MCAT Section Bank- Biology & Biochemistry

phosphatase

enzyme that removes a phosphate group from its substrate

According to the information in the passage, which statement best describes the function of the MR in response to stimulation of the sympathetic nervous system? The MR:

"Blood flow to the brain is regulated by changes in arterial vessel diameter in response to intravascular pressure, a process called the myogenic response (MR). The MR allows cerebral resistance vessels to achieve a state of partial constriction, from which they can dilate or further constrict to alter blood flow."

A) equalizes blood pressure in cerebral resistance arteries and the aorta.
B) moderates blood flow to the brain under high pressure.
C) enables cerebral resistance arteries to locally vasodilate.
D) redirects blood circulating in the brain to organs in the abdominal cavity.

B) moderates blood flow to the brain under high pressure. The answer to this question is B because the MR is a local response in resistance vessels that would slow blood flow to the brain to reduce the chance of tissue damage from high blood pressure resulting from activation of the sympathetic nervous system.

The polymerization of which structural component was analyzed in the experiment described in the passage? (Actin polymerization)
A) Microtubules
B) Microfilaments
C) Intermediate filaments
D) Thick filaments

B) Microfilaments The answer to this question is B because microfilaments are composed of actin.

Which amino acid is LEAST likely found in one of the transmembrane domains of GPCR43?
A) Asp
B) Gly
C) Trp
D) Phe

A) Asp The answer to this question is A because aspartate (Asp) has a negatively charged R group and is very hydrophilic.

Compared to untreated WT mice, antibiotic treatment of WT mice is likely to result in:

A) increased plasma butyrate levels.
B) decreased body mass.
C) increased volume of adipocytes.
D) decreased insulin sensitivity in adipocytes.

C) increased volume of adipocytes The answer to this question is C because if SCFA producing microbes were removed by antibiotic treatment, then GPCR43 would not be activated to reduce glucose uptake into adipocytes.

Which amino acid contains an unbranched alkyl side chain?
A) Ala
B) Ile
C) Leu
D) Val

A) Ala The answer to this item is A since alanine contains a methyl side chain, which is not considered a branched alkane.

The stereochemical designators α and β distinguish between:
A) enantiomers at an epimeric carbon atom.
B) enantiomers at an anomeric carbon atom.
C) epimers at an anomeric carbon atom.
D) epimers at a non-anomeric carbon atom.

C) epimers at an anomeric carbon atom. The answer to this item is C since the D versus L designation distinguishes between molecules with multiple chiral centers, but differ only in the configuration of the site known as the anomeric carbon atom.

Which ideal solution exhibits the greatest osmotic pressure?
A) 0.1 M MgCl2
B) 0.2 M NaCl
C) 0.2 M CaCl2
D) 0.5 M Glucose

C) 0.2 M CaCl2 The answer to this question is C because the solution listed has the greatest concentration of solute particles. CaCl2 dissociates into three ions, making the solution listed 0.6 M in solute particles.

Which event is directly mediated by a ligand-gated ion channel?

A) Release of Ca2+ from the sarcoplasmic reticulum of a muscle fiber to initiate muscle contraction
B) Influx of Na+ across the axon membrane of a somatic neuron during action potential propagation
C) Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane
D) Re-entry of Ca2+ back into the sarcoplasmic reticulum of a muscle fiber to end muscle contraction

C) Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane The answer to this question is C because the influx of Na+ across the motor end plate occurs when Na+ ion channels bind the ligand acetylcholine.

Southern blot

A Southern blot uses a restriction digest to differentiate between mutant and wild-type alleles. In order for a Southern blot to be useful, the mutation should either create or eliminate a restriction site, most of which are palindromes and 4 to 6 base pairs long. The mutation shown in this option is the only one that disrupts a palindromic sequence, AAGCTT. This sequence is the recognition sequence for HindIII.

What bond is cleaved by IN during the first reaction of integration?

"Upon binding, IN catalyzes cleavage of a GT dinucleotide from each 3 PRIME vDNA terminus"

A) P-C
B) P-P
C) P-H
D) P-O

D) P-O The answer to this question is D because in the first reaction of integration, integrase cleaves a dinucleotide from the 3' end of a strand of viral DNA.

A vDNA sequence encoding a protein is inserted into a host genome by IN. The protein is translated from the hypothetical mRNA sequence shown.

5'-GGCAACUGACUA-3'

Based on the passage, the segment of the original viral genome that encoded this protein had what nucleotide sequence?

5'-GGCAACUGACUA-3' The answer to this question is A because according to the passage, viral DNA integrated into a host cell genome by integrase would originate from a retrovirus. mRNA transcribed from retroviral DNA is either used to synthesize viral proteins, or used as the RNA genome for progeny viruses. Thus, the sequence of the nucleotide in the original viral genome will be the same as that of the transcribed mRNA.

Which experiment would best provide data to support the mechanism by which an ODN inhibits IN activity?

"ODNs are competitive inhibitors of vDNA for binding to IN."

A) Keep IN concentration constant and measure the initial velocity of the reaction at increasing vDNA concentrations in the presence and absence of the ODN.
B) Keep vDNA concentration constant and measure the initial velocity of the reaction at increasing concentrations of IN in the presence and absence of the ODN.
C) Keep ODN and IN concentration constant and measure initial velocity of the reaction at increasing concentrations of vDNA.
D) Keep IN and vDNA concentrations constant and measure initial velocity of the reaction at increasing concentrations the ODN.

A) Keep IN concentration constant and measure the initial velocity of the reaction at increasing vDNA concentrations in the presence and absence of the ODN. Competitive inhibition can be determined through rate experiments by applying the principles of the Michaelis-Menten equation. By keeping enzyme concentration constant, varying substrate concentration, and either including or excluding the inhibitor, the effect of the inhibitor on the Vmax and apparent Km of the reaction can be determined. A competitive inhibitor will increase the apparent Km and not affect the Vmax.

Which amino acid substitution for the conserved residue at position 64 is LEAST likely to affect the enzymatic function of IN?

Highly conserved residue = D64

A) Asparagine
B) Glutamate
C) Lysine
D) Valine

B) Glutamate The answer to this question is B because the conserved residue at position 64 in integrase is aspartate, which has a negatively charged side chain. Substitution with glutamate, which also has a negatively charged side chain, is least likely to affect integrase function.

An inactive tetramer of IN is expected to have approximately what molecular weight?

"IN is a 288-residue protein composed of three domains"

A) 16 kDa
B) 32 kDa
C) 64 kDa
D) 128 kDa

D) 128 kDa The answer to this question is D because the passage indicates that an integrase monomer is composed of 288 amino acids, which will have an approximate molecular weight of 32 kDa (the average molecular weight of an amino acid is 110 Da). Thus, a tetramer will have an approximate molecular weight of 128 kDa.

To determine if a small molecule acts like a LEDGIN with respect to IN, a researcher plans to incubate purified IN both with and without the small molecule and then perform a Western blot to detect IN in each sample. Under which condition(s) should the gel electrophoresis step be performed?
I. Denaturing
II. Reducing
III. Native

III only The answer to this question is C because if the intent is to confirm that a small molecule induces the formation of integrase tetramers from integrase dimers, it is necessary to visualize the proteins in their native state. Use of a denaturing agent will disrupt the interactions between monomers. Use of a reducing agent only will disrupt any disulfide bonds.

What is the dependent variable from the experiment shown in Figure 2?

X-axis = time
Y-axis = relative retinol fluorescence

A) STRA6/LRAT conditon
B) Retinol fluorescence
C) Time duration of assay
D) Untransfected cell membranes

B) Retinol fluorescence The answer to this question is B because the dependent variable is the variable that is being measured in the experiment.

The organ in which the holo-RBP complex forms also functions to:
(liver)

A) store bile
B) secrete glucagon
C) produce hydrochloric acid
D) detoxify drugs

D) detoxify drugs The answer to this question is D because according to the passage the holo-RBP complex forms in the liver and one of the function of the liver is drug detoxification.

In a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation). If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Note: Assume Mendelian inheritance patterns.)
A) 4/9
B) 1/2
C) 2/3
D) 3/4

A) 4/9 The answer to this question is A because given Mendelian inheritance patterns, a 3:1 ratio of F1 offspring means that the original crossed beetles are both heterozygotes, and the F1 offspring are 25% red (homozygous dominant), 50% red (heterozygous), and 25% brown (homozygous recessive). If two red F1 beetles are crossed and both red and brown beetles appear in the F2 generation, the F1 red beetles that were crossed must both be heterozygotes. The probability that, of the red F1 beetles, both were heterozygous is 2/3 × 2/3, or 4/9 (only red beetles were selected from and 2/3 of the red F1 beetles were heterozygous).

Hill coefficient

>1 cooperativity =1 no cooperativity <1 negative cooperativity

noncompetitive inhibitor

binds the enzyme and enzyme-substrate complex with the same affinity

In Figure 1, which statement best explains the data trend of an increase followed by a decrease in bacterial cell count?

A) The number of host cells in the sample increased until carrying capacity was reached.
B) Infection with VP is maximal at 5 hours of exposure.
C) VP replicates inside the host cell before causing host cell lysis.
D) Rate of VP entry into host cells increases then decreases over time.

