UW Biochemistry

how many Da in 1 amino acid

1 AA = 110 Da

protein degradation

1. Degradation by proteasomes -utilizes ubiquitin tags (ubiquitin-protein ligase recognizes them- destroys the protein- no lysosome is necessary here. Works via proteasome) 2. Degradation by lysosomes - secretory protein destroy by lysosome 3. Calcium-dependent enzymes

how many electrons do NADH carry

2; NADH is 2 electron carrier - FADH2 and UQH2 are also 2 electron carrier, but cytochrome c is 1 electron csrrier

Fermentation

A catabolic process that makes a limited amount of ATP from glucose without an electron transport chain and that produces a characteristic end product, such as ethyl alcohol or lactic acid. - the reduction of pyruvate to generate NAD+ for continued glycolysis under anaerobic conditions - convert pyruvate to lactate

Which of the following is most likely to be identified in SW-480 cells? A.An activating mutation in at least one K-ras allele B.Inactivating mutations in both K-ras alleles C.Decreased expression of K-ras protein D.Increased expression of hypoactive K-ras protein

According to the passage, the K-ras gene normally encodes a protein responsible for activating the signaling pathways that promote DNA replication (cell cycle S-phase) along with cell growth and survival. Mutated K-ras in SW-480 cells is most likely constantly inducing signaling pathways responsible for the characteristic uncontrolled growth, division, and survival of cancer cells. Therefore, K-ras is most likely a proto-oncogene with an activating mutation in at least one allele that turned it into an oncogene. choice C and D is wrong -> the mutated K-ras protein has either normal function with increased expression or hyperactive (not hypoactive) function with normal expression. correct answer is A

Hydrolysis

Breaking down complex molecules by the chemical addition of water

Based on Figure 1, which electron transport chain complex is inhibited by drug X? ** #6 on word doc A.Complex I B.Complex II C.Complex III D.Complex IV

Complex III facilitates consumption of UQH2 and cyt-Cox, producing UQ and cyt-Cred. If this complex is inhibited, the concentration of UQH2 will build as complexes I and II continue to produce it but Complex III fails to consume it. because cytochrome C is no longer being reduced, the reduced form will be depleted over time as complex IV continues to oxidize it. in summary,Reduced electron carriers upstream of the inhibited enzyme build up, and reduced electron carriers downstream are depleted. choice D is wrong -> if complex 4 is inhibited, reduced electron carriers would eventually build up b/c each will lose acceptors that will take its electrons correct answer is C

Acetylation of lysine residues in histones increases gene expression because: A.DNA is tightly bound to negatively charged amino acids on histones B.the carboxyl oxygen atoms in acetyl groups form hydrogen bonds with nitrogenous bases C.the salt bridges between charged amino acids and phosphate groups are disrupted D.lysine residues in histones associate with positively charged phosphate groups in DNA

DNA is predominately negatively charged d/t phosphate group it carries. histones forms salt bridge between positively charged amino acid and negatively charged phosphate group . this interaction cause histone bind tightly to DNA and prevent genes from being transcribed Acetylation of histones involves the transfer of acetyl groups from acetyl coenzyme A to positively charged amino groups on lysine or arginine residues. This modification disrupts salt bridges by reducing the positive charge on histones, which allows DNA to unwind and become more accessible to transcription machinery -> thus, acetylation increase gene expression choice A is wrong -> negatively charged amino acid residue will repel DNA choice B is wrong -> carboxyl oxygen atom form H bond with water choice D is wrong -> phosphate groups in DNA is negatively charged correct answer is C

parameter influence thermodynamic stability of the DNA duplex

DNA length -Longer DNA molecules take more time to both melt and reanneal. pH - Extreme changes in pH outside the physiological range lead to loss of hydrogen bonding and destabilize the DNA helix. Salt concentration (ionic strength) - High salt concentration of the solution increases double helix stability, but decreased salt concentration decreases stability.

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

Phosphorylation occurs at serine, threonine, and tyrosine residues, each of which contains a hydroxyl group in its side chain. Protein kinases catalyze the nucleophilic reaction in which the hydroxyl group on the amino acid side chain takes a phosphate group from ATP. To test the effect of post-translational modification, scientist replace with amino acid lack hydroxyl group correct answer is B

Based on the information in the passage, FCCP most likely causes a decrease in energy production by: A.decoupling the movement of protons down their concentration gradient from ATP synthase. B.carrying more protons into the intermembrane space than are pumped by the ETC alone. C.shifting the proton gradient further away from equilibrium than is achieved by the ETC alone D.causing a decrease in ETC activity, resulting in fewer available protons for ATP synthase

