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WorksheetsEOC practice with explained answers
Total questions: 24
Worksheet time: 48mins
Students modeled the changes in cells during mitosis, using paper plates, flat wooden sticks, cotton
swabs, and construction paper.
Which statement correctly uses the model to explain how mitosis maintains genetic continuity?
A. The chromosomes in cell 1 are the same as in cells 6 and 7.
B. Crossing-over occurs in cell 4, which increases the genetic diversity in cells 6 and 7.
C. When the nuclear membrane reforms in cell 5, each nucleus becomes diploid in number.
D. The independent assortment that is represented in cell 2 ensures that cell 3 has the correct number of
chromosomes.
Scientists recently developed a technique for synthesizing proteins outside of cells. These cell-free systems are used to produce pure proteins that can be used in many ways. One use is in the manufacturing of medicine.
Do cell-free systems the cost of producing some medicines so that more the medicine?
increase;
people can afford to buy
decrease;
people can afford to buy
increase; companies can profit from
decrease;
companies can profit from
Geographic isolation caused the separation of rainforest frog populations into a population in the
north and a population in the south. The separated populations later reconnected because the
climate became wetter and warmer, causing the rainforest to expand. When males from the north
mated with females from the south, the offspring failed to develop past the tadpole stage. When
males from the south mated with females from the north, the offspring developed more slowly than
the offspring of pairs of northern frogs. Based on these data, which event occurred while the two
populations of frogs were separated?
A. The two populations developed into new species.
B. The two populations mated with other species of frogs.
C. The two populations began a new method of reproduction.
D. The two populations had fewer offspring than before the separation.
Scientists studied a species of phytoplankton. These phytoplankton are an important part of marine
food webs and are major primary producers. As all organisms are regulated by their environment,
so are the phytoplankton. The scientists looked at the relationship between the populations of the
phytoplankton and viruses found in the same environment. They created three environments in a
laboratory setting to collect data on the numbers of phytoplankton and viruses over a period of time
and graphed the results once they were averaged, as shown below.
The scientists claimed that the growth and stability of the phytoplankton populations were affected by
the viruses. Using the information given, which TWO arguments support this claim?
A. The phytoplankton population was unable to absorb the light necessary for growth because the viruses
covered the surface of the water.
B. The phytoplankton population was affected by the viruses because the viruses were competitors for the
food sources in the environment.
C. As the phytoplankton population reached capacity, the number of viruses began to increase because
the phytoplankton were the hosts to the viruses and replicated the viruses’ genome.
D. The phytoplankton population was affected by the increase in the number of viruses in the environment
because the viruses used most of the carbon found in the environment.
E. The phytoplankton population decreased as the number of viruses increased because the cells of
the phytoplankton were destroyed as the viruses used them to increase the number of viruses in the
environment.
Today, the common type of banana we buy and eat is a Cavendish banana. They arose from chance
mutants that were produced sexually from wild banana plants. The Cavendish banana is infertile and
can only be produced by cloning from root shoots. Large commercial growers worldwide now plant
only the mutant type. Some information about both types of banana is recorded in the table.
Part A
Why are scientists warning that exclusively growing this mutant type by asexual reproduction presents
a serious disadvantage?
A. The loss of an adequate Cavendish banana seed supply could result in extinction of this variety.
B. The changes in characteristics from a parent plant to a clone will produce inconsistent plants that are
less healthy.
C. The lack of genetic variability among clones puts the whole species at increased risk of extinction
through a catastrophic disease or pest.
D. The increasing number of homologous sets of chromosomes with each successive generation of clones
will eventually result in widespread death of banana plants.
Today, the common type of banana we buy and eat is a Cavendish banana. They arose from chance
mutants that were produced sexually from wild banana plants. The Cavendish banana is infertile and
can only be produced by cloning from root shoots. Large commercial growers worldwide now plant
only the mutant type. Some information about both types of banana is recorded in the table.
Part B
Growers on large banana farms that supply food commercially have chosen to limit their plantings
exclusively to Cavendish banana plants. What advantage is likely cited by the growers for continued
planting of these asexually produced crops year after year?
A. Seedless cloned plants are not damaged by disease and pest organisms.
B. Successive generations of clones produce larger bananas and healthier plants.
C. The cloned banana plants rapidly adapt to extreme environmental changes due to their limited genetic
variation.
D. The bananas produced maintain consistent characteristics in quality, taste, and appearance from one
crop of clones to the next.
Bromothymol blue (BTB) is a pH indicator that is also used to detect carbon dioxide (CO2). BTB is blue
when pH is basic and CO2 is low. BTB is yellow when pH is acidic and CO2 is high. BTB is green when
pH is neutral. A group of students are planning to perform an investigation in which they will place
either a stalk of the aquatic plant elodea or a snail in a test tube that contains water with a neutral
pH of 7 and BTB. The students will also include a test tube that contains elodea and a snail. Observing
color change once the tubes have been placed under a growth light for several hours will allow the
students to answer which TWO of the following questions?
