Made by: Shahd A.Gaber
(Test 6-1)
Accepted classification systems of life do not include viruses. Although viruses possess
certain features of cellular organisms, including genetic material that codes for making
new viral particles, they cannot replicate (make copies of) themselves without first
infecting a living cell. Biologists agree that viruses originated from genetic material
called nucleic acid, but it is difficult to prove any single theory regarding how this
occurred. Three hypotheses of viral origin are presented here.
Coevolution Hypothesis
Some biologists argue that viruses evolved alongside other organisms over billions of
years. They suggest that simple molecules of ribonucleic acid (RNA), a nucleotide that
forms the genetic code for proteins, joined to form more complex sequences. These
RNA sequences developed enzyme-like abilities including the ability to self-replicate
and insert themselves into other nucleotide sequences. While some RNA sequences
became incorporated into membrane-bound cells, others were packaged inside proteins
as the first viral particles that could replicate after infecting cellular organisms (see
Figure 1).
Figure 1
Cellular Origin Hypothesis
Some biologists claim that nucleotide sequences within prokaryotic (non-nucleated)
and eukaryotic (nucleated) cellular organisms incorporated into a protein coating and
escaped from the cell as a viral particle. Initially, DNA or RNA nucleotide sequences
gained the code required for other cells to replicate them. Next, these sequences
associated with proteins to form an outer capsid. Finally, the virion (viral particle)
became capable of passing through the cell membrane and infecting other cells where it
Made by: Shahd A.Gaber
could be replicated. After the initial escape, viruses evolved independently from their
initial host and ultimately could infect either prokaryotic or eukaryotic cells.
Regressive Evolution Hypothesis
An alternative explanation of viral origin is that viruses evolved from cellular organisms.
Some cellular organisms, particularly certain bacteria, are obligate intracellular
parasites because they must infect a host cell in order to reproduce. Regressive
evolution suggests that some bacterial parasites gradually lost the structures required
for survival outside of a cell. The result was a virus particle containing only nucleotides,
a capsid (protein coating), and at times an outer membrane or envelope. This would
account readily for viruses that contain complex deoxyribonucleic acid (DNA) similar to
that found in bacteria and other cellular organisms (see Figure 2).
Figure 2
1. The development of which of the following is addressed in the passage by the
Coevolution Hypothesis, but NOT by the Regressive Evolution Hypothesis?
F. Self-replication
G. Capsid
H. Deoxyribonucleic acid
J. Cell membrane transit
Made by: Shahd A.Gaber
2. Supporters of all of the theories presented in the passage would agree with the
conclusion that the first viruses:
A. evolved from bacteria.
B. could self-replicate outside a cell.
C. were enclosed within a membrane.
D. contained nucleic acid.
3. The Coevolution Hypothesis does NOT provide an explanation for the earliest virus
particles possessing:
F. protein.
G. enzyme-like activity.
H. nucleotides.
J. DNA.
4. If the Cellular Origin Hypothesis is correct, which of the following conclusions can be
made about modern T4 DNA viruses, which infect Escherichia coli bacteria, and modern
PP7 RNA viruses, which infect Pseudomonas aeruginosa bacteria?
A. T4 and PP7 are more closely related to each other than to bacteria genetically.
B. T4 and PP7 are only distantly related genetically through a cellular organism.
C. T4 and PP7 both evolved from prokaryotic organisms.
D. T4 and PP7 both evolved from eukaryotic organisms.
5. The discovery of which of the following living organisms would provide the most
support for the Regressive Evolution Hypothesis?
F. Extracellular parasites with DNA resembling a known virus
G. Extracellular parasites with unique RNA nucleotide sequences
H. Intracellular parasites with DNA resembling a known virus
J. Intracellular parasites with unique RNA nucleotide sequences.
Made by: Shahd A.Gaber
6. Supporters of all the theories presented would agree with which of the following
conclusions about the origin of viruses?
A. Viral capsids contain a protein structure similar to the cell walls of modern
bacteria.
B. The first viruses did not originate before the first cellular organisms.
C. RNA viruses are more advanced than DNA viruses.
D. The first virus contained DNA and was surrounded by an envelope similar to a
cell membrane.
7. Which of the following questions is raised by the Coevolution Hypothesis, but is NOT
answered in the passage?
F. Why were some RNA sequences packaged into protein structures and others
incorporated into cell structures?
G. Why did obligate intracellular parasites lose their ability to survive outside of
cells?
H. How could two different types of cellular organisms account for the origin of
viruses?
J. How did virions develop the ability to pass through the cell membrane out of the
cell?
(Test 7-2)
Although many forms of bacteria are helpful for human health, they can also cause
illness and even death from severe infections. Antibiotics are a class of medicines used
to combat bacterial infections. Bacteriostatic activity inhibits bacteria cell division
and bactericidal activity kills bacterial cells. Both actions eliminate populations of
bacteria over time. Several classes of bacteriostatic and bactericidal antibiotics are
described in Table 1.
The effectiveness of several antibiotics against a bacterium known to cause common
skin infections was tested. Drugs were introduced to the bacterial culture by themselves
or in combination with sulfamethoxazole (forming SMX compounds). The effectiveness
of these antibiotics at eliminating the responsible bacterium is shown in Figure 1.
Made by: Shahd A.Gaber
Figure 1
Made by: Shahd A.Gaber
6. According to the information in
Table 1 and
Figure 1, what can be concluded about the use of sulfamethoxazole as an antibiotic for
common skin infections?
F. Using sulfamethoxazole 800 mg is ineffective as an antibiotic.
G. Increasing the dosage of sulfamethoxazole decreases its overall effectiveness as
an antibiotic.
H. As an antibiotic, the mechanism of action of sulfamethoxazole is unknown.
J. Compounding antibiotics with sulfamethoxazole increases their effectiveness
against common skin infections.
7. According to
Figure 1, if an investigator administered a sulfamethoxazole dose of 600 mg, 20% of the
original bacteria would remain after a treatment interval:
A. greater than 120 min.
B. between 90 and 120 min.
C. between 60 and 90 min.
D. between 30 and 60 min.
8. After treatment of a bacterial culture similar to that in the passage with 250 mg of
penicillin for 2 hours, the culture will probably contain:
F. less bacteria overall, but most will have survived.
G. less bacteria overall, and most will have been killed.
H. the same amount of bacteria overall, and most will have survived.
J. the same amount of bacteria overall, and most will have been killed.
9. Is the statement “antibiotics compounded with sulfamethoxazole are more effective
against common skin infections than when administered alone” supported by the
information shown in
Figure 1, and why?
A. No, because penicillin is more effective against a common skin infection
bacterium than sulfamethoxazole 400 mg.
B. No, because azithromycin is more effective against a common skin infection
Made by: Shahd A.Gaber
bacterium than SMX/azithromycin.
C. Yes, because sulfamethoxazole 800 mg is more effective against a common skin
infection bacterium than SMX/azithromycin.
D. Yes, because SMX/doxycycline is more effective against a common skin infection
bacterium than doxycycline.
10. According to the passage, the most effective antibiotic against bacteria is one that
results in the:
F. lowest percentage of bacterial elimination in the shortest treatment interval.
G. lowest percentage of bacterial elimination in the longest treatment interval.
H. greatest percentage of bacterial elimination in the shortest treatment interval.
J. greatest percentage of bacterial elimination in the longest treatment interval.
(Test 8-2)
The 4 different blood types in sheep are A, B, AB, and O. The blood type of an offspring
is determined by the blood types of its parents. Each parent contributes a single gene to
its offspring, forming a pair of genes. The genotype of an offspring refers to the
arrangement of the offspring's new gene formed by the combination of the parents' genes.
There are three possible alleles (forms) of this gene: the type-A blood allele (IA), the
type-B blood allele (IB), and the type-O blood allele (IO). Both IA and IB are dominant to IO,
and IO is recessive to IA and IB. This means that an individual with 1 IA and 1 IO will have
type-A blood, and an individual with one IB and one IO will have type-B blood. When an
individual has one IA and one IB allele, this individual will have type-AB blood, due to
the codominance of the IA and IB alleles.
Table 1
Blood Type
Possible Genotypes
A
IAIA or IAIO
B
IBIB or IBIO
AB
IAIB
O
IOIO
To explore the inheritance patterns of blood types in sheep, researchers conducted 4
analyses. In each analysis, male and female sheep of differing blood types were mated
and the resultant blood types of their offspring recorded.
Analysis 1
Made by: Shahd A.Gaber
One thousand males with type-O blood were mated with 1,000 females with type-AB
blood. The following blood types were observed in the offspring:
Type A: 50%
Type B: 50%
Analysis 2
Two hundred of the type-A offspring from Analysis 1 were mated with 200 type-O mates
from no previous experiment. The following blood types were observed in the offspring:
Type A: 50%
Type O: 50%
Analysis 3
One hundred of the type-A offspring from Analysis 1 parented children with 100 type-B
offspring from Analysis 1. The following blood types were observed in the offspring:
Type A: 25%
Type B: 25%
Type AB: 25%
Type O: 25%
Analysis 4
Twenty-five of the type-A offspring from Analysis 3 were mated with type-B mates with
Genotype IBIB who were not from any previous analysis. The following blood types were
observed in the offspring:
Type AB: 50%
Type B: 50%
7. The ratio of blood types containing at least one IA allele to the blood types containing
at least one IB allele produced in Analysis 3 was:
A. 1:00
B. 1:01
C. 2:01
D. 3:01
Made by: Shahd A.Gaber
8. An offspring whose blood type exhibits codominance has which of the following
genotypes?
F. IBIB
G. IBIO
H. IAIB
J. IAIO
9. To produce only offspring with AB blood, one would mate two sheep with which of the
following sets of genotypes?
A. IAIB × IAIB
B. IAIB × IOIO
C. IAIA × IBIB
D. IBIB × IAIO
10. In Analysis 3, the offspring used from Analysis 1 most likely had which of the
following genotypes?
F. IAIO and IBIB
G. IAIO and IBIO
H. IAIA and IBIB
J. IAIA and IBIO
11. Some or all of the offspring had 1 allele for type-O blood in Analyses:
A. 1 and 2 only.
B. 2 and 3 only.
C. 1, 2, and 4 only.
D. 1, 2, 3, and 4.
12. Suppose that 300 offspring were produced in Analysis 3. Based on the results, the
number of offspring with type-B blood produced in Analysis 3 would most likely have
been closest to:
F. 25
G. 50
Made by: Shahd A.Gaber
H. 75
J. 100
(Test 9-1)
Vasoconstriction involves a narrowing of blood vessels that could lead to poor blood
flow in the body if it persists over a long time. Ergotamine is a substance that can cause
vasoconstriction. When ergotamine is injected into a normal blood vessel,
vasoconstriction occurs quickly at the site of the injection (see Figure 1).
The diameter of the blood vessel at the site of vasoconstriction is less than the diameter
of the normal blood vessel, so blood flow has a higher velocity through this narrow site.
As a result, the blood pressure in the site of vasoconstriction is less than the blood
pressure in the normal blood vessel. Moreover, the higher the velocity of the blood flow
through the site of vasoconstriction, the lower the blood pressure at that site.
The percent change in blood pressure (% BP) can be defined as:
Blood vessel sections of similar diameters were isolated from laboratory rats and %
BP was measured over three experiments. When the researchers needed to create a site
of vasoconstriction for some of the experimental trials, they would inject ergotamine to
induce vasoconstriction within the blood vessel section.
Experiment 1
An artificial heart, which mimics a human's heartbeat, is used to move a constant
volume of 500 mL of blood with each beat through four blood vessel sections. These four
blood vessel sections were injected with the same amount of ergotamine, leading to sites
of vasoconstriction of the same diameter. The rate at which the blood is pumped was
Made by: Shahd A.Gaber
varied for the four different blood vessel sections, and the % BP values that resulted
were measured.
Table 1
Rate of artificial heart beat (beats per minute)
% BP
60
1.2
90
9.3
120
22.3
150
45.1
Experiment 2
The artificial heart used in Experiment 1 was then used to pump a constant volume of
500 mL of blood with each beat at a constant rate of 90 beats per minute through five
other blood vessel sections. These blood vessel sections were injected with different
amounts of ergotamine, resulting in sites of vasoconstriction with different diameters. The
% BP values were then measured.
Table 2
Diameter of site of vasoconstriction (cm)
0.4
0.6
0.8
1.0
1.2
% BP
40.3
18.6
9.3
4.6
2.5
Experiment 3
The artificial heart used in Experiment 1 was used to pump different volumes of blood
at a constant rate of 90 beats per minute through five blood vessel sections with the same
diameter at the site of vasoconstriction. The % BP values were then measured.
Table 3
Volume of blood pumped (mL)
% BP
400
8.4
450
8.8
500
9.3
550
9.7
600
10.2
Made by: Shahd A.Gaber
1. Under the conditions described for Experiment 3, a % BP of 9.0 would most likely be
obtained if the entering volume of blood equaled:
A. 350 mL.
B. 475 mL.
C. 550 mL.
D. 650 mL.
2. Based on the results of Experiment 1, if the rate of the artificial heart beat had been
less than 60 beats per minute, then the % BP would most likely have been:
F. less than 1.2.
G. between 1.2 and 9.3.
H. between 9.3 and 22.3.
J. greater than 22.3.
3. Which of the following is the most likely explanation for the results of Experiment 1?
As the rate of the artificial heart beat increases, % BP:
A. increases, because the velocity of blood through the site of vasoconstriction
increases.
B. increases, because the velocity of blood through the site of vasoconstriction
decreases.
C. decreases, because the velocity of blood flow through the site of vasoconstriction
increases.
D. decreases, because the velocity of blood flow through the site of vasoconstriction
decreases.
4. Consider blood flow through three regions of the same blood vessel, each of which
has a different diameter. The velocity of blood flow is measured in milliliters per minute
(mL/min) and the blood pressure is measured in millimeters of mercury (mmHg), and
their values for each of the blood vessel regions are shown in the following table:
Location
Velocity of blood flow (mL/min)
Blood pressure (mmHg)
A
500
31
B
1,000
29
C
900
30
Based on the information in the passage about blood flow, which of the following
diagrams best represents the relative diameters of the three blood vessel regions?
Made by: Shahd A.Gaber
F.
G.
H.
J.
5. Based on the results of Experiments 1 and 2, what was the diameter of the site of
vasoconstriction in the blood vessel section used in Experiment 3?
A. 0.4 cm
B. 0.6 cm
C. 0.8 cm
D. 1.0 cm
6. For the blood vessel sections used in Experiment 2 that had sites of vasoconstriction
with diameters of 0.4, 0.8, and 1.2 cm, which of the following graphs best displays the
comparison between blood pressure at each site of vasoconstriction and blood pressure
in the normal region of the blood vessel leading to the site of vasoconstriction?
Made by: Shahd A.Gaber
(Test 13-1)
The Citric cycle is an essential process used to transform carbohydrates, lipids, and
proteins into energy in aerobic organisms. If yeast is unable to produce succinate, it
cannot survive. The Citric cycle steps leading to the creation of succinate in yeast are
shown in Figure 1. Each step in this cycle is catalyzed by an enzyme, which is essential
to overcome the energy barrier between reactant and product. In the first step, Enzyme
1 is the enzyme, citrate is the reactant, and isocitrate is the product.
Figure 1
Experiment
A scientist grew four strains of yeast on several different growth media. Each strain was
unable to produce succinate because it lacked one of the enzymes required for the
Made by: Shahd A.Gaber
reaction pathway shown in Figure 1. Table 1 shows the results of the scientist's
experiment: "Yes" indicates that the strain was able to grow in the basic nutrition solution
(BNS) + the particular chemical. An undamaged strain of yeast would be able to grow in
the basic nutrition solution without any additional chemical. If a strain was able to grow in
a given growth medium, then it was able to produce succinate from the additional
chemical added to the basic nutrition solution.
If certain genes are damaged, the essential enzymes cannot be produced, which means
that the reactions that the enzyme catalyzes cannot go. Table 2 lists the genes
responsible for the enzymes in the steps of the Citric cycle leading to succinate
production in yeast. If an enzyme cannot be produced, then the product of the reaction
that enzyme catalyzes cannot be synthesized and the reactant in that reaction will
become highly concentrated. If a gene is damaged, then it is notated with a superscript
negative sign, as in Cat3-; if a gene is not damaged it is notated with a superscript
positive sign, as in Cat3+.
Table 2
Gene
Enzyme
Cat1
Enzyme 1
Cat2
Enzyme 2
Cat3
Enzyme 3
Cat4
Enzyme 4
1. Based on the information presented, the highest concentration of isocitrate would
most likely be found in which of the following yeasts?
A. Yeast that cannot produce Enzyme 1
B. Yeast that cannot produce Enzyme 2
C. Yeast that cannot produce Enzyme 3
D. Yeast that cannot produce Enzyme 4
Made by: Shahd A.Gaber
2. According to the information in the passage and Table 2, a strain of yeast that is
Cat1+ Cat2- Cat3- Cat4+ CANNOT produce:
F. Enzyme 1 and Enzyme 4.
G. Enzyme 3 and Enzyme 4.
H. Enzyme 2 and Enzyme 3.
J. Enzyme 1 and Enzyme 2.
3. Which of the following statements best describes the relationships between citrate,
isocitrate, and α-ketoglutarate as shown in Figure 1?
A. Isocitrate is a product of a reaction of α-ketoglutarate, and α-ketoglutarate is a
product of a reaction of citrate.
B. α-ketoglutarate is a product of a reaction of isocitrate, and isocitrate is a product
of a reaction of citrate.
C. α-ketoglutarate is a product of a reaction of citrate, and citrate is a product of a
reaction of isocitrate.
D. Citrate is a product of a reaction of isocitrate, and isocitrate is a product of a
reaction of α-ketoglutarate.
4. Strain X yeast was most likely unable to synthesize:
F. isocitrate from citrate.
G. α-ketoglutarate from isocitrate.
H. succinyl-CoA from α-ketoglutarate.
J. succinate from succinyl-CoA.
5. One of the growth media shown in Table 1 was a control that the scientist used to
demonstrate that all four strains of yeast had genetic damage that prevented the
reactions shown in Figure 1, the reactions which are responsible for the synthesis of
succinate. Which growth media was used as a control?
A. BNS
B. BNS + succinate
C. BNS + isocitrate
D. BNS + succinyl-CoA
Made by: Shahd A.Gaber
6. For each of the four strains of yeast, W-Z, shown in Table 1, if a given strain was able
to grow in BNS + succinyl-CoA, then it was also able to grow in:
F. BNS.
G. BNS + isocitrate.
H. BNS + α-ketoglutarate.
J. BNS + succinate.
F.
G.
H.
J.
Made by: Shahd A.Gaber
(Test 13-2)
Many viruses are known to persist more prevalently during certain times of the year. A
study of four relatively unknown viruses was conducted to examine their annual rate of
prevalence and mortality in a host population. A large survey was conducted of local
populations for the presence of antigen markers indicative of viral exposures to the four
virus types. Measurements were acquired monthly beginning in January of 2000 and
concluding two years later. All monthly measurements were averaged for comparison.
Figure 1 shows the incidence (cases per 1,000 individuals studied) of viral infections
attributed to each viral type over the duration of the study. Figure 2 shows the number of
deaths (per 1,000 individuals studied) attributed to virus A and D infections.
Figure 1
Made by: Shahd A.Gaber
Figure 2
7. According to Figure 1, the incidence of virus A is greatest during which season of the
year?
A. Spring (Mar-May)
B. Summer (Jun-Aug)
C. Fall (Sep-Nov)
D. Winter (Dec-Feb)
8. According to Figure 1, during April 2001, which virus was least prevalent in the
studied population?
F. Virus A
G. Virus B
H. Virus C
J. Virus D
9. In a previous study, a virologist claimed that the incidence of virus B has always
exceeded the incidence of virus C. As shown in Figure 1, the data for which of the
following months is inconsistent with the virologist's claims?
A. Jan-00
B. 1-Feb
C. 1-Aug
D. 1-Dec
10. According to Figure 1, the incidence of at least 3 of the viruses is most alike during
which of the following months?
F. Apr-00
G. Sep-00
H. 1-Nov
J. Jan-02
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11. During both years of the survey, in one month every year, 7 out of 1,000 individuals
died as a result of infection with virus A and 2 out of 1,000 individuals died as a result of
infection with virus D. According to Figure 2, these data most likely were obtained
during which of the following months?
A. January
B. March
C. May
D. October
Made by: Shahd A.Gaber
(Test 15-1)
Sylvatic, or jungle, Yellow Fever is caused by a virus transmitted by mosquitoes from
monkeys to humans. Figure 1 shows the life cycle of the mosquitoes who carry this
disease. These mosquitoes' eggs do not hatch unless there is enough water for the next
two stages of their life cycles. Yellow Fever is passed when an adult of these mosquitoes
first bites a monkey that is infected with the virus and then bites a human.
Figure 1
A study was done on a group of ecologists who went into a jungle where the monkeys
carrying the Yellow Fever virus live. These ecologists were divided into groups based on
how frequently they went into the jungle. The ecologists were tested monthly for Yellow
Fever. Figure 2 shows the number of new cases of Yellow Fever and the amount of rainfall
in the jungle. For each group, Table 1 shows the number of ecologists in each group,
number of mosquito bites, and percent of each group with Yellow Fever.
Made by: Shahd A.Gaber
Figure 2
1. Based on Figure 1, what is essential in maintaining the mosquito population?
A. Jungle
B. Water
C. Monkeys
D. Humans
2. Based on Table 1, the average percent of ecologists affected by the yellow fever
virus was closest to:
F. 20%.
G. 30%.
Made by: Shahd A.Gaber
H. 60%.
J. 80%.
3. Suppose additional data had been gathered for Table 1 about the number of
mosquito bites per month. Based on Figure 2 and Table 1, in which of the following
months would you expect to have the largest total of mosquito bites per month?
A. April
B. June
C. August
D. November
4. According to Figure 2, the amount of rainfall was different for each of the following
pairs of months EXCEPT:
F. May and December.
G. February and March.
H. January and October.
J. April and May.
5. Based on Table 1, as the number of trips into the jungle increased, the number of
monkeys seen:
A. increased only.
B. decreased only.
C. increased, then decreased.
D. varied with no consistency.
(Test 16-1)
Taraxicum, the common dandelion, can reproduce both through spreading seeds and
through vegetative reproduction. To spread its seeds, the dandelion grows seed pods
shaped like globes, in which the seeds are loosely attached to a central ball; each seed
grows a parachute-like tuft that lets it travel long distances on the wind (or when blown
upon by humans). In vegetative reproduction, a new dandelion stalk and leaves can
grow up from an existing root system. Two students discuss the spread of dandelion
populations.
Student 1
Made by: Shahd A.Gaber
In Taraxicum, vegetative reproduction and seed distribution make up the only means
of growing new plants. Each accounts for 50% of the growth of new dandelions.
Taraxicum grows throughout North America. In many places there is very little wind.
Therefore, Taraxicum must have a non-wind-based means of spreading itself. While
blowing dandelion seeds is a common pastime among humans, this human influence is
very recent in evolutionary terms; it is very unlikely that Taraxicum evolved to rely on
humans to distribute its seeds.
The way Taraxicum grows in a typical field shows that both vegetative reproduction and
seed distribution are at work. While seeds scatter over the whole field, the dandelions
tend to grow together in clumps. This suggests that individual seeds sprout the first new
dandelions, which then grow several more through vegetative reproduction.
Student 2
Seed distribution is the main way Taraxicum spreads itself. Without seed distribution,
there are very few new dandelions. Taraxicum does use vegetative reproduction, sending
new stalks from existing roots, but this is mainly to replace the above-ground plant if it
has been cut or eaten. This allows the plant to survive threats in the environment but does
not allow for the growth of new plants.
Plant studies show that plants which rely on vegetative reproduction to spread
themselves tend to have large, complex root networks or underground root
clusters. Taraxicum plants, however, each have a single large, deep taproot. This makes
them very difficult to uproot, but it also means that their roots do not spread out
underground, so any new plants growing from the roots would compete with each other
for sunlight. Even a slight breeze or the brush of a passing animal is enough to spread
dandelion seeds to a new area. Additionally, all known types of Taraxicum produce seed
globes. If half the new dandelions grew from vegetative reproduction, then a seedless
dandelion should not be at a competitive disadvantage and should be commonly
observed in the wild.
Experiment
The students proposed 3 trials using an introduced Taraxicum population in three fields
in a windy area where Taraxicum can naturally thrive (see Table 1).
Made by: Shahd A.Gaber
1. Suppose an experiment were performed in which several new Taraxicum plants were
planted in a field with their roots in glass jars and with plastic bags over the flowers.
Assuming that Student 1's hypothesis is correct, the number of new dandelions in the
field would most likely be what percent of the number in a control field?
F. 0%
G. 25%
H. 50%
J. 100%
2. Which of the following trials most likely provided the control group in the students'
experiment?
A. Trial 1, in which Taraxicum specimens are planted in the soil of a field with no
other Taraxicum plants
B. Trial 1, in which Taraxicum specimens are planted in large glass jars, which are
then buried in the soil of a field with no other Taraxicum plants
C. Trial 2, in which Taraxicum specimens are planted in the soil of a field similar to
that of Trial 1
Made by: Shahd A.Gaber
D. Trial 3, in which specimens are planted in large glass jars in a field similar to that
of Trial 1
3. Student 1 states that dandelions growing in clumps "suggests that individual seeds
sprout the first new dandelions, which then grow several more through vegetative
reproduction." Which of the following indicates why Student 2 believes this cannot be
true? Student 2 says:
F. Taraxicum tends to grow from a root network, while vegetative reproducers grow
from single roots.
G. Taraxicum tends to grow from a single root, while vegetative reproducers grow
from root networks.
H. Taraxicum has seeds that are attached loosely to the stem, a fact that suggests
they are not important to Taraxicum's reproductive strategy.
J. Taraxicum has seeds that are attached loosely to the stem, but vegetative
reproducers tend not to have seeds at all.
4. Student 2 would most likely agree with the statement that Taraxicum:
A. uses vegetative reproduction to compensate for windless environments.
B. improves its ability to survive by using vegetative reproduction to regenerate.
C. has evolved a dependency on humans to distribute its seeds.
D. tends to grow in clumps in fields to which it has spread itself.
5. With regard to the experiment described in the table, Students 1 and 2 would most
likely agree that the increase in the Taraxicum population would be greatest in a field
where:
F. neither plastic bags nor glass jars were used.
G. plastic bags were used, but not glass jars.
H. glass jars were used, but not plastic bags.
J. both plastic bags and glass jars were used.
Made by: Shahd A.Gaber
6. Suppose Trial 3 of the experiment were performed as described. Based on Student
1's hypothesis, the resulting population would be closest to what percentage of a control
population?
A. 0%
B. 25%
C. 50%
D. 100%
7. Suppose the 3 trials were performed as described. Student 2's hypothesis about the
way Taraxicum reproduces would be best supported if the number of new dandelions fit
which of the following patterns?
F. The field in Trial 3 had roughly the same number of dandelions as the field in Trial
1, both of which had fewer dandelions than the field in Trial 2.
G. The field in Trial 1 had more dandelions than the field in either Trial 2 or Trial 3,
while the fields in Trials 2 and 3 had roughly equal numbers of dandelions.
H. The field in Trial 2 had fewer dandelions than the field in Trial 3, which had more
dandelions than the field in Trial 1.
J. The field in Trial 3 had slightly fewer dandelions than the field in Trial 1, both of
which had many more dandelions than the field in Trial 2.
(Test 19-1)
Despite a global campaign since 1988 to eradicate poliomyelitis (polio), the virus that
causes this disease continues to be endemic in four countries. This polio virus, which
can exist as Type 1, Type 2, or Type 3, is most often transmitted through water that is
contaminated by human waste. People can be immunized from this virus with a highly
effective vaccine, which can be administered orally or by injection. Recent analyses of
polio virus transmission have focused on the four polio-endemic countries India,
Pakistan, Afghanistan, and Nigeria.
Study 1
In 2004, a temporary ban on polio vaccines was instituted in Nigeria in response to
concerns that they were contaminated. Researchers reviewed World Health
Organization (WHO) records to determine the number of Type 1 polio virus infections
that were reported in Nigeria in 2004 and tallied their findings by month (see Figure 1).
The World Health Organization has noted that in polio-endemic countries, official
records underestimate the number of people actually infected, because numerous
Made by: Shahd A.Gaber
infected individuals do not report their symptoms to clinics or rely on local therapists
who are not surveyed. In a polio-endemic country, for every person who has reported
an infection, as many as ten people may actually be infected in the local population.
Figure 1
Study 2
Although polio eradication efforts have been most consistent in the urban areas of polioendemic countries, these areas also have a high risk for a reemergence of polio,
especially when the large urban populations are exposed to water contaminated with
wastes that harbor the polio virus. In 2007, researchers analyzed the number of people
who reported infections with Type 3 polio virus in the five largest cities in India. These
cities were Mumbai in western India, New Delhi and Kolkata in northern India, and
Chennai and Hyderabad in southern India. The analysis was undertaken in the months
of June and August. June 2007 was chosen as a representative month for the dry
summer season in India, during which there was minimal rainfall. August 2007 was
chosen as a representative month for the wet monsoon season in India, during which
there was daily rainfall. The results of the findings are shown in Figure 2.
Made by: Shahd A.Gaber
Figure 2
1. According to Figure 1, the greatest increase in the number of reported polio infections
in Nigeria occurred between which two months?
A. January and February
B. February and March
C. April and May
D. November and December
2. It is estimated that for every person infected with the polio virus in an endemic
country, there are 200 people at risk for contracting the virus. Given the results of Study
1, how many people would have been at risk for becoming infected with the polio virus
in Nigeria in June 2004?
F. 80
G. 200
Made by: Shahd A.Gaber
H. 800
J. 16,000
3. Given the information in Figure 2, which of the following might explain the difference
in reported cases of polio in major Indian cities between June and August of 2007?
A. Water is more likely to become contaminated with polio-infected human waste in
periods of high rainfall.
B. Water is less likely to become contaminated with polio-infected human waste in
periods of high rainfall.
C. The polio virus infects more people in India during the summer and monsoon
seasons than during the autumn and winter seasons.
D. Those diagnosed with the polio virus in June are able to recover by August.
4. Which of the following hypotheses was most likely tested in Study 2?
F. The number of reported cases of polio infections varies significantly between
Nigeria and India.
G. Most cases of polio infections are not reported to medical authorities in India.
H. Poliomyelitis infections affect more people in certain regions in India than in other
regions.
