Review Questions

Question 1

Question At the end of his initial hospital stay, a few E. coli cells remained in the patient’s colon, even though he was taking antibiotics. These cells were most likely present because:

Answer 1

Chance mutations in the few E. Coli before the treatment made these cells and their descendants antibiotic-resistant Mutations which make cells drug resistant are very rare, but the few drug-resistant bacteria that do develop such mutations flourish when the nonresistant cells are killed by antibiotics.

Antibiotics are unlikely to have been the source of mutations

Question 2

In eukaryotes, oxidative phosphorylation occurs in the mitochondrion. The analogous structure used by bacteria to carry out oxidative phosphorylation is the:

Answer 2

Plasma membrane

The inner membrane of a mitochondrion is analogous to the plasma membrane of a prokaryote. The enzymes for oxidative phosphorylation are embedded in the inner membrane.

Question 3

Which of the following organelles most resembles the Golgi apparatus when an intact eukaryotic cell is viewed under the electron microscope?

Answer 3

The Smooth ER

Question 4

Applying a drug which blocks the absorption of NE into the adrenergic nerve terminal will result in:

Answer 4

Increased sympathetic activity

Blocking the reuptake of NE into the nerve terminal leaves more NE in the synapse where it can continue to stimulate the adrenergic receptors on the membrane of the tissue supplied by the neuron.

Question 5

If a cell’s membrane potential changes from –60mV to –70mV after treatment with an adrenergic drug, the NE receptor is most likely linked to:

Answer 5

A potassium channel

A change in membrane potential from –60mV to –70mV indicates that the cell membrane is most likely repolarizing to its resting potential of –70mV following an action potential. In a neuron, this is accomplished through the opening of K⁺ channels that release K⁺ from the cell to counterbalance the influx of Na⁺ that occurred during depolarization. Thus, D is the best answer.

Question 6

Capillaries in the kidney and elsewhere in the body maintain fluid homeostasis by balancing hydrostatic and osmotic pressures. Which of the following is the initial effect of a blood clot forming on the venous side of a capillary bed?

Net fluid flow in the direction of interstitial spaces will increase.

Net fluid flow in the direction of interstitial spaces will decrease.

Capillary osmotic pressure will increase.

Capillary osmotic pressure will decrease.

Answer 6

Net fluid flow in the direction of interstitial spaces will increase.

blood flows from arteries to capillaries and then to veins. If flow is blocked at the venous side, blood would accumulate in the capillaries. Thus, hydrostatic pressure would build up in the capillaries, causing a net increase in fluid flow into the interstitial spaces.

Question 7

The cell nucleus below contains the chromosomes of a sea urchin embryo at the two-cell stage.

Which of the diagrams below best represents the nucleus of an embryo at the 64-cell stage grown from this cell?

Question 8

Radioactively labeled uracil is added to a culture of actively dividing mammalian cells. In which of the following cell structures will the uracil be incorporated?

Chromosomes

Ribosomes

Lysosomes

Nuclear membrane

Answer 8

Ribosomes

Uracil is a component of RNA. Therefore, one would expect to find the radioactively labeled uracil in cell structures that contain RNA. B is the best answer because ribosomes contain rRNA and proteins.

Question 9

A resident of a famine area who appears undernourished and extremely emaciated has eaten only starches for the past 3 months. A urine analysis shows that a large amount of nitrogen is being excreted. This is most likely evidence of:

an abnormally high rate of glycogen breakdown in the liver.

breakdown of the body’s own structural proteins to provide energy.

utilization of the last remaining fat reserves to provide energy.

incomplete reabsorption of nitrogenous products due to kidney failure.

Answer 9

breakdown of the body’s own structural proteins to provide energy.

the nitrogen in the urine comes from breakdown of the body’s proteins.

Question 10

All of the following occur during normal inspiration of air in mammals EXCEPT:

elevation of the rib cage.

relaxation of the diaphragm.

reduction of pressure in the pleural cavity.

contraction of the external intercostal rib muscles.

Answer 10

Relaxation of the diaphragm

The diaphragm contracts during inspiration and pulls downward

The reduction of pressure in the pleural cavity is what causes air to move into the lungs

Question 11

What occurs during sympathetic activation?

Answer 11

the digestive system becomes less active; the heart beats faster and with greater force; the bronchi of the lungs relax to let in more air; and the pupils of the eyes dilate.

Question 12

Based on information in the passage, would the S or the PS division of the ANS be expected to produce a more rapid systemic (whole-body) response to a stimulus?

