biochem

Based on the information in the passage, which of the following best describes the regulation of p38α activity? p38alpha is activated by a covalent modification

Enzyme activity may be modulated by covalent modifications to the enzyme (ie, post-translational modifications) such as phosphorylation or glycosylation. Whether a covalent modification activates or inactivates an enzyme varies from one enzyme to another, and even from one modification to another within an enzyme. For example, phosphorylation at one amino acid residue may activate the enzyme, while phosphorylation at another may inactivate it (or decrease its activity). The passage states that p38α becomes active when phosphorylated. Therefore, p38α activity is regulated by a covalent modification that activates the enzyme.

Based on the passage, the SIRT4 protein most likely contains which of the following? A mitochondrial target signal

The pyruvate dehydrogenase complex (PDHC) catalyzes the decarboxylation of pyruvate to form acetyl-CoA, which then enters the citric acid cycle. This process occurs in the mitochondrial matrix, so the PDHC is found in this organelle. According to the passage, SIRT4 colocalizes with the PDHC and must also be found in the mitochondrial matrix. Therefore, SIRT4 most likely has a mitochondrial targeting signal to induce its transport to the mitochondria. Educational objective:In eukaryotes, different metabolic pathways occur in different subcellular compartments. The enzymes required for each pathway must move to the appropriate compartment, and N-terminal peptide sequences typically direct them to their correct destination. Proteins that colocalize with one another are found in the same compartment.

Based on Figure 2, which of the following is true of H6-G225D hemagglutinin?

H6-G225D hemagglutnin denatures faster than the other hemagglutinin forms tested at any given temperature. Educational objective:Increasing temperatures can cause proteins to unfold (denature) by disrupting the intermolecular forces that maintain secondary, tertiary, and quaternary structures. When the melting temperature of a protein is achieved, there are equal amounts of folded and unfolded proteins. A lower melting temperature indicates lower stability and a higher denaturation rate.

Which of the following could researchers add to restore oxygen consumption in the presence of drug X?

(Choice C) Ubiquinol provides the electrons necessary for complex III to produce cyt-Cred, which in turn can reduce O2. However, in the presence of drug X, complex III is inactive and cannot use electrons from ubiquinol to produce cyt-Cred no matter how much ubiquinol is added. Without cyt-Cred, oxygen consumption will not be restored. Educational objective:Each reduced electron carrier in the ETC facilitates reduction of the next acceptor in the chain. Depletion of one reduced electron carrier (ie, cyt-Cred), prevents reduction of the next in the chain (ie, oxygen). Adding more of a depleted carrier will reinstate activity until the carrier is depleted again.

Based on the passage, the functional RNase A recovered after the removal of denaturants and subsequent treatment with oxidizing agent displays which levels of protein structure? I. Primary II. Secondary III. Tertiary IV> Quaternary

(Number IV) Quaternary structure exists only for multimeric proteins. The passage states that RNase A is a monomeric protein. Therefore, functional RNase A enzymes do not consist of the multiple polypeptide chains required for quaternary structure. Educational objective: Proteins and enzymes require an intact structure to maintain their function. Protein structure can be divided into four levels. The fourth level, quaternary structure, exists only for multimeric proteins.

Which of the following expressions gives the number of peptides that were tested for their ability to reduce extracellular LDH activity?

3!For example, a dipeptide consisting of alanine and histidine can be arranged in 2! = 2 different ways (AH or HA). A tripeptide made of three different, known amino acids can be arranged in 3! = 6 different ways, a tetrapeptide can be arranged in 4! = 24 different ways, and so on. The passage states that researchers tested every possible sequence of tripeptides that contain one alanine, one glycine, and one histidine residue. In this case, n = 3, so 3! different combinations were tested.

On human receptors, neuraminidase action would lead to a decrease in what type of bond?

Educational objective:Glycosidic bonds are created by the linkage of a nucleophilic functional group, such as a hydroxyl or amino group, to the anomeric carbon of a sugar molecule. The bond is classified based on the assigned numbers of the linked carbons and on whether the anomeric carbon is in the α- or the β-conformation.

Which amino acid structure would most likely be hydroxylated by HIF-PHDs?

Educational objective:Hydroxylation reactions replace hydrogen atoms with hydroxyl (-OH) groups and can occur in amino acids with aliphatic R-groups. Aliphatic amino acids are nonpolar, hydrophobic molecules with R-groups consisting of chains of C-C bonds that are single, branched, or in nonaromatic rings

The Klenow fragment incorporates approximately 1 × 102 incorrect nucleotides for every 1 × 107 total nucleotides. If the exonuclease site removes 99% of the incorrect nucleotides while also removing 5% of the correctly incorporated nucleotides, which of the following statements describes nucleotide removal by the Klenow fragment?

Educational objective:Incorrect nucleotides added to the 3′ end of a new DNA strand by DNA polymerase can be removed by the 3′→5′ exonuclease of DNA polymerase.

Which conclusion about the effect of SIRT4 on blood glucose levels can be drawn from the data in Figure 3? SIRT4 faciliates glucose uptake by preventing insulin resistance.

Educational objective:Insulin is secreted in response to metabolic signals such as high blood glucose and induces glucose uptake by cells. Cells that are overexposed to insulin can develop resistance and take up less glucose than they normally would in the presence of insulin.

The effect of BHB on adipocyte lipolysis is an example of:

Educational objective:Negative feedback is a common mechanism for regulation of biochemical pathways. In negative feedback, the product of a biochemical pathway feeds backward and inhibits an earlier step of the product's synthesis pathway.

Which of the following statements regarding the binding of α-SNAP to syntaxin is best supported by the results shown in Figure 2?

Educational objective:On substrate binding curves, substrate-protein binding is approximately linear at low substrate levels. When proteins reach saturation, all binding sites are occupied and binding curves level off because the further addition of substrate does not increase binding.

The information in the passage suggests that the effects of LCA activity could be counteracted by which type of enzymatic activity? kinase activity

Educational objective:Phosphatases catalyze the dephosphorylation of their substrates by hydrolysis, producing inorganic phosphate. Kinases catalyze the phosphorylation of their substrates by transferring organic phosphate from a donor molecule, usually ATP. The two enzyme types often regulate biological processes by counteracting each others' effects.

If an eastern blot is used to detect phosphorylation, it will most likely detect modifications to which of the following amino acid residues?

Educational objective:Phosphorylation is a post-translational modification that most commonly occurs on the hydroxyl groups of serine, threonine, and tyrosine residues.

Wild-type syntaxin (Synt-WT) and mutant syntaxins containing a deletion of the N-terminal sequence (SyntΔA) or C-terminal sequence (SyntΔB) were separated and incubated with α-SNAP. After washing away unbound molecules, which of the following is the most likely result of protein immunoblotting for α-SNAP?

Educational objective:Western blot (protein immunoblotting) uses antibodies to detect specific proteins. Greater quantities of the protein of interest manifest as larger bands on the resultant blot.

Based on information given in the passage, stabilizing interactions are most likely to occur between phosphate groups on IP6 and which of the following amino acids lining the Tcd binding pocket?

Electrostatic interactions combine elements of both hydrogen bonding and ionic bonding. In proteins, it most commonly occurs between the anionic carboxylate of aspartic or glutamic acids and the cationic ε-amino group of lysine or the guanidinium group of arginine but can potentially involve any pair of oppositely charged chemical groups. Educational objective:Salt bridges are electrostatic attractions that contribute to the stability of folded proteins and interactions between proteins and ligands or substrates. Salt bridging combines elements of hydrogen bonding and ionic bonding, and occurs between oppositely charged moieties.