C) VP replicates inside the host cell before causing host cell lysis. The answer to this question is C because in Experiment 1, the host cells are exposed to VP for only two hours, then washed and incubated in VP-free media containing an antibiotic. Thus, the levels of intracellular bacteria (VP that has entered the host cell) measured at each time point represent the amount of VP in the host cells at that time point. The increasing levels at hours 3 and 4 suggest that VP is replicating inside the host cell. The sudden drop in intracellular bacterial levels is consistent with the hypothesis that the host cells are beginning to lyse, as the bacteria that had previously been inside the cell are rinsed away or killed by the antibiotic

The amino acid residue at position 61 in Rac is most likely:

"VopC contains a catalytic domain that irreversibly activates host cell GTPase Rac through the deamidation of the side chain of the residue at position 61."

A) glutamine
B) leucine
C) arginine
D) tyrosine

A) glutamine The answer to this question is A because according to the passage, VopC removes an amide from the side chain of the amino acid residue at position 61 in Rac via a deamidation reaction. Of the options, only glutamine contains a side chain with an amide group.

What is the independent variable in Experiment 1?

X-axis: hourly intervals in which sample of host cells were collected and rinsed
Y-axis: intracellular bacterial levels

A) Length of time host cells were exposed to the bacteria
B) Amount of cells collected in each sample
C) Time at which the cell samples were collected
D) Amount of intracellular bacteria measured at each time point.

C) Time at which the cell samples were collected The answer to this question is C because the independent variable in an experiment is that which is varied by the researcher in order to determine the effect (the dependent variable). In Experiment 1, the time at which cell samples are collected is the independent variable.

A Rac variant, in which the residue at position 61 was replaced with an alanine (Rac61A), was synthesized. Wild-type Rac and Rac61A were incubated separately with VopC. To obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of which compound?
A) CO2
B) NH3
C) O2
D) H2S

B) NH3 The answer to this question is B because a deamidation reaction releases NH3, thus, if NH3 is detected in the sample containing wild-type Rac but not Rac61A, this will provide data to support that VopC deamidates Rac at residue 61.

Why was the total amount of Rac measured in Experiment 2?

"To confirm that VopC activates Rac, a mutant strain of VP lacking the VopC gene was created..."

A) To confirm that the antibody used to detect GTP-Rac was working in each sample
B) To determine if the same amount of each sample was loaded on the gel
C) To act as a control for comparing rates of Rac activation by VopC
D) To determine if VopC affects Rac expression levels

D) To determine if VopC affects Rac expression levels The answer to this question is D because in order to determine how the presence or absence of VopC affects Rac activation, it is important to know if the total amount of Rac is affected by the strain being used. If the Rac expression is upregulated in one strain and not another, this will affect interpretation of the gel results.

adenine structure

guanine structure

cytosine structure

thymine structure

uracil structure

Isoelectric focusing of samples includes which step?:

A) Amplify the samples by PCR before running on the gel.
B) Establish a stable pH gradient in the gel before adding samples.
C) Denature the samples so the proteins unfold into a linear chain.
D) Treat the samples with a chemical that will add an overall negative charge to each protein.

B) Establish a stable pH gradient in the gel before adding samples. The answer to this question is B because according to the passage, visualization of the proteins in Experiment 2 is done through isoelectric focusing, which separates proteins based on isoelectric point. To do this, a stable pH gradient must be established in the gel.

When designing variant CREB327119, which of the following residues was most likely chosen as the substitute for residue 119 in CREB327WT?

"PKA phosphorylates CREB327 at serine 119. Variants of wild-type CREB327 (CREB327WT) that cannot be phosphorylated at serine 115 or 119 were created (CREB327115 and CREB327119, respectively)."

A) Tyrosine
B) Glutamine
C) Alanine
D) Threonine

C) Alanine The answer to this question is C because alanine is similar in size to serine, cannot be phosphorylated, and will not add a positive or negative charge to the protein. It is the best choice.

Which statement explains how two forms of CREB can be generated in cells?

"CREB exists as two isoforms in cells, one being CREB327."

A) Pre-mRNA is transcribed from one CREB allele versus the other.
B) CREB mRNA transcripts with different combinations of exons are generated.
C) Different post-translational modifications of CREB occur in the endoplasmic reticulum.
D) Pre-mRNA of various lengths are transcribed based on different termination sequences in the CREB gene.

B) CREB mRNA transcripts with different combinations of exons are generated. The answer to this question is B because CREB341 and CREB327 are isoforms. Different protein isoforms are synthesized from the same gene through alternative splicing, during which sections of the full transcript (both introns and exons) are spliced. Different combinations of exons can produce different protein isoforms.

Under certain conditions, PKA and GSK-3 have been shown to autophosphorylate. The control group used in Experiment 2 (Lane 1) was designed to account for this possibility. Given this, the control group most likely contained:

A) CREB327WT with ATP but no PKA or GSK-3.
B) PKA and GSK-3 with ATP but no CREB327WT.
C) PKA and GSK-3 with no ATP or CREB327WT.
D) CREB327WT with no ATP, PKA, or GSK-3.

B) PKA and GSK-3 with ATP but no CREB327WT. The answer to this question is B because the best control against the variable of enzyme autophosphorylation would be the enzymes alone without substrate. ATP must be included so that autophosphorylation would be possible. The substrate should be excluded so that it is clear that any phosphorylation detected is due to autophosphorylation and not phosphorylation of the substrate.

Based on the passage, FSHRs are found on cells of which type(s) of tissue?

"Follicle stimulating hormone (FSH) is a peptide hormone known to activate FSH G-protein coupled receptors (FSHRs) on ovarian cells. However, FSHRs have recently been found on osteoclasts, and their activation stimulates osteoclastogenesis and bone resorption."

I. Connective
II. Epithelial
III. Nervous

I and II only The answer to this question is C because the passage states that FSH regulates the release of estrogen from the ovaries and directly stimulates osteoclast activity. Ovarian cells are epithelial cells (II), and osteoclasts are connective tissue cells (I).

Kd

dissociation constant Higher affinity corresponds to lower Kd. Lower affinity corresponds to higher Kd.

Which statement best explains why the absorbance levels for FSH differ from those for FSHpep?

"A peptide consisting of a 13-residue sequence located within the receptor-binding domain of FSH was synthesized."

A) The FSH-Ab binds non-specifically to more than one site on FSHpep.
B) IgG competitively inhibits the binding of FSH-Ab to FSH.
C) The FSH-Ab cooperatively binds FSH.
D) The tertiary structure of FSH limits FSH-Ab binding interactions.

D) The tertiary structure of FSH limits FSH-Ab binding interactions. The answer to this question is D because the absorbance levels show that FSH-Ab binds FSHpep at higher levels than FSH. The difference between the two is that FSH is a fully folded protein, whereas FSHpep is a peptide sequence. Thus, the most likely explanation for why FSH-Ab binds FSH at lower levels than FSHpep is because the tertiary structure of FSH interferes with FSH-Ab binding.

Based on the FSHpep sequence, which amino acid substitution in the FSHR binding domain is most likely to have the greatest effect on reducing bone density loss in the presence of high levels of FSH?
A) K4R
B) R8D
C) V2L
D) Q12N

B) R8D The answer to this question is B because FSHR activation increases bone density loss. Therefore, to decrease loss, the substitution should disrupt binding. The R8D substitution replaces a positively charged residue with a negatively charged residue. This would most likely disrupt binding of FSH to the FSHR and the subsequent stimulation of osteoclast activity.

Which of the following reasons does NOT describe why FSHpep was included in the ELISA experiment?

A) To act as a positive control to confirm that the assay was functional
B) To provide a baseline against which to evaluate the affinity of FSH-Ab for FSH
C) To generate data to support that the FSH-Ab binds to the receptor-binding domain of FSH
D) To determine the amount of FSH-Ab that is most effective as a therapeutic treatment

D) To determine the amount of FSH-Ab that is most effective as a therapeutic treatment The answer to this question is D because the use of FSHpep would provide no useful information about the drug's therapeutic use, as the peptide does not exist in vivo; further studies using FSH would need to be performed to accomplish this. The use of the FSHpep acted as a positive control (A) to show that the assay was working as expected (the FSH-Ab was binding its target sequence), because the FSH-Ab was designed to bind the specific sequence of FSHpep. Its use provided a baseline to determine how the binding affinity of FSH-Ab for FSH compared to that of the peptide alone, which would be expected to be maximal (B). The use of FSHpep provided data to support that the binding of FSH-Ab to FSH, as observed in the assay, was most likely due to the binding of the antibody to the specific receptor-binding sequence and not another region of FSH (C).

Which statement describes a characteristic of FSH (peptide hormone)?

A) FSH does not require transport proteins to remain soluble in the bloodstream.
B) FSH enters the bloodstream by diffusing across the plasma membrane of endocrine cells.
C) FSH is synthesized from cholesterol.
D) FSH is derived from a single amino acid.

A) FSH does not require transport proteins to remain soluble in the bloodstream. The answer to this question is A because peptide hormones are hydrophilic and soluble in blood. Hormones that must bind transport proteins are steroid proteins, which are lipophilic.

Two gel electrophoresis analyses are performed on a sample of purified protein with unknown structure: SDS-PAGE (1 band appears) and SDS-PAGE under reducing conditions (2 bands appear). Which prediction about the protein is directly supported by these results? The protein:

A) is posttranslationally glycosylated.
B) contains a high proportion of charged residues.
C) contains no disulfide bonds.
D) is composed of multiple subunits.

