aamc self assesement biology

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Biological Sciences

Self-Assessment:

Biology Test

MCAT® is a program of the Association of American Medical Colleges

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Taking Your Test Offline

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Printing Guide

Use this printing guide as a reference to print selected sections of this test.

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Commitment Self-Assessment Click PAGES FROM radio to and enter

page 5 to 5

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pages 6 to 7

Periodic Table Click PAGES FROM radio to and enter

page 8 to 8

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pages 10 to 54

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Commitment Self-Assessment

How committed are you to completing this test and using the results to prepare for the MCAT?

1=Not committed , 2=Somewhat committed, 3=Committed

If answer is 1, --- Completing the test requires a commitment of time and energy. If you do not feel you can

commit the time to complete it, you may be better off waiting to take the test until you can commit the time. The

test must be completed to receive the feedback to guide your study.

If answer 2,--- It’s okay if you are unsure about your confidence to use the results to prepare for the MCAT. The

unknown can be daunting. However, it is important that you feel motivated to complete the test since you need

to answer all the questions to receive feedback. The Official MCAT® Self- Assessment Package will show your

relative strengths and weaknesses to help you determine in what areas you should focus your preparation.

The entire test will take a few hours to complete, but you don’t need to complete it all at once. It doesn’t matter

how long it takes you to finish the test, but you do need to finish to receive feedback!

If answer 3,-- You’ve taken an important step in preparing for the MCAT by committing your time and energy to

completing the Self-Assessments. It is okay if you don’t know all the answers. This time spent on preparation

and practice will help you figure out your relative strengths and weaknesses in the content of the MCAT so that

you can plan your study most effectively.

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Pre-test Confidence Self-Assessment

One of the factors that can influence both your preparation and performance on the actual MCAT exam is your

confidence. This questionnaire will help you assess your confidence on the topics in this section of the exam so

that you can use the information to decide where you should focus your study time. You will be asked to rate

your confidence again after completing the test to help you gauge how your experience with actual MCAT

questions influences your perception of your ability in these content areas so that you can decide if you were

overconfident, under confident or on target and why this may be.

Confidence: Using the 5-point scale below in the table, how confident are you are in your ability to perform

well on this section of the MCAT exam as well as for each content category?

MCAT®

is a program of the Association of American Medical Colleges

Test/Content

Categories

A=1-Not

Confident at

all

B=2-

Somewhat

Confident

C=3-

Moderately

Confident

D=4-Very

Confident

E=5-

Extremely

Confident

1. Biology Overall

2. Circulatory,

Lymphatic, and

Immune Systems

3. Digestive and

Excretory Systems

4. DNA and Protein

Synthesis

5. Enzymes and

Cellular Metabolism

6. Evolution

7. Generalized

Eukaryotic Cell

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8. Genetics

9. Microbiology

10. Muscle and Skeletal

Systems

11. Nervous and

Endocrine Systems

12. Respiratory System

13. Reproductive

System and

Development

14. Specialized

Eukaryotic Cells

and Tissues

15. Skin System

16. Eukaryotes

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1 H 1.0

Periodic Table of the Elements

2

He 4.0

3 Li 6.9

4 Be 9.0

5 B

10.8

6 C

12.0

7 N

14.0

8 O

16.0

9 F

19.0

10 Ne 20.2

11 Na 23.0

12 Mg 24.3

13 Al 27.0

14 Si

28.1

15 P

31.0

16 S

32.1

17 Cl 35.5

18 Ar 39.9

19 K

39.1

20 Ca 40.1

21 Sc 45.0

22 Ti

47.9

23 V

50.9

24 Cr 52.0

25 Mn 54.9

26 Fe 55.8

27 Co 58.9

28 Ni 58.7

29 Cu 63.5

30 Zn 65.4

31 Ga 69.7

32 Ge 72.6

33 As 74.9

34 Se 79.0

35 Br 79.9

36 Kr 83.8

37 Rb 85.5

38 Sr 87.6

39 Y

88.9

40 Zr 91.2

41 Nb 92.9

42 Mo 95.9

43 Tc (98)

44 Ru

101.1

45 Rh

102.9

46 Pd

106.4

47 Ag

107.9

48 Cd

112.4

49 In

114.8

50 Sn

118.7

51 Sb

121.8

52 Te

127.6

53 I

126.9

54 Xe

131.3 55 Cs

132.9

56 Ba

137.3

57 La* 138.9

72 Hf

178.5

73 Ta

180.9

74 W

183.9

75 Re

186.2

76 Os

190.2

77 Ir

192.2

78 Pt

195.1

79 Au

197.0

80 Hg

200.6

81 Tl

204.4

82 Pb

207.2

83 Bi

209.0

84 Po

(209)

85 At

(210)

86 Rn (222)

87 Fr

(223)

88 Ra

(226)

89 Ac† (227)

104 Rf

(261)

105 Db (262)

106 Sg

(266)

107 Bh

(264)

108 Hs

(277)

109 Mt (268)

110 Ds

(281)

111 Uuu (272)

112 Uub (285)

114

Uuq (289)

116

Uuh (289)

* 58 Ce

140.1

59 Pr

140.9

60 Nd

144.2

61 Pm (145)

62 Sm 150.4

63 Eu

152.0

64 Gd

157.3

65 Tb

158.9

66 Dy

162.5

67 Ho

164.9

68 Er

167.3

69 Tm 168.9

70 Yb

173.0

71 Lu

175.0

† 90 Th

232.0

91 Pa

(231)

92 U

238.0

93 Np (237)

94 Pu

(244)

95 Am (243)

96 Cm (247)

97 Bk

(247)

98 Cf

(251)

99 Es

(252)

100 Fm (257)

101 Md (258)

102 No

(259)

103 Lr

(260)

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Biological Sciences Self-Assessment: Biology Test

Number of Questions: 128

Approximate Time to Complete: 3-4 hours

Welcome to the Biological Sciences Self-Assessment: Biology

Test. The goal of this test is to analyze your knowledge in the

content of the MCAT. In order to obtain an accurate assessment

of your strengths and weaknesses, you must answer every

question. Because the test is lengthy, you are encouraged to

take breaks as needed.

Good luck!

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Passage I

Ectopic pregnancy is defined as the

development of a fertilized ovum outside the

uterine cavity. Most frequently, ectopic

development occurs in the fallopian tube (oviduct).

The first symptoms of tubal pregnancy are the same

as those of a normal early pregnancy. A positive

test result for the presence of the hormone human

chorionic gonadotropin (HCG) confirms

pregnancy, but does not indicate the location of the

pregnancy.

As the ectopic embryo begins to outgrow the

tiny fallopian tube, the woman may experience

lower abdominal discomfort and recurrent vaginal

bleeding. As rupture of the tube occurs or becomes

imminent, pain becomes severe, and the woman

may collapse due to internal hemorrhaging.

Treatment involves the immediate surgical removal

of the affected segment of the fallopian tube and

drainage of any blood that has accumulated in the

body cavity.

There are many causes of tubal pregnancy,

including abnormalities of the fallopian tube, the

zygote, and the endocrine system. For example,

diseases resulting in tubal infections (e.g.,

gonorrhea) may partially block a fallopian tube,

leading to ectopic pregnancy. The premature

breakdown of the protective acellular layer

surrounding the zygote may facilitate the

attachment of the zygote to the wall of the fallopian

tube rather than to the wall of the uterus. Finally,

altered hormone levels may delay ovulation and/or

inhibit ovum transport by decreasing the motility of

the tubal cilia.

Another factor associated with the development

of ectopic pregnancy is the use of contraceptives

such as the “morning-after pill” and the intrauterine

device. When these contraceptive methods fail, the

risk of developing an ectopic pregnancy increases

tenfold.

1. The one aspect of ectopic pregnancy common to

all the causes described in the passage is that the

zygote fails to:

A ) implant in the uterus.

B ) leave the ovary.

C ) reach the fallopian tube.

D ) begin its development.

2. A drug that increases the risk of a tubal

pregnancy is most likely to inhibit which one of

the following actions?

A ) Contraction of the uterus

B ) Secretion of follicle-stimulating hormone

C ) Onset of menstruation

D ) Transport of the ovum from ovary to uterus

3. From 4% to 10% of all maternal deaths in the

United States each year result from ectopic

pregnancy. The most likely cause of death in

these cases is:

A ) severe hormonal imbalance.

B ) loss of blood when the fallopian tube ruptures.

C ) infection in the region of the pregnancy.

D ) inadequate nutrition due to fetal use of maternal

nutrients.

4. Delayed ovulation, as a cause of tubal

pregnancy, would most likely be associated with

delayed secretion of which of the following

hormones?

A ) Progesterone

B ) Estrogen

C ) HCG

D ) Luteinizing hormone

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Passage II

Asthma is a disease of industrialized countries; rates

have doubled in the U.S. since 1980. The most life-

threatening asthmatic complication is lung

inflammation. This inflammatory response can be

triggered by exercise, respiratory viruses, or

environmental allergens, which stimulate

T lymphocytes to secrete cytokines that recruit B

lymphocytes and eosinophils to the airways.

Activated B lymphocytes secrete IgE, which

sensitizes mast cells to allergens. Activated mast cells

and eosinophils release histamine and small fatty

molecules called leukotrienes, respectively.

Leukotrienes function as chemoattractants for

granulocytic leukocytes and are potent constrictors of

bronchial smooth muscle, whereas histamine

functions as a vasodilator and can cause

microvascular endothelial cells to contract.

Current therapies such as steroids, antihistamines, and

bronchodilators treat the symptoms of the disease but

cannot prevent the onset and progression of an

asthmatic attack. Identifying points within the

inflammation cascade offers the opportunity to

develop more specific therapies to inhibit the process.

One therapeutic strategy would be to target a

particular subset of T lymphocytes known as T-helper

(TH) cells. TH1 cells secrete cytokines, such as

interferon-γ, and initiate cell-mediated responses that

eliminate cells infected with pathogens, such as

bacteria and viruses. TH2 cells secrete cytokines that

activate the inflammatory response and stimulate

antibody production. The activity of TH1 and TH2

cells are reciprocally regulated; the signal from one

cell type negatively regulates the activity of the other

cell type. Because overactivity of TH2 cells is

correlated with asthma, their inactivation would offer

a more effective treatment for this disease than is

currently available.

