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AP Biology Practice Exam

Section I: Multiple- Choice QuestionsComplete one hundred multiple- choice questions in 80 minutes, worth

60 percent of the total score.

BIOLOGY

SECTION I

Time— 80 minutes

100 Questions

Directions: Each of the questions or incomplete statements below is fol-

lowed by five suggested answers or completions. Select the one that is

best in each case.

1. During the evolutionary process, which of the following is the cor-

rect sequence of events?

A. Change in phenotype— change in genotype— speciation—

selection

B. Speciation— change in genotype— selection— change in phenotype

C. Speciation— selection— change in phenotype— change in genotype

D. Change in genotype— change in phenotype— selection—

speciation

E. Selection— speciation— change in phenotype— change in genotype

2. A lenticel is to a plant as a _______ is to an insect.

A. stoma

B. spiracle

C. trachea

D. mouth

E. siphon

AP BIology2

Questions 3– 6 refer to the following diagram.

3. Produce digestive enzymes.

A. I, III, and V

B. II, III, and V

C. III, IV, V, and VIII

D. I, III, V, and VII

E. II, III, and IV

4. Produce(s) glucagon.

A. II and IV

B. III

C. V

D. VIII

E. II and V

AP BIology PRACTICE EXAM 3

5. Absorbs lipids.

A. III

B. VII

C. VI

D. IV

E. VIII

6. Produce(s) chyme.

A. VIII

B. IV

C. VII and VIII

D. II and IV

E. III

7. A new biological structure has been discovered. It is larger than

most bacteria. Although primarily composed of protein, it is cov-

ered with a lipid- based membrane. Its genetic material is double-

stranded DNA, but the structure lacks RNA. The interior of the

structure has ribosomes and a few enzymes. What is this?

A. A virus

B. A protozoan

C. An alga

D. A bacterium

E. A prion

AP BIology4

8. Dinitrophenol (DNP) is a compound that allows hydrogen ions to

leak across a membrane. This leakage prevents the maintenance of

an H+ gradient. How would this impact a cell?

A. It would permit a greater growth rate, since the cell would no

longer have to spend energy pumping H+ across the membrane.

B. It would permit a greater growth rate, because the equalization

of H+ on both sides of a membrane would help the cell cope

with an acidic environment.

C. If the cell were a neuron, it would have no impact, because the

charge differential on the surface of the neuron membrane is

based on Na+ and K+, not on H+.

D. The cell would lose the bulk of its energy production because

of DNP’s interference with chemiosmosis.

E. It would slow the growth rate slightly, but the cell would com-

pensate by maintaining the function of proton pumps.

9. If you were to take 1.0 ml of a solution of HCl with a pH of 4.0

and add it to 9.0 ml of distilled water, what would be the pH of the

final solution?

A. The pH would remain unchanged.

B. The pH would rise to 5.5.

C. The pH would rise to 5.0.

D. The pH would be unmeasurable due to the amount of dilution.

E. The pH would rise to 7.0.

10. From where do the electrons used in photosynthesis originate?

A. Glucose

B. NAD

C. Photons

D. Chlorophyll

E. Water


11. What distinguishes mammalian from plant cell membranes?

A. Plant membranes contain chlorophyll, while mammalian mem-

branes do not.

B. Mammalian membranes contain cholesterol, while plant mem-

branes do not.

C. Plant membranes contain ion channels, while mammalian

membranes do not.

D. Mammalian membranes are attached to a cytoskeleton, while

plant cells are not.

E. Plant membranes contain cellulose, while mammalian cell

membranes do not.

12. In fungi, sporangiospores are found in bag- like structures. Ascospores

are also found in bag- like structures, but are always present in even

numbers. Why are these spore types not the same?

A. They are basically the same, but the sporangiospores are always

larger.

B. Ascospores are formed as a result of meiosis, while sporangio-

spores are a result of mitosis.

C. Sporangiospores are produced by sexual recombination, while

ascospores are produced asexually.

D. They are basically the same, but ascospores are always larger.

E. The structures are the same, but they were named differently

before they were determined to have the same function.

AP BIology6

13. One line of evidence for the evolutionary process is the identifica-

tion of homologous structures. Of the following pairs of character-

istics, which could be used in support of this mechanism?

A. The wings of birds compared to the wings of insects

B. The eyes of insects compared to the eyes of humans

C. The cardiovascular system of animals compared to the vascular

system of plants

D. The wings of birds compared to the legs of bats

E. The fins of whales compared to the hooves of horses

14. If a person contracts and survives an infectious disease such as mea-

sles, they tend not to get the disease again later in life. What cells are

involved in this protective process?

A. Platelets

B. Neurons

C. Lymphocytes

D. Neutrophils

E. Erythrocytes

Questions 15– 17 refer to the following simplified diagram.


15. The diagram represents a central process of catabolism. Identify the

process designated as (I).

A. Electron- transport chain

B. Light- independent reactions

C. TCA cycle

D. Glycolysis

E. Transcription

16. What element or compound is needed for the process designated as

(III) to occur?

A. Chlorine

B. Nitrogen

C. Oxygen

D. Potassium

E. Sodium

17. Which combination of processes takes place entirely within a cel-

lular organelle?

A. I only

B. III only

C. I and II

D. II and III

E. I, II, and III

AP BIology8

18. Animal phyla are broadly categorized into acoelomates, deuteros-

tomates, protostomates, and pseudocoelomates. Identify the phyla

that are classified as protostomates.

A. Platyhelminthes and cnidarians

B. Echinodermatans and chordates

C. Molluscans and arthropods

D. Nematodes

E. Poriferans

19. Some dinosaur bones were discovered fairly recently that still con-

tained significant amounts of the protein collagen in a non- fossilized

form. Enough material was available to determine its amino- acid

sequence. This allowed researchers to compare its sequence with the

sequences of collagen from currently existing animals. What group

of organisms did the dinosaur sequence most closely resemble?

A. Birds

B. Lizards

C. Amphibians

D. Rodents

E. Fish

20. A cell is composed of thousands of molecules that range in size from

very small to very large. Select the sequence of molecular size from

the smallest to the largest.

A. Carbon dioxide— phospholipid— DNA polymerase— ribosome

B. Phospholipid— DNA polymerase— ribosome— carbon dioxide

C. Phospholipid— carbon dioxide— ribosome— DNA polymerase

D. Carbon dioxide— DNA polymerase— phospholipid— ribosome

E. DNA polymerase— ribosome— carbon dioxide— phospholipid


Question 21 refers to the following graph.

21. A researcher was analyzing the life expectancy of a rodent found

on an island off the coast of California. She was able to date its age

based on the wear on its teeth. Her data is presented in the graph.

What kind of survivorship curve would this data represent?

A. A type I survivorship pattern

B. A blend of a type I and a type III survivorship pattern

C. A blend of a type II and a type III survivorship pattern

D. A type III survivorship pattern

E. A type II survivorship pattern

AP BIology10

Questions 22 and 23 refer to the narrative below and the options that

follow.

When restriction fragments produced from the same endonuclease are mixed under the proper conditions, they will spontaneously reanneal. However, they will not function properly within a cell until the nicks in the strands are repaired.

A. Taq DNA polymerase

B. Reverse transcriptase

C. DNA- dependent RNA polymerase

D. DNA ligase

E. Eco RI

22. Identify a restriction endonuclease.

23. What enzyme is necessary to repair the nicks?

24. Which relationship presents a condition that could be identified as

exploitative competition?

A. A woodpecker and a wren

B. A sturgeon and a minnow

C. A mouse and a mole

D. A lichen and a rock

E. Mistletoe and a tree


25. Inside which portion of a cell does translation take place?

A. The endoplasmic reticulum

B. The nucleus

C. The cytosol

D. The Golgi complex

E. The cell membrane

26. A woman has a colorblind father. Her son also has a colorblind

father. Colorblindness is caused by an X- linked recessive gene. If

only male offspring are considered in this case, what is the prob-

ability that her son will be colorblind?

A. 100 percent

B. 50 percent

C. 0 percent

D. 25 percent

E. 75 percent

27. What is the best way to describe the steps involved in converting

fatty acids to acetyl- CoA?

A. They are catabolic, because the free energy increases as a result

of the process.

B. They are metabolic, because the free energy increases as a result

of the process.

C. They are anabolic, because the free energy decreases as a result

of the process.

D. They are catabolic, because the free energy decreases as a result

of the process.

E. They are anabolic, because the free energy increases as a result

of the process.

AP BIology12

Questions 28– 30 refer to the diagram and the options that follow.

A. I, II, III, V, and VI

B. I, V, and VI

C. VI

D. II and III

E. IV and VI

28. Require or utilize both oxygen and carbon dioxide.

29. Classified as dermal tissue or structures.

30. Closely associated with transpiration.


31. Mast cells are leukocytes contained in tissues. They are filled with

granules that contain histamine. These cells are involved in allergies,

including hay fever, for which antihistamines are used as therapy.

What mechanism is the cause of the allergic reaction?

A. When an allergen binds to specific receptors on a mast cell, it

induces cellular degranulation and the release of histamine into

the tissues.

B. Seasonal allergens irritate the nasal mucosa and cause cellular

infiltration by mast cells.

C. Mast cells phagocytose allergens that induce cellular

proliferation.

D. Mast cells rove through tissues removing both histamines and

antihistamines that may be causing the allergy.

E. Antihistamines are antibodies that bind to histamine to pro-

duce a complex, which is then removed from the tissue by

mast cells.

32. The genome of the rabies virus is negative- sense RNA. Negative

sense means that the nucleic acid sequence is complementary to

mRNA. What enzyme must be present in order for this virus to

produce its viral proteins and replicate?

A. DNA- dependent DNA polymerase

B. RNA- dependent RNA polymerase

C. DNA ligase

D. DNA- dependent RNA polymerase

E. Restriction endonuclease

AP BIology14

Questions 33– 36 refer to the graph and narrative that follow.

The options below apply only to questions 33 and 34. The normal blood glucose level is 70– 150 mg/dl. The arrow indicates an injection of insulin only in someone who is diabetic.

A. I and II

B. III

C. I, II, and IV

D. IV

E. II

33. Which curve(s) indicate(s) probable diabetes?

34. Which curve(s) indicate(s) diabetic shock?


35. To what do you attribute the drop, then rise, of blood glucose levels

seen in curve II?

A. Very poor monitoring of blood glucose levels

B. Flaws in the instrumentation or methods used to determine the

blood glucose levels

C. Ingestion of additional glucose at about time 2 hours

D. A second injection of insulin at about time 2 hours

E. Sporadic release of insulin by the pancreas

36. The normal range is provided as 70– 150 mg/dl. What would be the

equivalent expressed as g/l?

A. 0.7– 1.5 g/l

B. 7.0– 150 g/l

C. 0.007– 0.015 g/l

D. 0.07– 0.15 g/l

E. 0.0007– 0.0015 g/l

AP BIology16

Questions 37 and 38 refer to the following narrative.

Erythroblastosis fetalis is a condition that affects a fetus in utero caused by an Rh incompatability with the mother.

37. What is the mechanism that produces this condition?

A. Antibodies from the fetus attack antibodies from the mother

found in fetal circulation.

B. Antibodies from the fetus attach to the red blood cells within

the maternal circulation.

C. Lymphocytes from the mother cross the placenta and attack

the red blood cells of the fetus.

D. Lymphocytes from the fetus cross the placenta and attack the

red blood cells in the mother.

E. Antibodies from the mother cross the placenta and attack the

red blood cells of the fetus.

38. This problem is caused only when a(n) __________ mother has a

child with phenotype _________.

A. Unsensitized Rh+; Rh–

B. Sensitized Rh– ; Rh–

C. Unsensitized Rh– ; Rh+

D. Unsensitized Rh– ; Rh–

E. Sensitized Rh– ; Rh+


39. What is the relationship between glucose and sucrose?

A. Glucose is sucrose with an extra – OH group at the 3′ position.

B. Sucrose is composed of two glucose molecules joined by a

glycoside bond.

C. A glucose molecule joined to a fructose molecule makes a mol-

ecule of sucrose.

D. Glucose is composed of two sucrose molecules joined by a

glycoside bond.

