1ST DIAGNOSTIC TEST
Directions: Each
set if lettered choices below refers to the numbered statements immediately
following it. Select the one lettered choice that best answers each question or
best fits each statement, and then fill I the corresponding oval on the answer
sheet. A choice may be used once, more than one, or not at all in each set.
Question 1-3
(A)
Mitochondria
(B)
Cytoplasm
(C)
Pyruvate
(D)
Lactic acid
(E)
Glucose
1.
Location of
cellular respiration in prokaryotes
2.
End product of
anaerobic metabolism in muscle cells
3.
Location of
glycolysis in eukaryotes
Question 4-6
(A)
Anaphase II
(B)
Metaphase I
(C)
Prophase II
(D)
Metaphase II
(E)
Prophase I
4.
Stage of
meiosis during which recommendation of genetic material occurs
5.
Stage of
meiosis during which pairs of homologous chromosomes align at the center of the
cell
6.
Stage of
meiosis during which sister chromatids are separated.
Question 7-9
(A)
Reasoning/insight
(B)
Imprinting
(C)
Classical
conditioning
(D)
Habituation
(E)
Instinct
7.
A simple kind
of learning involving loss of sensitivity to unimportant stimuli.
8.
Greese
recognize a ticking clock as “mother” if exposed to it during a critical period
shortly after hatching.
9.
Fish are given
food at the same time as a tap on their glass bowl and soon learn to approach
when a tap sounds even in the absence of food.
Question 10-12
(A)
Small intestine
(B)
Large intestine
(C)
Stomach
(D)
Esophagus
(E)
Mouth
10.
Structure where
most digestion and absorption of nutrient occurs
11.
Structure where
starch digestion first takes place
12.
Structure with
the lowest pH.
Directions: Each
of the questions or incomplete statements below is followed by five suggested
answers or completions. Some questions pertain to a set that refers to a
laboratory or experimental situation. For each question, select the one choice
that is the best answer to the question and then fill in the corresponding oval
on the answer sheet.
13.
Homologous
structures, which have similar underlying structure but may have different
functions, are formed by
(A)
Divergent
evolution
(B)
speciation
(C)
segregation
(D)
convergent
evolution
(E)
stabilizing
selection
14.
Hemoglobin is a
protein in red blood cells that binds and carries oxygen and some carbon
dioxide. Its affinity for oxygen changes as blood travels from the lungs to the
body tissues and back to the lungs again. One could expect hemoglobin to have.
(A)
A high carbon
dioxide affinity in the lungs and a low oxygen affinity in the tissues
(B)
A low carbon
dioxide affinity in the lungs and a high oxygen affinity in the tissues
(C)
A high oxygen
affinity in the lungs and a low oxygen affinity in the tissues
(D)
A low oxygen
affinity in the lungs and a high oxygen affinity in the tissues
(E)
A high oxygen
affinity in the lungs and a high carbon dioxide affinity in the tissues
15.
Which of the
following RNA sequences would be transcribed from the DNA sequence ATGCCTAGGAC?
(A)
TACGGATCCTG
(B)
UAGCGAUCCUG
(C)
AUGCCUAGGAC
(D)
UACGGAUCCUG
(E)
GCAUUCGAAGU
16.
Arthropods can
be characterized by all of the following EXCEPT.
(A)
A hard
exoskeleton
(B)
A water
vascular system
(C)
Jointed
appendages
(D)
Molting
(E)
Segmented body
17.
Which of the following
are functions of the kidney?
I.
Filtration of
blood to remove wastes
II.
Blood pressure
regulation
III. pH regulation
(A)
I only
(B)
I and II only
(C)
I and III only
(D)
II and III only
(E)
I, II and III
18.
In chickens,
the allele for long tail feathers (T) is dominant over the allele for short
tail feathers (t). If a pure-breeding long-tailed chicken (TT) mates with a
pure-breeding short-tailed chicken (tt), what percentage of their offspring (if
mated with the correct genotype) could give rise to chickens with short tails?
(A)
25%
(B)
50%
(C)
75%
(D)
100%
(E)
Unable to
determine from the information given
19.
All of the
following could be considered density dependent factors affecting population
growth EXCEPT.
(A)
Limited
nutrients
(B)
Climate
temperature
(C)
Build-up of
toxins
(D)
predation
(E)
limited water
20.
The best
definition of a species is
(A)
A group of
organisms that occupy the same niche
(B)
A population
that works together
(C)
A group of
organisms that can mate with each other
(D)
A population
that preys on other populations
(E)
A population in
which all members benefit from the association in some way
21.
Which of the
following contains blood poor in oxygen?
I.
Right ventricle
II.
Pulmonary vein
III. Pulmonary vein
(A)
I only
(B)
II only
(C)
III only
(D)
I and III only
22.
An organism
appears to be segmented worm. Upon observation it is determined that the
organism has a closed circulation, a mouth and an anus, and does NOT have an
exoskeleton. The organism most likely belongs to the pylulm.
(A)
mollusca
(B)
annelida
(C)
echinodermata
(D)
arthropoda
(E)
chordata
23.
Which of the
following sustances are produced by the light reactions of photosynthesis?
(A)
ATP and NADPH
(B)
ATP and glucose
(C)
NADPH and
glucose
(D)
ATP and NADH
(E)
NADPH and
glucose
24.
Consider the
following graph of substrate concentration vs product formation. Assume enzyme
concentration to be constant. Why does the graph level off at high substrate
concentrations?
Diagram
(A)
All the enzyme
is used up, and product formation cannot occur without is.
(B)
There is no
more substrate to be converted into product.
(C)
Substrate
concentration, exceeds enzyme concentration, and all active sites are
saturated.
(D)
The reaction
has run to completion.
(E)
An inhibitor
has been added, and it has slowed down the rate of product formation.
25.
A bird that
feeds on both insects and berries would be classified as a
I.
Primary
consumer
II.
Secondary
consumer
III. Tertiary consumer
(A)
I only
(B)
II only
(C)
III only
(D)
I & II only
(E)
II & III
only
26.
Which of the
following chemical formulas could represent a carbohydrate?
(A)
C6H6O6
(B)
C3H6O3
(C)
C6H12O3
(D)
C5H10O10
(E)
CH2O4
27.
A population of
birds lives in an area with many insects upon which it feeds. The insects live
inside trees, burrowing into the bark. Over many hundreds of years, the average
beak size in the bird population has increased. This is due to
(A)
increased
fitness of the birds, leading to speciation
(B)
decreased
fitness of the insects, allowing the birds to catch them more easily
(C)
increased
fitness of large-beaked birds, leading to evolution
(D)
decreased
fitness of small-beaked birds, leading to speciation
(E)
random mutation
and genetic recombination
28.
The location on
an enzyme where a substrate binds is called
(A)
binding site
(B)
reaction center
(C)
allosteric site
(D)
lock-and-key
model
(E)
active site
29.
Human cells
maintain concentration gradients across their plasma membranes, such that there
is a high sodium concentration outside the cell and a high potassium
concentration inside the cell. Suppose that within the cell membrane are sodium
“leak” channels. These channels would allow sodium to
(A)
move out of the
cell by simple diffusion
(B)
move into the
cell by simple diffusion
(C)
move out of the
cell by facilitated diffusion
(D)
move into the
cell by facilitated diffusion
(E)
move into the
cell by active transport
30.
The role of
decomposers in the nitrogen cycle is to
(A)
fix atmospheric
nitrogen into ammonia
(B)
incorporate
nitrogen into amino acids and organic compounds
(C)
convert ammonia
to nitrate, which can then be absorbed by plants
(D)
denitrify
ammonia, thus returning nitrogen to the atmosphere
(E)
release ammonia
from organic compounds, thus returning it to the soil
31.
All of the
following are true about the endocrine system EXCEPT.
(A)
it relies on
chemical messengers that travel through the bloodstream
(B)
it is a control
system that has extremely rapid effects on the body
(C)
the hormones
affect only certain “target” organs
(D)
it is involved
in maintaining body homeostasis
(E)
its organs
secrete hormones directly into the bloodstream, rather than through ducts
32.
Two organisms
live i close association with each other. One organism is helped by the
association, whereas the other is neither helped nor harmed. Which of the
following terms best describes this relationship?
(A)
Mutualism
(B)
Commensalism
(C)
Symbiosis
(D)
Parasitism
(E)
Predator-prey-relationship
33.
Cardiac output
(the amount of blood pumped out of the heart in one minute) and blood pressure
are directly proportiona. Which of the following graphs best depicts the
relationship between cardiac output and blood pressure?
Diagrams
Question 34-36 refer to the following diagram.
Diagram
34.
Location where
male haploid cells are produced
(A)
1
(B)
2
(C)
3
(D)
6
(E)
8
35.
Sticky
structure where pollen grains can attach and germinate
(A)
1
(B)
2
(C)
3
(D)
6
(E)
8
36.
Structure
which, when fertilized, develops into fruit
(A)
1
(B)
2
(C)
3
(D)
6
(E)
8
Question 37-38
Tropisms refer
to movements made by plants toward or away from certain stimuli. “Positive”
tropisms refer specifically to movements toward a stimulus, whereas “negative”
tropisms refer to movements made away from a stimulus.
37.
A plant growing
on the shady side of a building bends around the corner of the building toward
the sunlight. This is an example of
(A)
negative
geotropism
(B)
negative
phototropism
(C)
positive
phototropism
(D)
positive
hydrotropism
(E)
negative
hydrotropism
38.
The stem and
leaves of the plant grow up, away from the soil. This is an example of
(A)
negative
geotropism
(B)
positive
geotropism
(C)
negative
phototropism
(D)
positive
hydrotropism
(E)
negative
hydrotropism
Question 39-43 refer to the following diagram.
Diagram
39.
The hormone
labeled X in the diagram is often used in over the counter diagnostic tests to
determine when ovulation has occurred. This hormone is
(A)
estrogen
(B)
progesterone
(C)
FSH
(D)
LH
(E)
Testosterone
40.
Based on the
peak levels of hormone X, on what day of the cycle is ovulation most likely to
occur?
(A)
Day 21
(B)
Day 14
(C)
Day 12
(D)
Day 25
(E)
Day 28
41.
The hormone
labeled Y in the diagram is
(A)
progesterone,
secreted by the corpus luteum after ovulation has occurred
(B)
progesterone,
secreted by the ovary after ovulation has occurred
estrogen, secreted by the corpus luteum after ovulation has occurred
(C)
estrogen,
secreted by the ovary after ovulation has occurred
(D)
estrogen,
secreted by the follicle before ovulation occurs
42.
Immediately
after fertilization, the zygote begins to undergo rapid cell division. This
process is
(A)
Blastulation
(B)
Gastrulation
(C)
Neurulation
(D)
Implantation
(E)
cleavage
43.
From which of
the primary germ layers does the nervous system develop?
(A)
Endoderm
(B)
Mesoderm
(C)
Ectoderm
(D)
Enteroderm
(E)
Epidermis
Question 44-46
A barren, rocky community near a lake has virtually no vegetation
or animal life. After a period of approximately 75 years, the community boasts
a wide variety of flora and fauna, including deciduous trees, deer, and
raccoon.
44.
The process
which has taken place can best be described as
(A)
Progression
(B)
Succession
(C)
Evolution
(D)
Habitation
(E)
colonization
45.
The stable
community of deciduous trees and animals in known as the
(A)
Colonization
(B)
final community
(C)
climax
community
(D)
apex community
(E)
summit
community
(F)
composite community
46.
Usually the
first organisms to colonize rocky areas are lichen. These are known as the
(A)
primary
community
(B)
starter
community
(C)
colony
organisms
(D)
pioneer
organisms
(E)
settler
organisms
Question 47-50
Diuretics are
substances that help eliminate water from the body. The effects of various
substances were tested on several volunteers. All volunteers had a mass of 70
kg. They drank nothing for eight hours before the test and urinated just prior
to ingesting the test substance. The three substances (water, caffeine, and
salt) were tested on three separate days. The results are shown in the tables
below.
Table 1
|
Volunteer |
amount
caffeine ingested (in 100 ml
water) |
volume urine collected
after 1 hour |
|
A
|
50 mg |
302 ML |
|
B
|
100 mg |
492 ml |
|
C
|
150 mg |
667 ml |
|
D
|
200 mg |
863 ml |
Table 2
|
Volunteer |
amount sodium
chloride ingested (in 100 ml
water) |
volume urine collected
after 1 hour |
|
A
|
9 g |
201 ML |
|
B
|
1.8 g |
162 ml |
|
C
|
2.7 g |
125 ml |
|
D
|
3.6 g |
82 ml |
Table 3
|
Volunteer |
Volume water ingested |
volume urine collected
after 1 hour |
|
A
|
100 ml |
230 ML |
|
B
|
200 ml |
240 ml |
|
C
|
300 ml |
252 ml |
|
D
|
400 ml |
263 ml |
47.
Which of the
following substances could be classified as a diuretic?
I.
Caffeine
II.
Sodium
III. Water
(A)
I only
(B)
II only
(C)
I and II only
(D)
II and III only
(E)
I, II and III
48.
Which graph
best represents the change in urine volume when ingesting caffeine?