A

The S division, because its ganglia are interconnected

The S division, because it secretes norepinephrine

The PS division, because its ganglia are not interconnected

The PS division, because it secretes acetylcholine

Answer 12

The S division, because its ganglia are interconnected

Multiple pathways to ganglia lead to more rapid response of the sympathetic system. Multiple pathways are also more consistent with a systemic, rather than localized, response.

Question 13

Question

The ocular drug physostigmine inhibits acetylcholinesterase, the enzyme responsible for the hydrolysis of acetylcholine. Administration of physostigmine would be expected to cause the pupil to:

dilate, due to decreased acetylcholine levels.

dilate, due to increased acetylcholine levels.

constrict, due to decreased acetylcholine levels.

constrict, due to increased acetylcholine levels.

Answer 13

constrict, due to increased acetylcholine levels.

Administration of physostigmine, a drug that inhibits acetylcholinesterase, would increase acetylcholine levels. Acetylcholinesterase is the enzyme that breaks down acetylcholine, removing it from the synapse so that the neurotransmitter can no longer activate its receptor. Therefore, the amount of acetylcholine available at the synaptic junction increases. Constriction is due to increased, not decreased, acetylcholine levels.

Question 14

Which of the following media would most likely be used to grow virions in the laboratory?

A suspension of ribosomes and ATP

A suspension of human DNA

A nutrient broth

A tissue culture

Answer 14

A tissue culture

Viruses take advantage of the elaborate intracellular mechanisms of the host cell using them to make more virus particles. To do this they need intact host cells. Non-cellular media such as the nutrient broths and suspensions in choices A, B and C will not support culture of viruses.

Question 15

Most viral proteins are produced directly by:

translation of host nucleic acid.

translation of viral nucleic acid.

transcription of host nucleic acid.

transcription of viral nucleic acid.

Answer 15

translation of viral nucleic acid.

Question 16

The investigators concluded that the new hantavirus infects lung endothelial cells. Do the data support this conclusion?

Yes; viral replication was demonstrated in endothelial cells.

Yes; viral DNA was found associated with lung capillaries.

Yes; viral antigens were found associated with capillary endothelium.

No; viral antigens were found associated with lung alveoli.

Answer 16

Yes; viral antigens were found associated with capillary endothelium.

B is not the best answer because the gene sequences from known hantaviruses in Experiment 2 could have been hybridizing with RNA, not DNA, if hantaviruses have RNA genomes, and could have been hybridizing with viral sequences in lung tissue other than endothelium. C is the best answer because in Experiment 3, antibodies were binding viral antigens present in lung capillary walls or endothelium.

Question 17

In human females, mitotic divisions of oogonia that lead to formation of presumptive egg cells (primary oocytes) occur between:

Answer 17

Fertilization and birth only

All of the mitotic divisions that form primary oocytes occur prior to birth.

Question 18

The discovery that the amount of thymine equals that of adenine and the amount of guanine equals that of cytosine in a given cell provides supporting evidence that:

the Watson and Crick model of DNA is correct.

DNA is the genetic material.

the genetic code is universal.

the code for one amino acid must be a triplet of bases.

Answer 18

the Watson and Crick model of DNA is correct.

In the Watson and Crick model of DNA structure, the nitrogenous bases form hydrogen bonds with each other in a 1:1 ratio: guanine pairs with cytosine, and adenine pairs with thymine.

Question 19

Which of the following tissues have cells that are in direct contact with the external environment or elements of the external environment?

1. The lining of the reproductive tract
2. The lining of the respiratory tract
3. The lining of the gastrointestinal tract

Answer 19

All three

Tissues that are exposed to the external environment have mucosal membranes. All of the tissues listed contain an element of mucosa

Question 20

which of the following endocrine disorders would cause hypertension that could NOT be rectified by physiologically normal kidneys?

An excess of aldosterone

An excess of glucagon

A shortage of thyroxine

A shortage of insulin

Answer 20

An excess of aldosterone

Aldosterone is a hormone released by the adrenal glands. Physiologically normal kidneys respond to aldosterone by increasing the reabsorption of both sodium and water. This leads to an increase in blood volume and therefore blood pressure.

Question 21

What mechanism probably would be responsible for the increased urine output induced by hypertension according to Hypothesis B?

Increased blood flow to the bladder

Increased renal tubular reabsorption of solutes and water

Increased collecting duct permeability to water

Increased glomerular filtration rate

Answer 21

Increased glomerular filtration rate

An increase to the systemic blood pressure would initially increase the glomerular capillary blood pressure, which would increase the glomerular filtration rate. Without a corresponding increase in the rate of tubular reabsorption of water, this would lead to an increase in urine output

Question 22

If restriction of blood flow to the kidneys (by placing clamps on the renal arteries) resulted in an immediate but small increase in blood pressure, followed by the gradual development of severe hypertension, which hypothesis would these results best support?

