MCAT Biochemistry

Pyruvate dehydrogenase complex

- Pyruvate dehydrogenase (E1) catalyzes the redox-decarboxylation - Dihydrolipoyl transacetylase (E2) catalyzes the transfer of the acetyl group - Dihydrolipoyl dehydrogenase (E3) reforms the oxidized version of lipoamides Pyruvate decarboylation where pyruvate is converted into acetyl-CoA

Steps of signal transduction pathway

1) Release of appropriate signal molecule "primary messenger", 2) binding of the primary messenger to its receptor, 3) increase in intracellular concentration of secondary messengers, resulting in the amplification of the signal, 4) activation/inhibition of effectors, 5) Termination of pathway (regulated by dissociation of primary messengers, GTPase activity of G-proteins, and phosphatase) - Calcium serves as a secondary messenger in many signal transduction pathways and calmodulin is a regulatory protein that detects changes in calcium ion concentrations - Secondary messenger is used by all signal transduction pathways. Intracellular agents whose concentration can be amplified, causing an amplification of the initial signal

Site I and II are most likely lined with the side chains of which amino acids?

A. R and L (picked "D and E") Sites I and II recognize hydrophobic and anionic substrates. Thus, it's mostly likely that positively charged and hydrophobic side chains point into the binding pocket to make favorable interactions.

Introduction of which amino acid substitution would result in the largest decrease in the entropic penalty associated with a protein folding into its native conformation?

A: "Leu to Thr substitution at a surface-exposed site" Changing a surface-exposed hydrophobic residue for a more hydrophilic residue eliminates the entropic penalty associated with ordered water molecules around hydrophobic groups

Which of these ΔG° values most likely corresponds to a reversible step of glycolysis?

A: -1.7 KJ/mol Glycolysis is a catabolic process with a negative net charge in free energy. Reversible glycolytic steps are those that aren't committed steps as they don't represent large free energy change. In other words, the forward and reverse reactions of these processes have similar ΔG values, so they lack a strong tendency to move in one direction or the other. Thus, it is the choice with the ΔG° that is closest to zero that will be the most easily reversed.

Based on the information in Table 1, what is the ΔG° at 25°C for the hydrolysis of one mole of pyrophosphate into two phosphate groups?

A: -33.4 kJ/mol Reaction ΔG °' (kJ/mol at 25°C) ATP --> ADP + Pi -30.5 ADP --> AMP + Pi -33.4 ATP --> AMP + PP -30.5 The reaction for the hydrolysis of pyrophosphate (PP) into two phosphate groups (P) is: PP → 2 P Using the concept of Hess's law, we can combine the reactions in Table 1. This way, they give the desired reaction when added together, and summing the free energies of the reactions gives the desired ΔG° value. AMP + PP → ATP ΔG° = +30.5 kJ/mol ADP → AMP + P ΔG° = -33.4 kJ/mol ATP → ADP + P ΔG° = -30.5 kJ/mol PP → 2 P ΔG° = -33.4 kJ/mol

How many substituents within the ETC are capable of carrying more than two electrons at a time?

A: 0 There are 5 electron carriers involves in the ETC: NADH-Q reductase, ubiquinone, cytochrome reductase, cytochrome c, and cytochrome oxidase. Each carrier in the electron transport chain is only capable of carrying one or two electrons at a time

At pH = 7.3, what is the bond order of the shortest bond to oxygen in glycine?

A: 1.5 At physiological pH, the carboxyl group is deprotonated, and the negative charge is delocalized by resonance between the two oxygen atoms. As a result, the bond order of the shortest bond in glycine is not 2, but rather 1.5 - A bond order of 1 is a single bond

What are reasonable estimates of the optimal pH values for pepsin and chymotrypsin?

A: 1.5, 6.5 The stomach is a HIGHLY ACIDIC environment where HCl causes the pH to drop below 2. Any enzymes that operate here including pepsin must be able to function efficiently in such conditions. In contrast, chymotrypsin (small intestine) operates within the small intestine where its pH is higher due to the secretion of bicarbonate ion

What is the approximate molecular weight of active Dnmt3a?

A: 140 kDa (Picked 35) This is the answer because the enzyme is active as a homotetramer (which means that the molecular weight given by SDS-PAGE [35kDa] is 4 times less than the active molecular weight). This is due to the fact that the SDS-PAGE conditions denature the protein and eliminate quaternary structure

The cytochrome oxidase complex, encoded in part by COI, requires the transfer of electrons from how many molecules of cytochrome C in order to catalyze the formation of two molecules of water from molecular oxygen?

A: 4 The cytochrome c oxidase complex (complex IV) is the last enzyme of the electron transport chain. It receives one electron from each of four soluble cytochrome C molecules, transferring them to a single oxygen molecule and converting one O2 molecule into two molecules of water

A protein contains 4 disulfide bonds. In order to break these bonds researchers added a minimum of

A: 4 moles of NADH for each mole of protein Because each mole of NADH can reduce a mole of disulfide bonds. Since the protein has 4 disulfide bonds, 4 moles of NADH are needed

If the coding strand for a certain gene begins with 5' AGC CTT CGG CTG ACT GGC TGG, which of the following is a possible primer that researchers could use for reverse transcription PCR amplification?

A: 5' AGC CTT CGG CTG ACT GGC TGG 3' PCR uses DNA primers so we can eliminate answer choices that contain uracil (U). Recognize that the sequence given is the coding (sense) strand, not the template (antisense) strand. This means the mRNA that would be produced has the same sequence as the coding strand, except for the thymines which become oracles. So our mRNA will look like 5' AGC CUU... Now this mRNA is going to be reverse-transcribed into cDNA for use in PCR, so your cDNA will be 3' TCG GAA.. Remember how the primer will be complementary to this sequence: so it's going to be 5' AGC CTT..

In a particular biological system, 10 mM of substrate is present. If the initial velocity of the associated enzyme-catalyzed reaction is one-eighth of its Vmax, Km is equal to

A: 70 mM Solve this using the michaelis-menten equation

A scientist desires to use a single amino acid substitution to change the tertiary structure of a theoretical protein of interest. This scientist should consider:

A: A Cys -> Gly substitution Tertiary structure is driven by the tendency of hydrophobic residues to bury themselves inside the protein, away from aqueous environments and side chain interactions (i.e hydrogen bonding and disulfide bridges). Thus, we expect a substitution between amino acids that vary in polarity, charge, or sulfur content

Angiogenesis is a critical step in the development of tissue malignancies, as cancerous tissue is able to stimulate the growth of new vasculature to supply its unregulated cell division. Angiogenesis is mediated in part by the stimulatory action of VEGF, a transcription factor. Which of the following mutations would most likely be present in a cell displaying dramatically increased production of VEGF?

A: A mutation permitted decreased expression of Fas, a receptor that stimulates apoptosis when bound by Fas ligand If this receptor stimulates apoptosis, cancer cells that have progressed far enough to stimulate angiogenesis are unlikely to express it. If they did, they would be responsive to attack by the immune system.

What's the best explanation for the observed behavior of the E147D (compared to E147A) variant? The substitution results in

A: A repulsive interaction that reduces conformational stability and enzymatic activity With both side chains being negatively charged, aspartate has a slightly shorter chain and will be further from the active site. Since the resulting enzyme shows very low conformational stability, the best explanation is that a repulsive interaction develops reducing conformational stability, eliminating the possibility of catalysis

At excess levels, which of the following could serve as an activator of the TCA enzyme citrate synthase? A. ADP B. Citrate C. Succinyl-CoA D. NADH

A: ADP TCA is another name for the Krebs (citric acid) cycle. Increased concentration of ADP signals that more ATP is needed. Therefore, the citric acid cycle enzyme will be activated by the presence of ADP - Citrate and succinyl-CoA are intermediates of the citric acid cycle and NADH is the product of the citric acid cycle that signal to inhibit the cycle

In times of glucose deprivation, the body can obtain energy by utilizing its stores of fat. This involves the production of ketone bodies, the metabolism of which is relatively inefficient. One particular ketone body can be detected by smelling one's breath. Which product is responsible for yielding sweet-smelling breath and sweat?

A: Acetone Acetone is formed from the breakdown of acetoacetic acid. Up to 30% of this product can be excreted from the body in the urine or through mechanisms such as exhalation. Unlike the other ketone bodies, acetone cannot be converted back into acetyl-CoA.