J. The number of reported cases of polio infections in India is greatest during the
summer and least during the winter.
5. Polio-endemic countries are located in warm climates that harbor many mosquitoes.
Would the presence of mosquitoes directly affect the transmission of the polio virus?
A. Yes, because the polio virus is primarily transmitted through mosquitoes.
B. Yes, because the polio virus is primarily transmitted through human waste.
C. No, because the polio virus is primarily transmitted through mosquitoes.
D. No, because the polio virus is primarily transmitted through human waste.
Made by: Shahd A.Gaber
6. The comparison of reported polio infections in India in 2007, as shown in Figure 2,
indicates that relative to the number of people in Kolkata infected with polio in June, the
number of people infected with polio in Kolkata in August was approximately:
F. half as much.
G. the same.
H. twice as much.
J. ten times as much.
(Test 20-2)
A group of researchers performed the following study in order to investigate declines in
primarily carnivorous polar bear populations in the Arctic over a 10-year period.
Study
The researchers obtained previously collected data from several areas previously
identified as polar bear habitats. From this data, the researchers selected sixty 5 km × 5
km blocks that do not overlap with one another. The blocks were selected to fall into six
groups, each with a different set of conditions selected in order to conform to criteria for
listing animals as threatened species. Previous research has indicated that Arctic sea
ice and available food are among the factors which may affect polar bear populations.
Table 1 identifies each of the groups utilized in the study. Conditions other than the
ones listed were considered to be normal.
Table 1
Group
Conditions
1 These areas had significantly decreased populations of marine mammals
consumed by polar bears.
2 These areas had significantly increased populations of seaweed commonly
consumed by marine mammals.
3 These areas had been subject to excess thawing of Arctic sea ice.
4 These areas were subject to the same conditions as Groups 1 and 3.
5 These areas were subject to the same conditions as Groups 2 and 3.
6 Unaffected polar bear habitat.
Made by: Shahd A.Gaber
Data for each of the plots was collected, and the population density of polar bears was
calculated in terms of adult polar bears/km2. Table 2 shows the population density of the
blocks in Group 6.
Table 2
Area Label
Population density of Group 6 areas (polar bears/km2)
A
0.93
B
2.10
C
0.21
D
0.72
E
0.88
F
0.72
G
0.91
H
0.53
I
1.12
J
0.74
The data collected was analyzed to find the average population density ratio for each
group. The researchers defined the average population density ratio of a given group as
being equal to the result of the following expression:
Figure 1 shows the average population density ratio of Groups 1?5.
Figure 1
Made by: Shahd A.Gaber
6. Which of the following statements provides the best explanation for why the
researchers collected data for Group 6 in their study?
A. Group 6 provided data indicating the types of predators which most threaten
polar bears in their natural habitat.
B. Group 6 provided a standard by which the other groups could be compared in
order to determine how each set of conditions affected polar bear populations.
C. Group 6 provided a means by which the researchers could carefully identify and
select the conditions for the remaining five groups.
D. Group 6 provided a means of determining the greatest number of polar bears that
would be likely to survive in an area of 25 km2.
7. Which one of the following is a question that most likely explains why Group 2 areas
were included in the study?
F. Does an increase in the food source of their prey affect the population density of
polar bears?
G. If additional masses of seaweed were to be introduced to the Arctic, would polar
bears be increasingly omnivorous?
H. If additional masses of seaweed were to be introduced to the Arctic, would prey
population density increase?
J. Does an increase in the number of prey animals living in the same area as polar
bears affect the amount of Arctic ice?
8. Which of the following correctly ranks Groups 1-5 from the group where the
conditions are most conducive to polar bear population density in the study to the group
where the conditions are least conducive?
A. Group 1, Group 2, Group 3, Group 4, Group 5
B. Group 4, Group 3, Group 1, Group 5, Group 2
C. Group 2, Group 5, Group 1, Group 3, Group 4
D. Group 2, Group 1, Group 5, Group 3, Group 4
Made by: Shahd A.Gaber
9. Which of the following is most likely an organism that the researchers identified as
exhibiting a significantly decreased population when defining Group 1?
F. Snowy owl
G. Seal
H. Salmon
J. Polar bear
10. Synergy between two effects is said to exist when their combined effect is greater
than the sum of each effect considered separately. The study appears to be designed
such that the researchers can investigate possible synergy in which of the following two
groups?
A. Groups 1 and 2
B. Groups 1 and 4
C. Groups 4 and 5
D. Groups 1 and 3
11. Before performing their analysis of the data, the researchers developed four
different hypotheses. Each one of the four hypotheses below is supported by the results
of the study EXCEPT:
F. Declining prey populations have had some effect on polar bear populations.
G. The melting of Arctic sea ice has a greater effect on polar bear populations than
declining prey populations.
H. Declining prey populations have a greater effect on polar bear populations than
the melting of Arctic sea ice.
J. The melting of Arctic sea ice has had some effect on polar bear populations.
Made by: Shahd A.Gaber
(Test 22-2)
Bats of the family Vespertilionidae (Vesper bats) are commonly found in North America.
A guide for identifying Vesper bats found in Utah is presented in Table 1.
Students observed Vesper bats in a Utah nature reserve and recorded descriptions of
them in Table 2.
Made by: Shahd A.Gaber
7. Based on the given information, which of the following characteristics distinguishes
Bat IV from a Pipistrellus hesperus?
A. 4 mm and curved tragus
B. 15 mm long ears
C. 20 mm long forearm
D. Heavily furred uropatagium
8. Based on Table 1, Bats I and II share the same results through step:
F. 1
G. 5
H. 7
J. 9
9. Which of the following best describes the family Vespertilionidae?
A. Mammals
B. Protists
C. Lampreys
D. Birds
10. According to Table 1, Lasiurus cinereus and Lasiurus blossevillii could have all of
the following traits in common EXCEPT:
F. ears not separated at the base.
G. 35 mm long ears.
H. a heavily furred uropatagium.
J. 20 mm long ears.
11. Based on Table 1, which of the following is likely to be most genetically similar to
Bat II?
A. Lasiurus blossevillii
B. Idionycteris phyllotis
C. Lasionyceris noctivagans
D. Myotis volans
Made by: Shahd A.Gaber
(Test 22-4)
Pepsin is an enzyme in humans that catalyzes the digestion of proteins, like the milk
protein casein, into smaller subunits called peptides. Pepsin is active only in acidic
solutions.
The researchers prepared a solution of casein, a solution of anserine (a small peptide),
a solution of pepsin, and various buffer solutions (solutions maintaining a constant pH).
The following experiments were conducted using these solutions.
Experiment 1
Seven solutions were prepared in test tubes using a 5 mL solution buffered to pH 3.0.
Different amounts of casein, anserine, and pepsin solutions were added to each tube,
and then diluted to 10 mL with the buffer solution, so that the final pH in each test tube
would be 3.0. Each tube was incubated at a constant temperature for 15 minutes, and
then was monitored to determine whether there was any activity by pepsin (see Table
1).
Experiment 2
Seven solutions were prepared in test tubes according to the same procedure as in Trial
3 of Experiment 1, and each test tube was diluted with different buffer solutions of
varying pH (see Table 2).
Made by: Shahd A.Gaber
17. Pepsin is most likely to be found in which of the following organs?
A. Kidney
B. Heart
C. Stomach
D. Spinal cord
18. Suppose another trial had been performed in Experiment 2, and the results showed
a high level of pepsin activity. Which of the following would be the most likely pH of the
buffer solution used in this new trial?
F. 2
G. 4
H. 6
J. 8
19. Which of the following is the most likely reason that Trials 3 and 4 show high levels
of pepsin activity while Trial 5 shows no pepsin activity?
A. Pepsin activity is dependent on both casein and anserine.
B. Pepsin activity is blocked by anserine.
C. Pepsin is able to digest casein, but not anserine.
D. Pepsin is able to digest anserine, but not casein.
Made by: Shahd A.Gaber
20. According to the results from Experiment 1, which of the following trials are most
likely to contain undigested casein?
F. Trials 1, 3, 4, and 7 only
G. Trials 1, 5, 6, and 7 only
H. Trials 1 and 7 only
J. Trials 5, 6, and 7 only
21. The experimental conditions for Trial 3 are most similar to those for which of the
following trials?
A. Trial 9
B. Trial 11
C. Trial 13
D. Trial 14
22. According to the results from Experiments 1 and 2, which of the following best
explains the relationship between pepsin activity, pH, and temperature?
F. Pepsin digests proteins at a fast rate when the pH is greater than 4.0 and the
temperature is about 40°C.
G. Pepsin digests proteins at a fast rate when the pH is less than 4.0 and the
temperature is about 40°C.
H. Pepsin digests proteins at a fast rate when the pH is greater than 3.0 and the
temperature is about 30°C.
J. Pepsin digests proteins at a fast rate when the pH is less than 3.0 and the
temperature is about 30°C.
Made by: Shahd A.Gaber
(Test 23-4)
An ecological study measured the reflection of light by different algae types and water
samples. The study found that a water sample’s reflectance of light is determined by the
density of algae in it. As the density of algae in a water sample increases, the water
sample’s reflectance of light became more similar to the pure algae’s reflectance of light.
Table 1 lists the wavelength range of the visible spectrum and the wavelength ranges
of the colors of the visible spectrum.
Table 1
Color
Wavelength (nm)
Violet
380?430
Blue
430?500
Green
500?565
Yellow
565?585
Orange
585?630
Red
630?750
Figure 1 shows the relative reflectance of light by pure samples of water and three types
of algae versus the wavelength of light from 350 nm to 750 nm.
Made by: Shahd A.Gaber
Figure 2 shows the relative reflectance light of a sample of lake water versus the
wavelength of light from 350 nm to 750 nm.
Figure 2
17. Based on Table 1 and Figure 1, which color of light is most reflected by blue-green
algae?
A. Violet
B. Yellow
C. Red
D. Green
18. Autotrophic organisms, such as blue-green algae, absorb wavelengths using the
molecule chlorophyll. Chlorophyll is typically associated with which of the following
chemical reactions?
F. Binary fission
G. Condensation
H. Photosynthesis
J. Respiration
Made by: Shahd A.Gaber
19. According to Figure 1, at which of the following wavelengths does the amount of
light reflected by green algae exceed the amount of light reflected by diatoms?
A. 400 nm
B. 520 nm
C. 670 nm
D. 710 nm
20. Green algae is classified in which kingdom of organisms?
F. Animalia
G. Plantae
H. Fungi
J. Protista
21. Based on Figures 1 and 2, what type of algae has the greatest density in the lake
water sample?
A. Blue-green algae
B. Diatoms
C. Green algae
D. No algae are in the water sample.
(Test 23-6)
Haloarchaea are single-celled microorganisms that can use light to generate energy,
through a unique form of photosynthesis. To compare haloarchaeal photosynthesis with
plant photosynthesis and bacterial fermentation, researchers performed two experiments
in which they exposed plant haloarchaeal and bacterial cells to either red or green light.
The researchers measured the growth of these cells by measuring how much acid and
CO2 were produced; more production of these indicated more growth.
Experiment 1
Water containing salt and sucrose was added to eight large test tubes.
Next, phenolphthalein (a pH indicator that is colorless in the presence of acid and has a
pink color in its absence) was added to each large test tube. A smaller test tube was
then added, inverted, into each large test tube to collect CO2; if CO2 had been
produced, a gas bubble would appear in this smaller tube (see Figure 1).
Made by: Shahd A.Gaber
Figure 1
The large test tubes were capped, heated until the solutions were sterile, and then
cooled. Nothing was added to the first test tube (T1). Cells of the plant Rosa carolina were
added to the second test tube (T2), cells of the haloarchaea NRC-1 were added to the
third test tube (T3), and cells of the bacterium Bacillus anthracis were added to the fourth
test tube (T4). These four test tubes were exposed to red light, and incubated at 37°C for
48 hr. Then, the procedure was repeated with exposure to green light, using the four
remaining test tubes: T5 (no cells), T6 (plant cells), T7 (haloarchaeal cells), and T8
(bacterial cells). In Table 1, + means presence and – means absence.
Experiment 2
Some of the cells tested in Experiment 1 are thought to contain pigments that help them
absorb light. To determine whether these cells absorbed light to generate energy, cells
of the same species are exposed to red and green light in new test tubes. The
researchers measure the transmittance, or the amount of light that transmits through the
test tube. If the transmittance is low, then the cells in the test tube are assumed to
contain pigments that absorb most of the light to generate energy. If the transmittance is
high, then the cells are assumed to contain no pigment that could absorb light and
generate energy. Instead, most of the light passes through the test tube.
Made by: Shahd A.Gaber
28. In Experiment 1, which cell types grew in the presence of green light?
F. Plant cells only
G. Plant and bacterial cells only
H. Plant and haloarchaeal cells only
J. Haloarchaeal and bacterial cells only
29. Suppose that plant cells and haloarchaeal cells that are situated close to each other
do not interfere with each other's absorption of light and generation of energy. If a new
test tube containing both plant and haloarchaeal cells were prepared, what would be the
most likely results for Experiments 1 and 2?
A. - - High - - High
B. - + Low + - Low
C. + - Low - + High
D. + + High + + Low
30. Suppose that a scientist isolates a cell type that is one of the four cell types used in
Experiment 1. She finds that this cell type produces CO2 in the presence of red light.
She then tests the cell type in the presence of green light and finds that neither CO 2 nor
acid is produced. Based on the results of Experiment 1, the cell type is most likely the:
F. control with nothing added.
G. plant Rosa carolina.
Made by: Shahd A.Gaber
H. haloarchea NRC-1.
J. bacterium Bacillus anthracis.
31. What is the evidence from Experiments 1 and 2 that haloarchaea require green light
to generate energy?
A. In the presence of red light, haloarchaea show low transmittance of light and
produce acid.
B. In the presence of red light, haloarchaea show high transmittance of light and
produce no acid.
C. In the presence of green light, haloarchaea show low transmittance of light and
produce acid.
D. In the presence of green light, haloarchaea show high transmittance of light and
produce no acid.
32. Which of the following best illustrates the results of Experiment 1 for the plant Rosa
carolina in red light?
F.
G.
Made by: Shahd A.Gaber
H.
J.
33. Do the results of Experiment 1 support the hypothesis that haloarchaea and bacteria
use similar processes to generate energy?
A. Yes, because both haloarchaea and bacteria produce CO2 in the presence of
green light.
B. Yes, because both haloarchaea and bacteria produce CO2 in the presence of red
light.
C. No, because haloarchaea produce only acid in the presence of green light, while
bacteria produce acid and CO2 in both red and green light.
D. No, because neither haloarchaea nor bacteria produce CO2 in the presence of
either red or green light.
Made by: Shahd A.Gaber
(Test 23-7)
For most of the 20th century, scientists recognized two basic domains of living
organisms, prokaryotes and eukaryotes. The presence of nuclei and other membranebound organelles within the cell primarily distinguished eukaryotes from prokaryotes. The
possibility of revising this dichotomy resulted from the discovery of the Archaea,
organisms with unique cell membrane and ribosomal RNA (rRNA) structure. Cell
membranes are composed of phospholipids that have both water-insoluble and watersoluble subunits. Ribosomes are made of protein and rRNA and build new proteins within
the cell.
Two scientists in the 1990s debate whether organisms should be classified into two or
three domains.
2-Domain Hypothesis
The Archaea are prokaryotes because they lack intracellular membrane-bound
organelles. Although they are found in extreme and unusual environments, the gross
structure and life cycle of the Archaea are similar to prokaryotic bacteria. Like bacteria,
their cells are usually surrounded by a cell wall, and they reproduce asexually through
binary fission.
The structural and metabolic characteristics that are unique to the Archaea are not
significantly different from other prokaryotes to warrant their separation into a third
domain. Although the Archaea were distinguished very early on in the diversification of
life, today they remain appropriately defined by the original definition of prokaryote.
3-Domain Hypothesis
The Archaea are a distinct form of life requiring a revision of the previously held
dichotomy of prokaryote and eukaryote. Eukaryota should remain the same, but
prokaryotes should be split into Archaea and Bacteria because of significant differences
in genetics, structure, and metabolism.
Archaea as a domain is justified by detailed analysis. The genetic sequence of rRNA
in the Archaea is so distinct from prokaryotes and eukaryotes that these groups of
organisms likely diverged over 3 billion years ago. Archaea cell membranes contain more
rigid ether linkages instead of the ester linkages found in eukaryotes and bacteria. This
contributes to their survival in harsh environments. Finally, the Archaea are capable of
exploiting a wider range of energy sources compared to eukaryotes and bacteria.
Made by: Shahd A.Gaber
34. Which of the following statements is most consistent with the 3-Domain Hypothesis?
The time, in millions of years ago, when two groups of organisms diverge on the
evolutionary tree increases as the:
F. similarities between rRNA gene sequences increases.
G. differences between rRNA gene sequences increases.
H. number of ester linkages in the cell membrane increases.
J. number of ether linkages in the cell membrane decreases.
35. By referring to the observation that the newly discovered organisms do not have
membrane-bound organelles, the scientist supporting the 2-Domain Hypothesis implies
that these new organisms do not have which of the following structures?
A. Phospholipids
B. Ribosomes
C. rRNA
D. Nuclei
36. According to the passage, a similarity between eukaryotes and prokaryotes is that
both groups of organisms:
F. have ester linkages in their membranes.
G. contain membrane-bound organelles.
H. reproduce sexually.
J. are composed of cells.
37. According to the scientist who supports the 2-Domain Hypothesis, which of the
following is the strongest argument against using a 3-Domain classification?
A. rRNA does not exist in prokaryotes.
B. Ether linkages are found in the cell membranes of the Archaea.
C. The Archaea meet the primary definition of prokaryotic.
D. The Archaea synthesize proteins in the cell cytoplasm.
Made by: Shahd A.Gaber
38. It is shown that the Archaea have protein synthesis structures and mechanisms
more like eukaryotes than prokaryotes. This observation contradicts arguments stated
in which hypothesis?
F. The 2-Domain Hypothesis, because the discovery would show that the new
organisms and bacteria fundamentally differ in cellular metabolism.
G. The 2-Domain Hypothesis, because the discovery would show that the new
organisms and eukaryotes fundamentally differ in cellular metabolism.
H. The 3-Domain Hypothesis, because the discovery would show that the new
organisms and bacteria fundamentally differ in cellular metabolism.
J. The 3-Domain Hypothesis, because the discovery would show that the new
organisms and eukaryotes fundamentally differ in cellular metabolism.
39. The scientist who supports the 3-Domain Hypothesis implies that the 2-Domain
Hypothesis is weakened by which observation?
A. The Archaea have membrane-bound organelles.
B. Microscopes cannot accurately describe organisms.
C. The Archaea lack ester linkages in their cell membranes.
D. Eukaryotes are not related to the Archaea.
Made by: Shahd A.Gaber
40. Which of the following illustrations of a portion of a phospholipid cell membrane is
consistent with the description in the passage?
F.
G.
H.
J.
Made by: Shahd A.Gaber
(Test 24)
A corn seed, or kernel, is made up of pericarp, aleurone, and endosperm layers. Figure
1.1 shows the basic anatomy of a corn seed. The endosperm layer may be yellow or
white. The aleurone layer may be purple, red, or colorless. Unless the aleurone is
colorless, the color of the aleurone layer masks the color of the endosperm layer.
Figure 1.1
Endosperm color is determined by a single gene with two versions, or alleles. A corn
seed's specific combination of alleles, or genotype, determines the physical color of the
endosperm (phenotype). Aleurone color is determined by the interactions of three
independently assorting genes. A genotype that contains at least one aleurone color
allele R and one aleurone color inhibitor allele C will produce a purple aleurone. A
genotype that contains at least one R, one C, and two aleurone color modifier
alleles pp will produce a red aleurone. All other allele combinations will produce a
colorless aleurone.
Table 1.1 shows the phenotypes that result from all possible allele combinations for
each of the four corn seed color genes.
Made by: Shahd A.Gaber
TABLE 1.1 Corn Seed Cross
In a single ear of corn, each individual kernel is a separate seed representing an
independent outcome from the cross of the parental corn. This means that individual
kernels on the same ear of corn can have different genotypes and phenotypes.
Students in a biology class examined several ears of corn that resulted from three
different parental crosses. Students were told the parental phenotypes for each cross
and were instructed to count the number of kernels of each color present on each ear of
corn. Table 1.2 shows the students' kernel color data for each of the three crosses.
TABLE 1.2 Corn Seed Genetic Cross
Made by: Shahd A.Gaber
1. Which structure's color is only visible when the aleurone layer is colorless?
A. Pericarp
B. Tip cap
C. Cotyledon
D. Endosperm
2. It can most logically be inferred that the pericarp layer of a corn seed:
A. is colorless.
B. is beneath the aleurone and endosperm layers.
C. has the same phenotype as the endosperm layer.
D. is absent in most corn seeds.
3. According to Table 1.1, which trait has more than two alleles?
A. Aleurone color modifier
B. Aleurone color
C. Aleurone color inhibitor
D. Endosperm color
4. According to Table 1.1, how many unique kernel color phenotypes are possible?
A. Two
B. Four
C. Five
D. Nine
5. Based on the information in Table 1.1, a corn seed with the
genotype rrC'cPPyy would appear:
A. white.
B. purple.
C. yellow.
D. colorless.
Made by: Shahd A.Gaber
6. Based on the information in Table 1.1, which of the following genotypes would
produce a red kernel?
A. rrCCppyy
B. rrCCPpyy
C. RRCCppyy
D. RRCCPpyy
7. The term allele relationships describes how multiple alleles for the same gene
interact. Based on the information in Table 1.1, which statement accurately describes
the relationship between the alleles of the aleurone color modifier gene?
A. When both P and p are present, an intermediate phenotype is produced.
B. When P is present, the phenotype of p is masked.
C. When p is present, the phenotype of P is masked.
D. The relationship between P and p cannot be determined from the information in
the table.
8. Table 1.2 shows two different kernel colors on the same ear of corn. This is possible
because
A. different kernels have different parent plants.
B. some kernels do not have an aleurone layer.
C. each kernel only gets two of the four seed color genes.
D. each kernel represents a separate offspring.
9. If the genotype of the yellow parent in Cross 1 is rrCCppYY, which of the following
could be the genotype of the white parent?
A. rrCCppyy
B. rrccppYY
C. RrCCppYY
D. rrCCPPYY
Made by: Shahd A.Gaber
10. In Cross 2, two red parents are shown to produce yellow kernels. What is the most
likely explanation for this outcome?
A. A mutation occurring when the two parent plants were crossed resulted in a new
color phenotype.
B. The crossing of parent alleles resulted in some kernels with a colorless aleurone
phenotype.
C. The two parent plants for Cross 2 were incorrectly identified, resulting in
mismatched phenotypes.
D. One of the parent plants passed on a yellow allele instead of a red aleurone color
modifier allele.
11. The ratio of red to yellow kernels in Cross 2 is approximately:
A. 2:01
B. 3:02
C. 3:01
D. 4:01
12. To have the greatest probability of producing a yellow kernel, it would be most
appropriate to repeat Cross(es):
A. 1
B. 3
C. 1 and 3.
D. 1, 2, and 3.
13. Based on the relationships information in Table 1.1, what would be the outcome if
the yellow parent plant in Cross 3 were replaced with a white parent?
A. The ratio of purple to yellow kernels would increase.
B. The ratio of purple to yellow kernels would remain constant.
C. The resulting ears would contain purple and white kernels.
D. The resulting ears would contain purple, yellow, and white kernels.
Made by: Shahd A.Gaber
14. Corn seed color is considered a polygenic trait. Based on the information in the
passage, the term polygenic refers to a trait that
A. results from a single gene with multiple alleles.
B. can exhibit a variety of phenotypes over time.
C. has a phenotype that is influenced by multiple genes.
D. affects many different functions of an organism.
15. Most of the corn sold in grocery stores is yellow. This means that an ear of corn
seen at the grocery store possesses.
A. a different combination of genes than is shown in Table 1.1.
B. fewer color genes than the corn in the crosses shown here.
C. the same genotype as the yellow kernels produced in Cross 2.
D. a genotype that produces a colorless aleurone.
(Test 26)
Bacteria species are differentiated into two large groups, gram-positive and gramnegative, based on the properties of their cell walls. Peptidoglycan, a sugar-amino acid
polymer, is a structural component of the cells walls of both types of bacteria, though
the peptidoglycan layer is significantly thicker in gram-positive bacteria. Gram-negative
bacteria have an extra lipid bilayer, called the outer membrane, that surrounds the
entire cell. Figure 1.3 shows a structural comparison of the cell walls of gram-positive
and gram-negative bacteria.
Made by: Shahd A.Gaber
Figure 1.3
Gram staining is a technique used to identify to which group a particular bacteria
species belongs based on its ability to retain a dye when rinsed with a solvent. First, the
primary stain, crystal violet, is applied to the bacteria culture. An iodine solution is then
added to form a complex with the crystal violet inside the cells. A decolorizer (ethyl
alcohol or acetone) is added next. In gram-positive bacteria, the decolorizer dehydrates
and shrinks the thick peptidoglycan layer. This traps the large crystal violet-iodine
complex inside the cell, staining the cell purple. In gram-negative bacteria, the
decolorizer degrades the outer membrane. This prevents the thin peptidoglycan layer
from retaining the crystal violet-iodine complex, and the dye is washed out of the cell. A
counterstain (safranine or fuchsin) is then added to the culture, giving decolorized gramnegative cells a red color. The counterstain is lighter colored than the primary stain, so it
does not affect the outcome for gram-positive cells. After the staining procedure is
completed, the treated cells are examined under a microscope to determine their color,
thus identifying the group to which they belong.
Gram staining is typically the first test in a series of laboratory tests used to identify an
unknown bacteria sample. Table 1.6 is a dichotomous key of characteristics that can be
used to identify members of five common bacteria genera.
TABLE 1.6 Bacteria Dichotomous Key
Made by: Shahd A.Gaber
1. Which statement accurately describes a structural difference between gram-positive
and gram-negative bacteria?
A. Gram-positive bacteria have a thicker layer of peptidoglycan but lack an outer
membrane.
B. Both types of bacteria have a cell wall, but gram-negative bacteria lack a cell
membrane.
C. Gram-negative bacteria have an outer membrane instead of a peptidoglycan
layer.
D. The outer membrane is located beneath the peptidoglycan layer in gram-positive
bacteria.
2. Which structural feature is present in both gram-positive and gram-negative cells?
A. Porins
B. Lipoteichoic acid
C. Periplasmic space
D. Lipopolysaccharides
Made by: Shahd A.Gaber
3. Which of the following statements is most logically supported by the presence of
porins in gram-negative bacteria?
A. Cells walls are not permeable, so all substances entering a bacteria cell must
travel through porins.
B. The lipopolysaccharide and phospholipid bilayer is less permeable than
peptidoglycan.
C. In bacteria cells, a thicker peptidoglycan layer is more permeable than a thin
peptidoglycan layer.
D. Gram-negative bacteria transport larger molecules into their cells than do grampositive bacteria.
4. Which substance does not act as a tissue stain in the Gram staining technique?
A. Safranine
B. Crystal violet
C. Fuchsin
D. Ethyl alcohol
5. According to the Gram staining technique, a bacteria species is identified as gramnegative if its cells:
A. appear purple after the staining procedure.
B. have not been exposed to any stain.
C. appear colorless after the staining procedure.
D. appear red after the staining procedure.
6. In the Gram staining technique, which step must be performed before the addition of
the iodine solution?
A. Staining with safranine
B. Washing with acetone
C. Staining with crystal violet
D. Washing with ethyl alcohol
Made by: Shahd A.Gaber
7. Based on the information about the Gram staining technique, the most logical reason
for applying a counterstain is to:
A. intensify the appearance of gram-positive cells under a microscope.
B. prevent the primary stain from affecting gram-negative cells.
C. counteract the effects of the primary stain on gram-positive cells.
D. allow gram-negative cells to be seen more easily under a microscope.
8. Based on the information about the Gram staining technique, it is most reasonable to
expect a chain of which type of molecule to degrade in the presence of ethyl alcohol?
A. Lipids
B. Nucleotides
C. Sugars
D. Amino acids
9. Based on the information in Table 1.6, bacteria belonging to which genus would
appear purple after a Gram staining test?
A. Streptococcus
B. Escherichia
C. Pseudomonas
D. Enterobacter
10. In Table 1.6, Steps 2 and 3 list the same cell shape characteristics because:
A. gram-positive and gram-negative bacteria can both be rod- or sphere-shaped.
B. gram-positive bacteria can switch between rod and sphere shapes.
C. cell shape depends on the results of the bacteria's Gram staining test.
D. the cell shape of many gram-positive and gram-negative bacteria is unknown.
11. Of the five bacteria genera listed in Table 1.6, how many have a cell wall composed
of a thick peptidoglycan layer?
A. One
B. Two
Made by: Shahd A.Gaber
C. Three
D. Five
12. Based on the information in Table 1.6, which genera contains gram-negative, rodshaped bacteria that do not ferment lactose?
A. Pseudomonas
B. Enterobacter
C. Staphylococcus
D. Escherichia
13. Based on the information in Table 1.6, which characteristic is shared
by Pseudomonas and Enterobacter bacteria?
A. Gram-positive cells
B. Lactose fermentation
C. Use of citric acid as sole carbon source
D. Rod-shaped cells
14. A laboratory technician is examining a bacteria sample belonging to the
genus Escherichia under a microscope and notes that the sample remains colorless
after performing the Gram staining procedure. It is most reasonable to assume that an
error occurred during the:
A. application of the primary stain.
B. application of the counterstain.
C. decolorization of the cells.
D. bonding of iodine to the primary stain.
15. Since gram-negative bacteria are generally more resistant to antibiotics such as penicillin, Gram
staining can be used to inform appropriate antibiotic treatment for patients with bacterial infections. Based
on the information in Table 1.6, infections caused by bacteria belonging to which genera would be most
effectively treated with penicillin?
A. Staphylococcus and Streptococcus
B. Enterobacter and Escherichia
C. Staphylococcus and Enterobacter
D. Streptococcus and Escherichia
Made by: Shahd A.Gaber
(Test 27)
An organism's genetic information is stored within the nuclei of its cells as a set of
chromosomes. The number of chromosomes in a cell varies from species to species. In
some species, the number of chromosomes can vary between individuals. Table 2.1
lists the chromosome count for a variety of species.
TABLE 2.1 Species Chromosome Count
Source: http://en.wikipedia.org/wiki/List_of_organisms_by_chromosome_count.
