Download Biology I End-of-Course

Biology I End-Of-Course Practice Test Answer Key
Question 1
Reporting Category: Scientific Process
Benchmark: SC.BS.2.1 Explain how scientific advancements and emerging technologies have
influenced society
Answer Key: B
A judge allowed a DNA analysis to be entered as evidence in court.
Which of the following statements best explains why the use of DNA analysis is so effective in
establishing identity?
A. The procedure for DNA fingerprinting is an accurate series of simple techniques.
This answer is not correct. How simple or complex the technique is does not indicate
whether this procedure is a reliable way to identify individuals.
B. The probability of two people having similar DNA fingerprints is small.
This answer is correct. There are billions of base pairs within each individual’s DNA.
These base pairs form sequences, and the probability of two individuals having the same
sequence of base pairs is very slight. Therefore, the use of DNA fingerprinting to establish
identity can be used as evidence in a court of law.
C. The purpose of DNA fingerprinting is to match sections of base pairs.
This answer is not correct. The purpose of DNA fingerprinting is to identify individuals and
family relationships among individuals based on their DNA.
D. The percentage of identical DNA within a species is high.
This answer is not correct. Individuals within a species share a large percentage of their
DNA, but there is still enough difference among individuals for this to be a useful technique.
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Biology I End-Of-Course Practice Test Answer Key
Question 2
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.2 Explain the chemical reactions that occur in photosynthesis and
cellular respiration that result in cycling of energy
Answer Key: D
A researcher wants to measure the rate of photosynthesis in two different types of water plants
growing in two different aquariums under the same conditions.
Which measurement would give the most information about the rate of photosynthesis?
A. The intensity of light available to each aquarium
This answer is not correct. The amount of light available does affect the rate of
photosynthesis, but measuring the intensity of light would not provide the most information
on how quickly photosynthesis is occurring in each plant.
B. The temperature of water in each aquarium
This answer is not correct. The temperature of the water can change the rate of
photosynthesis, but measuring the water temperature would not provide the most
information on the rate of photosynthesis in each plant.
C. The level of dissolved nitrogen in each aquarium
This answer is not correct. The process of photosynthesis does not involve nitrogen.
D. The amount of oxygen produced in each aquarium
This answer is correct. Oxygen is produced during photosynthesis. Measuring the amount
of oxygen produced is the best way to measure the rate of photosynthesis.
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Biology I End-Of-Course Practice Test Answer Key
Question 3
Reporting Category: Structure and Function in Organisms
Benchmark: SC.BS.4.6 Explain the organization of life on Earth using the modern classification
system
Answer Key: D
The table below shows the classifications of four animals.
According to their classification, which of the following animals are most closely related?
A. Q and R
This answer is not correct. The most closely related organisms share the greatest number of
classification levels. Animals Q and R share only three classification levels, which is not the
greatest number of all the animal pairs.
B. S and T
This answer is not correct. The most closely related organisms share the greatest number of
classification levels. Animals S and T share only three classification levels, which is not the
greatest number of all the animal pairs.
C. Q and T
This answer is not correct. The most closely related organisms share the greatest number of
classification levels. Animals Q and T share only four classification levels, which is not the
greatest number of all the animal pairs.
D. R and S
This answer is correct. The most closely related organisms share the greatest number of
classification levels. Animals R and S share five classification levels, which is the greatest
number of all the animal pairs.
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Biology I End-Of-Course Practice Test Answer Key
Question 4
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.4 Explain how Mendel's laws of heredity can be used to determine the
traits of possible offspring
Answer Key: C
In pea plants, the allele for red flowers (R) is dominant, and the allele for white flowers (r) is
recessive. The Punnett square below shows the cross of two pea plants, each with red flowers.
According to the Punnett square, what percent of the offspring resulting from this cross will have
red flowers?
A. 25%
This answer is not correct. Because the allele for red flowers (R) is dominant, the offspring
with the Rr allele combination are also expected to develop red flowers, not just those with
the RR combination.
B. 50%
This answer is not correct. Because the allele for red flowers (R) is dominant, all offspring
with a copy of the R allele are expected to develop red flowers, not just those with the Rr
combination.
C. 75%
This answer is correct. Red flowers are dominant, and the offspring only need to inherit
one copy of the dominant (R) allele to develop red flowers. All the offspring with the RR and
Rr allele combinations are expected to develop red flowers. This is 75% of the offspring.
D. 100%
This answer is not correct. Because the allele for red flowers (R) is dominant, all offspring
with a copy of the R allele are expected to develop red flowers. Only 75% of the offspring
will receive at least one copy of the R allele. The other 25% are rr and are expected to
develop the recessive white flowers.
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Biology I End-Of-Course Practice Test Answer Key
Question 5
Reporting Category: Scientific Process
Benchmark: SC.BS.1.2 Design and safely implement an experiment, including the appropriate
use of tools and techniques to organize, analyze, and validate data
Answer Key: D
A student tests the effects of antibiotics on bacteria. She uses two petri dishes that contain a
solid agar on the bottom. Both dishes have nutrients in the agar, but the second dish also has
an antibiotic in the agar. She puts 0.1 mL of a bacterial solution into one petri dish and 0.2 mL of
the same bacterial solution into the second petri dish. She spreads out each solution over the
agar. She allows the bacteria to grow overnight. The pictures below show the student’s
experimental setup.
The next day, the student counts the number of colonies in each dish. Each colony appears as
a spot in the dish and contains millions of cells that grew from a single bacterium. The dish
without antibiotics has 190 colonies, and the dish with antibiotics has 200 colonies. She
concludes that antibiotics have little effect on bacteria.
Which statement describes what the student could do to improve her experimental procedure?
A. Add more of the antibiotic to the agar in the second dish.
This answer is not correct. Adding more antibiotic would not strengthen the conclusion,
because the amount of bacterial solution added to each dish still will not be the same.
B. Let the bacteria grow in the dishes for another day.
This answer is not correct. This would not necessarily change the results, because the initial
amount of bacterial solution added to each dish was not the same.
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Biology I End-Of-Course Practice Test Answer Key
C. Place the dish without the antibiotic in the refrigerator.
This answer is not correct. Placing the dish in the refrigerator adds another variable
(temperature). This would weaken the conclusion.
D. Put the same amount of bacterial solution into each petri dish.
This answer is correct. The student should ensure that the same amount of bacterial
solution is added to each dish. The experiment has two variables (the amount of bacterial
solution and the presence of antibiotics), which makes drawing a conclusion difficult.
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Biology I End-Of-Course Practice Test Answer Key
Question 6
Reporting Category: Scientific Process
Benchmark: SC.BS.1.3 Defend and support conclusions, explanations, and arguments based
on logic, scientific knowledge, and evidence from data
Answer Key: D
The graph shows changes in a population of shrimp over time.
Which conclusion is supported by this data?
A. An invasive species lives in this ecosystem.
This answer is not correct. The graph provides no information that would allow an individual
to make conclusions about the presence of an invasive species in this ecosystem.
B. Shrimp have a symbiotic relationship with another organism.
This answer is not correct. Because the graph only shows the population size of the shrimp,
a symbiotic relationship cannot be assumed or supported.
C. The food source for this population changes seasonally.
This answer is not correct. There is not enough information to conclude that the changes
correlate with seasonal changes.
D. This population of organisms lives in a stable environment.
This answer is correct. The graph shows periodic fluctuations in the population size of the
shrimp. There are no major increases or decreases in population size; therefore, the data
indicate that the population is living in a stable environment.
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Biology I End-Of-Course Practice Test Answer Key
Question 7
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.5 Describe the components and functions of a variety of macromolecules
active in biological systems
Answer Key: A
Carbohydrates are macromolecules used for energy in living organisms. Large carbohydrate
molecules are made of smaller building blocks called monosaccharides.
The arrangement of which three components is used to distinguish one monosaccharide from
another?
A. Carbon, hydrogen, and oxygen
This answer is correct. Monosaccharides all contain carbon, hydrogen, and oxygen. It is
the arrangement of these three components that makes one monosaccharide different from
another.
B. Glucose, fructose, and ribose
This answer is not correct. Glucose, fructose, and ribose are three examples of
monosaccharides.
C. Peptide, fatty acid, and purine
This answer is not correct. Peptide, fatty acid, and purine are all components of the other
three types of macromolecules.
D. Water, carbon dioxide, and nitrogen
This answer is not correct. Water, carbon dioxide, and nitrogen are essential for life
processes. However, they are not the components of carbohydrates.
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Biology I End-Of-Course Practice Test Answer Key
Question 8
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.3 Explain the structural properties of DNA and the role of DNA in heredity
and protein synthesis
Answer Key: D
The table shows which mRNA codons code for various amino acids.
Which amino acid sequence will be produced by translation of the mRNA sequence UAC UCU
ACC?
A. Asn – Pro – Thr
This answer is not correct. The amino acid sequence produced from the RNA sequence is
Tyr – Ser – Thr, not Asn-Pro-Thr.
B. Thr – Pro – Asn
This answer is not correct. The amino acid sequence produced from the RNA sequence is
Tyr – Ser – Thr, not Thr-Pro-Asn.
C. Thr – Ser – Tyr
This answer is not correct. The amino acid sequence produced from the RNA sequence is
Tyr – Ser – Thr, not Thr-Ser-Tyr.
D. Tyr – Ser – Thr
This answer is correct. The messenger RNA will code for the amino acid sequence
Tyr – Ser – Thr.
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Biology I End-Of-Course Practice Test Answer Key
Question 9
Reporting Category: Life and Environmental Sciences
Benchmark: SC.BS.4.4 Describe how homeostatic balance occurs in cells and organisms
The diagram shows a model of a cell. For proper cell function, sodium (Na+) and potassium (K+)
ions must be actively transported into and out of the cell. The arrows in the protein pump show
the direction of the net flow of each ion.
A. Place Na+ labels inside and outside the cell to illustrate the correct concentration gradient for
active transport to occur.
B. Place K+ labels inside and outside the cell to illustrate the correct concentration gradient for
active transport to occur.
C. Place the energy molecule that is required for active transport on top of the protein pump
structure.