5. According to the passage, what is the most

probable sequence of events after activation of T

lymphocytes by an allergen?

A ) IgE secretion → histamine release →

vasoconstriction

B ) B-cell activation → IgE secretion → mast-cell

activation

C ) IgE secretion → eosinophil activation →

leukotriene release

D ) Mast-cell activation → histamine release →

bronchodilation

6. The passage suggests that the most effective way

to prevent the onset and progression of an

asthmatic attack would be treatment with:

A ) a harmless bacterium that induces a strong TH1

response.

B ) IgE, antibodies with neutralizing activity.

C ) Zileuton, an inhibitor of leukotriene synthesis.

D ) colchicine to specifically prevent eosinophil

chemotaxis.

7. A person suffering an asthmatic attack often has

more difficulty exhaling than inhaling; the action

of leukotrienes on bronchial smooth muscle

contributes to this difficulty. As a result, what

primary effect do leukotrienes have on lung gas

exchange?

A ) No effect, because bronchial smooth muscle is not

found on alveoli

B ) An increased accumulation of O2, leading to

respiratory acidosis

C ) An increased accumulation of CO2, leading to

respiratory acidosis

D ) A decreased intake of O2, leading to metabolic

acidosis

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8. Leukotrienes are potent chemoattractants. What

other cells, in addition to eosinophils, would

probably respond to and be recruited by

leukotrienes to the inflammatory site?

A ) Erythrocytes

B ) Thrombocytes

C ) Neutrophils

D ) Myocytes

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Passage III

A healthy weight has been defined as a body mass

index (BMI) of 25 or less. BMI = w/h2, where w is

weight (kg) and h is height (m). Studies suggest that

genes account for about 40% of the factors that

determine BMI. Two genes affecting weight in mice

are related to leptin, a hormone that is released by fat

cells and required for maintaining normal weight.

One gene (designated ob) codes for leptin, and the

other gene (designated db) codes for a leptin receptor.

Stable weight is also believed to be regulated by

metabolic feedback loops linking the brain, fat cells,

the digestive tract, and muscles. Two hypotheses have

been proposed to explain the biological basis of

weight control.

Set Point Hypothesis

The brain regulates body weight just as a

thermostat maintains a constant room temperature.

The brain adjusts metabolism and behavior to

maintain a predetermined body weight. Genes also

influence the set point, which can increase with age—

but only to the extent dictated by inheritance. Diet

and exercise cannot reset the set point over the long

term.

Settling Point Hypothesis

Body weight is determined by the interaction of

two factors—metabolism and genes—with the

environment. Depending on genotype, various

metabolic feedback loops may allow weight to be

stabilized at a new level. Thus, in an environment

where high-calorie food is plentiful, individuals with

a genetic predisposition to obesity will tend to

become more overweight than those without such a

predisposition.

9. What type or class of chemical messenger

traveling in the blood would most probably link

the brain with the digestive tract and fat cells in the

control of body weight?

A ) Neurotransmitters

B ) Digestive enzymes

C ) Protein receptors

D ) Hormones

10. What do both the set point hypothesis and the

settling point hypothesis seek to explain?

A ) How multiple, interacting factors determine body

weight

B ) How individual factors acting alone influence

body weight

C ) How metabolism and the environment influence

body weight

D ) How the environment and behavior influence body

weight

11. Which hypothesis implies that a person can

deliberately alter his or her own body

maintenance weight?

A ) The set point hypothesis because a thermostat can

be reset

B ) The set point hypothesis because the set point can

change with age

C ) The settling point hypothesis because, with the

correct genotype, one’s metabolism may allow

weight to stabilize at a new level

D ) The settling point hypothesis because diet and

exercise cannot reset the set point

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12. One type of metabolic feedback loop that

influences weight control involves the regulation

of glucose levels in the blood. Which organ in the

digestive system participates in this regulation by

breaking down glycogen?

A ) Stomach

B ) Liver

C ) Pancreas

D ) Small intestine

13. Which gene would produce a product that acts

predominately on or in the cell in which it is

synthesized, ob or db?

A ) db because it encodes a hormone specific to fat

cells

B ) db because it encodes a hormone receptor

C ) ob because it encodes a hormone

D ) ob because it encodes a protein specific to fat cells

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Passage IV

The runny nose associated with common colds is

related to the release of either virus-induced histamine

or acetylcholine.The nasal mucosa contain receptors

for both histamine and acetylcholine.Activation of

either of these receptor types results in increased

secretion by nasal glands, producing a runny nose.

In an attempt to treat this condition, several new

drugs have been investigated.Drug A primarily blocks

histamine receptors but also partially blocks

acetylcholine receptors.Drug B blocks histamine

receptors but has no effect on the acetylcholine

receptors.

Eighteen subjects with severe common colds, whose

symptoms were judged to be identical, were randomly

assigned to three groups. Patients A-F in Group 1

received Drug A, Patients A-F in Group 2 received

Drug B, and those in Group 3 were treated with a

nondrug placebo.After 4 hours, patients reported their

own runny nose symptoms on a 5-point scale ranging

from a dry nose (1) to an excessively runny nose (5).

Table 1 Self-Reported Symptoms 4 Hours after

Treatment

Subject

Group 1

(Drug A)

Group 2

(Drug B)

Group 3

(Placebo)

A 2 3 1

B 1 4 4

C 3 2 5

D 2 3 4

E 1 3 5

F 1 3 4

14. In Group 3, the response of Patient A can best be

classified as a response most likely:

A ) not associated with the treatment.

B ) associated with histamine blocking only.

C ) associated with acetylcholine blocking only.

D ) associated with a combination of histamine and

acetylcholine blocking.

15. Based on Table 1, the individual who benefited

most from a specific blocking effect on the

histamine receptors only is:

A ) Subject B in Group 1.

B ) Subject C in Group 1.

C ) Subject B in Group 2.

D ) Subject C in Group 2.

16. The nasal mucosa cells responsible for the release

of excessive fluid during the common cold can

best be classified as:

A ) epithelial.

B ) connective.

C ) contractile.

D ) neurosecretory.

17. Based on the passage, which drug treatment

would hypothetically provide the maximum

reduction in nasal secretions?

A ) Antihistamine only

B ) Acetylcholine only

C ) Antihistamine and acetylcholine

D ) Antihistamine and acetylcholinesterase

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These questions are not based on a descriptive

passage and are independent of each other.

18. Which of the following describes a primary

function of the myelin sheath?

A ) It provides nutrients to motor neurons.

B ) It regulates synaptic vesicle discharge.

C ) It guides dendrite growth and branching.

D ) It increases the rate of conduction of action

potentials.

19. Which organ is involved in regulation of all of

the following: acid-base balance, blood pressure,

water balance, and removal of nitrogen wastes?

A ) Liver

B ) Spleen

C ) Kidney

D ) Large intestine

20. A sequence near the 3' end of bacterial 16S

ribosomal RNA (rRNA) base-pairs with a

sequence called the Shine–Dalgarno sequence in

the ribosome binding sites of prokaryotic

mRNAs. Given that the sequence in the

16S rRNA is

3' UCCUCCA 5'

what is the mRNA sequence of the Shine–

Dalgarno sequence that most strongly binds the

ribosome?

A ) 5' AGGAGGT 3'

B ) 5' AGGAGGU 3'

C ) 5' CUUCUUG 3'

D ) 3' AGGAGGU 5'

21. A hiker becomes lost and has no drinking water

for 2 days. At the end of this time, which of the

following changes in hormone production would

be expected to be significant in this individual?

A ) Decreased glucocorticoid secretion

B ) Decreased aldosterone secretion

C ) Increased insulin secretion

D ) Increased antidiuretic hormone secretion

22. Assuming that the vertebrates were all of

comparable size, which of the following

vertebrates would be expected to have the

strongest and heaviest bones?

A ) A land-dwelling mammal

B ) A water-dwelling mammal

C ) A flying bird

D ) An amphibian

23. The posttranslational modification of some of the

eukaryotic cell’s most abundant proteins is

thought to affect the ability of those proteins to

condense DNA into 30-nm fibers. Given this,

these proteins are most likely:

A ) tubulins.

B ) histones.

C ) transcription activators.

D ) DNA polymerase subunits.

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Passage V

Two major theories have been advanced to explain

why organisms age and die.

Theory I

The Genetic or Programmed Theory of Aging states

that aging is triggered by hormones and is an orderly

consequence of the genetically programmed processes

of growth, development, and differentiation.Life

spans of individuals in each species are finite,

species-specific, and vary little.Aging is thought to

improve the ability of the species to adapt to its

environment.

Evidence for Theory I:The difference in longevity

between fraternal twins is much greater than the

difference between identical twins.Also, cultured cells

of human embryonic connective tissue normally

double approximately 50 times before they die.For

example, when frozen at the 10th doubling and

thawed years later, the cells undergo 40 more

doublings before death.

Theory II

The Damage-Accumulation Theory of Aging states

that aging is nonadaptive and not genetically

programmed.Instead, aging results from random,

accumulated damage (to DNA, RNA, and proteins)

that is caused by free radical production within the

cells.This damage in turn leads to cellular changes

resulting in aging and death.

Evidence for Theory II:Metabolic rates of mammals

are directly proportional to the rate of generation of

free radicals.Dietary restriction, which decreases

metabolic rate, also increases the maximum life span

of rats from 125 to 185 weeks.The addition of

vitamins E and C (which react with free radicals and

render them harmless) to the feed of mice increases

the mice’s average life span.

24. Aging due to the production of free radicals can

occur by all of the following processes EXCEPT:

A ) absorption of ultraviolet radiation.

B ) production of partially reduced oxygen species

during normal metabolism.

C ) metabolic conversion of toxic chemicals such as

carbon tetrachloride (CCl4).

D ) consumption of excess quantities of vitamins E

and C.

25. Vitamin E is added to human connective tissue

cells in culture at the 30th doubling, and the

number of additional doublings before death is

counted. Theory II will be best supported if the

cells double an additional:

A ) 5 times.

B ) 10 times.

C ) 20 times.

D ) 40 times.

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26. Specific metabolic rates are 25 cal/g/day for

humans, 150 cal/g/day for rats, and 180 cal/g/day

for mice. When urinary output of a free-radical-

induced DNA damage product is plotted as a

function of metabolic rate in these 3 species,

which of the following graphs is most consistent

with Theory II, and best depicts the urine levels

of the damage product?

