MCB Final Exam
Glucagon-like peptide-1 (GLP-1) is a hormone peptide secreted by intestinal L cells. It is secreted in response to the nutrient detection in the small intestine. The resulting physiological response is the lowering of blood glucose levels. Which of the following is a potential mechanism for how GLP-1 causes this physiological response? A. stimulation of glucagon release by pancreatic a cells B. stimulation of insulin release by pancreatic b cells C. stimulation of glucagon release by pancreatic b cells D. stimulation of insulin release by pancreatic a cells E. downregulation of GLUT4
. stimulation of insulin release by pancreatic b cells
complete oxidation of this 20-carbon fatty acid to CO2 and water?
134 ATP
The graph on the right illustrates how temperature affects the relative absorbance of DNA at 260 nm. At which point (A, B, C, or D), the DNA is mostly in the double-helical form?
A - absorbance is increased when single stranded
Tyrosine has three groups with acid/base properties, and their pKa values are 2.20 (alpha-carboxyl group), 9.11 (alpha-amino group), and 10.1 (the side chain). Calculate its isoelectric point. A) (9.11 + 2.20) / 2 B) (9.11 + 10.1) /2 C) (2.20 + 10.1) / 2 D) (2.20 + 9.11 + 10.1) / 3
A) (9.11 + 2.20) / 2
The isoelectric point (pI) of the peptide is: A) 3.6 B) 6.1 C) 6.5 D) 7.0
A) 3.6
The technique illustrates in the diagram on the right is referred to as _________, which is very useful in _________. A) 3C: chromosome conformation capture; mapping TADs B) recombinant DNA technology; cloning a gene C) Sanger sequencing; determining the sequence of DNA D) None of the above
A) 3C: chromosome conformation capture; mapping TADs Chromosome Conformation Capture: a technique that allows the mapping of neighborhood info in chromosome - covalently joined, digest non-cross linked, new DNA = joint DNA fragments, sequence DNA and reveal identity of the two portions and detect neighborhood patterns ** useful in mapping topologically associated domains (TADs)
Using the pKa values provided in the table to estimate the isoelectric point of the following peptide. A) 5.6 B) 7.0 C) 8.5 D) 11.0
A) 5.6
Ten micrograms of carbonic anhydrase (MW = 30000 g/mol) in the presence of large excess of substrate exhibits a reaction rate of 6820 micromol/min. At [S] = 0.012 M, the rate is 3410 micromol/min. From this information, you can infer that the Vmax of this enzymatic reaction is __________, and Km is _______. A) 6820 micromol/min; 0.012 M B) 10000 micromol/min; 0.024 M C) 3410 micromol/min; 0.012 M D) Impossible to know based on the information
A) 6820 micromol/min; 0.012 M
Consider a double-stranded DNA with the sequence of the top strand being 5'-ATCGGCTCTTATC-3'. Which of the following can lead to an increase in its Tm? A) Adding a few extra CG base-pairs (i.e., increasing the length of the DNA) B) Changing its sequence to 5'-ATAAGCTCTTTTA-3' C) Changing its sequence to 5'-ATCGGCTT-3' D) None of the above
A) Adding a few extra CG base-pairs (i.e., increasing the length of the DNA)
The three highlighted functional groups in the figure are: A) Amine, carboxyl, amide B) Amine, hydroxyl, ester C) Thiol, carboxyl, amide D) Amide, carboxyl, ester
A) Amine, carboxyl, amide
Consider a reaction catalyzed by a Michaelis-Menten enzyme. Using twice as much enzyme will lead to A) An increase in Vmax and no change in Km B) An increase in both Vmax and Kcat C) An increase in both Vmax and Km D) No change in Vmax and an increase in Km
A) An increase in Vmax and no change in Km Vmax = Kcat [Et] Vo=Vmax[S]/Km+[S] if increasing Et, vmax is increased.
Like proteins, DNA can also absorb ultraviolet light. Heating a double-stranded DNA can lead to A) An increase in the absorbance of ultraviolet light B) A decrease in the absorbance of ultraviolet light C) No change in the absorbance of ultraviolet light D) A drastic decrease in the absorbance of ultraviolet light
A) An increase in the absorbance of ultraviolet light
Angiotensin II, endothelin, bradykinin, and vasopressin are four important peptides that control the tone of blood vessels in our body. These peptides have very different isoelectric points with 6.7 for angiotensin II, 9.4 for endothelin, 12.0 for bradykinin, and 10.9 for vasopressin. Among these four peptides, at pH 7, which of them comes out of a cation-exchange column the fastest? (Note that the solid support for cation-exchange column carries negative charges at pH 7). A) Angiotensin II B) Endothelin C) Bradykinin D) Vasopressin
A) Angiotensin II Cation exchange column binds positive charges, and therefore positive charges will take longer to leave the column. Angiotensin II will act as an acid and donate a proton, giving it a negative charge. It will be repelled form the negative resins and leave the column the fastest.
While many biochemists use the term "aspartate" and "aspartic acid" interchangeably, they do refer to different things. Which of these terms is true? A) Aspartate is the conjugate base of aspartic acid B) Aspartate is converted to aspartic acid upon increase in pH C) Aspartic acid is converted to aspartate upon decrease in pH D) All of the above
A) Aspartate is the conjugate base of aspartic acid
What does amino acid activation by amino acyl-tRNA synthetases have in common with ubiquitin activation by the ubiquitin activating protein (E1)? A) Both amino acids and ubiquitin are activated by adenylation followed by releasing AMP B) Both amino acids and ubiquitin are activated by ATP hydrolysis to generate ADP + Pi C) Both amino acids and ubiquitin are linked through thioester bonds to their substrates D) Both amino acids and ubiquitin are activated by GTP hydrolysis using specific protein factors E) Both amino acids and ubiquitin are linked through isopeptide bonds to their substrates
A) Both amino acids and ubiquitin are activated by adenylation followed by releasing AMP
Action of histone acetyl transferase (HAT) could lead to the formation of a binding site on histone that allows the recruitment of proteins containing this domain: _______________ A) Bromodomain B) Chromodomain C) Plant homeodomain D) SANT domain
A) Bromodomain * The PHD domain refers to plant homeodomain and it forms specific interaction with methylated lysine and it is often found in chromatin remodeling complex. * Bromodomain (recognize acetyl-Lys) * Chromodomain (recognize methyl-Lys) * TUDOR domain (recognize methyl-Lys) * SANT domain (recognize unmodified histone tails)
Methotrexate is a commonly used anti-cancer drug, which acts by inhibiting the enzyme dihydrofolate reductase (DHFR), an enzyme that is responsible for the regeneration of tetrahydrofolate from dihydrofolate. Notably, methotrexate has a structure similar to that of the folate. Thus, most likely methotrexate acts as a(n) __________ for DHFR. A) Competitive inhibitor B) Uncompetitive inhibitor C) Noncompetitive inhibitor D) Irreversible inhibitor
A) Competitive inhibitor
H2A.X is a histone variant that is important for A) DNA repair B) DNA replication C) centromere function D) Sliding camp binding
A) DNA repair H2A.X: Important for DNA repair; becomes phosphorylated upon DNA double-stranded break which helps to recruit DNA repair enzymes (signal). Without signal, cell would not recognize the break. CENP-A (an H3 variant): important for centromere function ** CENP-A allows the attachment of the mitotic spindle to the kinetochore - mainly found in centromere - allows centromere to bind to kinetochore
The segment of nucleic acid shown here is _________, and its sequence is ______________. A) DNA, 5'-GTAC-3' B) DNA, 5'-ATGC-3' C) DNA, 5'-CATG-3' D) RNA, 5'-GUAC-3'
A) DNA, 5'-GTAC-3'
A mixture of four proteins - G protein (92 kDa), GPCR (50 kDa), MAPK (35 kDa), and p53 (53 kDa) - is applied to a gel-filtration column; the protein that comes out of the column first (the earliest) is A) G protein B) GPCR C) MAPK D) p53
A) G protein * a type of column chromatography that separates proteins based on their size using size-exclusion beads; also called size-exclusion chromatography * larger molecules move more rapidly because they don't channel through the pores and therefore emerge from the column filter first * precisely when the molecule comes out depends on its molecular weight * quaternary structure stays intact (native state remains)
Ubp3 is a deubiquitinating enzyme that contains a histidine residue in its active site. In the process of catalysis, the side chain of histidine accepts a proton from the substrate. As such, which of the following catalytic strategies is in play here? A) General acid/base catalysis B) Covalent catalysis C) Metal ion catalysis D) Catalysis by approximation
A) General acid/base catalysis General acid/base catalysis * a non-water molecule plays the role of a proton donor or acceptor * in enzymes, residues with ionizable side chains such as His often serve this role * pH dependence of an enzyme is partially determined by the ionization of the catalytic group covalent catalysis * the active site contains a reactive group that becomes temporarily covalently attached to the substrate in the course of catalysis * eventually this covalent bond will be resolved catalysis by approximation * relevant when a reaction involves more than one substrate * substrates are brought together in favorable orientation metal ion catalysis * metal ions may facilitate the formation of a nucleophile * may serve as an electrophile
HDACs are enzymes that catalyze the removal of acetylation on histone proteins. Which of the following is a most likely relationship between HDAC and gene expression? A) HDAC likely acts as a gene regulator that impairs gene expression B) HDAC likely acts as a gene activator that enhances gene expression C) HDAC is likely very abundant in chromatin region with actively transcribed genes D) HDAC likely promotes the chromatin structure in a more open form
A) HDAC likely acts as a gene regulator that impairs gene expression
In the diagram of the first stage of glycolysis shown below, identify an enzyme that catalyzes a group transfer reaction. A) Hexokinase B) Phosphoglucose isomerase C) Fructose bisphosphate aldose D)Triose phosphate isomerase
A) Hexokinase
Hexokinase is a low Km enzyme while glucokinase is a high Km enzyme. We know that both can catalyze the conversion of glucose to glucose 6-phosphate. Which of the following is correct? A) Hexokinase has a higher affinity to glucose than glucokinase B) Hexokinase acts at a higher rate after one's blood glucose level increase to a much higher level C) Glucokinase has a higher affinity to glucose D) Converting glucose to glucose 6-phosphate by glucokinase mainly serves for glycolysis
A) Hexokinase has a higher affinity to glucose than glucokinase
A mixture of four proteins - Histone H1 (pI 10.7), p53 (pI 6.83), Ras (pI 6.33), and trypsinogen (pI 9.3) - is applied to a cation-exchange column (i.e., the matrix is composed of beads that contain negative charges) at pH 7. Which of the proteins will come out of the column last (the latest)? A) Histone H1 B) p53 C) Ras D) trypsinogen
A) Histone H1
Human IgG is composed of two heavy chains (~53 kD each) and two light chains (~23 kD each). These four chains are linked via inter-chain disulfide bonds. Which of the following is true regarding analyzing human IgG using SDS-PAGE? A) In the presence of beta-mercaptoethanol, two bands (~53 kD and ~23 kD) will show up on the gel B) In the absence of beta-mercaptoethanol, two bands (~53 kD and ~23 kD) will show up on the gel C) A single band with a size of ~76 kD will show up D) A single band with a size of ~106 kD will show up
A) In the presence of beta-mercaptoethanol, two bands (~53 kD and ~23 kD) will show up on the gel
As shown on the graph, increasing temperature results in an increased relative absorbance of DNA at 260 nm. Why it that? A) Increasing temperature leads to the formation of single stranded DNA, which has higher absorbance at 260 nm than double stranded DNA B) Increasing temperature promotes the formation of double stranded DNA, which has higher absorbance at 260 nm C) Increasing temperature removes protein contaminants D) Increasing temperature fragments DNA strand, and smaller fragments tend to have higher absorbance at 260 nm
A) Increasing temperature leads to the formation of single stranded DNA, which has higher absorbance at 260 nm than double stranded DNA hypochromism: * stacked bases in nucleic acid absorb less UV light than do unstacked bases * absorbance is increased when single stranded * 260 nm absorbance for DNA compared with 280 nm of proteins
The kinetic parameters of several isoenzymes are shown. Assuming the same isoenzyme concentration, which isoenzyme has the highest catalytic efficiency? A) Isoenzyme I B) Isoenzyme II C) Isoenzyme III D) Isoenzyme IV
A) Isoenzyme I Catalytic Efficiency: Kcat/Km Vmax = Kcat [ET]
The peptide shown below is a part of palmitoyl-pentapeptide-3, a component of antiaging creams. The sequence of this pentapeptide is __________________. If this peptide is going to form an alpha-helix structure, how many intra-strand backbone hydrogen bonds can it form? A) KTTKS; 1 B) KSSKT; 3 C) KSSKT; 1 D) KTTKS; 3
A) KTTKS; 1 A spiral shape constituting one form of the secondary structure of proteins, arising from a specific hydrogen-bonding structure. Hydrogen bonding occurs 4 residues apart. All groups except C-term and N-term are participating in H-bonding. That makes the two ends highly polar surfaces. INTERIOR backbone and side chains EXTEND OUT3.6 residues per turn height of each turn = 0.54 nm = 5.4 angstroms stabilized by H-bonding between the CO group of residue i and the NH group of residue i+4 HIGHLY restricted structure
The amino acid shown here is called _____________, and it has _______ chiral center(s) at __________. A) L-threonine; 2; carbon a and carbon b B) D-threonine; 2; carbon a and carbon b C) L-threonine; 1; carbon a D) L-serine; 2; carbon a and carbon b
A) L-threonine; 2; carbon a and carbon b
Lys and Arg are very abundant in histones, which are proteins that bind and organize DNA in the nucleus. What makes Lys and Arg very appropriate as the major components of histones? A) Lys and Arg have basic side chains that carry positive charges at physiological pH, thus they are very suitable in interacting with DNA B) The side chains of Lys and Arg have the buffering capacity to keep the pH around DNA to be neutral C) The side chains of Lys and Arg can absorb ultraviolet light, so they can prevent UV damage to DNA D) All of the above
A) Lys and Arg have basic side chains that carry positive charges at physiological pH, thus they are very suitable in interacting with DNA
For this question, refer to the genetic code chart on the right and the DNA sequence below. Assuming that the top strand of DNA is the coding strand, what amino acid sequence would you predict to be translated after transcription of the template strand? A) Met-Phe-Tyr-Cys B) Tyr-Lys-Met-Thr-Thr C) Ser-Ala-Val-Lys-His D) Ser-Arg-His-Phe-Val
A) Met-Phe-Tyr-Cys
Changes in pH can alter the charge status of protein molecules in aqueous solution. For example, as pH decreases, protein molecules in aqueous solution tend to become A) More protonated and more positively charged B) Less protonated and more negatively charged C) More stable and active D) None of the above
A) More protonated and more positively charged
Consider the peptide sequence, ILWANRMSHVLFAVEA, which amino acid residues would you expect to be on the solvent-exposed surface once it folds into its native conformation? A) N, R, S, H, E B) L, W, M, V, I C) N, W, S, H, L D) L, A, N, M, E
A) N, R, S, H, E
The figure shown on the right illustrates the relative position of an alpha helix to the heme group in hemoglobin. Which of the following statements is true regarding the two labeled proteins - position 1 and position 2 - of this helix? A) Position 1 is found in the R state of hemoglobin B) Position 2 is found in the R state of hemoglobin C) a shift from position 1 to 2 is induced by oxygen binding to the heme D) the shifting between these two positions is not influenced by oxygen binding
A) Position 1 is found in the R state of hemoglobin Hemoglobin: - an allosteric protein with quaternary structure - each subunit (4) has an active binding site - a RBC protein that carries oxygen from the lungs to the tissues - a tetramer consisting of two alpha subunits and two beta subunits (each subunit has a bound heme group) - hemoglobin is an allosteric protein that displays cooperativity in oxygen binding and release Oxygen binding changes the position of the iron atom. In deoxyhemoglobin: - the absence of oxygen binding - iron is positioned somewhat below the plane - proximal histidine bonded to 5th coordination site In oxyhemoglobin: - oxygen binding to the sixth coordination site - this oxygen binding changes the electron distribution around iron and lifts iron up by .4angstroms. - iron becomes effectively smaller - O2 exists in resonance with Fe2+, allowing the formation of superoxide ion with a negative charge on O and Fe3+. This is a highly reactive ion. Therefore, oxygen is negative charge state forms hydrogen bond with histidine to lock Fe3+ state, preventing Fe3+ O=O - reactive superoxide ion from leaking
Consider the properties of the following 4 proteins. If they are subjected to gel-filtration, which protein will elute out from the column last? A) Protein 1: MW 40 kD, pI 9.0 B) Protein 3: MW 70 kD, pI 7.0 C) Protein 2: MW 45 kD, pI 8.5 D) Protein 4: MW 100 kD, pI 5.0
A) Protein 1: MW 40 kD, pI 9.0 larger MW elute first
Which statement about eukaryotic versus prokaryotic gene regulation is NOT correct? A) Strong promoters in eukaryotes are generally fully active in the absence of regulatory proteins B) Access to eukaryotic promoters is restricted by the structure of chromatin C) Most regulation in eukaryotes is positive, involving activators and coactivators, rather than using negative regulation involving repressors as seen commonly in prokaryotes D) In eukaryotes, transcription and translation are separated by space and time, while in prokaryotes, transcription and translation happen simultaneously E) Larger and more multimeric proteins (complexes) are involved in regulation of eukaryotic genes than in regulation of prokaryotic genes
A) Strong promoters in eukaryotes are generally fully active in the absence of regulatory proteins
The data shown here were collected for a reaction catalyzed by the enzyme ABC. The same reaction can also be catalyzed by another enzyme called XYZ. The Km value of enzyme XYZ is 2 mM. Which of the following statements is likely to be true regarding the enzymes ABC and XYZ? A) The ABC enzyme binds substrate better than the XYZ enzyme B) The ABC enzyme has a higher Vmax than the XYZ enzyme C) The XYZ enzyme binds substrate better than the ABC enzyme D) Both enzymes are allosteric and do not follow Michaelis-Menten kinetics
A) The ABC enzyme binds substrate better than the XYZ enzyme Given the data, can be concluded that the Km value of ABC = 0.2 mM. Therefore, it has a higher affinity for the substrate.
We mentioned in class several times that triose phosphate isomerase (i.e., TIM) is one of the enzymes that have reached catalytic perfection. As such, we can assume that TIM has a very high value of A) The ratio of Kcat over Km B) The ratio of Km over Kcat C) Km D) Vmax
A) The ratio of Kcat over Km ** want Kcat to be very large and Km to be very small ** has to be the ratio that is ideal, not either of them individually
In the active site of lysozyme, there are two important chemical groups: -COO- (carboxylate) from the side chain of Asp52, and -COOH (carboxyl) from the side chain of Glu35. What sorts of catalytic strategies are executed by these two chemical groups? A) The side chain of Asp52 (with -COO-) performs covalent catalysis, and that of Glu35 (with -COOH) performs general acid catalysis B) The side chain of Asp52 (with -COO-) performs general base catalysis, and that of Glu35 (with -COOH) performs covalent catalysis C) Both side chains perform metal ion catalysis D) None of the above
A) The side chain of Asp52 (with -COO-) performs covalent catalysis, and that of Glu35 (with -COOH) performs general acid catalysis
Which of the following statements about functional tRNAs is correct? A) They contain many modified nucleotides. B) All of their nucleotides are in base-paired helical regions. C) They contain more than 1000 ribonucleotides. D) They have a terminal AAC sequence at their amino acid accepting end. E) They consist of two helical stems that are joined by loops to form a U-shaped tertiary structure.
A) They contain many modified nucleotides
Suppose you have recently purified a brand-new protein with a molecular weight of 250 kD, as measured by gel-filtration. Running this protein on a SDS-PAGE in the absence of beta-mercaptoethanol resulted in a single band with a molecular weight of 125 kD, while running this protein on a SDS-PAGE in the presence of beta-mercaptoethanol resulted in two bands with molecular weights of 100 kD and 25 kD, respectively. Which of the following is the most possible composition of this new protein? A) This protein is a tetramer: two long chains with a molecular weight of 100 kD, and two short chains with a molecular weight of 25 kD. Each long chain is connected with a short chain by a disulfide bond. B) This protein is a dimer: each subunit has a molecular weight of 125 kD C) This protein has a total of seven subunits: one with a size of 100 kD, and others have a size of 25 kD D) Not enough information to figure out the composition of this new protein: crystal structure is needed
A) This protein is a tetramer: two long chains with a molecular weight of 100 kD, and two short chains with a molecular weight of 25 kD. Each long chain is connected with a short chain by a disulfide bond.
Branched chain amino acids are essential amino acids for humans, meaning we have to obtain them from our diet. Which of the following is a group of branched chain amino acids? A) Val; Ile; Thr B) Ser; Lys; Ile C) Val; Pro; Leu D) Arg; Ser; Glu
A) Val; Ile; Thr
The sequence of the peptide shown in the question above is: A) YGGFL B) YGFL C) FGGYL D) YGGFI
A) YGGFL
What is a key function of the 23S rRNA in prokaryotic translation that is also true of the 28S rRNA in eukaryotic translation? A) catalyzes the peptide bond formation B) binds to mRNA C) forms the binding sites for the tRNA molecules D) recruits the translation initiation factors E) contains the anticodons to base pair with the codons
A) catalyzes the peptide bond formation
Many gene regulatory proteins contain specific DNA-binding motifs in their DNA-binding domains. In prokaryotes, what is the name of the most common DNA-binding motif, shown to the right? A) helix-turn-helix B) zinc finger C) leucine zipper D) beta sheet motif E) homeodomain
A) helix-turn-helix
A string of histidines (HHHHHH) is often added to the end of a protein because A) it serves as an affinity tag to allow easy purification of the protein B) it creates a unique active site for the protein to catalyze chemical reactions C) it generates a fusion protein that allows the protein to pass the lipid bilayer D) it adds positive charges to the protein so it can bind more tightly to an anion exchange column
A) it serves as an affinity tag to allow easy purification of the protein Inserting a piece of DNA that encodes a tag of choice to the gene that encodes a protein of interest. As a result, a fusion protein is produced that contains the entire protein of interest plus the tag. The added tag now allows easy purification and/or detection of the protein of interest. Histidine residues (HHHHHH) = "his tag" --> ligand = metal ions
Regulation of prokaryotic gene expression is mainly controlled by rapid responses to environmental signals. Which regulatory strategy below that we discussed in class is an example of inducible gene expression? A) lac repressor action on the lac operon B) lambda repressor action in controlling lambda lysogenic phase C) cro repressor action in controlling lambda lytic phase D) catabolite activator protein (aka CAP & CRP) action on the lac operon E) quorum sensing by biofilms
A) lac repressor action on the lac operon
Consider two peptides: peptide A has a pI of 4.1, and peptide B has a pI of 9.1. At which of the following conditions, both of them are capable of binding to a cation-exchange column? Note a cation-exchange column is packed with matrix materials that carry negative charges. A) pH = 2.1 B) pH = 4.1 C) pH = 9.1 D) pH = 10.1
A) pH=2.1 This way, both peptides will act as the base and accept a proton (become + charged) and bind to the negatively charged resins cation exchangers - in ion exchange chromatography: negatively charged resin, binds positively charged molecules (cations) and pushes negatively charged molecules quickly through
The information that can be gained from a Western blotting experiment include A) the presence of a protein of interest in the sample and the estimation of its molecular weight B) the abundance of a protein of interest in the sample and the estimation of its isoelectric point C) the presence of a specific DNA fragment in the sample D) the presence of a specific mRNA in the sample
A) the presence of a protein of interest in the sample and the estimation of its molecular weight This is because SDS PAGE is used in Western Blotting so it gives everything a negative charge, so not pI
Suppose that both enzymes ABC and XYZ are capable of catalyzing the reaction of phosphorylating glucose to produce glucose-6-phosphate. With glucose as a substrate, the enzyme ABC has a Km of 0.2mM and the enzyme XYZ has a Km of 10mM. With glucose as a substrate, the enzyme ABC A) the rate of ABC enzyme is less impacted by the fluctuation of blood glucose level B) ABC enzyme is a more efficient enzyme C) ABC enzyme has a lower affinity for glucose D) XYZ enzyme is a less efficient enzyme
A) the rate of ABC enzyme is less impacted by the fluctuation of blood glucose level
Out of the five regulatory strategies for fatty acid metabolism listed below, identify the regulatory strategy that would result in activation of fatty acid oxidation & inhibition of fatty acid synthesis. A. AMP activated protein kinase (AMPK) phosphorylation of acetyl CoA carboxylase B. Citrate inducing polymerization of acetyl CoA carboxylase C. Malonyl CoA inhibition of carnitine acyl transferase 1 D. Insulin signaling in response to high glucose levels E. Protein phosphatase 2A dephosphorylation of acetyl CoA carboxylase
A. AMP activated protein kinase (AMPK) phosphorylation of acetyl CoA carboxylase
The table to the right shows the four common "fad diets". These fad diets may lead to biochemical outcomes that are unfavorable. Which diet most likely could lead to ketoacidosis? A. Atkins diet B. Zone diet C. Ornish diet D. Weight Watchers E. None of these could lead to ketoacidosis.
A. Atkins diet
What does amino acid activation by amino acyl-tRNA synthetases have in common with ubiquitin activation by the ubiquitin activating protein (E1)? A. Both amino acids and ubiquitin are activated by adenylation followed by releasing AMP B. Both amino acids and ubiquitin are activated by ATP hydrolysis to generate ADP + Pi C. Both amino acids and ubiquitin are linked through thioester bonds to their substrates D. Both amino acids and ubiquitin are activated by GTP hydrolysis using specific protein factors E. Both amino acids and ubiquitin are linked through isopeptide bonds to their substrat
A. Both amino acids and ubiquitin are activated by adenylation followed by releasing AMP
What does amino acid activation by amino acyl-tRNA synthetases have in common with ubiquitin activation by E1? A. Both amino acids and ubiquitin are activated by adenylation followed by releasing AMP B. Both amino acids and ubiquitin are activated by ATP hydrolysis to generate ADP + Pi C. Both amino acids and ubiquitin are linked through thioester bonds to their substrates D. Both amino acids and ubiquitin are activated by GTP hydrolysis using specific protein factors E. Both amino acids and ubiquitin do not need to be activated for ligations to occur
A. Both amino acids and ubiquitin are activated by adenylation followed by releasing AMP Transfer RNAs: tRNAs are the adaptor molecules in protein synthesis - tRNA molecules react with specific amino acids in a reaction catalyzed by aminoacyl-tRNA synthetases - tRNA molecules also contain a template recognition site, called the anticodon- the anticodon, which consists of three bases, recognizes a complementary 3-base sequence in the mRNA called the codon Aminoacyl-tRNA attached the correct amino acid to the tRNA aminoacyl-tRNA synthetases catalyze the activation of amino acids by forming an ester linkage between the carboxyl group of the amino acid and either the 2' or 3' OH of the terminal adenosine of the tRNA, forming an aminoacyl-tRNA or charged tRNA. *** adenylation Amino acid + ATP --> aminoacyl-AMP + PPi --> aminoacyl-AMP + tRNA --> aminoacyl-tRNA + AMP overall reaction utilizes equivalent of 2 ATP because 2 phosphoanhydride bonds are broken Amino acid + ATP + tRNA + H20 --> aminoacyl-tRNA + AMP + 2Pi Three enzymes are required for ubiquitin (Ub) conjugation: 1. Ubiquitin-activating enzyme (E1) adenlyates Ub and transfers it to a cysteine residue on E1. - cell has lots of the same E1 and E2 but lots of different E3. 2. Ubiquitin-conjugating enzyme (E2) then transfers Ub to one of its own cysteine residues. ** puts it in position for E3 3. Ubiquitin-protein ligase (E3), using the E2-Ub complex as a substrate, transfers the Ub to the target protein. (ligase binding to ubiquitin and target protein and therefore must be specific) - E1 covalently attached ubiquitin by a thioester bond. E2 transfers the thioester linkage from E1 to itself. E2 binds to one of the active sites on E3. E3 recognizes the target and through ligation, transfers the ubiquitin on the thioester linkage onto the target protein using the C terminus of ubiquitin and the side chain of lysine on target protein. E2 is regenerated and will leave and the polyubiquitinated target protein is released. Ubiquitin targets the protein to the proteasome and it is recognized by the regulatory subunit of proteasome. It is then threaded through and chopped up, very ATP hydrolysis heavy. * very regulated process.
The most significant influence on why mRNA is processed differently in prokaryotes than eukaryotes is the fact that: A. Eukaryotes separate transcription and translation with a nucleus. B. Prokaryotic transcripts are often polycistronic. C. Eukaryotes are multicellular organisms. D. Eukaryotic transcripts are often monocistronic. E. Prokaryotes use formyl-methionine-tRNA as the initiator tRNA for translation.
A. Eukaryotes separate transcription and translation with a nucleus.
Which of the following features characterize the 26S proteasome? A. It consists of a 20S catalytic subunit and a 19S regulatory subunit B. It releases only free amino acids as products C. It digests ubiquitin along with the protein to which it is attached D. It contains no ATPase activity
A. It consists of a 20S catalytic subunit and a 19S regulatory subunit The proteasome digests ubiquitin-tagged proteins: The proteasome is a large molecular complex of proteolytic enzymes that degrades ubiquitinated proteins. The 26S proteasome consists of two components: a 20S catalytic subunit and a 19S regulatory subunit. The regulatory subunit binds to polyubiquitin, cleaves off intact ubiquitin, unfolds the condemned protein, and inserts it into the catalytic subunit. Ubiquitinated proteins are processed to peptide fragments from which the ubiquitin is subsequently removed and recycled. The peptide fragments are further digested to yield free amino acids, which can be used for biosynthesis reactions, most notably protein synthesis. The regulatory subunits recognize tagged protein.
Refer to the figures below, which depict a readout using quantitative PCR (qPCR). Which of the following statements is true of the value CT? A. It is the cycle number at which the fluorescence becomes detectable. B. It is directly proportional to the number of copies of the original template. C. It is the number of copies of the original template. D. All of the above are true.