You should use more than one Na+ label.
You should use more than one K+ label.
Only one energy molecule is needed.
Answer Key:
3 points
For this item, the response correctly:
 places more sodium (Na+) ions outside the cell than inside it
AND
 places more potassium (K+) ions inside the cell than outside it
AND
 places one or more energy (ATP) molecules on top of (or in the vicinity of) the protein pump
AND
 places no NADH molecules on top of (or in the vicinity of) the protein pump.
2 points
For this item, the response correctly:
 places more sodium (Na+) ions outside the cell than inside it
OR
 places more potassium (K+) ions inside the cell than outside it
AND
 places one or more energy (ATP) molecules on top of (or in the vicinity of) the protein pump
AND
 places no NADH molecules on top of (or in the vicinity of) the protein pump.
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Biology I End-Of-Course Practice Test Answer Key
1 point
For this item, the response correctly:
 places more sodium (Na+) ions outside the cell than inside it
OR
 places more potassium (K+) ions inside the cell than outside it
AND
 places one or more energy (ATP) molecules on top of (or in the vicinity of) the protein pump
AND
 places one or more NADH molecule on top of (or in the vicinity of) the protein pump.
0 points
The response is incorrect for all parts. The response does not have more sodium (Na+) ions
outside the cell than inside it and does not have more potassium (K+) ions inside the cell than
outside it, and does not have one or more energy (ATP) molecules on top of (or in the vicinity
of) the protein pump.
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Biology I End-Of-Course Practice Test Answer Key
Sample Student Answer:
Explanation of Correct Answer:
K+ ions are moving from the extracellular fluid into the cytoplasm. For active transport to be
occurring, the concentration of K+ ions inside the cell must be higher than outside the cell. Na+
ions are moving from the cytoplasm to the extracellular fluid. For active transport to be
occurring, the concentration of Na+ ions outside the cell must be higher than inside the cell. ATP
is the energy molecule required for active transport.
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Biology I End-Of-Course Practice Test Answer Key
Question 10
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.3 Explain how matter and energy flow through living systems and the
physical environment
Answer Key: C
The diagram below is an energy pyramid showing the amount of energy available at each
trophic level. Each level receives approximately 10% of the energy from the level below.
Which of the following statements best explains the difference in the amount of energy available
at each trophic level?
A. Higher trophic levels contain 90% more organisms than lower levels.
This answer is not correct. Since each trophic level receives approximately 10% of the
energy from the level below, this will severely limit the number of organisms at the higher
trophic levels.
B. Producers pass 90% of the energy from sunlight to the next level.
This answer is not correct. Each trophic level receives approximately 10% of the energy
from the level below.
C. Life functions at each trophic level consume 90% of the remaining energy.
This answer is correct. Since approximately 10% of energy is passed from one trophic
level to the next, approximately 90% of the energy must be consumed at each trophic level.
D. Organisms in lower trophic levels digest food 90% faster than those in the middle levels.
This answer is not correct. There is no indication in the energy pyramid regarding the rate of
food digestion.
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Biology I End-Of-Course Practice Test Answer Key
Question 11
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.2 Explain how cells are specialized into different tissues and organs
Answer Key: D
When a cyclist rides up a steep hill, the cyclist’s circulation and breathing rate increase, allowing
a greater amount of oxygen to reach the cyclist’s muscles. In order to generate energy from this
extra oxygen, skeletal muscle must contain a greater number of a certain cell part than other
tissues.
Which of the following cell parts is more numerous in skeletal muscle than in other tissues?
A. Golgi bodies
This answer is not correct. The Golgi bodies are responsible for modifying, packaging, and
transporting proteins through the cell.
B. Lysosomes
This answer is not correct. The lysosomes are responsible for the breakdown of ingested
substances, damaged organelles, and cell macromolecules.
C. Ribosomes
This answer is not correct. The ribosomes are responsible for protein synthesis.
D. Mitochondria
This answer is correct. The mitochondria are responsible for cellular respiration. During
cellular respiration, the energy molecule ATP is produced.
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Biology I End-Of-Course Practice Test Answer Key
Question 12
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.3 Differentiate between the processes of mitosis and meiosis
Answer Key: A
A student uses a microscope to observe cells in the root tissue of an onion. He concludes that
the cells are reproducing by mitosis.
Which hypothesis is supported by his conclusion?
A. The root tissue cells all have the same set of chromosomes.
This answer is correct. The cells created by mitosis are clones of each other and all have
the same chromosomes.
B. The root tissue cells each have a unique genetic make-up.
This answer is not correct. Each cell has the same genetic makeup because the cells were
created by mitosis―none is genetically unique.
C. The root tissue cells produce identical gametes.
This answer is not correct. Gametes are sex cells that are produced through meiosis.
D. The root tissue cells split to form leaf cells.
This answer is not correct. Root tissue cells that split through mitosis do not form cells from
other plant tissues, such as the leaf.
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Biology I End-Of-Course Practice Test Answer Key
Question 13
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.2 Explain the theory of natural selection
Answer Key: A
C. reinhardtii is a species of unicellular green algae that primarily produce energy for growth
through photosynthesis. However, when necessary, they can also produce energy from a
carbon source, which allows them to grow in total darkness. A scientist grows a population of
these algae in the dark and finds that after 600 generations, the algae population now grows
better in the dark than in the light.
Which statement best explains what has happened to the cells in the algae population?
A. The cells that were better adapted to growing in the dark reproduced more.
This answer is correct. Over the 600 generations of the algae culture, algae cells that were
better able to grow in the dark would survive better and reproduce more than the other algae
cells.
B. The cells that were better adapted to growing in the light got smaller in size.
This answer is not correct. This would not explain why the population grows better in the
dark.
C. They became contaminated with a species of algae that grow only in the dark.
This answer is not correct. It would not be likely that the algae population would be
contaminated by another species that could grow only in the dark.
D. They evolved into a different species that can grow only in the dark.
This answer is not correct. The introduction states that the population grows better in the
dark, not that it can grow only in the dark. Therefore, the algae still retain some ability to
grow in the light.
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Biology I End-Of-Course Practice Test Answer Key
Question 14
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.1 Describe different cell parts and their functions
Answer Key: C
The cytoplasm of muscle cells experiences an increase in CO2 levels and a decrease in pH
during heavy aerobic exercise.
The processes of which cell organelle are responsible for these changes?
A. Endoplasmic reticulum
This answer is not correct. The endoplasmic reticulum transports materials around the cell
and produces certain biological molecules. It does not directly affect the carbon dioxide
(CO2) levels or pH in the cell during exercise.
B. Golgi apparatus
This answer is not correct. The Golgi apparatus prepares material for export from the cell. It
does not directly affect the carbon dioxide (CO2) levels or pH in the cell during exercise.
C. Mitochondrion
