Biology Test Item Analysis

Biology Test Item Analysis
Office of Teaching and Learning 2017

New Test Item Types: The current on-line practice tests have the various types of items represented. The newer item types were added to the end of these tests in July. One new ability is an inline text choice. These are basically drop down boxes where students pick from a defined set of choices. They can be used within a sentence/paragraph structure (sort of like a fill in the blank) or within a table format. You can see an example in question 25 on the 5th grade practice test. Another thing that is new is the ability to combine multiple item types within the same question. You can see an example of that on the practice tests for biology test (#24). These multi-interaction items can be an avenue to provide depth to assessment questions, allowing us to glean more about the student’s thinking and understanding, in place of the human scored items which previously did some of that function. These items can have numerous parts and allow us to connect concepts. We also use simulations where students collect data to then use in responding to questions. This is not new but, with the practice tests now being interactive, students and teachers can now actually manipulate them. An example is found on biology (#16). Combining simulations with multi-interaction items provides a rich opportunity for future assessment development. The new types were not in operational use on the Spring 2017 tests, so the first released examples will come with the Spring 2018 released items next July. There will be no more human scored items on science assessments. That does not mean that all writing to address items is eliminated. All items will be machine scored.

Accelerated/Advanced
Heredity The diagram depicts the process of crossing over, which occurs between homologous chromosomes during gamete formation. What is the result of this process? An increase in offspring variation The deletion of amino acids in proteins An increase in mutations in genetic material The production of gametes that are genetically identical Content Standard: Cellular Genetics % Earned 0: , 40, 88 Strategies: Proficient Explain how sorting and recombination of genes in sexual reproduction and meiosis result in variations of traits in offspring. Accelerated/Advanced Investigate mapping genes using recombination frequency. Link crossing over to evolutionary benefits. Lessons and Resources: -use different models to show crossing over -Sordaria lab: -fruit fly lab simulation: -conduct a hands-on genetic mapping activity:

Scoring Guidelines: Rationale for Option A: Key – Crossing over during meiosis represented in the diagram leads to the formation of four unique chromatids thus increasing variation of traits. Rationale for Option B: This is incorrect. Chromosomes do not overlap during protein synthesis. Crossing over during meiosis represented in the diagram leads to the formation of four unique chromatids, thus increasing variation of traits. Rationale for Option C: This is incorrect. The variation in the gametes from crossing over is from the exchange of genetic material, not from genetic mutations. Rationale for Option D: This is incorrect. Crossing over during meiosis represented in the diagram leads to the formation of four unique chromatids. This exchange of genetic material during meiosis would not result in four identical gametes.

In a particular species of mice, a single gene (B or b) determines tail length. The short-tail allele is sex-linked and dominant. Shading in the pedigree below indicates that an individual has a short tail. The pedigree shows the pattern of inheritance of the short-tail allele over three generations of mice. Individual 6 is female. What is the genotype of individual 6? Heredity Content Standard: Cellular Genetics % Earned 0: , 40, 88 Strategies: Proficient Predict and interpret Mendelian and non-Mendelian (sex-linked traits) crosses from a pedigree. Accelerated/Advanced Given an incomplete pedigree, predict the genotypes and phenotypes of previous generations for Mendelian and non-Mendelian crosses. Explain the differences in possible outcomes for a somatic vs. sex-linked trait; and a sex-linked trait on the X chromosome vs. the Y chromosome. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – The female offspring receive one X- chromosome from the father. Since the father carries the dominant allele on his X-chromosome and the mother is homozygous recessive on both X- chromosomes, the female will be heterozygous dominant. This item requires the student to demonstrate an understanding of the genetic mechanisms of inheritance. In this scenario, the traits “B” (dominant for short-tailed) and “b” (recessive for long-tailed) are sex- linked. Since there is an affected female (XX), the traits must be linked to the sex chromosome X. Therefore, the genotype of “1” (affected female) must be either XBXB or XBXb, and the genotype for “2” (unaffected male) must be XbY. The offspring of “1” and “2” is “3” (unaffected female), so her genotype must be XbXb, and the genotype of “1” must therefore be XBXb. Since “6” is the female offspring resulting from the mating of “3” (XbXb) and the affected male, “4” (XBY), the genotype of “6” is XBXb because the XB must be donated by the male parent (“4”) and the Xb by the female parent (“3”).

Heredity Content Standard: Cellular Genetics % Earned 0: Strategies: Proficient Describe crossing over in meiosis. Accelerated/Advanced Model the process of meiosis, including all phases and mechanisms. Lessons and Resources:

Scoring Guidelines: Part A
Rationale for Option A: This is incorrect. Alleles separate in anaphase after crossing over has occurred. Rationale for Option B: This is incorrect. Spindle fibers attached to the chromatids pull the chromatids to opposite poles during anaphase, after crossing over has occurred. Rationale for Option C: This is incorrect. This is how translocation occurs. Rationale for Option D: Key – Crossing over, or the exchange of DNA, occurs between two chromatids on two different chromosomes that come in contact during prophase. This item requires the student to recall details about crossing over during meiosis. Crossing over during meiosis occurs between two homologous chromosomes when portions of their chromatids are exchanged. Crossing over provides additional genetic variability and is confirmed by conducting a dihybrid cross.

Scoring Guidelines: Part B
Rationale for Option A: This is incorrect. Cloning is the process of producing genetically identical individuals. Producing a genetic copy of an organism does not confirm that crossing over occurs. Rationale for Option B: Key – Test crosses involving two non-linked traits provide confirmation that alleles located on the same chromosome were inherited independently of one another. Rationale for Option C: This is incorrect. Scientists use gel electrophoresis to separate and analyze biological molecules based on their relative sizes and charges. Rationale for Option D: This is incorrect. Genetic engineering is a set of technologies used to change the genetic makeup of cells, including the transfer of genes within and across species boundaries. This item requires the student to recall details about crossing over during meiosis. Crossing over during meiosis occurs between two homologous chromosomes when portions of their chromatids are exchanged. Crossing over provides additional genetic variability and is confirmed by conducting a dihybrid cross.

Heredity Content Standard: Genetic Mechanisms and Inheritance % Earned 0: Strategies: Proficient Determine phenotypes and genotypes from a monohybrid incomplete dominance cross. Accelerated/Advanced Investigate Mendelian and non-Mendelian inheritance (e.g., dihybrid crosses, sex-linked traits, linkage) to compare predicted results to actual results (e.g., chi-square goodness of fit) Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – The data demonstrate incomplete dominance since the black genotype and white genotype combine to produce the intermediate blue genotype (BW) for the offspring. Rationale for Option B: This is incorrect. This set of data shows a situation where black feathers would show complete dominance over white feathers to produce offspring with black feathers. Rationale for Option C: This is incorrect. These results could not be expected based on incomplete dominance and black dominant to white. Rationale for Option D: This is incorrect. The student would not be able to conclude whether or not the inheritance of chicken feather color is incomplete dominance from these data. More information is needed for the student to make this conclusion. These data could be interpreted as an instance of complete dominance. This item requires the student to determine the resulting phenotypes produced by monohybrid crosses based on an example of incomplete dominance. The Punnett square crossing a black homozygous (BB) parent with a white homozygous (bb) parent would be: B B b Bb Bb If this were complete dominance, the heterozygous genotype (Bb) offspring would be black; however, with incomplete dominance, neither allele is dominant and both are expressed giving a blending of the traits resulting in an intermediate phenotype. In this item, the heterozygous genotype (Bb) produces a blue phenotype.

Heredity Content Standard: Genetic Mechanisms and Inheritance % Earned 0: Strategies: Proficient Determine phenotypes and genotypes for an epistatic dihybrid cross. Accelerated/Advanced Investigate Mendelian and non-Mendelian inheritance (e.g., dihybrid crosses, sex-linked traits, linkage) to compare predicted results to actual results (e.g., chi-square goodness of fit) Lessons and Resources:

Scoring Guidelines: For this item, a full-credit response includes:
Where top-left cell is A1, and bottom-right cell is D4: “Black” placed in cells A1, A2, A3, A4, B1, B3, C1, C2, C3, C4, D1, D3; AND “Yellow” placed in cells B2, B4, D2, D4 (1 point). This item requires the student to perform a dihybrid cross between two Labrador retriever dogs to determine the coat color of the offspring. The parent alleles are provided on the Punnett square, so the student performs the steps for a dihybrid cross to determine the coat color of the offspring as shown: BE Be BE Be BE BBEE BBEe BBEE BBEe Be BBEe BBee BBEe BBee bE BbEE BbEe BbEE BbEe be BbEe Bbee BbEe Bbee The black coat allele (B) is dominant. All of the dogs should have the potential of having a black coat, however scientists found an exception caused by the eumelanin gene. When the eumelanin gene is recessive for both alleles (ee), a yellow coat is produced regardless of the other coat color gene. Therefore, the potential for the dogs to exhibit a black coat phenotype remains with the dogs with the genotypes, BBEE, BBEe, BbEE or BbEe. The dogs with genotypes BBee and Bbee express the yellow coat phenotype.

