Homework Questions for Biochem final

Conversion of CO2 to organic compounds by a wide variety of living organisms is referred to as Choose one: A. carbon fixation.B. photosynthesis.C. photoablation.D. ossification.E. sequestration.

A: carbon fixation

What is the ultimate source of energy for life on Earth?Choose one:A. the SunB. fossil fuelsC. CO2D. Earth's core

A. the Sun

Which of the following occur as part of photosynthesis (pick two)?Choose one or more:A.Carbon is reduced.B.Water is oxidized.C.CO2 is generated.D.Glucose is consumed.

Carbon is reduced and water is oxidized

The electrons lost from chlorophyll photooxidation are replaced by the oxidation of water. How many electrons are generated from the oxidation of one water molecule?

2 electrons

Which of the following statements about redox reactions is correct? Choose one:A. A reduced compound becomes oxidized only when it acquires an electron from an oxidized compound.B. A reduced compound becomes oxidized only when it transfers an electron to an oxidized compound.C. A reduced compound has more carbon atoms than its oxidized counterpart.D. An oxidized compound has fewer protons than its reduced counterpart.

B: a reduced compound becomes oxidized only when it transfers an electron to an oxidized compound

The energy change in a system is related to the amount of heat transferred and work done. If there is no heat transferred, what is the magnitude of work in a system equal to?Choose one:A. total energy of the system (Einitial + Efinal)B. initial energy of the system (Einitial)C. energy change in the system (ΔE = Efinal - Einitial)D. final energy of the system (Efinal)

C. energy change in the system (E final - E initial)

Given the following reactions or reaction conditions, determine which are spontaneous.Choose one or more:A.ΔGº' > 0B.Keq = 10,000C.ΔH < 0 and ΔS > 0D.ΔGº' = +12.0 kJ/molE.The hydrolysis of ATP under standard conditions: ATP ⇔ ADP + Pi

Keq = 10,000 and H < 0 and S > 0 and the hydrolysis of ATP under standard conditions

Which of the following reactions is anabolic?Choose one or more:A.ATP ⇒ AMP + PPiB.GTP ⇒ GDP + PiC.glycerol + fatty acids ⇒ a triacylglycerolD.sucrose ⇒ fructose + glucose

glyceral + fatty acids -> a triacylglycerol

Protein folding can be represented by the following equation: unfolded ⇔ foldedIf the values of Δ𝐻∘ΔH∘ and Δ𝑆∘ΔS∘ for the folding of a protein are: Δ𝐻∘ΔH∘ = -270 kJ/mol Δ𝑆∘ΔS∘ = -790 J/mol·KWhat is the value of Δ𝐺∘ΔG∘ for folding at 25 °C?

-34.5 kj/mol

In degrees Celsius, at what temperature would the unfolding become favorable?

68.7 C

At standard conditions, the equilibrium lies toward products for a given biochemical reaction. What is known about the change in free energy?Choose one:A. ΔG°equals 0.B. ΔG°is unknown.C. ΔG°is less than 0.D. ΔG°is greater than 0.

C: G is less than 0

Given that the standard free energy change of ATP hydrolysis is -30.5 kJ/mol, what is the change in free energy in a cell at 37 °C where the ATP and Pi are both 10 mM and ADP is 10 uM?

-60.2 kJ/mol

What is the energy charge for the cell with concentrations of [ATP] = 1.000 mM, [ADP] = 10.00 uM, and [AMP] = 3.000 uM?

0.992

Calculate the EC (energy charge) given the following atypical adenylate concentrations for the cell containing the mutant adenylate kinase: ATP = 0.5 𝑚𝑀mM, ADP = 12.2 𝑚𝑀mM, AMP = 80 𝜇𝑀μM.

0.52

Mutations in adenylate kinase have led to a hyperactive enzyme that ultimately ends up elevating ADP levels in a cell. For cells without the hyperactive adenylate kinase mutation, how would you expect the EC to change?Choose one:A. The EC would increase.B. The EC would decrease.C. There is no way to predict how the EC would change.D. The EC would remain unchanged.

A: the EC would increase

Which of the following are properties of water?Choose one or more:A.Water is less dense as a solid than as a liquid.B.Water is a strong acid at high temperatures.C.Water is miscible with oil.D.Water is polar and an excellent solvent because of its hydrogen-bonding properties.E.Water freezes more readily when ionic solutes are dissolved in it.F.Water is liquid over a wide range of temperatures.

Water is less dense as a solid than as a liquid, water is polar and an excellent solvent because of its hydrogen-bonding properties, and water is liquid over a wide range of temperatures

Which best describes the pKa of a weak acid?Choose one:A. The pKa value depends on the concentration of the molecule present.B. The pKa value depends on the pH.C. pKa values change over the course of a titration.D. The pKa is an intrinsic property of a molecule in question and depends on how tightly bound the acidic H+ is.

D: The pKa is an intrinsic property of a molecule in question and depends on how tightly bound the acidic H+ is.

Given a solution composed of equal volumes of 0.05 M acetic acid and 0.1 M sodium acetate, determine if the following statements are true or false.