The human gluconeogenetic pathway can directly utilize all of the following molecules EXCEPT: A. oxaloacetate. B. pyruvate. C. acetyl-CoA. D. alanine.

A: Acetyl-CoA Acetyl-CoA can't be directly incorporated into gluconeogenesis, nor can it be converted to oxaloacetate in human cells - Oxaloacetate is an intermediate between pyruvate and phosphoenolpyruvate - Alanine can be converted to pyruvate in the liver, then fed directly into the pathway

Small intracellular G proteins, such as Ras and Rho, are regulated by a complex system of upstream signaling proteins. One such protein is called Son of Sevenless (SOS), which acts as a guanine nucleotide exchange factor (GEF) for Ras, promoting the ejection of GDP. Co-localization of Ras and SOS via the action of Grb2 will likely result in:

A: Activation of Ras G proteins are only active when GTP is bound, a state that necessitates the release of GDP. Because we are told that SOS promotes GDP release, it must activate Ras. - SOS itself is not a G protein; it simply promotes the release of GDP that has already been hydrolyzed and rests in the active site of the inactive Ras protein

Dnmt3a was purified using which type of column chromatography?

A: Affinity The answer is affinity because the use of histone tagging and a nickel column is a form of affinity chromatography

After the application of Compound Z, the hemoglobin in a blood sample is only able to transport oxygen at 50% efficiency. However, studies show that Compound Z does not attach to hemoglobin's oxygen-binding site. Assuming a direct interaction between Compound Z and hemoglobin, which mechanism is most likely responsible for this change in O2 transport efficiency?

A: Allosteric inhibition This question mentions that NO compound Z was found in the ligand-binding site. Inhibition of a protein's function through attachment to a position other than the active site is termed allosteric inhibition

A biological researcher observes that a certain enzyme displays a sigmoidal kinetic curve. He concludes that this compound must be:

A: An allosteric enzyme that participates in cooperative binding Sigmoidal curves are characteristic of enzymes that display cooperative binding. Through this mechanism, the enzyme's affinity for its substrate increases drastically once substrate has already bound to one/more subunits.

What is the best description of the catalytic mechanism of GK? Catalysis occurs through

A: An ordered mechanism in which a ternary complex is formed Passage states that both ATP and glycerol occupy the catalytic cleft, which means that a ternary complex is formed. Since glycerol binds first, it's an ordered mechanism as opposed to a random order mechanism where either substrate could be first to bind

When not bound to FMN, flavodoxin is best described as a: A. prosthetic group. B. cofactor. C. apoprotein. D. holoenzyme.

A: Apoprotein In its functional form, flavodoxin is bound non-covalently to FMN. When unbound to a prosthetic group (such as the cofactor FMN), the protein component is referred to as an apoprotein. - Prosthetic groups are tightly-bound organic molecules such as vitamins or inorganic molecules such as metal ions that are required for the hormonal or enzymatic function of a molecule. - Cofactor is a non-protein component of a molecule required for the molecule's biological activity. When a cofactor is tightly bound, it's often referred to as a prosthetic group. - Holoenzyme is the biologically active form of an enzyme where both the protein and necessary non-protein component are present. A protein component alone is referred to as an apoprotein.

Atorvastatin is a drug used to treat hypercholesterolemia. Which of the following is the most likely mechanism of action by which atorvastatin reduces serum cholesterol?

A: Atorvastatin inhibits HMG-CoA reductase, the rate-limiting enzyme in the cholesterol biosynthesis pathway. The rate-limiting step of a biosynthetic pathway is typically the best target for regulation, both exogenous and endogenous. Altering the activity of a rate-limiting enzyme produces the fastest and most pronounced change in metabolic output.

Which of the following must be true for a biochemical reaction at equilibrium? A. ΔGº' = 0 B. ΔG = 0 C. Keq = 0 D. Q = 0

A: B The gibbs free energy change (ΔG) is used to predict the spontaneity of a reaction, with negative values indicating that the reaction will proceed in the forward direction and positive values indicating that it will proceed in the reverse direction (from products to reactions). When the reaction has reached equilibrium, ΔG is 0. - ΔG° is the Gibbs free energy under *standard biological conditions*. Typically, this involves reactions that are not at equilibrium - An equilibrium constant Keq relates concentrations (or pressures for gases) in a mathematical expression in which the products are divided by the reactants, with all species raised to their stoichiometric coefficients. The only type Keq equals 0 is if the reaction never forms any products at all. For biochemical reactions at equilibrium, Keq can be very small, but it must be greater than zero. - The reaction quotient (Q) is calculated in the same manner as Keq, but it can be used for reactions at all stages, not just at equilibrium. When a process does reach equilibrium, Q = Keq, not zero

The products of the pentose phosphate pathway are linked to all of the following metabolic processes EXCEPT: A. glycolysis. B. fatty acid synthesis. C. glutathione reduction. D. blood pH regulation.

A: Blood pH regulation Regulation of blood pH involves the interconversion between water and carbon dioxide, carbonic acid, and bicarbonate; it is unrelated to the PPP

Glycolysis, among other vital functions, serves to help regulate blood sugar. One hour after a meal, what is the expected saturation status of hexokinase and glucokinase with glucose?

A: Both enzymes will be saturated Directly after a meal, sugars are absorbed into the blood, resulting in high plasma glucose levels. Hexokinase has a much higher affinity for glucose than its isozyme (glucokinase). However, at high glucose concentrations, both enzymes should be saturated with glucose

Conversion of pyruvate into glucose requires enzymes present in

A: Both the mitochondria and cytosol (I picked: "cytosol only") Conversion of pyruvate to glucose requires its initial conversion into oxaloacetate, catalyzed by pyruvate carboxylase in the mitochondria. OAA is then decarboxylated and phosphorylated. After transfer of OAA in the form of malate or PEP from the mitochondira, the remainder of gluconeogenesis takes place in the cytosol

Which of the four DNA bases contains the largest number of hydrogen bond acceptors when involved in a Watson-Crick base pair?

A: C (Picked G) Adenine contains 1 donor and 1 acceptor, thymine contains 1 donor and 1 acceptor, guanine contains 2 donors and 1 acceptor, and cytosine contains 1 donor and 2 acceptors

It has been noted that certain cancers are linked with high rates of aerobic metabolism. A researcher desires to track the metabolic rate in a series of cell cultures. This researcher should monitor the formation of which product?

A: CO2 The question mentions aerobic metabolism, during which carbon dioxide is produced as a waste product. Therefore, CO2 is a good choice to monitor as it's gaseous and easy to measure. - Ethanol, like lactic acid, is more closely associated with anaerobic pathways

An opiate overdose slows the respiratory drive, resulting in decreased gas exchange across the alveoli. This can damage tissues or cause death from

A: Cell damage resulting from reduced ATP levels If the respiratory rate slows, the transport of oxygen to tissues will decrease as well. In such conditions, cells cannot generate ATP through mitochondrial respiration, though they can still use anaerobic processes such as glycolysis and fermentation. Therefore, opiate overdose leads to a decrease in ATP concentration in the cells of vital organs, inhibiting their function.

Nicotinamide adenine dinucleotide (NAD) plays an important role in cellular respiration. In its reduced form, this species can transfer hydrogen to protein complexes that generate the H+ gradient required for ATP synthesis. NAD can be most specifically described as a

A: Coenzyme Coenzymes are a subset of cofactors that tend to bind LOOSELY to their associated enzymes. This type of molecule is also known for TRANSFERRING functional groups between species. NAD serves this function by donating its hydrogen to complex I.

How many ATPs does oxidation of caprylic acid (C8 saturated fatty acid) produce?

A: Complex oxidation of this acid yields 61 ATPs. The activated fatty acyl-CoA proceeds 4 rounds of B-oxidation (because C8) and is converted to 4 acetyl CoA. ♣ Each acetyl-CoA enters the krebs cycle which produces 1 GTP, 1 FADH2, and 3NADH. From here, 48 ATPs produced. 2) Complete oxidation of C8 fatty acid hives 3 FADH2 and 3 NADH from B-oxidation. Also the cleavage of the last 4 carbon compound gives 2 molecules of acetyl-CoA that enter the Krebs cycle, but neither undergo B oxidation. Thus, in total, we gained 61 ATPs.

The rate of actin polymerization in a solution was observed and plotted against the concentration of actin. What is the critical concentration of actin under these conditions?

