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the cell theory includes all of the following EXCEPT
What role does the active site Histidine play in thehydrolysis of a peptide bond catalyzed by chymotrypsin?
-Form part of the active site
-Donate a proton to histidine
-Initiatea nucleophilic attack on the carbonyl carbon of the substrate
-Form a transient covalent bond with the substrate
Ser, Thr, Tyr
Water is very abundant and very effective at competingfor the ionized functional groups. Therefore, ionized groups will hydrogen bond with water rather than formionic bonds with each other.
Theenzyme OMP decarboxylase was placed in a solution containing a highconcentration of urea and was found to rapidly lose its enzymaticactivity. Removal of the urearestored the enzyme’s activity.
A. What is urea most likely doing to disrupt the enzyme’s activity?
Causingthe protein to unfold by competing for H-bonds
The enzyme refolded back into its native state.
Muscle (because hexokinaseworks faster at 3.5 mM and therefore more glucose can be trapped in the cellsexpressing hexokinase).
Inglycolysis, the enzyme hexokinase converts glucose to glucose 6-phosphate. What two purposes does theinvestment of 1 ATP at this step serve?
If hexokinase was inhibited,glucose would not be trapped in the cell and stored.
Introduction of double bondsinto the fatty acyl chains of membrane phospholipids.
Human beings have approximately _____ fold more genes than the bacteria E. coli.
This organelle is similar in size and appearance to a bacterial cell.
The phosphoanhydride bonds in ATP are considered “high energy” because
Much more energy is released upon formation of hydrolytic productsthan is needed to break bonds in the reactants.
Alzheimer’s disease may be caused by aggregation of a protein fragment (Aß)extracellularly in large, abnormal structures called _________.
At a pH of 3.0, the product of [H+] x [OH-] will be ______
Which polysaccharide bond cannot be broken by mammalian enzymes that normally digestpolysaccharides?
What is the probability of finding the amino acid sequence VANDYWINS in aprotein?
What property is not a characteristic of enzymes?
For glucose, rotation around carbon 1 as the ring structure opens and closes generates
For a membrane protein where the polypeptide chain spans the thickness of themembrane bilayer (4 – 5 nm), about _____ hydrophobic amino acids are needed whenthe chain is in an α-helix, and about ______ are needed if the chain is in a β-strand.
In the Cn3D program for visualizing the 3D structure of proteins, what happens whenyou go to the Coloring Shortcut and choose “Charge”?
Charged residues turn red (for acidic residues) or blue (for basic residues)
The change in Gibbs free energy is described by the formula _______
∆G = ∆H - T∆S
What level of structure in proteins is held together by intermolecular R group interactions?
Lipid rafts are enriched in the lipids __________.
When human cells die, this lipid gets exposed on the outer leaflet of the plasmamembrane.
The approximately 200 different human cell types in adults are produced from ______cells through the process of ______.
By definition, a kilocalorie (kcal) is the amount of energy required to do what?
Mammals lack the enzyme that hydrolyzes cellulose. Yet many mammals are herbivores andthey eat grass and other plant material for nutrition. How can this be?
Bacteria living the guts of herbivores have enzymes that can degrade cellulose and aid in thedigestion.
The coloration patterns in Siamese Cats arise from a temperature-sensitive mutation. What is happening with the enzyme at themolecular level that explains this?
The enzyme has a mutation that perturbs folding at 37C (or normal bodytemperature), but does not perturb folding at the lower temperature. Tertiaryand quaternary structures are held together by weak bonds (H-bonds, etc.) thatcan be broken by increased temperature. The mutation likely reduces the numberof weak bonds that maintain the proper structure.
Describe how this sugar is metabolized in liver cells and why it is metabolized differently(what enzymes are different?)
Liver cells express glucokinase instead of hexokinase, and glucokinasecannot phosphorylate fructose. A different set of enzymes in the liver willphosphorylate and break fructose down into 3-carbon sugars. Fructose entersglycolysis at DHAP (dihydroxyacetone phosphate) and GAP (glyceraldehyde 3-phosphate).
In one step of glycolysis, glyceraldehyde 3-phosphate dehydrogenase catalyzes the additionof Pi to glyceraldehyde 3-phosphate. This reaction has a positive ∆G of +12 kcal/mol. How doesthe enzyme catalyze this energetically unfavorable reaction?
The addition of Pi is coupled to the oxidation of the aldehyde on glyceraldehyde 3-phosphate and transfer of high energy electrons from the substrate to NAD+ toform NADH. This oxidation reaction has a large negative ∆G of slightly morethan -12 kcal/mol. Therefore, the combined set of reactions catalyzed byglyceraldehyde 3-phosphate dehydrogenase has an overall negative ∆G.