Made by: Shahd A.Gaber
Ploidy is the number of sets of chromosomes present in the cell of an organism.
The monoploid number (x) is the number of chromosomes an organism has in one set.
In most species, a gamete (sex cell) contains one complete set of an organism's
chromosomes. The number of chromosomes in a gamete is referred to as
the haploid number (n). The fusing of two gametes into a zygote during sexual
reproduction produces somatic cells (body cells) containing two complete sets of
chromosomes. The total number of chromosomes in a somatic cell is referred to as
the diploid number (2n). In most species, the monoploid number (x) and the haploid
number (n) are the same.
Some species have more than two sets of chromosomes present in their cells, a
condition referred to as polyploidy. The somatic cells of triploid organisms have three
sets of chromosomes, for example, and tetraploids have four. In polyploidy organisms,
the term haploid is still used to describe the number of chromosomes in a gamete,
and diploid is used to describe the number of chromosomes in a somatic cell. However,
the monoploid number and the haploid number are not the same in a polyploidy
organism.
1. Based on the information in Table 2.1, which species does not exhibit variation in
chromosome numbers between individuals?
A. European honeybee
B. Swamp wallaby
C. Slime mold
D. Jack jumper ant
2. The first part of an organism's scientific name identifies the genus to which it belongs.
Which statement about the members of a genus is best supported by the information in
Table 2.1?
A. An organism's genus determines the number of chromosomes it has.
B. Organisms in the same genus tend to have similar chromosome counts.
C. No two organisms in the same genus can have the same number of
chromosomes.
D. Chromosome count can vary greatly between organisms in the same genus.
Made by: Shahd A.Gaber
3. Based on the information in Table 2.1, the relationship between diploid chromosome
count and organism complexity can best be described as exhibiting:
A. a direct correlation.
B. no correlation.
C. an inverse correlation.
D. a linear correlation.
4. To which kingdom does the organism exhibiting the greatest diploid number of
chromosomes in Table 2.1 belong?
A. Animalia
B. Plantae
C. Eubacteria
D. Protista
5. Which species has more chromosomes than a human but fewer chromosomes than a
dog?
A. Bombyx mori
B. Canis latrans
C. Ophioglossum reticulatum
D. Mus musculus
6. Cells from which pair of organisms have the same number of chromosomes in their
nuclei?
A. Horse and donkey
B. Zebra fish and pineapple
C. Earthworm and European honeybee
D. Oats and potato
7. A team of scientists have discovered three previously unknown insect species in the
Amazon rain forest. Which statement about the genetic information of these species is
best supported by the data in Table 2.1?
A. The largest species is most likely to have the highest number of diploid
chromosomes.
Made by: Shahd A.Gaber
B. There is no way to determine the diploid chromosome count of each species.
C. The three species are highly likely to have the same number of diploid
chromosomes.
D. It is not easy to predict the diploid chromosome count of each species.
8. Based on the information in the passage, which species produces gametes that each
contain 32 chromosomes?
A. Apis mellifera
B. Saccharomyces cerivisiae
C. Equus ferus caballus
D. Drosophila melanogaster
9. According to the information in Table 2.1, how many more total chromosomes does a
female European honeybee have than a male?
A. 1
B. 2
C. 16
D. 32
10. Based on the information in the passage, the total number of chromosomes in a
somatic cell is represented by which of the following terms?
A. n
B. x
C. 2x
D. 2n
11. Which statement about polyploidy is supported by the information in the passage?
A. The number of chromosomes varies among the somatic cells of a polyploid
organism.
B. The gametes and somatic cells of a polyploid organism contain the same number
of chromosomes.
C. The gametes of a polyploid organism contain more than one complete set of
chromosomes.
Made by: Shahd A.Gaber
D. The somatic cells of a polyploid organism contain too many chromosomes to be
considered diploid.
12. Which of the following organisms has four complete sets of chromosomes in its
somatic cells?
A. Alfalfa
B. Slime mold
C. Oats
D. Earthworm
13. Table 2.1 identifies the oat species Avena sativa as a hexaploid, containing six sets
of chromosomes. The numerical representation 2n = 6x = 42 describes the total number
of chromosomes in a somatic cell of this hexaploid species. How many chromosomes
does Avena sativa have in one set?
A. 6
B. 7
C. 21
D. 42
14. Which of the following correctly identifies the relationship between the diploid
number (2n), haploid number (n), and monoploid number (x) of Solanum tuberosum?
A. 2n is twice n, but 4 times x.
B. 2n is twice the sum of n and x.
C. 2n is the sum of n and x.
D. 2n is twice x, but 4 times n.
(Test 28)
The majority of scientists agree that global temperatures are rising, leading to a host of
climate changes that will produce significant worldwide effects over time. Still subject to
debate are the type and severity of effects that these climate changes will have on
various industries. Two scientists present their viewpoints regarding the effects of
climate change on agriculture in the United States.
Scientist 1
Made by: Shahd A.Gaber
Climate change is likely to have mixed effects on U.S. agriculture over time. Every crop
has a set of optimal conditions under which it grows and reproduces best. For many
crops, the growth rate increases as temperature increases, suggesting that the
progressive increase in average temperatures will have a beneficial effect on many
types of crops. On the other hand, a faster growth rate means less time for the seeds of
certain crops to mature, hindering their reproductive ability. Average temperatures will
eventually surpass the optimal growth temperature for some crops, causing their yields
to decline.
Crop yields also increase with carbon dioxide levels. The positive growth effect of
carbon dioxide can be suppressed, however, if the optimal growth temperature is
surpassed. The potential effects of climate change on other environmental conditions,
including soil moisture, nutrient levels, and water availability must be taken into account
as well.
Scientist 2
Agriculture in the United States will be adversely affected by climate change over the
next several decades. Many weeds, pests, and fungi thrive in warm, wet climates and
with increased levels of carbon dioxide. As average temperatures continue to increase
and these conditions become more widespread, the habitat ranges for these organisms
will spread northward. This will pose challenges to northern crops that have not
previously been exposed to certain competitors and pests.
The predicted increase in extreme weather events will also negatively impact crop
yields. An increase in the frequency of floods will destroy crops and potentially deter
farming along major waterways altogether. In areas in which drought conditions are
projected to become more common, a water supply capable of sustaining even modest
crop yields is a very real concern.
1. According to Scientist 1, how will a change in average temperature affect the growth
rates of crops?
A. As average temperature increases, all crops will begin to grow faster.
B. A change in average temperature will benefit some crops and harm others.
C. If average temperature changes too quickly, many crops will stop growing.
D. An increase in average temperature will hinder growth until crops adapt.
Made by: Shahd A.Gaber
2. If Scientist 1 is correct, which of the following trends will most likely occur over the
next several decades?
A. The agriculture industry will experience no significant change in crop yields.
B. The depletion of soil nutrients will cause yields of all crops to decline.
C. Crops with chemical defenses against pests will exhibit increased yields.
D. Crops with higher optimal growth temperatures will produce greater yields.
3. Which environmental change was discussed by Scientist 2, but not Scientist 1?
A. Elevated carbon dioxide levels
B. Increasing average temperatures
C. Limited water availability
D. Increased frequency of flooding.
4. Scientist 2 did not predict that climate change would cause an increase in which of
the following factors affecting crop yields?
A. Fungi
B. Pests
C. Seeds
D. Weeds
5. Based on the passage, the major difference between the opinions of Scientists 1 and
2 is that:
A. Scientist 2 does not predict any positive effects of climate change on agriculture.
B. Scientist 1 discusses the effects of increased temperature but not carbon dioxide.
C. Scientist 2 expects agriculture in southern areas to be unaffected by climate
change.
D. Scientist 1 focuses only on the effects of climate change on crop reproductive
rates.
Made by: Shahd A.Gaber
6. Scientist 2 states that high:
A. carbon dioxide levels will benefit crop yields.
B. carbon dioxide levels will lead to decreased crop yields.
C. average temperatures will improve crop yields.
D. average temperatures will hinder the growth of fungi.
7. An industry-wide increase in agricultural pesticide use over the next several decades
would support the opinion of:
A. Scientist 1.
B. Scientist 2.
C. both scientists.
D. neither scientist.
8. According to Scientist 1, what happens when a crop's optimal growth temperature is
surpassed?
A. The crop maintains growth at its maximum rate.
B. The crop continues to grow but at a reduced rate.
C. The crop experiences growth at an exponential rate.
D. The crop can no longer grow in that environment.
9. Based on the information in the passage, both scientists would agree with which of
the following statements?
A. The greatest threat posed by climate change to the U.S. agriculture industry is
the projected increase in extreme weather events.
B. Southern crops are better adapted than northern crops to withstand the effects of
elevated carbon dioxide levels associated with climate change.
C. The effects of climate change will have a greater negative impact on the
reproductive ability of crops than on their growth rate.
D. Increasing average temperatures associated with climate change will provide an
advantage to some organisms.
Made by: Shahd A.Gaber
10. It can be inferred that Scientist 1 believes elevated levels of carbon dioxide will
directly lead to crops with a(n):
A. shortened growing season.
B. higher optimal growth temperature.
C. decreased need for soil nutrients.
D. increased rate of photosynthesis.
11. Which of the following does Scientist 2 identify as potential competitors to northern
crops?
A. Invasive species of weeds
B. Newly introduced crop species
C. Other industries that use land
D. Migrating pest species
12. The hypothesis of Scientist 1 could best be tested by recording data over the next
decade on:
A. crop yields, average temperatures, and soil nutrient availability worldwide.
B. seed production, soil nutrient availability, and water availability worldwide.
C. seed production, carbon dioxide levels, and water availability in the United
States.
D. crop yields, average temperatures, and carbon dioxide levels in the United
States.
13. If Scientist 2 is correct, over time, the range of:
A. northern crops will become narrower.
B. southern crops will move farther south.
C. northern crops will overtake southern crops.
D. southern crops will remain constant.
Made by: Shahd A.Gaber
14. Assuming that increasing carbon dioxide levels cause average temperature to
increase, which graph best represents the relationship between carbon dioxide level
and crop yields, according to Scientist 1?
A.
Figure 2.1
B.
Figure 2.2
C.
Figure 2.3
Made by: Shahd A.Gaber
D.
Figure 2.4
(Test 29)
The leaves of green plants use the energy in sunlight to convert atmospheric carbon
into organic carbon through the reactions of photosynthesis. These reactions can be
summarized by the following equation:
Gas exchange between the leaf and the environment is an integral part of the
photosynthesis reactions. As carbon dioxide enters the leaf, the oxygen produced as a
by-product of photosynthesis is released into the environment in a 1:1 ratio. Enclosing a
leaf within a lighted chamber allows for the rate of this exchange, and therefore the rate
of photosynthesis, to be measured.
Students in a biology class used lighted chambers to measure the photosynthetic rate of
leaves from four common plant species: sunflower, water hyacinth, rhoeo, and pothos.
A leaf was placed inside the chamber, and a flow of air was introduced. Sensors within
the chamber recorded data on light intensity (LED irradiance), carbon dioxide
concentration, air temperature, and relative humidity.
The leaf was initially exposed to a constant light intensity of 300 μE/m 2/s to stimulate
photosynthesis. After this initial period, students incrementally increased the light
intensity to investigate the relationship between light intensity and photosynthetic rate.
Figure 2.5 shows the light intensity (LED irradiance) over time for a chamber containing
a water hyacinth.
Made by: Shahd A.Gaber
Figure 2.5
Source: "BISC 111/113: Introductory Organismal Biology," by Jocelyne Dolce, Jeff
Hughes, Janet McDonough, Simone Helluy, Andrea Sequeira, and Emily A. Bucholtz.
http://openwetware.org/wiki/Lab_5:_Measurement_of_Chlorophyll_Concentrations_and
_Rates_of_Photosynthesis_in_Response_to_Increasing_Light_Intensity.
Figure 2.6 shows the change in carbon dioxide concentration over time for a chamber
containing a water hyacinth.
Made by: Shahd A.Gaber
Figure 2.6
Source: "BISC 111/113: Introductory Organismal Biology," by Jocelyne Dolce, Jeff
Hughes, Janet McDonough, Simone Helluy, Andrea Sequeira, and Emily A. Bucholtz.
http://openwetware.org/wiki/Lab_5:_Measurement_of_Chlorophyll_Concentrations_and
_Rates_of_Photosynthesis_in_Response_to_Increasing_Light_Intensity.
Figure 2.7 shows the change in air temperature over time for a chamber containing a
water hyacinth.
Figure 2.7
Source: "BISC 111/113: Introductory Organismal Biology," by Jocelyne Dolce, Jeff
Hughes, Janet McDonough, Simone Helluy, Andrea Sequeira, and Emily A. Bucholtz.
http://openwetware.org/wiki/Lab_5:_Measurement_of_Chlorophyll_Concentrations_and
_Rates_of_Photosynthesis_in_Response_to_Increasing_Light_Intensity.
Figure 2.8 shows the change in relative humidity (RH) over time for a chamber
containing a water hyacinth.
Made by: Shahd A.Gaber
Figure 2.8
Source: "BISC 111/113: Introductory Organismal Biology," by Jocelyne Dolce, Jeff
Hughes, Janet McDonough, Simone Helluy, Andrea Sequeira, and Emily A. Bucholtz.
http://openwetware.org/wiki/Lab_5:_Measurement_of_Chlorophyll_Concentrations_and
_Rates_of_Photosynthesis_in_Response_to_Increasing_Light_Intensity.
Students performed 10 light-chamber trials with leaves from each of the four plant
species. The carbon dioxide concentration data was then used to calculate the
maximum carbon dioxide exchange rate for each leaf.
Table 2.2 shows the calculated and mean carbon dioxide exchange rates for each of
the four plant species.
TABLE 2.2 Carbon Dioxide Exchange Rates
1. The atmospheric carbon absorbed by green plants is in the form of:
A. carbon monoxide.
B. carbon dioxide.
C. carbohydrates.
D. water.
Made by: Shahd A.Gaber
2. Which molecule is formed as a by-product of the photosynthesis reactions?
A. Carbon dioxide
B. Glucose
C. Water
D. Oxygen
3. According to Figure 2.5, the initial photosynthesis-stimulating period lasted
approximately:
A. 5 minutes.
B. 20 minutes.
C. 50 minutes.
D. 80 minutes.
4. The slight increase in air temperature indicated in Figure 2.7 is most likely related to
the:
A. increasing light intensity as the study progressed.
B. peak in relative humidity at the 50-minute mark.
C. increase in CO2 concentration at the end of the study.
D. heat generated by the sensors in the light chamber.
5. Sensors within the lighted chamber monitor the presence of which chemical reactant
of the photosynthesis reactions?
A. Carbon dioxide
B. Oxygen
C. Sunlight
D. Glucose
6. Which graph represents the independent variable in the students' study?
A. Figure 2.6
B. Figure 2.7
C. Figure 2.5
D. Figure 2.8
Made by: Shahd A.Gaber
7. Based on the data in Figures 2.5 and 2.6, which light intensity causes a water
hyacinth leaf to absorb carbon dioxide at the fastest rate?
A. 0 μE/m2/s
B. 300 μE/m2
C. 100 μE/m2
D. 1,000 E/m2/s
8. The data in the table would best support the assertion that sunflower plants:
A. require less intense light than the other three species.
B. release more oxygen than the other three species.
C. are the fastest growing of the four species studied.
D. have the shortest life cycle of the four species studied.
9. According to Table 2.2, which plant showed the least variability across trials?
A. Water hyacinth
B. Pothos
C. Rhoeo
D. Sunflower
10. Based on the information in the passage, if the oxygen concentration within the
chamber had been recorded, its graph would most closely resemble which figure?
A. Figure 2.6
B. Figure 2.7
C. Figure 2.5
D. Figure 2.8
11. According to the data in Table 2.2, which plant species perform(s) photosynthesis at
a faster rate than pothos?
A. Sunflower only
B. Sunflower and water hyacinth
C. Rhoeo only
D. Rhoeo and water hyacinth
Made by: Shahd A.Gaber
12. Which of the following generalizations is supported by the data in Figures 2.5
through 2.8?
A. Photosynthesis occurs at a faster rate in a highly humid environment.
B. The rate of photosynthesis varies directly with air temperature.
C. The greater the light intensity, the faster the rate of photosynthesis.
D. The rate of photosynthesis depends on the level of carbon dioxide available.
13. Which of the following statements is supported by the data in Table 2.2?
A. The single leaf with the fastest gas exchange rate was from a sunflower plant.
B. The single leaf with the slowest gas exchange rate was from a water hyacinth
plant.
C. No two leaves from different species exhibited the same gas exchange rate.
D. No two leaves from the same species exhibited the same gas exchange rate.
14. Based on the data in Table 2.2, which plant could be expected to be most tolerant of
a low-light environment?
A. Rhoeo
B. Pothos
C. Sunflower
D. Water hyacinth
15. The passage states that the rates recorded in Table 2.2 represent the maximum
carbon dioxide exchange rates observed for each trial. Assuming that light intensity was
increased at the same intervals for each trial, at approximately which point during each
trial were the exchange rates recorded in the table most likely observed?
A. 30 minutes
B. 70 minutes
C. 10 minutes
D. 50 minutes
SUBMIT
Made by: Shahd A.Gaber
(Test 30)
Antigens occur on the surface of many cell types and provide a unique chemical
signature that allows the body to determine the cell's identity. Antibodies are proteins
that attack foreign substances that may pose an immune threat to the body. Antibodies
identify a substance as foreign by recognizing and binding to its surface antigens. Each
type of antibody is antigen-specific, attacking only one type of antigen.
Human blood is classified into different blood groups based on the presence of certain
antigens on the red blood cells. The most commonly used blood group system is ABO.
This system classifies blood into four groups (types) according to the presence or
absence of A and/or B antigens on the blood cells. Cells may contain A antigens only, B
antigens only, both A and B antigens, or neither antigen. Blood also contains antibodies
against the antigens that are absent from the red blood cells. For example, type A blood
contains A antigens and anti-B antibodies. Table 3.1 identifies the antigens and
antibodies present in each blood type.
TABLE 3.1 ABO Blood Types
Blood can also be classified as Rh-positive (Rh+) or Rh-negative (Rh-), based on the
presence or absence of a different antigen on the red blood cells. Table 3.2 identifies
whether the Rh antigen or antibody is present in each blood type.
TABLE 3.2 Rh Blood Types
The ABO and Rh blood group systems are combined to determine an individual's
medical blood type. Figure 3.1 illustrates the distribution of medical blood types in the
general population of the United States.
Made by: Shahd A.Gaber
Figure 3.1
Source: https://www.armydogtags.com.
Table 3.3 indicates the distribution of medical blood types by ethnicity in the United
States. The values listed represent the percentage of individuals within the given ethnic
group that exhibit each blood type.
TABLE 3.3 Blood Type Demographics
1. What is the total number of medical blood types possible for a human being?
A. Two
B. Four
C. Six
D. Eight
Made by: Shahd A.Gaber
2. The name of each ABO blood type is derived from the:
A. antibodies that are present in the blood.
B. antigens that are present on the red blood cells.
C. prevalence of each blood type in the general population.
D. antigens that are absent from the red blood cells.
3. According to the passage, antigens:
A. distinguish one cell type from another.
B. recognize and attack antibodies.
C. are only found on harmful cells.
D. block antibodies from attacking cells.
4. Rh + blood always contains:
A. Rh antigen.
B. anti-Rh antibodies.
C. A and B antigens.
D. anti-A and anti-B antibodies.
5. Blood containing anti-A and anti-Rh antibodies and B antigens would be identified as
which blood type?
A. A+
B. BC. ABD. B+
6. According to Figure 3.1, what percentage of the general population has type B blood?
A. 9%
B. 2%
C. 11%
D. 16%
Made by: Shahd A.Gaber
7. The least common blood type in the United States is type:
A. O+.
B. AB+.
C. B-.
D. AB-.
8. Based on the data in Table 3.3, which continent's population can be inferred to have
the greatest incidence of blood type B + ?
A. Asia
B. Europe
C. Africa
D. South America
9. In what percentage of the general population are A antigens present on red blood
cells?
A. 39%
B. 33%
C. 44%
D. 37%
10. The data in Table 3.3 support the statement that more than half of the:
A. Caucasian population has type O blood.
B. Hispanic population has type O+ blood.
C. general population with type O blood is Caucasian.
D. general population with type O+ blood is Hispanic.
11. An individual of African-American ethnicity has a greater chance of having a B+
blood type than:
A. the general population.
B. an A+ blood type.
C. an individual of Asian ethnicity.
D. an O+ blood type.
Made by: Shahd A.Gaber
12. Based on the information in Table 3.1, if an individual with an AB blood type
receives donated type A blood, the donated blood will cause:
A. the conversion of existing B antigens to A antigens, altering the individual's blood
type.
B. an immune reaction because the existing B antigens will attack the new A
antigens.
C. no immune reaction because the individual has no antibodies against the new
blood.
D. the individual's body to begin producing anti-A antibodies in response to the new
blood.
13. Blood type O- is often referred to as the "universal donor" because it can be
donated to any of the other blood types. This is because it has:
A. no antibodies to attack antigens.
B. no antigens to trigger an attack by antibodies.
C. both A and B antibodies to attack antigens.
D. both A and B antigens to prevent attack by antibodies.
14. An individual with blood type A- can safely receive a transfusion of which of the
following blood types?
A. A+ or AB. A- or ABC. A- or OD. O- or O+
15. The percentage of the Caucasian population that has blood type AB- is:
A. the same as the percentage for the Hispanic population.
B. less than the percentage for the African-American population.
C. equal to the percentage for the general population.
D. greater than the percentage for the general population.
Made by: Shahd A.Gaber
(Test 31)
Agarose gel electrophoresis is a technique in which an electric field is used to separate
fragments of DNA by size. Figure 3.2 illustrates a common setup of an electrophoresis
apparatus. A square of agarose gel is prepared and placed in a tray of buffer solution.
DNA in solution is loaded into small slits (wells) in the top of the gel. A solution of DNA
fragments of known length, called a DNA ladder, is loaded in the first well. DNA
samples to be studied are loaded in the remaining wells, and an electric current is
applied to the apparatus. Since DNA is negatively charged, the DNA molecules in the
wells travel toward the opposite, positive end of the gel. Smaller DNA fragments are
able to move through the gel more easily and thus move faster than longer fragments.
This causes the fragments to separate according to size as the procedure runs.
Comparison to the DNA ladder provides an estimate of the separated fragments' sizes.
Figure 3.2
In addition to fragment size, several factors can affect the rate of migration of DNA
fragments through the agarose gel. Table 3.4 provides a summary of the effects of
agarose gel concentration and voltage of the electric current.
TABLE 3.4 Factors Affecting Fragment Migration
Made by: Shahd A.Gaber
Table 3.5 identifies the agarose gel concentration needed for optimum resolution of
DNA fragments within various size ranges.
TABLE 3.5 Agarose Concentrations
One application of the gel electrophoresis technique is to identify the alleles an
individual carries for a particular gene. Although there may be multiple possible alleles
(versions) for a specific gene, each individual carries exactly two copies. When
subjected to electrophoresis, each allele separates out into a distinct band, allowing that
individual's pair of alleles to be identified. A single darker band indicates two copies of
the same allele.
Made by: Shahd A.Gaber
Figure 3.3 shows electrophoresis results for a gene with three possible alleles. Allele 2
is known to contain extra bases as compared to Allele 1. Allele 3 is known to be missing
bases as compared to Allele 1. DNA samples from 16 different individuals are loaded in
Lanes A through P. The sizes of the known fragments in the DNA ladder are listed
along the left.
Figure 3.3
1. In a standard gel electrophoresis procedure, the first well is generally reserved for:
A. a DNA ladder.
B. the buffer solution.
C. smaller DNA fragments.
D. the DNA sample of greatest interest.
2. Applying an electric current to the electrophoresis apparatus causes the DNA
fragments to travel:
A. toward the wells.
B. toward the cathode.
C. away from the anode.
D. away from the cathode.
3. According to the passage, the role of a DNA ladder is to:
A. propel the DNA fragments through the agarose gel.
B. provide an approximation of a DNA fragment's size.
C. identify the base sequence in a DNA fragment.
D. determine the total number of bases in a DNA fragment.
Made by: Shahd A.Gaber
4. Which of the following would be a disadvantage of running a gel electrophoresis at a
voltage of 6 V/cm?
A. Poor resolution of large DNA fragments
B. A short total run time for the procedure
C. A total run time of more than one day
D. Brittleness of the agarose gel
5. According to Table 3.5, as the concentration of agarose increases, the range of DNA
fragment sizes that can be resolved:
A. expands.
B. is constant.
C. shrinks.
D. expands, then shrinks.
6. Which combination of factors would provide the best results for DNA fragments of
0.5-0.7 kb?
A. 0.7% agarose and 0.5 V/cm
B. 1.0% agarose and 0.5 V/cm
C. 1.2% agarose and 5 V/cm
D. 1.5% agarose and 5 V/cm
7. When observing electrophoresis results, the largest DNA fragments will appear:
A. closest to the cathode.
B. closest to the anode.
C. as the largest bands.
D. as the smallest bands.
8. Based on the information in the passage, Allele 2 traveled through the agarose gel
faster than:
A. Allele 1 but slower than Allele 3.
B. neither Allele 1 nor Allele 3.
Made by: Shahd A.Gaber
C. Allele 3 but at the same rate as Allele 1.
D. both Alleles 1 and 3.
9. According to the passage, a single darker band, as seen in Lane E, most likely
indicates an:
A. error during the electrophoresis process.
B. error when collecting the DNA sample.
C. individual missing an allele due to mutation.
D. individual with two copies of the same allele.
10. What is the approximate size of Allele 1?
A. 3.0 kb
B. 1.0 kb
C. 8.0 kb
D. 0.5 kb
11. What is the most common allele combination represented in the DNA samples
shown in Figure 3.3?
A. Two copies of Allele 2
B. Allele 1 and Allele 2
C. Two copies of Allele 1
D. Allele 1 and Allele 3
12. Which of the following provides the best explanation for the result shown in Lane J?
A. Individual J carries Alleles 1 and 3.
B. Individual J carries Alleles 2 and 3.
C. Lane J contains the DNA ladder.
D. Sample J contains DNA from two individuals.
Made by: Shahd A.Gaber
13. Which allele combination is not represented in the DNA samples shown in Figure
3.3?
A. Allele 1 and Allele 3
B. Allele 1 and Allele 2
C. Two copies of Allele 3
D. Allele 2 and Allele 3
14. Which agarose concentration was most likely used in the electrophoresis in Figure
3.3?
A. 1.50%
B. 2.00%
C. 1.00%
D. 1.20%
15. Which combination of factors would cause the slowest migration of DNA fragments?
A. 0.5% agarose and 7 V/cm
B. 2.0% agarose and 7 V/cm
C. 2.0% agarose and 0.25 V/cm
D. 0.5% agarose and 0.25 V/cm
(Test 32)
A student wanted to test human reaction time to different stimuli to determine the
conditions that cause the fastest reaction. The student conducted three experiments to
test reaction time.
Experiment 1
The student used a computer program to record the time between the sounding of a
tone and the student pressing the spacebar on the keyboard. This process was
repeated 10 times per trial. The program then averaged the 10 response times to
produce an average for the trial. The student conducted three trials using a tone length
of 200 milliseconds (ms) and three trials with a tone length of 400 ms. Results are
shown in Table 3.6.
Made by: Shahd A.Gaber
TABLE 3.6 Experiment 1
Experiment 2
The student then used the same computer program to record the time between the
sounding of a tone or the appearance of an image on the screen and the student
pressing the spacebar. This process was repeated 10 times per trial, with the computer
again averaging the 10 response times for each trial. The student conducted three trials
using the tone as the stimulus and three trials using the image. Each stimulus lasted for
a duration of 400 ms. Results are shown in Table 3.7.
TABLE 3.7 Experiment 2
Experiment 3
The student repeated the previous experiment but alternated the stimulus (tone versus
image) with each trial. Results are shown in Table 3.8.
Made by: Shahd A.Gaber
TABLE 3.8 Experiment 3
1. In the three experiments, response time is measured as the time between:
A. exposures to two consecutive stimuli.
B. exposure to a stimulus and the subsequent response.
C. the registering of two consecutive responses.
D. the beginning and end of one trial.
2. The stimulus in Experiment 1 was the:
A. sounding of a tone.
B. appearance of a screen image.
C. pressing of the spacebar.
D. use of a computer program.
3. How do Experiments 2 and 3 differ?
A. Experiments 2 and 3 used different stimuli to test response times.
B. The length of exposure to the stimulus was greater in Experiment 2.
C. Experiment 3 included more trials than Experiment 2.
D. In Experiment 3, the type of stimulus was alternated with each trial.
Made by: Shahd A.Gaber
4. Based on the data in Table 3.7, the sense of hearing is:
A. twice as fast as sight.
B. more complex than sight.
C. not as readily testable as sight.
D. more acute than sight.
5. A stimulus duration of 400 ms was used during which experiment(s)?
A. Experiments 2 and 3 only
B. Experiment 1 only
C. Experiments 1 and 2 only
D. Experiments 1, 2, and 3
6. The fastest reaction time occurred in response to:
A. an auditory stimulus lasting 200 ms.
B. an auditory stimulus lasting 400 ms.
C. a visual stimulus lasting 400 ms.
D. a visual stimulus lasting 200 ms.
7. Based on the data in Table 3.6, what is the relationship between reaction time and
length of stimulus exposure?
A. Lengthening the stimulus improves reaction time.
B. A shorter stimulus produces the fastest reaction time.
C. Stimulus length has no measurable effect on reaction time.
D. A longer stimulus produces the slowest reaction time.
8. Scientists have found that it takes 20-40 ms for a visual signal to reach the brain.
Based on the data in Experiments 2 and 3, how long can an auditory signal be expected
to take to reach the brain?
A. 25-45 ms
B. 50-55 ms
C. 8-10 ms
D. 20-40 ms
Made by: Shahd A.Gaber
9. The data in Tables 3.7 and 3.8 best support the conclusion that alternating between
two stimuli:
A. increases the average response time for both stimuli.
B. improves auditory response time but not visual response time.
C. decreases the average response time for both stimuli.
D. improves visual response time but not auditory response time.
10. Scientists have found that a specific response time range exists for each particular
sense. Which of the following would be the range for auditory stimuli?
A. 140-160 ms
B. 180-200 ms
C. 150-170 ms
D. 125-145 ms
11. How many total responses were recorded during Experiment 2?
A. 10
B. 6
C. 60
D. 30
12. Which graph best represents the data collected during Experiment 3?
A.