E. Glucose is composed of a fructose and a sucrose molecule

joined by a glycoside bond.

Questions 40– 42 refer to the following diagram and narrative.

The lines represent three species within a single isolated ecosystem as observed by a team of qualified researchers.

AP BIology18

40. What relationship between two populations can you reasonably

deduce when comparing curves I and II?

A. The relationship is neutral.

B. There is strong evidence of competition but with resource

partitioning.

C. The population designated II appears to be an invasive species

that outcompetes the other.

D. The population designated I appears to be an invasive species

that could not survive in a new niche.

E. The population designated II appears to be a parasitic species

that uses the other as its host.

41. What relationship does the population designated as III reasonably

have with the other two populations?

A. The population designated III is a competitor of both of the

other populations.

B. The population designated III is a predator of population II but

not I.

C. The population designated III is a predator of population I but

not II.

D. The population designated III has a neutral relationship with

both of the other populations.

E. The population designated III is a predator of both populations

II and I.


42. To what might you reasonably attribute the loss of population III

within the ecosystem?

A. Population III was a prey of population I, which completely

consumed all its members within the ecosystem.

B. Population III was a prey of population II, which completely

consumed all its members within the ecosystem.

C. Population III was a predator of population II, but it could not

survive in the ecosystem.

D. Populations II and III both used population III as their prey,

completely consuming all its members within the ecosystem.

E. Population III was a predator of population I but it could not

survive in the ecosystem.

43. What differences are there between the cell membranes of eukary-

otes and prokaryotes?

A. Prokaryotic cell membranes are thinner and contain more proteins.

B. Eukaryotic cell membranes contain more proteins and larger

phospholipids.

C. Gases and water pass through a prokaryotic cell membrane at

greater rates than through those of eukaryotes.

D. Many functions that are compartmentalized within organelles of

eukaryotes are associated with the cell membranes of prokaryotes.

E. There are no significant differences in the structure or function

of the cell membranes.

AP BIology20

44. Which of the following best explains the origin of the predation

ability of Venus flytraps?

A. Only those plants with the ability to supplement the poor

nutrient content of their soils could survive.

B. Predation allowed the Venus flytrap to outcompete other spe-

cies because of its ability to move rapidly.

C. As consumers instead of producers, Venus flytraps have greater

access to better nutrients.

D. Predation allowed the Venus flytrap to collect a wider variation

of preformed organic materials.

E. Better nutrients allowed the Venus flytrap to grow faster than

other plants.

45. A type of fungus exists that lives symbiotically with plants. What is

that fungus?

A. Mycorrhizae

B. Lichens

C. Mushrooms

D. Morels

E. Truffles

46. Which of the following experiments first ascertained that the

nucleus possessed the ability to control phenotype?

A. Experiments with radiolabelled phages

B. Experiments with bacteria that identified DNA polymerase

C. Transformation experiments with Streptococcus pneumoniae

D. Identification of restriction endonucleases

E. Grafting experiments with Acetabularia


47. A father with type B blood and mother with type A blood have a

child. Their child, while in a biology lab at school, tests her blood

and discovers she has type O blood. Does she have any concerns

about her parentage?

A. Yes, because she should have type AB blood if they are her true

biological parents.

B. No, because type O blood is possible if her parents both had

genotypes AB.

C. No, because both of her parents could be heterozygous.

D. Yes, because both of her parents might be heterozygous.

E. No, because blood types A and B are codominant.

48. The HIV genome is roughly 10,000 nucleotides in size, while

the human genome compares at about 3,000,000,000 base pairs.

Reverse transcriptase has an error rate of about 1 in 20,000 bases.

What does this imply about the evolution of the virus?

A. The evolutionary rate is rather slow, because it is unlikely that

any one virus will contain a mutation.

B. The rate of evolution is extremely high, because every infected

cell will produce viruses with mutations.

C. The rate of evolution is rather slow, because the genome is

so small.

D. The rate of evolution is extremely high, because the human

genome is so comparatively large.

E. The evolutionary rate is slow, because the vast majority of

HIV mutations will prevent the virus from binding to human

cell receptors.

AP BIology22

Questions 49– 51 refer to the diagram

49. Which organ or tissue contains this structure?

A. Liver

B. Ovary

C. Lung

D. Kidney

E. CNS

50. Vasopressin affects the functioning of this structure. Which of the

following secretes this substance?

A. Posterior pituitary

B. Liver

C. Hypothalamus

D. Anterior pituitary

E. Adrenal glands


51. What product or process is this structure known for?

A. Production of urine

B. Production of LH and FSH

C. Feedback control of the autonomic nervous system

D. Gas exchange

E. Production of bile

52. “Human cells contain DNA sequences that match the sequences

found in bacterial 16S rDNA.” What is your response to this

statement?

A. The statement is untrue, because healthy human cells should

contain no bacterial DNA.

B. The statement is true, because mitochondria contain bacterial

DNA sequences.

C. The statement is untrue, because there is no such thing as

rDNA, only rRNA.

D. The statement is true, because the 16S sequences are part of

the human 18S sequences.

E. The statement is true, because there are many human sequences

that match 16S sequences.

53. The Na+/K+ pump uses a form of ________ _______ known as

_________.

A. passive transport; uniport

B. facilitated diffusion; antiport

C. active transport; symport

D. facilitated diffusion; uniport

E. active transport; antiport

AP BIology24

54. Many mammalian cell membrane receptors associated with the

immune system contain repetitive structures known as the immu-

noglobulin fold. Receptors containing even one of these folds are

considered members of the immunoglobulin superfamily. How

would you explain this common feature in so many molecules?

A. All of the genes that produce these proteins used to be part of

a much larger gene that fragmented in the past.

B. All of the genes in question are retrotransposons.

C. Most of the genes in question are introns that experienced

mutations that removed the splice sequence.

D. The fold was so successful as a signaling molecule that it encour-

aged its propagation.

E. The fold is repetitive because of gene duplication.


Assume you had a 10 m2 pond to study. You measured the amount of light energy bathing the water surface at 1,000 kcal.

55. About how much of that available energy would be expected to be

converted into producer biomass?

A. 100 kcal

B. 1,000 kcal

C. 2,000 kcal

D. 10 kcal

E. 200 kcal


56. Based on the energy captured above, about what percentage of that

would be transferred to higher trophic levels?

A. 1.65 percent

B. 16.5 percent

C. 33 percent

D. 3.3 percent

E. 66 percent

57. If a woman who suffers from rheumatoid arthritis becomes preg-

nant, the arthritis symptoms often subside temporarily during her

pregnancy. Not coincidentally, a woman who has once had genital

warts caused by the human papillomavirus (HPV) will find they

return temporarily during her pregnancy. What do these two obser-

vations have in common?

A. Both effects are caused by the production of FSH and LH dur-

ing pregnancy.

B. Both effects are caused by the reduction of progesterone and

estrogen during pregnancy.

C. Both are caused by the partial suppression of the immune sys-

tem during pregnancy.

D. Both effects are caused by the production of progesterone and

estrogen, and the drop in β- HCG levels, during pregnancy.

E. Both effects are caused by the production of oxytocin during

pregnancy.

AP BIology26

Questions 58 and 59 refer to the following diagram.

58. The diagram indicates an error in some process. What is that process?

A. Meiosis

B. Crossing over

C. Pleiotropy

D. Misogyny

E. Mitosis

59. Which of the following is true about the process as indicated?

A. It is the cause of sickle cell anemia.

B. It is the failure to get equal distribution of chromosomes during

mitosis.

C. It is associated with susceptibility to HIV infection.

D. It is an important part of the double- fertilization process.

E. It is the cause of the “superfemale” syndrome (XXX).

60. What do these two conditions have in common: webbed fingers and

toes in humans and the development of cancer cells?

A. Both are caused by excessive cell growth.

B. Both are caused by a loss of apoptosis.

C. Both are a result of exposure to carcinogenic compounds.

D. Both are caused by excessive fat- soluble vitamin levels.

E. Both are caused by infections from cancer- associated viruses.


61. Tobacco mosaic virus (TMV) is very common in the environment

and is frequently isolated from tobacco products. What health

implication does this present?

A. TMV is a cause of lung cancer.

B. TMV is associated with immune suppression.

C. TMV is associated with skin tumors.

D. There are no health implications; TMV cannot infect human

cells.

E. TMV is associated with genetic abnormalities within any cell

exposed to secondhand smoke.

Question 62 refers to the following image.

62. Which has the most phylogenetic relatedness to what is seen here?

A. Lycophytes

B. Monocots

C. Gingkos

D. Mosses

E. Ferns

AP BIology28

63. In addition to enzymes, at least one other organic molecule is known

to be capable of catalysis. What is that molecule?

A. DNA

B. Fatty acids

C. Cellulose

D. RNA

E. ATP

64. What is the primary source of the electrons used in the mitochondria?

A. Water

B. Glucose

C. ATP

D. Cellulose

E. Oxygen

65. Large populations of organisms are centered on deep sea hydrother-

mal vents. Which of the following is true about these populations?

A. The energy captured by these organisms is thermal in nature

and allows them to thrive in mixed populations.

B. Only a single species is found at any one vent, although the

species varies from vent to vent.

C. The species found at each site are usually evenly mixed between

producer plants and consumer microorganisms and animals.

D. Fungi and protozoans are usually not represented in these

populations.

E. Energy from mineral ions feed producers, which in turn feed

varieties of consumers.


66. A certain species of bird has two feather variations displayed by

males: a dull- colored one that blends into their preferred habitat

and a brightly colored one that is very obvious. The adult females

are dull colored. The predation rate for brightly colored adults is

three times higher than for the dull- colored adults. In spite of this,

the brightly colored plumage persists in the population. How would

you explain this?

A. Females prefer the brighter plumage and so mate with this

phenotype more often.

B. The dull plumage provides a selection advantage over the

bright plumage.

C. The adults with dull plumage require less energy investment to

ensure their survival.

D. Those brightly colored birds that do survive have more highly

developed survival skills than those that do not survive.

E. Dull- colored adult males are more lethargic than the brightly

colored birds.

AP BIology30

Questions 67– 69 refer to the diagram below and the options that follow.

A. VII

B. I and VI

C. II, III, IV, and V

D. II, V, and VII

E. III and VII

67. Meiosis takes place in which structure(s)?

68. If this flower relied on wind pollination, then this/these structure(s)

would be a lot more pronounced.

69. Aggregate fruit are produced from a single flower with multiple

numbers of these structures.


70. As has occurred innumerable times in the past, the global mean

temperature appears to be rising, at least in the short term. Which

of the following is an inevitable result of this change?

A. First frost dates will be delayed.

B. Birds will migrate shorter distances.

C. Oxygen levels in the oceans will rise.

D. Available fertile lands will decrease as desertification increases.

E. Animal life will be reduced worldwide.

71. The damming of the Nile River near Aswan in the 1960s introduced

a new disease to the area. While the water once flowed freely, stop-

ping its flow allowed new organisms to thrive in newly opened

niches. One of these new organisms causes a disease called schisto-

somiasis, which infects humans as part of its life cycle. What is the

taxonomic classification of this organism?

A. Arachnid

B. Helminth

C. Cephalopod

D. Bacterium

E. Cladoceran

72. Antigen- presenting cells communicate with effector cells of the

immune system. Which of the following mechanisms account for this?

A. Direct cell membrane- to- membrane contact

B. Quorum sensing

C. Release of highly specific steroidal hormones

D. Release of neurotransmitter- like substances

E. Induction of rapid ion flow through the membrane

AP BIology32

Questions 73– 75 refer to the following narrative.

Shigella is a bacterium that is very closely related to E. coli but is incapable of fermenting lactose (lactose negative). A gene for resistance to the antibiotic ampicillin (ampR) was inserted in between two of the structural genes of the lac operon and then inserted into Shigella.

73. What would be the effect of this insertion?

A. The cell would immediately be resistant to ampicillin.

B. The cell would be resistant to ampicillin only in the presence of

tryptophan.

C. The cell would produce the enzyme β- lactamase as soon as it

started to grow exponentially.

D. The cell would spontaneously utilize DNA- editing enzymes to

restore the original genotype.