Heredity Content Standard: Structure and Function of DNA % Earned 0: 73, 74 Strategies: Proficient Understand the function of DNA in a cell and cell differentiation. Accelerated/Advanced Describe cell differentiation and the implications on gene activity. The diagram shows four different cells from the same animal. Which statement about the cells is accurate? The four cells have identical DNA sequences. Cell 3 and Cell 4 have the longest DNA sequences because they have longer shapes. The four cells are different shapes because they come from different parent organisms. Cell 1 and Cell 2 have all of the same active genes even though they are from different body systems. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – The cells are from the same animal, so they would have the same DNA sequences. The cells look different due to the process of differentiation. “The many body cells in an individual can be very different from one another, even though they are all descended from a single cell and thus have essentially identical genetic instructions. Different genes are active in different types of cells, influenced by the cell’s environment and past history.”(AAAS) This item requires the student to recall that different cells from the same organism have the same DNA. The cells look different due to the process of differentiation. Different genes are active in the different types of cells. The cells look different due to the process of differentiation. Although the cells from the same organism have the same genetic information, different genes are active in the different types of cells.

Heredity Content Standard: Structure and Function of DNA % Earned 0: , 74 Strategies: Proficient Determine an amino acid sequence from a DNA/RNA base sequence. Explain how the sequence will change with a substitution or frameshift mutation. Accelerated/Advanced Given a real-world example, predict how a mutation in the DNA strand affected the resulting mRNA strand and how it lead to the faulty coding of the protein which produced the condition. Lessons and Resources:

Scoring Guidelines: For this item, a full-credit (2 point) response includes: “ACGTAACC” in the DNA box; AND “UGCAUUGG” in the mRNA box; OR “ACGTATCC” in the DNA box; AND “UGCAUAGG” in the mRNA box; OR “ACGTACCC” in the DNA box; AND “UGCAUGGG” in the mRNA box. For this item, a partial credit (1 point) response includes: “ACGTAACC” in the DNA box; OR “ACGTATCC” in the DNA box; OR “ACGTACCC” in the DNA box. This item requires the student to identify the resulting nucleotide in a strand of mRNA that results from a substitution mutation on the DNA strand at the sixth position. A substitution mutation means that the existing single nucleotide is substituted by another. Since this is a DNA strand, the existing guanine (G) could be substituted with either cytosine (C), thymine (T) or adenine (A). The resulting mRNA sequence would be constructed by entering the complementary nucleotide. In mRNA, thymine (T) becomes uracil (U). Therefore, the mRNA nucleotides guanine (G), cytosine (C), uracil (U) or adenine (A) pair with the DNA nucleotides. For example: DNA strand after the substitution ACGTAACC Resulting mRNA strand UGCAUUGG

Heredity Content Standard: Structure and Function of DNA % Earned 0: Strategies: Proficient Explain the steps of protein synthesis from DNA to translation. Accelerated/Advanced Design a model to explain the processes of transcription, translation, and protein synthesis. Lessons and Resources:

“DNA” placed in the top box; AND “mRNA” placed in the center box; AND “tRNA” placed in the bottom box (1 point). This item requires the student to accurately place where deoxyribonucleic acid (DNA) and the different forms of ribonucleic acid (RNA) reside in a graphical representation of a cell. In all eukaryotic cells, the nucleus contains the DNA. During the process of transcription, messenger RNA (mRNA) copies the genetic instructions from DNA in the nucleus. The mRNA is transferred out of the cell’s nucleus where it is decoded by transfer RNA (tRNA) in the cytoplasm during a process called translation. Translation occurs in the ribosomes located in the cytoplasm.

Heredity Content Standard: Structure and Function of DNA % Earned 0: Strategies: Proficient Determine an amino acid sequence from a DNA/RNA base sequence. Explain how the sequence will change with a substitution or frameshift mutation. Accelerated/Advanced Given a real-world example, predict how a mutation in the DNA strand affected the resulting mRNA strand and how it lead to the faulty coding of the protein which produced the condition. Lessons and Resources:

Scoring Guidelines: Part A
Rationale for Option A: This is incorrect. The correct sequence is Thr-Leu-Val. Rationale for Option B: Key – The correct sequence is Thr-Leu-Val. Rationale for Option C: This is incorrect. The correct sequence is Thr-Leu-Val. Rationale for Option D: This is incorrect. The correct sequence is Thr-Leu-Val. This item requires the student to analyze a DNA sequence using a codon chart to determine the amino acid sequence coded for by a specific strand of DNA and to identify the mRNA strand that the DNA strand coded for during transcription resulting in the amino acid sequence chosen in Part A. In Part A, the student first determines what mRNA strand is transcribed by the DNA strand. The compliment of DNA’s thymine is RNA’s uracil. Therefore, the DNA sequence …TGG-AAC-CAA… goes through the process of transcription to produce the mRNA strand …ACC-UUG-GUU…. The student uses the provided codon chart to determine which mRNA codon translates to which amino acid: the codon ACC translates to the amino acid threonine (Thr); the codon UUG translates to the amino acid leucine (Leu); and the codon GUU translates to the amino acid valine (Val). The resulting amino sequence is Thr-Leu-Val.

Heredity Content Standard: Structure and Function of DNA % Earned 0: Strategies: Proficient Determine an amino acid sequence from a DNA/RNA base sequence. Explain how the sequence will change with a substitution or frameshift mutation. Accelerated/Advanced Given a real-world example, predict how a mutation in the DNA strand affected the resulting mRNA strand and how it lead to the faulty coding of the protein which produced the condition. Part B: Which mRNA sequence was used to create the amino acid sequence you selected in part A? Lessons and Resources:

Scoring Guidelines: Part B
Rationale for Option A: This is incorrect. The correct sequence is ACC-UUG- GUU. Rationale for Option B: This is incorrect. Thymine is not present in mRNA. The correct sequence is ACC-UUG-GUU. Rationale for Option C: This is incorrect. The correct sequence is ACC-UUG- GUU. Rationale for Option D: Key – The correct sequence is ACC-UUG-GUU. This item requires the student to analyze a DNA sequence using a codon chart to determine the amino acid sequence coded for by a specific strand of DNA and to identify the mRNA strand that the DNA strand coded for during transcription resulting in the amino acid sequence chosen in Part A. In Part B, …ACC-UUG-GUU… is the mRNA sequence used to create the correct amino acid sequence in Part A.

Heredity Content Standard: Structure and Function of DNA % Earned 0: Strategies: Proficient Determine how DNA codes for proteins. Accelerated/Advanced Design a model to explain the processes of transcription, translation, and protein synthesis. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – This protein required a gene with at least 78 bases: 25 codons, one per amino acid, and a stop codon. Rationale for Option B: This is incorrect. Carbohydrates are not coded for by DNA or RNA. Rationale for Option C: This is incorrect. An mRNA with 20 codons has 60 bases, less than the 78 bases needed to code for a protein with 25 amino acids. Rationale for Option D: This is incorrect. An mRNA with 45 nucleotides has 45 bases, less than the 78 bases needed to code for a protein with 25 amino acids. This item requires the student to determine how DNA codes for proteins. DNA is used to code for mRNA during a process called transcription. During translation that takes place on ribosomes, each mRNA codon made up of 3 nitrogenous bases attaches to its complimentary tRNA anticodon with its amino acid attached. Therefore, the protein with 25 amino acids would be the longest strand of DNA with 78 bases: 1 start codon (3 bases) and 25 codons (3x25=75 bases).

Heredity A mutation occurs in the DNA base sequence GGG GAG TTA, resulting in the base sequence GGA GAG TTA. Use the information in the table to determine the effect of this mutation on the amino acid chain produced. What is the effect, if any, of this mutation on the amino acid chain produced? Proline is now lysine and leucine is unchanged. Proline is now lysine and asparagine is unchanged. Proline is unchanged and asparagine is unchanged. Proline is unchanged and leucine is now phenylalanine. Content Standard: Mutations % Earned 0: , 81 Strategies: Proficient Explain the possible outcomes when gene mutations occur in gametes or somatic cells. Accelerated/Advanced Explain the relative effects of different mutations in DNA on the operation of protein because of changes in sequence of amino acids. Differentiate germ cell and chromosomal mutations. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. This mutation does not cause a change to the amino acid sequence. Rationale for Option B: This is incorrect. This mutation does not cause a change to the amino acid sequence. Rationale for Option C: Key – This mutation does not cause a change to the amino acid sequence. Rationale for Option D: This is incorrect. This mutation does not cause a change to the amino acid sequence. This item requires the student to determine how a mutation in the DNA has affected the amino acid gene product. A codon, a 3-nitrogenous base sequence, codes for a specific amino acid. The table in this item, “Codons and Amino Acids,” lists only 7 of the 64 codon examples and 5 of the 20 amino acids. As evidenced in the table, a mutation in a codon may or may not affect the amino acid(s) produced.