A: Concentration where its polymerization rate is 0 With regard to actin, the critical concentration is the point at which no net polymerization/depolymerization occurs; this is the concentration where the length of the polymer is stable

Skeletal muscle is especially rich in phospholipids, of which 10-20% are arachidonic acid. Which of the following changes would lead to a decrease in the fluidity of the sarcolemma?

A: Converting some of the arachidonic acid present into its all-trans unsaturated geometric isomer Recall that the sarcolemma is the skeletal muscle plasma membrane. Cis double bonds tend to introduce "kinks" in phospholipids chains, decreasing the packing efficiency and increasing the fluidity of a membrane when compared to trans unsaturation. Converting some of the arachidonic acid present into its all-trans unsaturated geometric isomer would allow the hydrocarbon tails to pack more closely together, decreasing fluidity. - Increasing the temperature of the membrane leads to the transfer of thermal energy to phospholipids, increasing their kinetic energy and membrane fluidity in general - Shorter and more saturated phospholipids increase fluidity . Shorter phospholipids have a greater average velocity at the same temperature vs. larger phospholipids and have a smaller surface area with which to undergo stabilizing london dispersive attractions. Saturated fats pack very efficiently; less energy is required to obtain the same freedom of movement and fluidity found in an unsaturated fat. Thus, replacement of unsaturated fat by more saturated, shorter phospholipids would increase membrane fluidity.

Although PCR protocols almost exclusively employ heat to induce DNA denaturation, alternative denaturation methods are often used for Southern blotting. Which of these solutions would be LEAST effective when separating DNA strands to enable adherence to the blotting membrane? A: A mile sodium hydroxide solution B: A 6M urea solution C: A sodium dodecyl sulfate solution D: A mild Mg2+ solution

A: D Magnesium has a well-known STABILIZING effect on DNA (i.e the essential role of magnesium as a cofactor for DNA polymerase where magnesium coordinates with the pyrophosphate leaving group, stabilizing its transition into solution. - A: Most souther blotting recommends the use of dilute NaOH to denature DNA. The altered dielectric environment conferred by the negatively-charged hydroxide ions contributes to the disruption of the hydrophobic interactions between the strands. - B: Urea interferes with hydrogen bonding, disrupting the native double helix conformation

How can a scientist running an SDS-PAGE procedure assure that the difference in running distance on the gel is due to the length of the peptide and not its shape?

A: Detergent added to the running buffer SDS is a detergent (has BOTH a hydrophobic and hydrophilic end). This COATS the protein and DESTROYS hydrophobic interactions, largely denaturing all but primary structure. Thus, the protein moves through the gel in a linear fashion.o

Researchers wish to isolate the protein product of BRCA1, a gene involved in the development of breast cancer. They also want to know this protein's primary structure. Which analytical technique(s) should they use?

A: Edman degradation The Edman degradation is a technique used to sequence peptides by progressively removing amino acids - X-ray crystallography allows us to infer the 3D (secondary and tertiary) structure of the protein, but not its primary structure.

A particular protein largely lacks both secondary and tertiary structure. Which factor is mainly responsible for the resting state of this protein?

A: Entropy Hydrogen bonds, disulfide bonds, and dipole-dipole interactions produce the complex folding patterns in secondary and tertiary structure. However, if a protein doesn't have these elements of structure, it will adopt a state where its entropy is maximized.

Two separate gel electrophoresis analyses are performed on a sample of purified α enzyme and the following results are obtained: SDS-PAGE: 1 band SDS-PAGE + dithiothreitol: 3 bands Which prediction about the enzyme structure is most strongly supported by these observations?

A: Enzyme a contrails more than one subunit *Dithiothreitol (DTT)* is a reducing agent often used during SDS-PAGE to further denature proteins by reducing/cleaving disulfide linkages and breaking up the quaternary protein structure (oligomeric subunits). The presence of one band when not exposed to DTT and three bonds when exposed to DTT suggests that at least two disulfide linkages are present in the enzyme, and there linkages hold three separate subunits together.

Which of the following hormones promotes glycogen degradation while preventing glycogen synthesis via a G protein-coupled signal transduction pathway?

A: Epinephrine Epinephrine provides the body more access to glucose. Specifically, it activates a G protein-coupled signal transduction pathway that activates glycogen phosphorylase (enzyme involved in glycogen degradation) and inactivates glycogen synthase (enzyme involved in glycogen synthesis) - Somatostatin is a peptide hormone secreted by delta cells of the pancreas. It inhibits both insulin and gluten and its release is triggered by high glucose and amino acid levels

Most, but not all, human cells can utilize aerobic respiration to create ATP in the presence of oxygen. Which of these cell types cannot perform this function and must rely on glycolysis?

A: Erythrocytes Erythrocytes (red blood cells) do not contain a nucleus or mitochondria.

With unmanaged diabetes, weight loss can begin to occur, even if there is no change in eating habits and less exercise than usual. This can be attributed to:

A: Fat being broken down as glucose leaves in the urine Both type I and type II diabetes (untreated) come down to poor glucose uptake. If cells are unable to uptake glucose, they will signal for more glucose to be produced, assuming blood sugar is low. The excess glucose will eventually be flushed out of the system, but since the glucose is not being taken up by the cells to metabolize, glucose and ketone bodies will keep being produced and keep being flushed out of the system, which is incredibly damaging to the body tissues; one effect is weight loss, which operates very much like liposuction. Fat is being taken out and flushed out of the body rather than being "burned" metabolically first

A pharmaceutical compound targets and inhibits the carnitine shuttles used by cells. A human cell treated with this medication currently displays excess acetyl-CoA inside the mitochondria and insufficient cytosolic acetyl-CoA. What will be the most direct result of this circumstance? A. Fatty acid oxidation will be inhibited. B. Fatty acid synthesis will be inhibited. C. The Krebs cycle will be inhibited. D. Stores of fatty tissue will increase in mass.

A: Fatty acid synthesis will be inhibited While an inhibited carnitine shuttle causes many long-term issues, the question asks to consider only the most immediate effects. Fatty acid synthesis takes place in the cytosol and requires acetyl-CoA; thus, this process will be inhibited if cytosolic concentrations of this molecule are too low - Because the interior of the mitochondria contains many acetyl-CoA, fatty acid oxidation and krebs cycle relatively unaffected

A physiologist notices that whenever he becomes afraid, his blood sugar levels rise regardless of how recently he has eaten. How can this phenomenon be explained?

A: Fear promotes the release of high concentrations of epinephrine, which increases gluconeogenesis as part of the fight-or-flight response One goal of the fight-or-flight response is to increase the glucose that's available to the body for use. To do so, epinephrine triggers an increase in gluconeogenesis.

When an alcoholic abruptly stops consuming alcohol, he or she will experience withdrawal symptoms, among the most dangerous of which is uncontrolled electrical activity in the brain resulting in seizures. What is the most likely cause of seizures during withdrawal?

A: Fewer action potentials in neurons with GABA receptors (I picked "fewer action potentials in neurons with NMDA receptors). The question stem says that seizures are due to increased electrical (excitatory) activity in the brane. If a seizure occurs in someone who is used to having alcohol in their bloodstream and suddenly does not, this may be due to the sudden withdrawal of the inhibitory effects of alcohol in the brain.

In what order do electrons move through the ETC?

A: From carriers with lower reduction potential to carriers with higher reduction potential In the ETC, the movement of electrons between carriers is used to create a photon gradient across the inner mitochondrial membrane, which allows ATP synthase to generate ATP for the cell. For electrons to move properly through the chain, each carrier must "want" to be reduced more than the one before it, or the electron transfer would not be energetically favorable. The more a carrier wants to be reduced, the higher its reduction potential.

Which of the following peptides is most likely to form a covalent bonded dimer?

A: GHICEPN (I picked "AVTSYWP" that has the least ionic bond) The presence of cysteine indicates that the peptide could form a disulfide-linked dimer. We must know that covalent bond formation is based on the scientific model of disulfide bonds.

Of the amino acid sequences listed, which would incur the greatest entropic penalty when used to replace Ala-Ile-Trp in the buried interior of a protein?

A: Glu-Ser-Glu Entropic penalties are incurred when amino acids are exposed to an environment where they display poor solubility. The interior of a protein is generally hard far from the aqueous environment outside (hydrophobic residues are favored) while hydrophilic amino acids incur a penalty. Because glutamine, serine, and glutamic acid are all hydrophilic, it incurs the greatest entropic penalty

At pH 7, which of the following peptides will bind to an anion-exchange column and require the lowest concentration of NaCl for elution?