During cis-complex disassembly, what consequence of NSF function is most likely responsible for destabilization of the cis-complex? exposure of the zero ionic layer's charged amino acids to solvent molecules.

NSF destabilizes the cis-complex by breaking the leucine zipper seal (ie, unzipping). Without the watertight seal provided by the leucine zipper, the charged and polar residues of the zero ionic layer are exposed to water. Because water is polar, exposure of the charged and polar residues results in preferential binding to water and resultant destabilization of the cis-complex. Educational objective:Hydrophilic amino acids are typically found on the outside of proteins whereas hydrophobic amino acids exist in the protein interior and protein-protein interfaces. When polar or charged amino acids are exposed to water, they preferentially form hydrogen bonds with water molecules.

Before the carbon atoms in threonine can be incorporated into glucose, they must enter citric acid cycle

Scheme I shows that threonine can be converted to succinyl-CoA, which subsequently undergoes the steps of the cycle that convert it to malate. The malate is then able to undergo a series of reactions that eventually produce glucose. Because the reactions that convert succinyl-CoA to malate are part of the citric acid cycle, the carbon atoms in threonine must enter the citric acid cycle before they can be converted to glucose. Educational objective:Metabolic pathways can be identified by the molecules they use as substrates.

Which of the following statements correctly describes the activities of the metabolic enzymes found in the liver cells of the rats that were tested?

fructose-1,6-bisphosphatase is more active in the livers of rats that had food withheld for 3 days than for 1 day. The information in the passage shows that rats that had food withheld for 3 days converted threonine to glucose at a faster rate than rats that had food withheld for 1 day. Therefore, rats that had food withheld for 3 days were conducting gluconeogenesis, which occurs primarily in the liver, at a faster rate. To accomplish this, those rats must have upregulated the liver enzyme that controls gluconeogenesis—fructose-1,6-bisphosphatase. Accordingly, it is likely that fructose-1,6-bisphosphatase was more active in the livers of rats that had food withheld for 3 days than in rats that had food withheld for 1 day.

What is the total number of anomeric carbons in disaccharides that are linked by β-glycosidic bonds, such as lactose?

The passage describes a study of the sweetness of sugar-containing solutions, including those containing the disaccharides lactose and maltose, and the question asks for the total number of anomeric carbons in disaccharides linked by β-glycosidic bonds. The anomeric carbon in a monosaccharide is the carbon that has two bonds with oxygen. Each sugar residue in a disaccharide or polysaccharide has one anomeric carbon. Therefore, all disaccharides (regardless of configuration) have two anomeric carbons Educational objective:In a monosaccharide, the anomeric carbon is the one that has two bonds with oxygen. Because each monosaccharide has one anomeric carbon, disaccharides have two anomeric carbons. Monosaccharides

In the absence of SIRT4, the electrochemical potential across the inner mitochondrial membrane:

is not changed relative to mitochondra in the presence of SIRT4

If miR-277 reduces acyl-CoA dehydrogenase mRNA levels to 50% of control levels, which of the following results is most consistent with the expected qPCR result? (Note: Ct is the threshold cycle, the number of PCR cycles required to produce a prespecified amount of cDNA.)

miR-277 results in a Ct valve that is one cycle more than the number of cycels needed in control experiments.

If the unique properties of each amino acid in the EPN motif are necessary for carbohydrate binding, which of the following amino acids would best substitute for the asparagine residue? glutamine

The passage describes an EPN motif within a bactericidal lectin, and the question asks for the amino acid that would best replace the asparagine if the unique properties of each amino acid in the EPN motif are necessary for carbohydrate binding. Among the 20 most common amino acids in proteins, asparagine is one of the two with polar, amide-containing R groups. The other such amino acid is glutamine; therefore, glutamine would best replace asparagine in a similar motif. Educational objective:Of the 20 most common amino acids found in proteins, glutamine and asparagine are the only two that are polar and contain amide side chains.

Which finding, when combined with data presented in Table 1, would be most likely to lead researchers to conclude that fructokinase activity mediates the abnormal accumulation of liver triglycerides in sugar-drinking mice that have similar energy and fructose intake levels?

Energy and fructose consumption in the WT mice consuming the 10% sugar solution and the FK KO mice consuming the 30% sugar solution are not statistically different (ie, values in each group are within one standard error of each other). Therefore, abnormal triglyceride accumulation in the group with active fructokinase (ie, WT mice) would provide evidence that fructokinase activity mediates excess triglyceride accumulation in sugar-drinking mice that consume similar amounts of energy and fructose. Educational objective:Different tissues can exhibit different metabolic responses that are consistent with their normal roles in whole-body metabolism

Based on the passage, culturing wild-type and FH− cells in atmospheric oxygen conditions leads to: increased hydroxylation and ubiquitination of HIF-1alpha

According to the passage, HIF-1α should be present only in small quantities under normal oxygen conditions because it is hydroxylated by active HIF-PHDs. Hydroxylation makes HIF-1α a target for ubiquitination, an ATP-dependent process in which ubiquitin molecules are added to a protein substrate. The proteasome then recognizes the ubiquitinated protein (eg, HIF-1α) and degrades it to small peptides via proteolysis (ie, peptide bond cleavage). As a result, HIF-1α concentration is maximally decreased (or absent) under normoxia. In comparison, HIF-1α concentration is high during hypoxic conditions when HIF-PHDs are inactive. Educational objective:Proteins are tagged for degradation in a process known as ubiquitination, which causes the proteasome to recognize and degrade the marked protein via proteolysis (peptide bond cleavage). This system is responsible for degrading damaged or unnecessary proteins/enzymes.

Compared to the wild-type HA RNA transcript in H6N1, which of the following features is mostly likely found in the H6-P186L RNA transcript?

According to the passage, the H6-P186L subtype of H6N1 influenza A contains a CCC to CUC mutation in the viral RNA, indicating that a uracil base took the place of a cytosine base. Therefore, this position in the mutant RNA contains a ribose bonded through a glycosidic linkage to uracil instead of cytosine. Educational objective:RNA nucleotides are composed of a ribose sugar with a phosphate group bound to a nitrogenous base (eg, adenine, guanine, cytosine, and uracil). Adjacent sugars in the RNA backbone are bound to each other by phosphodiester bonds, and bases are linked to sugars by glycosidic bonds.

Some strains of Staphylococcus aureus modify their membranes by attaching lysine to phospholipids on the extracellular side. Given this information, which of the peptides tested is likely to be most effective against these strains?

As stated in the passage, most known antimicrobial peptides are positively charged. Lysine is also positively charged under physiological conditions, so bacteria coated with lysine can repel those antimicrobial peptides and therefore resist their activity. Negatively charged peptides, on the other hand, would be attracted to the positive charges of the lysine residues and therefore would interact strongly with the membrane. Of the peptides tested, Peptide 4 is the only one with a net negative charge at physiological pH, having several aspartate residues (D). It therefore has the best chance of disrupting lysine-coated membranes. Educational objective:Amino acids may have positively or negatively charged side groups. Because similar charges repel and opposite charges attract, membranes coated with a positively charged amino acid such as lysine repel positively charged molecules but interact strongly with negatively charged residues such as aspartate.

