Learning Objective Student Questions

Learning Objective Student Questions
2014 Review

Learning Objective 1.1 The student is able to convert a data set from a table of numbers that reflect a change in the genetic makeup of a population over time and to apply mathematical methods and conceptual understandings to investigate the cause(s) and effect(s) of this change

Learning Objective 1.1 Question 1 Table 1.1, Phenotypic Frequency over 60 years For a population of small forest trolls, the red allele is dominant to the purple allele. The habitat has no selection preference for either color. Assume during the time period represented in the graph, the small troll habitat was disrupted by a natural disaster. Describe an event that could account for the data in table 1.1. Explain how this event could account for the data in the table. Support your explanation using quantitative reasoning.

Learning Objective 1.1 Question 1 Answer Table 1.1, Phenotypic Frequency over 60 years For a population of small forest trolls, the red allele is dominant to the purple allele. The habitat has no selection preference for either color. Assume during the time period represented in the graph, the small troll habitat was disrupted by a natural disaster. Describe an event that could account for the data in table 1.1. Explain how this event could account for the data in the table. Support your explanation using quantitative reasoning. The natural disaster (fire, volcano, tsunami…) likely occurred between 1910 and 1925. This event likely wiped out a large quantity of the population and thus, the gene pool. This is an example of genetic drift, more specifically, the bottleneck effect. The remaining gene pool is different from the original. The data table shows us that the phenotypic frequencies changed between 1910 and 1925. We can find the change in allele frequency (and so change in gene pool) using the Hardy-Weinberg equation showing the values of p (frequency of dominant alleles) and q (frequency of recessive alleles) has changed significantly. p(1910) = 0.06, or 6% of the total alleles in the population are dominant in 1910. p(1925) = 0.86, or 86% of the total alleles in the population are dominant in 1925. q(1910) = 0.94, or 94% of the total alleles in the population are recessive in 1910. q(1925) = 0.14, or 14% of the total alleles in the population are recessive in 1925.

Learning Objective 1.1 Question 2 Use statistical analysis to ensure that the alleles present in 1925 are due to the natural disaster and not just random chance. There are 1,000 total alleles present in the population. Accepted probability on chi square chart is 0.05. For a population of small forest trolls, the red allele is dominant to the purple allele. The habitat has no selection preference for either color.

Learning Objective 1.1 Question 2 Confirm some or all of your conclusions for #1 by performing a chi-square test. The null hypothesis you are testing is that the observed and expected values are not significantly different from one another (because your expected values are calculated based on an assumption of Hardy-Weinberg equilibrium, this is the same as saying that the population is in H-W equilibrium for the genotype being tested). The critical value for the chi-square in this case is 3.841; if your calculated value of the chi-square is equal to or greater than that, the probability of the null hypothesis being correct (i.e., the probability of the population being in H-W equilibrium at that genotype) is 0.05 or less, and the null hypothesis is rejected.

1.1
The student is able to convert a data set from a table of numbers that reflect a change in the genetic makeup of a population over time and to apply mathematical methods and conceptual understandings to investigate the cause(s) and effect(s) of this change

Population of Dragons Ice dragons evolved from the same species fire dragons did except their breath gene was magically changed by a great wizard. The two dragon species dislike each other and are in constant battle with one another. They are evenly matched in strength and speed and have similar population sizes. The population of dragons per year is shown in the table above. Give a plausible explanation as to why the population changed so suddenly.

Population of Dragons Ice dragons evolved from the same species fire dragons did except their breath gene was magically changed by a great wizard. The two dragon species dislike each other and are in constant battle with one another. They are evenly matched in strength and speed and have similar population sizes. The population of dragons per year is shown in the table above. Give a plausible explanation as to why the population changed so suddenly. A solar flare caused a drastic increase in temperature that year, causing the ice dragons to weaken and die to rival fire dragons

Population of Dragons Ice dragons evolved from the same species fire dragons did except their breath gene was magically changed by a great wizard. The two dragon species dislike each other and are in constant battle with one another. They are evenly matched in strength and speed and have similar population sizes. The population of dragons per year is shown in the table above. Sheep are the main food source for dragons. A deadly virus infects and kills a majority of the sheep in the dragons’ habitat. Predict what will happen to the population of both fire and ice dragons.

Population of Dragons Ice dragons evolved from the same species fire dragons did except their breath gene was magically changed by a great wizard. The two dragon species dislike each other and are in constant battle with one another. They are evenly matched in strength and speed and have similar population sizes. The population of dragons per year is shown in the table above. Sheep are the main food source for dragons. A deadly virus infects and kills a majority of the sheep in the dragons’ habitat. Predict what will happen to the population of both fire and ice dragons. The population of fire dragons will drop drastically. The population of ice dragons will also drop but more gradually and not as much.

The graph to the left shows the distribution of light and dark colored phenotypes of a population of squirrel. They live in a forest with both white poplar trees and dark walnut trees. Name the type of distribution

The graph to the left shows the distribution of light and dark colored phenotypes of a population of squirrel. They live in a forest with both white poplar trees and dark walnut trees. Name the type of selection Disruptive Selection

The graph to the left shows the distribution of light and dark colored phenotypes of a population of squirrel. They live in a forest with both white poplar trees and dark walnut trees. A fire burns down a lot of the forest. Everything is covered in black ash and mostly walnut trees survive. Draw a graph of the expected population

Objective 1.4
Lauren Angello

Learning Objective 1.4 The student is able to evaluate data-based evidence that describes evolutionary changes in the genetic makeup of a population over time.

Learning Objective 1.4 Question 1 Describe two plausible reasons for a shift in phenotype.

Learning Objective 1.4 Answer 1 Describe two plausible reasons for a shift in phenotype. Genetic drift is plausible b/c if a lot of one phenotype of species dies or stops reproducing for any reason (natural disaster, ect.) it creates a higher frequency for another allele. Natural selection takes a hold of the population favoring one phenotype over another, causing a decline/increase for a given phenotype in a species.

Learning Objective 1.4 Question 2 Which type of genetic drift could this be an example of?

Learning Objective 1.4 Answer 2 Which type of genetic drift could this be an example of? The bottle neck effect because it may be due to an unusual force that drastically reduces numbers in a population.

Learning Objective 1.4 Question 3 Describe an event that could have occurred to cause this drastic shift in phenotype.

Learning Objective 1.4 Answer 3 Describe an event that could have occurred to cause this drastic shift in phenotype. Climate change, more snow favors light colored (albino) phenotype, causing all the brown bunnies to die. Invasive species that was introduced who could see brown bunnies more easily.

Learning Objective 1.4 Question 4 If the climate were to drastically shift again causing the albino bunnies to be seen more easily than the brown, what changes would occur in the population?

Learning Objective 1.4 Answer 4 If the climate were to drastically shift again causing the albino bunnies to be seen more easily than the brown, what changes would occur in the population? There would be a shift in the population again causing a decrease in albino and an increase in brown.

Phylogenetic Tree

Phylogenetic Tree Diagram

Phylogenetic Tree Question 1 According to this phylogenetic tree, annelids are most closely related to____ .

Phylogenetic Tree Question 1 According to this phylogenetic tree, annelids are most closely related to____ . Answer= Arthropods

Phylogenetic Tree Question 2 According to this phylogenetic tree, the clade that includes echinoderms and chordates is most closely related to the clade that includes____.

Phylogenetic Tree Question 2 According to this phylogenetic tree, the clade that includes echinoderms and chordates is most closely related to the clade that includes____. Answer=Molluscs

Phylogenetic Tree Question 3 According to this phylogenetic tree, cnidarians are most closely related to____.

Phylogenetic Tree Question 3 According to this phylogenetic tree, cnidarians are most closely related to____. Answer=Platyhelminths

Phylogenetic Tree Question 4 According to this phylogenetic tree, chordates are most closely related to____.

Phylogenetic Tree Question 4 According to this phylogenetic tree, chordates are most closely related to____. Answer=Echinoderms

Phylogenetic Tree How To Read The Phylogenetic Tree All of the Questions uses the same phylogenetic tree. The questions are all rather simply. First you must be able to read the tree. To read the tree it goes from right to left and it starts to branch off. Animals that have the least common would be furthest away and most common would be the closest to each other. Sub categories are for example where mollusca and arthropods are separated unlike proifera and cnidaria.

Concept: 1.5The student is able to connect evolutionary changes in a population over time to a change in the environment.

Question 1 Which of the following things is most consistent at causing populations to become better suited to their environments over the course of generations? Gene flow Mutation Natural selection Genetic drift

C. Natural Selection-the process where organisms better adapted to their environment tend to survive and produce more offspring. The theory is believed to be the main process that brings about evolution.

Question 2
All of the following options have a chance to influence gene frequencies in small populations rather than large populations, but which of the following options consistently requires a small population as a precondition to occur? Mutation Natural selection Random mating Genetic drift

D. Genetic Drift-is the change in the frequency of a gene variant (allele) in a population due to random sampling. When there are few copies of an allele, the effect of genetic drift is larger, and when there are many copies the effect is smaller.

Question 3
A population of birds has thrived on a small, secluded island for the past 2,000 years. There has been little to no human impact on this island whatsoever. However, over the past decade people have slowly come to this island to salvage its resources for the local industries on the nearest mainland. Deforestation and massive erosion has occurred on this island. (3 points possible) A). Name TWO possible effects/reactions the bird species could have. B). Name ONE adaption this bird species could develop over time in response to habitat loss.

General answer: A) Effects: threatened species, habitat loss, extinction Reactions: migration, adaption B) Adaption to other environmental conditions on the island

Question 4
Habitat loss is one of the leading human impacts on organisms and their environments. Of the following choices, chose TWO and reason how the effect population and how a population can adapt and evolve to meet these chosen changes. Impacts: Dredging, Draining, Bulldozing, Deforestation, Dams, Sewage Discharges, Mining, Resource Scavenging. (4 points possible, two for identification and two for accurate explanations)

General answer: Possible Explanations: Dams, fish have changed their migration patterns in response to the changes in stream flow. However, certain fish species have had to evolve new structures such as larger fins that allow them to propel themselves upstream in shallow conditions, as seen in some salmon species. Deforestation, animals have had to adapt to the ever changing conditions in forests as a result of habitat loss. Suburban expansion has resulted in massive habitat loss and as a result animals such as deer and bears have integrated themselves closer into human areas as a result of environment/ habitat capacity pressures.

Learning Objective #1.6
The student is able to use data from mathematical models based on the Hardy-Weinberg equilibrium to analyze genetic drift and effects of selection in the evolution of specific populations.

Learning Objective 1.6 Question 1 In a given population, 7,192 people have attached earlobes and 4,841 people have unattached earlobes. The allele for unattached earlobes is recessive. Assuming that the population is in Hardy-Weinberg equilibrium, what percentof the individuals in the population are heterozygous?

Learning Objective 1.6 Answer 1 In a given population, 7,192 people have attached earlobes and 4,841 people have unattached earlobes. The allele for unattached earlobes is recessive. Assuming that the population is in Hardy-Weinberg equilibrium, what percent of the individuals with attached earlobes are heterozygous? You can use A and a to represent the alleles for attached and unattached earlobes. The frequency of the aaphenotype is 4,841/(4,841+7,192)=.402. q2=.402 and q=.63.Becausep+q=1, p=.37. The frequency of the Aa genotype is equal to 2pq or 2(.63)(.37)=.47=47%

Learning Objective 1.6 Question 2 The delta-32 mutation, a recessive gene, gives humans protection from HIV infection. The allele frequency in a town in Sweden is 20%. What percent of the population have two copies of the gene and are therefore immune to HIV?

Learning Objective 1.6 Answer 2 The delta-32 mutation, a recessive gene, gives humans protection from HIV infection. The allele frequency in a town in Sweden is 20%. What percent of the population have two copies of the gene and are therefore immune to HIV? We know q=.2 because the allele frequency is 20%. Therefore p=.8. The question is asking for you to solve for q2 (homozygous recessive). q2=.22=.04=4%

Learning Objective 1.6 Question 3 The delta-32 mutation, a recessive gene, gives humans protection from HIV infection. The allele frequency in a town in Sweden is 20%. What percent of the population will be less susceptible to the disease since they are heterozygous?

Learning Objective 1.6 Answer 3 The delta-32 mutation, a recessive gene, gives humans protection from HIV infection. The allele frequency in a town in Sweden is 20%. What percent of the population will be less susceptible to the disease since they are heterozygous? We already solved for p and q. p is .8 and q is .2. Heterozygous would just be 2pq=2(.8)(.2)=.32=32%

Learning Objective 1.6 Question 4 The delta-32 mutation, a recessive gene, gives humans protection from HIV infection. The allele frequency in a town in Sweden is 20%. How would you expect the p and q values to change over time due to natural selection based on what you know about HIV.

Learning Objective 1.6 Answer 4 The delta-32 mutation, a recessive gene, gives humans protection from HIV infection. The allele frequency in a town in Sweden is 20%. How would you expect the p and q values to change over time due to natural selection based on what you know about HIV. The p value should decrease and the q value should increase because a gene that protects against HIV would be favorable and selected for, at least a little.

Learning Objective1.7The student is able to justify data from mathematical models based on the Hardy-Weinberg equilibrium to analyze genetic drift and the effects of selection in the evolution of specific populations.

Question 1 In a population of berry bushes, the allele for green berries is dominant to the allele for red berries. Red berries are usually selected against because animals can see the red berries against the green leaves. While being studied, a pesticide is sprayed on the bushes in year 3, changing the color of the leaves to red. Using what you know about genetic drift and evolution, justify the claim that the population is evolving.

In a population of berry bushes, green berries are dominant to red berries. Red berries are usually selected against because animals can see the red berries against the green leaves. While being studied, a pesticide is sprayed on the bushes in year 3, changing the color of the leaves to red. Using what you know about genetic drift and evolution, justify the claim that the population is evolving. The recessive phenotype frequency (q2) in year 2 was 0.13. Taking the square root of that to find the value of q, the frequency of the recessive allele, you get about 0.36, or 36% of the alleles in the population are recessive. The dominant allele frequency (p) would then be 0.64, or 64% of the alleles in the population are dominant, if you use the equation p+q=1. In year 3, q= 0.87 and p= 0.13. The change of the frequency of recessive alleles goes from 0.36 to 0.87, showing a dramatic change, indicating that evolution is occurring.

Question 2 In a population of berry bushes, the allele for green berries is dominant to the allele for red berries. Red berries are usually selected against because animals can see the red berries against the green leaves. In a population of 100 bushes, 25 have red berries and 75 have green berries. The next year, 28 had red berries and 72 have green berries. Using your knowledge of evolution and equilibrium, justify that the population is evolving or is not evolving.

In a population of berry bushes, the allele for green berries is dominant to the allele for red berries. Red berries are usually selected against because animals can see the red berries against the green leaves. In a population of 100 bushes, 25 have red berries and 75 have green berries. The next year, 28 had red berries and 72 have green berries. Using your knowledge of evolution and equilibrium, determine if the population is evolving. 25/100= 0.25, or 25% are the recessive phenotype., or q2 The frequency of the recessive allele , or q, would be 0.5. If p+q=1, then p would also be equal to 0.5. 28/100= 0.28, or 28% are the recessive phenotype , or q2,the next year. The frequency of the recessive allele, or q, would be approximately 0.53. If p+q=1, then p would equal 0.47. The allele frequencies changed slightly, however the change may be small enough to be considered insignificant as far as evolution purposes go.