A: HIPAGEATEKALRGD Anionic peptides bind to anion-exchange columns. The strength of the binding depends on the overall charge of the peptide. The answer choice has a charge of -1 (the other -5). Our answer would elute at a lower salt concentration that the one with -5.

During the synthesis of bile, cholesterol is conjugated by the addition of a carboxylic acid residue. Bile salts are thus able to solubilize lipids in aqueous digestive secretions because bile salts:

A: Have a polar region and a non-polar region, thus increasing the water-solubility of lipids Bile salts contain both polar elements and a hydrophobic cholesterol core. Because of this amphipathic structure, they contain both polar and non-polar regions, acting as a detergent and solubilizing lipids in aqueous digestive contents by interacting with both hydrophilic (lipid) and aqueous phases

G protein-coupled receptors are transmembrane proteins that fully span the lipid bilayer. Within the structure of such a protein, where is a region dense in acidic residues most likely to be located? I. On the cytosolic side II. On the extracellular side III. Within the bilayer itself

A: I and II Acidic amino acids have hydrophilic side chains that orient themselves to maximize contact with water and other polar substances. Therefore, both cytosol and extracellular fluid directly contact a polar, watery environment

A scientist exposes a yeast cell to a chemical that inhibits the synthesis of flippase proteins. He then uses 14C to track the eventual locations of newly synthesized lipids via radiolabeling. After the administration of the aforementioned inhibitor, where would the new, radioactive lipids be found? I. In the nuclear membrane II. In the cytosolic side of the cell membrane III. In the extracellular side of the cell membrane

A: I and II Newly synthesized lipids are originally inserted into the cytosolic leaflet of the ER membrane. The ER membrane is continuous with the nuclear membrane where lipid molecules move freely between structures. Synthesized lipids are transported from the ER to the cell membrane via vesicular transport. Lipids in the cytosolic leaflet of the ER are made to comprise the cytosolic layer of the membrane of a vesicle and later join the cytosol-facing side of the cell membrane. - III: Lipids begin on the cytosolic face of the ER. Flippases then "flip" them into the interior, lumen-facing leaflet. These lipids later become the inside layer of a vesicular membrane and join the exterior layer of the cell membrane when the vesicle merges with that structure. If flippase synthesis is inhibited, this will not occur. New lipids will thus be restricted to the interior face of the plasma membrane.

Which of these statements describe(s) the amino acid W at physiological pH? I. It has no net electrical charge. II. It is not a zwitterion. III. It is neither acidic nor basic.

A: I and III At physiological pH, a tryptophan (W) has 1 negatively charged carboxyl group and a positively charged amino group, totaling a net charge of 0. It is a neutral amino group.

Which of these serve as (an) appropriate response(s) to high blood sugar? I. Increased glycogenesis II. Increased rate of gluconeogenesis III. Downregulation of glycolysis

A: I only In response to elevated plasma sugar levels, the body will aim to either use the glucose or store it for later consumption. Therefore, the rates of both glycolysis and glycogenesis will increase

Dyneins, motor proteins that are associated with microtubules, play a vital role in the transport of cellular components. How can the intracellular movement of dyneins be described? I. Dyneins travel toward the center of the cell. II. Dyneins move toward the minus ends of their associated microtubules. III. Dyneins are involved in retrograde transport.

A: I, II, and III A dynein "walks" down its microtubule towards the minus end, a terminal typically oriented towards the central region of the cell. This act of traveling inward from the cell membrane is known as retrograde transport

You desire to measure the amount of antigen in a blood sample using a western blot. Which of these reagents or molecules will be necessary? I. Sodium dodecyl sulfate II. A reducing agent III. A detection antibody IV. An agarose gel

A: I, II, and III In a western blotting, the proteins in the original mixture are separated on an SDS-PAGE gel. Typically, in addition to SDS, a REDUCING agent is used to break disulfide bonds that may be present between cysteine residues. There, antibodies are used to detect particular species. - Generally, an agarose gel is more appropriate for DNA analysis than with proteins

According to Le Châtelier's principle, which glycolytic disturbances would force the metabolic system to reestablish equilibrium? I. Increased concentrations of glucose II. Decreased concentrations of pyruvate III. Increased production of hexokinase IV. Decreased availability of ATP

A: I, II, and IV - III is wrong because hexokinase is the enzyme. Enzymes are NEITHER reactants nor products and affect the KINETICS, not thermodynamics. Therefore, this change would not disrupt the equilibrium; it would simply affect the rate at which equilibrium is established

Thermal denaturation experiments can be used to follow the transition of double-stranded DNA into single-stranded DNA. Which of the following parameters affects the Tm of dsDNA in this experiment? I. pH of solution II. Ionic strength of solution III. Length of DNA strand

A: I,II,III (Picked I and III) Significant drops in pH result in protonation of hydrogen-bond acceptors, leading to a loss in base-pairing interactions. The presence of positive ions in solution (particularly Mg2+) leads to stabilization of the DNA folding via shielding of the repulsion between phosphate groups within the DNA backbone. The length of DNA also plays a role. Longer DNA strands are held together by more hydrogen bonds, meaning more energy is required to denature the dsDNA.

The enzyme adenylate kinase catalyzes the following reaction: ADP + ADP ↔ ATP + AMP The reverse process predominates when the cell is using ATP at a low rate. Which of these species likely downregulate(s) the activity of one or more glycolytic enzymes? I. AMP II. ATP III. ADP

A: II and III Because we know that low use of ATP causes the reverse reaction to predominate, we expect high concentrations of ATP to be found under these low-energy-use conditions. In contrast, more ATP and AMP should be present when the cell is in an energetically demanding state. However, ATP will then be used rapidly, leaving high concentrations of AMP and low ADP. Since this question asks about the downregulation of glycolysis, it implies that high ADP correlates iwth low energy demand (presence of ADP inhibits at least one glycolytic enzyme). Also, ATP limits its own production in a negative feedback.

The toxic cyanide anion (CN-) binds to cytochrome c oxidase and prevents electron transfer. Dosing of BP-CML cells with cyanide would result in: I. increased oxygen demand. II. reduced intracellular ATP. III. apoptosis or necrosis.

A: II and III Cytochrome c oxidase is an enzyme responsible for electron transfer in the ETC; therefore, inhibiting it would slow/stop the ETC. As a result, ATP production would be dramatically reduced because ETC is responsible for the majority of ATP production. The question states that cyanide is highly toxic; therefore, a large dose would cause cell death (apoptosis or necrosis) - I is false because inhibiting the ETC would reduce oxygen demand as electrons would no longer be flowing to oxygen, the final electron acceptor

Denaturation of whey proteins involves the loss of which levels of protein structure? I. Primary structure II. Tertiary structure III. Quaternary structure

A: II and III Denaturation is a process where proteins/nucleic acids lose the quaternary, tertiary, and secondary structures due to some external stress. Denaturation in this experiment involves breaking hydrogen bonds. Primary structure refers to the linear sequence of amino acids in a polypeptide chain. Primary structure is held together by covalent bonds, which are not broken by heating the proteins to the temperatures in this experiment. Side note: Disulfide bonds are stronger than hydrogen bonds, taking larger increases in temperature or acidity to be broken

Which statement(s) regarding the regulation of glycolysis is / are true? I. Each step of glycolysis can be directly up- or downregulated by the presence of certain coenzymes or products. II. Glycolysis and gluconeogenesis share all of the same enzymes, since one process is the exact reverse of the other. III. Glucose 6-phosphate allosterically inhibits Step 1 of glycolysis. IV. AMP allosterically activates Step 3 of glycolysis, but allosterically inhibits the final step of the glycolytic pathway.