E. The cell would be resistant to ampicillin only in the presence of

lactose.

74. If Shigella is normally lactose negative, from where did the lac

operon originate?

A. The bacterial genome was removed, the lac operon was added,

and then the genome was inserted back into the original cell.

B. The lac operon was present on a plasmid which was inserted

into the cell by transformation.

C. The bacterial genome was removed and replaced by a plasmid

containing the lac operon.

D. The lac operon developed in the cell by random mutation and

strong selection pressures.

E. The lac operon was derived from mutations of the tryptophan

operon which is normally present in Shigella.


75. What enzyme regularly used by genetic engineers would be vital for

the successful completion of this exercise?

A. DNA polymerase

B. Reverse transcriptase

C. Taq polymerase

D. β- galactosidase

E. Restriction endonuclease

76. Assume that you have been hired to work in a lab that tests plant

hormones for increasing crop yield. Your job today is to test spray a

large plot with a selected hormone to determine its effect in hybrid

vigor. Which of the following would you want to avoid using?

A. 2,4- D

B. IAA

C. Gibberellin

D. Cytokinin

E. Abscisic acid

77. Place the following vessels or tissues in the proper sequence from

containing the highest concentration of CO2 to the lowest.

A. Pulmonary veins— left atrium— tissues— right ventricle—

pulmonary arteries

B. Left atrium— tissues— right ventricle— pulmonary veins—

pulmonary arteries

C. Pulmonary arteries— right ventricle— tissues— left atrium—

pulmonary veins

D. Right ventricle— pulmonary arteries— pulmonary veins— left

atrium— tissues

E. Tissues— pulmonary veins— pulmonary arteries— left atrium—

right ventricle

AP BIology34

78. Individuals born with a condition known as phenylketonuria must

control their intake of phenylalanine or suffer developmental prob-

lems. Because of this they must avoid foods rich in what?

A. Proteins and lipids

B. Fats and oils

C. Proteins

D. Simple sugars

E. Complex carbohydrates and oils

Questions 79– 80 refer to the following table and narrative.

The genes that code for the human leukocyte antigen (HLA) antigens, also known as the major histocompatability complex (MHC) genes, are all located on human chromosome 6. Two brothers and a sister participated in a bone marrow donor registration, and their HLA typing results are given above.


79. What is the probability that, if they had a fourth sibling, that sibling

would match one of the three above?

A. 25 percent

B. 100 percent

C. 0 percent

D. 50 percent

E. 75 percent

80. What is your explanation for the single HLA- A result for sibling 3?

A. A gene- deletion event occurred during crossing over.

B. Nondisjunction occurred during meiosis.

C. A laboratory error caused the lost data; it should have a second

result.

D. A metabolic error prevented the synthesis of the second antigen.

E. Both parents were heterozygous for HLA- A1.

AP BIology36


81. Based on this monitoring of an individual infected with HIV at six

months, who was untreated during the time period indicated, how

would you identify what the three curves signify?

A. I = viral antigen in the blood; II = anti- HIV antibodies in the

blood; III = CD4+ cells in the blood

B. I = viral antigen in the blood; II = CD4+ cells in the blood; III =

anti- HIV antibodies in the blood

C. I = CD4+ cells in the blood; II = anti- HIV antibodies in the

blood; III = viral antigen in the blood

D. I = anti- HIV antibodies in the blood; II = viral antigen in the

blood; III = CD4+ cells in the blood

E. I = CD4+ cells in the blood, II = viral antigen in the blood; III =

anti- HIV antibodies in the blood


82. Where in a plant cell is a water molecule broken down to release

oxygen gas, electrons, and hydrogen ions?

A. In the inner matrix of the mitochondrion

B. In the outer membrane of the chloroplast

C. In the intermembrane space of the mitochondrion

D. In the thylakoid space of the chloroplast

E. In the intermembrane space of the chloroplast

83. What is the best measuring stick to use to determine the degree of

fitness in evolutionary selection?

A. The increase in lifespan resulting from selection

B. The increase in chromosomal number within the nucleus

C. Reproductive success

D. The number of available alleles for any one gene

E. The increase in geographical territory for any one species

84. Identify the general relative sizes of the listed structures from the

smallest to the largest.

I = bacterium

II = enzyme

III = nucleus

IV = chromosome

V = molecule of water

A. V— II— I— IV— III

B. II— V— IV— I— III

C. V— I— II— III— IV

D. II— V— I— IV— III

E. II— V— IV— III

AP BIology38


85. The graph displays the activity curves for three human enzymes.

Identify the curve designation with the enzyme.

A. I = pepsin; II = chymotrypsin; III = fibrin

B. I = chymotrypsin; II = fibrin; III = pepsin

C. I = fibrin; II = pepsin; III = chymotrypsin

D. I = pepsin; II = fibrin; III = chymotrypsin

E. I = fibrin; II = chymotrypsin; III = pepsin


86. Why do toadstools form fairy rings after a rain?

A. The ring reflects a pH gradient which is highest in the center of

the circle and diffuses outward. The size of the ring varies with

the alkalinity at the center.

B. The ring appears around the remnants of a tree stump that

provides nutrients at the perimeter of the buried wood.

C. The ring usually appears at the periphery of large buried boul-

ders or other dense material.

D. In actuality, the appearance of a fairy ring is usually just the

mind’s attempt to make a pattern out of random placements of

toadstools.

E. The ring indicates the intersection of a colony of subsurface

fungal hyphae expanding into a surrounding colony with a dif-

ferent mating type.


Hemophilia is a genetic disorder caused by an X- linked recessive allele. It is of historical interest because of the extensive familial intermarriages of the royal houses of Europe during the seventeenth to nineteenth centuries AD.

87. In evolutionary terms, why has this lethal allele continued to be

passed on?

A. Because of the founder effect

B. Because, while producing a bleeding disorder, it increases longevity

C. Because it only affects female carriers of the trait

D. Because the recessive allele provides a selection advantage

E. Because of disruptive selection

AP BIology40

88. Which of the following best describes this form of hemophilia?

A. This disorder is caused by reduced production of thrombocytes

in the bone marrow.

B. This disorder is caused by an increased production of antithrombin.

C. This disorder is caused by an absence of factor VIII in the blood.

D. This disorder is caused by a loss of the clotting ability of throm-

bocytes, which are present in normal numbers in the blood.

E. This disorder is caused by an increased level of anticoagulants

in circulation and in the tissues.

89. Numerous animals, notably reptiles, are capable of reproduction

and maintaining numbers when populations are composed of only

single sex members, even in isolated areas. How do you explain

this statement?

A. It is not possible and the statement is false.

B. When only females are present, hormonal signals will cause the

dominant female to develop male characteristics and engage in

sexual reproduction with the other females.

C. Meiosis within unfertilized eggs will arrest at a diploid stage and

undergo normal fetal development in an all-female population.

D. When only males are present, random males will become her-

maphroditic and lay eggs fertilized internally.

E. The sperm from males can be stored for years within a female,

allowing normal clutches of eggs to hatch years after males

have disappeared from a population.


90. An experiment is set up with a culture of algae undergoing normal

photosynthesis. A pulse of radioactive carbon, in the form of 14CO2,

is inserted into the culture at time zero. At five- second intervals,

some of the cells are removed by dropping them into boiling water

to halt metabolic processes. What is the first organic material that

would be identified as being radioactive during this process?

A. ATP

B. PGA

C. ADP

D. Glucose

E. Acetyl- CoA

91. Hydrilla is a vascular invasive plant species that grows under the

surface of lakes at a very high density during the summer but dies

back to its buried rhizomes during the winter. It has been linked

to oxygen depletion of the waters in which it is found, in spite of

the fact that it is photosynthetic. How can you account for this

seeming contradiction?

A. While photosynthetic, Hydrilla consumes oxygen at a faster

rate than it releases it because of its very high rate of growth.

B. Hydrilla’s nutrient density is so great that fish that feed upon it

proliferate beyond the carrying capacity of the lakes in which it

is found.

C. Because Hydrilla is vascular, it binds up large quantities of

oxygen in its biomass.

D. As an aquatic plant, Hydrilla requires large quantities of

oxygen to fill its air bladders.

E. When the vegetative structures die off in the winter, decom-

posing bacteria consume the oxygen during their rapid growth.

AP BIology42

Questions 92– 94 refer to the following graph.

92. Given that the curve designated II is FSH, identify the other three.

A. I = LH; III = progesterone; IV = estrogen

B. I = estrogen; III = LH; IV = progesterone

C. I = progesterone; III = LH; IV = estrogen

D. I = LH; III = estrogen; IV = progesterone

E. I = estrogen; III = progesterone; IV = LH

93. Identify the sources of the hormones.

A. Ovary = I and II; pituitary = III; hypothalamus = IV

B. Ovary = III and IV; pituitary = I and II

C. Ovary = I and II; pituitary = III and IV

D. Hypothalamus = I and II; pituitary = III and IV

E. Ovary = I and II; pituitary = IV; hypothalamus = III


94. Identify the day of follicle rupture.

A. About day 14

B. About day 10

C. About day 0

D. About day 21

E. About day 25

95. Which of the following observations support the fluid mosaic model

of membrane structure?

A. The capping of fluorescent antibodies specific for cell- surface

receptors

B. The presence of the cytoskeleton underneath the surface of the

membrane

C. The measurement of ions flowing through ion- gated channel

proteins by patch clamps

D. The pseudopodial movement of amoebas

E. The scaffolding and kinetic proteins associated with the move-

ment of cilia

96. Broca’s area, the portion of the brain associated with the motor con-

trol of speech, is localized in which portion of the brain?

A. The parietal lobe

B. The cerebellum

C. The occipital lobe

D. The frontal lobe

E. The lateral sulcus

AP BIology44

97. Which of the following means of reproduction is the most ancient?

A. Viviparity

B. Oviparity

C. Ovoviviparity

D. Placental viviparity

E. Viviparous matrotrophy

98. When carbohydrate monomers are joined together, the process is

identified as ___________ ______________, and when they are bro-

ken apart again it is called ___________.

A. hydrolytic bonding; degradation

B. hydrogen bonding; depletion

C. condensation synthesis; hydrolysis

D. polar synthesis; hydrolysis

E. condensation synthesis; anabolism

99. Some plants have the ability to twine around objects. What is this

ability to respond to physical contact with a growth response called?

A. Seismonastic movement

B. Phototropism

C. Thigmotropism

D. Somnambulism

E. Gravitropism


100. Many organisms have been used in research to help elucidate

biological processes. Which of the following is used exclusively in

developmental biology?

A. Drosophila melanogaster

B. Caenorhabditis elegans

C. Pisum sativum

D. Escherichia coli

E. Daphnia pulex

END OF SECTION I

AP BIology46

Section II: Free- Response QuestionsComplete four essay questions, with ten minutes of reading and plan-

ning time and an hour and a half of writing time. Each question is worth

10 percent of the total exam score.

BIOLOGY

SECTION II

Reading and planning time— 10 minutes

Writing time— 1 hour and 30 minutes

Directions: Answer all questions.

Your answers must be in essay form. Neither outlines alone nor diagrams

alone are sufficient. It is important that you read and understand each

question before you begin to write.

1. Cellular homeostasis must be maintained to sustain life and life pro-

cesses. In order to accomplish this, the cell must possess the ability

to operate with both positive and negative control mechanisms.

A. Identify and explain ONE positive and ONE negative control

mechanism involved in gene expression.

B. Identify and explain ONE positive and ONE negative control

mechanism involved in enzyme kinetics.

2. Organismal homeostasis must be maintained to sustain life and

insure the continuation of the species. In mammals, this mainte-

nance is often controlled by both neural and hormonal mechanisms.

A. Discuss the neural mechanism of the control of blood pressure.

B. Discuss the kidney’s role in maintenance of blood pressure.

C. Identify TWO disease processes caused by failure to control

blood pressure.


3. Speciation results from the application of selective pressures until

two non- interbreeding populations develop.

A. Identify and explain TWO prezygotic isolating mechanisms.

B. Identify and explain TWO postzygotic isolating mechanisms.

4. Plants maintain homeostasis but lack the pumps and muscles that

animals have in order to do so.