D) is composed of multiple subunits. The answer to this question is D because a reducing agent is used during SDS-PAGE to cleave disulfide bonds. The appearance of 1 band on SDS-PAGE without the reducing agent and 2 bands on SDS-PAGE with a reducing agent, suggests that at least one disulfide bond is present in the protein and that the disulfide bond(s) hold(s) two separate subunits of different masses together. This is the best response.

Which type of interaction does NOT contribute to the stabilization of the tertiary structure of a protein?
A) Disulfide bond
B) Phosphodiester bond
C) Hydrogen bond
D) Salt bridge

B) Phosphodiester bond The answer to this question is B because phosphodiester bonds link adjacent nucleotides in DNA. They do not contribute to the stabilization of protein structure.

Which type of inhibitor does NOT alter the KM/Vmax ratio of an enzyme?
A) Competitive
B) Uncompetitive
C) Noncompetitive
D) Mixed

B) Uncompetitive The answer to this question is B because uncompetitive inhibitors do not alter the slope of the Lineweaver-Burk plot, which is equal KM/Vmax.

An enzyme is more effectively inhibited by uncompetitive inhibitors when:

I. the substrate concentration is decreased.
II. the substrate concentration is increased.
III. the inhibitor concentration is increased.

II and III only The answer to this question is D because uncompetitive inhibitors bind their target enzymes only when the substrate is first bound to the enzyme. Since at higher substrate concentrations, the substrate-enzyme complex are more abundant, the uncompetitive inhibitor will work most effectively when the substrate concentration is the highest. Additionally, an increase in the inhibitor concentration results in increased enzyme binding and inhibition.

Which peptide sequence is most likely found in a transmembrane helix of a protein?
A) Ala-Ile-Phe-Val-Leu
B) Ala-Thr-Lys-Asn-Leu
C) Lys-Thr-Arg-Asn-His
D) Val-Thr-Pro-Tyr-Ser

A) Ala-Ile-Phe-Val-Leu The answer to this question is A because transmembrane helices are made up of mostly unbroken stretches of hydrophobic amino acids.

Which experimental approach would be LEAST effective to determine the localization of pRB within a cell?

"The retinoblastoma protein (pRB) has a primary role as a transcriptional coregulator of genes involved in several important cellular processes such as the cell cycle."

A) Fusion of a fluorescent tag to pRB
B) Use of a fluorescent probe that hybridizes to rb transcripts
C) Cellular fractionation to isolate different organelles
D) Pull down of pRB and identification of interacting proteins by mass spectrometry

B) Use of a fluorescent probe that hybridizes to rb transcripts The answer to this question is B because tagging the rb transcript would be the least helpful since this is expected to be found in the cytoplasm where the ribosomes are located.

Rhodamine 123 is a fluorescent dye that binds to polarized membranes and is used to label mitochondria. A cell line expressing basal levels of pRB or which form of induced pRB would most likely exhibit the lowest level of rhodamine 123 staining?

A) pRB induced
B) RB_N
C) RB_SP
D) RB_C

C) RB_SP The answer to this question is C because when apoptosis is induced by pRB, rhodamine 123 would no longer be able to label the mitochondria efficiently due to the loss of membrane potential. Figure 2 shows that the number of apoptotic cells is highest when an inducible copy of RB_SP is expressed, indicating that this would likely result in the lowest amount of mitochondrial labeling.

The caspase activator (cytochrome c) released during the mitochondrial apoptotic pathway primarily functions in which cellular process?

A) Proton pumping
B) Heme biosynthesis
C) Calcium signaling
D) Electron transport

D) Electron transport The answer to this question is D because cytochrom c is associated with the mitochondrial inner membrane, where it functions in the electron transport chain.

Based on the information in the passage, which protein domain of STAT3 is NOT predicted to play a role in its signaling?

A) Nuclear localization domain
B) Signal sequence domain
C) DNA binding domain
D) Protein binding domain

B) Signal sequence domain The answer to this question is B because based on the passage, STAT3 is a nuclear protein, which means that it requires a nuclear localization domain for nuclear translocation and a DNA binding domain for binding to regulatory regions of targeted genes. The passage also states that in addition to forming a homodimer, STAT3 associates with LEPRb/JAK2 complex which infers the presence of a protein binding domain within the STAT3 sequence. In contrast, signal sequence domains are protein domains required for proteins that are directed toward secretory pathways.

Which mechanism restricts the expression of leptin to adipocytes? Only adipocytes contain:

A) the ob gene.
B) a promoter for the expression of the ob gene.
C) enhancers for the expression of the ob gene.
D) nuclear factors for the expression of the ob gene.

D) nuclear factors for the expression of the ob gene. The answer to this question is D because among the listed options, nuclear factors are the only elements that vary in different cells and therefore can confer both temporal and spatial regulation of their target genes.

Which amino acid substitution within the consensus-binding site for STAT3 is LEAST likely to interfere with STAT3 binding?

A) Gln to Glu
B) Gln to Gly
C) Gln to Asn
D) Gln to Ala

C) Gln to Asn The answer to this question is C because a glutamine to asparagine substitution within the consensus-binding site for STAT3 (YXXQ) is the most conservative as both glutamine and asparagine are polar amino acids.

Which amino acids are most likely present at the dimerization interface of STAT3 proteins?

A) Polar amino acids
B) Hydrophobic amino acids
C) Positively charged amino acids
D) Negatively charged amino acids

B) Hydrophobic amino acids The answer to this question is B because the polar and charged amino acids most likely interact with water molecules in cytosol and would not be involved in protein-protein interactions. In contrast the side chains of hydrophobic amino acids are free and most likely participate in dimerization of STAT3.

The six different isoforms of LEPR are produced by using different:
A) exons of the LEPR gene.
B) isoforms of the LEPR gene.
C) promoters of the LEPR gene.
D) protein cleavage sites of LEPRb.

A) exons of the LEPR gene. The answer to this question is A because different isoforms of proteins are expressed from single genes through alternative splicing of exons of the primary transcript.

Which methods separate proteins based on their charge?

I. SDS-PAGE
II. Isoelectric focusing
III. Ion-exchange chromatography
IV. Affinity chromatography

A) I and II only
B) II and III only
C) I, II, and III only
D) II, III, and IV only

B) II and III only The answer to this question is B because isoelectric focusing separates proteins based on their isoelectric point (the pH at which the net charge of the protein is zero) and ion exchange chromatography separates proteins based on their net charge. In contrast SDS-PAGE separates proteins based on their mass and affinity chromatography separates proteins based on their interactions with specific ligands.

How many molecules of reduced electron carrier are generated during conversion of α-ketoglutarate to oxaloacetate in the citric acid cycle?
A) One
B) Two
C) Three
D) Four

C) Three The answer to this question is C because during conversion of α-ketoglutarate to oxaloacetate in the citric acid cycle 2 molecules of NADH and one molecule of FADH2 are generated.

Which enzyme is used both in gluconeogenesis and glycogenolysis?

A) Phosphoglucomutase
B) Glucose 6-phosphatase
C) Hexokinase
D) Glucokinase

B) Glucose 6-phosphatase The answer to this question is B because glucose 6-phosphatase catalyzes the final step of both gluconeogenesis and glycogenolysis

catalytic efficiency

efficiency = kcat/Km lower Km = higher efficiency

Changes to Kd

implies role in binding to protein

competitive inhibition

-binding occurs at the active site of the enzyme -no change in Vma

Based on the data presented in Figure 1, overexpression of PRR:
I. increases blood pressure in part through an angiotensin II-dependent pathway.

II. increases blood pressure in part through an angiotensin II-independent pathway.

III. decreases blood pressure in part through an angiotensin II-dependent pathway.

I and II only The answer to this question is D because compared to control, the PRR overexpression alone increased ROS (an indicator of blood pressure) levels even in presence of losartan (angiotensin II antagonist), indicating that PRR overexpression induces ROS formation in an angiotensin II-indepedent manner. Moreover, the addition of exogenous prorenin, a ligand for PRR, induces an increase in ROS formation which can be reversed by losartan, indicating its angiotensin-II-dependent nature.

Based on the data presented in Table 1, which amino acid residue of prorenin most likely interacts with the residue at position 201 of PRR?

"D201N - Kd = 0.441"

A) Alanine
B) Glutamate
C) Arginine
D) Glutamine

C) Arginine The answer to this question is C because substitution of aspartate (D) at position 201 with asparagine (N) results in a lower PRR affinity (higher Kd) towards prorenin. Since both asparagine and aspartate have equivalent carbon lengths, the decrease in affinity is most likely due to elimination of an ion pair. Among the possible options, only arginine could form an ion pair with aspartate.

Which enzyme was LEAST likely used in cloning of WT-PRR cDNA?
A) DNA polymerase
B) RNA polymerase
C) DNA ligase
D) Reverse transcriptase

B) RNA polymerase The answer to this question is B because while DNA polymerase (in DNA amplification), DNA ligase (in ligation of the cDNA to DNA vector) and reverse transcriptase (in reverse transcription of RNA to cDNA) are used during cDNA cloning, RNA polymerase is not used in cDNA cloning.

Compared to WT, what is the most likely effect of the W140L substitution on the stability of the PRR-prorenin complex?