Doubling the concentration of acetic acid and sodium acetate will double the pH capacity and adding acetic acid will decrease the pH of a solution

Which of the following correctly describe a feature of how enzymes function as reaction catalysts?Choose one or more:A.Most enzymes contain multiple protein subunits.B.Adding the suffix "-ase" to the end of a protein's name denotes that the protein is an enzyme.C.Enzymes function primarily by lowering the activation energy of a reaction in order to speed up the rate of the reaction.D.Some of the amino acids in an enzyme's active site play a direct role in lowering the activation energy of the given reaction.

All (.Most enzymes contain multiple protein subunits.B.Adding the suffix "-ase" to the end of a protein's name denotes that the protein is an enzyme.C.Enzymes function primarily by lowering the activation energy of a reaction in order to speed up the rate of the reaction.D.Some of the amino acids in an enzyme's active site play a direct role in lowering the activation energy of the given reaction.)

Which of the following mutations would most likely keep the transitions of T state to R state in hemoglobin unchanged, or similar to the transitions that occur in the native molecule? The three-letter codes for amino acids are used, and the numbers following the amino acids designate the amino acid position in the protein. By convention, the original, native amino acid is listed first, followed by the mutational change.Choose one:A. Tyr42→Phe42B. Tyr42→Gly42C. Asp94→Glu94D. Asp94→Ile94

C: Asp94 -> Glu 94

Isolate any one subunit of deoxyhemoglobin along with the heme cofactor. Observe the amino acid sequence of residues that likely interact with the heme cofactor, and determine how the cofactor interacts with the protein. The most logical means of interaction are both the hydrophobic interaction and also ________ bonding.Choose one:A. ionicB. hydrogenC. esterD. disulfide

B: hydrogen

Which of the following is the best explanation for why the heme group shifts from puckered to planar when O2 binds to hemoglobin?Choose one:The heme Fe 2+is bound to the proximal histidine in the F helix, and as the Fe2+ outer electrons are drawn towards the oxygen, the Fe2+is prevented from moving into the plane of the heme.The heme Fe 2+ is bound to the proximal histidine in the F-helix, and as the Fe2+ outer electrons are drawn towards the oxygen, the Fe2+ is pulled into the plane of the heme, pulling the F-helix along with it.The heme Fe 2+ is bound to the proximal histidine in the F-helix, and as the Fe2+ outer electrons are drawn away from the oxygen, the Fe2+is pulled into plane of the heme, pulling the F-helix along with it.The heme Fe 2+is bound to the proximal histidine in the F-helix, and as the Fe2+ outer electrons are drawn away from the oxygen, the Fe2+is pulled out of the plane of the heme, pulling the F-helix along with it.

B: The heme Fe 2+ is bound to the proximal histidine in the F-helix, and as the Fe2+ outer electrons are drawn towards the oxygen, the Fe2+ is pulled into the plane of the heme, pulling the F-helix along with it

Carbon monoxide (CO) is a colorless, odorless gas and competes for binding of O2 to hemoglobin. Which of the following best explains the mechanism by which CO-poisoning occurs?Choose one:While O2 can load and unload from hemoglobin reversibly in the lungs and in the tissues, CO binds so tightly to the Fe2+ in the heme group that it gradually replaces all of the O2 in RBCs, starving tissues of oxygen.CO sequesters all of the Fe2+ in the bloodstream so that it cannot be incorporated into heme groups.CO and O2 have the same affinity for Fe2+ in heme, but the carbon atom in CO is unable to draw the Fe2+ into the planar form, decreasing hemoglobin's affinity for O2.CO and O2 can both bind hemoglobin simultaneously, causing a push-pull mechanism that makes the planar heme formation unstable.

A: While O2 can load and unload from hemoglobin reversibly in the lungs and in the tissues, CO binds so tightly to the Fe2+ in the heme group that it gradually replaces all of the O2 in RBCs, starving tissues of oxygen.

A patient, rescued from a burning home, comes into the ER with suspected carbon monoxide poisoning. Which of the following would you consider to be the best treatment strategy for the patient?Choose one:Treat the patient with 100% oxygen (O2).Move the patient outside into fresh air.Treat the patient with carbon dioxide (CO2).Infuse the patient with iron.

A: treat with 100% oxygen

Would it be more difficult to treat a patient with CO poisoning at high altitude or low altitude and why?Choose one:High altitude, because there would be less partial pressure of O2 in the air, and O2 would have a harder time binding to the Fe2+ in hemoglobin.High altitude, because there would be a higher partial pressure of O2 in the air, and O2 would have a harder time binding to the Fe2+ in hemoglobin.Low altitude, because there would be less partial pressure of O2in the air, and O2 would have a harder time binding to the Fe2+ in hemoglobin.Low altitude, because there would be less partial pressure of O2 in the air, and O2 would have an easier time binding to the Fe2+ in hemoglobin.

A: High altitude, because there would be less partial pressure of O2 in the air, and O2 would have a harder time binding to the Fe2+ in hemoglobin.

positive allosteric effector of hemoglobin

O2

Fetal hemoglobin has a BLANK

affinity for 2,3-BPG than maternal hemoglobin due to its altered subunit. As a result, BLANK hemoglobin molecules are in the R state in fetal hemoglobin, and fetal hemoglobin has a BLANK affinity for O2 than maternal hemoglobin. This enables the mother to transfer oxygen to the developing fetus via her hemoglobin.

lower, more, higher

Carbonic anhydrase is an important enzyme necessary for transport of carbon dioxide from tissues to the lungs via red blood cells. Which of the following reactions is carbonic anhydrase responsible for?