The ∆G for movement of Na+ down itselectrochemical gradient into the cell is -3.1 kcal/mol at typicalintracellular and extracellular Na+ concentrations and a membrane potentialof -70 mV (negative inside, positive outside). If the membrane potential decreased to -20 mV, the ∆G would become
Electron transport complexes pump protons fromthe ______ to the _______?
3 Na+ ions bind and are pumped out of the cell(cytosol -> extracellular space)The terminal phosphate is transferred from ATPto the protein. (also ok… ATP ishydrolyzed leading to phosphorylation of the pump. ATP binding and phosphate transfer to the pumpdrives the E1 to E2 transition.)
2 K+ ions bind from theextracellular side and are pumped into the cytosol (extracellular space -> cytosol).
The phosphate is releasedfrom the pump, allowing it to relax back to the E1 conformation.
It pumps 3 Na+ ions out of the cell and 2 K+into the cell for each ATP hydrolyzed.
The net displacement of +1 charge per ATP hydrolyzedcreates an electrical gradient across the membrane (or a membrane potential). (The pump creates an electrical current bythe net displacement of charge across the membrane.)
To induce rotation, the ring subunits push against subunit ____ as each proton binds.
-It is the most abundant protein on Earth.
-It catalyzes both carbonfixation and photorespiration.
-It is soluble in the stroma ofthe chloroplast
-It is part of the Calvin cycle
What must a protein have in order to leave the cytosol and reside within a membrane-bound organelle?
A. recognizes ER signalsequences
B. causes a pause in translation of secretory proteins
C. facilitates transfer of newly synthesized proteinsto the ER translocon.
D. helps deliver ribosomes to the ER to form the roughER.
A nuclear localization signal of the type found inT-antigen is
1. Fall back toground state & emit heat to the environment.
2.Fall back to ground state & emit a photon of light that is lost to theenvironment. (Fluorescence)
3.Fall back to ground state & emit a photon that is captured by anotherchlorophyll (fluorescence resonance energy transfer - FRET).
4. The excited e- can be transferred to a series of electron acceptors in an e- transport chain to build up a proton gradient/reduce NADP (photooxidation)
(reduction means that an electron is added.)
NADP+ is the final electron acceptor of the electrons transport chain(ETC) Remoting electrons allows their continual flow down the ETC. This generates NADPH which is required as an elctron donor in the calvin cycle.
After P680 absorbs a photon of light and becomes oxidized to P680+, it is reduced backto P680 with an electron extracted from ________
TheRanGEF is localized to the nucleoplasm (by interaction with chromatin).
TheRanGAP is localized to the cytoplasm (because it lacks any sorting signal).
Fusion of this secretory vesicle with its target membranerequires both long fibrous and multisubunit _____ complexes, which are recruited to the correctmembrane by a small GTP binding protein in the ___ family. The final fusion of the membrane bilayers isdriven by a _____ in the vesicle (or VTC) membrane and a ______in the target membrane.
1. By inhibiting some of the KATP channels,sulfonylurea drugs will allow the beta cell membrane to depolarize more easilyas ATP levels rise
2. The depolarized membrane causes a calcium channel to open, increasing the cytosolic concentration of calcium
The mutation likely disrupts the ability of the LDLreceptor to bind to clathrin coat components in clathrin-coated pits. The LDLreceptor normally cycles between the endosomes and plasma membrane many times,removing an LDL particle from the serum every trip. The receptors in thepatient that could not undergo endocytosis would have a greatly diminishedcapacity to remove LDL from the bloodstream.
Dynein (Ciliary dynein) (Outer and inner dynein arms)
The patient would be sterile because he lacksdyneins that load onto the A microtubules. Thus, the cilia will not be able tobeat, and the sperm will be immotile.
The phospholipid bilayer of the red blood cell plasma membrane is supported by peripheralmembrane proteins called ________ that help prevent mechanical rupture.
Ankyrin and spectrin
The Na+/K+ ATPase:
A. pumps 3 Na+ out of the cell per ATP hydrolyzed.B. pumps 2 K+ into the cell per ATP hydrolyzed.
C. generates an electrochemical gradient.
D. can consume as much as 50% of the ATP in a cell.
For regulated secretion in beta cells of the pancreas and in neurons, opening of a______-gated ________ channel is required to drive exocytosis
Which product of the TCA cycle donates electrons to complex II of the electrontransport chain?
Which photosynthetic complex in the thylakoid membrane is most analogous to complexIII in the mitochondrial inner membrane?
Proteins are transported into this organelle using a gated transport mechanism and mosttypically using a T-antigen type of localization signal comprised of ~5 basic amino acids.
The nucleolus is the site of _________ synthesis.