Which of the following amino acid substitutions is most likely to interrupt the interactions of the zero ionic layer? Q53R on SNAP-25 (Sn1)

Back In this example, Q53R denotes the replacement of glutamine with arginine at position 53 on SNAP-25 (Sn1). As the passage suggests, the zero ionic layer is stabilized by attractive forces between the carbonyl groups (partial negative charge) on glutamine residues and the positively charged arginine on synaptobrevin. Substitution of glutamine with arginine results in a repulsive positive-positive ion interaction between the two arginine residues, destabilizing the interaction between the proteins in the complex. Educational objective:Amino acid substitutions are denoted by the original amino acid, the amino acid position, and the substituting amino acid. Amino acid substitutions that alter charge are likely to destabilize protein interactions, particularly if they result in repulsive forces between similarly charged amino acids.

Which of the following amino acid residues is LEAST likely to be present in the Tcd central region? His

Because any hydrophilic amino acid residues within the membrane would interact unfavorably with hydrophobic phospholipid tails, only membrane-spanning protein domains containing principally hydrophobic residues are thermodynamically stable. Histidine contains a basic, hydrophilic side chain. As such, histidine residues are most stable in aqueous environments. Of the residues listed, histidine would be the least likely to be found within a transmembrane domain. Educational objective:Transmembrane proteins contain hydrophobic transmembrane domains composed largely of nonpolar amino acids. They interact favorably with the hydrophobic tails of phospholipids in cellular membranes.

Which of the following metabolites is NOT produced in one of the metabolic pathways discussed in the passage?

Educational objective:Glycolysis occurs in the cytosol and ends with the synthesis of pyruvate. Pyruvate is transported into the mitochondria, where it is decarboxylated to form acetyl-CoA, which then enters the citric acid cycle. All steps of the citric acid cycle occur in the mitochondria and, unlike glycolysis, involve thioesters or multiple carboxylate groups.

Compared to proteins analyzed by BN-PAGE, proteins analyzed by native PAGE without Coomassie pre-treatment: may migrate in the opposite direction

Educational objective:Electrophoresis separates molecules by size, shape, and charge. In SDS-PAGE, sodium dodecyl sulfate coats proteins with a negative charge allowing all proteins to migrate to the anode. In native PAGE, anionic dyes like Coomassie serve a similar purpose without denaturing proteins.

Given that blood pH is 7.4, what is the approximate isopeptidase activity level of isoform 1 in blood expressed as a percentage of optimal activity? 50%

Educational objective:Enzyme activity is dependent on the ambient pH. Changes in pH can alter the protonation states of some amino acid side chains, which can alter their ability to stabilize enzyme structure and interact with substrates. The optimal pH of an enzyme must be determined experimentally.

During hypoxia, skeletal muscles have been shown to accumulate succinate and ADP molecules. Which of the following explains this result? anerobic conditions prevent mitochondrial regeneration of NAD+ and FAD

During hypoxia, less oxygen is available to accept electrons from the citric acid cycle; therefore, NAD+ and FAD are not regenerated in the mitochondria. Consequently, ADP concentrations increase (due to the absent proton motive force) as ATP production is significantly reduced. In addition, there is an accumulation of succinate and other citric acid cycle intermediates that require NAD+ or FAD for their oxidation reactions. Note that intermediates may not accumulate as they can leave the mitochondria (eg, citrate) or participate in side reactions like transamination (eg, α-ketoglutarate). Educational objective:The citric acid cycle uses oxidation reactions to convert NAD+ and FAD to NADH and FADH2, respectively. The movement of electrons from these high-energy electron carriers to electron transport chain (ETC) proteins drives phosphorylation of ADP to ATP. Regeneration of NAD+ and FAD in the ETC is only possible when oxygen is available to act as a final electron acceptor.

Why was a mild oxidizing agent added during the experiments? The bond between the sulfur and the hydrogen of a free thiol is a polar bond. Sulfur is more electronegative than hydrogen, and therefore it has a relatively high electron density around it. In contrast, because a disulfide bond is between two equally electronegative sulfur atoms, this bond is nonpolar. Each sulfur loses some electron density as electrons are equally shared between them. This loss of electron density is oxidation. Therefore, oxidizing agents were added to oxidize free thiolsto form disulfide bonds, which stabilizes the conformation of the protein.

Educational objective: Disulfide bonds (-‍S-‍S-‍) are formed by the oxidation of two nearby thiol (-‍SH) groups. Disulfide bonds form a covalent linkage that strongly stabilizes local protein structure.

Both isoforms of destabilase have approximately the same kcat values for isopeptidase activity at optimal pH, but Figure 1 shows that when 5 μL of each purified enzyme was provided with saturating levels of substrate, the isoforms had different levels of activity. What factor could explain this apparent discrepancy?

Educational objective:An enzyme's turnover number kcat is the number of reactions each enzyme catalyzes per second under saturating conditions. The maximum velocity Vmax of an enzyme-catalyzed reaction is equal to kcat multiplied by the concentration of enzyme [E]. Two enzymes with similar kcat values may have different Vmax values if one of the enzymes is present at a higher concentration than the other.

The following CD spectrum was obtained for P13: Based on this, the predominant secondary structure element in P13 is alpha helix

Educational objective:Circular dichroism (CD) spectroscopy is a technique that can be used to study the secondary structure of proteins. The CD spectrum of a given protein can be compared to standard spectra to elucidate the secondary structure features of the sample protein.

Compound 1 would most likely have what effect on the activity of LAR?

Educational objective:Competitive inhibitors typically bind enzymes at the active site, and therefore are usually similar in structure to the substrate. They have no effect on Vmax but increase the amount of substrate needed to achieve Vmax, resulting in an increased Km value.

Based on Figure 1, which electron transport chain complex is inhibited by drug X? complex III

Figure 1 shows that drug X causes an accumulation of UQH2 and its upstream electron carriers, NADH and FADH2. Drug X also causes a depletion of cyt-Cred. If complex III is inhibited, it can no longer consume UQH2 and can no longer produce cyt-Cred. The concentration of UQH2 builds as complexes I and II continue to produce it while complex III fails to consume it. As UQH2 levels increase, UQ levels fall, so eventually NADH and FADH2 will also accumulate as they have nothing to pass their electrons to. Because inhibition of complex III means cyt-Cox is no longer being reduced, cyt-Cred is depleted over time as complex IV continues to consume it. Therefore, drug X most likely inhibits complex III. Educational objective:Reduced electron carriers of the ETC are normally in a steady state, being consumed and regenerated as they move from one complex to another and back. Inhibiting a complex disturbs the steady state, causing an accumulation of upstream carriers and a depletion of downstream carriers.

Which statement best describes the structure of the avian receptor of influenza A shown in Figure 1? it is a glycoprotein with the anomeric carbon of sialic acid bound to nonreducing diaccharide.

Figure 1 shows that the influenza A receptor is a glycoprotein because the carbohydrate chain is linked to a protein through the amino acid asparagine. The terminal group of the chain consists of SA bound to two, linked sugar molecules (a disaccharide). Both sugars in the disaccharide are bound to another molecule at their anomeric carbons, so neither contains a hemiacetal or hemiketal. Therefore, the disaccharide is a nonreducing sugar, and the correct description of the receptor should state that it is a glycoprotein with SA bound to a nonreducing disaccharide. Educational objective:Reducing sugars contain free anomeric carbons that provide reducing power when they are oxidized. In linear form the anomeric carbon is an aldehyde or a ketone, and in cyclic form reducing sugars have hemiacetal or hemiketal configurations. Nonreducing sugars contain acetal or ketal structures in their cyclic forms.