This answer is correct. Mitochondria participate in metabolism, which is a process
whereby oxygen is used up and converted into carbon dioxide. Carbon dioxide (CO2) is a
weak acid that lowers the pH. During respiration, additional acids are produced that also
lower pH.
D. Nucleus
This answer is not correct. The nucleus has little direct effect on how much carbon dioxide
(CO2) is produced or how much the pH varies in a cell during exercise.
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Biology I End-Of-Course Practice Test Answer Key
Question 15
Reporting Category: Scientific Process
Benchmark: SC.BS.1.4 Determine the connection(s) among hypotheses, scientific evidence,
and conclusions
Answer Key: A
A scientist investigates the interaction of single-cell pond organisms S and T. He places
organism S in pond water in Container 1. He places organism T in pond water in Container 2.
He places both organisms in pond water in Container 3. The graphs show how the populations
of the organisms in the containers change over time.
What can the scientist conclude from this data about the interaction of organisms S and T?
A. Organisms S and T are neither benefited nor harmed by living together.
This answer is correct. Both Organism S and Organism T show similar population growth,
whether they are grown together or grown separately.
B. Organisms S and T both benefit from living together.
This answer is not correct. The data do not support this conclusion. In order to support it, the
data would have to show both populations of organisms increasing more when grown
together than when grown alone.
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Biology I End-Of-Course Practice Test Answer Key
C. Organism S benefits from living with organism T and organism T is harmed.
This answer is not correct. The data do not support this conclusion. In order to support it, the
data would have to show the population of Organism S increasing more when living with
Organism T than when living alone. The data would also have to show that Organism T
increases less when living with Organism S than when living alone.
D. Organism S benefits from living with organism T and organism T is unaffected.
This answer is not correct. The data do not support this conclusion. In order to support the
conclusion, the data would have to show the population of Organism S increasing more
when living with Organism T than when living alone.
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Biology I End-Of-Course Practice Test Answer Key
Question 16
Reporting Category: Scientific Process
Benchmark: SC.BS.1.8 Describe the importance of ethics and integrity in scientific
investigation
Answer Key: D
A scientist crosses a homozygous red-eyed fruit fly with a homozygous white-eyed fruit fly. He
returns to the lab several days later to record the phenotypes of the offspring in the F1
generation. While collecting his data, the scientist realizes that individuals in the F1 generation
have produced an F2 generation of offspring in the same test tube. Which statement explains
what the scientist should do next in his experiment?
A. Find similar fruit fly studies and use the data from those investigations.
This answer is not correct. Using the data from other studies would be unethical; the
investigation should be repeated to collect new data.
B. Finish collecting his data on all of the offspring and record it as the F1 generation.
This answer is not correct. Altering the data of an investigation is unethical; the investigation
should be repeated to collect new data.
C. Record half of the individuals as the F1 generation and the other half as the F2 generation.
This answer is not correct. Altering the data of an investigation is unethical. The
investigation should be repeated to collect new data.
D. Repeat the initial cross using a new pair of homozygous individuals.
This answer is correct. When there are mistakes that compromise the validity of an
investigation, the study should be repeated so that new data are collected.
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Biology I End-Of-Course Practice Test Answer Key
Question 17
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.5 Explain chromosomal mutations, their possible causes, and their
effects on genetic variation
Answer Key: D
A substitution of thymine with adenine in one DNA codon causes a particular disorder.
Which statement explains how the change in DNA leads to this disorder?
A. The deletion mutation prevents the production of the hemoglobin protein in the body.
This answer is not correct. This particular disorder is caused by a point mutation, not a base
deletion or frameshift mutation. The change in the base pair results in a different amino acid
being added to the protein chain.
B. The frameshift mutation prevents the production of several proteins found in the blood.
This answer is not correct. The same number of base pairs exists after the mutation;
therefore, the mutation is not a frameshift mutation but a point mutation. The substitution in
the base pair results in a different amino acid being added to the protein chain.
C. The insertion mutation causes extra hemoglobin proteins to attach to red blood cells.
This answer is not correct. This particular disorder is caused by a point mutation, not a base
insertion or frameshift mutation. The change in the base pair results in a different amino acid
being added to the protein chain.
D. The point mutation causes a different amino acid to be added to the hemoglobin protein.
This answer is correct. The substitution of the DNA base is a point mutation that leads to a
change in the conformation of the hemoglobin protein, causing some red blood cells to be
deformed.
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Biology I End-Of-Course Practice Test Answer Key
Question 18
Reporting Category: Scientific Process
Benchmark: SC.BS.1.9 Explain how scientific explanations must meet a set of established
criteria to be considered valid
Answer Key: C
A student is investigating the effect of temperature on the growth rate of two variations of a plant
species. One variation of the plant has broad leaves and the other has narrow leaves. The
student prepares to grow several plants of each variation at four different temperatures. She
then creates the data table shown to record the height of each plant at the end of the
experiment.
How can the student best improve the validity of the experiment?
A. Complete the experiment outside to better control the growing conditions.
This answer is not correct. While there are benefits to growing plants in their natural
environment, properly testing the hypothesis would require the student to control the
temperature at which the plants are grown.
B. Grow only one variation of the plant species to simplify the data analysis.
This answer is not correct. Experimenting on only one variation of the plant species would
not allow the student to compare the plants’ growth rates to their leaf shapes.
C. Record the heights of the plants daily throughout the length of the experiment.
This answer is correct. To calculate the growth rates of the two plant variations, the
student would need to record the heights of the plants throughout the experiment, not just at
the end of the experiment.
D. Reduce data collection by focusing on one temperature instead of several.
This answer is not correct. Having one temperature would not allow the student to
investigate how a change in temperature affects each plant's growth rate.
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Biology I End-Of-Course Practice Test Answer Key
Question 19
Reporting Category: Scientific Process
Benchmark: SC.BS.1.7 Revise, as needed, conclusions and explanations based on new
evidence
Answer Key: A
A student examines the information in the table. The student concludes that Organism W should
be placed at the base of the food web, to represent the feeding relationships in the marine
ecosystem. However, the student later learns that Organism W’s cells do not contain
chloroplasts.
Which conclusion would be appropriate, based on the new information?
A. Another organism in the marine ecosystem is a producer.
This answer is correct. The animal life in this ecosystem could not be sustained without
the existence of at least one producer. Therefore, another organism that is not listed must
be a producer.
B. Organism W is a secondary consumer in the marine ecosystem.
This answer is not correct. A secondary consumer eats primary consumers. There is no
behavioral information about Organism W to suggest it eats a non-producer organism.
C. The role of Organism T in the marine ecosystem changes during its life.
