AAMC Section Bank Biology

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Based on the information in the passage, which type of enzyme is most likely to suppress CBC?

A. Phosphorylase
B. Kinase
C. Phosphatase
D. Synthase

passage: Phosphorylation of LC20 is required to activate the myosin head, which binds to actin... this will later create a CBC
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Terms in this set (154)
phosphorylation: add phosphate to inorganic phosphate
kinase: add phosphate group from ATP to substrate
phosphatase: dephosphorylase substrate, remove phosphate group
synthase: link to molecules together via catalysis
lyase: catalyze cleavage without adding H2O
isomerase: catalyze conversion of isomers
ligase: bind 2 large biomolecules
hydrolase: catalyze cleavage with H2O addition
want to know the Mr function in sympathetic NS.

A- no, blood pressure is higher in aorta than in rest of body
B - yes, sympathetic NS cause increase BP
C - no, sympathetic NS causes vasoconstriction
D - no, blood flow is directed towards brain and skeletal muscles, NOT the organs.

most reasonable is B. due to high BP
thick filamentmyosinthin filamentactinWhat is the likely structure of the amino acid found at position 19 of LC20?passage states "which activates an enzyme that phosphorylates amino acid residue 1- of the myosin light chain" phosphorylates = OH! so D (serine) is the correct answeramino acid phosphorylationTHINK OF OH! serine, threoine, tyrosineWhich table shows the expected body weight of WT and Gpcr43-/- mice fed a HFD while housed for 16 weeks in a conventional (CONV) condition, or a germ-free (GF) condition where the gut does not become colonized? Passage: "microbial population...implicated the regulation of host metabolism, obesity & diabetes"yes, you got this right, just look at figure 1 for the answer. Since the independent variable is germ free (no gut microbiota) means that the WT would have more body weight because no metabolism can occur In figure 1: we see that GPCR has lower overall body weight compared to WT. Therefore we expect the table should correspond to that under convention for GF: as stated before, WT would have increased body weight. for GPCR, since it is able to regulate energy, we expect little to no change in the corresponding graphWhich amino acid is LEAST likely found in one of the transmembrane domains of GPCR43?transmembrane protein = hydrophobic! look for the amino acid that is polar aspartic acid is the only one polarhydrophobic amino acidsFAMILY VWCompared to WT mice, which experimental group of mice is most likely to remain lean when fed a HFD?if you look at figure 1, we see that without GPCR43, the body weight is significantly higher compared to GPCR43 present. Therefore we can conclude that GPCR leads to weight loss. C insulin: increase glucose -> fatter acetate: decrease glucose -> leaner also, we want to look at figure 1, where we see that if we compare GPCR (+) and WT, we see that GPCR is able to keep the body fat down. therefore we want GPCR to be overexpressed to remain "lean"Which conclusion about glucose uptake is best supported by the data in Figure 2?Figure 2. insulin vs. acetate insulin takes in glucose, acetate does the opposite. if there is insulin only there is high glucose. both insulin and acetate its lower in WT. this means that insulin is suppressed by acetate!! A look at the figure carefully!Insulin signaling results in the phosphorylation of the downstream target, Akt, which can be quantified by Western blot analysis. Which graphic shows the effect acetate administration will likely have on Akt phosphorylation levels in WAT and muscles of WT mice?insulin + = akt high. B is correct because in the present of insulin, Akt is high. but we also need to consider that acetate suppresses insulin. bc insulin is suppressed, so is Akt here.Compared to untreated WT mice, antibiotic treatment of WT mice is likely to result in:Antibiotic treatment means less gut microbiota look at figure 1, you see that WT on STD, the body weight is relatively low. antibiotic treatment would do the opposite, and make him fatter. (B wrong) increase adipocytes because if SCFA was removed by antibiotic treatment, then GPCR 43 would not be activated to reduce uptake into glucose. (C correct)adipocytesfat cellsBranched Amino AcidsLIV Leucine, Isoleucine, ValineThe stereochemical designators α and β distinguish between:epimers at an anomeric carbon atom alpha is a fish and fish goes down. (OH DOWN) beta (OH UP) when you think of alpha and beta for sterochemistry, think of anomers and the alpha and beta shit.Epimer vs Anomersepimers = different configuration in just 1 chiral carbon anomers = different configuration in the chiral, anomeric carbon when the molecule is in cyclic formepimerDiastereomers that differ at only one chiral center.anomerssubtype of epimers that differ at the anomeric carbon alpha = down beta = upWhich ideal solution exhibits the greatest osmotic pressure?easy question, you just dissociate the molecule into its ions then multiply C: 0.2 M CaCl2 (0.2) x (3 ions) = 6. (checked all the other answers already, and C has the greatest)Which event is directly mediated by a ligand-gated ion channel?Influx of Na+ across motor end plate resulting in the depolarization of the muscle fiber membrane!! this is where voltage gated Na+ channels open for depolarizationA single point mutation in a gene results in a nonfunctional protein. Individuals heterozygous for this mutation were identified using a Southern blot. Which pair of wild-type (WT) and mutant alleles most likely contains the mutation?Southern blot can only be useful if the mutation can either CREATE or ELIMINATE a restriction site (that are palindroms of 4 to 6 bp long) In this case, we see that AAGCTT is disrupted n A. Southern blot uses restriction enzymes to cut DNA sequences and only cut at these sequences. look at the answer choices and find the palindrome one.What bond is cleaved by IN during the first reaction of integration?passage "IN catalyzes cleavage of a GT dinucleotide from each 3' vDNA terminus" it cleaves a guanine-thymine. (nucleotide) therefore it cleaves its phosphodiester bond here. D is correct. (P-O)A vDNA sequence encoding a protein is inserted into a host genome by IN. The protein is translated from the hypothetical mRNA sequence shown. 5-GGCAACUGACUA-3 Based on the passage, the segment of the original viral genome that encoded this protein had what nucleotide sequence?5-GGCAACUGACUA-3 this has been translated. viral DNA integrated into host cell genome by integrating would originate from a retrovirus. mRNA is then transcribed from retrovirus and can either used to synthesize proteins or used as the RNA genome for progeny virusWhich experiment would best provide data to support the mechanism by which an ODN inhibits IN activity?"ODN are competitive inhibitors of vDNA" remember the lineweaver plots km: increase vm: no affect.Lineweaver Plot GraphsWhich amino acid substitution for the conserved residue at position 64 is LEAST likely to affect the enzymatic function of IN?D64 is aspartic acid. you want a polar, negative charge amino acid in place. so glutamic acid (E)An inactive tetramer of IN is expected to have approximately what molecular weight?Passage states: 288 residue protein you also want a TETRAMER (question stem) 4 units given that you know the molecular weight of protein (110kDa) 288 x 4 x 110 approx : 300 x 4 x 100 = 120000 Da = 120 kDa!Average molecular weight of amino acid110 kDaTo determine if a small molecule acts like a LEDGIN with respect to IN, a researcher plans to incubate purified IN both with and without the small molecule and then perform a Western blot to detect IN in each sample. Under which condition(s) should the gel electrophoresis step be performed? I. Denaturing II. Reducing III. NativeGel electrophoresis! LEDGIN = inhibitors that bind IN and shift its oligomerization equilibrium toward inactive tertramer. probably native, because you still want to know about its function. native will keep its natural/original function,denaturingdisrupts interaction between monomersreducingdisrupt disulfide bridges (created by cysteine)nativedoes nothing to proteins, in its native stateSDS non reducingeliminates quaternary structure by breaking hydrogen bond. It breaks into subunits (dimers into 2 and trimers into 3)SDS reducingeliminates quaternary & tertiary, broken disulfide bridgeSDS denaturingeliminates quaternary, tertiary, and secondary. only primary structure left.Retinol uptake was measured in untransfected and STRA6-transfected cells in the presence of retinol bound to either RBP or bovine serum albumin (BSA). Uptake of retinol bound to RBP was also measured in cells that were cotransfected with a small inhibitor RNA (siRNA) targeting STRA6. Which graphic shows the expected result of this experiment?Read the graph carefully. the binding protein STRA6 is on the X axis. so that means we are looking for the absence/presence of this molecule. In this case, untransfected is the control. The first two columns we expect the transfected column to be high, because there is an (-, absence) of the SiRNA, which is an inhibitor so we expect the levels to be high since there is no inhibitor. in the second column, BSA, we expect no uptake because BSA acts as a competitive inhibitor that prevents RBP from binding to retinol. This results in the levels to be low. in the last column, SiRNA is positive so there is inhibition, so we expect this level to be slightly lower than the first graph, but we still expect some uptake to happen.Which statement is supported by the data shown in Figure 1?Pay attention to the control and compare it with your results. Although LRAT, and CRBP lines are high, you still need to consider the STRA6 only line too. Since STRA6 line has a higher activity rate than the control, we still expect it to still produce activity. This means that STRA6 is not completely dependent on LRAT or CRBP for uptake. It can still uptake on its own, but a significantly lower rate BUT ITS STILL PRESENTWhat is the dependent variable from the experiment shown in Figure 