WT - Kd = 0.034
W140L = 0.163

A) Elimination of pi-stacking interaction decreases stability of the complex.
B) Elimination of an ionic interaction decreases stability of the complex
C) Addition of a hydrophobic interaction increases stability of the complex.
D) Addition of a hydrophobic interaction increases stability of the complex.

A) Elimination of pi-stacking interaction decreases stability of the complex. The answer to this question is A because based on results presented in Table 1; in W140L variant, the replacement of tryptophan (a hydrophobic, aromatic amino acid) by leucine (a hydrophobic but non-aromatic amino acid) results in a decrease in stability of the PRR-prorenin complex as indicated by the increase in Kd. This is most likely due to the elimination of a π-stacking interaction with the side chain of tryptophan.

Based on the passage, which metabolic pathways are upregulated during the transition from intrauterine to extrauterine environment?

"After depletion of the blood glucose pool, the initial source for energy replacement is from hepatic glycogen, which is sufficient to support basic metabolism for the first 12 hours. Because infants normally do not receive an adequate supply of glucose from feeding for the first days of life, hepatic glycogen stores are soon exhausted, and the glucose pool must be replenished from non-carbohydrate sources including fatty acids and glucogenic amino acids."

A) Glycogenesis followed by glycogenolysis
B) Glycogenolysis followed by gluconeogenesis
C) Gluconeogenesis followed by glycogenolysis
D) Gluconeogenesis followed by glycogenesis

B) Glycogenolysis followed by gluconeogenesis The answer to this question is B because the passage states that the initial source for energy replacement is from hepatic glycogen indicating upregulation of glycogen breakdown (glycogenolysis). The passage also states that after hepatic glycogen stores are exhausted the glucose pool must be replenished from non-carbohydrate sources implying upregulation of the gluconeogenic pathway.

After the depletion of hepatic glycogen in newborns, which compounds can be used as precursors to sustain the blood glucose level?

*Gluconeogenesis

I. Acetyl-CoA
II. Lactate
III. Oxaloacetate
IV. a-Ketoglutarate

II, III, and IV only The answer to this question is D because among the listed options, only lactate, oxaloacetate, and α-ketoglutarate are used as starting materials in gluconeogenesis.

Which event is NOT a likely outcome of glucagon binding to its receptor? Increase in:

A) GDP binding to Ga subunit of the G protein
B) adenylate cyclase activity
C) protein kinase A activity
D) cAMP generation

A) GDP binding to Ga subunit of the G protein The answer to this question is A because following glucagon binding to its receptor and activation of its coupled G protein, activities of the adenylate cyclase and the protein kinase A, and level of cAMP are all increased. In contrast activation of the G protein promotes the dissociation of bound GDP and its exchange for GTP on the α subunit.

Infusion of which peptide hormone will most likely prevent brain injury in newborn infants exposed to high glucose levels during their fetal development?
A) Insulin
B) Epinephrine
C) Glucagon
D) Cortisol

C) Glucagon The answer to this question is C because exposure to high glucose levels results in elevated levels of circulating insulin which prevents mobilization of endogenous glucose storage. Infusion of glucagon will result in mobilization of endogenous glucose storage thereby preventing hypoglycemia and brain injury.

Activation of which enzyme would support the metabolism of newborn infants during the first 12 hours?

"After the depletion of the blood glucose pool, the initial source for energy replacement is from hepatic glycogen, which is sufficient to support basic metabolism for the first 12 hours."

A) Glycogen phosphatase
B) Glycogen phosphorylase
C) Glycogen hydrolase
D) Glycogen kinase

B) Glycogen phosphorylase The answer to this question is B because the passage states that hepatic glycogen supports the basic metabolism of newborn during the first 12 hours. Glycogen phosphorylase is the enzyme that catalyzes the rate-limiting step in glycogen breakdown (glycogenolysis).

The initial source of energy replacement in the liver of newborn infants is formed by glycosidic bonds between glucose molecules through:

A) α(1→4) linkage linearly and β(1→6) linkage at branch points.
B) β(1→6) linkage linearly and α(1→4) linkage at branch points.
C) α(1→4) linkage linearly and α(1→6) linkage at branch points.
D) α(1→6) linkage linearly and α(1→4) linkage at branch points.

C) α(1→4) linkage linearly and α(1→6) linkage at branch points. The answer to this question is C because the glucose polymer in liver (glycogen) is formed by glycosidic bonds between glucose molecules through α(1→4) linkage linearly and α(1→6) linkage at branch point.

Which of the following compounds is NOT a gluconeogenic precursor or substrate?
A) Lactate
B) Glycerol
C) Oxaloacetate
D) Phosphogluconate

D) Phosphogluconate The answer to this question is D because while lactate, oxaloacetate and glycerol are gluconeogenic precursors, phosphogluconate is involved in the pentose phosphate pathway and is not a precursor or substrate in gluconeogenesis.

Which type of enzyme removes the chemical groups that are added to proteins by kinases?
A) Phosphorylase
B) Cyclase
C) Phosphatase
D) Acetylase

C) Phosphatase The answer to this question is C because while kinase catalyzes the addition of a phosphate group to a substrate, the removal of the phosphate groups are catalyzed by a phosphatase.

Under physiological conditions, increased activity of succinyl-coA synthetase will most likely result in:
I. increased levels of succinyl CoA
II. increased levels of succinate
III. increased levels of GTP

II and III only The answer to this question is D because increased activity of succinyl-CoA synthetase will result in greater levels of the reaction products, succinate and GTP. Succinyl-CoA is the substrate of the reaction and its levels will likely decrease with increased succinyl-CoA function.

The pentose phosphate pathway results in the generation of:
A) NADPH, which is used as a reductive agent in cellular respiratory processes.
B) NADPH, which is used as an oxidative agent in cellular respiratory processes.
C) NADH, which is used as a reductive agent in cellular respiratory processes.
D) NADH, which is used as an oxidative agent in cellular respiratory processes.

A) NADPH, which is used as a reductive agent in cellular respiratory processes. The answer to this question is A because the pentose phosphate pathway results in the generation of NADPH, which is utilized in reductive reactions.

The Vmax of an enzymatic reaction is altered by which types of inhibitors?
I. Competitive
II. Noncompetitive
III. Uncompetitive
IV. Mixed

II, III, and IV only The answer to this question is D because as observed in 1/V0 versus 1/ plot (Lineweaver-Burk plot), competitive inhibitors are the only inhibitors that do not alter the intercept on the y-axis of the plot and therefore the Vmax of the enzymatic reaction.

Which metabolic reaction is most likely affected by the treatment of cells with C75? The reaction that converts:

"mtKAS is inhibited by C75. mtKAS is also involved in the synthesis of lipoic acid (LA), a cofactor for several metabolic enzymes"

A) fructose-6-phosphate to fructose-1,6-bisphosphate
B) fumarate to malate
C) pyruvate to acetyl-CoA
D) oxaloacetate to phosphoenolpyruvate

C) pyruvate to acetyl-CoA

Based on the information from the passage, the deletion of mtKAS most likely results in decreased level of:

"The deletion of the gene encoding for mtKAS results in dissipation of the electrochemical gradient in mitochondria due to an increase in membrane permeability."

A) protons in the mitochondrial matrix
B) protons in the mitochondrial intermembrane space
C) electrons in the mitochondrial matrix
D) electrons in the mitochondrial intermembrane space

B) protons in the mitochondrial intermembrane space The answer to this question is B because the passage states that deletion of the gene encoding for mtKAS results in dissipation of the electrochemical gradient generated by the electron transport chain. The electron transport chain uses the free energy from redox reactions to pump protons from mitochondrial matrix to the intermembrane space thereby generating an electrochemical gradient across the inner mitochondrial membrane. Therefore any event that causes the dissipation of the electrochemical gradient across the inner mitochondrial membrane will result in decreased level of protons in mitochondrial intermembrane space.

In the pentose phosphate pathway, which enzyme catalyzes the production of 6-phosphogluconate?

A) Glucose 6-phosphate dehydrogenase
B) 6-Phosphogluconolactonase
C) Phosphogluconate dehydrogenase
D) Glycerol 3-phosphate dehydrogenase

A) Glucose 6-phosphate dehydrogenase The answer to this question is A because Glucose 6-phosphate dehydrogenase catalyzes the conversion of glucose 6-phosphate to 6-phosphogluconolactone in the pentose phosphate pathway.

A partial DNA sequence of the coding strand of a gene is shown.

5'-GACATGGACTCGCTA-3'

Which sequence corresponds to the mRNA for this DNA sequence?

A) 5'-GACAUGGACUCGCUA-3'
B) 5'-CUGUACCUGAGCGUA-3'
C) 5'-UAGCGAGUCCAUGUC-3'
D) 5'-AUCGCUCAGGUACAG-3'

A) 5'-GACAUGGACUCGCUA-3' The answer to this question is A because for any given gene, the nucleotide sequence of both the coding strand the mRNA are complementary to the sequence of the template strand.

Which technique CANNOT be used to analyze expression?
A) Western blotting
B) Northern blotting
C) Southern blotting
D) Reverse transcription PCR

C) Southern blotting The answer to this question is C because Southern blotting is a technique used to detect a particular sequence in a sample of DNA.