CO2 +H2O <-> HCO3- + H+

In heavy exercise, CO2 accumulates due to increased respiration. As a result:Choose one:blood [H+] increases and pH decreases.blood [H+] increases and pH increases.blood [H+] decreases and pH decreases.blood [H+] decreases and pH increases.

A: blood H+ increases and pH decreases

As the blood pH increases, the Bohr Effect says this will happen more easily in tissues:Choose one:O2 loadingO2 unloadingCO2 loadingCO2 unloading

B: O2 unloading

Student A is studying the structure of the K+ channel protein. She hypothesizes that it would be possible to modify the protein by making a conservative amino acid substitution to any one of the five residues that line the selectivity channel. The goal of the mutation would be to reduce the selectivity so that both desolvated K+ and Na+ could pass through the channel with roughly equal affinity. Which of the following correctly addresses student A's hypothesis?Choose one:A. Any mutation designed to alter the amino acid composition of the selectivity channel in such a way to increase the distance between tetramer-associated carbonyl oxygen atoms could strengthen the interaction with Na+.B. Any mutation designed to alter the amino acid composition of the selectivity channel in such a way to decrease the distance between tetramer-associated carbonyl oxygen atoms could strengthen the interaction with Na+. However, any change significant enough to be beneficial for Na+ selectivity is likely to restrict K+ passage. C. Any mutation designed to increase the negative charge of the selectivity channel could increase the favorable attraction for Na+. However, this change could lead to a decrease in the selectivity for K+.D. Any mutation designed to decrease the negative charge of the selectivity channel could increase the favorable attraction for Na+. However, any change significant enough to be beneficial for Na+ selectivity is likely to decrease the attraction for K+

B: Any mutation designed to alter the amino acid composition of the selectivity channel in such a way to decrease the distance between tetramer-associated carbonyl oxygen atoms could strengthen the interaction with Na+. However, any change significant enough to be beneficial for Na+ selectivity is likely to restrict K+ passage.

For which of the following types of membrane transport is the rate of transport directly proportional to the concentration of solute?Choose one or more:Active transport through a carrier.Passive facilitated diffusion through a channel.Passive facilitated diffusion through a carrier.Simple diffusion though a membrane.

B and D, simple diffusion through membrane and passive facilitated diffusion through a channel

If you were to deplete the cell of ATP, which of the following transport processes would continue for an extended period of time?Choose one:Primary active transport of H+ into the stomach lumen by the H+-K+ATPase.Primary active transport of Ca2+ into the sarcoplasmic reticulum by the SERCA pump.Secondary active transport concentrating I- in the thyroid gland by the Na+-I-symporter.Facilitated diffusion of water through aquaporins.

D: facilitated diffusion of water through aquaporins

Each subunit of the aquarporin tetramer can deliver up to 3 billion water molecules per second into the cell. A key Asn reside at the end of each alpha helix forms the constriction point. Based on the biochemistry of this selectivity area of the channel, which do you think best describes the association of the channel with water molecules?Choose one:No chemical binding occurs, only space restriction, which allows for almost 3 billion water molecules to pass each second.Hydrogen bonds form between the oxygen atoms in water and the Asn sidechain. Hydrogen bonds form between the oxygen atoms in water and the peptide backbone carbonyl groups.It is based on covalent bonding between water and the Asn residues.

B: hydrogen bonds form between the oxygen atoms in water and the Asn sidechhain

The constriction point is made up of two Asn residues that are Choose one:adjacent to each other in the primary sequence.the N- and C-termini of the polypeptide chain.the locations of subunit-subunit interaction when the tetramer forms.formed by the 3-D folding of the polypeptide chain, bringing two important amino acids in close proximity of each other.

D: formed by the 3-D folding of the polypeptide chain, bringing two important amino acids in close proximity of each other.

Thick filaments

connected to z disks via titin proteins, composed of myosin filaments, and bind ATP

thin filaments

contain tropomyosin, composed of actin monomers, composed of troponin complex, and bind calcium

In the sliding filament model of muscle contraction, the BLANK filaments are made of myosin protein and the BLANK filaments are made of actin and several other proteins. These two filaments slide past each other during muscle contraction, shortening the distance between BLANK This contraction is initiated by the binding of BLANK to troponin, and decreasing concentrations of this ion allow muscle relaxation.

thick, thin, z-disc proteins, Ca2+

Which of the following correctly describes the role of Ca2+ in the sarcomere?Choose one:Ca2+ binding to troponin C exposes thick filament binding sites on the thin filaments, allowing for the formation of crossbridges.Ca2+ binding to troponin C exposes thin filament binding sites on the thick filaments, allowing for the formation of crossbridges.Ca2+ binding to Titin allows for the formation of crossbridges in the sarcomere.Ca2+ binding to cAMP allows for the formation of crossbridges and the association of cAMP with thin filaments.

A: Ca2+ binding to troponin C exposes thick filament binding sites on the thin filaments, allowing for the formation of crossbridges.

Which aspect of the crossbridge cycle is sensitive to ATP-binding?Choose one:Crossbridge formationRelease of thick filaments from thin filamentsThe power strokeExposure of thick filament binding sites on thin filaments.