Under aerobic conditions, which of the following molecules would be expected to be present in significantly decreased levels in the livers of KO mice? succinyl CoA

Figure 1 shows that α-ketoglutarate is oxidized at a significantly lower rate in KO mice than in WT mice. Because all livers were treated with the same amount of α-ketoglutarate, this indicates that αKDH works more slowly in KO mice than in WT mice under identical experimental conditions. Therefore, the product of α-ketoglutarate oxidation, succinyl-CoA, will most likely be depleted in KO livers. Educational objective:The citric acid cycle facilitates several oxidative reactions, including two oxidative decarboxylation events that use isocitrate and α-ketoglutarate as substrates, respectively. Inhibition of a step in the cycle leads to accumulation of reactants and depletion of products.

Which of the following conclusions is most strongly supported by the data in the passage?

Figure 2 also shows that the Protein X Anap mutant displays a λmax of 420 nm when saturated with ligand. The conformational change upon ligand binding alters the environment of Anap such that its λmax now matches that of free Anap in ethyl acetate, which is much less polar than water. Therefore, Anap is likely positioned toward the more hydrophobic, buried core in ligand-bound Protein X. Together, these data support the conclusion that the Anap residue moves from a surface-exposed to a buried position upon ligand binding. Educational objective:Most soluble proteins have a hydrophilic surface and a hydrophobic core. Conformational changes that rearrange a residue from surface-exposed to buried also change the residue's environment from hydrophilic to hydrophobic.

Based on the data shown in Figure 2, Protein X can best be described as: a positively cooperative protein with a Hill coefficient greater than 1

Figure 2 is essentially a binding curve for Protein X that uses the emission wavelength as a reporter for ligand binding. Because the wavelength decreases as the fraction of Protein X increases, the binding curve appears inverted; however, the overall shape of the binding curve is still recognizable. At low concentrations of L, the curve is shallow, indicating low-affinity binding. The curve steepens as ligand-induced conformational changes cause high-affinity binding for subsequent ligands. The curve flattens again as the population of Protein X becomes saturated with ligand, forming the top of the sigmoidal curve. Because of the sigmoidal binding curve shown in Figure 2, Protein X can be described as a protein showing positive cooperativity and a Hill coefficient greater than 1. Educational objective:Positive cooperativity is indicated on a binding curve by a sigmoidal shape and a Hill coefficient greater than 1.

If a western blot showed P13 only in the 300 kDa band of the BN-PAGE gel, then the evidence best supports the conclusion that P13:

Focusing only on the single lane formed by the 300-kDa P13-containing complex, only one protein band is seen on the SDS-gel, indicating that only one peptide subunit (~13 kDa) contributed to the complex shown in the 300 kDa band of the BN-gel (Choice D). The large size difference noted before and after denaturation (ie, ~300 kDa vs. ~13 kDa) indicates that several copies of a single 13-kDa polypeptide join as a homo-multimer to form the 300 kDa complex. Educational objective:Polyacrylamide gel electrophoresis (PAGE) can be classified as native or denaturing. Native PAGE keeps protein structure and complexes intact whereas denaturing page (eg, SDS-PAGE) denatures structure. The output of native PAGE can be run as the input of SDS-PAGE to perform a two-dimensional PAGE experiment.

Antimicrobial peptides can be taken into cells by endocytosis. They are subsequently transported to the lysosome, where the pH decreases to approximately 5 and peptides are degraded. After lysosomal degradation, researchers found a tripeptide consisting of the last three amino acids of Peptide 2. What is the net charge of this tripeptide at lysosomal pH?

Histidine has a pKa of 6, so while it is deprotonated at cytosolic pH (7.4), it is protonated at lysosomal pH and carries a positive charge. Lysine and the N-terminus are also protonated and remain positively charged as the pH drops to lysosomal pH. Glutamate and the C-terminus have pKa values of ~4 and 2, respectively, so they remain deprotonated at lysosomal pH and are still negatively charged. The sum of the charges is therefore +1. Educational objective:Ionizable side chains and peptide termini can be charged. If the pKa of an ionizable group is higher than the surrounding pH, the group is protonated and therefore more positive (or less negative). The net charge of a peptide at a given pH is the sum of the charges of all the ionizable groups.

Which interpretation(s) is(are) consistent with the data in Table 1? The affinity of PGN with a 5-sugar saccharide is significantly less than that of PGN with a 35-sugar saccharide. The affinity of PGN with a 15-sugar saccharide is significantly greater than that of PGN with a 35-sugar saccharide. The affinity of PGN with a 15-sugar saccharide is significantly less than that of PGN with a 5-sugar saccharide.

I only The question asks for the interpretation most consistent with the data in table 1. A general rule for assessing statistical differences when p-values are not given is that groups are likely different if the mean ± 2 SEM ranges do not overlap. Statement I, stating that the affinity of PGN with a 5-sugar saccharide is significantly less than that of PGN with a 35-sugar saccharide, is therefore correct. lower Kd reflects higher affinity, and a higher Kd reflects lower affinity. Educational objective:When assessing binding interactions, a lower Kd reflects higher affinity and a higher Kd reflects lower affinity. As a general rule when exact p-values are not given, groups are statistically different if their mean ± 2 SEM ranges do not overlap

The tested tripeptide variants most likely have which of the following characteristics in common? I. Primary structure II. Isoelectric point III. Hydrophobicity

II and III only (Number I) Primary structure is defined as the sequence of amino acids in a polypeptide chain. The tripeptides tested are all made of the same three amino acids but arranged in different orders, so the tripeptides do not all have the same primary structure. Educational objective:Peptides are chains of amino acids linked together by peptide bonds. For short peptides that do not adopt higher levels of structure, properties such as isoelectric point and hydrophobicity generally depend only on which amino acids are present (composition) and not on the order in which the amino acids are arranged (primary structure)

Of the four original solutions studied, which of the following pairs of solutions contain epimers? I. Maltose and lactose II. Galactose and glucose III. Maltose and glucose IV. Lactose and galactose

II only The passage describes a study in which the sweetness of various sugar solutions is determined, and the question asks which pairs of solutions contain sugars that are epimers. Epimers are molecules that contain multiple chiral carbons and are identical apart from their configuration at a single chiral carbon. Of the sugar pairs provided, only glucose and galactose meet both these requirements Educational objective:Epimers contain more than one chiral carbon and are identical apart from their configuration at a single chiral carbon.

Following ingestion of a large amount of fructose, ATP consumption occurs more rapidly than ATP production in some cells, and the cytosolic ATP concentration of the small intestine and liver fall and then recover whereas those in skeletal muscle are relatively unchanged. Further analysis shows all three types of cells express fructose transporters. What mechanism is most likely responsible for these findings?

In most tissues (including muscle), allosteric feedback inhibition of hexokinase occurs when its products (ie, glucose-6-phosphate and fructose-6-phosphate) increase in concentration. This feedback inhibition limits ATP use for phosphorylation in the hexokinase reaction, thereby preventing the ATP depletion that might otherwise occur. In contrast, the fructokinase present in small intestine and liver cells does not undergo feedback inhibition and can therefore transiently deplete cellular ATP levels when fructose is abundant. The ATP levels eventually recover when later steps in glycolysis produce more ATP.

The peptide substrate (PS) is an inhibitor that binds HIF-PHDs outside the active site. The PS binds to both HIF-PHD and HIF-PHD-2-OG complexes with equal affinity. According to the passage, which graph would represent the effect of PS on HIF-PHD activity?