2?Retinol fluorescenceHill coefficientA measure of cooperative interaction between protein subunitshill coefficient = 0no cooperativityHill coefficient greater than 1positive cooperativity, binding of one ligand facilitates binding of subsequent ligands at other sites (increases affinity at remaining sites)Which statement regarding STRA6-mediated retinol uptake is best supported by the passage? Passage: "Retinal fluorescence, which is enhanced when retinol is bound to RBP, was measured over time"higher retinal fluorescence -> lower reuptake low retinal fluorescence -> higher reuptake if you're confused, just compare it with the control, most usually you're trying to do the opposite to what the control is showing since we are looking for reuptake, we are looking at the line that dips down the most. We see that STRA6 + LRAT 0.5 holo dips down the most, meaning "C" STRA6 becomes more dependent on LRAT for retinol release when the holo RBP to STRA6 molar ratio increases"negative cooperativityfirst binding event reduces affinity at remaining sitesFrom the data shown in Figure 2, which statement best explains the relationship between STRA6 and holo-RBP concentration in retinol release?need some key information from the passage. we are trying to understand the retinol release by looking at the graph. passage states that retinol flurescence is enhanced when retinol is bound to RBP so for retinol release, we expect the opposite (dec in fluorescene -> higher retinal release) looking at the answer choices, C makes the most sense because: STRA6 becomes more dependent on LRAT for retinol release (STRA6+LRAT line is lower than STRA6 only line). when holo-RBP STRA6 ratio increases -> we see that there is less retinol release.The organ in which the holo-RBP complex forms also functions to:found in the liver. it will detoxify!!liverproduces bile, detoxifysecrete glucagonalpha cells in the pancreasproduce HClparietal cells in the stomachIn a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation). If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Note: Assume Mendelian inheritance patterns.)4/9 What are the odds of crossing 2 hetereozygous F1 beetle? (Rr x Rr) = 2/3 What are the odds of having both red & brown beetles in F2? 2/3 x 2/3 = 4/9 *** Just break it up into simple punnet squares *** since it is asking for F2 generation, just break it up like that. AND = multiply OR = addThe diagram shows the size and position of the exons (numbered) and introns (lines) of a gene that codes for hypothetical Protein X, which can exist as two isoforms (either 16 or 17 amino acid residues long). Which technique can be used to determine if a sample of cells expresses both isoforms?1 amino acid consists of 3 residues. The isoforms NEED TO HAVE EXON 1 AND EXON 4. therefore the isoform 1 will have: exon 1, 2, 4, and isoform 2 will have exon 1, 3, 4. When you add the base pairs up and divide by 3. you will get either 16 or 17.isoformRelated by different proteins. They can be generated by alternative splicing of a gene.In Figure 1, which statement best explains the data trend of an increase followed by a decrease in bacterial cell count?VP replicates inside the host cell before causing host cell lysis *** all from the paragraph before the figure. VP bacteria were injected onto the host cell. Then, the sample was rinsed with ANTIBIOTIC that was effective against VP *** READ THE PARAGRAPH BEFORE FIGURE CAREFULLY, DRAW PIC AS NEEDED *** Figure 1 is measuring the lntracellular bacterial levels and we see that there is an increase then sharp decrease. during incubation, cells grew (increase) the antibiotics made VP to lysis (decrease)The amino acid residue at position 61 in Rac is most likely:passage states: deamidation of the side chain of the residue position 61 glutamine!A Rac variant, in which the residue at position 61 was replaced with an alanine (Rac61A), was synthesized. Wild-type Rac and Rac61A were incubated separately with VopC. To obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of which compound? Passage: "VopC contains a catalytic domain that irreversibly activates host cell GTPase Rac thru deamidation of a side chain residue at position 61"deamidation = amide group! looking for group that has NH3 therefore B is the correct answerWhy was the total amount of Rac measured in Experiment 2? Passage: "to confirm that VopC activates Rac, a mutant strain of VP lacking the VopC gene was created (VPvopC)"keyword: activates so we are thinking about transcription where things are activated on an mRNA. things will either be expressed or silence during this time. D! To determine if VopC affects Rac expression levelsUnder normal conditions in the absence of VP infection, what is the structure of the molecule that binds and activates Rac? Passage: "VopC contains a catalytic domain that irreversibly activates host cell GTPase RAC thru the deamidation"key word: GTPase!! you're looking for GTP (guanine, so a carbonyl group, 2 ring)The method used to visualize CREB327WT in Experiment 2 must include which step? Passage: "CREB327 is