B: release of thick filaments from thin filaments

Would a person go into rigor mortis sooner if they died running a marathon or sitting in a chair watching TV and why?Choose one:Running a marathon, because overall ATP levels in muscle would be low, and crossbridges would be unable to release.Running a marathon, because overall ATP levels in muscle would be high, and crossbridges would release quickly.Watching TV, because overall ATP levels in muscle would be low, and crossbridges would be unable to release.Watching TV, because overall ATP levels in muscle would be high, and crossbridges would release quickly.

A: Running a marathon, because overall ATP levels in muscle would be low, and crossbridges would be unable to release.

A low concentration of a first messenger binding its receptor leads to many large changes taking place inside the cell for all signal pathways. The large changes are due to a process calledChoose one:A. signaling specificity.B. protein kinase amplification.C. signaling compartmentalization.D. signal amplification.

D: signal amplification

Which of the following first messengers is unique because it is not water soluble and does not bind directly to a membrane receptor?Choose one:A. epinephrineB. estradiolC. acetylcholineD. nitric oxide

B: estradiol

The β2-adrenergic receptor is best characterized as which of the following?Choose one:none of the abovemostly β turnsmostly β sheetsmostly α helices

D: mostly alpha helices

Which binding site most likely represents the area of greatest structural change in response to ligand binding?Choose one:heterotrimeric G protein bindingDNA bindingSOS bindingcAMP binding

A: heterotrimeric G protein binding

Tasting involves many different cell-signaling processes that ultimately generate nerve signals transduced by membrane depolarization. Sweet tastes result in PIP2 hydrolysis, while salty tastes allow sodium ions to directly alter the membrane potential. What can you deduce about the signaling mechanisms for sweet and salty?Choose one or more:A.Sweet ligands bind to and close sodium channels, which starts a membrane potential.B.Sweet and salty signaling pathways use the same taste receptor molecule, but with different efficacy on the sweet and salty taste center in the brain.C.Sodium ions directly enter the cells, indicating the signal is transduced by an ion channel.D.Sweet utilizes the GPCR signaling pathway, activating phospholipase C.

C.Sodium ions directly enter the cells, indicating the signal is transduced by an ion channel.D.Sweet utilizes the GPCR signaling pathway, activating phospholipase C

More than 50% of the pharmaceuticals currently on the market target G protein-coupled receptor pathways. A number of medications meant for treating allergic reactions target the histamine receptor signaling pathway, initiated by histamine. The H1 histamine receptor is coupled to the Gq heterotrimeric G protein. As a new undergraduate researcher in the laboratory, which of the following strategies do you think would be most successful in your efforts to inhibit histamine signaling? More than one option may be correct.Choose one or more:activation of phosphodiesterase to enhance cAMP levelsactivation of phospholipase C β activityaddition of an IP3 receptor antagonist to prevent IP3 bindingaddition of H1 histamine receptor antagonist

C and D: addition of an IP3 receptor antagonist to prevent IP3 binding and addition of H1 histamine receptor antagonist

It is important to design an inhibitory strategy that has the smallest probability of inducing side effects. In essence, you want the treatment to be as specific as possible for the allergic reaction. Which of the following strategies has the best chance of avoiding off-target effects (like activation of glycogen synthesis) while preserving the health of the patient?Choose one:Include both a Gs inhibitor at the same time as a Gq inhibitor.Use a specific H1 receptor antagonist to prevent histamine binding.Inhibit production of IP3 by phospholipase C β.Add an inhibitor of phospholipase C β only when the patient has low blood sugar so that glucagon synthesis is not stimulated.

B: use a specific H1 receptor antagonist to prevent histamin binding

Rhodopsin is a well-characterized G protein-coupled receptor that binds retinal. Absorption of light by the retinal molecule allows important neuronal signals to be sent to your brain that enable vision. If you look at a bright light and quickly close your eyes, the retinal molecules quickly turn off that visual signal to the brain. Which of the following do you think is the best explanation for such an observation?Choose one:Termination of the visual signal is completely dependent on removal of rhodopsin from the plasma membrane.The G proteins, once activated, are separated into α and βγ subunits and this terminates the signal.Inhibition of the visual signal requires GTPase activating proteins (GAP) binding and GTP hydrolysis.Guanine nucleotide exchange factors, or GEFs, are slow to respond in order to reactivate the G proteins.

C: Inhibition of the visual signal requires GTPase activating proteins (GAP) binding and GTP hydrolysis.

For some diseases like cancer, therapeutic interventions targeting small G protein signaling pathways could be beneficial, especially those interventions that inhibit Ras signaling since, in many cancers, Ras is constitutively active, meaning it is always on. Which of the following might be the best approach to inhibiting growth and division of cancer cells expressing overactive Ras?Choose one:increasing MAP kinase activityinhibiting receptor tyrosine kinasesinhibiting Raf activityincreasing GRB2 activity

C: inhibiting Raf activity

Which of the following correctly describes how phosphorylation of PIP2 to generate PIP3 propagates the insulin receptor signal?Choose one:PIP3 is chemically inert, and must be dephosphorylated to PIP2 for downstream signaling to continue.PIP3 recruits Akt, a PH domain-containing protein, and PDK1 to the plasma membrane.PIP3 is degraded into IP3 and diacylglycerol.PIP3 binds to IRS-2, activating PI-3K.

B: PIP3 recruits Akt, a PH domain-containing protein, and PDK1 to the plasma membrane.