Diagrams
49.
The purpose of
ingesting the plain water (Table 3) was to
(A)
rehydrate the
volunteers
(B)
dissolve the
substances
(C)
act as a
control
(D)
flush out the
kidneys
(E)
act as a
positive test substance
50.
Based on the
results in table 2, if a volunteer were to ingest 4.5 g sodium chloride
dissolved in 100 ml water, what would be the approximate predicted urine volume
collected after one hour?
(A)
20 ml
(B)
30 ml
(C)
40 ml
(D)
50 ml
(E)
60 ml
Question 51-53 refer to the following information on heredity.
Hemophilia is a
disorder in which blood fails to clot. John, a male hemophiliac, marries Jane,
a normal woman, and together they have four children, two boys (Mark and Mike)
and two girls (Molly and Mary). None of the children display the symptoms of
hemophilia, Mark, Mike, Molly and Mary all marry normal individuals and have
children. None of Mark's or Mike's children, male or female, display symptoms
of hemophilia, but the sons of Molly and Mary all display symptoms of
hemophilia while the daughters of Molly and Mary do not.
51.
Which of the
following best explains the reason that Mark, Mike, Molly and Mary do not
display symptoms of hemophilia, even though their father, John, is a
hemophiliac?
(A)
Hemophilia is
an X-linked disorder, and John can pass on only his Y chromosome.
(B)
Hemophilia is
an X-linked disorder, and even though Molly and Mary received a hemophiliac X
chromosome from John, Jane gave them a normal X chromosome.
(C)
Hemophilia is a
Y-linked disorder and therefore cannot be displayed in females.
(D)
Hemophilia is a
Y-linked disorder and Mark and Mike must have received an X chromosome from
John.
(E)
Hemophilia is
an X-linked disorder, and even though Mark and Mike received a hemophiliac X
chromosome from John, Jane gave them a normal X chromosome.
52.
If one of
Mike's daughters marries a normal man, what is the probability that one of
their children will display symptoms of hemophilia?
(A)
0%
(B)
25%
(C)
50%
(D)
75%
(E)
100%
53.
Which of the
following individuals are heterozygous for hemophilia?
(A)
John, Mark and
Mike
(B)
Mark, Mike,
Molly and Mary
(C)
John and Jane
(D)
Molly and Mary
(E)
Mark and Mike
Question 54-57
A volunteer was
injected intravenously with several test substances to determine the effect of
each substance on normal body variables. The results are shown in Table 1.
Assume that enough time was allowed between injections so that the substances
do not interfere with one another
Table 1
|
Variable |
baseline
values |
values after
injecting substance A |
values after
injecting substance B |
values after
injecting substance C |
values after
injecting substance D |
|
Serum Ca++ |
2.3 mmol/L |
2.3 mmol/L |
3.0 mmol/L |
2.3 mmol/L |
2.3 mmol/L |
|
Serum Na+ |
135 mmol/L |
135 mmol/L |
136 mmol/L |
135 mmol/L |
147 mmol/L |
|
Serum glucose |
5.6 mmol/L |
3.3 mmol/L |
3.6 mmol/L |
7.4 mmol/L |
5.6 mmol/L |
54.
Based on the
information in Table 1, which of the following is most likely substance B?
(A)
Calcitonin
(B)
Insulin
(C)
Parathyroid
hormone
(D)
Glucagon
(E)
Aldosterone
55.
Based on the
information in Table 1, which of the following is most likely substance A?
(A)
Glucagon
(B)
Aldosterone
(C)
Calcitonin
(D)
Parathyroid
hormone
(E)
Insulin
56.
Under what
conditions might substance D be released normally?
(A)
soon after a
meal
(B)
When blood
pressure is low
(C)
Between meals
(D)
When there has
been limited intake of dietary calcium
(E)
when dietary
calcium is in excess
57.
All of the
following changes in variable values are significant EXCEPT
(A)
the change in
serum glucose when substance A is injected
(B)
the change in
serum Na+ when substance D is injected
(C)
the change in
serum Ca++ when substance B is injected
(D)
the change in
serum glucose when substance C is injected
(E)
the change in
serum Na+ when substance B is injected
Question 58-60
Three different cell types were observed under the microscope. The
observations are summarized in Table 1.
Table 1
|
Cell type |
Nucleus? |
Cell wall? |
Chloroplasts? |
|
A
|
No |
Yes |
No |
|
B
|
Yes |
Yes |
No |
|
C
|
Yes |
Yes |
Yes |
The three cell types were grown in separate cultures with plenty of
oxygen and nutrients available. Figure 1 shows their rates of growth. At Time
1, oxygen was no longer available to the cells.
Figure 1
58.
Based on the
information in Table 1, which of the following is the most likely
classification of cell Type A?
(A)
Fungi
(B)
Plant
(C)
Bacteria
(D)
Animal
(E)
protist
59.
Which of the
following equation is cell Type C able to run?
I.
C6H12O6
+ 6O2
II.
H2O
+ light
III. 6CO3 + 6H2O + ATP + NADPH
(A)
I only
(B)
II only
(C)
I and II only
(D)
II and III only
(E)
I, II and III
60.
Consider Figure
1. Which of the following statements best describes the reason for the difference
between the curves for cell Type B and cell Type C?
(A)
Cell type B is
unable to survive in the presence of oxygen, while cell Type C can ferment.
(B)
The products of
fermentation in cell Type C are toxic to the cells and they are dying.
(C)
Cell Type B is
an obligate aerobe while cell Type C is able to ferment.
(D)
Cell Type B is
a facultative anaerobe, while cell Type C is an obligate aerobe.
(E)
Cell Type C is
an obligate aerobe, while cell Type B is an obligate aerobe.
Directions: Each
of the questions or incomplete statements below is followed by five suggested
answers or completions. Some questions pertain to a set that refers to a
laboratory or experimental situation. For each question, select the one choice
that is the best answer to the question and then fill in the corresponding oval
on the answer sheet.
Question 61-64
61.
The driest of
all terrestrial biomes, characterized by low and unpredictable precipitation
(A)
Thundra
(B)
Taiga
(C)
Tropical rain
forest
(D)
Deciduous
forest
(E)
Desert
62.
Coniferous
forests, characterized by long, cold winters and short, wet summers
63.
Biome
characterized by great diversity of flora and fauna and high levels of
precipitation
64.
Northern areas,
characterized by permafrost, extremely cold temperatures and few trees
65.
Plants that
have true roots, stems and leaves, as well as flowers and seeds enclosed in
fruit, are classified as
(A)
bryophytes
(B)
tracheophytes
(C)
gymnosperms
(D)
angiosperms
(E)
endosperms
66.
Which of the
following indicates that animals have internal biological clocks?
(A)
A mouse kept in
constant darkness shows a daily rhythm of activity.
(B)
A rooster crows
whenever the sun rises in both winter and summer.
(C)
An owl kept in
constant light drifts away from a 24-hour cycle.
(D)
Some species ……
(E)
A squirrel
whose night and day are artificially reversed soon adapts to its schedule.
67.
Which of the
following correctly lists the phylogenic hierarchy?
(A)
Kingdom,
phylum, family, class, order, genus, species
(B)
Phylum, family,
order, class, kingdom, species, genus
(C)
Kingdom,
family, order, class, phylum, genus, species
(D)
Kingdom,
phylum, class, order, family, genus, species
(E)
Family,
kingdom, order, phylum, genus, class, species
68.
A rattlesnake
would be classified as a
(A)
Tertiary
consumer and a heterotroph
(B)
Secondary
consumer and an autotroph
(C)
producer and an
autotroph
(D)
producer and a
heterotroph
(E)
primary
consumer and a heterotroph
69.
At some point
in their development, chordates possess all of the following EXCEPT
(A)
A dorsal hollow
nerve cord
(B)
A notochord
(C)
Gill slits
(D)
postanal tail
(E)
An exoskeleton
Question 70-73
A population of birds (Population A) on a remote, isolated island
is studied to determine beak length. The resulting data are plotted in Figure
1.
Figure 1
Suppose that 200 years later, the beaks of the birds on the island
were again measured (Population B). The data, when plotted, yielded a graph as
in Figure?
Figure 2
70.
What is the
average beak length (in cm) of the birds in Figure 1?
(A)
30 cm
(B)
15 cm
(C)
5 cm
(D)
3 cm
(E)
4 cm
71.
What is the
most likely reason for the difference in distribution of beak lengths between
the data plotted in Figure 1 and the data plotted in Figure 1?
(A)
All birds with
beaks of 30 mm flew to a new island over the 200 year time span.
(B)
Birds with
beaks of 30 mm were selected against.
(C)
Predators
consume birds with beaks of 40 mm.
(D)
Birds with
beaks of 30 mm were selected for extinction.
72.
Suppose that it
researcher studying Population B found that birds with beaks of 20 mm were
unable to mate with birds that had 40 mm beaks. These two groups of birds would
now be classified as
(A)
occupying
different niches
(B)
separate
species
(C)
competitors
(D)
predators
(E)
separate
populations
73.
How would beak
length in the bird population change after another 200 year time span?
(A)
The average
beak length would return to 30 mm.
(B)
The average
beak length would shift to 40 mm.
(C)
The average
beak length would shift to 20 mm.
(D)
The differences
in beak length would more pronounced.
(E)
It is not
possible to determine how beak length might change.
Question 74-78
Acid rain is
formed after the burning of fossil fuels releases compounds containing nitrogen
and sulfur into the atmosphere. Sunlight and rain bring about chemical
reactions that convert these compounds into nitric acid and sulfur dioxide,
which combine with water droplets to form acid rain. Acid rain typically has a
pH of approximately 5.5.
The higher
acidity of soil and water affects many living organisms adversely. As the pH of
lake water falls, fish become ill and die. Table 1 shows the effects of pH on
the size of adult fish.
Table 1
|
pH of lake |
average
length of fish (cm) |
average mass
of fish (g) |
|
7.5 |
30 cm |
454 g |
|
7.0 |
28 cm |
450 g |
|
6.5 |
29 cm |
453 g |
|
6.0 |
25 cm |
401 g |
|
5.5 |
20 cm |
288 g |
|
5.0 |
17 cm |
127 g |
|
4.5 |
all fish dead |
all fish dead |
Mycorrhizal fungi, which form a mutualistic association with many
plant roots, are particularly sensitive to the effects of acid rain. These
fungi facilitate the absorption of water and nutrients by the plants: in turn,
the plants provide sugars and amino acids without which the fungi could not
survive.
74.
The effect of
acid rain on fish size is best represented by which of the following graphs?
Diagram
75.
The
relationship between mycorrhizal fungi and plants can best be described as one
in which
(A)
one partner
benefits from the association and the other partner is harmed
(B)
one partner
benefits from the association and the other partner is neither harmed nor
helped
(C)
one partner
prays upon the other partner
(D)
both partners
benefit from the association
(E)
neither partner
benefits from the association
76.
If the pH of
the soil were 7.0, what would be the effect on the mycorrhizal fungi and plants.
(A)
The fungi would
survive but the plant would be harmed.
(B)
The fungi would
harmed but the plant would survive.
(C)
The fungi would
be slightly harmed and the plant would be slightly harmed.
(D)
Neither the
fungi nor the plant would survive.
(E)
Neither the
fungi nor the plant would be harmed.
77.
What might be
the best strategy to prevent ecological damage due to acid rain?
(A)
Stock the lakes
with bigger fish so that they can resist the effects of the acid better
(B)
Reduce the
amount of fossil fuel that are burned.
(C)
Supply plants
with excess phosphorus and water
(D)
Supply fungi
with excess sugars and amino acids
(E)
Add alkalines
to soil and water to neutralize the acid.
78.
Fungi are
classified as
(A)
prokaryotic
decomposers
(B)
eukaryotic
producers
(C)
eukaryotic
decomposers
(D)
eukaryotic
autotrophs
(E)
prokaryotic
consumers
Question 79-80
The following
graphs show the growth of two closely related species of paramecia, both when
grown alone (Figure 1) and when grown together (Figure 2). Both species consume
bacteria as their food source and reproduce by binary fission as often as
several times a day.
79.
The data in
figure 2 indicate that
(A)
P. aurelia is
preying on P. caudata
(B)
P. aurelia is a
better competitor than P. caudata
(C)
P. aurelia and
P. caudate are in a symbiotic relationship
(D)
P. aurelia is a
parasite of P. caudate
(E)
P. aurelia grew
better when combined with P. caudata than it did when grown alone
80.
Paramecia are
members of the kingdom
(A)
fungi
(B)
Animalia
(C)
monera
(D)
protista
(E)
plantae
Directions: Each
of the questions or incomplete statements below is followed by five suggested
answers or completions. Some questions pertain to a set that refers to a
laboratory or experimental situation. For each question, select the one choice
that is the best answer to the question and then fill in the corresponding oval
on the answer sheet.
81.
All of the
following are true about RNA EXCEPT
(A)
it is single
standard
(B)
its bases are
adenine, thymine, guanine, and uracil
(C)
it has a sugar
phosphate backbone
(D)
its sugar is
ribose
(E)
it is found in
the both the nucleus and the cytoplasm of the cell
82.
