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MCAT The Princeton Review AAMC Practice Exam 2 - Biological and Biochemical, Exams of Biochemistry

MCAT The Princeton Review AAMC Practice Exam 2 - Biological and Biochemical Foundations of Living Systems

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MCAT The Princeton Review AAMC Practice
Exam 2 - Biological and Biochemical
Foundations of Living Systems
Passage 1 (Questions 1 - 4)
Ultraviolet radiation (UVR) causes skin cancer by producing
mutagenic cyclobutane pyrimidine dimers (CPDs) and (6-4)
photoproducts (PPs) in DNA. In mice and humans, the nucleotide
excision repair system (NER) recognizes, removes, and replaces
segments of DNA strands that contain CPDs and (6-4) PPs.
Figure 1 shows how mouse skin levels of XPA (a damage-
recognition and rate-limiting factor in NER) and clock regulatory
protein CRY1 fluctuate in accordance with the internal
physiological and metabolic time keeper, the circadian clock.
Figures 2 and 3 show the circadian rhythms of NER activity and
DNA replication in mouse skin, respectively.
1. The information in the passage suggests that in mice CRY1
most likely affects XPA by:
A. activating XPA protein activity.
B. activating translation of XPA-encoding transcripts.
C. repressing replication of the XPA-encoding gene.
D. repressing transcription of the XPA-encoding gene.
Answer
:
D. repressing transcription of the XPA-encoding gene
Reasoning
:
This is a Biology question that falls under the content category
“Transmission of genetic information from the gene to the
protein.” The answer to this question is D because Figure 1
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MCAT The Princeton Review AAMC Practice

Exam 2 - Biological and Biochemical

Foundations of Living Systems

Passage 1 (Questions 1 - 4)

Ultraviolet radiation (UVR) causes skin cancer by producing

mutagenic cyclobutane pyrimidine dimers (CPDs) and (6-4)

photoproducts (PPs) in DNA. In mice and humans, the nucleotide

excision repair system (NER) recognizes, removes, and replaces

segments of DNA strands that contain CPDs and (6-4) PPs.

Figure 1 shows how mouse skin levels of XPA (a damage-

recognition and rate-limiting factor in NER) and clock regulatory

protein CRY1 fluctuate in accordance with the internal

physiological and metabolic time keeper, the circadian clock.

Figures 2 and 3 show the circadian rhythms of NER activity and

DNA replication in mouse skin, respectively.

  1. The information in the passage suggests that in mice CRY

most likely affects XPA by:

A. activating XPA protein activity.

B. activating translation of XPA-encoding transcripts.

C. repressing replication of the XPA-encoding gene.

D. repressing transcription of the XPA-encoding gene.

Answer:

D. repressing transcription of the XPA-encoding gene

Reasoning:

This is a Biology question that falls under the content category

“Transmission of genetic information from the gene to the

protein.” The answer to this question is D because Figure 1

shows that XPA levels decrease as CRY1 levels increase. It is a

Scientific Reasoning and Problem Solving question because the

question involves determining which cellular process would lead

to a reduction in XPA levels.

  1. Which cells harvested from adult mice were most likely used as

the highly proliferative benchmark in the experiment that

generated the data shown in Figure 3?

A. Adipocytes

B. Cardiac muscle cells

C. Gastrointestinal epithelial cells

D. Neurons

Answer:

C. Neurons

Reasoning :

This is a Biology question that falls under the content category

“Structure and integrative functions of the main organ system.”

The answer to this question is C because the epithelial cells that

line the gastrointestinal tract are typically highly proliferative. It is

a Reasoning about the Design and Execution of Research

question because you must apply your knowledge of the different

cell types to predict which cell type would best suit the

experiment.

  1. After a section of a DNA strand containing a UVR-induced

lesion is removed and resynthesized, the newly synthesized

strand is rejoined to the remainder of the DNA strand by what

type of bond?

A. Disulfide

B. Hydrogen

C. Peptide

D. Phosphodiester

Answer :

D. Phosphodiester

Reasoning:

analysis of the effect of AlP on cytochrome oxidase (Figure 1).