Hypothesis A, because the clamps increased the vascular resistance to blood flow

Hypothesis A, because the clamps caused the kidneys to receive less blood

Hypothesis B, because the kidneys were responding to decreased glomerular blood pressure

Hypothesis B, because the volume of body fluids was probably decreasing

Answer 22

Hypothesis B, because the kidneys were responding to decreased glomerular blood pressure

The reduced flow of blood through the renal arteries due to the clamps would cause a decrease in glomerular blood pressure. The kidneys respond to this drop in pressure by activating the renin–angiotensin system of hormones. This increases the amount of sodium and water that is reabsorbed by the kidneys, therefore increasing blood volume and pressure.

Question 23

What is the primary process of the colon? (Large intestine)

Answer 23

The primary process that takes place in the colon is absorption of water

Question 24

Normally the immune system avoids attacking the tissues of its own body because:

a special intracellular process recognizes only foreign antigens.

the body does not make any antigens that the immune system could recognize.

it changes its antibodies to be specific only to foreign antigens.

it suppresses cells specific to the body’s own antigens.

Answer 24

it suppresses cells specific to the body’s own antigens.

The immune system is designed to attack foreign material in the body. It avoids attacking tissues of its own body because it suppresses cells that are specific to its own body’s antigens (surface molecules that would otherwise initiate an immune response).

Question 25

Question

An ulcer that penetrated the wall of the intestine would allow the contents of the gastrointestinal tract to enter:

Answer 25

The peritoneal cavity

Question 26

The finches observed by Darwin on the Galapagos Islands are an example of adaptive radiation. In order to set up conditions that would produce adaptive radiation, it would be necessary to place members of:

one species in one rapidly changing environment.

one species in several different environments.

several very similar species in the same environment.

several unlike species in one environment to compete for the same resources.

Answer 26

one species in several different environments.

Adaptive radiation involves the divergence of one species into multiple species over time, which can occur when subgroups of the original species are separated or isolated in different environments so that these subgroups evolve independently of one another.

Question 27

Hemophilia, a disease in which the time required for blood to clot is greatly prolonged, is determined by a sex-linked gene. Suppose a man with normal blood clotting marries a woman with normal blood clotting whose father was a hemophiliac. If this couple has three sons, what is the probability that hemophilia will be transmitted to all three of them?

1/8

1/4

3/8

1/2

Answer 27

1/8

The father only had one X chromosome to contribute and that X chromosome contained the hemophilia gene. The mother will pass on one X chromosome to her sons, either the X chromosome containing the normal (wild type) gene, or the X chromosome containing the hemophilia gene. The probability of a son receiving the hemophilia gene and being a hemophiliac is 1/2, because there is a 50-50 chance that this son received an X chromosome carrying the defective gene from his carrier mother. The cumulative probability of all three sons inheriting the hemophilia gene would be the product of the individual probabilities, or 1/2 × 1/2 × 1/2.

Question 28

Which participant in the electron transport chain has the greatest attraction for electrons?

Answer 28

Oxygen

the final electron acceptor of this chain, oxygen (C), has the greatest attraction for electrons.

Question 29

Postmenopausal women receiving estrogen and progesterone therapy will most likely experience which of the following side effects?

Breast tissue will atrophy.

Vaginal tissue will dry out.

Periodic menstruation will resume.

Lactation will be induced.

Answer 29

Periodic menstruation will resume.

Question 30

A man is treated with low doses of an estrogen analogue to destroy an estrogen-responsive adrenal tumor. Compared to an age-matched control (no estrogen treatment), this patient’s chances of developing osteoporosis will most likely be:

increased.

decreased.

approximately the same.

approximately the same, but the disease will appear at an earlier age.

Answer 30

approximately the same.

Administration of estrogen to men would then have no therapeutic effect on bone loss, since they should not be suffering from withdrawal of the hormone, having never had high levels. Men given estrogen should have the same chance of developing osteoporosis as a control population, answer choice C.

Question 31

Which of the following hormones is(are) directly required for spermatogenesis?

1. Luteinizing hormone (LH)
2. Follicle-stimulating hormone (FSH)
3. Inhibin
4. Testosterone

Answer 31

FSH and Testosterone only

Question 32

Uric acid enters the urine both through filtration and secretion in the kidney. The process of filtration of uric acid in the kidney takes place in the:

glomerulus.

loop of Henle.

distal convoluted tubule.

proximal convoluted tubule.

Answer 32

glomerulus

Filtrate is formed as fluid passes from the glomerular capillaries through the glomerular membrane into the Bowman’s capsule.

Question 33

Colchicine most likely relieves gout symptoms through what mechanism?

