AAMC FL2 bio/biochem

Question 1

When levels of CRY1 are high, levels of XPA are low.

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 1

D

One mechanism in which this could be achieved is through repressing transcription of XPA-encoding genes, which will lower XPA levels.

Question 2

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 2

c, gastrointestinal epithelial cells

• During adulthood, adipocytes are unlikely to undergo mitosis. While there is a limited replicative capacity of adipocytes under certain conditions, adipocytes typically respond to changing conditions by altering their size.
• Early in development, cardiac muscle cells are proliferative. However, as an organism reaches adulthood, mitotic capacity of such cells is decreased.
• Gastrointestinal epithelial cells exhibit mitotic activity throughout adulthood. Specifically, this epithelium possesses stem cells that continuously proliferate. As such, gastrointestinal epithelial cells may have been used by the researchers to assess proliferative activity.
• Neurons are post-mitotic cells that do not typically undergo cell division in adult mammals.

Question 3

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 3

D.Phosphodiester

The DNA backbone is formed by phosphodiester linkage between deoxyribonucleotide molecules.

Question 4

Answer 4

pyridines have longer name but are smaller molecule –> singe ring

only C has two molecules with single ring

Question 5

thymine dimer is made of

Answer 5

A cyclobutane pyrimidine dimer is a dimer of two pyrimidines as in the case of a thymine dimer which may be produced by UV damage to the DNA. This makes (C) correct.

thymine is a pyrimidine (large name, smaller thing) with a single ring

Question 6

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 6

look at the equation

Reaction 1 shows that 1 mole of phosphine is produced from 1 mole of AlP reacting with 3 moles of H+. According to Le Châtelier’s principle, adding more reactant, H+, will shift the equilibrium to the right toward products, leading to more phosphine production.

think about pH
(we know that we have higher H+)
The pH (acidity) of a solution is defined by pH = −log [H+], so a higher concentration of free H+ ions is associated with a lower pH (greater acidity).

Thus, adding a solution with a pH < 4 will result in the largest amount of phosphine production bc it has the most H+ and shifts to the right most. (A) is correct.

Question 7

The passage notes that phosphine reacts with sulfhydryl groups. Cysteine contains a sulfhydryl group on its sidechain, allowing it to react.

Based on the passage, which amino acid will most likely react with phosphine?

A.Met
B.Cys
C.Ser
D.Thr

Answer 7

cys

The passage notes that phosphine reacts with sulfhydryl groups. Cysteine contains a sulfhydryl group on its sidechain, allowing it to react.

Question 8

Based on the passage, AlP inhibits the electron transport chain function.

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 8

A.Glycolytic flux is increased after AlP treatment.

Based on the passage, AlP inhibits the electron transport chain function. Therefore, to compensate for the decreased ETC-mediated ATP, glycolytic flux must be increased.

following the dec ETC (induced from AIP), glycolysis must surge

Question 9

Why was it necessary for the researchers to determine the activity of the complexes independent of one another?

A.Complex stability is lost if the complexes are able to interact structurally.
B.The complexes have different cellular locations, and it is not feasible to isolate them together.
C.The complexes all use the same substrates, so their use must be monitored separately.
D.The reactions catalyzed by the complexes are coupled to one another.

Question 10

Why was it necessary for the researchers to determine the activity of the ETC complexes independent of one another?

A.Complex stability is lost if the complexes are able to interact structurally.
B.The complexes have different cellular locations, and it is not feasible to isolate them together.
C.The complexes all use the same substrates, so their use must be monitored separately.
D.The reactions catalyzed by the complexes are coupled to one another.

Answer 10

D. The reactions catalyzed by the complexes are coupled to one another.

Because the complexes function in a chain, inhibition of complexes I and II will affect the activity of Complex III and later on Complex IV.