In this scenario, the peptide substrate (PS) binds both HIF-PHD allosterically (outside the active site) and HIF-PHD-2-OG with equal affinity. By definition, the PS is considered a noncompetitive inhibitor, a molecule that binds both the enzyme and the enzyme-substrate complex (ES) with equal affinity. Noncompetitive inhibitors slow the reaction rate and decrease Vmax; however, they do not alter the affinity of the enzyme for its substrate, resulting in an unchanged KM. Accordingly, in the plot of reaction velocity V0 with increasing 2-OG cosubstrate shown in Choice D, the second curve shows that addition of the PS decreases Vmax but does not change KM compared to the first curve. Educational objective:Michaelis-Menten graphs plot the initial reaction velocity against substrate concentrations and can be used to determine KM and Vmax. Noncompetitive inhibitors allosterically bind the enzyme and enzyme-substrate complex with equal affinity. These inhibitors decrease Vmax but do not alter KM.

The absence of which of the following lipids is most likely to affect the assembly of viral envelopes? phosphatidylseine

Influenza A obtains its viral envelope from the plasma membrane of its host cell. However, the passage states that the composition of the envelope differs significantly from the plasma membrane. Unlike human membranes, the viral envelope contains mostly phosphatidylethanolamine and negatively charged glycerophospholipids. Therefore, phospholipids with phosphoethanolamine, glycerophospholipids with negatively charged head groups such as phosphoserine and phosphoinositol, and other negatively charged glycerophospholipids like diphosphatidylglycerol are needed for envelope formation. Of the choices presented, phosphatidylserine is the only negatively charged glycerophospholipid. Therefore, its absence is most likely to affect the assembly of the viral envelope. Educational objective: Phospholipids are composed of a hydrophilic polar head that contains a phosphate group and a hydrophobic tail with one or more fatty acids. They may contain a glycerol or a sphingosine backbone, and their properties depend on the head groups attached to the phosphate.

In Figure 2, the most likely reason that researchers added succinate was to stimulate: FADH2 formation by complex II

Oxygen is the final electron acceptor in the ETC and is consumed only if sufficient NADH and FADH2 are present. Increasing the rate of FADH2 synthesis increases the rate of electron transfer and therefore increases the rate of oxygen consumption. In Figure 2, researchers measured oxygen consumption to assess ETC output under various conditions. They most likely added succinate to stimulate formation of FADH2 (and the subsequent electron transfer) by Complex II. Educational objective:The electron transport chain (ETC) depends on the production of NADH and FADH2 to function. Complex II of the ETC is a flavoprotein (contains FAD) that is also known as succinate dehydrogenase. It is part of both the citric acid cycle and the ETC, and it oxidizes succinate to reduce FAD to FADH2. FADH2 is then oxidized back to FAD in the same enzyme, producing ubiquinol.

The side chain of the C-terminal amino acid residue in GAH is derived from which structure?

Peptides form when the C-terminus of one amino acid or peptide reacts with the N-terminus of another to form an amide bond. The newly formed peptide will have one N-terminus and one C-terminus. The amino acid residue closest to the N-terminus is the N-terminal residue, and the residue closest to the C-terminus is the C-terminal residue. Additional residues may be found between the N- and C-terminal residues. Educational objective:Amino acids consist of a side chain connected to a backbone structure that contains an amino group (N-terminus), an α-carbon, and a carboxyl group (C-terminus). Peptides are chains of amino acids that form when the amino group of an amino acid or peptide reacts with the carboxyl group of another amino acid or peptide. Like amino acids, peptides have an N-terminus and a C-terminus.

Which amino acid is LEAST likely to be found in normally functioning binding domains of t-SNARE and v-SNARE proteins? proline

Proline and glycine are often found in loops and linker regions between α-helices and β-sheets, particularly at sharp turns (eg, beta turns). Proline is unlikely to be found within an α-helix because it is structurally rigid and introduces a kink in the chain that is useful for sharp turns but disrupts α-helices. In addition, the backbone amide of a proline residue connects back to its side chain to form a ring structure and is therefore unable to participate as a hydrogen bond donor. Glycine has the smallest side group of all the amino acids (a hydrogen atom) and is quite flexible. Although ideal for sharp turns, this flexibility causes α-helices to be too "floppy" and disrupts the structure. Educational objective:α-Helices have 3.6 amino acids per turn and are stabilized by hydrogen bonds between carbonyl and amide groups. Of the amino acids, proline is least likely to be found within α-helices and is more often found in linker regions or at sharp turns due to its structural rigidity. Glycine also tends to disrupt helices due to its excessive flexibility.

Acetyl-CoA carboxylase (ACC) is an enzyme that converts acetyl-CoA to malonyl-CoA. Phosphorylation of ACC inactivates the enzyme. Based on this, how would the effect of phosphorylating ACC most likely compare to the effects of miR-277?

The ACC mutation opposes the effect of miR-277 on fatty acid oxidation. The question states that acetyl-CoA carboxylase (ACC) phosphorylation inactivates the enzyme. Phosphorylation of ACC can occur through the enzymes protein kinase A (PKA, activated through catabolic hormones like glucagon or epinephrine) or AMP-activated protein kinase (AMPK, activated by high AMP/low ATP levels). In both cases, these kinases signal a need for lipid catabolism to produce more ATP. Therefore, phosphorylation of ACC promotes fatty acid oxidation, which opposes miR-277-mediated inhibition of fatty acid oxidation.

How is the energy of the phosphate transfer from ATP to RCAN1 affected by the addition of active p38α?

The activation energy decreases but the Gibbs free energy is unaffected. The passage states that p38α is an enzyme that transfers a phosphate group from ATP to RCAN1 (ie, p38α is a kinase). Because p38α is an enzyme, it must decrease the activation energy of the reaction, allowing the reaction to proceed faster. However, the reactants are ATP and RCAN1, and the products are ADP and pRCAN1, regardless of whether p38α is present. Therefore, as with any catalyst, p38α does not affect the Gibbs free energy of the reaction. Educational objective:Enzymes are biological catalysts that increase the rate of a biological reaction without being consumed. As such, they decrease the activation energy of the reaction they catalyze but do not alter the Gibbs free energy.

Why did researchers test for the α-tubulin protein along with HIF-1α? α- tubulin serves as a loading control because it is expressed ubiquitously.

The most common loading controls are proteins necessary for baseline cellular function transcribed from housekeeping genes (eg, α- and β-tubulin proteins are structural/mobile cytoskeleton components). Based on this fact, α-tubulin was used as a loading control as it can be detected at consistent levels in all cell types and has a molecular weight different from HIF-1α. The consistent expression of α-tubulin in the gel under all conditions confirms that differences in HIF-1α expression were not a result of experimental errors in protein loading or detection. Educational objective:Loading controls normalize protein detection and ensure that protein loading is standardized across the gel. Proteins used as loading controls tend to be ubiquitously expressed and have consistent concentrations across all cell/tissue types. Housekeeping genes are the most common loading controls.

Given that RCAN1 noncompetitively inhibits calcineurin, which statement is true regarding the dissociation constants shown in Figure 1? Kd1=kd2

The passage and question state that RCAN1 noncompetitively inhibits calcineurin. Therefore, RCAN1 must have equal affinity for free calcineurin and the calcineurin-pRCAN1 complex. As a result, the two Kdvalues must be equal: Kd1 = Kd2.