A. Identify the elements associated with fluid flow within vascular

plants. Describe the mechanisms that produce the flow.

B. In order to maintain fluid flow, the vascular structure must be

protected. Explain TWO such protective mechanisms.

END OF EXAM

AP BIology48

Answers to the AP Biology Practice ExamSection I: Multiple- choice questionsAnSwER KEy

1. D

2. B

3. A

4. C

5. B

6. E

7. A

8. D

9. C

10. E

11. B

12. B

13. E

14. C

15. D

16. C

17. D

18. C

19. A

20. A

21. E

22. E

23. D

24. A

25. C

26. B

27. D

28. D

29. B

30. E

31. A

32. B

33. C

34. D

35. C

36. A

37. E

38. E

39. C

40. C

41. B

42. C

43. D

44. A

45. A

46. E

47. C

48. B


49. D

50. A

51. A

52. B

53. E

54. E

55. D

56. B

57. C

58. A

59. E

60. B

61. D

62. C

63. D

64. B

65. E

66. A

67. E

68. C

69. D

70. A

71. B

72. A

73. E

74. B

75. E

76. A

77. C

78. C

79. E

80. E

81. B

82. D

83. C

84. A

85. D

86. E

87. A

88. C

89. C

90. B

91. E

92. D

93. B

94. A

95. A

96. D

97. B

98. C

99. C

100. B

AP BIology50

AnSwER EXPlAnATIonS

1. D. This question is designed to evaluate your understanding of the

basic process of evolution.

Jean- Baptiste Lamarck proposed that physical changes that developed

in an individual would be passed on to its progeny. This could be inter-

preted to mean that if a person worked out a lot and became buff, that

extra toning would be passed on to his or her children. Charles Darwin,

on the other hand, proposed that selection was the process that deter-

mined what characteristics were passed on. In order for a characteristic to

appear, there must first be a mutation in the DNA (change in genotype).

Once the DNA has changed, the changed gene is expressed, modifying

what characteristic may be seen or measured (change in phenotype).

After the introduction of the new phenotype increases the diversity

within the population, environmental pressures then serve to select

which phenotype is more fit under those conditions. Accumulation of

a battery of new phenotypes eventually will cause divergence from the

original species genotype.

2. B. The purpose of this question is to determine if you can identify

respiratory structures on plants and insects.

You have to remember what a lenticel is. Lenticels are openings in

plants that permit gas exchange. You probably do remember the leaf

structure called a stoma (Greek for “mouth”) which is bounded by guard

cells. The stoma is a leaf lenticel. This allows you to eliminate option

A. A trachea is a tube that conducts air and is used for both human and

insect structures, just as a siphon conducts water in aquatic animals, but

it is not the opening that was asked for. You can now eliminate options

C and E. While a mouth is an equivalent mammalian structure to a

lenticel, in insects the mouth is not used for gas exchange. This then

eliminates option D. The last choice is the one you would know if you

remembered the respiratory system of an insect.

3. A. Questions 3– 6 assay your understanding of the human diges-

tive system.

First you should identify the labeled structures. (I) is the parotid


gland; (II) is the liver; (III) is the stomach; (IV) is the gall bladder; (V) is

the pancreas; (VI) is the large intestine; (VII) is the small intestine; and

(VIII) is the appendix. Question 3 then asks you to identify the struc-

tures that produce digestive enzymes. The parotid produces amylase and

bicarbonate, the stomach produces pepsinogen, and the pancreas pro-

duces pancreatic juice, which contains a number of enzymes including

trypsinogen and lipase. The liver produces bile, which is stored in the

gall bladder, but neither of these two produces digestive enzymes.

4. C. Glucagon in a hormone that helps regulate blood glucose levels.

It is released by the pancreas into the blood when glucose levels drop too

low. Glucagon stimulates the liver’s conversion of glycogen into glucose

and the production of insulin in the pancreas. As the glucose level rises

in the blood, the insulin stimulates the uptake of glucose by insulin-

dependent tissues.

5. B. One of the functions of bile is to aid in the absorption of lipids.

Once a bolus of food moves from the stomach into the small intestine,

its acidity is neutralized and lipase begins to break triglycerides into

monoglycerides and fatty acids. These compounds are then emulsified

and packaged into spontaneously forming micelles. These micelles are

then absorbed into the endothelial cells of the small intestine, where

their contents are then routed to the lymph system for transport to the

circulatory system.

6. E. Once food passes through the esophagus by peristalsis, it enters

the stomach through the cardiac sphincter. While there, it is bathed in

HCl and churned into a paste- like consistency. After about two hours

the food, now in the form of chyme, starts to move into the small intes-

tine through the pyloric sphincter.

7. A. The purpose of this question is to determine if you can use rea-

son to deduce the identification of a hypothetical biological entity.

All living organisms have a DNA- based genome, a membrane, ribo-

somes for translation, and RNA for transcription. Viruses may have a

DNA- based genome, may have a host- membrane derived envelope, may

be larger than some bacteria, may contain a stray ribosome from the

AP BIology52

host cell, and may have a couple of enzymes. The key characteristic

differences between a virus and a living organism are that viruses con-

tain either RNA or DNA, but not both, and while a virus may have an

enzyme or two, it lacks metabolic processes.

8. D. This question is designed to ascertain your understanding of the

functioning of mitochondria.

Chemiosmosis is the process whereby hydrogen ions are pumped

across a membrane. In the mitochondria, these proton pumps are pow-

ered by the energy extracted from electrons harvested from glycolysis

and the TCA cycle. Because of the gradient, the ions will rush back

through the membrane via a molecule of ATP synthase, thereby manu-

facturing ATP. If DNP is present, then the electron- transport system is

still functioning, and H+ ions are being pumped across the membrane,

but the leakage prevents the maintenance of the gradient required to

power ATP synthase. Thus, no energy is produced by the cell, and the

cell ceases to be able to maintain homeostasis.

9. C. The purpose of this question is to determine the extent of your

understanding of pH and biological systems.

pH is a measure of the concentration of hydronium ions (H3O+) in

water. Distilled water has a neutral pH of 7.0 (where the H3O+ con-

centration matches the OH– concentration).The scale is logarithmic, so

a tenfold drop in concentration raises the pH by one. Putting 1.0 ml

into 9.0 ml makes a total of 10.0 ml, so this represents a 1:10 dilu-

tion. Diluting an acidic solution makes it less acidic, so option A can be

removed from consideration. Making a pH 4.0 solution neutral would

require the addition of a basic solution, not just water, so E can also be

eliminated. A pH can always be measured, so option D can be removed.

A pH of 5.5 is halfway between 4.0 to 7.0 but really has no relevance.

Decreasing the concentration tenfold will raise the pH by 1.0.

10. E. This question is designed to determine your level of understand-

ing of the flow of energy in photosynthesis.

Photons are packets of energy, not matter like electrons, so option

C can be ignored. Photosynthesis is conducted by autotrophs to


manufacture organic compounds for the storage of energy (electrons),

not to use the compounds, so option A can be eliminated. Since NAD

must actually be reduced by the capture of an electron, it lacks one to be

the originator, so option B is out as well. Chlorophyll only captures the

energy of a photon, which it then transfers to an electron. The electrons

come from a molecule of water in the process of photolysis, which also

releases hydrogen ions and oxygen gas when the electrons are harvested.

11. B. The purpose of this question is to determine the extent of your

understanding of membrane structure.

Chlorophyll in plants is contained within the thylakoid membranes of

the chloroplast, not the cell membrane, so option A is wrong. Both plant

and mammalian cell membranes contain ion channels and are attached

to the cytoskeletons, so options C and D are also wrong. Plants cells are

surrounded with a cell wall of cellulose, not a cell membrane of cel-

lulose, so option E is wrong as well.

12. B. This question is based on one of the laboratory exercises and is

designed to see if you understand the differences between sexual and

asexual reproduction in fungi.

The ascomycetes produce their ascospores in the bag- like structures

called asci through the process of meiosis. This is why there is always an

even number of ascospores. Sporangiospores are asexually produced by

mitosis and thus are always present in larger, unpredictable numbers.

13. E. This question is designed to determine if you understand basic

mechanisms of evolution.

Analogous structures are those that have similar function but which

evolved separately within different groups of organisms. Homologous

structures are those that are evolutionarily linked by origin but may or

may not be similar in function. Options A, B, C, and D all posit analo-

gous structures.

14. C. The purpose of this question is to determine your understand-

ing of one of the mechanisms of the immune system.

The immune system has both innate and acquired mechanisms. The

purpose of immunizations against viral infections, bacterial infections,

AP BIology54

and toxins is to activate one of the acquired mechanisms called memory.

When a person gets an infection, his or her body responds with the

production of protective antibodies. These antibodies are produced by

a subset of agranulocytes known as plasma cells, which are terminally

differentiated B- lymphocytes, or B- cells. Some of these B- cells, before

becoming plasma cells, will instead enter an arrested interphase and

stick around in the circulatory and lymph systems for decades, waiting

to become reactivated if required at a later time. Neurons have no direct

role in immunity, so option B can be discarded. While both platelets

and erythrocytes are found in the blood, they are associated with blood

clot formation and oxygen transport respectively, not immunity against

viruses. Neutrophils are very short- lived cells only associated with

phagocytosis, so option D is also out.

15. D. Questions 15– 17 are designed to determine your understanding

of the major catabolic processes of cells.

First, you need to identify the processes described in the diagram.

Process (I) converts glucose to acetyl- CoA. The major portion of this

is glycolysis. Process (II) takes acetyl- CoA and harvests the high- energy

electrons contained in the cell, packing them onto NADH, and then

expels the waste gas carbon dioxide. This is the TCA cycle. Process (III),

if you have not guessed yet, is the electron- transport chain that takes

electrons and converts their energy into ATP. Neither options B nor E

are catabolic, and they can be discarded.

16. C. Identifying this as the electron- transport system is key to answer-

ing this question. Knowing that, what gas is required to act as the termi-

nal electron acceptor? Neither chlorine, nitrogen, potassium, nor sodium.

17. D. To answer this question you need to remember the processes

involved. Both the electron- transport system (III) and the TCA cycle (II)

take place entirely within the mitochondrion. Glycolysis, the part that

goes from glucose to pyruvate, takes place within the cytoplasm, while

pyruvate conversion to acetyl- CoA takes place within the mitochondrion.

18. C. This question helps to ascertain your understanding of the phy-

logenetic classification of animals.


The simplest animals have no body cavities (acoelomates). These

include the poriferans, cnidarians, and platyhelminthes, thus eliminat-

ing options A and E. At the other extreme, the most advanced animals

have a coelom developed from a digestive tube (deuterostomes). These

include the echinoderms and the chordates, thus eliminating option B.

The nematodes are pseudocoelomates, thus eliminating option D.

19. A. This is another question to ascertain your understanding of phy-

logenetic relationships.

This is actually a true discovery. Of the eight regions that were recov-

erable, three most closely identified the dinosaur protein as bird- like.

Guess what that means: dinosaur probably tasted like chicken!

20. A. This question is designed to apply your understanding of biomo-

lecular structure to derive relative sizes.

Carbon dioxide is a gas that consists of three atoms with a molecular

weight of 76 Da. A phospholipid consists of a phosphate group attached

to a glycerol head that is connected to two fatty acid chains of about 20

carbon atoms in length each for a total less than 30 carbon atoms; its

molecular weight is 700– 800 Da. DNA polymerase is an enzyme com-

plex of between 50 and 150 kDa, or about 2S. The bacterial ribosome

consists of about 55 proteins plus rRNA strands that total 70S.

21. E. This question tests your ability to interpret data presented in

a graph.

A type I survivorship pattern is characteristic of large animals that

have few offspring in which they invest significant nurturing to ensure

longer lives. A type II survivorship pattern is characteristic of small ani-

mals that have moderate numbers of offspring in which they invest some

nurturing and that present a fairly even death rate throughout their lives.

A type III survivorship pattern is characteristic of very small animals

which produce huge numbers of offspring and invest no nurturing. This

produces extremely high death rates early on. The curve presented is

not one normally seen in a textbook, which usually presents the data as

mortality vs. maximum life span. This data simply shows a level death

rate, one that is fairly constant throughout the lifespan of the rodent.