Question 3 In a population of blue-eyed and brown-eyed individuals, 30% of the individuals display the recessive phenotype for blue eyes. Assuming Hardy-Weinberg equilibrium, what percentage of the population has both alleles? Use the Hardy-Weinberg equation to help justify your answer. A) 30% B) 70% C) 49.5% D) 42.8%

In a population of blue-eyed and brown-eyed individuals, 30% of the individuals display the recessive phenotype for blue eyes. Assuming Hardy-Weinberg equilibrium, what percentage of the population has both alleles? Use the Hardy-Weinberg equation to help justify your answer. A) 30% B) 70% C) 49.5%, 30%= 0.3 (q2). q=0.55. p+q=1, p= 0.45. 2pq= 2*0.55*0.45= 0.495 (Because heterozygotes have one of each allele and are represented in the equation as 2pq). 0.495*100= 49.5%. D) 42.8%

Question 4 If there are 20 out of 100 people in a population that show the dominant phenotype, how many individuals in the population have at least one recessive allele. Use the Hardy-Weinberg equation to justify your answer. A) 80 B) 99 C) 27 D) 38

If there are 20 out of 100 people in a population that show the dominant phenotype, how many in the population have at least one recessive allele. Use the Hardy-Weinberg equation to justify your answer. A) 80 B) 99, 80 people would show recessive phenotype, which would make q2= 0.8. Then q would be 0.894. If p+q= 1, p would then equal 0.106. 2pq= 0.19. Because heterozygotes have at least 1 recessive allele, they will be included in the answer. 19+80= 99 individuals with at least one recessive allele. C) 27 D) 38

Learning Objective 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

Learning Objective 1.8 Question 1 A small population of toucans inhabit a small tropical island. One day, a hurricane hits the island and some of the finches are blown away to an uninhabited nearby island, resulting in a loss of genetic variation in the population. This is an example of Geographic isolation A cline Bottleneck effect Founder effect

Learning Objective 1.8 Question 1 Answer A small population of toucans inhabit a small tropical island. One day, a hurricane hits the island and some of the finches are blown away to an uninhabited nearby island, resulting in a loss of genetic variation in the population. This is an example of Geographic isolation A cline Bottleneck effect Founder effect *A natural disaster (hurricane) reduced the size of the population, thus resulting in a loss of genetic variation

Learning Objective 1.8 Question 2 Based on the previous question, if some environmental conditions were different than that of the previous island that the toucans lived on, what would eventually happen to the toucans that were blown away to the nearby island?

Learning Objective 1.8 Question 2 Answer Based on the previous question, if some environmental conditions were different than that of the previous island that the toucans lived on, what would eventually happen to the toucans that were blown away to the nearby island? Natural selection would eventually alter the genetic make up of the island to help the toucans have more reproductive success in the environmental conditions on the island.

Learning Objective 1.8 Question 3 For a population of aliens, green skin is dominant to purple skin. Astronauts capture this population of aliens and breed them. They think that the green skin looks gross, so they only wish to breed alien offspring with purple skin. What type of selection is this and how will it effect and what will happen to the skin color allele frequencies in the population of aliens?

Learning Objective 1.8 Question 3 Answer For a population of aliens, green skin is dominant to purple skin. Astronauts capture this population of aliens and breed them. They think that the green skin looks gross, so they only wish to breed alien offspring with purple skin. What type of selection is this and how will it effect the skin color allele frequencies in the population of aliens? This is artificial selection. Because purple skin is the desire trait, the allele frequency of green skin will decrease and the purple skin allele frequency will increase.

Learning Objective 1.8 Question 4 Based on the map, what can you infer about polar bears and grizzly bears? Polar bears will migrate farther south. Polar bears and grizzly bears will mate and produce hybrid offspring. Grizzly bears will migrate south back to their historic habitat. Both polar bears and grizzly bears will migrate east

Learning Objective 1.8 Question 4 Answer Based on the map, what can you infer about polar bears and grizzly bears? Polar bears will migrate farther south. Polar bears and grizzly bears will mate and produce hybrid offspring. Grizzly bears will migrate south back to their historic habitat. Both polar bears and grizzly bears will migrate east *The grizzly bear’s extended range of habitat is near the polar bear’s habitat, so they could possibly mate with each other

Katie Cummings

Learning Objective 1.10 Question 1 All of the following provide evidence for evolution except A. vestigial characters. B. Darwin’s finches. C. homologous characters. D. embryology. E. mutations.

Learning Objective 1.10 Question 1 Answer All of the following provide evidence for evolution except A. vestigial characters. B. Darwin’s finches. C. homologous characters. D. embryology. E. mutations. E – Mutations in and of themselves are not evidence for evolution, although they are necessary if evolution is going to occur.

Learning Objective 1.10 Question 2 Discuss TWO mechanisms of speciation that lead to the development of separate species from a common ancestor.

Learning Objective 1.10 Question 2 Answer Discuss TWO mechanisms of speciation that lead to the development of separate species from a common ancestor. Geographic isolation (or allopatric speciation) takes place when a population of one species becomes physically separated by some geographic barrier such as a river, mountain range, etc. Long-term isolation of two populations eventually leads to reproductive isolation. Sympatric speciation happens when new species arise as a result of reproductive isolation within the population range — for example, because of polyploidy or switching mating behaviors (fruit flies going from hawthorn to apple to lay eggs). Eventually the two populations are unable to interbreed. Reproductive isolation by prezygotic barriers, such as habitat, temporal, behavioral, mechanical, or gametic incompatibility. Reproductive isolation by postzygotic barriers (e.g., reduced hybrid viability or fertility) leads to speciation

Learning Objective Question 1.10 Question 3 Explain THREE methods that have been used to investigate the phylogeny of organisms. Describe a strength or weakness of each method.

Learning Objective Question 1.10 Question 3 Answer Explain THREE methods that have been used to investigate the phylogeny of organisms. Describe a strength or weakness of each method.

Learning Objective 1.10 Question 4 All of these are examples of random evolutionary processes except: A. An earthquake divides a single elk species into two populations, forcing them to no longer interbreed. B. A mutation in a flower plant results in a new variety. C. An especially long winter causes a group of migrating birds to shift their home range. D. A mutation results in a population of trees that spread their seeds more widely than their peers, causing their population to grow. E. A spider species declines in an area because individuals are consistently moving out of an old range into a new range.

Learning Objective 1.10 Question 4 Answer All of these are examples of random evolutionary processes except: A. An earthquake divides a single elk species into two populations, forcing them to no longer interbreed. B. A mutation in a flower plant results in a new variety. C. An especially long winter causes a group of migrating birds to shift their home range. D. A mutation results in a population of trees that spread their seeds more widely than their peers, causing their population to grow. E. A spider species declines in an area because individuals are consistently moving out of an old range into a new range. D - This is the only answer that shows evidence of natural selection, which is the nonrandom process by which evolution occurs. The two elk species splitting (answer A) is an example of allopatric speciation caused by a random factor (a geologic event). A mutation is also a random event (answer B; for example, if we had said that the new variety became the dominant allele in a population because it had an advantage over other variants, then that would be natural selection. A home range shift (answer C) is not evolution, but rather a behavioral change within an organism’s lifetime. Finally, a spider species declining in an area because individuals are slowly changing territory is an example of gene flow, which we know to be random process of evolution.

Learning Objective 1.12 The student is able to connect scientific evidence from many scientific disciplines to support the modern concept of evolution.

Learning Objective 1.12 Question 1 Discuss two mechanisms of speciation that lead to the development of two or more separate species from a common ancestor.

Learning Objective 1.12 Answer 1 Discuss two mechanisms of speciation that lead to the development of two or more separate species from a common ancestor. POSSIBLE ANSWERS: Geographic isolation (or allopatric speciation) takes place when a population of one species becomes physically separated by some geographic barrier such as a river, mountain range, etc. Long-term isolation of two populations eventually leads to reproductive isolation. Sympatric speciation happens when new species arise as a result of reproductive isolation within the population range. Eventually the two populations are unable to interbreed. Reproductive isolation by prezygotic barriers, such as habitat, temporal, behavioral, mechanical, or gametic incompatibility. Reproductive isolation by postzygotic barriers leads to speciation.

Learning Objective 1.12 Question 2 Explain three methods that have been used to investigate the phylogeny of organisms. Describe a strength and weakness of each method.

Learning Objective 1.12 Answer 2 Explain three methods that have been used to investigate the phylogeny of organisms. Describe a strength and weakness of each method. Fossils Strength: can determine time and extinct species. Weakness: not all species leave fossils behind. Morphology Strength: Homologous structures indicate evolutionary relationships. Weakness: Analogous structures are difficult to distinguish between. Embryology Strength: shows similarities in structure and pattern of development. Weakness: similarities may be lost in development into adult. Molecular Strength: shows a large number or traits, very accurate. Weakness: no data for extinct species; variation within species makes things difficult. Behavioral Strength: some may be genetic. Weakness: behavior may be culturally transmitted or learned.

Learning Objective 1.12 Question 3 All of the organisms show a similar pulley-shaped astralagus bone in the ankle except for the whale. Using these phylogenetic trees that show the relationship between the whale and six other mammals, identify which tree best represents the evolutionary relationship of these animals. Base your answer on the principle of parsimony and genomic information given on the previous slide.

Learning Objective 1.12 Answer 3 Tree II is the best choice to represent the evolutionary relationships between the whale and six other mammals. By looking at the genetic information, one can see the deer and cow, whale and hippo, and pig and peccary have similar sequences in common. But the camel had little to no common sequences, separating it from the rest. Because of the genetic similarities in certain species, they can be ordered with the camel as the out-group.

Learning Objective 1.12 Question 4 All of these are examples of random evolutionary processes except A: An earthquake divides a single elk species into two populations, forcing them to no longer interbreed. B: A mutation in a flowering plant results in a new variety. C: An especially long winter causes a group of migrating birds to shift their home range. D: A mutation results in a population of trees that spread their seeds more widely than their peers, causing them to overpopulate. E: A spider species declines in an area because individuals are consistently moving out of an old range and into a new range.

Learning Objective 1.12 Answer 4 D: A mutation results in a population of trees that spread their seeds more widely than their peers, causing them to overpopulate. D is the correct choice because it is the only option that shows evidence of natural selection, which is a nonrandom process. A is allopatric speciation, B is a random event, C is a behavioral shift of location randomly, and E is an example of gene flow out of a population, all which are random processes.

Learning Objective 1.13 The student is able to construct and/or justify mathematical models, diagrams or simulations that represent processes of biological evolution. By: Monica Headley

Learning Objective 1.13 Question 1 Based on the phylogenetic tree shown below, which of the following statements is true? A. Clownfish have four walking legs. B. Tortoises have hair. C. Red-Eyed Tree Frogs have both a vertebral column and an amnion. D. Amnions exist in both tortoises and lions. E. Sponges have a vertebral column.

Learning Objective 1.13 Answer 1 Based on the phylogenetic tree shown below, which of the following statements is true? D. Amnions exist in both tortoises and lions. A. is incorrect because only red-eyed tree frogs, tortoises, and lions have four walking legs. B. is incorrect because hair only develops in lions. C. is incorrect because amnions evolved after red-eyed tree frogs. E. is incorrect because sponges are the outgroup, meaning they don’t have any of the listed traits in the phylogenetic tree.

Learning Objective 1.13 Question 2 According to this phylogenetic tree, annelids are most closely related to _____________. Mollusca Arthropoda Nematoda Porifera Echinodermata

Learning Objective 1.13 Answer 2 According to this phylogenetic tree, annelids are most closely related to _____________. B. Arthropoda Molluscs are closely related to annelids, but annelids are closer related to arthropods because they are both derivatives of molluscs. Nematodes, porifera, and echinodermata are related to annelids, but they aren’t as close as arthropods.

Learning Objective 1.13 Question 3 Based on the character table below, sketch a phylogenetic tree.

Learning Objective 1.13 Answer 3 Based on the character table below, sketch a phylogenetic tree.

Learning Objective 1.13 Question 4 Please write your question here Create a cladogram illustrating the organisms and their traits listed above.

Learning Objective 1.13 Answer 4 Create a cladogram illustrating the organisms and their traits listed above.

The student is able to pose scientific questions that correctly identify essential properties of shared, core life processes that provide insights into the history of life on Earth

1.14 Question 1 Which group of organisms is the most prolific of all the organisms that have ever existed on Earth, with over a million existing species known and perhaps several million more to discover? Bacteria Vertebrates Fungi Arthropods

1.14 D Both bacteria and fungi have many species with many new ones to discover, but whenever you encounter a question asking about the group with “over a million species,” “the most diversity”, “the most diverse group”, and so on, think either arthropods or insects. There are more species of arthropods than all the rest of the species combined.

1.14 Question 2 A scientist places free strands of DNA, which contain a gene that codes for the protein allowing the metabolism of glucose, in a medium containing bacteria that can only survive on the sugar lactose. The scientist heat shocks the bacteria in CaCl2 and lets them recover before plating them in several petri dishes with only glucose as a nutrient source. After several days, there are no signs of bacterial growth in the glucose medium. All of the following are possible explanations for the results except which? All of the bacteria died from the heat shock treatment. The gene for glucose metabolism was not incorporated into any of the bacteria. The genes inserted into the bacteria didn’t have the code for the metabolism of glucose. The free strands of DNA were from a sheep so will not function in bacteria.

Question 2 Answer D Choices A, B, and C describe problems associated with the procedure or materials given. The correct answer is D. All DNA is composed of base pairs; it doesn’t matter what organism it comes from. DNA can even be manufactured from scratch and never exists in an organism.

1.14 Question 3 All of the following are considered post-transcriptional modifications that occur in the nucleus except 5’ capping with methylated guanines. The addition of a 3’ poly-adenine tail. The excision of introns from mRNA via spliceosome formation. mRNA attachment to polyribosomes.

Question 3 Answer D All of the choices listed describe posttranscriptional modifications, those that take place in the nucleus after transcription, except choice D. This takes place outside the nucleus during translation.

1.14 Question 4 Which of the following forms three hydrogen bonds when linked with a cytosine nucleotide? Uracil nucleotide Guanine nucleotide Transcription Splicing

Question 4 Answer B A guanine nucleotide forms three hydrogen bonds when linked with a cytosine nucleotide.

Learning Objective # 1.15 The student is able to determine the outcome of genes in relation to the inheritance of X-linked and sex-linked genes.

Learning Objective 1.15 Question 1 When Thomas Hunt Morgan crossed his red-eyed F₁ generation flies to each other, the F₂ generation included both red- and white-eyed flies. All the white-eyed flies were male. What was the explanation for this result? A) The gene involved is on the Y chromosome. B) The gene involved is on the X chromosome. C) The gene involved is on an autosome, but only in males. D) Other male-specific factors influence eye color in flies. E) Other female-specific factors influence eye color in flies.

Learning Objective 1.15 Answer 1 When crossed, red-eyed F₁ generation flies to each other, and the F₂ generation included both red and white-eyed flies. The remaining white-eyed flies were male. How is this possible? The gene involved is located on the X chromosome. This gene is only carried on the X chromosome, therefore, the male must have one x and one y, obtaining white eyes.

Learning Objective 1.15 Question 2 Why are males more often affected by sex-linked traits than females? A) Because male hormones (testosterone) often alter the effects of mutations on the X chromosome. B) Because female hormones (estrogen) often compensate for the effects of mutations on the X chromosome. C) Because X chromosomes in males have more mutations than X chromosomes in females. D) Because males are hemizygous for the X chromosome. E) Because mutations on the Y chromosome cause more the effects of X-linked mutations.

Learning Objective 1.15 Answer 2 Males are more often affected by sex-linked traits than females because males are hemizygous for the X chromosome. This means that males can have an X and a Y chromosome, unlike females with two X chromosomes. This increases the chance of becoming affected by a trait.

Learning Objective 1.15 Question 3 Green-Blue color blindness is a sex-linked recessive trait. Two people with normal color vision have a color-blind son. What are the genotypes of the parents? A) XcXc and XcYB) XcXc and XCY C) XCXC and XcYD) XCXC and XCY E) XCXc and XCY

Learning Objective 1.15 Answer 3 Green-Blue color blindness is a sex-linked recessive trait. Two people with normal color vision have a color-blind son. What are the genotypes of the parents? XCXc and XCY because the parents cannot be homozygous recessive because they are not affected. They have to be heterozygous because they cannot be homozygous dominant otherwise the son could not have been affected.

Learning Objective 1.15 Question 4 Duchenne muscular dystrophy is caused by a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin. How likely is it for a woman to have this conditionand why? A) Women can never have this condition.B) ½ of the daughters of an affected man could have this condition.C) 1/4 of the children of an affected father and a carrier mother could have this condition.D) Rarely would a woman have this condition.E) Only if a woman is XXX could she have this condition.