This answer is not correct. Organism T is the top consumer in this marine ecosystem. There
is no indication that Organism T changes its role during its lifetime.
D. This marine ecosystem does not have any producers.
This answer is not correct. The animal life in this ecosystem could not be sustained without
the existence of at least one producer.
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Biology I End-Of-Course Practice Test Answer Key
Question 20
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.1 Explain the theory of evolution and describe evidence that supports
this theory
Answer Key: D
Which similarity in two mammals provides the best evidence that the mammals are closely
related?
A. They have similar feeding mechanisms.
This answer is not correct. Similar feeding habits could be a result of convergent
evolution―adaptations evolving separately in unrelated organisms.
B. They have similar methods of processing ATP.
This answer is not correct. All animals have similar methods of processing ATP.
C. They live in similar habitats.
This answer is not correct. A given habitat is home for many unrelated organisms.
D. They share similar DNA base pair sequencing.
This answer is correct. Organisms that share similar DNA share a common ancestor.
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Biology I End-Of-Course Practice Test Answer Key
Question 21
Reporting Category: Scientific Process
Benchmark: SC.BS.1.6 Engage in and explain the importance of peer review in science
Answer Key: D
Corals are marine organisms that have symbiotic algae living in their tissues. The algae make
food using the energy from sunlight. Sometimes, “coral bleaching” occurs when the algae leave
the coral tissues. It is believed that different factors such as a change in temperature or
sediments blocking sunlight could cause this coral bleaching to occur.
A student plans to study the factors that could cause coral bleaching in the lab. The student
plans to grow corals in five different aquariums with different water temperatures and amounts
of light. The corals are collected from water that is 24 °C. The student’s setup is shown. During
a peer review, the student’s classmate says that it is good that the student has a control, but
advises him to change the setup before starting the experiment.
Which statement best describes how the student should change the experiment?
A. The student should expose Aquarium 3 to 25% light.
This answer is not correct. Aquarium 3 is the control in the study. It should not have the light
blocked.
B. The student should raise the temperature 2 degrees in Aquarium 3.
This answer is not correct. Aquarium 3 is the control in the study. It should not have the
temperature raised.
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Biology I End-Of-Course Practice Test Answer Key
C. The student should reduce the light and temperatures in the five aquariums.
This answer is not correct. The student should only test one variable at a time in an
aquarium. It would be best to manipulate the temperature or the light, not both.
D. The student should vary the temperatures and have 100% light in the five aquariums.
This answer is correct. The student should only test one variable at a time in an aquarium.
By only manipulating the temperature, the student could better test the effect of that
variable.
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Biology I End-Of-Course Practice Test Answer Key
Question 22
Reporting Category: Scientific Process
Benchmark: SC.BS.2.2 Compare the risks and benefits of potential solutions to technological
issues
Answer Key: C
Farmers often spray chemicals on their fields to kill weeds. These chemicals may also harm the
crops. Scientists have altered the DNA of some crops so that they will not be harmed by these
chemicals. However, some of the weeds can cross-breed with the crops.
Which statement describes a negative consequence of altering the DNA of the crops?
A. The amount of chemicals needed for the crop fields would increase.
This answer is not correct. This would not necessarily be a result of genetic engineering in
crop plants. The amount of chemicals needed may or may not increase.
B. The amount of crops grown in a field would decrease.
This answer is not correct. The crop yield would probably increase, with fewer crop plants
dying because of herbicide application.
C. The weeds could become resistant to the chemicals.
This answer is correct. When the weeds cross-breed with the genetically modified crop
plants, the weeds could also become resistant to the chemicals.
D. The weeds could become smaller than the crops.
This answer is not correct. The weeds becoming smaller than the crop plants would not be
considered a negative consequence.
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Biology I End-Of-Course Practice Test Answer Key
Question 23
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.4 Explain dynamic equilibrium in organisms, populations, and
ecosystems; explain the effect of equilibrium shifts
Answer Key: B
An ecologist studied a rare bird species within a regional wildlife preserve for several years. The
graph shows the data the ecologist collected on the birth rate, death rate, and emigration rate
for the bird species.
During which time period did the bird population experience the largest decline?
A. Between Years 0 and 1
This answer is not correct. The death rate and emigration rate are lower and the birth rate is
higher during this time period than during the period from Year 1 to Year 2.
B. Between Years 1 and 2
This answer is correct. The overall death and emigration rates are higher and the overall
birth rate lower during this time period than during any other shown on the graph. More birds
are leaving the ecosystem than are being replaced.
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Biology I End-Of-Course Practice Test Answer Key
C. Between Years 3 and 4
This answer is not correct. The death rate and emigration rate are lower and the birth rate is
higher during this time period than during the period from Year 1 to Year 2.
D. Between Years 4 and 5
This answer is not correct. The death rate and emigration rate are lower and the birth rate is
higher during this time period than during the period from Year 1 to Year 2.
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Biology I End-Of-Course Practice Test Answer Key
Question 24
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.1 Describe biogeochemical cycles within ecosystems
Answer Key: C
Which statement describes the potassium cycle?
A. Potassium is eliminated from organisms by respiration and stored in the atmosphere.
This answer is not correct. Potassium is mainly stored in the soil. It is removed through plant
harvest and runoff. Carbon is eliminated from animals through respiration.
B. Potassium is extracted from the atmosphere by animals and stored in the soil.
This answer is not correct. Potassium is mainly stored in the soil. Therefore, it cannot be
extracted from the atmosphere.
C. Potassium is leached from the soil by rain water and removed through crop harvest.
This answer is correct. Potassium is mainly stored in the soil. It is leached (removed/
washed away) from the soil through rain water. It is also removed from the soil when plants
that contain potassium are harvested.
D. Potassium is taken out of plants by bacteria and added back through decomposition.
This answer is not correct. Potassium is mainly stored in the soil. It is removed through plant
harvest and runoff. Carbon is removed from plants and added back to the atmosphere by
bacteria.
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Biology I End-Of-Course Practice Test Answer Key
Question 25
Reporting Category: Scientific Process
Benchmark: SC.BS.1.1 Describe how a testable hypothesis may need to be revised to guide a
scientific investigation
A student hypothesizes that feeding an adult goldfish more than once a day will make it grow
larger. He keeps three adult goldfish in separate 40-liter tanks at 20 °C for four weeks. He feeds
the first goldfish once a day, the second twice a day, and the third three times a day.
At the end of the experiment, the student concludes that more food does not make an adult
goldfish grow larger. He decides to test a new hypothesis and redesign the experiment.
A. Place a check mark next to one new hypothesis that the student could test.
B. According to the hypothesis you chose, set up a new experiment by placing labels in the
boxes for “Tank Size,” “Feeding Frequency,” and “Temperature.”