Researchers measured the levels of ATP, the rate of ATP

synthesis, and the rate of ATP hydrolysis in rat liver mitochondria.

It was found that AlP exposure resulted in a 65% decrease in ATP

levels and a 48% decrease in the rate of ATP synthesis. In

addition, the effect of AlP on the activity of three ETC complexes

was determined (Table 1). The activities of these complexes were

determined independently of each other.

  1. AlP exposed to an aqueous solution in which pH range will

result in the largest amount of phosphine production?

A. pH < 4

B. 4 < pH < 7

C. 7 < pH < 10

D. pH > 10

Answer:

A. pH < 4

Reasoning:

This is a General Chemistry question that falls under the content

category “Principles of bioenergetics and fuel molecule

metabolism.” The answer to this question is A because H+ is a

reactant, and the increase in the concentration of H+ at low pH

will favor product formation. It is a Scientific Reasoning and

Problem Solving question because it is necessary to take the

information from the chemical equation in the passage and

reason about how changing solution conditions will affect that

reaction.

  1. Based on the passage, which amino acid will most likely react

with phosphine?

A. Met

B. Cys

C. Ser

D. Thr

Answer:

B. Cys

Reasoning :

This is an Organic Chemistry question that falls under the content

category “Structure and function of proteins and their constituent

amino acids.” The answer to this question is B because the

passage states that phosphine reacts with sulfhydryl groups, and

the cysteine side chain contains a sulfhydryl group. It is a

Knowledge of Scientific Concepts and Principles question

because you must recall the structure of the cysteine side chain

and apply it to the reactivity described in the passage.

  1. When researchers determined the total cellular concentration of

ATP in AlP-exposed rat liver cells, they found the concentration to

be equal to the control value. Which conclusion about the

metabolic state of the cell is best supported by these data?

A. Glycolytic flux is increased after AlP treatment.

B. Glycolytic flux is decreased after AlP treatment.

C. Citric acid cycle flux is increased after AlP treatment.

D. Citric acid cycle flux is decreased after AlP treatment.

Answer:

A. Glycolytic flux is increased after AIP treatment

Reasoning:

This is a Biochemistry question that falls under the content

category “Principles of bioenergetics and fuel molecule

metabolism.” The answer to this question is A because ATP

production is the same in both control and AlP-exposed cells, and

the data in the passage show that mitochondrial ATP production

is decreased. This indicates that the flux through glycolysis is

increased, because this would be the major pathway for ATP

production once the electron transport chain is shut down. It is a

Scientific Reasoning and Problem Solving question because you

C. Lysosome

Reasoning:

This is a Biology question that falls under the content category

“Assemblies of molecules, cells, and groups of cells within single

cellular and multicellular organisms.” The answer to this question

is C because when a macrophage ingests foreign material, the

material initially becomes trapped in a phagosome. The

phagosome then fuses with a lysosome to form a

phagolysosome. Inside the phagolysosome, enzymes digest the

foreign object. Of the cell structures listed, the labeled

carbohydrate is most likely to be microscopically visualized within

a lysosome (phagolysosome). It is a Scientific Reasoning and

Problem Solving question because the question makes a

scientific prediction.

  1. Inhibition of phosphofructokinase-1 by ATP is an example of:

I. allosteric regulation.

II. feedback inhibition.

III. competitive inhibition.

A. I only

B. III only

C. I and II only

D. II and III only

Answer :

C. I and II only

Reasoning:

This is a Biochemistry question that falls under the content

category “Structure and function of proteins and their constituent

amino acids.” The

answer to this question is C because ATP, the

end product of glycolysis, downregulates through feedback

inhibition the activity of phosphofructokinase-1 by binding to a

regulatory site other than the active site of the enzyme (allosteric

regulation). In contrast, competitive inhibition involves competition

for binding to the active site. It is a Scientific Reasoning and

Problem Solving question because you need to reason about how

the product of glycolysis (ATP) inhibits a regulatory enzyme of

glycolysis (phosphofructokinase-1).

  1. The graph shows the average relative concentration of ions in

pond water and in the cytoplasm of green algae cells. Which

process moves chlorine ions into the cells of the green algae?