Learning Objective 1.15 Answer 4 Duchenne muscular dystrophy is caused by a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin. How likely is it for a woman to have this condition and why? Rarely would a woman have this condition. This is the result because it is not common for a male who has been affected to mate with a carrier of this gene.

1.15
Sean McLaughlin

What evidence can show these three domains all come from a common ancestor? (Hint: What are common in all cells)

What evidence can show these three domains all come from a common ancestor? (Hint: What are common in all cells) Basic features of all cells Plasma membrane Semifluid substance called cytosol (cell juice) Chromosomes (carry genes) Ribosomes (make proteins)

If Archaea is the only domain that has no photosynthetic organism, what could be understood about the means of getting energy by the common ancestor?

If Archaea is the only domain that has no photosynthetic organism, what could be understood about the means of getting energy by the common ancestor? The common ancestor would have to be a heterotroph and have to get its energy somehow. Chloroplast was not present in the common ancestor.

If Species 1 and Species 2 heterotrophs and Species 3 is a autotroph what is the derived characteristic?

If Species 1 and Species 2 heterotrophs and Species 3 is a autotroph what is the derived characteristic? The derived Characteristic is Chloroplast which separates the species in the beginnng

What is the theory for how the derived character is incorporated into the species?

What is the theory for how the derived character is incorporated into the species? The Endosymbiotic Theory

Learning Objective #1.16 The student is able to justify the scientific claim that organisms share many conserved core processes and features that evolved and are widely distributed among organisms today.

Learning Objective __1.16___ Question 1 If a population is in Hardy Weinberg equilibrium then: it is evolving to adapt to environmental changes B. the frequency of alleles is changing with each generation C. mutations, immigration, and selective mating are changing allele frequencies D. it is not evolving and allele frequencies remain the same with each generation E. Homozygous recessive individuals are less fit.

Learning Objective __1.16___ Answer 1 If a population is in Hardy Weinberg equilibrium then: it is evolving to adapt to environmental changes B. the frequency of alleles is changing with each generation C. mutations, immigration, and selective mating are changing allele frequencies D. it is not evolving and allele frequencies remain the same with each generation E. Homozygous recessive individuals are less fit. Due to the requirements of equilibrium: 1.) no mutations 2.) random mating 3.) no natural selection 4.) large population size 5.) no gene flow.

Learning Objective _1.16__ Question 2 For a population in Hardy-Weinberg equilibrium, the frequency of the recessive allele: increases with each generation until it reaches 50% B. decreases each generation until it reaches 25% C. remains the same in every generation. D. decreases due to negative selection pressure on homozygous recessive individuals. E. increases due to the occurrence of new mutations

Learning Objective _____ Answer 2 For a population in Hardy-Weinberg equilibrium, the frequency of the recessive allele: increases with each generation until it reaches 50% B. decreases each generation until it reaches 25% C. remains the same in every generation. D. decreases due to negative selection pressure on homozygous recessive individuals. E. increases due to the occurrence of new mutations Because the population is in equilibrium there is no change.

Learning Objective __1.16___ Question 3 DNA sequences can be used to determine the evolutionary relationships of species because… organisms with similar anatomy will develop similar DNA sequences by convergent mutations B. DNA sequences for proteins never change, so two species that have the same protein will have the same DNA sequences C. natural selection causes organisms that live in similar environmental conditions to have the same mutations in their DNA sequences D. mutations occur randomly in DNA at a steady rate, so the number of DNA difference is equivalent to the time since a pair of species that shared a common ancestor

Learning Objective _____ Answer 3 DNA sequences can be used to determine the evolutionary relationships of species because… A. organisms with similar anatomy will develop similar DNA sequences by convergent mutations B. DNA sequences for proteins never change, so two species that have the same protein will have the same DNA sequences C. natural selection causes organisms that live in similar environmental conditions to have the same mutations in their DNA sequences D. mutations occur randomly in DNA at a steady rate, so the number of DNA difference is equivalent to the time since a pair of species that shared a common ancestor Self explanitory

Learning Objective _1.16__ Question 4 The independent development of similarities between unrelated groups resulting from adaptation to similar environments is known as… founder effect B. kin selection C. competitive exclusion D. adaptive radiation E. convergent evolution

Learning Objective __1.16___ Answer 4 The independent development of similarities between unrelated groups resulting from adaptation to similar environments is known as… founder effect B. kin selection C. competitive exclusion D. adaptive radiation E. convergent evolution Definition

The student is able to justify the scientific claim that organisms share many conserved core processes and features that evolved and are widely distributed among organisms today.

Learning Objective 1.16 Question 1 Organisms that utilize cellular respiration, maximize its efficiency by the highly folded cristae. However, simpler organisms do not have organelles such as mitochondria. How can these organisms still acquire their energy?

Learning Objective 1.16 Answer 1 Many simpler cells contain highly folded plasma membranes which can serve as “separate” compartments for the organism to perform these cellular activities.

Learning Objective 1.16 Question 2 What structures are homologous? A) Bird’s wing & Bat’s wing B) Human’s arm & Whale’s fin C) Bat’s wing & Bee’s wing D) Frog’s leg & Zebra’s arm

Learning Objective 1.16 Answer 2 B) Human’s arm & Whale’s fin Homologous structures are essentially core structures that have retained their characteristic features but have evolved to serve different purposes. The other answers demonstrate analogous structures or structures that are not even related.

Learning Objective 1.16 Question 3 Which structure is found in all prokaryotic and eukaryotic organisms but has evolved to provide the most success to the organism? A) Mitochondria B) Endoplasmic Reticulum C) Chloroplast D) Plasma membrane

Learning Objective 1.16 Answer 3 D) Plasma membrane Prokaryotes utilize their plasma membrane to simulate an environment for various reactions. Prokaryotes and eukaryotes also share ribosomes and the cytosol. Prokaryotes do not contain membrane-bound organelles and thus the other answers are ruled out.

Learning Objective 1.16 Question 4 Why are analogous structures not that valuable in illustrating the features that are conserved and shared among many organisms?

Learning Objective 1.16 Answer 4 Analogous structures both serve the same function. However, the two structures have evolved totally independently of one another. Therefore they are not features that were initially conserved.

1.17
The student is able to pose scientific questions about a group of organisms whose relatedness is described by a phylogenetic tree or cladogram in order to (1) identify shared characteristics, (2) make inferences about the evolutionary history of the group, and (3) identify character data that could extend or improve the phylogenetic tree

1.17 1. Of the cladograms shown below, which one shows a different evolutionary history from the others?

1.17 Cladogram 3. In it, B and C are shown as sharing the most recent common ancestor, whereas in the others, C shares its most recent common ancestor with D.

1.17 2. Which derived characters are present in only one terminal taxon?

1.17 Antennae and spines

3.What’s the sister group to cows?

1.17 The answer is whales because they share a common ancestor.

4.At the top of the tree, a bracket marks the groups that are considered to belong to the reptiles. Would you consider the reptile group, as labeled, to be a true clade? If yes, why? If no, why not?

1.17 The answer is no. The reptile group doesn’t include all the descendants of the common ancestor. You’d have to add birds to make it a true clade.

Learning objective 1.17 Name 3 characteristics of birds?

Learning objective 1.17 Feathers, bipedalism, amniote egg, etc. These are the tick marks before birds are shown

Learning objective 1.17 Which specie(s) is an outgroup?

Learning objective 1.17 The least closely related organisms are lizards and snakes.

Learning objective 1.17 What characteristics separate the outgroup from the most evolved?

Learning objective 1.17 According to the cladogram, these traits are hair, mammary glands, and endothermy. Bipedalism, and parental care, is not observed by ALL mammals. This is why it is not shown before mammal evolution.

Learning objective 1.17 Identify the most recent and common ancestor of crocodiles, dinosaurs, and birds.

Learning objective 1.17 This is where they last joined before joining on the cladogram.

Learning Objective 1.18 The student is able to evaluate evidence provided by a data set in conjunction with a phylogenetic tree or a simple cladogram to determine evolutionary history and speciation.

Learning Objective 1.18 Question 1 In the above tree, assume that the ancestor had a long tail, ear flaps, external testes, and fixed claws. Based on the tree and assuming that all evolutionary changes in these traits are shown, what traits does a sea lion have? a) long tail, ear flaps, external testes, and fixed claws b) short tail, no ear flaps, external testes, and fixed claws c) short tail, no ear flaps, abdominal testes, and fixed claws d) short tail, ear flaps, abdominal testes, and fixed claws e) long tail, ear flaps, abdominal testes, and retractable claws

Learning Objective 1.18 Answer 1 d) short tail, ear flaps, abdominal testes, and fixed claws ‘d’ is the correct answer. Tracing up from the ancestor to sea lions, one sees that the only changes are in tail length and testes position. For the other traits, they have retained the ancestral condition.

Learning Objective 1.18 Question 2 In the above tree, assume that the ancestor was a herb (not a tree) without leaves or seeds. Based on the tree and assuming that all evolutionary changes in these traits are shown, which of the tips has a tree habit and lacks true leaves? a) Lepidodendron b) Clubmoss c) Oak d) Psilotum e) Fern

Learning Objective 1.18 Answer 2 a) Lepidodendron ‘a’ is the correct answer. Clubmosses are not trees. Oak (and yew) are trees but they have leaves. Psilotum lacks leaves, but it is not a tree. A fern has leaves and is not a tree.

Learning Objective 1.18 Question 3 Which of the four trees above depicts a different pattern of relationships than the others?

Learning Objective 1.18 Answer 3 Which of the four trees above depicts a different pattern of relationships than the others? ‘a’ is the correct answer. In all the other trees B is more closely related to D and E than is C. In ‘a,’ C is more closely related to D and E than is B.

Learning Objective 1.18 Question 4 Which of the trees below is false given the larger phylogeny above?

Learning Objective 1.18 Answer 4 Which of the trees below is false given the larger phylogeny above? ‘d’ is the correct answer. ‘d’ shows yeast being more closely related to plants than it is to animals. The true phylogeny (right) shows that yeast is more closely related to human than to any of the plants (green algae, lily, and fern).

Learning Objective 1.20 The student is able to analyze data related to questions of speciation and extinction throughout the Earth’s history.

Learning Objective 1.20 Question 1 The chart shows the mate choices of a Female European flycatcher. The pied flycatcher and collared flycatcher are closely related species. The female selects between her own species, pied, and collared males. Which of the following is most supported by the first graph (on the left)? a) All of the female European flycatcher choose to mate with the sympatric collared male because they share the same coat color, making them look nearly identical to each other. b) The female European flycatchers can tell the difference between the other species living amongst them, and thus, only mate with their own species. c) The female European flycatchers mate with both her own species and the other species. d) When choosing between mates outside of her own specie's living area, the female European flycatcher mates with both her own species and the other species. e) The female European flycatchers mate with the other sympatric species, leading to offspring that are hybrids of the two.

Learning Objective 1.20 Question 1 The chart shows the mate choices of a Female European flycatcher. The pied flycatcher and collared flycatcher are closely related species. The female selects between her own species, pied, and collared males. Which of the following is most supported by the first graph (on the left)? a) All of the female European flycatcher choose to mate with the sympatric collared male because they share the same coat color, making them look nearly identical to each other. b) The female European flycatchers can tell the difference between the other species living amongst them, and thus, only mate with their own species. c) The female European flycatchers mate with both her own species and the other species. d) When choosing between mates outside of her own specie's living area, the female European flycatcher mates with both her own species and the other species. e) The female European flycatchers mate with the other sympatric species, leading to offspring that are hybrids of the two. B. Answer choice a) is incorrect because none mate with the other species. Choice c) is true of the graph on the right, not on the left. Choice d) is also true of the graph on the left. Choice e) is incorrect. Choice b) is correct because sympatric means that the species live together. The female European flycatchers can tell the difference between the species since they live with them, and only mates with her own species.

Learning Objective 1.20 Question 2 The chart shows the mate choices of a Female European flycatcher. The pied flycatcher and collared flycatcher are closely related species. The female selects between her own species, pied, and collared males. Which of the following can be concluded about barriers to reproduction? a) No barriers to reproduction exist for the birds of allopatric population. b) No barriers to reproduction exist for the birds of sympatric population. c)The barriers to reproduction are equally as strong for the birds in both populations. d) Barriers to reproduction appear to be stronger in birds from allopatric populations than in birds from sympatric populations. e) Barriers to reproduction appear to be stronger in birds from sympatric populations than in birds from allopatric populations.

Learning Objective 1.20 Question 2 The chart shows the mate choices of a Female European flycatcher. The pied flycatcher and collared flycatcher are closely related species. The female selects between her own species, pied, and collared males. Which of the following can be concluded about barriers to reproduction? a) No barriers to reproduction exist for the birds of allopatric population. b) No barriers to reproduction exist for the birds of sympatric population. c)The barriers to reproduction are equally as strong for the birds in both populations. d) Barriers to reproduction appear to be stronger in birds from allopatric populations than in birds from sympatric populations. e) Barriers to reproduction appear to be stronger in birds from sympatric populations than in birds from allopatric populations. E. The female European flycatcher can tell the difference between the different species in the sympatric population, and thus no interbreeding occurs. These differences are harder to spot for the allopatric populations, and so reproduction between species is easier. Thus, the barriers are stronger in the sympatric populations than allopatric.

Learning Objective 1.20 Question 3 The chart shows the mate choices of a Female European flycatcher. The pied flycatcher and collared flycatcher are closely related species. The female selects between her own species, pied, and collared males. If all populations become allopatric, according to the data, which of the following can occur? a) Even if speciation occurs again and again, it will be still be impossible for the difference to be great enough for a new group of organisms to be formed. b) The reproductive behaviors in the sympatric population demonstrates the viability of the hybrid species. c) Extinction will never occur because all of these birds are considered the same species. d) If the hybrid species is viable and can reproduce, then over time, the original European, pied, and collared species may shrink and even become extinct. e) None of the above can occur.

Learning Objective 1.20 Question 3 The chart shows the mate choices of a Female European flycatcher. The pied flycatcher and collared flycatcher are closely related species. The female selects between her own species, pied, and collared males. If all populations become allopatric, according to the data, which of the following can occur? a) Even if speciation occurs again and again, it will be still be impossible for the difference to be great enough for a new group of organisms to be formed. b) The reproductive behaviors in the sympatric population demonstrates the viability of the hybrid species. c) Extinction will never occur because all of these birds are considered the same species. d) If the hybrid species is viable and can reproduce, then over time, the original European, pied, and collared species may shrink and even become extinct. e) None of the above can occur. D. When the little differences add up over many generations, leading to the hybrid species forming a new species entirely, then the old species may eventually die out if they become less common.

Learning Objective 1.20 Question 4 A Which branch point represents the most recent common ancestor of chordates and annelids? E B C D A B C D E

Learning Objective 1.20 Question 4 A Which branch point represents the most recent common ancestor of chordates and annelids? E B C D A B C D E C

Learning Objective 1.21 The student is able to analyze data related to questions of speciation and extinction throughout the Earth’s history.

Learning Objective _____ Question 1 Given the different coloration of bees, light brown and black. How could environment changes caused directional selection to either color of bees?

Learning Objective _____ Answer 1 If room allows, copy and paste your question here, and make the font smaller to fit. A climate change could caused a destruction of the habitat of light color or dark tree where the bees lives. The animals that only hunts light brown trees are endangered.

Learning Objective _____ Question 2 How could disruptive selection on mice lead to speciation?

Learning Objective _____ Answer 2 If room allows, copy and paste your question here, and make the font smaller to fit. The mice that were once closely related started to grew farther away due to selective pressure. Given time, the two mice might undergoes habitat isolation.

Learning Objective _____ Question 3 Given the following graph, what could be a possible reasons for the declined population of seals and polar bear to be endangered?

Learning Objective _____ Answer 3 If room allows, copy and paste your question here, and make the font smaller to fit. Their habitat being destroyed The meals are harder to find They are being hunted by predator Genetic mutation that causes diseases

Learning Objective _____ Question 4 Using the graph from the previous question, how could human interaction affect the animals to be endanger?