Place only one label in each box.
You may use a label more than once.
There may be more than one correct answer.
You do not need to use all the labels.
Answer Key:
2 points
For this item, the response correctly:
 places a check mark next to the “Goldfish grow larger in warmer water” hypothesis
AND
 designs a complete experimental setup that includes the same tank size volume and the
same feeding frequency for each tank, but a different temperature for each tank.
OR
 places a check mark next to the “Goldfish grow larger in more spacious environments”
hypothesis
AND
 designs a complete experimental setup that includes the same feeding frequency and the
same temperature for each tank, but a different volume for each tank size.
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Biology I End-Of-Course Practice Test Answer Key
1 point
For this item, the response correctly:
 places a check mark next to the “Goldfish grow larger in warmer water” hypothesis
AND
 designs a complete experimental setup that includes the same feeding frequency and the
same temperature for each tank, but a different volume for each tank size.
OR
 places a check mark next to the “Goldfish grow larger in more spacious environments”
hypothesis
AND
 designs a complete experimental setup that includes the same tank size volume and the
same feeding frequency for each tank, but a different temperature for each tank.
OR
 places a check mark next to the “Adult goldfish grow larger when fed more” hypothesis
AND
 designs a complete experimental setup that includes changing the tank size volume to 55
liters or 70 liters for each tank and/or changing the temperature to 18 °C or 22 °C for each
tank, and the same feeding frequency for each tank (i.e., changes tank size and/ or
temperature from the original experiment and the same feeding frequency).
0 points
The response is incorrect for all parts. The response does not select a new hypothesis to test
and does not design a complete experimental setup (two controlled variables and one
experimental variable).
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Biology I End-Of-Course Practice Test Answer Key
Sample Student Answer:
Explanation of Correct Answer:
In the student’s original experiment, he keeps tank size and tank temperature constant while
varying feeding frequency in order to determine whether feeding frequency will cause an adult
goldfish to grow larger. At the end of the original experiment, the student concludes that varying
feeding frequency does not make an adult goldfish grow larger, and he decides to test a new
hypothesis. A new hypothesis would be whether warmer tank temperature will cause an adult
goldfish to grow larger or a larger tank size will cause an adult goldfish to grow larger. To
properly set up an experiment to test whether warmer tank temperature will cause an adult
goldfish to grow larger, each tank must be a different temperature but the feeding frequency and
tank size must be kept constant. To properly set up an experiment to test whether a larger tank
size will cause an adult goldfish to grow larger, each tank must be a different volume but the
feeding frequency and tank temperature must be kept constant.
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Biology I End-Of-Course Practice Test Answer Key
Question 26
Reporting Category: Scientific Process
Benchmark: SC.BS.1.4 Determine the connection(s) among hypotheses, scientific evidence,
and conclusions.
Answer Key: C
A researcher conducts an experiment with fruit flies. He hypothesizes that the allele for white
eyes is dominant to the allele for red eyes. He crosses a red-eyed female with a white-eyed
male and repeats this pairing five times. All offspring from these pairs have red eyes.
What should the researcher do after obtaining these results?
A. Change his data to match his hypothesis.
This answer is not correct. It is not ethical to change data from an experiment and this would
be misleading.
B. Conclude that his data supported his hypothesis.
This answer is not correct. The results of the experiment do not support the hypothesis. If
the hypothesis were true, half or more of the offspring should have white eyes. The
researcher should not conclude that his hypothesis is correct.
C. Record his data and conclude that his hypothesis is not supported.
This answer is correct. Data should always be reported and interpreted so that a
conclusion can be made, even if the conclusion contradicts the hypothesis of the
experiment.
D. Revise his hypothesis to match his data.
This answer is not correct. It is not ethical to change a hypothesis so that it is supported by
the data from an experiment.
Page 34 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 27
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.3 Explain how matter and energy flow through living systems and the
physical environment.
Answer Key: B
In the ocean, most primary producers are found in the top zone of water that stretches from the
surface down to a maximum depth of 200 meters. In this zone, sunlight is plentiful, so the
growth of primary producers is mostly limited by the amount of nutrients available in the water.
The source of these nutrients is decaying plant and animal matter. Nutrients are released into
the water as this matter sinks to the bottom of the ocean.
According to the information above, which statement best explains why vertical mixing of water
is required for efficient cycling of matter through the marine ecosystem?
A. Bottom-dwelling consumers have difficulty reaching the zone where the primary producers
grow.
This answer is not correct. Bottom-dwelling consumers do not need access to primary
producers in the top zone of the ocean.
B. Decomposers act on dead consumers at levels deeper than where the primary producers
are found.
This answer is correct. The mixing of the deep water with the surface water brings
nutrients from decaying matter that has sunk to the bottom up to the surface, where primary
producers can use them.
C. Large consumers such as fish need to live at the same level as the decomposers on which
they feed.
This answer is not correct. Fish and other consumers do not need to live on the same level
as the decomposers.
D. Photosynthetic bacteria can produce nutrients only at levels that have limited amounts of
sunlight.
This answer is not correct. Photosynthetic bacteria need sunlight in order to grow and
produce nutrients.
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Biology I End-Of-Course Practice Test Answer Key
Question 28
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.4 Explain dynamic equilibrium in organisms, populations, and
ecosystems; explain the effect of equilibrium shifts
Answer Key: D
In 1935, the cane toad was introduced into Australia to control the population of crop-damaging
insects. The cane toad population has increased steadily and rapidly since its introduction.
Scientists are now trying to slow the growth of the cane toad population.
Which statement could explain the steady increase in Australia’s cane toad population?
A. The cane toad lays few eggs.
This answer is not correct. The cane toad lays many eggs. However, if it were true that the
cane toad lays few eggs, this would not suggest a sustained, rapid population increase.
B. The cane toad lives in few habitats.
This answer is not correct. The number of different habitats is less predictive of a sustained,
rapid increase than the abundance of those habitats. The cane toad prefers open grassland,
which is abundant in Australia.
C. The cane toad has few food sources.
This answer is not correct. The cane toad has many food sources. However, if it were true
that the cane toad has few food sources, this would not suggest a sustained, rapid
population increase.
D. The cane toad has few natural predators.
This answer is correct. The cane toad has few natural predators in Australia; thus, an
important limiting factor for its population growth is missing.
Page 36 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 29
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.3 Differentiate between the processes of mitosis and meiosis.
The stages of mitosis are shown.
Place the label naming the correct stage of mitosis in each blank box.


