Missed Questions - UWorld Biochemistry

A.Glutamine **deamidation = removal of amine to form a carboxylic acid or a ketone**** Glutamine/asparagine is the only ones that can deamidate to produce glutamic acid and aspartic acid**

Amino acid catabolism releases nitrogen in the form of ammonia. In the liver, the urea cycle prepares ammonia for excretion. Which amino acid could undergo deamidation to produce ammonia for the urea cycle? A.Glutamine B.Arginine C.Glycine D.Aspartate

A.Enzyme 1 only Because it shows a sigmoidal curve, enzyme 1 has positive cooperativity and therefore must have multiple active sites. "S" looking curve = multiple active sites

Researchers measured the activity of two enzymes that catalyze the same reaction, with results shown below. Which of the enzymes must have multiple active sites? A.Enzyme 1 only B.Enzyme 2 only C.Both enzymes D.Neither enzyme

B. B=PRUVATE

Which of the following molecules is a precursor to the citric acid cycle?

A.A noncovalent protein-protein interaction Antibodies bind their epitopes through noncovalent interactions such as hydrogen bonding and electrostatic attractions.*****

(The passage states that cell lysates were incubated with α-flag, an antibody that binds the flag peptide.) Which of the following most accurately describes the interaction between α-flag and flag-IRS4? A.A noncovalent protein-protein interaction B.A covalent protein-protein interaction C.A noncovalent protein-carbohydrate interaction D.A covalent protein-carbohydrate interaction

D.It would bind to the active site and increase the Km competitive inhibitors increase Km****

(in the presence of Compound 1, which was found to significantly reduce the activities of both LAR) Compound 1 would most likely have what effect on the activity of LAR? A.It would bind to an allosteric site and decrease the Km. B.It would bind to the active site and decrease the Vmax. C.It would bind to an allosteric site and increase the Vmax D.It would bind to the active site and increase the Km

D.It interacts with the negative charges of phosphate groups, stabilizing the transition state. ***The question states that Mg2+ is directly involved in TMPK-mediated catalysis, so Mg2+ must be involved in stabilization of the transition state.

A Mg2+ ion in the active site of TMPK is directly involved in catalysis. Given this, Mg2+ most likely plays which of the following roles in TMPK-mediated reactions? A.It binds negatively charged amino acids, stabilizing the tertiary structure of TMPK. B.It interacts with water molecules in the active site, facilitating phosphate hydrolysis. C.It binds anions, preventing them from competing with phosphate groups for the active site. D.It interacts with the negative charges of phosphate groups, stabilizing the transition state.

A.Incubate cardiomyocytes with pyruvate and measure oxygen consumption in the presence and absence of 5-AVAB. The scientists hypothesized that 5-AVAB inhibits fatty acid degradation (β-oxidation) in the mitochondria, and thereby decreases oxygen consumption To determine whether the ETC is directly affected by 5-AVAB, an alternative source of NADH and FADH2 must be provided. *** Pyruvate produces NADH and FADH2 by conversion to acetyl-CoA. Therefore, incubating cells with pyruvate and measuring oxygen consumption in the presence and absence of 5-AVAB would reveal whether 5-AVAB interferes with the ETC directly

A colleague suggested that 5-AVAB may interfere with the electron transport chain directly. Which experiment could provide evidence for the effect (if any) of 5-AVAB on the electron transport chain? A.Incubate cardiomyocytes with pyruvate and measure oxygen consumption in the presence and absence of 5-AVAB. B.Add an electron transport chain inhibitor to each set of cardiomyocytes and measure the change in oxygen consumption in the presence and absence of 5-AVAB. C.Increase the concentration of palmitate given to cardiomyocytes and measure oxygen consumption in the presence and absence of 5-AVAB. D.Increase the partial pressure of oxygen to which the cardiomyocytes are exposed and measure the rate of oxygen consumption in the presence and absence of 5-AVAB

B pi bonds interfere with B oxidation unless they are in a very specific position acetyl CoA = 2 carbons so you need 8 carbons total

A fatty acid undergoes complete β-oxidation to yield nine molecules of acetyl-CoA. No isomerization reactions occur. The structure of the original fatty acid was:

B.Hydroxyl basically asking which can be phosphorylated - hydroxyl is the only one ** Phosphorylation occurs at hydroxyl groups of serine, threonine, and tyrosine residues

A scientist proposed that phosphorylation of the R domain increases CFTR activity. Which of the following functional groups in the R domain could be removed by mutation to test this hypothesis in vivo? A.Amide B.Hydroxyl C.Thiol D.Carboxylic acid

D.Deoxyguanosine monophosphate Deoxyguanosine has the largest molecular weight of any nucleotide, so all other nucleotides will move faster through an agarose gel

A single-stranded DNA oligonucleotide composed of which of the following would move most slowly down an alkaline agarose gel during electrophoresis? A.Deoxyadenosine monophosphate B.Deoxythymidine monophosphate C.Deoxycytidine monophosphate D.Deoxyguanosine monophosphate

D.allosterically activates CFTR-mediated chloride transport Allosteric activators increase the ability of the protein to function,**** (A is wrong bc passage states PASSIVE transport, therefore you dont need energy from hydrolysis of ATP - active transport)

According to the information in the passage, which statement best describes the role of ATP in CFTR activity? ATP: (CTFR facilitates passive transport of chloride upon ATP binding) A.is hydrolyzed to provide the energy input necessary for ion transport. B.flows through CFTR in the opposite direction as chloride. C.inhibits chloride transport by occupying the channel. D.allosterically activates CFTR-mediated chloride transport

B.impaired ability to synthesize lipoic acid. The pyruvate dehydrogenase complex requires lipoic acid for activity. Because lipoic acid is a necessary cofactor of the PDHC, impaired lipoic acid production would result in decreased activity of the PDHC complex. (would be underexpression - A) (PDHC synthesizes acetyl-CoA, which enters the Krebs cycle (also known as the citric acid cycle). A deficient PDHC would decreaseacetyl-CoA concentration and downregulate its entry into the Krebs cycle. - C) (if lactic acid buildup = high rate of pyruvate fermentation - D)

An infant who experiences seizures and has lactic acid buildup is shown to have deficient pyruvate dehydrogenase complex (PDHC) activity. Genetic analysis shows that both the maternal and paternal alleles for all PDHC subunits are normal. These results could indicate: A.overexpression of one or more PDHC subunits. B.impaired ability to synthesize lipoic acid. C.upregulated acetyl-CoA entry into the Krebs cycle D.an unusually low rate of pyruvate fermentation.