The genome of the bacterium E. coli is considered to be “streamlined”, meaning that nearly all of theDNA encodes genes and there is little extra DNA. In contrast, human genomes are large and have lotsof DNA that does not encode genes, so-called junk DNA. Describe one possible evolutionary advantage for bacterial cells to keep their genome size small.
Describe one possible evolutionary advantage for human cells to having a large genome.
Chance of a single mutation hitting a vital nucleotide or gene is low
Describe one possible evolutionary disadvantage for human cells to maintain a large genome.
Longer time to replicateGreater chance of mutation
Humans inherit one set of genes from their father and one set from their mother therebyexpanding the size of the genome even more, i.e diploid. Describe one advantage to having two sets ofgenes in humans.
Inherited mutations on one set can be offset by normal gene in the other set
When an origin is first recognized and replication begins, two primers are made even thougheventually 4 strands will be synthesized at the two replication forks. Draw a picture of this event and thendescribe why only two primers are immediately made instead of 4?
Since replication can only proceed 5'-3', there is a leading and lagging strand. RNAprimers can be laid down on the leading strands right away but the helix must open upfurther before the lagging strand primers can be laid down.
If the sequence of the RNA in Telomerase is 5’-CCCGGUG-3’, what would be the sequence andpolarity of the telomeric DNA repeat? Would this be a prokaryotic or eukaryotic cell?
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. Broadly, how is the incredible amount of DNA in mammalian cells able to fit within the nucleus?
DNA is packaged into nucleosomes which consist of ~2 loops of DNA wound around a coregroup of 8 histones. The nucleosomes can be further compacted into heterochromatin and areeven more compact during mitosis
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. What is the term used to describe the state of the chromatin around the hsp70 gene in normal,unstressed cells?
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. What is the term used to describe the structure of the chromatin around the hsp70 gene in cellssubjected to stress or heat?
Euchromatin. Activation of the hsp70 gene will cause a rearrangement of the chromatinsurrounding the hsp70 gene allowing gene activation.
hsp70 is a gene that gets turned on during cell stress, particularly after exposure of cells to heat. What controls the change in the chromatin around the hsp70 gene in stressed versus unstressedcells?
The amino terminal tails of the 8 core histones in the nucleosome can be modified byenzymes that either acetylate or methylate the histones.
A chemical mutagen called EMS causes transfer of alkyl groups (carbon chains) to certain bases.Below is a base that contains an alkyl group (R) added by exposure to EMS. What is the base above to which the alkyl group was added?
A chemical mutagen called EMS causes transfer of alkyl groups (carbon chains) to certain bases.Below is a base that contains an alkyl group (R) added by exposure to EMS. How do you expect this modification to affect base pairing?
The oxygen that is now alkylated is normally double bonded making it a good hydrogenacceptor whereas with the addition of the alkyl group, it cannot participate in H-bonding.
Draw the structure of dCTP
For both structures, what base will pair across the DNA helix? Circle the atoms involved inpairing.
Circle three atoms on dCTP and two on the imino form
There are mutant forms of many proteins that are referred to as temperature sensitive mutations. Replication proceeded normally at the permissive temperature but for eachsituation, indicate what would be the specific effect on replication after a shift to the non-permissivetemperature.
A temperature sensitive helicase:
A temperature sensitive 3'-5' exonuclease activity (DNA Polymerase III)
A temperature sensitive 5'-3' exonuclease activity (DNA Polymerase I)
No ability to repair gaps between Okazaki fragments or repair mistakes after replication.
Lots of mutations.
Synthesis and rapid degradation of cyclinproteins are essential for proper cell cycle progression and checkpoint control. What will likely happento the replication status of a cell that suffers a mutation disabling the degradation machinery? Why?
Unregulated passage through the cell cycle due to loss of checkpoints controlled by cyclin-cdk complexes. Rapid proliferation, cancer.
After replication, a portion of one of the daughter chromosomes looks like the following with onemismatch in the middle (a T—G pairing). Assume that the top strand is the parental strand and that thisDNA is from a prokaryotic organism.
Describe the steps needed to repair the mismatch:
A repair endonuclease recognizes the mismatch
The endonuclease cleaves the daughter strand to expose a 3’-OH
DNA Polymerase I begins synthesis from the exposed 3’-OH using its 5’-3’ exonuclease
activity to remove the existing strand and its 5’-3’ polymerase activity to add new DNA
copying the parental strand
DNA ligase rejoins the strand
How would the repair machinery distinguish the parental strand from the new daughter strand?
What property of DNA Polymerase I is essential for this type of repair?
What is the most likely mechanism for how this mismatch arose?