A. It is the cycle number at which the fluorescence becomes detectable.
In eukaryotes, which polymerase is responsible for synthesizing most of the rRNA genes, including the 18S rRNA and 28S rRNA genes? A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III D. DNA polymerase a E. DNA polymerase d
A. RNA polymerase I
To the right is a typical promoter found in eukaryotic rRNA genes. This promoter MOST LIKELY directs the binding of transcription factors to recruit which RNA polymerase? A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III
A. RNA polymerase I rRNA is transcribed by Pol I mRNA is transcribed by Pol II tRNA is transcribed by Pol III
The phosphorylation of several RNA polymerase II serine residues on its C-terminal domain (CTD) has been identified as a key event in eukaryotic transcription initiation. Which of the following is one of the key enzymes responsible for this phosphorylation? A. TFIIH B. TFIIF C. TFIID D. TATA box binding protein (TBP) E. Histone kinase
A. TFIIH
What would happen to the electron transport chain (ETC) and oxidative phosphorylation pathway in the presence of excess NADH if the mitochondrial matrix were not closed, but opened up to the cytoplasm? A. The ETC complexes would function as normal, but no ATP would be made. B. The ETC and synthesis of ATP would continue as normal. C. The ETC complexes would transfer electrons from NADH to O2, but no protons would be pumped. D. The NADH would react directly with O2, generating excess heat. E. ATP would be made, but the ETC complexes would no longer function.
A. The ETC complexes would function as normal, but no ATP would be made.
What are the roles of the core RNA polymerase and sigma factor in prokaryotes? A. The core RNA polymerase recognizes the promoter, while the sigma factor performs the catalysis of RNA synthesis. B. The core RNA polymerase performs the catalysis of RNA synthesis, while the sigma factor recognizes the promoter. C. The core RNA polymerase, along with the sigma factor, only act to recognize the promoter. D. The core RNA polymerase, along with the sigma factor, only act to perform the catalysis of RNA synthesis. E. The core RNA polymerase, along with the sigma factor, work together to recognize the promoter as well as to perform the catalysis of RNA synthesis.
A. The core RNA polymerase recognizes the promoter, while the sigma factor performs the catalysis of RNA synthesis. Prokaryotic RNA Polymerase Structure RNA polymerase is composed of 5 subunits. Core enzyme: required for polymerization activity (catalysis: making RNA) - composed of 2alpha, 1beta, 1beta', and 1omega Sigma factor: required for correct initiation of transcription: binding to promoter (can associate and dissociate) - required to recognize the promoter Holoenzyme: composed of the core enzyme plus the sigma factor * The sigma subunit of bacterial RNA polymerase can dissociate from the core enzyme. * The sigma subunit of bacterial RNA polymerases helps polymerase find a site where transcription begins, participates in the initiation of RNA synthesis, and then dissociated from the rest of the enzyme. * The holoenzyme of bacterial RNA polymerases initiates transcription, while the core enzyme of bacterial RNA elongates RNA
Which of the following describes an accurate order of events in translation? A. The first (initiator) aminoacyl-tRNA molecule binds to the small subunit before the large subunit binds. B. Initiation of translation occurs only after the small subunit dissociates from the mRNA. C. Ribosomal translocation occurs before the first (initiator) aminoacyl-tRNA molecule binds to the small subunit. D. The aminoacyl-tRNA synthetase binds to the ribosome. E. The large subunit of the ribosome binds the mRNA before the small subunit.
A. The first (initiator) aminoacyl-tRNA molecule binds to the small subunit before the large subunit binds.
Which of the following is true of the mechanism of attenuation of the trp operon? A. The stalling of the ribosome at the adjacent Trp codons of the leader sequence inhibits the formation of the terminator structure. B. When Trp binds to the leader sequence, a riboswitch is triggered, terminating transcription prematurely. C. The trp repressor binds to the attenuator sequence instead of the operator sequence. D. The ribosome only stalls at the adjacent Trp codons when the levels of tryptophan in the cell are high. E. The leader peptide mechanism is a repressor protein but requires Trp to bind the repressor to dissociate from the DNA, using an inducible mechanism.
A. The stalling of the ribosome at the adjacent Trp codons of the leader sequence inhibits the formation of the terminator structure.
The sequence of a duplex DNA segment in a longer DNA molecule is: 5'-ATCGCTTGTTCGGA-3' 3'-TAGCGAACAAGCCT-5' When this segment serves as a template for RNA polymerase, it gives rise to a segment of RNA with the sequence 5'-UCCGAACAAGCGAU-3'. Which of the following statements about the DNA segment are correct? A. The top strand is the template strand. B. The top strand is the coding strand. C. The bottom strand is the antisense (-) strand. D. The bottom strand is the template strand. E. The top strand is the sense strand.
A. The top strand is the template strand.
Which of the following statements about functional tRNAs is correct? A. They contain many modified nucleotides. B. All of their nucleotides are in base-paired helical regions. C. They contain more than 1000 ribonucleotides. D. They have a terminal AAC sequence at their amino acid accepting end. E. They consist of two helical stems that are joined by loops to form a U-shaped tertiary structure.
A. They contain many modified nucleotides.
The catalytic headpiece of the ATP synthase enzyme is primarily composed of which subunits? A. a hexameric α,3β3 ring B. a circle of 8-12 c subunits C. the F0 subunit D. γ and ε subunits E. the a subunit
A. a hexameric α,3β3 ring
Insulin stimulates a signaling cascade that leads to the _______________ of the bifunctional protein phosphofructokinase 2/ fructose bisphosphatase 2, which results in the activation of ______________. A. dephosphorylation; glycolysis B. phosphorylation; glycolysis C. dephosphorylation; gluconeogenesis D. phosphorylation; gluconeogenesis E. ubiquination; glycolysis
A. dephosphorylation; glycolysis
Fructose can enter glycolysis at two distinct points, depending on the tissue. In adipose tissue, fructose is directly converted to __________, while in the liver, fructose is directly converted to ____________. A. fructose-6-phosphate; fructose-1-phosphate B. fructose-1-phosphate; fructose-6-phosphate C. fructose-6-phosphate; glyceraldehyde 3-phosphate (GAP) D. fructose-6-phosphate; dihydroxyacetone phosphate (DHAP) E. fructose-1-phosphate; glyceraldehyde 3-phosphate (GAP)
A. fructose-6-phosphate; fructose-1-phosphate
To what part of the target protein do molecular chaperones bind to in order to help the protein fold into its proper conformation (i.e., native fold)? A. hydrophobic patches on the polypeptide chain B. the N-terminus C. the C-terminus D. lysine residues E. ubiquitin proteins
A. hydrophobic patches on the polypeptide chain
Imagine that you just drank a large glass of milk on an otherwise empty stomach. For simplicity, imagine that the milk only contains lactose and nothing else (no glucose). Which of the following scenarios is most likely happening in your gut bacteria with regard to their lac operons? A. lac repressor bound to inducer, CAP bound near promoter, high (activated) lac operon transcription B. lac repressor bound to operator, CAP bound near promoter, no lac operon transcription C. lac repressor bound to inducer, CAP not bound, low (basal) level lac operon transcription D. lac repressor bound to operator, CAP not bound, no lac operon transcription
A. lac repressor bound to inducer, CAP bound near promoter, high (activated) lac operon transcription high lactose, low glucose
In glycolysis, what is the function of glyceraldehyde 3-phosphate dehydrogenase? A. oxidation of an aldehyde by NAD+ and formation of acyl-phosphate B. hydrolysis of glyceraldehyde 3-phosphate (GAP) C. oxidation of an alcohol to an aldehyde by NAD+ D. dehydration and dephosphorylation of glyceraldehyde 3-phosphate (GAP) E. removal of the 3-phosphoryl-group from glyceraldehyde 3-phosphate (GAP)
A. oxidation of an aldehyde by NAD+ and formation of acyl-phosphate
In class, we discussed how the COVID-19 virus, SARs-CoV-19, could be detected using a quantitative method for gene expression. Which method below is involved in the diagnostic test that we discussed? A. quantitative PCR (qPCR) B. DNA microarray (DNA chip) C. Exon skipping D. RNA inference (RNAi) E. Genome editing like CRISPR-Cas9
A. quantitative PCR (qPCR)
What sequence identifies a protein for translocation across the endoplasmic reticulum (ER)? A. signal sequence B. Shine-Dalgarno sequence C. poly-A sequence D. endoplasmic sequence E. cargo sequence
A. signal sequence
In order for a fatty acyl-CoA to be moved into the mitochondria for beta-oxidation, it must be A. transported by the carnitine-acylcarnitine translocase. B. activated by adenylation. C. reduced by NADH and FADH2 D. hydrolyzed into acetyl-coA fragments. E. attached to a lipid-soluble carrier protein called a lipoprotein.
A. transported by the carnitine-acylcarnitine translocase.
Diabetic ketosis is a dangerous condition in which ketone bodies build up in the blood. Identify the ketone bodies below.
B
Refer to the table below. Which small, noncoding RNA is BEST matched with its function? (note: the RNA names, abbreviations and sizes are correctly matched in all 4 options. Only the functions have been shuffled.)
B
Which of the following is an example of a ligation reaction?
B
Suppose this base-pair is present in a DNA double-helix structure. Which of the following labeled chemical groups are exposed to the major groove of DNA? A) #1, #2, #3, #4 B) #2, #3, #4, #5 C) #6, #7, #8, #9 D) #7, #8, #9, #10
B) #2, #3, #4, #5
Order the events that occur in one cycle of the polymerase chain reaction (PCR). #1: the reaction mixture is heated to 72°C #2: the double-stranded DNA is denatured to form single-stranded DNA #3: the temperature of the reaction mixture is lowered to allow the primer to anneal to the DNA template #4: the reaction mixture is heated to 94°C #5: Taq polymerase extends the primer. A) #1; #2; #3; #4; #5. B) #4; #2; #3; #1; #5. C) #1; #2; #4; #3; #5. D) #4; #2; #1; #3; #5.
B) #4; #2; #3; #1; #5.
The pKa values of the three acid/base groups in the amino acid here are 2.20, 9.11, and 10.01, respectively. The estimated pI value for this amino acid is A) (2.20 + 9.11 + 10.07) / 3 B) (2.20 + 9.11) / 2 C) (9.11 + 10.07) / 2 D) (2.20 + 10.07) / 2
B) (2.20 + 9.11) / 2 This is TYROSINE - a polar neutral molecule Neutral: PI= (pka1+pka2)/2 Basic: PI =(pkR+pk2)/2 Acidic: PI =(pkR+pk1)/2 where pKa1 = a-COOH where pKa2 = a-NH3+
Which of the following DNA sequence is the most compatible with the formation of nucleosome? Note: n means the sequence in parentheses repeats numerous times A) (ATCGTACTTA)n B) (ATTATCGGCG)n C) (TCAGAATCGA)n D) (ATCAATAATC)n
B) (ATTATCGGCG)n
20. Which of the following peptide segments is most likely to be part of a stable alpha helix at pH = 7? A) -Glu-Glu-Glu-Glu-Glu- B) -Glu-Leu-Ala-Lys-Phe- C) -Pro-Val-Thr-Pro-Trp- D) -Gly-Arg-Asn-His-Lys-
B) -Glu-Leu-Ala-Lys-Phe-
Histidine has a total of three ioizable groups: alpha-carboxyl (pKa 1.8 - pK1), side chain (pKa 6.0 - pKR), and alpha-amino (pKa 9.2 - pK2). At pH 6.0, the average charge of Histidine is about: A) 0.2 B) 0.5 C) 1.0 D) 1.5
B) 0.5
Based on the Lineweaver-burk plot shown in the figure, determine the value of Vmax. A) 1.04 microg/ml/min B) 2.04 microg/ml/min C) 3.04 microg/ml/min D) 4.04 microg/ml/min
B) 2.04 microg/ml/min
On average, each human gene has a size of 27,000 base-pairs. On average, how many histone H2A molecules are needed to package each human gene? A) 200 B) 270 C) 370 D) 1000
B) 270 Each nucleosome includes 200 nucleotide pairs. Need 27,000 / 200 *2 (because each histone has two H2A molecules)
The data in the table shown on the right were collected for a reaction catalyzed by enzyme ABC. Estimate the Km of this enzyme. A) 9x10^-6 mM B) 2x10^-5 mM C) 2x10^-4 mM D) 2x10^-3 mM E) cannot do it without my calculator
B) 2x10^-5 mM Km is the substrate concentration that is needed to achieve half of Vmax. Therefore, find Vmax on right = 280. Half of this is 140 which is on the chart with a Km of 2x10^-5.
The above peptide has an estimated pI of ____. A) 2.1 B) 3.6 C) 4.1 D) 5.1
B) 3.6 amino group - 8.0 glutamic acid - 4.1 carboxyl group - 3.1 +1-->0-->-1-->-2 3.1+4.1 / 2 = 3.6
Shown on the right is a DNA sequencing result. From this result, we know this piece of DNA has the following sequence: A) 5'-GACCTGACTGTA-3' B) 5'-ATGTCAGTCCAG-3' C) 5'-AAATTTCCCGGG-3' D) 5'-GGGCCCTTTAAA-3'
B) 5'-ATGTCAGTCCAG-3'
Shown on the right is a primer/template junction bound by a DNA polymerase. Which of the following correctly describes the sequence of template strand shown in this image? A) 5'-TTC-3' B) 5'-CTT-3' C) 5'-AAG-3' D) 5'-GAA-3'
B) 5'-CTT-3'
Based on the Lineweaver-burk plot shown to the right, determine the value of Vmax and Km Vmax Km A) 1 micromol/sec; 50 mM B) 50 micromol/sec; 2mM C) 0.02 micromol/sec 0.5 mM D) 2 micromol/sec; 50 mM
B) 50 micromol/sec; 2mM Km= -1/x-int Vmax = 1/y-int slope = Km/Vmax
A typical PCR reaction involves three different temperatures: 55°C, 72°C, and 95°C. Which of the following correctly explains the purpose of each temperature setting? A) 55°C: polymerization; 72°C: annealing primer to template; 95°C: generating single-stranded template B) 55°C: annealing primer to template; 72°C: polymerization; 95°C: generating single-stranded template C) 55°C: generating single-stranded template; 72°C: polymerization; 95°C: annealing primer to template D) 55°C: polymerization; 72°C: annealing primer to template; 95°C: inactivating polymerase
B) 55°C: annealing primer to template; 72°C: polymerization; 95°C: generating single-stranded template amplification of a specific DNA fragment by polymerase chain reaction (PCR): Step 1: heat to separate strands; denature (95 degrees, 30 s) Step 2: hybridization of primers (annealing, 50-60 degrees C, 30 s) Step 3: DNA synthesis from primers (elongation, 72 degrees Celsius)
Differences between eukaryotic and prokaryotic translation include that in eukaryotes, initiation factors bind to the _____________ on the mRNA, and the small ribosomal subunit binds to them, scanning the mRNA for the start codon. A) Shine-Delgarno sequence & ribosome binding site B) 7-methylguanine cap & poly-A tail C) Poly-A tail & Shine-Delgarno sequence D) 7-methylguanine cap & Shine-Delgarno sequence E) start codon & stop codon
B) 7-methylguanine cap & poly-A tail
Blood pH is maintained in a very narrow range of ___________, and this enzyme ___________ is very important in controlling blood pH. A ) 7.15 - 7.25; pyruvate decarboxylase B) 7.35 - 7.45; carbonic anhydrase C) 7.55 - 7.65; pyruvate decarboxylase D) 7.55 - 7.65; carbonic anhydrase
B) 7.35 - 7.45; carbonic anhydrase
Histidine is an amino acid with three titratable groups: an -NH3+ group (pKa=9.2), a -COOH group (pKa=1.8), and an imidazole (amine-like) group (pKa=6.0). The pI of Histidine is A) 5.5 B) 7.6 C) 3.9 D) 8.5
B) 7.6 (9.2+6)/2 = 7.6
In theory, maximally how many hydrogen bonds can each urea molecule form? A) 5 B) 8 C) 13 D) 20
B) 8
DNA polymerase catalyzes the reaction of adding nucleotides to the 3'-end of a primer strand, and dNTP is used by the enzyme. Under what condition, could a DNA polymerase use rNTP as a substrate and incorporate ribonucleotide into a DNA strand? A) A DNA polymerase that lacks proofreading exonuclease activity B) A DNA polymerase whose discriminator amino acids have been mutated to accommodate ribose C) A DNA polymerase that does not interact with sliding clamp D) A DNA polymerase that has a proofreading site too far from its polymerization site
B) A DNA polymerase whose discriminator amino acids have been mutated to accommodate ribose
Most likely, the diagram on the right illustrates A) A hydrogen bond formed between a base-pair B) A hydrogen bond formed between two adjacent beta-strands C) An ionic interaction between two adjacent alpha helices D) A hydrogen bond formed between two adjacent alpha helices
B) A hydrogen bond formed between two adjacent beta-strands
The sequence of this peptide is: A) AFDG B) AYDG C) AYEG D) GYEA
B) AYDG
The ADME property of a drug refers to these following four aspects of a drug's behavior: A) Absorption; Development; Metabolism; Efficiency B) Absorption; Distribution; Metabolism; Excretion C) Absorption; Distribution; Metabolism; Efficiency D) Absorption; Development; Metabolism; Excretion
B) Absorption; Distribution; Metabolism; Excretion ADME Properties Absorption (effectively) Distribution (specifically) Metabolism (slowly) Excretion (slowly) The concentration of a compound at its target site is affected by the extents and rates of ADME.
Histone proteins can undergo numerous post-translational modifications. Which of the following modifications decrease the DNA/histone interactions? A) Methylation and ubiquitination B) Acetylation and phosphorylation C) Phosphorylation and methylation D) Glycosylation and methylation
B) Acetylation and phosphorylation
The antibiotic ______________ inhibits prokaryotic transcription initiation by intercalating between the bases of the DNA double helix, preventing the DNA from being used as a template for RNA synthesis. A) Rifampicin B) Actinomycin D C) Streptomycin D) Erythromycin E) Puromycin
B) Actinomycin D
Shown here is the schematic of a general enzymatic pathway, in which numbered enzymes catalyze the conversion of lettered metabolites. Which of the following statements regarding regulatory interaction is correct? A) Inhibition of enzyme 1 by metabolite D is called homotrophic regulation B) Activation of enzyme 2 by metabolite B is called homotrophic regulation C) Activation of enzyme 3 by metabolite C is called heterotrophic regulation D) Inhibition of enzyme 3 by metabolite B is called feedback inhibition
B) Activation of enzyme 2 by metabolite B is called homotrophic regulation Homotrophic Effects *allosteric effects that occur when substrates (allosteric modulator) are bound to an enzyme on SUBSTRATE SITE *SAME AS SUBSTRATE SITE- causes a shift from T state to R state * effects shape of curve Heterotrophic Effects * allosteric effects that occur when non-substrate molecules are bound to an enzyme in a site other than the substrate active site * effects change the shifting of sinusoidal curve but not its shape * i.e. CTP can shift enzyme from R state to T state
Difference and similarity of ZFN, TALEN, CRISPR/Cas9: A) ZFN and TALEN use single-strand DNA to guide a nuclease, while CRISPR/Cas9 uses RNA B) All three methods are capable of guiding a nuclease to a specific locus in the genome C) ZFN and CRISPR/Cas9 use DNA-binding domains to guide a nuclease, while TALEN uses RNA D) TALEN and CRISPR/Cas9 use RNA to guide a nuclease, while ZFN uses DNA-binding domains
B) All three methods are capable of guiding a nuclease to a specific locus in the genome
The catalytic triad of trypsin, i.e., Ser-His-Asp, is responsible for speeding up the hydrolysis of the peptide bond on the carboxyl side of this residue ___________ on a substrate. In the course of catalysis, trypsin uses the side chain of this residue _____________ as a nucleophile and forms a temporary covalent bond with the substrate. A) Asp; Ser B) Arg/Lys; Ser C) Arg/Lys; Asp D) Asp/Glu; His
B) Arg/Lys; Ser
Four amino acids are shown below. Identify the ones drawn in the L configuration. A) Glutamine & Threonine B) Asparagine & Serine C) Isoleucine & Glutamate D) Leucine & Glutamate
B) Asparagine & Serine L-form of AMINO ACIDS - If H is pointing away and "CORN" is CCW - If H is pointing towards and "CORN" is CW D-form of AMINO ACIDS - If H is pointing away and "CORN" is CW - If H is pointing towards and "CORN" is CCW
30. DNA polymerase uses the strategy of metal ion catalysis to speed up the reaction. Two magnesium ions are coordinated by _________ residues in the palm region of the enzyme. The functions of these ions include ________________________________________________________. A) Histidine; forming hydrogen bonds with incoming nucleotide B) Aspartate; activating 3'-OH and positioning incoming nucleotide C) Glutamate; promoting base-pairing between incoming nucleotide and the template D) Asparagine; activating 3'-OH and breaking hydrogen bonds
B) Aspartate; activating 3'-OH and positioning incoming nucleotide The two metal ions are held in place by interactions with two highly conserved aspartate residues.
Many assumptions were used to simplify the derivation of Michaelis-Menten equation. Which of the following is one of the assumptions? A) At the very beginning of an enzymatic reaction, enzymes are saturated with substrates B) At the very beginning of an enzymatic reaction, the free substrate concentration is nearly the same as the initial substrate concentration C) Every substrate that binds enzyme will be converted to a product D) The reaction is set up in such a way that the concentration of enzyme is higher than that of substrate
B) At the very beginning of an enzymatic reaction, the free substrate concentration is nearly the same as the initial substrate concentration
Double-helical DNA can take on different forms, which include A form, B form, and Z form. The original double-helical structure proposed by Watson and Crick is right-handed with each base-pair nearly perpendicular to the axis of the helix. The structure was based on the X-ray diffraction data from fully hydrated DNA fiber. As such, the Watson/Crick model of double-helical model actually describes structure of DNA in A) A form B) B form C) Z form D) All of them
B) B form B form - right handed, most cellular DNA, anti, fully hydronated form A form - right handed, double stranded RNA, DNA/RNA hybrid, anti, dehydrated DNA Z-form - left handed, based on crystal structure of CGCGCG, anti or syn!
Due to hydrolytic attack, depurination (i.e., loss of purines) is a very frequent type of DNA damages. As a result, an AP (apurinic and apyrimidinic) site is produced in DNA. The pathway that repairs depurination is ______________. Without proper repair prior to next round of DNA replication, depurination can lead to this type of mutation _____________. A) Direct repair; deletion B) Base-excision repair; deletion C) Nucleotide-excision repair; insertion D) Mismatch repair; insertion
B) Base-excision repair; deletion
Which of the following statements is true regarding the two amino acids shown on the right? A) Both can be synthesized using pyruvate as a direct precursor B) Both can be synthesized using oxaloacetate as a direct precursor C) The one of the left but not the one of the right can be synthesized using pyruvate as a direct precursor D) Both are D-aspartate
B) Both can be synthesized using oxaloacetate as a direct precursor
CAF-1, i.e., chromatin assembling factor, and NAP-1, i.e., nucleosome assembling protein, are two important proteins for DNA replication in eukaryotes. Interestingly, CAF-1 is able to add newly translated H3/H4 tetramer to newly replicated DNA. How does CAF-1 achieve that? A) CAF-1 has a much higher affinity to newly synthesized DNA B) CAF-1 is able to interact with sliding clamp C) CAF-1 is able to interact with helicase D) CAF-1 is able to interact with topoisomerase I
B) CAF-1 is able to interact with sliding clamp
CAP-1 is a eukaryotic histone chaperone that is responsible for delivering H3/H4 tetramers to newly synthesized DNA to start the formation of a new nucleosome. Which of the following is a mechanism by which CAP-1 recognizes newly synthesized DNA? A) Newly synthesized DNA lacks methylation at the GATC sequence B) CAP-1 is able to interact with sliding clamp: naked DNA with sliding clamp is likely newly synthesized C) CAP-1 is able to interact with helicase: naked DNA with helicase attached is definitely newly synthesized D) CAP-1 is able to interact with RNase H, which helps it to recognize newly synthesized lagging strand
B) CAP-1 is able to interact with sliding clamp: naked DNA with sliding clamp is likely newly synthesized
A typical eukaryotic chromosome has at least 10 replication origins. What is one important mechanism cells use to prevent re-activation of an origin during each cell cycle to avoid replicating some DNA segment more than once? A) Cells use the length of telomere to count how many replication origins have been activated B) Cells restricts pre-RC assembly only in G1 phase C) Cells restricts pre-RC assembly and activation simultaneously in S phase D) Cells count how many histone molecules have been used to get accurate estimate of the origins being used
B) Cells restricts pre-RC assembly only in G1 phase
20. Nucleosome sliding can be facilitated by which of the following proteins? A) DNA ligase B) Chromatin remodeling complex C) Histone methyltransferase D) Histone H
B) Chromatin remodeling complex Nucleosome movement: nucleosome movement by sliding along a DNA molecule exposes sites for DNA binding proteins Chromatin remodeling complex: can use energy from ATP hydrolysis to catalyze changes in the structure of nucleosome - temporarily loosens interactions (now new information available for reading)
Remdesivir, compound 21, compound 34, and compound 101 are four competitive inhibitors that act on the RNA-dependent RNA polymerase in coronavirus. The Kd values of these four compounds are 1.1 nM (remdesivir), 0.2 nM (compound 21), 4 nM (compound 34), and 100 nM (compound 101). Based on these data alone, which drug candidate likely has the highest potency (implying the lowest dosage needed) when used as a drug in combating COVID-19? A) Remdesivir B) Compound 21 C) Compound 34 D) Compound 101
B) Compound 21
Damaged bases such as uracil/hypoxanthine are detected by ______________, and their repair does not need this enzyme __________________. A) uvrA/uvrB complex; DNA polymerase B) DNA glycosylases; Helicase C) uvrA/uvrB complex; Helicase D) DNA glycosylase; DNA polymerase
B) DNA glycosylases; Helicase
Adenylation, aka adenylylation or AMPylation, is the process of attaching an AMP to a chemical group in a molecule. Which of the enzymes/proteins important in DNA replication involves adenylation in the process of its action? A) DNA polymerase B) DNA ligase C) Helicase D) Clamp loader
B) DNA ligase
Why is methyl-C a hot spot for mutation in vertebrate genomes? A) Demethylase that removes methyl group in methyl-C is not efficient B) Deamination converts methyl-C to thymine, and glycosylase that repairs T:G base-pair is not efficient C) Methyl-C has a much higher tendency to undergo oxidation, and oxo-methyl-C is very hard to repair D) Methyl-C can base-pair with adenine, and C:A mispairing cannot be corrected
B) Deamination converts methyl-C to thymine, and glycosylase that repairs T:G base-pair is not efficient
The prosthetic group of hemoglobin and myoglobin is heme. In ____________, the central iron atom is displayed slightly out of the plane of the porphyrin ring system. The central iron has six bonds: four to nitrogen atoms in the porphyrin, one to a ___________ residue, and one to oxygen. A) Oxyhemoglobin; histidine B) Deoxyhemoglobin; histidine C) Oxyhemoglobin; cysteine D) Deoxyhemoglobin; cysteine
B) Deoxyhemoglobin; histidine
In Sanger sequencing, one most critical reagent is dideoxynucleotides. Which of the following is true regarding this reagent? A) Dideoxynucleotides provide sequence information because they add additional mass to DNA B) Dideoxynucleotides terminate DNA synthesis due to the lack of 3'-OH group C) Dideoxynucleotides terminate DNA synthesis due to the lack of 5'-OH group D) Dideoxynucleotides provide sequence information due to their incorporation into the 5'-end of a DNA strand
B) Dideoxynucleotides terminate DNA synthesis due to the lack of 3'-OH group
Biochemical standard free energy change is different from standard free energy change in which aspect? A) Distinct from standard free energy change, biochemical standard free energy change can be used to predict if a reaction can occur spontaneously in certain direction B) Distinct from standard free energy change, biochemical standard free energy change is measured under the condition that pH equals 7 C) Biochemical standard free energy change always has a larger value that of standard free energy change D) All of the above
B) Distinct from standard free energy change, biochemical standard free energy change is measured under the condition that pH equals 7
There are some resemblances between replication initiation in eukaryotes and prokaryotes. For example, the three key proteins required for prokaryotic replication initiation have equivalent/similar counterparts in eukaryotes. Which of the following is the most accurate pairing of these proteins? A) DnaA::helicase; DnaB::ORC; DnaC::Cdc6/Cdt1 B) DnaA::ORC; DnaB::helicase; DnaC::Cdc6/Cdt1 C) DnaA::Cdc6/Cdt1; DnaB::helicase; DnaC::ORC D) DnaA::ORC; DnaB::Cdc6/Cdt1; DnaC::helicase
B) DnaA::ORC; DnaB::helicase; DnaC::Cdc6/Cdt1 DnaA - initiator DnaB - helicase DnaC - Cdc/Cdt1
NHEJ and homologous recombination (HR) are two approaches for cells to repair ________________, and a key difference between these two is __________________. A) Pyrimidine dimer; NHEJ does not need ligase B) Double strand break; HR leads to accurate repair and NHEJ does not C) Chromosomal translocation; NHEJ needs the presence of sister chromatids D) Double strand break; HR needs topoisomerase and NHEJ does not
B) Double strand break; HR leads to accurate repair and NHEJ does not
Consider the following two drug candidates: drug 1 has a molecular weight of 900 Daltons, 7 hydrogen bond donors, 15 hydrogen bond acceptors, and a partition coefficient of 1; drug 2 has a molecular weight of 400 Daltons, 4 hydrogen bond donors, 4 hydrogen bond acceptors, and a partition coefficient of 2. According to Lipinski's rule, which of the following regarding the absorption property of these two drugs is correct? A) Both drugs have poor absorption B) Drug 1 likely has poor absorption C) Drug 2 likely has poor absorption D) Both drugs have good absorption
B) Drug 1 likely has poor absorption
Suppose that you want to clone your favorite gene into the vector pSAP shown here. Sequence analysis revealed that your favorite gene contains the following restriction enzyme sites: XhoI, NdeI, HindIII, SmaI, and SalI; the sequence upstream of your favorite gene contains the sites for EcoRI and HindIII; the sequence downstream of your favorite gene contains the sites for XhoI and KpnI. Which of the following enzyme(s) will you choose for your cloning project? A) HindIII and SmaI B) EcoRI and KpnI C) EcoRI and XhoI D) HindIII and NdeI
B) EcoRI and KpnI
A compound that resembles the transition state of a substrate is termed a transition-state analog. It has been demonstrated that a transition-state analog often acts as a more potent inhibitor for an enzyme than a compound that resembles the substrate itself. This observation supports which of the following hypotheses? A) Enzyme binds substrate better than the substrate in its transition state B) Enzyme binds substrate in its transition state better than its substrate C) Transition state is a stable state that can be isolated for analysis D) Transition state does not exist
B) Enzyme binds substrate in its transition state better than its substrate Once the substrate is in place, the enzyme changed conformation to bind it better and stabilize the "transition state." Enzymes have the HIGHEST AFFINITY for the substrate in its TRANSITION STATE.
