Preparation Manual: Section 4: Sample Selected-Response Questions

Section 4: Sample Selected-Response Questions
Life Science 7–12 (238)

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This section presents some sample exam questions for you to review as part of your preparation for the exam. To demonstrate how each competency may be assessed, sample questions are accompanied by the competency that they measure. While studying, you may wish to read the competency before and after you consider each sample question. Please note that the competency statements do not appear on the actual exam.

For each sample exam question, there is a correct answer and a rationale for each answer option. The sample questions are included to illustrate the formats and types of questions you will see on the exam; however, your performance on the sample questions should not be viewed as a predictor of your performance on the actual exam.

Domain I—Scientific Inquiry and Processes

Competency 001—The teacher understands how to select and manage learning activities to ensure the safety of all students and the correct use and care of organisms, natural resources, materials, equipment and technologies.

1. A hamster kept as a class pet had a litter of 14 babies. Once the babies were one week old, the students monitored their growth by weighing them every other day and recording their individual masses to the nearest gram. The students also used calculators to determine the average mass of all the babies each time they were weighed. Many of the students reported the average mass to hundredths of grams. Which of the following best explains why the averages should have been reported only to the nearest gram?

It is simpler to graph data points that are rounded to the nearest whole number.
A calculation cannot appear to be more accurate than the original measurement.
Rounding to the nearest whole number makes it easier to determine which students performed the most accurate measurements.
The sample size was too small to justify averages in hundredths of grams.

Enter to expand or collapse answer.Answer expanded: Option B is correct because significant figures rules state that there cannot be more significant figures in a calculation than in the data used for the calculation. Option A is incorrect because the ability to graph is irrelevant to the calculation of averages. Option C is incorrect because the question asks nothing about comparing the students’ results, and results can be compared equally well with more figures. Option D is incorrect because significant figures, rather than sample size, determine the number of figures in the averages.

2. In an effort to increase the number of students eating vegetables at lunch, the school cafeteria manager asked students to sample six different vegetables, with each vegetable presented both raw and cooked. The cafeteria personnel recorded the number of students who said that they would eat the vegetables if presented at lunch in each of the tested forms. Which of the following types of graph will most effectively show the data?

Scatterplot
Stem-and-leaf plot
Line graph
Bar graph

Enter to expand or collapse answer.Answer expanded: Option D is correct because each form of each vegetable is a discrete variable that can be indicated on the x-axis and form the basis for a bar. The absolute number of students who responded positively to each vegetable determines the height of each bar. To determine which vegetables are most likely to be consumed, the heights of the 12 bars can be easily compared. Option A is incorrect because there is no absolute x-axis variable to plot. Option B is incorrect because a stem-and-leaf plot is used to examine the distribution of a series of numerical observations within a single data set, such as the amount of one particular vegetable eaten by individual students. Option C is incorrect because there is no absolute x-axis variable to plot and so no points to connect into a line.

Competency 002—The teacher understands the nature of science, the process of scientific inquiry and the unifying concepts that are common to all sciences.

3. Researchers spent months in the field studying the behavior of monkeys in a particular population. In order to share their results with other scientists, the researchers should first do which of the following?

Perform controlled experiments with the monkeys in the field to support their observations.
Capture some of the monkeys to determine whether their behavior in the field is reproducible in a laboratory setting.
Submit a paper describing their observations to a scientific journal for review by their peers.
Invite an independent, unbiased group of scientists to observe the same population of monkeys.

Enter to expand or collapse answer.Answer expanded: Option C is correct because a major route by which scientific findings are shared is by publication in a peer-reviewed journal. Option A is incorrect because field observations are frequently impossible to test in controlled experiments. Option B is incorrect because wild animals brought into a laboratory generally do not behave as they do in the field. Option D is incorrect because although another group of scientists may observe the same population of monkeys, such a practice is not instrumental in sharing results with the greater scientific community.

Example of 2 species of Acetabularia green algae.

Species 1 has a cap with more of a webbed pedal feature, with a stalk and rhizoid. Species 2 has a similar stalk and rhizoid, however the cap has pedals growing individually.

4. Acetabularia is a genus of single-celled green algae that have a cap at one end that is attached to a stalk and a rootlike structure called a rhizoid. Two species are shown above. If the cap of a cell is removed, a new cap grows on the stalk and the detached cap degrades. If a stalk from species 2 is grafted onto a rhizoid from species 1, a new cap grows on the end of the stalk and resembles a species 1 cap. Similarly, grafting a species 1 stalk onto a species 2 rhizoid gives rise to a cap that resembles a species 2 cap. Which of the following is the best control for the grafting procedure?

Remove the cap and stalk from an individual of species 1 and observe whether the rhizoid regrows at the base of the stalk.
Remove the cap and stalk from an individual of species 1 and replace the same stalk to observe cap regrowth and morphology.
Remove the cap and stalk from an individual of species 1, replace both parts with the stalk from a third species of Acetabularia, and observe cap morphology.
Remove the cap and stalk from an individual of species 1, replace both parts with one stalk from each of species 1 and 2, and observe cap morphology.

Enter to expand or collapse answer.Answer expanded: Option B is correct because the procedure evaluates whether a cap and stalk can be removed from Acetabularia and the stalk replaced without a loss of algal viability or a change to the morphology, and whether the cap that regrows is identical to the original. Option A is incorrect because it involves no grafting and additionally tests the regrowth of the rhizoid rather than the regrowth of the cap and stalk. Option C is incorrect because it introduces the variable of an additional species. Option D is incorrect because it introduces the variable of two species’ stalks grafted onto a single rhizoid.