Prevention of uric acid diffusion through cell membranes

Inhibition of leukocyte phagocytosis of uric acid crystals

Inhibition of uric acid crystal formation

Maintenance of the pH optimum for PRPP synthetase

Answer 33

Inhibition of leukocyte phagocytosis of uric acid crystals

Phagocytosis requires that the cell change shape dramatically as it surrounds and engulfs large extracellular particles. Microtubules are one of the cytoskeletal elements that help determine cell shape. This function relies on the ability of the microtubules to disassemble and reorganize. The drug colchicine inhibits microtubule reorganization and would therefore inhibit phagocytosis of uric acid crystals by leukocytes.

Question 34

In the patient described in the passage, the likely genetic basis of the increased levels of uric acid is a mutation:

affecting an allosteric site of PRPP synthetase.

affecting the active site of PRPP synthetase.

in a promotor gene regulating the rate of transcription of the PRPP synthetase gene.

in a gene coding for a transcription factor for the PRPP synthetase gene.

Answer 34

affecting an allosteric site of PRPP synthetase.

Passage: Assays revealed that the patient had normal levels of PRPP synthetase, but the enzyme activity was three times normal levels in cultured cells.

The passage indicates that the patient produces the normal amount of PRPP synthetase, but its activity in vivo is 3 times the normal level. However, PRPP synthetase purified from the patient does not show this increased enzymatic activity in vitro. The fact that the enzyme binds the substrate and converts it to product at normal levels in vitro suggests that the active site of the enzyme has not been altered, but rather that an allosteric site on the enzyme has been affected. The mutant PRPP can most likely bind an intracellular molecule at an allosteric site, which changes the shape of the enzyme, enhancing its activity. This activity-enhancing molecule most likely is not present in the in vitro reaction mix.

Question 35

A drug that binds to tubulin molecules of plant cells and prevents the cells from assembling spindle microtubules would most likely cause the resulting plants or plant cells to have:

A

greater genetic variability than the parent plants.

more than two sets of chromosomes.

a stronger cell wall because of excess tubulin.

independent movement because of excess tubulin.

Answer 35

more than two sets of chromosomes.

Question 36

Cancer cells most likely have an abnormality in their:

DNA.

rRNA.

mitochondria.

lysosomes.

Answer 36

DNA

Question 37

Contraction of the diaphragm results in a:

more negative IPP and inspiration.

more negative IPP and expiration.

more positive IPP and inspiration.

more positive IPP and expiration.

Answer 37

more negative IPP and inspiration.

The diaphragm is a muscular partition between the abdominal and thoracic cavities. It is dome-shaped at rest curving up toward the lungs and heart. It flattens when it contracts during inspiration. Because it is anchored around its edges to the ribs and spine, when the diaphragm contracts the volume of the thoracic cavity increases. This decreases the pressure (IPP) in the pleural cavity between the thoracic wall and the lung. Fresh air flows in to equalize the pressure inflating the lung.

Question 38

The pancreas produces which of the following substances for the digestive system?

Bile salts

Emulsifier

Gastric juices

Proteolytic enzymes

Answer 38

Proteolytic enzymes

Question 39

Which statement correctly describes both PO₄^3– and TNBS?

Both TNBS and PO₄^3– are hydrophobic.

TNBS is hydrophobic and PO₄^3– is hydrophilic.

PO₄^3– is hydrophobic and TNBS is hydrophilic.

Both TNBS and PO₄^3– are hydrophilic.

Question 40

Question

Acetic acid and ethanol react to form an ester product as shown below.

In determining which reactant loses the –OH group, which of the following isotopic substitutions would be most useful?

• A.Replace the acidic H of acetic acid with D.
• B.Replace the alcoholic H of ethanol with D.
• C.Replace the carbonyl oxygen of acetic acid with O-18.
• D.Replace the hydroxyl oxygen of ethanol with O-18

Answer 40

D.Replace the hydroxyl oxygen of ethanol with O-18

1. Labeling the oxygen in ethanol with O-18 definitely determines which –OH is lost because 100% of the O-18 in radiolabeled ethanol will be retained in the ester, not the water. Thus, only acetic acid supplies –OH that is lost to form water.

Question 41

If 2-pentanol replaces 1-pentanol in the reaction shown in Figure 3, the rate of substitution is less because:

• A.the C–O bond in 2-pentanol is stronger than the C–O bond in 1-pentanol.
• B.there is a competing elimination reaction that slows the rate of substitution.
• C.there is more steric hindrance at the oxygen atom in 2-pentanol than in 1-pentanol, making protonation less likely.
• D.there is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-pentanol.