A: III Glucose-6-phosphate is the product of step 1. Therefore, when present in excess, it inhibits the forward reaction of the step through a negative feedback. - I: The glycolytic pathway has three major points of regulation, and ONLY three of its steps are strictly regulated - II: Glycolysis and gluconeogenesis do share most of the same enzymes. However, glycolysis is marked by three irreversible steps for which gluconeogenesis must use its own unique enzymes to catalyze the reverse reaction. - IV: AMP does allosterically activate Step 3 of glycolysis, but it does not inhibit the final step

At a pH of 6.0, which of the following amino acids would be found as neutral molecules? I. Arginine II. Aspartic acid III. Alanine IV. Phenylalanine

A: III and IV At a pH of 6, both alanine and phenylalanine have a protonated amino group, which carries a +1 charge. Both molecules will also have a deprotonated carboxyl group, which will be negative; this results in a net molecular charge of zero. Note that neither alanine nor phenylalanine has a charged R group. I: Arginine contains a basic side chain with a pKa of >12. At pH of 6, both its amino group and R group will be protonated (+2), while its carboxyl group is deprotonated, resulting in a net molecular charge of +1. II: Aspartic acid has an acidic side chain with a pKa of approximately 4. Under the conditions in the question, its amino group will be protonated (+), but its R group and carboxylic acid will be deprotonated (-2). Consequently, its overall charge will be -1.

Which of the following components, in addition to phospholipids, are likely to be found in whole or in part on the cytoplasmic face of a human plasma membrane? I. Glycoprotein II. Glycolipid III. Transmembrane protein IV. Phospholipase A2 V. Cholesterol

A: III and IV Transmembrane proteins are membrane-spanning integral proteins with cytosolic and ectoplasmic faces. Phospholipase A2 is both an extracellular and a peripheral protein. Peripheral proteins attach to integral membrane proteins and are anchored in the lipid bilayer via a hydrophobic domain. - Glycoproteins either occur as secreted extracellular proteins or as an extracellular segment of integral membrane proteins. They would not be found on the cytoplasmic face of the membrane. - Glycolipids are found on the ectoplasmic surface of all human cell membranes and extend from the phospholipid bilayer into the aqueous environment, where they act as recognition sites for chemicals and as cellular attachment sites.

Which of the following processes are NOT likely to contribute to hepatic glucose production by mice fed an HFD? I. Gluconeogenesis using TCA cycle intermediate as substrate II. Hepatic glycogen breakdown III. Gluconeogenesis using pyruvate as substrate IV. Gluconeogenesis using acetyl-CoA as substrate

A: IV Both the breakdown of stored liver glycogen and hepatic gluconeogenesis could contribute to hepatic glucose production in mice fed an HFD. Gluconeogenesis, which occurs in the liver during a period of fasting, produces glucose from non-carbohydrate carbon substrates such as pyruvate, glycerol, lactate, TCA cycle intermediates, and the carbon skeletons of glycogenic amino acids. However, acetyl-CoA cannot, when produced from the oxidation of even-chain fatty acids, serve as a substrate for gluconeogenesis. This is because the conversion of pyruvate to acetyl-CoA is irreversible; additionally, during the TCA cycle, the two carbons of acetyl-CoA are lost as two molecules of CO2 and are not integrated into the carbon skeletons of TCA cycle intermediates that can be directed into gluconeogenesis.

The internal region of the bacterial K+ channel contains a large amount of water. Additionally, crystallographic measurements indicate that its dimensions are sufficient to permit potassium ions to pass through most of the channel in a hydrated state until they reach the extracellular side. This feature most likely:

A: Increases the speed of channel conduction by reducing energetic penalty associated with dehydration Only a small section of the channel is narrow enough to strip the ions of their hydration shells. This reduces the energetic penalty associated with passing a charged particle through a hydrophobic environment, such as the interior of the bilayer.

Considering only the results for HERV-W transcription in schizophrenic samples (Figure 2), VPA is most likely to function as a(n) A. transcriptional repressor. B. promoter of heterochromatin formation. C. inhibitor of histone acetylase. D. inhibitor of histone deacetylase.

A: Inhibitor of histone deacetylase Figure 2 shows that VPA treatment significantly increases HERV-W transcription in samples taken from schizophrenic patients. In chromatin, DNA is wrapped around histone proteins; strong DNA-histone associations produce a tightly-packed chromatin structure that is difficult to access by transcriptional enzymes. In contrast, loose interactions between DNA and histone proteins promote transcription. Acetylated histones are less attracted to DNA than deacetylated histones; thus, an inhibitor of histone deacetylase would keep histone proteins acetylated, reducing DNA-histone attractions and promoting transcription.

Valproic acid dissociates in water to form its conjugate ion, valproate. Valproate is expected to associate most strongly with a protein rich in which of the following residues?

A: K As the conjugate base of valproic acid, valproate must be negatively charged. As such, it will associate most strongly with positive species. Of the choices given, only K (lysine) is a basic amino acid and thus will have a positively-charged side chain at moderate pH levels.

In humans, gluconeogenesis occurs in the

A: Kidneys Gluconeogenesis mainly takes place in the liber but does occur to a lesser extend in the kidneys * Recall that the liver plays an integral role in the maintenance of stable blood glucose levels

Phosphorylation is commonly used to modify regulatory enzymes because

A: Kinases are used to reversibly bind phosphate groups to polar amino acids Phosphorylation is advantageous because it is a REVERSIBLE process especially in regulatory proteins that must constantly be turned on/off. Phosphate groups may be attached to amino acids that have a terminal oxygen atom in their R group (i.e tyrosine) where its residues are at somewhere polar

Which of the following segments of amino acids would be most likely to be found in the membrane-spanning domain of the sodium channel in a nerve axon?

A: LIV (Picked: EVE) The inside of the membrane is hydrophobic, so the membrane-spanning domain of a protein will consist of more hydrophobic amino acid residues.

The side chain of which amino acid can form a bond that's similar to a peptide bond?

A: Lysine Lysine side chain can form isopeptide bonds by reacting with a carboxylic acid group, which is the same way that peptide bonds are formed.

Consider a mixture of lysine, histidine, and aspartate in a buffered solution at pH 4.0. You perform an isoelectric focusing procedure on these amino acids. What would be the order of the amino acids in terms of migration distance from greatest to smallest?

A: Lysine > Histidine > Aspartate You should know that lysine and histidine are basic amino acids, but histidine has a pI around 7 since it is a common residue used to transfer protons in enzymes. Lysine has a pI in the basic range. You should recognize aspartate as an acidic amino acid that would have a pI in the acidic range. At pH = 4.0, we are closest to the pI of aspartate, then histidine, and finally lysine. This can be reasoned without knowing the actual values. Here are the actual pI values for the amino acids: pI of lysine: 9.74 pI of histidine: 7.59 pI of aspartate: 2.77 Migration Distance: Lysine > Histidine > Aspartate

Aspartic acid, lysine, and glutamine undergo cation-exchange chromatography. How should they be ordered from longest to shortest retention time?

A: Lysine, glutamine, aspartic acid Cation-exchange chromatography features a column with a negatively-charged stationary phase. This phase interacts with the positive lysine molecules while repelling aspartic acid due to its negative carboxylate side chain. Therefore, lysin elutes last while aspartic acid travels the fastest In cation-exchange, (+) molecules are purified via binding to an anionic solid phase

One method by which cells drive nonspontaneous reactions is to couple them to other processes that involve high-energy compounds. All of the following are species that contain high-energy bonds EXCEPT: A.ATP. B.NAD+. C.creatine phosphate. D.acetyl-CoA.

A: NAD+ ATP and acetyl CoA are similar because they contain high-energy phosphate or thioester bonds. In contrast, NAD+ is the oxidized version of NADH (electron carrier). While NADH may be a high energy bond, NAD+ itself represents a low-energy state.

Two cysteine residues undergo a reaction to form a disulfide bond. Which of these reactions can be coupled to this process?

A: NAD+ -> NADH The formation of a disulfide bond involves oxidation; therefore, it can be coupled with a reduction reaction.

Which phases of the pentose phosphate pathway should be activated to produce the greatest amount of nucleotides and NADPH, respectively?

A: Non-oxidative phase, oxidative phase NADPH is produced only in the oxidative phase of the pentose phosphate pathway. Therefore, the answer to the second part of this question must be the oxidative phase. Remember that the oxidative phase is irreversible, while the non-oxidative phase is reversible. Ribulose 5-phosphate, a product of both the oxidative and non-oxidative pathways, can be converted to ribose 5-phosphate, which in turn can be used to make nucleotides. Therefore, both the oxidative and non-oxidative phases could be utilized for nucleotide production. In the oxidative phase, a carbon dioxide molecule is lost when converting the six-carbon glucose 6-phosphate to the five-carbon ribulose 5-phosphate. However, no carbon dioxide is lost in the non-oxidative phase. Thus, the non-oxidative phase yields more moles of ribulose 5-phosphate per mole of glucose 6-phosphate than the oxidative phase.