Competency 003—The teacher understands the history of science, how science impacts the daily lives of students and how science interacts with and influences personal and societal decisions.

5. Prior to the 1940s and 1950s, it was generally accepted that heritable genetic material was composed of protein. It was known that chromosomes contain both DNA and proteins, the latter of which appeared to provide the complexity required by genetic material. Experiments performed in the 1940s and 1950s confirmed that genetic material was actually DNA by demonstrating both its heritability and its importance in dictating the phenotype of an organism. Which of the following is best demonstrated by the scenario described?

Hypotheses should not be proposed before knowing all the facts.
Experimental data obtained before the middle of the twentieth century are unreliable.
Scientific theories are subject to change in light of new data.
A theory must be absolutely supported by data prior to its publication.

Enter to expand or collapse answer.Answer expanded: Option C is correct because the experiments to determine what molecule contained genetic material took place over many years and continuously built on prior experiments. As technology improved and as more data were generated, the theory describing the nature of heritable genetic material changed. Option A is incorrect because hypotheses are based on logical analysis of available data and then are further tested to be eventually accepted, discarded, or modified. Option B is incorrect because much experimental data obtained early in the twentieth century have proven to be very reliable. In many cases, further experimentation with more modern techniques has upheld previous data and conclusions. Option D is incorrect because theories based on experiments that have received peer reviews are deemed to be logical and worthy of further investigation and are published with supporting data in scientific journals. This allows the theories to be tested further by the same or many other investigators.

6. Individuals in many parts of the world are typically immunized against tuberculosis (TB), but immunization against TB in the United States is rare. Which TWO of the following are the most likely reasons why TB immunization in the United States is rare?

Individuals worry about possible unintended effects of vaccines.
The incidence of TB is relatively low in the United States.
A TB skin test cannot be used to diagnose disease in an immunized individual because the results are always positive.
The vaccine is prohibitively expensive in the United States.

Enter to expand or collapse answer.Answer expanded: Options B and C are correct because, in contrast to some other countries, there is a relatively insignificant number of TB infections in the United States, and the medical community wants to preserve skin testing as a quick and inexpensive way to test for TB. Once an individual is immunized, the skin test will be positive, requiring a more expensive test, such as a chest x-ray. Option A is incorrect because there is no evidence that the vaccine is associated with any unintended effects. Option D is incorrect because the vaccine is used in other countries that are financially less well off than the United States is, so the cost of the vaccine is not a problem in this country.

Domain II—Cell Structure and Processes

Competency 004—The teacher understands the structure and function of biomolecules.

7. A virus infects only some types of cells in certain animal species. Which of the following is of greatest importance in determining the cell types that are infected?

The structures of a ligand on the surface of a virus and a receptor on the surface of a cell
The ratio between the concentration of virus and the number of cells of a particular type
The pH and temperature environment around different cell types
The availability to viruses of products secreted by different cell types

Enter to expand or collapse answer.Answer expanded: Option A is correct because the specificity of interaction between the amino acids of the ligand on the surface of a virus and the amino acids and carbohydrates of a receptor molecule on the surface of a cell determines both host range and tissue-type specificity of virus binding. Option B is incorrect because the ratio of virus to cell number might influence the number of cells infected, but it will not determine which cells are infected. Option C is incorrect because pH and temperature should be relatively constant in the body, and these factors are unlikely to alter viral specificity. Option D is incorrect because products secreted by the cells are unlikely to affect viral specificity.

8. Which of the following is a property of carbon that makes this element basic to the chemistry of life?

Carbon has many isotopes, two of which are stable.
Carbon atoms are arranged in different ways in different crystalline structures.
A carbon atom has four valence electrons and forms four covalent bonds with other atoms.
A carbon atom has an equal number of protons and electrons.

Enter to expand or collapse answer.Answer expanded: Option C is correct because the atom’s ability to make four covalent bonds allows it to form large, complex molecules. Option A is incorrect because the number of isotopes does not influence carbon’s role in the chemistry of life. Option B is incorrect because carbon’s ability to form different kinds of crystalline structures does not influence carbon’s role in the chemistry of life. Option D is incorrect because, while this statement is true, the number of protons is typically identical to the number of electrons in any uncharged atom.

Competency 005—The teacher understands that cells are the basic structures of living things and have specialized parts that perform specific functions.

An eukarytotic membrane with X labeling the insides of the two cells nucleus.

9. A model of a eukaryotic cell membrane is shown above. Which of the following is the most likely function of the structures labeled X?

Facilitating the exchange of O₂ and CO₂ by the cell
Catalyzing the synthesis of molecules secreted by the cell
Promoting the entry of hormones into the cell
Enabling the movement of ions across the cell membrane

Enter to expand or collapse answer.Answer expanded: Option D is correct because the structures labeled X appear to be integral membrane proteins with channels running through them. Ions in solution are hydrated by water and are too large to pass between membrane lipids. They require channel proteins, with different proteins specific to each type of ion. Option A is incorrect because O₂ and CO₂ can pass between membrane lipids. Option B is incorrect because molecules secreted by a eukaryotic cell are synthesized by cellular organelles rather than by channel proteins. Option C is incorrect because most protein hormones bind to cell membrane receptors and do not enter the cell, while lipid hormones simply pass between the membrane lipids to enter.

10. Which of the following properties best distinguishes prokaryotic and eukaryotic cells from viruses?

Prokaryotes and eukaryotes divide by mitosis, whereas viruses do not.
Prokaryotes and eukaryotes synthesize proteins on ribosomes, whereas viruses do not.
Prokaryotes and eukaryotes are surrounded by a lipid bilayer, whereas viruses are not.
Prokaryotes and eukaryotes have genomes of DNA, whereas viruses have genomes of RNA.