Answer 41

D.there is more steric hindrance at the 2-position of 2-pentanol than at the 1-position of 1-pentanol.

1. At a secondary carbon, there is more crowding from the adjacent carbon chains, which makes nucleophilic substitution slower.

Question 42

Question

Which of the following will decrease the percentage ionization of 1.0 M acetic acid, CH₃CO₂H(aq)?

• A.Chlorinating the CH₃ group
• B.Diluting the solution
• C.Adding concentrated HCl(aq)
• D.Adding a drop of basic indicator

Answer 42

C.Adding concentrated HCl(aq)

1. HCl is a strong acid that will increase the amount of H⁺ in solution and thus decrease the percentage of CH₃CO₂H that ionizes.

Question 43

When the current in the micro-robot’s circuit increases and the resistance of the circuit remains constant, the voltage of the circuit:

• A.decreases.
• B.increases.
• C.is constant.
• D.is zero.

Answer 43

B.increases.

1. This option is correct because according to Ohm’s law, current is equal to voltage divided by resistance. If current increases and resistance is constant, then voltage increases as well.

Question 44

Knowing that the speed of light in the vitreous humor is 2.1 × 10⁸ m/s, what is the index of refraction of the vitreous humor? (Note: The speed of light in a vacuum is 3.0 × 10⁸ m/s.)

• A.0.7
• B.1.4
• C.2.1
• D.3.0

Answer 44

B.1.4

The index of refraction of a medium is equal to the ratio of the speed of light in vacuum to the speed of light in the medium, thus in this case is equal to (3.0 × 10⁸ m/s)/(2.1 × 10⁸ m/s) = 1.4.

Question 45

What amino acids can be phosphorylated?

Answer 45

Serine, threonine, and tyrosine

Question 46

Compared to the wild-type LipA, what is the change in net charge in variant XI at pH 7?

R33G, K112D, M134D, Y139C, I157M

• A.+4
• B.+3
• C.−3
• D.−4

Answer 46

D.−4

Question 47

By what factor is the proton concentration increased in the experiments shown in Figure 1B from those shown in Figure 1A?

pH 7 → pH 4

• A.2
• B.10
• C.200
• D.1000

Answer 47

1000

Difference in pH = 3

pH = -log [H+]

3 = - log = 10³ = 1,000

x = - log = 10^x

Question 48

Question

Each of the following equations shows the dissociation of an acid in water. Which of the reactions occurs to the LEAST extent?

A.HCl + H2O → H3O+ + Cl−

B.HPO42− + H2O → H3O+ + PO43−

C.H₂SO₄ + H₂O → H₃O⁺ + HSO₄⁻

D.H₃PO₄ + H₂O → H₃O⁺ + H₂PO₄⁻

Answer 48

B.HPO4^2- + H2O → H3O+ + PO4^3-

1. HPO₄^2– has a high negative charge and so dissociation of it will occur to the least extent.

Question 49

In the above figure, an object O is at a distance of three focal lengths from the center of a convex lens. What is the ratio of the height of the image to the height of the object?

• A.1/3
• B.1/2
• C.2/3
• D.3/2

Answer 49

½

1. This is correct because the ratio of the image height to the object height is equal to the ratio of the lens-image distance to the object-lens distance. According to the thin lens equation, the distance between the lens and the image is ​​. The ratio sought is then equal to​ ​.

Question 50

Emission of protons is during _____

Emission of electrons is during ____

Emission of neutrons is during _____

Answer 50

Beta+ decay

Beta- decay

Nuclear fission

Question 51

Which of the following most likely will occur if a homogeneous catalyst CANNOT be separated from the products at the end of a reaction?

A.The catalyst will become heterogeneous.
B.The products will be contaminated.
C.The reaction will not occur.
D.The reaction rate will speed up.

Answer 51

The products will be contaminated.

1. If a suitable separation technique is not found, the product will be contaminated by the catalyst.

Question 52

How will the rate of a catalyzed reaction be affected if the solid catalyst is finely ground before it is added to the reaction mixture?

A.The rate will be faster because a greater mass of catalyst will be present.
B.The rate will be faster because a greater surface area of catalyst will be exposed.
C.The rate will be slower because the fine catalyst particles will interfere with product formation.
D.The rate will remain the same because the mass of catalyst will be the same.

Answer 52

The rate will be faster because a greater surface area of catalyst will be exposed.

1. Grinding a heterogeneous catalyst increases the amount of catalyst available to the reaction and therefore increases its rate.