Learning Objective _____ Answer 4 If room allows, copy and paste your question here, and make the font smaller to fit. The pollution from factories and vehicles cause chemical reaction ( ex. Chloride and bromide) in the ozone and cause it to start depleting. The sunlight is heating up the earth more. The ozone was a mechanism that protect the earth from being overly heated by sun. The increasing amount of temperature is destroying the polar bear and seal habitats.

The student is able to use data from a real or simulated population(s), based on graphs or models of types of selection, to predict what will happen to the population in the future.

Learning Objective 1.22 Question 1 The following graph shows the frequency of a certain species of wood mouse by fur color. If a new species of bird was introduced that does a majority of its hunting under darker, fallen trees, how would the population of wood mice be affected?

Learning Objective 1.22 Answer1 The following graph shows the frequency of a certain species of wood mouse by fur color. If a new species of bird was introduced that does a majority of its hunting under darker, fallen trees, how would the population of wood mice be affected? The introduction of the bird species would likely lead to a reduction of individuals with lighter fur. This selection pressure would lead to a higher frequency of individuals with darker fur, and directional selection will occur.

Learning Objective 1.22 Question 2 Assume that the graph below, representing a mouse population, began as a normal distribution curve. What type of selection could lead the population to have to a distribution of phenotypes represented by this graph?

Learning Objective 1.22 Answer 2 Using the same graph as the previous question, what type of selection could lead to a distribution of individuals relative to a particular phenotype that is similar to the mouse population in this graph. A particular population would need to undergo disruptive selection in order to have the population exhibit a particular phenotype at both extremes of the spectrum.

Learning Objective 1.22 Question 3 Due to environmental contamination, the surrounding environment is changing for the dark peppered moths. The following chart shows the number of moths captured with respect to the color phenotype. Using the data given, predict what will happen to the population of moths by year 15 if the trend continues.

Learning Objective 1.22 Answer 3 Due to environmental contamination, the surrounding environment is changing for the dark peppered moths. The following chart shows the number of moths captured with respect to the color phenotype. Using the data given, predict what will happen to the population of moths by year 15 if the trend continues. If this trend continues, it is probable that the number of observed light colored moths will continue to approach zero, and may possibly cease to exist in the population due to the directional selection that is occurring.

Learning Objective 1.22 Question 4 At year 10 on the graph, the local government issues a mandate that should drastically cut down on the air pollution in the area. The drastic change is expected to get the trees in the area back to their normal, lighter color within a 10 year period. What changes would be expected to happen to the population of moths that inhabit the trees at the end of this new 10 year period?

Learning Objective 1.22 Answer 4 At year 10 on the graph, the local government issues a mandate that should drastically cut down on the air pollution in the area. The drastic change is expected to get the trees in the area back to their normal, lighter color within a 10 year period. What changes would be expected to happen to the population of moths that inhabit the trees at the end of this new 10 year period? If the trees begin to return to their original state, so should the population of moths. In this case, the moths will no longer be selected for their darker color, but now will be selected for their lighter color as the trees will return to their original lighter color. So by the end of the new 10 year period, the moth population should resemble the original population at year one.

Learning Objective 1.23 The student is able to justify the selection of data that address questions related to reproductive isolation and speciation

Learning Objective 1.23 Question 1 A certain species of bird on Herman Island is known for its short and thin beak and for using the insects within trees as its main food source. After a large storm in 1966, half of this bird’s population was blown away into the O’Neil Isles. Forty years later, researchers visit the isles and notice that this certain population of birds all mostly have long and thick beaks. It is deemed a possibility that speciation has occurred. Name this type of speciation and explain one possible environmental factor that caused the change in phenotype.

Learning Objective 1.23 Answer 1 A certain species of bird on Herman Island is known for its short and thin beak. After a large storm in 1966, half of this bird’s population was blown away into the O’Neil Isles. Forty years later, researchers visit the isles and notice that this certain population of birds all mostly have long and thick beaks. It is deemed a possibility that speciation has occurred. Name this type of speciation and explain one possible environmental factor that caused the change in phenotype. This is an example of allopatric speciation. A possible environmental factor is that the trees in the O’Neil isles have thicker bark. It would be more difficult for the birds to get to their necessary food source so their beaks would be thicker to break the bark and longer to reach the insects.

Learning Objective 1.23 Question 2 Researchers would like to prove that speciation had occurred in the previous example. Identify one way that they can prove this through an experiment.

Learning Objective 1.23 Answer 2 Researchers would like to prove that speciation had occurred in the previous example. Identify one way that they can prove this through an experiment. The researchers could reintroduce the bird population on Herman Island with the population on the O’Neil isles. They can try to induce mating between birds of the two populations but if they are incompatible, then speciation occurred. They could also compare the DNA nucleotide sequences between birds of the two populations to look for significant differences.

Learning Objective 1.23 Question 3 A population of black beetles and a population of brown beetles used to be part of one uniform population. The habitat they lived in was quite large until interference by humans cut down the area by half. This created a lot less breeding grounds for the beetles. As a result, half the population continued to mate during the spring as before while the other half had to adapt and mate during the fall. Due to the varied mating schedules, inbreeding between the black and brown beetles is rare. What kind of speciation does this indicate?

Learning Objective 1.23 Answer 3 A population of black beetles and a population of brown beetles used to be part of one uniform population. The habitat they lived in was quite large until interference by humans cut down the area by half. This created a lot less breeding grounds for the beetles. As a result, half the population continued to mate during the spring as before while the other half had to adapt and mate during the fall. Due to the varied mating schedules, inbreeding between the black and brown beetles is rare. What kind of speciation does this indicate? This indicates parapatric speciation because it came about due to a change in the habitat. The two populations still inhabit the same area but the change caused them to develop distinct characteristics (the difference in mating schedules.)

Learning Objective 1.23 Question 4 The genetic differences between the brown and black beetles continue to increase over the years. What might happen to the offspring of two beetles that are largely genetically different? Would it be possible for a new hybrid species to emerge?

Learning Objective 1.23 Answer 4 The genetic differences between the brown and black beetles continue to increase over the years. What might happen to the offspring of two beetles that are largely genetically different? BEST ANSWER (may be others): The offspring would most likely be sterile and not able to reproduce with other beetles. As a result, a hybrid species could not be sustained.

Objective 1.24 : The student is able to describe speciation in an isolated population and connect it to change in gene frequency, change in enviroment, natural selection, and/or genetic drift

Objective 1.24 Which of the following is true about sympatric speciation? A) Sympatric speciation has never been seen in nature. B) Sympatric speciation can occur in only a single generation. C) Sympatric speciation is always initiated by the geographic isolation of two populations. D) Sympatric speciation is imaginary. E) All of the above.

Objective 1.24 Answer: BAlthough sympatric speciation usually takes place over many generations, it can occur in a single generation in plants.

A group of mermaids exits in which 70% of the individuals have green fins and the remaining 30% have purple fins. Due to the disastarous oil spills caused by humans, the population was spilt into two, seperating Ariel from her sisters. Coincidently, the split left Ariel's group with 90% green fins and the other with 90% purple fins. Over time, the two populations become different species, one of which is entirely composed of green tailed mermaids and the other completely purple finned. What speciation has taken place? A) No Speciation B) Sympatric Speciation C) Allopatric Speciation D)Parapatric Speciation

Objective 1.24 Answer: CAllopatric speciation, the most common form of speciation, occurs when populations of a species become geographically isolated.

Objective 1.24 A population of grass plants live at the edge of a mining area. Close to this area, the soil has become infused with a high concentration of metals, making it more difficult for plants to grow there. The grass plants that grow in these areas develop more slowly than those that grow farther away from the mines and they flower later in the season. As a result, two different populations of grass arise that experience temporal reproductive isolation and evolve into different species. What type of speciation is occuring? A) No speciation B) Sympatric Speciation C) Allopatric Speciation D) Parapatric Speciation

Objective 1.24 Answer: DParapatric speciation occurs when populations are seperated not only a geographical barrier, but by an extreme change in habitat. Each population develops distinct characteristics and lifestyles. Reproductive isolation is temporal or behavior, rather than geographic.

Objective 1.24 A flowering plant usually produces seeds that are diploid. In one generation, however, this plant produces several seeds that have four copies of each chromosome, also called tertaploids. These seeds germinate and mate with each other to create more tetrapolid offspring. What type of speciation is most likely to have occured? A) No Speciation B) Sympatric Speciation C) Allopatric Speciation D) Parapatric Speciation

Objective 1.24 Answer: BSympatric speciation occurs when populationsof a species that share the same habitat become reproductively isolated from each other. This usually occurs through polyploidy, in which an offspring will be produced with twice the normal number of chromosomes.

The student is able to describe a model that represents evolution within a population.

Learning objective 1.25 question 1

Learning objective 1.25 Answer 1 D

Learning objective 1.25 Answer 2 B

Learning objective 1.25 Answer 3

Learning objective 1.25 question 4 (question on the next slide)

Learning objective 1.25 Answer 4

The student is able to evaluate given data sets that illustrate evolution is an ongoing process.

Learning Objective 1.26: Question 1 Fur Color in Mice Population A population of mice exhibit either black fur or white fur phenotypes. Use the data table above to select which event is the most likely to have happened. The white furred mice lived longer than black furred mice. The black furred mice were able to hide from predators more easily than white furred mice. The black furred mice were killed by a disease that the white furred mice were resistant to.

Learning Objective 1.26: Question 1 Answer b) The black furred mice were able to hide from predators more easily than white furred mice. Their black fur allowed them to hide in dark places while white furred mice could easily be seen and caught by predators. Since the black furred mice lived they were able to reproduce and pass on their genes.

Learning Objective 1.26: Question 2 Fur Color in Mice Population In 2013, a lethal virus infected the mice population. The fur color gene was linked to the gene which carried resistance to the virus. Which is the most probable combination? The black fur allele is linked with the allele for resistance to the virus. The white fur allele is linked with the allele for resistance to the virus. The white fur allele is linked with the allele for NO resistance to the virus. The genes are not linked.

Learning Objective 1.26: Question 2 Answer b) The white fur allele is linked with the allele for resistance to the virus. This means that the mice with white fur are not infected by the virus and remain healthy. The white furred mice are able to live and reproduce.

Learning Objective 1.26: Question 3 Fur Color in Mice Population All mice are now resistant to the virus. What will most likely happen to the percentages of white furred and black furred mice in the future? The black furred mice will all die because of pollution. The percentage of black furred mice will slowly increase, while the percentage of white furred mice will slowly decrease. The percentages will remain the same because the population is at equilibrium. The virus will return and kill all the white furred mice.

Learning Objective 1.26: Question 3 Answer b) The percentage of black furred mice will slowly increase, while the percentage of white furred mice will slowly decrease. This is because the black fur will still be an advantage for hiding from predators. More black furred mice will live and reproduce while more white furred mice are caught. Slowly there will be more black furred mice than white furred mice.

Learning Objective 1.26: Question 4 Height in Tree Population Based on the data table above, what would you expect to happen to the percentages of tall and short trees if they are planted close to one another? The percentage of short trees will increase while the percentage of tall trees decreases. The percentage of tall trees will increase while the percentage of short trees decreases. The percentages will remain the same.

Learning Objective 1.26: Question 4 Answer b) The percentage of tall trees will increase while the percentage of short trees decreases. The tall trees will block out sunlight from the short trees. If the short trees do not receive sunlight, they will die.

The student is able to describe a scientific hypothesis about the origin of life on Earth

What experiment stimulated the conditions thought at the time to be present on early Earth? Learning Objective 1.27; Question 1 E. Nirenberg and Matthaei experiment F. Hershey-Chase experiment G. Miller-Urey experiment H. Ernest Rutherford experiment I. Robert Millikan experiment

Learning Objective 1.27; Answer 1 Answer: G What experiment stimulated the conditions thought at the time to be present on early Earth? G. The Miller-Urey experiment proved that the hypothesis on primitive Earth favored chemical reactions that created more complex compounds, which in turn, would lead to complex life on Earth. E. Nirenberg and Matthaei experiment F. Hershey-Chase experiment G. Miller-Urey experiment H. Ernest Rutherford experiment I. Robert Millikan experiment

What is abiogenesis? And what do most scientists believe about this process and its relation to Earth? Learning Objective 1.27; Question 2

Learning Objective 1.27; Answer 2 Answer: What is abiogenesis? And what do most scientists believe about this process and its relation to Earth? Abiogenesis is the process by which life arises form non-living matter such as simple organic compounds (oxygen, hydrogen, etc.). Most scientists believe Earth went through this process, that life originated spontaneously on Earth. Everything literally popped out of nowhere. Well, not really. But sort of.

Learning Objective 1.27; Question 3 The early earth was a harsh environment and present day organisms that could possibly have survived that type of environment are A. Archaebacteria B.Early plants called blue-green algae C.Protobionts D.Eukaryotic organisms

Learning Objective 1.27; Answer 3 Answer: A The early earth was a harsh environment and present day organisms that could possibly have survived that type of environment are A. Archaebacteria are known to survive in extreme conditions, which would make them perfect to live in earth’s early enviroment. A. Archaebacteria B. Early plants called blue-green algae C. Protobionts D. Eukaryotic organisms

Learning Objective 1.27; Question 4 Miller and Urey's experiments proved that A. life evolved on earth from inanimate chemicals B. complex organic molecules can form spontaneously under conditions that probably existed on the early earth C. bacteria were the first type of living organism to appear on the earth D. RNA can act as an enzyme and assemble new RNA molecules from RNA templates

Learning Objective 1.27; Answer 4 Miller and Urey's experiments proved that B. Miller and Urey’s experiment proved that the origin of life on earth came from the spontaneous reactions between simple compounds that led to more complex compounds. A. life evolved on earth from inanimate chemicals B. complex organic molecules can form spontaneously under conditions that probably existed on the early earth C. bacteria were the first type of living organism to appear on the earth D. RNA can act as an enzyme and assemble new RNA molecules from RNA templates

Learning Objective 1.28 The student is able to evaluate scientific questions based on hypotheses about the origin of life on Earth

Learning Objective __1.28___ Question 1 How did the volcanic eruptions contribute to the origin of life?

Learning Objective __1.28___ Answer 1 How did the volcanic eruptions contribute to the origin of life? Volcanic eruptions released thick water vapor and compounds such as carbon dioxide, ammonia, nitrogen, hydrogen, and more which provided materials to build organic molecules.

Learning Objective __1.28___ Question 2 What were the early conditions of earth? Filled with rocks and ice The atmosphere was filled with thick water vapor Had volcanic eruptions that released compounds All of the above

Learning Objective __1.28___ Answer 2 What were the early conditions of earth? D) all of the above Hypotheses suggest that bombardments of huge chunks of rocks and ice formed the planet during the creation of the solar system. Furthermore, volcanoes were active and constantly erupted to release substances like gases and water vapor.

Learning Objective __1.28___ Question 3 Which of the following is a characteristic of vesicles in early earth conditions? a) able to reproduce, have simple metabolism, and able to maintain internal chemical environments b) able to produce enzymes to aid metabolism c) are made of RNA sequences

Learning Objective __1.28___ Answer 3 Which of the following is a characteristic of vesicles in early earth conditions? a) able to reproduce, have simple metabolism, and able to maintain internal chemical environments These characteristics are all required for life to occur and vesicles were the basis of the creation of life.

Learning Objective __1.28___ Question 4 What is the benefit of reproduction in protocells?

Learning Objective __1.28___ Answer 4 What is the benefit of reproduction in protocells? Able to pass on genetic information to produce more proto cells and continue life

Learning Objective #1.29 The student is able to describe the reasons for revisions of scientific hypotheses of the origin of life on Earth.

Learning Objective 1.29 Question 1 Several scientific laboratories across the globe are involved in research concerning the origin of life on Earth. Which of these questions is currently the most problematic and would have the greatest impact on our understanding if we were able to answer it? A) How can amino acids, simple sugars, and nucleotides be synthesized abiotically? B) How can RNA molecules catalyze reactions? C) How did RNA sequences come to carry the code for amino acid sequences? D) How could polymers involving lipids and/or proteins form membranes in aqueous environments? E) How can RNA molecules act as templates for the synthesis of complementary RNA molecules?