Enter to expand or collapse answer.Answer expanded: Option B is correct because viruses do not contain ribosomes to synthesize proteins but are dependent on the ribosomes of the prokaryotes and eukaryotes they infect for synthesis of viral proteins. Option A is incorrect because only eukaryotes divide by mitosis; neither prokaryotes nor viruses do so. Option C is incorrect because enveloped animal viruses are surrounded by a lipid layer that is usually derived from the plasma membrane of the cells they infect. Naked capsid viruses have no such lipid layer. Option D is incorrect because only some viruses have genomes of RNA; many viruses, including all bacteriophages, have genomes of DNA.

Competency 006—The teacher understands how cells carry out life processes.

11. The major source of energy for cellular respiration in plants is provided by which of the following?

Sunlight
CO₂ and water
Nutrients absorbed from the soil
Sugars synthesized by photosynthesis

Enter to expand or collapse answer.Answer expanded: Option D is correct because plants use photosynthesis to produce three-carbon sugars that can be converted to glucose. The energy stored in glucose is then released by glycolysis in the cytosol and the citric acid cycle in the mitochondria of plant cells. Option A is incorrect because sunlight provides energy for photosynthesis but not for cellular respiration. Option B is incorrect because CO₂ and water are not energy sources. CO₂ is released during cellular respiration and fixed in photosynthesis; H₂O is produced during cellular respiration and provides a source of electrons for noncyclic photophosphorylation in photosynthesis. Option C is incorrect because no nutrients that are absorbed from the soil provide energy for cellular respiration.

12. In eukaryotes which of the following molecules is produced in anaerobic cellular respiration but not in aerobic cellular respiration?

Lactic acid
ATP
Pyruvic acid
H₂O

Enter to expand or collapse answer.Answer expanded: Option A is correct because lactic acid is a product of fermentation, which is a common form of anaerobic cellular respiration. Option B is incorrect because ATP is produced in both anaerobic and aerobic respiration. Option C is incorrect because pyruvic acid is the final breakdown product of glucose in glycolysis, the initial metabolic pathway of both anaerobic and aerobic respiration in eukaryotes. Option D is incorrect because H₂O is released when oxygen is reduced during oxidative phosphorylation of aerobic respiration.

Competency 007—The teacher understands how specialized cells, tissues, organs, organ systems and organisms grow and develop.

13. Which of the following is most likely to trigger uncontrolled cell division and tumor formation?

A mutation of a cyclin protein that interferes with the binding of cyclins and cyclin-dependent kinases (Cdk)
A mutation to a cell surface receptor that causes continuous signaling in the absence of ligand binding
An increase in the cytosolic concentration of second messengers such as cAMP and Ca²⁺
An increase in the rate of cellular respiration and ATP production

Enter to expand or collapse answer.Answer expanded: Option B is correct because binding of a ligand such as a cellular growth factor to a cell surface receptor is the normal way by which intracellular signaling is triggered. Mutations to such receptors often induce conformational changes to the receptors that cause the receptors to initiate signaling in the absence of ligand binding. Option A is incorrect because blocking interaction between cyclins and cyclin-dependent kinases blocks cell cycle progression and cell division. Option C is incorrect because cAMP and Ca²⁺are ubiquitous second messengers, so increasing their concentrations is likely to cause so many metabolic problems in a cell that the cell will die. Option D is incorrect because increased cellular respiration and ATP production are likely to be a consequence of, rather than a cause of, increased cell division.

14. The drugs used to treat some types of cancer are toxic to rapidly dividing cells. Which of the following best explains why patients whose bone marrow has been damaged by such drugs often receive bone marrow transplants?

Bone marrow must be dense so that the bones can support the mass of an individual.
Bone marrow contains stem cells that give rise to the red and white blood cells of an individual.
Bone marrow contains adipose cells that store fat as a source of energy for an individual.
Bone marrow must have cells around which to lay down a calcium-storing matrix in an individual.

Enter to expand or collapse answer.Answer expanded: Option B is correct because the hematopoietic stem cells that give rise to red blood cells and white blood cells are found in the bone marrow. These cells divide rapidly and so are typically killed by drugs that are used to kill rapidly dividing cancer cells. Option A is incorrect because the density and strength of bone are derived from collagen fibers and deposits of mineral salts in the compact bone that surrounds the softer marrow. Option C is incorrect because the adipose cells in the bone marrow store an insignificant amount of energy. Option D is incorrect because calcium is not stored in the bone marrow.

Domain III—Heredity and Evolution of Life

Competency 008—The teacher understands the structures and functions of nucleic acids in the mechanisms of genetics.

15. If a single base pair is deleted from the coding region of gene X, which of the following describes a possible effect of the deletion on protein X?

Protein X will be of normal sequence, because three sequential base pairs must be deleted to affect the protein sequence.
Protein X will be of normal sequence, because one of every three base pairs is a wobble base that can be deleted without altering protein sequence.
Protein X will differ from the normal sequence by the single amino acid coded in the region of the deletion.
Protein X will be significantly shorter than the normal length, because a stop codon will be introduced downstream of the deletion.

Enter to expand or collapse answer.Answer expanded: Option D is correct because each codon specifying an amino acid is three bases, so the loss of a single base pair in the DNA will alter the entire amino acid coding sequence past the deletion (that is, will alter the translation reading frame). Because three codons of the genetic code are translation stop codons rather than amino acid–dictating codons, one of the altered codons may be a stop codon, causing a shorter protein to be produced. Option A is incorrect because the deletion of one base pair alters the translation reading frame. Option B is incorrect because only the third base of many codons is a wobble base whose alteration does not alter the amino acid encoded, and the question does not indicate that the third base is the one that is deleted. Additionally, the third base of some codons is not a wobble base, so changing the third base of such codons does alter which amino acid is encoded. Option C is incorrect because the deletion of one base pair alters the entire translation reading frame and not just a single amino acid.