A tautomeric shift that resulted in Tenol:G
A) DNA polymerases proofread newly synthesizedDNA
B) Repair endonucleases nick the DNA to expose3’-OH groups
C) DNA Polymerase I has 5’ to 3’ exonucleaseactivity
D) Theparental DNA is recognized due to methylation
What would youpredict would happen to the error rate of DNA replication if the equilibriumconstant between tautomeric forms was smaller by two orders of magnitude (i.e.100x smaller)? What if the equilibriumconstant was larger by two orders of magnitude?
What would happen toa rapidly dividing cell that lost the main gene that methylates DNA. Why?
No ability to distinguish betweenthe parental and daughter DNA strands so that postreplication repair will havea 50/50 chance of repairing damaged DNA correctly. Thus, more mutations.
Predict the effecton DNA replication for a prokaryoticcell that lost the gene for telomerase.
Why is it important that prokaryotic transcripts terminateat specific positions whereas termination is seemingly random in eukaryotes.
More sequence information contained within the mRNA itselfis required to initiate translation in prokaryotes than in eukaryotes.
What sequences are these and why is more informationneeded in prokaryotes?
1. The Shine-Dalgarno sequence which pairs witha region in 16S rRNA in the small subunit of the ribosome.
2. The start codon which pairs with a methioninetRNA.
Because prokaryoticmRNAs are often polycistronic, the sequence information in the RNA itself is neededto guide the ribosome to the correct AUG start sites.
More sequence information contained within the mRNAitself is required to initiate translation in prokaryotes than in eukaryotes. What allows eukaryotic ribosomes to initiate properly andefficiently without such extensive sequence information in the mRNA?
The Cap structure, a7methyl G residue added to the 5' end of the mRNA via a 5'-5' bond.
Describe the effects on transcription and explain an increase in thedistance between the -10 and -25 elements for sigma.
A mutation thatchanges the promoter from TATA to TGTG
TBP will not bindwith high affinity so the transcription rate will decrease; will still occur at VERY low rates
Effect on transcription of an insertion of100 nucleotides between the TATA box and the normal start site.
No effect, the promoters for tRNA genes arewithin the gene itself.
Atriplet of nucleotides that specifies a single amino acid
How many codons are there in the genetic code?
How can a smaller number of tRNAs pair with all the codons?
7-methyl guanosine triphosphate. This is the modified nucleotide that formsthe cap structure that is added in a 5'-5' bond to the first base of mRNAtranscripts
the creation of alternative reading frames when translating an mRNA.
During transcription of DNA to RNA in prokaryotes:
an RNA polymerase must first bind to a promoter sequence.
Since the two strands of DNA are complementary, for any given gene:
the ribosomal small subunit and tRNAmet are joined by the large subunit over the start codon.
the site where TBP binding leads to recruitment of RNA polymerase II.
Each of the 3 potential reading frames of an mRNA are used to make protein (T/F)
Initiation of translation in prokaryotes involves binding of the sigma factor to a promoter (T/F)
In a prokaryotic ribosome, catalysis of peptide bonds is driven by the ______subunit, whereas identifying the start site is driven by the ___________ subunit.During elongation, each in-coming aminoacyl-tRNA binds to the __-site of the ribosome,whereas the growing peptide chain is held on the tRNA in the _P__-site.
In bacteria, the protein called the _____________ associated with theRNA polymerase is principally responsible for binding to the promoter DNA.
-Occurs between the third base of the codon and the first base of the antibodon
-Is necessary because there are more codons than tRNAs
Hutchinson-Gilford Progeria Syndrome (HGPS)
A very rare, fatal, premature aging syndrome caused by amutation in the lamin A that produces the farnesylatedaberrant lamin A protein, progerin.
Causes premature aging and acclerated atherosclerosis leading to early death.
Hormone-Induced Activation and Inhibition of Adenylyl Cyclase in Adipose Cells
Role of β-arrestin in GPCR desensitization andsignal transduction
Cell replication is controlled by a complex network of signaling pathways that integrate signals from the extracellular environment with intracellular cues about cell size and developmental program.
S, G2, and mitosis in the absence of growth factors.
1. Cohesins are loaded onto chromosomes during G1
2. Closely behind the replication fork cohesins are converted into cohesive molecules
3. Cohesins are retained only at the centromere region by the end of metaphase
1. matter - membranes and macromolecules
2. energy – ATP and enzyme couplin
3.information - nucleic acids
o Enzymeforces an energy releasing rxn to give up its energy to an energy gaining rxt
Howcome it wasn’t chloroplast first to be engulfed by early bacteria and then mitochondria ?
o Ifchloroplast was taken in before the mitochondria then the oxygen will destroythe cell
o Mitochondriais needed to avoid being poisoned by oxygen brought in by the chloroplast
1.'M' Mitotic phase (division phase)
2. Interphase (G1, Synthesis, G2)