The addition of FCCP provides an alternate pathway to carry protons across the membrane from high concentration to low concentration. However, protons that enter the mitochondrial matrix by this means will not interact with ATP synthase, and therefore will not drive ATP synthesis. This phenomenon is called "decoupling." Because fewer protons are being used to drive ATP synthase, the rate of ATP synthesis will decrease. choice B is wrong -> carry more H to intermembrane space will actually increase ATP production, b/c more H will flow through ATP synthase down to mitochondria matrix correct answer is A

During prolonged fasting, which of the following liver enzymes has upregulated activity? A.Glycogen synthase B.Pyruvate carboxylase C.Glucokinase D.Phosphofructokinase

The first step in gluconeogenesis is carboxylation of pyruvate to form oxaloacetate. This step bypasses the irreversible pyruvate kinase reaction and is catalyzed by the enzyme pyruvate carboxylase choice C and D are wrong -> they participate in glycolysis correct answer is B

Thermophiles are microorganisms that thrive at temperatures above 40 °C. Which of the following characteristics LEAST accurately describes an enzyme isolated from a thermophile? A.Its optimal catalytic activity occurs above 40 °C.[4%] B.It is resistant to thermal denaturation.[21%] C.It has a relatively low catalytic rate at room temperature.[9%] D.It loses tertiary structure at temperatures below 37 °C.[63%]

Thermophilic enzymes must be resistant to thermal denaturation so that they can maintain their tertiary structures at the high temperatures of their natural environment Although enzymes from thermophiles typically operate slowly below 37 °C, this is most likely due to inflexibility of the active site, not denaturation. Accordingly, thermophilic enzymes most likely maintain their tertiary structure below 37 °C because molecular motion at this temperature is too slow to break intramolecular bonds. correct answer is D

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

Vmax = k_cat*[enzyme] if Kcat are the same for both isoform, it's likely isoform 2 is expressed in higher concentration so its activity is higher than isoform 1 correct answer is D

condensation reaction

a chemical reaction in which two or more molecules combine to produce water or another simple molecule

Each of the following molecules can be converted to oxaloacetate by a single enzymatic reaction EXCEPT: A.pyruvate B.malate C.alanine D.aspartate

alanine can be deaminated to pyruvate, which can then be converted to oxaloacetate, and this process will require 2 enzymatic steps instead of 1 choice A is wrong -> enzyme pyruvate carboxylase is used to converted pyruvate to oxaloacetate in gluconeogenesis => CO2 and ATP participate in the rxn choice B is wrong -> enzyme malate dehydrogenase is used to convert malate to oxaloacetate is the final step of citric acid cycle => NADH is produced choice D is wrong -> In amino acid degradation, amino acids are converted to α-keto acids by transamination, in which an amino group is transferred from an amino acid to α-ketoglutarate to synthesize glutamate (Choice D). The α-keto acid derivative of aspartate is oxaloacetate. correct answer is C

lipoprotein lipase (LPL)

an enzyme that hydrolyzes triglycerides passing by in the bloodstream and directs their parts into the cells, where they can be metabolized for energy or reassembled for storage. - necessary for the metabolism of chylomicrons and VLDL (very-low-density-lipoproteins)

Enzyme: low vs high temp

at high temperature -molecular motion speed up - intermolecular bonds that maintain tertiary structure are broken - protein denature that the shape of enzyme active sites are destroyed and halting catalytic activity at low temperature - molecular motion slow down - active site becomes inflexible - rxn rate slow down b/c active sites can't change shape to accommodate the transition site

An enzyme generates a Lineweaver-Burk plot with a slope of m. If the enzyme's Michaelis constant Km is known, which expression yields the maximum velocity Vmax of the enzyme-catalyzed reaction? A.Km/m[51%] B.Km × m[30%] C.m/Km[13%] D.Km + m[4%]

be careful with the axis if L-B plot x-axis is 1/S and y-axis is 1/V, while m is the slope: 1/V = (Km/Vmax)(1/S) + 1/Vmax m = (1/Vmax)/(1/Km) = Km/Vmax so, V = Km/m correct answer is A

mixed inhibitor

bind to ES complex and enzyme active site - effect on Km vary -> may decrease, increase, and no change on Km, depend on whether they bind to E (increase Km) or ES (decrease Km) - decrease Vmax

noncompetitive inhibitor

bind to ES complex and enzyme active site equally - no effect on Km - decrease Vmax

competitive inhibitor

bind to enzyme active site and prevent substrate from binding - increase Km -> higher substrate concentration is required to reach Vmax - no effect on Vmax

uncompetitive inhibitor

bind to enzyme-substrate complex, preventing enzyme from converting substrate to product - decrease Km -> inhibitor binds to ES complex decrease amount of ES in the solution and shift equilibrium toward more ES formation. This causes a decrease in the amount of substrate needed to achieve ½Vmax - decrease Vmax