Question 53

BaCrO₄(s) Ba²⁺(aq) + CrO₄^2–(aq)

exists in a saturated aqueous solution of BaCrO₄. Dissolution of Na₂CrO₄ in a saturated aqueous BaCrO₄ solution would:

• A.have no effect on the position of this equilibrium.
• B.shift this equilibrium left.
• C.shift this equilibrium right.
• D.shift this equilibrium first right and then left.

Answer 53

shift this equilibrium left.

1. Dissolution of Na₂CrO₄ would introduce the common ion, CrO₄^2–, which would reduce the solubility of BaCrO₄ due to the common ion effect.

Question 54

What volume of a 0.120 M CaI₂ solution would contain 0.078 mol of the solute?

• A.35.0 mL
• B.65.0 mL
• C.350 mL
• D.650 mL

Answer 54

one needs to take the number of moles and divide by the solution concentration in molarity: 0.078 mol × 1 L/0.120 mol = 0.65 L = 650 mL.

Question 55

Which nucleotide pairing(s) would be recognized by the MMR system during DNA replication?

1. dTMP and dCMP
2. dGMP and dAMP
3. dAMP and dTMP

• A.I only
• B.III only
• C.I and II only
• D.I, II, and III

Answer 55

I and II only

1. The passage indicates that MMR system recognizes the pair mismatch. Both options I and II represent mismatches, thus they will be recognized by the MMR system.

Question 56

The information in the passage best supports the conclusion that Intron 8 of HSP110 most likely contains which of the following?

• A.Stop codon
• B.Splice acceptor site
• C.HSP110 gene promoter
• D.Partial coding sequence of HSP110

Answer 56

Splice acceptor site

1. Based on the passage, intron 8 does not contain a stop codon as translation ends with exon 10.
2. It is most likely that intron 8 contains a splice acceptor site that allows exon 9 to be omitted upon mutation.
3. The promoter is usually not located in an intron.
   
   1. The intron does not code for a gene, an exon does.

Question 57

According to the information in the passage, HSP110 is most likely which type of protein?

HSP110 can bind structurally similar heat shock proteins and functions to facilitate proper protein folding and to reduce levels of nonfunctional protein aggregates

A. Adhesion

B.Chaperone

C.Clathrin

D.Enzyme

Answer 57

Chaperone

1. The passage states that HSP110 is involved in protein folding. Adhesion proteins act in binding with other cells or with the extracellular matrix, not protein folding.
2. The passage states that HSP110 is a heat shock protein, which facilitates proper protein folding and inhibits the formation of nonfunctional protein aggregates. Proteins that perform this function are chaperone proteins.
3. Clathrin functions in formation of vesicles for intracellar trafficking, not in protein folding.
   
   1. Based on the passage, HSP110 is involved in protein folding, not enzymatic catalysis of specific biochemical reactions.
      1.

Question 58

According to the information in the passage, HSP110 is most likely which type of protein?

HSP110 can bind structurally similar heat shock proteins and functions to facilitate proper protein folding and to reduce levels of nonfunctional protein aggregates

A. Adhesion

B.Chaperone

C.Clathrin

D.Enzyme

Answer 58

Chaperone

1. The passage states that HSP110 is involved in protein folding. Adhesion proteins act in binding with other cells or with the extracellular matrix, not protein folding.
2. The passage states that HSP110 is a heat shock protein, which facilitates proper protein folding and inhibits the formation of nonfunctional protein aggregates. Proteins that perform this function are chaperone proteins.
3. Clathrin functions in formation of vesicles for intracellar trafficking, not in protein folding.
   
   1. Based on the passage, HSP110 is involved in protein folding, not enzymatic catalysis of specific biochemical reactions.
      1.

Question 59

Pellagra also results from a deficiency of nicotinamide, which is synthesized from tryptophan. Nicotinamide nucleotides are neither oxidized nor reduced during which step of cellular respiration?

• A.Glycolysis
• B.Chemiosmosis
• C.Citric acid cycle
• D.Electron transport chain

Answer 59

Chemiosmosis

1. During glycolysis NAD⁺ is reduced to form NADH.
2. Chemiosmosis is the only step of cellular respiration where NAD⁺ is neither reduced to form NADH, nor is NADH oxidized to form NAD⁺.
3. During the citric acid cycle, NAD⁺ is reduced to form NADH.
   
   1. During the electron transport chain, NADH is oxidized to form NAD⁺.
      1.

Question 60

Question

Epilepsy may result in motor seizures due to massive synchronous firing of neurons in a small area of the cerebral cortex (the epileptic focus). Excitation spreads from the focus, involving an increasingly larger area of the cortex. A drug for the treatment of epilepsy would be most effective if it caused which of the following changes in the epileptic focus?