phosphorylated by glycogen synthase kinase-3 (GSK-3) at serine 115. PKA phosphorylates CREB327 at serine 119."Keyword: WT and PKA pka we think of pH, WT we think control "CREB327 WT cannot be phosphorylated at serine 115 or 119 were created" dependent variable: "CREB327 phosphorylation by GSK-3 at serine 11.5. B: establish a stable pH gradient in the gel before adding samplesWhich conclusion about the phosphorylation of CREB327WT is supported by the data in figures 1 and 2?think about the independent and dependent variable independent: serine 119 dependent: phosphorylation D: phosphorylation of CREB327WT by GSK 3 can only occur after phosphorylation with PKA ***just look at the figures and confer***When designing variant CREB327119, which of the following residues was most likely chosen as the substitute for residue 119 in CREB327WT?CREB327 119 has serine (polar, phosphorylation) therefore we are looking for a control that doesn't have OH nor anything funky Alanine!Which statement explains how two forms of CREB can be generated in cells? Passage: "Second messenger cAMP activates protein kinase A (PKA), which can then activate nuclear transcription factor cAMP response-element binding protein (CREB). CREB exists as two isoforms in cells, one being CREB327"Keyword: second messenger, nuclear transcription factor, isoforms we are dealing with transcription!! isoforms: formed by splicing exons and introns; alternative splicing B: CREB mRNA transcripts with different combination of exons are generatedisoform definitionRelated by different proteins. They can be generated by alternative splicing of a gene.Which statement about CREB327 phosphorylation is supported by the data presented in Table 1? Passage: "Second messenger cAMP activates protein kinase A (PKA), which can then activate nuclear transcription factor cAMP response-element binding protein (CREB)."second messenger activates PKA -> activates nuclear transcription factor so, that tells us that PKA is the "iniatitor" to have other things activated so, in fig 1 we see that PKA activates CREB, and it's even more activated with GSK-3 + PKA! so: C - phosphorylation by PKA can partially activate CREB327Under certain conditions, PKA and GSK-3 have been shown to autophosphorylate. The control group used in Experiment 2 (Lane 1) was designed to account for this possibility. Given this, the control group most likely contained Passage: "CREB327WT was incubated with radiolabeled ATP and either PKA, GSK-3, or sequentially with PKA then GSK-3. Isoelectric focusing and autoradiography were used to detect phosphorylation (Figure 2)."all the wells have ATP. purpose of experiment: to see if PKA and GSK can phosphorylate on its own. B: PKA and GSK 3 with ATP but with no CREB327 WTBased on the passage, FSHRs are found on cells of which type(s) of tissue? I. Connective II. Epithelial III. Nervous Passage: "Follicle stimulating hormone (FSH) is a peptide hormone known to activate FSH G protein-coupled receptors (FSHRs) on ovarian cells. However, FSHRs have recently been found on osteoclasts, and their activation stimulates osteoclastogenesis and bone resorption."Keyword: ovarian cells (epithelial) & bone (connective) answer: connective and epithelialEpithelial cellsskin cells that cover the outside of the body and line the internal surfaces of organs example: digestive tract, bladder, air sacs, urinary system, reproductive systemWhich conclusion is supported by the data presented in Figure 2? passage:FSH activates FSHR on ovarian cell & bones estrogen inhibits these effects, but high FSH can lead to bone loss. independent: specificity of FSH ab for FSH pep or FSH dependent: intensity of fluoresence, increase absorbance for increased binding. high absorbance = high binding FSH has lower binding A: The Kd for binding of FSHpep by FSH ab is lower than the Kd for binding of FSH by FSH abHigh Kdlow affinity for binding Kd = dissociation constantWhich statement best explains why the absorbance levels for FSH differ from those for FSHpep? Passage: An antibody (FSH-Ab) was then designed to bind the specific sequence of FSHpep. The antibody was also able to bind FSH.FSH has lower absorbance level compared to FSH pep, means that there is lower binding. There is no indication that there was cooperative or inhibition happing because passage says FSH ab was specific to FSHpep. therefore the only answer that makes sense is D D: The tertiary structure of FSH limits FSH-ab binding interactionsBased on the FSHpep sequence, which amino acid substitution in the FSHR binding domain is most likely to have the greatest effect on reducing bone density loss in the presence of high levels of FSH? Passage: "high levels of FSH, despite normal levels of estrogen, are associated with bone density loss."FSHR activation increases bone density loss. we need to decrease loss. (the opposite effect) therefore we need a substitution that disrupts binding R8D changes from + to - therefore it has most disruptive binding!Which of the following reasons does NOT describe why FSHpep was included in the ELISA experiment?