The ETC performs oxidative phosphorylation through a series of four protein complexes (complex I-IV) which are involved in transferring electrons from NADH and FADH2 to molecular oxygen. Complex I establishes the proton gradient by pumping hydrogen ions from the matrix into the intermembrane space. Complex II acts as a second entry point to the ETC where FADH2 enters. Complex III receives electrons from both complex I and complex II and passes them to complex IV, which reduces oxygen

Question 11

A large carbohydrate is tagged with a fluorescent marker and placed in the extracellular environment around a macrophage. The macrophage ingests the carbohydrate via phagocytosis. Which cellular structure is most likely to be fluorescently labeled upon viewing with a light microscope soon after phagocytosis?

A.Nucleus
B.Golgi apparatus
C.Lysosome
D.Endoplasmic reticulum

Answer 11

C, lysosome

During phagocytosis, large molecules from the extracellular environment are internalized into an organelle referred to as a phagosome. Phagosomes can fuse with lysosomes, which contain hydrolytic enzymes. This allows for the digestion of contents in the phagosome, such as carbohydrate molecules.

Question 12

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 12

C. I and II only

I. Phosphofructokinase is the main regulatory enzyme of glycolysis and, as such, it is regulated both allosterically as II. well as by a negative feedback loop by the final glycolytic product, ATP.

Option III, competitive inhibition, is not true since ATP is not competing with the substrate for the active site but is the substrate of PFK-1.

Phosphofructokinase-1 (PFK-1) catalyzes the irreversible conversion of fructose-6-phosphate and ATP into fructose-1,6-bisphosphate and ADP during glycolysis. At high ATP concentrations, ATP saturates the active site. Thus, ATP is available to bind to the allosteric site of PFK-1, which inhibits PFK-1. This is an example of allosteric regulation. Option I is true. Eliminate (B) and (D). Moreover, ATP is a product of glycolysis, so inhibition of PFK-1 by ATP prevents unnecessary production of ATP through glycolysis. This is an example of feedback inhibition. Option II is also true. (C) is correct.

Question 13

Answer 13

Active transport

According to the graph, the concentration of chloride ions is greater in cytoplasm than in pond water. Spontaneously, chloride ions will want to travel down the concentration gradient, from an area of higher concentration to an area of lower concentration. Moving chloride ions into the cells of the green algae is against the concentration gradient and will require energy. Active transport uses chemical energy such as ATP to move ions across a cell membrane from an area of lower concentration to an area of higher concentration. Thus, (C) is correct.

Question 14

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.
B.It would shorten the refractory period.
C.It would prolong the action potential.
D.It would prevent depolarization.

Answer 14

In response to a stimulus, a somatic motor neuron fires an action potential. Depolarization occurs because voltage-gated sodium ion channels open and allow sodium ions to diffuse into the cytoplasm. The membrane potential of a neuron becomes highly positive because of the influx of sodium ions. Once the threshold is reached, sodium ion channels close and voltage-gated potassium channels open, allowing potassium ions to diffuse out during repolarization. Hyperpolarization occurs when the membrane potential becomes more negative than the resting potential, so the neuron cannot transmit another action potential during this refractory period until the membrane potential equals the resting potential.

Since dendrotoxin blocks voltage-gated potassium channels, exposure to dendrotoxin will lead to a prolonged action potential and the cell will not be able to repolarize and transmit another action potential. (C) is correct.

Question 15

The passage notes that Wnt proteins have an isoelectric point of 9

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 15

B

The passage notes that Wnt proteins have an isoelectric point of 9 and therefore they are charged positively at physiological pH range.

so pH < pI will mean the protein is pos charged (low pH)

pH > pI will mean the protein is neg charged (high pH)

Question 16

Secretory proteins are produced by __ on the _______ (ER) and folded in the lumen of the ____ ER.

Answer 16

Secretory proteins are produced by ribosomes on the rough endoplasmic reticulum (ER) and folded in the lumen of the rough ER.

Question 17

isoelectric point, pI

Answer 17

The isoelectric point (pI) is the pH at which the number of negatively charged groups (—COO−) and the number of positively charged groups (—NH3+) are roughly the same and the molecule is electrically neutral. At pH lower than the pI, the molecule will be positively charged as the amino groups become more protonated, similarly at pH higher than pI, the molecule becomes negatively charged as the carboxyl group undergoes deprotonation.