Ethidium bromide is a commonly used dye for staining DNA. It is very important to be careful when working with ethidium bromide because it could damage DNA. Which of the following statements regarding ethidium bromide is correct? A) Ethidium bromide is an alkylating agent, which can chemically modify guanine B) Ethidium bromide is an intercalating agent, which can slip into the space between the base-pairs C) Ethidium bromide is an oxidizing agent, which can oxidize adenine D) Ethidium bromide binds backbone of DNA, which causes double-strand breaks
B) Ethidium bromide is an intercalating agent, which can slip into the space between the base-pairs
Etoposide is a commonly used chemotherapy agent in treating cancer. Its molecular target is topoisomerase II. One mechanism by which inhibiting topoisomerase II by etoposide can kill cancer cells is A) Topoisomerase II is only expressed in cancer cells and is essential for the viability of cancer cells B) Etoposide inhibits topoisomerase II in such a way that leads to a large accumulation of double-strand breaks C) Etoposide inhibits topoisomerase II in such a way that leads to the activation of telomerase D) Etoposide inhibits topoisomerase II in such a way that causes degradation of histone octamers
B) Etoposide inhibits topoisomerase II in such a way that leads to a large accumulation of double-strand breaks
BFP can be excited by violet light to emit blue light, while GFP can be excited by blue light to emit green light. In cells that express two fusion proteins: BFP-protein X and GFP-protein Y, which of the following is most likely to happen if protein X and protein Y interact in the nucleus? A) Exciting the cells with violet light leads to emission of blue light in the nucleus B) Exciting the cells with violet light leads to emission of green light in the nucleus C) Exciting the cells with blue light leads to emission of violet light in the nucleus D) Exciting the cells with blue light leads to emission of violet light in the cytoplasm
B) Exciting the cells with violet light leads to emission of green light in the nucleus
Without a proper telomere, inside of our body, which of the following enzymes would most likely cause harm to our chromosomes? A) DNA polymerase B) Exonuclease C) Endonuclease D) Trypsin
B) Exonuclease In order to be accurately replicated and segregated, linear chromosomes need 3 important sequence elements: telomere, replication origin, centromere * required for maintenance and replication! Telomere: Contain repeated nucleotide sequences that enable the ends of chromosomes to be replicated (and ensure that they are properly replicated). Also protect the end of the chromosomes from being mistaken by the cell as broken DNA molecule in need of repair. Telomere FUNCTION: 1) they protect eukaryotic chromosomes 2) without telomeres, the ends of chromosomes would be "repaired," leading to chromosome fusion and massive genome instability 3) they solve the end-replication problem (requires telomerase) 4) telomeres are thought to be the "clock" that regulates how many times an individual cell can divide. Telomeric sequences shorten each time the DNA replicates 5) telomerase is turned off in somatic cells 6) once the telomere shrinks to a certain level, the cell no longer divide. Its metabolism slows down, it ages, and dies. 7) cancer cells: telomerase always on to ensure constant dividing
Why does DNA synthesis always occur by extending the primer strand from the 3'-end? A) Only 3'-end of the primer strand is able to bind Mg++ B) Extending the primer strand from the 5'-end is not compatible with proofreading C) Extending the primer strand from the 3'-end prevents the erroneous usage of rNTP D) Phosphodiester linkage cannot be formed at the 5'-end of the primer strand
B) Extending the primer strand from the 5'-end is not compatible with proofreading 3'-5' direction of DNA chain growth is NOT optimal for proofreading. This is because hypothetical 3'-5' strand growth occurs and the 5' end is produced, if one nucleotide is removed by proofreading would thus prevent incoming correct dNTP from proceeding as no high-energy bond would be cleaved. (phosphate group not properly oriented)
RecA, RecBCD, and Rad51 are proteins important for homologous recombination. The function of RecA and Rad51 is A) Serving as a scaffolding protein for the recruitment of RecBCD B) Facilitating invasion of double-stranded DNA by single-stranded DNA C) Serving as a helicase to generate single-stranded DNA D) Stimulating the ligase activity
B) Facilitating invasion of double-stranded DNA by single-stranded DNA Recombination requires a host of enzymes and other proteins. RecBCD is an enzyme complex with both helicase and nuclease activities. Helicase and nuclease activities of RecBCD degrade the DNA. On reaching a chi (crossover hotspot instigator) sequence, nuclease activity on the strand with the 3' end is suppressed. The other strand continues to be degraded, generating a 3'-terminal single-stranded end. Degraded the 5' end. Likely successful in generating single strand region, allowing for more homologous recombination. RecA (pro) / Rad51 (euk) catalyzes DNA synapsis: Step 1: RecA intertwines the DNA single strand and the DNA duplex in a sequence-independent manner. Step 2: The DNA single strand searches the duplex for homologous sequence. *** if not successful, dissociates. Step 3: Stand invasion - the single strand forms base-pairs with the complementary strand in the duplex (heteroduplex)
Which of the following correctly describes how a histone octamer is assembled? A) H1 forms a tetramer with H2A, H2B, and H3; two H2A/H2B dimers then join in B) H3 and H4 forms a tetramer; two H2A/H2B dimers then join in C) H2A and H2B forms a tetramer; two H1 homodimers then join in D) H2A and H2B forms a tetramer; two H3/H4 dimers then join in
B) H3 and H4 forms a tetramer; two H2A/H2B dimers then join in
HAT is often linked to transcriptional activation. This is because A) HAT increases ubiquitination of histone octamers B) HAT elevates the level of histone acetylation, which helps to open up chromatin structure C) HAT is responsible for serine phosphorylation, a mark that is associated with gene expression D) HAT has the intrinsic ability in activating transcription factors
B) HAT elevates the level of histone acetylation, which helps to open up chromatin structure
DNA primase is associated with ________ in the prokaryotic replication fork but with ____________ in the eukaryotic replication fork. A) DNA polymerase I; DNA polymerase alpha B) Helicase; DNA polymerase alpha C) DNA polymerase II; helicase D) Helicase; DNA polymerase epsilon
B) Helicase; DNA polymerase alpha Primase associates with ____ in bacteria but with ____ in mammals: - helicase in bacteria - DNA polymerase in mammals
Which of the following statements provides the most accurate explanation for the capability of hemoglobin as an oxygen transporter molecule? A) Hemoglobin is a protein with high molecular weight B) Hemoglobin has a quaternary structure with four cooperative oxygen binding sites C) Hemoglobin has a much higher affinity to oxygen than myoglobin D) Hemoglobin is a protein that can easily form a polymer to facilitate oxygen transport
B) Hemoglobin has a quaternary structure with four cooperative oxygen binding sites
This amino acid ______ contains an imidazole group on its side chain, which allows it to function as a proton shuttle in the active site of the enzyme. A) Cysteine B) Histidine C) Asparagine D) Tryptophan
B) Histidine
The chemical structure shown on the right is the side chain of ____ which can function as a ____ in the active side of an amino acid. A) Proline; hydrogen bond acceptor B) Histidine; proton acceptor C) Tyrosine; hydrogen bond acceptor D) Histidine; proton donor
B) Histidine; proton acceptor
Barrier DNA sequences can block the spread of reader-writer and thus separate neighboring chromatin domain. Which of the following proteins could potentially act as a barrier protein if it can bind specifically to the barrier DNA sequence? A) G proteins B) Histone demethylase (HDM) C) HAT D) Protein kinase
B) Histone demethylase (HDM) Barrier DNA: sequences block the spread of reader-writer and thereby separating neighboring chromatin domains ** stops spread of code ** either physical barrier protein or HDM/HDT that can serve to erase newly added code to stop the spread
Why are histone proteins so basic? A) Histone proteins carry a large number of glutamate and aspartate residues, which are negatively charged at physiological pH and thus basic B) Histone proteins carry a large number of lysine and arginine residues, which give those proteins very high isoelectric points and make them very suitable for wrapping DNA C) Histone proteins are very small and easy to learn D) All five histone proteins are present in every living organism
B) Histone proteins carry a large number of lysine and arginine residues, which give those proteins very high isoelectric points and make them very suitable for wrapping DNA
Which of the following is the most accurate description of the End-of-Replication problem? A) In replicating circular DNA, the replication machinery cannot determine the exact ending point of DNA replication B) In replicating linear DNA, the 3'-end of template DNA cannot be completely replicated C) In replicating linear DNA, the 5'-end of template DNA cannot be completely replicated D) In replicating linear DNA, replication forks from adjacent replication bubbles interfere with each other
B) In replicating linear DNA, the 3'-end of template DNA cannot be completely replicated
Shorten the distance between polymerization site and proofreading site on DNA polymerase will A) Increase the rate of DNA replication B) Increase the fidelity (i.e., accuracy) of DNA replication C) Decrease the fidelity of DNA replication D) Have no effect on either the rate or the fidelity of DNA replication
B) Increase the fidelity (i.e., accuracy) of DNA replication shortening the distance between polymerization site will increase the fidelity (i.e. accuracy) of DNA replication but slow down the process (i.e. fidelity increases, rate decreases)
Increasing pH from 7 to 10 can lead to a drastic decrease in the double-helical form of DNA. Which of the following is the most reasonable explanation for this phenomenon? A) Increasing pH allows the formation of more hydrogen bonds between A and T B) Increasing pH impairs the formation of proper hydrogen bonds between G and C C) Increasing pH leads to circulation of single-stranded DNA D) Increasing pH leads to severe degradation of DNA
B) Increasing pH impairs the formation of proper hydrogen bonds between G and C
Which of the following amino acids have two chiral centers and can potentially form four stereoisomers? A) Serine and Threonine B) Isoleucine and Threonine C) Tyrosine and Tryptophan D) Isoleucine and Tryptophan
B) Isoleucine and Threonine
The value of ______ is used for describing the catalytic efficiency of an enzyme, and its upper limit is set by _____. A) Kcat; Vmax B) Kcat/Km; K1 C) Kcat; K1 D) Kcat./Km; Vmax
B) Kcat/Km; K1 Catalytic Efficiency: Kcat/Km - the ratio Kcat/Km combines information about how fast bound substrate is processed and how well the substrate is captured - can be used to compare the efficiency of different enzymes as well as substrate preferences of the same enzyme Upper limit of Kcat/Km: K1, rate constant of ES formation. The value of K1 has a diffusion limit around 10^8-10^9. Reaching the upper limit of Kcat/Km: catalytic perfection!
Refer to the diagram of common fatty acids found in plants. Which of the fatty acids listed would you predict to have the lowest melting temperature? A) Oleic acid (18:1) B) Linolenic acid (18:3) C) Palmitic acid (16:0) D) Erucic acid (22:1) E) Steric acid (18:0)
B) Linolenic acid (18:3)
When scientists use mass spectrometry to identify a protein, they often digest the protein with trypsin. Which of the following is the best rationale for this? A) Breaking an intact protein into fragments is important for assessing the purity of the protein sample. B) Mass information from both the intact protein and tryptic fragments ensures unique identification of the protein from the database C) The presence of large basic residues in the middle of a protein impairs its analysis by mass spectrometry D) Trypsin treatment increases the mass of the protein
B) Mass information from both the intact protein and tryptic fragments ensures unique identification of the protein from the database Protein identification by mass spec: * measure the movement of ions in the electric/magnetic field in vacuum * mass/charge ratio (m/z)- movement of ions determines this ratio precisely- knowing Z allows us to then deduce mass. *rarely would two proteins have the same pattern after digested with trypsin
Glutamine is a common amino donor for many biosynthetic reactions. Enzymes that catalyze reactions involving glutamine as the amino donor often contain a molecular tunnel. Which of the following is the most reasonable explanation for the presence of molecular tunnel in those enzymes? A) glutamine is very unstable, thus a molecular tunnel helps to retain it in the active site of the enzyme B) NH3 generated from glutamine as an intermediate needs to be protected from leaking out C) cellular concentration of glutamine is extremely low, thus a tunnel is needed to enrich it D) all of the above are true
B) NH3 generated from glutamine as an intermediate needs to be protected from leaking out
For haploid yeast cells (i.e., only one set of chromosomes are present), double-strand breaks are repaired by which of the following pathways? A) Homologous recombination B) NHEJ (non-homologous end joining) C) Direct repair D) Nucleotide excision repair
B) NHEJ (non-homologous end joining) ** another strand is not present for replication Homologous recombination VS. NHEJ: Nonhomologous end joining: not an accurate repair ** possible gene manipulation = gene deletion/inactivation Homologous recombination: if another copy is close by from replication ** possible gene manipulation = gene replacement if donor DNA nearby
There are some similarities and functional equivalences between proteins important for initiating DNA replication in prokaryotes and eukaryotes. The eukaryotic equivalents of DnaA, DnaB, and DnaC proteins are: _____, ______, and _______, respectively. A) Dna alpha; Dna beta; Dna gamma B) ORC; helicase; Cdc6/Cdt1; C) ORC; helicase; Abf1 D) Cdc6; Cdt1; helicase
B) ORC; helicase; Cdc6/Cdt1;
DNA polymerase is equipped with some elegant mechanisms to ensure correct incoming nucleotide is selected to be incorporated to the growing primer strand. Nevertheless, it still makes mistakes occasionally. What is one major source of the error that is made by DNA polymerase? A) Occasionally DNA polymerase may mistakenly use dNDP as substrate B) Occasionally DNA polymerase may be tricked by the rare tautomeric form of the bases C) Occasionally DNA polymerase may not be able to distinguish dATP from dCTP D) Occasionally DNA polymerase may not be able to distinguish dGTP from TTP
B) Occasionally DNA polymerase may be tricked by the rare tautomeric form of the bases
Which of the following keto-acid is a direct precursor for the synthesis of aspartate in our body? A) Pyruvate B) Oxaloacetate C) Alpha-ketoglutarate D) None of the above
B) Oxaloacetate
Polymerase switch refers to the phenomenon that synthesis of the first 20 or so nucleotides of each Okazaki fragment is catalyzed by one polymerase, and synthesis of the rest is catalyzed by a different polymerase. Specifically, the first polymerase in this switch is ________, and this phenomenon occurs in this organism ___________. A) Polymerase I; E. coli B) Polymerase alpha; dog C) Polymerase II; Salmonella D) Polymerase epsilon; mice
B) Polymerase alpha; dog
Which of the following residues has the most restricted phi angle? A) Gly B) Pro C) Ser D) Leu
B) Pro Exceptions for phi/psi conformations Proline: the cyclic chain of Pro limits its range of phi values to around -60 degrees making it the most conformationally restricted residue. === MOST RESTRICTED Glycine: much less sterically hindered; its permissible range of phi/psi covers a larger area of the Ramachandran diagram. At Gly residues, polypeptide chains often assume confirmations that are forbidden to other residues. === LEAST RESTRICTED
Which of the following correctly describes the order of events in determining protein structure using X-ray crystallography? A) Growing crystal; purifying protein; diffracting crystal using X-ray; solving the structure B) Purifying the protein; growing crystal; diffracting crystal using X-ray; solving the structure C) Purifying the protein; solving the structure; growing crystal; diffracting crystal using X-ray D) Growing crystal; diffracting crystal using X-ray; purifying protein; solving the structure
B) Purifying the protein; growing crystal; diffracting crystal using X-ray; solving the structure
Replicating linear DNA has a so-called "end replication problem". Among the four highlighted regions on the linear DNA, which regions will be impossible to be replicated completely without telomerase? A) Regions 1 & 2; B) Regions 2 & 3; C) Regions 3 & 4; D) Regions 4 & 1
B) Regions 2 & 3; The end replication problem: - without extension, no place to add primer to copy last portion of lagging strand ** only applies to linear DNA - replicate the ends of lagging strand template can be tricky - the telomerase: a protein-RNA complex that carries an RNA template - telomerase extends 3' end of linear chromosome and through reverse transcriptase uses RNA to direct synthesis of DNA
Which of the following peptides has the highest absorbance of ultraviolet light with a wavelength of 280 nm? A) SYLGKMNCSTYGQNTY B) SWGKWNYSTLYWGQN C) SFGKLNQKYLSTEGWNL D) TWKLSMCYKHISELNQS
B) SWGKWNYSTLYWGQN Tyrosine (Y) Tryptophan (W)
The catalytic triad of trypsin is composed of these three residues ___________, and this one ________ serves as a nucleophile and forms a temporary covalent bond with its substrate. A) Ser-His-Asp; His B) Ser-His-Asp; Ser C) Thr-His-Glu; Thr D) Thr-His-Glu; His
B) Ser-His-Asp; Ser
Trypsin contains a catalytic triad composed of ________, and the side chain of this residue _______ acts as a nucleophile in the course of catalysis. A) Ser/Thr/Tyr; Tyr B) Ser/His/Asp; Ser C) Thr/His/Asp; Thr D) Ser/His/Asp; Asp
B) Ser/His/Asp; Ser Catalytic triad of chymotrypsin, trypsin, and elastase: Referred to as the serine proteases; Use a highly reactive serine residue in the active site to break peptide bond. Trypsin, a protease with a highly reactive serine residue in its active site, is an example of using the strategy of covalent catalysis, because a transient covalent bond is formed between trypsin (the enzyme) and its protein substrate in the course of catalysis. The serine residue in the active site of trypsin is highly reactive because it is located close to a histidine residue, which can pull a proton from its side chain. The proton transfer to Histidine is further facilitated by an aspartate residue that is adjacent to Histidine. Ser-His-Asp is a catalytic triad present in many proteases. During catalysis, the negatively charged and highly reactive Serine side chain attacks the peptide bond of the protein substrate and forms a transient covalent bond with the carbonyl carbon, which leads to the breakage of the peptide bond. The covalent linkage between the enzyme (Trypsin) and the protein substrate is resolved when a molecule of water comes and splits the bond to displace the enzyme.
Which of the following statements is true regarding serine proteases and cysteine proteases? A) Serine proteases selectively cleave the peptide bond after a serine residue, while cysteine proteases cleave the peptide bond after a cysteine residue B) Serine proteases have a serine in the active site, while cysteine proteases have a cysteine reside in the active site C) Serine proteases include trypsin, while cysteine proteases include chymotrypsin D) All of the above are correct
B) Serine proteases have a serine in the active site, while cysteine proteases have a cysteine reside in the active site
Aspirin can transfer an acetyl group to a ________ residue on ___________, which leads to an inhibition of the biosynthesis of inflammatory mediator ____________________. A) Cysteine; cyclooxygenase; cytokine B) Serine; prostaglandin H2 synthase; prostaglandin H2 C) Lysine; cyclooxygenase; PGH2 D) Serine; guanylate cyclase; cGMP
B) Serine; prostaglandin H2 synthase; prostaglandin H2
Which of the following proteins make DNA polymerase extraordinarily processive? A) Helicase B) Sliding clamp C) SSB D) Ligase
B) Sliding clamp
Deamination of methylated cytosine (methyl-C) converts a C::G base-pair to A) A::G base-pair B) T::G base-pair C) U::G base-pair D) C::A base-pair
B) T::G base-pair
The two structures shown here were formed from the double-stranded DNA with the same sequence. Which of the following regarding them is correct? A) The DNA shown on the left has the same linking number as the one shown on the right B) The DNA shown on the left has the same twisting number as the one shown on the right C) The DNA shown on the left has the same writhe number as the one shown on the right D) The DNA shown on the right is positively supercoiled (meaning a positive writhe number)
B) The DNA shown on the left has the same twisting number as the one shown on the right Lk=Tw+Wr
Biochemical standard free energy change is unique in A) The [H+] is set as 1 M B) The acidity is set at pH 7 C) The temperature is set as 37°C D) All of the above
B) The acidity is set at pH 7
Despite being very accurate in selecting correct nucleotide for polymerization, DNA polymerase still makes mistake. Which of the following is a major factor that causes DNA polymerase to make mistakes in synthesizing DNA? A) The cellular concentration of rNTP is too high B) The existence of rare tautomers of bases C) The lacking of 5'-exonuclease activity of DNA polymerase D) The existence of uracil in the cells
B) The existence of rare tautomers of bases Tautomers: * Isomers that can interconvert by exchanging the location of a proton. * Major source of mistakes by DNA polymerase during DNA synthesis.
As organism gets more complex, the density of genes in its genome often decreases. This is because A) More complex organisms have less number of genes B) The gene size increases and the intergenic space increases as organism gets more complex C) More complex organisms have smaller size of genome D) More complex organisms typically do not contain introns in their genes
B) The gene size increases and the intergenic space increases as organism gets more complex
Generally speaking, anti-parallel beta-sheet is a more stable structure than parallel beta-sheet. The correct explanation for this observation is: A) it is easier to form a compact hydrophobic core in anti-parallel beta-sheet B) The inter-strand hydrogen bonds are stronger in anti-parallel beta-sheet C) There are more ionic interactions in anti-parallel beta-sheet D) The loops linking the strands in anti-parallel beta-sheet have strong structural stabilizing effects
B) The inter-strand hydrogen bonds are stronger in anti-parallel beta-sheet
According to an early model proposed by Pauling and Corey, DNA has three intertwined chains, with the phosphates near the fiber axis, and the bases on the outside. Why is this model "unsatisfactory"? A) The bases on the outside will make them not accessible to proteins B) The phosphates near the axis will repel each other C) Three intertwined chains tend to make a very unstable structure D) None of the above
B) The phosphates near the axis will repel each other
Transversion is a type of mutation in which a purine has been changed to a pyrimidine. Base oxidation without repair is a common cause of such mutation. In particular, oxidation of guanine is responsible for a large portion of transversion mutation. Why does oxidation of guanine lead to transversion mutation? A) The product of guanine oxidation, i.e., 8-oxo-guanine, can base-pair with thymine, resulting in G:T mispairing B) The product of guanine oxidation, i.e., 8-oxo-guanine, can base-pair with adenine, resulting in G:A mispairing C) Oxidation of guanine inhibits the enzymes required for nucleotide excision repair D) Oxidation of guanine inhibits the enzymes required for mismatch repair
B) The product of guanine oxidation, i.e., 8-oxo-guanine, can base-pair with adenine, resulting in G:A mispairing
In an experiment, a scientist treated cells with or without X-rays and analyzed the cell extracts using western blotting with a primary antibody that specifically recognizes phosphorylated H2A.X. What was the most likely purpose of such an experiment? A) The scientist was checking if X-ray treatment impairs the viability of the cells B) The scientist was checking if X-ray treatment leads to DNA strand breaks C) The scientist was checking if X-ray treatment leads to H1 degradation D) The scientist was checking if X-ray treatment causes cancer
B) The scientist was checking if X-ray treatment leads to DNA strand breaks H2A.X important for DNA repair; becomes phosphorylated upon DNA double-stranded break which helps to recruit DNA repair enzymes (signal). ** Without signal, cell would not recognize the break.
Biological function of many gene regulatory proteins relies on their ability to recognize and bind to a specific DNA sequence. Typically, these proteins bind to the major groove of DNA. Why is that? A) Major groove has more charged groups that can form ionic interactions B) The sequence of DNA can be easily distinguished by the unique chemical groups exposed in the major groove by each base-pair C) Major groove has extensive contact with histone octamer D) All of the above
B) The sequence of DNA can be easily distinguished by the unique chemical groups exposed in the major groove by each base-pair
Protein concentration can be estimated based on its absorbance at 280 nm. This is because: A) The side chain of histidine and phenylalanine absorbs UV light at 280 nm B) The side chain of tryptophan and tyrosine absorbs UV light at 280 nm C) The side chain of arginine and lysine absorbs UV light at 280 nm D) The peptide bonds absorb UV light at 280 nm
B) The side chain of tryptophan and tyrosine absorbs UV light at 280 nm
Each DNA polymerase has a pair of residues (Arg and Gln) in its palm region that are essential for its function in replicating genetic material. Why is this pair of residues important? A) These two residues can position metal ions in correct places in the active site B) These two residues play an important role in ensuring correct base-pairing between incoming nucleotide and the template C) These two residues help generating single-stranded primer strand for proofreading D) These two residues are important for hydrolyzing ATP to provide energy for the enzyme
B) These two residues play an important role in ensuring correct base-pairing between incoming nucleotide and the template Arg & Gln are two residues in the palm region of DNA polymerase and they are important in checking if a correct base pair is formed in the active site
Suppose you have a drug candidate with a very low value of Kd to its intended target in the body. Which of the following statements regarding this drug is true? A) This drug should have no problem to survive clinical trials because it is very potent B) This drug has very high affinity to its target C) The binding site for this drug on the target must be different from its active site D) The target of this drug must be an enzyme
B) This drug has very high affinity to its target The dissociation constant, Kd: * a measure of the strength of the interaction between the drug candidate and the target R: receptor L: ligand RL: the receptor ligand complex Kd = [R][L]/[RL] Kd equals the concentration of free ligand at which 1/2 of the binding sites are occupied. * A lower Kd means tighter binding which means a larger potency and a higher affinity
Drug ABC is a drug with a molecular weight of 151 Da and a logP value of 0.91. The structure of drug ABC is shown on the right. According to Lipinski's rule, A) This drug has poor absorption if taken orally because logP value is too small B) This drug should have reasonable absorption rate if taken orally C) This drug has too few H-bond acceptors, and thus its absorption should be poor D) The molecular weight of this drug is too small, and thus its absorption should be poor
B) This drug should have reasonable absorption rate if taken orally Lipinski's Rules: poor absorption is likely when... 1) the molecular weight is greater than 500 Da 2) the number of H-bond donors is greater than 5 3) the number of H-bond acceptors is greater than 10 4) the partition coefficient (logP) is greater than 5 (too hydrophobic --> too many aggregations --> difficult to pass lipid bilayer)
Which of the following is a group of amino acids with polar (neutral) side chain? A) Val; Pro; Ser B) Thr; Asn; Cys C) Asp; Glu; Lys D) Gly; Met; Cys
B) Thr; Asn; Cys
What is one reason that you would want to use a quantitative PCR (qPCR) assay? A) To silence a particular gene B) To determine how much of a particular gene is expressed compared to another gene C) To determine when and where a particular gene is expressed D) To compare gene expression across hundreds of genes under a certain condition E) To determine the specific DNA binding site of a particular gene regulatory protein
B) To determine how much of a particular gene is expressed compared to another gene qPCR is used to determine how much of a particular gene is expressed compared to another gene. qPCR measures how much individual mRNA is expressed compared to a housekeeping gene. DNA microarrays allow the determination of the expression pattern of many genes
Both the U1 snRNA and the U6 snRNA bind to the 5' splice site during the splicing mechanism, albeit at different times. Refer to the figures on the right that show how U1 and U6 independently bind to the 5' splice site. If the splice site sequence was mutated from 5'-GUAAGU-3' to 5'-GGAACC-3', would both U1 and U6 still be able to bind? A) Yes, both U1 and U6 would bind with the same affinity as before the mutation. B) U1 would bind, but with a lower affinity; U6 would likely not bind at all. C) U6 would bind, but with a lower affinity; U1 would likely not bind at all. D) No, neither U1 nor U6 will be able to bind after the mutation. E) Yes, both U1 and U6 would bind, but with lower affinities.
B) U1 would bind, but with a lower affinity; U6 would likely not bind at all.