A.29% Table 2 shows that 2746 participants had type A blood and 174 had type AB blood, for a total of 2920 participants who expressed the A-antigen. Therefore, 2920 participants expressed GalNAcyl transferase. The total number of participants was 10,000, so approximately 29% of the participants in the study expressed GalNAcyl transferase.**

Approximately what percentage of the population measured in Table 2 expresses GalNAcyl transferase? A.29% B.27% C.11% D.9%

C.Male ME patients produce more glutamate in their liver cells than female patients. males have more protein catabolism than females so must produce more amino acids

Assuming that 3-methylhistidine is the only indicator of protein catabolism, which observation would best support the difference in the metabolism of amino acids in male and female patients with ME? A.Male ME patients undergo fewer transamination reactions than female patients. B.Female ME patients excrete more ammonia in their urine than male patients. C.Male ME patients produce more glutamate in their liver cells than female patients. D.Female ME patients produce more α-keto acids in their muscles than male patients.

C.Km increases to 25 μM. Right shift = Km increases If approaches 0 = volume increased

Based on Figure 2, 250 μM acivicin has what effect on the Km of hGGT? A.Km decreases to 5 μM. B.Km stays constant at 13 μM C.Km increases to 25 μM. D.Km increases to 50 μM.

A.It increases the turnover number. increases Kcat (c- wrong bc opposite, as slope gets steeper Kcat/Km ratio would decrease)

Based on Figure 2, Compound 1 activates hGGT in which of the following ways? A.It increases the turnover number. B.It decreases the Michaelis constant. C.It increases the kcat/Km ratio. D.It decreases the maximum velocity

D.Phosphorylation of PDH by PDK1 They didnt even measure PDH mRNA! SILLY

Based on Figure 2, the availability of ATP in PBMCs is most likely affected by which process? A.Overexpression of PDH mRNA B.Removal of lipoic acid cofactors by SIRT4 C.Upregulation of PPARα transcription factors D.Phosphorylation of PDH by PDK1

D.7 Kcat = Vmax/[E] For CDC45: Kcat = 2.8/0.01 = 280 For LAR Kcat = 2/0.05 = 40 RATIO: 280/40 = 7***

Based on Table 1, the turnover number (kcat) of CD45 exceeds that of LAR by what factor? A.1.5 B.2 C.5 D.7

C.Amino acid sequence The passage states that the 15 globular domains of CI-MPR fold similarly. Therefore, they most likely all have similar amino acid sequences. (a - Post-translational modifications are non-amino acid additions to proteins, such as phosphorylation or glycosylation. Table 1 shows that some CI-MPR domains have no glycosylation sites, some have one, some two, and some three. Therefore, the domains are not similar in their post-translational modifications.)

Based on information in the passage and the data in Table 1, the 15 domains of CI-MPR are most likely similar to each other in which of the following ways? A.Post-translational modifications B.Affinity for their ligands C.Amino acid sequence D.Primary ligand

D.glutamate LOOK at the freaking picture!!!

Based on information in the passage, NMR data indicated electrostatic interactions involving all of the following amino acids EXCEPT: A.lysine. B.aspartate. C.arginine. D.glutamate

B.Fewer binding events occur on BAZ2B as the experiment progresses All it is is accessing binding and what ITC specifically does which is binding!

Based on the information in the passage, why does the size of the ITC peaks decrease over the course of each experiment in Figure 2? A.Increasing H3 concentration pushes the equilibrium toward dissociation. B.Fewer binding events occur on BAZ2B as the experiment progresses C.Less H3 is present in each subsequent injection as the titration progresses. D.Negative cooperativity exhibited by BAZ2B causes binding to decrease as H3 is added.

D.E. coli expressed isoform 2 at higher levels than isoform 1 Figure 1 shows that 5 μL of isoform 2 has substantially higher activity than the same volume of isoform 1. The best explanation for this observation is that isoform 2 was present at a higher concentration than isoform 1. This could occur if the E. coli bacteria used to produce each isoform expressed isoform 2 at higher levels than isoform 1.

Both isoforms of destabilase have approximately the same kcat values at optimal pH, but Figure 1 shows that when 5 μL of each purified enzyme was saturated with substrate, isoform 2 had substantially higher activity. What factor could explain this apparent discrepancy? A.The affinity column bound isoform 1 more tightly than isoform 2. B.Under saturating conditions, only isoform 2 operated at Vmax. C.Isoform 1 became denatured at the optimal pH of isoform 2. D.E. coli expressed isoform 2 at higher levels than isoform

B.Phosphorolysis of glycogen **when DAB is added (column 3), we see the same pathways are active since both glycogen and lactate change in the same direction as without it **DAB is acting as an inhibitor of one of the forward steps in the process (likely the rate determining step** so B) D/C DONT HAPPEN

DAB most likely inhibits which reaction? A.Phosphorylation of glucose B.Phosphorolysis of glycogen C.Export of lactate from cells D.Hydrolysis of glycogen

D.Gluconeogenesis and glycolysis The process of carrying lactate from the muscle to the liver and moving regenerated glucose from the liver back to muscles is called the Cori cycle, which connects gluconeogenesis and glycolysis.