Question 18

According to the passage and Figure 1, β-catenin activates transcription factors for Wnt target genes when the Wnt signaling pathway is turned on

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 18

C

Since transcription factors are found in the nucleus, β-catenin must transport from the cytoplasm into the nucleus. For a protein to gain entry into the nucleus, it must possess a nuclear localization signal (NLS), an amino acid sequence that identifies a protein for nuclear transport. Therefore, β-catenin most likely has a nuclear localization sequence. (C) is correct.

Question 19

“In the absence of Frizzled activation, CK1 and GSK3 sequentially phosphorylate β-catenin, which targets it for ubiquitination”

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 19

In the absence of Frizzled activation, β-catenin is phosphorylated and ubiquitinated.

Ubiquitination marks proteins for degradation by a proteasome.

this means its cut into small segments (A)

Question 20

The middle region, as opposed to either end, of each of the seven α-helical domains of Frizzled is most likely to contain a high proportion of which type of amino acid residue?

A.Nonpolar
B.Polar uncharged
C.Positively charged
D.Negatively charged

Answer 20

A.Nonpolar

The passage notes the seven α-helical domains are located within the membrane and therefore most likely to contain nonpolar and thus hydrophobic amino acids.

Question 21

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 21

B

Figure 1 shows the formation of a single band in gel electrophoresis under the native and denaturing conditions, indicating Protein X is one unit. Under the denaturing/reducing conditions, Protein X consists of two subunits, as indicated by the formation of two bands in gel electrophoresis. Reducing agents such as dithiothreitol (DTT) are used to break protein-protein disulfide linkages. Thus, the formation of two bands in gel electrophoresis under the reducing conditions indicates disulfide linkages between the Protein X subunits. Disulfide linkages can only be formed by cysteines as they contain thiol (—SH) groups. Thus, (B) is correct.

Question 22

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.

“The passage notes that upon binding to hydrophobic surface residues, ANS exhibits increased fluorescence.”

Answer 22

Figure 2 shows that ANS fluorescence when DPC is added to ANS + Protein X is greater than ANS fluorescence of ANS + Protein X or ANS alone. The increased fluorescence can be inferred as the result of a conformational change in Protein X, leading to more ANS binding to hydrophobic amino acids that are exposed in the presence of DPC. Thus, (C) is correct.

so DPC must somehow expose the enzymes

Question 23

Which interpretation(s) is(are) consistent with the observations in the passage?

Surface amino acids of Protein X are mostly hydrophilic in aqueous solution.
Surface amino acids of Protein X are mostly hydrophilic in presence of DPC.
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 23

I and III

Based on data shown in Figure 2, Protein X has only a few surface hydrophobic residues in aqueous solution, but as evidenced by increased ANS binding, addition of DPC results in increased surface hydrophobic residues.

Question 24

What event will most likely occur if Protein X is inserted into the inner membrane of mitochondria?

A.The citric acid cycle will cease to function.
B.The electron transport chain will cease to function.
C.The proton gradient across the inner membrane will dissipate.
D.The pH of the intermembrane space will decrease.

The gradual release of dye from liposome when Protein X is added

Answer 24

C.

The gradual release of dye from liposome when Protein X is added indicates that the protein can form a channel, thus making the mitochondrial inner membrane permeable, which causes the dissipation proton gradient across the membrane.

Insertion of a channel across the mitochondrial inner membrane will most likely increase, not decrease, the pH of intermembrane space by the dissipation of the proton gradient across the membrane.

Protein X can form a channel across the mitochondrial inner membrane, but this will not directly interfere with the functioning of the citric acid cycle, as the citric acid cycle occurs within the mitochondrial matrix OR the functioning of the electron transport chain, as the electron transport chain can bypass the channel

Question 25

The enzyme encoded by the glyceraldehyde-3-phosphate dehydrogenase (GAPDH), 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 25

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

The reversible interconversion of glyceraldehyde-3-phosphate and 1,3-bisphosphoglycerate during glycolysis and gluconeogenesis is catalyzed by the enzyme glyceraldehyde-3-phosphate dehydrogenase, which is encoded by the tdh2 gene.

