. D. The mRNA is synthesized in the nucleus, is translated on a bound ribosome, then is sent to the ER and modified. A transport vesicle carries the protein to the Golgi apparatus. After modification in the Golgi, another transport vesicle carries the protein back to the ER. FRQ Question: Organelles are found in both prokaryotic and eukaryotic cells. a. Compare and contrast prokaryotic and eukaryotic cells. b. List and describe the basic function of three organelles. c. Describe how these organelles work together to allow the animal cell to function. A eukaryotic cell has membrane-enclosed organelles, the largest of which is usually the nucleus. A prokaryotic cell is usually smaller and does not contain a nucleus or other membrane-enclosed organelles. Some organelles are animal or plant specific, but these SEVEN organelles are found in both types of cell: the nucleus, vacuole, mitochondria, the Golgi apparatus, cytoskeleton, peroxisomes, and the Endoplasmic reticulum (ER). The nucleus includes the nuclear envelope, the nucleolus, and chromatin, and is the information center of the cell. A vacuole is a membrane-bound sac that contains hydrolytic enzymes that aid in intracellular digestion and the release of cellular waste products. There is a central vacuole in plant cells. Mitochondria aid in energy capture and transformation. The Golgi apparatus, consisting of a series of flattened membrane sacs (calledcisternae), synthesizes and packages small molecules for transport in vesicles, and produces lysosomes (described later). The cytoskeleton reinforces the cell’s shape. The ER consists of a smooth and a rough ER. The rough ER gets its name from its surface containing ribosomes (small structures containing rRNA and protein) and is where protein synthesis occurs and contributes to intracellular transport. Ribosomes can be free ribosomes (suspended in the cytosol) or bound ribosomes (attached to the outside of the ER or nuclear envelope). The smooth ER synthesizes lipids. Peroxisomes are organelles that aid in metabolism and produce hydrogen peroxide. Plant cells include chloroplasts, organelles that contribute to energy capture and conversion for photosynthesis. Animal cells contain lysosomes, whichaid in intracellular digestion, apoptosis, and recycling cell material. Centrosomes are organelles where the cell’s microtubules are initiated, but in animal cells these organelles contain centrioles (function unknown). Plant cells also contain a cell wall, with plasmodesmata (channels) throughout, that help keep the cell’s shape and protect the cell from mechanical damage. (next slide has visuals)

  33. Organelle Interactions Now that we know the functions of the organelles, let’s look at how they interact… The nucleus houses the most of the cell’s DNA and the ribosomes use information from the DNA to make proteins. Vesicles move from the ER to the Golgi. Vesicles form and leave Golgi, carrying specific proteins to other locations or to the plasma membrane for secretion. Vesicles also transport certain proteins back to the ER. These vesicles give rise to lysosomes and vacuoles. Lysosomes are then available for fusion with another vesicle for digestion. Transport vesicles carry proteins to the plasma membrane for secretion. The Plasma membrane expands by fusion of the vesicles and the proteins are secreted from the cell. This makes up the endomembrane system. In addition to this system, mitochondria (animal cells) and chloroplasts (plant cells) contain DNA and change energy from one form to another.