MCAT Section Bank- Biology & Biochemistry - Subjecto.com

MCAT Section Bank- Biology & Biochemistry

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phosphatase

enzyme that removes a phosphate group from its substrate

According to the information in the passage, which statement best describes the function of the MR in response to stimulation of the sympathetic nervous system? The MR:

"Blood flow to the brain is regulated by changes in arterial vessel diameter in response to intravascular pressure, a process called the myogenic response (MR). The MR allows cerebral resistance vessels to achieve a state of partial constriction, from which they can dilate or further constrict to alter blood flow."

A) equalizes blood pressure in cerebral resistance arteries and the aorta.
B) moderates blood flow to the brain under high pressure.
C) enables cerebral resistance arteries to locally vasodilate.
D) redirects blood circulating in the brain to organs in the abdominal cavity.

B) moderates blood flow to the brain under high pressure. The answer to this question is B because the MR is a local response in resistance vessels that would slow blood flow to the brain to reduce the chance of tissue damage from high blood pressure resulting from activation of the sympathetic nervous system.

The polymerization of which structural component was analyzed in the experiment described in the passage? (Actin polymerization)
A) Microtubules
B) Microfilaments
C) Intermediate filaments
D) Thick filaments

B) Microfilaments The answer to this question is B because microfilaments are composed of actin.

Which amino acid is LEAST likely found in one of the transmembrane domains of GPCR43?
A) Asp
B) Gly
C) Trp
D) Phe

A) Asp The answer to this question is A because aspartate (Asp) has a negatively charged R group and is very hydrophilic.

Compared to untreated WT mice, antibiotic treatment of WT mice is likely to result in:

A) increased plasma butyrate levels.
B) decreased body mass.
C) increased volume of adipocytes.
D) decreased insulin sensitivity in adipocytes.

C) increased volume of adipocytes The answer to this question is C because if SCFA producing microbes were removed by antibiotic treatment, then GPCR43 would not be activated to reduce glucose uptake into adipocytes.

Which amino acid contains an unbranched alkyl side chain?
A) Ala
B) Ile
C) Leu
D) Val

A) Ala The answer to this item is A since alanine contains a methyl side chain, which is not considered a branched alkane.

The stereochemical designators α and β distinguish between:
A) enantiomers at an epimeric carbon atom.
B) enantiomers at an anomeric carbon atom.
C) epimers at an anomeric carbon atom.
D) epimers at a non-anomeric carbon atom.

C) epimers at an anomeric carbon atom. The answer to this item is C since the D versus L designation distinguishes between molecules with multiple chiral centers, but differ only in the configuration of the site known as the anomeric carbon atom.

Which ideal solution exhibits the greatest osmotic pressure?
A) 0.1 M MgCl2
B) 0.2 M NaCl
C) 0.2 M CaCl2
D) 0.5 M Glucose

C) 0.2 M CaCl2 The answer to this question is C because the solution listed has the greatest concentration of solute particles. CaCl2 dissociates into three ions, making the solution listed 0.6 M in solute particles.

Which event is directly mediated by a ligand-gated ion channel?

A) Release of Ca2+ from the sarcoplasmic reticulum of a muscle fiber to initiate muscle contraction
B) Influx of Na+ across the axon membrane of a somatic neuron during action potential propagation
C) Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane
D) Re-entry of Ca2+ back into the sarcoplasmic reticulum of a muscle fiber to end muscle contraction

C) Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane The answer to this question is C because the influx of Na+ across the motor end plate occurs when Na+ ion channels bind the ligand acetylcholine.

Southern blot

A Southern blot uses a restriction digest to differentiate between mutant and wild-type alleles. In order for a Southern blot to be useful, the mutation should either create or eliminate a restriction site, most of which are palindromes and 4 to 6 base pairs long. The mutation shown in this option is the only one that disrupts a palindromic sequence, AAGCTT. This sequence is the recognition sequence for HindIII.

What bond is cleaved by IN during the first reaction of integration?

"Upon binding, IN catalyzes cleavage of a GT dinucleotide from each 3 PRIME vDNA terminus"

A) P-C
B) P-P
C) P-H
D) P-O

D) P-O The answer to this question is D because in the first reaction of integration, integrase cleaves a dinucleotide from the 3′ end of a strand of viral DNA.

A vDNA sequence encoding a protein is inserted into a host genome by IN. The protein is translated from the hypothetical mRNA sequence shown.

5′-GGCAACUGACUA-3′

Based on the passage, the segment of the original viral genome that encoded this protein had what nucleotide sequence?

5′-GGCAACUGACUA-3′ The answer to this question is A because according to the passage, viral DNA integrated into a host cell genome by integrase would originate from a retrovirus. mRNA transcribed from retroviral DNA is either used to synthesize viral proteins, or used as the RNA genome for progeny viruses. Thus, the sequence of the nucleotide in the original viral genome will be the same as that of the transcribed mRNA.

Which experiment would best provide data to support the mechanism by which an ODN inhibits IN activity?

"ODNs are competitive inhibitors of vDNA for binding to IN."

A) Keep IN concentration constant and measure the initial velocity of the reaction at increasing vDNA concentrations in the presence and absence of the ODN.
B) Keep vDNA concentration constant and measure the initial velocity of the reaction at increasing concentrations of IN in the presence and absence of the ODN.
C) Keep ODN and IN concentration constant and measure initial velocity of the reaction at increasing concentrations of vDNA.
D) Keep IN and vDNA concentrations constant and measure initial velocity of the reaction at increasing concentrations the ODN.

A) Keep IN concentration constant and measure the initial velocity of the reaction at increasing vDNA concentrations in the presence and absence of the ODN. Competitive inhibition can be determined through rate experiments by applying the principles of the Michaelis-Menten equation. By keeping enzyme concentration constant, varying substrate concentration, and either including or excluding the inhibitor, the effect of the inhibitor on the Vmax and apparent Km of the reaction can be determined. A competitive inhibitor will increase the apparent Km and not affect the Vmax.

Which amino acid substitution for the conserved residue at position 64 is LEAST likely to affect the enzymatic function of IN?

Highly conserved residue = D64

A) Asparagine
B) Glutamate
C) Lysine
D) Valine

B) Glutamate The answer to this question is B because the conserved residue at position 64 in integrase is aspartate, which has a negatively charged side chain. Substitution with glutamate, which also has a negatively charged side chain, is least likely to affect integrase function.

An inactive tetramer of IN is expected to have approximately what molecular weight?

"IN is a 288-residue protein composed of three domains"

A) 16 kDa
B) 32 kDa
C) 64 kDa
D) 128 kDa

D) 128 kDa The answer to this question is D because the passage indicates that an integrase monomer is composed of 288 amino acids, which will have an approximate molecular weight of 32 kDa (the average molecular weight of an amino acid is 110 Da). Thus, a tetramer will have an approximate molecular weight of 128 kDa.

To determine if a small molecule acts like a LEDGIN with respect to IN, a researcher plans to incubate purified IN both with and without the small molecule and then perform a Western blot to detect IN in each sample. Under which condition(s) should the gel electrophoresis step be performed?
I. Denaturing
II. Reducing
III. Native

III only The answer to this question is C because if the intent is to confirm that a small molecule induces the formation of integrase tetramers from integrase dimers, it is necessary to visualize the proteins in their native state. Use of a denaturing agent will disrupt the interactions between monomers. Use of a reducing agent only will disrupt any disulfide bonds.

What is the dependent variable from the experiment shown in Figure 2?

X-axis = time
Y-axis = relative retinol fluorescence

A) STRA6/LRAT conditon
B) Retinol fluorescence
C) Time duration of assay
D) Untransfected cell membranes

B) Retinol fluorescence The answer to this question is B because the dependent variable is the variable that is being measured in the experiment.

The organ in which the holo-RBP complex forms also functions to:
(liver)

A) store bile
B) secrete glucagon
C) produce hydrochloric acid
D) detoxify drugs

D) detoxify drugs The answer to this question is D because according to the passage the holo-RBP complex forms in the liver and one of the function of the liver is drug detoxification.

In a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation). If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Note: Assume Mendelian inheritance patterns.)
A) 4/9
B) 1/2
C) 2/3
D) 3/4

A) 4/9 The answer to this question is A because given Mendelian inheritance patterns, a 3:1 ratio of F1 offspring means that the original crossed beetles are both heterozygotes, and the F1 offspring are 25% red (homozygous dominant), 50% red (heterozygous), and 25% brown (homozygous recessive). If two red F1 beetles are crossed and both red and brown beetles appear in the F2 generation, the F1 red beetles that were crossed must both be heterozygotes. The probability that, of the red F1 beetles, both were heterozygous is 2/3 × 2/3, or 4/9 (only red beetles were selected from and 2/3 of the red F1 beetles were heterozygous).

Hill coefficient

>1 cooperativity =1 no cooperativity <1 negative cooperativity

noncompetitive inhibitor

binds the enzyme and enzyme-substrate complex with the same affinity

In Figure 1, which statement best explains the data trend of an increase followed by a decrease in bacterial cell count?

A) The number of host cells in the sample increased until carrying capacity was reached.
B) Infection with VP is maximal at 5 hours of exposure.
C) VP replicates inside the host cell before causing host cell lysis.
D) Rate of VP entry into host cells increases then decreases over time.