A )

B )

C )

D )

27. To examine the effects on life span of

undernutrition without malnutrition, one group of

just-weaned rats was fed every day, and a second

group was fed every other day. Which of the

following survival curves for the 2 groups is

consistent with the evidence for Theory II?

A )

B )

C )

D )

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Passage VI

Most mammalian cells contain the Na+, K

+-

ATPase enzyme (the sodium pump).The sodium

pump is responsible for regulating Na+ and K

+

gradients across the cell membrane through the

transportation of K+ into and Na

+ out of the cell.

The sodium pump has been studied by using

reconstituted erythrocytes.Reconstituted erythrocytes

are formed by bathing erythrocytes in distilled water

under controlled conditions.These cells swell,

forming pores that release cytoplasmic proteins and

ions into the distilled water.If the swollen cells are

then placed in a solution isotonic to normal cells, ions

or proteins within the isotonic solution will

equilibriate with the water inside the swollen

cells.The swollen cells will also shrink, resealing the

plasma membrane.Therefore, ions from the isotonic

solution will be trapped within the reconstituted

erythrocytes.

Experiment 1

Reconstituted erythrocytes were prepared

containing a mixture of several ions.The reconstituted

cells were then placed in solutions containing either

KCl, NaCl, NH4Cl, or RbCl.The rates of hydrolysis of

ATP were recorded.Table 1 lists the results of this

experiment.

Experiment 2

Several reconstituted erythrocytes were prepared,

each containing ADP, Pi, and KCl, but not NaCl.The

erythrocytes were placed in a solution containing

NaCl, but not KCl.ATP was formed inside the

erythrocytes.

Based on experiments similar to these, researchers

have proposed the following overall equation for the

sodium pump.

3 Na+

(inside) + 2 K+

(outside) + ATP4–

+ H2O → 3

Na+

(outside) + 2 K+

(inside) + ADP3–

+ Pi 2–

+ H+

Reaction A

Table 1

Extracellular environment* Erythrocyte contents

NH4Cl RbCl KCl NaCl

N

H

-

N

N

H

N

-

-

H

N

N

L

-

L

L

KCl

NaCl

NH4Cl

RbCl

ATP

ATP

ATP

ATP

Mg2+

Mg2+

Mg2+

Mg2+

N

N

-

N

N

N

N

-

-

N

N

N

N

-

N

N

KCl

NaCl

NH4Cl

RbCl

ATP

ATP

ATP

ATP

no

Mg2+

no

Mg2+

no

Mg2+

no

Mg2+

N

N

-

N

N

N

N

-

-

N

N

N

N

-

N

N

KCl

NaCl

NH4Cl

RbCl

no

ATP

†

no

ATP

†

no

ATP

†

no

ATP

†

Mg2+

Mg2+

Mg2+

Mg2+

* N = no ATP hydrolysis; L = low rate of ATP

hydrolysis; H = high rate of ATP hydrolysis

† ATP was contained in the extracellular contents,

rather than in the erythrocytes.

28. The sodium pump would be most active in cells

of which of the following structures?

A ) Veins

B ) Loop of Henle

C ) Lungs

D ) Bone marrow

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29. When erythrocytes are placed in distilled water,

the volume of each erythrocyte increases because

the:

A ) gradient of ions causes water to enter the cells.

B ) contractile filaments of the cytosol open pores in

the plasma membrane.

C ) sodium pump transports sodium out of the

erythrocytes more rapidly than normal.

D ) erythrocyte’s DNA produces degradative

enzymes.

30. A student postulated that the sodium pump

directly causes action potentials along neurons.

Is this hypothesis reasonable?

A ) No; action potentials result in an increased

permeability of the plasma membrane to sodium.

B ) No; the myelin sheaths of neurons prevent

movement of ions across the plasma membranes of

the neurons.

C ) Yes; sodium is transported out of neurons during

action potentials.

D ) Yes; action potentials are accompanied by the

hydrolysis of ATP.

31. Based on Reaction A, if all the energy produced

from glycolysis were used to remove Na+ from a

cell, how many molecules of Na+ would be

removed per molecule of glucose?

A ) 3

B ) 6

C ) 9

D ) 12

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Passage VII

Sarah, a scientist from New Orleans, takes two-week

vacations to different locations every year to

experience new sports.

One year she went to the Caribbean Sea to learn skin

diving.Although she was in excellent physical

condition from daily swimming in the ocean, she

noticed that the first time she went diving, she

experienced an elevated pulse and ventilation rate.By

the third time she went diving, her heart and breathing

rate were no longer elevated.By the end of the two

weeks, her skin had become darker.

Another year she went skiing on snow in the

mountains of Colorado.Again, she noticed that the

first time she went skiing, her heart and ventilation

rate were faster than usual.Although it was not as

elevated by the end of the first week, her heart and

breathing rates were still higher than usual.She also

noticed that her appetite and caloric intake were

considerably greater during her skiing vacation

compared with her diving vacation.However, she

noticed that her body weight did not change

significantly.

Sarah calculated the actual work that she performed

skiing and diving.There was not enough difference in

the work performed to account for the observed

difference in appetite; although the physical work of

diving and skiing was approximately equal and she

ate more calories during the skiing trip, she did not

gain any weight.

On a third vacation, Sarah had a serious accident

while playing sports.

32. The prolonged increase in heart and breathing

rates during the snow skiing trip was probably a

result of:

A ) activation of the sympathetic autonomic nervous

system by the new experience.

B ) activation of the parasympathetic autonomic

nervous system by the new experience.

C ) hypoxia caused by insufficient blood hemoglobin

concentration to supply oxygen for exercise at the

low oxygen pressure found at high altitudes.

D ) depressed core body temperature (hypothermia)

caused by exposure to cold temperatures at high

altitudes.

33. Control of heart rate, muscle coordination, and

appetite is maintained by the:

A ) hypothalamus, cerebrum, and brain stem,

respectively.

B ) brain stem, hypothalamus, and cerebrum,

respectively.

C ) cerebellum, hypothalamus, and brain stem,

respectively.

D ) brain stem, cerebellum, and hypothalamus,

respectively.

34. The initial increase in heart and breathing rates

during the skin diving trip was probably a result

of:

A ) activation of the sympathetic autonomic nervous

system by the new experience.

B ) activation of the parasympathetic autonomic

nervous system by the new experience.

C ) hypoxia caused by the inability of her blood

hemoglobin concentration to supply sufficient

oxygen for the strenuous exercise of swimming at

sea level.

D ) elevated core body temperature caused by

swimming in warm tropical waters.

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35. During the initial skin diving session, when her

heart and breathing rates were increased, Sarah

noticed that she produced more urine than usual.

This was most probably a result of:

A ) increased blood pressure caused by her excitement

or anxiety.

B ) reduced blood pressure caused by her excitement

or anxiety.

C ) absorption of water from the ocean.

D ) inability to cool the skin through evaporative

water loss.

36. Sarah noted that her skin blood vessels were

usually constricted to conserve body heat in the

cold environment of the mountains. However,

her skin blood vessels would occasionally dilate

for short periods of time. What would be the

most probable physiological purpose for this

periodic vasodilation?

A ) Maintain normal skin tone

B ) Maintain sufficient oxygenation of cells

C ) Reduce excessive blood pressure

D ) Maintain normal muscle tone

37. After Sarah’s accident, her attending physician

detected the protein myoglobin in her urine.

What type of injury is consistent with this

observation?

I. Broken bone

II. Damaged muscle

III. Damaged kidney

A ) I only

B ) III only

C ) I and III only

D ) II and III only

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Passage VIII

The Ames test is often used in the initial screening

of suspected carcinogenic compounds because it

provides a good indication of the mutagenic

characteristics of many chemicals. The test uses

special strains of the bacterium Salmonella

typhimurium that are nutritional mutants; they also

lack a mechanism for DNA repair.

When grown on a medium that lacks the amino

acid histidine, the Salmonella test strains do not

survive even though wild-type Salmonella grow well

on this medium. During the Ames test, the suspected

carcinogen is added to a histidine-deficient growth

medium. If the chemical is a mutagen, some of the

bacteria will back-mutate, and a visible colony will

form. The usefulness of the Ames test can be

improved when the growth medium contains rat-liver

enzymes.

Figure 1 illustrates several Ames tests performed

on the air from 3 different cities.

Figure 1

Figures adapted from Peter Flessel, Yi Y. Yang, Kuo-In Chang, and

Jerome J. Wesolowski, Ames Testing for Mutagens and Carcinogens in

Air. © 1987 by the Division of Chemical Education, American

Chemical Society.

38. Cancer cells most likely have an abnormality in

their:

A ) DNA.

B ) rRNA.

C ) mitochondria.

D ) lysosomes.

39. Why is the Ames test for mutagens used to test

for carcinogens?

A ) Salmonella transform mutagens into carcinogens.

B ) Most mutagens are also carcinogens.

C ) Salmonella contain oncogenes.

D ) Salmonella’s RNA distinguishes between

carcinogens and mutagens.

40. Which of the following best explains why

bacterial colonies formed on Plate IV in Figure

1?

A ) The air contained mutagens.

B ) The agar contained mutagens.

C ) Spontaneous mutations occurred.

D ) The DNA repair system became activated.

41. The passage indicates that when Salmonella have

back-mutated, they:

A ) contain a pigment that makes them visible.

B ) are capable of synthesizing histidine.

C ) will metabolize the carcinogen in the presence of

light.

D ) lack histidine in their proteins.

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Passage IX

Autoimmune diseases result when lymphocytes from

the immune system attack the body’s own tissues.

This is normally prevented by the body’s ability of

self-tolerance; that is, the immune system

“recognizes” the body’s own tissues and forms very

few lymphocytes that act against them. Autoimmune

diseases may affect any type of body tissue.

Two hypotheses have been advanced to explain how

autoimmunity develops.

Hypothesis 1

Most of the body’s self-tolerance is generated within

a few months of birth, when the body is processing T

and Blymphocytes. Identical groups (clones) of

circulating lymphocytes remain inactive until they

encounter their specific antigens, after which they

proliferate. During this time, the process of clonal

deletion destroys any newly formed groups of

lymphocytes that might attack the body’s own tissues.

If clonal deletion of such lymphocytes does not occur

or is hindered, these lymphocytes will incorrectly

recognize a specific body tissue as foreign or non-self,

and begin to destroy it.