Figure 3.4
Made by: Shahd A.Gaber
B.
Figure 3.5
C.
Figure 3.6
D.
Figure 3.7
13. The student wants to test how varying the length of exposure to a visual stimulus
affects response time. The best way to do this is to repeat:
A. all three experiments using visual stimuli only.
B. Experiment 1, replacing the tone with an image.
C. Experiment 3, using a visual stimulus only.
D. Experiment 2, using a stimulus duration of 200 ms.
Made by: Shahd A.Gaber
14. What was the slowest auditory response time recorded during the three
experiments?
A. 199 ms
B. 152 ms
C. 142 ms
D. 158 ms
15. Based on the data from the three experiments, what can be done to improve
response time?
A. Alternate exposure to two different stimuli.
B. Decrease the duration of each exposure to a stimulus.
C. Repeat exposure to the same stimulus.
D. Increase the number of stimuli used at one time.
(Test 35)
Shebay Park has been the site of ongoing population dynamics studies since the 1960s.
Consisting of a group of isolated islands, the park provides ecologists with a unique,
closed ecosystem in which to analyze the relationship between predator and prey
populations. Figure 4.5 illustrates the food web for the Shebay Park ecosystem.
Figure 4.5
Ecological research in the park has focused mainly on the predator-prey relationship
between the jaguar and peccary (a type of pig) populations. In addition to the typical
selective pressures each species exerts on the other, scientists have observed specific
events over the years that have affected population sizes. The inadvertent introduction
of feline leukemia by humans in the late 1980s severely reduced the jaguar population.
Made by: Shahd A.Gaber
In 2004, the severest winter on record and an outbreak of ticks did the same to the
peccary population. Figure 4.6 compares the annual population sizes for both species
observed between 1968 and 2012.
Figure 4.6
1. Shebay Park is considered a closed ecosystem because:
A. organisms cannot easily migrate in from other ecosystems.
B. population sizes within the ecosystem do not fluctuate.
C. scientists have never had the opportunity to study the ecosystem.
D. predator-prey is the only type of relationship that exists in the ecosystem.
2. According to the food web in Figure 4.5, peccary can be categorized as which type of
consumer?
A. Scavengers
B. Herbivores
C. Carnivores
D. Omnivores
Made by: Shahd A.Gaber
3. According to Figure 4.6, what has been the maximum size of the jaguar population
since 1968?
A. 20
B. 60
C. 50
D. 10
4. The peccary population reached its smallest size in which year?
A. 2006
B. 2003
C. 1995
D. 1988
5. It can be inferred that the 13-year trend in the peccary population that began after
1990 was largely influenced by a sharp decline in:
A. the jaguar population caused by disease.
B. cactus growth caused by disease.
C. the jaguar population during a severe winter.
D. cactus growth during a severe winter.
6. According to Figure 4.5, how many secondary consumer species are present in the
Shebay Park ecosystem?
A. 0
B. 1
C. 6
D. 2
7. Organisms that compete for many of the same resources within an ecosystem are
said to occupy similar niches. Based on the information in Figure 4.5, which populations
occupy a niche most similar to that of the peccary population?
A. Nutria and squirrel
B. Fox and jaguar
Made by: Shahd A.Gaber
C. Duck and fox
D. Nutria and duck
8. Based on the data in Figure 4.6, a sharp decline in a population's size most
commonly occurs in response to:
A. a sharp increase in another population's size.
B. an event that reduces individuals' immediate survival.
C. a parallel decline in the size of other populations.
D. an event that limits individuals' reproductive ability.
9. Ecologists believe that an increase in parasites is partially responsible for the shift in:
A. the peccary population after 2004.
B. the peccary population before 2004.
C. the jaguar population after 1990.
D. the jaguar population before 1990.
10. Which of the following statements is best supported by the information in the
passage?
A. Predation is the single greatest factor affecting peccary population size.
B. Food availability is the single greatest factor affecting peccary population size.
C. Peccary population size varies independently of the predator population size.
D. Predation is one of several factors that impact the size of the peccary population.
11. Based on Figure 4.5, which population is least likely to be affected by a change in
the peccary population?
A. Aquatic plants
B. Squirrel
C. Cacti
D. Duck
Made by: Shahd A.Gaber
(Test 36)
An invasive species is a species that is not native to an ecosystem and whose
introduction has harmful environmental, economic, and/or human health effects.
Eichhornia crassipes (water hyacinth) is an invasive species of floating aquatic weed
found in freshwater waterways in tropical and temperate regions worldwide. It is highly
tolerant of fluctuations in water level, nutrient availability, pH, and temperature. This
allows it to grow rapidly and outcompete native aquatic plant species for resources.
Dense floating mats of E. crassipes further alter aquatic communities by reducing
dissolved oxygen levels and access to light. Decomposing matter from E.
crassipes mats increases sediment deposition in waterways.
Ecological studies have shown that the growth of a plant can be influenced by
competition with different species of neighboring plants. A group of scientists carried out
the following studies to determine the effects on the growth of E. crassipes when paired
with three other, more benign, aquatic weed species.
Study 1
Scientists collected growth data on E. crassipes mats in the Kagera River in Tanzania.
Scientists marked off 1 square meter (m2) sample areas containing E. crassipes alone
and in combination with three other aquatic weeds common to the Kagera River.
To determine the effects of the other three weed species on E. crassipes growth,
scientists analyzed five growth parameters. Fresh weight was determined by removing
and immediately weighing 10 E. crassipes plants from each area. Plant height was
measured from the base of the plant to the tip of the tallest leaf. The total number of E.
crassipes plants within a sample area was recorded as plant density, which was then
multiplied by fresh weight to determine total biomass. The number of leaves per plant
was also recorded. Table 5.1 lists the averages for each growth parameter for E.
crassipes growing alone and in combination with the three other aquatic weed species.
TABLE 5.1 Kagera River Data
Made by: Shahd A.Gaber
Source:
http://www.academicjournals.org/ijbc/fulltext/2011/August/Katagira%20et%20al.htm.
Study 2
Scientists transplanted young E. crassipes, Commelina sp., Justicia sp., and V.
cupsidata plants from the Kagera River to a greenhouse. In the greenhouse, E.
crassipes potted alone and in combination with the other three weed species were
allowed to grow in water from the Kagera River for four months. At the end of the fourmonth growth period, the parameters of fresh weight, plant height, and leaves per plant
were all determined by the same methods used in Study 1.
Table 5.2 lists the averages for each growth parameter for E. crassipes growing in the
greenhouse alone and in combination with the other aquatic weed species.
TABLE 5.2 Greenhouse Experiment Data
Source:
http://www.academicjournals.org/ijbc/fulltext/2011/August/Katagira%20et%20al.htm.
Made by: Shahd A.Gaber
1. According to the passage, species identified as invasive are always:
A. aggressively growing plants.
B. disruptive to an ecosystem.
C. introduced by humans.
D. economically profitable.
2. According to the passage, water hyacinths upset freshwater ecosystems by doing all
of the following except:
A. increasing sediment deposition in waterways.
B. outcompeting native plants for resources.
C. altering the pH of aquatic environments.
D. limiting aquatic organisms' access to sunlight.
3. Which weed combination was tested in Study 1 but not Study 2?
A. Water hyacinth alone
B. All four aquatic weeds together
C. Water hyacinth with V. cupsidata
D. Water hyacinth with Justicia sp.
4. In Study 1, plant density was measured as:
A. the total number of E. crassipes plants in 1 m2.
B. the total number of weed plants in 1 m2.
C. fresh weight divided by water volume in 1 m2.
D. fresh weight divided by plant volume in 1 m2.
5. Which weed combination serves as the control group in Study 1?
A. E. crassipes with Justicia sp.
B. E. crassipes with all three other weeds
C. E. crassipes with V. cupsidata
D. E. crassipes alone
Made by: Shahd A.Gaber
6. In Study 1, V. cupsidata caused the greatest reduction in:
A. all E. crassipes growth parameters.
B. E. crassipes fresh weight only.
C. all growth parameters except fresh weight.
D. E. crassipes height and density only.
7. Based on the data in Table 5.1, which weed exerts the least competitive pressure
on E. crassipes?
A. Justicia sp.
B. V. cupsidata
C. Commelina sp.
D. The combination of all three weeds.
8. In Table 5.2, Commelina sp. and Justicia sp. are both shown to have:
A. a stronger effect on fresh weight than V. cupsidata.
B. no effect on E. crassipes plant height.
C. the same effect on fresh weight as V. cupsidata.
D. a positive effect on E. crassipes plant height.
9. In Study 2, the water hyacinths grown alone exhibited a greater average:
A. number of leaves than in Study 1.
B. plant height than in Study 1.
C. fresh weight than in Study 1.
D. total biomass than in Study 1.
10. Total biomass was not included as a growth parameter in Table 5.2 because:
A. plant density was not measured in Study 2.
B. the fresh weight values recorded in Table 5.2 were too low.
C. the plants used in Study 2 had no biomass.
D. total biomass is not a good indicator of plant growth.
Made by: Shahd A.Gaber
11. Which of the following statements is supported by the data collected in both
studies?
A. V. cupsidata has the most negative effect on water hyacinth growth.
B. Commelina sp. has a positive effect on water hyacinth growth.
C. Water hyacinth growth is not affected by the presence of other weed species.
D. Justicia sp. has no effect on water hyacinth growth.
12. Based on the data in Table 5.2, the most significant impact of growing E.
crassipes in combination with other weeds in a greenhouse environment appears to be
the production of:
A. shorter plants.
B. lighter plants.
C. fewer leaves per plant.
D. fewer plants.
13. The greatest advantage of the experimental design in Study 2 is that scientists were
able to:
A. choose on which weed species to focus their observations.
B. record data more frequently than could be done at the Kagera River.
C. control for other environmental factors that may affect plant growth.
D. obtain more precise measurements for each of the growth parameters.
14. According to Table 5.1, the presence of all three competitor weeds within the same
square meter appears to have:
A. a greater effect on E. crassipes fresh weight than the presence of any single
competitor weed.
B. an effect approximately equal to the sum of the effects of each single competitor
weed on fresh weight.
C. a lesser effect on E. crassipes fresh weight than the presence of any single
competitor weed.
D. an effect approximately equal to the mean of the effects of each single competitor
weed on fresh weight.
Made by: Shahd A.Gaber
15. Ecologists have found that introducing a competitor to an ecosystem is sometimes
more effective in reducing an unwanted population than introducing a predator. Based
on the results of this pair of studies, increasing the presence of which of the following
species can be predicted to best reduce the water hyacinth population?
A. Justicia sp.
B. V. cupsidata and Commelina sp.
C. Commelina sp. and Justicia sp.
D. V. cupsidata
(Test 38)
The scientific classification of organisms provides information about the relative level of
relatedness between species. Biologists use a hierarchical grouping system to classify
organisms into various taxa (groups) based on shared physiological, developmental,
and genetic characteristics. Table 5.5 identifies the scientific classification of five
common species.
TABLE 5.5 Taxonomic Classification
Biologists use a phylogenetic tree to illustrate the evolutionary history of related species.
In a typical tree, currently living species called extant taxa are listed along the right.
Moving to the left, the point at which two or more extant taxa meet is called a node. A
node indicates an ancestral taxon, or a common ancestor shared by the extant taxa.
Horizontal line length in a phylogenetic tree indicates relative divergence time, an
estimation of how long ago the extant taxa are thought to have diverged into separate
species. Figure 5.6 shows a phylogenetic tree of the species listed in Table 5.5.
Made by: Shahd A.Gaber
Figure 5.6
1. The phylogenetic tree in Figure 5.6 identifies evolutionary relationships between
which type of organisms?
A. Mammals
B. Arthropods
C. Invertebrates
D. Amphibians
2. To which family does Panthera pardus belong?
A. Mustelidae
B. Canidae
C. Felidae
D. Carnivora
Made by: Shahd A.Gaber
3. All organisms in Table 5.5 are members of the same:
A. genus.
B. order.
C. species.
D. family.
4. Canis latrans is the scientific name of which organism?
A. American badger
B. Gray wolf
C. Leopard
D. Coyote
5. According to Figure 5.6, the European otter is most closely related to which species?
A. Panthera pardus
B. Canis lupus
C. Taxidea taxus
D. Canis latrans
6. According to Figure 5.6, how many common ancestors does Panthera pardus share
with Taxidea taxus?
A. Three
B. One
C. Two
D. Four
7. According to Figure 5.6, which pair of species have the most recent divergence time?
A. Lutra lutra and Panthera pardus
B. Canis latrans and Lutra lutra
C. Taxidea taxus and Lutra lutra
D. Canis lupus and Canis latrans
Made by: Shahd A.Gaber
8. Based on the information in the passage and Figure 5.6, Taxidea taxus would be
considered:
A. an extant taxon.
B. a node.
C. an ancestral taxon.
D. an order.
9. Which of the following inferences can be made about the seven-level classification
system used in Table 5.5?
A. Organisms classified in the same kingdom are classified in the same phylum.
B. Organisms within the same class share a common kingdom and phylum.
C. Organisms that share a common order cannot be classified in the same family.
D. Organisms within the same family must share a common genus and species.
10. The scientific classification of the lynx is shown in Table 5.6. With which species
would the lynx share the most recent ancestor?
TABLE 5.6 Lynx Classification
A. Lutra lutra
B. Canis latrans
C. Panthera pardus
D. Taxidea taxus
Made by: Shahd A.Gaber
11. How many taxonomic levels does the lynx have in common with the gray wolf ?
A. One
B. None
C. Two
D. Four
12. Based on the information in Table 5.5, it can be predicted that the common ancestor
shared by all five species belonged to which taxon?
A. Mustelidae
B. Carnivora
C. Canidae
D. Felidae
13. A clade is the taxonomic term for a grouping composed of all the descendants of a
single ancestral taxon. According to Figure 5.6, which of the following groupings would
not constitute a clade?
A. Canis latrans and Canis lupus
B. Taxidea taxus and Lutra lutra
C. Canis latrans, Canis lupus, and Lutra lutra
D. Canis latrans, Canis lupus, Taxidea taxus, and Lutra lutra
14. The wolverine (Gulo gulo) belongs to the family Mustelidae. Which of the following
assumptions can be made about the wolverine?
A. It is most closely related to the American badger.
B. It shares the most genetic similarity with the European otter.
C. It belongs to the same family as the gray wolf.
D. It belongs to the same order as the coyote.
15. Which of the following statements is best supported by the information in Figure
5.6?
A. The Canidae taxon diverged from the Mustelidae taxon more recently than from
the Felidae taxon.
B. Canidae, Mustelidae, and Felidae all diverged into separate taxa at the same
Made by: Shahd A.Gaber
time.
C. The Mustelidae taxon diverged from the Felidae taxon more recently than from
the Canidae taxon.
D. The Felidae taxon first diverged from the Canidae taxon and then from the
Mustelidae taxon.
(Test 39)
A gene is composed of a series of exon and intron segments. Exons are the coding
regions of a gene, the segments that contain the instructions for building a protein. A
gene's exons are connected by noncoding regions, or introns.
To build a protein, the cell must first transcribe the gene into messenger RNA (mRNA).
Then a process called RNA splicing removes the noncoding introns and connects all of
the exons to produce an mRNA transcript that can be used to build the protein.
Tropomyosins are a family of proteins that help maintain the cytoskeleton structure in all
cells and support the contraction of muscle cells. In the late 1980s, a group of scientists
discovered that the alpha-tropomyosin (α-TM) gene can code for several different
tropomyosin proteins within different tissues of the same organism.
Figure 6.1 shows the structure of the seven mRNA transcripts identified as the result of
the scientific study. In each transcript, each box represents an exon. Each transcript
was found to be a product of the same α-TM gene.
Figure 6.1
Made by: Shahd A.Gaber
Scientists continue to study the α-TM gene as a model of alternative splicing, in which
mRNA transcripts containing different combinations of exons can lead to the production
of different proteins. Figure 6.2 shows the structure of the α-TM gene, which is
composed of 12 exons connected by 11 introns.
Figure 6.2
Each exon in a gene codes for a specific series of amino acids in the corresponding
protein. The complete α-TM gene codes for a protein composed of 284 total amino
acids. Table 6.1 shows the series of amino acids coded by each of the 12 exons in the
α-TM gene.
TABLE 6.1 Alpha-Tropomyosin Exon
1. To produce a tropomyosin protein, which of the following steps must occur first?
A. The introns are removed from the α-TM mRNA.
B. Exons are alternatively spliced to code a specific tropomyosin.
C. The α-TM gene is transcribed into mRNA.
D. Amino acids are arranged based on the α-TM mRNA sequence.
Made by: Shahd A.Gaber
2. According to the passage, each mRNA transcript in Figure 6.1 is produced from:
A. the same gene.
B. multiple genes.
C. an independent gene.
D. the same tissue.
3. In Figure 6.1, what is the maximum number of exons present in an mRNA transcript?
A. 10
B. 7
C. 9
D. 11
4. Constitutive exons are present in all mRNA transcripts of a gene and are thought to
be integral in the proteins' basic structure. Which of the following exons appears to be
constitutive?
A. Exon 3
B. Exon 7
C. Exon 12
D. Exon 4
5. Alternatively spliced exons (ASEs) are those that only appear in certain mRNA
transcripts. Which of the following cell types appears to have the least number of ASEs?
A. Myoblast
B. Brain
C. Nonmuscle/fibroblast
D. Smooth muscle
6. Which exons do not appear in any of the same mRNA transcripts?
A. Exons 10 and 12
B. Exons 7 and 11
C. Exons 2 and 3
D. Exons 3 and 11
Made by: Shahd A.Gaber
7. Two types of muscle tissues-skeletal and cardiac-are both striated. Based on Figure
6.1, how do the α-TM mRNA transcripts of skeletal and cardiac muscle tissues differ?
A. One transcript contains a greater total number of exons.
B. The exons present in one transcript are absent in the other.
C. One contains Exon 2, while the other contains Exon 3.
D. Each transcript contains a different final exon.
8. The total number of exons in the α-TM gene is:
A. unknown.
B. 12
C. variable.
D. 11
9. Based on the data in Table 6.1, which α-TM exon codes for the longest sequence of
amino acids?
A. Exon 8
B. Exon 4
C. Exon 11
D. Exon 6
10. Which α-TM mRNA transcript is missing amino acids 258-284?
A. Hepatoma
B. Myoblast
C. Smooth muscle
D. Brain
11. Based on Table 6.1, which mRNA transcript contains a repeated sequence of amino
acids?
A. Striated muscle
B. Nonmuscle/fibroblast
C. Smooth muscle
D. Brain
Made by: Shahd A.Gaber
12. A myoblast is an embryonic cell that can differentiate into a muscle cell. Based on
Figure 6.1, which of the following happens to the α-TM mRNA transcript when a
myoblast differentiates into a smooth muscle cell?
A. Exon 10 is added.
B. Exon 3 is replaced by Exon 2.
C. Exon 12 is replaced by Exon 11.
D. Exon 10 is removed.
13. Untranslated regions (UTRs) are sequences that exist at the beginning and end of
every mRNA transcript. Instead of coding for amino acids, UTRs regulate the
expression of the transcribed gene. In the α-TM mRNA, Exons 1 and 12 both contain
UTRs. Based on the data in Table 6.1, which other exon contains a UTR?
A. Exon 5
B. Exon 8
C. Exon 11
D. Exon 3
14. The passage states that in addition to their function in all cells, tropomyosins also
support contraction in muscle cells. It can be inferred that this extra function is related to
which of the following sequences of amino acids?
A. Amino acids 39-80
B. Amino acids 81-125
C. Amino acids 258-284
D. Amino acids 1-38
15. A hepatoma is a tumor that forms within the liver. Based on Figure 6.1, it can be
inferred that tumor formation may correlate to a loss of which exon?
A. Exon 2
B. Exon 10
C. Exon 11
D. Exon 7
Made by: Shahd A.Gaber
(Test 40)
Over the past several decades, scientists have seen a rapid decline in honeybee
populations worldwide. In an effort to boost population sizes, the European Union
recently instituted a temporary two-year ban on neonicotinoids, a class of pesticides
thought to be harmful to honeybees.
Two scientists present their viewpoints regarding the value of instituting a similar ban in
the United States.
Scientist 1
A short-term ban on the class of pesticides called neonicotinoids is a viable option that
should seriously be considered by the United States. Studies have found neonicotinoid
concentrations in pollen and nectar that can be lethal to pollinators. Although research
has not identified a direct link between neonicotinoids and a reduction in honeybee
populations, recent studies suggest that these pesticides may increase honeybees'
susceptibility to parasites and diseases. The health of honeybee populations directly
affects the agriculture industry and the overall ecosystem. Twenty-three percent of
crops grown in the United States are pollinated by honeybees. Some crops, such as
almonds, apples, onions, and carrots, are pollinated almost exclusively by honeybees.
The reproductive rates of these crops vary directly with the availability of honeybees.
Many of the plants that make up the base of the food web in the natural ecosystem also
rely on these pollinators. Because the honeybee's role as pollinator is so pervasive, any
measures that have the potential to support the health of honeybee populations should
be taken.
Scientist 2
Honeybees are important pollinators for both natural ecosystems and the agriculture
industry, and the health of their populations should be monitored closely. Instituting a
ban on neonicotinoids, however, is unnecessary. Based on current research, the
benefits of neonicotinoid use to the agriculture industry outweigh the threat to honeybee
health. Though the exact causes are difficult to identify, researchers attribute the
decrease in honeybee populations in recent years to weather, environmental stress,
disease, and varroa mites. Environmental stressors include nectar and water that is
scarce or of poor quality and exposure to pesticides, although researchers have found
the latter to have the weakest correlation to honeybee loss of all stressors. Therefore, a
ban on neonicotinoids will not be an effective approach for improving the health of
honeybee populations. A more effective method should address varroa mites and
disease, the greatest known threats to honeybee health.
Made by: Shahd A.Gaber
1. According to the passage, neonicotinoids are a type of:
A. parasite.
B. pollinator.
C. pesticide.
D. pathogen.
2. According to Scientist 1, neonicotinoids:
A. have been directly linked to declines in honeybee populations.
B. affect honeybees by increasing their vulnerability to parasites.
C. provide agricultural benefits that outweigh the risk to honeybees.
D. are the greatest threat to honeybee health in the United States.
3. Scientist 1 identifies all of the following crops as being highly dependent on
pollination by honeybees except:
A. cherries.
B. almonds.
C. carrots.
D. apples.
4. The major difference between the two scientists' viewpoints is that:
A. Scientist 1 believes honeybee populations should be saved, while Scientist 2
believes humans should not interfere with honeybee populations.
B. Scientist 1 believes all threats to honeybee health should be addressed, while
Scientist 2 believes that efforts should focus on the greatest threats to these
populations.
C. Scientist 1 believes honeybee populations are declining in the United States,
while Scientist 2 believes that honeybee populations are stable.
D. Scientist 1 believes neonicotinoids are harmful to honeybees, while Scientist 2
believes neonicotinoids do not pose any threat.
Made by: Shahd A.Gaber
5. According to Scientist 1, which of the following graphs best represents the
relationship between honeybees and producers in an ecosystem?
A.
Figure 6.3
B.
Figure 6.4
C.
Figure 6.5
Made by: Shahd A.Gaber
D.
Figure 6.6
6. Which of the following does Scientist 2 identify as the greatest threats to honeybee
populations in the United States?
A. Varroa mites and disease
B. Neonicotinoids and weather
C. Disease and nectar quality
D. Water and nectar scarcity
7. Which factor affecting honeybee health was discussed by Scientist 2 but not by
Scientist 1?
A. Pesticides
B. Disease
C. Parasites
D. Water quality
8. It can be inferred that Scientist 1 believes honeybees' most important role in natural
ecosystems is to:
A. act as a host for varroa mites.
B. provide a food source for birds.
C. transfer pollen between plants.
D. compete with other bee species.
Made by: Shahd A.Gaber
9. According to Scientist 2, which graph best represents the relationship between
neonicotinoid exposure and honeybee health?
A.
Figure 6.7
B.
Figure 6.8
C.
Figure 6.9
D.
Figure 6.10
Made by: Shahd A.Gaber
10. A doubling of the average honeybee population size in Europe over the next five
years would support the opinion of:
A. both scientists.
B. neither scientist.
C. Scientist 1.
D. Scientist 2.
11. Based on the information in the passage, both scientists would support efforts to:
A. institute a one-year ban on neonicotinoids in the United States.
B. improve disease and parasite prevention in honeybee populations.
C. reduce private consumer use of pesticides near honeybee habitats.
D. monitor changes in the size of honeybee populations without interfering.
12. According to Scientist 1, approximately what proportion of the agriculture industry in
the United States is dependent on honeybees?
A.
B.
C.
D.
13. If Scientist 2 is correct, which of the following trends is most likely to be seen if a
ban on neonicotinoids is enacted in the United States?
A. Honeybee populations will continue to decline at the preban rate.
B. Honeybee populations will begin to increase at a rapid rate.
C. Honeybee populations will continue to decline but at a slower rate.
D. Honeybee populations will begin to increase at a moderate rate.
14. If Scientist 2 is correct, it can be inferred that honeybee health is most strongly
affected by:
A. seasonal conditions.
B. resource availability.
Made by: Shahd A.Gaber
C. human interference.
D. biotic factors.
(Test 41)
Corals build the habitat that is the home for the fish and other marine species that live
on the reef. The corals grow by creating aragonite forms of calcium carbonate cups in
which the polyp sits. Figure 6.11 identifies the anatomy of a coral polyp.
Figure 6.11
Source: http://oceanservice.noaa.gov/education/kits/corals/media/supp_coral01a.html.
Millions of photosynthetic algae, called zooxanthellae, reside inside polyp tissues. They
serve as an energy source for corals as well as providing the coloration for which corals
are known.
Current research indicates that increasingly acidic waters may be to blame for the
decline in coral populations. Oceans absorb atmospheric carbon dioxide. Table 6.2
depicts changes to ocean chemistry and pH estimated using scientific models
calculated from surface ocean measurement data.
Made by: Shahd A.Gaber
TABLE 6.2 Ocean Chemistry and pH
Some coral become less successful at reproducing sexually in acidic waters. Studies
also show links between ocean acidification and coral bleaching. Figure 6.12
summarizes the physiological responses of marine organisms to biological ocean
acidification experiments done by various scientists.
Figure 6.12
Made by: Shahd A.Gaber
1. Zooxanthenellae would logically inhabit which part of a coral?
A. Stomach
B. Basal plate
C. Outer epidermis
D. Stinging nematocyst
2. The data in Table 6.2 indicates that as the concentration of carbon dioxide in the
water rises:
A. the pH decreases and the balance shifts toward bicarbonate instead of
carbonate.
B. the pH increases and the carbonate ion concentration increases.
C. both the pH and the bicarbonate concentration decrease.
D. the pH increases and the balance shifts toward carbonate instead of bicarbonate.
3. Based on the information in Table 6.2, what conclusions can be drawn about ocean
chemistry?
A. Future emissions of carbon dioxide are less likely to significantly impact ocean
chemistry over time.
B. Increased atmospheric carbon dioxide will have little impact on the concentration
of carbonate ions.
C. Chemical changes in oceans are a result of the water absorbing atmospheric
carbon dioxide produced by human activities.
D. Ocean acidification is an unpredictable response that is unlikely to be linked to
human activities that increase the atmospheric concentration of carbon dioxide.
4. Factors that might impact the data found in Table 6.2 include:
I. seasonal changes in temperature.
II. variations in photosynthesis.
III. runoff from rivers.
IV. fluctuations in respiration.
Which of these is likely to account for fluctuations in the geographic pH of ocean
waters?
Made by: Shahd A.Gaber
A. II
B. III
C. I and IV
D. II and III
5. The saturation horizon is a natural boundary in seawater, above which calcium
carbonate (CaCO3) can form and below which it dissolves. Which species from Figure
6.12 most likely lives below the saturation horizon?
A. Corals
B. Gastropods
C. Crustaceans
D. Calcareous algae
6. Calcifying organisms that produce the calcite form of calcium carbonate, such as
foraminifera, can be less vulnerable to acidification than those constructed with
aragonite structures, such as corals. Which of these provides a logical explanation for
these findings?
A. Aragonite is more soluble than calcite.
B. Calcite is more soluble than aragonite.
C. Aragonite saturation is farther from the surface of oceans.
D. Calcite saturation is farther from the surface of oceans.
7. Based on the information in Figure 6.12, decreased fertilization affects corals as well
as:
A. echinoderms.
B. bivalves.
C. crustaceans.
D. calcareous algae.
Made by: Shahd A.Gaber
8. In Figure 6.12, the most significant data with regard to the health of marine
ecosystems is:
A. the decline in coral calcification.
B. the rise in crustacean calcification.
C. the declining metabolism of bivalves.
D. the decreased larval survival in gastropods.
9. According to Figure 6.12, which species appear to be most affected by ocean
acidification?
A. Echinoderms
B. Gastropods
C. Calcareous algae
D. Crustaceans
10. Which of the following terms best describes the relationship between zooxanthellae
and corals?
A. Parasitic
B. Codependent
C. Symbiotic
D. Mutualistic
11. Corals have several features that help them survive in the shallow ocean. Which
part of a coral's anatomy may protect against fluctuating environmental changes such
as temperature?
A. Stomach
B. Nematocysts
C. Basal plate
D. Outer epidermis
12. Some corals can reproduce in a variety of ways. Which of these methods would
produce the most diverse offspring?
A. Coral fragments regenerate to form new coral.
B. Adult coral sprouts tiny buds to form new coral.
Made by: Shahd A.Gaber
C. Adult coral divides and both pieces grow new coral.
D. Coral eggs join with coral sperm to form new coral.
13. Which of the following is most likely to result from declines in a coral polyp's
zooxanthellae population?
A. Bleaching
B. Hyperpigmentation
C. Increased thermal tolerance
D. Accelerated growth of nematocysts.
14. Scientists design an experiment in an attempt to predict the effect increasingly
acidic seawater will have on coral reproduction. They may use the following in the
experiment:
- Aquarium tanks
- Seawater
- Tap water
- Corals
- Carbon dioxide bubbles
Which experimental design will allow the scientists to investigate their hypothesis fairly
and produce high-quality data for analysis?
A. Use two tanks filled with seawater and corals. Add carbon dixiode bubbles to one
tank.
B. Use two tanks filled with tap water and corals. Add carbon dioxide bubbles to one
tank.
C. Use two tanks filled with corals. Add tap water to one tank and seawater to the
other.
D. Use three tanks filled with carbon dioxide bubbles. Add tap water and seawater
to each tank.