Based on the findings of this passage, it is most likely that BPA, DEHP, and DPB interfere with endocrine function by interacting with:

A: Nuclear receptors The passage provides data about pubertal anomalies, primordial follicle loss, polycystic ovary disease, and tumor development. All of these reflect long-term developmental patterns associated with the function of steroid hormones rather than peptide hormones. Steroid hormones exert their effects by binding NUCLEAR RECEPTORS. Therefore, it's likely that a substance that interferes with the effects of steroid hormones would do so by interacting with nuclear receptors - Peptide hormones typically react with membrane-bound receptors

A description of the structures of 4 proteins is shown in the table 1. Protein 1: 32 kDa monomer 2. Protein 2: Disulfide-linked homodimer comprised of 19kDa monomers 3. Protein 3: Homotrimer comprised of 25 kDa monomers 4. Protein 4: Homodimer comprised of 38 kDa monomers Which protein has the highest electrophoretic mobility in SDS-PAGE under non-reducing conditions?

A: Protein 3 (picked protein 2) Because in an SDS-PAGE gel that's run under non-reducing conditions, protein 1, 3, and 4 will run as monomers. Protein 2 will run as a dimer because the disulfide bonds between Cys residues are not reduced. The running masses are: A=32, B=38, C=25, D=38. As the smallest one, protein 3 will have the greatest electrophoretic mobility.

A protein with which properties will most likely have the largest negative net charge at pH 7?

A: Protein that binds to an anion-exchange column at pH 7 and requires a high concentration of NaCl for elution Largest negative net charge implies the presence of a large quantity of negatively charged amino acids, allowing the protein to bind tightly to the column. A high concentration of NaCl would be required for elution

In the lab, the technique of immunohistochemistry is used for

A: Providing a detailed and visual report on the protein expression within a tissue Immunohistochemistry uses antibodies that are specific to proteins to determine which antigens are expressed in a particular region of tissue.

Why does RNA electrophoresis not require SDS?

A: RNA molecules already have a net negative charge proportional to the number of nucleotides SDS detergent is used to ensure that polypeptides have a negative charge proportional to the length of the molecule. The phosphate backbone on nucleic acids provides a net negative charge on the molecule. This allows the nucleic acids to migrate towards the positively charged anode, with the gel acting as a sieve to separate sequences based on size

Nucleic acids that can catalyze biological reactions include:

A: RNA only While we typically think of enzymes are protein-based (most are), some forms of RNA do perform enzymatic functions, molecules known as ribozymes - There is no enzyme composed of DNA

According to the classic Watson-Crick model of base pairing, the two complementary nucleobases comprising a pair are aligned such that they lie in the same plane, perpendicular to the long axis of the helix. However, recent applications of NMR spectroscopy have revealed substantial deviations from this ideal model. In some cases, the two nucleobases may transiently lie in two planes that are almost orthogonal. The angle between the two complementary bases is termed the propeller angle. A base pair with a large propeller angle will:

A: Reduce the overall melting temperature of the DNA duplex due to reduced hydrophobic contact The HYDROPHOBIC effect is the PRINCIPLE contributor to DNA stability. An increased propeller angle will reduce the total hydrophobic contact between the base pairs and the adjacent bases in the stack. This will weaken the attractions holding the DNA together and less heat will need to enter the system to melt the helix. - Hydrogen bonding constitutes a SECONDARY contribution to the overall stability of the helix

In which phase of the cell cycle would it be reasonable to expect the highest activity of the pentose phosphate pathway?

A: S The pentose phosphate pathway produces ribose 5-phosphate, a compound that plays an important role in nucleic acid synthesis. During the S phase, DNA replication is ongoing, and the demand for nucleic acids is expected to be very high.

When performing experiments to measure the Kcat of an enzyme, the substrate concentration should be

A: Saturating Kcat is used to describe the rate-limiting step of catalysis under saturating conditions of substrate

To determine a protein's thermodynamic stability, chemical denaturation studies can be performed. Assuming that only the native and unfolded states can be observed under experimentally available conditions, what's the most likely shape of the curve for the dependence of the fraction of folded protein upon denaturant concentration?

A: Sigmoidal (Picked "hyperbolic") Because the unfolding of proteins is a cooperative process; it's marked by sigmoidal curves.

Cefixime displays what inhibitive behavior on PBP?

A: Suicide inhibition According to the passage, cephalosporins bind irreversibly to the active site of PBP. Suicide inhibition occurs when an enzyme binds the inhibitor (structurally a substrate analogue) and forms an IRREVERSIBLE complex through a covalent bond.

The proline in the PTS domain of mucin has a unique cyclical side chain and cannot form a standard α-helix. As a result, it is often used when mucin needs to make a tight turn in the polypeptide chain. This is likely to support the generation of which mucin structure?

A: Tertiary (picked "secondary") Turning or bending of a polypeptide chain is the tertiary structure of a protein - Secondary structure is the formation of B-pleated sheets or a-helices; proline does not aid in the generation of secondary structure

Why does polymerization occur much more rapidly at the (+) end at an actin polymer than at the (-) end

A: The (+) ends of actin molecules display stronger intermolecular interactions than the (-) ends While the rate of polymerization/depolymerization depends on the surrounding actin concentration, the (+) end grows much more quickly. (+) actin is bound to ATP while (-) actin is bound to ADP. When the (+) actin polymerizes, ATP is cleaved to become ADP; ADP at (=) end interacts much weakly, promoting slower polymerization - At the (-) end of a microtubule, depolymerization is prevented by anchoring of that end to an MTOC (Microtubule-organizing centers). All microtubules originate from MTOC.

In the process used to create cDNA, the ssDNA produced by reverse transcriptase can be made into dsDNA through treatment with DNA polymerase I. In this case, no primer is required because

A: The 3' end of the DNA strand often loops back onto itself, providing a primer for the enzyme The 3' end of ssDNA often loops back onto itself due to complementary base pairing. This serves as a primer for DNA polymerase I, providing the free 3' OH group necessary for the enzyme to initiate synthesis. - Primase is responsible for synthesizing primers, not DNA polymerase I - DNA polymerase I does require a primer to begin the replication process

A researcher compares two antibodies that recognize the same antigen, even though they are made by different animal species. How will these antibodies differ?

A: The antibodies will have different constant regions Antibodies of different isotypes (IgA, IgG) differ in their constant regions as they are made by separate species. The constant region is recognized by other immune systems to further stimulate the response to a pathogen. - two antibodies CAN recognize the same antigen

By coupling an unfavorable reaction to the hydrolysis of ATP:

A: The coupled reactions become exergonic overall The goal of coupling is to make the total of the two reactions exergonic in nature. A reaction that typically would occur may be paired with one that is so favorable that it may promote both processes. Gibbs free energy change of each individual reaction remains unaffected. - The activation energy of each energy remains unchanged.

The ability for certain lipid-based compounds to remain liquid at room temperature depends on their ratio of saturated to unsaturated fats. What molecular reasoning might explain this phenomenon?

A: The double bonds in unsaturated fats introduce kinks in the structure, allowing more fluidity and reducing the likelihood of solidification Unsaturated fats give cell membranes greater fluidity and functionality, while lowering the freezing temperature of the membrane as a whole.

What is likely to be the relative rate of fatty acid metabolism if the concentration of insulin in the bloodstream is very high?

A: The fatty acid metabolism will be low Insulin is an inhibitor of lipolysis. The enzyme inactivates hormone-sensitive lipase, an enzyme responsible for the hydrolysis of triacylglycerides to free fatty acids

The conversion between glucose and pyruvate strongly favors the formation of pyruvate, and yet the gluconeogenic pathway is able to utilize several shared enzymes with glycolysis to crease glucose from pyruvate. It can do this because

A: The formation of glucose, fructose-6-phosphate, and PEP push the equilibrium to favor gluconeogenesis Production of PEP, glucose, and fructose-6-phosphate by gluconeogenesis-specific enzymes bypassing irreversible steps of glycolysis push the equilibrium of reversible enzymes that function both in glycolysis and gluconeogenesis in the direction of glucose production - The drives of glycolysis and gluconeogenesis are reliant on changes in enthalpy through breaking/forming high-energy bonds rather than entropic changes

Ethanol is used as a temporary treatment for poisoning with ethylene glycol (ethane-1,2-diol). The efficacy of alcohol as an antidote is best explained by:

A: The high affinity of ethanol for ADH and the toxicity of glycolic acid Ethanol and ethylene glycol are both alcohols, so both can be processed by ADH in the liver. Since ethanol act as an antidote, it blocks the toxic effects of ethylene glycol. If ethanol has a high affinity for ADH, it could act like a competitive inhibitor and prevent the conversion of ethylene glycol to its aldehyde and carboxylic acid. Alcohol's high affinity for ADH keeps ethylene glycol from being converted to glycolic acid, a toxic metabolite. Etylene glycol itself is not very toxic.