Which of the following distinguishes ubiquitination from acetylation? A) Ubiquitination is a type of post-translational modifications, but acetylation is not B) Ubiquitination involves the formation of an isopeptide bond, but acetylation does not C) Acetylation occurs on histone proteins, while ubiquitination does not D) Acetylation occurs on lysine residues, while ubiquitination occurs on arginine residues
B) Ubiquitination involves the formation of an isopeptide bond, but acetylation does not Ubiquitylation: Ubiquitination is the biochemical process in which proteins are marked by ubiquitin, a 76 amino acid protein. Ubiquitin can be conjugated to other proteins such as histone by its carboxyl end to a -Lys group (this forms an isopeptide bond)
Several proteins are relevant to the infection of our body by SARS-CoV-2. Specifically, the spike proteins on the surface of the virus contact ACE2 proteins on the surface of our epithelial cells, which then leads to endocytosis and internalization of the virus into our cells. Inside our cells, the virus replicates its genome using its RNA-dependent RNA polymerase and produces viral proteins using the translational machinery of the host. The newly translated viral proteins are further processed by viral M2 protease to be fully functional. Suppose that you are in charge of using the novel mRNA approach to develop a vaccine for SARS-CoV-2, which of the following would be the most ideal strategy? A) Using mRNA for ACE2 protein as vaccine candidate B) Using mRNA for the spike protein as vaccine candidate C) Using mRNA for RNA-dependent RNA polymerase as vaccine candidate D) Using mRNA for M2 protease as vaccine candidate
B) Using mRNA for the spike protein as vaccine candidate
Assuming that you are investigating whether a tumor suppressor p53 protein is present in a particular cell line. Which of the following techniques will be the most suitable for your purpose? A) Southern blotting B) Western blotting C) Northern blotting D) x-ray crystallography
B) Western blotting
Suppose that you have a very sensitive and highly specific antibody that recognizes the spike protein from SARS-CoV-2 virus. Which of the following methods will allow you to detect if your patient may contain the SARS-CoV-2 virus? A) Native gel electrophoresis of the samples from the patient B) Western blotting analysis of the samples from the patient C) Anion exchange chromatography of the samples from the patient D) Gel filtration analysis of the samples from the patient
B) Western blotting analysis of the samples from the patient S -> D N -> R O O W -> P
In comparison to A and B forms of DNA, Z-DNA is very unique in A) Z-DNA is only present in viruses B) Z-DNA is left-handed and can have nucleotides in syn-conformation C) Z-DNA is right-handed and all nucleotides are in syn-conformation D) Z-DNA has a much lower GC content
B) Z-DNA is left-handed and can have nucleotides in syn-conformation
The difference between ZFN/TALEN and CRISPR/Cas9 is A) ZFN/TALEN primarily generate single-strand break, while CRISPR/Cas9 generates double-strand break B) ZFN/TALEN employs DNA-binding protein, while CRISPR/Cas9 uses guide RNA C) ZFN/TALEN only works in prokaryotes, while CRISPR/Cas9 only works in eukaryotes D) All of the above
B) ZFN/TALEN employs DNA-binding protein, while CRISPR/Cas9 uses guide RNA
Imagine that you just consumed delicious potato chips. For simplicity, let's say that the chips ONLY contain triacylglycerols and amylose. Which of the following enzymes would you expect to be active to help you digest your tasty snack? A) a-amylase, lactase, & sucrase B) a-amylase, pancreatic lipases, & maltase C) a-amylase, pancreatic lipases, & proteases D) maltase, pancreatic lipases, & proteases E) pancreatic lipases, lactase, and maltase
B) a-amylase, pancreatic lipases, & maltase
Taking sildenafil would lead to A) an inhibition of guanylate cyclase, a decrease in cGMP, and a reduced smooth muscle relaxation B) an inhibition of phosphodiesterase 5, an increase in cGMP, and an enhanced smooth muscle relaxation C) an activation of guanylate cyclase, an increase in cGMP, and an elevated smooth muscle relaxation D) an inhibition of phosphodiesterase 5 and an elevation of blood pressure
B) an inhibition of phosphodiesterase 5, an increase in cGMP, and an enhanced smooth muscle relaxation
Consider the atoms highlighted in the base-pair shown on the right. If this base-pair is present in DNA, the atoms that form N-glycosidic bonds are ___________, and the atoms that are found in the major groove of DNA are _____________. A) atoms 1 & 2; atoms 3 & 4 B) atoms 1 & 4; atoms 2 & 3 C) atoms 3 & 4; atoms 5 & 6 D) atoms 1 & 4; atoms 5 & 6
B) atoms 1 & 4; atoms 2 & 3 N-glycosidic: bond between base and sugar
In Anfinsen's elegant demonstration that amino acid sequence determines 3-D structure, beta-mercaptoethanol was used to ___________, and urea was used to __________________. A) add negative charges; solubilize the protein B) break the disulfide bonds; denature the protein C) denature the protein; break the disulfide bonds D) solubilize the protein; add negative charges
B) break the disulfide bonds; denature the protein
Which factor is NOT known to be involved in transcription initiation in eukaryotes? A) DNA helicase activity B) construction and use of a RNA primer C) formation of an open complex from a closed complex D) protein binding of specific proteins to specific DNA sequences E) protein phosphorylation
B) construction and use of a RNA primer
The tertiary structure of ribonuclease is stabilized by four pairs of disulfide bonds. In the Anfinsen experiment, two chemical reagents - urea and beta-mercaptoethanol - were used. The function of urea is to ________ and the function of beta-mercaptoethanol is to _______. Between these two reagents, this one _______ should be removed first to ensure correct refolding of the protein. A) denature the protein; oxidize the protein; B-mercaptoethanol B) denature the protein; reduce the disulfide bonds; urea C) reduce the disulfide bonds; denature the protein; B-mercaptoethanol D) solubilize the protein; reduce the disulfide bonds; urea
B) denature the protein; reduce the disulfide bonds; urea The tertiary structure of ribonuclease is stabilized by four pairs of disulfide bonds. In the Anfinsen experiment, two chemical reagents - urea and beta-mercaptoethanol - were used. The function of urea is to denature the protein and the function of beta-mercaptoethanol is to reduce the disulfide bonds. Between these two reagents, urea should be removed first to ensure correct refolding of the protein. Add urea then B-mercaptoethanol to change active state to denatured reduced state. Urea - destroys hydrophobic effects and thus denatures the protein (uncoils). B-mercaptoethanol - SH (thiol group) can react with disulfide bonds and leads to its reduction, becoming free cysteine. Return protein to its native state by first removing urea to ensure folding pattern is correct before allowing disulfide bonds to reform. If B-me is removed prior to urea, a wrong pair of cys-residues may form a disulfide bond, which could prevent the protein from then refolding and functioning correctly.
DNA polymerase requires Mg2+ ions as an essential cofactor. The complex of DNA polymerase and Mg2+ is called an/a ________, while DNA polymerase lacking Mg2+ is called an/a _______. A) apoenzyme; holoenzyme B) holoenzyme; apoenzyme C) prosthetic group; apoenzyme D) holoenzyme; prosthetic group
B) holoenzyme; apoenzyme Holoenzyme = Enzyme + Cofactor * if cofactor is removed, then this would constitute an apoenzyme
Deamination of cytosine occurs due to ____________, while pyrimidine dimer forms due to __________. A) oxidation of DNA; ionization irradiation B) hydrolytic attack by water; exposure to ultraviolet light C) oxidation of DNA; exposure to ultraviolet light D) hydrolytic attack by water; ionization irradiation
B) hydrolytic attack by water; exposure to ultraviolet light
Both deamination of cytosine and depurination are caused by _____________ on DNA, and both can be repaired by ______________________. A) oxidative damage; nucleotide excision repair B) hydrolytic attack; base excision repair C) uncontrolled methylation; nucleotide excision repair D) hydrolytic attack; nucleotide excision repair
B) hydrolytic attack; base excision repair many factors can cause accidental lesions in DNA Types of DNA Repair Systems 1) base excision repair - abnormal bases (eg. uracil) ** no helicase needed a - recognize incorrect base b- hydrolyze base c- hydrolyze abasic nucleotide using endonuclease d- fill in gap with DNA polymerase e- close DNA ends with DNA ligase ** The DNA glycosylases travel along DNA using base-flipping to evaluate the status of each base and remove the damaged base. 2) nucleotide excision repair - DNA lesions that cause large structural changes (eg. pyrimidine dimers) 3) mismatch repair - mismatches 4) homologous recombination and NHEJ - double-strand breaks 5) direct repair - goes to lesion directly and corrects it ***only repair method that does not require DNA synthesis - Photoreactivation repair (direct cleavage of pyrimidine dimer into its component bases using the reusable DNA photolyase) - demethylase (direct removal of methyl groups from methylated bases using the one-and-done methyltransferase)
A major driving force for a polypeptide chain to fold into its native conformation is ______, which is distinct from many other non-covalent interactions, because it is largely __________. A) forming disulfide bonds; non-specific B) hydrophobic effect; entropy driven and not electrostatic in nature C) hydrogen bonding; directional D) Ionic interaction; sensitive to the aqueous environment
B) hydrophobic effect; entropy driven and not electrostatic in nature
Post-translational modifications (PTMs) are important mechanisms for protein regulation. Shown below is an example of such PTMs. This PTM is a _____________, which occurs very frequently in this protein ___________. A) cyclic serine; tumor suppressors B) hydroxyproline; collagen C) cyclic threonine; collagen D) hydroxyhistidine; oncoproteins
B) hydroxyproline; collagen Hydroxyproline During PTM, you can add a hydroxyl group to the side chain of proline. Hydroxyproline is very common in the 3-helices protein called collagen joined together through H-bonding. Therefore, the extra OH group allows for additional H-boding which is important for strong structure in collagen (found in connective tissue). Vitamin C can help catalyze this reaction.
The switch between life cycles in lambda phage is controlled largely by the protein levels of lambda repressor and cro protein. When lambda phage switches from lysogenic to lytic phase: A) lambda repressor levels increase and cro protein levels decrease B) lambda repressor levels decrease and cro protein levels increase C) both lambda repressor and cro protein levels decrease D) both lambda repressor and cro protein levels increase E) both lambda repressor and cro protein levels stay the same
B) lambda repressor levels decrease and cro protein levels increase
Like many cellular proteins, histones can undergo ubiquitination, a post-translational modification in which a small protein called ubiquitin is typically linked to a(n) ____________ residue on a substrate protein via the formation of a(n)_________________. A) alanine; hydrophobic interaction B) lysine; isopeptide bond C) arginine; amide bond D) cysteine; disulfide bond
B) lysine; isopeptide bond
Cystine is formed via A) Reduction of disulfide bond B) Oxidation of the side chain of two cysteines C) Dimerization between a cysteine and a tyrosine D) Dimerization between a cysteine and a methionine
B) oxidation of the side chain of two cysteines
The following reactions illustrate how acetaminophen is metabolized in our body. The step from acetaminophen to N-acetyl-p-benzoquinone imine is classified as ___________________, and this reaction is catalyzed by ______________________. A) oxidation; hydrolyase B) oxidation; p450 enzyme C) conjugation; trypsin D) conjugation; p450 enzyme
B) oxidation; p450 enzyme Drug metabolism: the body's two main pathways for defending against foreign compounds 1) Oxidation by cytochrome P450 in the liver 2) Conjugation by glutathione, glucuronic acid, and sulfate in the liver *** these drugs already had functional groups to be conjugated. ** functional groups may act as tags for kidneys to recognize that it needs to be removed
Succinylcholine is a fast-acting, short-duration muscle relaxant that is used when a tube is inserted into a patient's trachea. Within seconds of the administration of succinylcholine, the patient experiences muscle paralysis. Succinylcholine is hydrolyzed by blood-serum cholinesterase. Paralysis lasts until the succinylcholine is hydrolyzed by the serum cholinesterase, usually several minutes later. Some patients have a mutant form of the serum cholinesterase that displays a Km of 10 mM, rather than the normal 1.4 mM. What will be the likely effect of this mutation on the patient? A) patients with the mutation will clear the drug at a higher rate than normal people B) patients with the mutation will clear the drug at a lower rate than normal people C) patients with the mutation will clear the drug at the same rate than normal people D) patients with the mutation will stay paralyzed for a much shorter period of time
B) patients with the mutation will clear the drug at a lower rate than normal people
Polylysine (i.e., a peptide with all lysine residues) and polyglutamate (i.e., a peptide with all glutamate residues) are two examples of synthetic polypeptides that can assume an alpha-helical structure under appropriate conditions. Identify which two synthetic polypeptides can form an alpha-helix at the indicated pH values. A) polylysine at pH 2 and polyglutamate at pH 7 B) polylysine at pH 13 and polyglutamate at pH 2 C) polylysine at pH 7 and polyglutamate at pH 13 D) polylysine at pH 6 and polyglutamate at pH 12
B) polylysine at pH 13 and polyglutamate at pH 2
DNA polymerase uses two Mg++ to speed up the polymerization reaction. The function of one Mg++ is to ____________________, and the function of the other Mg++ is to _______________________________. A) produce a free 3'-OH group on the primer strand; stabilize the structure of discriminator amino acids B) produce a powerful nucleophile; position the incoming dNTP and to stabilize the resulting -PP C) bind the deoxyribose component of dNTP; bind the phosphate component of dNTP D) bind the ribose component of rNTP; bind the phosphate component of rNTP
B) produce a powerful nucleophile; position the incoming dNTP and to stabilize the resulting -PP Two metal ions (Mg++) bound to DNA polymerase catalyze nucleotide addition: * The two metal ions are held in place by interactions with two highly conserved aspartate residues. * Metal ion A primarily interacts with the 3'-OH, resulting in reduced association between the O and the H. This produces a powerful nucleophile * Metal ion B interacts with the incoming dNTP to position it for nucleophilic attack and to stabilize the resulting pyrophosphate.
The molecule shown on the right is _______________, which is a _______________. A) anti-adenine; nucleotide B) syn-adenosine; nucleoside C) anti-guanine; nucleotide D) syn-guanosine; nucleotide
B) syn-adenosine; nucleoside
DNA polymerases alpha, epsilon, and gamma functions to synthesize _________, ___________, and _________, respectively. A) leading strand; lagging strand; damaged DNA B) the first 20 nucleotides of each Okazaki fragment; leading strand; mitochondrial DNA C) leading strand; the first 20 nucleotides of each Okazaki fragment; the first 20 nucleotides of leading strand D) lagging strand; leading strand; Okazaki fragment
B) the first 20 nucleotides of each Okazaki fragment; leading strand; mitochondrial DNA Major types of DNA polymerase in eukaryotes: DNA polymerase alpha - initiator polymerase primase subunit - synthesizes the RNA primer DNA polymerase unit - adds stretch of about 20 nucleotides to the primer DNA polymerase Beta - DNA repair DNA polymerase delta - primary enzyme of DNA synthesis (lagging) DNA polymerase epsilon - primary enzyme of DNA synthesis in leading strand DNA polymerase gamma - mitochondrial polymerase
In bacteria, which are paired by a Shine-Dalgarno sequence? A) the 30S subunit with the 50S subunit of the ribosome B) the mRNA sequence just upstream from the start codon with the 16S rRNA embedded in the small subunit of the ribosome C) a termination codon with a release factor D) the mRNA sequence just downstream from the start codon with the 23S rRNA embedded in the large subunit of the ribosome E) the first codon of the mRNA with its matching tRNA anticodon
B) the mRNA sequence just upstream from the start codon with the 16S rRNA embedded in the small subunit of the ribosome Initiation in bacteria begins at least 25 nucleotides downstream of the 5' end of the mRNA, at a Shine-Dalgarno sequence (aka ribosome binding site, rbs), which is a purine-rich sequence approx. 10 base pairs upstream of the start site that interacts with the 16S rRNA to correctly position the AUG codon in the P site. Two kinds of interactions determine where protein synthesis starts: 1) the pairing of mRNA bases with the 3' end of 16S rRNA 2) the pairing of the initiator codon on mRNA with the anticodon of an initiator tRNA molecule
The main reasons that eukaryotic mRNA transcripts are modified at the 5' and 3' ends are: A) so that the promoter can be recognized and to facilitate termination of transcription B) to protect the ends of the transcripts from exonuclease degradation and to facilitate initiation of translation C) to ensure that the CTD of RNA polymerase is phosphorylated and that the polymerase can escape the promoter region D) to facilitate the export of the mature mRNA and to guide the removal of introns E) so that the two ends of mRNA remain associated with each other as the mRNA exits the nucleus and is transported to the rough endoplasmic reticulum
B) to protect the ends of the transcripts from exonuclease degradation and to facilitate initiation of translation
SANT domain recognizes _________________, while chromodomain recognizes ________________. A) acetylated lysine; methylated arginine B) unmodified histone tail; methylated lysine C) methylated lysine; acetylated lysine D) methylated lysine; unmodified histone tail
B) unmodified histone tail; methylated lysine
UV exposure can lead to the formation of pyrimidine dimer in DNA, which can interfere with normal DNA replication and transcription. Which of the following is a correct order of events in repairing pyrimidine dimer in bacteria? A) Photolyase binds to pyrimidine dimer; urvC enzyme cuts the damaged strand; helicase removes the damaged strand; DNA synthesis occurs and ligase completes the process B) uvrA/uvrB complex binds to pyrimidine dimer; uvrC cuts the damaged strand; uvrD removes the damaged strand; DNA synthesis occurs and ligase completes the process C) Photolyase binds to pyrimidine dimer; uvrA/uvrB cuts the damaged strand; DNA synthesis occurs and ligase completes the process D) DnaA binds to pyrimidine dimer; DnaB cuts the damaged strand; DnaC removes the damaged strand; DNA synthesis occurs and ligase completes the process
B) uvrA/uvrB complex binds to pyrimidine dimer; uvrC cuts the damaged strand; uvrD removes the damaged strand; DNA synthesis occurs and ligase completes the process
SARS-CoV-19 and HIV are both plus strand (sense) RNA viruses. However, HIV is a retrovirus, while SARS-CoV-19 is not a retrovirus. Outline the path of the "central dogma" for HIV inside its host. A. + RNA à - RNA à mRNA à protein B. + RNA à cDNA à mRNA à protein C. + RNA à mRNA à protein D. + RNA à - RNA à cDNA à protein E. + RNA à - RNA à cDNA à mRNA à protein
B. + RNA à cDNA à mRNA à protein
Which of the following is an allosteric inhibitor of phosphofructokinase both in the muscle & in the liver? A. NADH B. ATP C. AMP D. ADP E. fructose 1,6-bisphosphate
B. ATP
Which electron transport complex is unable to generate sufficient free energy to power ATP synthesis? A. Complex I B. Complex II C. Complex III D. Complex IV E. Complex V
B. Complex II
Which of the following statements about an E. coli ribosome is correct? A. It is composed of two spherically symmetrical subunits. B. Its large subunit, 50S, contains the 23S catalytic rRNA. C. It has a sedimentation coefficient of 80S. D. It has two small subunits, one housing the A site and the other the P site. E. It has a mass of approximately 270 kDa, one-fifth of which is RNA (the rest is protein).
B. Its large subunit, 50S, contains the 23S catalytic rRNA.
Visceral fat tissue secretes various adipokines in response to levels of the fat stores. Which hormone below is an apdiokine that acts as a "lipostat" in that it reports to the brain that the fat levels are sufficient? A. Insulin B. Leptin C. Ghrelin D. Adiponectin E. Glucagon-like peptide 1
B. Leptin
When bacteria produce mammalian proteins, complementary DNA to mature mRNA (cDNA) is used rather than genomic DNA. Which of the following is the best explanation? A. Bacteria have polyribosomes, but mammals don't. B. Mammalian DNA contain introns that bacteria do not have the mechanisms to remove. C. Most eukaryotic gene promoters do not function in bacteria. D. The polycistronic nature of bacterial genes precludes efficient expression of mammalian DNA. E. A different genetic code is used for mammalian genes than in bacteria.
B. Mammalian DNA contain introns that bacteria do not have the mechanisms to remove.
The Lambda repressor (cI) has the lowest affinity and the Cro protein (cro) has the highest affinity for which of the following binding sites? A. OR1 B. OR3 C. OR2 D. PRM E. PR
B. OR3
Which of the following applies only to prokaryotic proteins? A. Some proteins are synthesized on polyribosomes. B. Protein synthesis is coupled with mRNA synthesis C. Secreted proteins have signal sequences in their N-terminal regions. D. Some proteins require help from molecular chaperones to fold properly. E. Some proteins are covalently modified after translation.
B. Protein synthesis is coupled with mRNA synthesis
We discussed a variety of ribozymes in this unit. Out of the choices below, which enzymatic activity is NOT due to a ribozyme? A. Peptidyl transferase center B. RNA polymerase C. RNase P D. Spliceosome E. Processing of eukaryotic rRNA genes
B. RNA polymerase
What are the roles of the core RNA polymerase and sigma factor in prokaryotes? A. The core RNA polymerase, along with the sigma factor, only act to recognize the promoter. B. The core RNA polymerase performs the catalysis of RNA synthesis, while the sigma factor recognizes the promoter. C. The core RNA polymerase recognizes the promoter, while the sigma factor performs the catalysis of RNA synthesis. D. The core RNA polymerase, along with the sigma factor, only act to perform the catalysis of RNA synthesis. E. The core RNA polymerase, along with the sigma factor, work together to recognize the promoter as well as to perform the catalysis of RNA synthesis.
B. The core RNA polymerase performs the catalysis of RNA synthesis, while the sigma factor recognizes the promoter.
Which of the following statements about functional tRNAs is correct? A. All of their nucleotides are in base-paired helical regions. B. They contain many modified nucleotides. C. They contain more than 1000 ribonucleotides. D. They have a terminal 5' AAC 3' sequence at their amino acid accepting end. E. They consist of two helical stems that are joined by loops to form a U-shaped tertiary structure.
B. They contain many modified nucleotides.
Which of the following statements about eukaryotic mRNAs is true? A. They are often polycistronic. B. They result from extensive processing of their primary transcripts before serving as translation components. C. They usually have poly(A) tails at their 5' ends. D. They have a cap at their 3' ends. E. They are often encoded by contiguous segments of template DNA.
B. They result from extensive processing of their primary transcripts before serving as translation components. ** cap is 5' ** poly(A) is 3'
Both the U1 snRNA and the U6 snRNA bind to the 5' splice site during the splicing mechanism, albeit at different times. Refer to the figures on the right that show how U1 and U6 independently bind to the 5' splice site. If the splice site sequence was mutated from 5'-GUAAGU-3' to 5'-GGAAGU-3', would both U1 and U6 still be able to bind? A. Yes, both U1 and U6 would bind with the same affinity as before the mutation. B. U1 would bind, but with a lower affinity; U6 would likely not bind at all. C. U6 would bind, but with a lower affinity; U1 would likely not bind at all. D. No, neither U1 nor U6 will be able to bind after the mutation. E. Both U1 and U6 would bind, but with a lower affinity.
B. U1 would bind, but with a lower affinity; U6 would likely not bind at all.
Which of the following enzymes are the key regulatory sites in the citric acid cycle? A. malate dehydrogenase & a-ketoglutarate dehydrogenase B. a-ketoglutarate dehydrogenase & isocitrate dehydrogenase C. succinyl CoA synthetase & malate dehydrogenase D. succinate dehydrogenase & succinyl CoA synthetase E. isocitrate dehydrogenase & succinate dehydrogenase
B. a-ketoglutarate dehydrogenase & isocitrate dehydrogenase
Which of the following answers completes the sentence correctly? The wobble hypothesis: A. accounts for the conformational looseness of the amino acid acceptor stem of tRNAs that allows sufficient flexibility for the peptidyl-tRNA and aminoacyl- tRNA to be brought together for peptide-bond formation. B. accounts for the ability of some anticodons to recognize more than one codon. C. explains the occasional errors made by the aminoacyl-tRNA synthetases. D. explains the oscillation of the peptidyl-tRNAs between the A and P sites on the ribosome. E. assumes steric freedom in the pairing of the first (5') nucleotide of the codon and the third (3') nucleotide of the anticodon.
B. accounts for the ability of some anticodons to recognize more than one codon. assumes steric freedom in the pairing of the third(3') nucleotide of the codon and the first(5') nucleotide of the anticodon.
What tissue or organ is the primary storage site for triacylglycerols in animals? A. brain B. adipose C. kidney D. heart E. muscle
B. adipose
Which of the following molecules has a higher phosphoryl-transfer potential than ATP? A. pyrophosphate B. creatine phosphate C. 2,3-bisphosphoglycerate D. glucose-1-phosphate E. glucose-6-phosphate
B. creatine phosphate
The first stage of catabolism is: A. the citric acid cycle (Kreb's cycle) B. digestion of large macromolecules to smaller metabolic fuels like amino acids, fatty acids, and monosaccharides C. production of pyruvate from monosaccharides or amino acids D. production of acetyl CoA from pyruvate or fatty acids E. production of polymers from monomers
B. digestion of large macromolecules to smaller metabolic fuels like amino acids, fatty acids, and monosaccharides
The poly A binding proteins (PABP) bound to the poly(A) tail and eIF4E bound at the 5' cap interact through their binding to: A. Cap-binding complex (CBC) B. eIF4G C. eIF-2 D. GTP E. snRNA
B. eIF4G
DNA microarrays can be used for evaluating ___________, starting with RNA and using the enzyme ________ to synthesize cDNA and simultaneously incorporate nucleotides labeled with fluorescent dyes. A. gene expression; RNA polymerase II B. gene expression; reverse transcriptase C. gene expression; taq DNA polymerase (thermostable DNA polymerase) D. gene regulation; dicer E. gene regulation; RNA-induced silencing complexes (RISCs)
B. gene expression; reverse transcriptase
In which of the following metabolic conversions is ATP "consumed" (i.e., invested) during glycolysis?
B. glucose à glucose-6-phosphate
Mitochondria: A. have a porous inner membrane and nonporous outer membrane. B. have a higher pH inside the matrix than the inner membrane space during active electron transport. C. have an outer membrane that is composed of lipids and protein electron transport complexes. D. generate ATP for the cell under anaerobic and aerobic conditions. E. are where all the central metabolic pathways we discussed in class take place.
B. have a higher pH inside the matrix than the inner membrane space during active electron transport.
The DNA-binding motif shown to the right is called a ______________ motif, and its recognition helix binds directly in the __________ groove of the DNA. A. basic leucine zipper, major B. helix-turn-helix, major C. zinc finger, major D. basic leucine zipper, minor E. helix-turn-helix, minor
B. helix-turn-helix, major
Molecular protein chaperones: A. degrade proteins that have folded incorrectly B. help new proteins fold correctly and repair incorrectly folded proteins C. are found in the nucleus and aid in folding of RNA D. are only present in cells that are exposed to high temperatures E. only work through hydrophobic interactions and do not need ATP hydrolysis to function
B. help new proteins fold correctly and repair incorrectly folded proteins
Glucose is converted to ______ in skeletal muscle under anaerobic conditions. A. acetyl-CoA B. lactate C. ethanol & CO2 D. glycogen E. alanine
B. lactate
Which of the following enzymes are found in the glyoxylate cycle but not in the citric acid cycle? A. malate synthase and malate dehydrogenase B. malate synthase and isocitrate lyase C. isocitrate lyase and aconitase D. citrate synthase and aconitase E. malate dehydrogenase and citrate synthase
B. malate synthase and isocitrate lyase
What is the major fuel source for ATP production that is used in a 200-meter sprint? A. liver glycogen B. muscle phosphocreatine C. muscle glycogen D. adipose tissue fatty acids E. liver produced ketone bodies
B. muscle phosphocreatine
Different mature eukaryotic mRNAs can be obtained from the same gene by: A. terminating transcription at alternative start codons. B. splicing the pre-mRNA with different exons. C. adding the 7-methylguanosine cap at different sites. D. alternate folding of mRNA. E. editing the DNA in the genome by ZFNs, TALENs, or CRISPR-Cas9.
B. splicing the pre-mRNA with different exons.
What protein is used as a tag to identify proteins that are targeted for destruction? A. proteasome B. ubiquitin C. destructin D. degradin E. cyclin
B. ubiquitin
The Cori cycle facilitates the rapid production & use of glucose simultaneously to facilitate rapid muscle contraction because: A. your muscle cells will be performing gluconeogenesis, and your liver cells will be performing glycolysis B. your muscle cells will be performing glycolysis, and your liver cells will be performing gluconeogenesis C. both your muscle cells and liver cells will be performing glycolysis D. both your muscle cells and liver cells will be performing gluconeogenesis E. your brain will start burning fatty acids so that all the glucose can go to your muscles
B. your muscle cells will be performing glycolysis, and your liver cells will be performing gluconeogenesis
Eukaryotic gene expression can be regulated in many ways that can be classified into 6 main categories as shown below. Which step along the pathway would you expect miRNA to play an important role in regulation of gene expression?
C ***RNA Interference ** degradation control microRNAs (miRNAs) are small regulatory RNAs encoded by the genome that suppress translation: - a larger initial transcript, usually a product of RNA Pol II, is cleaved to yield the mature miRNA - the mature miRNAs bind to members of the argonaute family of proteins in order to exert their regulatory roles - miRNA-aronaute assemble to form RNA inducing silencing complexes (RISCs) - RISC complex binds to complement in target mRNA - if miRNA is perfect complement, target mRNA is cleaved resulting in the mRNA not being translated - if miRNA is only partial complement, translation is blocked, and mRNA processing is trapped in processing bodies (P bodies)
Consider a fibrous protein that is composed of two alpha helices. The residues present in this protein would most likely be clustered on which region of the Ramachandran plot (A, B, C, or D)?
C a map that displays the allowed polypeptide backbone conformation (allowed = no clash of residues) - the backbone conformation of any particular residue in a protein can be described by a point on a map with coordinates of phi and psi - favorable: sterically allowed phi/psi angles - allowed: more crowded but acceptable phi/psi angles - not allowed: local steric clash
19. Which of the following nucleotide sequences is most compatible with the formation of nucleosome? Note: n means that the sequence gets repeated numerous times A) (ATCGTAGCAC)n B) (ACTGCTAGTG)n C) (ATTATCGCGC)n D) (GCAATTCGTA)n
C) (ATTATCGCGC)n
The enzyme XYZase has a single active site. Given that Vmax for XYZase is 40 mmol /second, and the total enzyme present in the reaction is 40 micromol, determine the turnover number. Note that 1 mmol = 1,000 micromol. A) 200 per second B) 500 per seond C) 1,000 per second D) 2,000 per second
C) 1,000 per second Kcat = Vmax / [ET] Kcat = (40*10^-3)/(40*10^-6) = 1000
Consider an enzyme-catalyzed reaction that follows Michaelis-Menten kinetics, what will be the initial velocity (V0) when the initial substrate concentration is 1 mM, Km equals 9 mM, and Vmax is 100 mmol/min? A) 1 mmol/min B) 5 mmol/min C) 10 mmol/min D) 25 mmol/min
C) 10 mmol/min Vo=Vmax[S]/Km+[S] Vo = (100*1) / (9+1) = 10
Given that Vmax for ABCase is 2.6 mmol · L-1 · S-1, and [ET] = 1.3 μmol · L-1, determine its turnover number. Note that ABCase has a single active site, and 1 mmol = 1000 μmol. A) 500 per second B) 1000 per second C) 2000 per second D) 2600 per second
C) 2000 per second Kcat = turnover number Kcat = Vmax / [ET]
The size of human genome is _________ in a haploid cell. Suppose every nucleosome covers 200 base-pairs of DNA. How many nucleosomes are present in a typical diploid human cell? A) 2 billion base-pairs; 10 million B) 3.2 billion base-pairs; 16 million C) 3.2 billion base-pairs; 32 million D) 6.4 billion base-pairs; 64 million
C) 3.2 billion base-pairs; 32 million
Which of the following can most likely contribute to epigenetic regulation? A) Several nucleotides in a region of DNA have been mutated B) A stretch of nucleotides in DNA have been deleted C) A chromatin region becomes heavily methylated, resulting in silencing of genes in that region D) A gene is split into two chromosomes
C) A chromatin region becomes heavily methylated, resulting in silencing of genes in that region A mutation alters the nucleotide sequence in DNA. An epigenetic change does not alter the DNA sequence, but can be inherited by daughter cells.