  34. Answer Key- LO 4.4 MC Question: The endomembrane system is a complicated system involving multiple organelles. Which of the following correctly describes a the movement through this system of a protein that functions best in the ER? A. The mRNA is synthesized in the ER, translated on a free ribosome to the Golgi, and then a transport vesicle carries the mRNA back to the ER for modification. The mRNA is then sent to the nucleus. B. The mRNA is synthesized in the Golgi, is translated on a free ribosome to the ER for modification, and then a transport vesicle carries the mRNA back to the Golgi. C. The mRNA is synthesized in the nucleus, is translated on a free ribosome, then is sent to the ER and modified. A transport vesicle carries the protein to the Golgi apparatus. After modification, another transport vesicle carries the protein back to the ER. D. The mRNA is synthesized in the nucleus, is translated on a bound ribosome, then is sent to the ER and modified. A transport vesicle carries the protein to the Golgi apparatus. After modification in the Golgi, another transport vesicle carries the protein back to the ER. The question requires background about the difference between bound and free ribosomes, the functions of each organelle in the endomembrane system, and the relationship between the different organelles in the system. First, the student must understand that the order of the endomembrane system, with the specified protein from the question, is: nucleus, ER, Golgi, back to the ER (since the protein functions best in the ER). This eliminates A and B. Between C and D, D is the only answer choice that has the correct order and mentions a bound ribosome. Understanding that this protein requires a bound ribosome is important because bound ribosomes are attached to the rough ER. FRQ: Organelles are found in both prokaryotic and eukaryotic cells. a. Compare and contrast prokaryotic and eukaryotic cells. b. List and describe the basic function of three organelles. c. Describe how these organelles work together to allow the animal cell to function. a. A eukaryotic cell has membrane-enclosed organelles, the largest of which is usually the nucleus. A prokaryotic cell is usually smaller and does not contain a nucleus or other membrane-enclosed organelles. Prokaryotes contain a single circular chromosome, while eukaryotes contain multiple linear chromosomes. b. Three animal cell organelles are the Golgi apparatus, the nucleus, and the ER. The Golgi apparatus, consisting of a series of flattened membrane sacs (called cisternae), synthesizes and packages small molecules for transport in vesicles, and produces lysosomes. Thenucleusincludes the nuclear envelope, the nucleolus, and chromatin, and is the information center of the cell. The ER consists of a smooth and a rough ER. The rough ER gets its name from its surface containing ribosomes (small structures containing rRNA and protein) and is where protein synthesis occurs and contributes to intracellular transport. Ribosomes can be free ribosomes (suspended in the cytosol) or bound ribosomes (attached to the outside of the ER or nuclear envelope). The smooth ER synthesizes lipids. c. The nucleus houses the most of the cell’s DNA and the ribosomes use information from the DNA to make proteins. Vesicles move from the ER to the Golgi. Vesicles form and leave Golgi, carrying specific proteins to other locations or to the plasma membrane for secretion. Vesicles also transport certain proteins back to the ER. These vesicles give rise to lysosomes and vacuoles. Lysosomes are then available for fusion with another vesicle for digestion. Transport vesicles carry proteins to the plasma membrane for secretion.

  35. LO 1.24: The student is able to describe speciation in an isolated population and connect it to change in gene frequency, change in environment, natural selection and/or genetic drift.SP 7.2: The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. Learning Log/FRQ-style Question: Explain the difference between gradualism and punctuated equilibrium in the context of speciation. Use diagrams to enhance your answer. Which of the two models best describes a species in which a change in appearance occurred all at once?Is one model more likely to fit under sympatric speciation or allopatric speciation? Explain your answer. Images in slides from ib.bioninja.com.au and palaeos.com Multiple Choice Question:Which of the following is an accurate example of allopatric speciation? A) A mutation causing longer digits on a salamander causes the divide between ground-dwelling salamanders and tree-dwelling salamanders. B) Chimpanzees and humans have similar body proportions, so it is likely they share a common ancestor. C) A lake splits into two separate bodies of water and the population of yellow perches divides and is unable to mate with each other successfully. D) Due to behavioral isolation, blue-footed boobies will not mate with red-footed boobies. Explanation: Speciation is when a new species originates due to evolution. It is important to understand speciation because it is the basis for biological diversity. Isolation by any means in a species causes what was once a single species to evolve and become two separate ones that are unable to mate and/or reproduce with one another due to biological differences or not being in the same place or time when it is possible to do so. This can also help scientists and researchers to understand the overall evolution of species and how they originated. Two types of speciation include allopatric speciation and sympatric speciation. Allopatric speciation is when a geographic barrier isolates a population and blocks gene flow, which is defined as additions or subtractions to a population's gene pool due to the movement of fertile individuals or gametes. Examples of this type of speciation would be an emerging river separating a field or a mountain being created and dividing a piece of land. Sympatric speciation is when intrinsic factors such as chromosomal changes or non-random mating alter the gene flow. There is no geographic change whatsoever. Examples of this would include many plants such as oats or cotton. Both of these ways can lead to new species being created due to the gene make-up of a certain species divide because of environmental or internal changes. Natural selection favors the genes that produce the most favorable traits to help organisms survive in a certain type of environment. Genetic drift is defined as unpredictable fluctuations in allele frequencies from one generation to the next because of a population's finite size. This means that when allele frequencies shift, it can also lead to eventual speciation because what traits are prevalent in a certain population can change due to genetic drift.