Which one or more of the following statements is true about the complex structure above?Choose one or more:The GDP is held in place by the ends of three α helices.A β strand forms the active site of the Gqα protein.RGS2 competes for binding to the GTP binding site of Gqα.RGS2 does not participate in the active site chemistry needed to facilitate GTP hydrolysis.

A and D: the GDP is held in place by the ends of the three alpha helices and RGS2 does not participate in the active site chemistry needed to facilitate GTP hydrloysis

Based on the activation state of the RGS2-Gqα protein complex shown in the virtual model, which of the following would you expect to happen next?Choose one:GDP will be replaced with GTP.Gα will reassociate with Gβγ.Gqα will reassociate with its G protein-coupled receptor.Gqα will interact with downstream effector molecules.

B: Galpha will reassociate with G by

Long polypeptides, when folded into tertiary structures, can bring two amino acids within the polypeptide close together even though they are far apart in sequence. In the Gqα protein, the amino terminus and carboxyl terminus are found in close proximity. What would you say is an accurate description of this part of the protein?Choose one:Gqα begins and ends with α helices that come in close proximity when folded.Gqα begins and ends with β strands that come in close proximity when folded.Gqα has an α helix at one end and a β strand at the other that come in close proximity when folded.The Gqα polypeptide has little secondary structure at its ends that instead can be described as loose coils.

C: Gqa has an alpha helix at one end and a beta strand at the other that come in close proximity when folded

Both epinephrine (a tyrosine derivative) and glucagon (a peptide hormone) increase glucose export from the liver into the bloodstream. Each ligand binds a different receptor, but both lead to an activation of PKA. How does this happen?Choose one:A. Each activated receptor interacts with a unique GTP binding protein that activates PKA.B. The receptors for the two ligands interact before binding to the heterotrimeric G proteins.C. Both receptors bind and activate the same Gα subunit, Gsα, which indirectly leadsto PKA activation.D. Both receptors bind directly to and activate PKA after the Gsα protein dissociatesfrom the receptor.

C. Both receptors bind and activate the same Gα subunit, Gsα, which indirectly leadsto PKA activation.

The wild-type Ras gene normally has the codon GGT, coding for glycine at position 12. Which of the mutations shown in the above figure is least likely to affect Ras protein function?Choose one:GAT, AspGTT, ValGCT, AlaCGT, Arg

C: GCT, Ala

Which of the following best describes a death domain?Choose one:A. the active site portion of cysteine-aspartate proteases (caspases)B. the TRADD-FADD complex including procaspase 8C. the cytoplasmic tail of the TNF receptor, which functions as a protein-protein interaction moduleD. the proteolytic target site for CASP8 by CASP3

C: the cytoplasmic tail of the TNF receptor, which functions as a protein-protein interaction module

One subtype of breast cancer involves the human epidermal growth factor receptor 2 gene (EGF receptor). One in every five breast cancers has a mutation in this gene. Understanding that this is a growth factor receptor gene, which of the answer choices best describes how this type of cancer develops?Choose one:A. Interactions of the mutated receptor, which activates protein kinases, will not impact tumor growth because the ligand cannot bind to the receptor.B. Mutations that block the receptor from functioning will block cell growth, indicating that this is a loss-of-function gene.C. Mutations in the intracellular domain of the receptor will cause Akt to bind without further signaling and regulation of cell apoptotic processes.D. Mutations that activate the kinase portion of the receptor result in a receptor that is constantly phosphorylated. This causes constitutive activation of downstream signaling and the resulting cell growth and proliferation.

D. Mutations that activate the kinase portion of the receptor result in a receptor that is constantly phosphorylated. This causes constitutive activation of downstream signaling and the resulting cell growth and proliferation.

Long-term activation by nuclear receptors differs from the more transient membrane receptor signaling for what reason?Choose one:A. It takes longer for the steroid to diffuse through the cell, while membrane receptor ligands do not require translocation into the cell.B. Long-term activation occurs by nuclear receptors activating sets of genes, while membrane receptors involve small second messenger molecules responding quickly, as their signaling occurs by activation of kinases, phosphatases, and other enzymes already expressed in the cell. C. Receptor-mediated signaling takes longer to reverse the phosphorylation caused by kinases, while nuclear protein expression is much faster.D. Steroid receptors are antagonistic to most GPCR signaling.

B. Long-term activation occurs by nuclear receptors activating sets of genes, while membrane receptors involve small second messenger molecules responding quickly, as their signaling occurs by activation of kinases, phosphatases, and other enzymes already expressed in the cell.

Steroid receptors

bind to inverted base repeats due to the orientation of the receptor dimer, these receptors typically bind cholesterol-derived hormones and bind as head-to-head homodimers

metabolite receptors

bind to direct repeat DNA sequences, form head to tail dimers, and ligands are often dietary nutrients including vitamins lipids and derivatives of amino acids

Association of glucocorticoids with their DNA binding sites is essential for the regulation of inflammatory response genes. Which one or more of the following might disrupt normal association of glucocorticoids with their DNA binding sites and be cause for concern?Choose one or more:zinc deficiency in a patientexcess serum glucocorticoid levelstaking prednisone for an asthma attackgenetic mutation in the GRE

A and D zinc deficiency in a patient and genetic mutation in the GRE

What kinds of chemical bonds do you identify as being critical for the formation of this quaternary complex?Choose one or more:ionic bondshydrogen bondscovalent bondsglycosidic bonds

A and B, ionic and hydrogen

Which of the following best describes the location of the coordinating cysteines in the glucocorticoid receptor DNA binding element?Choose one:Each zinc ion is held in place by cysteines located in both β sheets and α helices, but not random coils.Each zinc ion is held in place by cysteines located in random coils only.Each zinc ion is held in place by cysteines located in α helices and random coils, but not the β sheets.Each zinc ion is held in place by cysteines located in α helices only.