During nucleic acid synthesis, the Klenow fragment incorporates an incorrect nucleotide, which is subsequently removed. Compared with the action of the polymerase region of the Klenow fragment, its exonuclease region most likely acts:

The passage describes the Klenow fragment of DNA polymerase I and asks where its exonuclease site acts relative to the site of action of the fragment's polymerase. The fact that each DNA strand is synthesized in a 5′→3′ direction means that the polymerase successively adds new nucleotides to the 3′ end of the new DNA strand. Exonucleases act at the ends of DNA strands, and a 3′→5′ exonuclease acts at the 3′ end (ie, the growing end) of a new DNA strand. In other words, a 3′→5′ exonuclease acts in the opposite direction of a polymerase to remove nucleotides that the polymerase recently added to the growing strand. Therefore, the 3′→5′ exonuclease acts at the same site as the polymerase but in the 3′→5′ direction. Educational objective:DNA polymerases possess several different enzyme activities. A 3′→5′ exonuclease proofreads and helps make corrections at the 3′ end of a growing DNA strand.

The researchers concluded that curcumin inhibition of NF-κB binding to target gene sequences was linearly related to the curcumin concentration. Do the data support the researchers' conclusion?

The passage presents results from an experiment in which NF-κB binding to DNA was studied, and the question asks if the data support the conclusion that NF-κB binding was linearly related to curcumin concentration. The loading controls established a range over which band intensities are linearly related to NF-κB binding; the other lanes show band intensities within this range. Table 1 shows that for every 20 μM of curcumin added, band intensity decreases by approximately 0.25 units. Therefore, the decrease in intensity is directly proportional to increasing curcumin concentration (ie, it is linear), so the data support the researchers' conclusion. Educational objective:Standards can be used to establish the relationship between variables over a particular range, which can be used to better interpret measured values in experimental samples.

Based on the information in the passage, FCCP most likely causes a decrease in energy production by

The passage states that FCCP can passively transport protons across membranes. Therefore, the addition of FCCP provides an alternative to ATP synthase by which protons can cross the membrane down their concentration gradient (ie, from high concentration to low concentration). Because the proton concentration is normally higher outside the mitochondrial matrix than inside it, FCCP transports protons into the matrix. However, protons that enter the mitochondrial matrix by this means do not interact with ATP synthase, so they do not drive ATP synthesis; this phenomenon is called "decoupling." Because this decoupling results in fewer protons going through ATP synthase, the rate of ATP synthesis decreases in the presence of FCCP. Educational objective:ATP synthesis is driven by protons crossing the inner mitochondrial membrane through ATP synthase. When protons are transported across the membrane by means that bypass ATP synthase, the energy released in the process cannot be used to produce ATP. This phenomenon is known as "decoupling" because proton transfer is no longer coupled to ATP synthesis.

Which of the following conclusions about protein folding is best supported by the information in the passage?

The passage states that RNase A was fully denatured until only the amino acid sequence (ie, primary structure) remained intact. Nevertheless, RNase A was able to refold to a functional state (ie, regain higher levels of structure) without an external energy source. Based on this evidence, it can be concluded that the higher levels of RNase A structure are encoded by its primary structure due to the dependency of protein structure levels. Educational objective:Protein structure is subdivided into four levels. The three higher levels of protein structure are dependent on the proper configuration of the lower levels of protein structure.

Which of the following statements correctly characterizes the bond that is broken during the reaction catalyzed by the IP6-dependent Tcd protease? the bond restricts molecular rotation

The passage states that Tcd undergoes autoprocessing in the form of intramolecular cleavage; in other words, the Tcd protein cleaves itself. In addition, the question asks about a reaction catalyzed by protease activity. Therefore, the reaction must be the cleavage of a peptide bond, which has restricted rotation. Educational objective:The peptide bond is a resonance-stabilized, planar structure exhibiting partial double-bond character. Rotation about the bond axis is restricted.

Which of the techniques described in the passage would be best for probing the effect of a mutation on the affinity of a transcription factor for its ligand? southwestern blot

The passage states that southwestern blots detect protein-DNA interactions. Therefore, of the available choices, the southwestern blot would be best for determining the effect of a mutation on binding affinity between a TF (a protein) and its ligand (DNA). Educational objective:Transcription factors are proteins that bind DNA and either increase or decrease transcription levels of one or more genes.

Compared to livers obtained from WT mice, KO livers are likely to show: decreased transmination of leucine

The passage states that αKIC is the α-keto derivative of leucine, meaning that it is produced by leucine transamination. The box plot in Figure 1 shows that αKIC is oxidized at a slower rate in the livers of KO mice than in the livers of WT mice. The resulting accumulation of αKIC shifts the equilibrium of the transamination reaction toward the production of leucine (reactant) rather than its consumption and degradation. Therefore, the transamination of leucine is likely to be decreased in the livers of KO mice.

Based on the passage, hepatocytes (ie, liver cells) that are producing BHB are also most likely to have: downregulated phosphofructokinase-2 (PFK2) activity

The question asks about the activity of enzymes in hepatocytes (ie, liver cells) that are producing BHB. The liver produces ketone bodies like BHB under glucose-poor (ie, fasting) conditions. During fasting, glucagon stimulates the liver to produce and export metabolic fuels, which include both BHB (through β-oxidation) and glucose (through glycogenolysis and gluconeogenesis). To regulate the glycolysis-gluconeogenesis equilibrium, glucagon regulates the activity of the bifunctional enzyme phosphofructokinase-2/fructose-2,6-bisphosphatase (PFK-2/FBPase-2). This enzyme regulates the level of fructose 2,6-bisphosphate (F2,6BP), which is an allosteric stimulator of glycolysis. Under fasting conditions in which BHB is also being produced, liver cells use gluconeogenesis to provide fuel to other cells, so glycolysis in these cells would be detrimental to the organism overall. Therefore, phosphofructokinase-2 activity is most likely downregulated in these cells to downregulate glycolysis. Educational objective:Ketone bodies are produced under fasting conditions. In the fasting state, glucagon signals the phosphorylation and regulation of various enzymes to increase the production of glucose through both glycogenolysis and gluconeogenesis.

The second reaction shown in Figure 1 involves an enzyme that transfers coenzyme A from succinyl-CoA to acetoacetate. If this enzyme were used in the citric acid cycle in place of succinyl-CoA synthetase, the overall yield of high-energy molecules in one round of the citric acid cycle would be: decreased by one NTP (either ATP or GTP)

The question asks for the change in yield of the citric acid cycle (TCA cycle) if the succinyl-CoA → succinate reaction shown in Figure 1 replaced the traditional succinyl-CoA synthetase reaction. In the traditional succinyl-CoA synthetase-catalyzed reaction, the free energy released after hydrolyzing the high-energy thioester bond of succinyl-CoA is used to form an NTP (either ATP or GTP, depending on the organism) by substrate-level phosphorylation. (Note: This reaction is the reverse of a ligase reaction). The alternative pathway shown in Figure 1 is a transfer reaction. The coenzyme A molecule is simply transferred from succinyl-CoA to acetoacetate without consumption or production of an NTP. Because no NTP is made in the alternate pathway, but an NTP is made in the traditional pathway, the yield of that round of the TCA cycle is decreased by one NTP molecule. Educational objective:In the citric acid cycle, the conversion of succinyl-CoA to succinate by succinyl-CoA synthetase is accompanied by the production of GTP (or ATP) by substrate-level phosphorylation. If succinyl-CoA is converted to succinate by an alternative enzyme or pathway, the NTP yield of that round of the citric acid cycle will change.

Which of the following enzymes involved in the small intestinal metabolism of fructose is part of the glycolytic pathway?