Each plasmid has several features that are important for its function. For the feature highlighted by Box1, which of the following phrases best describes its function? A) A region with multiple restriction sites for insertion of a DNA fragment B) The site where DNA replication can begin C) A gene that allows cells to grow when exposed to certain conditions D) A region that DNA is prioritized for repair
C) A gene that allows cells to grow when exposed to certain conditions amp: marker gene; a gene that allows cells to grow when exposed to certain conditions polylinker: a region with multiple restriction sites for insertion of a DNA fragment Ori: origin of replication; the site where DNA replication can start
ADME property of a drug refers to its performance in terms of A) Administration, dependence potential, mechanism understood, and excretion B) Administration, distribution, mechanism understood, and expenses C) Absorption, distribution, metabolism, and excretion D) Absorption, dependence potential, metabolism, and expenses
C) Absorption, distribution, metabolism, and excretion
Both salicylic acid and acetylsalicylic acid (i.e., aspirin) can block biosynthesis of prostaglandin H2 by inhibiting prostaglandin H2 synthase. However, salicylic acid is a competitive inhibitor while acetylsalicylic acid is an irreversible inhibitor for this enzyme. Why is that? A) Acetylsalicylic acid binds to a site distinct from the active site of the enzyme B) Acetylsalicylic acid forms a covalent bond with the substrate of the enzyme C) Acetylsalicylic acid transfers its acetyl group to a serine residue in the active site of the enzyme D) Acetylsalicylic acid has a much higher affinity to the enzyme than salicylic acid
C) Acetylsalicylic acid transfers its acetyl group to a serine residue in the active site of the enzyme
The following diagram illustrates the relationship among holoenzyme, cofactor, and apoenzyme. Which of the statements is a correct description of the diagram? A) Holoenzyme (1) + cofactor (2) make an apoenzyme (3) B) Cofactor (1) + holoenzyme (2) make an apoenzyme (3) C) Apoenzyme (1) + cofactor (2) make a holoenzyme (3) D) Holoenzyme (1) + apoenzyme (2) make a cofactor (3)
C) Apoenzyme (1) + cofactor (2) make a holoenzyme (3)
Metal ions are coordinated to this residue ___________ in the active site of DNA polymerase. The functions of these metal ions include _____________________. A) Histidine; activating 5'-OH and stabilizing pyrophosphate B) Alanine; activating a water molecule and promoting H-bonding between the bases C) Aspartate; activating 3'-OH and stabilizing pyrophosphate D) Glutamine; promoting nucleophilic attack by 3'-OH
C) Aspartate; activating 3'-OH and stabilizing pyrophosphate
Refer to the sugar structures on the right. The sugar labeled ______ is sucrose, while the sugar labeled ______ is glucose. A) A, E B) B, C C) B, E D) D, E E) E, C
C) B, E
Which of the following is a reason that rNTP is not used by DNA polymerase in DNA synthesis? A) The cellular concentration of rNTP is too low as compared to dNTP B) rNTP is only present in the cytoplasm and not in the nucleus C) Binding of rNTP to the DNA polymerase is not optimal for its incorporation into primer strand D) Only dNTP can base-pair with the nucleotide on the DNA template strand
C) Binding of rNTP to the DNA polymerase is not optimal for its incorporation into primer strand Steric constraints prevent DNA polymerase from using rNTP precursors, despite the much higher concentration: - the extra -OH would lead to misalignment - dNTP always used as DNA substrate - discriminator amino acids check to make sure it really is dNTP
Compare and contrast base-excision repair and nucleotide-excision repair: A) Both repair require photolyase and demethylase B) Both repair require glycosylase and AP endonuclease C) Both repair require DNA synthesis and ligase D) Both repair require topoisomerase I to generate a single-strand break
C) Both repair require DNA synthesis and ligase
Aspartate transcarbamoylase (ATCase) is composed of 6 regulatory subunits and 6 catalytic subunits. One important regulator of ATCase is CTP, and interestingly CTP does not resemble either of its two substrates. How does CTP inhibit ATCase? A) CTP forms a complex with the substrates of ATCase B) CTP binds to the catalytic subunits of ATCase and keeps the enzyme in the R state C) CTP binds to the regulatory subunits of ATCase and keeps the enzyme in the T state D) CTP binds to the interface of regulatory and catalytic subunits to prevent the binding of substrates
C) CTP binds to the regulatory subunits of ATCase and keeps the enzyme in the T state
GPR41 is an important drug target. Two lead candidates that can beneficially modulate the activity of GPR41 have been identified. Among them, candidate ABC has a Kd of 1.4 nM, and candidate DEF has a Kd of 140 nM. Which of the following statements regarding these two candidates is most likely to be correct? A) Candidate ABC has to be a competitive inhibitor of GPR41 B) Candidate DEF has to be a noncompetitive inhibitor of GPR41 C) Candidate ABC has a higher affinity to GPR41, and it would be a more potent drug than candidate DEF D) Candidate ABC has a lower affinity to GPR41, and it would be a less potent drug than candidate DEF
C) Candidate ABC has a higher affinity to GPR41, and it would be a more potent drug than candidate DEF
Drugs can be metabolized in our liver. For example, ibuprofen can be oxidized and then conjugated to glutathione. What purpose(s) does conjugating glutathione to ibuprofen serve? A) Conjugating glutathione to ibuprofen enhances the pain killing effect of the drug B) Conjugating glutathione to ibuprofen helps retaining the drug in our body C) Conjugating glutathione to ibuprofen increases the solubility of the drug and facilitates its excretion D) Conjugating glutathione to ibuprofen enhances the drug's ability to pass blood-brain barrier
C) Conjugating glutathione to ibuprofen increases the solubility of the drug and facilitates its excretion
2,3-BPG is an important regulator for hemoglobin. The three curves on the graph represent oxygen binding behavior of hemoglobin of normal blood (termed Hb-normal), hemoglobin of blood from individual adapted to high altitudes (termed Hb-high altitude), and hemoglobin stripped of 2,3-BPG (termed Hb-no-BPG). Which of the following is a correct match? A) Curve 1: Hb-normal; Curve 2: Hb-high altitude; Curve 3: Hb-no-BPG B) Curve 1: Hb-no-BPG; Curve 2: Hb-high altitude; Curve 3: Hb-normal C) Curve 1: Hb-no-BPG; Curve 2: Hb-normal; Curve 3: Hb-high altitude D) Curve 1: Hb-high altitude; Curve 2: Hb-normal; Curve 3: Hb-no-BPG
C) Curve 1: Hb-no-BPG; Curve 2: Hb-normal; Curve 3: Hb-high altitude
In the picture of an uncharged tRNAAla to the right, this letter ______is where you would find the "wobble" position, while this letter _______ is where the amino acid would be covalently linked. A) A ; B B) B ; A C) D ; A D) A ; D E) C ; A
C) D ; A
ATP has a very high phosphoryl transfer potential for many reasons. Which statement below is NOT such a reason? A) Charge repulsion higher before ATP hydrolysis B) Resonance stabilization of ATP hydrolysis products C) Decrease in entropy after ATP hydrolysis D) Stabilization by hydration after ATP hydrolysis E) Less hydrogen bond donors and acceptors before ATP hydrolysis
C) Decrease in entropy after ATP hydrolysis
In addition to making DNA polymerase more processive, sliding clamp has other important functions, and one of these functions is: A) Directing the assembly of pre-RC to the origin B) Acting as a licensing factor to recruit helicase to the origin C) Directing the loading of H3/H4 tetramer to newly synthesized DNA D) Directing the binding of telomerase to the telomere region of DNA
C) Directing the loading of H3/H4 tetramer to newly synthesized DNA Roughly equal distribution of parental H3-H4 tetramers to daughter DNA molecules: - H2A-H2B dimers completely dissociated from parent histone octamer - H3-H4 stay bound to DNA- chromatin needs to be rereplicated after DNA synthesis - histone chaperones restore the full complement of histones to daughter molecules ** NAP-1 loading H2A-H2B dimer randomly redelivered ; load dimers ** CAF-1 loading newly synthesized H3-H4 tetramer delivers new tetramer to newly synthesized naked DNA and knows based on which strand is interacting with the sliding clamp
DnaA, DnaB, and DnaC are proteins that are important for this organism _________ to start DNA replication. Among these three, this one ______________ binds DNA in an ATP-dependent manner, while this one __________________ keeps this one _________________ inactive. A) Yeast; DnaA; DnaB; DnaC B) E. coli; DnaC; DnaB; DnaA C) E. coli; DnaA; DnaC; DnaB D) Yeast; DnaA; DnaB; DnaC
C) E. coli; DnaA; DnaC; DnaB Bacterial proteins: DnaA - initiator protein DnaB - helicase DnaC - helicase loading protein *** loads helicase and keeps it in the inactive form, preventing helicase from binding to wrong state in the active form **** only in prokaryotic systems
In prokaryotic translation elongation, what two GTP-bound factors are required to efficiently elongate the peptide chain by one amino acid? A) IF1 & IF2 B) eIF4E & eIF4G C) EF-Tu & EF-G D) EF1 & EF2 E) release factors
C) EF-Tu & EF-G
Suppose you have a DNA fragment you would like to insert into the pSAP plasmid shown in the question above. The fragment has PstI and EcoRI restriction endonuclease sites near the 5' end and HindIII and Smal restriction endonuclease sites near the 3' end. Choose the one or two best restriction endonucleases to digest both the DNA fragment and pSAP. A) EcoRI only B) PstI and HindIII C) EcoRI and HindIII D) EcoRI and SmaI
C) EcoRI and HindIII
As shown in the table below, the SIR2 protein complex is one of the histone modifying enzymes. Most likely, which of the following functions could the SIR2 complex have in an organism? A) Promoting gene expression B) Recruiting RNA polymerase II to a gene C) Enabling gene silencing and possibly heterochromatin formation D) Facilitating the binding of H3 variant to centromere
C) Enabling gene silencing and possibly heterochromatin formation
15. Iron toxicity occurs in patients in which iron levels are high. In patients suffering from this type of toxicity, which of the following scenarios would you predict to be happening? A diagram of the two mRNA transcripts is shown below. Ferritin mRNA Transferrin Receptor mRNA A) Transferrin receptor will be translated efficiently. B) Iron response proteins (IRPs) will not contain iron in their binding sites. C) Endonucleases will cleave transferrin receptor mRNA, causing the mRNA to be rapidly degraded. D) Iron response proteins (IRPs) will be bound to iron response elements (IREs) on the 3' UTR of transferrin receptor mRNA, allowing translation of transferrin receptor. E) Iron response proteins (IRPs) will be bound to iron response elements (IREs) on the 5' UTR of ferritin mRNA, preventing the translation of ferritin.
C) Endonucleases will cleave transferrin receptor mRNA, causing the mRNA to be rapidly degraded.
Which of the following catalyzes the activation of trypsin from trypsinogen? A) Renin B) π-chymotrypsin C) Enteropeptidase D) Salt
C) Enteropeptidase - enteropeptidase produced in small amounts to convert just enough trypsinogen to trypsin and then trypsin itself can convert trypsinogen to trypsin
RecA is a prokaryotic protein, and its eukaryotic equivalent is Rad51. Which of the following biological processes require RecA/Rad51? A) Separating two sister chromatids B) Repairing pyrimidine dimer C) Facilitating DNA synapsis for homologous recombination D) None of the above
C) Facilitating DNA synapsis for homologous recombination
Which of the following statements about fetal hemoglobin is correct? A) Fetal hemoglobin has higher affinity for 2,3-BPG B) The four subunits of fetal hemoglobin do not show any cooperativity C) Fetal hemoglobin 2 alpha and 2 gamma subunits and a higher affinity for O2 than adult hemoglobin D) Fetal hemoglobin has 2 beta and 2 gamma subunits and a lower affinity for O2 than adult hemoglobin
C) Fetal hemoglobin 2 alpha and 2 gamma subunits and a higher affinity for O2 than adult hemoglobin Fetal hemoglobin - fetal hemoglobin must bind oxygen when the mother's hemoglobin is releasing oxygen - in fetal hemoglobin, the Beta chain is replaced with a gamma chain - the fetal alpha2gamma2 hemoglobin does NOT bind 2,3-BPG as tightly as adult hemoglobin binds 2,3-BPG - the fetal red cells have higher affinity for oxygen - two histidine residues have been replaced with serine residues in the hemoglobin structure of fetal Hb and therefore, there is less of an attraction to bind to 2,3-BPG, so this interaction is less tight - fetal Hb has a LOWER affintiy for 2,3-BPG and a HIGHER affinity for oxygen O2 flows from maternal oxyhemoglobin to fetal deoxyhemoglobin
FRET refers to __________________, and this technique allows scientists to examine _____________. A) Fluorescence resonance energy transfer; if a molecule is present in the cell B) Fluorescence in situ hybridization; if a DNA sequence is present in the cell C) Fluorescence resonance energy transfer; if two molecules are in close proximity and interact D) Far right energy transfer; if a molecule can emit fluorescence light
C) Fluorescence resonance energy transfer; if two molecules are in close proximity and interact
Pre-RC is assembled in __________, activated in ______________, and this enzyme _______ is required for its activation. A) M phase; G1 phase; topoisomerase II B) G1 phase; S phase; helicase C) G1 phase; S phase; Cdk (cyclin-dependent protein kinase) D) S phase; S phase; Cdk (cyclin-dependent protein kinase)
C) G1 phase; S phase; Cdk (cyclin-dependent protein kinase) DNA replication initiation in eukaryotes: G1: - ORC (origin recognition complex) binds along with Cdc6 and Cdt1 - Cdc6 is helicase loading and keeps helicase inactive - Cdt1 always bound to origin unless that piece needs to be replicated - pre-RC (pre-replicative complex) = ORC + Cdc6 + Cdt1 + helicase assembly of pre-RC occurs ONLY in the G1 PHASE S: Cdk (cycling dependent protein kinase) is active in the S phase and is a mediated phosphorylation - phosphorylates Cdc6 which promotes Cdt1 to also fall off - now helicase = active and begins melting DNA Activation of the pre-RC occurs ONLY during S PHASE
Below are the promoter regions for 5 gene sequences in E. coli. Knowing that the standard promoter sequence is 5' TTGATA 3' for the -35 region and 5' TATGAT 3' for the -10 region, which gene below has the weakest promoter? A) Gene 1 B) Gene 2 C) Gene 3 D) Gene 4 E) Gene 5
C) Gene 3
Screening cDNA library is more beneficial than screening genomic library in identifying a gene of interest. This is because A) It is easier to generate a cDNA library than a genomic library B) cDNA library typically has a larger size than genomic library C) Genomic library often contains clones with non-functional gene fragment D) DNA clones in cDNA library are circular, and they are easier to replicate
C) Genomic library often contains clones with non-functional gene fragment
The palm region of DNA polymerase contains several highly conserved residues that are important for its function. Those include a pair of aspartate residues, a glutamine residue, and an arginine residue. Which of them are important for the confirmation of correct base-pairing between incoming dNTP and the template? A) All of them B) Aspartate and glutamine C) Glutamine and arginine D) The two aspartates
C) Glutamine and arginine Palm region of DNA polymerase: The palm region is composed of a beta sheet and contains the primary element of the catalytic site. - this region binds two divalent metal ions that alter the chemical environment around the correctly-based paired dNTP and the 3'-OH of the primer - this region also monitors the base pairing of the most recently added nucleotides Arginine & Glutamine are two residues in the palm region of DNA polymerase and they are important in checking if a correct base pair is formed in the active site.
The sequence of the dipeptide shown on the right is _______, and the bond # ____ has a partial double bond feature and thus cannot freely rotate. A) FL, #3 B) HK, #4 C) HR, #3 D) RH, #3
C) HR, #3
Thalassemia is a disease that is caused by an imbalanced production of hemoglobin. Which of the following is correct regarding thalassemia? A) There are two types of thalassemia, one due to the insufficient production of alpha chain and the other due to the insufficient production of gamma chain. B) Insufficient production of alpha chain leads to aggregation of beta chains, which cause the loss of red blood cells. C) Hemoglobin composed entirely of beta chains has a higher affinity to oxygen, and that is why it is also called hemoglobin H. D) alpha-thalassemia is more prevalent than beta-thalassemia.
C) Hemoglobin composed entirely of beta chains has a higher affinity to oxygen, and that is why it is also called hemoglobin H. Thalassemia - an imbalanced production of hemoglobin chains can cause this - an inherited blood disorder, with fewer healthy RBCs and less functional hemoglobin than normal - alpha-thalassemia: the alpha chain is not produced in sufficient quantity. Form Hemoglobin H (HbH), which is composed of only the beta chains and binds oxygen with high affinity - beta-thalassemia: the beta chain is not produced in sufficient quantity. Without the beta chain, the alpha chains can form insoluble aggregates inside immature RBCs, thereby leading to the loss of the RBC *** more prevalent because our body only codes for two beta genes, instead of 4 genes like for alpha. ***beta is a less common gene and therefore more likely to be missing making beta-thalassemia more prevalent
Myoglobin and hemoglobin are two important oxygen-binding proteins. Which of the following is a correct statement regarding the structure and function of these two proteins? A) Myoglobin has a higher affinity to oxygen and serves to transport oxygen from lung to tissues B) Myoglobin is a monomeric protein that serves to carry CO2 from tissues to lung C) Hemoglobin has quaternary structure and displays cooperative oxygen binding and release D) Hemoglobin is composed of four identical subunits
C) Hemoglobin has quaternary structure and displays cooperative oxygen binding and release Myoglobin: * a single polypeptide chain consisting of mainly alpha helices arranged to form a globular structure * has a single bound heme * the binding of oxygen by myoglobin is NOT cooperative * myoglobin binds oxygen in MUSCLE CELLS * has high affinity for oxygen; main function is to store oxygen for later use Hemoglobin: * an allosteric protein with quaternary structure * each subunit (4) has an active binding site * a RBC protein that carries oxygen from the lungs to the tissues * a tetramer consisting of two alpha subunits and two beta subunits (each subunit has a bound heme group) * hemoglobin is an allosteric protein that displays cooperativity in oxygen binding and release
Shown on the right (the top panel) is MCS (multiple cloning sites) of a plasmid, which is composed of a set of restriction enzyme recognition sites that allow researchers to insert their DNA fragment into the plasmid. Based on the restriction map for GPA1 gene as well as its flanking region, decide the restriction enzyme(s) that you can use to insert (aka clone) GPA1 gene to the above plasmid. A) KpnI and XbaI B) KpnI and SacII C) HindIII and BamHI D) HindIII and XbaI
C) HindIII and BamHI
Many drugs can be conjugated to glutathione, a tripeptide (gamma-Glu-Cys-Gly), in liver. Typically, this conjugation event serves to A) Enhance the potency of the drug B) Facilitate the effective distribution of the drug to brain C) Increase the solubility of the drug and facilitate its excretion D) Increase the molecular weight of the drug and impair its removal from the body
C) Increase the solubility of the drug and facilitate its excretion Drug Metabolism: Body's two main pathways, for defending against foreign compounds 1) oxidation by cytochrome P450 enzymes in the liver 2) conjugation by glutathione, glucuronic acid, and sulfate in the liver
Which of the following conditions shifts the oxygen binding curve of hemoglobin (solid line in the graph) to the right? A) Decreased hydrogen ions B) Decreased CO2 C) Increased 2,3-BPG D) Decreased exercise
C) Increased 2,3-BPG Right shift = decreased affinity for oxygen; more willing to unload oxygen in tissues hemoglobin high altitude - count of 2,3-BPG increases in high altitudes - right shift, lower oxygen affinity and more effective unloading - slightly less loading but BEST unloading and therefore overall more effective transport of oxygen from lungs to tissue H+ and carbon dioxide promote the release of oxygen - carbon dioxide and H+, produced by actively respiring tissues, enhance oxygen release by hemoglobin - the stimulation of oxygen release by carbon dioxide and H+ is called the Bohr effect -important bc CO2 and H+ in high concentration in tissues when exercising, therefore need O2 transport to be more effective at these times
Consider the kinetics data for each enzyme in the presence and absence of its inhibitor, select the statement that is correct. A) Inhibitor A binds to a regulatory site of carbonic anhydrase B) Inhibitor B is a structural mimic of the substrate for chymotrypsin and binds to its active site C) Inhibitor C only binds to the enzyme-substrate complex D) All of the above
C) Inhibitor C only binds to the enzyme-substrate complex Competitive Inhibition: Inhibitor binds to the active site, therefore increasing the amount of [S] needed to need Vmax constant. Therefore, competitive inhibitors raise Km and leave Vmax unchanged. Uncompetitive Inhibitors: Inhibitor binds to the UNITED ES complex. It depletes [ES], therefore shifting the reaction to the right and increasing the enzyme's affinity for the substrate but doing this all at a slower rate. Therefore, both Km and Vmax decrease. These values decrease to the same extent, and the parent line and inhibitor line remain parallel. Noncompetitive Inhibitor: Inhibitor can bind to either E or ES, therefore Km is not changed but Vmax decreases.
One mutation that occurs on the beta chain of hemoglobin is the replacement of a glutamic acid residue at position 6 to a valine. What is the consequence of such a mutation? A) It can lead to destabilization of the beta chain, which can trigger beta-thalassemia B) It can lead to an imbalance between the alpha and gamma chain of hemoglobin C) It can lead to the polymerization of hemoglobin molecules that are in deoxy states D) It increases the affinity of hemoglobin to oxygen
C) It can lead to the polymerization of hemoglobin molecules that are in deoxy states
Sildenafil is able to stimulate smooth muscle relaxation because A) It inhibits the biosynthesis of prostaglandins B) It inhibits the proliferation of smooth muscle cells C) It inhibits the enzyme that catalyzes hydrolysis of cGMP to GMP D) It inhibits the activity of myosin
C) It inhibits the enzyme that catalyzes hydrolysis of cGMP to GMP Sildenafil: * a case of drug development by serendipity * increases smooth muscle relaxation * Because sildenafil is a structural mimic of cGMP, it can function as an inhibitor of phosphodiesterase 5 allowing cGMP to increase and therefore increase muscle relaxation * phosphodiesterase 5 breaks down cGMP in the body
Remdesivir (RDV) is a potent inhibitor for RNA-dependent RNA polymerase, an essential enzyme for the replication of coronaviruses. The triphosphate form of the inhibitor (RDV-TP) competes with its natural counterpart ATP for the synthesis of RNA (note that ATP is a natural substrate for RNA synthesis, along with TTP, CTP, and GTP). How would RDV-TP change the kinetics of its target enzyme? A) It will decrease the Vmax of the enzyme B) It will decrease the Kcat of the enzyme C) It will increase the Km of the enzyme D) It will not affect the kinetics of the enzyme
C) It will increase the Km of the enzyme Competitive inhibitor: Vmax = constant Km = increased
Based on the linweaver-burk plot shown, estimate the Km and Vmax. A) Km = 0.59 μmol · min-1; Vmax = 0.44 mM B) Km = 0.59 μmol · min-1; Vmax = 0.44 mM C) Km = 0.44 mM; Vmax = 0.59 μmol · min-1 D) Km = -0.44 mM; Vmax = 0.59 μmol · min-1
C) Km = 0.44 mM; Vmax = 0.59 μmol · min-1
Based on the double-reciprocal plot for the enzyme inhibition shown on the right, which of the following is true? A) Line 1 contains an uncompetitive inhibitor B) Line 2 contains an uncompetitive inhibitor C) Line 1 contains a noncompetitive inhibitor D) Line 2 contains a noncompetitive inhibitor
C) Line 1 contains a noncompetitive inhibitor Competitive Inhibitor: Competitive inhibitors bind directly to an enzyme's active site and, in doing so, decrease the affinity of enzyme for it substrate. In other words, competitive inhibitors result in an increase in the Km value of an enzyme for its substrate. Competitive inhibitors do not change the enzyme's Vmax. Uncompetitive Inhibitor: Uncompetitive inhibitors bind to the enzyme-substrate complex, thereby decreasing an enzyme's maximal rate of catalysis and increasing its affinity for its substrate. Thus, uncompetitive inhibitors both decrease Vmax of an enzyme and decrease its Km. As [ES] continues to become inhibited, the K1 reaction will shift to the right, pushing more E+S to turn to ES, thereby, increasing enzyme affinity for substrate (and decreasing Km). Noncompetitive Inhibitor: Noncompetitive inhibitors bind to either an allosteric site on the enzyme or to the enzyme-substrate complex. In either case, the result is a decrease in the enzyme's Vmax and no change to its Km (both [E] and [ES] depleted and therefore Km is constant, but the rate of enzymatic reaction obviously must decrease).
OMP decarboxylase (ODCase), is an enzyme that catalyzes decarboxylation of OMP (the molecule on the left side of the reaction shown) to form UMP (the molecule on the right side of the reaction shown) and CO2. According to enzyme classification, ODCase should be classified as A) Oxidoreductase B) Hydrolase C) Lyase D) Ligase
C) Lyase A) Oxidoreductase: catalyze oxidation-reduction reactions that involve the TRANSFER OF ELECTRONS B) Hydrolase: catalyze hydrolysis reactions by transfer of functional groups to water; INTRODUCE WATER TO BREAK BONDS C) Lyase: catalyze reactions in which functional groups are added to double bonds or double bonds are formed via removal of functional groups D) Ligase: catalyze the formation of C-C, C-S, C-O, and C-N bonds by condensation reactions coupled to ATP cleavage. JOINING of molecules. Need hydrolysis of the diphosphate bond in ATP as energy input Isomerase: catalyze the rearrangement of bonds within a single molecule to yield an isomer Transferase: catalyze the movement of a functional group from one molecule to another
XIC is a locus on X-chromosome, and its gene product is XIST. If a mutation has occurred on the XIC locus of an X-chromosome that prevents it from expressing XIST, what would be a functional consequence? A) Other genes from that X-chromosome are not able to be expressed B) If the other X-chromosome of the individual has the same mutation, that individual would live longer C) Most likely that X-chromosome would not be inactivated D) All of the above
C) Most likely that X-chromosome would not be inactivated An individual X chromosome can be completely inactivated by heterochromatin formation: XIC (x-inactivation center) transcribed XIST (x inactivation specific transcript) which spreads along an x chromosome - need to silence an X gene so girls don't have double ** dosage compensation XIC: refers to x chromosome inactivation center, a locus on X-chromosome XIST: refers to x-inactivation specific transcript, a non-coding RNA that directly binds X-chromosome and initiates its inactivation
MutL, MutS, and MutH are three bacterial proteins required for mismatch repair. What are their specific functions? A) MutL serves to recognize mismatch; MutS serves to recognize GATC sequence; MutH serves to recruit AP endonuclease B) MutL serves to recognize single-strand break; MutS serves to recruit exonuclease; MutH serves to attract DNA polymerase C) MutL serves to distinguish new strand from template strand by looking for GATC sequence; MutS serves to identify mismatch and bind to it; MutH has endonuclease activity and cuts the newly synthesized strand identified by MutL D) MutL serves to methylate GATC sequence; MutS serves to search for GATC sequence; MutH serves to cut the GATC sequence
C) MutL serves to distinguish new strand from template strand by looking for GATC sequence; MutS serves to identify mismatch and bind to it; MutH has endonuclease activity and cuts the newly synthesized strand identified by MutL
Which of the following correctly describes mismatch repair? A) MutH binds the mismatch, MutS identifies the new strand, and MutL seals DNA B) MutS recognizes the mismatch, MutH identifies the new strand, and MutL hydrolyzes the backbone C) MutS binds the mismatch, MutL identifies the new strand, and MutH breaks the new strand D) Both prokaryotes and eukaryotes need all three proteins (MutH, MutL, MutS) for mismatch repair
C) MutS binds the mismatch, MutL identifies the new strand, and MutH breaks the new strand Mismatch repair (strand specific): - detecting and binding a mismatched base pair - removing the lesion from the newly synthesized strand but NOT from the parental strand - must be able to distinguish two strands - repairing DNA synthesis - enzymes identify the recently copied DNA strand, snip out a portion of the DNA strand that contains the mismatched nucleotides and resynthesize the segment correctly Special proteins required: MutS: detecting and binding a mismatched base pair (Scanning for the mismatch) MutL: looking for GATCH sequence that is close to distinguish the newly synthesized strand from the template strand (looking for GATC) MutH: nicks the unmethylated GATC sequences (hydrolyzing the phosphodiester linkage) prokaryotes: MutS recuits MutL which recuits MutH which gets activated and cuts eukaryotes: MutS and MutL only
Suppose the base-pair shown on the right is present in double-stranded DNA. From left to right, the chemical groups located in the major groove that form a code to distinguish this base-pair from three others are: A) H, N, O, and NH2 B) N, NH2, and O C) N, O, NH2, and H D) N, C, O, and H
C) N, O, NH2, and H
In DNA, the bond that joins purine and deoxyribose is referred to as _____________, which could have __________ conformation. A) Phosphodiester bond; trans B) Phosphodiester bond; cis C) N-glycosidic bond; either anti or syn D) N-glycosidic bond; syn only
C) N-glycosidic bond; either anti or syn
Which of the following sequences is the most compatible with the formation of an alpha helix? A) ATTSACYLENGPSD B) ASKDNPRGITKNDG C) PLKACEYQLEAAMR D) LKACPDKNGCRITK
C) PLKACEYQLEAAMR Amino acids rarely found in alpha helix include: Pro, (side chain has an N with three H bonds and not likely to participate in H bonding) Ile, Thr, Val (branching leads to lots of clashing) Ser, Asn, Asp (polar groups near main chain may compete for H bonding) Gly (putting a highly flexible structure into a highly organized structure will lead to a large entropy cost)
Consider a fibrous protein that is composed of two alpha helices. The residues present in this protein would most likely be clustered on which region of the Ramachandran plot? A) Region A B) Region B C) Region C D) Region D
C) Region C
MutL/MutS/MutH proteins are important for which of the following? A) Repairing damaged base B) Repairing double strand breaks C) Repairing mismatched base-pair D) Repairing pyrimidine dimer
C) Repairing mismatched base-pair
Which of the following processes involves adenylation (aka AMPylation, or adenylylation)? A) Adding a new nucleotide to the 3'-end of a DNA strand B) Removing a newly added nucleotide by 3'-exonuclease C) Sealing the two Okazaki fragments by ligase D) Adenylation never occurs in nature
C) Sealing the two Okazaki fragments by DNA ligase seal together the okazaki fragments (requires ATP) He Likes The Clover Helicase, Ligase, Top 2, Clamp loader all require energy input to function
Even though any sequence of DNA can potentially be wrapped around histone octamer to form a nucleosome, some sequence is preferred. Which of the following is a preferred sequence for the formation of nucleosome? A) Sequence that has high GC content B) Sequence that has high AT content C) Sequence that alternates between A:T- and G:C-rich region with a periodicity of 5 base-pair D) There is no such sequence
C) Sequence that alternates between A:T- and G:C-rich region with a periodicity of 5 base-pair Site of contact between histones and DNA: * the majority of H-bonds are between the histone proteins and the oxygen atoms in the phosphodiester backbone near the MINOR groove of the DNA * nucleosomes prefer to bind bent DNA * 146 base pairs, 14 turns, therefore 14 contact sites between DNA and octamer * A:T base pairs have a tendency to bind toward the minor groove and G:C base pairs have the opposite tendency. * Sequences that alternate between A:T and G:C rich sequences with periodicity of ~5bp will act as preferred nucleosome-binding sites.