Compound 1 would most likely have what effect on the activity of LAR? A.It would bind to an allosteric site and decrease the Km B.It would bind to the active site and decrease the Vmax C.It would bind to an allosteric site and increase the Vmax D.It would bind to the active site and increase the Km.

compound 1 and substrate have similar structure, so it's very likely it will compete with substrate by binding to active site -> compound 1 is competitive inhibitor => d/t the present of competitive inhibitor, amt of substrate needed is increased to achieve Vmax, so Km will increase and Vmax will not change correct answer is D

Which of the following most accurately describes the interaction between α-flag and flag-IRS4? A.A noncovalent protein-protein interaction B.A covalent protein-protein interaction C.A noncovalent protein-carbohydrate interaction D.A covalent protein-carbohydrate interaction

each antibody contains a variable region that binds to a chemical structure known as epitope in an antigen - > epitope is a sequence of amino acid w/n a protein or peptide -> antibodies bind to epitope through noncovalent interactions such as hydrogen bond or electrostatic attraction according to the passage α-flag is an antibody that binds to flag peptide, so flag is epitope (was this portion of flag-IRS4 to which the antibody noncovalently bound) correct answer is A

Activity assays were performed to determine the best purification method for a cell-free extract of Protein X. The results of the assays are shown below. ** plot is #4 in word doc Based on the data, what was the highest purification yield of Protein X? A.0.5% B.25% C.50% D.75%

extract are cells composed of protein, lipids, and carbohydrates, and to isolate a specific protein, use different methods, such as gel electrophoresis, precipitation, and column chromotagraphy -> the effectiveness of purification can be measured by performing activity assays that measure the rate of the reaction catalyzed by the enzyme of interest. Protein purity is represented as specific activity (u/mg) **y-axis on the plot**, which is a ratio of the activity of the target protein and the total protein obtained in a purification step. -> specific activity increases during purification The yield achieved by a purification step is determined by calculating the ratio of total activity of the purified protein to the unpurified extract. - yield = total activity of purified protein / total activity of unpurified extract to find the yield, need to calculate total activity of purified protein and extract total activity = total protein (mg) * specific activity (u/mg) total activity of purified protein = 200*1 = 200 u total activity of extract = 800*0.5 = 400 u yield=200/400 = 0.5 correct answer is C

Hormone Sensitive Lipase (HSL)

hydrolyze triacylglycerols to produce fatty acid and glycerols - stimulate by low insulins level, high epinephrine and cortisol level

what happened during fasting

in early stage of fasting, liver degrades glycogen to produce glucose -> glycogenolysis in prolonged fasting, glycogen stores are depleted, liver synthesizes glucose from smaller recursor molecules -> gluconeogenesis => use the same enzyme from glycolysis, except glucokinase, phosphofructokinase, and pyruvate kinase => glucose 6-phosphatase and fructose 1,6-bisphosphatase bypass irreversible steps

affinity columns

isolate molecules that bind a particular ligand.

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

k_cat = Vmax / [enzyme] Kcat for CD45 = 2.8/0.01 = 280 s^-1 Kcat for LAR = 2/0.05 = 40 s^1 CD45/LAR = 280/40 = 7 correct answer is D

turnover number (kcat)

measure the ability of a folded enzyme to catalyze a reaction -the number of reactions catalyzed per second per active site in solution

Proteins A, B, C, and D are dimers. They were run on a reducing SDS gel, shown below: #1 word doc What is the order of molecular weights of the proteins in their native state, from largest to smallest? A.Protein D → Protein C → Protein B → Protein A[5%] B.Protein C → Protein D → Protein A → Protein B[52%] C.Protein D → Protein B → Protein C → Protein A[33%] D.Protein C → Protein A → Protein D → Protein B[8%]

on reducing SDS gel, SDS causes protein to unfold and break disulfide bonds that link subunit together. subunits of different sizes are separated - homomultimer: if subunits are identical, they will migrate in the same distance and appears as a single band - heteromultimer: if subunits are different, they will migrate different distance according to molecular weight and appear as separate bands Band A (homomultimer) = 30*2=60 Band B (heteromultimer) = 30 + 25 = 55 Band C (homomultimer) = 40*2 = 80 Band D (heteromultimer) = 40 + 25 = 65 C >D >A > B correct answer is B

what kind of process is the first step of glycogenolysis

phosphorolysis; convert glycogen to glucose-1-phosphate with glycogen phosphorylase

Studies on the catalytic activity of hexokinase provided the following data: ** #7 on word doc A.Experiment I B.Experiment II C.Experiment III D.Experiment IV

protein denature can happen at high temperature and in the present of urea -> decrease rxn rate and small amt of folded protein production experiment 2 has the highest rxn rate, that indicates large amt of folded protein is produced protein unfolded -> lost fxn correct answer is B