C) VP replicates inside the host cell before causing host cell lysis. The answer to this question is C because in Experiment 1, the host cells are exposed to VP for only two hours, then washed and incubated in VP-free media containing an antibiotic. Thus, the levels of intracellular bacteria (VP that has entered the host cell) measured at each time point represent the amount of VP in the host cells at that time point. The increasing levels at hours 3 and 4 suggest that VP is replicating inside the host cell. The sudden drop in intracellular bacterial levels is consistent with the hypothesis that the host cells are beginning to lyse, as the bacteria that had previously been inside the cell are rinsed away or killed by the antibiotic

The amino acid residue at position 61 in Rac is most likely:

"VopC contains a catalytic domain that irreversibly activates host cell GTPase Rac through the deamidation of the side chain of the residue at position 61."

A) glutamine
B) leucine
C) arginine
D) tyrosine

A) glutamine The answer to this question is A because according to the passage, VopC removes an amide from the side chain of the amino acid residue at position 61 in Rac via a deamidation reaction. Of the options, only glutamine contains a side chain with an amide group.

What is the independent variable in Experiment 1?

X-axis: hourly intervals in which sample of host cells were collected and rinsed
Y-axis: intracellular bacterial levels

A) Length of time host cells were exposed to the bacteria
B) Amount of cells collected in each sample
C) Time at which the cell samples were collected
D) Amount of intracellular bacteria measured at each time point.

C) Time at which the cell samples were collected The answer to this question is C because the independent variable in an experiment is that which is varied by the researcher in order to determine the effect (the dependent variable). In Experiment 1, the time at which cell samples are collected is the independent variable.

A Rac variant, in which the residue at position 61 was replaced with an alanine (Rac61A), was synthesized. Wild-type Rac and Rac61A were incubated separately with VopC. To obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of which compound?
A) CO2
B) NH3
C) O2
D) H2S

B) NH3 The answer to this question is B because a deamidation reaction releases NH3, thus, if NH3 is detected in the sample containing wild-type Rac but not Rac61A, this will provide data to support that VopC deamidates Rac at residue 61.

Why was the total amount of Rac measured in Experiment 2?

"To confirm that VopC activates Rac, a mutant strain of VP lacking the VopC gene was created…"

A) To confirm that the antibody used to detect GTP-Rac was working in each sample
B) To determine if the same amount of each sample was loaded on the gel
C) To act as a control for comparing rates of Rac activation by VopC
D) To determine if VopC affects Rac expression levels

D) To determine if VopC affects Rac expression levels The answer to this question is D because in order to determine how the presence or absence of VopC affects Rac activation, it is important to know if the total amount of Rac is affected by the strain being used. If the Rac expression is upregulated in one strain and not another, this will affect interpretation of the gel results.

adenine structure

guanine structure

cytosine structure

thymine structure

uracil structure

Isoelectric focusing of samples includes which step?:

A) Amplify the samples by PCR before running on the gel.
B) Establish a stable pH gradient in the gel before adding samples.
C) Denature the samples so the proteins unfold into a linear chain.
D) Treat the samples with a chemical that will add an overall negative charge to each protein.

B) Establish a stable pH gradient in the gel before adding samples. The answer to this question is B because according to the passage, visualization of the proteins in Experiment 2 is done through isoelectric focusing, which separates proteins based on isoelectric point. To do this, a stable pH gradient must be established in the gel.

When designing variant CREB327119, which of the following residues was most likely chosen as the substitute for residue 119 in CREB327WT?

"PKA phosphorylates CREB327 at serine 119. Variants of wild-type CREB327 (CREB327WT) that cannot be phosphorylated at serine 115 or 119 were created (CREB327115 and CREB327119, respectively)."

A) Tyrosine
B) Glutamine
C) Alanine
D) Threonine

C) Alanine The answer to this question is C because alanine is similar in size to serine, cannot be phosphorylated, and will not add a positive or negative charge to the protein. It is the best choice.

Which statement explains how two forms of CREB can be generated in cells?

"CREB exists as two isoforms in cells, one being CREB327."

A) Pre-mRNA is transcribed from one CREB allele versus the other.
B) CREB mRNA transcripts with different combinations of exons are generated.
C) Different post-translational modifications of CREB occur in the endoplasmic reticulum.
D) Pre-mRNA of various lengths are transcribed based on different termination sequences in the CREB gene.

B) CREB mRNA transcripts with different combinations of exons are generated. The answer to this question is B because CREB341 and CREB327 are isoforms. Different protein isoforms are synthesized from the same gene through alternative splicing, during which sections of the full transcript (both introns and exons) are spliced. Different combinations of exons can produce different protein isoforms.

Under certain conditions, PKA and GSK-3 have been shown to autophosphorylate. The control group used in Experiment 2 (Lane 1) was designed to account for this possibility. Given this, the control group most likely contained:

A) CREB327WT with ATP but no PKA or GSK-3.
B) PKA and GSK-3 with ATP but no CREB327WT.
C) PKA and GSK-3 with no ATP or CREB327WT.
D) CREB327WT with no ATP, PKA, or GSK-3.

B) PKA and GSK-3 with ATP but no CREB327WT. The answer to this question is B because the best control against the variable of enzyme autophosphorylation would be the enzymes alone without substrate. ATP must be included so that autophosphorylation would be possible. The substrate should be excluded so that it is clear that any phosphorylation detected is due to autophosphorylation and not phosphorylation of the substrate.

Based on the passage, FSHRs are found on cells of which type(s) of tissue?

"Follicle stimulating hormone (FSH) is a peptide hormone known to activate FSH G-protein coupled receptors (FSHRs) on ovarian cells. However, FSHRs have recently been found on osteoclasts, and their activation stimulates osteoclastogenesis and bone resorption."

I. Connective
II. Epithelial
III. Nervous

I and II only The answer to this question is C because the passage states that FSH regulates the release of estrogen from the ovaries and directly stimulates osteoclast activity. Ovarian cells are epithelial cells (II), and osteoclasts are connective tissue cells (I).

Kd

dissociation constant Higher affinity corresponds to lower Kd. Lower affinity corresponds to higher Kd.

Which statement best explains why the absorbance levels for FSH differ from those for FSHpep?

"A peptide consisting of a 13-residue sequence located within the receptor-binding domain of FSH was synthesized."

A) The FSH-Ab binds non-specifically to more than one site on FSHpep.
B) IgG competitively inhibits the binding of FSH-Ab to FSH.
C) The FSH-Ab cooperatively binds FSH.
D) The tertiary structure of FSH limits FSH-Ab binding interactions.

D) The tertiary structure of FSH limits FSH-Ab binding interactions. The answer to this question is D because the absorbance levels show that FSH-Ab binds FSHpep at higher levels than FSH. The difference between the two is that FSH is a fully folded protein, whereas FSHpep is a peptide sequence. Thus, the most likely explanation for why FSH-Ab binds FSH at lower levels than FSHpep is because the tertiary structure of FSH interferes with FSH-Ab binding.

Based on the FSHpep sequence, which amino acid substitution in the FSHR binding domain is most likely to have the greatest effect on reducing bone density loss in the presence of high levels of FSH?
A) K4R
B) R8D
C) V2L
D) Q12N

B) R8D The answer to this question is B because FSHR activation increases bone density loss. Therefore, to decrease loss, the substitution should disrupt binding. The R8D substitution replaces a positively charged residue with a negatively charged residue. This would most likely disrupt binding of FSH to the FSHR and the subsequent stimulation of osteoclast activity.

Which of the following reasons does NOT describe why FSHpep was included in the ELISA experiment?

A) To act as a positive control to confirm that the assay was functional
B) To provide a baseline against which to evaluate the affinity of FSH-Ab for FSH
C) To generate data to support that the FSH-Ab binds to the receptor-binding domain of FSH
D) To determine the amount of FSH-Ab that is most effective as a therapeutic treatment

D) To determine the amount of FSH-Ab that is most effective as a therapeutic treatment The answer to this question is D because the use of FSHpep would provide no useful information about the drug’s therapeutic use, as the peptide does not exist in vivo; further studies using FSH would need to be performed to accomplish this. The use of the FSHpep acted as a positive control (A) to show that the assay was working as expected (the FSH-Ab was binding its target sequence), because the FSH-Ab was designed to bind the specific sequence of FSHpep. Its use provided a baseline to determine how the binding affinity of FSH-Ab for FSH compared to that of the peptide alone, which would be expected to be maximal (B). The use of FSHpep provided data to support that the binding of FSH-Ab to FSH, as observed in the assay, was most likely due to the binding of the antibody to the specific receptor-binding sequence and not another region of FSH (C).

Which statement describes a characteristic of FSH (peptide hormone)?

A) FSH does not require transport proteins to remain soluble in the bloodstream.
B) FSH enters the bloodstream by diffusing across the plasma membrane of endocrine cells.
C) FSH is synthesized from cholesterol.
D) FSH is derived from a single amino acid.

A) FSH does not require transport proteins to remain soluble in the bloodstream. The answer to this question is A because peptide hormones are hydrophilic and soluble in blood. Hormones that must bind transport proteins are steroid proteins, which are lipophilic.

Two gel electrophoresis analyses are performed on a sample of purified protein with unknown structure: SDS-PAGE (1 band appears) and SDS-PAGE under reducing conditions (2 bands appear). Which prediction about the protein is directly supported by these results? The protein:

A) is posttranslationally glycosylated.
B) contains a high proportion of charged residues.
C) contains no disulfide bonds.
D) is composed of multiple subunits.