Which of the following residues are rarely found in an alpha helix? A) Glu, Arg, Lys B) Tyr, Phe, Leu C) Ser, Val, Thr D) Phe, Trp, Tyr
C) Ser, Val, Thr Amino acids rarely found in alpha helix include: Pro, Ile, Thr, Val Ser, Asn, Asp Gly
The antibiotic ______________ inhibits prokaryotic translation by competing with fMet for initiation as well as causing the misreading of mRNA. A) Tetracycline B) Chloramphenicol C) Streptomycin D) Erythromycin E) Puromycin
C) Streptomycin
Which basal transcription factor binds FIRST to the promoter region of RNA polymerase II synthesized genes? A) TFIIA B) TFIIB C) TFIID D) TFIIE E) TFIIH
C) TFIID
In deriving the Michaelis-Menten equation, it was assumed that A) the concentration of initial substrate is much lower than that of enzyme B) the enzyme is quickly saturated by the substrate so ES concentration stays at a steady state C) The enzyme will be either free (E) or bound to the substrate (ES) D) The rates of forward and reverse reactions are the same so ES concentration does not change
C) The enzyme will be either free (E) or bound to the substrate (ES) 1. ES quickly comes to a steady state, so [ES] is constant. --- 2. [S]>>>[Et], so the fraction of S that binds to E is negligible, and [S] is constant are early time points. --- 3. at early time points, where initial velocity is measured, [P] approx = 0.
Consider the image shown on the right. Which of the following is true? A) The left half illustrates chromatin region with no H1 B) The right half illustrates chromatin region with heavily methylated histones C) The left half illustrates chromatin region with low level of acetylated histones D) All of the above are correct
C) The left half illustrates chromatin region with low level of acetylated histones Prominent covalent modifications found on histones: Acetyl lysine - neutralizes lysine on histone which weakens DNA and histone interactions Methyl lysine - preserves positive charge of lysine on histone and tends to enhance DNA and histone interactions Phosphorylating serine - introduce an additional negative charge to serine on histone, weakening DNA and histone interaction
Histone H1 can be very important in regulating gene expression. In a chromatin region that contains actively transcribed genes, one would expect which of the following about histone H1? A) The level of histone H1 in that region is likely to be high B) The lysine residues of H1 in that region are heavily methylated C) The level of histone H1 in that region is likely to be low D) Histone H1 binds very tightly to the histone octamers found in that region
C) The level of histone H1 in that region is likely to be low
Chromatin remodeling complexes often have ATPase activity. Which of the following would likely to happen if you make a mutation to disrupt the ATPase activity of these complexes? A) The mutant complexes would promote ejecting of histone octamer from the nucleosome B) The mutant complexes would facilitate sliding of DNA around the histone octamer C) The mutant complexes would fail to hydrolyze ATP and fail to remodel chromatin structure D) The mutant complexes would gain the ability to methylate DNA
C) The mutant complexes would fail to hydrolyze ATP and fail to remodel chromatin structure
Consider the circular DNA molecules shown on the right, which of the following statements is true? A) The one on the left has a larger linking number than the one on the right B) The one on the left has the same twisting number as the one on the right C) The one on the left has a negative writhing number D) The one on the right has a negative writhing number
C) The one on the left has a negative writhing number DNA Supercoiling: Lk = Tw + Wr
Shown on the right are the three known possible conformations a double-stranded DNA could adopt. Which of them is left-handed and contains purine nucleotide in syn conformation? A) The one on the left B) The one in the middle C) The one on the right D) All double-stranded DNA molecules are right-handed
C) The one on the right The one of the left is B. The one in the middle is A. The one on the right is Z. B form - right handed, most cellular DNA, anti, fully hydronated form A form - right handed, double stranded RNA, DNA/RNA hybrid, anti, dehydrated DNA Z-form - left handed, based on crystal structure of CGCGCG, anti or syn!
ATCase is an example of allosteric enzyme that can exist as either R state or T state. Note that R state is the highly active form of the enzyme and T state is the less active form of the enzyme. Typically, in the absence of substrate, ATCase predominantly exists in the T state. Suppose a mutation caused the ATCase to always stay in the R state. How would this mutation affect the relationship between the ratio of the reaction (V) and substrate concentration ([S])? A) The plot of V vs [S] would be shaped like an S, sigmoidal B) The enzyme would be less active because it is a mutant version C) The plot of V vs [S] would likely follow Michaelis-Menten kinetics as no more T/R transitions D) None of the above
C) The plot of V vs [S] would likely follow Michaelis-Menten kinetics as no more T/R transitions * instead of being sigmoidal like it would be with cooperative binding, it would no longer exhibit this behavior
CENP-A is a variant of histone H3, and it is important for the proper function of centromere. Hypothetically, what would be the consequence for a mutant that lacks CENP-A? A) The ends of the mutant's chromosome would be destroyed by nuclease in the cell B) The mutant DNA cannot be properly replicated C) The sister chromatids of the mutant cannot be separated properly during cell division D) All of the above
C) The sister chromatids of the mutant cannot be separated properly during cell division
Biochemical standard free energy change differs from standard free energy change in that A) The standard conditions of biochemical standard free energy charge include 1 M of H2O B) Biochemical standard free energy change is always larger than that of standard free energy change C) The standard conditions of biochemical standard free energy change include pH at 7.0 D) The standard conditions of biochemical standard free energy change include [H+] = 1 M
C) The standard conditions of biochemical standard free energy change include pH at 7.0
Val, Leu, Ile, and Thr are considered as essential amino acids from humans, meaning we have to get them from our diet. Structurally, what is a common feature shared by those four amino acids? A) They all have a polar side chain B) They all have a non-polar side chain C) They all have a branched side chain D) They all have an acidic side chain
C) They all have a branched side chain
Consider the circular DNA shown on the right. Which of the following statements is true regarding this circular DNA? A) The linking number of this DNA is larger than its twisting number B) The writhe number of this DNA is larger than its linking number C) This circular DNA is negatively supercoiled D) The linking number of this DNA is less than the sum of its twisting number and writhe number
C) This circular DNA is negatively supercoiled Twisting number (Tw): the number of times one strand completely wraps around another strand Linking number (Lk): the number of times that a strand of DNA winds in the right-handed direction around the helical axis when the axis lies in a plane Writing number (Wr): the number of times the long axis of the double helix crosses over itself or is wound in a cylindrical manner (- for right-handed) DNA Supercoiling: Lk = Tw + Wr
Which of the following is true regarding the graph shown on the right? A) This is a typical isoelectric focusing, and the protein C has the highest pI value B) This is a typical SDS-PAGE, and the protein A has the highest molecular weight C) This is a 2-D gel, and the protein B has the highest molecular weight D) This is a 2-D gel, and the protein C has both the highest pI and molecular weight
C) This is a 2-D gel, and the protein B has the highest molecular weight
Which of the following statements is true regarding the following peptide? A) This is a tetrapeptide, and it can absorb light at 280 nm B) This is a tripeptide, and it cannot absorb light at 280 nm C) This is a pentapeptide, and it can absorb light at 280 nm D) This is a pentapeptide, and it carries two net positive charges at pH 1
C) This is a pentapeptide, and it can absorb light at 280 nm Tryptophan (Trp) and tyrosine (Tyr) side chains can absorb light at 280 nm
Mass spectrometry analysis of tryptic fragments (i.e., fragments generated from trypsin digestion) instead of the intact protein is often used to identify a protein. Which of the following is the most likely reason for this? A) It is impossible to accurately measure the molecular weight of an intact protein by mass spectrometry B) It is more convenient to measure the molecular weight of tryptic fragments because they are smaller C) Two proteins could have exactly the same molecular weight, but it is less likely they have tryptic fragments with the same molecular weight D) All of the above
C) Two proteins could have exactly the same molecular weight, but it is less likely they have tryptic fragments with the same molecular weight
Suppose that you want to solve crystal structure of the spike proteins found on the surface of SARS-CoV-2 virus. Your research mentor has provided you some cell extracts that contain His-tagged spike proteins. Which of the following is the most suitable method for purifying the spike proteins from this point on? A) Running the cell extracts on SDS-PAGE B) Running the cell extracts on a 2-dimensional gel C) Using metal-ion beads and running affinity chromatography D) Running gel filtration chromatography
C) Using metal-ion beads and running affinity chromatography Affinity tag examples: * Histidine residues (HHHHHH) = "his tag" --> ligand = metal ions * Flag peptide (DYKDDDDK) = "flag tag" --> ligand = flag antibody * Maltose binding protein --> ligand = maltose * Protein A --> ligand = IgG EX: if a protein contains a "his-tag" then it can tightly bind to metal ions
What separates x-ray crystallography from NMR and cryo-EM in determining protein structure is A) X-ray crystallography is the only method that can achieve atomic resolution B) X-ray crystallography is the only method that does not require purified protein C) X-ray crystallography is the only method that relies on the formation of protein crystal D) X-ray crystallography is the only method that can determine structure of membrane proteins
C) X-ray crystallography is the only method that relies on the formation of protein crystal Methods to determine protein structure 1. xray crystallography 2. NMR spectroscopy 3. electron microscopy (flaw: cannot achieve atomic resolution)
Consider the structure shown below. In what kind of elements can this structure most likely be found? A) a coiled-coil B) a transmembrane helix C) a parallel beta-sheet D) DNA
C) a parallel beta-sheet A) a coiled-coil (this is an alpha helix) B) a transmembrane helix C) a parallel beta-sheet D) DNA
Acetyl CoA is a key metabolic intermediate. Coenzyme A (CoA) is an activated carrier of what type of molecule? A) phosphate B) electrons C) acyl groups D) aldehyde E) carboxylic acid
C) acyl groups
The pKa values for histidine's protonatable groups are 1.82, 6.00, and 9.17. The pKa value of 9.17 corresponds with histidine's _____ group; the pI of histidine equals _______. A) -COOH group; 7.59 B) R group; 6.00 C) amino group; 7.59 D) amino group; 3.91
C) amino group; 7.59 (9.7+6)/2 = 7.59
In general, an anti-parallel beta sheet is slightly more stable than a parallel beta sheet. This is because A) an anti-parallel beta sheet always contains a higher content of hydrophobic residues B) an anti-parallel beta sheet always has more strands C) an anti-parallel beta sheet allows the formation of stronger hydrogen bonds D) all of the above
C) an anti-parallel beta sheet allows the formation of stronger hydrogen bonds Anti-parallel: neighboring hydrogen-bonded polypeptide chains run in opposite directions H-bonding is not uniform H-bonding is collinear and therefore very strong Parallel: all of the N-termini of successive strands are oriented in the same direction uniform H-bonding; not collinear and therefore weak
All amino acids have both alpha-amino group and alpha-carboxyl group. But this amino acid _______ also has a beta-carboxyl group, and this amino acid _____ also has an epsilon-amino group. A) glutamic acid; arginine B) glutamine; arginine C) aspartic acid; lysine D) asparagine; lysine
C) aspartic acid; lysine
In SDS-PAGE, the function of SDS is to A) increase the size of the protein, so it can migrate slower in the electric field B) reduce the disulfide bonds present in the protein C) denature the protein and give protein extensive negative charges D) strengthen the hydrophobic interactions and keeping the subunits together
C) denature the protein and give protein extensive negative charges
What is the name of the sugar depicted on the right? A) glucose B) fructose C) galactose D) mannose E) maltose
C) galactose
Sucrose is a disaccharide composed of which two monosaccharides? A) glucose and galactose B) galactose and fructose C) glucose and fructose D) glucose and mannose E) two molecules of glucose
C) glucose and fructose
The molecule shown on the right is: A) adenosine triphosphate B) deoxy-adenosine triphosphate C) guanosine triphosphate D) deoxy-guanosine triphosphate
C) guanosine triphosphate
What type of interaction would you expect between serine and aspartic acid in a tertiary structure? A) hydrophobic interaction B) salt bridge C) hydrogen bond D) disulfide bond
C) hydrogen bond
Histone H1 is very unique, because A) it is the only histone protein that has more Asp/Glu than Lys/Arg B) it can form a dimer with any other histone proteins C) it binds to the linker DNA and facilitates the formation of 30 nm fiber D) it is always acetylated
C) it binds to the linker DNA and facilitates the formation of 30 nm fiber H1: linker Histone bound to the DNA where the double helix enters and leaves the nucleosome core (linker Histone)
As an inhibitor for the shikimate pathway, glyphosate (roundup) is able to kill weeds because A) it blocks the synthesis of branched chain amino acids such as Ile and Val B) It blocks the conversion of alpha-ketoglutarate to glutamate C) It blocks the synthesis of aromatic amino acids such as Phe and Trp D) It stimulates human beings to pull the weeds out of their yards
C) it blocks the synthesis of aromatic amino acids such as Phe and Trp
Suppose a lactose-intolerant individual, such as Dr. J, forgot to take her lactase supplement prior to eating a large bowl of mint chocolate chip ice cream. For simplicity, assume that the ice cream contains only lactose and no other sugars (i.e., no glucose is present). Which of the following scenarios is most likely happening in her gut bacteria with regard to the bacteria's lac operons? A) lac repressor and catabolite activator protein (CAP) are bound to DNA, resulting in no lac operon transcription B) lac repressor is bound to DNA, but catabolite activator protein (CAP) is not bound, resulting in no lac operon transcription C) lac repressor is not bound to DNA, but catabolite activator protein (CAP) is bound to DNA, resulting in high lac operon transcription D) neither the lac repressor or the catabolite activator protein (CAP) is bound to DNA, resulting in low level lac operon transcription
C) lac repressor is not bound to DNA, but catabolite activator protein (CAP) is bound to DNA, resulting in high lac operon transcription high lactose, low glucose
Small interfering RNAs (siRNAs) and microRNAs (miRNAs) are noncoding RNAs with important roles in gene regulation. Both can be used in the biochemical assay referred to as RNA interference (RNAi). What is the main difference between siRNA and miRNA? A) siRNA is derived from the nucleus, while miRNA is derived from single-stranded RNA B) miRNA regulates gene expression by preventing transcription, while siRNA regulates gene expression by preventing translation C) miRNA is derived from the nucleus, while siRNA is derived from single-stranded RNA D) siRNA regulates gene expression by preventing transcription, while miRNA regulates gene expression by preventing translation E) miRNA binds to mRNA to prevent translation, while siRNA binds to the protein sequence, tagging the protein for degradation
C) miRNA is derived from the nucleus (product of mRNA), while siRNA is derived from single-stranded RNA (made in the lab)
Select the correct grouping of the following amino acids: asparagine, methionine, arginine, aspartate A) polar neutral: Met; non-polar: Asn; positively charged: Arg; negatively charged: Asp B) polar neutral: Asp; non-polar: Met; positively charged: Asn; negatively charged: Arg C) polar neutral: Asn; non-polar: Met; positively charged: Arg; negatively charged: Asp D) polar neutral: Arg; non-polar: Asn; positively charged: Met; negatively charged: Asp
C) polar neutral: Asn; non-polar: Met; positively charged: Arg; negatively charged: Asp
Which of the following can be achieved by an enzyme in the course of catalyzing a chemical reaction? A) increasing the probability of product formation B) shifting the position of the reaction equilibrium C) preferential binding and thus stabilizing the transition state D) all of the above
C) preferential binding and thus stabilizing the transition state
One major source of mis-pairing in DNA synthesis is A) rNTP B) ATP C) rare tautomer of the four bases D) All of the above
C) rare tautomer of the four bases
Chromatin structure has to be dynamic, because otherwise A) histone molecules would be excluded from the nucleus B) DNA would be more susceptible for hydrolytic attack C) the information carried on DNA would be hard to access D) DNA will be positively supercoiled
C) the information carried on DNA would be hard to access
t-loop refers to A) the loop formed by single-stranded DNA between helicase and DNA polymerase on lagging strand B) the loop formed by the 30 nm chromatin fibers C) the loop formed in the end of the linear chromosome D) the loop formed by negative supercoiling of DNA
C) the loop formed in the end of the linear chromosome telomere: - repetitive DNA sequences at the ends of all eukaryotic chromosomes - they contain thousands of repeats of the six-nucleotide sequence, TTAGGG (in human) or TTGGGG (in ciliate) - together with telomere binding proteins, they form a unique t-loop at the ends of each linear chromosome to protect the ends. The presence of t-loop protects the linear chromosome and prevents chromosome fusion. - together with telomerase, telomere can help solve the end replication problem associated with the replication of linear DNA
What is used to target a protein for degradation by the proteasome? A) phosphorylation B) glycosylation C) ubiquitination D) acetylation E) methylation
C) ubiquitination
Using the following sequences, derive the consensus sequence of the E. coli promotor:
C. -35 TTGATA & -10 TATGAT
Knowing that the standard promoter consensus sequence in E. coli is 5' TTGATA 3' for the -35 region and 5' TATGAT 3' for the -10 region, which gene below would you predict has the weakest promoter? A. -35: 5' TTGATA 3' ; -10: 5' TATGAT 3' B. -35: 5' TAGATA 3' ; -10: 5' TATTAT 3' C. -35: 5' TAAAGTT 3' ; -10: 5' TTTATT 3' D. -35: 5' TTGGTT 3' ; -10: 5' TAGGTT 3' E. -35: 5' TTCATT 3' ; -10: 5' TATTAT 3'
C. -35: 5' TAAAGTT 3' ; -10: 5' TTTATT 3'
The catalytic rRNA in eukaryotic ribosomes that catalyzes the peptidyl transferase activity is called: A. 18S rRNA B. 5.8S rRNA C. 28S rRNA D. 5S rRNA E. 16S rRNA
C. 28S rRNA
Which of the mRNA messages would represent the longest open reading frame, starting with a start codon and ending with a stop codon? A. 5' AUGUUUUCAUAAAGUUGA 3' B. 5' AUGUAAUCAUGAAGUUAA 3' C. 5' AUGUUUUCACAAAGUUGA 3' D. 5' AUCUAAUCAUGAAGUUAA 3' E. 5' AUGUAGUCAUUAAGUUAG 3'
C. 5' AUGUUUUCACAAAGUUGA 3'
Which of the following statements about the chemiosmotic theory is true? A. It requires a mitochondrial membrane embedded with "purple protein" (bacteriorhodopsin). B. The ATP synthase has no significant role in the theory. C. A proton-motive force drives the ATP synthase to produce ATP. D. Energy is coupled through a transmembrane electron gradient. E. It explains how ATP energy is used to create a pH gradient.
C. A proton-motive force drives the ATP synthase to produce ATP.
Out of the descriptions below, which one describes a correct function of a prokaryotic translation factor? A. IF1 moves the peptidyl-tRNA from the A to the P site B. IF2 recognizes stop codons C. EF-Tu delivers aminoacyl-tRNA to the A site D. EF-G delivers fMet-tRNAfMet to the P site E. RF prevents the combination of the 50S
C. EF-Tu delivers aminoacyl-tRNA to the A site
The antibiotic ______________ inhibits prokaryotic transcription initiation by binding to the RNA polymerase in the channel that is normally occupied by the newly formed RNA-DNA hybrid and blocking the exit of the nascent RNA. A. Actinomycin D B. Erythromycin C. Rifampicin D. Streptomycin E. Puromycin
C. Rifampicin
The picture to the right shows the TATA binding protein (TBP). What is its role? A. The TATA binding protein is a prokaryotic transcription factor that binds to bacterial RNA polymerase and the -10 & - 35 promoter to initiate transcription. B. The TATA binding protein is a eukaryotic transcription factor that binds to the promoter and helps to recruit other transcription factors and RNA polymerase I to initiate transcription. C. The TATA binding protein is a eukaryotic transcription factor that binds to the promoter and helps to recruit other transcription factors and RNA polymerase II to initiate transcription. D. The TATA binding protein is a eukaryotic transcription factor that binds to the promoter and helps to recruit other transcription factors and RNA polymerase III to initiate transcription.
C. The TATA binding protein is a eukaryotic transcription factor that binds to the promoter and helps to recruit other transcription factors and RNA polymerase II to initiate transcription.
The sequence of a duplex DNA segment in a longer DNA molecule is: 5'-ATCGCTTGTTCGGA-3' 3'-TAGCGAACAAGCCT-5' When this segment serves as a template for E. coli RNA polymerase, it gives rise to a segment of RNA with the sequence 5'-UCCGAACAAGCGAU-3'. Which of the following statements about the DNA segment are correct? A. The top strand is the coding strand. B. The bottom strand is the antisense strand. C. The top strand is the template strand. D. The top strand is the sense strand
C. The top strand is the template strand. TOP: Template (antisense) BOTTOM: Coding (sense)
Which of the following statements about eukaryotic mRNAs is true? A. They are often polycistronic. B. They usually have poly(A) tails at their 5' ends. C. They result from extensive processing of their primary transcripts before serving as translation components. D. They usually have a cap at their 3' ends. E. They are often encoded by contiguous segments of template DNA.
C. They result from extensive processing of their primary transcripts before serving as translation components.
Imagine that you just consumed delicious French fries. For simplicity, let's say that the fires ONLY contain triacylglycerols and amylose. Which of the following enzymes would you expect to be active to help you digest your tasty snack? A. a-amylase, lactase, & sucrase B. a-amylase, pancreatic lipases, & maltase C. a-amylase, pancreatic lipases, & proteases D. maltase, pancreatic lipases, & proteases E. pancreatic lipases, hormone-stimulated lipases, and maltase
C. a-amylase, pancreatic lipases, & proteases
Excess alcohol consumption leads to a buildup of which two key molecules? A. glucose & lactate B. glucose & acetyl CoA C. acetyl CoA & NADH D. NADH & FADH2 E. lactate & FADH2
C. acetyl CoA & NADH
What type of pathway can be either anabolic or catabolic depending on the energy conditions of the cell? A. amphipathic B. amphiprotic C. amphibolic D. amphoteric
C. amphibolic
α-Amylase: A. removes glucose residues sequentially from the reducing end of starch. B. breaks the internal α-1,6 glycosidic bonds of starch. C. breaks the internal α-1,4 glycosidic bonds of starch. D. cleaves the α-1,4 glycosidic bond of lactose. E. can hydrolyze cellulose in the presence of an isomerase.
C. breaks the internal α-1,4 glycosidic bonds of starch.
Which of the following is considered a product of the citric acid cycle? A. acetate B. a-ketoglutarate C. carbon dioxide D. NAD+ E. pyruvate
C. carbon dioxide
Translation is an RNA-directed process whereby amino acids are specified by ________ on the mRNA, which are complementary to ________ on aminoacyl tRNAs. A. anticodons; codons B. amino acids; codons C. codons; anticodons D. codons; amino acids E. amino acids; anticodons
C. codons; anticodons
In the rho-independent transcription termination signal in prokaryotes, the GC-rich palindromic region forms a ______________ in the RNA, which results in the _____________________. A. Rho protein; ATP-dependent unwinding of the RNA-DNA hybrid helix B. Rho protein; RNA polymerase to pause transcription and dissociate from the DNA C. hairpin stem-loop structure; RNA polymerase to pause transcription and dissociate from the DNA D. stretch of U oligos; RNA polymerase to pause transcription and dissociate from the DNA E. hairpin stem-loop structure; ATP-dependent unwinding of the RNA-DNA hybrid helix
C. hairpin stem-loop structure; RNA polymerase to pause transcription and dissociate from the DNA
Alpha helices are often found in DNA binding proteins such as gene regulatory proteins because they: A. are the only secondary feature found in proteins. B. often can form cross-beta filaments that fit perfectly in the major groove of DNA. C. have the appropriate diameter to fit into the major groove of DNA. D. have the appropriate diameter to fit into the minor groove of DNA. E. are the only secondary structure that are capable of forming both heterodimers and homodimers.
C. have the appropriate diameter to fit into the major groove of DNA. The DNA-binding motif from the DNA binding protein binds to the DNA usually by inserting an alpha helix into the major groove of DNA.
A common feature of translation in both eukaryotes and prokaryotes is: A. hydrolysis of ATP to facilitate translocation of the ribosome. B. a conserved ribosomal binding site on the mRNA directs the small subunit to the start codon using direct rRNA and mRNA base-pairing. C. hydrolysis of GTP to promote accommodation of the charged aminoacyl-tRNA (AA-tRNAAA). D. an E site that can contain a tRNA covalently bound to a growing polypeptide. E. initiation requires formyl methionine to bind to the start codon in the P site of the small ribosomal subunit.
C. hydrolysis of GTP to promote accommodation of the charged aminoacyl-tRNA (AA-tRNAAA).
Suppose a lactose-intolerant individual, such as Dr. J, forgot to take her lactase supplement prior to eating a large bowl of vanilla ice cream. For simplicity, assume that the ice cream contains only lactose and no other sugars (i.e., no glucose is present). Which of the following scenarios is most likely happening in her gut bacteria with regard to the bacteria's lac operons? A. lac repressor bound to operator, CAP bound near promoter, no lac operon transcription (repressed) B. lac repressor bound to operator, CAP not bound, no lac operon transcription (repressed) C. lac repressor bound to inducer, CAP bound near promoter, high (activated) lac operon transcription D. lac repressor bound to inducer, CAP not bound, low (basal) level lac operon transcription E. lac repressor bound to inducer, CAP bound near promoter, no lac operon transcription (repressed)
C. lac repressor bound to inducer, CAP bound near promoter, high (activated) lac operon transcription glucose = low lactose = high The lac operon regulates lactose metabolism in bacteria: An example of regulated transcription is the control of the enzymes for lactose metabolism in bacteria. The gene for beta-galactosidase, which metabolizes lactose, is minimally transcribed unless lactose is present. In the presence of lactose, the genes for B-galactosidase as well as two other enzymes - a permease and thiogalactoside transacetylase- are expressed. The lac repressor is an example of negative regulation, requiring a ligand to induce gene expression: negative, inducible * The lac repressor inhibits transcription unless induced by allolactose not to bind to the DNA. Ligand binding can induce structural changes in regulatory proteins. This is an example of inducible repression. Low Lactose:- repressor bound to operator site prevents transcription of operon High Lactose:- repressor-inducer complex does not bind to DNA - lets go of DNA exposing promoter sequence - transcription of operon can occur 1) Lactose absent, repressor bound to operator, operon repressed: In the absence of lactose, the repressor remains bound to the operator, and RNA polymerase is therefore prevented from moving down the lac operon and transcribing its genes. 2) Lactose present, repressor not bound to operator, operon depressed: In the presence of lactose, the repressor is converted to its inactive form, which does not bind to the operator. RNA polymerase can therefore move past the operator and transcribe the lacZ, lacY, and lacA genes into a single mRNA. Transcription can be stimulated by proteins that contact RNA polymerase: Expression of the lac operon can also be stimulated by the catabolite activator protein (CAP), also called the cAMP receptor protein (CRP), when not under glucose (catabolite) repression. High glucose: Glucose (catabolite) repression Glucose levels regulate cAMP levels - not made when high levels of glucose. Low glucose: cAMP is made when glucose levels are low and this activates transcription, binding DNA to allow better binding for RNA polymerase Summary: the lac operon is an example of both negative (involves a repressor) and positive (involves an activator) regulation Inducible regulation - when lactose is low, repressor bound to operator resulting in inhibition (no expression of gene) - when lactose is high, repressor dissociates, permitting transcription, basal expression (low level expression - if activator not bound because glucose levels are high then only low levels of expression) Repressible regulation - when glucose is high, CAP (CRP) is not bound and transcription is dampened (glucose or catabolite repression) ** not enough cAMP, basal expression (low level expression) - when glucose is low, cAMP is high and CAP (CRP) is bound, resulting in activation - activated expression * when glucose is low and lactose is high, cAMP can bind = activated expression
The primary raw materials for gluconeogenesis are:
C. lactate, glycerol and alanine
Which of the following coenzymes is an activated carrier of acyl groups and also functions as a "flexible swinging arm" when it transfers the reaction intermediate from one active site of the pyruvate dehydrogenase complex to the next? A. FAD B. NAD+ C. lipoamide D. thiamine pyrophosphate (TPP) E. coenzyme A
C. lipoamide
Operons can be defined as units of DNA containing ________ gene(s) under control of _________ promotor(s). A. one; one B. one; multiple C. multiple; one D. multiple; multiple E. no; no
C. multiple; one
Coenzyme A is derived from this vitamin: ___________, and coenzyme A is an activated carrier for this functional group: ____________. A. riboflavin; acyl groups B. niacin; electrons C. pantothenic acid; acyl groups D. pantothenic acid; CO2 E. niacin; phosphate
C. pantothenic acid; acyl groups
The proteasome is a large multi-subunit ATP-dependent protease that degrades proteins that have been modified by the attachment of ________. A. PEST sequences B. N-terminal signal sequence C. polyubiquitin (a chain of at least 4 linked ubiquitin molecules) D. acetyl or methyl groups E. cyclin degradation boxes
C. polyubiquitin (a chain of at least 4 linked ubiquitin molecules)
Which of the following is an RNA molecule that can be catalytic (i.e., a ribozyme)? A. mRNA B. tRNA C. rRNA D. miRNA E. siRNA
C. rRNA
From the choices below, which proteins act in a coordinated manner with nuclear hormone receptors to mediate gene expression? A. coreceptors B. cohormones C. receptor-transcription mediators D. cotranscription factors E. coactivators
C. receptor-transcription mediators Nuclear hormone receptors are an elegant example of transcription initiation regulation: - Steroid hormones are powerful regulatory molecules that control gene expression. ** steroids are small molecules that bind nuclear hormone receptors to have an effect - Nuclear receptors bind to specific regions of DNA called steroid response elements. When steroids bind to nuclear hormone receptors, there will be a conformational change and expose a site for a coactivator / repressor to then bind and modify chromatin in some way.