Question 26

What is the best experimental method to analyze the effect of tdh2 gene (PROTEIN) deletion on the rate of histone acetylation? Comparing histone acetylation in wild-type and Δtdh2 cells by:

A.Western blot
B.Southern blot
C.Northern blot
D.RT-PCR

Answer 26

Histone acetylation is a post-translational modification where an acetyl group is transferred to a lysine residue in a histone molecule. Acetylation of histone molecules increases their molecular weight.

A method often used to identify proteins based on molecular weight is western blot. A western blot is a technique in which protein samples can be separated on the basis of molecular weight and then detected using specific antibodies. Thus, the rate of histone acetylation due to the deletion of the tdh2 gene can be analyzed by performing the western blot technique with proteins extracted from the wild-type and Δtdh2 cells. (A) is correct.

_____________
(B) and (C) are incorrect because a Southern blot is a technique used to separate and identify DNA, while a northern blot is used to separate and detect RNA. Neither of these techniques can analyze histone acetylation, which is a modification of the histone protein. (D) is incorrect because RT-PCR is the process of amplifying a DNA strand obtained from its corresponding RNA using reverse transcriptase and does not identify proteins.

Question 27

Which experimental approach(es) can be used to analyze the effect of ROS on the lifespan of yeast? Comparing the lifespans of:

I. wild-type yeast versus yeast lacking antioxidant enzymes
II. wild-type yeast versus yeast overexpressing antioxidant enzymes
III. yeasts growing in the presence or absence of hydrogen peroxide
A.I only
B.II only
C.II and III only
D.I, II, and III

Answer 27

All three of the variables indicated in options I, II, and III affect ROS levels in yeast, and therefore any of them can be used to study the role of ROS in influencing lifespan in yeast.

Reactive oxygen species (ROS) include hydroxyl radicals, superoxides, and peroxides. Build up of ROS results in oxidative stress in cells, which can cause many harmful effects including DNA damage and apoptosis. Antioxidant enzymes can deactivate oxygen free radicals to protect cells from ROS-induced damage. Thus, comparing the lifespan of wild-type yeasts with the lifespan of yeast lacking antioxidant enzymes can be used to analyze the effect of ROS on yeast lifespan. Option I is true. Eliminate (B) and (C). A similar comparison with yeast overexpressing antioxidant enzymes can also be used to analyze the effect of ROS on yeast lifespan. Option II is true. Eliminate (A). (D) is correct, so option III is also true. Since hydrogen peroxide (H2O2) is endogenous ROS, cells growing in the presence and absence of hydrogen peroxide (H2O2) can also be used to analyze the effect of ROS on the lifespan of yeast.

Question 28

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 28

A. The collecting duct is a renal structure responsible for the concentration of urine as it passes from individual nephrons to a minor calyx. The process of concentrating urine is regulated by aquaporin proteins that mediate water reabsorption from the urine. Specifically, the hormone vasopressin is released during periods of dehydration, which increases expression of aquaporins in the collecting ducts.

Question 29

aquaporins in the proximal convoluted tubule and vasopressin

Answer 29

Aquaporins in the proximal convoluted tubule mediate water reabsorption. However, expression of aquaporins in this region is relatively constant and is not regulated by vasopressin.

Question 30

bowmans capsule

Answer 30

Bowman’s capsule is the initial site where contents that have been filtered from the blood, such as water, are located.

Question 31

Aquaporins are expressed along the apical membrane of the ____ loop of Henle.

Answer 31

Aquaporins are expressed along the apical membrane of the descending, not ascending, loop of Henle.

Vasopressin does not regulate aquaporin expression in the loop of Henle.

Question 32

aquaporins in the ascending loop are where?