Heredity Some bacteria have developed a resistance to antibiotics through random mutations. One such mutation allows bacteria to produce enzymes that inactivate the antibiotics when they enter the bacteria. Which statement describes how this mutation occurs? The mutation occurs when nucleic acids are replicated. The mutation occurs in chloroplasts during photosynthesis. The mutation occurs in mitochondria during cellular respiration. The mutation occurs when molecules move across the cell membrane. Content Standard: Mutations % Earned 0: , 81 Strategies: Proficient Explain the process of DNA replication and that mutations occur when it does not copy correctly. Accelerated/Advanced Examine the effects of different mutations (e.g., substitution or insertion/deletion resulting in a frameshift) in DNA and the effect it has on the resulting protein. Explain how gene mutations can be passed onto offspring. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – Enzymes are proteins produced from information coded in nucleic acids. A mutation that occurs during DNA replication can lead to the production of new enzymes. Rationale for Option B: This is incorrect. Photosynthesis converts light energy into stored chemical energy; it does not cause mutations in DNA to occur. Rationale for Option C: This is incorrect. Cellular respiration produces usable energy in the form of ATP from stored chemical energy in sugars; it does not cause mutations in DNA to occur. Rationale for Option D: This is incorrect. The movement of molecules in and out of cells does not cause mutations in DNA to occur. This item requires the student to communicate an understanding of the molecular basis of inheritance by selecting an explanation for how mutations occur. A genetic mutation is passed on to any cell that replicates from the parent cell through its DNA by mitosis. Only if the genetic mutation occurs in a gamete through meiosis will the mutation be passed on to the offspring.

A farmer wants to know which plants come from two given adult plants. Using gel electrophoresis, the farmer tests the same specific section of DNA of the two adult parent plants. Then, he tests the same section of DNA on six individual young plants to determine their relatedness to the two adult parent plants. The electrophoresis gel results are shown. Heredity Content Standard: Modern Genetics % Earned 0: , 63 Strategies: Proficient Recognize which tests are used to compare DNA of individuals. Explain how offspring receive alleles from each parent cell during meiosis. Accelerated/Advanced Interpret data from a real-world scenario involving biotechnology. Identify one young plant that is an offspring of the adult parents, and explain how the results support your choice. Describe how the results of the electrophoresis gel from the adult parent and young individual plants demonstrate the movement of chromosomes during meiosis. Lessons and Resources:

Scoring Guidelines: 2 points The response includes:
• a correct identification of a young plant and a correct explanation of results supporting the young plant identification; AND • a correct description of how the electrophoresis gel results from the parent and young plants demonstrate the movement of chromosomes during meiosis. 1 point The response includes: • a correct identification of a young plant and a correct explanation of results supporting the young plant identification; OR This item requires the student to analyze an electrophoresis gel of a set of parent plants and six young plants to determine which of the young plants are from the parents by comparing the DNA banding. In meiosis, each parent contributes 50% of their genetic material to create gametes. The student is looking for offspring with components from both parents. Y1, Y3, Y4 and Y6 receive one band from P1 and one band from P2, providing evidence that they are offspring of P1 and P2. A total of four young plants (Y1, Y3, Y4 and Y6) are possible offspring of the given parent plants, and each of these young plants shows differing genotypes. Since all four young plants have at least one band from each parent but display differing genotypes, segregation occurred during meiosis.

Heredity Content Standard: Modern Genetics % Earned 0: , 63 Strategies: Proficient Recognize electrophoresis results can be used to compare relationships between individuals. Accelerated/Advanced Interpret data from a real-world scenario involving electrophoresis. Lessons and Resources:

“Unrelated” placed below “Rat B”; AND “Identical Sibling” placed below “Rat C”; AND “Sibling or parent” placed below “Rat D” (1 point). This item requires the student to infer genetic relationships between individuals by comparing DNA banding produced through gel electrophoresis. A rat is a diploid organism receiving 50% of its genetic material from each of its parent’s gametes. By analyzing the bands created during gel electrophoresis, identical matches, parents and siblings can be identified. The appearance of similar bands provides evidence for the relationships between the rats. Rat B is unrelated because there are no similar banding patterns compared to Rat A. Rat C is identical to Rat A because the banding is identical. This means that Rat A and C are identical siblings. Rat D has 3 of the 6 bands that are identical to Rat A, which indicates that Rat D could be a parent or sibling.

Heredity Content Standard: Modern Genetics % Earned 0: , 63 Strategies: Proficient Explain that electrophoresis results can be used to compare traits (alleles) between individuals. Accelerated/Advanced Use biotechnology (electrophoresis) to interpret real-world data and draw conclusions. Lessons and Resources:

“DNA Sample #2” selected (1 point). This item requires the student to analyze the results of a gel electrophoresis to classify a sample. The evidence is in the table, “Gel Electrophoresis Results,” which compares the known base sample labeled “Resistant DNA Sample” with the unknown DNA samples labeled “DNA Sample [#1, #2 #3].” Through visual analysis of the gel electrophoresis results, it can be determined that “DNA Sample #2” is the only sample that matches the known “Resistant DNA Sample.”

Heredity Content Standard: Modern Genetics % Earned 0: , 63 Strategies: Proficient Describe the process involved in genetic engineering. Accelerated/Advanced Given a genetic engineering scenario, make and justify a conclusion about how the technology used changed the genetic makeup of the cells. How could genetic engineering be used to produce a more successful crop in a hot, dry climate? non-essential DNA could be removed from the genomes of the crop plants. DNA from plants adapted to dry areas could be added to the genomes of the crop plants. RNA from a variety of nonagricultural plants could be added to the genomes of the crop plants. mRNA transcripts of genes from dry weather crops could be added to the fertilizer used on the crop plants. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. The removal of non-essential DNA would not increase the crops’ ability to deal with hot, dry climates. Rationale for Option B: Key – This would help the current plant exhibit traits of the plants that better tolerate a dry climate. Rationale for Option C: This is incorrect. RNA from non-crop plants would not give the crop any advantages in hot, dry climates. Rationale for Option D: This is incorrect. mRNA may be absorbed from the fertilizer, but it has no long-term ability to increase the crops’ output in hot, dry climates. This item requires the student to identify a way that genetic engineering can be used to produce better crops. Through the process of genetic engineering, segments of DNA are replaced in an organism’s genetic composition (DNA) to modify a physical trait, which provides the desired benefit to the engineered organism. In this specific question, the desired trait is drought resistance. This item assesses the student’s ability to identify a way that genetic engineering can be used to produce more successful crops.

Evolution Content Standard: Diversity of Life % Earned 0: , 89, 74, 53 Strategies: Proficient Using a cladogram, analyze and sort organisms by shared and derived characteristics. Accelerated/Advanced Given organisms and shared/derived characteristics, construct a cladogram. Lessons and Resources: NOVA Cladogram simulations (click through all screens)

Scoring Guidelines: For this item, a full-credit (2 point) response includes: Only the segmented organism with no antenna in the first box; AND Only the segmented organism with antenna in the second box AND only the segmented organism with antenna and wings in the third box; OR Only the segmented organism with antenna and wings in the second box AND only the segmented organism with antenna in the third box; AND Traits arranged as shown below (or “wings” at the top second from the right blue dot or in the top right box as long as it is below or in the box that contains the organism with antenna and wings). This item requires the student to distinguish the differences in features of different organisms, completing a cladogram for these organisms, and labeling the development of certain traits on the cladogram. The student must rationalize the evolutionary development and correctly place the different organisms in the correct sequence. The organism given has four legs and a single body segment. The next logical step would be the segmentation of body parts, and the “segments” label placed on the bottom blue dot. This is followed by either the segmented organism with antennae, or the winged, segmented organism with antennae. The label “antennae” is placed on the blue dot to the right and above the “segments” label. Lastly, either the remaining segmented organism with antennae, or the winged, segmented organism with antennae is placed in the far-right box. The label “wings” is placed on the blue dot under the winged organism. The unsegmented organism with six legs does not fit the evolutionary development pattern and is not used.

Evolution Content Standard: Diversity of Life % Earned 0: , 89, 74, 53 Strategies: Proficient Determine evolutionary relationships based on comparisons of amino acid sequences. Accelerated/Advanced Use amino acid sequence codes to determine the number of differences between organisms and place the organisms correctly on a cladogram. Lessons and Resources:

“B” placed in top box and “C” placed in 2nd-from-top box OR “C” placed in top box and “B” placed in 2nd-from-top box; AND “E” placed in the box directly below “D”; AND “A” placed in the bottom box (1 point). This item requires the student to determine evolutionary relationships based on comparisons of amino acid sequences. The more differences found in amino acid sequences, the more distantly related the organisms are to each other. The evidence is contained in the table, “Differences in 113-Amino Acid Sequence.” Organism A and the fact that it does not have any differences is the base organism, which is annotated on a single lineage. Organisms B and C each differ from Organism A by 11 amino acids and are therefore closely related to each other and A. Organism E has the greatest number of amino acid differences and is thus the furthest removed from Organism A.

Evolution Content Standard: Diversity of Life % Earned 0: , 89, 74, 53 Strategies: Proficient Determine evolutionary relationships based on comparisons of DNA sequences. Accelerated/Advanced Use DNA sequences to determine the number of differences between organisms and place the organisms correctly on a phylogenetic tree. Lessons and Resources:

“Diphyllodes” placed in the top box; AND “Manucodia” placed in the middle box; AND “Empidonax” placed in the bottom box (1 point). This item requires the student to use DNA sequence data to construct a cladogram of evolutionary relationships. The more differences found in the nitrogen base sequence, the more distantly related the organisms are to each other. The evidence is contained in the table, “DNA Sequences of Avian Families.” The Avian family of Ptiloris is the base organism, which is annotated on a single lineage. The Avian families Vireo and Diphyllodes differ by 4 nitrogenous bases and are therefore the closest relatives to Ptiloris. The Avian family Manucodia differs from Ptiloris by 6 nitrogenous bases and is thus the next closest relative. Finally, the Avian family of Empidonax is the most distant relative with 10 different nitrogenous bases.