16. For which of the following reasons is it better to clone cDNA rather than genomic DNA into bacteria if the bacteria are to synthesize a eukaryotic protein of the correct sequence?

There are discrepancies between the genetic codes of prokaryotes and eukaryotes.
The transcription factors of prokaryotes and eukaryotes differ.
Bacteria cannot process mRNAs to remove introns.
Bacteria cannot modify proteins in the same way that eukaryotes can.

Enter to expand or collapse answer.Answer expanded: Option C is correct because cDNA, a double-stranded DNA copy of an mRNA, does not contain introns, but genomic DNA does. Bacteria do not have the molecules necessary to process pre-mRNAs and remove the introns before translation. The only way to translate a protein of correct amino acid sequence is to clone a cDNA that lacks introns. Option A is incorrect because the genetic codes of prokaryotes and eukaryotes are essentially identical. Option B is incorrect because transcription factors are irrelevant to pre-mRNA processing. Option D is incorrect because it refers to protein modifications, but the difference between a cDNA and genomic DNA is the lack or presence, respectively, of introns.

Competency 009—The teacher understands the continuity and variations of traits from one generation to the next.

Table outlining test content and subject weighting by domain.
Genotype	Percent of Progeny
RrTt	39%
Rrtt	11%
rrTr	13%
rrtt	39%

17. In a particular species of fly, gene R can be found in two allelic forms: R, which is dominant, and r, which is recessive. Similarly, gene T can be found in two allelic forms: T, which is dominant, and t, which is recessive. A female fly of genotype RrTt is crossed with a male fly of genotype rrtt. The progeny are examined with respect to genes R and T, and their genotypes are determined. The results are shown in the table. Which of the following is most consistent with the data?

Genes R and T are on different chromosomes that exhibit independent assortment.
Genes R and T are linked on a single chromosome.
Genes R and T are both on the X chromosome.
Genes R and T are on different chromosomes that do not assort independently.

Enter to expand or collapse answer.Answer expanded: Option B is correct because the expected data, if genes R and T are completely unlinked, should show 25% of each genotype of progeny. The genotype of the gamete from the male fly can only be rt and, if the genes are unlinked, there is a 25% chance of gametes with each of the following genotypes being produced by the female fly: RT, Rt, rT, rt. The data suggest that (1) the female fly, with genotype RrTt, is the result of a pairing between RT and rt gametes and (2) the R and T or r and t alleles are close enough on the chromosome that relatively little crossing-over occurred between them during formation of gametes by the female fly. Option A is incorrect because 25% of each genotype of progeny is expected if the genes are on different chromosomes and assorting independently. Option C is incorrect because there is no information to suggest that male and female progeny differ in their expression of the genes, as would be the case with X-linkage. Option D is incorrect because different chromosomes do assort independently of each other.

18. Siamese cats are white at birth, but as they mature, they become tan with much darker extremities, including the ears, nose, legs, and tail. Which of the following is the most likely explanation of the color pattern observed in adult Siamese cats?

An allele of a gene required for pigment synthesis encodes a temperature-sensitive enzyme.
An allele of a gene required for pigment synthesis is triggered by sex hormones.
A gene required for pigment synthesis is located in mitochondrial DNA and is only expressed in cells of the extremities.
A gene required for pigment synthesis in cells is activated by increased light exposure.

Enter to expand or collapse answer.Answer expanded: Option A is correct because the extremities of a cat are cooler than the core, and if an enzyme that is required for pigment synthesis or deposition has a different conformation and exhibits greater activity at a lower temperature, then only the extremities of an adult cat will be dark. Option B is incorrect because it is highly improbable that sex hormones will activate a gene associated with pigment deposition only in cells of the extremities. Option C is incorrect because most mitochondrial genes encode products required for mitochondrial function and are not expected to differentially affect pigment production or deposition. Option D is incorrect because other parts of the animal are exposed to light to the same degree as the extremities.

Competency 010—The teacher understands the theory of biological evolution.

19. Scientists have observed that when two species of stickleback fish occur in a single lake, they have distinct morphologies and niches. The species differ in size and shape and in the number and length of the gill rakers that are involved in food–particle retention. One species feeds in open water, and the other feeds at the lake bottom. When only one species of stickleback inhabits a lake, the species exhibits an intermediate morphology and niche. This scenario illustrates which of the following phenomena?

Müllerian mimicry
Character displacement
Facilitation
Competitive exclusion

Enter to expand or collapse answer.Answer expanded: Option B is correct because character displacement refers to the tendency for traits to be more divergent in sympatric populations of two species than in allopatric populations of the same species. Option A is incorrect because Müllerian mimicry refers to a resemblance between two or more unpalatable species that generally have a common predator, but the stickleback fish described differ in appearance and occupy different niches that probably have different predators. Option C is incorrect because facilitation refers to the positive effect of one species on the survival and reproduction of another species, but the two stickleback species have separate niches. Option D is incorrect because competitive exclusion refers to the greater success of one species when two species compete for the same resources, but the two stickleback species have separate niches and do not compete.

20. Which of the following scenarios best illustrates disruptive selection?

Female widowbirds preferentially mate with male widowbirds with longer tails.
Water boatmen occur in three distinct shades of brown; when a population includes all three shades, predatory fish eat a disproportionately large number of the most common form and reduce its frequency in the population.
Guppies that are preyed on by fish that prefer large, mature guppies reproduce at a smaller size and younger age than more‑rapidly growing guppies that are preyed on by fish that prefer small, juvenile guppies.
Under drought conditions, a population of cactus finches includes individuals with short, deep bills and others with long, shallow bills but few with bills of intermediate shape, as are common with normal rainfall.