C: Each zinc ion is held in place by cysteines located in α helices and random coils, but not the β sheets.

Which of the following cell lysis techniques uses a high-frequency sound to disrupt the membrane?Choose one:A. sonicationB. French pressC. needle shearD. enzymatic digestion

A: sonication

An important step in the purification of a protein is salting out, which requires the addition of increasing amounts of a saturating salt, like ammonium sulfate, to a protein solution. Which of the following correctly characterizes the salting-out process?Choose one:Salting out causes aggregation of insoluble protein conjugates and a purified protein pellet upon centrifugation. This is not ideal as it results in a nonfunctional protein.Salting out causes aggregation of insoluble protein conjugates and a purified protein pellet upon centrifugation. The resulting protein retains its 3-D conformation and is fully functional.Salting out decreases the solubility of insoluble protein conjugates and the protein stays in solution. This is not ideal as it results in a nonfunctional protein.Salting out increases the solubility of insoluble protein conjugates and the protein stays in solution. The resulting protein retains its 3-D conformation and is fully functional.

B: Salting out causes aggregation of insoluble protein conjugates and a purified protein pellet upon centrifugation. The resulting protein retains its 3-D conformation and is fully functional.

Now that we have salted out our protein, we need to get rid of the excess salt before moving on. Dialysis is a great way to achieve this outcome. Your lab mate gives you a dialysis bag with a pore size of 25 kDa; the protein you are trying to purify is 30 kDa. Is your lab mate helping or hurting your efforts?Choose one:Helping; the pore size of the dialysis bag is smaller than the protein, so the protein will remain in the bag and the salt will flow out into the surrounding solution.Helping; the pore size of the dialysis bag is larger than the protein, so the salt will remain in the bag and the protein will flow out into the surrounding solution.Hurting; the pore size of the dialysis bag is smaller than the protein, so the salt will remain in the bag and the protein will flow out into the surrounding solution.Hurting; the pore size of the dialysis bag is larger than the protein, so the protein will remain in the bag and the salt will flow out into the surrounding solution.

A: Helping; the pore size of the dialysis bag is smaller than the protein, so the protein will remain in the bag and the salt will flow out into the surrounding solution.

Which of the following most accurately describes luciferase?Choose one:Luciferase is a monomeric protein with N- and C-termini in close proximity to each other.Luciferase is a dimeric protein with N- and C-termini on opposing sides of the molecule.Luciferase is a monomeric protein with N- and C-termini on opposing sides of the molecule. Luciferase is a dimeric protein with N- and C-termini in close proximity to each other.

C: Luciferase is a monomeric protein with N- and C-termini on opposing sides of the molecule.

The active site of luciferase is bound by a luciferin analog, oxyluciferin, in the model above. Zoom in on the active site and choose the most accurate description of luciferase secondary structures participating in binding active site formation.Choose one:three β strands, an α helix, and a β turntwo β strands and one α helixthree β strands and a β turntwo β strands, two α helices, and a β turn

A: three β strands, an α helix, and a β turn

Lactate dehydrogenase (LDH) is a tetrameric enzyme present in the cytosol of cells. It catalyzes the reversible reaction that converts pyruvate to lactate, and it is highly expressed in muscle cells. LDH participates in the clearing of lactate when it accumulates following heavy exercise. Which of the following strategies would be most effective in purifying LDH for further enzyme assays?Choose one:fixing muscle cells, immunostaining with fluorescent antibody to LDH, and obtaining microscopic imagesion exchange chromatography, followed by SDS-PAGE and Western blot with anti-LDH antibodydifferential centrifugation of muscle cell extract, salting out, followed by column chromatography2-D isoelectric focusing of muscle cell extracts, followed by tandem mass spectrometry

C: differential centrifugation of muscle cell extract, salting out, followed by column chromatography

Now that you have isolated LDH enzyme from muscle cells, you perform an activity assay to assess purity. In the muscle cell extract, your specific activity for LDH is 2.5 units/mg protein and in your column chromatography samples it is 12 units/mg. What is the fold purification of your new LDH sample?Choose one:4.8300.219.5

A: 4.8

The mass-to-charge ratios of denatured proteins are equivalent for different mass proteins. However, the cross-linked nature of the acrylamide media can limit migration through the polymer matrix. Gels with less cross-linked acrylamide (low % SDS gels) will do which of the following?Choose one:A. separate proteins as in size-exclusion chromatography, with the smaller proteins migrating most slowly and the larger proteins migrating farther through the gelB. separate larger proteins at the expense of smaller proteins, which will not resolve wellC. allow the negative charge contributed by the smaller SDS molecules to more easily migrate through a low-percent acrylamide because of the increase in negative chargeD. separate the smaller proteins based on the percentage gel at the expense of larger proteins