You should use each label only once.
Each blank box should be filled with a label.
Answer Key:
3 points
For this item, a full-credit response includes
 the prophase label in the first box
AND
 the metaphase label in the second box
AND
 the anaphase label in the third box.
2 points
For this item, a partial-credit response includes
 the prophase label in the first box
AND
 the metaphase label in the second box
OR
 the anaphase label in the third box.
OR



the anaphase label in the third box
AND
the metaphase label in the second box
OR
the prophase label in the first box.
OR



the metaphase label in the second box
AND
the prophase label in the first box
OR
the anaphase label in the third box.
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Biology I End-Of-Course Practice Test Answer Key
1 point
For this item, a partial-credit response includes
 the prophase label in the first box
OR
 the metaphase label in the second box
OR
 the anaphase label in the third box.
Sample Student Answer:
Explanation of Correct Answer:
The prophase is the first stage of cell division where the chromosomes begin to coil and the
spindle apparatus begins to form as centrosomes move apart. The metaphase is the stage
where the chromosomes become aligned on a plane. The anaphase is the stage where the
chromatids separate and move toward opposite ends of the cell.
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Biology I End-Of-Course Practice Test Answer Key
Question 30
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.4 Explain how Mendel's laws of heredity can be used to determine the
traits of possible offspring
Answer Key: C
A type of rodent can have white or brown fur. The allele for white fur is recessive to the allele for
brown fur.
What is the genotype of an individual with white fur?
A. Heterozygous dominant
This answer is not correct. A heterozygous individual would have one white fur allele and
one brown fur allele. Since the brown fur allele is dominant, the organism would have brown
fur.
B. Heterozygous recessive
This answer is not correct. A heterozygous individual would have one white fur allele and
one brown fur allele. Since the brown fur allele is dominant, the organism would have brown
fur.
C. Homozygous recessive
This answer is correct. White fur is controlled by a recessive allele which means that an
individual must have two copies of the recessive allele in order to have white fur.
D. Homozygous dominant
This answer is not correct. A homozygous dominant individual would have brown fur
because it would have two brown fur alleles.
Page 39 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 31
Reporting Category: Unity and Diversity
Benchmark: SC.BS.5.5 Explain chromosomal mutations, their possible causes, and their
effects on genetic variation
Answer Key: B
Answer Key (Part II): C
Part A
A wild sheep is observed to have a number of physical ailments. A geneticist takes a cell
sample from the sheep and produces the karyotype shown to investigate the cause of these
ailments. The geneticist labels four parts of the karyotype as K, L, M, and N.
What is the genetic cause of the sheep’s physical ailments, based on information shown in the
karyotype?
A. The sheep did not receive any dominant alleles from its parents.
This answer is not correct. Karyotypes do not reveal information about the alleles for
different traits.
B. The sheep did not receive the correct combination of chromosomes from its parents.
This answer is correct. This diploid sheep has three of chromosome 14; it should only
have two - one from each parent.
C. The sheep is infected by a virus that is removing essential genes from its DNA.
This answer is not correct. The karyotype does not contain information about viral infections.
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Biology I End-Of-Course Practice Test Answer Key
D. The sheep experienced a new selection pressure, which caused multiple insertion
mutations.
This answer is not correct. Selection pressures do not cause mutations to occur.
Part B
Which part of the karyotype provides evidence to support your choice in part A?
A. K
This answer is not correct. The bent structure of the chromosome is not a sign of disease or
genetic problems.
B. L
This answer is not correct. The small size of these chromosomes is not a sign of disease or
genetic problems.
C. M
This answer is correct. A third copy of a chromosome in diploid organisms is a common
cause of genetic disease and associated physical ailments.
D. N
This answer is not correct. The X/Y chromosome pair shows that this sheep is a male.
Page 41 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 32
Reporting Category: Scientific Process
Benchmark: SC.BS.1.8 Describe the importance of ethics and integrity in scientific
investigation
A teacher presents her students with a passage about ethics in research.
Select each sentence in the passage that is an example of an unethical research practice.
Ethics Passage
A research group submits a paper for publication that describes a new air filter they made. They
included a detailed procedure so that others could replicate their data. Their methods section
states that there were ten trials completed. The group did five trials and duplicated their data.
The results section of the paper discusses positive outcomes, including a large decrease in the
level of pollutants in the air. The group did not include any of their negative results because they
felt that more people would read positive results. They left out data showing how easily the filter
mildewed and rusted.
The reviewers of the article asked for additional data, which the authors had not collected. The
authors used data from another group’s study without proper citation.
The reviewers of the paper noticed the unethical behavior of the authors and rejected the paper.
Page 42 of 61
Biology I End-Of-Course Practice Test Answer Key
Answer Key:
For this item, a full-credit response includes