Evolution Content Standard: Diversity of Life % Earned 0: , 89, 74, 53 Strategies: Proficient Identify cladograms that correctly model evolutionary relationships. Accelerated/Advanced Given a scenario, make and justify a conclusion about evolutionary mechanisms using a cladogram. Some orders of land and marine mammals are closely related. Molecular data recently collected from dolphins, camels, deer, and hippopotamuses have shown that hippopotamuses are the closest living relatives of dolphins. Which cladogram models this evolutionary relationship? Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – This is the only cladogram that shows hippopotamuses and dolphins with a common ancestor not shared by other groups. Rationale for Options B, C, & D: This is incorrect. This diagram shows that camels are the closest living relatives of dolphins. In this cladogram, hippopotamuses and dolphins do not share an exclusive common ancestor. This item requires the student to identify the cladogram that correctly models the evolutionary relationship that hippopotamuses are the closest living relatives of dolphins. For the hippopotamus and dolphin to be closely related, they would have to share a common ancestor. Therefore, the correct response is the cladogram that models the hippopotamus and the dolphin having the most recent common ancestor.

Stimulus for Next 3 Questions:

Evolution Content Standard: Diversity of Life % Earned 0: Strategies: Proficient Identify cladograms that correctly model evolutionary relationships. Explain how geographical isolation and gene flow influence the evolution of a species. Accelerated/Advanced Given a scenario, make and justify a conclusion about evolutionary mechanisms using a cladogram. Lessons and Resources:

Scoring Guidelines: From LEFT to RIGHT:
“L” and “N” placed in the first and second boxes OR “N” and “L” placed in the first and second boxes (1 point); AND “O” and “P” placed in the third and fourth boxes OR “P” and “O” placed in the third and fourth boxes AND “M” placed in the fifth box (1 point). This item requires the student to determine how geographical isolation and gene flow affects the evolution of a species. The unlabeled salamander subpopulation is provided in the last box on the right. From the left, the cladogram has boxes 1 and 2 that represent closely-related sister groups, boxes 3 and 4 that represent closely-related sister groups, and box 5 that is the outgroup to sister groups represented by boxes 3 and 4. Looking at the map of California in the stimulus, the two salamander subpopulations L and N in the western part of the state are closely related as they diverged and were separated from the others by the Central Valley. Therefore, salamander subspecies L and N are entered in boxes 1 and 2, in either order. On the eastern side of the state, there are salamander subspecies M, O, and P. Salamander subspecies M has the larger geographic area and was the first to diverge. Therefore, salamander subspecies M is entered in box 5. Salamander subspecies O and P diverged from salamander subspecies M and are entered in boxes 3 and 4 in either order to complete the cladogram.

Evolution Content Standard: Diversity of Life % Earned 0: Strategies: Proficient Given a real-world example, predict and provide evidence to support how a population has responded to environmental changes. Accelerated/Advanced Given data and/or a scenario, make and justify a conclusion about evolutionary mechanisms in a population; analyze data and provide evidence for the selection of an evolutionary mechanism that has occurred in a given population. Lessons and Resources:

“L” and “N” placed individually in the boxes (1 point); OR “M” and “O” placed individually in the boxes (1 point); OR “O” and “P” placed individually in the boxes (1 point). This item requires the student to determine how the environment affects the mechanism of gene flow, the transfer of genes between populations. For gene flow to occur between the salamander subpopulations, there must be a means for members of the subpopulations to reproduce. The species originally were separated by a large lake, and then later by the Central Valley. Therefore, with the geographical barrier that the lake and later the valley created, it is not possible for the subpopulations L or N to have gene flow with subpopulations M, O, or P. Also, due to distance and the presence of subpopulation O, subpopulations M and P do not have an opportunity for interbreeding. Only subpopulations L and N, M and O, and O and P have an opportunity for gene flow to occur along the common terrain joining their subpopulations.

Evolution Content Standard: Diversity of Life % Earned 0: Strategies: Proficient Given a real-world example, predict and provide evidence to support how a population has responded to environmental changes. Accelerated/Advanced Given data and/or a scenario, make and justify a conclusion about evolutionary mechanisms in a population; analyze data and provide evidence for the selection of an evolutionary mechanism that has occurred in a given population. Which two salamander subpopulations would be least likely to reproduce successfully if brought together? Salamanders L and M Salamanders L and N Salamanders P and N Salamanders P and O Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. These two salamanders are not as geographically isolated or as evolutionarily divergent as P and N. Rationale for Option B: This is incorrect. These two salamanders can experience gene flow and are likely to be able to reproduce successfully. Rationale for Option C: Key – These two species are geographically isolated and have been diverging for the longest time, from two already unique subpopulations. Rationale for Option D: This is incorrect. These two salamanders can experience gene flow and are likely to be able to reproduce successfully. This item requires the student to determine how the environment affects the mechanism of gene flow, the transfer of genes between populations. For gene flow to occur between the salamander subpopulations, there must be a means for members of the subpopulations to interbreed. The species originally were separated by a large lake, and later by the Central Valley. The subpopulations P and N are not only separated by the geographical barrier, but they have also been diverging and genetically isolated for the longest period.

Evolution Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Describe the effects of reproductive isolation. Accelerated/Advanced Given data and/or a scenario, make and justify a conclusion about evolutionary mechanisms in a population. From an evolutionary perspective, what is the significance of reproductive isolation developing between two populations within a species? Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. Speciation between two populations means that the lineage splits, resulting in the divergence of a new (or second) species from the ancestral species, not the extinction of the ancestral species. Rationale for Option B: Key – Reproductive isolation restricts the gene flow between populations which results in speciation. Rationale for Option C: This is incorrect. Reproductive isolation over time results in speciation, which changes the type of competition from intra-species to interspecies, not the amount of competition between individuals or populations. Rationale for Option D: This is incorrect. Whether a mutation is favorable or unfavorable determines whether it results in greater fitness, not the mutation rate. This item requires the student to explain the significance of reproductive isolation developing between two populations within a species. Reproductive isolation restricts the gene flow between populations, which results in speciation. Reproductive isolation does not increase competition between individuals or populations, but it changes the type of competition from intra-species to interspecies. Speciation results in the divergence of a new (or second) species from the ancestral species, not the extinction of the ancestral species. Mutations occur in all species regardless of population size. Whether a mutation is favorable or unfavorable determines whether the trait caused by the mutation results in greater fitness, not the mutation rate.

Evolution Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Understand the basis for the Hardy- Weinberg Equilibrium Equation and know the conditions. Accelerated/Advanced Explain why a population is or is not in Hardy-Weinberg equilibrium based on given data or conditions (e.g., no mutation is occurring, mating is random). Lessons and Resources:

Scoring Guidelines: For this item, a full-credit (2 point) response includes: “Mating is random.” AND “No lizards migrate in or out of the population.” AND “Natural selection is not occurring in the population.” AND no other statements are selected in part A (1 point); AND ONLY “in a population” in the first box AND “remain constant” in the second box of part B (1 point). This item requires the student to determine that a population is not evolving by identifying what conditions illustrate the population is in Hardy-Weinberg equilibrium and what happens to allele frequencies when those conditions are met. The Hardy-Weinberg equilibrium, also known as the Hardy-Weinberg principle, model, theory or law, states that the allele and genotype frequencies will remain constant from generation to generation in the absence of other evolutionary influences. It is used to measure how observed genotype frequencies differ from frequencies predicted by the equation. Therefore, the first, third and fifth options, “Mating is random.”, “No lizards migrate in or out of the population.”, and “Natural selection is not occurring in the population.”, respectively, are selected in Part A. To be in Hardy-Weinberg equilibrium, the population to be studied must be large; therefore, the second option, “The lizard population is small.” is invalid. Also, the fourth option, “A new mutation occurs within the population.” is invalid since evolutionary influences like mutations cannot occur in a population that is in equilibrium. In Part B, the Hardy-Weinberg equilibrium only applies to populations and not individuals.

Evolution How does natural selection affect populations? By creating variation individuals that allows them to better adapt to changing environmental conditions By increasing the rate of beneficial mutations which allow individuals to increase their rate of reproduction By reducing variation between individuals in a population thereby making the population more homogeneous and likely to survive By acting on variations between individuals that make some better adapted to their environment and more likely to survive and reproduce Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Identify how natural selection affects populations. Accelerated/Advanced Given data and/or a scenario, make and justify a conclusion about evolutionary mechanisms in a population; especially in regards to frequency of genes. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. Natural selection does not affect populations by creating variation in individuals. Rationale for Option B: This is incorrect. Natural selection does not affect populations by increasing the rate of beneficial mutations in individuals. Rationale for Option C: This is incorrect. Natural selection does not affect populations by reducing variation in individuals in a population and thereby making the population more homogenous and likely to survive. Rationale for Option D: Key – Natural selection affects populations by allowing individuals with certain variations of traits to have a selective advantage over individuals with different variations of those traits. This item requires the student to identify how the process of natural selection affects populations. Natural selection is the natural process by which only the organisms that are best adapted to their environment survive and transmit their genetic characteristics through sexual or asexual reproduction to succeeding generations, increasing the frequency of those characteristics within the population.