B: separate larger proteins at the expense of smaller proteins, which will not resolve well

X-ray crystallography

nearly any sized protein can be used, dynamic changes not easily captured, poorly ordered domains will have multiple arrangements averaging out the signal

NMR spectroscopy

requires isotopic labing of elements, smaller proteins are more amenable, large proteins will reorient slowly in solution averaging out the signal

You have discovered a small organic compound that you think will cause a significant shift in the peptide loop covering the active site of the enzyme, inhibiting the protein's function. Which approach(es) would be most appropriate to test for this hypothesis?Choose one or more:A.X-ray crystallographyB.Affinity chromatographyC.ELISAD.NMR spectroscopy

Xray crystallography and NMR

An important drug design strategy is to map the structure of a large macromolecular complex. You have a concentrated solution of this complex, which is quite large. Which of the following methods has recently emerged as having the highest probability of helping you map the structure?Choose one:nuclear magnetic resonance spectroscopyX-ray crystallographycryo-electron microscopymass spectrometry

C: Cryo-electron microscopy

When trying to determine the shape of a protein that is structurally flexible, meaning it can adopt multiple conformations, which of the following methods should you explore further?Choose one:cryo-electron microscopynuclear magnetic resonance spectroscopyX-ray crystallographymass spectrometry

NMR

The molecular structure of an immunoglobulin Fab fragment is illustrated above; it is bound to an antigen (in yellow). The positioning of the antigen binding site is important for the functioning of the immune system. Take a careful look at the placement of the antigen by manipulating the molecule. Which of the following statements correctly describes the antigen binding region?Choose one:The antigen makes contact with both the VL and VH variable domains.The antigen is held in place by the VH variable domain only.The antigen makes contact with the constant domains only.The antigen is held in place by the variable domain only.

A: the antigen makes contact with both the Vl and Vh variable domains

Which antibody, typically made using a hybrid between tumor cells and mouse B cells, produces a single antibody that recognizes a single epitope on an antigenic molecule, instead of several antibodies each binding multiple epitopes on a single antigen? (Choose the specific name of this antibody.)Choose one:A. monoclonal antibodyB. monocellular antibodyC. polyclonal antibodyD. antigen-specific antibody

A: monoclonal antibody

You are an undergraduate research student working in a laboratory that studies breast cancer cells. You are asked by the lab manager to determine whether Protein X is overexpressed in breast cancer cells. Which of the following would be the most appropriate strategy for determining this?Choose one:Use affinity chromatography to separate the normal cell Protein X from the cancer cell Protein X.Combine the normal and cancer cell samples and perform isoelectric focusing to determine relative abundance of Protein X.Perform SDS-PAGE and Western blot to demonstrate relative abundance of Protein X in normal cells and breast cancer cells.Determine an A280 value for both normal cells and breast cancer cells to determine total protein concentrations.

C: Perform SDS-PAGE and Western blot to demonstrate relative abundance of Protein X in normal cells and breast cancer cells.

From your previous experiment, you were unable to confirm that the relative abundance of Protein X was any different in cancer cells versus normal cells. The lab manager seems convinced that Protein X is the main cause of the cancer cell phenotype, but you are not so sure of that! They ask you to compare the cellular localization of Protein X in normal cells to cancer cells. How should you proceed?Choose one:Prepare fixed slides of normal cells and cancer cells, immunostain with antibody to Protein X, and then add a secondary fluorescent antibody that will bind to the primary antibody. View under a fluorescence microscope.Incubate normal cell lysates with red fluorophore and cancer cell lysates with green fluorophore, and perform 2-D PAGE.Prepare cell lysates from both normal cells and cancer cells, and perform an immunoprecipitation with antibody to Protein X. Then use a Western blot to detect differences.Perform column chromatography on lysates from normal cells and cancer cells

A: Prepare fixed slides of normal cells and cancer cells, immunostain with antibody to Protein X, and then add a secondary fluorescent antibody that will bind to the primary antibody. View under a fluorescence microscope.

Your lab manager was right! There is a difference in the localization: in normal cells, Protein X is located in the cytosol, but in cancer cells, Protein X seems to be in the nucleus. Now we have to confirm this biochemically. Which of the following seems like the best secondary approach?Choose one:Fractionate the normal cells and cancer cells using differential centrifugation to generate cytosolic and nuclear fractions. Then perform SDS-PAGE and Western blot using antibody specific for Protein X.Fractionate the cytosol and nucleus from the rest of the cell and perform fluorescence microscopy on the organelles using primary antibody to Protein X and fluorescence secondary antibody.Fractionate the normal cells and cancer cells using column chromatography to generate cytosolic and nuclear fractions. Then perform SDS-PAGE and Western blot using antibody specific for Protein X.Fractionate the normal cells and cancer cells using differential centrifugation to generate cytosolic and nuclear fractions. Then perform tandem mass spectrometry to determine the amino acid composition of Protein X.

A: Fractionate the normal cells and cancer cells using differential centrifugation to generate cytosolic and nuclear fractions. Then perform SDS-PAGE and Western blot using antibody specific for Protein X.