“The group did five trials and duplicated their data.” selected
AND
“The group did not include any of their negative results because they felt that more
people would read positive results.” selected
AND
“They left out data showing how easily the filter mildewed and rusted.” selected
AND
“The authors used data from another group’s study without proper citation.” selected.
Sample Student Answer:
Explanation of Correct Answer:
Duplicating data and reporting that twice as many trials were run is an unethical practice. It is
also unethical to not include any negative results, no matter the effect on readership. In this
case, the data that was left out had negative implications and should have been included in the
research paper. It is unethical to use data from another person’s research study without proper
citation.
Page 43 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 33
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.1 Describe biogeochemical cycles within ecosystems
The table shows several biogeochemical cycles.
Select the boxes to match each cycle with the part of Earth that the cycle occurs in.
Answer Key:
Explanation of Correct Answer:
Plants photosynthesize carbon dioxide from the air into glucose that is stored inside the plant.
When the plant decays in the soil, carbon dioxide is released again into the atmosphere. Carbon
dioxide also is found dissolved in water where marine plants photosynthesize it into glucose.
Nitrogen gas can be found in the air. Plants process nitrogen found in the soil and nitrogen can
be found in water in the form of nitrates as a result of fertilizer runoff. Phosphorus can be found
in soil and plants uptake the phosphorus. Phosphorus can enter the water due to runoff.
Page 44 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 34
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.5 Describe the components and functions of a variety of macromolecules
active in biological systems
Answer Key: B, D, E
The diagram shows a particle moving into a cell membrane.
Select all of the lettered particles in the diagram.
A. Amino acid
This answer is not correct. Although proteins are composed of amino acids, an amino acid is
not shown in the diagram.
B. Carbohydrate
This answer is correct. The carrier within the cell membrane helps to facilitate the diffusion
of glucose into the cell.
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Biology I End-Of-Course Practice Test Answer Key
C. Nucleic acid
This answer is not correct. Neither DNA nor RNA is shown in the diagram.
D. Phospholipid
This answer is correct. The phospholipid bilayer comprises the cell membrane.
E. Protein
This answer is correct. Proteins within the phospholipid bilayer facilitate the diffusion of
glucose into the cell.
F. Water
This answer is not correct. Although water can be found both inside and outside cells, this
diagram does not show or label a water molecule.
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Biology I End-Of-Course Practice Test Answer Key
Question 35
Reporting Category: Organisms and the Environment
Benchmark: SC.BS.3.1 Describe biogeochemical cycles within ecosystems
Answer Key: A
Which of the following is a gas that soil bacteria convert into a form that plants can use?
A. Nitrogen
This answer is correct. Soil bacteria are an active part of the nitrogen cycle and convert
nitrogen into a form that can be used by plants.
B. Carbon dioxide
This answer is not correct. Carbon is released, not converted, as dead organisms are
broken down.
C. Oxygen
This answer is not correct. Plants convert carbon dioxide into oxygen.
D. Water vapor
This answer is not correct. Water vapor is a stage in the water cycle. Bacteria are not a part
of the water cycle.
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Biology I End-Of-Course Practice Test Answer Key
Question 36
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.1 Explain the theory of evolution and describe evidence that supports
this theory
Answer Key: D
The skeletal structures of a crocodile leg and a mouse leg are shown.
What can be hypothesized about crocodiles and mice, based on the skeletal anatomy of these
two legs?
A. They have long life spans.
This answer is not correct. Similarities in bone structure provide no information about the life
spans of these species.
B. They have similar diets.
This answer is not correct. Similarities in bone structure provide no information about these
species' diets.
C. They move at a similar speed.
This answer is not correct. Similarities in bone structure provide no information about the
speed at which a species moves.
D. They share a common ancestor.
This answer is correct. The similar skeletal structures are homologous and are evidence
that the crocodile and mouse have a common ancestor.
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Biology I End-Of-Course Practice Test Answer Key
Question 37
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.4 Explain how Mendel's laws of heredity can be used to determine the
traits of possible offspring
Answer Key: C
A female carrier of colorblindness and a male who is not colorblind produce a son.
Given the fact that colorblindness is a recessive X-linked trait, what is the chance that the son
will be colorblind?
A.
0%
This answer is not correct. Since all males inherit an X chromosome from their mothers, the
son does have a chance of inheriting the recessive allele.
B.
25%
This answer is not correct. The son will inherit one of the mother’s X chromosomes, so he
will have a 50% chance of being colorblind.
C.
50%
This answer is correct. Since the mother is a carrier, only one of her X chromosomes will
have the recessive allele. Since the son will receive a copy of one of her X chromosomes,
he has a 50% chance of being affected by the recessive trait.
D. 100%
This answer is not correct. Since one of the mother’s X chromosomes does not have the
colorblind allele, there is a chance that the son will not be colorblind.
Page 49 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 38
Reporting Category: Structure and Functions in Organisms
Benchmark: SC.BS.4.4 Describe how homeostatic balance occurs in cells and organisms
Two diagrams of a carrier protein and cell membrane are shown.
A. In the Active Transport diagram, place an appropriate concentration of molecules in the
“Concentration” boxes on each side of the cell membrane to show active transport.
Then, place an arrow in the “Net flow” box to show the direction of active transport
corresponding to these concentrations.
B. In the Passive Transport diagram, place an appropriate concentration of molecules in
the “Concentration” boxes on each side of the cell membrane to show passive transport.
Then, place an arrow in the “Net flow” box to show the direction of passive transport
corresponding to these concentrations.


There may be more than one correct answer.
Place only one object in each box.
Answer Key:
2 points
For this item, a full-credit response includes


in the Active Transport diagram, a low molecule concentration on one side of the
membrane AND a high molecule concentration on the other side of the membrane AND
an arrow pointing from the low molecule concentration to the high molecule
concentration (1 point)
AND
in the Passive Transport diagram, a low molecule concentration on one side of the
membrane AND a high molecule concentration on the other side of the membrane AND
an arrow pointing from the high molecule concentration to the low molecule
concentration (1 point).
Page 50 of 61
Biology I End-Of-Course Practice Test Answer Key
1 point
For this item, a partial-credit response includes


in the Active Transport diagram, a low molecule concentration on one side of the
membrane AND a high molecule concentration on the other side of the membrane AND
an arrow pointing from the low molecule concentration to the high molecule
concentration (1 point)
OR
in the Passive Transport diagram, a low molecule concentration on one side of the
membrane AND a high molecule concentration on the other side of the membrane AND
an arrow pointing from the high molecule concentration to the low molecule
concentration (1 point).
Sample Student Answer:
Explanation of Correct Answer:
Molecules move from a lower concentration to a higher concentration during active transport.
Molecules move from a higher concentration to a lower concentration during passive transport.
Page 51 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 39
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.5.1 Explain the theory of evolution and describe evidence that supports
this theory
The allele for brown fur is dominant to the allele for white fur in two different rabbit populations.
Each rabbit population lives in this snowy ecosystem.
Place an immigrating organism in the response box in the Snowy Ecosystem that would provide
a selective pressure favoring the recessive fur allele.