Evolution A biologist conducting a long-term field study noticed there were two types of mice in a valley. One type of mouse had webbed toes, and one type of mouse lacked webbed toes. A dam was built in the valley turning the land into a wetland. As the years passed, all mice living in the valley were found to have webbed toes. No mice were found with non-webbed toes. Which statement best explains the observed change in the mouse population? Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Determine how variations within populations in a changing environment can lead to evolution; define natural selection. Accelerated/Advanced Formulate and revise explanations for gene flow and sexual selection based on real-world problems. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. This is a misconception that a physical characteristic (e.g., webbed toes) is controlled by the environment. The trait must naturally occur in the population through genetic variation or mutation, and then it must provide a survival advantage allowing individuals with the trait to reproduce more successfully, impacting the population. Rationale for Option B: This is incorrect. This is a misconception that a physical characteristic was grown due to change in the environment (e.g., webbing). The trait must naturally occur in the population through genetic variation or mutation, and then it must provide a survival advantage allowing individuals with the trait to reproduce more successfully, impacting the population. Rationale for Option C: Key – The trait (webbed toes) naturally occurred in the population and provided a survival advantage allowing individuals with the trait to reproduce more successfully, impacting the population. Rationale for Option D: This is incorrect. This is a misconception that change in the environment (e.g., increased water levels) affects the mutation rate. The trait must naturally occur in the population through genetic variation or mutation, and then it must provide a survival advantage allowing individuals with the trait to reproduce more successfully, impacting the population. This item requires the student to determine the effect of environment on the survival of organisms through the process of natural selection. The process of natural selection requires a trait to exist within a population. It is the exhibition of this trait that provides the advantage that allows individuals with the trait to reach reproductive age. These individuals then produce more offspring than the individuals without the trait, thus increasing the frequency of the trait within the population. In this example, webbed toes offer an advantage in the changing environment, allowing those mice to out-reproduce the mice without the trait.

Evolution In most species, scientists have observed that related individuals share about 80% of the same genes. Cheetahs today, however, are known to share 99% of the same genes. What does this lack of genetic diversity imply about the history of the cheetah population? Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Describe how reproductive isolation can result in a bottleneck. Accelerated/Advanced Given data and/or a scenario, make and justify a conclusion about how bottlenecks and founder effect can be influenced by evolutionary mechanisms in a population. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. Mutations would add to the genetic diversity if the mutation occurred in a gamete. The mutation would not be passed to the offspring if it occurred in a somatic cell. Rationale for Option B: Key – Today’s cheetah underwent a sharp decline in population, creating a genetic bottleneck. Today’s cheetah genetic pool is based on these relatively few survivors. This bottleneck created a decrease in genetic diversity within the current cheetah population. Rationale for Option C: This is incorrect. By the definition of a species, cheetahs cannot interbreed with other species of large cats. If possible, this would add to the genetic diversity rather than subtract from it. Rationale for Option D: This is incorrect. Sharing a common ancestor would have no effect on the cheetah’s genetic diversity. This item requires the student to recall factors that limit the genetic diversity within a population. Today’s cheetahs underwent a sharp decline in population, creating a genetic bottleneck. Today’s cheetah genetic pool is based on relatively few survivors. This bottleneck created a decrease in genetic diversity within the current cheetah population.

Evolution Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Given scenarios with selective pressure on a population, explain how a change in the ecosystem can affect the change in available alleles. Discuss incomplete dominance. Accelerated/Advanced Describe how ecosystems and the gene pool can change as geological or biological conditions vary. Predict the differences between Mendelian and non-Mendelian scenarios. Lessons and Resources:

“No change” placed next to “B allele”; AND “No change” placed next to “b allele” (1 point). This item requires the student to use knowledge of incomplete dominance to predict the changes in allele frequencies in a population. Since the heterozygous condition of medium brown (Bb) is being selected by predation, there is a 1:1 reduction in the alleles. Therefore, the relative frequencies would not change.

Evolution Rabbits were introduced to Australia in the 1800s. They rapidly overpopulated because they had few natural predators in the area. To control their population, scientists introduced a rabbit-specific virus into the population and their numbers greatly decreased. However, after several generations, the rabbit population began to increase again. Which statement explains the new increase in the number of rabbits? Content Standard: Mechanisms % Earned 0: 82, 66, 45, 79, 57, , 71 Strategies: Proficient Explain natural selection so students understand how genetic variation within a species can enable individuals with advantageous characteristics to survive. Accelerated/Advanced Discuss examples of real-world examples of a trait in a population giving that population the ability to survive to reproductive age and the implications it has for competing species. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. According to the given information, rabbits were introduced to Australia; therefore, there are no native rabbit species with which to breed. Rationale for Option B: Key – Rabbits with the advantageous characteristic of being immune or partially immune to the virus would be able to survive and reproduce, passing this advantage to their offspring. The population of rabbits would begin to increase again. Rationale for Option C: This is incorrect. An organism cannot “learn” to survive a disease or infection. Rationale for Option D: This is incorrect. Organisms cannot change their reproductive cycles at will. This item requires the student to understand how genetic variation within a species can enable individuals with advantageous characteristics to survive. The process of natural selection requires a trait (e.g., resistance to a virus) to exist within a population. It is the exhibition of this trait that provides the advantage that allows individuals with the trait to reach reproductive age. These individuals then produce more offspring than the individuals without the trait, which increases the frequency of the trait within the population. In this example, the viral resistance offered an advantage in the changing environment, allowing those rabbits to out-reproduce the rabbits without the trait.

Evolution Content Standard: Mechanisms % Earned 0: 23 Strategies: Proficient Explain natural selection so students understand how genetic variation within a species can enable individuals with advantageous characteristics to survive. Accelerated/Advanced Discuss examples of real-world examples of a trait in a population giving that population the ability to survive to reproductive age and the implications it has for competing species. Lessons and Resources:

In the “New Population” box, more than 2 brown dots AND less than 8 green dots (1 point); AND “Natural selection” selected (1 point). This item requires the student to demonstrate in Part A how phenotype frequencies change due to environmental pressures and in Part B identify the evolutionary mechanism. Due to changes in the environment, the brown frogs become better camouflaged, so the brown phenotype will increase in frequency from the original population (greater than 2) due to less predation caused by better camouflage. The green phenotype will decrease from the original population (less than 8) in frequency due to increased predation caused by being less camouflaged and more visible. In Part B, natural selection is the environmental mechanism evidenced by changes in the environment that gives a survival advantage of one phenotypic trait over another phenotypic trait.

Diversity of Life Content Standard: Ecosystems % Earned 0: 64, 74, 61, 66, 55, 91 Strategies: Proficient Given population graphs or charts containing data, determine the changes to an ecosystem, including carrying capacity. Accelerated/Advanced Given population graphs or charts containing data, analyze current and past data to predict and provide evidence to support their predicted changes in an ecosystem. Lessons and Resources:

Scoring Guidelines: For this item, a full-credit (2 point) response includes: For the Key: “dashed red line” placed next to Lake Trout AND “solid blue line” placed next to Sea Lamprey (1 point); AND For the graph: “red triangle” placed at or above 50 AND “blue square” placed at or below 25 (1 point). This item requires the student to evaluate a graph to determine the potential outcome of an invasive species (Sea Lamprey) on a native species (Lake Trout) in the Great Lakes. The red dashed line represents the lake trout. At year 50, the number of lake trout should be increasing (above 50) because the population is recovering with the decline of the sea lamprey. The solid blue line represents the sea lamprey, and with a successful chemical control program, this population would decrease (below 50) at year 50.

Diversity of Life Content Standard: Ecosystems % Earned 0: 64, 74, 61, 66, 55, 91 Strategies: Proficient Understand the parts of the logistic (carrying capacity) graph. Accelerated/Advanced Understand concepts relating to population growth and carrying capacity in regards to both exponential and logistic growth models. Determine what will happen to a population over time given data. Lessons and Resources:

“2” placed in the top box; AND “4” placed in the middle box; AND “3” placed in the bottom box (1 point). This item requires the student to analyze a population vs. time graph. The student must apply the concepts relating to population growth and carrying capacity. Position 1 is a horizontal line depicting a stable population at equilibrium. The upward/positive slope of the line at Position 2 depicts exponential growth as the population responds to some favorable change in the ecosystem. As the carrying capacity of the ecosystem is reached at Position 3, the slope of the line decreases and begins to fluctuate as a new equilibrium is established. The ecosystem experiences a negative influence at Position 4, as the slope decreases and another new equilibrium is established, denoted by the line fluctuations to the right of Position 4.

Diversity of Life Content Standard: Ecosystems % Earned 0: 64, 74, 61, 66, 55, 91 Strategies: Proficient Describe how carrying capacity causes changes in ecosystems over time. Accelerated/Advanced Use mathematical models to explain carrying capacity and homeostasis within ecosystems; predict changes in the ecosystem. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. This curve indicates that something else is being introduced to the environment that eventually lowers the size of the population. Rationale for Option B: Key – As the number of individuals in a population exceeds the carrying capacity, the population size falls below the carrying capacity and eventually fluctuates around the carrying capacity. Rationale for Option C: This is incorrect. A population size will not start at a number above zero. Rationale for Option D: This is incorrect. This growth curve suggests that carrying capacity does not change a population's ability to grow exponentially. This item requires the student to communicate an understanding of how carrying capacity affects change in ecosystems over time. Carrying capacity is the maximum population size that resources in an environment can sustain for that specific population. A population will increase in number using a logistical curve until the carrying capacity is reached. Since environments will naturally fluctuate, the carrying capacity will also fluctuate around the theoretical limit marked by the dashed line.