The primary action of steroid hormones binding to nuclear hormone receptors is to A. facilitate export of mRNA out of the nucleus B. induce splicing of introns C. recruit coactivators to modify local chromatin D. degrade mRNA E. repress transcription
C. recruit coactivators to modify local chromatin - Steroid hormones are powerful regulatory molecules that control gene expression. ** steroids are small molecules that bind nuclear hormone receptors to have an effect - Nuclear receptors bind to specific regions of DNA called steroid response elements. When steroids bind to nuclear hormone receptors, there will be a conformational change and expose a site for a coactivator / repressor to then bind and modify chromatin in some way.
In the picture of an amino acid, what region of the amino acid would become covalently attached to a tRNA? A. region B B. region A C. region D D. region C E. All of the regions are equally probable
C. region D
Regulation of the trp operon involves ALL statements below except which of the following? A. controlling the amount of polycistronic mRNA at the level of transcription termination B. the sequential and coordinate production of five enzymes of tryptophan metabolism from a single mRNA C. the sequential and coordinate production of five enzymes of tryptophan metabolism from five different mRNAs produced in equal concentrations D. the production of transcripts of different sizes, depending on the level of tryptophan in the cell
C. the sequential and coordinate production of five enzymes of tryptophan metabolism from five different mRNAs produced in equal concentrations Many amino acid biosynthesis genes are regulated at the transcriptional and translational levels: The trp operon of bacteria contains a gene regulatory region and the coding sequences for five enzymes in the tryptophan biosynthetic pathway. The Trp operon is controlled by two mechanisms: Trp repressor (only binds to DNA when Trp is present) and attenuation.
In glycolysis, there is one enzyme that is considered "kinetically perfect" (aka "catalytically perfect"). What is the name of that enzyme?
C. triose phosphate isomerase
Shown on the right is a plot of the log of the molecular weights of proteins against their relative mobility on SDS-PAGE. Which circled protein (A, B, C, or D) has the smallest number of SDS molecules bound to it?
D
TIM, i.e., triose phosphate isomerase, has a molecular weight of 28.0 kD and an isoelectric point of 6.01. On a two-dimensional gel, most likely TIM will be located in which circled region of the gel?
D
For an enzymatic reaction that follows Michaelis-Menten kinetics, if [S] = 1/9 Km (i.e., one-ninth of Km), the V0 equals ___________; if [S] = 9 Km (i.e., 9 times of Km), the V0 equals __________. A) 0.4 Vmax; 0.6 Vmax B) 0.3 Vmax; 0.7 Vmax C) 0.2 Vmax; 0.8 Vmax D) 0.1 Vmax; 0.9 Vmax
D) 0.1 Vmax; 0.9 Vmax Vo=Vmax[S]/Km+[S]
The peptide shown on the right has _____ peptide bonds, and it sequence is _________. A) 3; Glu-Leu-Ser B) 3; Glu-Leu-Thr C) 2; Asp-Ile-Ser D) 2; Glu-Leu-Ser
D) 2; Glu-Leu-Ser
Consider a simple enzyme-catalyzed reaction that follows Michaelis-Menten kinetics; what will be the initial velocity (Vo) when the initial substrate concentration is nine times of the value of Km (i.e., 9 Km) and Vmax is 1000 mmol/min? A) 100 mmol/min B) 200 mmol/min C) 400 mmol/min D) 900 mmol/min
D) 900 mmol/min Vo=Vmax[S]/Km+[S]
Which is a simplifying assumption that allows the derivation of Michaelis-Menten equation? A) ES quickly comes to steady state, so [ES] is constant B) [S] >> [ET], so the fraction of S that binds to E is negligible, and [S] is constant at early time points C) At early time points, where initial velocity is measured, [P] is about 0 D) All of the above
D) All of the above
Which of the following catalytic strategies is clearly used by lysozyme? A) General acid catalysis B) General base catalysis C) Covalent catalysis D) All of the above
D) All of the above
Which of the following leads to an increase in the Tm of DNA? A) an increase in the cation concentration B) an increase in the GC% of the DNA C) an increase in the length of the DNA D) All of the above
D) All of the above
Which of the following describes the properties of a peptide bond? A) Peptide bond has partial double bond feature and it cannot freely rotate B) Most peptide bonds found in proteins are trans C) Six atoms around peptide bonds (two alpha carbons, C=O, and NH) are on the same plane D) All of the above
D) All of the above * The peptide bond is rigid and planer. * Partial double bond feature -- shorter bond length -- no free rotation -- six atoms on a plane * Almost all peptide bonds in proteins are trans
Lysozyme uses a combination of catalytic strategies, which include A) General acid catalysis B) Covalent catalysis C) General base catalysis D) All of the above
D) All of the above Asp52: covalent catalysis Glu 35: general acid and general base catalysis 1. Binding: upon binding to lysozyme, non-covalent interactions between the enzyme and the substrate lead to a conformational change of one sugar unit in the substrate, which puts the substrate in a less stable state. This strategy can be referred to as "ground state destabilization of the substrate." 2. Binding also brings the sugar bond on the substrate to be split close to the two catalytic residues in lysozyme: Glu35 and Asp52. Due to the difference in their neighboring microenviornment, the side chain of Asp52 exists predominantly in its conjugated based form (-COO-) and the side chain of Glu35 exists predominantly in the acid form (-COOH). Crystal structures of lysozyme indicate that there are more hydrophobic residues surrounding Glu35, thus effectively raising the pKa value of its side chain. 3. The negatively charged side chain of Asp52 (-COO-) acts as a nucleophile and forms a transient covalent bond with the C in the sugar (covalent catalysis), whereas the side chain of Glu35 (-COOH) donates a proton (acts as a general acid so this is acid catalysis) to the O that is linked to the C1 carbon. As a result, the sugar bond linking sugars D and E is broken. 4. Now water comes into play. The now negatively charged side chain of Glu35 polarizes water and makes it a better nucleophile in attacking the C1 carbon and the side chain of Asp52. As a result, OH- from water joins the C1 carbon, and the H+ from water joins the side chain of Glu35 (acts as a proton acceptor, so this is general base catalysis). 5. After all these steps, a sugar bond joining sugar units D and E is hydrolyzed, and enzyme is returned to its original state. Facilitate the hydrolysis of polysaccharides
Which of the following is true regarding the enzyme shown on the right? A) This protein can use the energy from ATP hydrolysis to change its conformation B) The hole in the middle of this protein only allows single-stranded DNA to pass C) The action of this enzyme generates a product that needs to be protected by SSB D) All of the above
D) All of the above Helicase; hexamer
Which of the following factors increases the Tm of DNA? A) A higher salt concentration B) A higher GC content C) An increased length D) All of the above
D) All of the above Tm: melting temperature the temperature at which half the helical structure is lost - cooperative - double strand --> single strand high temp = high absorbance lower temp = low absorbance Tm is affected by several factors: - concentration of ions in the solution: especially cations because the two phosphates are (-) charged, so presence of (+) charge to stabilize that would increase Tm - DNA sequence: increase GC bonds increases Tm because 3H bonds = stronger IMFs - Length of DNA: increase length, increase bonds needed to break, increase Tm
Which of the following is a type of post-translational modifications? A) Hydroxylation of a proline residue on collagen B) Phosphorylation of a serine residue on a signaling protein C) Methylation of a lysine residue on histone proteins D) All of the above are considered post-translational modifications
D) All of the above are considered post-translational modifications
Which of the following sets of amino acids have three ionizable groups? A) Ala, Pro, Asn B) Asn, Asp, Gln C) Asp, Ser, Trp D) Asp, Cys, Tyr
D) Asp, Cys, Tyr ** three ionizable groups means that the R group must be ionizable
The molecule shown on the right is called _____. At pH 7, in theory, this molecule can maximally bond with this many ______ water molecules. A) Glutamine; 10 B) Glutamine; 13 C) Asparagine; 10 D) Asparagine; 13
D) Asparagine; 13
Which of the following techniques allows the identification of topologically associating domains (TADs) within an interphase chromosome? A) Western blot B) Southern blot C) FISH D) Chromosome conformation capture
D) Chromosome conformation capture FISH analysis: * a FISH analysis using a different mixture of fluorochromes for marking the DNA of each chromosome, detected with seven color channels in a fluorescence microscope, allows each chromosome to be distinguished in 3D space. * The first step is to prepare short sequences of single-stranded DNA that match a portion of the gene the researcher is looking for. These are called probes. * The next step is to label these probes by attaching one of a number of colors of fluorescent dye. * The next step is to let the probe bind to the complementary strand of DNA (called hybridization), wherever it may reside on a person's chromosomes. When a probe binds to a chromosome, its fluorescent tag provides a way for researchers to see its location. * FISH provides researchers with a way to visualize and map the genetic material in an individual's cells, including specific genes or portions of genes. This may be used for understanding a variety of chromosomal abnormalities and other genetic mutations. * USEFUL FOR GENETIC ANALYSIS OF A PATIENT * The basic elements of FISH are a DNA probe and a target sequence. The probe is complementary to a region on target. * Before hybridization, the DNA probe is labeled to contain a fluorophore. The labeled probe and the target DNA are denatured. Combining the denatured probe and target allows the annealing (forming of a double-stranded structure) of complementary DNA sequences. Chromosome Conformation Capture: a technique that allows the mapping of neighborhood info in chromosome - covalently joined, digest non-cross linked, new DNA = joint DNA fragments, sequence DNA and reveal identity of the two portions and detect neighborhood patterns
The titration curve for an amino acid is shown below. Most like, the amino acid is ______. A) Alanine B) Histidine C) Glutamate D) Cysteine
D) Cysteine **basic group
Sanger sequencing refers to ____________________, using this essential reagent ________________. A) protein sequencing; SDS B) RNA sequencing; chain-terminating nucleotide C) DNA sequencing; urea D) DNA sequencing; chain-terminating nucleotide
D) DNA sequencing; chain-terminating nucleotide Sanger sequencing
To examine if two proteins interact in live cells, you fused one protein with blue fluorescence protein, which can be excited by violet light and emits blue light; you fused another protein with green fluorescence protein, which can be excited by blue light and emits green light. Which of the following results would support the hypothesis that these two proteins interact? A) Exciting cells with blue light leads to the emission of green light only B) Exciting cells with violet light leads to the emission of blue light only C) Exciting cells with blue light leads to the emission of violet light D) Exciting cells with violet light leads to the emission of green light
D) Exciting cells with violet light leads to the emission of green light
A technique in which a fluorescence-labeled DNA probe is used to locate a gene in a chromosome is called: A) Southern blotting B) Northern blotting C) FRET D) FISH
D) FISH a FISH analysis using a different mixture of fluorochromes for marking the DNA of each chromosome, detected with seven color channels in a fluorescence microscope, allows each chromosome to be distinguished in 3D space. FISH provides researchers with a way to visualize and map the genetic material in an individual's cells, including specific genes or portions of genes. This may be used for understanding a variety of chromosomal abnormalities and other genetic mutations.
Which macromolecule below is NOT considered a ribozyme? A) spliceosome B) large subunit of ribosome C) group I introns D) FMN riboswitch
D) FMN riboswitch
The O-helix is present in this region ________ of DNA polymerase, and its Lys and Arg are responsible for interacting with this part ____________ of the incoming nucleotide. A) Palm; base B) Thumb; phosphates C) Fingers; base D) Fingers; phosphates
D) Fingers; phosphates DNA polymerase "grips" the template and the incoming nucleotide when a correct base pair is made using finger region: correct base pairing between incoming nucleotide and the template DNA triggers a conformation change in DNA polymerase: a 40 degree rotation of one of the helices in the finger domain called the O-helix * in the open conformation, the O-helix is distant from the incoming nucleotide; when the polymerase is in the closed conformation, this helix moves and makes several important interactions with the incoming dNTP - a tyrosine makes stacking interactions with the base of the dNTP - the two charged residues (Lys & Arg) associate with the triphosphase - the combination of these interactions positions the dNTP for catalysis mediated by the two metal ions bound to the DNA polymerase
Which of the following enzymes/proteins is not required for repairing pyrimidine dimer? A) Nuclease B) Helicase C) Ligase D) Glycosylase
D) Glycosylase During nucleotide excision repair of damaged DNA, enzymes unwind DNA, cut out a section on one strand that contains damaged DNA, and resynthesize the section with the correct DNA sequence. The UvrA dimer binds with a UvrB dimer and forms a complex that is able to detect DNA damage. uvrC has nuclease activity that cuts out the damaged section. uvrD has helicase activity that can unwind. DNA polymerase (synthesize) plus DNA ligase (reseal fragments) resynthesize the correct DNA sequence.
In forming a nucleosome, which of the following proteins first contact the naked DNA? A) Histone H1 B) H2A/H2B dimer C) H3/H4 dimer D) H3/H4 tetramer
D) H3/H4 tetramer
During the assembly of histone octamer, which of the following sub-structures initiates DNA binding to form a nucleosome? A) H2A/H2B dimer B) H3/H4 dimer C) H2A/H2B tetramer D) H3/H4 tetramer
D) H3/H4 tetramer The assembly of a histone octamer: H3-H4 functional unit = tetramer (Formed when two H3-H4 dimers join together) H2A - H2B functional unit = dimer x 2 (2 dimers individually join the established tetramer) = histone octamer
According to Lipinski's rule, a drug candidate would have poor absorption if it has this property: A) A molecular weight of 200 Daltons B) Having four hydrogen bond acceptors C) Having four hydrogen bond donors D) Having a log(P) value of 7
D) Having a log(P) value of 7
Both hexokinase and glucokinase can catalyze the reaction that converts glucose to glucose-6-phosphate. Notably, the Km of hexokinase for glucose is 0.1 mM, while the Km of glucokinase for glucose is 10 mM. Which of the following statements is correct regarding these two enzymes? A) Glucokinase has a higher affinity for glucose than hexokinase B) Glucokinase is likely to be present in every cell to process glucose for glycolysis C) Hexokinase is a less effective enzyme than glucokinase D) Hexokinase is likely to be responsible for a vital process that is largely independent of the fluctuating glucose level in the blood
D) Hexokinase is likely to be responsible for a vital process that is largely independent of the fluctuating glucose level in the blood
Which of the following can potentially function as a barrier protein to prevent the spread of histone code? A) Histone acetyltransferase B) G protein-coupled receptor C) Albumin D) Histone deacetylase
D) Histone deacetylase
Genome editing is a very hot and promising area in molecular biology. ZFN, TALEN, and CRISPR-Cas9 are three important strategies commonly used in this exciting area, and all of them make use of DNA repair pathways in the host. Specifically, the following host repair pathways are very important for genome editing to work: A) Mismatch repair pathway B) Base-excision repair pathway C) Nucleotide-excision repair pathway D) Homologous recombination and nonhomologous end-joining (NHEJ)
D) Homologous recombination and nonhomologous end-joining (NHEJ)
Which of the following organisms has the least dense genome? A) Bacteria B) Yeast C) Fruit Fly D) Human
D) Human
Which of the following non-covalent interactions is crucial for driving membrane formation and protein folding? A) Ionic interaction B) Hydrogen bond C) Van der Waals interaction D) Hydrophobic interaction
D) Hydrophobic interaction
Consider the three amino acids shown in the figure. From left to right, these three amino acids are A) L-alanine (S); L-valine (S); L-cysteine (S) B) L-alanine (S); L-leucine (S); L-cysteine (R) C) D-glycine (R); D-valine (R); D-cysteine (S) D) L-alanine (S); L-valine (S); L-cysteine (R)
D) L-alanine (S); L-valine (S); L-cysteine (R)
Arginine-tRNA synthetase has three substrates: arginine, tRNA, and ATP. Specifically, it catalyzes the reaction that joins arginine and tRNA facilitated by the energy input from ATP hydrolysis. Substrates: ATP + arginine + tRNA; Products: AMP + diphosphate + arginyl-tRNA Accordingly, arginine-tRNA synthetase should be classified as A) Oxidoreductase B) Transferase C) Lyase D) Ligase
D) Ligase
An enzyme that catalyzes the reaction shown below should be classified as: A) Hydrolase B) Isomerase C) Ligase D) Lyase
D) Lyase
The enzyme that catalyzes the reaction shown here should be classified as: A) Oxidoreductase B) Isomerase C) Transferase D) Lyase
D) Lyase note the movement of the double bond
When bacteria produce mammalian proteins, complementary DNA (cDNA) is used rather than genomic DNA. Which of the following is the best explanation of why cDNA is used over genomic DNA? A) The polycistronic nature of bacterial genes precludes efficient expression of mammalian DNA. B) Bacteria have polyribosomes, while mammals don't have polyribosomes. C) Most eukaryotic promoters don't function in bacteria. D) Mammalian DNA contain introns that bacteria don't have mechanisms to remove. E) Many mammalian proteins require post-translational modifications and processing that bacteria do not have the capability to perform.
D) Mammalian DNA contain introns that bacteria don't have mechanisms to remove.
Which statement that best explains the role of 2,3-BPG in oxygen transport from mother to fetus? A) Maternal hemoglobin-bound oxygen will tend to move to fetal hemoglobin because maternal hemoglobin binds an additional 2,3-BPG per tetramer when it enters placental circulation B) In the placental circulation, fetal hemoglobin will load up on oxygen as 2,3-BPG dissociates from maternal hemoglobin and binds to fetal hemoglobin C) Fetal hemoglobin will extract oxygen from oxygenated maternal hemoglobin because 2,3-BPG enhances oxygen binding to fetal hemoglobin D) Maternal hemoglobin-bound oxygen will tend to move to fetal hemoglobin because fetal hemoglobin has a lower affinity for 2,3-BPG, an allosteric inhibitor of oxygen binding
D) Maternal hemoglobin-bound oxygen will tend to move to fetal hemoglobin because fetal hemoglobin has a lower affinity for 2,3-BPG, an allosteric inhibitor of oxygen binding
E. coli, budding yeast, fruit fly, and mice are four very commonly used model organisms in biomedical research. In case you are unclear, E. coli belongs to prokaryotes, and yeast is single-cell eukaryote. Among these four model organisms, which one has the least number of genes in every million base-pairs of its genome? A) E. coli B) Budding yeast C) Fruit fly D) Mice
D) Mice
47. ______________ is commonly used as an oxidizing agent in catabolic pathways, while ______________ is commonly used as a reducing agent in anabolic pathways. A) NAD+ ; NADH B) FAD ; FADH2 C) NADP+ ; NADPH D) NAD+ ; NADPH
D) NAD+ ; NADPH
Adenosine is a ________, and it is composed of ________. A) Nucleotide; adenine, ribose, and one phosphate B) Nucleoside; adenine and deoxyribose C) Nucleotide; adenylate, ribose, and one phosphate D) Nucleoside, adenine and ribose
D) Nucleoside, adenine and ribose
Similar amino acids are sometimes attached to the incorrect amino-acyl tRNA synthetase. Suppose Phe was attached to tRNATyr, creating Phe-tRNATyr. Choose the correct equation for editing Phe-tRNATyr by the tyrosinyl-tRNA synthetase that takes place during the proofreading step: A) Phe-tRNATyr + Tyr Tyr-tRNATyr B) Phe-tRNATyr + Phe Phe-tRNAPhe C) Phe-tRNATyr Tyr-tRNATyr D) Phe-tRNATyr tRNATyr + Phe E) Phe-tRNATyr tRNATyr + Tyr
D) Phe-tRNATyr tRNATyr + Phe
Examine the promoters shown below. These promoters MOST LIKELY direct transcription factors to recruit which polymerase for RNA synthesis? A) bacterial RNA polymerase B) RNA polymerase I C) RNA polymerase II D) RNA polymerase III E) DNA polymerase
D) RNA polymerase III
DNA synthesis needs a primer, which is eventually removed from the final product by the action of this enzyme: A) DNA polymerase alpha B) Helicase C) Ligase D) RNase H
D) RNase H
In prokaryotes, the enzyme that processes the broken ends of DNA generated from a double-stranded break is _______, and the resulting single-stranded DNA with a 3'-end is bound with ______ to facilitate DNA synapsis. A) RecA; RecBCD B) MutS; RecA C) DnaA; DnaB/DnaC D) RecBCD; RecA
D) RecBCD; RecA
A small generic section of the primary structure of an alpha helix is given by -residue 1-residue 2- residue 3-residue 4-residue 5-residue 6-residue7-residue 8- Which residue's backbone forms a hydrogen bond with the backbone of the residue 2? A) Residue 3 B) Residue 4 C) Residue 5 D) Residue 6
D) Residue 6
Which of the following chromatin binding domains has a higher affinity to nucleosomes built from newly translated and unmodified histone proteins? A) Chromodomain B) Bromodomain C) TUDOR domain D) SANT domain
D) SANT domain
How are viruses related to the presence of pseudogenes in our genome? A) A major source of pseudogene in our genome stems from integration of virus genes B) In ancient times, viruses had more genes than humans C) Viruses have more compact genomes, so their genes can be easily integrated into its host genome D) Some viruses carry reverse transcriptase that could reverse-transcribe host mRNA to DNA
D) Some viruses carry reverse transcriptase that could reverse-transcribe host mRNA to DNA
How does the shape of a competitive enzyme inhibitor determine its mechanism of inhibition? A) The inhibitor has a different shape than the substrate. The inhibitor is bound by the active site of the enzyme along with the substrate, preventing the catalyzed reaction. B) The inhibitor has a similar shape to the substrate. The inhibitor is bound by the active site of the enzyme along with the substrate, preventing the catalyzed reaction. C) The inhibitor has a different shape than the substrate. The inhibitor is bound by an allosteric site on the enzyme, preventing the substrate from being bound by the active site. D) The inhibitor has a similar shape to the substrate. The inhibitor is bound by the active site of the enzyme, preventing the substrate from being bound by the active site.
D) The inhibitor has a similar shape to the substrate. The inhibitor is bound by the active site of the enzyme, preventing the substrate from being bound by the active site.
Pyruvate, oxaloacetate, and alpha-ketoglutarate are the direct precursors of the following amino acids, respectively. A) alanine; asparagine; glutamine B) glycine; aspartate; glutamate C) serine; glutamine; asparagine D) alanine; aspartate; glutamate
D) alanine; aspartate; glutamate
Maintaining correct pH is very important for living systems. Which of the following molecules is important for pH control in living systems? A) Proteins, especially those with high content of histidine residues B) Dihydrogen phosphate and hydrogen phosphate C) Bicarbonates D) all of the above
D) all of the above
Which of the following is true regarding ATCase? A) ATCase should be classified as a transferase B) ATCase does not follow Michaelis-Menten kinetics C) ATCase catalyzes an important reaction in pyrimidine biosynthesis D) all of the above
D) all of the above - an example of an allosteric enzyme - ATCase combines two reactants (carbonyl phosphate and aspartate) to become N-carbamoylaspartate (transferase) - N-carbamoylaspartate is used for synthesis of CTP- CTP itself is a regulator for this reaction - ATCase can be inhibited by CTP- as [CTP] increases, there is less need for the reaction to continue - therefore, CTP inhibits because it wants to avoid all the intermediate steps when there is already enough CTP end product= FEEDBACK INHIBITION ATCase is made up of a total of 12 polypeptide chains- two catalytic trimers (= six active sites) - three regulatory dimers (= six regulatory sites) ATCase has two distinct conformation states of ATCase revealed by PALA binding: 1. T-state inactive (Tense) 2. R-state active (Relaxed) - active sites has low affinity for substrate when in the T state because of its tight conformation - when PALA binds to the active site, the enzyme can expand and active site is now in perfect position to bind a substrate = high affinity for substrate - when CTP binds to regulatory site, regulatory sites have a high affinity for CTP and CTP binds and stabilizes the enzyme in the inactive T state and this is how CTP inhibits ATCase - the binding of substrates to the enzyme drives ATCase into its R state -the binding of inhibitors to the enzyme drives ATCase into its T state - ATCase displays sigmoidal kinetics - cooperativity: in some MULTI SUBUNIT enzymes, ligand binding to one subunit can affect the properties of an adjacent subunit by changing its conformation - cooperativity: multiple active sites that cooperate with one another (ex: shift from T state to R state) - in T state, beginning stretch of sigmoidal curve, low affinity for substrates, so it would be very hard for a substrate to bind to the active site (hence it takes a lot of [S] to bind) - once some [S] are bound, influences the rest of the active sites to grab substrates (hence share rise) = cooperativity
Which of the following statements accurately describe the blood buffering system in humans? A) The blood buffering system utilizes the H2CO3/HCO3- conjugate acid/base pair B) The blood buffering system maintains the pH of blood near 7.4 C) The blood buffering system is facilitated by the enzyme carbonic anhydrase D) All of the above
D) all of the above carbonic acid-bicarbonate buffer system - very important for the pH control of blood and body fluids - the blood buffering system maintains the pH of blood near 7.4 - the blood buffering system utilizes the H2CO3/HCO3- conjugate acid/base pair dihydrogen phosphate-hydrogen phosphate system - plays a major role in controlling intracellular pH because it has pKa value near 7.0 and it is abundant in cells - contributes to the buffering in the cytoplasm - abundant IN THE CELL proteins (as a buffer system) - proteins contain many weakly acidic or basic groups and some of these have pKa values near 7.0. - Because proteins are abundant both in cells and in body fluids, pH buffering is very strong. - example: histidine side chain in proteins contributes to the cytoplasmic buffering
Aspirin inhibits its target cyclooxygenase by covalently modifying a serine residue in the active site of the enzyme. Accordingly, aspirin should be classified as A) a competitive inhibitor B) an uncompetitive inhibitor C) a noncompetitive inhibitor D) an irreversible inhibitor
D) an irreversible inhibitor
What are the two most important metabolic fuels? A) carbohydrates and proteins B) proteins and nucleic acids C) lipids and proteins D) carbohydrates and lipids E) lipids and nucleic acids
D) carbohydrates and lipids
Which of the following functions is not a purpose of metabolism? A) extract chemical energy from substances obtained from the external environment B) form and degrade the biomolecules of the cell C) convert exogenous foodstuffs into building blocks and precursors of macromolecules D) equilibrate extracellular substances and the biomolecules of the cell E) assemble the building-block molecules into macromolecules
D) equilibrate extracellular substances and the biomolecules of the cell
Hemoglobin is a protein with a total of four subunits, each with a molecular weight of 16 kDa. The isoelectric point of hemoglobin is 6.8. A related protein, myoglobin, is a protein with a single polypeptide chain, which has a molecular weight of 16.7 kDa and an isoelectric point of 6.87. Which of the following methods can best resolve a mixture of hemoglobin and myoglobin? A) SDS-PAGE B) Isoelectric focusing C) 2-dimensional gel D) gel-filtration
D) gel-filtration A) SDS-PAGE: Anions of SDS bind to main chains of polypeptides at a ratio of one SDS anion for every two amino acid residues (tight binding). Consequently, all proteins carry a large net negative charge that is roughly proportional to the mass of the protein. The rate of migration is strictly determined by the mass of the proteins (smaller ones travel faster). If you mix a protein in its native state with this "ionic detergent," it will denature the protein and elongate it. Smaller protons will move faster towards the positive electrode of the set up. B) Isoelectric focusing: Separating proteins based on their isoelectric points. Requires a unique type of gel. When the pH of the gel regions equals the pI of a protein, the protein STOPS migrating in an electric field because it carries NO net charge. When the pH of the gel region does NOT equal pI of a protein, the protein continues migrating in an electric field until it reaches the region where pH=pI. C) 2-dimensional gel: Iso-electric focusing and SDS-PAGE can be combined to allow high resolution separation of complex protein mixtures. FIRST separation: isoelectric focusing (separation based on charge) SECOND separation: SDS PAGE (separation based on size) This order cannot be switched because once proteins are treated with SDS, then they all have the same charge. D) gel-filtration: A type of column chromatography that separates proteins based on their size using size-exclusion beads; also called size-exclusion chromatography. Larger molecules move more rapidly because they don't channel through the pores and therefore emerge from the column filter first. Can estimate the molecular weight of molecules, especially for a protein with more than one subunit. Precisely when the molecule comes out depends on its molecular weight. Quaternary structure stays intact (native state remains).