The function of
the Golgi appraratus is to
(A)
package and
store proteins for secretion
(B)
synthesize
proteins
(C)
function in
cellular respiration
(D)
help the cell
expel waste
(E)
digest foreign
substances
83.
A eukaryotic
cell that has a cell wall but lacks chloroplasts would be classified as a
(A)
moneran
(B)
chordate
(C)
plant
(D)
fungus
(E)
bacteria
84.
All of the
following could give rise to new species EXCEPT
(A)
Variations in
antler size between male and female reindeer
(B)
An earthquake
that physically separates a population of lizards into two separate groups
(C)
Divergent
evolution
(D)
Evolution of a
population of cats such they can no longer mate with their ancestors
(E)
A massive flood
that separates a population of frogs onto opposite sides of a large lake
85.
A retrovirus requires
which of the following enzymes in order to integrate its genome with its host's
genome?
(A)
Host’s RNA
polymerase
(B)
RNA dependent
RNA polymerase
(C)
DNA dependent
RNA polymerase
(D)
DNA dependent
DNA polymerase
(E)
RNA dependent
DNA polymerase
86.
Which of the
following groups have the most in common with one another?
(A)
Members of the
same kingdom
(B)
Members of the
same genus
(C)
Members of the
same phylum
(D)
Members of the
same class
(E)
Members of the
same family
87.
Which of the
following individuals is the LEAST fit in evolutionary terms?
(A)
A 45 year old
male with a terminal disease who has fathered three children
(B)
A 20 year old
man who has fathered one children
(C)
A 35 year old
woman with four children
(D)
A healthy 4
year old child
(E)
A 25 year old
woman with one child, who has a tubal ligation to prevent future pregnancies
Question 88-92
Most bacteria
can be grown in the laboratory on agar plates containing glucose as their only
carbon source. Some bacteria require additional substances, such as amino acids,
to be added to the growth medium. Such bacteria are termed auxotrophs. These
bacteria are denoted by the amino acid they require followed by a “-” in superscript
(e.g., arg). Bacteria that do not require that particular amino acid can be
indicated by a “+” in superscript.
Different
strains of bacteria were grown on several plates containing a variety of nutrients.
Figure q shows the colonies (numbered) that grew on each plate. The supplements
in each plate are indicated.
Figure 1
In a second
experiment, Colony 1 was mixed with soft agar and spread over a plate so that
an even lawn of bacteria grew. Bacterial lawns appear cloudy on agar plates. A
single drop of an unknown organism was placed in the center of the bacterial
lawn and after 24 hours, a clear area known as a “plaque” appeared at that
spot. The clear area continued to expand at a slow rate. Although new colonies
could be grown from samples taken from the lawn, attempts to grow new colonies
from samples taken from the plaque area were unsuccessful.
88.
Referring to
Figure 1, what is the genotype of Colony 3?
(A)
arg+, leu+,
pro+
(B)
arg+, leu-,
pro+
(C)
arg+, leu+,
pro-
(D)
arg-, leu-,
pro+
(E)
arg-, leu-,
pro-
89.
Is Colony 1 an autotroph?
(A)
Yes, it is able
to grow in the presence of the three amino acids being tested
(B)
Yes, it can
only grow if glucose is present
(C)
No, it is able
to grow in the absence of glucose.
(D)
No, it is able
to grow in the absence of any additional amino acids.
(E)
The data
available are insufficient to determine the answer.
90.
Which structures
could be observed in a sample of Colony 2?
I.
Nuclei
II.
Ribosome
III. Mitochondria
(A)
I only
(B)
II only
(C)
I and III only
(D)
II and III only
(E)
I, II and III
91.
If a liquid
culture medium containing glucose, leucine, and proline was inoculated with
Colony 4, would bacterial growth be observed?
(A)
No, Colony 4 is
an arginine auxotroph (arg).
(B)
No, Colony 4
cannot, grow in the presence of leucine.
(C)
Yes, Colony 4’s
genotype is leu-, pro-.
(D)
Yes, Colony 4
requires only glucose to grow.
(E)
The data
available are insufficient to make a prediction.
92.
What is the
most likely reason for the clearing (the plaque) in the lawn of bacteria in the
second experiment?
(A)
The unknown
organism-is bacterial Colony 2, and these bacteria are eating the 1acteria from.
Colony 1 forming the lawn.
(B)
The unknown organism
is a virus that is infecting the bacteria and causing them to lyse (killing
them).
(C)
The drop placed
iii the center of the lawn contained a strong acid that destroyed the bacteria
at that spot
(D)
Bacteria are very
delicate and the disturbance caused them to die.
(E)
The unknown
organism began producing threonine, which is toxic to Colony 1
Question 93-96
In 1910, a
small town on the East Coast of the United States relied primarily on
agriculture to support its economy. In the mid-1930s, a steel mill was built,
and the economy shifted from being agriculturally supported to being
industrially supported. The steel mill released a lot of smog and soot into the
air, which collected on the bark of trees in a wooded area near the outskirts
of town. Over a period of ten years the bark gradually darkened, then
maintained a constant dark color.
A variety of
animals and insects lived in the wooded area. In particular, a certain species
of moth served as the primary food source for a population of birds. The moths
lay their eggs in the bark of the trees and, thus, must spend a fair amount of
time sitting on the tree trunks. Table 1 presents data on the moth population.
Table 1
|
Year |
% white moths |
% black moths |
|
1910 |
95 |
5 |
|
1920 |
95 |
5 |
|
1930 |
95 |
5 |
|
1940 |
50 |
50 |
|
1950 |
20 |
80 |
|
1960 |
5 |
95 |
93.
The wings of
the moths and the wings of the birds are both used for flight (similar
functions); however, their underlying structures are very different. Moth wings
and bird wings are hut classified as
(A)
homologous structures
(B)
autologous
structures
(C)
divergent strictures
(D)
analogous structures
(E)
emergent
structures
94.
What is the
most likely explanation for the shift in the percentage of black moths in the
population?
(A)
The white moths
no longer blended, the Color of the tree bark and, thus were selected for.
(B)
The black moths
blended better with the color of the tree bark and, thus were selected for.
(C)
The black moths
blended better with the color of the tree bark and, thus were selected against.
(D)
The white moths
blended better with the color of the tree bark and, thus were selected against.
(E)
The black moths
did not blend with the color of the tree bark and, thus were selected against.
95.
If a seed from
one of the trees was planted in an area far from the steel mill, what color
would the bark of the tree be?
(A)
Black, because
the parent tree had black bark
(B)
White, because
the gene causing black bark was mutated due to environmental pollution
(C)
Black, because
the gene causing white bark was mutated due to environment pollution
(D)
White, because
the black bark was an acquired characteristic and is there passed on to progeny
(E)
The color of
the bark is not able to be determined.
96.
Birds track
their prey visually, whereas bats rely on sonar to locate their food. If the
bird population were replaced with a bat population in 1940, what would eb ratio
of white moths to black moths?
(A)
95% white, 5%
black
(B)
80% white, 20%
black
(C)
50% white, 50%
black
(D)
20% white, 80%
black
(E)
5% white, 90%
black
Question 97-100
Dialysis tubing
is a semipermeable membrane. It allows small molecules, such as water, to pass
through easily, while larger molecules, such as sucrose, are restricted.
Movement of molecules across the tubing is due to concentration gradients. In
an experiment designed to study osmosis, several pieces of dialysis tubing were
filled with sucrose solutions of varying concentration and placed in beakers
containing distilled water. The rate and direction of water movement was
determined weighing the bags before and after placing them in the distilled
water. The data are recorded below.
Table 1
|
Tube number |
Tube contents
(breaker contents) |
Mass (g) 0 minutes |
Mass (g) 15 minutes |
Mass (g) 30 minutes |
Mass (g) 45 minutes |
Mass (g) 60 minutes |
|
1
|
Distilled
water (distilled water) |
22.3 g |
22.4 g |
22.2 g |
22.3 g |
22.3 g |
|
2
|
100% sucrose
(distilled water) |
24.8 g |
25.3 g |
25.7 g |
26.4 g |
26.9 g |
|
3
|
40% sucrose
(distilled water) |
25.1 g |
26.3 g |
27.5 g |
28.9 g |
29.6 g |
|
4
|
Distilled
water (40% sucrose) |
22.7 g |
21.3 g |
20.5 g |
19.8 g |
18.7 g |
97.
Why does the
mass of Tube 3 increase while the mass of Tube 4 decreases?
(A)
Water moving into
Tube 3 and sucrose is moving into Tube 4.
(B)
Water is moving
into Tube 4, and sucrose is moving into Tube 3
(C)
Water is moving
into Tube 3, and water is moving out of Tube 4.
(D)
Sucrose is
moving into Tube 3, and sucrose is moving out of Tube 4.
(E)
Sucrose is
moving out of Tube 3, and water is moving out of Tube 4.
98.
Why does the
mass of Tube 1 remain relatively unchanged throughout the experiment?
(A)
The dialysis
tubing in Tube 1 is defective and does not allow water to cross.
(B)
There is no
concentration gradient to drive the movement of sucrose.
(C)
The dialysis
tubing broke, allowing the tube contents to mix with the beaker contents.
(D)
There is no
concentration gradient to drive the movement of water
(E)
The experiment
failed to record the date properly.
99.
Which of the
following graphs best illustrates the relationship between Tube 2 and Tube 3?
Diagram
100.
Cell membranes
are also semipermeable, allowing water but not other substances to cross
easily. A red blood cell placed in a 0.9% NaCl solution will neither swell nor
shrivel. Based on this knowledge, and the information presented in Table 1,
what would happen to a red blood cell placed in a 20% NaCI solution?
(A)
Water would be
drawn out of the cell and the cell would swell.
(B)
Water would be
drawn into the cell and the cell would swell.
(C)
Water would be
drawn out of the cell and the cell would shrivel.
(D)
Water would be
drawn into the cell and the cell would shrivel.
(E)
No change would
occur to the cell.
Answers and
Explanations
|
Question |
Answer |
Explanation |
|
1
|
B |
Prokaryotes
(bacteria, monerans) have no membrane-bound organelles, so all reactions and
processes occur in the cytoplasm. You should have been able to eliminate
choice C, A and e because they did not describe locations. |
|
2
|
D |
When
muscle cells run out of oxygen and switch to anaerobic metabolism (glycolysis
only) to make ATP, the end product is lactic acid. Yeast can also switch to
anaerobic metabolism; their end product is ethanol. You should have been able
to eliminate choices A and B because they are not products. |
|
3
|
B |
Eukaryotes
possess organelles and as such divide the location of their cellular
processes among them. Glycolysis occurs in the cytoplasm, whereas the Krebs
cycle and electron transport occur in the mitochondria. As in Question 1, you
should have eliminated choices C, D and E. |
|
4
|
E |
Recombination
occurs when the homologous chromosomes are paired and crossing over can take place.
This occurs during prophase I of meiosis. Remember that recombination occurs
during prophase, and this would help you eliminate choices A, B and D. |
|
5
|
B |
After
prophase I, the homologous chromosomes remain paired and align at the center
of the cell, on the “metaphase plate.” (The prefix meta means “middle.” Use
this fact to help you elimate choices A, C and E.) During metaphase II, the
individual, unpaired chromosomes align at the cell center. |
|
6
|
A |
During
meiosis, the chromosomes remain replicated (i.e., remain as two jointed
sister chromatids) for the entire first set of divisions. The whole point to
the second set of meiotic divisions is to separate the sister chromatids.
This takes place during and phase II. |
|
7
|
D |
Habituation
involves becoming accustomed to certain stimuli that are not harmful or
important. For example, if you walk down the hallway and a friend jumps out
at you and you get scared, that is a normal reaction to a startling stimulus.
However, if this happens every time you walk down the hallway, you get
accustomed to it and no longer are startled. You have become habituated to
the stimulus. Note that for Questions 7-9, you just had to know the basic
definitions of these types of learning. Most of the classification type
questions are like that. |
|
8
|
B |
Some
animals do not have an instinctive sense for who their mother is and will
bond with any object they are exposed to during a certain time period after
their birth. The “imprints” on their minds, and thereafter, even if exposed
to their real mother, they will still treat the object as Mom. |
|
9
|
C |
Conditioning
involves the association of and response to one stimulus with a second,
different stimulus. The best example is Ivan Pavlov’s dog. He rang a bell
when he fed them, and the dogs salivated in response to the food. Soon, all
he had to do was ring the bell, and the dogs would salivate, even in the
absence of food. |
|
10
|
A |
Most
digestion and absorption occur in the small intestine. A very small amount of
digestion (starch only) takes place in the mouth, and a very small amount of
digestion takes place in the stomach (acid hydrolysis of food and some
protein digestion). As with Questions 7-9, Question 10-11 require the same
type of knowledge - memorization of basic facts. |
|
11
|
E |
Saliva
contains the enzyme amylase, which breaks down starch. |
|
12
|
C |
Cells
in the stomach secrete hydrochloric acid, which keeps the pH of the stomach
around 1-2, the other regions of the digestive tract maintain a fairly
neutral pH. |
|
13
|
A |
Divergent
evolution occurs when the same ancestral organism is placed into different
environments and must then adapt to function in these different environments.