Enter to expand or collapse answer.Answer expanded: Option D is correct because disruptive selection is a form of natural selection in which individuals with extreme phenotypes survive and reproduce with greater success than do individuals with intermediate phenotypes. Option A is incorrect because the scenario illustrates sexual selection. Option B is incorrect because the scenario illustrates frequency-dependent selection, in which the frequency of different phenotypes is maintained at approximately equal levels by the reduction of the most common phenotype, in this instance because of the greater likelihood of being eaten. Option C is incorrect because the scenario illustrates directional selection in which there is a shift, under certain predator conditions, toward more-rapidly maturing guppies that can reproduce before being eaten.

Competency 011—The teacher understands evidence for evolutionary change during Earth�s history.

21. Recent evidence suggests that the segmented body plan of several protostome phyla arose independently in the phyla, rather than from a common ancestor. The evidence suggests convergent evolution of segmentation in Annelida (segmented worms), Arthropoda (arthropods), and Onychophora (velvet worms). Which of the following most likely provided the evidence?

Analyses of ribosomal RNA (rRNA) sequences and mitochondrial DNA sequences
Studies of newly uncovered fossils from several geologic periods
Comparison of interactions among cells in different body segments
Comparison of chromosome numbers among organisms in the three phyla

Enter to expand or collapse answer.Answer expanded: Option A is correct because comparing the similarities or dissimilarities of rRNA sequences and mitochondrial DNA sequences in different organisms has been shown to be a reliable way of analyzing possible relationships. Such technologies were not available in the early days of phylogeny construction. Option B is incorrect because discovery of a few fossils in these rather diverse phyla is unlikely to demonstrate the origin of segmentation. Option C is incorrect because interactions among cells might provide data on cell signaling, but the interactions will not provide data on the origin of segmentation. Option D is incorrect because chromosome numbers provide no indication about the similarities or dissimilarities of gene sequences or the evolution of segmentation.

22. Which of the following adaptations is most directly linked to the diversity of angiosperms?

The variation in photoperiods among species
The presence of vascular tissue
The ability to reproduce asexually
The ability to perform photosynthesis

Enter to expand or collapse answer.Answer expanded: Option B is correct because the evolution of vascular tissue first allowed plants to colonize land. The evolution of flowers and the ability to transport water through the vascular system then allowed angiosperms to adapt to a wide range of terrestrial biomes. Option A is incorrect because the variation in photoperiods is not linked to the diversity of angiosperms. Option C is incorrect because only a fraction of angiosperms reproduce asexually through such means as the production of runners. Option D is incorrect because all plants perform photosynthesis, not just angiosperms.

Domain IV—Diversity of Life

Competency 012—The teacher understands similarities and differences between living organisms and how taxonomic systems are used to organize and interpret the diversity of life.

23. Which of the following adaptations is more likely to be enhanced in a desert mammal than in a mammal living in the northern boreal forest?

Ability to burrow into the ground
Avoidance of predators
Ability to retain water
Complex mating behavior

Enter to expand or collapse answer.Answer expanded: Option C is correct because water is far more plentiful in the northern boreal forest than in the desert. Desert mammals have evolved mechanisms for conserving water that include more efficient water extraction from the oxidation of macromolecules in their food and production of more concentrated urine. Option A is incorrect because some species of mammals in both regions burrow into the ground. Option B is incorrect because mammals in both regions try to avoid predators. Option D is incorrect because mammals in both regions exhibit complex mating behaviors.

24. Which of the following best distinguishes the structure of bacteria from the structure of archaeans?

Genomes of bacteria are located in nucleoids, whereas those of archaeans are located in nuclei.
Cell walls of bacteria contain peptidoglycan, whereas those of archaeans lack peptidoglycan.
Chromosomes of bacteria are circular, whereas those of archaeans are linear.
Chlorophyll in bacteria is located in infoldings of the plasma membrane, whereas chlorophyll in archaeans is located in chloroplasts.

Enter to expand or collapse answer.Answer expanded: Option B is correct because the presence or absence of peptidoglycan is a major structural distinction between bacteria and archaeans. Option A is incorrect because both bacteria and archaeans have nucleoids; nuclei are a characteristic of eukarya only. Option C is incorrect because the chromosomes of both bacteria and archaeans are generally circular. Option D is incorrect because some bacteria perform photosynthesis using chlorophyll embedded in infoldings of the plasma membrane, but neither chlorophyll nor photosynthesis has been demonstrated in any archaeans.

25. Which of the following pieces of information would be most useful to include in a dichotomous key for students identifying Texas wildflowers?

Ribosomal RNA (rRNA) sequences
Germination times
Petal colors
Chromosome numbers

Enter to expand or collapse answer.Answer expanded: Option C is correct because a dichotomous key is typically used to allow visual identification of plants or animals. Selection from a series of features such as leaf shape, petal number, and petal color allow an individual to identify a wildflower. Options A, B, and D are incorrect because rRNA sequences, germination times, and chromosome numbers do not aid in visual identification of an organism and are thus not included in dichotomous keys.

Competency 013—The teacher understands that, at all levels of nature, living systems are found within other living systems, each with its own boundaries and limits.

26. Which of the following best describes the mechanism by which vaccination provides protection from pathogens?

The lymphatic system carries pathogens more rapidly to lymphoid organs after pre‑exposure to a similar vaccine.
The vaccine remains in the body and competes with the real pathogen for cells to infect.
Memory lymphocytes are produced that rapidly respond upon subsequent encounter with a pathogen resembling a vaccine.
There is an increase in the number of phagocytic cells that can find and ingest pathogens resembling a vaccine.