Learning Objective 1.29 Answer 1 Several scientific laboratories across the globe are involved in research concerning the origin of life on Earth. Which of these questions is currently the most problematic and would have the greatest impact on our understanding if we were able to answer it? A) How can amino acids, simple sugars, and nucleotides be synthesized abiotically? B) How can RNA molecules catalyze reactions? C) How did RNA sequences come to carry the code for amino acid sequences? D) How could polymers involving lipids and/or proteins form membranes in aqueous environments? E) How can RNA molecules act as templates for the synthesis of complementary RNA molecules? We would be able to gain more knowledge on the origin of amino acids, one of the building blocks of life.

Learning Objective 1.29 Question 2 List six main sources of evolutionary evidence.

Learning Objective 1.29 Answer 2 List six main sources of evolutionary evidence. Fossil record Comparative anatomy Comparative biology Comparative embryology Biogeography Molecular biology

Learning Objective 1.29 Question 3 Which of the following is the correct sequence of events in the origin of life? I. formation of protobiontsII. synthesis of organic monomers III. synthesis of organic polymers IV. formation of DNA-based genetic systems A) I, II, III, IV B) I, III, II, IV C) II, III, I, IV D) II, III, IV, I

Learning Objective 1.29 Answer 3 Which of the following is the correct sequence of events in the origin of life? I. formation of protobiontsII. synthesis of organic monomers III. synthesis of organic polymers IV. formation of DNA-based genetic systems A) I, II, III, IV B) I, III, II, IV C) II, III, I, IV D) II, III, IV, I The synthesis of organic monomers led to the synthesis of polymers. Protobionts formed as a result of the organic polymers, and thus, DNA-based genetic systems were formed.

Learning Objective 1.29 Question 4 What 3 domains of science can be used to explain the transition of Earth into its state of being able to sustain life?

Learning Objective 1.29 Answer 4 What 3 domains of science can be used to explain the transition of Earth into its state of being able to sustain life? Geology – fossil record Chemistry – Miller-Urey experiment Molecular – homologies and what common ancestor exists

Learning Objective 1.31 The student is able to evaluate the accuracy and legitimacy of data to answer scientific questions about the origin of life on Earth.

Learning Objective 1.31 Question 1 1. What are the four essential steps for life to emerge on Earth?

1. What are the four essential steps for life to emerge on Earth? 1. First: An abiotic (nonliving) synthesis of Amino Acids and Nucleic Acids must occur. Second: Monomers must be able to join together to form more complex polymers using energy that is obtained from the surrounding environment. Third: The RNA/DNA molecules form and gain the ability to reproduce and stabilize by using chemical bonds and complimentary bonding. Fourth: The evolution of a protobiont (means “first life form”) membrane of phospholipids and proteins to keep the “cell” intact.

Learning Objective 1.31 Question 2 2. What is Abiotic synthesis? Who discovered it? What can be used as evidence to support this claim?

2. What is Abiotic synthesis? Who discovered it? What can be used as evidence to support this claim? 2. Abiotic Synthesis is the preparation of a compound, often of biological relevance, without the use of biological agents such as enzymes or nucleic acids. Stanley Miller and Harold Urey were the two scientists that conducted an experiment, in 1953, supporting the ideas of abiotic synthesis. The experiment took the inorganic gases hypothesized to have been in early Earth’s atmosphere, H2, CH4, NH3, and water vapor and formed a variety of organic molecules, such as amino acids and oils. These organic molecules are the building blocks of all life forms. The energy source for the formation was an electrical spark that was to represent lightning in the early atmosphere. The gases hypothesized to be present in the early Earth’s atmosphere came from data associated with analyzing active volcanoes eruptions. Volcanoes would have been very active and present in the early Earth time period, as well as still today. These organic molecules would have accumulated in the Earth’s early oceans over millions of years and thus making possible the formation of more complex structures.

Learning Objective 1.31 Question 3 3. What is planetary habitability?

3. What is planetary habitability? 3. Planetary habitability is the measure of a planet's or a natural satellite's potential to develop and sustain life. Life may develop directly on a planet or satellite or be transferred to it from another body, a theoretical process known as panspermia. As the existence of life beyond Earth is currently uncertain, planetary habitability is largely an extrapolation of conditions on Earth and the characteristics of the Sun and Solar System

Learning Objective 1.31 Question 4 4. Old rocks are rare on Earth, mostly poorly preserved because of metamorphism, deformation and inadequate exposure, and so signs of primordial life are hard to find and even more difficult to interpret. What is one method scientists use to study these early rock forms?

4. Old rocks are rare on Earth, mostly poorly preserved because of metamorphism, deformation and inadequate exposure, and so signs of primordial life are hard to find and even more difficult to interpret. What is one method scientists use to study these early rock forms? 4. In order to understand fossils, it is useful to learn how they formed. Taphonomy, is the study of the conditions under which plants, animals, and other organisms become altered after death and sometimes preserved as fossils.

The student is able to evaluate the accuracy and legitimacy of data to answer scientific questions about the origin of life on Earth.

Learning Objective 1.31 ; Question 1 Use the cladogram and data table to answer the following question. Which pair of organisms have the least in common? Shark and Gorilla Shark and Tiger Lamprey and Human Lizard and Gorilla

Learning Objective 1.31 ; Answer 1 C. In this example the Lamprey is being used as the common ancestor. This means that the characteristics being observed would not be present in this organism. The Human contains all of the characteristics while the Lamprey contains none, meaning the Human and the Lamprey would have the least in common.

Learning Objective 1.31 ; Question 2 If it was discovered that there was evidence that gorillas still had a functional tail, how could that change the data? Tigers and Gorillas would have the same amount of differences and would be grouped together on the cladogram Gorillas and Humans would now have the same amount of differences and would be grouped together on the cladogram Gorillas would have the same amount of differences as the lizards and they would be grouped together on the cladogram Use the cladogram and data table to answer the following question.

Learning Objective 1.31 ; Answer 2 A. Would be correct because of the fact that the gorilla still has a tail, that would mean that the gorilla and the tiger would have the same amount of different characteristics from the lamprey. Since they have the same amount of differences gorillas would get grouped with the tigers in the cladogram.

Learning Objective 1.31 ; Question 3 Use the cladogram to answer the following question. What characteristics do all the organisms besides the hagfish share? Lungs Claws or nails Jaws Fur; mammary glands

Learning Objective 1.31 ; Answer 3 C. In this example the outgroup organism is the hagfish. The characteristic closest to this group but not included in this group would be the characteristic that would exist in all the organisms that come after that unless it is replaced with another adaptation. That would mean that having jaws would be a common characteristic in all the organisms except the outgroup.

Learning Objective 1.31 ; Question 4 Use the graph to answer the following question During which 10 hour period was the population increasing most rapidly? a. 0 - 10 hours b. 10 - 20 hours c. 20 - 30 hours d. 30 - 40 hours e. 40 - 50 hours

Learning Objective 1.31 ; Answer 4 A. From 0-10 hours there was an increase in micro- organisms from 0-400, while in the other 10 hour increments there was only an increase of 200 micro-organisms before it reached its carrying capacity and leveled-off. So from 0-10 hours the population increased most rapidly.

Learning Objective 1.31 The student is able to evaluate the accuracy and legitimacy of data to answer scientific questions about the origin of life on Earth.

Learning Objective 1.31; Question 1 Explain how the “organic soup” model serves as scientific evidence for the rise of complex organic molecules and simple cell-like structures.

Learning Objective 1.31; Answer 1 The “organic soup” model states that the hypothesized primitive atmosphere contained inorganic precursors from which organic molecules could have been synthesized through natural chemical reactions catalyzed by the input of energy. As a result, these molecules served as monomers for the formation of more complex molecules, including amino acids and nucleotides, which would have the ability to replicate, store, and transfer information. This answer makes the most sense because simple cell-like structures would need molecules like nucleotides so the cells could divide and pass on their genetic information.

Learning Objective 1.31; Question 2 Life on earth is carbon based. Similar molecules could be formed with Potassium Aluminum Iron Silicon

Learning Objective 1.31; Answer 2 The correct answer is D) Silicon. This is the correct answer because silicon, like carbon, needs four electrons to fill its outer shell and could, therefore, form similar compounds.

Learning Objective 1.31; Question 3 Which of the following pieces of evidence most strongly supports the common origin of all life on Earth? All organisms require energy All organisms use essentially the same genetic code All organisms reproduce All organisms show heritable variation All organisms have undergone evolution

Learning Objective 1.31; Answer 3 The answer is B) All organisms use essentially the same genetic code. This is the correct answer because genes and DNA essentially make up the organism, indicating that organisms with a similar genetic code have a similar genetic make up and therefore originated from a common ancestor.

Learning Objective 1.31; Question 4 What evidence strengthens the hypothesis that chloroplasts could have been photosynthetic prokaryotes and mitochondria could have been aerobic prokaryotes?

Learning Objective 1.31; Answer 4 The fact that chloroplasts are the organelles responsible for photosynthesis in plants leads to the supposition that before symbiosis they were autotrophic prokaryotes. For the reason that mitochondria are the center of the aerobic cellular respiration, the powerhouse of the eukaryotic cell, it is supposed that they were once aerobic prokaryotes. The endosymbiotic hypothesis to explain the emergence of aerobic and autotrophic eukaryotic beings is strengthened further by the following evidence: chloroplasts as well as mitochondria have their own DNA, similar to bacterial DNA; chloroplasts and mitochondria reproduce asexually by binary division, like bacteria do; both organelles have ribosomes and synthesize proteins.

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The student is able to justify a scientific claim that free energy is required for living systems to maintain organization, to grow or to reproduce, but that multiple strategies exist in different living systems

Learning Objective 2.2 Question 1 Name 2 explanations of how free energy is used to maintain organization, growth and reproducing?

Learning Objective 2.2 Answer 1 Name 2 explanations of how free energy is used to maintain organization, growth and reproducing? Endothermy (maintaining homeostasis) Ectothermy (maintaining body temperature) Elevated floral temperatures in some plant species Seasonal reproduction in animals and plants Life-history strategies (biennial, reproductive diapause) size of organism versus metabolic rate energy storage

Learning Objective 2.2 Question 2 Choose the pair of terms that correctly completes this sentence: Catabolism is to anabolism as _______ is to _______. entropy; enthalpy exergonic; spontaneous exergonic; endergonic work; energy free energy; entropy

Learning Objective 2.2 Answer 2 Choose the pair of terms that correctly completes this sentence: Catabolism is to anabolism as _______ is to _______. entropy; enthalpy exergonic; spontaneous exergonic; endergonic work; energy free energy; entropy

Learning Objective 2.2 Question 3 Food chains are sometimes short because only a single species of herbivore feeds on each plant species. most producers are inedible. most of the energy in a trophic level is lost as it passes to the next higher level. predator species tend to be less diverse and less abundant than prey species. local extinction of a species causes extinction of the other species in its food chain.

Learning Objective 2.2 Answer 3 Food chains are sometimes short because only a single species of herbivore feeds on each plant species. most producers are inedible. most of the energy in a trophic level is lost as it passes to the next higher level. predator species tend to be less diverse and less abundant than prey species. local extinction of a species causes extinction of the other species in its food chain.

Learning Objective 2.2 Question 4 Describe the use of biological processes to offset entropy and maintain or increase order?

Learning Objective 2.2 Answer 4 Describe the use of biological processes to offset entropy and maintain or increase order? coupling of cellular processes that increase entropy with those that decrease entropy energy input must exceed free energy lost to maintain order and power cells ATP--> ADP coupled with other processes Krebs Cycle Glycolysis Calvin cycle Fermentation

Learning Objective 2.3 The student is able to predict how changes in free energy availability affect organisms, populations and ecosystems.

Learning Objective 2.3 Question 1 Cellular respiration uses glucose and oxygen, which have high levels of free energy, and releases CO2 and water, which have low levels of free energy. Is cellular respiration spontaneous or not? Is it exergonic or endergonic? What happens to the energy released from glucose?

Learning Objective 2.3 Answer 1 Cellular respiration uses glucose and oxygen, which have high levels of free energy, and releases CO2 and water, which have low levels of free energy. Is cellular respiration spontaneous or not? Is it exergonic or endergonic? What happens to the energy released from glucose Cellular respiration is a spontaneous and exergonic process. The energy released from glucose is used to do work in the cell or is lost as heat.

Learning Objective 2.3 Question 2 Some night-time partygoers wear glow-in-the-dark necklaces. The necklaces start glowing once they are “activated,” which usually involves snapping the necklace in a way that allows two chemicals to react and emit light in the form of chemiluminescence. Is the chemical reaction exergonic or endergonic. Explain your answer.

Learning Objective 2.3 Answer 2 Some night-time partygoers wear glow-in-the-dark necklaces. The necklaces start glowing once they are “activated,” which usually involves snapping the necklace in a way that allows two chemicals to react and emit light in the form of chemiluminescence. Is the chemical reaction exergonic or endergonic. Explain your answer. The reaction is exergonic because it releases energy –in this case, in the form of light.

Learning Objective 2.3 Question 3 Describe the forms of enerygy found in an apple as it grows on a tree, then falls, then is digested by someone who eats it.

Learning Objective 2.3 Answer 3 Describe the forms of enerygy found in an apple as it grows on a tree, then falls, then is digested by someone who eats it. The apple has potential energy in its position hanging on the tree, and the sugars and other nutrients it contains have chemical energy. The apple has kinetic energy as it falls from the tree to the ground. Finally, when the apple is digested and its molecules broken down, some of the chemical energy is used to do work, and the rest is lost as thermal energy

Learning Objective 2.3 Question 4 What is the second law of thermodynamics and how does it help explain the diffusion of a substance across a membrane?

Learning Objective 2.3 Answer 4 What is the second law of thermodynamics and how does it help explain the diffusion of a substance across a membrane? The second law of thermodynamics: Every energy transfer or transformation increases the entropy of the universe. When concentrations of a substance on both sides of a membrane are equal, the distribution is more random than when they are unequal. Diffusion on a substance to a region where it is initially less concentrated increases entropy, making it an energetically favorable process as described by the second law

Learning Objective 2.4 The student is able to predict how changes in free energy availability affect organisms, populations and ecosystems.

Question 1 Which of the following statements concerning energy transformations is true? A) Increases in entropy reduce usable energy. B) Energy may be created during transformation. C) Potential energy increases with each transformation. D) Increases in temperature decreases total amount of energy available.

Answer 1 A) Increases in entropy reduce usable energy Entropy The ratio of usable and unusable energy in a given system. Higher entropy represents a greater % of unusable energy, where as Lower entropy represents a greater % of usable energy.

Question2 What happens to the free energy availability of a certain bird species if another bird species is introduced and occupies the same niche. What will happen to the population size of the original bird species?

Answer 2 The two species will compete for resources and if the second bird species is stronger then original species will lose free energy availability and the population size will decrease.

Question 3 What are some examples of disruptions that can cause a decrease in free energy availability in ecosystems?

Answer 3 Natural Disaster Competition Invasive species Disease Human impact

Question 4 Where do the tropics get their abundance of free energy?

Answer 4 The plentiful plant life offers a lot of free energy for primary consumers and their ecosystems.

Learning Objective 2.5 The student is able to construct explanations of the mechanisms and structural features of cells that allow organisms to capture, store or use free energy.

Learning Objective 2.5 Question 1 Which of the following does NOT occur during the Calvin cycle? A) release of oxygen B) regeneration of the CO2 acceptor C) carbon fixation D) oxidation of NADPH E) consumption of ATP

Learning Objective 2.5 Answer 1 Which of the following does NOT occur during the Calvin cycle? A) release of oxygen B) regeneration of the CO2 acceptor C) carbon fixation D) oxidation of NADPH E) consumption of ATP A, the release of oxygen The Calvin cycle uses the chemical energy of ATP and NADPH to reduce CO2 to sugar. One molecules of G3P exits the cycle per 3 CO2 molecules fixed and is converted to glucose or other organic molecules.