  36. Multiple Choice Question:Which of the following is an accurate example of allopatric speciation? A) A mutation causing longer digits on a salamander causes the divide between ground-dwelling salamanders and tree-dwelling salamanders. B) Chimpanzees and humans have similar body proportions, so it is likely they share a common ancestor. C)A lake splits into two separate bodies of water and the population of yellow perches divides and is unable to mate with each other successfully. D) Due to behavioral isolation, blue-footed boobies will not mate with red-footed boobies. C is the correct answer because the population is affected by an environmental change that causes the two sides to evolve independently of each other, which is what allopatric speciation is based off of. (Think “allo-” means “other,” so they evolve in anotherplace.) Learning Log/FRQ-style Question: Explain the difference between gradualism and punctuated equilibrium in the context of speciation. Use diagrams to enhance your answer. Which of the two models best describes a species in which a change in appearance occurred all at once? Explain how both models fit into allopatric speciation and sympatric speciation. Gradualism and punctuated equilibrium have to do with the rate of speciation. Gradualism has to do with a species slowly evolving over a long period of time in which slow changes take place from generation to generation. Each adaptation changes the physiology of the organism and the organisms selected for that trait will survive, while the organism to do not have that trait will not. This results in gradual, or slow, changes rather than a major change all at once. Punctuated equilibrium has to do with a species experiencing a major change after a long period of stasis. Instead of small changes over time, a species will change after it separates from the parent species, but will do so very little after that initial point. The punctuated model best describes a species that has undergone a sudden morphological change after a long time of staying the same. In allopatric speciation, the species that becomes environmentally isolated but remains similar to the original population only changing slightly over time would be gradualism. This would mean that neither population becomes extremely morphologically different after separating at first. A species that experiences allopatric speciation in a punctuated equilibrium model would be a population that is divided, and one side of the population dramatically changes in the next generation in order to accommodate the environmental factors and differences they experience. A species that slowly changes over time and after many generations is unable to reproduce with others in the original population would have experienced sympatric speciation in a gradualism model. With the same type of speciation in a punctuated equilibrium model, one part of the population would experience a dramatic morphological change that would break away from the parent species. All of these situations would result in a new species entirely, whether that new species is in the same or a different location, or if it took place over a long or short period of time.

  37. Learning Objective 1.32: The student is able to justify the selection of geological, physical, and chemical data that reveals early Earth conditions. Science Practice 4.1: The student can justify the selection of the kind of data needed to answer a particular scientific question. Multiple Choice: Which of the following statements about the conditions of Earth’s beginning and the scientific evidence that proves that condition is TRUE? The cyanobacteria’s bacteria that metabolized dihyrogen sulfide mutated to use water rather than dihydrogen sulfide, which proves that the Earth’s atmosphere in early times had an abundance of dihydrogen sulfide. The cyanobacteria’s bacteria that metabolized dihyrogen sulfide mutated to use water rather than dihydrogen sulfide, which proves that the Earth’s atmosphere in early times had an abundance of water. The percent of common amino acids in all organisms proves that the theory of evolution is invalid. Monomers combined to make polymers which indicates that there were great amounts of volcanic activity in Earth’s early years. Explanation: At Earth’s beginning, there was very little oxygen, and a lot of water, methane, carbon monoxide, and carbon dioxide. There was no ozone layer which led to high levels of UV radiation, and lots of meteorite bombardment. There was a lot of lightening and volcanic activity. Through absolute and relative dating of fossils and rocks, and the knowledge that the environment was too hostile for life until 3.9 byo, we can conclude that the origin of life occurred between 3.5 and 3.8 billion years ago. The oldest rocks were found in Greenland 3.8 billion years ago, and the oldest fossils were dated to 3.5 byo. In 1953, the Miller-Urvey experiment was conducted to see if Earth favored reactions that formed organic compounds from inorganic compounds, which was a hypothesis made in 1920 by Oparin and Haldane. Through repeated experiments, they concluded that all monomers necessary for life (amino acids, all the components of nucleotides, sugars, lipids, etc.) could be formed from inorganic compounds. This gave evidence that living matter can come from lifeless matter, and that this is how the earth was formed since it started out with no living matter but had an abundance of free energy and barely any oxygen. These monomers connected to make polymers as Syndey Fox proved by dripping monomers onto hot sand, clay, or rocks to see if they would connect. These connected monomers created protienoids which are poly peptides created by abiotic means. These protienoids gave way to protobionts which are abiotically produced molecules surounded by a membrane. We believe that eukaryotes originated through endosymbiosis which is when molecules absorb one another and live inside of each other. This is proven by the similarities between bacteria and mitochondria & chloroplasts. They both reproduce by binary fission, have small, circular genomes, and very similar DNA sequences , along with many others. Current endosymbioticrelatationships also provide evidence… it had to start sometimes, so it could have been then. How similar all organisms DNA sequences/amino acid composure is a sign that we all are evolved from one organism and so we conclude that there is a common genetic code. There is some evidence that RNA was the first genetic material because through ribozymes, RNA can self replicate and the RNA molecules best suited for their environment replicate their DNA and reproduce so natural selection has an impact. The origin of photosynthesis tells us what the atmosphere was like during early times because the cyanobacteria used to have a bacteria that metabolized dihydrogen sulfide, and that bacteria mutated to use water. Water was abundant in the atmosphere at that time so the cyanobacteria used it in photosynthesis. The oxygen that was released from the photosynthesis reacted with dissolved iron and formed oxide precipitate and the saturated water was released into the atmosphere. FRQ: Explain the process of the origin of life on Earth and the scientific evidence that supports these theories. Begin with no life, and end with the formation of eukaryotes. Make sure that you include information about how natural selection contributed and cite the names of any scientists who contributed experimental data to help solidify this theory.