Hypothesis 2

In addition to effector lymphocytes (such as helper T

cells and cytotoxic T cells) that selectively attack and

destroy antigens, the body contains other lymphocytes

(suppressor T cells) that prevent this destruction by

selectively limiting the action of the effector cells.

Normally, a regulatory balance is maintained between

effector and suppressor T cells. However, when this

balance is disturbed (for example, by loss or

inactivation of suppressor-cell clones), an

autoimmune disease may result.

42. The human body never develops self-tolerance to

the proteins of the cornea. According to

Hypothesis 1, one reason for this might be that

the corneal proteins:

A ) never circulate in the body fluids; therefore, they

are never exposed to lymphocytes.

B ) are not part of a living tissue; therefore,

development of self-tolerance is unnecessary.

C ) are adequately protected by tears and other

external barriers to antigens.

D ) are protected by suppressor T cells rather than by

clonal deletion.

43. According to the normal mechanism of self-

tolerance described in the passage, the body will

respond to each antigen it encounters by

activating:

A ) either B or T lymphocytes, but not both.

B ) lymphocytes against the antigen, if the antigen is

from the body’s own tissues.

C ) all clones of lymphocytes that have not been

destroyed by clonal deletion.

D ) a clone of lymphocytes specific for that antigen.

44. According to Hypothesis 2, the normal balance

between effector cells and suppressor cells

specific for a certain tissue will most likely be

disturbed if the tissue is injected with:

A ) cells for that tissue obtained from an identical

twin.

B ) cells from another tissue to which the tissue has

already been exposed.

C ) a foreign substance that cross-reacts with cells of

that tissue.

D ) a foreign substance that does not cross-react with

cells of that tissue.

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45. Does acceptance of the mechanism of self-

tolerance described in Hypothesis 1 rule out

acceptance of the mechanism described in

Hypothesis 2?

A ) Yes; if clonal deletion occurs, no self-reactive

lymphocytes will be left for suppressor T cells to

act upon.

B ) Yes; Hypothesis 1 deals with the formation of self-

tolerance and Hypothesis 2 deals with its

maintenance.

C ) No; suppressor T cell formation can only occur

after clonal deletion has occurred.

D ) No; clones of self-reactive lymphocytes not

destroyed by clonal deletion may be controlled by

suppressor T cells.

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46. Rates of endocytosis vary from cell type to cell

type. What cell would be predicted to have the

highest rate of endocytosis?

A ) A macrophage

B ) An erythrocyte

C ) An osteoblast

D ) A neuron

47. After the gall bladder is removed from a patient,

the patient will most likely have reduced ability

to digest:

A ) protein.

B ) starch.

C ) sugar.

D ) fat.

48. Which statement below most accurately describes

the roles of the proteins actin and myosin during

muscular contraction?

A ) Both actin and myosin shorten, causing the muscle

tissue to which they are attached to contract.

B ) Both actin and myosin catalyze the reactions that

result in muscle contraction.

C ) Actin molecules are disassembled by myosin,

leading to a shortening of muscle sarcomeres.

D ) Bridges between actin and myosin form, break,

and re-form, leading to a shortening of muscle

sarcomeres.

49. It was observed that when a mother baboon died,

her infant was cared for by the mother’s sibling.

A biologist explained that this behavior would

increase the chance that the sibling’s genes would

be passed on to the next generation. This

conclusion is most likely based on the fact that

the:

A ) infant’s mother was unfit to live in the

environment.

B ) infant would probably have survived without the

sibling’s help.

C ) sibling had many of the same genes as the infant’s

mother.

D ) population of baboons would increase in number.

50. The antibiotic penicillin has the effect of

inhibiting the production of the chemical

peptidoglycan. Therefore, penicillin is likely to

be most effective in treating infection by:

A ) viruses.

B ) bacteria.

C ) fungi.

D ) protozoa.

51. Actin filaments within cells can be identified

experimentally by the use of a labeled molecule

that binds specifically to actin and not to other

cell substances. Which of the following would be

best to use as the labeled molecule?

A ) ATP

B ) Myosin

C ) Albumin

D ) Myoglobin

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52. A wound that penetrates the rib cage and lets air

into the right pleural cavity stops air flow into the

right lung because the:

A ) lung cannot be expanded.

B ) rib cage cannot be expanded.

C ) diaphragm cannot be lowered.

D ) air dries and stiffens the lung.

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Passage X

Toxic shock syndrome (TSS) is an illness

characterized by the rapid onset of high fever,

hypotension, and a rash that results in skin

desquamation (separation of cell layers). It affects at

least three organ systems. In the early 1980s, an

increased risk of TSS was associated with use of

high-absorbency tampons. Some high-absorbency

brands of tampons were removed from the market,

and warning labels were required for all remaining

brands. Although reported cases of TSS decreased

markedly at that time, significant menstrual and

nonmenstrual cases of TSS continue to occur.

The two bacteria that cause TSS are Staphylococcus

aureus and Streptococcus pyogenes. Most studies of

these pathogens have focused on the effects of the

protein toxins they produce. Chemical and

biological/immunological tests indicate that these

toxins are superantigens.

Superantigens differ from other proteins in their

antigenic nature; they do not stimulate T lymphocytes

in the immune system in the same manner that

conventional protein antigens do. Superantigens

bypass a processing step normally performed by

antigen-presenting cells, and also differ from normal

antigens by binding to T cells outside the standard

antigen binding site. Because this unique type of

binding activates approximately 20% of the T

lymphocytes, as opposed to 1 in 100,000 T cells

activated by conventional antigenic stimulation,

superantigens are considered nonspecific stimulators.

Negative effects of nonspecific stimulation by

superantigens occur because the activation of so many

T cells causes the release of massive levels of

cytokines. This increased cytokine release is probably

responsible for many of the acute problems seen in

TSS, and also in some autoimmune diseases such as

arthritis, multiple sclerosis, and rheumatic fever.

53. Staphylococcus and Streptococcus bacteria cause

problems in acute infections such as toxic shock

syndrome primarily by:

A ) multiplying to produce large numbers of bacteria.

B ) stimulating exaggerated immune responses.

C ) causing autoimmune reactions.

D ) inhibiting metabolic enzymes with toxins.

54. According to the passage, superantigens increase

the number of activated T cells over activation

levels observed with conventional antigens by a

factor of:

A ) 20.

B ) 5000.

C ) 20000.

D ) 100000.

55. In addition to the skin and circulatory systems,

which of the following organ systems is most

likely to be affected by TSS?

A ) The musculoskeletal system

B ) The digestive system

C ) The lymphatic system

D ) The respiratory system

56. If the dose of Streptococcus Strain A required to

cause infection is 1 x 105 bacteria and that of

Streptococcus Strain B is 5 x 104

bacteria, which

of the following statements describes the relative

potencies of these strains?

A ) Strain A is five times as potent as Strain B.

B ) Strain A is one-fifth as potent as Strain B.

C ) Strain A is twice as potent as Strain B.

D ) Strain A is half as potent as Strain B.

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Passage XI

Joe and Steve were firefighters. Joe was an

experienced veteran in the fire department, whereas

Steve was a 23-year-old new member. Although Joe

was 60 years old, he kept his muscles in excellent

physical condition and only 8% of his weight was

body fat. However, his long firefighting career had

had serious effects on his health. At the age of 25, Joe

had been trapped in a burning house and had suffered

severe burns over 50% of his body, which resulted in

massive scarring of his skin. As a consequence of

long-term inhalation of smoke, Joe also had an early-

stage emphysema – a disease in which the elastic

tissue of the lungs loses its ability to recoil after it is

stretched. Although Steve weighed only half as much

as Joe, Steve had a higher percentage of body fat

(15% of his weight). However, he was in excellent

physical condition. Steve and Joe had identical vital

signs: a resting heart rate of 60 beats per minute and

blood pressure of 125/70 mmHg.

Joe and Steve were assigned to a team sent to fight a

major fire in an industrial warehouse. Fighting a big

fire is often a frightening experience. Because this

was Steve’s first major fire, it was especially

frightening for him. After the fire was extinguished,

an inspection of the building revealed that a chemical

storage container had ruptured, possibly exposing Joe

and Steve to a hepatotoxic agent, which could damage

the liver.

57. Joe’s body might have a greater tendency to

overheat during strenuous work than Steve’s

body would, because:

A ) older males have a higher basal metabolic rate.

B ) Steve has a greater percentage of body fat.

C ) Joe’s scarred skin would reduce evaporative

cooling.

D ) Joe has more skin surface area relative to his body

volume.

58. Damage to the liver would most directly affect

the production of:

A ) digestive enzymes.

B ) antidiuretic hormone.

C ) new blood cells.

D ) bile salts.

59. Assuming the lungs are fully perfused (meaning

fully permeated with blood), which factor is least

likely to influence the oxygenation of blood in

the pulmonary circulation?

A ) Rate and depth of breathing

B ) Hemoglobin concentration of the blood

C ) Blood pressure in the pulmonary artery

D ) Surface area of the alveoli

60. If both Steve and Joe performed the same work

tasks, which of the following statements would

describe their individual energy consumption?

A ) Steve’s body would consume more energy

because of his lesser weight.

B ) Steve’s body would consume less energy because

of the greater basal metabolic rate in younger

people.

C ) Joe’s body would consume more energy because

of his greater weight.

D ) Joe’s body would consume the same amount of

energy as Steve’s because the basal metabolic

rates are equal.

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61. If Steve’s blood pressure had increased

significantly more than Joe’s increased (other

factors being equal) while they worked, what

difference in their urinary system function would

be expected?

A ) Joe’s glomerular filtration rate would increase

more than Steve’s would.

B ) Steve’s reabsorption rate of glomerular filtrate by

the peritubular capillaries would be lower than

Joe’s would be.

C ) Steve’s reabsorption rate per milliliter of

glomerular filtrate by the peritubular capillaries

would be higher than Joe’s would be.

D ) Steve’s glomerular filtration rate would increase

more than Joe’s would.

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Passage XII

Researchers have isolated G proteins, a new family of membrane-associated proteins. These proteins are

believed to regulate all or most of the intracellular signaling systems operating across the plasma membrane,

including those involving hormones and action potentials. Figure 1 depicts how this mechanism is hypothesized

to function.