Enter to expand or collapse answer.Answer expanded: Option C is correct because vaccination produces clones of lymphocytes specific to the introduced material, and many of the lymphocytes are memory cells that are activated much more rapidly than naïve cells upon subsequent encounter with a pathogen similar to the material in the vaccine. Option A is incorrect because vaccination has no effect on the rate at which material is carried through the lymphatic system. Option B is incorrect because the vaccine is typically completely cleared from the body within a few weeks. Option D is incorrect because phagocytic cell populations do not expand upon pre‑exposure to a vaccine, and the cells exhibit no memory.

27. Mutation of a protein specific to which of the following cell types is most likely to interfere with the transport of sugars in angiosperms?

Sclereids
Tracheids
Collenchyma cells
Sieve-tube elements

Enter to expand or collapse answer.Answer expanded: Option D is correct because sieve-tube elements are the cells that conduct sugars and other organic nutrients in the phloem of angiosperms. The sieve-tube elements are alive but lack most organelles, as this leaves maximal space for nutrient passage through the cells. Option A is incorrect because sclereids are cells with very thick secondary cell walls that provide support to a plant or to its products, such as seed coats, and are often dead at maturity. Option B is incorrect because tracheids are cells of the xylem that extend in the direction of water movement and possess pits to allow water movement among the cells. Option C is incorrect because collenchyma cells are flexible cells that are alive at maturity and provide support to a plant, especially to new growth.

Competency 014—The teacher understands the processes by which organisms maintain homeostasis.

28. In the flipper of a dolphin, each artery is surrounded by several veins. Which of the following is the best explanation for the arrangement of blood vessels?

Cool blood in the veins receives heat from arterial blood as the venous blood returns to the animal’s core.
Oxygen-poor blood in the veins absorbs oxygen from the artery as the blood returns to the animal’s core.
The artery is protected from trauma by the surrounding veins.
Flipper movement ensures that the rate of blood flow in the veins equals the rate of flow in the artery.

Enter to expand or collapse answer.Answer expanded: Option A is correct because this is an example of a countercurrent heat exchanger. To conserve body heat, venous blood that has cooled in the flipper and is returning to the body core absorbs heat from arterial blood directed to the flipper. Option B is incorrect because there is no gas exchange between blood in the artery and the veins. Option C is incorrect because veins do not have strong muscular walls and thus are too soft to provide protection for an artery. Option D is incorrect because while flipper movement does help blood flow in veins, it does not ensure a flow rate equal to that of the flow in the artery, and it does not explain the particular arrangement of the vessels.

29. Which THREE of the following best explain how small mammals can survive long periods of hibernation without ingesting food or water?

The production of enzymes that hydrolyze the triglycerides of stored fat in the mammals increases.
The pH of the mammals’ blood significantly decreases, except for blood that circulates to the brain.
The heart rate and breathing rate of the mammals decrease.
The core temperature of the mammals decreases.

Enter to expand or collapse answer.Answer expanded: Options A, C, and D are correct because all describe physiological processes that occur during hibernation. There is an increase in the production of enzymes that hydrolyze the triglycerides of the fat stores accumulated by the animal before the start of hibernation. Additionally, the heart rate, breathing rate, and core temperature all decrease in association with a decrease in metabolic rate during hibernation. Option B is incorrect because a decrease in the pH of the blood during hibernation would have an adverse effect on the cells of the animal.

Competency 015—The teacher understands the relationship between biology and behavior.

30. Which of the following best demonstrates that some behaviors involve both an innate and a learned component?

A pheromone produced by a queen honeybee attracts workers to the queen in the hive.
The females of some cichlid fish species hold fertilized eggs in their mouths until the eggs hatch.
Male weaver birds build similar but not identical nests, and individual birds modify their technique from one nest to the next.
A rat that presses a lever in a box and routinely receives a food reward will continue to press the lever.

Enter to expand or collapse answer.Answer expanded: Option C is correct because a bird initially uses instinct to build a nest that is generally characteristic of the species in terms of location and structure. However, there must also be a learned component to nest building because the structure of the nests can improve and become more elaborate with each subsequent nest built by birds of certain species, as exemplified by the nests built by male weaver birds. Option A is incorrect because attraction to a pheromone is strictly innate. Option B is incorrect because a cichlid does not learn to hold eggs and fry in its mouth but is exhibiting an innate behavior. Option D is incorrect because pressing a lever with the expectation of a reward is an example of operant conditioning, in which a behavior may be changed by a consequence; lever-pressing behavior by the rat is reinforced by a food reward.

Domain V—Interdependence of Life and Environmental Systems

Competency 016—The teacher understands the relationships between abiotic and biotic factors of terrestrial and aquatic ecosystems, habitats and biomes, including the flow of matter and energy.

31. The primary productivity of coral reefs is substantially greater than the primary productivity of the open ocean. Which of the following best explains the high primary productivity of coral reefs?

Corals grow in relatively deep water with a low salt concentration and a high concentration of dissolved CO₂.
Corals grow in warm water that contains a high plankton count and has a high concentration of nutrients.
Corals grow in shallow, cold water with abundant sunlight and a high concentration of dissolved O₂.
Corals grow in relatively shallow, nutrient-poor water that is easily penetrated by sunlight.