  38. FRQ Answer: The Miller-Urvey experiment was conducted to see if Earth favored reactions that formed organic compounds from inorganic compounds, which was a hypothesis made earlier by Oparin and Haldane. Through repeated experiments, they concluded that all monomers necessary for life (amino acids, all the components of nucleotides, sugars, lipids, etc.) could be formed from inorganic compounds. This gave evidence that living matter can come from lifeless matter, and that this is how the earth was formed since it started out with no living matter but had an abundance of free energy and barely any oxygen. These monomers connected to make polymers as Syndey Fox proved by dripping monomers onto hot sand, clay, or rocks to see if they would connect. These connected monomers created protienoids which are poly peptides created by abiotic means. These protienoids gave way to protobionts which are abiotically produced molecules surounded by a membrane. Natural selection then comes in to play because protobionts best suited for their enviroment could reproduce and create others who were best suited for their enviroment. Coacervate is a stable protobiont droplet that self assembles when a suspension of macromolecules are shaken. Since we know that macromolecules can be created in earth’s hostile enviroment due to the findings of Miller and Urvey, we know that protobionts can also be formed. We believe that eukaryotes originated through endosymbiosis which is when molecules absorb one another and live inside of each other. This is proven by the similarities between bacteria and mitochondria & chloroplasts. They both reproduce by binary fission, have small, circular genomes, and very similar DNA sequences , along with many others. Current endosymbioticrelatationships also proof that it could have been responsible for forming eukaryotes so early because it had to have started at some point in the history of the Earth, so why not in the very beginning? Answer Key • Multiple Choice Question and Answer: Which of the following statements about the conditions of Earth’s beginning and the scientific evidence that proves that condition is TRUE? • The cyanobacteria’s bacteria that metabolized dihyrogen sulfide mutated to use water rather than dihydrogen sulfide, which proves that the Earth’s atmosphere in early times had an abundance of dihydrogen sulfide. • The cyanobacteria’s bacteria that metabolized dihyrogen sulfide mutated to use water rather than dihydrogen sulfide, which proves that the Earth’s atmosphere in early times had an abundance of water. • The percent of common amino acids in all organisms proves that the theory of evolution is invalid. • Monomers combined to make polymers which indicates that there were great amounts of volcanic activity in Earth’s early years. B is the correct answer. The other options are all wrong.