AP BIology56

22. E. Questions 22 and 23 are included to determine your under-

standing of a basic genetic- engineering procedure.

Restriction endonucleases are produced by bacteria as a defense

mechanism against bacteriophage attack. Each endonuclease cuts double-

stranded DNA at a specific sequence and usually cuts unevenly across

the two strands, producing “sticky ends.” These restriction enzymes are

identified by the first letter of the genus name and the first two letters

of the species name of the bacterium that produced the enzyme. These

three letters are followed by an additional unique identifier. Two such

common enzymes are Eco RI (for E. coli restriction enzyme I) and Hin

dIII (Haemophilus influenzae restriction enzyme dIII). Options A and C

can be eliminated because they are polymerases, as can B, which is also

called RNA- dependent DNA polymerase. Options D can be eliminated

because it is a ligase, not an endonuclease.

23. D. When DNA from any source is cut with the same restriction

enzyme, producing restriction fragments, then every sticky end is equally

complementary with every other sticky end. However, the cut in the

DNA backbone, produced by the endonuclease, is still present. This nick

in each strand must be rejoined, or ligated, by the enzyme responsible

for DNA repair caused by UV irradiation, DNA ligase.

24. A. This question is included to determine your understanding of

interspecies competition.

Sturgeons are primarily bottom feeders, while minnows prefer more

open water near shorelines. Since they rarely interrelate, their interac-

tion is considered neutral; therefore option B can be discarded. The same

is true for option C. A rock is nonliving and simply serves as a substrate

for lichen attachment, and therefore no interspecies competition is

involved. Thus option D can be eliminated. While mistletoe is a parasite

on trees, this is not considered competition, as different resources are

involved (the tree draws from the soil, the mistletoe from the tree).

Woodpeckers are able to retrieve food from under the surface of wood,

while wrens are restricted to the surface only. The woodpecker is thus

able to exploit a food supply that the wren cannot.


25. C. This question is designed to allow you to indicate your level of

understanding of organization within a cell.

The nucleus is the location of the cell’s genome and the site of ribo-

somal subunit assembly, but not translation. Option B is thus elimi-

nated. The Golgi complex, also called the Golgi apparatus, is the site of

protein modification and packaging, but not protein synthesis. Option D

is thus eliminated. No translation takes place on the cell membrane, so

option E is also eliminated. While the endoplasmic reticulum (ER) may

have newly synthesized proteins extruded into its lumen, no ribosomes

are present within the ER, although they sometimes are on its cytosolic

surface. Ribosomes are only present in the cytosol, which is the site of

all protein synthesis.

26. B. This question is here to determine your understanding of com-

mon inheritance patterns.

X- linked recessive means that the trait will be expressed in women

only if they are homozygous for the trait, but males will always express

the trait if they have it because they will be hemizygous. The mother is

heterozygous, because her father was afflicted and yet she is not. The

father is hemizygous for the trait, because he is afflicted. Constructing a

Punnett square shows that there is a 50 percent chance of a child’s being

female, and these possibilities are not to be considered. Of the two pos-

sible male combinations one is afflicted, one is unafflicted. Therefore,

of possible male offspring, one-half will be afflicted and one-half will

be unafflicted.

27. D. The purpose of this question is to determine the extent of your

understanding of free energy in a cell.

Metabolism consists of all cellular processes used to process energy.

Catabolism consists of those reactions and pathways that break down large

molecules into smaller ones and release the energy therein contained.

Anabolism consists of those reactions and pathways that construct larger

molecules from smaller ones and store energy in the resulting additional

chemical bonds. Fatty acids are the major energy- containing substruc-

tures of triglycerides (lipids) that are broken down by b- oxidation to

AP BIology58

produce acetyl- CoA, which, in turn, feeds into the TCA cycle, which

produces 11 ATP molecules from every acetyl- CoA fed in.

28. D. Questions 28– 30 are based on one of the lab exercises and

require you to interpret the diagram based on your understanding of leaf

structure and function.

There are two aspects to this question: which cells respire and which

cells photosynthesize. The ground tissue cells in the leaf do both. Dermal

tissue cells lack chloroplasts to allow light to penetrate into the leaf, so

they only respire. Vascular tissue cells do neither. The structure desig-

nated as (I) is cuticle and nonliving, so any answer including it can be

discarded; out go options A and B. The structures designated as (V) and

(VI) are dermal, so options C and E can be ignored.

29. B. Dermal tissues or structures of the leaf are those associated with

protection. This includes the cuticle (I), dermal cells (V), and stomata (VI).

30. E. Transpiration is the process during which water is allowed to

evaporate within the leaf and escape through the stomata. By acting like

a sink, the vanishing water draws more water (and nutrients) up from

the roots through the vascular system. Therefore both stomata (VI) and

xylem (included in IV) are involved in transpiration.

31. A. This question is designed to get you to deduce the mechanisms

involved in allergic reactions based on your understanding of the immune

system and cell- signaling pathways.

Mast cells contain membrane surface receptors that attach to a cer-

tain class of antibody called IgE. The presence of these allergen- specific

antibodies on the receptors sensitizes the mast cells to those allergens.

When the allergen appears, it attaches to the antibody, which changes

conformation and initiates a signal- transduction pathway. This results in

an influx of Ca2+ ions into the cell, which causes the release of histamine

into the tissue.

32. B. This question determines if you understand the mechanism of

translation and can apply that understanding.

In order for proteins to be produced, the coding must be brought to

the ribosome in the form of mRNA, even that required for viral proteins.


If the rabies virus genome is complementary to the mRNA, it must have

a mechanism for converting its complementary code into mRNA code.

Retroviruses do this by converting their RNA genome into DNA with

reverse transcriptase, then back into mRNA by normal transcription.

However, reverse transcriptase is not one of your choices. Options C and

E have nothing to do with transcription so can be removed as choices.

The enzymes in options A and D are directly involved in replication and

transcription directly, but do not solve the problem. This type of virus

comes packaged with its own RNA- dependent RNA polymerase inside

its capsid so it can immediately transcribe its complementary code into

mRNA as soon as it uncoats in the host cell.

33. C. Questions 33– 35 determine your abilities to interpret a graph

and apply your interpretation in order to explain the physiologic feed-

back mechanism.

This first question is rather easy. You know that diabetes is the con-

dition of having elevated blood glucose levels. The normal values are

provided in the question, so you know diabetes would be a condition

when the value exceeds 150 mg/dl. The only curve that reflects a normal

resting value is III.

34. D. Diabetic shock occurs when excess insulin is in circulation. This

excess causes too much glucose to be moved from the blood into insulin-

sensitive cells. When the glucose level drops to the point where no cell

can get any glucose, then you are in insulin shock. While curves (II)

and (IV) both indicate excess insulin was administered, only the latter

remains critically low.

35. C. The curve designated (II) indicates an initial diabetic condition,

which means an insulin injection was administered at the time indicated

by the arrow. The glucose level plummeted, sending the person into

insulin shock. However, insulin shock can be treated by either removing

insulin from the blood (and it’s too late for that) or by increasing the

glucose level back up to the normal range. This is commonly done by

ingesting glucose in the form of a candy bar.

36. A. This is a question that simply requires shifting decimal places

AP BIology60

and understanding metric measurements. It also demonstrates why blood

values are often expressed per deciliter, as opposed to the more com-

monly used per milliliter: because it requires fewer decimals. Converting

milligrams to grams shifts the decimal three positions to the left, and

converting deciliters to liters shifts the decimal one position to the right.

37. E. Questions 37 and 38 are designed to ascertain your under-

standing of the genetics and disease mechanism of a fairly common

perinatal problem.

Erythroblastosis fetalis is caused when an Rh– mother becomes preg-

nant with a second or subsequent Rh+ child. During the birth of an

Rh+ child, an Rh– mother gets exposed to the Rh factor antigens on the

infant’s red blood cells. Consequently, she starts to produce antibodies

against the Rh factor. If she later becomes pregnant with an Rh– child,

the preformed antibodies will not damage the baby’s factorless red blood

cells. However, if she later becomes pregnant with an Rh+ child, her

antibodies will cross the placenta and attack the red blood cells of her

child. If severe enough, this can result in the death of the fetus.

38. E. Inheritance of the Rh blood group antigen is by simple domi-

nance. A homozygous recessive mother lacking the antigen has to have

been exposed to the antigen during an earlier birth. The problem only

arises when a subsequent child inherits at least one copy of the gene that

produces the antigen.

39. C. This question is designed to see how well you understand the

relationships between two very common carbohydrates.

Glucose is central to the cellular metabolic pathways of most cells

and is manufactured within plants. However, it is primarily sucrose that

serves as the carbohydrate that transports energy in the sap of plants.

Combining a molecule of glucose with a molecule of fructose produces

a molecule of the disaccharide sucrose.

40. C. Questions 40– 42 are included to determine your grasp of the

mechanisms of competition with an ecosystem.

By looking at the graph you can see that only one population, desig-

nated I, was present at the beginning of the collection of data. The other


two appear to have been introduced into the ecosystem, either acciden-

tally or intentionally. When comparing the curves designated I and II,

you can see an inverse relationship, i.e., as the numbers of II increase, the

numbers of I decrease, and vice versa. This suggests either that they are

direct competitors or that population II uses population I as either prey

or hosts. This latter is made less likely because when population I drops

to zero, population II continues to rise. Thus, it appears that population II

is a successful invasive species that completely outcompetes population I.

41. B. Population III arrives on the scene around week 45 and initially

prospers. At the same time, the invasive species numbers start to drop

off while population I starts to recover, making it appear that the drop

in population II was due to the introduction of population III. If that was

true, then it appears that population III was not capable of succeeding in

this new environment, and its numbers started to drop off, restoring the

growth of population II at the expense of population I. It appears likely

that population III was a natural predator of population II but could not

survive in this new environment.

42. C. If the understanding presented in the answer to question 41 is

correct, and it certainly appears to be the most reasonable of the choices,

then population I died out as a result of being outcompeted by popula-

tion II. That the loss of population I was not due to either predation or

parasitism by population II is indicated by the latter still increasing in

numbers following the disappearance of population I. This most likely

means that population III failed independently of either I or II.

43. D. This question is included to determine your level of understand-

ing of cell membrane function.

Essentially, the basic structure of eukaryotic and prokaryotic cell mem-

branes is the same. They are the same thickness, eliminating option A. The

chemistry doesn’t change from type to type, eliminating options B and C.

While option E is tempting, the role of the prokaryotic cell membrane is

more general and broader than that of the more specialized eukaryote.

44. A. This question allows you to demonstrate your ability to discern

selective pressures in evolutionary processes.

AP BIology62

Venus flytraps live naturally only in areas with nitrogen- poor soils.

They grow slowly and within a rather limited range, so option E is out.

If the mechanisms that produced speciation were based on their speed,

as in option B, or the richness of their food, as in option C, or a wider

variation in nutrient forms, as in option D, then they would have a

much wider range and would be found in much larger numbers. It is

only because of their ability to survive in a rather hostile environment

that they exist at all.

45. A. The purpose of this question is to determine if you understand

and can identify symbiotic relationships.

Truffles, morels, and mushrooms are all sexual reproductive structures

of fungi, so options C, D, and E are eliminated immediately. Lichens are

fungal symbiotes, but with algae, not plants.

46. E. This question seeks to let you identify historical experiments

leading to the discovery that DNA contained genetic information.

Actually, the question is almost a giveaway. Option A identifies

phages, as in bacteriophages, as in bacteria, which don’t have nuclei. So

option A can be discarded, as can B and C. Option D involves restriction

enzymes, but these all come from bacteria, as they serve to protect them

from phages. Even if you don’t remember Acetabularia as a large nucle-

ated cell, it is the only remaining choice.

47. C. This question allows you to demonstrate your understanding of

ABO blood group inheritance patterns.

Blood types A and B are determined by the presence of the A or

B allele, respectively. Blood type O is detected if neither A nor B is

present. A person with type A or B blood may be homozygous (AA or

BB) or heterozygous (AO or BO). Option A can be discarded because

type AB blood is not a certainty, with only a 25 percent probability.