During anaerobic exercise, the Cori cycle connects which two metabolic pathways? A.Glycogenolysis and the citric acid cycle B.The urea cycle and gluconeogenesis C.The pentose phosphate pathway and glycogenesis[ D.Gluconeogenesis and glycolysis

A.Oxaloacetate + GTP → GDP + CO2 + PEP During decarboxylation, CO2 is removed from oxaloacetate (a reactant), yielding CO2 as a product of the reaction.

During gluconeogenesis, the enzyme phosphoenolpyruvate carboxykinase (PEPCK) catalyzes production of phosphoenolpyruvate (PEP) by decarboxylating oxaloacetate and transferring a phosphate group from GTP. What is the balanced equation for this reaction? A.Oxaloacetate + GTP → GDP + CO2 + PEP B.Oxaloacetate + GTP + PEPCK → GDP + 2 CO2 + PEP C.Oxaloacetate + GTP → GDP + 2 Pi + PEP D.Oxaloacetate + GTP + PEPCK → GDP + Pi + PEP

C.I, II, and III only primary protein structure is maintained by peptide bonds. Secondary structure (α-helices and β-pleated sheets) is maintained by hydrogen bonds. Tertiary structure (three-dimensional conformation) and quaternary structure are maintained by van der Waals forces, hydrogen bonds, ionic bonds, and disulfide bonds.

During protein folding, which of the following interactions help stabilize secondary, tertiary, or quaternary structure? I. Hydrogen bonds II. Ionic bonds III. van der Waals forces IV. Amide bonds A.I and II only B.III and IV only C.I, II, and III only D.I, III, and IV only

B.25%

Escherichia coli bacteria containing only 15N-labeled DNA were grown in media containing only 14N nucleotides. What percentage of double helices would be composed of one 15N strand and one 14N strand after three generations? A.12.5% B.25% C.50% D.100%

B.3.3 × 10−3 units/μg activity=0.30 nmol/30 min=0.01 units 0.01 units/ 3 µg=3.3×10−3 units/ µg

Given that 1 unit of activity equals 1 nmol of substrate converted per minute, what was the specific activity of purified TMPK toward dTMP in Experiment 1? (Note: Assume the rate of phosphorylation is constant throughout the experiment.) (According to the passage, 0.30 nmol of dTMP became phosphorylated over the course of the 30-minute assay.) A.1.7 × 10−3 units/μg B.3.3 × 10−3 units/μg C.5.0 × 10−3units/μg D.1.0 × 10−2 units/μg

B.50% Figure 1 shows that for isopeptidase activity, both isoforms of destabilase have an optimal pH of approximately 6. Isoform 1 converts substrate to product at a rate of ~800 ng/min at optimal pH. As pH increases above 6, isopeptidase activity decreases. The graph shows that at the pH of blood (7.4), the isopeptidase activity diminishes to ~400 ng/min. Therefore, at pH 7.4, isoform 1 exhibits isopeptidase activity that is 50% of its optimal activity (400/800).

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? A.40% B.50% C.80% D.100%

C.4 kDa The average molecular weight of an amino acid is 110 Da. The molecular weight of a polypeptide can be estimated by multiplying the number of amino acids in the polypeptide by 110. 626-591 = 35 amino acids 35 x 110 = 3860 Da ~ 4 kd

If a frameshift mutation changes the number of amino acids in a protein from 591 to 626, the difference between molecular weights of the wild-type and mutant proteins would be closest to: A.1 kDa B.2 kDa C.4 kDa D.8 kDa

A.Increased cytosolic NADPH. fatty acid synthesis occurs in the cytosol so increased cytosolic NADPH indicates fats are going through pentose phosphate pathway energy for fatty acid synthesis comes from: ATP, NADPH, and acetyl-CoA, respectively. If fatty acid synthesis is inhibited in ME patients, the conversion of NADPH to NADP+ will occur less frequently, and cytosolic NADPH will build up.

If lower phospholipid levels in ME patients are due to inhibition of fatty acid synthesis, researchers would most likely observe which of the following? A.Increased cytosolic NADPH. B.Decreased mitochondrial acetyl-CoA. C.Decreased mitochondrial fumarate D.Increased cytosolic lysine

B.4 Because gluconeogenesis consumes 6 ATP equivalents and glycolysis produces 2 ATP equivalents, a net total of 4 ATP equivalents are consumed.

If one glucose molecule is converted to pyruvate in the muscle and then the resulting pyruvate molecules are converted back to glucose in the liver, what is the net number of ATP equivalents consumed? A.2 B.4 C.6 D.8

D.I, III, and V only in order from glycogen to lactate need these steps 1. Glycogen to glucose via glycogenolosis 2. Glucose to pyruvate via glycolysis 3. pyruvate to lactate via fermentation THEREFORE we know V must be true and IV isnt true so need to make decision between I. Glucose-1-phosphate to glucose-6-phosphate AND II. Glucose to glucose-6-phosphate II comes from dietary glucose****** so WRONG. We need glucose that came from glycogenolysis so therefore needs to be I. Glucose-1-phosphate

If the proposed mechanism in the passage is correct, which of the following metabolic reactions must occur for astrocytes to convert glycogen into a fuel for neurons? (passage states glycogen --> lactate) I. Glucose-1-phosphate to glucose-6-phosphate II. Glucose to glucose-6-phosphate III. 3-phosphoglycerate to 2-phosphoglycerate IV.pyruvate to acetyl-CoA V. Pyruvate to lactate A.I, II, and III only B.II, III, and V only C.I, IV, and V only D.I, III, and V only