The question asks for the glycolytic enzyme that participates in fructose metabolism in small intestinal cells. Of the choices given, only enolase, which catalyzes the formation of phosphoenolpyruvate from 2-phosphoglycerate, is an enzyme of glycolysis. Educational objective:Glycolysis is the series of cytosolic reactions in which a glucose molecule is converted to two pyruvate molecules, resulting in the net production of two ATP molecules and two NADH molecules.

According to the Henderson-Hasselbalch equation, the pH range over which the EPN motif would go from being approximately 99% protonated to approximately 1% protonated is equal to:

The question asks for the pH range over which the EPN motif would go from being approximately 99% protonated to approximately 1% protonated, based on the Henderson-Hasselbalch equation. Only the glutamate (Glu) residue in the EPN motif has an ionizable R group, and the effect of the pH on the deprotonated-to-protonated ratio is logarithmic. Applying the Henderson-Hasselbalch equation to glutamate yields: pH=pKaGlu+logdeprotonated Glu[protonated Glu] Because the logarithm of 99 (corresponding to 1% protonation) is approximately 2 and the logarithm of 199 (corresponding to 99% protonation) is approximately −2, the correct range is glutamate's R-group pKa plus or minus 2 pH units. Educational objective:The Henderson-Hasselbalch equation relates a chemical species pKa to the pH of its environment. Each unit change in a number's log10 corresponds to an order-of-magnitude change in the number itself.

If the researchers used the form of galactose most common in nature (ie, d-galactose), the sequence of stereochemical configurations from carbon 2 to carbon 5 must be:

The question asks for the sequence of stereocenter configurations in the most common naturally occurring form of galactose (ie, ᴅ-galactose). For a sugar to be of the ᴅ form, the stereocenter farthest from the anomeric carbon must have the R configuration (note that this only applies to sugars and not to all chiral molecules). Therefore, the sequence of R, S, S, R stereocenter configurations is the only one that could possibly be that of d-galactose. Educational objective:ᴅ- and ʟ-sugars have opposite R and S configurations at each stereocenter. The stereocenter farthest from the anomeric carbon is used to classify the configuration of a sugar, and will be in the R configuration for a ᴅ-sugar and in the S configuration for an ʟ-sugar.

Each of the blotting techniques mentioned in the passage most likely requires which of the following steps? separation of molecules by gel electrophoresis

The question asks for the step that is most likely required for each blotting technique mentioned in the passage. Of the given answer choices, separation of molecules by gel electrophoresis is the only choice that is common to all the blotting techniques mentioned in the passage.

Which of the following reasons does NOT describe why threonine was labeled with carbon-14 atoms?

The question asks which reason does NOT describe why labeled threonine was used. Chemical separation of molecules requires significant differences in chemical properties that can be exploited (eg, different interactions with mobile or stationary phases in chromatography columns). Carbon-14 essentially has the same chemical properties as any isotope of carbon; therefore, its incorporation into threonine does not alter the overall chemical properties of threonine or its downstream metabolites. Consequently, incorporation of carbon-14 into these metabolites does not allow them to be separated from unlabeled metabolites by chemical means. Educational objective:Radioactive isotopes of atoms are commonly incorporated into molecules to track their progress through metabolic pathways and to monitor specific compounds. Different isotopes of an atom have essentially the same chemical properties.

Which solution(s) will likely be rated as sweetest after the addition of glucosidase enzymes to each solution? (Note: Assume solution sweetness is determined equally by intrinsic sugar sweetness and sugar concentration.) lactose and maltose equally

The question describes a study of sugar solution sweetness and asks for the solution(s) likely to be rated as sweetest if the experiment is repeated with the addition of glucosidase enzymes to the solutions. Glucosidases are enzymes that break the glycosidic bonds in sugar polymers. Glucosidase treatment would double the sugar concentrations of the disaccharide solutions by converting them to monomers whereas the sugar concentration in the original sugar monomer solutions would be unaffected by glucosidase treatment. Because the total concentrations of glucose and/or galactose (two equally sweet monosaccharides) would be twice as high in the disaccharide solutions as in the monosaccharide solutions, the lactose and maltose solutions would be sweetest after glucosidase treatment. Educational objective:Glucosidases are enzymes that facilitate the breaking of glycosidic bonds in sugar polymers (ie, disaccharides and polysaccharides).

The first step of the autoprocessing reaction mechanism described in the passage involves modification of a thiol in the active site by an amino acid functioning as a general base. This action is likely to enhance protease activity by:

The question states that the mechanism begins with the modification of a thiol group. Cysteine is the only amino acid with a thiol (-SH) in its R group. The question further states that the thiol is modified by an amino acid (histidine, in this case) acting as a general base. Bases are proton acceptors; therefore, histidine must deprotonate (not protonate) cysteine. Educational objective:Cysteine and serine frequently act as nucleophiles in enzyme-catalyzed reactions, but they are relatively weak nucleophiles on their own. Therefore, the mechanisms of these reactions often include a deprotonation step to enhance nucleophilicity.

How is the enzymatic regulation of HIF-PHDs by fumarate and succinate best described? fumarate is a more potent competitive inhibitor than succinate.

Therefore, fumarate is a more potent competitive inhibitor as it results in a greater increase in the KM value of 2-OG (ie, greater right shift) than does succinate, given equal inhibitor concentrations. Fumarate occupies the HIF-PHD active site for longer periods than succinate, effectively reducing the enzyme's affinity for 2-OG. Educational objective:The x-intercept on a Lineweaver-Burk plot represents −1/KM, and the y-intercept represents 1/Vmax. Competitive inhibitors bind reversibly to an enzyme's active site and block substrate access, causing an increase in KM but no change in Vmax.

Dephosphorylation of pRCAN1 by calcineurin most likely causes which of the following? downregulation of T-cell activation

These statements indicate that calcineurin inhibition helps suppress the immune response by decreasing T-cell activation. Dephosphorylation of pRCAN1 most likely leads to inhibition of calcineurin by producing the inhibitor RCAN1. This would result in inability to dephosphorylate NFAT, which would lead to downregulation of T-cell activation.

What was the purpose of the measurements performed after glucosidase treatment?

To determine whether glucosidase treatment can modulate the sweetness of reducing sugar solutions, as established in the first experiment. The passage describes measurements of sugar solution sweetness, initially in the absence of glucosidase treatment and, in a second set of measurements, after glucosidase treatment. Each of the solutions in the study contains a reducing sugar, and the question asks for the purpose of the second set of measurements with glucosidase treatment. Therefore, the second set of measurements allows the researchers to determine whether glucosidase treatment alters the sweetness of the reducing sugar solutions as established in the first experiment. Educational objective:Reducing sugars are those with a free anomeric carbon (ie, one not involved in a glycosidic linkage).

If the mitochondrial membrane potential decreased, which of the following changes to the metabolic pathway in Scheme I would best help restore the membrane potential?

To increase the mitochondrial membrane potential in a cell, more reduced cofactors must enter the ETC. The citric acid cycle produces 3 NADH and 1 FADH2 molecules per round. Therefore, increasing the activity of the cycle could provide the needed cofactors to increase the mitochondrial membrane potential. Educational objective:The electric potential across the mitochondrial membrane arises as protons are pumped from the mitochondrial matrix into the intermembrane space by the electron transport chain. The citric acid cycle provides the reduced electron carriers needed for the electron transport chain and therefore helps control the membrane potential.