Thus the same original structures evolve separately and “diverge” from one
another. Examples of homologous structures are the arm of a man, the wing of
a bat, and the flipper of a whale. All have the same basic bone structure but
vastly different functions. The opposite of divergent evolution is convergent
evolution, in which vastly different organisms are placed into the same
environment and must adapt to perform similar functions with different
structures. Convergent evolution produces analogous structures, examples of
which are the wings of bats, the wings of butterflies. Speciation is often
the result of divergent evolution, not the cause of it |
|
14
|
C |
The
job of the blood is to carry oxygen from the lungs, where it is plentiful, to
the tissues, where it is not. Thus hemoglobin should have a high affinity for
oxygen in the lungs so it can bind oxygen (choices A, B and D could be
eliminated) and a low oxygen affinity in the tissues (so it can release the
oxygen where it is needed). The reverse is true for carbon dioxide.
Hemoglobin has a high carbon dioxide affinity in the tissues and a low carbon
dioxide affinity in the lungs. |
|
15
|
D |
In
RNA, the base thymine (T) is replaced with uracil (U), so choice A can be
immediately eliminated. Further, A will always pair with U, and G will always
pair with C. The only choice that has the bases paired correctly is choice D.
|
|
16
|
B |
Choices
A, C, D and E all describe characteristics of the phylum arthropoda. A water
vascular system is a characteristic of the phylum echinodermata, the “spiny
skinned” animals such as sea stars and sea urchins. Their water vascular
system ends in tube feet that play a role in locomotion and feeding. Don’t
forget your LEAST/EXCEPT/NOT technique of circling the word “EXCEPT” and
drawing a vertical line through the answer choices to help you remember to
choose the incorrect statement. |
|
17
|
E |
The
kidney’s primary role is to filter blood to remove wastes (statement I is
true and choice D is eliminated), but it is also involved to a fair extent in
blood pressure regulation (through renin and aldosterone, so II is true and
choices A and C are eliminated) and in pH regulation through excretion of
hydrogen ions, so III is true and choice B is eliminated). |
|
18
|
D |
If
a pure-breeding long-tailed chicken (TT) mates with a pure-breeding
short-tailed chicken (tt), all of their offspring (the F1 generation) will
have the genotype Tt (and have long tails). Thus all of them, if mated with
the correct genotype (Tt or tt), could produce offspring with short tails.
Draw some quick Punnett squares to prove it to yourself. |
|
19
|
B |
Density-dependent
factors are those that get more significant as the size of the population
increases. Limited nutrients and water, toxic waste build-up, and predation
are all issues that are of greater concern to a large population that to a
small one. Only choice B, climate temperature, is not more worrisome to a
large group than to a small one. It, will affect all populations equally,
regardless of their size. Remember your LEAT/EXCEPT/NOT technique. |
|
20
|
C |
Two
populations are considered separate species when they are so different from
one another that they can no longer mate and produce viable offspring. Thus,
organisms that can mate with each other must be of the same species. |
|
21
|
E |
Blood
that is poor in oxygen returns from the body to the right side of the heart
(I is true, so choices B and C are eliminated), then travels through the
pulmonary artery (III is true, so choices A and D are eliminated) to get to
the lungs, where it picks yup oxygen again. This oxygen-rich blood returns to
the left side of the heart through the pulmonary vein (II is false) and is
pumped back out to the body through the aorta. |
|
22
|
B |
The
characteristics described are those of the phylum annelida, the best example
of which is the earthworm. Mollusks have external shells (snails),
echinoderms and arthropods have exoskeletons (sea stars, crustaceans,
insects), and chordates have endoskeletons and, in any case, are not worms. |
|
23
|
A |
The
light reactions of photosynthesis convert solar energy to usable energy in
the form of ATP (choices C and E are eliminated) and NADPH (a reduced
electron carrier). The ATP and NADPH (i.e. energy) produced during these
reactions are used later during the Calvin cycle to fix carbon dioxide into
carbohydrates, like glucose. Because glucose is a product of the
light-independent reactions, choice B could be eliminated. Remember that
NADPH belongs with photosynthesis to eliminate choice D. |
|
24
|
C |
When
the concentration of substrate far exceed the concentration of enzyme
(remember, the question states that enzyme concentration is assumed to be
constant), all the enzyme active sites are saturated with substrate, and the
product is being formed at its maximum rate. The only way to increase product
formation at this point is to increase the concentration of the enzyme. Note
that enzymes should not be used up in the course of the reaction (A is
wrong). Furthermore, product formation is still occurring, just at a stable
rate (B and D are wrong). There is no reason to assume an inhibitor has been
added; the rate of product formation remains constant. |
|
25
|
D |
Berries
are plant products (i.e. primary producers), thus any organism that eats
berries is a primary consumer, on an herbivore (I is true, choices B, C and E
can be eliminated). Notice at this point that the only remaining choices (A
and D) do not contain option III, therefore, option III is false. Secondary
consumers, carnivores and omnivores (e.g. birds) eat primary consumers (e.g.
bugs), thus II is also true, and choice A can be eliminated. Terriary
consumers are carnivores (e.g. cats) that eat other carnivores (e.g. birds,
secondary consumers). III is false, as we saw earlier. |
|
26
|
B |
Carbohydrate
have the general molecular formula C6H2nOn,
as in glucose, which is C6H12O6. The only formula that fits this rule is
choice B. |
|
27
|
C |
A
change in a population that occurs over a long period of time is evolution.
This alone is a good tip-off that choice C, the only choice that mentions
evolution, is correct. Speciation has not occurred, only a change in the
characteristic of the birds, thus choices A and D can be eliminated. Any
change in the fitness of the insects would change the characteristics of the
insect population, not the bird population (B is wrong), and random mutation
would not produce a specific, directed effect (E is wrong). Birds with large
beaks had greater fitness because they could more easily obtain food, thus
they had an advantage over birds with smaller beaks, which dies out as tie
passed. |
|
28
|
E |
The
other choices do not describe substrate binding sites on an enzyme. |
|
29
|
D |
Break
this question down one piece at a time. First, the question states that there
is a high sodium concentration outside the cell. This means sodium wants to
move into the cell (where concentration is lower). Choices A and C can be
eliminated because they state that sodium would move out of the cell. Active
transport requires ATP, and because there is no ATP involved, choice E can be
eliminated. Last, because sodium moves across the membrane with the help of a
channel, it is moving by facilitated diffusion (choice B is eliminated). |
|
30
|
E |
Decomposers
take organic material and break it down into its individual compounds, thus
returning these compounds back to the earth. The other processes listed are
carried out by other organisms: nitrogen fixing bacteria (choice A),
heterotrophs and autotrophs (choice B), and other soil bacteria (choice C and
D) |
|
31
|
B |
The
endocrine system is a body control system, but it is NOT rapid. The fastest
hormone in the body is adrenaline, and even that takes a few seconds,
compared the nervous system’s milliseconds. Most hormones operate in the
minutes to hours range. The other choices regarding the endocrine system are
all true. Remember your LEAST/EXCEPT/NOT technique! |
|
32
|
B |
This
symbiotic relationship describes commensalism. In mutualism both partners
benefit, in parasitism and predator prey relationships one partner benefits
while the other is harmed, and symbiosis is just a general term used to
describe close living arrangements. |
|
33
|
D |
Relationships
that are directly proportional have linear graphs with a positive slope. |
|
34
|
B |
Male
haploid cells (pollen grains or microspores are produced on the anther (#2 in
the diagram), which is at the tip of the filament (#3 in the diagram). |
|
35
|
A |
Pollen
grains stick to the stigma (#1 in the diagram), which is supported by the
style (#4 in diagram). |
|
36
|
E |
The
pollen fertilizes the ovule (female haploid cells, or megaspores, # 6 in the
diagram); once fertilized, the ovary (#8 in the diagram) develops into a
fruit. |
|
37
|
C |
The
prefix photo refers to light; because the light is growing toward light we
can eliminate choices A, D and E. “Positive” means growing toward, so choice
B can be eliminated. |
|
38
|
A |
The
prefix geo refers to the earth, or soil’ thus we can eliminate choices C, D
and E. because the plant is growing away from the earth, this is a negative
tropism and we can eliminate choice B. |
|
39
|
D |
A
surge in LH is what causes ovulation and is measured in the ovulation
prediction kits. Estrogen and progesterone affect the uterus, not the ovary,
and FSH causes development of a follicle (A, B and C are wrong). Testosterone
is a male hormone (E is wrong). |
|
40
|
BH |
Because
we know hormone X is peaking at ovulation time, a quick look at the graph
shows hormone X peaking at about Day 14 of the cycle. |
|
41
|
A |
The
rise in hormone Y occurs after ovulation (choice E is wrong) and coincides
with formation of the corpus luteum (choices B and D are wrong). The primary
hormone secreted by the corpus luteum is progesterone. |
|
42
|
E |
Rapid
cell division after fertilization is known as cleavage. Blastulation is the
formation of a hollow ball of cells (A is wrong), gastrulation is formation
of the three primary germ layers (B is wrong), neurulation is development of
the nervous system (C is wrong), and implantation is when the morula (solid
ball of cells) burrows into the uterine lining (D is wrong). |
|
43
|
C |
You
should remember the prefixes anto, meso, and endo for the primary germ layers
and therefore eliminate choices D and E right away. The ectoderm forms the
skin, hair, nails, mouth lining and nervous system. The mesoderm forms
muscle, bone, blood vessels, and organs (B is wrong). The endoderm forms
inner linings and glands (A is wrong). |
|
44
|
B |
The
development of a thriving ecosystem from a barren area is known as
succession. Note the evolution usually has a much longer time frame than
succession. |
|
45
|
B |
The
climax community is the final, stable community in secession. The key word
here is “stable.” That should tip you off that this is the end of the
process, or the “climax.” |
|
46
|
D |
The
first organisms to colonize a barren area are known as the pioneer organisms.
|
|
47
|
A |
Diuretics
help eliminate water (i.e., increase urine production) from the body. From
the data tables, the only substance that increases urine production
significantly is caffeine. Don’t forget the I, II, III technique here; even
just knowing that option I is true allows you to eliminate two (choices B and
D) of the five choices. |
|
48
|
A |
Again,
from the data tables, there is a directly proportional (i.e., linear)
relationship between the amount of caffeine ingested and the volume of urine
produced. As caffeine ingested and the volume of urine produced. As caffeine
consumption increases, so des urine volume. The only graph that shows this
relationship is choice A. |
|
49
|
C |
Because
the caffeine and the sodium chloride were dissolved in water, plain, water
was consumed as a control, to make sure the effects seen were due to the
added substances and not the water. Questions about experimental controls
come up fairly frequently on the SAT Biology E/M. Subject Test; make sure you
know the definition for a control and how to spot in the experiment. |
|
50
|
C |
From
Table 2, an increase in sodium chloride of 0.9 g results in a decrease in
urine volume of approximately 40 ml. when 3.6 g sodium chloride are ingested,
82 ml urine is produced; thus if 4.5 g sodium chloride were to be ingested,
the expected urine volume would be 40 ml less, approximately 40 ml. |
|
51
|
B |
As
soon as you see “hemophilia,” you should be thinking “X-linked disorder.”
Then use the technique for avoiding the temptation trap, which is
particularly dangerous here, because the passage and the questions are
confusing, and it’s very tempting to just guess blindly. Resist! Take the
paragraph apart piece by piece, sentence by sentence. Write out genotypes as
you read through and construct Punnett squares to help you see probability.
Out of this family, the only members that express this condition are males.