A. Osmosis

B. Diffusion

C. Active transport

D. Facilitated diffusion

Answer :

C. Active Transport

Reasoning:

This is a Biology question that falls under the content category

“Assemblies of molecules, cells, and groups of cells within single

cellular and multicellular organisms.” The

answer to this question

is C because in order to maintain a higher concentration of

chlorine ions inside the cell, the ions must be moved into the cell

against their concentration gradient, which requires energy. This

process is active transport. In the other processes, ions would

move along their concentration gradient, either with or without the

help of transport proteins. It is a Knowledge of Scientific Concepts

and Principles question because it addresses assumed

knowledge of membrane transport processes.

  1. Dendrotoxin from the mamba snake blocks voltage-gated

potassium channels in somatic motor neurons that regulate

skeletal muscle contraction. In what way would initial exposure to

dendrotoxin affect the ability of a somatic motor neuron to

propagate an electrical signal in response to a stimulus?

A. It would inhibit the initiation of an action potential.

domains. Additional proteins in this pathway include the enzyme

casein kinase-1 (CK1), glycogen synthase kinase-3 (GSK3), and

β-catenin, which activates expression of Wnt target genes. In the

absence of Frizzled activation, CK1 and GSK3 sequentially

phosphorylate β-catenin, which targets it for ubiquitination. When

Frizzled is activated, CK1 and GSK3 activity is inhibited, and Wnt

target genes are transcribed (Figure 1). In healthy adult cardiac

tissue, the Wnt signaling pathway is silent; however, it is

reactivated by cardiac tissue injury.

  1. Based on the passage, which statement describes Wnt

proteins?

A. They are composed of multiple subunits.

B. They have a positive charge.

C. They are synthesized in the smooth endoplasmic reticulum.

D. They fold into their tertiary structure in the cytoplasm.

Answer:

B. They have a positive charge

Reasoning:

This is a Biochemistry question that falls under the content

category “Structure and function of proteins and their constituent

amino acids.” The answer to this question is B because based on

the passage, Wnt proteins are a family of secretory proteins with

isoelectric points around 9, implying that they are positively

charged at physiological pH. A is incorrect because there is no

information in the passage to support this response. C is incorrect

because secretory proteins are synthesized in the rough

endoplasmic reticulum. D is incorrect because folding of secretory

proteins occurs in the rough endoplasmic reticulum. It is a

Scientific Reasoning and Problem Solving question because it

requires you to estimate the charge of Wnt proteins, based on the

value of their isoelectric points as described in the passage.

  1. Based on the passage, β-catenin most likely has:

A. multiple subunits.

B. very few disulfide bonds.

C. a nuclear localization sequence.

D. a high proportion of surface-exposed nonpolar residues.

Answer:

C. a nuclear localization sequence

Reasoning:

This is a Biology question that falls under the content category

“Transmission of genetic information from the gene to the

protein.” The answer to this question is C because according to

Figure 1, β-catenin activates transcription factors for Wnt target

genes. As transcription factors are found in the nucleus, β-catenin

must enter the nucleus. Proteins that are translocated into the

nucleus usually contain a nuclear localization sequence. It is a

Scientific Reasoning and Problem Solving question because it

requires you to bring together assumed knowledge and passage-

based observations to make a conclusion.

  1. In the absence of Frizzled activation, β-catenin is covalently

modified and:

A. bound by a proteasome to initiate degradation into short

peptides.

B. translocated into the Golgi body for secretion through

exocytosis.

C. engulfed by a lysosome where it is hydrolyzed by proteases.

D. stored in vesicles until the signaling pathway is activated.

Answer:

A. bound by a proteasome to initiate degredation into short

peptides

Reasoning:

This is a Biology question that falls under the content category

“Transmission of genetic information from the gene to the

Protein X , its cDNA was cloned and expressed. The purified

protein was analyzed by gel electrophoresis under native,

denaturing, and denaturing/reducing conditions as shown in

Figure 1.