The electron transport chain results in the generation of: A.protons, which produce an electrochemical gradient that drives phosphorylation of ADP B.protons, which reduce NAD+ in a coupling reaction that drives phosphorylation of ADP C.electrons, which produce an electrical gradient that drives phosphorylation of ADP D.electrons, which oxidize NADH in a coupling reaction that drives phosphorylation of ADP

protons are produced in a series of redox rxns in oxidative phosphorylation. excess energy can pump protons (H+) from the mitochondrial matrix to the intermembrane space. This establishes both a concentration and a charge gradient (collectively called an electrochemical gradient) across the membrane. The gradient drives the movement of protons from the intermembrane space (higher concentration) back into the matrix (lower concentration) through the ATP synthase complex - the movement of protons down electrochemical gradient provides energy needs to ATP synthase to produce ATP from ADP B is wrong - it's oxidation of NADH in electron transport chain, not reduction C and D are wrong - electrons are not generated; they are transferred from one chemical species to the other in sucssessful coupling rxns correct answer is A

An enzyme was assessed for its ability to cleave three substrates with similar structures, Compounds 1-3, as shown. **** #2 on word doc Based on the graph, which statement best describes the interaction of the enzyme with the compounds? The enzyme: A.is strongly inhibited by Compounds 1 and 3.[23%] B.is allosterically activated by Compound 2.[6%] C.has high specificity for Compound 2.[66%] D.has a high Hill coefficient for Compounds 1 and 3.[3%]

read the question carefully! it says Compound 1 and 3 are substrates, not inhibitor. even though their structures are similar to Compound 2, their Km is high, which means their affinity to active site is low. Some enzymes have high specificity for their substrates. These enzymes catalyze only reactions with a particular substrate and do not act on other molecules with similar, but not identical, structure. -> For example, the α-amylase enzyme cleaves the glycosidic bonds in starch but not cellulose, even though the two molecules differ only in the orientation of their glycosidic bonds. A high Hill coefficient indicates cooperativity and would result in a sigmoidal curve. Neither Compound 1 nor Compound 3 shows sigmoidal behavior. correct answer is C

Which of the following characteristics do the flag (DYKDDDDK) and myc (EQKLISEEDL) tags have in common? They have the same: A.number of nonpolar residues. B.binding interactions with α-flag C.number of acidic residues D.net charge at physiological pH

review amino acids Myc has net charge -3: 3 glutamates (E) = -3 1 asparate (D) = -1 1 lysine (K) = +1 Flag has net charge -3: 5 asparate (D) = -5 2 lysine (K) = +2

Certain RNA molecules can fold into structures known as hairpins, as depicted below. Which of the following RNA hairpins is the most stable? * see 3. in word doc. answer choices are all graphs

the hairpin contains the most bond will be the most stable is they have the same number of G-C bonds. Since choice C and D have the same number of G-C bonds, C will be more stable b/c their are 8 pair of bonds in its hairpin while D has 7 pair. correct answer is C

An enzyme kinetics assay reveals that Vmax = 100 nmol/s and Km = 10 μM. For this assay, what substrate concentration is required for a reaction rate Vo of 75 nmol/s? A.20 μM B.30 μM C.50 μM D.75 μM

use Michaelis-Menten equation to calculate Vo: Vo = Vmax*[S]/Km + [S] plug in all the number to find [S]: 75 = 100*S/10+S 750+75S=100S 25S=750 S=30 mM Correct answer is B

What are the physical states of wild-type (WT) and SW-480 K-ras DNA at 55°C? A.100% of SW-480 DNA is double-stranded and 50% of WT DNA is single-stranded B.100% of SW-480 DNA is single-stranded and 50% of WT DNA is single-stranded C.50% of SW-480 DNA is double-stranded and 100% of WT DNA is double-stranded D.50% of SW-480 DNA is single-stranded and 100% of WT DNA is double-stranded.

when double strand DNA is exposed to increasing temperature, double strands will be separated into 2 single strands -> The melting temperature (Tm), depicted as the midpoint of the melting curve, is achieved when 50% of double-stranded DNA has become single-stranded In Figure 2, both WT K-ras and SW-480 K-ras DNA have a maximal fluorescence of 0.04 at 20°C, which signals that all DNA molecules are double-stranded and bound by LC Green Plus. The fluorescence of WT K-ras DNA decreases to 0.02 at 55°C, which is the Tm of WT K-ras DNA and signifies that 50% of the WT DNA has melted and become single-stranded while the other 50% is double-stranded. However, mutant K-ras DNA from SW-480 cells reacts differently at 55°C; the fluorescence decreases to a value near zero, indicating that 100% of mutant K-ras DNA is single-stranded. correct answer is B