D.D-galactose with sugar enantiomers ONE IS D***, OTHER IS L*** Therefore since fucose is L, we are looking for something with D, and ribulose is 5 carbons so also no so has to be D-galactose

Ignoring the anomeric carbon, the fucose shown in Figure 1 is the dehydroxylated enantiomer of: A.L-xylulose. B.D-ribulose. C.L-glucose. D.D-galactose

C.FADH2 synthesis by Complex II. Complex II of the ETC is a flavoprotein (contains FAD) that is also known as succinate dehydrogenase. They most likely added succinate to stimulate FADH2 synthesis (and subsequent electron transfer) by Complex II. (A wrong bc complex I consumes NADH, does not synthesize it)

In Figure 2, the most likely reason that researchers added succinate was to stimulate: A.NADH synthesis by Complex I. B.GTP synthesis by succinyl-CoA synthetase. C.FADH2 synthesis by Complex II. D.ubiquinol synthesis by fumarase.

D.Oxaloacetate Fatty acids must be activated with coenzyme A followed by carnitine to enter the mitochondrial matrix. Activation requires ATP hydrolysis. Although oxaloacetate is an intermediate in the citric acid cycle, gluconeogenesis, and transport of acetyl-CoA from the mitochondria into the cytosol for fatty acid synthesis, it is not involved in transport of fatty acids from the cytosol into the mitochondria.

In order to transport long-chain fatty acids from the cytosol to the mitochondria, three of the following molecules are required. Which one is NOT necessary? A.Coenzyme A B.ATP C.Carnitine D.Oxaloacetate

D.Covalent bonds between carbon and nitrogen atoms Because reflectins do not fold, they only participate in primary structure, characterized by the covalent carbon-nitrogen linkages of peptide bonds***

In the absence of ACh, which of the following bond types best characterizes the structure of reflectins? (reflectins have no secondary or tertiary structure) A.Hydrogen bonds between backbone carbonyls and amides B.Covalent bonds between sulfur atoms C.Hydrogen bonds between R group alcohols D.Covalent bonds between carbon and nitrogen atoms

C.is not changed relative to mitochondria in the presence of SIRT4. In each case, O2 was consumed at the same rate in both WT and KO livers, indicating that the ETC functions normally in KO mice. If ETC functions normally, there is no change in the electrochemical gradient across the inner mitochondrial membrane occurs. (Isocitrate is a precursor to α-ketoglutarate in the citric acid cycle. Because α-ketoglutarate is consumed more slowly in KO livers, isocitrate is also likely to be consumed more slowly, and will accumulate.)

In the absence of SIRT4, the electrochemical potential across the inner mitochondrial membrane: A.is increased relative to mitochondria in the presence of SIRT4. B.is decreased relative to mitochondria in the presence of SIRT4. C.is not changed relative to mitochondria in the presence of SIRT4. D.is increased relative to mitochondria with SIRT4 only when Cyt-C is added

B.2 NADH is a two-electron carrier.

In the electron transport chain, four electrons are required to fully reduce one O2molecule to two H2O molecules. Given this information, how many NADH molecules are required? A.1 B.2 C.3 D.4

C.Healthy individuals, because insulin stimulates the transport of acetyl-CoA to the cytosol pyruvate dehydrogenase complex is activated by insulin since it is part of a well fed state*** (hormone sensitive lipase = triggers release of fatty acids from triglycerides in a starvation condition to allow beta-oxydation ***controlled by glucagon, not insulin)

Lipids serve as energy stores for healthy patients and patients with ME. If both groups were fed the same diet, would the rate of lipid synthesis after a meal be faster in healthy individuals or in patients with ME? A.Patients with ME, because the release of glucagon activates acetyl-CoA carboxylase in the liver B.Patients with ME, because more chylomicrons are broken down in the bloodstream C.Healthy individuals, because insulin stimulates the transport of acetyl-CoA to the cytosol D.Healthy individuals, because fewer fatty acids are being generated by hormone-sensitive lipase

reannealing is faster in shorter carbon chains, physiological pH, and increases salt concentration

Prior to inserting the STR sequences into K-ras DNA, the researchers perform MCA analysis using only the double-stranded STR sequences. Which sequence from Table 1 would have the fastest reannealing time in this experiment? A.Sequence 1 B.Sequence 2 C.Sequence 3 D.Sequence 4

C.contains a secondary amine in its backbone.

Proline differs from the other standard amino acids in that only the former: A.has a negative charge in its side chain. B.forms peptide bonds that are planar. C.contains a secondary amine in its backbone. D.is commonly found in flexible loop regions of folded protein

A.heteromultimer that formed in the oxidizing environment of the endoplasmic reticulum. The fact that a reducing gel is required to break the multimer indicates that disulfide bonds are present. They most likely formed in the oxidizing environment of the endoplasmic reticulum.**** reducing environment breaks the bonds, oxidizing environment forms them

Researchers isolate a protein that runs as a single band on a nonreducing SDS-PAGE gel but as two distinct bands on a reducing gel. The protein is most likely a: A.heteromultimer that formed in the oxidizing environment of the endoplasmic reticulum. B.heteromultimer that formed in the reducing environment of the cytosol. C.homomultimer that formed in the oxidizing environment of the endoplasmic reticulum. D.homomultimer that formed in the reducing environment of the cytosol.

B at saturating conditions!! So affecting enzyme concentration is going to affect the rate!! enzyme concentration alters Vmax**

The CSL substrates from Table 1 were tested in vitro at saturating substrate concentration in solutions containing 10 pM or 100 pM Hip1 enzyme. The relative fluorescence of these substrates was then measured. Which of the following graphs best reflects the expected results?