This is a tip-off for X-linked disorders, which are more common in males
because they have only a single X chromosome. (In any case, you should
remember the two most common X-linked disorders; hemophilia and color
blindness). John’s genotype is YXh. He passed his Y chromosome to
Mark and Mike; they also received a normal X from Jane, thus they do not have
hemophilia, nor can they pass it on to their kids. Molly and Mary received Xh
from John but also received a normal X from Janem thus they are carriers of
hemophilia but do not display its symptoms. |
|
52
|
A |
Because
Mike does not carry the gene for hemophilia (see above), he cannot pass it on
to his children, and they in turn cannot pass it on to their children. |
|
53
|
D |
Mark
and Mike does not carry the gene for hemophilia (see solution to 51 above),
thus we can eliminate choices A, B, and E. Jane is normal, so choice C is
eliminated as well. |
|
54
|
C |
From
Table 1, substance B caused on increase in serum calcium levels, which is the
effect that insulin has on the body. |
|
55
|
E |
Again,
from Table 1, substance A caused on increase in serum calcium levels, which
is the effect that insulin has on the body. |
|
56
|
B |
Substance
D causes an increase in serum sodium, which is the effect aldosterone has on
the body. Aldosterone is released when blood pressure is low, because excess
sodium will have the effect of causing water retention, which will increase
blood volume, which will increase blood volume, which will increase blood
volume, which will increase blood pressure. (Note: Even if you did not know
this, you should have been able to eliminate the other choices.) |
|
57
|
E |
The
change in serum sodium after injection of Substances B is insignificant. All
other choices cause significant change from the baseline values of the
variable being measured. Don’t forget the LEAST/EXCEPT/NOT technique. |
|
58
|
C |
Cell
Type A has no nucleus. The only organisms that do not have nuclei are bacteria
(kingdom Monera). |
|
59
|
E |
Cell
Type C has a nucleus, a cell wall, and chloroplasts and therefore most likely
comes from a plant. Equations II and II are the equations for photosynthesis
and would occur in plants (choices A, B, and C are eliminated), and equation
I is the equation for cellular respiration, which also occurs in plants
(choice D is eliminated). |
|
60
|
D |
This
is a great question to do some answer predicting on. At Time I the oxygen was
removed from the cultures and cell Type C died. Clearly it is an obligate
aerobe. Thus we can eliminate choices A, B, and C. Because cell Type B was
growing well in the presence of oxygen, it cannot be an obligate anaerobe,
thus choice E is eliminated. Cell Type B must be a facultative anaerobe,
suing oxygen when kit is available and fermenting when oxygen is not
available. The decrease in growth of cell Type B after Time I is most likely
because energy is produced during fermentation than during aerobic
metabolism. |
|
61
|
E |
These
are the characteristics of desert. |
|
62
|
B |
These
are the characteristics of taiga. |
|
63
|
C |
These
are the characteristics of tropical rain forest. |
|
64
|
A |
These
are the characteristics of tundra. |
|
65
|
D |
Flowering
plants are angiosperms. Gymnosperms are confiders (naked seed plants; C is
wrong), bryophytes are mosses (A is wrong), and tracheophytes are non-seed
producing plants (ferns; B is wrong). “Endosperm” is not a classification for
plants. |
|
66
|
A |
If
an organism’s environment remains absolutely constant, and that organism
still exhibits regular rhythms of activity there must be some internal
“clock” that keeps it on schedule (C and E are wrong). Roosters vary the time
of their crow as the sun varies the time it rises (B is wrong). ‘The magnetic
field has nothing to do with internal clocks (D is wrong). |
|
67
|
D |
Remember:
“King Philip came Over From Germany – so?” or make up you own! |
|
68
|
A |
Rattlesnakes
are clearly heterotrophs (only photosynthetic organisms are autotrophs), so
we can eliminate choices B and C. Rattlesnakes are carnivores, not producers
(D is wrong) or primary consumers (herbivores, so E is wrong). |
|
69
|
E |
Remember
your LEAST/EXCEPT/NOT technique. There are four features present in al
chordates – dorsal nerve cords (choice A), notochords (choice B), gill slits
(choice C), and postnatal tails (choice D). Note that some of these features
may be found only in embryonic or larval stages. The “wrong” answer is choice
E _ note all chordates (for example, sea squirts and lancelets) have a bony
endoskeleton. |
|
70
|
D |
The
question asks for the average beak length in cm, but the graph gives it in
mm. Average beak length is 30 mm. 10 mm = 1 cm; therefore, 30 mm = cm. read the questions carefully! |
|
71
|
B |
Clearly
the birds with 30 mm beaks were not surviving too well. There is no reason to
assume they flew to another island; remember, they are on a remote, isolated
island. There may not be another island near enough to fly to (Asis wrong).
If predators consumed birds with 20 mm or 40 mm beaks they would not be the
prevalent populations in Figure 2 (C and D are wrong), and if birds with 30 m
beaks were selected for, the population would
not have been divided (E is wrong). |
|
72
|
B |
The
defining characteristic for speciation is an inability to interbreed. |
|
73
|
E |
Just
because we have some information about how the population change in the last
200 years, it doesn’t tell us how it may change in the next 200 years. It
would depend on how the environment changed during that time period. |
|
74
|
D |
As
the acidity increases (pH goes down), the average mass of the fish decreases.
How-ever, it does not decrease linearly; rather, it says constant for a
while, then gradually drops off, then rapidly drops off as acidity becomes
severe. The best representation of this is choice D. |
|
75
|
D |
The
plant benefits by easier availability of water and phosphorus, and the fungi
benefit by receiving amino acids and sugars. Another term for this type of
relationship is mutualism. |
|
76
|
E |
pH
7.0 is neutral, thus neither the plant nor the fungi would be harmed. |
|
77
|
B |
The
best way to prevent damage from acid rain would be to prevent its formation
by reducing the burning of the fossil fuels that cause it. There is no
guarantee that bigger fish will resist the acidity any better than smaller
fish (A is wrong), supplying plants with sugars and amino acids will not help
them overcome the effects of acid soil (D is wrong), and, even if alkalines
will neutralize acid, they might be just as harmful to the environment (E is
wrong)! |
|
78
|
C |
Fungi
are euakryotes (A and E are eliminated), and they are not photosynthetic (B
and D are eliminated). |
|
79
|
B |
Clearly
P. Aurelia can compete better and get more food that P. caudate; thus it will
grow while P. caudate is competed to extinction. Choice A is highly unlikely
because the good source the paramecia prefer is bacteria, not each other.
This is not a symbiotic relationship but a competitive one (C and D are
eliminated), and the data contradict choice E. |
|
80
|
D |
Paramecia
are single-celled eukaryotes, members of kingdom Protista. |
|
81
|
B |
RNA
bases do not include thymine; they are adenine, guanine, cytosine, and
uracil. All other statements about RNA are correct. Remember the
LEAST/EXCEPT/NOT technique! |
|
82
|
A |
Ribosomes
synthesize protein (B is wrong), mitochondria function in respiration (C is
wrong), vacuoles help expel waste (D is wrong), and lysosomes function in
digestion (E is wrong). |
|
83
|
D |
Choices
A and E are prokaryotic and can be eliminated. Chordates have no cell walls
(B is wrong), and plants have chloroplasts (C is wrong). |
|
84
|
A |
The
defining characteristic for speciation is an inability to interbreed. Choices
B, C, D, and E could all ultimately produce two different populations that
lack the ability to mate. Choice A would not lead to an inability to mate.
Remember this is a LEAST/EXCEPT/NOT questions. |
|
85
|
E |
A
retrovirus has an RNA genome, so its polymerase must be able to read RNA
(DNA-dependent choices C and D can be eliminated). Furthermore, retroviruses
go through the lysogenic life cycle, and so must insert their genome into
their host’s genome. Because all other organisms have a DNA genome, a DNA
copy of the viral (RNA) genome must be synthesized. A DNA polymerase is
needed (choices A and B can be eliminated). Note that the host’s RNA
polymerase is DNA dependent, the same answer as choice C. |
|
86
|
B |
The
members that have the most in common with one another are the members near
the bottom of the hierarchy. Of the choices given, genus is the closest to
the bottom of the hierarchy. |
|
87
|
D |
Anyone
who has produced offspring has demonstrated their fitness. Regardless of how
healthy a child is, he has not yet produced offspring to prove his fitness.
Use the LEAST/EXCEPT/NOT technique. |
|
88
|
C |
The
only plate that Colony 3 cannnot grow on is Plate C, which lacks proline.
Thus colony 3 requires proline to row and is a proline auxotroph (pro). This
eliminates choices A, B, and D. Choice E is incorrect because Colony 3 does
not require arginine or leucine to grow; it can grow just fine in the absence
of these amino acids, as is indicated on plates A and B. |
|
89
|
D |
Colony
1 can grow on any of the plates, thus it does not require any additional
amino acids. Auxotrophs require additional supplements to their growth media.
|
|
90
|
B |
Because
these are bacterial colonies, they would not have any membrane-bound
organelles, and thus no nuclei (choices A, C, and E can be eliminated) or
mitochondria (choie D can be eliminated). Bacteria do have ribosomes to
synthesie proteins. |
|
91
|
A |
It
really helps to predict an answer BEFORE you look at the answer choices.
Sometimes looking at the choices first can confuse your thinking and lead you
to a trap, but if you have an idea of the correct answer before you look at
the choices, you will be less tempted. Colony 4 cannot grow in the absence of
arginine as is evidenced by Plate B. thus, because the liquid medium does not
contain arginine, no bacterial growth would be observed. |
|
92
|
B |
Again,
try to predict an answer first. Clear spots lawns of bacteria are due to
infection by viruses that cause lysis of the bacteria. Even if this was not
obvious to you, you should have been able to eliminate the other choices.
There is no reason to assume Colony 2 would prey on Colony 1 (A is wrong);
strong acid would lyse and destroy the Bacteria immediately, not after 24
hours (C is wrong); bacteria are not delicate, they can grow just about
anywhere, under any conditions (D is wrong); and there is no data to support
fact that threonine may be toxic to the bacteria or that the “unknown
organism” was producing it. |
|
93
|
D |
Structures
with similar functions but different underlying structures are the result of
vastly different organisms being placed into similar environments and having
to adapt to the same stresses with different starting materials. These are
termed “analogous structures” and are the result of convergent evolution.
(See also Questions # 13). |
|
94
|
B |
Because
the percentage of black moths is increasing, they must be selected for. This
eliminates C and E. because the percentage of white moths is decreasing, they
must be selected against, eliminating choice A. choice D is wrong because
white moths would not blend better against dark tree bark. |
|
95
|
D |
The
original parent had white bark. The change in bark color is the result of an
accumulation of soot on the tree. This is an acquired characteristic and
would not be passed on to offspring. Seedlings that grew far from the plant
would not be exposed to soot in the air and would not experience
discoloration of their bark. |
|
96
|
C |
The
reason the percentage of black moths increased was because they were no
longer visible against the now darkened bark of the trees. Because the white
moths were more easily seen by the birds, their numbers declined. However,
because bats rely on sonar to locate prey instead of vision, darker coloring
would not give the moths any advantage, and the population percentages would
stay at the point they were at when the birds were replaced by bats, a
fifty-fifty split. |
|
97
|
C |
Sucrose
cannot cross the dialysis membrane, so it cannot cause any effect on the mass
of the tubes. This eliminates choices A, B, D, and E. |
|
98
|
D |
Because
movement across the membrane relies strictly on concentration gradients, the
fact that there is no gradient in Tube 1 would prevent the movement of water
into or out of the tube thus there would be no change in miss. |
|
99
|
B |
The
gradient in Tube 3 is much larger than the gradient in Tube 2, thus water
would be expected to enter more rapidly. This is confirmed by the data in
Table 1. A linear graph should show two lines with positive slopes, and the
slope of Tube 3’s line should be greater than the lope of Tube 2’s line. |
|
100
|
C |
The
fact that the cell does not swell or shrivel in 0.9% NaCl implies that there
is no concentration gradient. A cell in a 20% NaCl solution would experience
similar stresses to a dialysis tube filled with water sitting in a beaker
filled with a more concentrated solution, such as Tube 4. The data indicate
that Tube 4 lost mass, thus water exited the tube, and the same would happen
to the red blood cell. |
2ND DIAGNOSTIC TEST
1.
Sexually
reproducing organisms show greater variation than a sexually reproducing ones
because
(A)
they exhibit
fewer mutations
(B)
they exhibit a
greater mutation rate
(C)
a sexually
reproducing organisms do not have internal membranes.
(D)
their alleles
recombine
(E)
they are larger
2.
Several species
of rhododendron are growing in the same area. All of the plants are capable of
hybridization, but none ever do because some of the plants produce pollen in
early June while others produce pollen in late June. This best describes an
evolutionary process known as
(A)
Survival of the
fittest
(B)
Overpopulation
(C)
Reproductive
isolation
(D)
Artificial
selection
(E)
Stabilizing
selection
3.
All of the
following contribute to variation in a population EXCEPT
(A)
Mutation
(B)
Isolation
(C)
Sexual
reproduction
(D)
Conjugation
(E)
Genetic drift
4.
Oxygen released
by plants comes from
(A)
Air
(B)
Carbon dioxide
(C)
Glucose
(D)
Chlorophyll
(E)
Water
5.
All of the
following are mammals Except
(A)
tiger
(B)
ape
(C)
kangaroo
(D)
blue jay
(E)
duck-billed
platypus
6.
Which of the
following is a density-independent factor?
(A)
Disease
(B)
Famine
(C)
Floods
(D)
Predation
(E)
Increase in
toxins in the environment
7.
A gene pool in
a population of jackrabbits in a field remained constant for many generations.
The most probable reason for this stable gene pool it that
(A)
no migration
occurred in a large population with random mating and no mutation
(B)
no migration
occurred in a small population with random mating and no mutation
(C)
no migration
occurred in a large population with nonrandom mating and no mutation
(D)
there was much
migration into and out of the large population, but matting was random and
there were few mutations.
(E)
the population
was small with no mutations, no migrations, and nonrandom mating.
8.
All of the
following about plasma membrane structure and function arc correct EXCEPT.
(A)
All plasma
membranes have the identical composition and structure
(B)
Diffusion of
gases across a membrane require that the membrane be moist
(C)
Facilitated diffusion
is an example of passive transport.
(D)
proteins serve
as membrane channels.
(E)
plasma
membranes contain receptors that are specific for the molecules they uptake
9.
Which of the
following exhibits internal fertilization, external development of the embryo,
few eggs, and much parenting?
(A)
Mammals
(B)
Amphibians
(C)
Reptiles
(D)
Birds
(E)
Fish
10.
A solution with
a pH of 5 is ____ times more acidic than a solution a pH of 7.