Answer 32

Instead, aquaporins are positioned along the basolateral membrane of the ascending loop of Henle. Vasopressin does not regulate aquaporin expression in the loop of Henle.

Question 33

What are the primary myelin-forming cells in the peripheral nervous system?

A.Microglia
B.Astrocytes
C.Schwann cells
D.Oligodendrocytes

Answer 33

C.

Schwann cells are the primary myelin-producing cells in the peripheral nervous system. Specifically, myelin insulates axons, allowing for the rapid propagation of action potentials.

(Oligodendrocytes are the primary myelin-producing cell in the central nervous system, not the peripheral nervous system)

Question 34

Answer 34

Question 35

A prion is best described as an infectious:

A.prokaryote.
B.transposon.
C.protein.
D.virus.

Answer 35

C.protein

Question 36

Assume that S. typhi immediately enters the bloodstream from the small intestine. Of the following, which would be the first major organ that bloodborne S. typhi would encounter?

A.Stomach
B.Pancreas
C.Large intestine
D.Liver

Answer 36

Deoxygenated blood carrying nutrients and toxins absorbed during digestion travels from the small intestines to the liver via the hepatic portal vein.

(This allows for nutrient-rich blood to be balanced and detoxified before it is delivered to the rest of the body.)

The liver then breaks down complex molecules for storage or passes the products into the circulating blood for further utilization or excretion. Therefore, S. typhi entering the bloodstream from the small intestine would first encounter the liver. (D) is correct.

(A) and (B) are incorrect because the stomach and pancreas are upstream of the small intestine. Food moves out of the stomach into the small intestine, and the pancreas delivers digestive enzymes to the small intestine.

Within the gastrointestinal tract, the large intestine comes immediately after the small intestine. However, once nutrients are absorbed from the small intestine into the blood, the blood is not routed to the large intestine.

Question 37

According to the passage, “the causative agent of typhoid fever is the bacterium Salmonella typhi.” Antibodies disrupt essential processes of the bacteria such as protein synthesis, which lead to slowed growth and/or death.

Researchers have noted that chloramphenicol (a commonly used antibiotic) is becoming less effective in treating typhoid fever. The best explanation for this observation would be selection:

A.against chloramphenicol in ΔF508 heterozygotes.
B.against chloramphenicol in wild-type homozygotes.
C.for chloramphenicol resistance in populations of S. typhi.
D.for ΔF508 CFTR, which cannot bind chloramphenicol.

Answer 37

ecreased efficacy of antibiotic chloramphenicol in treating typhoid fever suggests the increased selection for S. typhi that are resistant to chloramphenicol (i.e., S. typhi that are resistant to chloramphenicol are those that survive). The emergence of pathogens developing resistance against the damaging effects of an antibiotic after repeated exposure is antibiotic resistance. (C) is correct.

Question 38

CF can be caused by two copies of the ΔF508 allele, however, ΔF508 heterozygotes only contain one ΔF508 allele. Therefore, ΔF508 heterozygotes are not affected by CF. However, since S. typhi infection utilizes wild-type CFTR for host infection, individuals possessing the ΔF508 allele will have decreased S. typhi infection, and therefore be more resistant against typhoid fever.

Which of the following best describes the phenotype of an individual who is heterozygous with one ΔF508 and one wild-type CFTR allele?

A.More susceptible to typhoid fever than wild-type homozygotes and has CF
B.More susceptible to typhoid fever than ΔF508 homozygotes and does not have CF
C.More resistant to typhoid fever than ΔF508 homozygotes and has CF
D.More resistant to typhoid fever than wild-type homozygotes and does not have CF

Answer 38

D

More resistant to typhoid fever than wild-type homozygotes and does not have CF

Question 39

releasing factors vs neuromodulators

Answer 39

releasing factors directly stimulate release of a hormone on their own, while neuromodulators either enhance or reduce the action of a releasing factor.

Question 40

NPY enhances the ability of GnRH to increase LH levels. GnRH achieves this through activation of its receptor

Scientists are hopeful that NPY can be used in combination with GnRH to treat certain cases of female infertility. Individuals with a deficiency in what receptor system would be most likely to benefit from such a treatment?