  39. LO 4.5 The student is able to construct explanations based on scientific evidence as to how interactions of sub cellular structures provide essential functions.SP 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. • Explanation: In addition to the plasma membrane at its outer surface, eukaryotic cells have extensive elaborately arranged internal membranes that partition the cell into compartments. These compartments are the various organelles that provide different local environments that facilitate specific metabolic functions, so incompatible processes can go on simultaneously inside the same cell. Ribosomes are particles made of ribosomal RNA and protein. These two components of the ribosome interact to carry out protein synthesis by translating the genetic material to produce specific polypeptides. Ribosomes, however, are unlike most other organelles in the fact that they are not enclosed in a membrane. Ribosomes can synthesize proteins in two cytoplasmic locations, either suspended in the cytosol (free ribosomes) or attached to the endoplasmic reticulum (bound ribosomes). The rough ER containing the bound ribosomes serves to compartmentalize the cell and plays a role in intracellular transport. The ER is formed by a network of membranous sacs and tubes. The smooth ER allows for the synthesis of lipids, metabolism of carbohydrates, and detoxification. The Golgi apparatus is a membrane bound structure that consists of a series of flattened membrane sacs, known as cisternae. The Golgi is involved in the synthesis and packaging of materials for transport in vesicles and also allows for the production of lysosomes. The mitochondria are the sites of cellular respiration and specialize in the transformation of energy from sugars and other fuels to ATP. The mitochondria is comprised of a double membrane that aid in the formation of the concentration gradient that activates ATP synthase to produce ATP. The inner membrane of the mitochondria is folded to allow for increased surface area to aid in the mitochondria’s efficiency, these folds are known as cristae. Chloroplasts are also specialized organelles found in algae and higher plants that capture energy through photosynthesis. Chloroplasts have double membranes that contain interconnected sacs called thylakoids. These thylakoids are stacked to from grana and produce ATP and NADPH, which fuel carbon fixing reactions of the Calvin cycle. Chloroplasts contain chlorophylls that cause the green color of plants and function as the light trapping molecules in photosynthesis. Lysosomes are membrane enclosed sacs that contain hydrolytic enzymes that aid in intracellular digestion, recycling of organic cell material and apoptosis. Vacuoles are membrane bound sacs that also play a role in intracellular digestion and the release of cellular waste products. • MC Question • A protein that is synthesized in the rough endoplasmic reticulum is eventually incorporated into the cell’s plasma membrane. However, a researcher observes that the protein that was integrated into the membrane is slightly different than the protein made in the ER. The protein was most likely altered in the _____ before being incorporated. • A) Plasma membrane • B) lysosomes • C) Chloroplasts • D) Golgi Apparatus • FRQ Question • Imagine you are a scientist studying the liver, an organ that has the main function of bile production along with blood filtration. You have prepared a slide of liver cells ready to be observed under a microscope for analysis. Hypothesize which organelle is most likely to be highly concentrated within the cell and provide an explanation as to why this organelle is abundant.

  40. MC Question • A protein that is synthesized in the rough endoplasmic reticulum is eventually incorporated into the cell’s plasma membrane. However, a researcher observes that the protein that was integrated into the membrane is slightly different than the protein made in the ER. The protein was most likely altered in the _____ before being incorporated. • A) Plasma membrane (membrane does not alter proteins and only contains the integral proteins already altered) • B) lysosomes (lysosomes specialize in digestion of intracellular materials) • C) Chloroplasts (Chloroplasts aid in energy transformation, not transformation of proteins) • D) Golgi Apparatus (Golgi Apparatus is involved in the alteration of proteins for transportation) • FRQ Question • Imagine you are a scientist studying the liver, an organ that has the main function of bile production along with blood filtration. You have prepared a slide of liver cells ready to be observed under a microscope for analysis. Hypothesize which organelle is most likely to be highly concentrated within the cell and provide an explanation as to why this organelle is abundant. • The liver cell most likely contains a large concentration of Smooth Endoplasmic Reticulum. The smooth ER plays a vital role in detoxification along with the synthesis of lipids and metabolism of carbohydrates. Within the liver the smooth ER provides the synthesis of enzymes that help detoxify certain compounds such as alcohol. Another possibility for an abundant organelle is the high concentration of lysosomes. Lysosomes also aid in the breakdown of toxic chemicals by containing the enzymes produced to aid in digestion or detoxification. Since the liver is the site of detoxification and filtration, organelles specialized in the breakdown of toxic chemicals would be most abundant within the cells.