- act as positive control to confirm that assay was functional - to provide baseline against which to evaluate the affinity of FSH-Ab for FSH - to generate data to support that the FSH-ab binds to receptor binding domain of FSHWhich statement describes a characteristic of FSH? Passage: Follicle stimulating hormone (FSH) is a peptide hormone known to activate FSH G protein-coupled receptors (FSHRs) on ovarian cells.Peptide hormone! they are water soluble therefore they do not need transport proteins to be in the bloodstream steroid hormones need that. so APeptide Hormone Characteristics- water soluble (hydrophilic) - binds to membrane bound receptor, triggers 2nd messenger systemSteroid Hormone Characteristics- Fat soluble (hydrophobic) - bind to carrier proteins for transport in bloodstream - diffuses thru cell membrane & bind to receptors in cytosol & mitochondriaHow do hydrophobic hormones get into the bloodstream?using carrier proteins for transport they are then diffused thru the cell membrane & bind to receptors in cytosol & mitochondria this includes steroid hormonesTyrosine Hormone Characteristics- either water or fat soluble - includes thyroid hormones (fat soluble)Two gel electrophoresis analyses are performed on a sample of purified protein with unknown structure: SDS-PAGE (1 band appears) and SDS-PAGE under reducing conditions (2 bands appear). Which prediction about the protein is directly supported by these results? The protein:contains multiple subunits!! they were dissociated by the reducing agent as they has a disulfide bridge (caused by cysteine) between the subunitsWhich type of interaction does NOT contribute to the stabilization of the tertiary structure of a protein?Tertiary structures: (interactions bw R groups of amino acids) - hydrophobic interactions - van der waals - disulfide bridge (cysteine) - hydrogen bonding (OH, N) - salt bridge (bw + and -) not phophodiester bonds bc that's in DNAWhich type of inhibitor does NOT alter the KM/Vmax ratio of an enzyme?Uncompetitive inhibitor!! Km: reduced Vm: reduced just think about the lineweaver plot graphs!An enzyme is more effectively inhibited by uncompetitive inhibitors when: I. the substrate concentration is decreased. II. the substrate concentration is increased. III. the inhibitor concentration is increased.uncompetitive inhibitors Km: reduced Vm: reduced uncompetitive inhibitors can only function with an enzyme-substrate complex. Therefore we want a higher substrate concentration III yes because if there are more inhibitors, enzyme cannot bind II yes because if there are too much substrates it won't functionWhich peptide sequence is most likely found in a transmembrane helix of a protein?transmembrane = hydrophobic!! amino acids that are hydrophobic, nonpolar F, A, M, I, L, Y, V, W Ala - Ile - Phe - Val - LeuIn order to determine the effect of pRB on tumor growth, tumor cells containing no pRB (Rb-/-), basal levels of pRB (basal pRB), or additional induced amounts of pRB (induced pRB), were injected into mice. Tumor cell growth was measured at eleven days post graft. Which graphic shows the expected result of this experiment? Passage: pRB may also pay a role in mitochondrial apoptosisEasy question!! just identify the independent, dependent, & control control: no PRB (Rb-/-) independent: basal pRB & induced pRB dependent: tumor growth C!Which statement is best supported by the data shown in Figure 1? Passage: "Cells with basal or induced pRB expression for 24 h were treated with the pro-apoptotic factor TNFα for 48 h" "The mitochondrial apoptosis pathway involves the activation of the BAX and BAK proteins"TNFα: increases apoptosis! BAX and BAK proteins = facilitate apoptosis! then look at fig 1 where Bak-/- and Bax -/- the Bax - we see that there is no change where Bak - we see that there is change. This means that Bax is needed!!Which experimental approach would be LEAST effective to determine the localization of pRB within a cell? Passage: "The retinoblastoma protein (pRB) has a primary role as a transcriptional coregulator of genes involved in several important cellular processes such as the cell cycle."keyword: localization of pRB (wants to know where pRB is) transcript is not needed to locate proteins. If you look through the answer choices you see that: A tags the protein (makes sense). C isolates their organelles (since pRB play a role in cellular processes, it makes sense) D: pull down assay (measure protein-protein interactions; measure the affinity for protein), identification of interacting proteins using mass spect, this will help locate! B: does not make sense, because the transcript of mRNA is not representative of where the protein is located, the mRNA will travel to the ribosomes for translation. ~ it will continue to move.Rhodamine 123 is a fluorescent dye that binds to polarized membranes and is used to label mitochondria. A cell line expressing basal levels of pRB or which form of induced pRB would most likely exhibit the lowest level of rhodamine 123 staining?Look at figure 2 "lowest level of rhodamine 123 