C.3 A glycosidic bond is a bond between the anomeric carbon of a carbohydrate and any other biological molecule, including proteins, lipids, nucleotides, and other carbohydrates. therefore since three bonds with O-R*** there is three

The D-galactose moiety of the A-antigen participates in how many glycosidic bonds? A.1 B.2 C.3 D.4

B.Replacement of nucleotides at the 3′ end of the growing strand However, the passage states that the Klenow fragment (KF) enzyme described in the experiment does not have 5′-3′ exonuclease activity. KF can only proofread DNA in the 3′-5′ direction on the template strand, so only errors at the 3′ end of the growing strand can be repaired.

The Klenow fragment used in the experiment would be able to perform which of the following repair processes? A.Excision of thymine dimers at the 5′ end of the growing strand B.Replacement of nucleotides at the 3′ end of the growing strand C.Correction of mismatched nucleotides in the middle of a completed strand D.Removal of damaged bases from the middle of the template strand

A.The x-intercept would be left-shifted and the slope would be decreased. A decrease in Km corresponds to a left-shift in the x-intercept because as Km decreases, the magnitude of −1/Km increases, moving it farther away from the origin. Therefore, the x-intercept for isoform 2 will be left-shifted relative to isoform 1. Vmax is equal to kcat × [E], where [E] is the total concentration of enzyme. Because both enzyme isoforms have approximately the same kcat, they will also have similar Vmax values if both isoforms are tested at the same [E] Relative to the slope of the plot for isoform 1, the numerator (Km) of isoform 2 is decreased but the denominator (Vmax) remains constant, yielding a smaller Km/Vmax ratio. Therefore, the plot for isoform 2 will have a decreased slope compared to isoform 1.

The Lineweaver-Burk plot shown above depicts the glycosidase activity of isoform 1. If both isoforms were tested at equal enzyme concentrations, which of the following would describe the appearance of the plot for isoform 2 relative to isoform 1? (According to the passage, isoforms 1 and 2 of destabilase have approximately the same kcat values, but isoform 2 has a lower Km.) A.The x-intercept would be left-shifted and the slope would be decreased. B.The slope would be decreased but the x-intercept would be unaffected. C.The x-intercept would be right-shifted but the slope would be unaffected. D.The slope would be increased and the x-intercept would be right-shifted.

A.assessing the precision of the results. In this scenario, an experiment was repeated multiple times. This allowed researchers to determine whether repeated trials yield similar results. Therefore, the purpose of the repetition was to assess the precision of the result

The affinity of a protein for its ligand was measured by isothermal titration calorimetry. The experiment was repeated multiple times to mitigate error by: A.assessing the precision of the results. B.verifying the calibration of the instruments. C.assessing the validity of the experiment. D.verifying the accuracy of the results

C.dephosphorylated and consumes F2,6BP. enzyme phosphorylation activates F2,BPase and inactivates PFK2 since passage states that the PFK-2 domain of the bifunctional enzyme is active (and the F2,6BPase domain is inactive) when it is dephosphorylated. Therefore, when glycolysis is active, the PFK-2/F2,6BPase enzyme is most likely dephosphorylated and produces F2,6BP

The information in the passage and the data in Figure 1 support the conclusion that when glycolysis is active, the bifunctional PFK-2/F2,6BPase enzyme is: (The passage states that the PFK-2 domain of the bifunctional enzyme is active (and the F2,6BPase domain is inactive) when it is dephosphorylated. ) A.phosphorylated and consumes F2,6BP. B.phosphorylated and produces F2,6BP. C.dephosphorylated and consumes F2,6BP. D.dephosphorylated and produces F2,6BP

D.ATP binds both free pyruvate kinase and the enzyme-substrate complex, but it favors free enzymes Km increases, Vmax decreases = mixed inhibition, therefore binds both free pyruvate kinase and the enzyme-substrate complex, but it favors free enzymes (Km increases = indicates that it binds the free enzyme with higher affinity)

The information in the passage and the data in Figure 1 support which of the following conclusions about ATP as an inhibitor of the R510Q enzyme? A.ATP can bind only free pyruvate kinase. B.ATP binds free pyruvate kinase and the enzyme-substrate complex with equal affinity. C.ATP can bind only the enzyme-substrate complex. D.ATP binds both free pyruvate kinase and the enzyme-substrate complex, but it favors free enzymes

A.The citric acid cycle fumarate and malate both increased****

The intermediates of which metabolic pathway were shown to increase in muscles due to exercise? A.The citric acid cycle B.Fermentation C.Glycolysis D.Glycogenolysis

A.to demonstrate that GAH does not directly inhibit LDH. They found that LDH activity was constant regardless of the amount of GAH used, and therefore they were able to demonstrate that GAH does not directly inhibit LDH activity but instead most likely prevents LDH leakage.

The purpose of measuring purified LDH activity with various amounts of GAH was most likely: A.to demonstrate that GAH does not directly inhibit LDH. B.to determine baseline extracellular LDH activity. C.to show that GAH is an effective LDH antagonist. D.to characterize the GAH-LDH binding site

C.II and III only 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.*** All tripeptides tested consist of the same three amino acids (A, G, and H), so they contain the same functional groups. Therefore, they most likely have the same hydrophobicity All possible arrangements of the tripeptide have approximately the same net charge at any given pH, giving them the same isoelectric point

The tested tripeptide variants most likely have which of the following characteristics in common? (each contained one alanine, one glycine, and one histidine residue.) I. Primary structure II. Isoelectric point III. Hydrophobicity A.I and II only B.I and III only C.II and III only D.I, II, and III

A.gradual deprotonation as reflectins migrated from anode to cathode. Arginine will be (+) at physiological pH and neutral at high pH therefore when placed near pH of 7, it will be + and migrate towards the cathode the low-pH end of the gradient is placed near the anode and the high-pH end is placed near the cathode therefore since at low pH side (more positive-cathode), once current is applied they will move towards high pH (more negative-anode) In order to become more negative, need to be deprotonated***

To generate Figure 2, reflectins were loaded onto the low-pH (pH = 7) side of a gel with a pH gradient, and a current was applied. This resulted in: A.gradual deprotonation as reflectins migrated from anode to cathode. B.gradual deprotonation as reflectins migrated from cathode to anode. C.gradual protonation as reflectins migrated from anode to cathode. D.gradual protonation as reflectins migrated from cathode to anode

Figure 1 shows that α-ketoglutarate is oxidized at a significantly lower rate in KO mice than in WT mice. Therefore, the product of α-ketoglutarate oxidation, succinyl-CoA, will most likely be depleted in KO livers.