Option B can be discarded because the explanation is wrong. Option E

can be discarded because the answer is irrelevant. While the reasoning

for option D is correct, the answer is wrong, so D is also thrown out. If

the parents are heterozygous, then any child has a 25 percent chance of

being homozygous for blood type O (OO).


48. B. The purpose of this question is to determine your level of under-

standing of basic evolutionary mechanisms.

Human DNA polymerase has an error rate of about one in 1 × 10– 7

bases. HIV replicates with an error rate of one in 2 × 10– 4 bases, or about

1000 times greater. This means that, on average, one of every two viruses

made will be mutations. While the vast majority will be noninfective or

nonproductive, the mutation rate ensures that the viral population will

constantly change. Option E is wrong because, while the explanation is

true, the numbers produced greatly speed the process. Options C and D

have the relationships reversed, i.e., the larger the genome, the greater,

not the less, the likelihood of error. Option A is wrong because, on aver-

age, 50 percent of all viruses produced will contain mutations.

49. D. Questions 49– 51 test your ability to identify the structures and

function of the urinary system.

The round globular structure (glomerulus) connected to the long

inverted loop (loop of Henle) should provide the key to identifying this

as a nephron, the functional unit of the kidney. The parallel tubes on

the left are renal arterioles and venules, and the tube on the right is the

collecting tubule.

50. A. If you know that vasopressin is another name for antidiuretic

hormone, then this helps, although even if you didn’t know this you

could narrow down the choices. While the liver does secrete a few hor-

mones, they are associated with growth and metabolism, not with the

kidney, so you can ignore option B. The hormones produced by the

hypothalamus and the adrenals are primarily developmental or growth

hormones, so options C and E can be ignored as well. To distinguish

between the anterior and posterior pituitary, just remember that “the

posterior deals with the posterior” (urine, uterine contractions).

51. A. Even if you didn’t recognize the nephron from the image, you

can eliminate some choices here. You can see this is not the pituitary,

so you can eliminate option B. Because of the branching, you can elim-

inate the axons of neurons from being present, so option C can also

be removed from consideration. While the glomerulus resembles an

AP BIology64

alveolus, the latter are always in clusters, so option D is out. The liver

has no such organized structure, so option E is similarly gone.

52. B. This question delves into your ability to identify evidence for

the endosymbiosis theory as a part of evolution.

The 70S bacterial ribosome consists of two subunits identified as

30S and 50S. The smaller subunit contains a strand of rRNA identified

as 16S. The rRNA is made by transcription in almost the exact same

way as mRNA, except the “gene” being transcribed is identified as 16S

rDNA. In humans, the 80S functional ribosomal subunit equivalents

found in the cytosol are 40S and 60S, with the rRNA in the smaller

subunit being 18S. However, the bacteria- sized mitochondria contain

their own self-replicating DNA and their own 16S rRNA as part of their

70S ribosome.

53. E. This question ascertains your understanding of and your ability

to identify the membrane transport mechanisms used by cells.

The question consists of two parts that need to be in agreement. If

you remember that this pump requires the expense of ATP, and thus the

pump is active transport, then you can eliminate options A, B, and D. If

you remember that the ions flow in opposite directions simultaneously,

and thus the pump is antiport, then you can eliminate options A, C, and

D. Combining the two leaves only one option remaining.

54. E. This question is here to determine if you can recognize a mecha-

nism involved in evolution.

Option A can be ignored as it begs the issue: if a gene is fragmented

into repetitive parts, where did the repetitive parts come from? Option

B won’t hold water, because there are hundreds of thousands of ret-

rotransposons, while repetitive immunoglobulin sequences number in

the scores. Option C would not explain how the repetition came about

within introns. Option D is a poor choice because the concept is rather

Lamarckian, with phenotype driving genotype.

55. D. Questions 55 and 56 are designed to allow you to demonstrate

your understanding of energy transfer through an environment.

Light bathes the Earth constantly. Actually, a rather small percentage


is captured by photosynthesis, only about 1 percent. Therefore, if 1,000

kcal were available, only 10 kcal would end up in trophic level 1 bio-

mass. While you might remember the exact percentage, the important

fact to remember is that the percentage captured is a very small amount

compared to the amount available.

56. B. While question 55 dealt with the amount of energy captured

from the available sunlight by consumers, this question deals with how

much of that which was captured is passed on to consumers at higher tro-

phic levels. About 15 percent of producer biomass is converted to trophic

level 2; about 10 percent of that would be passed on to level 3. Thus, of

the original 10 kcal, 15 percent (or 1.5 kcal) would be transferred to level

2, and 10 percent of that (or 0.15 kcal) would be transferred to level 3,

for a total 16.5 percent transfer to higher trophic levels.

57. C. This question is designed to assay your level of understanding of

the role of hormones during pregnancy and your ability to reason from

two observations.

When a woman becomes pregnant, the production of LH and FSH

are suppressed, as they are associated with menstruation; thus option A

is removed from consideration. When a pregnancy begins, the process

is supported by the production of b- HCG, progesterone, and estrogen;

thus options B and D are no longer to be considered. Oxytocin is asso-

ciated with perinatal and postnatal situations: the induction of labor

with uterine contractions and lactation. Thus option E is eliminated.

Rheumatoid arthritis is an autoimmune reaction caused by the infiltra-

tion of immune active cells into joints which causes damage. Thus, a

suppression of the immune system will reduce the attacks on the joints.

Genital warts are caused by infection with HPV, which the immune

system eventually suppresses, but if the immune system is suppressed,

the warts can reappear. When a woman becomes pregnant, her immune

response is slightly suppressed to prevent her from immunologically

rejecting the fetus.

58. A. This question determines if you can recognize the possible con-

sequences of a genetic abnormality occurring during meiosis.

AP BIology66

During meiosis, sister chromatids in synapsis, as indicated in the top

of the image, would undergo two sequential separation events. The cor-

rect process would be indicated if one chromosome of the original tetrad

remained in each nucleus. If this was mitosis, then only one division

would have been involved, so ignore option E. Crossing over occurs

before the reduction divisions shown, so ignore option B. Pleiotropy

deals with gene expression, not chromosomal separation, so option C

is also to be ignored. And it is a good thing that you worked on your

vocabulary so you immediately recognized that, although one of the

resulting cells may have lacked an “X” chromosome, this drawing has

nothing to do with hatred of women.

59. E. Nondisjunction is when the chromosomes separate unequally

during meiosis, a condition that would then be passed on to every cell

of a zygote that was produced by affected gametes. Knowing this allows

you to eliminate options A, B, and D. While you may not be able to

completely eliminate option C with what you know, the last choice

clearly indicates an extra “X” chromosome.

60. B. This question is designed to see if you are capable of applying

your understanding of the immune system to identify common elements.

During gestation, fingers and toes are initially all connected together

by a layer of skin. However, at some point the cells joining these struc-

tures receive a signal to undergo apoptosis, or genetically programmed

cell death. The loss of these cells then allows the independent func-

tioning of these digits. All cells have the genes for the expression of

apoptosis. Cancer cells are allowed to proliferate only because of a loss

of apoptosis expression.

61. D. The purpose of this question is to allow you to demonstrate

your understanding of cell- surface receptors in viral transmission.

TMV is a rather common infector of many plant species, and people

who smoke or use tobacco products are often excluded from working

in plant nurseries for this reason. However, as with bacteriophages, just

because they are there doesn’t mean you are doomed. TMV lacks the

ability to attach to and infect human cells.


62. C. This question is designed to get you to use your understanding

of plant anatomy to group phylogenetically.

What is pictured is a pine cone from a conifer, which is classified

within the gymnosperms, or vascular plants with naked seeds. Mosses

are nonvascular plants, so option D is out. Ferns and lycophytes are vas-

cular but do not produce seeds, so you can also eliminate options A and

D from consideration. Monocots are a subset of angiosperms, or flower-

ing plants, so they can also be eliminated. Of course, it would help if you

remembered that gingkos are also gymnosperms.

63. D. This is a question designed to get you to analyze probable func-

tion based on known structure.

Enzymes are able to act as biologic catalysts because of the near infi-

nite variety of shapes which they are able to assume. Fatty acids are long

linear polymers of carbon and vary only occasionally as double bonds

appear, so out goes option B. Cellulose is a similar long polymer, but of

glucose, not just carbon, so option C is also out. DNA is very regular in

its structure as well, always in a long antiparallel double helix, so option

A is removed. ATP is too small, so you also get to eliminate option E.

Only RNA has the structural flexibility to become catalytic, in which

case it is known as ribozyme.

64. B. This question is included to get you to express your understand-

ing of energy flow within a cell.

Electrons are transported to the electron system within the mitochon-

drion by molecules of NADH. NADH picks up most of its electrons

from the TCA cycle as it processes molecules of acetyl- CoA, and the

rest of them from the reactions that produce pyruvate. The electrons in

both acetyl- CoA and pyruvate come with them as they are processed

and released from a molecule of glucose. While cellulose is polyglucose,

it is not used as an energy source by plants, only as a structural polymer.

Therefore option D can be removed from consideration. ATP is an energy

carrier, not an electron source, so ignore option C as well. The electron

transport system discards its energy- drained electrons onto a molecule of

oxygen to produce water, so options A and E are not correct either.

AP BIology68

65. E. This one deals with your understanding of energy flow in an

ecosystem not dependent on light.

Light cannot penetrate to these deep-sea vents so the populations

that arise are based on chemical energy sources. Although the water

temperature reaches several hundred degrees Celsius, it cannot boil

because of the high pressure. At that temperature, however, the water

easily dissolves the minerals from the crust, which immediately forms

large clouds of precipitate that produce the classic “black smoker” struc-

ture. Bacteria are able to harvest electrons from these precipitates and

grow to large numbers. They, in turn, are fed upon by protozoan and

larger planktonic organisms, which then support large animals such as

tube-worms and large crabs. Organisms cannot feed on heat so option

A is not true. Ecosystems always form from cross- dependent organisms

within a food web, so options B and D are not correct. Since no light can

penetrate to these depths there are no plants, so ignore option C.

66. A. This question centers on your ability to identify a basic evolu-

tionary selection pressure.

Under completely equal conditions a phenotype that dies off three

times faster than another phenotype would quickly disappear from the

gene pool, so there must be some selective advantage to having bright

feathers. Options B, C, and D are probably true but do not explain the

survival of the bright plumage. Option E would actually probably have

the opposite effect of what is being observed. If females mated three

times more often with colorful males than with dull males, then the

bright phenotype would most likely be able to survive within the popu-

lation in spite of the increased predation, simply because they would be

more reproductively successful.

67. E. Questions 67– 69 help you demonstrate your understanding of

the structure and function of a flower.

Meiosis is the process in which gametes are formed. In the case of

flowers, this means the formation of eggs and sperm. The eggs are pro-

duced in ovules and the sperm, as packaged in pollen, are produced in

the anther.


68. C. Flowers are pollinated by different means. Many are fertilized

by animals that are attracted to the flower by color or scent. Others,

which do not bother to invest their energy in producing large, bright,

or attractive flowers, instead invest in producing prodigious amounts of

pollen to be cast to the winds. To do this, they only need lots of pollen-

producing stamens as well as large stamens.

69. D. Simple fruit, like peaches, are formed from a single ovary.

Multiple fruit, such as pineapples, are formed from the fused ovaries

of multiple flowers. Accessory fruit, such as seen in the strawberry, are

formed from tissues not derived from the ovary. Aggregate fruit, such

as seen in blackberries and apples, are formed from multiple carpels in a

single flower.

70. A. This question is here to determine your ability to discriminate

the possible from the probable when it comes to climate change.

If global mean temperatures rise, then the oceans become warmer

and, as a result, are able to dissolve less oxygen, not more. This

eliminates option C from consideration. While a warming planet will

probably accelerate desertification, it will also open land previously

uninhabitable due to an excessively cold climate. This means the

amount of land available for cultivation will remain about the same,

so option D is out. Warm climes do not necessarily mean reduced ani-

mal diversity, as can be seen in the tropical rain forests, so option E

is removed as well. While a change in migration patterns is hard to

predict, it is more likely that flight lengths will increase, not decrease.