staining" apoptosis would effect the ability of rhodamine 123 to label the mitochondria from loss of membrane potential Therefore higher the apoptosis rate -> lower level of rhodamine 123 staining RB_SP has the highest percent apoptosisThe caspase activator released during the mitochondrial apoptotic pathway primarily functions in which cellular process? Passage: "mitochondrial apoptosis pathway... release of caspase activator cytochrome C"electron transport chain!! keyword: mitochondria = ETC, cytochrome C = ETCcytochrome Chelp carry electrons from one complex of the integral membrane protein to anotherCells were incubated with the indicated amounts of recombinant pRB only, or recombinant pRB in the presence of BAX and BAK. Following centrifugation, cell membranes were disrupted but organelles remained intact. Levels of cytochrome c in the supernatant (SN) and pellet fractions were analyzed by Western blot. Which graphic shows the expected result?control: recombinant pRB only independent: BAX & BAK presence in recombinant pRB purpose: cell membrane disrupted but organelle intact dependent: cytochrome C level & pellet fractions we expect cytochrome C to have a presence in pellet. we expect cytochrome C to have none in SN bc of the disruption of mitochondria (organelle), no longer have normal functionBased on the information in the passage, which protein domain of STAT3 is NOT predicted to play a role in its signaling? A. nuclear localization domain B. signal sequence domain C. DNA binding domain D. protein binding domainA: "nuclear factor" B: none C: "nuclear factor" D: "homodimer" nuclear factor: requires nuclear localization domain for nuclear translocation & DNA binding domain homodimer: protein made up of 2 identical polypeptide chains (protein binding!!)Which amino acid substitution will most likely result in upregulation of leptin signaling? Passage: "SOCS3 binds to Y985 within the LEPRb, thereby blocking recruitment of STAT3 to the LEPRb/JAK2 complex."SOCS3 is blocking LEPRb from happening, therefore we need to have a substitution to prevent this so changing Y985 to Y985F would be the best so SOCS3 can no longer bind and try to inhibit it.Which mechanism restricts the expression of leptin to adipocytes? Only adipocytes contain: Passage: "Leptin is encoded by the ob gene and is primarily expressed in adipocytes in response to feeding." A: ob gene B: promotor for ob gene C: enhancer for ob gene D: nuclear factors for ob genewhat is the only thing that can be found on adipocytes? A: all cells contain the same DNA therefore the ob gene is most likely not exclusive to only adipocytes. B & C: Promotors and enhancers are DNA sequences. They are also not unique to adipocytes. D: nuclear factors are expressed on the ob gene, which are basically transcription factors. This makes sense because nuclear factors are the only elements that are different in cells. nuclear factors = different in cells!!Which amino acid substitution within the consensus-binding site for STAT3 is LEAST likely to interfere with STAT3 binding? Passage: YXXQglutamine and asparagine are alike, as they are both polar. They differ as glutamine as an extra carbon in the chain. Gln to Asn Q to NWhich amino acids are most likely present at the dimerization interface of STAT3 proteins? Passage: "Ligand binding causes LEPRb to undergo dimerization and to bind to Janus-activated kinase 2 (JAK2), resulting in JAK2 autophosphorylation at Y570 and its activation"Y570 Y (tyrosine) is hydrophobic, nonpolar. so hydrophobic amino acidsdimerizationpairing of two receptor-hormone complexesThe six different isoforms of LEPR are produced by using different:recall definition of isoforms: Related by different proteins. They can be generated by alternative splicing of a gene. they are expressed via exons! so A. Exons of the LEPR geneResearchers purified the wild-type and a variant form of a 140 kDa protein (Protein X). The researchers performed a native gel and an SDS gel electrophoresis under non-reducing condition and observed the following electrophoretic patterns.SDS takes proteins then denature them and separate them. Gel B is native gel because the protein appears to be in its native form. Gel A, looking at WT, we see that it separates into 50kDa and 20kDa. this means that the proteins has been denatured and separated into 2 bands. However, they do not add up to the original mass of 140 kDa (in the passage) therefore we need to multiply these bands by 2. (50x2)+(20x2)=140 kDa. This means that protein X has 4 components, meaning it is a tetramer. Disulfide bridges are present and they contain covalent bonds. so A, B, C are all true. D is wrong because the largest subunit is actually 50.SDS Pageseparates proteins by massNative Pagemaintains the protein's shape separates by charge and massIsoelectric focusing-proteins separated based on isoelectric point (pI) -electric charge is applied to the sample - protein migrates and stops when it reaches its isoelectric point(stops b/c charges cancel out)ion exchange chromatographyseparates by chargeaffinity