Under aerobic conditions, which of the following molecules would be expected to be present in significantly decreased levels in the livers of KO mice? A.Succinyl-CoA B.ATP C.Isocitrate D.Cytochrome C

B.thermodynamically unstable but kinetically stable Anabolic processes such as DNA and protein synthesis require energy input (usually from ATP or GTP) but form kinetically stable bonds. A reaction that is thermodynamically favorable but slow involves bonds that are thermodynamically unstable (prone to breaking) but kinetically stable (break slowly).

Under physiological conditions, peptide bond formation and degradation both require enzymes, but only formation requires coupling to GTP hydrolysis. Based on this information, peptide bonds under physiological conditions are: A.both thermodynamically and kinetically stable. B.thermodynamically unstable but kinetically stable C.thermodynamically stable but kinetically unstable D.both thermodynamically and kinetically unstable

C.causes a decrease in F2,6BP synthesis. glucagon is opposite of insulin** insulin increases, therefore glucagon must decrease it

What effect does glucagon most likely have on F2,6BP synthesis in the liver? Glucagon: A.does not alter F2,6BP synthesis. B.causes an increase in F2,6BP synthesis. C.causes a decrease in F2,6BP synthesis. D.causes F2,6BP to be synthesized and degraded in a cycle

C.0.25 mg/mL **analyze all parts of graph** Of the GAH concentrations tested, 0.25 mg/mL is the lowest concentration that yields a p-value of less than 0.05 (p = 0.009). Therefore, 0.25 mg/mL is the lowest tested concentration that produces a significant reduction in extracellular LDH activity.

What is the lowest tested concentration of GAH that produces a significant reduction in extracellular LDH activity when 5 mM H2O2 is used? A.0.0125 mg/mL B.0.0625 mg/mL C.0.25 mg/mL D.1 mg/mL

D.The lower affinity for the substrate *dotted is final* -Vmax decreases and Km increases therefore, lower affinity

What stress-induced change in Figure 2 provides evidence that mechanical stress alters the binding properties of the GLUT transporters? A.The left-shift of Km B.The downward-shift of Vmax C.The decreased catalytic efficiency D.The lower affinity for the substrate

B.Condensation condensation joins two molecules and release h20 in the process**

What type of reaction does porphobilinogen synthase catalyze? A.Hydrolysis B.Condensation C.Deamination D.Methylation

B.K132N wild-type and K132N have similar Vmax and Km values*****

Which HMBS mutant variant is LEAST likely to cause acute intermittent porphyria? A.C73R B.K132N C.R167W D.V215E

D.Proline As mentioned in the passage, the interacting protein domains are α-helical. Proline and glycine are often found in loops and linker regions between α-helices and β-sheets, particularly at sharp 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.

Which amino acid is LEAST likely to be found in normally functioning binding domains of t-SNARE and v-SNARE proteins? A.Asparagine B.Tryptophan C.Threonine D.Proline

D.Nonpolar aliphatic nonpolar would be in membrane not on surface Nonpolar residues tend to be found buried within the protein or in the hydrophobic environment of a membrane, where they can avoid water. Therefore, nonpolar aliphatic residues are the least likely to be found on the surface of a CI-MPR globular domain.

Which class of amino acids is LEAST likely to be found on the surface of CI-MPR globular domains? A.Positively charged B.Polar uncharged C.Negatively charged D.Nonpolar aliphatic

B.SIRT4 facilitates glucose uptake by preventing insulin resistance. Cells that are overexposed to insulin can develop resistance and take up less glucose than they normally would in the presence of insulin. Therefore since still high insulin and high glucose, SIRT4 must be preventing insulin resistance

Which conclusion about the effect of SIRT4 on blood glucose levels can be drawn from the data in Figure 3? A.SIRT4 causes insulin secretion, leading to decreased glucose uptake. B.SIRT4 facilitates glucose uptake by preventing insulin resistance. C.SIRT4 increases insulin sensitivity, causing increased blood glucose. D.SIRT4 reduces insulin secretion, inducing increased glucose uptake.

D.I and III only Each of the assumptions of Michaelis-Menten holds true only during the initial phase of the reaction, before substrate is depleted or product accumulates - why I is correct** To ensure that ES formation does not significantly impact [S], the total concentration of enzyme in solution should be much smaller than any substrate concentration tested (Number III).** (II is wrong bc enzyme concentration must be much smaller than Km, not larger.)

Which experimental procedure(s) must scientists use to determine Vmax and Km of an enzymatic reaction using the Michaelis-Menten model? They must ensure that: I. they only measure the initial reaction rate for each substrate concentration. II. total enzyme concentration is much greater than the Km of the reaction. III. each initial substrate concentration tested is much greater than enzyme concentration. A.I only B.III only C.I and II only D.I and III only

D.GL6 double bond makes more fluid, but also does SHORTER carbon chain so GL6

Which lipid in Table 3 would make the greatest contribution to membrane fluidity? A.GL2 B.GL3 C.GL5 D.GL6

D.PanZ binds AcCoA followed by PanD to form a ternary complex Ternary complexes form when three distinct molecules, at least one of which is a protein, bind to each other to form a single system. They can form through random or ordered mechanisms. (a-wrong PanD will not bind PanZ or undergo cleavage unless AcCoA is bound to PanZ first.)