(A)
1/10
(B)
1/100
(C)
10
(D)
100
(E)
1,000
11.
Vitamins are
essential for normal cell function. They are important because they
(A)
function as an
energy source
(B)
are hormones
(C)
directly assist
in the normal conduction of impulses
(D)
resist pH
changes
(E)
enable enzymes
to function normally
12.
Tendons connect
________ to ________: ligaments connect ________ to ________.
(A)
bone to bone;
bone to muscle
(B)
bone to muscle
bone to bone
(C)
bone to bone;
muscle to muscle
(D)
muscle to
muscle; bone to bone
(E)
ligaments to
bone; tendons to 6 ones
13.
Food chains never
consist of more than 4 or 5 trophic levels: The reason for this is
(A)
energy is lost
along the food chain
(B)
there are fewer
primary consumers in the world than secondary consumers
(C)
producers are
always shown on the bottom of any food chain
(D)
pioneer organisms
compete with consumers
(E)
all of the
above are correct
14.
Here is a sketch.
All of the fol1owin processes produce this molecule EXCEPT the
Diagram
(A)
Calvin cycle
(B)
Krebs cycle
(C)
electron
transport chain
(D)
light dependent
reactions
(E)
glycolysis
15.
A black animal
as crossed with a white animal an all the offspring are black. Which pattern of
inheritance nr likely at work?
(A)
Law of
dominance
(B)
Law of
segregation
(C)
Incomplete dominate
(D)
Codominance
(E)
Sex linked
inheritance
16.
According to
the best scientific evidence about evolution, species descended from a common
ancestor.
(A)
Slowly and
gradually by the accumulation of many small changes
(B)
rapidly through
divergent evolution alone
(C)
rapidly through
mutation alone
(D)
in spurts of
relatively rapid change
(E)
because they
needed to adapt to a changing environment or they
17.
A black h is crossed
with a white rooster and only gray offspring result If two of these ray
offspring are mated, what is the chance of hatching a white offspring
(A)
0%
(B)
25%
(C)
50%
(D)
75%
(E)
100%
18.
Farmers have
successfully bred Brussels sprouts, broccoli, kale, and cauliflower from the
mustard plant. This demonstrates
(A)
convergent
evolution.
(B)
coevolution
(C)
adaptive
radiation
(D)
natural
selection
(E)
artificial
selection
Questions 19-21
(A)
imprinting
(B)
Classical
condition
(C)
Fixed action
(D)
Altruism
(E)
Operant
condition
19.
Geese hatching
follow the first thing they see
20.
Innate, highly
stereotypical behavior, which, once begun, is continued to completion no matter
how useless
21.
Trial and error
learning
22.
Scientists
believe chat the giraffe original had a short neck that has grown long over
time. The most likely explanation of this is which of the following?
I.
Natural
selection
II.
Adaptive
radiation
III. Divergent evolution
(A)
I only
(B)
I and II only
(C)
I and III only
(D)
II and III only
(E)
I, II, and III
23.
Which of the
following are most closely
I.
Acer rubrum
II.
Acer sucte
III. P.seudorricon nubrum
(A)
I and II
(B)
II and III
(C)
I and II
(D)
All are closely
related.
(E)
It cannot be
determined using only scientific names.
Question 24-26.
Refer to this drawing of the eye.
Diagram
24.
Identify the
structure that change to allow different amounts of light to enter the eye.
25.
Identify the
structure that absorb light and sends nerve impulses to the brain.
26.
Identify the
retina
Question 24-26.
Five beakers are used in an experiment about osmosis. Each beaker
contains 50 mL of sucrose solution of varying concentrations: 0.2 M 0.4 M 0.6
M, 0.8-M, 1.0 M. Pieces of fresh potato (each 10.0 g in mass) are cut up
weighed, and plated into the beakers. After 12 hours the potatoes are crefu1ly
removed from each beaker and weighed again. Sec the data in the table below.
|
Beaker |
Concentration
of sucrose solution |
Mass of
Potato at Time Zero |
Mass of
Potato After 12 Hours |
|
1
|
1.0 M |
10.0 g |
8.2 g |
|
2
|
1.8 M |
10.0 g |
9.4 g |
|
3
|
0.6 M |
10.0 g |
3.8 g |
|
4
|
0.4 M |
10.0 g |
11.5 g |
|
5
|
0.2 M |
10.0 g |
13.5 g |
27.
In this
experiment
(A)
water flowed
into the potato only
(B)
water flowed
out of the potato
(C)
sucrose flowed
into the potato only
(D)
sucrose flowed
both into and out of the potato
28.
given the
result of this experiment what is the molarity (concentration) with the potato
cells?
(A)
Less than 0.2 M
(B)
Less than 0.4 M
but greater than 0.2 M
(C)
Less than 0.6 M
but greater than 0.4 M
(D)
Less than 0.8 M
but greater than 0.6 M
(E)
Greater than 0.8
M
29.
The results of
this experiment give support to the theory that
(A)
water diffuses
down a gradient
(B)
water can be
actively transported against a gradient
(C)
solutes will
diffuse from high concentration to low concentration
(D)
living cells
respond in different ways to the same condition
(E)
potato ceil
respond differently of mother living cells
Question 36-39
Refer to this graph of an impulse passing across a neuron.
Diagram
36.
The impulse is
passing
37.
The
sodium-potassium pump is responsible for pumping ions across the membrane
38.
A steep
gradient of sodium and potassium ions exists at the axon membrane
39.
An impulse
cannot pass
Question 40-41
A study of a small farm in Michigan was carried out in 2004. A
variety of organisms to live there, including meadow voles, grasshoppers
spiders, birds, and mice. The farmer retired and moved away, leaving the land
to grow wild.
40.
The meadow
voles, grasshoppers, spiders, mice, and other organisms, along with the soil,
minerals, and water make up a(n)
(A)
Ecosystem
(B)
Population
(C)
Community
(D)
food chain
(E)
desert biome
41.
The study of
the farm revealed the population size of the different species of animals
during the summer months of June July, and August. The results are recorded in
the table below
|
|
Number of
Organisms |
||
|
Species of
animals |
June |
July |
August |
|
Spiders |
850 |
300 |
550 |
|
Grasshoppers |
1,600 |
4,600 |
4,000 |
|
Mice |
275 |
225 |
250 |
|
Birds |
95 |
80 |
90 |
Which is
correct about the data collected from June to August?
(A)
Only the spider
population changed to any extent
(B)
The population
of mice increased as summer went or
(C)
The population
of grasshoppers remained constant:
(D)
The population
of birds remain fairly constant
(E)
Both the
population of spiders and mice remained constant.
42.
What will most likely
occur if the farm is sold and the fields are allowed to grow wild?
(A)
The plant will
change, but the animals will stay the same.
(B)
They animals
will change, but the plants will stay the same
(C)
Neither the
plant not the animals will change because the climate will not change.
(D)
Both animals
and plants will change.
(E)
All will slowly
die out because they will not be adapted to the new environment.
43.
Although
several different species of birds inhabit the farm, competition between these
birds rarely occurs. The best explanation for this lack of competition is that these
birds
(A)
share food with
each other
(B)
have a limited
red supply of food
(C)
live in
different ecological niches
(D)
are closely
related
(E)
have
experienced mutations in their DNA that prevent them from competing
44.
Humans eat corn
and eat other vegetables. Humans also eat beef from cattle that were corn fed.
In those cases, cattle and humans occupy which of the following trophic levels?
(A)
Producer and
primary consumer
(B)
Primary
consumer and secondary consumer
(C)
Secondary
consumer and tertiary consumer
(D)
Tertiary
consumer and quaternary consumer
(E)
Both are
primary
Questions 45-47
(A)
Tundra
(B)
Marine biome
(C)
Desert
(D)
Temperate
deciduous forest
(E)
Tropical rain
forest
45.
Only 4 percent
of the land surface but accounts for 20 percent of Earth’s food production
46.
Provides most
of Earth’s food and oxygen
47.
Consists of
trees that drop their leaves in winter
Question 48-51
Refer to the diagram of a flower
Diagram
48.
Site where the
pollen germinates
49.
Sire of sperm
production
50.
Becomes the
fruit
51.
Becomes the
seed
Questions 52-54
The table shows a Series of Four fruit fly experiment breeding normal
and vestigial wing flies. The flies in each cross could be either homozygous heterozygous
for wing trait.
|
Parents |
offspring |
|||
|
Cross |
Female |
Male |
Normal wing |
Vestigial wing |
|
1
|
Normal Wing |
Vestigial wing |
95 |
0 |
|
2
|
Vestigial wing |
Normal wing |
105 |
0 |
|
3
|
Normal Wing |
Normal Wing |
76 |
23 |
|
4
|
Normal Wing |
Vestigial wing |
56 |
51 |
52.
The trait for
vestigial wings is most likely
(A)
Autosomal
dominant
(B)
Autosomal
recessive
(C)
Sex linked
dominant
(D)
Sex-linked
recessive
(E)
It cannot be
determined
53.
What is the
most likely genotype of the female normal wing in cross 4?
(A)
Nn
(B)
NN
(C)
Nn
(D)
X-X
(E)
X-X-
54.
What is the
most likely genotype of the male normal wing in cross 3?
(A)
NN
(B)
Nn
(C)
nn
(D)
X-X
(E)
X-X-
55.
Which is the
true about the karyotype below?
Diagram
(A)
It shows a
normal female.
(B)
It shows a
normal male.
(C)
It shows a
person who suffers from Down syndrome.
(D)
It shows person
who suffers from a gene mutation.
(E)
It shows a
person who suffers from a condition that result from nondisjunction.
Questions 56-58
Refer to this sketch of prokaryotic DNA as it commonly undergoes
replication and transcription simultaneously
56.
If 1 is
adenine, then A must be
(A)
Guanine
(B)
Cytosine
(C)
Thymine
(D)
Adenine
(E)
uracil
57.
If 1 is
adenine, then A must be
(A)
Guanine
(B)
Cytosine
(C)
Thymine
(D)
Adenine
(E)
Uracil
58.
In what way
would the ocess shown be different in a eukaryotic cell?
(A)
Eukaryotic
cells & not carry out transcription.
(B)
Eukaryotic
cells do not carry out replication.
(C)
Eukaryotic
cells do not carry out both transcription and replication.
(D)
Eukaryotic
cells carry our both processes, but they do not occur at the same time.
(E)
Eukaryotic
cells carry oat both these processes in the Golgi body.
59.
A fungus
infection affected nearly all-the oak trees in a particular that the coloration
of the bark turned almost black. Scientists studying the diseased trees
discovered that a moth population that inhabited the forest changed from being
light brown to being almost black. Which of the following would best explain
that color change of the moth population?
(A)
The moths.
developed darker wings to blend in with the trees.
(B)
The. fungus
infected the moths as well as the oak tree.
(C)
The almost
black moths within the population were the only ones to survive once
(D)
The trees
darkened because of the fungus infection.
(E)
The moths were
the first to change color, which caused the trees to darken.
(F)
The fungus
caused mutations to occur in the moths as well as in the oak trees
60.
According to
scientific evidence, the age of Earth is closest to
(A)
400 years old
(B)
4,000 years old
(C)
400,000 years
old
(D)
4,000,000 years
old
(E)
4 billion years
old
61.
All of the
following are true of organisms classified in the domain Archaea EXPECT.
(A)
One example is
E. coli, the organism that lives in the human gut
(B)
they can thrive
in environment with very high temperature
(C)
they can thrive
in environment with high salt concentrations
(D)
they have no
intrnal membranes
(E)
their DNA an
contain introns
62.
Factors that
influence population density include which of the following?
I.
Predation
II.
Interspecies
coopetition
III. Interspecies competition
(A)
I only
(B)
II only
(C)
1 and III only
(D)
II and III only
(E)
I, II, and III
63.
The human
population today can best described as
(A)
Declining
(B)
growing
linearly
(C)
growing
exponentially
(D)
at the carrying
capacity
(E)
fluctuating
seasonally
64.
All of the are
true of K-strategists EXCEPT
(A)
Intensive
parenting
(B)
reproduce only
once or twice
(C)
example humans
(D)
large young
(E)
slow maturation
65.
Lamprey eels
attach to the skin of certain tout and absorb nutrients from the body of the trout.
Which symbols best represent this relationship?
(A)
(+/+)
(B)
(+/0)
(C)
(+/–)
(D)
(–/+)
(E)
(–/0)
Answers and
Explanations
|
Question |
Answer |
Explanation |
|
1
|
B |
Prokaryotes
(bacteria, monerans) have no membrane-bound organelles, so all reactions and
processes occur in the cytoplasm. You should have been able to eliminate
choice C, A and e because they did not describe locations. |
|
2
|
D |
When
muscle cells run out of oxygen and switch to anaerobic metabolism (glycolysis
only) to make ATP, the end product is lactic acid. Yeast can also switch to
anaerobic metabolism; their end product is ethanol. You should have been able
to eliminate choices A and B because they are not products. |
|
3
|
B |
Eukaryotes
possess organelles and as such divide the location of their cellular
processes among them. Glycolysis occurs in the cytoplasm, whereas the Krebs
cycle and electron transport occur in the mitochondria. As in Question 1, you
should have eliminated choices C, D and E. |
|
4
|
E |
Recombination
occurs when the homologous chromosomes are paired and crossing over can take place.