Enter to expand or collapse answer.Answer expanded: Option D is correct because corals primarily live around islands and along the edges of some continents in a zone of water that exhibits great clarity. The abundant sunlight promotes the growth of photosynthetic algae that live in a symbiotic relationship with the corals, a relationship that contributes to the great productivity of the reefs. Most corals live in warm water, and all appear to be very sensitive to a high concentration of nutrients and have in fact declined in areas exposed to sewage and fertilizer runoff. Option A is incorrect because most corals do not live in deep water, where the Sun does not penetrate, and deep water is unlikely to be low in salt and high in CO₂ . Option B is incorrect because high nutrient concentrations that often promote plankton growth kill corals. Option C is incorrect because the vast majority of corals grow in warm water, and a very few grow in cool—but not cold—water.

32. If the energy available from all secondary consumers to the next trophic level in a terrestrial food web is 90 kcal / m² / year, then the net primary productivity of the food web is most likely closest to

0.9 kcal / m² / year
9 kcal / m² / year
900 kcal / m² / year
9,000 kcal / m² / year

Enter to expand or collapse answer.Answer expanded: Option D is correct because 90 kcal / m² / year is 1% of 9,000 kcal / m² / year. It is estimated that approximately 10% of the energy at any trophic level is available to the next trophic level. If the net primary productivity of the food web is closest to 9,000 kcal / m² / year, then the energy available to primary consumers (herbivores) is approximately 900 kcal / m² / year, and the energy available to secondary consumers is approximately 90 kcal / m² / year. Options A, B, and C are incorrect because 90 kcal / m² / year is not 1% of the value in any of those options.

Competency 017—The teacher understands the interdependence and interactions of living things in terrestrial and aquatic ecosystems.

Biomass pyramind

The longest layer in the first layer with is labeled as primary producers, the second layer is labeled primary consumers, the third layer is labeled secondary consumers, the fourth and last layer is labeled tertiary consumers.

33. Which of the following best explains why each level of a standing crop biomass pyramid such as that shown above is so much smaller than the level immediately below it?

Organisms at higher levels die more rapidly than organisms at lower levels do.
Inefficient energy transfers between levels support fewer organisms at higher levels.
Organisms at higher levels typically require more nutrients per unit mass than organisms at lower levels do.
Recycling of organic matter and essential chemical elements occurs less efficiently at higher levels.

Enter to expand or collapse answer.Answer expanded: Option B is correct because trophic efficiencies generally range from 5% to 20%. Energy is lost between trophic levels through respiration and feces as well as through energy in the lower trophic level that is not consumed by organisms in the next trophic level. Option A is incorrect because organisms at higher trophic levels often live as long as or longer than organisms at lower trophic levels. Option C is incorrect because organisms at higher trophic levels typically do not require more nutrients per unit mass than organisms at lower trophic levels do. Option D is incorrect because recycling occurs with about the same efficiency at all trophic levels, but there is simply less energy to recycle at higher levels.

34. Truffles are fungi that grow underground. Truffles produce chemicals with an aroma that attracts pigs and helps them detect the truffles. The pigs sometimes eat the truffles they find. Which of the following best describes why the relationship between a pig and a truffle can be considered mutualistic?

The pig eats the truffle and gains chemical energy from it.
The pig eats the truffle, and the truffle remnants benefit scavengers in the soil.
The pig eats the truffle, and the pig’s feces contain digested components that decomposers return to the soil.
The pig eats the truffle and spreads fungal spores as it digs the truffle out of the ground.

Enter to expand or collapse answer.Answer expanded: Option D is correct because both the pig and truffle benefit: the pig gains a source of chemical energy when it eats the truffle, and the spores of the truffle are dispersed by the pig, facilitating the growth of more truffles. Option A is incorrect because this activity benefits the pig but not the truffle. Options B and C are incorrect because both activities benefit the pig and other organisms but do not directly benefit the truffle.

Competency 018—The teacher understands the relationship between carrying capacity and changes in populations and ecosystems.

35. Assuming that no emigration, immigration, or catastrophic event occurs, which of the following graphs best illustrates a human population that is likely to decrease in size over the next 25 years?

The left side of the graph represents the male population starts on the 0 age point on the y axis and 10 percentage point on the x axis. The line curves up to the 85 year old age mark. The right side of the graph, which represents the female population, starts at the 0 mark on the x axis and at 85 year old age mark on the y axis. It curves down to 0 years of age on the y axis and 10 percentage on the x axis.
The left side of the graph represents the male population starts on the 0 age point on the y axis and 10 percentage point on the x axis. The line increases up coming to a more round shapt at the 85 year old age mark. The right side of the graph, which represents the female population, starts at the 0 mark on the x axis and at 85 year old age mark on the y axis. From it's semi-round beginning, the line declines to 0 years of age on the y axis and to 10 percentage on the x axis.
The left side of the graph represents the male population starts on the 0 age point on the y axis and between the 2 and 4 percent point on the x axis. The line increases up to the 85 year old age mark with a slight left curve in to staying within that 2 to 4 percent range. The right side of the graph, which represents the female population, slightly curved to the right side starts at 85 age mark on the y axis and at 0 percent mark on the x axis. It declines to 0 years of age on the y axis and stays between 0 and 2 percent on the x axis.
The left side of the graph represents the male population starts on the 0 age point on the y axis and between the 2 and 4 percent point on the x axis. The line increases up to the 85 year old age mark with a slight right curve in to staying within that 2 to 4 percent range. The right side of the graph, which represents the female population, starts at 85 age mark on the y axis and at 0 percent mark on the x axis. It declines to 0 years of age on the y axis and stays between 0 and 2 percent on the x axis.