Which of the following best describes the ADC activation mechanism? A.AcCoA binds to the PanD-PanZ complex to induce cleavage. B.PanD, PanZ, and AcCoA bind in a random order to form a ternary complex. C.A binary complex composed of PanD and AcCoA cleaves PanZ. D.PanZ binds AcCoA followed by PanD to form a ternary complex

D.Aconitase converts citrate into isocitrate through an intermediate. The conversion of citrate into isocitrate by aconitase and the transformation of fumarate into malate by fumarase are stereospecifc reactions.***** B is wrong bc fumarase converts fumarate into only L-malate and is a hydration reaction, not a condensation reaction

Which of the following correctly describes the stereospecific formation of a citric acid cycle intermediate? A.Citrate synthase consumes ATP to generate citrate from oxaloacetate and acetyl-CoA. B.Fumarase converts fumarate into both L- and D-malate through a condensation reaction. C.Succinyl-CoA synthetase consume GTP to transform succinyl-CoA to succinate. D.Aconitase converts citrate into isocitrate through an intermediate.

D.Formation of the glycosidic linkage between uracil and the ribose sugar just changing a base!! therefore there will be a glycosidic linkage between uracil and the ribose only the base is being replaced rather than the entire nucleotide

Which of the following correctly identifies a reaction involved in the formation of the P186L mutation in H6N1 RNA? A.Replacing a carbonyl group in uracil with an amine group B.Cleavage of a glycosidic linkage between a deoxyribose sugar and cytosine[ C.Formation of a phosphoester bond between 3′ and 5′ carbon of adjacent sugars D.Formation of the glycosidic linkage between uracil and the ribose sugar

A.Adding Ru1,5BP and measuring CO2 consumption over time to test for proper protein folding, you should just look to see that it is functioning properly

Which of the following experiments, performed in a closed system, could confirm that GroEL/ES successfully facilitated proper rubisco folding? A.Adding Ru1,5BP and measuring CO2 consumption over time B.Running rubisco on an SDS gel to assess its dimerization C.Observing a decrease in the concentration of inorganic phosphate D.Measuring the Kd of GroEL/ES interactions with Mg2+ over time

A.3! For a peptide of known composition, in which each amino acid is different from every other amino acid, the number of possible arrangements is n!, where n is the number of amino acids in the peptide.

Which of the following expressions gives the number of peptides that were tested for their ability to reduce extracellular LDH activity? A.3! B.3^3 C.3^20 D.20^3

D The passage also states that PanD is the inactive form of the enzyme ADC and cannot decarboxylate L-aspartate. Enzymes in their inactive forms, referred to as zymogens or proenzymes, must be altered by covalent modifications to become active. In the case of PanD, proteolytic cleavage (proteolysis) is used to activate the proenzyme. Because PanD is inactive, it displays no enzymatic activity at any pH (Choice A). Only Choice D illustrates both the correct optimal pH of ADC and the inactivity of PanD.

Which of the following graphs accurately predicts the aspartate decarboxylation activity of PanD and ADC between pH 3 and pH 9?

A.The sb domain facilitates ubiquitination of IRS4, targeting it to the proteasome. ***Proteins with ubiquitin tags are typically targeted to the PROTEASOME for destruction. not lysosome

Which of the following is the most likely mechanism by which the sb region of Asb-4 decreases IRS4 levels? A.The sb domain facilitates ubiquitination of IRS4, targeting it to the proteasome. B.The sb domain induces IRS4 expression by entering the nucleus to bind DNA. C.The sb domain prevents degradation of IRS4 by inhibiting proteases. D.The sb domain becomes ubiquitinated, causing IRS4 to enter the lysosome

A.It is a pentofuranose. Ribose is a pentose (five-carbon sugar) that links to the nitrogenous base through a glycosidic bond. Ribose is a five-carbon sugar (pentose) that must adopt the furanose form to be incorporated into the nucleotide triphosphates. In this form, it can be classified as a pentofuranose.

Which of the following most accurately describes the ribose component of a nucleotide triphosphate? A.It is a pentofuranose. B.It is a pentopyranose. C.It is a hexofuranose. D.It is a hexopyranose

D.Allosteric inhibition of fructose-1,6-bisphosphatase catalysis Fructose-2,6-bisphosphate allosterically activates phosphofructokinase-1 and inhibits fructose-1,6-bisphosphatase, which stimulates glycolysis and reduces gluconeogenesis.*** Inhibiting gluconeogenesis in the liver leads to increased net glucose catabolism.

Which of the following regulatory mechanisms helps increase net glucose catabolism in the liver after a meal? A.Inhibition of hexokinase by glucose-6-phosphate B.Allosteric suppression of phosphofructokinase-1 by ATP binding C.Hormonal suppression of fructose-2,6-bisphosphate synthesis D.Allosteric inhibition of fructose-1,6-bisphosphatase catalysis

A.I only (the lipids that are primarily responsible for energy storage are triacylglycerides) Sphingolipids are structural lipids that help influence the fluidity and curvature of biological membranes* The long hydrocarbon chain of the sphingosine head group cannot be readily hydrolyzed, and sphingolipids are not a primary means of energy storage.