Figure 1 Intracellular signaling by G proteins

According to this hypothesis, extracellular signals such as some hormones bind to specific receptors on the

surface of the plasma membrane. This binding activates the receptor that then binds to a G protein embedded in

the lipid bilayer of the plasma membrane. This, in turn, causes the G protein to release guanosine diphosphate

(GDP) and bind guanosine triphosphate (GTP). The G protein with bound GTP interacts with various enzymes

or proteins in the plasma membrane. The end result is either activation or inactivation of the enzyme, depending

on the specific system. The G protein is then inactivated, which “turns off” the initial steps of the signaling

system.

Adenylate cyclase, the enzyme that synthesizes the intracellular signal cyclic AMP (cAMP), is believed to be

activated by this mechanism.

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62. According to the passage, all of the following

events must occur in order to activate a G protein

EXCEPT:

A ) binding of the G protein to the receptor.

B ) binding of the extracellular signal to the receptor.

C ) displacement of GDP by GTP on the G protein.

D ) phosphorylation of GTP.

63. Which of the following best explains the role of

GTP in controlling the function of G proteins?

A ) GTP is required for binding the G protein to the

hormone receptor.

B ) GTP alters the conformation of the G protein

allowing interaction with the enzyme.

C ) Hydrolysis of GTP acts as a source of energy to

activate the G protein.

D ) Hydrolysis of GTP is required for activation of the

membrane-bound enzyme.

64. Binding of G proteins to a membrane-bound

enzyme was observed to inactivate the signaling

system. Is this observation consistent with the

information presented in the passage?

A ) No, because G proteins are believed to activate

signaling systems

B ) No, because G proteins should have been activated

by binding to the signal-receptor complex

C ) Yes, because the effect of G proteins is dependent

upon the specific signaling system involved

D ) Yes, because binding of G proteins to the

membrane-bound enzyme results in the hydrolysis

of GDP, which inactivates the signaling system

65. Which of the following, if found to be true,

would best refute the hypothesis that a

membrane-bound enzyme is activated by a G

protein?

A ) The enzyme can be activated in the absence of

bound GTP.

B ) The enzyme is activated only when hormone is

present.

C ) The enzyme cannot be activated when GDP is

bound to a G protein.

D ) The enzyme is always found in the activated state.

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Passage XIII

Stem cells are immature cells that continually replace

cells having short life spans, such as those of the skin

and blood. When stem cells divide, one of the

daughter cells remains a stem cell and the other

differentiates to produce a mature cell. A pluripotent

stem cell can replicate often, but can only produce a

few cell types. A totipotent stem cell cannot replicate

as often as a pluripotent cell, but can differentiate into

many cell types. A committed stem cell is committed

to producing a particular specialized cell line, such as

leukocytes.

Hematopoietic stem cells (HSCs) are located in the

bone marrow and have the potential to differentiate

into blood cells. HSCs can also be pluripotent,

totipotent, or committed; totipotent cells can replace

all the types of blood cells of the immune system.

Current investigations into the mechanism governing

HSC differentiation have revealed two properties

common to all stem cells. First, receptors on stem

cells respond to hormones that regulate the production

of different types of blood cells. Second, stem cells

grown in cultures lacking naturally occurring support

cells differentiate randomly.

Two hypotheses describe different mechanisms

governing differentiation. The deterministic view

holds that external signals, such as hormones, direct

stem cell differentiation. The stochastic view

maintains that differentiation into various cell types

occurs randomly.

66. HSCs are vital to human body function because

blood cells:

A ) typically have short life spans.

B ) are not stored in the body.

C ) constantly leave the body through the urinary and

digestive systems.

D ) constantly differentiate into other types of cells.

67. Which of the following observations supports

both the deterministic and stochastic views?

A ) When hormone X is presented to a culture of

HSCs, only erythrocytes are formed.

B ) When stem cells initially express hormone

receptors, they do so randomly.

C ) The receptors on HSCs are expressed in response

to external signals from surrounding bone marrow.

D ) Genes within the HSCs are turned on according to

internal signals early in fetal development.

68. Which of the following observations supports the

deterministic view?

A ) Cultured stem cells develop hormone receptors

when exposed to naturally occurring support cells.

B ) When cultured stem cells are exposed to

hormones, they divide more rapidly.

C ) When cultured stem cells are reintroduced into the

body, they continue random differentiation.

D ) When genes for erythrocytes are introduced into

cultured stem cells, erythrocytes are formed.

69. Damaged or destroyed bone marrow can be

replaced with transplanted tissue. If the main goal

of such a transplant is to replace all blood cell

types, the transplanted tissue should contain

which of the following cell types?

A ) Pluripotent HSCs, because they differentiate

stochastically and therefore would replicate the

fastest

B ) Pluripotent HSCs, because they differentiate

stochastically and the body could therefore signal

which types of cells it needs

C ) Totipotent HSCs, because they can differentiate

stochastically and the random differentiation

would therefore produce the most types of blood

cells

D ) Totipotent HSCs, because they can differentiate

deterministically and the cells would therefore

have the greatest ability to differentiate in

response to the new host

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70. In almost all vertebrates, when the optic cup fails

to develop in the embryo, the lens also fails to

form. This constitutes evidence that:

A ) the process of neurulation follows gastrulation.

B ) the eye develops early in vertebrate

morphogenesis.

C ) cells may induce neighboring cells to differentiate.

D ) cell differentiation is an “all or none”

phenomenon.

71. When viewing an X ray of the bones of a leg, a

doctor can tell if the patient is a growing child,

because the X ray shows:

A ) cartilaginous areas in the long bones.

B ) bone cells that are actively dividing.

C ) the presence of haversian cells.

D ) shorter-than-average bones.

72. The different antigenic blood types are inherited

through allelic genes. The actual molecular

difference between two blood types is in the

carbohydrate that is attached to a common

molecular backbone. The best explanation for

how genes determine blood type, therefore, is

that each gene:

A ) produces RNA that is converted to a specific

carbohydrate.

B ) codes for a carbohydrate instead of a protein.

C ) codes for an enzyme that attaches the type-specific

carbohydrate.

D ) contains a specific carbohydrate as part of the

nucleic-acid structure.

73. Which of the following changes would NOT

interfere with the repeated transmission of an

impulse at the vertebrate neuromuscular

junction?

A ) Addition of a cholinesterase blocker

B ) Addition of a toxin that blocks the release of

acetylcholine

C ) An increase in acetylcholine receptor sites on the

motor end plate

D ) Addition of a substance that binds to acetylcholine

receptor sites

74. During the repolarization phase of an action

potential in a neuron, which of the following is

generally true of the voltage-gated channels that

cause that action potential?

A ) The voltage-gated K+ channels are closed, and the

voltage-gated Na+ channels are closed.

B ) The voltage-gated K+ channels are open, and the

voltage-gated Na+ channels are open.

C ) The voltage-gated K+ channels are closed, and the

voltage-gated Na+ channels are open.

D ) The voltage-gated K+ channels are open, and the

voltage-gated Na+ channels are closed.

75. Antidiuretic hormone (ADH) acts to decrease

urine output by increasing the water permeability

of the walls of:

A ) the glomerulus.

B ) Bowman’s capsule.

C ) the loop of Henle.

D ) the distal tubule and collecting duct.

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Passage XIV

The study of human disease has revealed many details

and raised many questions about the genetic basis of

cellular physiology. The study of neurofibromatosis

type I (NF1) suggests that defects in a single gene

cause the various clinical features that characterize

this disease, which include skeletal abnormalities,

learning disabilities, and benign and malignant

tumors. This array of clinical features is difficult to

reconcile with the idea of defects in a single gene.

Studies of DNA mutations in NF1 patients have

revealed few mutations–too few to explain all the

features of this genetically dominant disease.

At least four alternative mRNA transcripts are

expressed from the single NF1 gene. Each mRNA

transcript is expressed differently in different tissues

and at different developmental stages. Some

investigators propose that changes in the types of NF1

transcript may drive cellular differentiation, whereas

others propose that cellular differentiation causes

changes in the type of NF1 transcript expressed. In

either case, epigenetic (developmental) events

evidently affect the expression of the NF1 gene.

Mistakes in RNA processing might contribute to the

disease phenotype.

The events involved in NF1 gene expression probably

are not unique to this gene. Many genes may undergo

a similarly complex series of events, which ultimately

regulate the amount and composition of protein

expressed from a particular DNA sequence.

76. The mechanisms that regulate gene expression

are:

A ) simple because they are contained within the cell.

B ) complex and occur at many levels within the cell.

C ) complex and affect only DNA.

D ) simple because all genes are regulated in the same

way.

77. Changes in the type of NF1 transcript expressed

will cause changes in the type of NF1:

A ) protein synthesized by the ribosomes.

B ) protein transcribed by the ribosomes.

C ) gene passed to the offspring of those affected.

D ) gene within cells of those affected.

78. Epigenetic modulation of gene expression is most

likely to be important in evolutionary terms

because it allows:

A ) multiple proteins to be encoded by a single gene.

B ) multiple genes to encode the same protein.

C ) the posttranslational modification of defective

proteins.

D ) more variation at the DNA level.

79. The passage suggests that the expression of

disease genes probably is important in regulating

normal cellular physiology because:

A ) altered expression of disease genes leads to

disease.

B ) the expression of disease genes leads to disease.

C ) disease genes are frequent targets for mutation.

D ) gene mutations frequently lead to disease.

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Passage XV

Scientists have hypothesized that mitochondria evolved from aerobic heterotrophic bacteria that entered and

established symbiotic relationships with primitive eukaryotic anaerobes.Many structural and functional

similarities between mitochondria and present-day bacteria support this hypothesis.They are approximately the

same size, reproduce by similar means, and contain non-histone-bound DNA.They contain the tRNAs,

ribosomes, etc., necessary for transcription and translation, and they show some similarities in base sequences

of rRNAs.

In addition, the inner membranes of mitochondria have enzymes and transport systems similar to those on the

plasma membranes of bacteria.One similar system is the electron transport system (ETS).Electron transport in

both mitochondria and bacteria is accomplished using three large protein complexes, each composed of multiple

polypeptides (Figure 1).

Figure 1 Electron transport across inner mitochondrial membrane

Hydrogen atoms and electrons donated from NADH are passed between components of the electron transport

chain and eventually reduce oxygen to form water.This chain of events creates both a pH gradient and an

electrical potential across the membrane.The protons are thought to move down the pH gradient, interacting

with the enzyme ATP synthetase.This results in the production of ATP from ADP and phosphate.