Learning Objective 2.5 Question 2 The immediate energy source that drives ATP synthesis by ATP synthase during oxidative phosphorylation is the A) oxidation of glucose and other organic compounds B) affinity of oxygen for electrons C) transfer of phosphate to ADP D) flow of electrons down the electron transport chain E) H+ concentration across the membrane holding ATP synthase

Learning Objective 2.5 Answer 2 The immediate energy source that drives ATP synthesis by ATP synthase during oxidative phosphorylation is the A) oxidation of glucose and other organic compounds B) affinity of oxygen for electrons C) transfer of phosphate to ADP D) flow of electrons down the electron transport chain E) H+ concentration across the membrane holding ATP synthase E, H+ concentration across the membrane holding ATP synthase In the ETC, electron transfer causes protein complexes to move H+ ions to the intermembrane space. This causes a H+ gradient. H+ diffuses back into the matrix through ATP synthase, phosphorylating ADP in chemiosmosis.

Learning Objective 2.5 Question 3 (a) Identify the pigment (chlorophyll a or bacteriorhodopsin) used to generate the absorption spectrum in each of the graphs above. Explain and justify your answer.

Learning Objective 2.5 Answer 3 (a) Identify the pigment (chlorophyll a or bacteriorhodopsin) used to generate the absorption spectrum in each of the graphs above. Explain and justify your answer. Identify BOTH pigments: Graph 1 = bacteriorhodopsin AND graph 2 = chlorophyll a Explain that an organism containing bacteriorhodopsin appears purple because the pigment absorbs light in the green range of the light spectrum and/or reflects violet or red and blue light. The reflected red and blue light appears purple. Explain that an organism containing chlorophyll a appears green because the pigment absorbs light in the red and blue ranges of the light spectrum and/or reflects green light.

Learning Objective 2.5 Question 4 In an experiment, identical organisms containing the pigment from Graph II as the predominant light-capturing pigment are separated into three groups. The organisms in each group are illuminated with light of a single wavelength (650 nm for the first group, 550 nm for the second group, and 430 nm for the third group). The three light sources are of equal intensity, and all organisms are illuminated for equal lengths of time. Predict the relative rate of photosynthesis in each of the three groups. Justify your predictions.

Learning Objective 2.5 Answer 3 In an experiment, identical organisms containing the pigment from Graph II as the predominant light-capturing pigment are separated into three groups. The organisms in each group are illuminated with light of a single wavelength (650 nm for the first group, 550 nm for the second group, and 430 nm for the third group). The three light sources are of equal intensity, and all organisms are illuminated for equal lengths of time. Predict the relative rate of photosynthesis in each of the three groups. Justify your predictions.

Learning Objective 2.7 Students will be able to explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination.

Learning Objective 2.7 Question 1 Cell size is limited by the surface area to volume ratio of the cell membrane. Discuss why cell size is limited by this ratio.

Learning Objective 2.7 Answer 1 Cell size is limited by the surface area to volume ratio of the cell membrane. Discuss why cell size is limited by this ratio. A higher ratio of surface to volume allows for greater space for solutes to move in and out of cells. Cells must maintain homeostasis and in order to do this must eliminate wastes, ingest nutrients and maintain osmotic and ion balances. As a cell grows larger this ratio diminishes, limiting cell size.

Learning Objective 2.7 Question 2 Describe two adaptations that increase surface area in organisms.

Learning Objective 2.7 Answer 2 Describe two adaptations that increase surface area in organisms. Alveoli Convoluted membranes in chloroplasts and mitochondria Root hairs Villi and microvilli Endoplasmic reticulum

Learning Objective 2.7 Question 3 Describe the processes by which small polar and small nonpolar molecules can cross cell membranes according to their concentration gradient and give an example of each type of molecule.

Learning Objective 2.7 Answer 3 Describe the processes by which small polar and small nonpolar molecules can cross cell membranes according to their concentration gradient and give an example of each type of molecule. Small polar molecules cross by facilitated diffusion and require membrane channels. Water is an example of a small polar molecule that crosses through channels called aquaporins. Ions are polar species that cross the membrane via protein channels. Small nonpolar molecules can freely diffuse across the lipid bilayer. Seed/seedcoat Examples of small nonpolar molecules include CO2, O2, and steroid hormones.

Learning Objective 2.7 Question 4 Simple cuboidal epithelial cells line the ducts of certain human exocrine glands. Various materials are transported into or out of the cells by diffusion. (The formula for the surface area of a cube is 6 × S2, and the formula for the volume of a cube is S3, where S = the length of a side of the cube.) Which of the following cube-shaped cells would be most efficient in removing waste by diffusion?

Learning Objective 2.7 Answer 4 A. You could actually calculate the surface to volume ratio of these cells but the easiest way is to just realize that it takes 8 of the choice A boxes to make the volume of 1 choice B box.

Learning Objective 2.7 Students will be able to explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination.

100µm 200µm 300µm 1000µm During microscopic examination of human tissue samples, a student observed the following at different magnifications. In some cases, individual cells were clearly visible (A and B). In others, the magnification was too low to clearly visualize individual cells although the dark patches of nuclei are visible in fig C. FIGURE A. FIGURE B. FIGURE C. FIGURE D.

Learning Objective __2.7_ Question 1 a. Identify the image that contains a cell or cells with the lowest surface area to volume ratio. Explain your reasoning and provide a sample calculation to illustrate this.

Learning Objective __2.7_ Answer 1 Fig. B. The field of view is 200 microns, the cell is therefore about 100 microns across. The closest to this is fig. A in which the cells are approximately 20-30 microns. Assuming the cells are all roughly the same shape, the cell with the largest volume will have the lowest surface area to volume ratio. Many possible answers – check for accuracy in calculation. e.g. for a spherical cell in which the radius is r, the volume (v) is 4/3πr3 (check answer) The surface area (sa) is 4πr2 . Ratio is sa/v. Accept answers for rectangular and cuboidal cells. Student must compare two cell sizes and show how sa:v ratio drops as size decreases.

Learning Objective __2.7_ Question 2 b. Identify an image that shows a tissue that is ideally suited for the exchange of materials with the environment. Justify your response and suggest the role this tissue may have in the organism.

Learning Objective __2.7__ Answer 2 Fig. C or D are appropriate responses. These tissues indicate a large surface area/volume ratio. Their structure facilitates exchange of material with the environment, maximizing the area through which substances can move into the cell while minimizing the distance these substances have to move to perfuse the cell fully. Fig. C is alveolar tissue and fig. D is jejunal villi.

Learning Objective __2.7__ Question 3 c. Explain how surface area to volume ratio can place a limit on the maximum size of a cell.

Learning Objective _2.7_ Answer 3 Rapid exchange of materials such as oxygen, carbon dioxide, and glucose with the environment is crucial for effective metabolism and the maintenance of homeostasis. A low sa:v ratio in a large cell in effect decreases the speed at which materials can fully perfuse and leave the cell. This makes exchange slow, increasing concentration of waste products and decreasing the rate at which nutrients can enter. A cell beyond a particular size is therefore unable to operate basic life processes because of this limitation.

Learning Objective _2.7_ Question 4 d. Describe how you could model that the rate of exchange of materials is affected by surface area to volume ratio.

Learning Objective _2.7_ Answer 4 ANSWERS MAY VARY Responses are acceptable if the model and measurement method seem reasonable. For example, create blocks of agar/phenolphthalein/base of different sizes. Immerse in acid for set length of time. Retrieve blocks, cut open and measure distance of diffusion. Diffusion of the acid in smaller block will fill the block indicating a correlation between higher sa:v ratio and an object being fully perfused quicker. Note that in this method, the movement of acid is the same in each block, limiting how much of a larger block can be filled in a given time.

Learning objective 2.9
The student is able to represent graphically or model quantitatively the exchange of molecules between an organism and its environment, and the subsequent use of these molecules to build new molecules that facilitate dynamic homeostasis, growth and reproduction.

Learning Objective 2.9 Question 1 Mr. Dumbledore loves papayas. Papaya plants require much water which they get from the rain in the tropical forest. Mr. Dumbledore gets a papaya plant from the tropical forest and plants it in his garden in Hogwartz.Hogwarts is in a temperate grassland. A temperate grassland rains a little bit less on average compared to a rainforest. Make a graph comparing the rate of photosynthesis of a papaya plant in a temperate grassland and a tropical forest.

Learning Objective 2.9 Question 1 answer The top line is of the one in a tropical forest. The bottom line is the temperate grassland. The papaya plant needs water because they provide electrons for the photosystems. They also ionize NADP to NADPH which would be used in the calvin cycle. A less efficent calvin cycle means less G3P. Many plant products are produced from G3P.

Learning Objective 2.9 Question 2 What is the factor which is limiting the rate of photosynthesis at point X on the graph?

Learning Objective 2.9 Question 2 answer What is the factor which is limiting the rate of photosynthesis at point X on the graph? Answer: In any question on limiting factors, the factor on the X axis remains a limiting factor for as long as the graph continues to rise; in this case the point at which it levels off. At this point photosynthesis may still be limited by other factors such as temperature or carbon dioxide concentration.

Learning Objective 2.9 Question 3 What evidence on the graph indicates that chlorophyll a is not the only pigment involved in photosynthesis?

Learning Objective 2.9 Question 3 answer What evidence on the graph indicates that chlorophyll a is not the only pigment involved in photosynthesis? Answer : The rate of photosynthesis is high at wavelengths when the absorption by the pigment is low. or The rate of photosynthesis is high at wavelengths such as 450 nm and 650 nm where the absorption is low.

Learning Objective 2.9 Question 4 What evidence from the graph supports the statement that GP can be converted to RuBP in the light?

Learning Objective 2.9 Question 4 answer What evidence from the graph supports the statement that GP can be converted to RuBP in the light? Answer: The GP concentration decreases as the RuBP increases.

Learning Objective 2.10 The student is able to use representations and models to pose scientific questions about the properties of cell membranes and selective permeability based on molecular structure.

Learning objective 2.10; Question 1 Membranes are essential components of cells. a. Identify THREE macromolecules that are components of the plasma membrane in an eukaryotic cell and draw a model shows the macromolecules in the cell membrane.

Learning objective2.10; Answer 1 A. Three macromolecules that are in the cell membrane are lipids, proteins and carbohydrates. Lipids are the main component of the cell membrane. It is composed of a phospholipid bilayer which is composed of lipids as stated in the name. The bilayer controls the permeability of the membrane. Proteins are the transport vessels that bring molecules in or out of the cell. The transport proteins are located embedded in the cell membrane. The last macromolecule is the carbohydrates that are located on the peripheral of the cell membrane. The carbs allow for cell to cell recognition and signaling. B. A basic structure of he cell membrane will be drawn showing the lipid bilayer of the membrane. The transport protein will be obvious and the glycolipids will be drawn on the exterior of the membrane.

Learning Objective 2.10; Question 2 Which of the following types of molecules are the major structural components of the cell membrane?

Learning Objective 2.10; Answer 2 A. phospholipids and cellulose B. nucleic acids and proteins C. proteins and cellulose D. phospholipids and proteins E. glycoproteins and cholesterol D. phospholipids and proteins

Learning Objective 2.10; Question 3 All of the following are functions of integral membrane proteins- except?

Learning Objective 2.10; Answer 3 A. enzyme synthesis B. active transport C. cell adhesion D. hormone reception E. cytoskeleton attachment A. enzyme synthesis

Learning Objective 2.10; Question 4 When human skin is submerged in water for an extended period of time it shrivels. Explain why this occurs and draw a diagram showing the movement of solutes and/or water.

Learning objective 2.10; Answer 4 The human skin shrivels because it is placed in a hypotonic solution. The inside of the cell membranes is isotonic with the normal surroundings. When in solely water the solutes diffuse through the membrane into the water outside. This causes the skin to shrivel up due to the loss of solutes and molecules through the cell membrane via diffusion. Skin Solutes Water

Learning Objective _2.11____ Answer 1 All of the following are methods of moving materials across cell membranes along a concentration gradient EXCEPT a. exocytosis b. facilitated diffusion c. osmosis d. active transport

Answer: D; Active transport requires ATP because materials are going against their concentration gradients. Another way to solve this questions is to consider which three words are most similar, and pick the most different one.

Learning Objective _2.11____ Question 2 Which of the following statements are not true about the sodium potassium pump? A) The ions from the intracellular fluid plus an ATP molecule bind to the carrier protein on the inside of the cell membrane. B) ATP is broken down into ADP and potassium to supply the energy. C) The carrier protein changes shape as it transports ions from the intracellular fluid to the extracellular fluid. D) The ions from inside the cell are transported across the cell membrane. E) The ions are then released into the extracellular fluid.

Learning Objective _2.11____ Answer 2 Which of the following statements are not true about the sodium potassium pump? A) The ions from the intracellular fluid plus an ATP molecule bind to the carrier protein on the inside of the cell membrane. B) ATP is broken down into ADP and potassium to supply the energy. C) The carrier protein changes shape as it transports ions from the intracellular fluid to the extracellular fluid. D) The ions from inside the cell are transported across the cell membrane. E) The ions are then released into the extracellular fluid ATP does not break down to potassium first of all. And second, ATP is not broken down to supply the energy.

Learning Objective _2.11____ Question 3 If a plant cell’s ΨP = 2 bars and its ΨS = -5.5 bars, what is the resulting Ψ?

Learning Objective _2.11____ Answer 3 . If a plant cell’s ΨP = 2 bars and its ΨS = -5.5 bars, what is the resulting Ψ? Ψ = ΨP + ΨS = 2 bars + (-5.5 bars) = -3.5 bars

Learning Objective _2.11____ Question 4 A sample of cells is placed in a salt solution. The cells shrink and the membrane is distorted. Relative to the cell, the solution is probably: A. isotonic. B. hypotonic. C. osmotic. D. hypertonic.

Learning Objective _____ Answer 4 A sample of cells is placed in a salt solution. The cells shrink and the membrane is distorted. Relative to the cell, the solution is probably: A. isotonic. B. hypotonic. C. osmotic. D. hypertonic. Net movement of water is going outside of the cell and in a hypertonic solution it is normal for cells to decrease and cell membranes to become distorted.

Learning Objective 2.12 The student is able to use representations and models to analyze situations or solve problems qualitatively and quantitatively to investigate whether dynamic homeostasis is maintained by the active movement of molecules across membranes.

Learning Objective 2.12 Question 1 Assuming the membrane is only permeable to water, explain the movement of water molecules over time?

Learning Objective 2.12 Answer 1 What will happen to the water molecules and dissolved salts over time? Because the solute cannot move across the semipermeable membrane, only water can. The water molecules, which are randomly moving around, will go from the higher concentration to the lower concentration. In this case, they will move from left to right. This process of water diffusing from an area of high concentration to an area of low concentration is called osmosis.

Learning Objective 2.12 Question 2 Paramecium and other protists that live in hypotonic environments have cell membranes that limit water uptake, while those living in isotonic environments have membranes that are more permeable to water. What water regulation adaptations might have evolved in protists in hypertonic habits such as Great Salt Lake? In habitats with changing salt concentration?

Learning Objective 2.12 Answer 2 Paramecium and other protists that live in hypotonic environments have cell membranes that limit water uptake, while those living in isotonic environments have membranes that are more permeable to water. What water regulation adaptations might have evolved in protists in hypertonic habits such as Great Salt Lake? In habitats with changing salt concentration? Protists that live in hypertonic habitats such as Great Salt Lake might have evolved a lack of aquaporins, this would limit water loss. In a hypertonic solution, the net movement of water would be out of the cell. In habitats of changing salt concentration, they might have evolved contractile vacuoles that pump water out of the cell because they can change their rate of pumping out water depending on the concentration of the habitat.

Learning Objective 2.12 Question 3 The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at 27 degrees Celsius. Round your answer to the nearest hundredth.