A. Do both elodea and snails require oxygen to survive?
B. Does photosynthesis performed by elodea remove CO2 from the water?
C. Does cellular respiration occur at a higher rate than photosynthesis in the tube with only elodea?
D. Do snails respire faster when placed in a tube with elodea?
The ribosome of the bacterium E. coli includes the ribosomal protein L4 (rpl4). The rpl4 gene carries
the instructions for making rpl4 protein. Which of the following arguments provides support for the
claim that E. coli has a common ancestor with all other organisms?
A. Every organism depends on proteins to carry out essential cellular processes. Ribosomes are needed
by all organisms to synthesize proteins such as rpl4.
B. Every organism possesses in its ribosome a protein that is similar to rpl4. This protein has an amino
acid sequence that is similar to the sequence of E. coli’s rpl4.
C. Every organism contains a structure that is similar to a ribosome. This structure helps convert the
instructions from the rpl4 gene into amino acids.
D. Every organism has proteins made of amino acids. The code for amino acids is the same in E. coli
because the instructions for amino acids come from the DNA. DNA contains the same components in
all organisms.
Heterozygous; codominance;
Partly expressed
Heterozygous; incomplete dominance;
Partly expressed
Heterozygous; codominance;
completely expressed
Homozygous; codominance;
Partly expressed
A male and female have a child that has three copies of chromosome 18. Although both parents are
unaffected, their doctor claims that the disorder associated with having an extra chromosome 18 is
the result of a chromosomal mutation in cells that carry inherited material. Which argument supports
this claim?
A. A mutation occurred when crossing over caused chromosome 18 to be replicated twice during meiosis,
allowing one parent to donate two copies of chromosome 18 to the child.
B. A nondisjunction mutation was caused by the improper separation of the genetic material during
meiosis, allowing the gamete of one parent to donate two copies of chromosome 18 to the child.
C. A substitution mutation during replication allowed the genetic material of chromosome 18 to
replace the genetic material of a nearby chromosome, causing the child to have three copies of
chromosome 18.
D. An insertion mutation during replication allowed the genetic material of chromosome 18 to be inserted
into the genetic material of another chromosome, causing three copies of chromosome 18 to be
made.
red-backed shrike;
highly;
highest;
natural selection;
increasing
reed-warbler shrike;
highly;
highest;
natural selection;
increasing
brambling shrike;
poorly;
highest;
natural selection;
increasing
red-backed shrike;
highly;
highest;
genetic drift;
increasing
Which of the following cladograms BEST represents these data?
A genome is the complete set of genetic material present in a species. Genome analysis usually includes the study of both the structure and the function of genes. Analysis of the structure involves determining the sequence of all genes in the chromosome of individuals, and analysis of the function involves determining how the genes are expressed in an individual
Part A
Analysis of genome structure and function helps researchers understand how the products of genes interact during the of individuals of the species
Part B
Understanding the structure and function of the genomes of species helps researchers determine
amino acids;
mutation;
how to prevent genetic mutations
proteins; development;
how species are related to each other
nucleic acids;
evolution;
why some species' traits don't change over time
proteins;
mutation;
why some genes are passed on and other are not
The model shows part of a process that uses tRNA.
Which description explains the role of the tRNA in the process shown in this model?
A. The tRNA delivers amino acids to the ribosome so that they can be added to the developing peptide.
B. The tRNA recognizes the stop codon of a developing peptide so that no new amino acids are added.
C. The tRNA signals the release of the peptide from the ribosome once all of the amino acids have been
added.
D. The tRNA scans the developing peptide to make sure that the sequence of the amino acids matches
the mRNA.
The diagram represents a model of how bacteria become resistant to an antibiotic, allowing bacteria
to survive treatment.
Which BEST explains how the indicated step in the model allows bacteria to develop resistance?
A. Genetic mutations that promote resistance occur.
B. The bacteria are infected by viruses that confer resistance.
C. Alleles for antibiotic resistance become dominant over recessive alleles.
D. A portion of the genetic material is re-replicated, allowing for resistance.
Hypotonic -> Hypertonic
Hypertonic -> Hypotonic
Isotonic -> Hypertonic
Isotonic->Hypotonic
Hemoglobin is a protein found in the red blood cells of vertebrates and in the plasma of many
invertebrates. The function of this protein is to transport oxygen throughout the body and to bring
carbon dioxide back to be expelled from the organism. If the amino acid sequence of the protein is
altered, the mutated protein is not as efficient at carrying oxygen as is the normal hemoglobin. Which
argument is supported by this information?
A. The mutated hemoglobin protein can still carry carbon dioxide to be expelled from the organism.
B. Hemoglobin must be a simple molecule because it is found in both vertebrates and invertebrates.
C. Structural changes of hemoglobin affect its ability to carry oxygen, indicating that the shape of a
protein is important to its function.