• A.An increase in the neuron-firing threshold
• B.An increase in extracellular Na⁺ concentration
• C.A decrease in axon–membrane permeability to negative ions
• D.A decrease in the length of the depolarization stage

Answer 60

An increase in the neuron-firing threshold

1. An increase in the threshold would make it more challenging for a neuron to fire an action potential. This will reduce the frequency of downstream neuron depolarizations, resulting in relief of the symptoms.

Question 61

Question
When concentrated urine is being produced, in which of the following regions of the kidney will the glomerular filtrate reach its highest concentration?

A.Proximal convoluted tubule
B.Distal convoluted tubule
C.Cortical portion of the collecting duct
D.Medullary portion of the collecting duct

Answer 61

Medullary portion of the collecting duct

1. The collecting duct is the final structure in which water reabsorption occurs, which concentrates filtrate. The medullary portion of the collecting duct is the last portion of the tubules where reabsorption can occur. In the portion of the tubule that follows, there will be no more reabsorption. Thus, the medullary portion of the collecting duct contains the most concentrated glomerular filtrate that will correspond to the urine.

Question 62

Which approach does NOT measure the activity of the Na+K+ ATPase?

A.Measuring the rate of ATP hydrolysis
B.Measuring the free energy of the ion transport
C.Measuring the rate of ADP production
D.Measuring the change in ion concentration within the liposome

Answer 62

Measuring the free energy of the ion transport

1. Free energy does not correlate with enzyme activity because it is NOT a measure of reaction rate/activity.

Question 63

Upon activation, p65 and cRel control the level of IL-6 mRNA by:

The NF-κB proteins p65 and cRel are transcription factors that can be continuously activated when mutations occur to proteins upstream in the signaling pathway

A.binding RNA.
B.binding DNA.
C.replicating RNA.
D.replicating DNA.

Answer 63

Binding DNA

1. p65 and cRel are transcription factors and regulate the expression of other genes by binding to the promoter or the enhancer of the gene located on the DNA.

Question 64

Considering the structure of STN, what is the most likely mechanism for its entry into the cell?

A.Active transport
B.Receptor mediated endocytosis
C.Diffusion directly through the membrane
D.Passage through an ion channel

Answer 64

Diffusion directly through the membrane

1. The structure of STN shows that it is a hydrophobic molecule. These types of molecules pass through the membrane by simple diffusion.

Question 65

According to the passage, what is most likely correct about the genes that are expressed as a result of the constant activation of the NF-κB pathway?

• A.They cause disruption of the mitochondria.
• B.They contain a p65/cRel binding site in their promoter region.
• C.They have accumulated mutations that alter function.
• D.They bind to STN in the cytoplasm.

Answer 65

They contain a p65/cRel binding site in their promoter region.

1. Based on the passage, activation of the NF-κB pathway results in translocation of the transcription factor subunits p65/cRel in the nucleus, where they regulate the expression of downstream target genes. Expression regulation is possible because the downstream target genes contain in their promoter or enhancer, DNA regions that are able to interact with p65 and cRel.

Question 66

It has been shown that P-gp exhibits ATPase activity and is localized in cholesterol-rich domains within the membrane.

What is the most likely location of P-gp within the plasma membrane?

• A.Associated with lipids on the cytoplasmic side only
• B.Associated with lipids on the extracellular side only
• C.Peripheral to the plasma membrane
• D.Within a lipid raft

Answer 66

Within a lipid raft

1. The passage states the P-gp is found in cholesterol rich domains, pointing to lipid rafts.

Question 67

Question
Enzymes alter the rate of chemical reactions by all of the following methods EXCEPT:

A.co-localizing substrates.
B.altering local pH.
C.altering substrate shape.
D.altering substrate primary structure.

Answer 67

altering substrate primary structure.

1. The enzyme needs to co-localize with the substrates to be able to modify the rate of chemical reactions.
2. It is one of the enzyme functions to change the pH of the environment so that the reaction can occur.
3. The enzyme will alter the shape of the substrate to modify the rate of chemical reactions.
   
   1. Altering the primary structure of the substrate is not a way for the enzyme to modify the rate of chemical reaction.

Question 68

Ornithine (R = –CH₂CH₂CH₂NH₂) is an amino acid that is found in cells, but not incorporated into proteins.

Which of the following statements gives the most fundamental reason why ornithine is unlikely to be found in proteins synthesized in vivo?

• A.There is no codon for it in the standard genetic code.
• B.It cannot form a peptide bond.
• C.It is not available in the diet.
• D.It has a net positive charge in aqueous solution.

Answer 68

There is no codon for it in the standard genetic code.

Question 69

However, cP-450 does not metabolize all toxins at the same rate. For example, although cP-450 usually metabolizes barbiturates, alcohol acts as a competitive inhibitor of barbiturate metabolism.