Learning Objective 2.12 Answer 3 The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the water potential at 27 degrees Celsius. Round your answer to the nearest hundredth. Water potential=Solute Potential + Pressure Potential=-7.48 bars Solute potential=-(1)(0.3M)(0.0831 liter bars/mole K)(27+273=300°K)=-7.48 bars Pressure potential=0 because it is an open beaker.

Learning Objective 2.12 Question 4 Extensive irrigation in arid regions causes salts to accumulate in the soil. Explain why increased soil salinity might be harmful to crops. Suggest ways to minimize damage. What costs are attached to your solutions?

Learning Objective 2.12 Answer 4 Extensive irrigation in arid regions causes salts to accumulate in the soil. Explain why increased soil salinity might be harmful to crops. Suggest ways to minimize damage. What costs are attached to your solutions? Increased salinity might be harmful to crops because it would make the soil hypertonic to the plant’s root cells, thus causing them to loose water. A solution could be to apply excess water and allow the water to drain into the ocean or river. The costs of this would be the increased water consumption and the cost of draining the water away.

Learning Objective 2.14 The student is able to use representations and models to describe differences in prokaryotic and eukaryotic cells.

Learning Objective 2.14 Question 1 The eukaryotic cell is considered the more “sophisticated” cell when compared to the prokaryotic cell. Why is this?

Learning Objective 2.14 Answer 1 The eukaryotic cell is considered the more “sophisticated” cell when compared to the prokaryotic cell. Why is this? The eukaryotic cell contains internal membranes the facilitate the cellular process. The eukaryotic cell’s organelles complete the task of intracellular metabolic processes. Prokaryotes lack internal membranes and organelles.

Learning Objective 2.14 Question 2 Create a diagram/chart highlighting the structural differences between the eukaryotic cell and the prokaryotic.

Learning Objective 2.14 Answer 2 Create a diagram/chart highlighting the structural differences/similarities between the eukaryotic cell and the prokaryotic.

Learning Objective 2.14 Question 3 A B Which is cell is eukaryotic? Which is prokaryotic? Explain.

Learning Objective 2.14 Answer 3 Which is cell is eukaryotic? Which is prokaryotic? Explain. Cell A is Eukaryotic, notice the many organelles and mainly the nucleus. Cell B is Prokaryotic, The lack of the nucleus and simplistic structure prove it is prokaryotic.

Learning Objective 2.14 Question 4 Why do prokaryotes lack the membrane bound organelles that eukaryotes posses?

Learning Objective 2.14 Answer 4 Why do prokaryotes lack the membrane bound organelles? The prokaryotes do not need membrane bound organelles because they carry out metabolic processes in the cytoplasm.

Learning Objective 2.18 The student can make predictions about how organisms use negative feedback mechanisms to maintain their internal environments.

Learning Objective 2.18 Question 1 The endocrine system incorporates feedback mechanisms that maintain homeostasis. Which of the following demonstrates negative feedback by the endocrine system? (A) During labor, the fetus exerts pressure on the uterine wall, inducing the production of oxytocin, which stimulates uterine wall contraction. The contractions cause the fetus to further push on the wall, increasing the production of oxytocin. (B) After a meal, blood glucose levels become elevated, stimulating beta cells of the pancreas to release insulin into the blood. Excess glucose is then converted to glycogen in the liver, reducing blood glucose levels. (C) At high elevation, atmospheric oxygen is more scarce. In response to signals that oxygen is low, the brain decreases an individual’s rate of respiration to compensate for the difference. (D) A transcription factor binds to the regulatory region of a gene, blocking the binding of another transcription factor required for expression.

Learning Objective 2.18 Answer 1 The endocrine system incorporates feedback mechanisms that maintain homeostasis. Which of the following demonstrates negative feedback by the endocrine system? (B) After a meal, blood glucose levels become elevated, stimulating beta cells of the pancreas to release insulin into the blood. Excess glucose is then converted to glycogen in the liver, reducing blood glucose levels. This system represents a negative feedback loop. The pancreas responds to excess glucose with the release of insulin, which ultimately reduces glucose levels.

Learning Objective 2.18 Question 2 Identify ONE type of hormone pathway that demonstrates a negative feedback loop in order to maintain homeostasis. Describe the pathway and identify the hormones involved.

Learning Objective 2.18 Answer 2 Identify ONE type of hormone pathway that demonstrates a negative feedback loop in order to maintain homeostasis. Describe the pathway and identify the hormones involved. The pathway that regulates glucose levels, composed of the pancreas and liver, is a negative feedback loop that maintains a homeostatic concentration of glucose in the blood. When blood glucose levels rise too much, the beta cells of the pancreas release insulin, which causes body cells and liver to take up more glucose, reducing blood glucose levels. If the blood gulcose concentration drops too much, alpha cells in the pancreas stimulate the release of the hormone glucagon. Glucagon causes the liver to break down glycogen and return glucose into the blood. This system maintains a certain blood glucose level.

Learning Objective 2.18 Question 3 Several diseases can emerge from the dysfunction of feedback systems. Name one and describe what failure occurred in the given system to lead to that disease.

Learning Objective 2.18 Answer 3 Several diseases can emerge from the dysfunction of feedback systems. Name one and describe what failure occurred in the given system to lead to that disease Diabetes is a disease that emerges from increased levels of glucose concentration in the blood. Type 1 results from an inadequate production of insulin, and type 2 results from target cells failure to react to insulin and take in glucose.

Learning Objective 2.18 Question 4 An example of antagonistic hormones controlling homeostasis is Thyroxineand parathyroid hormone in calcium balance. Insulin and glucagon in glucose metabolism Progestinsand estrogens in sexual differentiation. Epinephrine and norepinephrine in fight-or-flight responses. Oxytocin and prolactin in milk production. Progestinsand estrogens in sexual differentiation.

Learning Objective 2.18 Answer 4 An example of antagonistic hormones controlling homeostasis is B. insulin and glucagon in glucose metabolism Insulin an glucagon perfom two opposite tasks in the regulation of glucose in the body. Glucagon, when broken down by the liver, releases more glucose into the blood. Conversely, when insulin is is present, it triggers the uptacke of glucose by body cells, ultimately decreasing blood glucose levels. These two hormones have opposite affects but work together to prevent BCL from getting too high or low.

The student is able to justify the selection of the kind of data needed to answer scientific questions about the relevant mechanism that organisms use to respond to changes in their external environment.

Learning Objective 2.21 1. What are some ways that animals respond to changes in external environments?

Learning Objective 2.21 Hibernation and migration in animals Taxis and kinesis in animals Shivering and sweating in humans Nocturnal and diurnal activity; circadian rhythms

Learning Objective 2.21 2. What are some ways plants and fungi respond to changes in external environments?

Learning Objective 2.21 Photoperiodism and phototropism in plants Sexual reproduction in fungi

Learning Objective 2.21 3. What are some various mechanisms that organisms use for obtaining nutrients and eliminating wastes?

Learning Objective 2.21 Gas exchange in aquatic and terrestrial plants Digestive mechanisms: good vacuoles, gastrovascular cavities, one-way digestive systems Respiratory systems of aquatic and terrestrial animals Nitrogenous waste production and elimination in aquatic and terrestrial animals

Learning Objective 2.21 4. Which system coordinates the body’s response to changes in its internal and external environment?

Learning Objective 2.21 Nervous System

Learning Objective 2.25 Question 1 Explain how skin functioned in homeostasis, and if characteristic of the common ancestor of many animals.

Learning Objective 2.25 Question 1 Explain how skin functioned in homeostasis, and if characteristic of the common ancestor of many animals. Skin acts as a nonspecific defense barrier that prevents pathogens from entering the body. Preventing infection helps maintain homeostasis, and through natural selection, animals without skin died off from infection from pathogens.

Learning Objective 2.25 Question 2 A species of wolf is separated from its original habitat and placed in either a cold environment or a warm environment. Determine which wolf would have thick fur and which would have thin fur, and explain the homeostatic significance of this.

Learning Objective 2.25 Question 2 A species of wolf is separated from its original habitat and placed in either a cold environment or a warm environment. Determine which wolf would have thick fur and which would have thin fur, and explain the homeostatic significance of this. The wolves in the cold environment would have thick fur and those in the warm would have thin. Due to natural selection, wolves with thin fur in the cold environment died off due to the cold. Wolves with thick fur in the warm environment died off from overheating.

Learning Objective 2.25 Question 3 A few individuals from the common ancestor of bears contain a mutated gene that prevents glucagon from functioning properly. He is dead, and all the others with the same mutation are dead too. Explain why this mutation is not the wildtype trait of bears today.

Learning Objective 2.25 Question 3 A few individuals from the common ancestor of bears contain a mutated gene that prevents glucagon from functioning properly. He is dead, and all the others with the same mutation are dead too. Explain why this mutation is not the wildtype trait of bears today. That mutation prevents glycogen in the liver from being converted into glucose, leading to low blood sugar. Because of natural selection, the organisms with this died, and the trait was not passed on to bears today.

Learning Objective 2.25 Question 4 A few individuals from the common ancestor of bears contain a mutated gene that prevents glucagon from functioning properly. However, this gene also activates heightened awareness in these organisms. Today, there are two species of bears that have this mutated gene and bears without it. Explain how this is possible.

Learning Objective 2.25 Question 4 A few individuals from the common ancestor of bears contain a mutated gene that prevents glucagon from functioning properly. However, this gene also activates heightened awareness in these organisms. Today, there are two species of bears that have this mutated gene and bears without it. Explain how this is possible. Speciation occurred between the bears with and without the gene due to some sort of barrier. The bears with the mutation were able to kill prey more easily with their heightened awareness, and thus keep blood sugar high from eating a lot. The other bears are able to regulate their blood sugar normally.

Lauren Angello

Learning Objective 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information.

Learning Objective 3.1 Question 1 What did Hershey and Chase discover?

Learning Objective _____ Answer 1 What did Hershey and Chase discover? They discovered that the radioactively phosphorus labeled DNA was injected into the bacteria and transformed.

Learning Objective 3.1 Answer 2 Name one conclusion made from Griffith’s experiment.

Learning Objective 3.1 Question 2 Name one conclusion made from Griffith’s experiment. ANSWER: Living bacteria acquired genetic information from dead bacteria - particularly the instructions for making capsules, thus transforming the naked bacteria into incapsulated bacteria. OR The Transforming agent was discovered to be DNA. DNA was isolated and added to live naked bacteria, and they were transformed into the incapsulated kind.

Learning Objective 3.1 Question 3 Examine the pedigree from a family with a genetic disease and answer questions 3 and 4. 3. Does this pedigree indicate autosomal dominant, recessive or sex-linked type of inheritance? Give reasons for your choice.

Learning Objective 3.1 Answer 3 3. Does this pedigree indicate autosomal dominant, recessive or sex-linked type of inheritance? Give reasons for your choice. This pedigree is consistent with autosomal dominnat inheritance with reduced penetrance. X-linked inheritance is excluded because of male to male transmission (II:1 to III:1).

Learning Objective 3.1 Question 4 Examine the pedigree from a family with a genetic disease and answer questions 3 and 4. 4. Individual II.3 requested genetic counseling. What is the probability that her child would be affected. Explain why.

Learning Objective 3.1 Answer 4 4. Individual II.3 requested genetic counseling. What is the probability that her child would be affected. Explain why. II:3 is obviously transmitting the disease although she is unaffected. The chance that her children would be heterozygotes is 50%. However, only some of these would be affected, depending on the penetrance. In the aggbove pedigree, three out of five heterozygotes were affected, equal to 60%. This is only a rough measure of the penetrance because of the small number of individuals involved. Thus II:3 has a risk of 60% of 50% (=30%) that her offspring would be affected, and 20% risk that her normal offspring are in fact carriers of the abnormal gene.

Learning Objective #3.2 The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of heritable information

Learning Objective 3.2 Question 1 In Griffith’s experiment, he found that when he put a rough strain (harmless) into a mouse and mixed it with a heat killed smooth strain (harmless), the mouse dies. What did he learn from this experiment?

Learning Objective 3.2 Answer 1 In Griffith’s experiment, he found that when he put a rough strain (harmless) into a mouse and mixed it with a heat killed smooth strain (harmless), the mouse dies. What did he learn from this experiment? He learned that there was a transforming factor. Something was being transferred from the dead smooth strain to the live rough strains that was transforming them into a virulent bacteria. (Remember that the smooth strain-when it’s not heat killed- kills the mouse)

Learning Objective 3.2 Question 2 The Hershey-Chase used bacteriaphages. They labeled radioactive sulfur in one experiment, infected the bacteria, and saw that the phages produced in the bacteria contained no radioactivity. In the other experiment they labeled radioactive phosphorous, infected the bacteria, and saw that the phages produced in the bacteria contained radioactivity. Why was it important that they used sulfur and phosphorous?

Learning Objective 3.2 Answer 2 The Hershey-Chase used bacteriaphages. They labeled radioactive sulfur in one experiment, infected the bacteria, and saw that the phages produced in the bacteria contained no radioactivity. In the other experiment they labeled radioactive phosphorous, infected the bacteria, and saw that the phages produced in the bacteria contained radioactivity. Why was it important that they used sulfur and phosphorous? Sulfur is not found in DNA, and phosphorous is not found in protein. By using these, they were able to definitively say that DNA is the genetic material, not proteins, as many people at that time thought.

Learning Objective 3.2 Question 3 Chargaff did experiments involving the four bases of DNA. What did he find?

Learning Objective 3.2 Answer 3 Chargaff did experiments involving the four bases of DNA. What did he find? He found that the amount of T is equal to the amount of A and the amount of G is equal to the amount of C.

Learning Objective 3.2 Question 4 Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA? A) the diameter of the helix B) the rate of replication C) the sequence of nucleotides D) the bond angles of the subunits E) the frequency of A vs. T nucleotides

Learning Objective 3.2 Answer 4 Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA? A) the diameter of the helix B) the rate of replication C) the sequence of nucleotides D) the bond angles of the subunits E) the frequency of A vs. T nucleotides A. The diameter of the helix. Franklin found using X-ray diffraction that the diameter of the helix is about 2nm.

The student is able to describe representations and models illustrating how genetic information is translated into polypeptides

Learning Objective 3.4 Question 1 Eukaryotic cell What process occurs from A to B? DNA A DNA replication Termination Translation Transcription B mRNA C Polypeptide

Learning Objective 3.4 Question 1 Answer Eukaryotic cell What process occurs from A to B? DNA A DNA replication Termination Translation Transcription B mRNA Transcription is the process by which the information in a DNA sequence is copied into a complementary RNA sequence. C Polypeptide

Learning Objective 3.4 Question 2 During RNA splicing, which parts of the gene will be included in the mRNA ?

Learning Objective 3.4 Question 2 Answer During RNA splicing, which parts of the gene will be included in the mRNA ? Exon 1, Exon 2, Exon 3 *During RNA Splicing, introns are removed and exons are spliced together

Learning Objective 3.4 Question 3 Which site is the P site? C A B

Learning Objective 3.4 Question 3 Answer Which site is the P site? A B C The P site holds the tRNA that is attached to the growing polypeptide chain C A B

Learning Objective 3.4 Question 4 What is the complementary RNA sequence of the following DNA sequence? T-C-T-A-G-T-C-T-A-C-A-T-C

Learning Objective 3.4 Question 4 Answer What is the complementary RNA sequence of this DNA sequence? T-C-T-A-G-T-C-T-A-C-A-T-C A-G-A-U-C-A-G-A-U-G-U-A-G

Sydni LaPlaga AP Biology

Learning Objective 3.6 The student can predict how a change in a specific DNA or RNA sequence can result in changes in gene expression.

Learning Objective 3.6; Question 1 A nucleotide deletion in DNA replication Causes one amino acid of the protein to be incorrect Causes all of the amino acids of the protein to be incorrect Causes the amino acids inserted after the deletion to be incorrect Causes the amino acids inserted before the deletion to be incorrect Has no effect on the resulting protein

Learning Objective 3.6; Answer 1 Answer: c) causes the amino acids inserted after the deletion to be incorrect. All of the amino acids inserted after the deletion will be incorrect, as al of the groupings of codons will be changed.