71. B. This question is here to allow you to use reasoning skills to iden-

tify a taxonomic group.

The building of the Aswan Dam reduced the river flow and allowed

a certain snail to migrate into the still waters. This snail serves as a host

for a trematode called Schistosoma. After infecting the snail, the fluke

changes form into a cercaria, which then infects humans wading in the

water by burrowing into the skin and migrating into the liver. There it

matures, mates, and lays eggs that are expelled from the body in feces or

urine, depending on the species. The key to this question is the term “life

AP BIology70

cycle.” Bacteria do not have life cycles, only binary fission, so exclude

option D from your consideration. Similarly, arachnids do not metamor-

phose, so option A can be excluded. Since you had a lab on the physiol-

ogy of the circulatory system using Daphnia as a model, and of course

you remember that water fleas are members of the order Cladocera,

then you can exclude option E as well. Option C, which includes squid

and octopi, can obviously be removed. Helminthes include tapeworms

(cestodes), roundworms (nematodes), and flukes (trematodes).

72. A. This question is designed to ascertain your level of understand-

ing of cell communications.

Steroidal hormones are generally less specific than protein hormones,

and the immune system relies heavily on cell- surface receptors. Option

C can be discarded. Quorum sensing is bacterial, so option B is out.

Neurotransmitter release actually precludes cell- to- cell contact, so

option D is out. A change in ion flow through a membrane is a result of

cell signaling, not a form of it, so out goes option E as well.

73. E. Questions 73– 75 evaluate your ability to deduce the result of a

mutation based on your understanding of the regulation of gene expres-

sion and molecular biology techniques.

The lac operon is the archetype of the inducible bacterial gene. Under

normal conditions, lactose is not available to bacterial cells as a food

source because it is a rarely encountered carbohydrate. Consequently,

the operon that codes for the production of enzymes that metabolize

lactose is repressed. Under conditions of starvation and the appearance

of lactose, cells that possess this gene will be capable of using the newly

available lactose. The sugar enters the cell and binds to the repressor at

the operator region of the operon. The repressor then releases its hold on

the DNA, de- repressing (inducing) the operon. RNA polymerase then

binds to the promoter region, passes through the operator region, and

then transcribes the three structural genes that code for enzymes needed

to catabolize lactose. If the ampR gene was inserted into the sequence,

the polycistronic mRNA would be translated into the required enzymes

for lactose use, as well as the β- lactamase enzyme coded by ampR.


However, the ability to resist the effects of ampicillin would only be seen

when the operon was induced by the presence of lactose. Thus options A

and C are incorrect and can be discarded. Option B makes reference to

tryptophan, but that is the archetypal repressible gene, not the inducible

one referred to here, so it can be discarded. Proofreading enzymes would

only work if there were a template available, and that would require a

diploid condition, so option D is out.

74. B. The evolution of a specific gene by random mutations, even

under tremendous selection pressure, would take an exceedingly lengthy

period of time, so option D is removed from consideration. The lac

operon and the tryptophan operon have nothing in common, so option

E is removed as well. If the genome of the bacterium was removed and

not replaced, how could the cell survive? Option C is also eliminated.

And, as tempting as option A might be, the bacterial genome is huge in

molecular terms and extremely difficult to manipulate as a whole, which

is why working with much smaller plasmids is so successful.

75. E. Three of these options are fairly easy to deal with. Option D is an

enzyme encoded within the lac operon, but is not a necessary part of the

exercise. Options A and C are both DNA polymerases, and the exercise

deals with transcription, not replication. Option B presents a transcrip-

tion enzyme, but not the one needed in this exercise. The enzyme that is

necessary here is the one that allows researchers to cut- and- paste DNA.

76. A. This question is here to assess your understanding of the func-

tions of various plant hormones.

IAA is an auxin that is responsible for promoting cell elongation.

Cytokinin stimulates cell division and retards senescence. Abscisic acid

induces dormancy. While you may want to avoid these because it is

gibberellin (option C) that you specifically are dealing with, the one you

really want to stay away from is 2,4- D, because, if you used it, it would

kill all your test plants. It is a herbicide.

77. C. The purpose of this question is to determine your level of under-

standing of the flow of oxygen through the body.

Both oxygen and carbon dioxide flow freely through cell membranes

AP BIology72

in the direction of the concentration gradient. However, the direction of

the gradient throughout the body is reversed for the two gasses. While

the oxygen level is highest in the atmosphere, thus in the lung alveoli,

it is lowest in the tissues. This means the correct sequence for O2 is

(alveoli)— pulmonary veins— left atrium— (left ventricle)— (aorta)—

(arteries)— (arterioles)— tissue capillaries— (venules)— (veins)— (vena

cava)— (right atrium)— right ventricle— pulmonary arteries— (back to

the alveoli). This question, however, is based on the reverse process of

ridding the body of CO2, which is highest in the tissues and lowest in the

pulmonary veins.

78. C. This question is included to determine your level of understand-

ing of biological molecule structure.

There are four broad categories of biological molecules: carbo-

hydrates, lipids, proteins, and nucleic acids. Almost everything within

cells is based on these or hybrid structures. Phenylalanine is an amino

acid, and proteins are polymers of amino acids. A person with phenylke-

tonuria must minimize his or her intake of proteins and does not need to

exclude any of the other groups.

79. E. Questions 79 and 80 test your ability to recognize simple inheri-

tance patterns from what appears to be complex data but is not really

that complex.

Each HLA antigen has numerous alleles, all codominant. Each sibling

received one of their mother’s chromosomes and one of their father’s,

each of which has the HLA- A, HLA- B, HLA- C, and HLA- DR loci.

Because these genes are closely linked, they are commonly inherited

together. So the question is simply “What is the probability of a sibling’s

having the same gene as another sibling?” The answer is 25 percent. The

question asks, however, the probability of a fourth sibling’s matching

any one of the previous three, and for that you simply add up the prob-

ability of matching the first, or second, or third.

80. E. You will notice that when you compare the results for siblings

1 and 2, there are four results for the antigens present for HLA- B, - C,

and - DR, but only three for HLA- A. While this might be due to gene


deletion (option A) or lab error (option C), these are not likely and are

not necessarily explanations. Option B can be discarded, because this

would have caused the loss of all four antigens coded on the missing

chromosome, not just one. And option D can be discarded, because if it

affected one antigen, it would have affected all. The simplest answer is

that both parents have an allele for HLA- A1, and sibling 3 just happens

to be homozygous for that allele.

81. B. This question focuses on your understanding of the physiologi-

cal response to an infection involving an immune- suppressing virus.

The incubation period for HIV is about two weeks. After this two-

week period, someone infected exhibits flu- like symptoms for a week to

10 days, after which the symptoms subside. The virus is replicating dur-

ing this time and reaches a detectable level, as measured by viral antigen

level in the blood, at about the same time. Antibodies specific for HIV are

manufactured simultaneously but, because they are neutralizing viruses,

they do not reach detectable levels until about six weeks post-infection.

As the antibody levels rise, they are successful in clearing the blood of

any detectable viruses. The infection stays at this asymptomatic stage for

years. Eventually the virus starts to reduce the number of CD4+ T- helper

cells within the blood. As the antibody levels continue to decrease, the

virus starts to replicate at an uncontrollable rate, reducing the CD4+ cell

count to below 200 cells/mm3. When this threshold is crossed, the per-

son is classified as having AIDS. While treatment with antiretrovirals can

keep the virus suppressed, which means infection with HIV is treatable,

the infection remains incurable, with a 100 percent fatality rate.

82. D. This question is included to determine your level of understand-

ing of the structure and function of the chloroplast.

The energy of light is used in the process of photolysis as described

in the question. The electrons then flow through the electron- transport

mechanism embedded in the thylakoid membrane within the chloro-

plast. Within two complexes, photosystems I and II, energy from addi-

tional photons is harvested and transferred to the electrons, which are

eventually used to reduce NADP+ + H+ to NADPH. The H+ released

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from photolysis then accumulates within the thylakoid space to drive

the synthesis of ATP by ATP synthase.

83. C. This question helps determine your level of understanding of a

very basic evolutionary concept.

The term “fitness” simply means the ability of an organism to sur-

vive and pass on its genes to its descendents. Lifespan is a rather poor

measure, because some very successful organisms that have been around

for millions of years have lifespans measured in hours, so option A is a

poor choice. Neither the number of chromosomes (bacteria lack them,

and yet they are very successful) nor the number of alleles has any value

in rating success, so out also go options B and D. Many animals once

covered the planet but are no longer with us, so option E is also a poor

choice. The only true way to measure fitness is by the ability to hang in

there during the millennia, which requires reproductive success.

84. A. This question attempts to clarify your understanding of the

molecular scales involved in measuring the cell.

Water surrounds everything associated with life, including enzymes,

so it is the smallest. Therefore you can discard options B, D, and E

because they do not list it first. Bacteria contain hundreds of different

enzymes, so an enzyme must be smaller than a bacterium, eliminating

option C. And yes, a bacterium is generally smaller than a chromosome.

85. D. This question measures your ability to apply what you know

about enzyme activity and your understanding of the digestive and circu-

latory systems to the interpretation of data presented in graphical form.

The curve designated I indicates the enzyme in question that has an

optimal enzyme activity at a pH of around 2. Now, where in the human

body do you find a pH that low? Correct, in the stomach. So enzyme I

is an enzyme secreted in the stomach. What about the curve designated

III with an optimal pH of about 8? Where do you find a pH that high?

Right again, in the duodenum, right after the release of bile from the gall

bladder. So enzyme III is a digestive enzyme secreted by the liver. While

curve II would fit just about any human enzyme having an optimal activ-

ity near physiological pH (7.4), that is especially true in the blood. If


you remember that pepsin is secreted as pepsinogen in the stomach, you

can eliminate options B, C, and E. If you remember that chymotrypsin

(the “chyme” is a clue) is associated with protein digestion in the small

intestine, then you can eliminate options A, B, and E. There is only

one option not eliminated with this reasoning. You do not even need to

remember that fibrin is associated with blood- clot formation to get this

one right.

86. E. This question centers on your ability to discern the implications

of the ring based on your understanding of fungal sexual reproduction.

Toadstools and mushrooms are both the basidiospore- bearing struc-

tures formed when the subsurface hyphae of two different mating types

intersect with each other. At each location where such contact is made

the thallus arises out of the soil. A ring indicates that a colony of one

mating type started growing in the middle of another mating type and

the ring shows the intersection of the mating types.

87. A. Questions 87– 89 focus on your understanding of inheritance

and gene expression.

The easiest way to approach this question is by discarding the obvi-

ously wrong answers first. By remembering that hemophilia is a lethal

disorder, you understand that if a person bleeds to death it can hardly

increase longevity, so options B and D can be discarded. By remembering

that a person is afflicted if they inherit two recessive alleles (if female) or

one (if male) you can discard option C, because it is wrong. Disruptive

selection is when the selective pressures remove the mean and encour-

age the extremes; unfortunately, the extreme here is death, so option E

is out as well. While the usual examples of the founder effect center on

geographic separation, here the separation is social.

88. C. While all of the options are plausible, and some actually produce

known morbidity, in this case the disorder is caused by a lack of factor

VIII in the blood. It is the only cause of hemophilia that is sex linked.

89. C. The purpose of this question is to determine your level of under-

standing of a form of asexual reproduction known as parthenogenesis.

First of all, some reptiles in an all- female populations can lay eggs that

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hatch normally, so option A is incorrect. While sperm can be stored for

later use in many animal species, the phenomenon does not occur in

reptiles, so option E is out. While some animals such as mollusks and

fish can change gender, it does not happen in reptiles. Parthenogenesis

involves the ability of some cells to interrupt the meiotic process early,

leaving some diploid cells to continue as a form of zygote. This does,

however, produce a population that continues to be only female.

90. B. This question is designed to determine your understanding of

the Calvin- Benson cycle.

In the 1940s this very experiment was conducted by Melvin Calvin

and Andrew Benson using the newly developed techniques of radio-

labeling. While you might be tempted to think the technique would

not work because the boiling water would destroy the radiolabeled

materials, this is not the case, as only proteins are so labile. And, since

you remember the Calvin- Benson cycle, you remember that RuBP

fixes atmospheric CO2 to form two molecules of PGA. This is the

technique they used to figure that out because PGA was where the 14C

first showed up.