D. Normal hemoglobin must be a larger molecule than the mutated hemoglobin since it has sufficient
space to attach to and carry both oxygen molecules and carbon dioxide molecules.
Nutrient sources;
Presence of Algae
Presence of Algae;
Water Temperatures
Nutrient sources;
Water Temperatures
Wind and water currents; Presence of the Algae
Reduce fertilizer runoff into rivers and oceans; Reduce overfishing in rivers and oceans
Reduce the temperature of wastewater release into rivers and oceans; Reduce fertilizer runoff into rivers and oceans
Reduce overfishing in rivers and oceans; Reduce the use of nonrenewable energy sources
Reduce the use of nonrenewable energy sources; Reduce overfishing in rivers and oceans
Tropic latitudes experience less seasonal changes in climate; There is a large amount of annual precipitation in tropic latitudes; Regions near the equator have a longer growing season for plants;
There is a large amount of annual precipitation in tropic latitudes; There is less land in temperate and polar latitudes; Regions near the equator have a longer growing season for plants;
Regions near the equator have a longer growing season for plants; There is a large amount of annual precipitation in tropic latitudes
There is less land in temperate and polar latitudes; Regions near the equator have a longer growing season for plants; There is a large amount of annual precipitation in tropic latitudes
Claim 1:
Homologous chromosomes seperate incorrectly, producting daughter cells with unequal amounts of genetic material; Daughter cells have too much or too little genetic material
Claim 2:
Daughter Cells are all haploid but are not genetically identical;
Homologous chromosomes exchange genetic material during meiosis
Claim Claim 1:
Homologous chromosomes seperate incorrectly, producting daughter cells with unequal amounts of genetic material; Daughter cells have too much or too little genetic material
Claim 2:
Daughter Cells are all haploid but are not genetically identical;
Deletions during replication are passed to daughter cells
Claim 1:
Homologous chromosomes seperate incorrectly, producting daughter cells with unequal amounts of genetic material; Daughter cells have too much or too little genetic material
Claim 2:
Heritable mutations caused by environmental factors result in the loss of entire chromosomes;
Homologous chromosomes exchange genetic material during meiosis
Claim 1:
Homologous chromosomes seperate incorrectly, producting daughter cells with unequal amounts of genetic material; Daughter cells have too much or too little genetic material
Claim 2:
Daughter Cells are all haploid but are not genetically identical;
Heritable mutations caused by environmental factors result in the loss of the entire chromosomes.
A group of researchers gathered data from 40 forest sites and determined that forests with many different species of trees are loss affected by drought (long periods of dry conditions) than forests with few or only one tree species. Since different species of trees have traits that cause them to use and move water in different ways, they are not equally impacted by dry conditions.
has thick bark;
has deep roots;
has large leaves
produces fleshy fruit;
sheds and grows new leaves annually;
has needles instead of large; flat leaves
produces fleshy fruit; has thick bark; has large leaves
has thick bark; has deep roots; has needles instead of large, flat leaves
Part A
Individual II1 must be a male and must have the trait (shaded) since he has offspring with the trait and the
female II2 does not have the trait.
Individual II8 must be a female and must have the trait (shaded) since she has offspring with the trait and
the male II7 does not have the trait.
Individual II4 must be a female without the trait because the offspring do not have the trait.
Individual III4 must be the remaining male without the trait. This is confirmed since neither of his parents
have the trait.
Part B
50% gg
Part A
Individual II1 must be a male and must have the trait (shaded) since he has offspring with the trait and the
female II2 does not have the trait.
Individual II8 must be a female and must have the trait (shaded) since she has offspring with the trait and
the male II7 does not have the trait.
Individual II4 must be a female without the trait because the offspring do not have the trait.
Individual III4 must be the remaining male without the trait. This is confirmed since neither of his parents
have the trait.
Part B
50% GG
Part A
Individual II1 must be a male and must have the trait (shaded) since he has offspring with the trait and the
female II2 does not have the trait.
Individual II8 must be a female and must have the trait (shaded) since she has offspring with the trait and
the male II7 does not have the trait.
Individual II4 must be a female without the trait because the offspring do not have the trait.
Individual III4 must be the remaining male without the trait. This is confirmed since neither of his parents
have the trait.
Part B
75% Gg
Part A
Individual II1 must be a male and must have the trait (shaded) since he has offspring with the trait and the
female II2 does not have the trait.
Individual II8 must be a female and must have the trait (shaded) since she has offspring with the trait and
the male II7 does not have the trait.
Individual II4 must be a female without the trait because the offspring do not have the trait.
Individual III4 must be the remaining male without the trait. This is confirmed since neither of his parents
have the trait.
Part B
25% gg
A;
at
B;
after
C;
at
D;
after