Glucose and BHB can each be converted to two acetyl-CoA molecules. How many NAD+ molecules (both cytosolic and mitochondrial) are consumed by conversion of glucose to two acetyl-CoA and by conversion of BHB to two acetyl-CoA, respectively?

four NAD+ consumed per glucose, one NAD+ consumed per BHB

Based on Figure 2, residues H759 and H757 of wild-type TcdA and TcdB, respectively, may function to: mediate an interaction in the flap region that obstructs the active site in the absence of IP6

Unlike wild-type Tcd, autoprocessing occurs in H759A and H757A mutants even in the absence of IP6. This behavior supports the conclusion that in wild-type proteins, an interaction between the flap region and the active site is abolished when histidine is mutated to alanine in the mutants. Therefore, H759 and H757 of wild-type Tcd proteins likely help mediate an interaction that is necessary for the flap region to obstruct the active site in the absence of IP6. Educational objective:Allosteric regulation occurs when an allosteric effector binds to the allosteric site and alters catalytic activity in the active site. Mutations that result in the instability of normal conformations can disrupt allosteric regulation.

Which conclusion about the effect of the H759A and H757A mutations is best supported by Figure 2? Tcd mutant protease activity is less dependent on IP6 than is wild type activity

When 0.1 mM IP6 is added, the extent of Tcd hydrolysis increases substantially in mutant proteins but cannot be further increased by adding more IP6, demonstrating that only a small amount of IP6 is required to reach full activity in mutants. In contrast, wild-type autoprocessing steadily increases with increasing IP6 concentration over the range tested. Therefore, the protease activity of TcdA H759A and TcdB H757A is less dependent on IP6 than the activities of their wild-type counterparts.

In certain bactericidal lectins that are similar to the one that was tested, carbohydrate binding is mediated by an X-P-X motif, in which X refers to amino acids that replace E and N in the EPN motif. The best explanation for proline's presence in both motifs is that:

a secondary amino group is required for favorable positioning of hte neighboring amino acids. Educational objective:As the only secondary amino acid, proline imparts structural rigidity to polypeptide regions in which it is found.

After removal of an incorrectly incorporated nucleotide by the Klenow fragment exonuclease, the correct nucleotide is incorporated. Compared to the incorrectly incorporated nucleotide, the chances of the correctly incorporated nucleotide encountering the exonuclease site of the Klenow fragment will most likely be

decreased Educational objective:For a growing DNA strand to have the proper shape, it must have the correct base pairing. An incorrectly paired nucleotide alters the shape of a growing DNA strand in a way that increases the likelihood that the nucleotide will encounter the 3′→5′ exonuclease and be removed. The question asks about the relative chances of an incorrect nucleotide or its correct replacement encountering the exonuclease site of the Klenow fragment. As shown in Figure 1, incorrect base pairing alters the geometry of a new DNA strand and directs it on a new trajectory, often toward the 3′→5′ exonuclease site. Replacement of the incorrect nucleotide with the correct one restores the normal geometry of the strand, making it less likely to encounter the 3′→5′ exonuclease. Therefore, the chances of the correctly incorporated nucleotide encountering the exonuclease site will be decreased relative to those for the incorrectly incorporated nucleotide.

Which of the following most likely explains why fructose was NOT detected in portal vein blood prior to ingestion of the sugar solution in Experiment 2?

elevated glucagon levels during fasting promoted release of stored glucose but not fructose

In liver cells, a molecule of fructose requires 2 ATP molecules to become phosphorylated and enter glycolysis as 2 molecules of glyceraldehyde-3-phosphate. In muscle cells, 1 molecule of fructose is initially phosphorylated by hexokinase, entering glycolysis directly as fructose-6-phosphate. Compared to net ATP production from fructose in muscle cells, net glycolytic ATP production from fructose in liver cells will:

not change Educational objective:During glycolysis of 1 glucose molecule, 2 ATP molecules are invested to form intermediates that will eventually yield 4 ATP molecules, representing a net production of 2 ATP molecules in the pathway. The question asks for the net glycolytic ATP production from fructose in the liver cells relative to that in the muscle cells. The number of ATP molecules invested in the formation of GAP is the same (ie, 2) when fructose is phosphorylated by fructokinase as when fructose (or glucose) is phosphorylated by hexokinase, and the energy payoff phase in both scenarios is identical to that occurring when glucose is metabolized in glycolysis. Therefore, the net glycolytic ATP production will not change when fructose is metabolized via fructokinase rather than by hexokinase.

Which of the following describes the high-energy cofactors produced as the threonine carbons are converted from succinyl-CoA to malate?

one GTP molecule and one FADH2 molecule are produced. he question asks for the high-energy molecules produced as succinyl-CoA is converted to malate. Two steps within this process produce such molecules: conversion of succinyl-CoA to succinate produces GTP, and conversion of succinate to fumarate produces FADH2. The conversion of fumarate to malate does not produce any high-energy molecules. Therefore, the given pathway produces one GTP molecule and one FADH2 molecule. Educational objective:The citric acid cycle produces three NADH molecules, one GTP molecule, and one FADH2 molecule per round. Each molecule is produced in a distinct step of the cycle, and a given subset of cycle steps will always produce the same set of molecules.

One of the challenges of antimicrobial peptides is their short physiological half-life. To solve this problem, some researchers have begun investigating synthetic antimicrobial peptides composed of D-amino acids. Why might these peptides have an increased half-life in the body relative to natural antimicrobial peptides?

proteases cann't act on peptides made of D- amino acids Educational objective:Amino acids have chiral centers and can exist in two conformations, designated L and D. Almost all natural amino acids are synthesized in the L-conformation. As a result, enzymes essentially work exclusively with L-amino acids and do not affect D-peptides.

Given that the mutated oligonucleotide lacks the sequence needed to bind NF-κB, what was the most likely purpose of including the mutated oligonucleotide probe in the experiment? The mutated oligonucleotide:

served to confirm that DNA binding by NF-kB is specific to the proposed binding sequence The passage describes an experiment in which an EMSA was performed, and the question asks for the likely purpose of the mutated oligonucleotide probe lacking the consensus NF-κB binding site. The mutated oligonucleotide probe is a target for nonspecific NF-κB binding (ie, interactions between NF-κB and DNA of a sequence other than the binding sequence). Nonspecific binding should not occur if the radioactive probe is specific for NF-κB (ie, the mutated oligonucleotide probe serves as a negative control to verify that a result that should not occur, does not occur). Educational objective:Mutations can influence interactions of other molecules with DNA. Negative and positive controls help verify that an experimental result is due to the independent and dependent variables under study rather than other factors. A negative control is an experimental condition in which no response to or signal from an experimental intervention is expected.

The Lineweaver-Burk plot shown above depicts the glycosidase activity of isoform 1. Based on Figure 1 and the information in the passage, which of the following would describe the appearance of the plot for isoform 2 glycosidase activity at pH 7 relative to isoform 1?

the x-intercept would be left-shifted, and the slope would be decreased.

Why were the positions of the oligonucleotide complexes with NF-κB measured in both the presence and absence of antibodies to NF-κB subunits?

to confirm presence or absence of specific subunits in the complexes Typically, antibody-epitope interactions are highly specific, meaning the antibody will not bind to other molecules. As such, each antibody added is expected to bind one and only one subunit of NF-κB. Antibody binding to NF-κB subunits alters the sizes of complexes containing those subunits. Therefore, bands that shift in the presence of an antibody (as shown in Figure 2) confirm the presence of the corresponding subunit in the complex. Educational objective:Antibody binding shifts the migration of a molecule in a gel, thereby confirming the presence of the molecule targeted by that antibody.