D) is composed of multiple subunits. The answer to this question is D because a reducing agent is used during SDS-PAGE to cleave disulfide bonds. The appearance of 1 band on SDS-PAGE without the reducing agent and 2 bands on SDS-PAGE with a reducing agent, suggests that at least one disulfide bond is present in the protein and that the disulfide bond(s) hold(s) two separate subunits of different masses together. This is the best response.

Which type of interaction does NOT contribute to the stabilization of the tertiary structure of a protein?
A) Disulfide bond
B) Phosphodiester bond
C) Hydrogen bond
D) Salt bridge

B) Phosphodiester bond The answer to this question is B because phosphodiester bonds link adjacent nucleotides in DNA. They do not contribute to the stabilization of protein structure.

Which type of inhibitor does NOT alter the KM/Vmax ratio of an enzyme?
A) Competitive
B) Uncompetitive
C) Noncompetitive
D) Mixed

B) Uncompetitive The answer to this question is B because uncompetitive inhibitors do not alter the slope of the Lineweaver-Burk plot, which is equal KM/Vmax.

An enzyme is more effectively inhibited by uncompetitive inhibitors when:

I. the substrate concentration is decreased.
II. the substrate concentration is increased.
III. the inhibitor concentration is increased.

II and III only The answer to this question is D because uncompetitive inhibitors bind their target enzymes only when the substrate is first bound to the enzyme. Since at higher substrate concentrations, the substrate-enzyme complex are more abundant, the uncompetitive inhibitor will work most effectively when the substrate concentration is the highest. Additionally, an increase in the inhibitor concentration results in increased enzyme binding and inhibition.

Which peptide sequence is most likely found in a transmembrane helix of a protein?
A) Ala-Ile-Phe-Val-Leu
B) Ala-Thr-Lys-Asn-Leu
C) Lys-Thr-Arg-Asn-His
D) Val-Thr-Pro-Tyr-Ser

A) Ala-Ile-Phe-Val-Leu The answer to this question is A because transmembrane helices are made up of mostly unbroken stretches of hydrophobic amino acids.

Which experimental approach would be LEAST effective to determine the localization of pRB within a cell?

"The retinoblastoma protein (pRB) has a primary role as a transcriptional coregulator of genes involved in several important cellular processes such as the cell cycle."

A) Fusion of a fluorescent tag to pRB
B) Use of a fluorescent probe that hybridizes to rb transcripts
C) Cellular fractionation to isolate different organelles
D) Pull down of pRB and identification of interacting proteins by mass spectrometry

B) Use of a fluorescent probe that hybridizes to rb transcripts The answer to this question is B because tagging the rb transcript would be the least helpful since this is expected to be found in the cytoplasm where the ribosomes are located.

Rhodamine 123 is a fluorescent dye that binds to polarized membranes and is used to label mitochondria. A cell line expressing basal levels of pRB or which form of induced pRB would most likely exhibit the lowest level of rhodamine 123 staining?

A) pRB induced
B) RB_N
C) RB_SP
D) RB_C

C) RB_SP The answer to this question is C because when apoptosis is induced by pRB, rhodamine 123 would no longer be able to label the mitochondria efficiently due to the loss of membrane potential. Figure 2 shows that the number of apoptotic cells is highest when an inducible copy of RB_SP is expressed, indicating that this would likely result in the lowest amount of mitochondrial labeling.

The caspase activator (cytochrome c) released during the mitochondrial apoptotic pathway primarily functions in which cellular process?

A) Proton pumping
B) Heme biosynthesis
C) Calcium signaling
D) Electron transport

D) Electron transport The answer to this question is D because cytochrom c is associated with the mitochondrial inner membrane, where it functions in the electron transport chain.

Based on the information in the passage, which protein domain of STAT3 is NOT predicted to play a role in its signaling?

A) Nuclear localization domain
B) Signal sequence domain
C) DNA binding domain
D) Protein binding domain

B) Signal sequence domain The answer to this question is B because based on the passage, STAT3 is a nuclear protein, which means that it requires a nuclear localization domain for nuclear translocation and a DNA binding domain for binding to regulatory regions of targeted genes. The passage also states that in addition to forming a homodimer, STAT3 associates with LEPRb/JAK2 complex which infers the presence of a protein binding domain within the STAT3 sequence. In contrast, signal sequence domains are protein domains required for proteins that are directed toward secretory pathways.

Which mechanism restricts the expression of leptin to adipocytes? Only adipocytes contain:

A) the ob gene.
B) a promoter for the expression of the ob gene.
C) enhancers for the expression of the ob gene.
D) nuclear factors for the expression of the ob gene.

D) nuclear factors for the expression of the ob gene. The answer to this question is D because among the listed options, nuclear factors are the only elements that vary in different cells and therefore can confer both temporal and spatial regulation of their target genes.

Which amino acid substitution within the consensus-binding site for STAT3 is LEAST likely to interfere with STAT3 binding?

A) Gln to Glu
B) Gln to Gly
C) Gln to Asn
D) Gln to Ala

C) Gln to Asn The answer to this question is C because a glutamine to asparagine substitution within the consensus-binding site for STAT3 (YXXQ) is the most conservative as both glutamine and asparagine are polar amino acids.

Which amino acids are most likely present at the dimerization interface of STAT3 proteins?

A) Polar amino acids
B) Hydrophobic amino acids
C) Positively charged amino acids
D) Negatively charged amino acids

B) Hydrophobic amino acids The answer to this question is B because the polar and charged amino acids most likely interact with water molecules in cytosol and would not be involved in protein-protein interactions. In contrast the side chains of hydrophobic amino acids are free and most likely participate in dimerization of STAT3.

The six different isoforms of LEPR are produced by using different:
A) exons of the LEPR gene.
B) isoforms of the LEPR gene.
C) promoters of the LEPR gene.
D) protein cleavage sites of LEPRb.

A) exons of the LEPR gene. The answer to this question is A because different isoforms of proteins are expressed from single genes through alternative splicing of exons of the primary transcript.

Which methods separate proteins based on their charge?

I. SDS-PAGE
II. Isoelectric focusing
III. Ion-exchange chromatography
IV. Affinity chromatography

A) I and II only
B) II and III only
C) I, II, and III only
D) II, III, and IV only

B) II and III only The answer to this question is B because isoelectric focusing separates proteins based on their isoelectric point (the pH at which the net charge of the protein is zero) and ion exchange chromatography separates proteins based on their net charge. In contrast SDS-PAGE separates proteins based on their mass and affinity chromatography separates proteins based on their interactions with specific ligands.

How many molecules of reduced electron carrier are generated during conversion of α-ketoglutarate to oxaloacetate in the citric acid cycle?
A) One
B) Two
C) Three
D) Four

C) Three The answer to this question is C because during conversion of α-ketoglutarate to oxaloacetate in the citric acid cycle 2 molecules of NADH and one molecule of FADH2 are generated.

Which enzyme is used both in gluconeogenesis and glycogenolysis?

A) Phosphoglucomutase
B) Glucose 6-phosphatase
C) Hexokinase
D) Glucokinase

B) Glucose 6-phosphatase The answer to this question is B because glucose 6-phosphatase catalyzes the final step of both gluconeogenesis and glycogenolysis

catalytic efficiency

efficiency = kcat/Km lower Km = higher efficiency

Changes to Kd

implies role in binding to protein

competitive inhibition

-binding occurs at the active site of the enzyme -no change in Vma

Based on the data presented in Figure 1, overexpression of PRR:
I. increases blood pressure in part through an angiotensin II-dependent pathway.

II. increases blood pressure in part through an angiotensin II-independent pathway.

III. decreases blood pressure in part through an angiotensin II-dependent pathway.

I and II only The answer to this question is D because compared to control, the PRR overexpression alone increased ROS (an indicator of blood pressure) levels even in presence of losartan (angiotensin II antagonist), indicating that PRR overexpression induces ROS formation in an angiotensin II-indepedent manner. Moreover, the addition of exogenous prorenin, a ligand for PRR, induces an increase in ROS formation which can be reversed by losartan, indicating its angiotensin-II-dependent nature.

Based on the data presented in Table 1, which amino acid residue of prorenin most likely interacts with the residue at position 201 of PRR?

"D201N – Kd = 0.441"

A) Alanine
B) Glutamate
C) Arginine
D) Glutamine

C) Arginine The answer to this question is C because substitution of aspartate (D) at position 201 with asparagine (N) results in a lower PRR affinity (higher Kd) towards prorenin. Since both asparagine and aspartate have equivalent carbon lengths, the decrease in affinity is most likely due to elimination of an ion pair. Among the possible options, only arginine could form an ion pair with aspartate.

Which enzyme was LEAST likely used in cloning of WT-PRR cDNA?
A) DNA polymerase
B) RNA polymerase
C) DNA ligase
D) Reverse transcriptase

B) RNA polymerase The answer to this question is B because while DNA polymerase (in DNA amplification), DNA ligase (in ligation of the cDNA to DNA vector) and reverse transcriptase (in reverse transcription of RNA to cDNA) are used during cDNA cloning, RNA polymerase is not used in cDNA cloning.