15. Changes in the biological processes in the surface ocean water affect deeper
portions of the ocean because:
A. habitats at deeper levels depend on dissolved oxygen occurring at the surface.
B. organisms living at lower ocean levels rely mainly on products created by
organisms at shallow levels.
Made by: Shahd A.Gaber
C. the pH of organisms in shallow waters is altered and becomes non-nutritious to
deep-water organisms.
D. the calcification of shallow-water organisms provides an additional layer of
protection that prevents predation.
(Test 43)
In 1906, the Kaibab Plateau in northern Arizona was declared a federal game refuge by
President Theodore Roosevelt. Before this time, the Kaibab was home to mule deer,
cattle, sheep, and a variety of predators. The approximately 4,000 Rocky Mountain deer
were an important source of food for the wolves, coyotes, bears, mountain lions, and
bobcats that lived on the Kaibab and competed with sheep, horses, and cattle for the
limited grass resources of the plateau.
When the game refuge was created, all deer hunting was banned in an attempt to
protect the "finest deer herd in America." In 1907, the U.S. Forest Service began to
exterminate the natural predators of the deer. With the deer freed from the checks and
balances of predators, the population began to multiply. By the early 1920s, scientists
estimated that there were as many as 100,000 deer on the plateau.
Sheep and cattle were also banned from the Kaibab. Signs of overgrazing were
everywhere, and disease began to attack the crowded deer population. Hunting was
reopened, but it was not enough to reduce the number of deer. Some estimate that as
many as 60,000 deer starved to death in the winters of 1925 and 1926.
Two scientists exchange views about "The Kaibab Deer Incident: A Long-Persisting
Myth."
Scientist A
The Kaibab Plateau should be a lesson to everyone about the disruption of the
predator-prey relationship. This is a classic example of predator control hurting the very
species that the wildlife biologists are attempting to help. If the predators had not been
removed from the Kaibab Plateau, the deer population would have grown under normal
conditions and would not have been subjected to the cruel fate of starvation and
disease. This is a moral case that should be heeded by all biologists when considering
predator control and presented to biology students in their studies of predator checks in
population dynamics.
Scientist B
Predator removal is only a small part of the disaster on the Kaibab and has been
grossly overdramatized. The deer population on the plateau grew rapidly because of the
increase in food supply after the removal of competitive species. With no sheep and
cattle to compete with for grazing, the environment could readily support more deer.
Made by: Shahd A.Gaber
The increased food supply allowed the population to grow quickly and to fall just as
quickly due to the density-dependent factors of starvation and disease. In fact, data
about the peak total number of deer on the plateau are unreliable, and there may have
only been 30,000. The factors are more complex than early ecologists believed, but the
lesson is still valuable.
1. Which of the following pieces of information would Scientist A use to support his
claim?
A. Before 1906, the Kaibab Plateau had already been overgrazed by the herbivores
in the area.
B. It is estimated that between 1907 and 1939, 816 mountain lions, 20 wolves,
7,388 coyotes, and more than 500 bobcats were killed.
C. The U.S. Forest Service reduced the number of livestock grazing permits.
D. In 1924, a committee formed to oversee the situation recommended that all
livestock not owned by local residents be removed immediately.
2. Which of the following reflects evidence presented by both Scientist A and Scientist B
about the deer situation on the Kaibab Plateau?
A. Competition among herbivores was reduced due to restrictions on grazing.
B. The food chain was disrupted when secondary consumers were reduced.
C. Starvation and disease reduced the herd during the winters of 1925 and 1926.
D. Human intervention in the predator population was the cause of the upsurge in
the deer population.
3. Which statement would LEAST likely be attributed to Scientist B?
A. Data about the deer herd are unreliable and inconsistent, and the factors that
may have led to an upsurge are hopelessly confounded.
B. Conclusions that have been made about the Kaibab are based on the maximum
estimate and evolved by unjustified tampering with original data.
C. This is a classic example of how the effects of disruption of the predator-prey
relationship can be seen plainly.
D. The reduction in sheep alone from 1889 to 1908 might have totaled 195,000.
Made by: Shahd A.Gaber
4. The following statements have been made by biologists to describe the Kaibab
Plateau situation. On which statement would Scientists A and B be likely to agree?
A. The plateau represents the unforeseen and disastrous possibilities of ignorant
interference in natural communities.
B. The Kaibab is a classic example of what happens when people set out to protect
prey from their enemies" (sometimes only to preserve them for their human ones) by
killing the predators."
C. Man is the most destructive predator alive.
D. This situation is a well-documented example of what can happen when predators
are removed from an ecosystem.
5. The views of Scientist A:
A. minimize the role of the bounty placed on predators.
B. emphasize the lack of competition for resources.
C. show a more balanced view of the problem by taking into account all factors that
led to the increase in population.
D. are likely to be used by someone trying to illustrate the dangers of removing a
species from the food chain.
6. Scientist A would be most likely to support:
A. the introduction of non-native species into an area where there are no natural
predators.
B. controlled hunting of predators to protect endangered species.
C. future efforts to reorganize natural ecosystems through human intervention.
D. the view that predators help preserve ecosystems.
7. Scientists A and B tend to agree on:
A. the role of an increase in grass abundance in the increase of the deer population.
B. the role that disease and starvation played in reducing the population.
C. the role of predation in the increase of the deer population.
D. the data to be used to represent the situation on the Kaibab.
Made by: Shahd A.Gaber
8. Which of the following facts would support the view of Scientist B regarding the cause
for rapid increase in the deer population?
A. Coyotes were hunted in the thousands.
B. Starvation and disease were rampant from 1924 to 1926.
C. Hunters killed 674 deer in 1924.
D. Sheep and cattle were banned during this time period.
9. Lethal reduction of midsized mammal predators that target duck nests is a method
used to increase the duck population available for sport hunters. Which of the following
statements would Scientist A most likely make regarding this practice?
A. Hunting will keep the duck population from increasing unchecked and limit
growth.
B. The removal of mammals such as foxes and skunks will disrupt other areas of the
food chain, such as the population of mice.
C. The duck population will have greater nesting success as a result of reduced
predatory concerns.
D. Other waterfowl will enjoy the benefits of less predation.
10. On which of the following conclusions would Scientists A and B agree?
A. Human intervention in natural ecosystems is a necessary step to protect
populations.
B. Caution should be taken when creating an ecological situation that favors a single
species.
C. Humans can make a change involving a single species with little or no effect on
other species in the area.
D. The Kaibab Plateau does not offer any lessons applicable to modern-day issues
in ecology.
Made by: Shahd A.Gaber
(Test 49)
Phosphorus is an essential nutrient that can negatively affect water quality, primarily by
promoting excessive plant and algae growth. When this occurs, plants and animals that
live in the water are affected by the reduced sunlight and lower oxygen levels that
develop as organic matter decomposes. For humans, algal blooms lead to a reduction
in the quality of drinking water, a decrease in the use of the water source for
recreational activities, and a decline in property value along waterfront areas.
Lakes are often classified according to their trophic state, which indicates their biological
productivity. The least productive lakes are called oligotrophic. Bodies of water
classified as oligotrophic are typically cool and clear, have relatively low nutrient
concentrations, and provide excellent drinking water. The most productive lakes are
called eutrophic and are characterized by high nutrient concentrations that result in algal
growth, cloudy water, and low dissolved oxygen levels. Table 8.6 shows the phosphorus
levels that are found in lakes with different trophic classifications.
TABLE 8.6
Lake managers collected data over approximately two decades in four different
ecological areas of a large lake. Each area had a different target phosphorus level
based on the natural ecological factors of the area, as indicated in Figures 8.7 through
8.10. For proper lake health, the level must be at or below the target amount. Levels in
excess of the target amount lead to an imbalance in nutrient flow.
Made by: Shahd A.Gaber
Figure 8.7
Figure 8.8
Figure 8.9
Made by: Shahd A.Gaber
Figure 8.10
1. In which area of the lake did the scientists fail to attain the target phosphorus level
in any of the years of the study?
A. Bay Area 1
B. North Lake
C. Bay Area 2
D. South Lake
2. The target for Bay Area 2 falls in the trophic category of:
A. oligotrophic.
B. mesotrophic.
C. eutrophic.
D. hypereutrophic.
3. In 2011, flooding occurred in the region surrounding the lake. This caused:
A. all areas of the lake to exceed their target phosphorus levels.
B. all areas of the lake to meet the standard for target phosphorus level.
C. all but one of the lake areas to exceed the recommended phosphorus levels.
D. two of the four lake areas to exceed the target phosphorus level.
4. In 2005, ecologists managing the lake began a concentrated effort to reduce
agricultural runoff. This appears to have had the greatest effect on phosphorus levels in:
A. Bay Area 1.
B. North Lake.
C. Bay Area 2.
D. South Lake.
5. In 2002, the South Lake area had a trophic classification of:
A. oligotrophic.
B. mesotrophic.
Made by: Shahd A.Gaber
C. eutrophic.
D. hypereutrophic.
6. For how many years of the study was Bay Area 2 found to be mesotrophic?
A. 0
B. 3
C. 9
D. 19
7. During which of the following years did Bay Area 1 have a trophic classification of
mesotrophic?
A. 1992
B. 1995
C. 2009
D. 2011
8. Which of the following best describes the range in phosphorus levels in the North
Lake over the 21-year period?
A. 32 μg/L
B. 62 μg/L
C. 25 μg/L
D. 45 μg/L
9. Which lake had the narrowest range of phosphorus levels over the time period of the
study?
A. Bay Area 1
B. North Lake
C. Bay Area 2
D. South Lake
Made by: Shahd A.Gaber
10. During how many years did Bay Area 1 meet or surpass the standard for proper
lake health?
A. 4
B. 10
C. 10
D. 15
11. What was the difference in phosphorus concentration between Bay Area 1 and
North Lake in 1993?
A. 9 μg/L
B. 41 μg/L
C. 50 μg/L
D. 59 μg/L
12. A neighboring lake was tested in 2005 and found to have a phosphorus level of 10
μg/L. It was most likely taken from a body of water similar to:
A. Bay Area 1.
B. North Lake.
C. Bay Area 2.
D. South Lake.
13. Which of the following statements is best supported by the information in the
passage and figures?
A. When studying the North Lake area, one would expect to find cool, clear water
and high oxygen levels.
B. None of the target levels for any of the lake areas fell in the eutrophic category.
C. The management of phosphorus levels does not have a positive impact on
humans who use the lake for recreation.
D. During the 21-year period of the study, none of the lakes could be classified as
hypereutrophic.
Made by: Shahd A.Gaber
14. The category of lake classification appearing most frequently in the North and South
Lake areas was:
A. oligotrophic.
B. mesotrophic.
C. eutrophic.
D. hypereutrophic.
15. According to the information provided in the passage, which of the areas was the
most likely to have the lowest dissolved oxygen levels in 2010?
A. Bay Area 1
B. North Lake
C. Bay Area 2
D. South Lake
(Test 59)
Photosynthesis is the process by which organisms such as green plants, algae, and
cyanobacteria convert light energy from the sun to chemical energy stored in the bonds
of carbohydrates. Chlorophyll is a pigment employed by many autotrophic organisms to
absorb the various wavelengths of visible light from the sun for use in photosynthesis. A
variety of photosynthetic pigments exist; they are specifically adapted for absorbing
different ranges of the visible light spectrum and reflecting others. The absorption
spectrum of chlorophyll and accessory pigments can be obtained through
spectrophotometry and later used to gain insight into plant growth, determine the
abundance of photosynthetic organisms in fresh- or saltwater, and evaluate water
quality.
The data in Figure 11.1 and Table 11.2 were collected by students measuring the
absorption spectra of three commonly encountered photosynthetic pigments.
Made by: Shahd A.Gaber
Figure 11.1
TABLE 11.2
1. Which of the following statements represents a valid assessment of the data?
A. Chlorophyll a and b absorb the most green light.
B. Neither chlorophyll a, b, nor the carotenoids absorb light in the wavelengths of
425 to 475 nm.
C. Chlorophyll a and b have the most reflection in the wavelengths of 525 to 625
nm.
D. Carotenoids absorb the most light in the red portion of the spectrum.
Made by: Shahd A.Gaber
2. Using the information in Table 11.2, which of the following wavelengths of light would
phycocyanin, a pale blue-colored accessory pigment, reflect most?
A. 400 nm
B. 460 nm
C. 550 nm
D. 750 nm
3. Based on Figure 11.1, which of the following wavelengths of visible light would be
absorbed to promote the most photosynthetic activity in green plants?
A. 400 nm
B. 440 nm
C. 550 nm
D. 625 nm
4. Based on the information in Figure 11.1 and Table 11.2, which pigment has the
highest relative absorbance in the red portion of the spectrum?
A. Chlorophyll a
B. Chlorophyll b
C. Carotenoids
D. All pigments absorb the same amount of light in this portion of the spectrum.
5. Which of the following statements about the relative absorbance levels of the
pigments is most accurate?
A. Chlorophyll a absorbs 8 times as much light as chlorophyll b at 450 nm.
B. Carotenoids behave similarly to chlorophyll a and b at wavelengths greater than
550 nm.
C. Chlorophyll b has the lowest relative absorbance of violet light.
D. Carotenoids absorb approximately 60% as much light as chlorophyll b at 450 nm.
Made by: Shahd A.Gaber
6. Which pigment would absorb the most violet light at a wavelength of 425 nm?
A. Chlorophyll a
B. Chlorophyll b
C. Carotenoids
D. All pigments absorb violet light equally.
7. Which wavelength represents the maximum absorbance of chlorophyll b?
A. 425 nm
B. 450 nm
C. 515 nm
D. 650 nm
8. From the information in the passage, one can conclude that chlorophyll appears
green to the human eye because:
A. wavelengths of light between 550 and 600 nm are highly absorbed.
B. wavelengths in the green portion of the visible spectrum are absorbed.
C. wavelengths in the green portion of the visible spectrum are reflected.
D. wavelengths in the red portion of the visible spectrum are reflected.
9. Phycoerythrin is a photosynthetic pigment that is found in marine algae. It has
absorption peaks at 495 nm and 566 nm and reaches its lowest values over 600 nm.
Based on this information, one would expect phycoerythrin to:
A. appear very similar to chlorophyll a and b to the naked eye.
B. reflect green light and absorb red light.
C. reflect red light and absorb green light.
D. absorb red and green light equally.
Made by: Shahd A.Gaber
(Test 62)
Staphylococcus aureus (S. aureus) is a bacterium found on the skin of 25% of healthy
individuals with no adverse effects. However, when transferred to food products, the
toxins it can produce have been known to cause food-borne illness, particularly in
cooked and cured meats. It is important to determine conditions that promote the growth
of pathogenic species of bacteria and apply that knowledge in food preparation and
handling. Two groups of students set out to determine the ideal growth conditions for S.
aureus. In each group, bacteria were inoculated in a nutrient broth and allowed to grow
over a period of time.
When analyzing the growth of bacteria in a liquid medium, an increase in turbidity
corresponds to an increase in the bacteria growing in suspension. Because the bacterial
cells scatter light, spectrophotometry can be used to determine bacterial growth.
Changes in the logarithmic absorbance scale on the spectrophotometer correspond to
changes in the number of cells, and a growth curve can be plotted by graphing the
absorbance readings from the spectrophotometer versus time. The rate of growth can
be determined by the slope of the lines, and the type of growth occurring at a given time
can be determined by the shape of the curve.
Group 1
The students in Group 1 set out to determine whether temperature affected the growth
rate of S. aureus in a nutrient-rich medium. They tested S. aureus by inoculating a flask
of nutrient-rich broth with a pH of 6 at the following temperatures: 3°C, 20°C, 37°C,
45°C, and 60°C. The spectrophotometer absorbance readings are shown in Table 11.4.
TABLE 11.4
Made by: Shahd A.Gaber
Group 2
Students in Group 2 planned to determine the effect of pH on the growth of S. aureus.
They wanted to test growth at a range of values from highly acidic (low pH) to highly
basic/alkaline (high pH). The cultures were grown at a temperature of 37°C, and the pH
levels of each culture were as follows: 3, 5, 6, 7, and 9. Table 11.5 shows their data.
TABLE 11.5
Made by: Shahd A.Gaber
1. (A) Which of the following is a valid assessment of the data from Group 1?
A. As temperature increases, the growth rate of S. aureus increases proportionally.
B. S. aureus grows optimally at a temperature of 37°C.
C. As temperature decreases, so does the growth rate of S. aureus.
D. S. aureus exhibits exponential growth at a temperature of 37°C.
2. The stationary phase of growth is entered as the nutrients in the medium begin to run
out and the growth of bacteria changes the conditions in the flask. Cell division slows,
and the turbidity ceases to increase because the overall population remains unchanged.
When did the bacteria in Group 1/Trial 3 likely begin the stationary phase of growth?
A. 97 minutes
B. 148 minutes
C. 600 minutes
D. 1,470 minutes
3. Which of the following statements is most accurate?
A. S. aureus responds well to increases in pH above 7.
B. pH factor does not have an effect on the growth of S. aureus.
C. S. aureus is viable at pH levels below 2.
D. pH is a control for Group 1 and an experimental variable for Group 2.
4. Generation time is the amount of time it takes for the bacterial population to double.
Some bacteria, such as E. coli, have a doubling time of 20 minutes under ideal
conditions, while other bacteria may take days to double their population size. Which of
these is most likely the approximate generation time for the bacteria in Group 1/Trial 3
during the first 100 minutes of the experiment?
A. 6 minutes
B. 29 minutes
C. 75 minutes
D. 100 minutes
Made by: Shahd A.Gaber
5. Which of the following does NOT represent a controlled variable for both groups?
A. The nature of the nutrient media
B. The temperature of incubation
C. The strain of S. aureus used
D. The wavelength of light set on the spectrophotometer.
6. Which of the following is a valid assessment of the data in Table 11.5?
A. S. aureus has a faster growth rate at pH 7 than at pH 5.
B. S. aureus achieves a higher overall turbidity at pH 5 than at pH 7.
C. S. aureus achieves maximum growth at pH levels below 4.
D. S. aureus has a faster growth rate at pH 5 than at pH 7.
7. The data for Group 2/Trial 5 indicate that:
A. at this high level of acidity, S. aureus bacteria cannot grow.
B. at this high level of alkalinity, S. aureus bacteria cannot grow.
C. at this low level of alkalinity, S. aureus bacteria cannot grow.
D. at this high level of alkalinity, S. aureus bacteria thrive.
8. Which of the following trials were run under the same conditions for Groups 1 and 2?
A. Group 1/Trial 1 and Group 2/Trial 1
B. Group 1/Trial 2 and Group 2/Trial 4
C. Group 1/Trial 3 and Group 2/Trial 3
D. Group 1/Trial 5 and Group 2/Trial 5
9. If scientists wanted to do further testing on the growth of S. aureus to inform food
processing and handling decisions, such as the effect of salinity, what conditions would
be best as the controls for their test?
A. 37°C and pH 6
B. 37°C and pH 5
C. 20°C and pH 6
D. 45°C and pH 7
Made by: Shahd A.Gaber
10. Which of the following graphs most closely resembles the shape of a graph that
could be drawn for Group 1/Trial 3?
A.
Figure 11.9
B.
Figure 11.10
C.
Figure 11.11
D.
Figure 11.12
Made by: Shahd A.Gaber
11. Chemical inhibitors such as sodium benzoate are often used as food preservatives
because of their ability to retard, although not completely inhibit, bacterial growth. If a
study were done on the effects of sodium benzoate on the growth of S. aureus at 37°C,
one would expect to see results similar to those of:
A. Group 2/Trial 1.
B. Group 1/Trial 5.
C. Group 2/Trial 2.
D. Group 1/Trial 3.
12. What is the independent variable for both Group 1 and Group 2?
A. Time
B. pH
C. Temperature
D. The type of bacteria being tested.
13. Meat in a delicatessen tested positive for the presence of bacteria, and this bacteria
was to be identified using spectrophotometry. Which of the following test results would
most clearly indicate that the bacteria found was S. aureus?
A. The bacteria died at temperatures above 55°C.
B. The bacteria experienced rapid growth at temperatures below 20°C.
C. The bacterial generation time was determined to be approximately 30 minutes at
37°C.
D. The bacteria showed no increase in turbidity when incubated in an ice bath.
14. Which of the following statements is the best explanation for what occurred in Group
2/Trial 2 between 10 and 24 hours?
A. The bacterial population grew exponentially.
B. The birth and death rate of bacterial cells were relatively equal during this time.
C. The bacteria were dying more rapidly than new bacteria could be generated.
D. No new bacteria were generated or died during this time.
Made by: Shahd A.Gaber
(Test 67-1)
More than 5 percent of Americans have asthma, a chronic disease that affects the
airways and lungs, causing shortness of breath, wheezing, and sometimes death. In the
United States, rates for asthma have steadily increased, nearly doubling during the past
20 years. There is no cure for asthma. Two researchers discuss factors that cause
individuals to develop asthma.
Researcher 1
There has long been an association between the allergen Dermatophagoides
pteronyssinus (dust mites) and asthma. Evidence for a causal relationship has been
supported by bronchial challenge studies and avoidance experiments. Studies have
shown that exposure in the child’s own house was the primary determinant of
sensitization. Research from around the world has provided evidence about other
indoor allergens, specifically cats, dogs, and the German cockroach. These studies
showed that perennial exposure to allergens was an important cause of inflammation in
the lungs and associated nonspecific bronchial hyperreactivity. Children are being
exposed to more perennial allergens now than ever before. Houses are built more
tightly and are better insulated and have more furnishings and fitted carpets. In addition,
children are spending more time indoors. This increased exposure to allergens,
including dust mites, has led to increased sensitization, and more cases of asthma.
Since assays for total serum IgE (immunoglobulin E) became available, it has been
clear that patients with asthma have, on average, higher total IgE than patients with hay
fever or no allergy. Recent work on patients hospitalized for asthma has suggested that
the interaction between rhinovirus and allergy occurs predominantly among patients
with total IgE > 200 IU/ml. Thus, the different properties of allergens could influence
both the prevalence and severity of asthma.
Researcher 2
It is widely accepted that air pollution exacerbates asthma. For example, when traffic
controls were put in place during the 1996 Summer Olympic Games in Atlanta, Georgia,
morning peak traffic counts declined by 23 percent. This in turn lowered ozone (O3)
concentrations by 13 percent, carbon monoxide (CO) by 19 percent, and nitrogen
dioxide (NO2) by 7 percent. Associated with these declines in ambient air pollution were
drops in Medicaid-related emergency room visits and hospitalizations for asthma (down
42 percent), asthma-related care for health maintenance organizations (down 44
percent), and citywide hospitalizations for asthma (down 19 percent). Despite such
striking relationships between exposure to air pollution and asthma aggravation, air
pollution has not been regarded as a cause of the disease. Increasingly, however,
recent studies have been suggesting that air pollution may, indeed, be a cause of
asthma.
Made by: Shahd A.Gaber
The Children’s Health Study (CHS) followed 3,535 children with no lifetime history of
asthma for five years. During that period 265 reported a new physician diagnosis of
asthma. Analysis of CHS data has shown that children living in communities with high
ozone levels developed asthma more often than those in less polluted areas. The
hypothesis that ozone might cause asthma is reinforced by a study of 3,091
nonsmoking adults aged 27 to 87 years who were followed for 15 years. The results of
this study showed that 3.2 percent of the men and 4.3 percent of the women reported
new doctor-diagnosed asthma. The researchers concluded that there was a connection
between ozone concentration and development of asthma.
1. If ozone levels decrease nationwide, Researcher 2 would expect to see what change
in asthma rates?
A. An increase in the prevalence of asthma
B. A decrease in the prevalence of asthma
C. No change in the prevalence of asthma
D. First a decrease and then an increase in the prevalence of asthma
2. Which of the following statements would Researcher 1 agree with?
F. Asthma rates are lower in rural areas.
G. Men are more likely to have asthma than women.
H. People who have pets are more likely to have asthma.
J. Asthma rates are related to the quality of air.
3. Researcher 1 would most likely agree with which of the following statements about
IgE?
A. People who have IgE levels of 400 IU/ml have a high chance of having severe
asthma.
B. People who have IgE levels of 100 IU/ml have a high chance of having severe
asthma.
C. Most people who have asthma have low levels of IgE; less than 200 IU/ml.
D. There has been no connection made between IgE levels and the prevalence of
asthma.
Made by: Shahd A.Gaber
4. If the prevalence of asthma in the United States continues to increase, Researcher 1
would likely cite which of the following as a solution to the problem?
F. People need to spend less time outside.
G. Houses need to be better insulated.
H. People need to be given supplements to increase their IgE levels.
J. Fans need to be added to houses to allow more circulation and to bring more
outside air into the house.
5. Researchers 1 and 2 would both agree with which of the following statements?
A. Asthma rates are likely to decline over the next 20 years.
B. Air pollution and high IgE levels are the two leading causes of asthma.
C. Women are more likely to develop asthma than men.
D. Measures can be taken to lower a person’s risk of developing asthma.
6. If Researcher 2 is correct, which of the following graphs would best represent the
relationship between CO concentrations and cases of asthma?
F.
G.
H.
J.
Made by: Shahd A.Gaber
7. Researcher 2 would most likely agree with which of the following statements
regarding the prevalence of asthma 20 years ago?
A. There was a higher prevalence of asthma 20 years ago because there was less
pollution.
B. There was a lower prevalence of asthma 20 years ago because there were
higher ozone levels and less pollution.
C. There was a lower prevalence of asthma 20 years ago because there was less
pollution and lower ozone levels.
D. There was a lower prevalence of asthma 20 years ago because people spent
more time outside.
(Test 68-1)
Pertussis, commonly known as whooping cough, is a highly infectious disease of the
respiratory tract caused by bacteria. The disease spreads by direct contact with
secretions from the nose or throat of an infected person, or by breathing in the air when
an infected person coughs. Pertussis most easily passes between people in the initial
stage of illness, but it can be spread at any time during the course of the illness. Figure
1 depicts the course of pertussis from exposure to recovery.
Figure 1
Made by: Shahd A.Gaber
The number of reported cases of pertussis from 1974 through 2004 is depicted in Figure
2.
Figure 2
The number of reported cases of pertussis in 2004 by age group is shown in Figure 3.
Figure 3
1. A new vaccine for pertussis was introduced in 1991, which claimed to be more
effective than the previous vaccine. Does the data in Figure 2 support this claim?
A. No. Rates of pertussis increased after 1991.
B. No. Rates of pertussis remained the same after 1991.
C. Yes. Rates of pertussis decreased after 1991.
D. Yes. Rates of pertussis increased after 1991.
2. If a person was experiencing a cough, was medicated, and was still contagious, he or
she would be in what stage of the disease?
F. Incubation
G. Catarrhal
H. Paroxysmal
J. Convalescent
Made by: Shahd A.Gaber
3. What is the maximum number of weeks that pertussis can be transmitted?
A. 1
B. 3
C. 8
D. 12
4. Doctors hypothesized that because of their immature immune systems young
children are the most susceptible to pertussis. Does the data from 2004, in Figure 3,
support this theory?
F. No. Pertussis mostly affects the elderly.
G. No. Ten- to fourteen-year-olds were the most likely to contract pertussis.
H. Yes. Infants had the highest rate of pertussis.
J. Yes. Seven thousand young children contracted pertussis.
5. Which of the following statements could be a plausible explanation for increased
incidence of pertussis?
A. Fewer infants are being vaccinated for pertussis, and are therefore contracting
the disease.
B. Regulations on reporting pertussis are more lax now, so the numbers are
inaccurate.
C. There are more blood transfusions performed now, increasing people’s risks of
contracting pertussis.
D. Vaccine immunity wanes after 5–10 years, so more young adults are succumbing
to pertussis.
6. Which of the following statements about pertussis is NOT true?
F. The disease has an incubation period ranging from 1 to 3 weeks.
G. Symptoms are similar to those of a common cold.
H. The disease is a virus that cannot be treated with antibiotics.
J. Pertussis can be communicated via particles left in the air after a person coughs.
Made by: Shahd A.Gaber
(Test 69-1)
The most common type of red-green color perception defect is caused by a mutation on
the X chromosome. X-linked red color blindness is a recessive trait. The eggs of the
mother will contain either a normal X chromosome (XR) or an X chromosome with the
mutation (Xr) causing red-green color blindness. The sperm of the father will contain the
normal X chromosome, the X chromosome with the mutation, or the Y chromosome.
Females heterozygous (one normal gene, one mutated gene) for this trait have normal
vision. The color perception defect manifests itself in females only when it is inherited
from both parents. By contrast, males inherit their sole X chromosome from their
mothers, and become red-green color blind when this one X chromosome has the color
perception defect.
Genotype refers to the combination of alleles that an individual has for a gene. Table 1
lists the possible genotypes for red-green color perception and their corresponding
effects on vision.
Table 1
Students interviewed members of the Allen family to investigate the inheritance of redgreen color blindness.
Figure 1 displays the family tree of the Allen family.
Figure 1
Study 1
Students interviewed Barbara and John and learned that Barbara is red-green color
blind, while John is not. Based on this information, the students deduced that their four
children have the following genotypes:
Made by: Shahd A.Gaber
Figure 2
Study 2
Students assumed that Liz would have children with a man who has normal vision, and
using a Pundit square calculated all of the possible genotypes for their children:
Figure 3
Study 3
Students assumed that David would have children with a woman who has red-green
color blindness, and using a Pundit square calculated all of the possible genotypes for
their children:
Figure 4
1. A female who has normal vision but is a carrier of red-green color blindness must
have which of the following genotypes?
F. XR XR
G. XR Xr
H. Xr Xr
J. XR Y
2. For a couple to produce only red-green colorblind children, regardless of the child’s
sex, they would need to have which of the following pairs of genotypes?
A. XR XR and Xr Y
B. Xr Xr and XR Y
C. XR Xr and Xr Y
D. Xr Xr and Xr Y
3. All of the male offspring exhibited red-green color blindness in Studies:
F. 1 and 3.
G. 1, 2, and 3.
H. 2 and 3.
J. 1 and 2.
Made by: Shahd A.Gaber
4. Suppose that the students who worked on Study 2 are correct, and that Liz has
children with a man who has normal vision. If they have 8 children, how many of them
would be expected to have red-green color blindness?