Which metabolic process below is an example of catabolism? A) DNA synthesis B) RNA synthesis C) protein synthesis D) glycolysis E) gluconeogenesis
D) glycolysis
In the rho-independent transcription termination signal in prokaryotes, the GC-rich palindromic region forms a ______________ in the RNA, which results in the _____________________. A) Rho protein; ATP-dependent unwinding of the RNA-DNA hybrid helix B) hairpin stem loop structure; ATP-dependent unwinding of the RNA-DNA hybrid helix C) stretch of U oligos; RNA polymerase to pause transcription D) hairpin stem loop structure; RNA polymerase to pause transcription E) Rho protein; RNA polymerase to pause transcription
D) hairpin stem loop structure; RNA polymerase to pause transcription
Many types of noncovalent interactions are electrostatic in nature; one notable exception is A) Hydrogen bond B) Van der Waals interaction C) Ionic interaction D) Hydrophobic interaction
D) hydrophobic interaction
The control point for most eukaryotic gene regulation occurs at: A) initiation of translation B) RNA processing C) mRNA degradation D) initiation of transcription E) protein degradation
D) initiation of transcription
Aspirin is a drug that can inhibit the enzyme COX1 by covalently modifying a residue close to the active site of COX1. Thus, aspirin should be classified as this type of inhibitor for COX1: . A) Non-competitive inhibitor B) competitive inhibitor C) uncompetitive inhibitor D) irreversible inhibitor
D) irreversible inhibitor
The most significant influence on why prokaryotic mRNA is essentially not processed at all, while eukaryotic mRNA is highly processed is the fact that: A) prokaryotic genes are often polycistronic B) prokaryotic mRNA does not need to have a 3' poly-A tail C) eukaryotes are multicellular, while prokaryotes are unicellular D) prokaryotes simultaneously synthesize RNA and protein E) eukaryotic genes have introns, while prokaryotic genes don't
D) prokaryotes simultaneously synthesize RNA and protein
In addition to the four nitrogen atoms of the porphyrin, the iron in the heme group present in hemoglobin is also coordinated to an atom in A) the carboxyl terminus of the alpha chain of hemoglobin B) the amino terminus of the alpha chain of hemoglobin C) the amino terminus of the beta chain of hemoglobin D) the proximal histidine residue on hemoglobin
D) the proximal histidine residue on hemoglobin Fe2+ can form 6 bonds: 4 with nitrogen, one with oxygen, and one with a histidine residue
DNA topoisomerase II enzyme has ___________ catalytic ____________ residue(s), and this enzyme is able to generate a _____________ on ____________________. A) one; serine; single-strand break; T segment B) one; tyrosine; single-strand break; T segment C) two; serine; double-strand break; G segment D) two; tyrosine; double-strand break; G segment
D) two; tyrosine; double-strand break; G segment DNA topoisomerase: relieve helical winding and DNA tangling DNA Top I: single strand break to solve winding problem Type I DNA topoisomerase: * solves the "winding problem" that arises during DNA replication * type 1 DNA topoisomerase with tyrosine at the active site and can attack phosphodiester linkage 1. one end of the DNA double helix cannot rotate relative to the other end 2. DNA topoisomerase covalently attaches to a DNA phosphate, thereby breaking a phosphodiester linkage in one DNA strand (single strand break, a nick) 3. the two ends of the DNA double helix can now rotate relative to each other, relieving accumulated strain * acts like a reversible endonuclease 1. the original phosphodiester bond energy is stored in the phosphotyrosine linkage, making the reaction reversible 2. spontaneous reformation of the phosphodiester bond regenerates both the DNA helix and the DNA topoisomerase DNA Top II: double strand break to solve tangling problem (requires ATP input) DNA topoisomerase II (dimer) * untangles DNA, scissors * two circular DNA double helices are interlocked * G segment - gate * T segment - transport * double stranded break
DNA topoisomerase II has two catalytic __________ residues, which attack the backbone of ________ to allow the resolving of interlocking DNA segments. A) serine; T segment B) cysteine; G segment C) threonine; T segment D) tyrosine; G segment
D) tyrosine; G segment
Which is NOT a way that prokaryotes can regulate their gene expression? A) use of alternative sigma factors B) use of riboswitches C) use of activators and/or repressors that bind directly to the DNA D) use of alternative splice sites or alternative poly-A sites E) use of operons to regulate related genes at once
D) use of alternative splice sites or alternative poly-A sites
Which of the following compounds contains a "high-energy" bond and is used to produce ATP by substrate-level phosphorylation in glycolysis?
D. 1,3-bisphosphoglycerate
Starting with lactose, and using the glycerol 3-phosphate shuttle, how may ATP equivalents would you generate upon complete oxidation of lactose to CO2 and water? A. 2 B. 10 C. 30 D. 60 E. 106
D. 60
Differences between eukaryotic and prokaryotic translation include that in eukaryotes, the small ribosomal subunit recognizes the ______________ rather than a _____________ on the mRNA, as it happens in prokaryotes. A. Shine-Delgarno sequence; ribosome binding site B. 7-methylguanine cap; poly-A tail C. Poly-A tail; Shine-Delgarno sequence D. 7-methylguanine cap; Shine-Delgarno sequence E. first AUG; stop codon
D. 7-methylguanine cap; Shine-Delgarno sequence Transcription and processing (5' cap, splicing, poly A tail) are coordinated by the phosphorylated CTD of RNA pol II: 1) acquisition of 5' 7-methylguanylate cap, 2) splicing introns, and 3) acquisition of poly(A) tail
Which of the following events occur in the presence of moderate levels of λ repressor? A. λ repressor binds to OR1 and OR2, but not OR3. B. The λ repressor bound to OR1 blocks access to the promoter on the right side of the operator sites, repressing transcription of Cro. C. The λ repressor bound to OR2 stimulates transcription of the λ repressor promoter. D. All of the above.
D. All of the above.
Refer to the figure to the right. What method do you surmise would have generated the result shown? This method is used to determine where on the DNA sequence a specific DNA-binding protein will bind. A. Chromatin immunoprecipitation (ChIP) B. Quantitative polymerase chain reaction (qPCR) C. DNA microarrays (DNA chips) D. DNA footprinting E. RNA interference (RNAi)
D. DNA footprinting DIDN"T talk about this
Which of the following prokaryotic regulatory proteins serves as both an activator and a repressor of prokaryotic gene expression? A. Lac repressor B. Catabolite activator protein (CAP, aka CRP) C. Tryptophan repressor D. Lambda repressor E. Cro protein
D. Lambda repressor bacteriophage lambda has two alternative infection modes, the lytic pathway and the lysogenic pathway. Regulatory proteins control which pathway is followed. Lytic cycle: lyse bacteria and kill cell 1) phage attaches to host cell and injects DNA 2) phage DNA circularizes and enters the lytic cycle or lysogenic cycle. 3) new phage DNA and proteins are synthesized and assembled into virions 4) cell lyses, releasing phage virions Lysogenic Cycle: virus integrates its DNA into bacteria genome to be replicated 1) phage attaches to host cell and injects DNA 2) phage DNA circularizes and enters the lytic cycle or lysogenic cycle 3) phage DNA integrates within the bacterial chromosome by recombination, becoming a prophage 4) lysogenic bacterium reproduces normally; many cell divisions 5) occasionally, the prophage may excise from the bacterial chromosome by another recombination event, initiating a lytic cycle. Lambda repressor binding to OR1 represses Cro expression, binding to OR2 activates lambda expression (repressor), and binding to OR3 represses lambda expression (repressor). A circuit based on lambda repressor and Cro form a genetic switch: The lambda repressor binds to operators cooperatively, with the highest binding affinity for OR1 (OR1>OR2>OR3). Once it binds to OR3, it will stop making more cI. When lambda repressor level low, lambda repressor gene expression is stimulated by OR2 binding. When lambda repressor levels are high, lambda repressor gene expression is blocked because there is already enough. When something changes (virus detects damaged DNA?) RecA (protease) cleaves right between lambda repressor, leaving OR3, OR2, then OR1, allowing all lytic proteins to be synthesized. The lambda repressor blocks the transcription of all genes in the lambda genome except the gene that encodes the repressor. *****Express lambda repressor only (cI) - lysogenic cI activates cI and inhibits cro, favoring lysogenic pathway ** cI binds to OR1 and OR2 Cro = controlled of other genes Express lots of viral proteins, and make virus Cro inhibits cI, allowing cro to be expressed and favors the lytic pathway ** Cro binds to OR3
Which of the following statements is FALSE concerning gene regulation in eukaryotes? A. Deamination of specific RNA nucleotides in a process known as RNA editing is a mechanism of altering the protein product encoded by some genes. B. miRNAs are used to target specific mRNAs for degradation, preventing or slowing down translation. C. RNA interference by siRNAs can be a mechanism of defense against viral genes. D. Most related genes under transcriptional control are clustered together in operons and are regulated by a single promoter and a single operator. E. DNA methylation of cytosine residues in major grooves of DNA can silence genes and are usually located in CpG islands that are "hot spots" for mutations.
D. Most related genes under transcriptional control are clustered together in operons and are regulated by a single promoter and a single operator.
Specific combinatorial control of transcription in eukaryotes is facilitated by all of the following EXCEPT: A. specific interactions between transcription factors and specific DNA sequences. B. given regulatory proteins can have different effects depending upon the neighboring proteins with which they are associated. C. specific interactions between transcription factors that don't specifically bind DNA and transcription factors that do bind DNA. D. RNA polymerase directly binding the promoter, recruiting transcription factors to the start site. E. chromatin remodeling, allowing transcription factors to recruit another transcription factor.
D. RNA polymerase directly binding the promoter, recruiting transcription factors to the start site.
Out of the RNA molecules below, which type of RNA would you expect to be the most abundant in a cell? A. Small nuclear RNA (snRNA) B. MicroRNA (miRNA) C. Small nucleolar RNA (snoRNA) D. Ribosomal RNA (rRNA) E. Transfer RNA (tRNA)
D. Ribosomal RNA (rRNA)
Similar amino acids are sometimes attached to the incorrect amino-acyl tRNA synthetase. Suppose serine was attached to tRNAThr, creating Ser-tRNAThr. Choose the correct equation for editing Ser-tRNAThr by the threonyl-tRNA synthetase that takes place during the proofreading step: A. Ser-tRNAThr + Thr Thr-Thr-tRNAThr B. Ser-tRNAThr + Ser Ser-Ser-tRNAThr C. Ser-tRNAThr Thr-tRNASer D. Ser-tRNAThr tRNAThr + Ser E. Ser-tRNAThr tRNAThr + Thr
D. Ser-tRNAThr tRNAThr + Ser
What is the committed step in fatty acid synthesis? A. Binding of the fatty acyl group to the acyl carrier protein B. The reduction reaction catalyzed by NADPH C. The formation of acetoacetyl-ACP D. Synthesis of malonyl CoA by acetyl CoA carboxylase E. Transport of an activated fatty acid from the cytosol into the mitochondrion
D. Synthesis of malonyl CoA by acetyl CoA carboxylase
What is the correct order of general transcription factor assembly at an RNA polymerase II promoter? A. TFIIA à TFIIB à TFIID à TFIIE à TFIIF à TFIIH B. TFIID à TFIIA à TFIIB à TFIIE à TFIIF à TFIIH C. TFIID à TFIIH à TFIIB à TFIIA à TFIIF à TFIIE D. TFIID à TFIIA à TFIIB à TFIIF à TFIIE à TFIIH E. TFIIA à TFIIB à TFIIH à TFIIF à TFIIE à TFIID
D. TFIID à TFIIA à TFIIB à TFIIF à TFIIE à TFIIH
What is the correct order of general transcription factor assembly at an RNA polymerase II promoter? A. TFIID à TFIIAà TFIIB à TFIIE à TFIIH à TFIIF B. TFIIA à TFIIB à TFIID à TFIIF à TFIIE à TFIIH C. TFIIF à TFIID à TFIIA à TFIIB à TFIIE à TFIIH D. TFIID à TFIIA à TFIIB à TFIIF à TFIIE à TFIIH E. TFIIH à TFIID à TFIIA à TFIIB à TFIIE à TFIIF
D. TFIID à TFIIA à TFIIB à TFIIF à TFIIE à TFIIH
Which of the following BEST explains the "wobble" hypothesis? A. The genetic code is ambiguous in that each codon can specify more than one amino acid. B. The anticodon can pair with any part of the corresponding codon. C. Inosine in the codon can pair up with A, U, or C in the anticodon. D. The 5' base of the anticodon can make non-Watson-Crick hydrogen bonds with several different bases at the 3' position of the codon. E. The genetic code is not degenerate in that there is only one codon for each amino acid.
D. The 5' base of the anticodon can make non-Watson-Crick hydrogen bonds with several different bases at the 3' position of the codon.
What is the role of the TATA binding protein (TBP) in transcription? A. The TATA binding protein is a prokaryotic transcription factor that binds to bacterial RNA polymerase and the -10 & -35 promoter to initiate transcription. B. The TATA binding protein is a eukaryotic termination factor that binds to the termination signal on mRNA and helps to recruit the cleavage enzyme and poly A polymerase to terminate transcription. C. The TATA binding protein is a eukaryotic elongation factor that binds to the phosphorylated CTD on RNA polymerase II and helps to recruit the capping enzymes and spliceosome to elongate transcription. D. The TATA binding protein is a eukaryotic transcription factor that binds to the promoter and helps to recruit other transcription factors and RNA polymerase II to initiate transcription. E. The TATA binding protein is a eukaryotic elongation factor that binds to the spliceosome to induce alternative splicing, RNA editing, and alternative poly-A tails.
D. The TATA binding protein is a eukaryotic transcription factor that binds to the promoter and helps to recruit other transcription factors and RNA polymerase II to initiate transcription.
Which of the following statements about the poly(A) tails that are found on most eukaryotic mRNAs are correct? A. They are added as preformed polyriboadenylate segments to the 3' ends of mRNA precursors by an RNA ligase activity. B. They are encoded by stretches of polydeoxythymidylate in the template strand of the gene. C. They are added by RNA polymerase II in a template-independent reaction using ATP as the sole nucleotide substrate. D. They are added by poly(A) polymerase using ATP as the sole nucleotide substrate. E. They are cleaved from eukaryotic mRNAs by a sequence-specific endoribonuclease that recognizes the RNA sequence AAUAAA.
D. They are added by poly(A) polymerase using ATP as the sole nucleotide substrate. pre-mRNA acquires a poly(A) tail on the 3' end The cleavage signal (its terminator) resides in the transcript itself. Specific proteins (termination factors) bind to specific sequences on the transcript to mark where pre-mRNA will be cleaved. - A specific endonuclease cleaves the RNA downstream of AAUAAA (cleavage signal) - Poly (A) polymerase then adds about 250 adenylate residues - Poly A binding proteins (PAB) binds to the poly (A) tail to "mark" the 3' end of mRNA and protect it. - Poly A tail protects mRNA from RNases (exonucleases)
The exchange of fatty acids and triacylglycerols between the liver and adipose tissue is an ongoing process that helps maintain metabolic homeostasis. What is the name of this process? A. Cori cycle B. Glycerogenesis pathway C. Gluconeogenesis D. Triacylglycerol cycle E. Glucose-alanine cycle
D. Triacylglycerol cycle
Expression of a(n) ________ alters gene transcription in E. coli by activating transcription of specialized heat stress response genes, nitrogen metabolism genes, and/or flagella genes. A. attenuator B. lac repressor C. lambda repressor D. alternative sigma factor E. autoinducer
D. alternative sigma factor
In the ρ-independent transcription termination signal, the GC-rich palindromic region forms a ______________ in the RNA, which results in the _____________________. A. Rho protein, ATP-dependent unwinding of the RNA-DNA hybrid helix B. hairpin stem loop structure, ATP-dependent unwinding of the RNA-DNA hybrid helix C. stretch of U oligos, RNA polymerase to pause transcription D. hairpin stem loop structure, RNA polymerase to pause transcription
D. hairpin stem loop structure, RNA polymerase to pause transcription Two main types of termination of RNA synthesis in bacteria: - elongation continues until a termination signal is detected - the simplest stop signal is the transcribed product of a segment of palindromic DNA - the other type involves a protein, Rho - both types require the transcribed product of the stop signals encoded by the DNA to terminate transcription** signals are within the DNA transcript in both types Rho-Independent (Intrinsic) - palindromic, G-C rich loop, string of U's (which means string of A's in DNA) - string of U-A interactions is a weaker interaction than the stem-loop position, allowing RNA to fall off polymerase- uses stem loop secondary structure Rho-dependent - Rho, helicase - climbs up transcript and once it hits the hybrid, it will unwind it using ATP hydrolysis, forcing RNA to leave polymerase
What enzyme below would MOST LIKELY have a role in repression of eukaryotic gene expression? A. histone kinase B. histone demethylase C. histone acetylase D. histone deacetylase
D. histone deacetylase
What is the expected outcome in a liver cell after exposure to insulin? A. activation of fructose-1,6-bisphosphatase B. allosteric inhibition of protein phosphatase 1 C. phosphorylation of phosphofructokinase-2/fructose-2,5-bisphosphatase D. increased cytoplasmic concentration of fructose-2,6-bisphosphate E. activation of phosphoenolpyruvate carboxykinase (PEPCK)
D. increased cytoplasmic concentration of fructose-2,6-bisphosphate
Which of the following is the primary mode of regulation for the pyruvate dehydrogenase complex? A. feedback inhibition by citrate B. feed-forward activation by glucose C. allosteric activation by ATP and NADH D. inhibition by phosphorylation E. allosteric inhibition by ADP and CO2
D. inhibition by phosphorylation
When glucose levels are high, the hormone __________ is secreted from the pancreas, which activates a signaling cascade in target tissues (such as the muscle and adipose tissue) to allow this receptor _________ to move to the membrane for glucose uptake. A. glucagon, GLUT2 B. insulin, GLUT2 C. glucagon, GLUT4 D. insulin, GLUT4 E. insulin, GLUT1
D. insulin, GLUT4
What prosthetic group is present in complexes I, II, and III of electron transport? A. Copper B. flavin mononucleotide (FMN) C. coenzyme Q D. iron-sulfur cluster E. heme
D. iron-sulfur cluster
The antibiotic ______________ inhibits both prokaryotic and eukaryotic protein synthesis because it binds in the tRNA binding sites on the ribosomes. A. tetracycline B. chloramphenicol C. streptomycin D. puromycin E. erythromycin
D. puromycin
The irreversible glycolytic step catalyzed by pyruvate kinase is bypassed in gluconeogenesis by the action of these two enzymes: A. pyruvate carboxylase & glucose 6-phosphatase B. glucose 6-phosphatase & fructose 1,6-bisphosphatase C. malate dehydrogenase & fructose 1,6-bisphosphatase D. pyruvate carboxylase & phosphoenolpyruvate carboxykinase E. phosphoenolpyruvate carboxykinase & malate dehydrogenase
D. pyruvate carboxylase & phosphoenolpyruvate carboxykinase
The antibiotic ______________ inhibits prokaryotic transcription initiation by binding to the RNA polymerase in the channel that is normally occupied by the newly formed RNA-DNA hybrid and blocking the exit of the nascent RNA. A. actinomycin B. streptomycin C. erythromycin D. rifampicin E. puromycin
D. rifampicin
What citric acid cycle enzyme is responsible for catalyzing the reaction that is a substrate-level phosphorylation? A. malate dehydrogenase B. fumarase C. aconitase D. succinyl-CoA synthetase E. alpha-ketoglutarate dehydrogenase complex
D. succinyl-CoA synthetase
The main reasons that eukaryotic mRNA transcripts are modified at the 5' and 3' ends are: A. so that the promoter can be recognized and to facilitate termination of transcription B. to ensure that the CTD of RNA polymerase is phosphorylated and that the polymerase can escape the promoter region C. so that the two ends of mRNA remain associated with each other as the mRNA exits the nucleus and is transported to the rough endoplasmic reticulum D. to protect the ends of the transcripts from exonuclease degradation and to facilitate initiation of translation E. to facilitate the export of the mature mRNA and to guide the removal of introns
D. to protect the ends of the transcripts from exonuclease degradation and to facilitate initiation of translation
What is the role of the E2 enzyme in the ubiquitin system? A. acetylate ubiquitin for activation B. activate ubiquitin by adenylation C. add a poly Lys tag on the target protein that ubiquitin recognizes D. transfer ubiquitin thioester linkage from E1 to E2 E. form thioester linkage between E1 and ubiquitin
D. transfer ubiquitin thioester linkage from E1 to E2
Which of the following can be used as a metabolic control mechanism? A) Enzyme compartmentation B) Action of hormones C) Covalent modification of an enzyme D) Regulation of enzyme degradation E) All of the above
E) All of the above
Catabolic pathways are always paired with anabolic pathways. Why? A) Catabolic pathways build up new molecules and anabolic break down molecules. B) Anabolic pathways require redox reactions to operate, while catabolic pathways require ATP to operate. C) Both require ATP to operate. D) Both require redox reactions to operate. E) Catabolic pathways break down molecules, producing ATP that is used for anabolic pathways.
E) Catabolic pathways break down molecules, producing ATP that is used for anabolic pathways.
What tissue or organ are the primary storage sites for glycogen in animals? A) brain & adipose tissue B) adipose tissue & liver C) kidney & muscle D) heart & muscle E) muscle & liver
E) muscle & liver
If the E. coli RNA polymerase was missing the _________ subunit, the RNA polymerase would bind the DNA with ____________ specificity. A) alpha; greater B) rho; less C) sigma; greater D) rho; greater E) sigma; less
E) sigma; less
Differential mRNA processing in eukaryotes, such as in using alternative splice sites or alternative poly-A sites, may result in: A) a shift in the ratio of mRNA product from two adjacent genes. B) attachment of the poly-A tail to the 5' end of an mRNA. C) inversion of certain exons in the final mRNA. D) the production of the same protein from two different genes. E) the production of two or more different proteins from a single gene.
E) the production of two or more different proteins from a single gene.
Refer to the figure on the right, which we discussed in class. Which of the following statements about apolipoprotein B (apo B) is true? A. The apo B-48 form is formed by the proteolytic cleavage of the primary (apo B-100) translation product. B. Apo B-48 and apo B-100 are formed in the same tissues at the same time. C. Apo B-48 arises from the expression of a form of the gene for apo B-100 that has been shortened by nonhomologous recombination. D. The transcript of the apo B-100 gene is spliced to remove a segment and form apo B-48. E. A specific nucleotide in the apo B-100 transcript is altered, thereby creating a stop codon in the mRNA, resulting in an mRNA encoding apo B-48.
E. A specific nucleotide in the apo B-100 transcript is altered, thereby creating a stop codon in the mRNA, resulting in an mRNA encoding apo B-48.
If there were a mutation in TFIIH causing it to lose its kinase activity, which of the following scenarios would you expect? A. RNA polymerase II would not bind to the other basal (general) transcription factors B. Mediator would fail to bind the initiation complex C. The CTD of RNA polymerase II would be phosphorylated at a higher rate than usual D. Chromatin remodeling complexes and other histone code readers would not be recruited to the DNA E. Initiation of transcription would be inhibited due to basal (general) transcription factors failure to dissociate from the initiation complex (a phenomenon called promoter escape)
E. Initiation of transcription would be inhibited due to basal (general) transcription factors failure to dissociate from the initiation complex (a phenomenon called promoter escape)
Concerning the translational control of iron metabolism in animals, which of the scenarios below would you expect to find when there are low levels of iron in the blood? A. Endonucleases will cleave transferrin receptor mRNA, causing the mRNA to be rapidly degraded. B. Iron response proteins (IRPs) will contain iron in their binding sites. C. Ribosomes will be bound to IRE elements on the 5' UTR of ferritin mRNA, preventing translation of ferritin. D. Ferritin, the iron storage protein, will be translated efficiently. E. Iron response proteins (IRPs) will be bound to IRE elements on the 3' UTR of transferrin receptor mRNA, allowing efficient translation of transferrin receptor.
E. Iron response proteins (IRPs) will be bound to IRE elements on the 3' UTR of transferrin receptor mRNA, allowing efficient translation of transferrin receptor.
Refer to the figure depicting five common fatty acids. Which of the fatty acids would you except to be the MOST fluid (i.e., have the lowest melting temperature)? A. Myristic acid B. Palmitic acid C. Stearic acid D. Oleic acid E. Linoleic acid
E. Linoleic acid
What sequence identifies a protein for translocation across the endoplasmic reticulum (ER)? A. Shine-Delgarno sequence B. Poly-A sequence C. TATA box sequence D. Promoter sequence E. N-terminal signal sequence
E. N-terminal signal sequence
What is NOT required for RNA synthesis, but is required for DNA synthesis? A. rNTPs B. DNA template C. Mg2+ or Mn2+ D. RNA polymerase E. Primer
E. Primer
Which enzyme transfers electrons from a membrane-soluble carrier to a water-soluble carrier in the electron-transport process? A. ATP synthase B. cytochrome c oxidase C. NADH-Q oxidoreductase D. succinate-Q reductase E. Q-cytochrome c oxidoreductase.
E. Q-cytochrome c oxidoreductase.
An infant who obtains nourishment from milk and who has galactosemia (i.e., has a deficiency in galactose-1-phosphate uridyltransferase) is unable to convert: A. galactose-1-phosphate to glucose-6-phosphate B. glucose-6-phosphate to galactose-1-phosphate C. galactose to galactose-1-phosphate D. glucose-1-phosphate to galactose-1-phosphate E. galactose-1-phosphate to glucose-1-phosphate
E. galactose-1-phosphate to glucose-1-phosphate
What is the main difference between small interfering RNA (siRNA) and microRNA (miRNA)? A. miRNAs bind to RNA-induced silencing complexes (RISCs); siRNAs do not. B. siRNAs recognize mRNA by base-pairing with it; miRNAs do not. C. miRNAs are processed by Dicer; siRNAs are not. D. siRNAs silence genes; miRNAs activate genes. E. miRNAs are encoded by the host genome; siRNAs are not.
E. miRNAs are encoded by the host genome; siRNAs are not.
Which of the following are common mechanisms used by bacteria to regulate their metabolic pathways? A. control of the expression of genes at the level of transcription initiation B. utilization of specific and distinct sigma factors to suit the environmental condition C. formation of altered enzymes by the alternative splicing of mRNAs D. deletion and elimination of genes that specify enzymes E. options A & B are both common mechanisms used by bacteria
E. options A & B are both common mechanisms used by bacteria
Which citric acid cycle is also a key intermediate in gluconeogenesis? A. citrate B. fumarate C. a-ketoglutarate D. succinate E. oxaloacetate
E. oxaloacetate
The different sensitivities to ________ revealed that eukaryotes have three distinct RNA polymerases for synthesis of mRNA, rRNA, tRNA, and small RNAs. A. diphtheria toxin B. ricin toxin C. actinomycin D. cycloheximide E. α-amanitin toxin
E. α-amanitin toxin
Iron deficiency anemia is a common type of anemia in which iron levels are low. In patients suffering from this type of anemia, which of the following scenarios would you predict to be happening? A. Ferritin, the iron storage protein, will be translated efficiently. B. Iron response proteins (IRPs) will contain iron in their binding sites. C. Endonucleases will cleave transferrin receptor mRNA, causing the mRNA to be rapidly degraded. D. Iron response proteins (IRPs) will be bound to iron response elements (IREs) on the 3' UTR of transferrin receptor mRNA, preventing translation of transferrin receptor. E. Iron response proteins (IRPs) will be bound to iron response elements (IREs) on the 5' UTR of ferritin mRNA, preventing the translation of ferritin.
Iron response proteins (IRPs) will be bound to iron response elements (IREs) on the 5' UTR of ferritin mRNA, preventing the translation of ferritin. Genes associated with iron metabolism are translationally regulated in animals: Iron transport and storage must be carefully regulated in animals to prevent oxidative damage by iron. Ferritin and transferrin receptor are translationally controlled. ** always transcribed and regulated at the transcriptional level IRE: iron response element IRP: IRE-binding protein High Iron: make Ferritin (iron storage) Low Iron: make Transferrin Receptor (iron transport) IRON STARVATION: - IRP bound to 5' UTR on ferritin mRNA - prevents ribosome from translating - IRP bound to 3' UTR on transferrin receptor mRNA - ribosome can translate, transferrin receptor made, this is what you want (will help transport iron to parts of body) EXCESS IRON: - Iron binds to the IRP preventing it from binding to the 5' UTR on ferritin mRNA - allows for translation and ferritin is made (allows for iron storage when conditions are in excess) - Iron binds to the IRP preventing it from binding to the 3' UTR on the transferrin receptor mRNA and this UTR has endonucleolytic cleavage, leading to mRNA degradation due to lack of protection and no transferrin receptor made
repeat about U1 and U6
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What would happen to a mutated ATP synthase enzyme where the proton binding aspartate residue on the c subunits was mutated to an alanine? A. The enzyme would make ATP as normal in the presence of a proton gradient. B. The enzyme would not make ATP in the presence of a proton gradient. C. The enzyme would make ATP without the need for a proton gradient. D. The rotor would rotate in response to a proton gradient, but no ATP would be made. E. The ATP synthase will reverse direction and hydrolysis ATP readily in response to the proton gradient.
The enzyme would not make ATP in the presence of a proton gradient.
What is the chemical effect of oligomycin on aerobic metabolism? A. The flow of electrons from Cyt a-a3 to oxygen is blocked B. The flow of electrons from NADH to CoQ is blocked C. The transport of ATP out of and ADP into the mitochondria is blocked D. The proton transfer through F0 of ATP synthase is blocked, blocking the phosphorylation of ADP E. Oxidative phosphorylation is uncoupled from electron transport and all the energy is lost as heat
The proton transfer through F0 of ATP synthase is blocked, blocking the phosphorylation of ADP
How many introns and exons are shown in the images to the right (both images show the same DNA segment hybridized to the same mRNA segment). A. 2 introns, 3 exons B. 3 introns, 4 exons C. 3 introns, 2 exons D. 4 introns, 3 exons E. 3 introns, 3 exons
don't need to know
The common diabetic type II treatments are shown in the table to the right. Monitoring fatty acid metabolism in a diabetic patient before and after administration of metformin would show a decrease of _____________ in the liver and an increase in ______________ in the skeletal muscle.
fatty acid synthesis; fatty acid oxidation
Which of the following metabolic conversions is considered to be the rate-determining step of glycolysis, and thus a major control point of glycolysis?
fructose-6-phosphate --> fructose-1,6-bisphosphate
What reaction is catalyzed by aldolase?
reversible cleavage of Fructose-1,6-bisphosphate (F-1,6-BP) to DHAP and GAP
n the picture of an uncharged tRNAAla, this letter ______is where you would find the "wobble" position, while this letter _______ is where the amino acid would be covalently linked. repeat
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Similar amino acids are sometimes attached to the incorrect amino-acyl tRNA synthetase. Suppose Phe was attached to tRNATyr, creating Phe-tRNATyr. Choose the correct equation for hydrolytic editing of Phe-tRNATyr by the tyrosine amino acyl-tRNA synthetase that takes place during the proofreading step: A. Phe-tRNATyr + Tyr à Tyr-tRNATyr B. Phe-tRNATyr à Tyr-tRNATyr C. Phe-tRNATyr à tRNATyr + Phe D. Phe-tRNATyr + Phe à Phe-tRNATyr E. Phe-tRNATyr à tRNATyr +Tyr
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