A.GnRH
B.LH
C.NPY
D.LH and NPY

Answer 40

A

According to the passage, NPY enhances the ability of GnRH to increase LH levels. GnRH achieves this through activation of its receptor. Individuals experiencing infertility as caused by deficiencies of the GnRH receptor would likely benefit from increased functioning of the GnRH receptor. Since NPY does this, NPY and GnRH treatment would likely be an effective treatment for infertility caused by deficiencies with the GnRH receptor.

Question 41

Scientists have hypothesized that NPY is necessary for the generation of the preovulatory LH surge, a hormonal event that triggers ovulation. Which of the following findings best supports this hypothesis?

A.When the actions of NPY are blocked in female rats, the LH surge and ovulation do not occur.
B.NPY release from the hypothalamus increases just prior to the preovulatory LH surge.
C.NPY can enhance GnRH-stimulated LH secretion in female rats during the preovulatory period.
D.NPY has no effect on GnRH-stimulated LH secretion in male rats.

Answer 41

A When the actions of NPY are blocked in female rats, the LH surge and ovulation do not occur.

To demonstrate that NPY is necessary for the generation of a preovulatory LH surge and subsequent ovulation, one would need to block the actions of NPY. Specifically, if one were to observe that the LH preovulatory surge and subsequent ovulation failed to occur following NPY blockade, then this would indicate that NPY is required for these events to occur.

Question 42

NPY alone does not significantly alter LH levels.

Answer 42

A

Estrogen levels can be enhanced by LH. However, according to the passage, NPY alone does not significantly alter LH levels. Therefore, treatment with NPY alone will be unlikely to alter LH levels, which would keep estrogen levels relatively consistent.

Question 43

The passage mentions that inhibition of endosomal acidification can block EboV infection.

Based on the passage, CatB and CatL most likely act on EGP in which of the following cellular compartments to facilitate membrane fusion?

A.Endoplasmic reticulum
B.Golgi apparatus
C.Endosomes
D.Cytosol

Answer 43

C

The passage mentions that inhibition of endosomal acidification can block EboV infection. This suggests that EboV is internalized through the process of endocytosis, in which endosomes form. In such an endosome, CatL and CatB can target EGP, allowing for fusion of the viral membrane with the endosomal membrane. This permits entry of EboV into the cytoplasm.

Question 44

According to the passage EGP is a transmembrane viral glycoprotein. Paragraph 3 states that CatB and CatL are proteases that, when inhibited, greatly reduce the infection of EGP-containing viruses.

Based on the passage, CatB or CatL or both would be expected to have which of the following effects, if any, on EGP?

A.No effect
B.Reduction of enzyme activity
C.Formation of protein dimers
D.Digestion into smaller protein fragments

Answer 44

D

CatB and CatL are proteases that target EGP, which is a glycoprotein. Proteases degrade proteins, such that CatB and CatL will likely digest EGP into smaller fragments.

Question 45

The precursor of EGP is translated from a transcript that has had one nontemplated nucleotide added to the open reading frame. This change does not create or eliminate a stop codon. Compared with the protein sGP, which is produced from the unedited transcript, EGP most likely has the same primary:

A.amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence.
B.carboxy-terminal sequence as sGP, but a different primary amino-terminal sequence.
C.sequence as sGP except that EGP has one additional amino acid.
D.sequence as sGP except that EGP has one less amino acid.

Answer 45

A

Codons before the inserted nucleotide will not be affected, but every codon downstream of the inserted nucleotide will be altered. Since proteins are synthesized from the N-terminus to the C-terminus, the amino terminal sequence before the inserted nucleotide in sGP will be unaffected, but the carboxy-terminal sequence after the inserted nucleotide will be changed. So, EGP will likely have a different carboxy-terminal sequence than sGP. This matches (A).
(B) is incorrect because proteins are synthesized from N-terminus to C-terminus, so the amino-terminal sequence will be the same but the carboxy-terminal sequence will be different. (C) is incorrect because in order to add an amino acid to the primary sequence without causing a frameshift, a complete codon of three nucleotides must be added to the mRNA transcript, not a single nucleotide as described in the question. Similarly, removal of an amino acid would require the removal of a complete codon of three nucleotides, making (D) incorrect as well.