The passage states that "Scientists utilizing SDS-PAGE often run into difficulties with proteins that have a greater hydrophobic content— typically those produced by organisms in contexts where surfactants are regularly encountered." Which of the following is a reasonable explanation of why this might be the case?

A: The negative-charged sulfate "head" on the end of sodium dodecyl sulfate cannot easily penetrate hydrophobic regions of the protein to disrupt non-covalent interactions, inducing higher variability in the ratio of bound SDS to the target protein molecule. Since SDS denatures proteins and coats them in negative charge, it's the bound SDS that we care about

The sequence of a typical HERV is most likely to contain a gene that codes for:

A: The production of an RNA-dependent DNA polymerase The passage states that HERVs, or human endogenous retroviruses, arose through the integration of retroviral material into the genome. It also mentions that HERVs are able to perform reverse transcription, a process that requires the enzyme reverse transcriptase. This enzyme catalyzes the production of DNA from an RNA template. Reverse transcriptase thus must have DNA polymerase activity, since it builds a new DNA strand; it is also RNA-dependent, since it reads an RNA template. - The production of an RNA-dependent RNA polymerase; this term describes a polymerase that forms RNA from an RNA template, which do not exist in eukaryotes (they are found in viruses) - The production of a DNA-dependent RNA polymerase; this describes a polymerase that creates RNA from a DNA template.. This happens during regular transcription

A researcher is looking to isolate a protein from a sample of crude tissue lysate. Following a properly conducted western blot, the researcher is unable to obtain any conjugation and the blot yields no signal for the protein of interest. Immunoprecipitation conducted on the sample using the same monoclonal antibody under neutral conditions returns a pure sample of the desired protein; however, results are less successful under mildly alkaline conditions. Which of these statements is a possible explanation for these results?

A: The protein of interest is a dimer, and the antibody binds at the junction between two monomers. Western blotting includes a denaturation step that separates any bound monomers. Immunoprecipitation under alkaline conditions produces the same dissociation but would not cause the proteins monomers to unfold. The best results would yield from immunoprecipitation in a neutral environment. If the antibody must bind to amino acid residues that reside on adjacent monomers, it will be unable to do so if the protein is denatured or if the two subunits are separated.

A protein from the urine of a chronically ill patient was isolated and run on an SDS-PAGE gel. The result displayed on the gel was a long smear instead of a crisp band. These results can most likely be attributed to the fact that: A. the protein was partially degraded. B. SDS denatured a portion of the primary structure. C. the protein being analyzed was composed of multiple subunits of different molecular weights. D. multiple proteins of distinct molecular weights were present in the sample.

A: The protein was partially degraded Degradation would break the protein into a very large number of lighter-colored bands, which would appear to blend together on the polyacrylamide gel - C and D would result in multiple well-defined bands, not a single smear

A certain metabolic process in the liver produces NADH as a part of the process. If this process is up-regulated, which of the following effects associated with gluconeogenesis is most likely to follow?

A: The rate of gluconeogenesis in the liver will decrease (I picked "plasma glucose concentration will increase significantly") When NADH increases, availability of both OAA and pyruvate decrease. Thus, the rate of gluconeogenesis decreases too

Which of the following best explains what would result if CoQ accepted only one electron instead of two?

A: There would be a free radical This question requires us to think about what will happen if only one electron and H+ atom are added to ubiquinone. We know from question 3 that the electrons are added to one of the ketones. If only one is added, then the pi electrons from the ketone bond will be transferred into the ring, pushing the pi electrons from one of the double bonds onto the adjacent single bond, pushing the pi electrons from the other ketone bond to the oxygen and a H+ will be grabbed from solution to create an alcohol. Normally, when two electrons are added, the original ketone oxygen will have a negative charge as well and grab a H+. However, if only one electron is added, there will be 5 valence electrons on that oxygen and it will be a free radical. The resulting molecule is called a semiquinone.

Consider a particular amino acid sequence taken from an enzyme: Pro-Leu-Asp-Cys-Arg-Ser-Tyr-His-Ser-Gly-Cys-Asp This catalyst is physiologically inactive until it is exposed to stomach acid. After exposure, it can be used to react with and activate digestive zymogens. How can this process be explained?

A: This sequence is part of an enzyme with a low optimal pH Every enzyme has a specific temperature and pH at which it functions most effectively. Outside of these conditions, the enzyme catalyzes its reactions less rapidly or become entirely denatured. This enzyme is said to function in the presence of stomach acid (pH around 2) - Disulfide linkage is cleaved EXCLUSIVELY via an oxidation/reduction reaction

Maxine is researching ways to reduce the activity of an enzyme implicated in a number of diseases. She is attempting to engineer an antagonist molecule that will competitively inhibit this disease-causing catalyst. To do so, Maxine should create a molecule that most closely resembles the

A: Transition state of the enzyme-catalyzed reaction Because enzymes increase the rate of reaction by reducing activation energy, they have the highest affinity for (and best stabilizes) the transition states. Therefore, a transition state analogue can be a potent inhibitor.

Acidic and basic amino acids can be freely found in the bloodstream. How are these molecules involved in buffering the plasma at physiological pH?

A: Under low-pH conditions, acidic amino acids can decrease H+ concentration Amino acids can buffer a solution either by gaining/releasing H+ depending on their abilities and the condition of the environment. At physiological pH, acidic amino acids are negatively changed. If H+ concentration increase, pH decrease, and acidic proteins pick up some excess protons to re-establish a healthy level

The figure below shows the substrate binding curves of two enzymes, A and B. Consider Enzyme B at a point below Vmax. As substrate concentration is increased and an allosteric activator is added, how will reaction rate and Vmax be affected?

A: Vmax will remain constant, while the reaction rate will increase Vmax is NOT dependent on substrate concentration, and it does not change upon addition of an activator. Vmax increase ONLY IF more enzyme were added. However, increased substrate concentration does raise the reaction rate.

If a student wished to detect a particular protein in a sample containing polypeptides of various molecular weights, which biotechnological technique would be most useful?

A: Western blotting Western blotting is a technique in which proteins are run on a gel using electrophoresis, transferred to a nitrocellulose or similar membrane, and finally washed with an antibody marker. It is typically used to determine the identity of particular proteins present in a sample and can sort these proteins by molecular weight.

Uncoupling agents, such as 2,4-dinitrophenol, operate by facilitating proton diffusion across the mitochondrial membrane, dissipating the proton gradient. After administration of a concentrated solution of 2,4-dinitrophenol to an actively respiring muscle cell, the overall rate of ATP synthesis:

A: Will decrease by over 50% without immediately halting the electron transport chain Uncoupling agents dissipate the proton gradient that's generated by the four complexes in the ETC. The effect is analogous to poking holes in the inner membrane; thus, protons can't be sequestered outside the matrix. Without the proton gradient, no flow of protons can occur through ATP synthase to regenerate ATP. Since most ATP in cells produced via the action of ATP synthase, ATP production will significantly decrease; however, the ETC will continue. It will simply no longer be "coupled" to ATP production. - the chain has no reason to immediately stop. Instead, it continually pumps protons out of the matrix. But due to the uncoupler, these protons enter again without traveling through ATP synthase - This uncoupling may work beneficially! Normally the electrochemical gradient has a large amount of potential energy used to regenerate ATP. When the gradient is uncoupled, the energy is instead "wasted" as heat.

In comparison to the transmembrane region of ASIC1, the finger region

A: Would promote a less negative change in entropy when immersed in water The transmembrane region of a protein is the portion that spans the lipid bilayer. As such, it's largely hydrophobic (nonpolar). Nonpolar molecules or parts of molecules do not mix water. When immersed in aqueous solution, they cause surrounding water molecules to form an ordered structure known as the salvation layer around them. This causes a large decrease in the entropy of the water. The passage tell us that the finger region extends far from the transmembrane region and plays a role in the sensing of [H+] in the nearby environment. We conclude that the finger region protrudes into the extracellular fluid and be polar. Polar molecules do mix well with water (especially those who can hydrogen bond). Therefore, water molecules form a less ordered salvation layer around them and the change in entropy is less negative.