Enter to expand or collapse answer.Answer expanded: Option C is correct because the size of the child-bearing population in 25 years will be smaller than the size of the current child-bearing population. Option A is incorrect because the population is heavily weighted toward individuals who will be of child-bearing age in 25 years. Option B is incorrect because there are more individuals who will be of child-bearing age in 25 years than are currently of child-bearing years. Option D is incorrect because the number of individuals of child-bearing age is fairly constant for the next 25 years and is greater than the current older population; the population will probably grow slowly over the next 25 years.

36. The reproductive strategies of organisms are often described as being primarily r‑selected or K‑selected. Organisms with r‑selected reproductive strategies often have shorter life expectancies and produce many more offspring than organisms with K‑selected reproductive strategies do. Which of the following is more likely to be true of a population of organisms with a K‑selected reproductive strategy than of a population with an r‑selected reproductive strategy?

The population size is near the carrying capacity of the environment.
The population lives in an unstable environment.
The population lives in an environment with a temperate climate.
The population is made up of organisms that are keystone species.

Enter to expand or collapse answer.Answer expanded: Option A is correct because the population size of organisms with a K‑selected reproductive strategy is typically close to the carrying capacity of the environment. Option B is incorrect because organisms with a K‑selected reproductive strategy typically live in a stable environment. Option C is incorrect because organisms with a K‑selected reproductive strategy are found in a variety of climates. Option D is incorrect because organisms with a K‑selected reproductive strategy need not be keystone species.

Domain VI—Science Learning, Instruction and Assessment

Competency 019—The teacher understands research-based theoretical and practical knowledge about teaching science, how students learn science and the role of scientific inquiry in science instruction.

37. Which of the following procedures is most likely to engage and motivate students to learn about regulation of cell growth?

Playing a game in which teams of students compete to see who can correctly answer the most questions about organelle function in cell growth
Directing students to Internet sites on which they can watch animations of cells as they grow and divide
Having students read short biographies of scientists who are engaged in research about cell-growth regulation
Having students use a microscope to examine cultured cells before and after they add substances to the cells that affect cell growth and division

Enter to expand or collapse answer.Answer expanded: Option D is correct because students who use microscopes to observe the effects of adding substances to cells are likely to be engaged and motivated to learn more about factors that can affect cell growth. Option A is incorrect because competing to see who can answer the most questions about organelle function is less engaging than observing cells through a microscope. Option B is incorrect because watching animations is passive rather than engaging. Option C is incorrect because reading biographies of scientists is also passive.

38. Students in a biology laboratory class will work in groups using the enzyme catalase to study factors that affect the activity of enzymes. Which of the following strategies by the teacher is the most effective inquiry-based learning approach?

Ask the students first to develop a hypothesis related to catalysis and then to design and perform an experiment with catalase to investigate the hypothesis.
Give the students several hypotheses about factors that affect catalase activity and help them to design a single experiment that addresses all the hypotheses.
Give the students a protocol that has been used to investigate catalysis by catalase and ask them to carefully follow and then evaluate the protocol.
Tell the students to read research papers about catalase and its activity, then have them select one paper and attempt to reproduce the methods and results.

Enter to expand or collapse answer.Answer expanded: Option A is correct because asking the students to develop a hypothesis as well as perform an experiment that they design to investigate the hypothesis is a learning approach that is entirely inquiry based. Option B is incorrect because giving the students hypotheses and helping them to design an experiment does not provide much opportunity to develop independent-inquiry skills. Option C is incorrect because giving the students a protocol to follow and evaluate provides little opportunity to develop independent-inquiry skills. Option D is incorrect because reproducing published methods to obtain published results does not provide much opportunity to develop independent-inquiry skills.

Competency 020—The teacher knows how to monitor and assess science learning in laboratory, field and classroom settings.

39. Students perform a laboratory exercise to investigate how cell membranes regulate the movement of materials into and out of cells. They expose both live cells and artificial membranes to several different solutions and monitor the effects of the solutions on the cells and membranes. Which of the following methods will best allow the teacher to assess understanding by each student of the significance of the results?

After completing the exercise, have the students write individual reports in which they use their results to propose mechanisms by which various materials cross cell membranes.
As they perform the exercise, have the students work in pairs and discuss the procedure and data with each other.
After completing the exercise, select several volunteers to stand up and summarize the significance of the results to the rest of the class.
After completing the exercise, give the students a multiple-choice quiz with questions about the procedure and the results they obtained.

Enter to expand or collapse answer.Answer expanded: Option A is correct because the ability to write a report and, based on the experimental results, propose mechanisms by which materials, including some the student did not test, cross cell membranes suggests that a student truly understands the significance of the results. Option B is incorrect because it is difficult to determine how much the students understand if they are only talking with one another. Option C is incorrect because the process does not assess the understanding of all students but only of those few who volunteer to speak to the class. Option D is incorrect because a quiz on the procedure and results, particularly a multiple-choice quiz, does not allow the students to describe the broader significance of the results.

40. Which of the following is the best example of a summative assessment?

Asking students who have presented independent experiments at a science fair to describe their science fair experiences to their classmates
Asking students at the end of a physiology unit to identify and give the function of structures identified by flags in dissected organisms
Asking students who worked in assigned groups for an activity to evaluate the work of the other members of the group
Asking students at the beginning of a unit on ecology to describe the biological basis for succession and niche partitioning

Enter to expand or collapse answer.Answer expanded: Option B is correct because testing student knowledge at the end of a unit or course is a summative assessment. Option A is incorrect because asking students to describe science fair experiences does not directly assess knowledge or skills. Option C is incorrect because asking students to evaluate the work of other group members is a peer assessment. Option D is incorrect because assessing prior knowledge at the beginning of a unit is typically part of a formative assessment.

Section 4: Sample Selected-Response Questions Life Science 7–12 (238)