Diversity of Life Which species in the food web would be subject to the greatest biomagnification of pesticides applied to the grasses? Deer Grasses Grasshoppers Hawks Content Standard: Ecosystems % Earned 0: 71 Strategies: Proficient Given a real-world example, predict and provide evidence to support how a population has responded to environmental changes (e.g. biomagnification). Accelerated/Advanced Create models to explain how toxins increase in concentration as you go up the food chain. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. The deer do not consume as many types of organisms as hawks, whose prey consume the toxins. Rationale for Option B: This is incorrect. The grasses have the toxins; it does not biomagnify in them. Rationale for Option C: This is incorrect. Grasshoppers are exposed to the toxins, but at lower levels than hawks. Rationale for Option D: Key – Hawks consume the greatest number of organisms exposed to the toxins. This item requires the student to evaluate a food web to determine the potential outcome of introduced pesticides in an ecosystem on the organisms that live there. Biomagnification affects organisms at the top of the food web greater than those organisms at lower trophic levels. When an herbivore eats a producer contaminated by a fat-soluble substance in the environment, such as a pesticide or herbicide, the contaminant is retained in the herbivore’s cells. When that herbivore is then consumed by a carnivore or omnivore, the contaminants are retained and increase in concentration within the tissues of the consumer. This process is known as biomagnification. When the apex predator, the hawk for this item, eats these consumers, it is affected by these contaminants in concentrations greater than the consumers at the lower trophic levels.

Diversity of Life Content Standard: Ecosystems % Earned 0: 78 Strategies: Proficient Given graphs or charts containing data, determine the changes to an ecosystem, including inter- and intra-species competition. Accelerated/Advanced Given population graphs or charts containing data, analyze current and past data to predict and provide evidence to support their predicted changes in an ecosystem. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. The data do not support any inferences about the level of competition within any of the three bird species. Rationale for Option B: This is incorrect. Even though the bird preference for worm size shows three distinct size ranges, this does not mean the worms are of different species. Rationale for Option C: This is incorrect. Since worms at all three ranges are equally targeted by the birds, there is no evidence to show that small and large worms will become more frequent in the population. Rationale for Option D: Key – If worms of all size ranges are available, birds will eat the worms they prefer and won't compete with each other over worms in other size ranges. This item requires the student to evaluate a graph showing inter-species competition. The graph depicts three different bird species that prey on worms of different sizes. There is some overlap of worm size preference, but only at the extreme margins. If there is a significant population in a range of sizes, interspecies competition will be minimal. Only if a specific worm size becomes unavailable, will the affected species seek a different size worm, creating an inter-species competition issue.

Diversity of Life Content Standard: Ecosystems % Earned 0: 93 Strategies: Proficient Describe how carrying capacity causes changes in ecosystems over time. Accelerated/Advanced Use mathematical models to explain carrying capacity and homeostasis within ecosystems; predict changes in the ecosystem. Lessons and Resources:

Scoring Guidelines: For this item, a full-credit response includes
Only “Increase” selected for “Deer population size during time B”; AND Only “Increase” selected for “Carrying capacity of the habitat during time B”; AND Only “At Equilibrium” selected for “Carrying capacity of the habitat during time C” (1 point). This item requires the student to evaluate changes in an ecosystem and depict how carrying capacity is affected by available resources. A stable deer population at carrying capacity and equilibrium is expanded by adding neighboring land to a protected forest preserve. This addition of resources, habitat and food allows the deer population to expand as well as establish a new carrying capacity and equilibrium.

Diversity of Life Content Standard: Classification Systems % Earned 0: 35, 66, 49, 49, 48 Strategies: Proficient Interpret sequence percentages of a shared gene and document evolutionary relationships among species. Accelerated/Advanced Given gene sequence tables, complete cladograms to determine relationships among organisms. Lessons and Resources:

Scoring Guidelines: The species are ordered from left to right 5, 2, 3, 4, 1. This item requires the student to interpret the sequence percentages of a gene shared by five different species and to then use this sequence to document the evolutionary relationships among the species. The more similar the gene sequences, the more closely related the species. Therefore, these similarities can be used to determine the order of the evolutionary relationships. Species 1, which has the highest similarities (4 – 100%), will be entered in the first box either on the right side or left side of the cladogram. Second, depending on the starting point, Species 4, which has the next highest similarities (4 – 99%), will be entered in the second box. Third, Species 3, which has the third highest similarities (3 – 99% and 1 – 98%), will be placed in the third box. Fourth, Species 2, which has the fourth highest similarities (1 – 98%, 2 – 97% and 1 – 93%), will be entered into the fourth box. Finally, Species 5, which has the lowest similarities (93%, 90%, 87% and 81%), will be entered in the last box.

Scientists construct the cladogram shown, of plants in the genus Clusia, based on evidence collected from DNA analysis. Which conclusion is supported by the cladogram? Diversity of Life Content Standard: Classification Systems % Earned 0: 35, 66, 49, 49, 48 Strategies: Proficient Interpret cladograms to determine how species are related. Accelerated/Advanced Use DNA sequences from a table to construct a cladogram showing relatedness of species and delineating DNA similarities. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. While C. valerioi and C. rosea are more closely related to each other than to C. cupulata, C. cylindrica is more closely related to C. rosea since they are nested more closely in the cladogram that was constructed based on DNA sequences. Rationale for Option B: Key – C. rosea and C. cylindrica are nested the most closely in the cladogram that was constructed based on DNA sequences, so they would be the most closely related of the species, based on the cladogram. Points Possible: 1 See Alignment for more detail. 113 Rationale for Option C: This is incorrect. C. cupulata is the least closely related to the other three species in the cladogram based on the links presented in the cladogram. The general positioning of C. cupulata at the top does not indicate relatedness of the species. Rationale for Option D: This is incorrect. All four of the species come from the same genus Clusia. C. rosea and C. cylindrica are nested the most closely in the cladogram that was constructed based on DNA sequences, so they would be the most closely related of the four species. This item requires the student to interpret a cladogram to determine the relatedness of species. The cladogram was constructed based on DNA sequences. C. rosea and C. cylindrica are nested the most closely and share a common ancestor, so they would be the most closely related of the species based on their placement in the cladogram.

Diversity of Life Content Standard: Classification Systems % Earned 0: 35, 66, 49, 49, 48 Strategies: Proficient Interpret and complete a cladogram. Accelerated/Advanced Use conceptual knowledge to interpret and explain cladogram concepts. Create a cladogram using real-world data. Lessons and Resources:

“Species X” placed in the 2nd box from the left; AND “Species Y” placed in the 3rd OR 4th box from the left (1 point). This item requires the student to place organisms in an existing cladogram based on the traits of the organisms. The evidence used to complete the cladogram is contained in the table. Species X exhibited the physical traits “Exoskeleton,” “Compound Eyes” and “Mandibles.” Species Y exhibited the same physical traits as Species X with the addition of “Antennae.” The student must trace the emergence of the traits in order to properly position the species in the cladogram.

Diversity of Life The cladogram shows the evolutionary relationships among four species. One of the species in the cladogram is labeled as Species X. Click on the blank box or boxes to select the species that are members of the smallest clade that includes Species X. Content Standard: Classification Systems % Earned 0: 67 Strategies: Proficient Interpret and complete a cladogram. Accelerated/Advanced Use conceptual knowledge to interpret and explain cladogram concepts. Create a cladogram using real-world data. Lessons and Resources:

The bottom two boxes selected (1 point). This item requires the student to interpret the evolutionary relationships between four species using a cladogram. A clade includes an ancestor and all descendants of that ancestor. To select the box or boxes within the cladogram that identify the clade that would include species X, they would have to be more recent. Since the root species would be located at the top, the bottom two boxes are selected in this cladogram.

Diversity of Life Content Standard: Classification Systems % Earned 0: 45 Strategies: Proficient Interpret graphs or data (separation temperature) to explain how closely related two species are. Accelerated/Advanced Use conceptual knowledge to interpret and explain new technologies to determine species’ genetic similarities.

Scoring Guidelines: Part A Rationale for Option A: This is incorrect. These species have one of the greatest differences in hybridization temperatures, and so are two of the least related species. Rationale for Option B: This is incorrect. These species have a moderate difference in hybridization temperatures, so they are neither the most nor the least closely related. Rationale for Option C: Key – Species 2 and 4 have the least difference in hybridization temperatures, so their DNA sequences are the most alike and they are the most closely related. Rationale for Option D: This is incorrect. These species have a moderate difference in hybridization temperatures, so they are neither the most nor the least closely related. This item requires the student to evaluate differences in DNA separation temperature that can be used to determine how closely related two species are. DNA hybridization is a technique that measures the degree that two species are genetically similar. This is accomplished by heating the two DNA strands and measuring the temperature in which they separate. The closer the separation temperatures, the greater number of base pair similarities there are and more genetically related the two species are. Therefore, similar temperatures are needed to separate DNA from species that share more genetic information. Also, species that have diverged from a common ancestor more recently will also share more complementary base pairs. Part A requires the student to identify the two species that are closely related based upon the difference in separation temperatures.