80. According to the hypothesis described in the

passage, the bacteria that entered primitive

eukaryotic cells were able to carry out which of

the following functions that the primitive

eukaryotic cells could NOT?

A ) Glycolysis

B ) Krebs cycle and electron transport

C ) Cell division

D ) Transcription and translation

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81. Most proteins in present-day mitochondria are

made by cytoplasmic ribosomes from mRNA

transcribed from nuclear genes. Can this fact be

reconciled with the hypothesis described in the

passage?

A ) Yes; the transfer of genes from symbionts to the

eukaryotic nucleus could have occurred during the

last billion years of evolution.

B ) Yes; this difference from bacteria is unimportant,

because the many similarities between bacteria

and mitochondria provide sufficient evidence in

favor of the hypothesis.

C ) No; the fact that mitochondrial proteins are made

in the cytoplasm is convincing evidence that

mitochondria do not have a bacterial origin.

D ) No; because bacteria can make all their own

proteins and mitochondria cannot, this disproves

the hypothesis.

82. The chemical gramicidin inserts into membranes

and creates an artificial pathway for proton

movement. Based on Figure 1, if mitochondria

are treated with gramicidin, the rate of ATP

synthesis will most likely:

A ) increase, because of increased proton movement

back into the mitochondria.

B ) decrease, because of a decreased rate of hydrogen-

atom donation by NADH.

C ) decrease, because the proton gradient will rapidly

reach equilibrium.

D ) not be altered, because sufficient protons will

remain between the membranes to generate ATP.

83. Which of the following pieces of evidence most

strongly supports the hypothesis of mitochondrial

origin described in the passage?

A ) Mitochondria have fewer genes than typical

bacterial cells have.

B ) Mitochondria contain hundreds of different

enzymes.

C ) The diameters of mitochondria and typical

present-day bacteria are approximately equal.

D ) Nitrogen-fixing bacteria live symbiotically inside

the cells of present-day plants.

84. To support the symbiotic hypothesis presented in

the passage, mitochondria should be similar to

bacteria in which of the following ways?

A ) They should use 80S ribosomes.

B ) They should be incapable of binary fission.

C ) They should have circular DNA.

D ) They should be capable of anaerobic respiration.

85. The chemical valinomycin inserts into

membranes and causes the movement of K+ into

the mitochondria. Based on Figure 1, if

mitochondria are treated with valinomycin, the

rate of ATP synthesis in the mitochondria will

most likely:

A ) decrease, because the K+ will compete with

protons at the active site on ATP synthetase.

B ) decrease, because movement of K+ into the

mitochondrial compartments will disrupt proton

movement into the intermembrane space.

C ) increase, because the net positive charge in the

mitochondria will cause increased movement of

protons into the intermembrane space.

D ) increase, because the additional positive charge

will further activate ATP synthetase.

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Passgage XVI

An AIDS infection is especially dangerous because

the AIDS virus attacks the cells of the immune

system.A primary target of the virus is the CD4

lymphocyte (helper T cell). Helper T cells produce

substances that trigger the maturation of B

lymphocytes and CD8 lymphocytes (killer T cells).

During the infection of a helper T cell, gp120 proteins

of the viral coat first bind to the CD4 antigens on the

cell membrane.The viral coat then fuses with the cell

membrane, and the RNA-containing core of the virus

is dumped into the cell.Viral RNA is used as a

template to produce DNA with the help of the enzyme

reverse transcriptase, several copies of which are also

contained in the viral core.The viral DNA is then

incorporated into the chromosomes of the helper T

cell.At a later time, the viral DNA will be activated

and used to make new viral particles, resulting in the

destruction of the helper T cell.

One approach to the treatment of AIDS infections is

to interfere with the binding of the virus to the helper

T cell.This can be done by producing antibodies that

bind to the gp120 protein on the viral

surface.However, there are several difficulties with

this approach.First, because of the high mutation rate

of the gp120 protein, the most antigenic region of the

protein is extremely variable in structure.Second, the

binding region of gp120 does not readily stimulate

antibody production because the region is well

shielded by sugar molecules.Third, the gp120 protein

has a very strong affinity for the CD4 antigen that

must be overcome by any antibody produced against

gp120.

A second approach to AIDS treatment is to interfere

with the function of reverse transcriptase by

producing nucleotides that lack the hydroxyl group on

the 3′ carbon.These nucleotides will be preferentially

incorporated into a growing DNA chain by reverse

transcriptase, but not by the DNA polymerase of the

host cell, which is much more specific than the viral

enzyme.Because no subsequent nucleotides can be

added to the viral DNA chain, the chain will be

terminated.The drug AZT, which has an azide (N3)

group at the 3′ carbon, has been shown to interfere

with reverse transcriptase function and to prolong the

lives of AIDS patients.

86. When an AIDS virus has been incorporated into a

CD4 cell, but has NOT yet been replicated, the

viral genetic information is located in the CD4

cell’s:

A ) mitochondria.

B ) endoplasmic reticulum.

C ) nucleus.

D ) ribosomes.

87. The direction of information flow in the process

catalyzed by the enzyme reverse transcriptase is

the reverse of:

A ) DNA replication.

B ) RNA synthesis.

C ) protein synthesis.

D ) carbohydrate synthesis.

88. Some antibiotics work by interfering with the

function of bacterial (but not eukaryotic)

ribosomes. Such antibiotics are NOT effective in

fighting viruses because:

A ) viral ribosomes are too similar to eukaryotic

ribosomes.

B ) viral ribosomes are protected by the viral coat.

C ) viral ribosomes are too small to bind to any drug.

D ) viruses ordinarily lack ribosomes.

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89. Which of the following evolutionary mechanisms

most likely explains the presence in humans of

CD4 receptors on the helper T cells that bind to

the gp120 proteins of the AIDS virus?

A ) Coevolution (development of a series of reciprocal

adaptations that benefited both virus and host)

B ) Convergent evolution (development of

resemblances between virus and host after they

entered the same environment)

C ) Natural selection favoring chance mutation(s) of

the virus

D ) Natural selection favoring chance mutation(s) of

the host

90. AZT is effective for treating AIDS because it is

missing a hydroxyl group on the 3′ carbon, a

normal site for binding between:

A ) a phosphate and a sugar.

B ) a sugar and a nitrogenous base.

C ) a phosphate and a nitrogenous base.

D ) 2 complementary nitrogenous bases.

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Passage XVII

The ability of hemoglobin (Hb) to exchange O2 is

represented by the O2 equilibrium curve, a graph

depicting the saturation of deoxygenated Hb by O2 at

different pressures of O2 (Figure 1).

Figure 1

O2 equilibrium curve for human fetal (F) and

maternal (M) blood.

Figure adapted from Arthur C. Guyton, M.D., Textbook of Medical

Physiology, Sixth Edition. © 1981 by W.B. Saunders company.

The P50 is the O2 pressure at which the Hb binds

to 50% of the amount of O2 to which it can bind

under buffered standard conditions; the P50 is

inversely related to O2 affinity.A probe for dissolved

O2 is used to measure the amount of O2 present in the

sample, and corrections are made for the ambient

barometric pressure and water vapor.Because a color

change occurs when Hb changes its conformation

from the relaxed (oxygenated) to the tense

(deoxygenated) state, saturation of Hb may be

measured by spectrophotometry.

The Hb molecule is composed of 4 iron-

containing heme molecules to which the O2 molecules

attach and a globin portion consisting of 4 large

polypeptide chains (2 alpha and 2 beta chains) that

determine the binding affinity of the Hb for O2.One

heme molecule is associated with each of the 4

polypeptide subunits.The subunits can bind and

release O2 molecules separately, but the entire

tetramer cooperatively “clicks” into either a relaxed

or a tense conformation at a point where it has 2 or 3

O2 doublets bound to the hemes.At the P50 in a given

organism, individual Hb molecules may have 4, 3, 2,

1, or even zero O2 molecules bound to them, resulting

in a mean of 2.0 for the organism.

Several other ligands, such as carbon monoxide

(CO) and nitrous oxide (N2O), have a very high

affinity for Hb and are able to bind almost irreversibly

to Hb in place of O2.Once these ligands bind, the

molecule is fixed in the relaxed state.

Genetic variations leading to production of

different structural types of Hb in different species, or

in different life-cycle stages of the same species,

result in differences in P50 values.For example, larger

concentrations of Hb molecules with variations in the

beta chains typically cause organisms to have lower

P50 values.For this reason, the P50 of the human fetus

is lower than that of the mother (Figure 1), promoting

fetal ability to compete for O2 at the placenta.On the

other hand, some species, such as cats, mice, and

deer, have very high P50 values.

The P50 is also acutely sensitive to physiological

conditions.For example, O2 affinity is decreased by

decreasing pH and increasing temperature, as in

strenuous exercise; this lowered O2 affinity serves to

off-load O2 to the tissues.

91. The fetus is better able to compete for O2 at the

placenta because, compared to maternal Hb, fetal

Hb has an increased ability to:

A ) off-load O2 at lower O2 concentrations.

B ) off-load O2 at higher O2 concentrations.

C ) bind O2 at lower O2 concentrations.

D ) bind O2 at higher O2 concentrations.

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92. High P50 values, such as those of cats, mice, and

deer, are adaptive for these animals because they

enable the animals to:

A ) compete for O2 more efficiently than other animals

in the environment.

B ) be very active.

C ) live under conditions of high atmospheric

pressure.

D ) live under conditions of high environmental

temperature.

93. Cats have a much higher P50 reading than

humans. This means that, compared to human

Hb, cat Hb can:

A ) transport more O2.

B ) transport less O2.

C ) off-load O2 more readily under the same

conditions.

D ) off-load O2 at a lower partial pressure of O2.

94. Which phrase correctly describes the level of a

gas in fetal blood relative to its level in the

maternal blood in the placenta?

A ) CO–poisoned.

B ) CO2–depleted.

C ) O2–deprived.

D ) O2–enriched.

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Passage XVIII

The testes have both an endocrine and an exocrine

portion. The exocrine portion consists of the tightly

coiled seminiferous tubules. Before puberty, the

seminiferous tubules contain only spermatogonia and

Sertoli cells. Beginning at puberty, each

spermatogonium will undergo a series of mitotic and

meiotic divisions, called spermatogenesis, that result

in the production of mature spermatozoa (Figure 1).