This occurs during prophase I of meiosis. Remember that recombination occurs
during prophase, and this would help you eliminate choices A, B and D. |
|
5
|
B |
After
prophase I, the homologous chromosomes remain paired and align at the center
of the cell, on the “metaphase plate.” (The prefix meta means “middle.” Use
this fact to help you elimate choices A, C and E.) During metaphase II, the
individual, unpaired chromosomes align at the cell center. |
|
6
|
A |
During
meiosis, the chromosomes remain replicated (i.e., remain as two jointed
sister chromatids) for the entire first set of divisions. The whole point to
the second set of meiotic divisions is to separate the sister chromatids.
This takes place during and phase II. |
|
7
|
D |
Habituation
involves becoming accustomed to certain stimuli that are not harmful or
important. For example, if you walk down the hallway and a friend jumps out
at you and you get scared, that is a normal reaction to a startling stimulus.
However, if this happens every time you walk down the hallway, you get
accustomed to it and no longer are startled. You have become habituated to the
stimulus. Note that for Questions 7-9, you just had to know the basic
definitions of these types of learning. Most of the classification type
questions are like that. |
|
8
|
B |
Some
animals do not have an instinctive sense for who their mother is and will
bond with any object they are exposed to during a certain time period after
their birth. The “imprints” on their minds, and thereafter, even if exposed
to their real mother, they will still treat the object as Mom. |
|
9
|
C |
Conditioning
involves the association of and response to one stimulus with a second,
different stimulus. The best example is Ivan Pavlov’s dog. He rang a bell
when he fed them, and the dogs salivated in response to the food. Soon, all
he had to do was ring the bell, and the dogs would salivate, even in the
absence of food. |
|
10
|
A |
Most
digestion and absorption occur in the small intestine. A very small amount of
digestion (starch only) takes place in the mouth, and a very small amount of
digestion takes place in the stomach (acid hydrolysis of food and some
protein digestion). As with Questions 7-9, Question 10-11 require the same
type of knowledge - memorization of basic facts. |
|
11
|
E |
Saliva
contains the enzyme amylase, which breaks down starch. |
|
12
|
C |
Cells
in the stomach secrete hydrochloric acid, which keeps the pH of the stomach
around 1-2, the other regions of the digestive tract maintain a fairly
neutral pH. |
|
13
|
A |
Divergent
evolution occurs when the same ancestral organism is placed into different environments
and must then adapt to function in these different environments. Thus the
same original structures evolve separately and “diverge” from one another.
Examples of homologous structures are the arm of a man, the wing of a bat,
and the flipper of a whale. All have the same basic bone structure but vastly
different functions. The opposite of divergent evolution is convergent
evolution, in which vastly different organisms are placed into the same
environment and must adapt to perform similar functions with different
structures. Convergent evolution produces analogous structures, examples of
which are the wings of bats, the wings of butterflies. Speciation is often
the result of divergent evolution, not the cause of it |
|
14
|
C |
The
job of the blood is to carry oxygen from the lungs, where it is plentiful, to
the tissues, where it is not. Thus hemoglobin should have a high affinity for
oxygen in the lungs so it can bind oxygen (choices A, B and D could be
eliminated) and a low oxygen affinity in the tissues (so it can release the
oxygen where it is needed). The reverse is true for carbon dioxide.
Hemoglobin has a high carbon dioxide affinity in the tissues and a low carbon
dioxide affinity in the lungs. |
|
15
|
D |
In
RNA, the base thymine (T) is replaced with uracil (U), so choice A can be
immediately eliminated. Further, A will always pair with U, and G will always
pair with C. The only choice that has the bases paired correctly is choice D.
|
|
16
|
B |
Choices
A, C, D and E all describe characteristics of the phylum arthropoda. A water
vascular system is a characteristic of the phylum echinodermata, the “spiny
skinned” animals such as sea stars and sea urchins. Their water vascular
system ends in tube feet that play a role in locomotion and feeding. Don’t
forget your LEAST/EXCEPT/NOT technique of circling the word “EXCEPT” and
drawing a vertical line through the answer choices to help you remember to
choose the incorrect statement. |
|
17
|
E |
The
kidney’s primary role is to filter blood to remove wastes (statement I is
true and choice D is eliminated), but it is also involved to a fair extent in
blood pressure regulation (through renin and aldosterone, so II is true and
choices A and C are eliminated) and in pH regulation through excretion of
hydrogen ions, so III is true and choice B is eliminated). |
|
18
|
D |
If
a pure-breeding long-tailed chicken (TT) mates with a pure-breeding
short-tailed chicken (tt), all of their offspring (the F1 generation) will
have the genotype Tt (and have long tails). Thus all of them, if mated with
the correct genotype (Tt or tt), could produce offspring with short tails.
Draw some quick Punnett squares to prove it to yourself. |
|
19
|
B |
Density-dependent
factors are those that get more significant as the size of the population
increases. Limited nutrients and water, toxic waste build-up, and predation
are all issues that are of greater concern to a large population that to a
small one. Only choice B, climate temperature, is not more worrisome to a
large group than to a small one. It, will affect all populations equally,
regardless of their size. Remember your LEAT/EXCEPT/NOT technique. |
|
20
|
C |
Two
populations are considered separate species when they are so different from
one another that they can no longer mate and produce viable offspring. Thus,
organisms that can mate with each other must be of the same species. |
|
21
|
E |
Blood
that is poor in oxygen returns from the body to the right side of the heart
(I is true, so choices B and C are eliminated), then travels through the
pulmonary artery (III is true, so choices A and D are eliminated) to get to
the lungs, where it picks yup oxygen again. This oxygen-rich blood returns to
the left side of the heart through the pulmonary vein (II is false) and is
pumped back out to the body through the aorta. |
|
22
|
B |
The
characteristics described are those of the phylum annelida, the best example
of which is the earthworm. Mollusks have external shells (snails), echinoderms
and arthropods have exoskeletons (sea stars, crustaceans, insects), and
chordates have endoskeletons and, in any case, are not worms. |
|
23
|
A |
The
light reactions of photosynthesis convert solar energy to usable energy in
the form of ATP (choices C and E are eliminated) and NADPH (a reduced
electron carrier). The ATP and NADPH (i.e. energy) produced during these
reactions are used later during the Calvin cycle to fix carbon dioxide into
carbohydrates, like glucose. Because glucose is a product of the
light-independent reactions, choice B could be eliminated. Remember that
NADPH belongs with photosynthesis to eliminate choice D. |
|
24
|
C |
When
the concentration of substrate far exceed the concentration of enzyme
(remember, the question states that enzyme concentration is assumed to be
constant), all the enzyme active sites are saturated with substrate, and the
product is being formed at its maximum rate. The only way to increase product
formation at this point is to increase the concentration of the enzyme. Note
that enzymes should not be used up in the course of the reaction (A is
wrong). Furthermore, product formation is still occurring, just at a stable
rate (B and D are wrong). There is no reason to assume an inhibitor has been
added; the rate of product formation remains constant. |
|
25
|
D |
Berries
are plant products (i.e. primary producers), thus any organism that eats
berries is a primary consumer, on an herbivore (I is true, choices B, C and E
can be eliminated). Notice at this point that the only remaining choices (A
and D) do not contain option III, therefore, option III is false. Secondary
consumers, carnivores and omnivores (e.g. birds) eat primary consumers (e.g.
bugs), thus II is also true, and choice A can be eliminated. Terriary
consumers are carnivores (e.g. cats) that eat other carnivores (e.g. birds,
secondary consumers). III is false, as we saw earlier. |
|
26
|
B |
Carbohydrate
have the general molecular formula C6H2nOn,
as in glucose, which is C6H12O6. The only formula that fits this rule is
choice B. |
|
27
|
C |
A
change in a population that occurs over a long period of time is evolution.
This alone is a good tip-off that choice C, the only choice that mentions
evolution, is correct. Speciation has not occurred, only a change in the
characteristic of the birds, thus choices A and D can be eliminated. Any
change in the fitness of the insects would change the characteristics of the
insect population, not the bird population (B is wrong), and random mutation
would not produce a specific, directed effect (E is wrong). Birds with large
beaks had greater fitness because they could more easily obtain food, thus
they had an advantage over birds with smaller beaks, which dies out as tie
passed. |
|
28
|
E |
The
other choices do not describe substrate binding sites on an enzyme. |
|
29
|
D |
Break
this question down one piece at a time. First, the question states that there
is a high sodium concentration outside the cell. This means sodium wants to
move into the cell (where concentration is lower). Choices A and C can be
eliminated because they state that sodium would move out of the cell. Active
transport requires ATP, and because there is no ATP involved, choice E can be
eliminated. Last, because sodium moves across the membrane with the help of a
channel, it is moving by facilitated diffusion (choice B is eliminated). |
|
30
|
E |
Decomposers
take organic material and break it down into its individual compounds, thus
returning these compounds back to the earth. The other processes listed are
carried out by other organisms: nitrogen fixing bacteria (choice A),
heterotrophs and autotrophs (choice B), and other soil bacteria (choice C and
D) |
|
31
|
B |
The
endocrine system is a body control system, but it is NOT rapid. The fastest
hormone in the body is adrenaline, and even that takes a few seconds,
compared the nervous system’s milliseconds. Most hormones operate in the
minutes to hours range. The other choices regarding the endocrine system are
all true. Remember your LEAST/EXCEPT/NOT technique! |
|
32
|
B |
This
symbiotic relationship describes commensalism. In mutualism both partners
benefit, in parasitism and predator prey relationships one partner benefits
while the other is harmed, and symbiosis is just a general term used to
describe close living arrangements. |
|
33
|
D |
Relationships
that are directly proportional have linear graphs with a positive slope. |
|
34
|
B |
Male
haploid cells (pollen grains or microspores are produced on the anther (#2 in
the diagram), which is at the tip of the filament (#3 in the diagram). |
|
35
|
A |
Pollen
grains stick to the stigma (#1 in the diagram), which is supported by the
style (#4 in diagram). |
|
36
|
E |
The
pollen fertilizes the ovule (female haploid cells, or megaspores, # 6 in the
diagram); once fertilized, the ovary (#8 in the diagram) develops into a
fruit. |
|
37
|
C |
The
prefix photo refers to light; because the light is growing toward light we
can eliminate choices A, D and E. “Positive” means growing toward, so choice
B can be eliminated. |
|
38
|
A |
The
prefix geo refers to the earth, or soil’ thus we can eliminate choices C, D
and E. because the plant is growing away from the earth, this is a negative
tropism and we can eliminate choice B. |
|
39
|
D |
A
surge in LH is what causes ovulation and is measured in the ovulation
prediction kits. Estrogen and progesterone affect the uterus, not the ovary,
and FSH causes development of a follicle (A, B and C are wrong). Testosterone
is a male hormone (E is wrong). |
|
40
|
BH |
Because
we know hormone X is peaking at ovulation time, a quick look at the graph
shows hormone X peaking at about Day 14 of the cycle. |
|
41
|
A |
The
rise in hormone Y occurs after ovulation (choice E is wrong) and coincides
with formation of the corpus luteum (choices B and D are wrong). The primary
hormone secreted by the corpus luteum is progesterone. |
|
42
|
E |
Rapid
cell division after fertilization is known as cleavage. Blastulation is the
formation of a hollow ball of cells (A is wrong), gastrulation is formation
of the three primary germ layers (B is wrong), neurulation is development of
the nervous system (C is wrong), and implantation is when the morula (solid
ball of cells) burrows into the uterine lining (D is wrong). |
|
43
|
C |
You
should remember the prefixes anto, meso, and endo for the primary germ layers
and therefore eliminate choices D and E right away. The ectoderm forms the
skin, hair, nails, mouth lining and nervous system. The mesoderm forms
muscle, bone, blood vessels, and organs (B is wrong). The endoderm forms
inner linings and glands (A is wrong). |
|
44
|
B |
The
development of a thriving ecosystem from a barren area is known as
succession. Note the evolution usually has a much longer time frame than
succession. |
|
45
|
B |
The
climax community is the final, stable community in secession. The key word
here is “stable.” That should tip you off that this is the end of the
process, or the “climax.” |
|
46
|
D |
The
first organisms to colonize a barren area are known as the pioneer organisms.
|
|
47
|
A |
Diuretics
help eliminate water (i.e., increase urine production) from the body. From
the data tables, the only substance that increases urine production
significantly is caffeine. Don’t forget the I, II, III technique here; even
just knowing that option I is true allows you to eliminate two (choices B and
D) of the five choices. |
|
48
|
A |
Again,
from the data tables, there is a directly proportional (i.e., linear)
relationship between the amount of caffeine ingested and the volume of urine
produced. As caffeine ingested and the volume of urine produced. As caffeine
consumption increases, so des urine volume. The only graph that shows this
relationship is choice A. |
|
49
|
C |
Because
the caffeine and the sodium chloride were dissolved in water, plain, water
was consumed as a control, to make sure the effects seen were due to the
added substances and not the water. Questions about experimental controls
come up fairly frequently on the SAT Biology E/M. Subject Test; make sure you
know the definition for a control and how to spot in the experiment. |
|
50
|
C |
From
Table 2, an increase in sodium chloride of 0.9 g results in a decrease in
urine volume of approximately 40 ml. when 3.6 g sodium chloride are ingested,
82 ml urine is produced; thus if 4.5 g sodium chloride were to be ingested,
the expected urine volume would be 40 ml less, approximately 40 ml. |
|
51
|
B |
As
soon as you see “hemophilia,” you should be thinking “X-linked disorder.”
