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

The full length test may be taken online, printed and taken offline, or a combination of both methods.

If you started a test online, the answer sheet provided at the end of this printout does not include the answers you entered online. Your online answers will appear on the online answer sheet used to submit your answers for scoring.

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Printing Guide

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

To print, click the PRINTER icon located along the top of the window and enter one of the following options in the PRINT RANGE section of the print dialog window:

To Print Enter Print Range Options

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Click ALL radio button

Commitment Self-Assessment Click PAGES FROM radio to and enter page 5 to 5

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Periodic Table Click PAGES FROM radio to and enter page 8 to 8

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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,--- Its 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 dont need to complete it all at once. It doesnt matter how long it takes you to finish the test, but you do need to finish to receive feedback!

If answer 3,-- Youve 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 dont 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?


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-, 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 agebut 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 factorsmetabolism and geneswith 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, ones 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 ShineDalgarno 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 ShineDalgarno 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 cells 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 mices 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 ) erythrocytes 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 Sarahs 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 ) Salmonellas 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 bodys own tissues. This is normally prevented by the bodys ability of self-tolerance; that is, the immune system recognizes the bodys 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 bodys 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 bodys 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 bodys 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 mothers sibling. A biologist explained that this behavior would increase the chance that the siblings genes would be passed on to the next generation. This conclusion is most likely based on the fact that the:

A ) infants mother was unfit to live in the environment.

B ) infant would probably have survived without the siblings help.

C ) sibling had many of the same genes as the infants 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 Steves 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. Joes body might have a greater tendency to overheat during strenuous work than Steves body would, because:

A ) older males have a higher basal metabolic rate. B ) Steve has a greater percentage of body fat. C ) Joes 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 ) Steves body would consume more energy because of his lesser weight.

B ) Steves body would consume less energy because of the greater basal metabolic rate in younger people.

C ) Joes body would consume more energy because of his greater weight.

D ) Joes body would consume the same amount of energy as Steves because the basal metabolic rates are equal.

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61. If Steves blood pressure had increased significantly more than Joes increased (other factors being equal) while they worked, what difference in their urinary system function would be expected?

A ) Joes glomerular filtration rate would increase more than Steves would.

B ) Steves reabsorption rate of glomerular filtrate by the peritubular capillaries would be lower than Joes would be.

C ) Steves reabsorption rate per milliliter of glomerular filtrate by the peritubular capillaries would be higher than Joes would be.

D ) Steves glomerular filtration rate would increase more than Joes 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 ) Bowmans 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 mutationstoo 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 cells:

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 ) COpoisoned. B ) CO2depleted. C ) O2deprived. D ) O2enriched.

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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 bloodtestis 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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