The fluorescent molecule 1-anilinonaphthalene-8-sulfonic acid

(ANS), which exhibits increased fluorescence upon binding to

hydrophobic surface residues of proteins, was used to probe for

conformational changes in Protein X in the presence of the

detergent dodecyl phosphocholine (DPC) (Figure 2).

To investigate the mechanism of Protein X pathogenicity,

researchers filled phospholipid-bound liposomes (micelles) with a

fluorescent dye. Addition of Protein X to the liposomes caused a

gradual release of the fluorescent dye from the liposomes (Figure

  1. Subunits of Protein X are linked covalently by bonds between

the:

A. thiol groups of methionine residues.

B. thiol groups of cysteine residues.

C. hydroxyl groups of serine residues.

D. hydroxyl groups of threonine residues.

Answer:

B. thiol groups of cysteine residues

Reasoning :

This is a Biochemistry question that falls under the content

category “Structure and function of proteins and their constituent

amino acids.” The answer to this question is B because Figure 1

shows that reducing agents separate subunits of Protein X. This

indicates that subunits of Protein X are linked together by disulfide

bonds, which implicate the thiol groups of cysteine residues. It is a

Knowledge of Scientific Concepts and Principles question

because it requires knowledge of the involvement of the thiol

groups of cysteine residues in the formation of disulfide bonds in

proteins.

  1. Based on the results in Figure 2, what effect does DPC have

on the hydrophobic amino acids in Protein X?

A. DPC phosphorylates these amino acids.

B. DPC hydrolyzes these amino acids.

C. DPC exposes these amino acids.

D. DPC suppresses these amino acids.

Answer:

C. DPC exposes these amino acids

Reasoning:

This is a Biochemistry question that falls under the content

category “Structure and function of proteins and their constituent

amino acids.” The answer to this question is C because Figure 2

shows that in the presence of DPC, Protein X has increased

fluorescence, indicating that it has adopted a conformation that

exposes hydrophobic residues on its surface. It is a Scientific

Reasoning and Problem Solving question because it requires

you to evaluate the information presented in Figure 2 to explain

the scientific basis of an observation.

  1. Which interpretation(s) is(are) consistent with the

observations in the passage?

I. Surface amino acids of Protein X are mostly hydrophilic in

aqueous solution.

II. Surface amino acids of Protein X are mostly hydrophilic in

presence of DPC.

III. Surface amino acids of Protein X are mostly hydrophobic in

presence of DPC.

A. I only

B. II only

C. III only

D. I and III only

Answer:

D. I and III only

Researchers investigated the roles of energy metabolism and

histone acetylation level in regulating the lifespan of yeast.

Experiment 1

Cells with deletion of the hst3 gene (Δ hst3 ), which encodes a

histone deacetylase; deletion of the rtt gene (Δ rtt ), which encodes

an acetyltransferase; or the double deletions of both genes

hst3 , Δrtt ) were prepared, and the lifespan for each mutant was

compared with the lifespan of wild-type cells (Figure 1).

Experiment 2

Cells with single deletion of the tdh2 gene (Δ tdh2 ), which encodes

glyceraldehyde-3-phosphate dehydrogenase (GAPDH), or double

deletions of the hst3 and tdh2 genes (Δ hst3 , Δ tdh2 ) were

prepared. The lifespan for each mutant was compared with the

lifespan of the Δ hst3 mutant and wild-type cells (Figure 2).

  1. The enzyme encoded by the tdh2 gene catalyzes the

reversible conversion of:

A. 3-phosphoglycerate to 1,3-bisphosphoglycerate.

B. glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate.

C. fructose-1,6-bisphosphate to glyceraldehyde-3-phosphate.

D. 2-phosphoglycerate to 3-phosphoglycerate.

Answer:

B. glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate

Reasoning :

This is a Biochemistry question that falls under the content

category “Principles of bioenergetics and fuel molecule

metabolism.” The answer to this question is B because GAPDH,

the enzyme encoded by the tdh2 gene, is a glycolytic enzyme that

catalyzes the reversible conversion of glyceraldehyde-3-

phosphate to 1,3-bisphosphoglycerate. It is a Knowledge of

Scientific Concepts and Principles question because it requires

knowledge of glycolytic enzymes and their substrates and

products.