Which of the following statements accurately describes the sphingolipids in Table 3? I. They provide structure to biological membranes II. They are primarily used for energy storage III. They can be hydrolyzed to produce two fatty acids A.I only B.III only C.I and II only D.II and III only

C.Conversion of succinyl-CoA to succinate

Which of the following steps in the citric acid cycle does NOT produce reduced electron carriers? A.Conversion of isocitrate to α-ketoglutarate B.Conversion of α-ketoglutarate to succinyl-CoA C.Conversion of succinyl-CoA to succinate D.Conversion of succinate to fumarate

C.OH group on the 2′ carbon of the pentose sugar If the pentose has a hydroxyl group at the 2′ carbon, it is ribose; if a hydrogen is present at the same position, it is deoxyribose. DNA replication only utilizes nucleotides that contain deoxyribose sugars.*** so would need on H on 2' carbon

Which of the following would NOT be present in the structure of the dFTP nucleotide used for replication? A.Bond to phosphate on the 5′ carbon of the pentose sugar B.Glycosidic bond on the 1′ carbon of the pentose sugar C.OH group on the 2′ carbon of the pentose sugar D.H group on the 5′ carbon of the pentose sugar

C.H3 acetylation of histone residues reduces its positivity, which is also the reason it has reduced interactions with DNA leading to euchromatin formation - D wrong H3 is most positive out of all of the choices

Which of the proteins mentioned in the passage has the highest isoelectric point? A.WT BAZ2B B.10M+ BAZ2B C.H3 D.H3K14ac

B.The equilibrium concentrations of 2-phosphoglycerate and 3-phosphoglycerate Because ΔG° does not change in the presence of a catalyst, the equilibrium concentrations of products and reactants also do not change. Therefore, the addition of phosphoglycerate mutase does not change the equilibrium concentrations of 2-PG or 3-PG. Catalysts increase reaction rates by facilitating formation of a more stable transition state between reactants and products. This enhanced stability decreases the activation energy Ea necessary for the molecules to react, allowing them to reach the energy threshold more often at any given temperature. As a result, the reaction rate increases in both the forward and reverse directions, resulting in an increased rate constant as calculated by the Arrhenius equation (Choices A, C, and D).

Which parameter would NOT be affected by the addition of phosphoglycerate mutase? A.The stability of the transition state between 3-phosphoglycerate and 2-phosphoglycerate B.The equilibrium concentrations of 2-phosphoglycerate and 3-phosphoglycerate C.The activation energy barrier between 3-phosphoglycerate and 2-phosphoglycerate D.The rate constant for conversion of 2-phosphglycerate to 3-phosphoglycerate

A.Isoprene → monoterpene → squalene → cholesterol

Which series depicts the order in which the precursors of steroid hormones are synthesized? A.Isoprene → monoterpene → squalene → cholesterol B.Cholesterol → isoprene → monoterpene → squalene C.Monoterpene → isoprene → squalene → cholesterol D.Isoprene → cholesterol → monoterpene → squalene

C.TMPK can only phosphorylate pyrimidine nucleotides that have two carbonyls.

Which statement about TMPK specificity is supported by Figure 1? A.TMPK can modify the monophosphate form of nucleotides but not the diphosphate form. B.TMPK can only interact with nucleotides that have methyl groups. C.TMPK can only phosphorylate pyrimidine nucleotides that have two carbonyls. D.TMPK can phosphorylate dNMPs but not NMPs

C.The effector is an inhibitor that decreases both Km and Vmax. Therefore, the allosteric effector is an inhibitor that decreases both Vmax and Km. **REMEMBER 1/VALUE

Which statement about the allosteric effector is correct? A.The effector is an activator that increases both Vmax and Km. B.The effector is an activator that increases Vmax but decreases Km. C.The effector is an inhibitor that decreases both Km and Vmax. D.The effector is an inhibitor that increases Km but decreases Vmax.

A.It is a glycoprotein with the anomeric carbon of sialic acid bound to a nonreducing disaccharide non-reducing sugar = no free anomeric carbons!!!

Which statement best describes the structure of the avian receptor of influenza A shown in Figure 1? A.It is a glycoprotein with the anomeric carbon of sialic acid bound to a nonreducing disaccharide B.It is a glycolipid with three reducing sugars bound to each other through their anomeric carbons C.It is a glycoprotein with sialic acid bound to the anomeric carbon of a reducing disaccharide D.It is a glycolipid with three nonreducing sugars in which only one bond contains an anomeric carbon

D.An oxidoreductase catalyzes the transfer of electrons to break the bond Disulfide bonds form in proteins when two cysteine residues are oxidized (lose electrons) to form cystine. They must be reduced (gain electrons) for the bond to break. Enzymes that catalyze oxidation-reduction reactions (transfer of electrons) are collectively known as oxidoreductases.

Which statement correctly describes the cleavage of protein disulfide bonds by the enzyme thioredoxin? A.A protease breaks the bond by catalyzing the addition of water across it. B.An isomerase catalyzes a rearrangement of functional groups to break the bond. C.A phosphorylase breaks the bond by catalyzing the addition of a phosphate group. D.An oxidoreductase catalyzes the transfer of electrons to break the bond

C.Glucose 6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase Glucose 6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase are the only oxidoreductases in the pentose phosphate pathway;***** These enzymes generate the majority of the NADPH in a cell.

Which two enzymes of the pentose phosphate pathway catalyze production of NADPH? A.Ribulose 5-phosphate epimerase and ribose 5-phosphate isomerase B.Transketolase and transaldolase C.Glucose 6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase D.Lactonase and aldolase

C.I and II only (The double mutant does not cause infection in either of the two target species of influenza A. It is not necessary as a control because it has the same effect as no virus at all.)

Which viruses would provide the necessary controls for an experiment designed to determine whether H6-G225D can infect human cells but not avian cells? I. WT-H6N1 that attaches only to avian tracheal tissue II. H1N1 that attaches only to human tracheal tissue III. V190D and G225D double mutant, which binds neither α-2,6- nor α-2,3- sialic acids A.I and III only B.II and III only C.I and II only D.I, II, and III