chromatographyseparates by affinity for ligandcolumn chromatographypolar elutes last nonpolar elutes firstHow many molecules of reduced electron carrier are generated during conversion of α-ketoglutarate to oxaloacetate in the citric acid cycle?3! NAD+ -> NADH FAD+ -> FADH2 NAD+ -> NADH just know: Can I (nad-> nadh) keep (nad-> nadh) selling (gdp -> gtp) sex (fad-> fadh2) for (h2o) money (nad->nadh) officerWhich enzyme is used both in gluconeogenesis and glycogenolysis?Glucose 6 Phosphatase!catalytic efficiency equationKcat/Km Vmax = Kcat [E]Based on the data presented in Table 1, which amino acid residues of PRR are involved in binding to prorenin? I. Residue 109 II. Residue 140 III. Residue 201 IV. Residue 269Your WT is your control. You're trying to see which amino acid residues changes the Kd value the most. In this case we see that 140, 201, and 269 has the most fluctuations in Kd values! 109 residue showed little changes.Which Lineweaver-Burk plot most likely exhibits the inhibition of prorenin by compound C17XA? Passage: "a substrate analog which binds to the active site of activated prorenin has been used to inhibit the catalytic activity of prorenin."competitive inhibitor! look for plot that has 1/[v] intersectsBased on the data presented in Figure 1, overexpression of PRR: I. increases blood pressure in part through an angiotensin II-dependent pathway. II. increases blood pressure in part through an angiotensin II-independent pathway. III. decreases blood pressure in part through an angiotensin II-dependent pathway.angiotensin II receptor antognist (losartan) ROS: level of reactive oxygen species Scroll down and look at figure 2 lol. It shows you that in the presence of PRR, there is an increase of ROS (oxygen level that is responsible for hypertension aka high blood pressure. Angiotensin I independent: look at protenin lorsatan graph, we see an increase despite the antagonist blocking angiotensin. Angiotensin II dependent: increase! I and IIBased on the data presented in Table 1, which amino acid residue of prorenin most likely interacts with the residue at position 201 of PRR?Amino acids interacting with each other via: ionic bonds (charged amino acids) hydrogen bonds (polar amino acids), hydrophobic side chains create van der Waals interactions. D201N: aspartic acid replaces with glutamine. Aspartic acid is negative & polar; Glutamine is neutral & polar. WE ARE LOOKING FOR A POLAR MOLECULE TO FORM A HYDROGEN BOND (preferably basic/positive polar molecule so it can create an ion bond with aspartic acid) A: Alanine is hydrophobic, wrong. B: glutamate: negative, polar, wrong. C: arginine: basic (positive), polar YES. D: glutamine: neutral, polar. NO. know your bonds, and amino acid characteristics!!! charge and polarity.Ways amino acids can interact with each other:ionic bond: charged amino acids hydrogen bond: polar amino acids van der waals interaction: hydrophobic side chainRNA primaselays down RNA primer on unwounded DNADNA polymerasesynthesize nucleotides from the RNA primer placed in DNA replicationRNA polymerasecopy DNA sequence into an RNA sequence during transcriptionreverse transcriptionsynthesis of DNA from an RNA templateCompared to WT, what is the most likely effect of the W140L substitution on the stability of the PRR-prorenin complex?W140L from tryptophan to leucine no ring structure elimination of pi stacking interaction -> decreases stability of complexBased on the passage, which metabolic pathways are upregulated during the transition from intrauterine to extrauterine environment? "For instance, the sudden disruption of glucose supply from maternal blood is immediately followed by the activation of metabolic pathways involved in maintenance of glucose homeostasis."disruption of glucose supply glycogenlysis! (breakdown of glycogen into glucose) gluconeogenesis (pyruvate into glucose)After the depletion of hepatic glycogen in newborns, which compounds can be used as precursors to sustain the blood glucose level? I. Acetyl-CoA II. Lactate III. Oxaloacetate IV: α-Ketoglutaratesustain blood glucose level = gluconeogenesis!! so what is used? II, III, IVGPCR sequence1. ligand bind to GPCR 2. GPCR goes thru conformational change 3. alpha subunit exchange GDP to GTP 4. alpha subunit dissociates and regulate target proteins 5. target protein relays signal as 2nd messenger & triggers cascade response 6. GTP is hydrolyzed to GDP, everything returns to original place, sequence is repeatedInfusion of which peptide hormone will most likely prevent brain injury in newborn infants exposed to high glucose levels during their fetal development? passage: "This particular hormonal environment inhibits the rate of glucose release from the liver, and unless steps are taken to supply exogenous glucose or to alter the endocrine environment so endogenous stores can be mobilized, the infant of a diabetic mother is at great risk for brain injury"peptide hormone, water soluble!! glucose glucagon!! (increases glucose levels) -> more glucoseActivation of which enzyme would support the metabolism of newborn infants