Place only one object in the box.
Answer Key:
For this item, a full-credit response includes

only the “Fox” in the Immigrating Organism box (1 point).
Sample Student Answer:
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Biology I End-Of-Course Practice Test Answer Key
Explanation of Correct Answer:
Brown fur is a dominant trait in fur color, but the environment for the rabbits is white. The fox is a
predator of rabbits and the white fox has an advantage of being able to hide in its surroundings.
This would lead to brown rabbits being more obvious to the foxes and therefore preyed upon
more often than the white rabbits. The white fur recessive allele would then be favored.
Page 53 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 40 Simulation
Students are investigating cell behavior in different concentrations of starch solutions. To do
this, the students construct cell models using a dialysis tube and different concentrations of
starch solutions.
They investigate both the concentration inside the cell model and the concentration outside the
cell model. At the end of the investigation, the students conduct an iodine test to test for the
presence of starch.
Using the lowest number of trials, set up an experiment to:
A. Determine the solution that is isotonic to a given solution in a dialysis tube.
B. Determine the solution that is hypertonic to a given solution in a dialysis tube.
C. Determine the solution that is hypotonic to a given solution in a dialysis tube.
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Biology I End-Of-Course Practice Test Answer Key
Explanation of How to Use Simulation:
A student can choose one of four Solutions from the “Solution Dialysis Tube” or “Solution in
Beaker” drop-down menus:
 Distilled Water

7% Starch Water
 14% Starch Water
 20% Starch Water
After choosing the solution for the tube and another for the beaker, the student presses the
green “Start” button. An animation of the activity plays and the output table populates with the
information shown, depending on the solutions chosen.
Dialysis Tube
Solution
Distilled Water
Beaker Solution
Distilled Water
Distilled Water
7% Starch Water
Distilled Water
14% Starch Water
Distilled Water
20% Starch Water
7% Starch Water
Distilled Water
7% Starch Water
7% Starch Water
7% Starch Water
14% Starch Water
Observations
No change in mass or
volume of the cell
model. Beaker tests
negative for starch.
The mass and volume
of the cell model
decrease. Beaker
tests positive for
starch.
The mass and volume
of the cell model
decrease. Beaker
tests positive for
starch.
The mass and volume
of the cell model
decrease. Beaker
tests positive for
starch.
The mass and volume
of the cell model
increase. Beaker tests
negative for starch.
No change in mass or
volume of the cell
model. Beaker tests
positive for starch.
The mass and volume
of the cell model
Iodine Test Results
for Dialysis Tube
Orange yellow; tests
negative for starch.
Orange yellow; tests
negative for starch.
Orange yellow; tests
negative for starch.
Orange yellow; tests
negative for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
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Biology I End-Of-Course Practice Test Answer Key
7% Starch Water
20% Starch Water
14% Starch Water
Distilled Water
14% Starch Water
7% Starch Water
14% Starch Water
14% Starch Water
14% Starch Water
20% Starch Water
20% Starch Water
Distilled Water
20% Starch Water
7% Starch Water
20% Starch Water
14% Starch Water
20% Starch Water
20% Starch Water
decrease. Beaker
tests positive for
starch.
The mass and volume
of the cell model
decrease. Beaker
tests positive for
starch.
The mass and volume
of the cell model
increase. Beaker tests
negative for starch.
The mass and volume
of the cell model
increase. Beaker tests
positive for starch.
No change in the
mass or volume of the
cell model. The
beaker tests positive
for starch.
The mass and volume
of the cell model
decrease. The beaker
tests positive for
starch.
The mass and volume
of the cell model
increase. Beaker tests
negative for starch.
The mass and volume
of the cell model
increase. Beaker tests
negative for starch.
The mass and volume
of the cell model
increase. Beaker tests
negative for starch.
No change in the
mass and volume of
the cell model. Beaker
tests positive for
iodine.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
Black; tests positive
for starch.
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Biology I End-Of-Course Practice Test Answer Key
Example of Student Answer:
The student selects 7% Starch Water for the Solution in Dialysis Tube input and runs trials using
Distilled Water, 7% Starch Water, and 14% Starch Water as Solution in Beaker input.
Explanation of Correct Answer:
A solution is hypertonic to another solution when the solution concentration in the beaker is
higher than the solution concentration on the inside of the dialysis tubing. Water will rush into
the dialysis tubing causing it to swell in volume. A solution is hypotonic to another solution when
the solution concentration in the beaker is lower than the solution concentration on the inside of
the dialysis tube. Water will flow out of the membrane tubing into the beaker and the dialysis
tube will shrink. A solution is isotonic to another solution when the solution concentration inside
the dialysis tube is the same as the solution concentration in the beaker. The iodine test turns
black when starch is present, and this is used by the student to determine that only water is
moving across the membrane and not starch. The minimum number of trials is three to verify a
solution that is isotonic, hypertonic, and hypotonic to the solution in the dialysis tube.
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Biology I End-Of-Course Practice Test Answer Key
Question 41 Simulation
Students are investigating the concept of tonicity in the laboratory. To do this, the students use
two types of cells and four concentrations of salt solutions.
They observe the behavior of cells under a microscope and record their observations in the data
table.
Explanation of How to Use Simulation:
The student can choose one of two cell types from the drop-down menu:
 Red Blood
 Elodea
The student can choose one of four solutions from the drop-down menu:
 Distilled Water

1% NaCl

5% NaCl
 20% NaCl
After choosing a cell type and a solution, the student presses the green “Start” button. An
animation of the process plays and the output table populates with the information shown,
depending on the cell type and solution chosen.
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Biology I End-Of-Course Practice Test Answer Key
Cell Type
Red Blood
Solution
Distilled Water
Red Blood
1% NaCl
Red Blood
Red Blood
Elodea
6% NaCl
20% NaCl
Distilled Water
Elodea
1% NaCl
Elodea
6% NaCl
Elodea
20% NaCl
Observations
The cells get larger and some
cells burst.
The cells do not change in
size.
The cells get smaller.
The cells shrivel.
The cell walls expand and the
cell membranes expand.
The cell walls keep their
shape and the cell
membranes do not change in
size.
The cell walls keep their
shape, the cell membranes
get smaller, and the
chloroplasts get closer
together.
The cell walls keep their
shape, the cell membranes
get smaller and shrink, and
the chloroplasts get really
close together.
Page 59 of 61
Biology I End-Of-Course Practice Test Answer Key
Question 42
Reporting Category: Diversity, Genetics, and Evolution
Benchmark: SC.BS.4.4 Describe how homeostatic balance occurs in cells and organisms
Answer Key: B, D, E, F
Select all of the statements that are conclusions that are supported by the data collected in the
investigation.
A. An Elodea cell is healthiest when it is placed in an isotonic salt solution.
This answer is not correct. The cell membrane has a large gap in an isotonic solution. Plant
cells are healthiest when placed in hypotonic solutions so that they are turgid.
B. Elodea and blood cells both have approximately the same salt concentration inside them.
This answer is correct. Both cell types have similar salt concentrations inside the cell. This
is evidenced by the cells not changing in the same concentration of salt water.
C. Hypertonic solutions cause all cell types to burst.
This answer is not correct. Neither Elodea nor red blood cells burst in hypertonic solutions.
They would shrivel in a hypertonic solution.
D. The concentration of salt inside of an Elodea cell decreases when it is placed into a
hypotonic salt solution.
This answer is correct. A hypotonic solution has less solute than the inside of the cell. This
causes water to rush inside the cell and the concentration inside of the cell to decrease.
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Biology I End-Of-Course Practice Test Answer Key
E. The concentration of the salt outside a blood cell increases if the solution is hypotonic to the
cell.
This answer is correct. A hypotonic solution has less solute than the inside of the cell. This
causes water to rush into the cell, increasing the concentration of the solution outside the
cell.
F. Water leaves a blood cell when it is placed in a hypertonic salt solution.
This answer is correct. A hypertonic salt solution has a higher concentration of solute
outside the cell. Water leaves the cell to maintain homeostasis.
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