Scoring Guidelines: Rationale for First Option: This is incorrect. Species that share more common ancestors, and are thus more closely related, will have more bonded complementary base pairs, and so higher temperatures will be needed to break more hydrogen bonds. Rationale for Second Option: Key – Species that are more genetically similar, like species 2 and 4, will have fewer unbonded base pairs when their DNA is hybridized, which is demonstrated by the low differences in the temperatures at which they separate. Rationale for Third Option: This is incorrect. More differences in base pairs between species that are less closely related decrease the temperatures needed to separate the hybrid DNA strands and cause a greater difference in temperature separation. Rationale for Fourth Option: Key – A greater number of complementary base pairs between two species suggests they are more closely related; therefore, the amount of heat needed to separate the DNA strands would be similar between hybridized and un-hybridized strands This item requires the student to evaluate differences in DNA separation temperature that can be used to determine how closely related two species are. DNA hybridization is a technique that measures the degree that two species are genetically similar. This is accomplished by heating the two DNA strands and measuring the temperature in which they separate. The closer the separation temperatures, the greater number of base pair similarities there are and more genetically related the two species are. Therefore, similar temperatures are needed to separate DNA from species that share more genetic information. Also, species that have diverged from a common ancestor more recently will also share more complementary base pairs. Part B requires the student to identify the two statements that provide evidence for their selection in Part A.

Cells Which type of macromolecule stores energy and provides thermal insulation for the body? Lipid Protein Nucleic acid Carbohydrate Content Standard: Cell Structure and Function % Earned 0: 73, 44, 70, 75, 73, 63 Strategies: Proficient List and define the four types of macromolecules found within the cell. Accelerated/Advanced Describe the essential functions of cells that involve chemical reactions utilizing water and carbohydrates, proteins, lipids and nucleic acids. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – Lipids have high energy bonds and are used to build the plasma membrane and insulate the body. Rationale for Option B: This is incorrect. While proteins perform a number of functions in the body, they do not provide thermal insulation. Rationale for Option C: This is incorrect. Nucleic acids contain genetic information that codes for protein synthesis. They do not provide thermal insulation or store energy. Rationale for Option D: This is incorrect. Carbohydrates do store energy but are not used to insulate the body. This item requires the student to identify the essential functions of macromolecules found within the cell. Lipids play a role in long-term storage of energy and provide thermal insulation for the body. Proteins are responsible for most of the work that occurs in the cell. Nucleic acids, DNA and RNA, store information that is used to make proteins. Carbohydrates provide an immediate source of energy.

Cells Which structural feature of the cell membrane allows molecules such as oxygen and carbon dioxide to diffuse into and out of the cell? The cell membrane contains protein molecules. The cell membrane contains cholesterol molecules. The cell membrane is made up of two layers of phospholipids. The cell membrane is anchored to the cytoplasm by the cytoskeleton. Content Standard: Cell Structure and Function % Earned 0: 73, 44, 70, 75, 73, 63 Strategies: Proficient Describe the function of organelles involved in transport and protein modification. Accelerated/Advanced Demonstrate how cell parts are specialized for the transport of materials, energy transformation, protein building, waste disposal, information feedback and movement. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. Small gas molecules do not require a protein or channel to pass through the cell membrane. Rationale for Option B: This is incorrect. Small molecules do not require cholesterol molecules to pass through the cell membrane. Rationale for Option C: Key – The phospholipid bilayer architecture allows for the passage of small molecules like oxygen, carbon dioxide and water in and out of the cell interior. Rationale for Option D: This is incorrect. The cytoskeleton does not contribute to the passage of particles across the cell membrane. This item requires the student to identify the structural feature that provides the cell membrane the ability to allow small molecules to diffuse across it. The phospholipid bilayer is the universal component of all cell membranes. This membrane functions as the barrier between the cell’s cytoplasm and the cell’s outside environment. This phospholipid bilayer allows only some substances to pass through, such as water, oxygen and carbon dioxide, while keeping other substances out. This property is called selective permeability or semi-permeability.

Cells Content Standard: Cell Structure and Function % Earned 0: 61 Strategies: Proficient Describe the function of organelles involved in transport, enzyme production and protein modification. Accelerated/Advanced Demonstrate how cell parts are specialized for the transport of materials, energy transformation, protein building, waste disposal, information feedback and movement. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – Cells that produce and secrete large amounts of enzymes (and other proteins) will have higher than normal amounts of endoplasmic reticulum, the organelle set where those proteins are produced and packaged for distribution. Rationale for Option B: This is incorrect. The nuclear envelope surrounds the nucleus, and is not involved in enzyme secretion. Rationale for Option C: This is incorrect. The cytoskeleton provides support and structure to the cell and is not directly involved in enzyme secretion. Rationale for Option D: This is incorrect. Animal cells do not have cell walls, and cell walls do not contribute to increased enzyme secretion. This item requires the student to determine the function of cell organelles. Chief cells located in the stomach release enzymes required for food digestion. This requirement for increased enzyme production would necessitate the cell to have an abundance of endoplasmic reticulum. The endoplasmic reticulum is responsible for the folding of protein molecules and the transport of synthesized proteins. The other three selections that include nuclear envelope, cytoskeleton, and cell wall, have other functions and are not involved in enzyme production.

The presence of which structure helps classify an organism as eukaryotic? nucleus cell wall flagellum chromosome Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, 67 Strategies: Proficient Identify prokaryotic and eukaryotic cells. Accelerated/Advanced Explain the major differences between prokaryotic and eukaryotic cells including differences in cellular structure, mode of reproduction, DNA, size, and metabolism. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: Key – Eukaryotic cells contain membrane- bound organelles such as the nucleus. Prokaryotes do not. Rationale for Option B: This is incorrect. The cell wall provides structure for a cell and is found in both eukaryotes and prokaryotes. Rationale for Option C: This is incorrect. The flagellum, a structure used for movement, may be found in both eukaryotes and prokaryotes. Rationale for Option D: This is incorrect. The chromosome is a bundle of DNA and protein and is found in both eukaryotes and prokaryotes. This item requires the student to identify the structural feature of eukaryotic cells. Eukaryotic cells have a nucleus. Prokaryotic cells have a nucleoid rather than a nucleus. Both eukaryotes and prokaryotes have cell walls, flagella and chromosomes.

Cells A student is asked to draw four cells. Which drawing is a representation of a prokaryotic cell? Content Standard: Cell Structure and Function % Earned 0: 73, 44, 70, 75, 73, 63 Strategies: Proficient Identify prokaryotic and eukaryotic cells. Accelerated/Advanced Explain the major differences between prokaryotic and eukaryotic cells including differences in cellular structure, mode of reproduction, DNA, size, and metabolism. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. This diagram is representative of a cell from a fungus. Rationale for Option B: This is incorrect. This diagram shows an animal cell. Rationale for Option C: Key – This cell has a cell wall but lacks internal membrane-bound compartments (organelles). Rationale for Option D: This is incorrect. This diagram shows a plant cell. This item requires the student to identify which cell is a prokaryotic cell. Since options A, B and D have membrane-bound organelles, option C must be a prokaryote.

Cells Which pair of organelles works together to give structure and support in animal cells? Content Standard: Cell Structure and Function % Earned 0: 73, 44, 70, 75, 73, 63 Strategies: Proficient Describe how cell components work together to perform the functions of the cell. Accelerated/Advanced Discuss the implications to the cell when specific organelles do not perform properly. Lessons and Resources:

Scoring Guides: Rationale for Option A: This is incorrect. Cell walls are found in plant cells, not animal cells. Rationale for Option B: This is incorrect. The Golgi complex does not aid in cell support. Rationale for Option C: Key – The cell membrane and cytoskeleton work together to provide support and structure for animal cells. Rationale for Option D: This is incorrect. The endoplasmic reticulum is not part of a cell structure. This item requires the student to identify which organelles give structure and support to animal cells. The only pair of organelles listed that gives both structure and support in animal cells is the cell membrane and cytoplasm. The cell wall, although it does provide support, is only present in plant cells. The Golgi complex and endoplasmic reticulum do not have a support function.

Cells A group of students studied four different cell specimens under a microscope and recorded information about each cell in this table. Content Standard: Cell Structure and Function % Earned 0: 73, 44, 70, 75, 73, 63 Strategies: Proficient Use data tables to identify prokaryotic and eukaryotic cells. Accelerated/Advanced Use data to explain the major differences between prokaryotic and eukaryotic cells including differences in cellular structure, mode of reproduction, DNA, size, and metabolism. Which cell is a prokaryote? Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. The characteristics of this cell are typical of a plant cell. Rationale for Option B: This is incorrect. The characteristics of this cell are typical of an animal cell. Rationale for Option C: Key – The characteristics of this cell are typical of a prokaryote. Rationale for Option D: This is incorrect. The characteristics of this cell are typical of a fungus. This item requires the student to distinguish a prokaryotic cell from other cell types based on its features. Prokaryotic cells are the most ancient life forms on Earth. Although they do have DNA used to transfer genetic information, the DNA is found in a nucleoid (nucleus-like). Prokaryotes do not have membrane-bound organelles, such as a true nucleus or chloroplasts. A true nucleus and membrane-bound organelles are traits of eukaryotes.

Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, 67 Strategies: Proficient Explain that enzymes have optimal temperatures and pH. Plan and conduct an experiment to determine the effects of external factors on cellular functions. Accelerated/Advanced Explain how the cellular environment affects the mechanical operation of an enzyme. Lessons and Resources:

Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, 67 Strategies: Proficient Explain how most cells function within a narrow range of temperature and pH. Accelerated/Advanced Explain how the cellular environment affects the mechanical operation of an enzyme. Lessons and Resources:

Scoring Guidelines (parts A & B):
Part A Rationale for Option A: This is incorrect. While changes in temperature can affect the shape of the enzyme, the data collected do not provide information about how the shape of this enzyme may have changed at different temperatures. Rationale for Option B: Key – Based on the data to be collected, scientists want to measure the enzyme activity at different temperatures. Rationale for Option C: This is incorrect. Substrate concentration is not changing in this experimental design. Rationale for Option D: This is incorrect. The data table does not include information about the appearance of the active site at different temperatures. Part B Rationale for Option A: This is incorrect. The data and the experimental design do not investigate activation energy, which will likely be lowest at the ideal temperature range of this enzyme. Rationale for Option B: This is incorrect. Enzymes do not adapt. Rationale for Option C: This is incorrect. This may be true, but this experiment is comparing how the enzyme operates at different temperatures. Rationale for Option D: Key – The bacterium lives in hot springs and, therefore, the enzymes it produces will likely operate best within that temperature range.

Scoring Guidelines (parts A & B) Continued:
This item requires the student to analyze a table developed by scientists to collect experimental data. The student then selects a question that could be answered based upon the data being collected. Finally, the student selects a reasonable hypothesis for the study by comparing the environmental data and the data that is being collected by the scientists. The scientists are studying an enzyme that was extracted from a bacterium that lives in a hot springs environment ranging from 50 – 80 degrees Celsius. The scientists develop a table to collect data using six different cultures at six different temperatures. The table then has spaces to record the reaction rates at four different time intervals. Therefore, the scientists are studying in part A how temperature affects the reaction rates of the enzyme. Since the bacteria have evolved to survive in a hot springs environment where the temperature ranges from 50 – 80 degrees Celsius, the reasonable hypothesis in part B would be that the optimum reaction rate of this enzyme would be found in the cultures grown within this range, i.e., tubes 4 or 5. Since these two parts are linked, the student can only get partial credit, 1 point, for correctly answering part A. One point cannot be awarded for a correct answer in part B only, since part B is linked to part A.

Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, Strategies: Proficient Describe how every cell is covered by a membrane that controls what can enter and leave the cell. In all but quite primitive cells, a complex network of proteins provides organization and shape. Within the cell are specialized parts for the transport of materials, energy transformation, protein building, waste disposal, information feedback and movement. Accelerated/Advanced Design an experiment to examine a cell in hyper, hypo, and isotonic solutions. Lessons and Resources:

A number greater than 3 in the top box; AND A number less than 3 in the bottom box (1 point). This item requires the student to infer and communicate the behavior of water in an onion cell in different salt solution concentrations based on observations. The image of the cell in a 3% salt solution is in an isotonic state (iso = equal), which means that the salt concentration is the same inside and outside the cell and that the water concentration is the same. The upper cell is in a hypertonic (hyper = above) state. The concentration of salt is lower within the cell, which equates to the water concentration being higher within the cell. Since the water concentration is higher inside the cell than outside the cell, water molecules diffuse from within the cell to the outside, causing the cell to shrink within the cell wall. The lower cell is in a hypotonic (hypo = under) state. The concentration of salt is higher within the cell, which equates to the water concentration being lower within the cell. Since the water concentration is lower inside the cell than outside the cell, water molecules diffuse from outside the cell to the inside, causing the cell to expand against the cell wall.

Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, 67 Strategies: Proficient List the products of anaerobic fermentation in yeast cells and animal cells. Accelerated/Advanced Conduct experiments to determine the optimal temperature/conditions for the production of CO2 in yeast cells during anaerobic respiration. Lessons and Resources:

For Part A: “Ethanol” and “CO2” placed individually in the boxes (1 point); AND For Part B: “35º C” placed in the box (1 point). This item requires the student to identify the products of fermentation in an oxygen-free environment and to identify the optimal temperature for fermentation based on the data provided. Fermentation is an anaerobic reaction where glucose in an aqueous solution results in energy, ethanol and carbon dioxide.

Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, 67 Strategies: Proficient Describe regulation of the cell environment. Accelerated/Advanced Explain how every cell is covered by a membrane that controls what can enter and leave the cell; that a complex network of proteins provides organization and shape. Describe specialized parts for the transport of materials, energy transformation, protein building, waste disposal, information feedback and movement. In addition to these basic cellular functions, most cells in multicellular organisms perform some specific functions that others do not. Lessons and Resources:

“Channel” placed in the left box; AND “Phospholipids” placed in the center box; AND  “ATP Pump” placed in the right box (1 point). This item requires the student to determine the correct membrane transport mechanism for different substances given the identity and concentration gradient of each substance. The “Channel” (protein channel) allows large-sized molecules, such as glucose, to cross. Since glucose is water soluble, it is transported through the protein channel with the concentration gradient (high concentration to low concentration) without the expenditure of energy. The “Phospholipids” (phospholipid layer) diffuse oxygen across the membrane without expenditure of energy using the concentration gradient (high concentration to lower concentration). The “ATP Pump” allows large-sized molecules such as sodium to be transported against the concentration gradient (low concentration to high concentration), requiring energy through expenditure of ATP.

Identify two cell structures involved in the function described. Place a structure number into each of the blank boxes. You do not need to use all the numbers. Place only one number in each blank box. Cells Content Standard: Cellular Processes % Earned 0: 56, 75, 69, 64, 79, 67 Strategies: Proficient Describe the function of organelles involved in protein modification and transport. Accelerated/Advanced Demonstrate how cell parts are specialized for the transport of materials, energy transformation, protein building, waste disposal, information feedback and movement. Lessons and Resources:

A “2” and a “6” placed individually in the boxes (1 point). This item requires the student to identify plant cell structures that perform various functions. Plant Cells are eukaryotes, which exhibit membrane-bound organelles. The numbered organelles shown are: “1,” a chloroplast that contains chlorophyll and the site of photosynthesis; “2,” the Golgi body, which is responsible for the modification, sorting and packaging of proteins; “3,” a vacuole responsible for temporary storage; “4,” the outermost layer of a plant cell, which is the cell wall, responsible for rigid structure and support; “5,” the nucleus, which is the control center of the cell, housing the DNA; “6,” the endoplasmic reticulum, which is responsible for the transport of proteins and materials; and “7,” the mitochondria, which is responsible for the conversion of glucose to ATP and is the “energy plant” of the cell. Mitochondria have their own heredity information and can self-replicate.

Cells Content Standard: Cellular Processes % Earned 0: Strategies: Proficient Describe the similarities and differences between photosynthesis and chemosynthesis. Accelerated/Advanced Design a model to represent the processes of photosynthesis and chemosynthesis. Lessons and Resources:

Scoring Guidelines: Rationale for Option A: This is incorrect. Both of these metabolic processes require a multitude of enzymes. Rationale for Option B: Key – Chemosynthesis harvests the energy from inorganic compounds to make organic, energy-rich compounds like sugar, and does not require sunlight. Rationale for Option C: This is incorrect. Chemosynthesis requires inorganic compounds from the environment as sources of energy. Rationale for Option D: This is incorrect. Both of these processes produce sugars. This item requires the student to recall the difference between chemosynthesis and photosynthesis. Both chemosynthesis and photosynthesis use inorganic compounds found in the environment as reactants. Enzymes are utilized to control the reactions that produce the sugars needed to produce energy. The difference between the two processes is chemosynthesis can take place in environments completely absent of sunlight.

Cells Content Standard: Cellular Processes % Earned 0: Strategies: Proficient Describe regulation of the cell environment. Accelerated/Advanced Explain how every cell is covered by a membrane that controls what can enter and leave the cell; that a complex network of proteins provides organization and shape. Describe specialized parts for the transport of materials, energy transformation, protein building, waste disposal, information feedback and movement. In addition to these basic cellular functions, most cells in multicellular organisms perform some specific functions that others do not. Lessons and Resources:

Part A, labels placed Left to Right: “H2O out,” “H2O in,” “H2O in/out” (1 point); AND Part B, labels placed Left to Right: “Shrivel,” “Burst,” “No Effect” (1 point). This item requires the student to show the process and result of osmosis, the movement of water across a membrane, on a cell. In Part A, the student identifies the direction of water movement, and then in Part B, shows the effect the movement has on the red blood cell. The red blood cell on the left has a lower concentration of solute (salt), therefore a higher concentration of water than its environment. Water moves (high to low) out of the cell into the environment causing the cell to shrivel. The red blood cell in the center has a higher concentration of solute (salt), therefore a lower concentration of water than its environment. Water moves (high to low) into the cell from the environment causing the cell to expand and burst. The red blood cell on the right has an equal concentration of solute (salt) and water as its environment. Water moves in and out of the cell at an equal rate into the environment having no effect on the cell.

Biology Test Item Analysis

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