  41. LO 3.15- The student is able to explain deviations from Mendel’s model of the inheritance of traits.SP 6.5- The student can evaluate alternative scientific explanations. Explanation: Deviations from Mendel’s model of inheritance/ Punnett Square is shown through deviations from predicted ratios like in the chi square technique (and Chi Square Lab with the corn kernels), sex-linked chromosomes, and genes that are not passed through nucleic DNA. From the Chi Square lab we did during Unit 4, we counted the kernels on the corn and then calculated the expected outcome of the number of kernels. The number of kernels counted and the number of kernels expected were definitely different from one another and they never matched, but they were close. Genes also deviate from Mendel’s laws by being linked a sex-linked genes on the sex chromosomes or having an extra chromosome like Trisomy 21/Down Syndrome. Sex-linked usually means that the allele is located on the X chromosome. Mothers and fathers pass an X-linked allele onto their sons or daughters. If a sex-linked allele is recessive, daughters show the phenotype of the allele when they are homozygous and the sons only show it when it is on their one X chromosome. More males are affected by sex linked-genes because they only need one infected X chromosome to display that phenotype. Some examples of sex-linked alleles are Duchenne’s Muscular Dystrophy, Hemophilia, and Color Blindness. There is also other types of DNA that are inherited that lie outside of Mendel’s rules because he only focused on nuclear DNA. There is also DNA in the chloroplasts and the mitochondria. Chloroplasts and mitochondria are randomly assorted to gametes and daughter cells; therefore, they do not follow Mendel’s Punnett Square and his rules of Independent Assortment and Segregation. In animals, mitochondrial DNA is transmitted by egg so mitochondrial traits are maternally inherited. M.C. Question: A scientist finds that a group of rabbits that spawns multiple generations. Some have black and white striped fur while the others have a normal solid coat of white. Determine which of these statements are true with the pedigree of the rabbits provided. (See slide 2 for pedigree) A) The traits for both the white and black striped fur and the solid coat of white fur are on the same chromosome. B) The black and white striped fur is a result of co-dominance. C) The trait for black and white striped fur is a X-linked recessive trait. D) The trait for black and white striped fur is an autosomal recessive trait. E)There is not enough information to determine how the gene is inherited. Pedigree for Rabbits (multiple generations) Black- Striped Rabbits White- Normal White Coated Rabbits Learning Log/FRQ-style Question: i. Explain Mendel’s Laws of Independent Assortment and Segregation and then explain TWO ways of inherited genes diverging from Mendel’s predicted ratios. Justify your answer. ii. Genetics are also determined by some alleles that are located on the sex chromosomes. Discuss sex-linked transmission and then provide an example of a sex-linked trait. • Slide 1 of 2

  42. ANSWER KEY-LO 3.15 A scientist finds that a group of rabbits that spawns multiple generations. Some have black and white striped fur while the others have a normal solid coat of white. Determine which of these statements are true with the pedigree of the rabbits provided. (See slide 2 for pedigree) A) The traits for both the white and black striped fur and the solid coat of white fur are on the same chromosome. B) The black and white striped fur is a result of co-dominance. C) The trait for black and white striped fur is a X-linked recessive trait. D) The trait for black and white striped fur is an autosomal recessive trait. E)There is not enough information to determine how the gene is inherited. Explanation: A is not correct because linked traits that are on the same chromosome are autosomal traits and are not sex-linked. B is not correct because the black and white striped fur is not a result of codominance because in the population of bunnies the scientist has there are no black bunnies in it so there are no black fur genes in the gene pool. C is correct because most of the sons of the parent bunnies get the black and white striped fur because it is X-linked. It is recessive because the daughter bunnies need to have two of the black and white striped fur gene on the X chromosome to show that version of fur. D is not correct because it is not an autosomal trait because it deviates from the Law of Segregation. i. Explain Mendel’s Laws of Independent Assortment and Segregation and then explain TWO ways of inherited genes diverging from Mendel’s predicted ratios. Justify your answer. ii. Genetics are also determined by some alleles that are located on the sex chromosomes. Discuss sex-linked transmission and then provide an example of a sex-linked trait. i. The Law of Independent Assortment discusses alleles aligning then separating independently during gamete formation when genes for two traits are on homologous chromosomes. This takes place during Metaphase I. The Law of Segregation discusses that each allele in a pair segregates into a different gamete during gamete formation. This takes place during Anaphase I. One way inherited genes diverge from Mendel’s predicted ratios is by genes being on a sex chromosome. Sex-linked traits mostly occur in males because most are on the X chromosome and only one infected X chromosome is needed for a male to be infected. Another way genes differ from Mendel’s laws is there is always a chance for mutation in the genome during transcription and translation. Mendel doesn’t take into account that part of gene can be changed during gene replication during gamete formation. ii. Sex-linked transmission occurs during the deciding factor of whether a parent’s offspring is a boy or a girl. Sex-linked traits are usually X-linked. Fathers pass X-linked allele to daughters (XX) and mothers pass on X-linked alleles to either a son or daughter. SRY is the determining region of the Y. Without SRY, gonads develop into ovaries, and with SRY the gonads develop in the testes. If X-linked alleles are recessive, females show phenotype when they are homozygous recessive. While males on the other hand show the phenotype when hemizygous, making males more affected by sex-linked traits. An example of a sex-linked trait would be Hemophilia, which is sex-linked recessive. If a person is a Hemophiliac, they do not produce enough protein needed for blood clotting; therefore, there is no clotting. • Slide 2 of 2

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