Question 46

Which molecule is NOT formed during the citric acid cycle?

A.Malate
B.Succinate
C.α-Ketoglutarate
D.Phosphoenolpyruvate

Answer 46

D.Phosphoenolpyruvate

Question 47

Under anaerobic conditions, how many net molecules of ATP are produced by the consumption of 5 moles of glucose?

A.3 × 1024
B.6 × 1024
C.9 × 1024
D.1.2 × 1025

Answer 47

In anaerobic respiration, the pyruvate generated at the end of glycolysis is converted to lactate, which does not yield additional ATP. Glycolysis uses two ATP to produce four ATP, so anaerobic respiration results in the net production of 2 moles of ATP per 1 mole of glucose. Accordingly, 5 moles of glucose would produce 5 mol glucose × 2 mol ATP/1 mol glucose = 10 moles of ATP.

There are 6 × 10&23 molecules in 1 mole, so 10 moles of ATP equals 10 mol ATP × 6 × 10^23 molecules/1 mole = 60 × 10^23 molecules or 6 × 10^1 × 10^23 = 6 × 10^24 molecules. (B) is correct

Question 48

In humans, eggs and sperm are most similar with respect to:

A.cell size.
B.genome size.
C.the time required for development.
D.the numbers produced by a single individual.

Answer 48

genome size

Both human sperm and egg cells are haploid, meaning that they contain one copy of each chromosome. This contrasts with most somatic cells, which are diploid, meaning they contain two copies of each chromosome.

• egg cell is much larger than a sperm cell. Unequal cytokinesis during oogenesis distributes ample cytoplasm to one daughter cell, resulting in one viable egg and a variable number of polar bodies.
• Oogenesis occurs once per month from puberty to menopause. One cycle of spermatogenesis results in four functional sperm since cytoplasmic divisions are equal. Approximately 3 million sperm are produced per day starting from puberty.

Thus, the human egg and sperm vary in their cell size and development but are similar with respect to their genome size. (B) is correct.

Question 49

In humans, the characteristic tissue of which of the following organs is NOT derived from mesoderm?

A.Brain
B.Heart
C.Kidney
D.Skeletal muscle

Answer 49

A, The brain is derived from ectoderm, not mesoderm.

Question 50

what does the mesoderm develop into?

Answer 50

The mesoderm is the middle layer of the three primary germ layers formed during embryonic development. This layer gives rise to the musculoskeletal, circulatory, and most of the excretory systems, as well as the gonads. Thus, the heart, kidney, and skeletal muscle are all derived from the mesoderm.

Question 51

Given that secretory lysosomes form normally in the melanocytes and CTLs of ashen and dilute mice, the data in Table 1 best support the conclusion that melanosome secretion and lytic granule secretion differ in that the secretion of lytic granules does NOT require:

A.Rab27a.
B.myosin Va.
C.microtubules.
D.fusion of these secretory lysosomes with the plasma membrane.

Answer 51

According to Table 1, mice with inactivation of myosin Va still exhibit killing by CTLs, suggesting that lytic granules can still be secreted. Mice with inactivated myosin Va, however, exhibit partial albinism, demonstrating that melanosomes are not secreted in these mice. This suggests that melanosome secretion requires myosin Va, while lytic granule secretion does not.

Question 52

Based on the passage, myosin Va most likely directly binds:

A.tubulin.
B.actin.
C.melanin.
D.Rab27a.

The passage indicates that myosin Va binds to microfilaments.

Answer 52

According to paragraph 2, Rab27a connects to myosin Va, which is a motor protein on microfilaments. Since microfilaments are made up of actin monomers, myosin Va most likely directly binds to actin. Thus, (B) is correct.