In the example of the alcoholic in the passage, the presence of alcohol caused death because the alcohol:

• A.denatured the cP-450.
• B.inhibited the cP-450.
• C.decreased the concentration of cP-450.
• D.increased the concentration of cP-450.

Answer 69

inhibited the cP-450.

Question 70

The lung cells of heavy smokers would be expected to have greatly increased concentrations of cP-450 and:

• A.DNA sequences that code for cP-450.
• B.mRNA sequences that code for cP-450.
• C.rRNA that process cP-450.
• D.tRNA that are specific for cysteine.

Answer 70

mRNA sequences that code for cP-450.

1. cP-450 DNA is present in the genome, and its levels do not change.
2. The levels of cP-450 mRNA should be elevated in smokers as they are almost directly related to the levels ofcP-450 protein produced to get rid of the toxins.
3. The levels of ribosomal RNA (rRNA) should not change in the epithelial cells of smokers.
   
   1. tRNA levels specific for cysteine are not related to the production of cP-450 protein.

Question 71

A student hypothesized that EC growth might be affected by the DNA from circulating erythrocytes. Is this student’s hypothesis reasonable?

A.No; the DNA in circulating erythrocytes is needed to help transport O2 through the capillaries.
B.No; circulating erythrocytes do not contain DNA.
C.Yes; DNA is responsible for cell division in most cells.
D.Yes; circulating erythrocytes carry DNA nutrients through the capillaries.

Answer 71

No; circulating erythrocytes do not contain DNA.

1. Mature erythrocytes are enucleated cells that do not contain DNA.

Question 72

Based on Figure 1 and the passage, which of the following cells are most important in the exchange of O₂ between the blood and the surrounding tissues?

• A.Pericytes
• B.Endothelial cells
• C.Smooth-muscle cells
• D.Fibroblasts

Answer 72

1. Pericytes are located between the endothelial cells and the matrix and they do not have direct contact with circulating blood.
2. Endothelial cells are the cells that are in direct contact with blood and the surrounding matrix so these are the cells that play the most important role in gas exchange.
3. Smooth-muscle cells are not in direct contact with blood, thus they are least likely to play a critical role in gas exchange between blood and the surrounding tissues.
4. Fibroblasts are supportive cells that are not in direct contact with the circulating blood, thus they are not likely to play a role in gas exchange between blood and the surrounding tissues.

Question 73

Which of the following best describes the bond that would form between the following two nucleotides if they were located adjacent to each other as shown in a single strand of DNA?

• A.A bond between the phosphate of the thymine and the phosphate of the adenine
• B.A bond between an oxygen in the thymine base and a nitrogen in the adenine base
• C.A bond between the phosphate of the thymine and the sugar of the adenine
• D.A bond between the phosphate of the adenine and the sugar of the thymine

Answer 73

A bond between the phosphate of the adenine and the sugar of the thymine

Question 74

Normally, a hypothalamic factor stimulates the release of adrenocorticotropic hormone (ACTH) from the pituitary gland. In a patient with Addison’s disease, the secretion of the hypothalamic factor will:

• A.be lower than normal.
• B.be higher than normal.
• C.be unchanged.
• D.increase before disease onset and decrease thereafter.

Answer 74

Higher than normal

1. Release of ACTH from the pituitary is regulated by negative feedback. In normal conditions, high levels of circulating glucose and other stressors activate the production of corticotropin-releasing hormone (CRH) from the hypothalamus. CRH will stimulate the pituitary gland to release ACTH which will trigger cortisol release from the adrenal cortex. Finally, the presence of high levels of circulating cortisol will inhibit CRH secretion (negative feedback) thus closing the loop. In Addison’s disease, the circulating levels of ACTH will be higher than normal because the factor that triggers the inhibition of CRH production, high cortisol levels, is absent or low.

Question 75

If a patient with Addison’s disease is given too high a replacement dose of glucocorticoids, the effect over time will be an increase in:

A.muscle mass.
B.muscle weakness.
C.red blood cell count.
D.heart rate.

Answer 75

muscle weakness.

1. High levels of circulating glucocorticoids will increase protein degradation in various tissues, muscles included. A direct consequence of protein degradation in muscles is muscle weakness.

Question 76

During the production of insulin, the translated polypeptide is cleaved into the mature form and secreted from the cell. The cleavage most likely takes place in which of the following locations?

• A.Nucleus
• B.Ribosomes
• C.Endomembrane system
• D.Cytoplasm

Answer 76

1. The endomembrane system is the portion of the cells that is in charge of modifying proteins that will be secreted. Thus, it is most likely that insulin cleavage will occur in the endomembrane system.