Compared to WT, what is the most likely effect of the W140L substitution on the stability of the PRR-prorenin complex?

WT – Kd = 0.034
W140L = 0.163

A) Elimination of pi-stacking interaction decreases stability of the complex.
B) Elimination of an ionic interaction decreases stability of the complex
C) Addition of a hydrophobic interaction increases stability of the complex.
D) Addition of a hydrophobic interaction increases stability of the complex.

A) Elimination of pi-stacking interaction decreases stability of the complex. The answer to this question is A because based on results presented in Table 1; in W140L variant, the replacement of tryptophan (a hydrophobic, aromatic amino acid) by leucine (a hydrophobic but non-aromatic amino acid) results in a decrease in stability of the PRR-prorenin complex as indicated by the increase in Kd. This is most likely due to the elimination of a π-stacking interaction with the side chain of tryptophan.

Based on the passage, which metabolic pathways are upregulated during the transition from intrauterine to extrauterine environment?

"After depletion of the blood glucose pool, the initial source for energy replacement is from hepatic glycogen, which is sufficient to support basic metabolism for the first 12 hours. Because infants normally do not receive an adequate supply of glucose from feeding for the first days of life, hepatic glycogen stores are soon exhausted, and the glucose pool must be replenished from non-carbohydrate sources including fatty acids and glucogenic amino acids."

A) Glycogenesis followed by glycogenolysis
B) Glycogenolysis followed by gluconeogenesis
C) Gluconeogenesis followed by glycogenolysis
D) Gluconeogenesis followed by glycogenesis

B) Glycogenolysis followed by gluconeogenesis The answer to this question is B because the passage states that the initial source for energy replacement is from hepatic glycogen indicating upregulation of glycogen breakdown (glycogenolysis). The passage also states that after hepatic glycogen stores are exhausted the glucose pool must be replenished from non-carbohydrate sources implying upregulation of the gluconeogenic pathway.

After the depletion of hepatic glycogen in newborns, which compounds can be used as precursors to sustain the blood glucose level?

*Gluconeogenesis

I. Acetyl-CoA
II. Lactate
III. Oxaloacetate
IV. a-Ketoglutarate

II, III, and IV only The answer to this question is D because among the listed options, only lactate, oxaloacetate, and α-ketoglutarate are used as starting materials in gluconeogenesis.

Which event is NOT a likely outcome of glucagon binding to its receptor? Increase in:

A) GDP binding to Ga subunit of the G protein
B) adenylate cyclase activity
C) protein kinase A activity
D) cAMP generation

A) GDP binding to Ga subunit of the G protein The answer to this question is A because following glucagon binding to its receptor and activation of its coupled G protein, activities of the adenylate cyclase and the protein kinase A, and level of cAMP are all increased. In contrast activation of the G protein promotes the dissociation of bound GDP and its exchange for GTP on the α subunit.

Infusion of which peptide hormone will most likely prevent brain injury in newborn infants exposed to high glucose levels during their fetal development?
A) Insulin
B) Epinephrine
C) Glucagon
D) Cortisol

C) Glucagon The answer to this question is C because exposure to high glucose levels results in elevated levels of circulating insulin which prevents mobilization of endogenous glucose storage. Infusion of glucagon will result in mobilization of endogenous glucose storage thereby preventing hypoglycemia and brain injury.

Activation of which enzyme would support the metabolism of newborn infants during the first 12 hours?

"After the depletion of the blood glucose pool, the initial source for energy replacement is from hepatic glycogen, which is sufficient to support basic metabolism for the first 12 hours."

A) Glycogen phosphatase
B) Glycogen phosphorylase
C) Glycogen hydrolase
D) Glycogen kinase

B) Glycogen phosphorylase The answer to this question is B because the passage states that hepatic glycogen supports the basic metabolism of newborn during the first 12 hours. Glycogen phosphorylase is the enzyme that catalyzes the rate-limiting step in glycogen breakdown (glycogenolysis).

The initial source of energy replacement in the liver of newborn infants is formed by glycosidic bonds between glucose molecules through:

A) α(1→4) linkage linearly and β(1→6) linkage at branch points.
B) β(1→6) linkage linearly and α(1→4) linkage at branch points.
C) α(1→4) linkage linearly and α(1→6) linkage at branch points.
D) α(1→6) linkage linearly and α(1→4) linkage at branch points.

C) α(1→4) linkage linearly and α(1→6) linkage at branch points. The answer to this question is C because the glucose polymer in liver (glycogen) is formed by glycosidic bonds between glucose molecules through α(1→4) linkage linearly and α(1→6) linkage at branch point.

Which of the following compounds is NOT a gluconeogenic precursor or substrate?
A) Lactate
B) Glycerol
C) Oxaloacetate
D) Phosphogluconate

D) Phosphogluconate The answer to this question is D because while lactate, oxaloacetate and glycerol are gluconeogenic precursors, phosphogluconate is involved in the pentose phosphate pathway and is not a precursor or substrate in gluconeogenesis.

Which type of enzyme removes the chemical groups that are added to proteins by kinases?
A) Phosphorylase
B) Cyclase
C) Phosphatase
D) Acetylase

C) Phosphatase The answer to this question is C because while kinase catalyzes the addition of a phosphate group to a substrate, the removal of the phosphate groups are catalyzed by a phosphatase.

Under physiological conditions, increased activity of succinyl-coA synthetase will most likely result in:
I. increased levels of succinyl CoA
II. increased levels of succinate
III. increased levels of GTP

II and III only The answer to this question is D because increased activity of succinyl-CoA synthetase will result in greater levels of the reaction products, succinate and GTP. Succinyl-CoA is the substrate of the reaction and its levels will likely decrease with increased succinyl-CoA function.

The pentose phosphate pathway results in the generation of:
A) NADPH, which is used as a reductive agent in cellular respiratory processes.
B) NADPH, which is used as an oxidative agent in cellular respiratory processes.
C) NADH, which is used as a reductive agent in cellular respiratory processes.
D) NADH, which is used as an oxidative agent in cellular respiratory processes.

A) NADPH, which is used as a reductive agent in cellular respiratory processes. The answer to this question is A because the pentose phosphate pathway results in the generation of NADPH, which is utilized in reductive reactions.

The Vmax of an enzymatic reaction is altered by which types of inhibitors?
I. Competitive
II. Noncompetitive
III. Uncompetitive
IV. Mixed

II, III, and IV only The answer to this question is D because as observed in 1/V0 versus 1/[S] plot (Lineweaver-Burk plot), competitive inhibitors are the only inhibitors that do not alter the intercept on the y-axis of the plot and therefore the Vmax of the enzymatic reaction.

Which metabolic reaction is most likely affected by the treatment of cells with C75? The reaction that converts:

"mtKAS is inhibited by C75. mtKAS is also involved in the synthesis of lipoic acid (LA), a cofactor for several metabolic enzymes"

A) fructose-6-phosphate to fructose-1,6-bisphosphate
B) fumarate to malate
C) pyruvate to acetyl-CoA
D) oxaloacetate to phosphoenolpyruvate

C) pyruvate to acetyl-CoA

Based on the information from the passage, the deletion of mtKAS most likely results in decreased level of:

"The deletion of the gene encoding for mtKAS results in dissipation of the electrochemical gradient in mitochondria due to an increase in membrane permeability."

A) protons in the mitochondrial matrix
B) protons in the mitochondrial intermembrane space
C) electrons in the mitochondrial matrix
D) electrons in the mitochondrial intermembrane space

B) protons in the mitochondrial intermembrane space The answer to this question is B because the passage states that deletion of the gene encoding for mtKAS results in dissipation of the electrochemical gradient generated by the electron transport chain. The electron transport chain uses the free energy from redox reactions to pump protons from mitochondrial matrix to the intermembrane space thereby generating an electrochemical gradient across the inner mitochondrial membrane. Therefore any event that causes the dissipation of the electrochemical gradient across the inner mitochondrial membrane will result in decreased level of protons in mitochondrial intermembrane space.

In the pentose phosphate pathway, which enzyme catalyzes the production of 6-phosphogluconate?

A) Glucose 6-phosphate dehydrogenase
B) 6-Phosphogluconolactonase
C) Phosphogluconate dehydrogenase
D) Glycerol 3-phosphate dehydrogenase

A) Glucose 6-phosphate dehydrogenase The answer to this question is A because Glucose 6-phosphate dehydrogenase catalyzes the conversion of glucose 6-phosphate to 6-phosphogluconolactone in the pentose phosphate pathway.

A partial DNA sequence of the coding strand of a gene is shown.

5′-GACATGGACTCGCTA-3′

Which sequence corresponds to the mRNA for this DNA sequence?

A) 5′-GACAUGGACUCGCUA-3′
B) 5′-CUGUACCUGAGCGUA-3′
C) 5′-UAGCGAGUCCAUGUC-3′
D) 5′-AUCGCUCAGGUACAG-3′

A) 5′-GACAUGGACUCGCUA-3′ The answer to this question is A because for any given gene, the nucleotide sequence of both the coding strand the mRNA are complementary to the sequence of the template strand.

Which technique CANNOT be used to analyze expression?
A) Western blotting
B) Northern blotting
C) Southern blotting
D) Reverse transcription PCR

C) Southern blotting The answer to this question is C because Southern blotting is a technique used to detect a particular sequence in a sample of DNA.

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