Then use the technique for avoiding the temptation trap, which is
particularly dangerous here, because the passage and the questions are
confusing, and it’s very tempting to just guess blindly. Resist! Take the
paragraph apart piece by piece, sentence by sentence. Write out genotypes as
you read through and construct Punnett squares to help you see probability.
Out of this family, the only members that express this condition are males.
This is a tip-off for X-linked disorders, which are more common in males
because they have only a single X chromosome. (In any case, you should
remember the two most common X-linked disorders; hemophilia and color
blindness). John’s genotype is YXh. He passed his Y chromosome to
Mark and Mike; they also received a normal X from Jane, thus they do not have
hemophilia, nor can they pass it on to their kids. Molly and Mary received Xh
from John but also received a normal X from Janem thus they are carriers of
hemophilia but do not display its symptoms. |
|
52
|
A |
Because
Mike does not carry the gene for hemophilia (see above), he cannot pass it on
to his children, and they in turn cannot pass it on to their children. |
|
53
|
D |
Mark
and Mike does not carry the gene for hemophilia (see solution to 51 above),
thus we can eliminate choices A, B, and E. Jane is normal, so choice C is
eliminated as well. |
|
54
|
C |
From
Table 1, substance B caused on increase in serum calcium levels, which is the
effect that insulin has on the body. |
|
55
|
E |
Again,
from Table 1, substance A caused on increase in serum calcium levels, which
is the effect that insulin has on the body. |
|
56
|
B |
Substance
D causes an increase in serum sodium, which is the effect aldosterone has on
the body. Aldosterone is released when blood pressure is low, because excess
sodium will have the effect of causing water retention, which will increase
blood volume, which will increase blood volume, which will increase blood
volume, which will increase blood pressure. (Note: Even if you did not know
this, you should have been able to eliminate the other choices.) |
|
57
|
E |
The
change in serum sodium after injection of Substances B is insignificant. All
other choices cause significant change from the baseline values of the
variable being measured. Don’t forget the LEAST/EXCEPT/NOT technique. |
|
58
|
C |
Cell
Type A has no nucleus. The only organisms that do not have nuclei are bacteria
(kingdom Monera). |
|
59
|
E |
Cell
Type C has a nucleus, a cell wall, and chloroplasts and therefore most likely
comes from a plant. Equations II and II are the equations for photosynthesis
and would occur in plants (choices A, B, and C are eliminated), and equation
I is the equation for cellular respiration, which also occurs in plants
(choice D is eliminated). |
|
60
|
D |
This
is a great question to do some answer predicting on. At Time I the oxygen was
removed from the cultures and cell Type C died. Clearly it is an obligate
aerobe. Thus we can eliminate choices A, B, and C. Because cell Type B was
growing well in the presence of oxygen, it cannot be an obligate anaerobe,
thus choice E is eliminated. Cell Type B must be a facultative anaerobe,
suing oxygen when kit is available and fermenting when oxygen is not
available. The decrease in growth of cell Type B after Time I is most likely
because energy is produced during fermentation than during aerobic
metabolism. |
|
61
|
E |
These
are the characteristics of desert. |
|
62
|
B |
These
are the characteristics of taiga. |
|
63
|
C |
These
are the characteristics of tropical rain forest. |
|
64
|
A |
These
are the characteristics of tundra. |
|
65
|
D |
Flowering
plants are angiosperms. Gymnosperms are confiders (naked seed plants; C is
wrong), bryophytes are mosses (A is wrong), and tracheophytes are non-seed
producing plants (ferns; B is wrong). “Endosperm” is not a classification for
plants. |
|
66
|
A |
If
an organism’s environment remains absolutely constant, and that organism
still exhibits regular rhythms of activity there must be some internal
“clock” that keeps it on schedule (C and E are wrong). Roosters vary the time
of their crow as the sun varies the time it rises (B is wrong). ‘The magnetic
field has nothing to do with internal clocks (D is wrong). |
|
67
|
D |
Remember:
“King Philip came Over From Germany – so?” or make up you own! |
|
68
|
A |
Rattlesnakes
are clearly heterotrophs (only photosynthetic organisms are autotrophs), so
we can eliminate choices B and C. Rattlesnakes are carnivores, not producers
(D is wrong) or primary consumers (herbivores, so E is wrong). |
|
69
|
E |
Remember
your LEAST/EXCEPT/NOT technique. There are four features present in al
chordates – dorsal nerve cords (choice A), notochords (choice B), gill slits
(choice C), and postnatal tails (choice D). Note that some of these features
may be found only in embryonic or larval stages. The “wrong” answer is choice
E _ note all chordates (for example, sea squirts and lancelets) have a bony
endoskeleton. |
|
70
|
D |
The
question asks for the average beak length in cm, but the graph gives it in
mm. Average beak length is 30 mm. 10 mm = 1 cm; therefore, 30 mm = cm. read the questions carefully! |
|
71
|
B |
Clearly
the birds with 30 mm beaks were not surviving too well. There is no reason to
assume they flew to another island; remember, they are on a remote, isolated
island. There may not be another island near enough to fly to (Asis wrong).
If predators consumed birds with 20 mm or 40 mm beaks they would not be the
prevalent populations in Figure 2 (C and D are wrong), and if birds with 30 m
beaks were selected for, the population would
not have been divided (E is wrong). |
|
72
|
B |
The
defining characteristic for speciation is an inability to interbreed. |
|
73
|
E |
Just
because we have some information about how the population change in the last
200 years, it doesn’t tell us how it may change in the next 200 years. It
would depend on how the environment changed during that time period. |
|
74
|
D |
As
the acidity increases (pH goes down), the average mass of the fish decreases.
How-ever, it does not decrease linearly; rather, it says constant for a
while, then gradually drops off, then rapidly drops off as acidity becomes
severe. The best representation of this is choice D. |
|
75
|
D |
The
plant benefits by easier availability of water and phosphorus, and the fungi
benefit by receiving amino acids and sugars. Another term for this type of
relationship is mutualism. |
|
76
|
E |
pH
7.0 is neutral, thus neither the plant nor the fungi would be harmed. |
|
77
|
B |
The
best way to prevent damage from acid rain would be to prevent its formation
by reducing the burning of the fossil fuels that cause it. There is no
guarantee that bigger fish will resist the acidity any better than smaller
fish (A is wrong), supplying plants with sugars and amino acids will not help
them overcome the effects of acid soil (D is wrong), and, even if alkalines
will neutralize acid, they might be just as harmful to the environment (E is
wrong)! |
|
78
|
C |
Fungi
are euakryotes (A and E are eliminated), and they are not photosynthetic (B
and D are eliminated). |
|
79
|
B |
Clearly
P. Aurelia can compete better and get more food that P. caudate; thus it will
grow while P. caudate is competed to extinction. Choice A is highly unlikely
because the good source the paramecia prefer is bacteria, not each other.
This is not a symbiotic relationship but a competitive one (C and D are
eliminated), and the data contradict choice E. |
|
80
|
D |
Paramecia
are single-celled eukaryotes, members of kingdom Protista. |
|
81
|
B |
RNA
bases do not include thymine; they are adenine, guanine, cytosine, and
uracil. All other statements about RNA are correct. Remember the
LEAST/EXCEPT/NOT technique! |
|
82
|
A |
Ribosomes
synthesize protein (B is wrong), mitochondria function in respiration (C is
wrong), vacuoles help expel waste (D is wrong), and lysosomes function in
digestion (E is wrong). |
|
83
|
D |
Choices
A and E are prokaryotic and can be eliminated. Chordates have no cell walls
(B is wrong), and plants have chloroplasts (C is wrong). |
|
84
|
A |
The
defining characteristic for speciation is an inability to interbreed. Choices
B, C, D, and E could all ultimately produce two different populations that
lack the ability to mate. Choice A would not lead to an inability to mate.
Remember this is a LEAST/EXCEPT/NOT questions. |
|
85
|
E |
A
retrovirus has an RNA genome, so its polymerase must be able to read RNA
(DNA-dependent choices C and D can be eliminated). Furthermore, retroviruses
go through the lysogenic life cycle, and so must insert their genome into their
host’s genome. Because all other organisms have a DNA genome, a DNA copy of
the viral (RNA) genome must be synthesized. A DNA polymerase is needed
(choices A and B can be eliminated). Note that the host’s RNA polymerase is
DNA dependent, the same answer as choice C. |
|
86
|
B |
The
members that have the most in common with one another are the members near
the bottom of the hierarchy. Of the choices given, genus is the closest to
the bottom of the hierarchy. |
|
87
|
D |
Anyone
who has produced offspring has demonstrated their fitness. Regardless of how
healthy a child is, he has not yet produced offspring to prove his fitness.
Use the LEAST/EXCEPT/NOT technique. |
|
88
|
C |
The
only plate that Colony 3 cannnot grow on is Plate C, which lacks proline.
Thus colony 3 requires proline to row and is a proline auxotroph (pro). This
eliminates choices A, B, and D. Choice E is incorrect because Colony 3 does
not require arginine or leucine to grow; it can grow just fine in the absence
of these amino acids, as is indicated on plates A and B. |
|
89
|
D |
Colony
1 can grow on any of the plates, thus it does not require any additional
amino acids. Auxotrophs require additional supplements to their growth media.
|
|
90
|
B |
Because
these are bacterial colonies, they would not have any membrane-bound organelles,
and thus no nuclei (choices A, C, and E can be eliminated) or mitochondria
(choie D can be eliminated). Bacteria do have ribosomes to synthesie
proteins. |
|
91
|
A |
It
really helps to predict an answer BEFORE you look at the answer choices.
Sometimes looking at the choices first can confuse your thinking and lead you
to a trap, but if you have an idea of the correct answer before you look at
the choices, you will be less tempted. Colony 4 cannot grow in the absence of
arginine as is evidenced by Plate B. thus, because the liquid medium does not
contain arginine, no bacterial growth would be observed. |
|
92
|
B |
Again,
try to predict an answer first. Clear spots lawns of bacteria are due to
infection by viruses that cause lysis of the bacteria. Even if this was not
obvious to you, you should have been able to eliminate the other choices.
There is no reason to assume Colony 2 would prey on Colony 1 (A is wrong);
strong acid would lyse and destroy the Bacteria immediately, not after 24
hours (C is wrong); bacteria are not delicate, they can grow just about
anywhere, under any conditions (D is wrong); and there is no data to support
fact that threonine may be toxic to the bacteria or that the “unknown
organism” was producing it. |
|
93
|
D |
Structures
with similar functions but different underlying structures are the result of
vastly different organisms being placed into similar environments and having
to adapt to the same stresses with different starting materials. These are
termed “analogous structures” and are the result of convergent evolution.
(See also Questions # 13). |
|
94
|
B |
Because
the percentage of black moths is increasing, they must be selected for. This
eliminates C and E. because the percentage of white moths is decreasing, they
must be selected against, eliminating choice A. choice D is wrong because
white moths would not blend better against dark tree bark. |
|
95
|
D |
The
original parent had white bark. The change in bark color is the result of an
accumulation of soot on the tree. This is an acquired characteristic and
would not be passed on to offspring. Seedlings that grew far from the plant
would not be exposed to soot in the air and would not experience
discoloration of their bark. |
|
96
|
C |
The
reason the percentage of black moths increased was because they were no
longer visible against the now darkened bark of the trees. Because the white
moths were more easily seen by the birds, their numbers declined. However,
because bats rely on sonar to locate prey instead of vision, darker coloring
would not give the moths any advantage, and the population percentages would
stay at the point they were at when the birds were replaced by bats, a
fifty-fifty split. |
|
97
|
C |
Sucrose
cannot cross the dialysis membrane, so it cannot cause any effect on the mass
of the tubes. This eliminates choices A, B, D, and E. |
|
98
|
D |
Because
movement across the membrane relies strictly on concentration gradients, the
fact that there is no gradient in Tube 1 would prevent the movement of water
into or out of the tube thus there would be no change in miss. |
|
99
|
B |
The
gradient in Tube 3 is much larger than the gradient in Tube 2, thus water
would be expected to enter more rapidly. This is confirmed by the data in
Table 1. A linear graph should show two lines with positive slopes, and the
slope of Tube 3’s line should be greater than the lope of Tube 2’s line. |
|
100
|
C |
The
fact that the cell does not swell or shrivel in 0.9% NaCl implies that there
is no concentration gradient. A cell in a 20% NaCl solution would experience
similar stresses to a dialysis tube filled with water sitting in a beaker
filled with a more concentrated solution, such as Tube 4. The data indicate
that Tube 4 lost mass, thus water exited the tube, and the same would happen
to the red blood cell. |