A. 1
B. 2
C. 4
D. 8
5. The ratio of Barbara and John’s offspring with normal vision to offspring with redgreen color blindness is:
F. 1:2
G. 2:1
H. 1:1
J. 3:1
6. Barbara’s parents could have had which of the following pairs of genotypes?
A. Xr Xr and XR Y
B. XR Xr and Xr Y
C. XR XR and Xr Y
D. XR XR and XR Y
(Test 71-1)
Ever since Darwin proposed his theory of natural selection in 1859, biologists have
regarded the gene as the sole unit of inheritance, and the discovery of DNA in the
twentieth century only served to reinforce this view, known as “hard inheritance.” The
idea that environmental factors could produce heritable changes in an organism without
altering the organism’s DNA (as occurs in gene mutation due to radiation exposure, for
example), known as “soft inheritance,” had long been written off as an impossibility.
“Inheritance of acquired traits” became a biological fallacy associated with preDarwinians like Jean-Baptiste Lamarck and fanatical Soviet scientists like Trofim
Lysenko. But the very recent discovery of what has been termed epigenetics—biological
mechanisms that leave DNA unchanged but can alter the ways in which individual
genes express themselves—has opened the door on the possibility of “soft inheritance”
once again, although many biologists remain skeptical.
Made by: Shahd A.Gaber
Scientist 1
Although study of the so-called “epigenome” is worthwhile and possesses the potential
to answer questions about a host of problems from diabetes to cancer, it would be
premature to call true “soft inheritance” a reality, at least in animal species. Isolated
situations in which environmental factors could affect an organism’s immediate
descendants—a mother with a zinc deficiency producing children and grandchildren
with weakened immune systems, for example—had already been documented without
anyone ever suggesting that there was more to heritability than the gene. Immediate
successive generations may exhibit observable effects, but the bloodline always reverts
to the true expression of its DNA. A map with dust on it may be hard to read, but the
information on the map remains unchanged, and it is simply a matter of how long it
takes for a strong wind to blow away the dust.
Scientist 2
Someone who is determined never to see “true soft inheritance” will never see it, but
that doesn’t change the fact that it occurs, and indeed has been occurring all along
without our realizing it. Epigenetic mechanisms may not alter DNA, but they possess the
ability to turn genes “on” or “off” for the duration of an organism’s bloodline, at least until
such time as the genes are affected by another epigenetic mechanism. There is already
compelling evidence, for example, that tobacco smoking poses risks not only to the
individual smoker, but to all of his or her descendants. Imagine a three-way light bulb
with three possible brightness settings being put into a three-way lamp on which one of
the settings is broken. The fact that the bulb still possesses three settings is immaterial
if it is permanently set into a broken lamp. It will continue to “express itself” in a limited
way. And the same thing can happen to a gene. If environmental factors can
permanently alter the way in which an unaltered gene expresses itself down along a
bloodline, resulting in observable effects in the members of that bloodline, then what
can we possibly call this other than true soft inheritance?
1. According to the information in the passage, which of the following might constitute
an example of true soft inheritance?
A. inherited gene damage due to a parent’s heavy radiation exposure
B. a susceptibility to diabetes as the result of a malnourished grandparent
C. a left-handed parent producing predominantly left-handed offspring
D. a parent who loves books teaching her children to love books
Made by: Shahd A.Gaber
2. Who first proposed the theory of epigenetics?
F. Charles Darwin
G. Trofim Lysenko
H. Scientist 2
J. The information is not included in the passage.
3. In the offspring of which of the following organisms might Scientist 1 currently be
willing to concede the documented existence of true soft inheritance?
A. a rhinoceros
B. a radiation-exposed human being
C. an apple tree
D. a zinc-deficient human being
4. Scientist 2 would most likely disagree with the suggestion that genes:
F. are the primary units of inheritance.
G. act in isolation to produce traits.
H. possess the ability to express themselves.
J. are made up of stretches of DNA.
5. Scientist 1 would be most likely to predict that any ill effects on the grandchildren of a
tobacco smoker:
A. have nothing to do with the epigenome.
B. are probably psychological rather than biological in origin.
C. must be the result of tobacco use having altered the smoker’s DNA.
D. will cease to be expressed in the bloodline given enough time.
6. Histones are proteins that attach themselves to sequences of DNA and can alter the
expression of genes depending on whether they are acetylated or methylated. Given
this information, it would appear that histones are analogous to the:
F. sole unit of inheritance.
G. zinc-deficient mother mentioned by Scientist 1.
Made by: Shahd A.Gaber
H. map mentioned by Scientist 1.
J. lamp mentioned by Scientist 2.
7. A study of what would be most likely to resolve the disagreement between Scientist 1
and Scientist 2?
A. the length of time that epigenetic mechanisms tend to remain in a bloodline
B. the unpublished work of Jean-Baptiste Lamarck
C. a comparison of cancer rates among identical twins
D. the inner workings of three-way light bulbs
(Test 71-2)
Figure 1 shows the average sleep pattern of a child, Figure 2 shows the average sleep
pattern of a young adult, and Figure 3 shows the average sleep pattern of an elderly
person. At Stage 0, the person is awake. As sleep moves from Stage 1 to Stage 4, it
grows progressively deeper. REM sleep, commonly associated with dreaming, is
predominant in the final third of a sleep cycle.
Figure 1
Figure 2
Made by: Shahd A.Gaber
Figure 3
8. According to Figure 2, a young adult who has been asleep for 5 hours will most likely
be in which sleep stage?
F. Stage 0
G. Stage 1
H. Stage 3
J. Stage 5
9. Based on the information in the passage, a child will wake up how many times during
a 7-hour stretch of sleep?
A. 7
B. 2
C. 1
D. 0
10. Based on the data presented in Figures 1, 2, and 3, which of the following
conclusions can be properly drawn?
F. As people age, they wake up more frequently during the night.
G. As people age, they wake up less frequently during the night.
H. As people age, they spend more time in deep sleep.
J. As people age, they spend more consecutive time in each sleep stage.
Made by: Shahd A.Gaber
11. Based on the information in the passage and in Figure 2, at which of the following
hours into a sleep interval will a young adult be most likely to experience REM sleep?
A. 0
B. 1
C. 2
D. 7
12. At which of the following hours of sleep will a child most likely be in the deepest
sleep?
F. 2
G. 3
H. 5
J. 7
SUBMIT
(Test 73-2)
Blood type is a hereditary trait. The type is established by the genes inherited from the
mother and father. The ABO system is widely accepted as the best blood classification
system. In the ABO system, there are four types of blood: A, B, AB, and O. The
combination of inherited genes is known as the genotype and the actual blood type is
known as the phenotype. The genes ensure that only the blood cells of the proper blood
type remain in the body.
Table 1
Made by: Shahd A.Gaber
The Rh (+/–) factor is inherited separately from the blood type. It is possible to have the
Rh+ phenotype yet still carry the recessive gene for Rh–. The Rh+ (R) is the dominant
gene and Rh– (r) is recessive. Table 2 shows the Rh phenotypes resulting from the
various genotypes inherited from parents.
Table 2
The surface of every red blood cell is covered with proteins. Rh factor and blood type
determine the proteins and the compatibility of donated blood as shown in Table 3.
Table 3
6. According to Table 1, parents with blood types O and B can only produce offspring
with which blood types?
A. B or O
B. A or B
C. A or O
D. A
7. Parents with which blood types could produce offspring with AB+ blood?
F. O+ and A–
G. A– and B+
Made by: Shahd A.Gaber
H. AB– and AB–
J. B+ and B+
8. A person who can donate blood to anyone could have parents with which of the
following blood types?
A. A+ and AB–
B. AB– and AB–
C. B– and B–
D. AB+ and AB+
9. List all of the blood types possible for the offspring of parents of blood types A+ and
O+.
F. A+ or O+
G. A+, A–, O–, or O+
H. O– or O+
J. AB+, A–, or O–
10. The genes that determine blood type are also responsible for:
A. Rh factor.
B. controlling the types of cells in the blood.
C. controlling blood volume.
D. creating proteins on white blood cells.
(Test 74-1)
In a study of the effects of Ritalin and Adderall on children with ADHD, subjects were
given one of four possible doses of medication. Their behavior in social and academic
settings was then monitored and rated. The four possible doses were placebo (P),
Ritalin given once in the morning (R1), Ritalin given twice daily (R2), or Adderall given
once in the morning (A1).
The results for each group were averaged. Figure 1 shows the average behavioral
rating (on a scale of 0–15, with 0 meaning no undesirable behavior) at various time
periods throughout the day. Figure 2 shows the percentage of children who
demonstrated side effects at a moderate or severe level on at least one day.
Made by: Shahd A.Gaber
Figure 1
Figure 2
1. Based on Figure 1, during which of the following time periods was the average
behavior rating most similar for the four groups of children?
F. 8:15–9:25
G. 9:25–10:35
H. 10:35–11:45
J. 3:35–4:45
Made by: Shahd A.Gaber
2. A scientist claimed that children given one dose of Adderall daily would exhibit fewer
behavior problems than children given either one or two doses of Ritalin daily. During
which of the following time periods shown in Figure 1 are the results inconsistent with
this claim?
A. 9:25–10:35
B. 10:35–11:45
C. 2:35–3:35
D. 3:35–4:45
3. According to Figure 2, for the group given Ritalin twice daily, the percentage of
children who experienced an adverse side effect was greatest for which side effect?
F. Dull
G. Headache
H. Withdrawn
J. Appetite loss
4. According to Figure 1, which dose of medication was the least successful in
controlling children’s behavior problems from 3:35–4:45?
A. P
B. R1
C. R2
D. A1
5. Suppose four groups of children were given one of the four possible medication
regimens. Between 12:15 and 1:25, one group had a behavior rating of 5. Which
medication regimen was most likely given to them?
F. P
G. R1
H. R2
J. A1
Made by: Shahd A.Gaber
6. Assume that an ideal medication is one that has the fewest side effects, yet is most
effective. Based on the data provided, which dose of medication is the most ideal?
A. P
B. R1
C. R2
D. A1
(Test 80-2)
Scientists studying sucrose examined oranges and lemons to determine how the two
fruits form and synthesize sucrose. Studies were conducted both on extractions from
the fruits and on small, intact fruits.
Study 1
Mature lemon and orange fruits were obtained and then juiced by hand. Formation of
fructose was determined using two portions of the fruits: the juice, and the particulate
sediment. The results are shown in Table 1.
Table 1—Formation of Sucrose by Fructose Reaction
Preparation
Sucrose Synthesized (μ moles)
Oranges
Juice
Particulate sediment 0.12
0.9
Lemons
Juice
Particulate sediment 0.4
0.17
Study 2
Scientists found lemons of varying ages from the same fruit grove. Each piece of fruit
was cut into sections, and various tissue samples were isolated for testing. The three
tissue types tested were the flavedo (colored, outer layer of the peel), the albedo (white
peel layer), and the vesicle (fleshy part of the fruit). The results of the formation of
sucrose in lemons are shown in Figure 1.
Made by: Shahd A.Gaber
Figure 1
Study 3
Whole fruits (each weighing approximately 5 g) were tested and then cut into sections to
measure sucrose activity in the various tissues. Figure 2 displays the results.
Figure 2
6. Based on Figure 1, which type of lemon tissue forms the most sucrose?
A. The albedo tissue in a large lemon
B. The albedo tissue in a small lemon
C. The flavedo tissue in a large lemon
D. The vesicles in a small lemon
7. According to the results of Study 2, as fruit size increased, the sucrose found in the
vesicles:
F. increased only.
G. increased, then decreased.
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H. decreased only.
J. decreased, then increased.
8. Based on the results of Study 3, the largest difference in sucrose activity was found
in:
A. oranges between the flavedo tissue and the albedo tissue.
B. oranges between the albedo tissue and the vesicle tissue.
C. lemons between the flavedo tissue and the vesicle tissue.
D. lemons between the flavedo tissue and the albedo tissue.
9. Based on the results of Study 1, which of the following is most accurate about the
formation of sucrose?
F. Orange juice is more effective at forming sucrose than lemon juice.
G. Orange juice is less effective at forming sucrose than lemon juice.
H. Lemon juice and orange juice are equally effective at forming sucrose.
J. Lemon juice forms less sucrose than lemon particulate sediment.
10. Assume Study 2 was repeated using oranges instead of lemons. Based on the
information presented in Figures 1 and 2, and assuming that the fruits used in Study 3
were 30 mm in diameter, which of the following would most likely be the sucrose
formation (as a percent of total activity) in the albedo tissue of the oranges?
A. 0
B. 3
C. 9
D. 20
11. Based on the results of Study 2, which statement most accurately summarizes the
formation of sucrose in lemons?
F. The fleshy part of the lemon formed the majority of the sucrose.
G. The colored outer layer of the peel does not form any sucrose.
H. The part of the lemon that forms the most sucrose is dependent on the size of the
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lemon.
J. More sucrose is formed in the white peel layer of a lemon than anywhere else.
(Test 81-2)
Since carbon is essential to life on Earth, understanding the global carbon cycle can
provide valuable information. Through the process of photosynthesis, plants on land or
algae living in water convert carbon dioxide into organic products. Through the process
of respiration, organisms convert these organic compounds into energy and release
carbon dioxide. Figure 1 displays the global carbon cycle.
Figure 1
Made by: Shahd A.Gaber
A significant amount of carbon is released through the burning of fossil fuels. Some of
this carbon is stored in the atmosphere, while a significant portion is stored in ocean
waters and sediments. Carbon in the ocean is described as dissolved inorganic carbon,
particulate inorganic carbon, particulate organic carbon (POC), and dissolved organic
carbon (DOC). Particulates are defined as those larger than 0.2 micrometers, whereas
dissolved matter is that smaller than 0.2 micrometers (see Figure 2).
Figure 2
Chromophoric dissolved organic matter (CDOM) is the colored portion of DOC. It is
colored because it intensely absorbs violet and blue light. Figure 3 displays DOC and
CDOM absorption in the tropical North Atlantic Ocean.
Made by: Shahd A.Gaber
Figure 3
6. Based on Figure 1, how much carbon is transmitted from the ocean to the
atmosphere?
F. 91.9 Pg C/yr
G. 90.6 Pg C/yr
H. 918 Pg C/yr
J. 700 Pg C/yr
7. According to Figure 2, which of the following is NOT considered a dissolved carbon?
A. Proteins
B. Salts
C. Algae
D. Viruses
8. Based on Figure 3, which of the following statements best describes the relationship
between ocean depth and amount of DOC?
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F. As the ocean depth increased, the amount of DOC remained fairly constant, then
decreased steadily, then began to increase again.
G. As the ocean depth increased, the amount of DOC steadily decreased.
H. As the ocean depth increased, the amount of DOC remained fairly constant, then
decreased steadily, then became fairly constant again.
J. As the ocean depth increased, the amount of DOC steadily increased.
9. Suppose a fragment of organic carbon was found in the ocean and measured
approximately 0.01 micrometers. It would mostly likely be featured in Figure 2 by which
of the following?
A. Gases
B. Protozoa
C. Bacteria
D. Small polymers
10. At an ocean depth of 100 m, there is approximately how much CDOM, according to
Figure 3?
F. 0.25
G. 0.12
H. 40
J. 70
11. Which of the following fragments, if found in the ocean, would be described as
dissolved inorganic carbon?
A. Bacteria
B. Protozoa
C. Viruses
D. Polymers
(Test 82-1)
Made by: Shahd A.Gaber
Three substances have been shown to inhibit respiration in various organisms:
hydrocyanic acid, hydrogen sulfide, and carbon monoxide. An experiment was
conducted to determine the effect that each of these substances has on the respiration
of the green alga Chlorella. Figure 1 shows the results.
Figure 1
Researcher 1 argued that glucose would have a contradictory effect on the alga, so that
if the Chlorella were suspended in a solution containing 1 percent glucose, there would
be less of an effect on respiration. Figure 2 displays the results of the experiment.
Figure 2
Researcher 2 hypothesized that light may play an active part in respiration of Chlorella,
so an experiment was done to measure the effect of successive periods of light and
darkness on the respiration of cells suspended in carbon monoxide and in nitrogen. The
results are shown in Figure 3.
Made by: Shahd A.Gaber
Figure 3
1. The theories of the two researchers are similar in that both researchers believe that:
F. glucose will increase the respiration of Chlorella.
G. HCN and H2S have no effect on the respiration of Chlorella.
H. an additional factor will have an effect on the respiration of Chlorella.
J. CO and N also affect the respiration of Chlorella.
2. According to Figure 1, Chlorella would typically consume about how much oxygen
after half an hour?
A. 10 mm3
B. 18 mm3
C. 22 mm3
D. 35 mm3
3. If Researcher 1 was correct about the effect of glucose on the respiration of Chlorella,
then based on the information in Figure 2, Researcher 1 would most likely predict
that Chlorella suspended in a solution containing 2 percent glucose and HCN would
consume how much oxygen after 20 minutes?
F. 15 mm3
G. 25 mm3
H. 30 mm3
J. 40 mm3
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4. Does the data in Figure 3 support Researcher 2’s hypothesis?
A. Yes. It shows that light is necessary for the respiration of Chlorella.
B. Yes. It shows that equal respiration occurs in the darkness and in light.
C. No. More respiration of Chlorella occurred in the darkness.
D. No. Varying the amount of light did not yield significantly different rates of
respiration.
5. Suppose a third researcher studied the effect of glucose and CO on the respiration
of Chlorella, using the same conditions and methods as Researcher 1. After 20 minutes
the results showed that the sample suspended in CO had consumed 50 mm 3 of oxygen,
while the sample suspended in a mixture of CO and glucose had consumed 60 mm 3 of
oxygen. How would this experiment most likely affect the researchers’ viewpoints?
F. It would weaken Researcher 2’s viewpoint only.
G. It would weaken Researcher 1’s viewpoint only.
H. It would strengthen both researchers’ viewpoints.
J. It would have no effect on either researcher’s viewpoint.
6. According to Figure 2, which substance had the LEAST effect on respiration
of Chlorella after 30 minutes?
A. HCN
B. H2S
C. Glucose + HCN
D. Glucose + H2S
7. According to Researcher 2, experiment findings indicate the following is FALSE
regarding the relationship between Chlorella cells suspended in carbon monoxide
and Chlorella cells suspended in nitrogen EXCEPT:
F. Chlorella cells suspended in carbon monoxide consume more oxygen.
G. Chlorella cells suspended in nitrogen consume less oxygen.
H. Chlorella cells suspended in carbon monoxide and nitrogen consume the same
amount of oxygen.
J. Chlorella cells suspended in nitrogen consume more oxygen.
Made by: Shahd A.Gaber
(Test 83-1)
Table 1 lists two genes found in Sesamum indicum (sesame), the possible alleles of
each gene, and the possible genotypes for each gene.
Table 1
Table 2 lists various sesame genotypes and the phenotype associated with each
genotype. Each gene affects only one of the phenotype traits listed.
Table 2
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Table 3 lists four sesame plant crosses, the genotypes of the parents, and the percent
of offspring that had each phenotype for the traits listed in Table 2. In each cross, each
parent donated one allele to each offspring at each gene.
Table 3
1. Based on Table 2, which of the 2 genes affects leaf texture?
A. P
B. L
C. Neither
D. Both
2. Based on Table 2, a sesame plant with 2 recessive alleles for each of the 2 genes will
have which of the following phenotypes?
F. 1 pod and normal leaves
G. 3 pods and normal leaves
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H. 3 pods and wrinkled leaves
J. 1 pod and wrinkled leaves
3. Based on the information provided, all of the offspring of Cross 3 had 3 pods because
each received:
A. allele p from its female parent and allele p from its male parent.
B. allele L from its female parent and allele l from its male parent.
C. allele l from its female parent and allele L from its male parent.
D. allele P from its female parent and allele P from its male parent.
4. In Cross 3, what percent of the offspring had genotype pp?
F. 0%
G. 25%
H. 50%
J. 100%
5. Based on the information provided, a sesame plant with 3 pods and normal leaves
could have which of the following genotypes?
A. PPll
B. PPLL
C. ppll
D. ppLl
(Test 83-2)
To determine the effect of ambient ultraviolet radiation (UVR) on
the chlorophyll production, growth rate, and cell death of Antarctic phytoplankton,
scientists conducted experiments on phytoplankton communities exposed to natural
levels of solar radiation.
Experiments were conducted during three different periods of time. Experiment 1 took
place February 1–6, Experiment 2 February 7–12, and Experiment 3 February 13–20. A
meteorological station automatically recorded solar radiation for the duration of the
study (see Figure 1).
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The experiments involved sampling surface seawater, placing them in bottles
submersed in incubators, and exposing them to either natural solar radiation (UVR) or
radiation filtered to exclude ultraviolet radiation (PAR).
Figure 1
Every two days, duplicate samples were taken from each location to determine
chlorophyll amounts and phytoplankton cell death (see Table 1 and Figure 2). Net
population growth rates were calculated for diatoms and flagellates from the cell
abundances obtained at the beginning and end of the experiments (see Figure 3).
Table 1
Figure 2
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Figure 3
6. Based on the data in Table 1, how much chlorophyll was measured in Experiment 3
for the sample exposed to UVR after 4 days?
F. 1
G. 1.3
H. 3
J. 4
7. Based on the information presented in Table 1, UVR seems to have what effect on
the amount of chlorophyll measured?
A. UVR seems to inhibit the production of chlorophyll.
B. UVR seems to stimulate the production of chlorophyll.
C. UVR seems to have no effect on the production of chlorophyll.
D. It cannot be determined what the effect of UVR is on chlorophyll.
8. According to the data shown in Figure 3, which cells showed the least growth in
Experiment 3?
F. Diatoms shielded from UVR
G. Diatoms exposed to UVR
H. Flagellates shielded from UVR
J. Flagellates exposed to UVR
9. Based on the information presented in Table 1, if Experiment 1 had been continued
for another 2 days, the amount of chlorophyll measured in the samples that were not
exposed to UVR would most likely be closest to what amount?
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A. 0
B. 17
C. 3
D. 36
10. Based on the results displayed in Figure 2, it can be assumed that ultraviolet
radiation (UVR) has what effect on the life of the diatoms and flagellates found in
phytoplankton?
F. The presence of UVR leads to fewer dead diatoms and more dead flagellates.
G. The presence of UVR leads to more dead diatoms and fewer dead flagellates.
H. The presence of UVR leads to fewer dead cells for both diatoms and flagellates.
J. The presence of UVR leads to more dead cells for both diatoms and flagellates.
(Test 84-1)
Ecological indicators are used to assess the condition of the environment, to provide
early warning signals of changes in the environment, or to diagnose the cause of
environmental problems. Some important criteria for selecting an indicator include:
1. The indicator should provide an accurate picture of what it is supposed to indicate,
such as “ecosystem health.”
2. The indicator should respond quickly to environmental changes.
3. The indicator should be easy to monitor.
4. The responses of the indicator in one or a few locations should indicate the state of
the ecosystem in a larger area.
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It has been suggested that seabirds are useful indicators. Two researchers discuss the
effectiveness of using seabirds as indicators.
Researcher 1
Seabirds are valuable indicators because they are top predators in their ecosystem.
Seabird populations and reproduction rates are regulated by prey abundance, and will
therefore reflect environmentally induced changes in prey availability. The effects of
reductions in prey resources, such as declining numbers of seabirds, are usually very
rapid, due to the short food chain, which is another advantage.
A few species of small fishes are important species of the ecosystem, having a major
effect on the ecosystem as a whole. Seabirds that eat mainly these key fish are good
indicators of the ecosystem in general.
Additional demographic parameters that can easily be monitored in seabirds are
population size, duration of foraging trips, and changes in body mass and offspring
growth rate, all of which are useful environmental monitors.
Overall, seabirds are a cost-effective, useful, and meaningful indicator of environmental
changes in ocean ecosystems.
Researcher 2
There are several reasons why seabirds may not be suitable as indicators of the
impacts of environmental changes. First of all, not all marine ecosystems follow a topdown food chain. Some marine food webs are dynamic and can alternate between
bottom-up and topdown. In addition, change in seabird numbers due to food scarcity
usually has a lag of several months or even years. Because of this seabirds are not
suitable as indicators of the food supply in all ecosystems.
In general, the effect of environmental change on seabird populations may take many
years to become apparent. Its effect is complex and involves many physical and
biological processes. This is a severe drawback to using seabirds as indicators. Another
drawback is the handling effect. It is not fully understood what effect handling seabirds
has on them. There is some evidence that bands placed on penguins to track them can
reduce both breeding success and survival rates. With seabirds that are tagged and
handled, it is difficult to be sure that observed effects are actually due to changes in the
environment and not other factors.
The effect of environmental changes on seabirds is complex and it is not clear what the
best way to monitor seabirds is. Therefore, seabirds are not a good choice for
indicators.
1. Based on Researcher 1’s discussion, demographic factors that can be easily
monitored in seabirds include all of the following EXCEPT:
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A. population size.
B. changes in offspring growth rate.
C. mortality rates.
D. length of foraging trips.
2. Sardines are plentiful in Chiriqui Bay, and their presence affects the entire ecosystem
of the bay. Kingfisher birds eat mostly sardines and because their numbers are
declining, scientists are concerned about the environmental health of Chiriqui Bay. Is
this concern consistent with the viewpoint of Researcher 1?
F. No, because Researcher 1 states that changes in the food chain are not
adequate indicators of the environmental health of a body of water.
G. No, because Researcher 1 states that only changes in body mass and offspring
growth rate are valuable indicators of environmental changes.
H. Yes, because Researcher 1 states that the effect of environmental changes on
seabirds will be slow due to the short food chain in a marine ecosystem.
J. Yes, because Researcher 1 states that seabirds that consume key fish in an
ecosystem are valuable indicators of the ecosystem in general.
3. Which researcher believes that handling seabirds may affect the validity of the results
gathered by monitoring them?
A. Researcher 1, because that researcher states that changes in prey resources are
very quick, and therefore results gathered will be valid.
B. Researcher 1, because that researcher states that overall, seabirds are a useful
indicator of environmental changes.
C. Researcher 2, because that researcher states that placing bands on penguins
has been shown to reduce their survival rates.
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D. Researcher 2, because that researcher states that it is clear that handling
seabirds will always reduce their breeding rates.
4. A study found that within two months of an oil spill in a bay, there was a drastic
reduction in the number of small forage fish found in the water, and that the number of
seabirds in the area was drastically reduced as well. Which researcher would most
likely use this study to support his or her viewpoint?
F. Researcher 1, because it would demonstrate how quickly environmental changes
impact prey resources and seabird behavior.
G. Researcher 1, because it would demonstrate how important seabirds are for the
overall ecosystem.
H. Researcher 2, because it would demonstrate how quickly environmental changes
impact prey resources and seabird behavior.
J. Researcher 2, because it would demonstrate how important seabirds are for the
overall ecosystem.
5. Based on Researcher 2’s discussion, seabirds would not be effective indicators
because they do not meet which criteria of a good indicator?
A. Point 1 only
B. Points 1 and 2 only
C. Points 1, 2, and 3 only
D. Points 2 and 3 only
6. Which of the following graphs is most consistent with Researcher 1’s view on the
relationship between prey abundance and seabird population?
F.
G.
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H.
J.
7. Researcher 2 cites all of the following as reasons why seabirds may not be suitable
as indicators EXCEPT:
A. the impact of food scarcity on seabird numbers may not be apparent for several
months or years.
B. it is not known how handling seabirds might impact them.
C. it is expensive to use seabirds as indicators.
D. it is uncertain how to best monitor seabirds.
(Test 85-2)
Scientists conducted a series of experiments to determine the effect of changes in
temperature on the intertidal rocky shore crab, Petrolisthes granulsus (P. granulosus).
Several fitness-related traits, such as body size and reproductive capacity in P.
granulosus individuals from three sites were compared after the crabs were exposed to
various temperatures. In addition, metabolic rate experiments were conducted to
determine the energetic cost associated with crab exposure to high temperatures.
Table 1 displays the temperatures (°C) used in the experiments. Thermal category TC1
refers to the maximum temperature registered at each site, while TC2 refers to the
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average of maximum temperatures recorded every day. The Control thermal category is
defined as the average temperature experienced by the crabs during the acclimation
period.
Table 1
Figure 1 displays the average size (cephalothorax length) of P. granulosus in the 3
populations, while Figure 2 displays the average egg volume in the 3 populations.
Figure 1
Figure 2
Figure 3 shows the average standard metabolic rates (SMR) of male crabs from the 3
populations exposed to 3 thermal categories.
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Figure 3
7. According to Table 1, crabs from which site were exposed to the highest
temperatures during the experiments?
A. Iquique
B. Coquimbo
C. Concepcion
D. Crabs at all of the locations were exposed to the same maximum temperatures.
8. Based on the information in Figure 2, the average eggs of crabs from the Coquimbo
site were found to be closest to which size?
F. 0.1 mm
G. 0.17 mm
H. 0.2 mm
J. 0.25 mm
9. According to Figure 3, adult male crabs from which category were found to have the
highest average standard metabolic rates?
A. TC1 from Concepcion
B. Control from Coquimbo
C. TC1 from Iquique
D. TC2 from Concepcion
10. Based on the information provided, which of the following best describes the effect
that average maximum temperature has on the average size of the crabs?
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F. As the average maximum temperature increased, the average size of the crabs
stayed the same.
G. As the average maximum temperature increased, the average size of the crabs
increased.
H. As the average maximum temperature decreased, the average size of the crabs
decreased.
J. As the average maximum temperature decreased, the average size of the crabs
increased.
11. Suppose the scientists had measured the average egg volume of crabs in the
control groups. Based on the information in Table 1 and Figure 2, the average egg
volume of crabs at the Concepcion site would have been closest to:
A. 0.05 mm.
B. 0.1 mm.
C. 0.2 mm.
D. 0.3 mm.
12. The scientists started this experiment with the theory that crabs exposed to higher
temperatures would develop higher standard metabolic rates. Do the results of the
experiment support this theory?
F. Yes. The crabs at Iquique in the thermal category TC2 were exposed to the
highest temperatures and were found to have the highest SMR.
G. Yes. At all locations the crabs from thermal category TC1 were found to have the
highest SMR.
H. No. At all locations the crabs from thermal category TC1 were found to have the
lowest SMR.
J. No. The results were mixed. At two locations the crabs from thermal category
TC1 were found to have the highest SMR, but at one location the crabs from the control
group had the highest SMR.
(Test 86-2)
Made by: Shahd A.Gaber
Recombination of genes is usually associated with the sexual reproduction of cells, or
meiosis. However, it can also occur when cells that undergo asexual reproduction, or
mitosis, need to be repaired, such as after radiation exposure. This repair process,
known as homologous recombination, aligns two copies of the same double strand of
DNA, one with the error and one without. As seen in Figure 1, correct genes are
transplanted from the correct strand to the one with errors (genes with errors are
represented with a *).