91. E. This question is included to determine if you are able to deduct

the correct answer based on your understanding of simple metabolism.

The amount of oxygen released by photosynthesis is greater than that

consumed by organic organisms. If that weren’t the case, there would be

no residual oxygen in the atmosphere for us, so you can exclude options

A and C. Even if Hydrilla had air bladders, they would contain air, not

pure oxygen, so option C can be excluded as well. While option B gets

closer to the actual mechanism, it is not fish that consume the oxygen, it

is the bacteria that come later.

92. D. Questions 92– 94 ascertain your understanding of the female

reproductive cycle and its control by hormones, and your ability to inter-

pret data in a graphical form.

The key here is to use two benchmarks on the graph: menstruation

starting on day 0 and the FSH peak on day 14. Remembering that FSH

and LH are tied together also helps a lot. Given that, you can see that


curve I represents LH, which eliminates options B, C, and E. While you

could select randomly from the last two options and possibly improve

your score, remembering that the level of progesterone rises during

the latter parts of the secretory phase of the uterine cycle gives you

the answer.

93. B. Deducing the hormones from the previous question helps to

answer this one, but remembering the major glands of the endocrine

system and the hormones they secrete helps a lot as well. In addition,

remembering that the hypothalamus controls the pituitary by neural

mechanisms and secretes nothing on its own allows you to eliminate

options that include it: A, D, and E. Since the pituitary provides most

of the coordinating hormones, and the spike around day 14 for both LH

and FSH synchronizes both the uterine and ovarian cycles, then you can

conclude that option C is incorrect.

94. A. The key here is to connect the mechanism of follicle rupture

with the more common term “ovulation.” The surge of FSH (follicle

stimulating hormone) triggers ovulation. The surge of LH (luteiniz-

ing hormone) supports the formation of the hormone- secreting corpus

luteum. The date of ovulation usually falls between the thirteenth and

sixteenth day from the onset of menses.

95. A. This question is designed to determine your level of understand-

ing of basic membrane structure.

The fluid mosaic model is based on the concept of proteins float-

ing in a hydrophobic interface between the two aqueous environments

found inside and outside the cell. It has nothing to do with motility,

eliminating options D and E, nor with the movement of substances

through the membrane, eliminating option C. While the cytoskeleton

does provide for cell structure, it has no fluid qualities. When antibod-

ies bind to membrane proteins scattered over the surface, the cell will

move and collect those bound receptors together into one region of the

cell surface. This process can be observed in a UV microscope, as the

fluorescent material all congregate into one spot, forming a cap on one

portion of the cell.

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96. D. This question focuses on your understanding of the organization

and function of the brain and the central nervous system.

Speech is a fine motor skill, so option B can be ignored. A sulcus is

a division between regions of the brain, not a functional unit of it, so

option E can be removed from consideration. The occipital lobe, at the

back of the brain, is associated with vision, so option C is out. Most of

the parietal lobe is associated with sensory interpretation, so option A

is out. The frontal lobe is best associated with memory, which includes

how to speak.

97. B. This question concentrates on your ability to extrapolate

between what you know and what you probably don’t know.

So many words, so little time. You may know that these all deal with

the means of bearing young in animals. But even if you didn’t, an under-

standing of classical root words and suffixes would be of great benefit

here. The root vivi- (as in vivacious or vivisection) means “alive.” The

root ov- (as in ovary or ovum) means “egg.” And the modifier par- means

“to hold” or “to bear.” Knowing this you can determine the answer.

Viviparity means bearing young alive. Oviparity means bearing young

in eggs. Ovoviviparity means having young in eggs that hatch internally,

producing a live birth. Placental viviparity is what humans have: live

birth from a placenta. While you may get hung up on the last term, just

remember the question: which is the oldest? In evolutionary terms, ani-

mal life developed from multicellular organisms producing fertilization

of eggs outside the body; therefore eggs came first, and it doesn’t matter

what matrotrophy means.

98. C. This question is designed to reveal the depth of your knowledge

of basic organic chemical reactions.

Carbohydrate monomers are also known as simple sugars. Sugars con-

nect together to form a glycoside bond. These bonds are formed when

– OH groups on adjacent molecules expel a molecule of water and join

together through the remaining atom of oxygen. When H2O is removed

from a molecule, it is called condensation; therefore this mechanism

is called condensation synthesis. Knowing this allows you to discard


options A, B, and D. Of the remaining two options, anabolic is an adjec-

tive meaning “building up,” the opposite of what you are looking for.

99. C. This question ascertains your recollection of the repertoire of

plant responses.

The root word photo- refers to light, so out goes option B. The root

grav- you can readily link to gravity, so out goes option E. The root

seismo- , as in seismograph, can be linked to motion, so you can also elim-

inate option A. The Greek root somn- means “sleep” and ambu- means

moving, so this term means “sleep walking,” which is not a plant option.

The Greek root thigmo- means “touch,” and trop- means “growth,” so the

word means “growth in response to what it is touching.”

100. B. This question helps to determine your level of understanding of

the organisms used in biologic research.

While we use the binomial system of identifying organisms to the genus

and species level, we also frequently use shortened forms which everyone

involved is just supposed to know. Sometimes just using the first letter

of a genus name is common, so E. coli is generally more recognizable

than the full genus name. Since this is a bacterium, it cannot be used in

learning about cellular differentiation, so option D can be ignored. You

should recognize at least the genus name Daphnia, which was used in

your lab on physiology, not on gene expression, so you can skip option E.

While Drosophila might be tempting, and has been used in learning gene

expression, its strength has been in understanding inheritance patterns,

so option A is out of the running. Pisum you probably remember from

when you studied Mendel because this was the variety of pea he used

in his research, so eliminate option C as well. Although the last one is a

mouthful, you may well recognize C. elegans, the nematode that lives for

less than three weeks and has exactly 959 cells as an adult.

Section II: Free- Response QuestionsPlease note that the answers provided here are only typical of what

might be acceptable. Many free- response questions have several possible

answers. Just remember that is important for you to answer each portion

AP BIology80

of the question reasonably. Note also that the answers need not be lengthy,

just complete. It is helpful to use the correct terminology in answering

these questions, as this is commonly rewarded in the scoring rubric. You

do not need to provide introductory or concluding paragraphs.

1. A. One example of positive control of gene expression is the induc-

ible operon in bacteria, of which the lac operon is the best example.

Under normal conditions, lactose is not available to the cell and the

operon is repressed. However, under periods of starvation when lactose

becomes available, the lactose binds to the repressor and causes it to

release its hold on the operator region of the DNA. This allows the RNA

polymerase to bind at the promoter region, pass through the operator

region, and transcribe the genes needed for the use of lactose as an energy

source. The resulting mRNA is translated into the needed proteins.

One example of negative control of gene expression is the repres-

sible operon in bacteria, of which the trp operon is the best example.

Under normal conditions of cell growth the amino acid tryptophan is

required for the manufacture of proteins. Consequently, the gene is nor-

mally expressed and the enzymes needed for tryptophan synthesis are

transcribed and the mRNA is translated. However, under conditions of

no growth, additional tryptophan is not needed. Under these conditions,

excess tryptophan binds to an inactive repressor, causing it to become

active and bind at the trp operon’s operator region. This represses the

operon and shuts down tryptophan synthesis.

B. One positive control mechanism of enzyme activity is the

requirement for a cofactor or coenzyme. Cofactors and coenzymes

bind at an enzyme’s allosteric site, causing a conformational change in

the enzyme structure. This change converts the enzyme from an inac-

tive to an active form, allowing the enzyme to catalyze the reaction in

which it participates.

One negative control mechanism of enzyme activity is the pres-

ence of either competitive or noncompetitive inhibitors. A competitive

inhibitor interferes with the ability of reactants to enter the active site


of the enzyme. The more competitive inhibitor there is in solution, the

slower the effective rate of enzyme activity. A noncompetitive inhibitor

binds at an allosteric site and changes the conformation of the enzyme,

changing it from an active to an inactive form. The more noncompeti-

tive inhibitor there is in a solution, the slower the effective rate of the

enzyme until saturation is reached, at which point additional inhibitor

produces no increasing effect.

2. A. Blood pressure in mammals is controlled by the autonomic ner-

vous system which consists of two parts, the sympathetic and the para-

sympathetic nervous systems. Most organs function only within a narrow

range of blood pressure, so both excessively high and excessively low pres-

sures must be avoided. The autonomic nervous system controls the tension

of the smooth muscles surrounding the arteries and veins. Increasing the

vascular pressure is simply a matter of relaxing or increasing the muscle

tension. Relaxation of the muscles increases the lumen diameter, increas-

ing the volume available for the blood, and drops the pressure. Increasing

the muscle tension decreases the lumen diameter, decreasing the volume

available for the blood, and raising the blood pressure.

B. Blood pressure can also be controlled by altering the fluid bal-

ance between the tissues and blood by controlling the Na+ level in the

blood, which is controlled significantly by the readsorption into the

blood from the urine within the kidneys. The more Na+ is in the blood,

the more water will be drawn into the blood to maintain proper Na+

levels, which increases blood volume and, in turn, raises blood pressure.

When blood pressure drops, the liver secretes angiotensin into the blood,

which eventually becomes a vasoconstrictor, raising blood pressure, and

which also stimulates the secretion of aldosterone. Aldosterone from the

adrenal glands promotes the readsorbtion of Na+ into the blood, also

raising blood pressure.

C. Chronic high blood pressure damages the glomerulus in the kid-

neys and causes kidney failure. A weakened area of a blood vessel may

burst if subjected to excessive blood pressure.

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3. A. One mechanism of prezygotic isolation is habitat isolation. This

occurs when two populations become separated geographically or by

habitat choice. An example of habitat isolation is species restrictions to

certain levels of a rain forest which prevent interactions between similar

organisms. Another mechanism of prezygotic isolation is behavioral iso-

lation. This occurs when mating takes place only after specific recogni-

tion of the opposite sex of the same species. A good example of this is

the flashing light patterns of fireflies, unique for each species.

B. One mechanism of postzygotic isolation is hybrid sterility. This

occurs when two species can mate but the offspring are infertile. An

example of this is the mating of a horse with a donkey, which produces

an infertile mule. Another mechanism is gamete isolation. This occurs

when the sperm of one species cannot physically fertilize the egg of

another. An example of this is when the sperm of one species cannot

survive in the reproductive structures of another.

4. A. Vascular plants have to transport water and nutrients from the

soil to the leaves sucrose from the leaves back to the rest of the plant.

The vascular tissues that accomplish this are xylem and phloem, respec-

tively. Xylem is composed of vessel elements and tracheids. Phloem is

composed of sieve- tube members and companion cells.

Water and nutrients are transported to the leaves by transpiration.

Water within the soil enters into the root, flowing between the cells

until it reaches the Casparian strip, at which point it flows within cells

until it reaches the xylem. This produces some pressure that pushes the

water up from the roots. Additionally, evaporation within the leaves,

controlled by the guard cells of the stomata, removes water from the top

of the water column, which, because of the cohesive characteristic of

water, pulls the water up as well.

Once photosynthesis has taken place and sucrose synthesized, it is

transported from the leaves to the rest of the plant by a mechanism

identified as the pressure- flow model. Sugar is loaded into the phloem

by active transport. The increase in sugar concentration then pulls water


into the phloem by osmosis. This increases the pressure within the tube,

forcing the sugar (or sap) to flow downward. At the other (sink) end of

the plant, tissues are pulling sugar out of the phloem, and water follows

out as well. This decreases the pressure, sucking the sap downward.

B. Both xylem and phloem are structurally supported by adjacent

parenchymal cells of the ground tissue system. This is required because a

break in either system would eliminate vital fluid flow through that sys-

tem. Additionally, these vessels and supporting structures are organized

into strengthened vascular bundles interspersed in the pith. A second

protective structure includes the dermal tissue system that strengthens

the plant from the outside. The woody bark protects the vascular cam-

bium from damage, and the cork cambium protects the vital tissues by

filling in fissures caused by the increase in girth during secondary growth.

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