Question 53

Secretory lysosomes are classified as lysosomes because secretory lysosomes have some functional components in common with conventional lysosomes. Given this, secretory lysosomes most likely contain:

A.ribosomes.
B.Krebs cycle enzymes.
C.RNA and DNA polymerases.
D.degradative enzymes that function at low pH.

Answer 53

D
Lysosomes are organelles with an acidic lumen that is enriched with digestive enzymes. This is important for the hydrolysis of macromolecules and infectious agents. Consequently, it is likely that secretory lysosomes also have these structures.

Question 54

Melanosomes most likely move along microtubules that originate in and radiate from the:

A.centrosome.
B.kinetochores.
C.Golgi apparatus.
D.microfilaments under the plasma membrane.

Answer 54

A - Centrosomes are organelles that serve as the primary microtubule organizing center of a cell. As such, melanosomes most likely move along microtubules produced by and radiating from, the centrosome.

Question 55

centrosome vs centromere

Answer 55

Centrosomes are organelles that serve as the primary microtubule organizing center of a cell. As such, melanosomes most likely move along microtubules produced by and radiating from, the centrosome.

The centromere appears as a constricted region of a chromosome and plays a key role in helping the cell divide up its DNA during division (mitosis and meiosis). Specifically, it is the region where the cell’s spindle fibers attach

Question 56

kinetechores

Answer 56

Kinetochores are proteins positioned at the centromere of a chromosome. Kinetochores can attach microtubules, but do not produce microtubules.

Question 57

golgi apparatus and MT

Answer 57

The Golgi apparatus is an organelle often positioned near the centrosome, which produces microtubules. However, the Golgi apparatus is not the primary site for the generation of microtubules.

Question 58

lytic granules are generally released from CTLs when the T-cell receptors on these cells bind specifically to:

A.viral antigens presented on the surface of virus-infected cells.
B.growth factors secreted by helper T lymphocytes.
C.B-cell receptors on activated B lymphocytes.
D.constant regions of secreted antibodies.

Answer 58

A

CTLs are effector T-cells produced in response to activation of a naive T-cell by an antigen presenting cell. T-cell receptors of CTLs bind to antigen present on the surface of an infected cell and release lytic granules which in turn activate apoptosis of the infected cell. Thus, (A) is correct.

Question 59

Based on the passage, Foxp3 affects ErbB2 expression in noncancerous mammary epithelium most likely by directly:

A.inhibiting synthesis of ErbB2 mRNA.
B.stimulating synthesis of ErbB2 mRNA.
C.inhibiting synthesis of ErbB2 protein.
D.stimulating synthesis of ErbB2 protein.

The passage notes that Foxp3 binds to the ErbB2 promoter

Answer 59

A.

The passage notes that Foxp3 binds to the ErbB2 promoter, which controls transcription of the ErbB2 gene. However, expression of ErbB2 increases when the Foxp3 binding site is deleted. This indicates that Foxp3 functions to inhibit synthesis of mRNA for ErbB2, which is produced from transcription.

Question 60

Given that the tumorigenicity of a certain mouse mammary tumor cell line is dependent on ErbB2-mediated intracellular signaling, mice injected with variants of this cell line that have which of the following modifications would be most likely to survive the longest?

A.The Foxp3 genes deleted
B.One of the ErbB2 genes amplified
C.A Foxp3-expressing plasmid introduced
D.The Foxp3 binding sites deleted in the promoter of one of the ErbB2 genes

Foxp3 binds to the ErbB2 promoter. However, deletion of this binding site leads to increased expression of ErbB2.

Answer 60

Foxp3 binds to the ErbB2 promoter. However, deletion of this binding site leads to increased expression of ErbB2.

Binding of Foxp3 to the ErbB2 promoter would, therefore, be associated with decreased expression of ErbB2. Compared to cancerous mammary epithelium, there is lower expression of ErbB2 in noncancerous mammary epithelium. Therefore, introduction of a Foxp3-expressing plasmid would reduce ErbB2 levels. This would decrease cancer progression and increase survival.