Learning Objective 3.6; Question 2 If a frameshift mutation causes a stop codon to be inserted into the DNA sequence The resulting protein will not be affected The phenotype will change but not the genotype The resulting protein will be too short and non-functional The resulting protein will be too long and non-functional

Learning Objective 3.6; Answer 2 Answer: c) the resulting protein will be too short and non-functional. The resulting protein will stop being made before it is complete. It will be too short and non-functional.

Learning Objective 3.6; Question 3 A mutation that causes a change in a single nucleotide in DNA Will have no affect on the resulting protein Changes the corresponding nucleotide in mRNA, resulting in a different codon Causes the codon to be correct, but the anticodon to be incorrect Caused protein synthesis to stop

Learning Objective 3.6; Answer 3 Answer: b) changes the corresponding nucleotide in mRNA, resulting in a different codon. A change in a single nucleotide in DNA changes the corresponding nucleotide in mRNA, resulting in a different codon that codes for a different amino acid.

Learning Objective 3.6; Question 4 A mutation that changes a single nucleotide can result in a different amino acid being added into a protein. True False

Learning Objective 3.6; Answer 4 Answer: a) True. A change in one nucleotide in DNA gives a corresponding change in mRNA which will then code for an incorrect amino acid.

Learning Objective 3.7 The student can make predictions about natural phenomena occurring during the cell cycle

Learning Objective 3.7 Question 1 Predict the results of a mutation that allows a cell to move past checkpoint G1 even though the cell has not grown sufficiently

Learning Objective 3.7Answer 1 Predict the results of a mutation that allows a cell to move past checkpoint G1 even though the cell has not grown sufficiently . The cell may get stopped a the G2 or M phase. Because it has not grown all the way, it may be to small to survive and will probable go through apoptosis.

Learning Objective 3.7 Question 2 What would happen if cell division was not controlled during the cell cycle?

Learning Objective 3.7Answer 2 What would happen is cell division was not controlled during the cell cycle? There would be unrestricted over growth of cells. This would cause the formation of a tumor. Uncontrolled cell growth ultimately causes cancer.

Learning Objective 3.7 Question 3 Paclitxel is a chemotherapy drug used to treat a variety of cancers. It inhibits the assembly and disassembly of microtubules. Based on this information, which checkpoint in the cells cycle is it affecting

Learning Objective 3.7Answer 3 Paclitxel is a chemotherapy drug used to treat a variety of cancers. It inhibits the assembly and disassembly of microtubules. Based on this information, which checkpoint in the cells cycle is it affecting It is affecting M checkpoint because microtubule assembly occurs during mitosis

Learning Objective 3.7 Question 4 How does the drug from the previous question inhibit the growth of cancer?

Learning Objective 3.7 Answer 4 How does the drug from the previous question inhibit the growth of cancer? It inhibits the growth of cancer because it stops the cell at checkpoint M, so cell division is not completed.

Learning Objective 3.8 The student can describe the events that occur in the cell cycle

Learning Objective 3.8 Question 1 The cell cycle is fundamental to the reproduction of eukaryotic cells. (a) Describe the phases of the cell cycle.

Learning Objective 3.8 Answer 1 Interphase → Prophase → (Prometaphase) → Metaphase → Anaphase → Telophase → Cytokinesis Interphase –-Chromatin dispersed in nucleus; nuclear envelope and nucleoli are intact and functional; DNA is replicated here. • G1, G2—C-ell growth. • S---DNA replication. • Mitosis---Nuclear division. • Prophase: Chromosomes begin to condense from chromatin; spindle apparatus assembled. • Prometaphase---Nuclear envelope disperses, nucleoli disperse, chromosomes connect to spindle apparatus fibers and begin to show motility. • Metaphase---Chromosomes reach maximum condensation and align on metaphase plate/plane. • Anaphase---Two-chromatid chromosomes split into two daughter chromosomes; chromosomes move to opposite poles of the spindle apparatus. • Telophase---chromosomes disperse back to chromatin form, nuclear envelope reassembles, nucleoli reassemble. • Cytokinesis---if this occurs, it is normally coordinated with telophase; cell division.

Learning Objective 3.8 Question 2 Explain the role of THREE of the following in mitosis or cytokinesis. (a) Kinetochores (b)Microtubules (c) Motor proteins (d) Actin filaments

Learning Objective 3.8 Answer 2 Kinetochores: Located in centromeres of condensed chromosomes; microtubule attachment sites necessary for chromosome positioning and movement. • Microtubules: Fundamental structural element of the spindle apparatus; framework on which chromosome motility is generated; define axis of division and cytokinesis. • Motor proteins: In Kinetochores, move chromosomes during mitosis, including anaphase separation; involves kinesins and dyneins. OR In animal cell cleavage furrow, generate force to pinch cell in two; involves myosins. • Actin filaments: Assemble under the membrane at the cytokinesis site; interact with myosin motor proteins to generate force to pinch cell in two; also interact with astral microtubules of the spindle to position the spindle apparatus in the cell.

Learning Objective 3.8 Question 3 Describe how the cell cycle is regulated and discuss ONE consequence of abnormal regulation.

Learning Objective 3.8 Answer 3 Regulation: Action of MPF and CDKs in checkpoint regulation Contact inhibition of mitosis Hormones; growth factor control of cell cycle activity Consequences of Abnormalities: • Uncontrolled cell proliferation, as in cancer spreading rapidly and randomly • Apoptosis—programmed cell death • Non-disjunction, aneuploidy or broken chromosomes from abnormal spindle events

Learning Objective 3.8 Question 4 Which is TRUE of the cell cycle? The timing of cell division is controlled by cyclins and CDK’s. A characteristic of cancer cells is density-dependent inhibition. The cell cycle is controlled solely by signals external to the cell. The cell cycle is controlled solely by internal signals.

Learning Objective 3.8 Answer 4 The timing of cell division is controlled by cyclins and CDK’s. Explanation:the timing of the cell cycle responds to external and internal cues and to fluctuations in levels of cyclins and CDK’s. Normal cells stop dividing when crowded. This phenomenon is called contact inhibition or density-dependent inhibition. Cancer cells are characterized by uncontrolled growth.

Shreya Modi

Learning Objective 3.10 The student is able to represent the connection between meiosis and increased genetic diversity necessary for evolution.

Learning Objective 3.10; Question 1 If a horticulturist breeding gardenias succeeds in having a single plant with a particularly desirable set of traits, which of the following would be her most probable and efficient route to establishing a line of such plants? Backtrack through her previous experiments to obtain another plant with the same traits. Breed this plant with another plant with much weaker traits. Clone the plant asexually to produce an identical one. Add nitrogen to the soil of the offspring of this plant so the desired traits continue.

Learning Objective 3.10; Answer 1 C Cloning the plants would be the most efficient route because it is guaranteed that they will have the same traits.

Learning Objective 3.10; Question 2 If a cell has completed the first meiotic division and is just beginning meiosis II, which of the following is an appropriate description of its contents? It has half the amount of DNA as the cell that began meiosis. It has half the chromosomes but twice the DNA of the originating cell. It has one-fourth the DNA and one-half the chromosomes as the originating cell. It is identical in content to another cell from the same meiosis.

Learning Objective 3.10; Answer 2 A This is because meiosis II is not preceded by chromosomal replication.

Learning Objective 3.10; Question 3 Which of the following happens at the conclusion of meiosis I? Homologous chromosomes are separated. The chromosome number per cell is conserved. Sister chromatids are separated. The sperm cells elongate to form a head and a tail end.

Learning Objective 3.10; Answer 3 A During anaphase I, the homologous pairs separate and are pulled to opposite poles of the cell. This process is called disjunction, and it accounts for a fundamental Mendelian law; independent assortment.

Learning Objective 3.10; Question 4 Independent assortment of chromosomes is a result of the random distribution of the sister chromatids to the two daughter cells during anaphase II. the random and independent way in which each pair of homologous chromosomes lines up at the metaphase plate during meiosis I. the relatively small degree of homology shared by the X and Y chromosomes. the random nature of the fertilization of ova by sperm.

Learning Objective 3.10; Answer 4 B The homologous pairs line up at the metaphase plate during metaphase I and each pair attaches to a separate spindle fiber by its kinetochore.

Learning Objective # 3.11 The student is able to evaluate evidence provided by data sets to support the claim that heritable information is passed from one generation to another generation through mitosis, or meiosis followed by fertilization.

Learning Objective ___3.11__ Question 1 A boy suffered from a skin condition that included thickening of the skin and the formation of loose spines that were periodically occurring. This “porcupine man” married and had six sons. All six sons had the same condition. He also had several daughters, all of whom were unaffected. What might you concluded about the location of the abnormal gene?

Learning Objective ___3.11__ Answer 1 A boy suffered from a skin condition that included thickening of the skin and the formation of loose spines that were periodically occurring. This “porcupine man” married and had six sons. All six sons had the same condition. He also had several daughters, all of whom were unaffected. What might you concluded about the location of the abnormal gene? This condition is most likely located on the Y chromosome. It is located on the Y chromosomes because the female are unaffected and have two X chromosomes, while the male have an X and a Y and are affected.

Learning Objective ___3.11__ Question 2 There are no carriers for Huntington’s Disease- . Is Huntington’s disease caused by a dominant or recessive trait? *The shaded shapes represent the male and females affected by the disease.

Learning Objective ___3.11__ Answer 2 There are no carriers for Huntington’s Disease- . Is Huntington’s disease caused by a dominant or recessive trait? *The shaded shapes represent the male and females affected by the disease. Dominant because if it was recessive, more offspring in the third generation would not have been affected.

Learning Objective __3.11___ Question 3 What is the genotype of e? *The shaded individuals are homozygous recessive.

Learning Objective __3.11___ Answer 3 What is the genotype of e? Heterozygous because it is an offspring of a homozygous recessive and B has to be heterozygous, therefore so does E.

Learning Objective ___3.11__ Question 4 *Affected individuals are represented by the open shape and the trait is recessive. What is the genotype of B-5?

Learning Objective _3.11____ Answer 4 What is the genotype of B-5? ww because it is affected by the trait which is recessive although they are not related to the first generation.

The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring.

3.12 Question 1 In dragonflies, the allele for gray body g is dominant to the allele for black body b. Additionally, the allele for straight wings is dominant to straight wings. However, the observed ration does not meet the expected 9:3:3:1 ratio. Which of these is the most likely reason for the observed Having the black allele produces a protein that suppresses the straight wing gene There was mutation in all the offspring Mate choice influences genes from an embryonic stage The genes were linked

3.12 Answer D) The genes for gray bodies and straight wings are linked genes and therefore are inherited together. The same goes for the black body gene and the curly wing gene.

3.12 Question 2 If the genes are in fact linked, then why are there any hybrids in the observed offspring? The genes are linked; however, due to recombination during meiosis there is a variation in alleles The genes are also affected by random X inactivation and as a result few offspring are hybrids. Being a hybrid has very low survivor ability I lied about the genes being linked

3.12 Answer 2 A) Although a majority of the genes while inherited as a link, chromosomes still undergo crossing over during meiosis and as a result offspring have the possibility of being a hybrid although it is much smaller.

3.12 Question 3 Green dominant to yellow. Tall dominant to short. Fred performed a dihybrid cross between a heterozygote for both alleles and a purebred homozygous recessive. The results are shown above. Name the pattern of inheritance that allows this phenomenon. Mitosis Epigenetic Independent Assortment Codominance

3.12 Answer 3 C) Due to independent assortment there is a 9:3:3:1 ratio. This means regardless of the other allele the allele in question will line up independently of other during metaphase.

3.12 Question 4 Trisomy-21 is a disorder caused by an error during the passage of the genes from the parent to offspring. Which of these could be the error? A) Insertion during prophase I B) Nondisjunction in mitosis C) Failure to separate during meiosis D) Frameshift error

3.12 Answer 4 C) Because of the chromosomes failure to separate during this process, the individual could receive a gamete that has an abnormal number of chromosomes.

3.13 Luke Elwell
The student is able to pose questions about ethical, social or medical issues surrounding human genetic disorders.

3.13 1. Free Response: Explain how a single base-pair mutation in DNA can alter the structure and, in some cases, the function of a protein.

3.13 Define mutation; change in bases: A, C, G or T. Describe type of mutation: duplication, frameshift, nonsense, deletion, substitution (point mutation). Describe central dogma: DNA  RNA  protein. Describe process of central dogma: transcription  translation. Translation of codons: 3 nucleotides  1 amino acid. Redundancy in genetic code: 64 combinations: 20 amino acids (or can result in “stop” codon). Describe altered protein structure: primary, secondary, tertiary, quaternary. Describe protein function change: active site conformation, oxygen binding. Describe structural change: hydrophobic/hydrophilic interactions, disulfide bonds, R-group interactions, hydrogen bonds.

3.13 2. Free Response: Explain, using a specific example, the potential consequences of the production of a mutant protein to the structure and function of the cells of an organism.

3.13 Type of change: dominant, recessive. Changed protein -> changed trait/character/function (gain or loss of function). Description of example (any trait). Description of protein structure or example after change. Description of function after change. Elaboration with sickle: mutation/effect in organism, Glu -> Val, etc. Heterozygotic advantage (resistance to malaria).

3. When homologous chromosomes fail to separate during meiosis, this is termed A. cross-over. B. RFLP. C. linked genes. D. disjunction. E. nondisjunction

3.13 Nondisjunction By definition

3.13 4. in a family the father has some disease, and the mother is normal. When they have kids, only the daughter get the disease. None of the sons do. What kind of disease is this? Sex linked dominant Sex linked recessive Autosomal recessive Autosomal dominant

3.13 Sex link dominant Father only gives it to the daughters and no boys so it is sex linked. It overrides the X chromosome from the mother which means its dominant, thus sex linked dominant

3.14
3.14
The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data sets.

3.14 Question 1 The Panda Bear has a colored trait that can be shown on the ears as White ears (W) and Black ears (w). What is the probability of have any panda with White ears if you cross a pair of heterozygous Pandas?

3.14 Answer – 75%

3.14 Question 2 If two parents with white ears from the F1 generation mated what would the odds be of having an offspring with black ears?

3.14 Answer – 25%

3.14 Question 3 What is the phenotypic ratio outcome of a dihibrid cross between two heterozygotes? 9:6:3:1 9:9:3:3 3:9:6:1 9:3:3:1

3.14 Answer - D

3.14 Question 4 A cross between an individual homozygous dominant for a trait and an individual homozygous recessive for the same trait would result in what kind of outcome. 4:1 1:0 1:1 3:1

3.14 Answer - B

Concept: 3.32The student is able to describe basic chemical processes for cell communication shared across evolutionary lines of descent.
By, Kristin Jones

Concept: 3.32
Question 1
Every minute of every day, the billions of cells in our brains send and receive signals that influence everything from the memories we form to the emotions we feel. Upon receiving new information, a nerve cell transmits an electrical signal, triggering the release of chemicals called neurotransmitters at special locations called synapses. These chemicals act as messengers, passing along instructions that switch nearby cells on or off. What system does this form of communication use?

Concept: 3.32 Answer: the nervous system

Concept: 3.32
Question 2
Identify the indicated parts of the neuron:

Concept: 3.32 Answer:

Concept: 3.32
Question 3
Every second of every day billions of brain cells called neurons transmit signals to each other. And they do it at trillions of junctions called synapses. It is a fast process that is vital to everything the brain does such as learning, memorizing, planning, reasoning, and movement. When just one part of the process breaks down, brain disorders and nervous system diseases can result. Name and describe two diseases that can result to a breakdown in the nervous system.

Concept: 3.32 Answer:Diseases: schizophrenia and Alzheimer’s disease Descriptions: Schizophrenia is a mental disorder characterized by a breakdown in thinking and poor emotional responses.Alzheimer’s disease is the most common form of dementia that resutls in loss of memory and function.

Question 4
When a signal is released by a neurotransmitter, that neurotransmitter was initially received by which of the following of that neuron: Synapse Postsynaptic cell Axon Presynaptic cell

Concept: 3.32 Answer: d). Presynaptic cell

Learning Objective Student Questions