Figure 1
The activities of some genes have been found to promote homologous recombination
(HR). In an experiment to quantify the genetic control over HR, 4 scientists measured
the frequency of HR per hour over a 24-hour period in isolated connective tissue cells
from rats placed in growth media. They then lysed the cells, separated out the entire
protein content, and used gel electrophoresis to count the amount of protein present in
the cells (see Figure 2).
Made by: Shahd A.Gaber
Figure 2
These scientists noticed that only a few specific proteins appeared to be responsible for
promoting HR, and labeled the genes encoding them as W, X, Y, and Z. They
engineered cells to express combinations of two active genes and recorded the HR.
They then analyzed the DNA content of the lysed cells and calculated distances
between four genes that encoded the relevant proteins (see Table 1).
Table 1
Genes HR
Distance between genes (centimorgans)
(events per hour)
W and X 75
20
X and Y 125
30
W and Z 60
15
Each of the 4 scientists then proposed individual models for the positions of the genes
they studied, taking into account the findings in Table 1. Each model shows where
genes may be located along a strand of DNA (see Figure 3). Each model correctly
assumes that the lengths of the genes are insignificant compared to the length of the
DNA.
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Figure 3
A final experiment showed that rat connective tissue cells in which genes W and Y were
active had an HR frequency of 45 times per hour.
6. All 4 models agree on the distance between which of the following pairs of genes?
F. Genes W and X
G. Genes W and Y
H. Genes X and Z
J. Genes Y and Z
7. According to Figure 2, if some of the connective tissue cells had a protein content of
3,500 molecules per cell, the HR of these cells is most likely closest to which of the
following?
A. 50 events per hour
B. 100 events per hour
C. 150 events per hour
D. 200 events per hour
8. If Scientist 2's model is correct and an additional gene, Gene V, is 10 centimorgans
from Gene X and 15 centimorgans from Gene Z, then Gene V is most likely between:
F. Genes W and X.
G. Genes W and Z.
H. Genes X and Y.
J. Genes X and Z.
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9. The result of the final experiment studying the distance between Genes W and Y is
consistent with models proposed by which of the following scientists?
A. Scientists 1 and 3
B. Scientists 1 and 4
C. Scientists 2 and 3
D. Scientists 3 and 4
10. Based on the information provided, HR would occur when connective tissue cells
are exposed to:
F. growth media.
G. sexual reproduction.
H. asexual reproduction.
J. X-rays.
11. Which scientist's model proposes that Genes Y and Z are separated by 65
centimorgans?
A. Scientist 1's
B. Scientist 2's
C. Scientist 3's
D. Scientist 4's
12. Genes A and B are separated by 10 centimorgans on a chromosome. An organism
has alleles A and B* on 1 chromosome and alleles A* and B on the homologous
chromosome. If a single HR event occurred between these 2 genes as shown in Figure
1, the genotype of Genes A and B for the 2 chromatids involved in the crossover would
be:
F. AB and AB.
G. AB and A*B*.
H. A*B and AB*.
J. A*B* and A*B*.
SUBMIT
Made by: Shahd A.Gaber
(Test 88-1)
Simple diffusion (SD) is the process by which an uncharged solute in water migrates
directly across an uncharged membrane, while facilitated diffusion (FD) is the process
by which a charged or polar solute travels through a channel or transporter that crosses
the membrane. Figure 1 illustrates how two solutes can diffuse, one by SD and one by
FD.
Figure 1
Solutes that cross a membrane by SD or by FD show different rates of flow across a
membrane, also known as flux. As a solute crosses a membrane by SD, the flux follows
a linear pattern over time, with smaller solutes having the greatest increase in flux over
time. As a solute crosses a membrane by FD, the flux follows a logarithmic pattern,
leveling off at a maximum flux since there are only a limited number of channels or
transporters through which the solute can travel.
Experiment 1
One scientist introduced five different solutes of the same concentration to similar
membranes at a constant temperature. The molecular masses of these solutes are
shown in Table 1.
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Table 1
Solute Molecular mass
(amu)
#1
160
#2
800
#3
2,000
#4
10,000
#5
40,000
This scientist then measured the time it took for the solute to reach equilibrium, which is
a state of equal concentration of the solute on both sides of the membrane. The results
are shown in Figure 2.
Figure 2
Experiment 2
Mixtures of solutes are subsequently introduced near three different membranes with
different properties. The results of these three trials are presented in Figure 3.
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Figure 3
1. Based on the results of Experiments 1 and 2, Mixture #3 is likely to consist of which
solutes from Experiment 1 ?
A. Solute #1 only
B. Solutes #1 and #3 only
C. Solutes #3 and #5 only
D. Solutes #2, #4, and #5 only
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2. In Experiment 1, which solute spends the least amount of time flowing across the
membrane before reaching equilibrium?
F. Solute #1
G. Solute #2
H. Solute #3
J. Solute #4
3. Based on the results of Experiments 1 and 2, which of the following ranks Solute #3,
Solute #4, and Mixture #2 in order of smallest to largest average molecular mass?
A. Solute #3, Solute #4, Mixture #2
B. Solute #4, Mixture #2, Solute #3
C. Mixture #2, Solute #3, Solute #4
D. Mixture #2, Solute #4, Solute #3
4. In Experiment 1, on average, did molecules of Solute #3 or molecules of Solute #4
more easily diffuse across the membrane?
F. Solute #3, because it has a larger molecular mass.
G. Solute #3, because it has a smaller molecular mass
H. Solute #4, because it has a larger molecular mass.
J. Solute #4, because it has a smaller molecular mass.
5. In which mixture is the molecular mass most likely less than 160 amu ?
A. Mixture 1
B. Mixture 2
C. Mixture 3
D. Neither Mixture 1, 2, or 3
6. How does the number of molecules in 1 gram of Solute #1 compare with the number
of molecules in 1 gram of Solute #5 ? The number of molecules in 1 gram of Solute #1
is:
F. less, because Solute #1 has a larger molecular mass than Solute #5.
G. less, because Solute #1 has a smaller molecular mass than Solute #5.
Made by: Shahd A.Gaber
H. more, because Solute #1 has a larger molecular mass than Solute #5.
J. more, because Solute #1 has a smaller molecular mass than Solute #5.
(Test 88-2)
The term "evolution" is often used in the context of biological changes in organism
populations over time, but it can also be applied to the change in the chemical
composition of the Earth's atmosphere. The hypotheses of two studies claim that this
chemical evolution has altered the types of chemicals found in the atmosphere between
the early stages of Earth's existence and the present day.
Study 1
Based on the hypothesis that volcanic eruptions were the source of gases in the early
Earth's atmosphere, scientists recreated four model volcanic eruptions in closed
chambers, each containing different percentages of the same volcanic particulate
matter. They then observed the gases in the air above this model over time. The
percent composition of this air after 1 day, when the air achieved a steady state of
constant gas concentrations, is represented in Table 1.
Since the experiment provided only a suggestion of the gas levels in the early Earth's
atmosphere, the scientists then analyzed the amount of trapped gases in sediment
layers, which indicate the changing atmospheric levels of gases over billions of years.
The data collected on O2 and H2O vapor are presented in Figure 1.
Study 2
A separate study used the same volcanic models as in Study 1, but it hypothesized that
the scientists in Study 1 underestimated the amount of H2 in the early Earth
atmosphere. They proposed a different composition of gases, highlighting an increased
H2 level in the atmosphere, also represented in Table 1. Based on these new data, the
scientists proposed an alternative graph for the changing atmospheric levels of O2 and
H2O vapor, also shown in Figure 1.
Made by: Shahd A.Gaber
Figure 1
Made by: Shahd A.Gaber
7. According to the results of Study 2, between 4 and 3 billion years before the present
day, the percent composition of O2 in the atmosphere:
A. increased only.
B. increased, then decreased.
C. decreased only.
D. decreased, then increased.
8. According to the results of Study 1, the percent composition of H2O vapor in the
atmosphere decreased most rapidly over what period of time?
F. Between 2.5 and 2 billion years ago
G. Between 2 and 1.5 billion years ago
H. Between 1.5 and 1 billion years ago
J. Between 1 and 0.5 billion years ago
9. Suppose that the actual early Earth atmosphere had a high H2 composition of 42%.
Based on Study 2, is it likely that the corresponding H2S and N2 compositions of this
atmosphere were each 3%?
3% H2S 3% N2
A. Yes Yes
B. Yes No
C. No Yes
D. No No
10. Suppose that in a new trial in Study 2, the percent composition of H2 in the
atmosphere was set at 33%, and the percent composition of N2 was found to be 2%.
The percent composition of H2O vapor in this trial would most likely be:
F. greater than 40%.
G. greater than 35% and less than 40%.
H. exactly 35%.
J. greater than 30% and less than 35%.
Made by: Shahd A.Gaber
11. Consider an early Earth environment that featured microorganisms. Based on the
results of Study 2, is it more likely that aerobic organisms (those that require O 2 to
survive) or anaerobic organisms (those that do not require O2 to survive) would have
existed on Earth 4 billion years ago?
A. Aerobic organisms, because of the high H2O level 4 billion years ago
B. Aerobic organisms, because of the low O2 level 4 billion years ago
C. Anaerobic organisms, because of the high H2O level 4 billion years ago
D. Anaerobic organisms, because of the low O2 level 4 billion years ago
12. According to Study 2, how long did it take the H2O vapor level to decrease to 75% of
its composition 4 billion years before the present day?
F. 500 million years
G. 1 billion years
H. 1.5 billion years
J. 2 billion years
(Test 89-2)
he apoptotic index (AI) for a group of dividing cells is calculated as follows:
Figure 1 shows the AI for a culture of fibroblast cells as a function of the surrounding
concentration in parts per million (ppm) of a cell toxin.
*
Figure 1
One thousand actively dividing fibroblast cells in culture were studied. Figure 2 shows
the distribution of the cells in each of the stages of the dividing cell cycle.
Made by: Shahd A.Gaber
Figure 2
Electron micrographs were taken of the fibroblasts in culture. Figure 3 shows an
example of cells in each of the 4 stages of the dividing cell cycle. Although the cells are
not arranged in the sequence of the cell cycle, each stage is shown only once.
Made by: Shahd A.Gaber
Figure 3
7. Which cell in Figure 3 is most likely in the stage of the cell cycle during which
cytokinesis is occurring as mitosis nears completion?
A. Cell 1
B. Cell 2
C. Cell 3
D. Cell 4
8. Based on Figure 1, of the fibroblast cells that are surrounded by a toxin concentration
of 90 ppm, the percent that are in apoptosis most likely is represented by which of the
following ranges?
F. Less than 0.5%
G. Between 0.5% and 0.6%
H. Between 0.6% and 0.7%
J. Greater than 0.7%
Made by: Shahd A.Gaber
9. Which of the following cells in Figure 3 is most likely in the first stage of the actively
dividing cell cycle?
A. Cell 1
B. Cell 2
C. Cell 3
D. Cell 4
10. According to Figure 2, how did the number of fibroblast cells in stage G 2 compare
with the number of cells in stage S? The number in G2 was approximately:
F. 2 times as great as the number in S.
G. 3 times as great as the number in S.
H.
J.
as great as the number in S.
as great as the number in S.
11. Based on Figure 2, of the fibroblast cells that were in the actively dividing cell cycle,
the proportion that were in G1 is closest to which of the following?
A.
B.
C.
D.
SUBMIT
Made by: Shahd A.Gaber
(Test 90-1)
A polymorphism is the persistent occurrence of different appearances for a particular
trait in a species. All humans have slight differences in their genotypes (genetic code)
that result in different phenotypes (observable characteristics). Genetic polymorphisms
are persistent variations in gene sequences at a particular location in chromosomes,
such as those accounting for different blood types. Variations that cannot be observed
with the naked eye require techniques such as capillary electrophoresis (the separation
of genetic or protein material based on charge characteristics using an electric field).
The label on a vial of blood from a hospital patient was lost. The sample just tested
positive for a disease of the blood protein hemoglobin that is very common in the
hospital population. The sample was traced to a room with 4 patients who were
subsequently tested to determine the source of the initial vial.
Tests and Results
Smears of the blood from the unidentified patient (P) and from the 4 newly tested
patients (1–4) were observed under the microscope for the appearance of the blood
cells. Results are shown in Table 1.
Table 1
Patient Blood smear findings
P Sickle cells
1 Target cells
2 Sickle cells
3 Normal blood cells
4 Sickle cells
Serum was isolated from the blood of Patient P and from Patients 1–4 and placed in
separate tubes. A buffer was added to each vial to establish a pH of 8.6. One at a time,
samples from each tube were injected into the capillary electrophoresis device set at 7.5
kilovolts (kV) to separate the types of hemoglobin present into peaks. The hemoglobin
proteins composing a peak had similar charge characteristics. Figure 1 shows the
peaks that resulted from all 5 samples.
Made by: Shahd A.Gaber
Figure 1
1. Are the data in Table 1 consistent with the hypothesis that Patient 4 and Patient P are
the same person?
F. Yes; Patient 4 has the same blood cell appearance as Patient P.
G. Yes; Patient 4 has different blood cell appearance as Patient P.
H. No; Patient 4 has the same blood cell appearance as Patient P.
J. No; Patient 4 has different blood cell appearance as Patient P.
Made by: Shahd A.Gaber
2. What is the most likely reason that the serum samples were treated with a buffer to
bring pH to 8.6 ?
A. Hemoglobin protein breaks down at that pH.
B. All bacteria and viruses are destroyed at that pH.
C. Capillary electrophoresis separation of hemoglobin functions best at that pH.
D. Capillary electrophoresis separation of hemoglobin does not function at that pH.
3. Sickle cell anemia is caused by certain hemoglobin genotype combinations of 3
different alleles. The HbA allele is responsible for normal hemoglobin, the HbS allele is
responsible for one variant that results in sickle cells, and the HbC allele is responsible
for a different variant also resulting in sickle cells. Based on Table 1, the genotype of
Patient 4 could be which of the following?
I. HbA HbA
II. HbA HbS
III. HbA HbC
F. II only
G. I or III only
H. II or III only
J. I, II, or III
4. According to Figure 1, the pattern of protein peaks produced by serum from Patient P
most closely resembles the pattern produced by the serum sample from:
A. Patient 1.
B. Patient 2.
C. Patient 3.
D. Patient 4.
5. Based on Figure 1, the hemoglobin proteins in which of the following 2 peaks were
most likely closest in charge characteristic?
F. W and X
G. W and Z
H. X and Y
J. X and Z
Made by: Shahd A.Gaber
6. During the capillary electrophoresis, all the hemoglobin proteins started with some
quantity of charge before migrating from left to right in Figure 1. Therefore, the proteins
resulting in peaks furthest to the left must have been the most:
A. negative, as opposite charges attract each other.
B. negative, as opposite charges repel each other.
C. positive, as opposite charges attract each other.
D. positive, as opposite charges repel each other.
(Test 92-2)
The force per unit area resulting from the separation of solutions of different
concentrations by a selectively permeable membrane is called osmotic pressure.
Molecules, including water, have a tendency to move from regions of high concentration
to regions of low concentration. Selectively permeable membranes act as filters, only
allowing molecules below a certain threshold size to pass through. Osmotic pressure is
the pressure required to stop water from moving across such a membrane from a region
of high to low water concentration.
Cupric ions (Cu2+) and glucose were dissolved separately in equal volumes of water to
make two solutions. The glucose solution was more dilute, meaning that it had a higher
percentage of water molecules than the cupric ion solution. Of the three molecules used
for the solutions, water is the smallest and glucose is the largest. Water and glucose
solutions are colorless while cupric ion solutions are blue. However, mixing glucose and
cupric ions results in a red solution.
Experiment 1
A U-shaped tube contains a selectively permeable membrane, dividing it into equal
halves. Glucose solution is poured in the left and an equal volume of cupric ion solution
is poured in the right. Over 2 hours, the water level fell on the left and rose on the right.
At this time, the left-sided solution was red and the right-sided was blue.
Experiment 2
Cupric ion solution is poured in the left and an equal volume of pure water is poured in
the right. Over 2 hours, the water level fell on the right and rose on the left. At this time,
both sides of the tube contained blue-colored solutions.
Experiment 3
Glucose solution is poured in the left and an equal volume of pure water is poured in the
right. Over 2 hours, the water level fell on the right and rose on the left. At this time, both
sides of the tube contained colorless solutions.
Made by: Shahd A.Gaber
8. Albumin molecules do not pass through the selectively permeable membrane used in
Experiments 1–3 and form clear solutions in water. If Experiment 2 were repeated, but
the left side was filled with an albumin solution, the solution levels would:
A. fall on the left and rise on the right, resulting in a left-sided red solution and rightsided clear solution.
B. fall on the right and rise on the left, resulting in red solutions on both sides.
C. fall on the left and rise on the right, resulting in red solutions on both sides.
D. fall on the right and rise on the left, resulting in clear solutions on both sides.
9. In Experiments 1 and 2, cupric ion particles were able to move:
F. through the membrane into both the glucose solution and pure water.
G. through neither membrane into neither the glucose solution nor the pure water.
H. only through the membrane separating it from the glucose solution.
J. only through the membrane separating it from pure water.
10. In Experiments 2 and 3, what did the left side of the U-tube contain at the start of the
experiment?
Experiment 2 Experiment 3
A. Cupric ion solution Pure water
B. Glucose solution Pure water
C. Cupric ion solution Glucose solution
D. Glucose solution Cupric ion solution
11. In Experiment 1, if the selectively permeable membrane allowed cupric ions,
glucose, and water molecules all to pass, how would the results have differed, if at all?
F. The water level would have fallen on the right and risen on the left.
G. A red color would have appeared on both sides of the U-tube.
H. A blue color would have appeared on both sides of the U-tube.
J. The same results would have been observed.
Made by: Shahd A.Gaber
12. After watching Experiment 1 only, an observer asserted that since the left-sided
solution ended up red, cupric ions must be bigger than water molecules. Is this a valid
assertion?
A. No; the results show only that cupric ions and water molecules are smaller than
glucose molecules.
B. No; the results show only that cupric ions and water molecules are larger than
glucose molecules.
C. Yes; the results show that water molecules but not cupric ions can pass through
the selectively permeable membrane.
D. Yes; the results show that both water molecules and cupric ions can pass
through the selectively permeable membrane.
13. In Experiment 1, before the molecules began to move relative to the semipermeable membrane, the appearance of the right-sided solution in the U-tube was:
F. clear.
G. blue.
H. red.
J. purple.
SUBMIT
(Test 93)
Amino acids are considered the building blocks of protein in the body. Amino acids
combine with each other to form chains called peptides, which then combine to form
proteins. The human body requires twenty different amino acids, whose combinations
produce every essential protein in the body. When amino acids form peptides,
the residue is what is left after the amino acid sheds a molecule of water (a hydrogen
ion from one end and a hydroxide ion from the other end). The reaction rate is the factor
by which the protein is able to build itself up through the combination of peptides.
Figures 1-3 show the effects that changes in temperature, water volume, and residue
concentration have on the rate of reaction when Amino Acids A and B are present.
Figure 4 shows the effects that changes in the concentrations of Amino Acids A and B
have on the rates of reaction in solutions of the same concentration.
Made by: Shahd A.Gaber
Figure 1
Figure 2
Made by: Shahd A.Gaber
Figure 3
Figure 4
Made by: Shahd A.Gaber
1. According to Figure 2, Amino Acid A has the highest reaction rate at a water volume
closest to:
A. 0 mL.
B. 4 mL.
C. 12 mL.
D. 20 mL.
2. Based on the data presented in Figure 2, at approximately which of the following
water volumes does Amino Acid A have the same reaction rate as Amino Acid B?
F. 30 mL
G. 40 mL
H. 50 mL
J. 60 mL
3. A researcher claims that the reaction rate of Amino Acid B is dependent on both
residue concentration and amino acid concentration. Do the data in Figures 3 and 4
support this claim?
A. No, the reaction rate is dependent on the amino acid concentration, but not on
the residue concentration.
B. No, the reaction rate is not dependent on either the residue concentration or the
amino acid concentration.
C. Yes, the reaction rate is dependent on both the residue concentration and the
amino acid concentration.
D. Yes, the reaction rate is dependent on the residue concentration, but not on the
amino acid concentration.
4. A researcher claims that under the conditions used to determine the data for Figure
4, the reaction rate for Amino Acid A at any given concentration will always be greater
than the reaction rate for Amino Acid B at the same concentration. Do the data support
this conclusion?
F. No, Amino Acid A has a lower reaction rate at all given residue concentrations
tested.
G. No, Amino Acid A has a lower reaction rate at all given amino acid
concentrations tested.
Made by: Shahd A.Gaber
H. Yes, Amino Acid A has a higher reaction rate at all given residue concentrations
tested.
J. Yes, Amino Acid A has a higher reaction rate at all given amino acid
concentrations tested.
5. The figure below shows the relative reaction rates for alanine, an amino acid found in
DNA, and glycine an amino acid found in the muscles.
Based on this figure, one would best conclude that compared to the water volume at the
peak reaction rate of amino acids in DNA, the water volume at the peak reaction rate of
amino acids in the muscles:
A. is higher.
B. is lower.
C. is the same.
D. cannot be measured.
SUBMIT
Made by: Shahd A.Gaber
(Test 97)
Aphids are small plant-eating insects known to feed on rosebushes. In the cultivation of
roses, certain pesticides are often applied when the presence of aphids is detected.
However, sometimes the flowers that are treated with the pesticides are not as vibrant
or fragrant as those that did not receive the pesticide treatment. Two experiments were
conducted to study the effects of certain pesticides on rosebushes.
Experiment 1
A gardener filled 125 pots with Soil Type 1. No pesticide was added to the soil in 25
pots. The other pots were divided into four groups of 25 and the soils in each group
were treated with 5, 15, 25, or 35 parts per million (ppm) of either Pesticide A or
Pesticide B. All other factors were held constant. Fully grown rosebushes with buds but
no flowers were planted after the pesticide was placed in the soil. After 30 days the
rosebushes were uprooted, sun-dried, and the total number of petals produced by the
bushes was counted. The results are shown in Table 1.
Experiment 2
Experiment 1 was repeated with 100 pots of Soil Type 1 and 100 pots of Soil Type 2.
The same pesticide doses and type and number of rosebushes were used. All other
factors were held constant. After 30 days the rosebushes were uprooted and weighed.
The results are shown in Table 2.
Made by: Shahd A.Gaber
Information on the composition of the two soil types used is given in Table 3.
1. Which of the following sets of rosebushes served as the control in Experiment 1?
F. Rosebushes grown in soil with no pesticide added
G. Rosebushes grown in soil treated with 15 ppm of Pesticide A
H. Rosebushes grown in soil treated with 15 ppm of Pesticide B
J. Rosebushes grown in soil treated with 35 ppm of Pesticide A
2. Which of the following, if true, best explains why the pesticides were applied to the
soil as opposed to being placed directly on the rosebushes?
A. Pesticides are never applied to the plants when treating aphids or other pests.
B. Aphids are not affected when a pesticide is applied directly to the soil.
C. The experiments were testing how water levels affect growth patterns.
D. Rosebushes generally die when pesticides are directly applied to them.
Made by: Shahd A.Gaber
3. Assume that there is a direct correlation between plant weight and the number of
petals on the flowers. If a rosebush was grown in Soil Type 2, one would predict that the
number of petals would be lowest under which of the following conditions?
F. Pesticide B at 35 ppm.
G. Pesticide A at 35 ppm.
H. Pesticide B at 25 ppm.
J. Pesticide A at 15 ppm.
4. Assume that a rosebush was grown in soil treated with varying doses of a third
pesticide (Pesticide C). Based on the results of the experiments, what prediction, if any,
about the effect of Pesticide C on the growth of this rosebush can be made?
A. Pesticide C would have no impact on the growth of the rosebushes.
B. Pesticide C would interfere with the growth of these rosebushes by making them
smaller.
C. Pesticide C would interfere with the growth of these rosebushes by making them
less fragrant.
D. No prediction can be made on the basis of the results.
5. The results of Experiment 2 indicate that, at every pesticide dose, average plant
weight was lowest under which of the following conditions?
F. Pesticide B and Soil Type 1
G. Pesticide A and Soil Type 1
H. Pesticide B and Soil Type 2
J. Pesticide A and Soil Type 2
SUBMIT
Made by: Shahd A.Gaber
(Test 98)
Dopamines serve as enhancers or catalysts (a substance that initiates or increases the
rate of impulses during a chemical reaction, but is not depleted during the process) to
certain reactions involved in the activity of human thought. The dopamine intropin is
involved in the stimulation of the neurotransmitters in the brain when thought is initiated.
A student investigated the effects of dopamine activity on a specific neurotransmitter.
Experiment 1
To each of 10 test tubes, 7 milliliters (mL) of a peptide (a neurotransmitter) solution was
added. Two mL of an intropin solution was added to each of Tubes 1-9. Tube 10
received 2 mL of water without intropin. The tubes were then stirred at a constant rate in
water baths at various temperatures and incubated (heated) from 0 to 15 minutes (min).
At the end of the incubation period, 0.3 mL of NaCl solution was added to each tube.
The NaCl stopped the reaction between the intropin and the peptide. The precipitates,
solids formed in a solution during a chemical reaction, which in this case were caused
by the reaction of NaCl and the peptide, were removed from the tubes and dried. The
masses of the precipitates, in milligrams (mg), were measured to determine the relative
amount of enhancer that remained in the tube. The results are shown in Table 1.
Experiment 2
Peptide solution (8 mL) was added to an additional 8 test tubes to which 2 mL of
intropin solution was then added. The tubes were incubated at 10 degrees Celsius and
stirred at a constant rate for 15 min. The effect of acidity on the neurotransmitter was
observed by varying the acidity levels (using the pH scale). The relative amount of
neurotransmitter present in each tube was determined in the same manner as
Made by: Shahd A.Gaber
Experiment 1, by adding NaCl solution to each test tube. The results are in shown in
Table 2.
1. In Experiment 1, which of the following conditions allowed for the large amount of
precipitate in Tube 1?
A. Lack of intropin.
B. Higher temperature.
C. Lack of water.
D. Shorter incubation period.
2. In which of the following ways did the designs of Experiments 1 and 2 differ?
F. A larger volume of the peptide solution was used in Experiment 2 than in
Experiment 1.
G. The temperature was held constant in Experiment 1 and varied in Experiment 2.
H. No NaCl was added after incubation in Experiment 2, but it was in Experiment 1.
J. The remaining fluid level was measured in Experiment 1 but not in Experiment 2.
Made by: Shahd A.Gaber
3. Which of the following hypotheses about the effects of pH on intropin activity is best
supported by the results of Experiment 2? As the pH of the solutions increases from 2 to
13, the effectiveness of intropin:
A. increases only.
B. decreases only.
C. increases, then decreases.
D. remains the same.
4. Suppose that NaCl had been added immediately to Tube 5 with no incubation period.
Based on the results from Experiment 1, the best prediction about the amount of
precipitate formed would be:
F. 4.1 mg.
G. 3.5 mg.
H. 2.1 mg.
J. 1.4 mg.
5. According to Table 1, which of the following combinations of water bath temperature
and incubation time yielded the greatest amount of precipitate?
A. 25°C, 5 min
B. 25°C, 10 min
C. 35°C, 5 min
D. 35°C, 10 min
6. According to the results of both experiments, one can predict that the LEAST amount
of precipitate would be formed if tubes were incubated for 15 min under which of the
following conditions?
F. 20°C at pH of 2.0
G. 20°C at pH of 6.0
H. 30°C at pH of 2.0
J. 30°Cat pH of 6.0
SUBMIT
Made by: Shahd A.Gaber
(Test 99)
Several scientists considered some different environmental factors and their influence
on the growth of certain bacteria. The following experiments used Salmonella bacteria
to measure the effect of pH levels, nutrients, and temperature on the number of bacteria
produced within a given time period.
Experiment 1
A known quantity of Salmonella bacteria was placed in each of 3 Petri dishes with the
same nutrient concentration at the same temperature. The pH level of each nutrient
concentration in each dish was varied according to Table 1. On the pH scale, 7
represents neutral, values less than 7 indicate an acid, and values greater than 7
indicate a base. The lids of the Petri dishes were replaced after the bacteria were added
and the dishes were left alone. After 6 hours, the percent growth of Salmonella bacteria
was recorded (Table 1).
Experiment 2
A known quantity of Salmonella bacteria was placed in each of 3 Petri dishes with
different nutrient concentrations in the form of organic compounds. The temperature
and pH level (neutral 7) were held constant in each sample. The lids of the Petri dishes
were replaced after the bacteria were added and the dishes were left alone. After 6
hours, the percent growth of Salmonella bacteria was recorded (Table 2).
Made by: Shahd A.Gaber
Experiment 3
A known quantity of Salmonella bacteria was placed in each of 3 Petri dishes at
different temperatures. The pH level and nutrient concentrations were held constant.
The lids of the Petri dishes were replaced after the bacteria were added and the dishes
were left alone. After 6 hours, the percent growth of Salmonella bacteria was recorded
(Table 3).
1. According to Table 1, what might best contribute to the growth
of Salmonella bacteria?
A. A pH level above 9
B. A pH level below 5
C. A pH level near 7
D. A pH level near 5
Made by: Shahd A.Gaber
2. According to the results of the three experiments, which combination of the three
factors studied would be expected to produce the highest percent growth?
F. pH level of 5, organic compound in Dish 2, temperature of 40°C
G. pH level of 7, organic compound in Dish 2, temperature of 10°C
H. pH level of 5, organic compound in Dish 1, temperature of 40°C
J. pH level of 9, organic compound in Dish 1, temperature of 90°C
3. Which of the following conclusions is strengthened by the results of Experiment 1?
A. Salmonella bacteria reproduce most efficiently in an acidic environment.
B. Salmonella bacteria reproduce most efficiently in a neutral environment.
C. Salmonella bacteria cannot reproduce in a basic environment.
D. Salmonella bacteria cannot reproduce in an acidic environment.
4. Bacteria will generally reproduce until all of the nutrients available have been
depleted. How could Experiment 2 be altered to maximize the length of time that
bacteria will reproduce?
F. Change the observation time from 6 hours to 12 hours.
G. Regularly re-supply each group of bacteria with unlimited nutrients.
H. Increase the rate of growth by decreasing the pH levels.
J. Do not test the effect of different nutrient combinations on growth.
5. Which of the following was the independent variable in Experiment 3?
A. pH level
B. temperature
C. organic compound
D. growth of bacteria
6. The experiments recorded the percent growth that occurred over a 6-hour period.
Bacteria often reproduce at a rate that drastically varies from one stage to the next. The
best way to study the different stages of growth would be to record the percent growth:
F. after 2 hours only.
G. after 4 hours, then again after 6 hours.
Made by: Shahd A.Gaber
H. after 8 hours only.
J. every 15 minutes for 3 hours.
(Test 105)
The common grackle is one of the most abundant species of bird in North America.