The Sertoli, or “nurse” cells, provide nutrients for the

developing sperm. In addition, the Sertoli cell

membranes form tight junctions, establishing a

blood–testis barrier that protects developing sperm

from potentially toxic bloodborne substances, such as

proteins and polar compounds.

Figure 1

The endocrine portion of the testes consists of the

Leydig cells located between the seminiferous

tubules. The Leydig cells secrete testosterone, an

important male hormone. Testosterone acts on the

Sertoli cells to promote maturation of sperm; it also

controls the development and maintenance of male

sexual organs and secondary sexual characteristics.

Both the exocrine and endocrine functions of the

testes are controlled by hormones from the

hypothalamus and the pituitary (Figure 2).

Gonadotropin-releasing factor (GnRF) from the

hypothalamus stimulates the pituitary to synthesize

and release follicle-stimulating hormone (FSH) and

luteinizing hormone (LH). FSH acts directly on the

Sertoli cells to promote and maintain

spermatogenesis. LH acts on the Leydig cells to

stimulate the production of testosterone. Testosterone

in turn regulates testicular activity by inhibiting

GNRF release from the hypothalamus and LH release

from the pituitary. Inhibin, produced by the Sertoli

cells, inhibits FSH release.

Figure 2

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95. Which of the following hormones is(are) directly

required for spermatogenesis?

I. Luteinizing hormone (LH)

II. Follicle-stimulating hormone (FSH)

III. Inhibin

IV. Testosterone

A ) IV only

B ) I and IV only

C ) II and IV only

D ) I, II, and III only

96. Which of the following statements correctly

describes the distinction between the exocrine

and endocrine portions of the testis?

A ) The exocrine portion secretes only peptides; the

endocrine portion secretes only steroids.

B ) The exocrine portion releases its products into

ducts; the endocrine portion releases its products

into the blood.

C ) The exocrine portion secretes only cellular

elements; the endocrine portion secretes only

chemical substances.

D ) The exocrine portion is the target tissue for the

products of the endocrine portion.

97. A male taking excess testosterone may become

infertile because of reduced spermatogenesis.

According to Figure 2, this could result directly

from:

A ) an increase in inhibin concentration.

B ) a reduction in inhibin concentration.

C ) a reduction in FSH concentration.

D ) a reduction in LH concentration.

98. The cell type in the male reproductive system that

is most analogous to the female ovum is the:

A ) spermatogonium.

B ) primary spermatocyte.

C ) spermatid.

D ) spermatozoon.

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99. Of the following tissues, which is NOT derived

from embryonic mesoderm?

A ) Circulatory

B ) Bone

C ) Dermal

D ) Nerve

100. What is an alternative to sexual reproduction?

A ) Isogamy

B ) Hermaphroditism

C ) Pseudohermaphroditism

D ) Parthenogenesis

101. Double-stranded DNA can adopt one of three

helical conformations depending on the

nucleotide makeup of the molecule and the

amount of hydration. The nucleotide base pairs

in a DNA helix are arranged like steps in a

spiral staircase. Each one is rotated a few

degrees from the previous base pair.

Table 1 Average Helical Twist between

Adjacent Nucleotide Pairs (Mean and Standard

Deviation, in Degrees)

Conformation Helical twist

A 33.1 +6

B 35.9 +4

Z 29.9 +1

In investigating the properties of a strand of

DNA, researchers determined that there were 12

nucleotide base pairs for every complete 360o

turn of the helix. The conformation of the DNA

strand was:

A ) A, not B or Z.

B ) B, not A or Z.

C ) Z, not A or B.

D ) A or B, not Z.

102. Which of the following correctly pairs a cellular

process with the location in which that process

occurs in a prokaryotic cell?

A ) Transcription, cytoplasm

B ) ATP synthesis, mitochondria

C ) Degradation of macromolecules, lysosomes

D ) Modification of carbohydrates on transmembrane

proteins, Golgi complex

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103.

The diagram above shows the pattern of

inheritance of a certain disease. Females are

represented by circles, males by squares.

Individuals that exhibit the disease are

represented by shaded circles or squares. What

is the most likely method of inheritance of this

disease?

A ) Autosomal dominant

B ) Autosomal recessive

C ) Sex-linked dominant

D ) Sex-linked recessive

104. Translation of antibody proteins in eukaryotic

cells is associated with what organelle?

A ) Nucleus

B ) Mitochondrion

C ) Endoplasmic reticulum

D ) Golgi apparatus

105. Which of the following tissues have cells that

are in direct contact with the external

environment or elements of the external

environment?

I. The lining of the reproductive tract

II. The lining of the respiratory tract

III. The lining of the gastrointestinal tract

A ) I and II only

B ) I and III only

C ) II and III only

D ) I, II, and III

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Passage XIX

Several species of Drosophila have X-linked e and f

genes that affect the sex ratios of individuals’

offspring. However, the genes only affect sex ratios if

they are brought close together by an inversion of one

arm of the X chromosome (Figure 1).

Figure 1 Standard and inverted X chromosomes

An XsY male is standard: he sires equal numbers of

sons and daughters. An XiY male expresses the sex

ratio trait: he sires only daughters. Total reproductive

output is not affected; XsY males and XiY males sire

equal numbers of offspring.

If none of the Xi-bearing genotypes (XiY, XiXi, or

XiXs) is selected against, then the frequency of Xi is

expected to increase to 100%, unless other genes act

to suppress expression of e and f.

Occasionally, XiY males sire viable but sterile sons of

normal appearance. Genetic analyses show that all

these sons are XO, inheriting their X chromosome

from their mother and lacking a Y chromosome.

106. If the e and f genes are expressed, the Xi

chromosome will be prevented from reaching

100% frequency if selection pressures cause

which of the following to be true?

A ) XsXs flies have the lowest fitness of any genotype.

B ) XsXs flies have the highest fitness of any genotype.

C ) XiY flies and XsY flies have equal fitness.

D ) XiXs flies and XsXs flies have equal fitness.

107. A virgin female Drosophila mates and produces

34 daughters and 38 sons. Eighteen of these

sons sire only daughters, while the remainder

sire approximately equal numbers of daughters

and sons. What are the genotypes of the original

female and the male with whom she mated?

A ) XiXs and XsY

B ) XiXs and XiY

C ) XiXi and XsY

D ) XsXs and XiY

108. Which of the following statements best explains

why Xi has the potential to increase to 100%

frequency in gene pools that contain it?

A ) XiXs flies have the highest fitness of any genotype.

B ) XiXi flies tend to migrate and introduce the Xi

chromosome into new populations.

C ) XiXi flies pass X chromosomes to all their

offspring, but XsXs flies pass their X chromosomes

to only half their offspring.

D ) XiY flies pass their X chromosome to all their

offspring, but XsY flies pass their X chromosome

to only half their offspring.

109. In a laboratory population of Drosophila, all the

males are XsY. Among the females, 15% are

XiXi, 50% are XiXs, and 35% are XsXs.

Assuming random mating, what proportion of

male flies in the next generation will be XiY?

A ) 12%

B ) 30%

C ) 40%

D ) 65%

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110. If all genotypes are equally fit and if there are

no genetic modifiers of the sex ratio trait, what

will be the ultimate fate of a population in

which 50% of the X chromosomes are currently

Xi and 50% are Xs?

A ) Extinction

B ) Stable population size, with a predominance of

females

C ) Stable population size, with all individuals

producing a 50:50 sex ratio

D ) Stable population size, with some individuals

producing an excess of females and some

producing an excess of males

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Passage XX

Bone remodeling (involving the opposing

processes of bone deposition and resorption) is

regulated by several factors, including the

homeostasis of Ca2+

concentrations in the bone cells

and in their extracellular environment. Ca2+

concentrations in turn are partially controlled by the

levels of 2 hormones, parathyroid hormone (PTH)

and calcitonin.

Two major kinds of bone cells are involved in

bone remodeling: osteoblasts, which are responsible

for forming new bone tissue during growth and repair,

and osteoclasts, which are active during bone

resorption. Mature bone cells, or osteocytes, are

osteoblasts that have lost their ability to produce new

bone tissue, and therefore are not involved in bone

remodeling.

Experiments have shown that PTH binds to

receptors on the plasma membrane of osteoclasts,

triggering a series of reactions that lead to a transient

increase in the intracellular concentration of free Ca2+

ions. The experiments also show that the increase in

intracellular free Ca2+

concentration is strongly

correlated with stimulation of bone resorption. This

indicates that PTH release increases the rate of bone

remodeling. Conversely, calcitonin decreases the rate

of bone remodeling by inhibiting bone resorption and

promoting bone deposition.

To study the effects of PTH and calcitonin on

intracellular free Ca2+

concentrations, changes in Ca2+

concentrations in living osteoclasts were measured.

Several tetracarboxylate compounds that fluoresce

(absorb radiation from ultraviolet light and emit it at a

wavelength different from the incident light) in the

presence of free Ca2+

ions were used as Ca2+

indicators. The indicator compounds are esterified

before use, allowing them to enter cells more easily.

Experiment 1 proceeded as follows:

Step 1:

Osteoclasts were suspended in media containing

an esterified indicator and placed in a light-proof

chamber.

Step 2:

Different concentrations of Ca2+

were present in

the media bathing the cells. A standardized amount of

ultra-violet (UV) light was continuously transmitted

to the cell cultures at all concentrations.

Result:

The cells emitted light continuously, with an

intensity proportional to their intracellular free Ca2+

concentrations.

Step 3:

PTH was then added to the media.

Result:

The intensity of the light emitted by the cells

increased, even in the absence of Ca2+

in the media

bathing the cells.

Experiment 2 was identical to Experiment 1

except that calcitonin was added instead of PTH. The

intensity of the light emitted by the cells increased

rapidly at first, then slowly decreased.

111. Which of the following would NOT be a reason

for continuous remodeling of bone in healthy

adults?

A ) To adjust the strength of the bones in response to

stress

B ) To increase the total mass of bone in the body

C ) To heal bone injuries, such as fractures

D ) To prevent the bones from becoming brittle

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112. In Experiment 2, the intensity of emitted light

changed when calcitonin was added because

calcitonin causes:

A ) osteoclasts to absorb more UV light.