  1. Based on Figure 1, how do the deletions of the hst3 or

the rtt genes affect the lifespan of yeast?

A. The deletion of the hst3 gene compensates for the ∆ rtt -induced

decrease in lifespan.

B. The deletion of the hst3 gene has no effect on the ∆ rtt -induced

increase in lifespan.

C. The deletion of the rtt gene compensates for the ∆ hst3 -induced

decrease in lifespan.

D. The deletion of the rtt gene has no effect on the ∆ hst3 -induced

decrease in lifespan.

Answer:

D. The deletion of the rtt gene has no effect on the ∆hst-3 induced

decrease in lifespan

Reasoning:

This is a Biology question that falls under the content category

“Processes of cell division, differentiation, and specialization.” The

answer to this question is D because according to data shown in

Figure 1, compared to wild type cells, the ∆ hst3 cells exhibit a

decrease in lifespan. Furthermore, the deletion of rtt gene in

hst3 cells has no effect on the cells lifespan. It is a Scientific

Reasoning and Problem Solving question because you need to

evaluate data shown in Figure 1 and reason about how deletion of

genes affects the phenotypes in yeast.

  1. What is the best experimental method to analyze the effect

of tdh2 gene deletion on the rate of histone acetylation?

Comparing histone acetylation in wild-type and Δ tdh2 cells by:

A. Western blot

B. Southern blot

C. Northern blot

D. RT-PCR

Answer:

A. Western blot

  1. Vasopressin regulates the insertion of aquaporins into the

apical membranes of the epithelial cells of which renal structure?

A. Collecting duct

B. Proximal tubule

C. Bowman's capsule

D. Ascending loop of Henle

Answer:

A. Collecting duct

Reasoning:

This is a Biology question that falls under the content category

“Structure and functions of the nervous and endocrine systems

and ways in which these systems coordinate the organ systems.”

The answer to this question is A because vasopressin regulates

the fusion of aquaporins with the apical membranes of the

collecting duct epithelial cells. It is a Knowledge of Scientific

Concepts and Principles question because it requires knowledge

of hormone function in the excretory system.

  1. What are the primary myelin-forming cells in the peripheral

nervous system?

A. Microglia

B. Astrocytes

C. Schwann cells

D. Oligodendrocytes

Answer :

C. Schwann cells

Reasoning:

This is a Biology question that falls under the content category

“Structure and functions of the nervous and endocrine systems

and ways in which these systems coordinate the organ systems.”

The answer to this question is C because Schwann cells are the

myelin-forming cells in the peripheral nervous system. It is a

Knowledge of Scientific Concepts and Principles question

because it assesses knowledge of the different types of glia within

the nervous system.

  1. Osmotic pressure Π is given by the relation:

Π = iMRT

where i is the van’t Hoff factor, M is the concentration of

solute, R is the gas constant, and T is the temperature. The

osmotic pressure of sea water is approximately 24 atm at 25°C.

What is the approximate concentration of salt in sea water

(approximated by NaCl with i = 2)? (Note: Use R = 0.

L•atm/mol•K.)

A. 0.25 M

B. 0.50 M

C. 0.75 M

D. 1.0 M

Answer:

B. 0.50 M

Reasoning:

This is a General Chemistry question that falls under the content

category “Assemblies of molecules, cells, and groups of cells

within single cellular and multicellular organisms.” The answer to

this question is B since algebraic manipulation of the relation

gives M = Π/iRT = 24 atm / (2 × 0.08 L•atm/mol•K × 300 K) = 24 /

(2 × 24) mol/L = 0.5 M. It is a Scientific Reasoning and Problem

Solving question since you must manipulate a scientific formula

algebraically to solve a problem.

  1. A prion is best described as an infectious:

A. prokaryote.

B. transposon.

C. protein.

D. virus.

Answer:

C. protein

Reasoning :

This is a Biology question that falls under the content category

“The structure, growth, physiology, and genetics of prokaryotes

and viruses.” The answer to this question is C because a prion is