Applied biochemistry is used in which of the following industries?Choose one or more:A.agricultureB.biotechnologyC.clinical diagnosticsD.pharmaceuticals

ALL

Yeast produces alcohol during glucose metabolism, and this is how beer and wine are made. Bakers also use yeast in order to produce CO2, which makes dough rise. Which of the following is the most reasonable explanation of why people cannot get drunk from eating bread?Choose one:A. When baking bread, nearly all residual ethanol evaporates out of the bread, just as most of the water evaporates out of the bread.B. Bread bakers use a specific type of yeast that doesn't produce ethanol, and a different type of yeast for beer and wine making.C. During bread making, the cook adds a certain ingredient that deactivates alcohol dehydrogenase.D. During bread making, the cook adds a certain ingredient that deactivates pyruvate decarboxylase.

A. When baking bread, nearly all residual ethanol evaporates out of the bread, just as most of the water evaporates out of the bread.

Within a biomolecule, a carbon atom that is attached to four other atoms is said to have tetrahedral bond geometry around that carbon atom, and all bond angles are expected to be approximatelyChoose one:A. 104.5°.B. 120.0°.C. 90.00°.D. 109.5°.

D: 109.5

Considering the figure shown below (Figure 1.24 in the textbook), select two of the cellular processes that depend on interstrand base pairing between the nucleotide bases adenine and uracil. Choose one or more:A.mRNA translated into proteinB.DNA replicationC.DNA double helix formationD.mRNA synthesis (DNA transcription)

A and D, Translation and Transcription

Sections of DNA, called genes, must be transcribed into mRNA prior to translation into proteins. Considering the following short section of DNA, select the correct mRNA sequence that would be transcribed from this DNA.Coding strand: 5'-ATT CGG GGG CGA AAA CTT-3'Template strand: 3'-TAA GCC CCC GCT TTT GAA-5'Choose one:A. 5'-ATT CGG GGG CGA AAA CTT-3'B. 5'-TAA GCC CCC GCT TTT GAA-3'C. 5'-AUU CGG GGG CGA AAA CUU-3'D. 5'-UAA GCC CCC GCU UUU GAA-3'

C. 5'-AUU CGG GGG CGA AAA CUU-3

Which of the following strands of DNA would have the highest melting temperature? The melting temperature of double-stranded DNA is the temperature at which the double helix of the DNA will disassociate due to the hydrogen bonds between the base pairs disassociating.Choose one:A. 5'-ATT CGC CCG GCT-3'3'-TAA GCG GGC CGA-5'B. 5'-TAC TTA ATG CTT-3'3'-ATG AAT TAC GAA-5'C. 5'-ATC TTG TAT AAA-3'3'-TAG AAC ATA TTT-5'D. 5'-ATG CCG GCC GGG-3'3'-UAC GGC CGG CCC-5'

A. 5'-ATT CGC CCG GCT-3'3'-TAA GCG GGC CGA-5'

Throughout the study of molecular evolution and speciation, many examples of gene duplications have been uncovered. Within humans, two genes that have evolved from one another following a gene duplication are called BLANKs , and when comparing one of these genes in humans to its counterpart in chimpanzees, you are observing BLANK .

paralogs, orthologs

Which of the following three terms includes the other two?Choose one:A. homologousB. orthologousC. paralogous

A: homologous

Which of the following protein changes is most likely a result of the CF genetic mutation for the form of CF that is described here?Choose one:A. A polar amino acid is substituted by another polar amino acid.B. A negatively charged amino acid is substituted by a non-polar amino acid.C. A positively charged amino acid is substituted by a non-polar amino acid.D. A non-polar amino acid is substituted by another non-polar amino acid.

C. A positively charged amino acid is substituted by a non-polar amino acid.

These days, treatment of CF symptoms has allowed patients to survive much longer than in prior decades. Based on the information presented in the intitial passage, which of the following have likely become increasingly more common for CF patients? Choose one or more:A.lung transplantB.inhaled antibioticsC.gene therapy

ALL

Of all the different bond types found in proteins, which has partial double bond character?Choose one:A. A hydrogen bondB. An ionic bondC. A peptide bondD. A disulfide bond

C: peptide bond

What is a major consequence of the partial double bond character?Choose one:A. Greatly reduced rotational freedom compared to the φ (phi)/ψ (psi) torsion anglesB. Greatly increased rotational freedom compared to the φ (phi)/ψ (psi) torsion anglesC. Minimal difference in rotational freedom compared to the φ (phi)/ψ (psi) torsion anglesD. Unrestricted rotational freedom

A. Greatly reduced rotational freedom compared to the φ (phi)/ψ (psi) torsion angles

Which of the following best describes the relationship between D-amino acids and L-amino acids?Choose one:A. They are mirror images of each other.B. They are counterclockwise to each other.C. They are oppositely charged to each other.D. They are chemically equivalent.

A: they are mirror images of each other

Charged amino acids

Aspartate, Glutamate, Lysine, Arginine, histidine

Aromatic amino acids

phenylalanine, tyrosine, tryptophan

Which of the following amino acids in a protein would have an affinity for water?Choose one or more:leucinephenylalaninetyrosinethreonine

tyrosine and threonine

Which of the following amino acids is most likely to be found buried on the inside of a protein?Choose one or more:alaninelysineaspartatevaline

alanine and valine

Due to its partial double bond character, the peptide bond restricts rotation, constraining atoms to lie within the same peptide plane. Rotation in the peptide backbone is around the two bonds that flank adjacent BLANK atoms, between which a peptide plane is formed.

Ca