B ) osteoclasts to absorb less UV light.

C ) Ca2+ in the medium to emit light.

D ) the intracellular concentration of free Ca2+ to

change.

113. An advantage of adding the tetracarboxylate

indicators in esterified forms to cell suspensions

is that esterification renders carboxylates:

A ) more polar, so they can enter the cell membrane

more easily.

B ) more polar, increasing their permeability in lipids.

C ) less polar, so they can dissolve into solution more

easily.

D ) less polar, so they are more lipophilic.

114. In Experiment 1, how was the location of Ca2+

most likely changed by the addition of PTH?

A ) It moved into the osteoclast cytoplasm from the

extracellular medium.

B ) It moved into the osteoclast cytoplasm from

intracellular membranes and organelles.

C ) It moved out of the osteoclast cytoplasm and into

the extracellular medium.

D ) It moved out of the membranes and organelles of

the osteoclasts and into the extracellular medium.

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Passage XXI

Humans in hot, dry environments regulate their

body temperatures by circulatory adjustment

(vasodilation of cutaneous blood vessels) and

evaporative cooling (increased sweat gland

secretions). However, during severe heat stress and

dehydration, temperature regulatory mechanisms

(particularly sweating) may upset physiological

homeostasis.

Excessive sweating may upset homeostasis by

impairing water and salt regulation. During

dehydration, the kidneys may reduce their urinary

output from the normal level of 1.0-1.5 L H2O/day to

as little as 0.5 L H2O/day, and renal salt excretion

may decline to near zero. Excessive sweating also

reduces the volume of blood available for delivering

O2 to the internal body tissues.

Vertebrates that have evolved in deserts are better

adapted than humans for maintaining homeostasis in

hot, dry environments. When severely dehydrated,

humans can produce urine that is 4 times as

concentrated as plasma. However, camels can more

than double, and kangaroo rats can more than triple,

the urine-concentrating capacity of humans. Reptiles,

which lack sweat glands, maintain homeostasis by

means of lower metabolic rates and scaly, relatively

impermeable integuments.

115. When the environmental temperature is 33° C,

vasodilation of cutaneous blood vessels helps to

regulate the body temperature of a human by:

A ) slowing blood flow through the skin.

B ) maintaining an even distribution of heat

throughout the body.

C ) radiating excess body heat into the environment.

D ) preventing needed body heat from being lost to the

environment.

116. When the environmental temperature is 45° C,

which of the following organisms will have the

highest body temperature?

A ) Human

B ) Kangaroo rat

C ) Camel

D ) Lizard

117. Kidney failure during severe dehydration is

most likely due to:

A ) inadequate blood volume for effective filtration.

B ) inability to produce sufficient urine.

C ) buildup of salts in the distal tubules.

D ) increased body temperature.

118. People who are born without sweat glands are

likely to die of heat stroke in the tropics. This

indicates that, under tropical conditions, the

human body may:

A ) gain, rather than lose, heat by evaporation.

B ) gain, rather than lose, heat by radiation.

C ) need to use different mechanisms than in

temperate zones to maintain body temperature.

D ) be better able to regulate body temperature than

under temperate conditions.

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Passage XXII

Saccharomyces cerevisiae is a unicellular fungus that has both a haploid and a diploid phase. Cells in the

haploid phase divide by mitosis. When 2 haploid cells of opposite mating type fuse, they form a diploid cell

that can also divide by mitosis. Under certain environmental conditions, a diploid cell will undergo meiosis to

produce 4 haploid cells. The following experiments were conducted to investigate 2 temperature-sensitive

mutations (M1 and M2). These mutants replicated normally at 20° C, but did not replicate at 35° C.

Experiment 1

In order to determine the effect of the mutations the researchers grew cells at 20° C and 35° C in the

presence of radioactive precursors that could be incorporated into RNA or protein. The amounts of radioactive

RNA and protein that accumulated in cells incubated at the 2 temperatures were monitored. The graphs in

Figure 1 summarize the results for the temperature-sensitive strains and the wild-type strain (+).

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Figure 1

Experiment 2

Haploid wild-type cells were fused with haploid M1 mutant cells; the phenotype of the resulting diploid

cells (M1/+) was wild-type. Similarly, the fusion of a haploid wild-type and a haploid M2 mutant cell resulted

in wild-type phenotype.

Experiment 3

Haploid cells containing the M1 mutation were fused to haploid cells containing the M2 mutation. The

resulting diploid cells (M1 +/+ M2) exhibited the wild-type phenotype. When these diploid cells were induced

to undergo meiosis, 1,000 haploid cells were produced. A random selection of 200 of these cells revealed that

152 haploid cells exhibited the mutant phenotype, and 48 haploid cells exhibited the wild-type phenotype.

119. Which of the following results could support the

hypothesis that the M1 mutation affects a

different gene product than does the M2

mutation?

A ) Neither the M1 nor the M2 mutant incorporated

radioactive protein when grown at 35° C.

B ) The M1 +/+ M2 diploid has a wild-type

phenotype.

C ) The M1 mutation reverts to + at a different rate

than the M2 mutant reverts to +.

D ) The M1 mutation is found at a different frequency

in natural populations than is the M2 mutation.

120. What information would be necessary to

determine if the M2 mutation is recessive to the

wild-type allele?

A ) The frequency of the M2 mutation in an untreated

culture

B ) The ratio of wild-type to M2 mutants in an

untreated culture

C ) The phenotype of the M2/+ diploid

D ) The location of the M2 mutation on the

chromosomes

121. Based on the information in Figure 1, a student

concluded that the wild-type cells synthesized

more RNA at 35° C that at 20° C. Which of the

following assumptions must be true for this

conclusion to be valid?

A ) The ratio of labeled to unlabeled RNA precursors

is the same at both temperatures.

B ) The amount of protein in the cells is the same at

both temperatures.

C ) The degradation of RNA is slower at 35° C than at

20° C.

D ) The number of DNA molecules is greater at 35° C

than at 20° C.

122. If the M1 mutation is produced by a change in a

single gene, the expected phenotype ratio for the

diploid’s (M1/+) products of meiosis would be:

A ) 4 wild-type:0 mutant.

B ) 3 wild-type:1 mutant.

C ) 2 wild-type:2 mutant.

D ) 1 wild-type:3 mutant.

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123. Suppose that an extract from a muscle cell

contains only the following: all the enzymes of

the glycolytic pathway, including the enzyme

that converts pyruvate to lactate; phosphate and

other salts; NAD+and ADP. When the extract is

incubated anaerobically and glucose is

introduced, neither pyruvate nor lactate is

produced. What must be added in order for

pyruvate to be made?

A ) O2

B ) ATP

C ) NADH

D ) Acetyl-coenzyme A

124. What is the net volume of fresh air that enters

the alveoli each minute, assuming that the

breathing rate is 10 breaths/min, the tidal

volume is 800 mL/breath, and the nonalveolar

respiratory system volume (dead space) is 150

mL?

A ) 65 mL

B ) 95 mL

C ) 6500 mL

D ) 7850 mL

125. In the human "knee-jerk" reflex, the knee is

struck and the lower leg jerks forward. Which of

the following represents the complete pathway

that the nerve impulse travels in effecting this

response?

A ) Sensory neuron, motor neuron

B ) Sensory neuron, brain, motor neuron

C ) Sensory neuron, associative neuron, brain,

associative neuron, motor neuron

D ) Sensory neuron, associative neuron, motor neuron,

associative neuron, motor neuron

126. Which of the following hormones is LEAST

directly regulated by the anterior pituitary?

A ) Cortisone

B ) Epinephrine

C ) Progesterone

D ) Thyroxin

127. Which of the following structures is derived

from the same germ cell layer as the heart?

A ) Eye

B ) Bone

C ) Spinal cord

D ) Liver

128. The mineral component of human bone is a salt

that consists primarily of all of the following

EXCEPT:

A ) calcium.

B ) phosphate.

C ) potassium.

D ) hydroxyl groups.

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Post-test Confidence Self-Assessment

Congratulations!

You have completed a test of The Official MCAT® Self-Assessment Package.

Now that you have completed the test, reconsider your confidence rating of your knowledge of the content in

this test. Has your confidence level changed after answering the questions? Why might this be? Both your initial

and revised confidence ranking for each content category will be displayed in the analytic summary.

MCAT®

is a program of the Association of American Medical Colleges

Test/Content

Categories

A=1-Not

Confident at

all

B=2-

Somewhat

Confident

C=3-

Moderately

Confident

D=4-Very

Confident

E=5-

Extremely

Confident

1. Biology Overall

2. Circulatory,

Lymphatic, and

Immune Systems

3. Digestive and

Excretory Systems

4. DNA and Protein

Synthesis

5. Enzymes and

Cellular Metabolism

6. Evolution

7. Generalized

Eukaryotic Cell

8. Genetics

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9. Microbiology

10. Muscle and Skeletal

Systems

11. Nervous and

Endocrine Systems

12. Respiratory System

13. Reproductive

System and

Development

14. Specialized

Eukaryotic Cells

and Tissues

15. Skin System

16. Eukaryotes

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Pre-test Confidence Self-

Assessment

1 (A) (B) (C) (D) (E)

2 (A) (B) (C) (D) (E)

3 (A) (B) (C) (D) (E)

4 (A) (B) (C) (D) (E)

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16 (A) (B) (C) (D) (E)

Biology Questions

1 (A) (B) (C) (D)

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66 (A) (B) (C) (D)

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116 (A) (B) (C) (D)

117 (A) (B) (C) (D)

118 (A) (B) (C) (D)

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119 (A) (B) (C) (D)

120 (A) (B) (C) (D)

121 (A) (B) (C) (D)

122 (A) (B) (C) (D)

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126 (A) (B) (C) (D)

127 (A) (B) (C) (D)

128 (A) (B) (C) (D)

Post-test Confidence Self-

Assessment

1 (A) (B) (C) (D) (E)

2 (A) (B) (C) (D) (E)

3 (A) (B) (C) (D) (E)

4 (A) (B) (C) (D) (E)

5 (A) (B) (C) (D) (E)

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12 (A) (B) (C) (D) (E)

13 (A) (B) (C) (D) (E)

14 (A) (B) (C) (D) (E)

15 (A) (B) (C) (D) (E)

16 (A) (B) (C) (D) (E)

aamc self assesement biology

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