When two male grackles encounter each other, there is often a threat display (a loud,
abrasive call). The dominant male usually forces the submissive male to cower and
eventually fly away. A biologist conducted two experiments to determine the rank in
aggression in male grackles. In the experiments described below, five adult male birds
were placed together in a cage and their interactions were observed and recorded.
Experiment 1
To determine what factors might affect aggressiveness, the biologists recorded the
sequence in which the birds were placed in the cage, their weight, their ages, and the
number of calls each grackle made during the experiment. In addition, the birds were
ranked according to their aggressiveness toward each other, from most aggressive (1)
to least aggressive (5). The results are shown in Table 1.
Experiment 2
The male grackles were placed back into the cage in the same sequence as in
Experiment 1. The results of all aggressive encounters (number of calls) between pairs
of birds were recorded. A bird was declared a "winner" if it forced the other bird, the
"loser," to flee from the encounter. Table 2 shows the results of the interactions between
the birds. There were no draws (ties) observed.
Made by: Shahd A.Gaber
Table 3 summarizes the results of all the encounters for each bird.
1. Which of the following generalizations about the relationship between body weight
and rank is consistent with the experimental results?
F. The heaviest bird will be the most dominant.
G. The heaviest bird will be the most submissive.
H. Body weight has no effect on rank.
J. The lightest bird will be the most dominant.
2. It was suggested that the more dominant a male grackle is, the more likely it is to
mate. Accordingly, one would predict, based on win-loss records, that the grackle with
the highest likelihood of mating would be:
A. Grackle B.
B. Grackle D.
C. Grackle C.
D. Grackle A.
Made by: Shahd A.Gaber
3. A sixth grackle, whose body weight was 330 grams and whose age was 24 months,
was added to the experimental cage. It was observed that the bird called a total of 10
times during the experiment. Based on the results of Experiment 1, what would be the
rank of the sixth grackle in terms of its aggressiveness?
F. 3
G. 4
H. 5
J. 6
4. According to the results of Experiments 1 and 2, which of the following factors is (are)
related to the dominance of one male grackle over other males?
I. Age
II. Body weight
III. Number of calls
A. I and II only
B. I and III only
C. II only
D. III only
5. One can conclude from the results of Experiment 2 that Grackle C and Grackle A had
a total of how many encounters with each other?
F. 5
G. 20
H. 25
J. 30
6. A criticism of this study is that the order that the grackles were placed in the cage
may have affected the aggressiveness of each bird. The best way to refute this criticism
would be to:
A. randomize the order before starting the experiments.
B. repeat the experiments several times with different orders each time.
C. place the birds in the cage in order of their age.
D. place the birds in the cage in order of their weight
Made by: Shahd A.Gaber
(Test 106)
In nature, different types of organisms often form symbiotic (mutually beneficial)
relationships with each other. One such example of this is between certain types of
fungi and plants; this relationship is known as a mycorrhiza. The association provides
the fungus with food through access to sugars from photosynthesis in the plant. In
return, the plant gains the use of the fungi's surface area to absorb mineral nutrients
from the soil. It is believed that without the assistance of fungi, these plants would not
be able to absorb crucial nutrients, including phosphates, from the soil. Two
experiments were performed to study the effect that the plant-fungi relationship has on
plant growth.
Experiment 1
For 6 weeks, several specimens of three different types of plants, selected from among
four different types of plants, were grown in a greenhouse. The average growth in
inches of each type of plant was recorded every two weeks. The soil used for the plants
was treated to remove any trace of fungi to establish expected growth without the plantfungi association. The results are shown in Table 1.
Experiment 2
In this experiment, several specimens of four different types of plants were grown in a
greenhouse for six weeks, and the average growth in inches of each type of plant was
recorded every two weeks. This time, however, untreated soil that contained fungi was
used. The results are shown in Table 2.
Made by: Shahd A.Gaber
Information on the plant types used is given in Table 3.
1. The results of Experiment 1 indicate that during what time frame did all of the plant
types studied experience the greatest increase in growth rate?
F. 0-2 weeks
G. 2-4 weeks
H. 4-6 weeks
J. Cannot be determined from the given information.
2. A plant from which climate type was NOT studied in Experiment 1?
A. Prairie
B. Tropical forest
C. Northern forest
D. All climate types were studied in Experiment 1.
Made by: Shahd A.Gaber
3. Based on the results of Experiment 1, which plant type experienced the most total
growth between Week 2 and Week 6?
F. Plant Type 1
G. Plant Type 3
H. Plant Type 4
J. Each plant type experienced the same total growth.
4. Based on the experiments, on the growth of which plant type did the presence of
fungi in the soil have the greatest effect?
A. Plant Type 1
B. Plant Type 3
C. Plant Type 4
D. The fungi had the exact same effect on all plant types.
5. Based on the results of Experiments 1 and 2, which of the following statements is
most accurate?
F. The presence of fungi has little or no impact on plant growth.
G. Removing fungi from soil can help to increase growth in some plants.
H. The presence of certain fungi in the soil increases plant growth.
J. Fungi cannot survive in local greenhouses.
Made by: Shahd A.Gaber
(Test 108)
The order Lepidoptera includes butterflies and moths. Table 1 is a key for identifying
some Lepidoptera in North America.
Table 2 describes 4 Lepidoptera that were seen in North America.
Made by: Shahd A.Gaber
1. Table 1 is used to identify animals that belong to which of the following groups?
F. Birds
G. Reptiles
H. Insects
J. Mammals
2. Based on the information provided, the Lepidoptera listed in Table 1 that is most
closely related to Agraulis vanillae most likely has which of the following characteristics?
A. Fuzzy body.
B. Wings with orange and black markings.
C. Hindwings with pronounced spots.
D. Green marbling on underside of wings.
3. Based on Table 1, which of the following traits of Lepidoptera Y indicates that it is
NOT a Speyeria coronis?
F. Slim body.
G. Yellow upper side of wings, with markings.
H. Triangular, silver markings on the underside of wings.
J. Round, elongated silver markings on the underside of wings.
4. The results from Table 1 for Lepidoptera W and Lepidoptera Z first diverge at which
of the following steps?
A. Step 1.
B. Step 3.
C. Step 7.
D. Step 8.
Made by: Shahd A.Gaber
5. According to Table 1, Automeris io and Antheraea polyphemus both have which of
the following traits?
F. Pronounced spot on hindwings.
G. Round markings on the underside of body.
H. A wingspan of 5-8 cm.
J. Slim bodies.
(Test 110)
Clinical research has become an important element in the development of modern
medicine. Perhaps one of the most widely-debated issues in today's clinical research is
the use of placebos, treatments believed to be biologically ineffective but used anyway
for psychological or experimental purposes; an example of a placebo is a sugar pill,
which contains no medication. In the realm of clinical research, placebos are used to
establish a control group within the pool of research participants. A certain percentage
of research patients are administered the test treatment, and another percentage is
administered a placebo treatment. Patients are not informed of which percentage they
are a part.
For various reasons, the use of placebos in clinical research is a controversial issue.
Two scientists debate whether the use of placebos is a good or bad practice in
research.
Scientist 1
Placebos are an important aspect of clinical research for many reasons. Not only do
they establish a control group for the test treatment in question, but they also help
address the issue of mind over matter, which is an important issue when working
towards treatment for a particular illness. The body is a powerful life force, with natural
recuperative abilities. A placebo encourages such recuperation.
Placebos also address the psychological aspect of illness. Because patients are
unaware of whether they are receiving treatment or a placebo, the possibility of
receiving treatment often provides patients with a psychological boost. The use of
placebos addresses the question of whether a person's positive attitude may be
important in recovery from illness. As a result, the placebo effect-a change in the
patient's condition due to the idea of treatment, rather than its biological effectivenessmay be a measurable change in behavior as a result of the belief in treatment.
For both their physical and psychological benefits, placebos should be used in clinical
research.
Made by: Shahd A.Gaber
Scientist 2
There are many reasons why placebos shouldn't be used in clinical research. For
example, placebos encourage deception in the doctor-patient relationship. Because this
relationship is crucial to the confidence of both the doctor and the patient, and therefore
the overall success of the patient's involvement in a study, placebos not only deceive
patients, but can also have an adverse affect on research results. Placebos also violate
patients' autonomy, or their right to choose treatment. While they can choose to be
involved in a study, patients are unable to select their own course of treatment because
it chosen for them.
Some argue that the placebo is worth its implementation in order to evaluate for the
occurrence of the placebo effect. However, such action may skew the results of the
study. For example, placebo-related changes could be over-estimated. Different
illnesses, by definition, will react differently to the placebo. For example, in the instance
of chronic pain or mood disorders, it's possible for patients to show spontaneous
improvement. The placebo effect can also result from contact with doctors or a
respected professional. Patients are vulnerable to their environment, which significantly
affects the psychological results of the placebo.
Due to its capability to skew research results, the placebo shouldn't be used in clinical
research.
1. Which of the following is most consistent with the reasons supporting the use of
placebos in clinical research? Many patients administered a placebo during a study:
F. found it very difficult to trust their various medical professionals.
G. found it much easier to deal with their illness due to the support of their doctors
and medical team.
H. experienced a heightening in their overall confidence and willingness to beat their
illnesses.
J. enjoyed no physical improvements while participating in the study.
2. According to Scientist 1's viewpoint, the placebo effect often reveals in the patient:
A. a negative change in behavior leading to a worsening of the patient's condition.
B. a positive change in behavior leading to recovery from the illness.
C. a negative change in behavior leading to a loss of faith in doctors.
D. no discernable change in behavior.
Made by: Shahd A.Gaber
3. According to the passage, both Scientists agree that:
F. the implementation of the placebo is worthwhile for the evaluation of the placebo
effect.
G. the use of placebos can cause the placebo effect in patients.
H. few patients experience any sort of a response to placebos.
J. placebos are valuable for both their physical and psychological effects on the
body.
4. According to Scientist 2's viewpoint, which of the following observations provides the
strongest argument against using placebos in clinical research?
A. The fact that patients sometimes deceive researchers
B. The danger of adverse reactions to sugar pills
C. The potential to skew the results of the research
D. The possibility that patients might choose ineffective treatments.
5. An evaluation of several placebo-using studies found that those patients who were
involved were not only very trusting of their doctors and medical teams, but they were
also more willing to communicate the various effects the treatment was having on them.
This finding contradicts evidence stated in which viewpoint?
F. Scientist 1's viewpoint, because the patients had a positive relationship with their
doctors.
G. Scientist 1's viewpoint, because use of the placebos discouraged the medical
team from obtaining accurate results.
H. Scientist 2's viewpoint, because placebos are a critical component in any clinical
study.
J. Scientist 2's viewpoint, because the placebo studies encouraged rather than
discouraged communication between doctors and patients.
6. Scientist 1's viewpoint implies that Scientist 2's viewpoint would be weakened by
which of the following observations?
A. All patients in a control group recover less quickly than the patients receiving
medical treatment.
B. Some patients seem to benefit from choosing their own treatments.
C. A patient who learns that she has been given a placebo becomes ill.
Made by: Shahd A.Gaber
D. A patient who is given a sugar pill develops new symptoms due to an allergic
reaction to the pill.
7. Which of the following assumptions about the use of placebos is implicit in Scientist
1's viewpoint?
F. Placebos are more cost-effective than other research tools.
G. Placebos are significantly more difficult to administer than real medications.
H. Real experimental medications are often dangerous to patients.
J. The use of placebos is safe for most patients.
(Test 115)
Asian soybean rust (ASR) is a disease caused by the fungus Phakospora pachyrhizi.
ASR spreads by windborne spores that infect soybean leaves. As rust lesions mature,
they produce thousands of additional spores. Over time, large spore loads build up
within fields and across large geographical areas. In 2004, this disease was detected in
nine states in the American southwest, and by 2005 it had invaded several other states.
ASR can drastically reduce crop yields in areas where it commonly occurs, so
monitoring and application of preventive measures such as fungicide will likely be
necessary.
Certain fungicides have been tested for their effectiveness against ASR. These
fungicides are listed in Table 1. The simplest classification of fungicides divides them
into three categories: contact, locally systemic, and systemic. Properties of these
fungicide categories are given in Table 2.
Made by: Shahd A.Gaber
ASR infections generally begin in the lower leaf canopy where humidity is higher and
leaves stay wet for longer periods. For this reason, the lower soybean leaf canopy is the
primary spray target. Both upper and lower leaf surfaces must be sprayed. Coverage as
dense as 400 spray droplets per square inch is considered ideal.
The different properties of fungicide types have important implications for spray
application. Contact and locally systemic fungicides require better spray coverage than
systemic fungicides. Contact fungicides, because they do not penetrate the plant tissue,
are more easily washed off the leaf by rain. This results in a shorter residual control
period and more frequent re-application of the fungicide.
Tests have shown that fungicide effectiveness varies based on the soybean growth
stage at which the fungicide is applied. Figure 1 identifies some of the different stages
of soybean growth. Soybean leaves can be infected at any time with ASR. However,
research has shown that the most critical time to protect soybean plants with fungicides
is from the R1 through R3 growth stages. Fungicide applications should not be initiated
after the R4 growth stage (seed development and mature plant).
Made by: Shahd A.Gaber
Figure 1
1. According to the passage, which of the following fungicides should be reapplied
frequently?
F. boscalid
G. azoxystrobin
H. pyraclostrobin
J. tebuconazole
2. Based on the information provided, during which stage of growth will the application
of chlorthalonil be least effective?
A. R1
B. R3
C. R4
D. Chlorthalonil should not be applied during any growth stage.
3. A student claimed that, "Application of a systemic fungicide will only prevent the
growth of fungi if applied after Growth Stage 4." Does the passage support this claim?
F. No; systemic fungicides are active only on the surface of the leaf.
G. No; fungicides are most effective when applied between Growth Stage 1 and
Made by: Shahd A.Gaber
Growth Stage 3.
H. Yes; systemic fungicides are most effective when applied during Growth Stage 4,
but not before.
J. Yes; soybean plants can only be infected with ASR late in their development.
4. According to the passage, if fewer than 400 spray droplets per square inch
of tebuconazole were applied during Growth Stage 4, the chances that the soybean
plants would become infected with ASR would most likely:
A. not be affected.
B. decrease only.
C. decrease, then increase.
D. increase only.
5. Equal amounts of azoxystrobin, boscalid, and myclobutanil were applied to three
different soybean plants during Growth Stage 3. After 24-hours, each of the plants was
sprayed with water. Based on the data, which of the following represents the order, from
least effective to most effective, of the fungicides' likelihood of preventing ASR?
F. myclobutanil, azoxystrobin, boscalid.
G. azoxystrobin, boscalid, myclobutanil.
H. boscalid, azoxystrobin, myclobutanil.
J. myclobutanil, boscalid, azoxystrobin.
6. According to Figure 1, during which of the following stages in the growth of a soybean
plant should fungicide NOT be applied?
A. Beginning bloom
B. Full pod
C. Full bloom
D. Beginning pod
Made by: Shahd A.Gaber
(Test 116)
Echinoderms are defined as any of a variety of invertebrate marine animals
characterized by a hard, spiny covering or skin. They have attracted much attention due
to their extensive fossil record, ecological importance, and bizarre body forms. Most
echinoderms are extinct, but many living representatives still exist. All living
echinoderms have an internal skeleton and a central cavity, but the outward
appearance can vary significantly. For example, starfish and brittle stars have arms that
extend from a central disk; sea lilies have a central stalk, or stem, and resemble flowers;
sea cucumbers are wormlike and tend to burrow.
The ways in which echinoderms move and feed are as diverse as their body forms.
Table 1 lists certain echinoderms and their methods of locomotion (movement) and
feeding.
Table 2 includes examples of echinoderm habitats around the world.
Made by: Shahd A.Gaber
1. The echinoderm shown below is most likely a:
F. sea lily.
G. starfish.
H. sea cucumber.
J. brittle star.
2. According to Table 1 and Table 2, crinoids can be found feeding on plankton:
A. near the shore.
B. in deep ocean trenches.
C. in offshore mud and ooze.
D. on the deep-sea floor.
Made by: Shahd A.Gaber
3. Based on the data provided in the passage, sea cucumbers most likely burrow in
order to:
F. locate food.
G. avoid worms.
H. move offshore.
J. shed their spines.
4. Suppose scientists discover a new echinoderm that uses its tube feet to move across
the deep-sea floor as it hunts for prey. This newly discovered echinoderm can most
likely be classified as a(n):
A. crinoid.
B. asteroid.
C. ophiunoid.
D. holothurian.
5. A student hypothesized that large populations of sea cucumbers could greatly alter
the physical and chemical composition of the sea floor. Is this hypothesis supported by
the data in the passage?
F. Yes; sea cucumbers often prey upon commercially important organisms, such as
oysters.
G. Yes; sea cucumbers feed by swallowing sediment, extracting organic matter, and
ejecting the remainder.
H. No; sea cucumbers cannot burrow into the sediment, so will not affect the
composition of the sea floor.
J. No; sea cucumbers do not have a viable method of locomotion.
Made by: Shahd A.Gaber
(Test 117)
Students wanted to test the effects of nutrition on the growth of guinea pigs. Two
experiments were conducted using different feeds and vitamin supplements. For both
experiments, four groups of 10 guinea pigs each were given a different type of feed over
an 8-week period. Each group received the same quantity of food and was provided
with fresh water daily. The guinea pigs were measured and weighed weekly. The
guinea pigs in each group had an average starting weight of 50 grams (g) and an
average starting length of 20 centimeters (cm).
Experiment 1
Group 1 was fed a high-protein feed (Feed P).
Group 2 was fed a grain-based feed with vitamin supplements (Feed Q).
Group 3 (control group) was fed a grain-based feed without supplements (Feed R).
Group 4 was fed a grain-based feed without supplements plus
fruits and vegetables (Feed S).
The results and average measurements are recorded in Table 1 below.
Experiment 2
Group 5 was fed a high-protein feed plus fruits and vegetables (Feed V).
Group 6 was fed a grain-based feed with vitamin supplements plus fruits and
vegetables (Feed W).
Group 7 (control group) was fed a grain-based feed without supplements (Feed X).
Group 8 was fed a grain-based feed without supplements plus fruits only (Feed Y).
The results and average measurements are recorded in Table 2 below.
Made by: Shahd A.Gaber
1. Based on the results of the experiments, which feed resulted in the greatest weight
gain?
A. Feed P.
B. Feed S.
C. Feed V.
D. Feed Y.
2. Based on the results of Experiment 1, the guinea pigs in the group that was fed a
grain-based feed with vitamins gained how much weight, on average, during each week
of the experiment?
F. 29 grams.
G. 11 grams.
H. 10 grams.
J. 4 grams.
3. If the students added vitamin supplements to Feed V for a new group (Group 9), what
might the result be after 8 weeks?
A. The guinea pigs in Group 9 would weigh less than those in Group 5.
B. The guinea pigs in Group 9 would weigh less than those in Group 6.
C. The guinea pigs in Group 9 would have a greater average length than those in
Group 5.
D. The guinea pigs in Group 9 would have a shorter average length than Group 6.
Made by: Shahd A.Gaber
4. Which of the following statements is true, according to Table 2?
F. Feed W produces guinea pigs that are almost twice as long as those in the
control group.
G. Feed V produces guinea pigs that weigh three times as much as those in the
control group.
H. Feed Y produces guinea pigs with the greatest average length.
J. Feed X produces guinea pigs similar to those produced by Feed Y.
5. From the results of the experiments the students would hypothesize that the guinea
pigs in Groups 3 and 4 are similar because:
A. the control group was fed larger quantities of food.
B. the fruits and vegetables given in Experiment 1 did not have a very high
nutritional value.
C. neither group received enough high-protein food.
D. the vitamin supplements given in Experiment 2 were more potent than those
given in Experiment 1.
6. According to the passage, the guinea pigs in which of the following groups showed
the least overall growth?
F. Group 8.
G. Group 7.
H. Group 4.
J. Group 1.
(Test 120)
Turf grasses are used throughout the United States in many suburban lawns. Kentucky
bluegrass is the most common type of turf grass used in the northern part of the United
States. To keep lawns green and healthy, many homeowners apply fertilizer up to five
times a year. Inorganic fertilizers are becoming more popular, and contain three
common elements - nitrogen, phosphorous, and potassium - for the development of
plant color, strength, and health. Most turf grass lawns do not use all of the nutrients
provided in the fertilizer, which means that much of the nitrogen, phosphorous, and
potassium remains in the soil. When water enters the soil, it accumulates a portion of
Made by: Shahd A.Gaber
the excess nitrogen from the soil. This water, now termed leachate, flows into
surrounding waterways. The leaching of high concentrations of nitrogen into natural
waterways can throw off the environmental equilibrium of the aquatic ecosystem, often
resulting in an increase in plant growth that can have a negative impact on the native
fish populations.
A study was performed to examine the degree of nitrogen leaching in Kentucky
bluegrass turf; 2 one-acre plots of turf were compared. The scientists conducting the
study relied completely on natural rainwater to irrigate the test plots. Each plot received
fertilizer applications containing different levels of nitrogen two times per week during
the months of April and September for 5 years. The plots had a 5% slope to facilitate
leaching; leachate was collected in one-liter jugs. The leachate collected from each plot
was measured for nitrogen concentration.
Plot A received a low nitrogen application: 98 kilograms of N per acre from 2000 to
2004. Plot B received an initially high nitrogen application: 245 kilograms of N per acre
from 2000-2003. In the last year of the study, the amount of nitrogen in the fertilizer was
decreased to 196 kilograms of N per acre for Plot B. Table 1 shows the average
nitrogen concentration in milligrams per liter (mg/L) in the leachate collected from each
plot during each year. Figure 1 shows the percent concentration of nitrogen in the
leachate.
Made by: Shahd A.Gaber
1. According the passage, as the amount of nitrogen in the fertilizer increased, the
average amount of nitrogen in the leachate:
F. decreased only.
G. increased only.
H. decreased for several years, then increased.
J. increased for several years, then decreased.
2. Based on the data in Table 1 and Figure 1, one can conclude that when fertilizer with
a low nitrogen concentration is applied, native fish populations in surrounding
waterways will most likely:
A. remain stable.
B. be reduced by 5%.
C. be completely decimated.
D. not have enough food.
3. It was determined that during times of heavy rain, more nitrogen was leached from
the soil. Based on the results of the study, which year most likely had times of heavy
rain in April and September?
F. 2000.
G. 2001.
Made by: Shahd A.Gaber
H. 2003.
J. None.
4. According to the Environmental Protection Agency, average nitrogen levels in
leachate must be less than 10 mg/L to be safe for the environment. Based on this
standard and the results of the study, which of the following fertilizer applications is
considered safe?
A. 196 kilograms of N per acre.
B. 98 kilograms of N per acre.
C. 245 kilograms of N per acre.
D. None of the tested applications is safe.
5. In 2005, it was found that average nitrogen levels in the leachate from Plot B were
8.2 mg/L. The data from the study supports which of the following conclusions?
F. Kentucky bluegrass should not be used for lawns in suburbs near a public
waterway.
G. Once high-nitrogen fertilizer has been applied to a suburban lawn, nitrogen levels
in the leachate will remain high, even if low-nitrogen fertilizer is later applied.
H. Following the application of low-nitrogen fertilizers, it will take more than one year
to reach safe nitrogen levels in leachate from suburban lawns previously fertilized with
high-nitrogen fertilizer.
J. The measurable concentration of nitrogen in leachate from suburban lawns will
always be within the range considered safe by the Environmental Protection Agency, as
long as irrigation is kept to a minimum.
Made by: Shahd A.Gaber
(Test 123)
Predation is an interaction between individuals of 2 species in which one is harmed (the
prey), and the other is helped (the predator). Predation can occur among plants and
animals as well as between plants and animals. Some biologists contend
that herbivores, or plant eaters, are predators. Table 1 indicates some characteristics
and examples of certain predators.
Predation is very important in maintaining a natural balance in any given ecosystem.
For example, without predators, prey populations tend to grow exponentially. Without
prey, predator populations tend to decline exponentially. Predators consume individual
members of the prey population, thereby controlling the overall numbers in the
ecosystem. The number of prey consumed depends on the number of prey present as
well as the number of predators present. The rate of change in the number of prey is a
function of the birth of new prey minus the death of other prey, due either to predation or
other causes. The death rate is assumed to depend on the number of prey available
and the number of predators. The rate of change in the number of predators is a
function of the births of new predators-which depends on the number of prey-minus the
death of some predators.
Over long periods of time, predator and prey tend to balance each other out. This is
called the predator-prey cycle. Prey numbers will increase when predator numbers
decrease. When the number of prey reaches a certain point, predators will start to
increase until they eat enough prey to cause a decline in prey numbers. When this
happens, the number of predators will begin to decrease because they can't find
enough prey to eat, and the cycle will begin again. Figure 1 represents an example of
a predator-prey cycle.
Made by: Shahd A.Gaber
Figure 1
1. Based on information in the passage and in Table 1, an herbivore is:
F. a predator only.
G. both a parasite and a predator.
H. prey only.
J. both a predator and prey.
2. According to information in the passage, the number of prey consumed in an
ecosystem is dependent on:
A. the natural balance of the ecosystem.
B. the total number of predators that die because of predation.
C. the type of parasites available in the ecosystem.
D. the number of predators present and the number of prey present.
3. Based on Figure 1, during the first year, predator numbers:
F. were inversely related to prey numbers.
G. were directly related to prey numbers.
H. were equal to prey numbers.
J. were always greater than prey numbers.
Made by: Shahd A.Gaber
4. Studies have shown that a certain species of deer will only eat a specific type of plant
found in the deer's natural habitat, and nothing else. Is this finding supported by the
information in the passage?
A. No, because a deer is an herbivore, which means it eats all plants.
B. No, because a deer is a carnivore and does not eat plants.
C. Yes, because a deer is an herbivore, and herbivores can be selective eaters.
D. Yes, because a deer is a prey animal, so it must use caution when eating.
5. Based on Figure 1, during which year were the greatest number of prey animals
available?
F. 1
G. 2
H. 3
J. 4
SUBMIT
(Test 125)
Sea anemones look like plants, but they actually are predatory animals. They are
invertebrates, which means that they do not have a skeleton. To protect themselves,
they will attach to firm objects on the sea floor, such as rock or coral.
Sea anemones can alter their body shape according to changes in their environment.
For example, when ocean currents are strong, the sea anemone will reduce its internal
volume in order to decrease the surface area that is exposed to the current. Sea
anemones are dependent on water flow for food and nutrients and also for assistance in
eliminating waste.
Most anemones share a symbiotic relationship with marine algae called zooxanthellae.
These are photosynthetic organisms whose waste products are a food source for the
sea anemone. The sea anemone also enjoys a mutualistic relationship with the clown
fish. This fish is immune to the stinging tentacles of the sea anemone, and it helps the
anemone by actually cleaning the tentacles. The cleaning process yields food for the
clown fish, while it remains protected from potential predators by the sea anemone's
stinging tentacles.
Figure 1 shows a cross-section of portions of the internal anatomy of a sea anemone.
Made by: Shahd A.Gaber
Figure 1
1. According to Figure 1, the sea anemone's mouth is located:
A. below the pharynx.
B. at its center.
C. near its base.
D. within the sphincter muscle.
2. According to information in the passage, the sea anemone benefits from the
presence of:
F. both the clown fish and zooxanthellae.
G. the clown fish only.
H. zooxanthellae only.
J. neither the clown fish nor zooxanthellae.
3. Which of the following statements about the sea anemone is supported by the
passage? The sea anemone most resembles:
A. a clown fish.
B. a flower.
C. marine algae.
D. a rock.
Made by: Shahd A.Gaber
4. Suppose that a strong storm stirred up the water in which a sea anemone was living.
The sea anemone's response would most likely be:
F. to expose itself to the strong current.
G. to seek the protection of a clown fish.
H. to reduce its internal volume.
J. to detach itself from the seafloor.
5. As shown in Figure 1, the part of the sea anemone's anatomy that connects its mouth
to its gastrovascular cavity is the:
A. oral disk.
B. tentacle.
C. pharynx.
D. sphincter muscle.
(Test 126)
Compost is the name given to a mixture of decaying leaves and other organic material.
This mixture is often used as fertilizer. Several students designed experiments to test
various types of soil, and various combinations of soil and compost on plant growth.
Experiment 1
The students dug a soil sample from an empty field next to the school. They put soil into
4 different clay pots, and mixed in various amounts of compost so that the volume of
soil mixture was the same in each pot. They then planted the same number of radish
seeds (4) in each pot. The soil/compost mixtures for each pot are shown in Table 1.
Made by: Shahd A.Gaber
The clay pots were placed next to each other on a windowsill and watered at the same
time each day. The students took care to ensure that the pots each received the same
amount of sunlight and water each day. After 2 weeks, the students began recording the
growth of the radish plants. They continued recording this data for two more weeks. The
results are shown in Table 2.
Experiment 2
The students repeated Experiment 1, with the following changes; each pot contained a
different soil type, and no compost was used. This experiment was begun at the same
time as Experiment 1. The results of Experiment 2 are shown in Table 3.
1. Based on the results of Experiment 2, which soil type yielded the most overall growth
after 28 days?
F. Sand
G. Potting soil
H. Soil from the field near the school
J. Mixture of sand and potting soil
Made by: Shahd A.Gaber
2. Based on the results of Experiment 1, which soil/compost mixture yielded the
greatest average plant height after the first 2 weeks?
A. 4
B. 3
C. 2
D. 1
3. Experiment 2 was different from Experiment 1 in that none of the clay pots:
F. were watered during the first 2 weeks.
G. contained any compost.
H. contained any soil.
J. were placed on the windowsill.
4. The results of Soil Type 3 in Experiment 2 and Pot Number 4 in Experiment 1 were
almost identical. This is most likely because:
A. the same amount of compost was used.
B. the plants were allowed to grow for 2 more weeks.
C. the pots were the same size.
D. the same type of soil was used.
5. In Experiment 2, how many seeds were planted in each clay pot?
F. 4
G. 14
H. 21
J. Cannot be determined.
6. According to the results of Experiment 1, what percentage of compost yielded the
highest average number of leaves?
A. 100%
B. 75%
C. 50%
D. 25%