Would a person with diabetes be at a greater risk of developing ketoacidosis?

A: Yes, because lack of insulin means lack of glucose in liver cells, halting the Krebs cycle and causing an overabundance of acetyl-CoA, resulting in ketogenesis and ketoacidosis. This question is asking us to remember what changes occur in a patient with diabetes and how those might contribute or not to ketoacidosis. Diabetes patients cannot produce insulin and might have high blood glucose and low levels of glucose in the cells. Thus, liver cells will lack the machinery to undergo the Krebs cycle but will still be able to obtain plenty of acetyl CoA via beta-oxidation of fatty acids. This will cause ketogenesisand ketoacidosis.

If the amino acids leucine and valine are run through the same ion-exchange chromatograph

A: it would not be possible to determine which peak corresponds to which amino acid. Ion-exchange chromatography separates compounds on the basis of charge. Depending on the stationary phase, either cations or anions will have longer retention times. In this situation, however, both compounds have nonpolar side chains. Thus, using a charge-based chromatography would be ineffective at separating them.

The diverse category of eukaryotic non-coding RNA includes:

A: microRNA, snoRNA, and tRNA, but not mRNA or hnRNA Non-coding RNA is never translated into a peptide product. Both mRNA and hnRNA (=pre-mRNA) are coding molecules.

Much effort has been exerted to optimize the timing and temperature ranges of the PCR thermal cycle, thus maximizing desired DNA yield and minimizing error. Poor yield in a PCR reaction could most reasonably be attributed to which of the following phenomena? A. Taq polymerase lacks 5' to 3' exonuclease activity. B. Magnesium stabilizes DNA double helices. C. Primers may potentially polymerize to create long single-stranded repeats. D. Insoluble DNA aggregates often form during the cooler phases of the thermal cycle.

Answer: B Although magnesium is essential for PCR to proceed, it can coordinate strongly with the double helix and prevent complete DNA denaturation, reducing the amount of substrate available for DNA polymerase and reduces yield. - A: Taq polymerase lacks 3' -> 5' exonuclease activity - B: While primers may dimerize, they do not spontaneously ligate in an end-to-end fashion and form long single-stranded repeats - D: The cooling phases of PCR generally proceed at ~68 degrees, which is still well above physiological temperature. If DNA aggregation doesn't occur in living cells, it should not occur during a PCR.

Which of the following is NOT a difference between a cDNA library and a genomic library? A. A genomic library is larger than a cDNA library. B. A genomic sequence contains both coding and non-coding sequences, whereas a cDNA library includes only coding sequences. C. A cDNA sequence is difficult to express in a prokaryotic system, whereas a genomic sequence can be conveniently expressed in a prokaryotic system. D. A genomic library includes promoters, but a cDNA library does not.

Answer: C A cDNA library includes only the coding regions; therefore, it can be easily expressed in prokaryotes via their transcriptional machinery. However, expression of an entire genome in prokaryotes is hard for many reasons including the fact that there is no splicing mechanism in prokaryotes to remove the non-coding regions. - A genomic library includes the entire genome of an organism. In contrast, a cDNA library is derived from the extracted mRNA. Thus, a genomic library is larger than a cDNA library - Since the genomic library includes the entire genome, it includes both coding and non-coding regions. A cDNA library includes only the coding regions because the non-coding regions were spliced out during the post-transcriptional modification of mRNA

Many processes in living cells produce free radicals. All of these molecules can perform an antioxidant function in vivo EXCEPT: A: Uniquinone B. Vitamine E C. NADH D. FAD

Answer: D FAD is the oxidized form of the electron carrier FADH2 so it does not act as a reducing agent (electron donor) in vivo The other three are biological reducing agents.

A researcher isolates an active enzyme from the digestive tract. She has grounds to believe that the zymogen precursor to the enzyme: A. has a larger mass than the active enzyme. B. is structurally distinct from the active enzyme. C. contains a high number of positively charged amino acid residues. D. more than one of the above are possible.

Answer: D Zymogens ("proenzyme") require covalent modification via HYDROLYSIS, significant structural rearrangement, or BOTH before they can function as active enzymes. A- correct; the precursor is likely cleaved into two or more fragments while only one is the active enzyme. Therefore, the zymogen is likely larger than the enzyme. B- correct; structural rearrangements often occur as a zymogen or proenzyme is activated C- false

DCCD is a chemical that blocks the proton pore of ATP synthase. If treated with DCCD, which of the following is most likely to decrease in the actively respiring mitochondria of adult rat cardiac cells? A. H+ concentration within the intermembrane space B. ADP concentration within the matrix C. Oxygen consumption D. Chemiosmotic gradient across the inner membrane

C is correct. If, because of DCCD, protons are unable to return to the matrix through the ATP synthase proton pore, the proton gradient of the intermembrane space will increase, making translocation of additional protons into this space by the ETC increasingly less favorable. This will reduce the rate of electron transport through, and oxygen consumption by, the ETC. This conclusion is consistent with choice C only. A: Following treatment, H+ concentration within the intermembrane space is likely to increase. B: ATP synthesis by the ATP synthase F1 depends on proton movement through the F0 proton pore. Absent this movement, ATP synthesis will decline and ADP concentration in the matrix will increase. D: The chemiosmotic gradient refers to the dual electrical and chemical gradient established across the inner mitochondrial matrix by electron transport. In the absence of a pathway for protons to return down this gradient into the matrix, the chemiosmotic gradient would be expected to increase.

Lactose intolerance

Cells use lactose to produce ATP via lactic acid fermentation. The build up of lactate and lactose in intestine (from lactose intolerance) will cause water to flow out of the cells and into lumen, causing diarrhea - High concentration of galactose can lead to: 1) lethargy, 2) liver environment and damage, 3) delayed mental development

Glucose transporters

Glucose molecules are large and polar; they can't simply diffuse across the membrane. Thus, the cells of our body need "glucose transporters" to uptake glucose from the blood. Glut-1: Found in all cells of the body Glut-2: Typically in liver and pancreas Glut-3: Dendrites and neurons Glut-4: Muscle and adipose tissue Glut-5: Small intestine cells

Malate-aspartate shuttle

In cardiac muscle cells and liver cells, the malate-aspartate shuttle is used to transport NADH molecules into the matrix of mitochondria (2.5 ATP per NADH) It translocates electrons produced during glycolysis across the semipermeable inner membrane of the mitochondrion for oxidative phosphorylation in *eukaryotes*. These electrons enter the electron transport chain of the mitochondria via reduction equivalents to generate ATP. The shuttle system is required because the mitochondrial inner membrane is impermeable to NADH, the primary reducing equivalent of the electron transport chain. To circumvent this, malate carries the reducing equivalents across the membrane.

Reversed-phase chromatography

In reversed-phase chromatography, the stationary phase of the column is hydrophobic and the mobile phase is hydrophilic. Thus, polar molecules will interact more strongly with the mobile phase and elute more quickly (a shorter retention time) than nonpolar species - conventional chromatography is the opposite of a reversed-phase technique

Size-exclusion chromatography

In size-exclusion chromatography, the beads in the column are permeated by small pores. While small proteins can fit through these pores and trapped in the beads, large molecules can't; therefore, larger proteins elute relatively quickly. - Size-exclusion chromatography does not assess proteins in terms of charge (it would be ion-exchange chromatography) - Size-exclusion chromatography is commonly used to PURIFY proteins based on size

Lipid rafts

Lipid rafts surround transmembrane proteins and serve signaling purposes. They tend to be comprised of long, saturated lipids that strongly adhere to each other and to their associated protein.

Glycerol 3-phosphate shuttle

Predominantly used by skeletal muscle cells which allows them to quickly regenerate NAD+ from NADH (by-product of glycolysis) and synthesize ATP (1.5 ATP per NADH) Found in animals, fungi, and plants

Energy charge

The ratio of ATP to AMP; low energy charge indicates the need of glycolysis and production of ATP

Which of the following amino acids CANNOT exist as a β-amino acid? A. Arg B. Pro C. Tyr D. Gly

β amino acids have their amino group bonded to the β carbon rather than the α carbon. Glycine is the only amino acid without a β carbon, which means that making β-glycine is not possible.