during the first 12 hours? Passage: "After the depletion of the blood glucose pool, the initial source for energy replacement is from hepatic glycogen, which is sufficient to support basic metabolism for the first 12 hours."keyword: glycogen so its gluconeolysis! breakdown of glycogen into glucose gluconeolysis RLS is glucose phosphorylase!rate limiting step of gluconeolysisglucose phosphorylaseThe initial source of energy replacement in the liver of newborn infants is formed by glycosidic bonds between glucose molecules through:alpha 1 4 linkage (linear) alpha 1 6 linkage (branched)Which of the following compounds is NOT a gluconeogenic precursor or substrate? A. lactate B. glycerol C. oxaloacetate D. PhosphogluconateDGluconeogenic precursors- lactate - pyruvate - glycerol - glucogenic amino acidsglucogenic amino acidan amino acid that can be converted to glucose via gluconeogenesis all amino acids except leucine & lysineketogenic amino acidAn amino acid that can be converted to acetyl CoA for the synthesis of free fatty acids. leucine lysineWhich type of enzyme removes the chemical groups that are added to proteins by kinases?Phosphatase!!!Under physiological conditions, increased activity of succinyl-CoA synthetase will most likely result in: I. increased levels of succinyl-CoA. II. increased levels of succinate. III. increased levels of GTP.II and III just used the mnemonic again can I keep selling sex for money officerThe pentose phosphate pathway results in the generation of:NADPH which is used as a reductive agent in cellular respiratory processes reducing agent: The agent loses electrons. Oxidizing agent: The agent gains electrons. The pentose phosphate pathway goes from G6P to Ribulose 5 Phosphate, during which NADP is being reduced to NADPH.Mixed inhibitors- bind at other site that isn't the active site - does NOT prevent substrate from binding. can bind to either enzyme alone or enzyme-substrate complex; has either competitive or uncompetitive effects; all lower Vmax to some extentWhich conclusion about the data shown in Figure 1 is valid?Look at the two graphs. In graph A, we see changes in cell viability values at 12 hours. In graph B we see changes in relative MMP around 2 hours. This means that changes in MMP will cause changes in cell viability. (since MMP changes occurred first, this means that MMP affects cell viability) MMP -> cell viability is affected so A! change in MMP likely affect cell viabilityIn a follow-up experiment, researchers treated cells with 50 µM of C75 in the presence or absence of the free radical scavenger N-acetyl cysteine (NAC) and then measured cell viability. The most likely rationale for this experiment was to test whether:Anti oxidants like NAC, glutathione, and vitamin E act as reducing agents to neutralize ROS from reacting with things like DNA and proteins. so B! reactive oxygen species play a role in mediating the effects of C75Which control experiments should be included in the design of Experiment 2 to validate the results shown in Figure 2? I. Evaluating the protein levels of FASN and mtKAS following siRNA treatment II. Evaluating the siRNAs specificity for FASN and mtKAS by assessing the cellular levels of unrelated proteins III. Evaluating the siRNAs specificity by showing that a non-specific siRNA has no effects on the protein levels of FASN and mtKASALL!! I: Does siRNA work in this experiment, yes or no? II: Does siRNA impact external factors, that we don't want? III: Does siRNA impact this stuff or only siRNA A and B that we care about, yes or no? This shows research controls.Which metabolic reaction is most likely affected by treatment of cells with C75? The reaction that converts:pyruvate to acetyl CoApyruvate oxidationpyruvate -> pyruvate dehydrogenase -> acetyl coalipoic acidcofactor for pyruvate dehydrogenase pyruvate dehydrogenase in acetyl coa oxidation pyruvate -> acetyl coaBased on the information from the passage, the deletion of mtKAS most likely results in decreased level of:proton in the mitochondrial intermembrane spaceETC haspumps H+ from matrix (inside mitochondria) to intermembrane space (outside) using ATP & electrons from metabolic processes - high H+ intermembrane space - low H+ mitochondrial matrixWhat are the effects of increasing mRNA and protein levels?Upregulation (positive feedback) of transcription!! DNA -> RNA -> ProteinWhat are the effects of increasing protein levels only?Upregulation (positive feedback) of translation! DNA -> RNA -> ProteinSuccinate -> _____ -> MalateFumarate!In the pentose phosphate pathway, which enzyme catalyzes the production of 6-phosphogluconolactone?Glucose 6 Phosphate Dehydrogenase! rate limiting step tooA partial DNA sequence of the coding strand of a gene is shown. 5-GACATGGACTCGCTA-3 Which sequence corresponds to the mRNA for this DNA sequence?Template Strandstrand that mRNA copies directly one that is mRNA is attached to antisense (-) transcribe as normalCoding StrandmRNA DOESN'T copy from directly one that mRNA ISN'T attached to sense (+) only change T to U on mRNADNA STRAND EXPLAINED