Exam 4 Study Set

Match the bacterial DNA replication component on the left with its primary function. A. Unwind the DNA at OriC B. Unwind the DNA double helix C. Synthesize an RNA primer D. Maximize processivity of DNA Pol E. Remove supercoils ahead of fork F. Prevent DNA strand reannealing (a)____Clamp protein (b) ___ Topoisomerase (c)____DNA helicase (d)____Single-strand binding protein e)____Primase (f)____ DNA A protein

(a)__D__Clamp protein (b) _E__ Topoisomerase (c)__B__DNA helicase (d)__F__Single-strand binding protein (e)__C__Primase (f)__A__ DNA A protein

Based on the figure below, cleavage followed by rejoining at which sites would lead to no recombination or loss of genetic information? -1 and 2 -1 and 3 -2 and 3 -1 and 4

-1 and 3 FEEDBACK: In a Holliday junction, four different strands of DNA come together and how the strands are cleaved and rejoined determines if a recombination event occurs. No recombination or a crossover and recombination event occurs when cleavage happens across the junction. Cleavage at adjacent sites would result in broken strands and a loss of genetic information.

If the following mRNA was added to a cell-free translation system, how many unique protein sequences would be generated?ACCACCACCACCACCACCACCACCACCACCACCACCACCACC -14 -7 -3 -2

-3

In an iteration of the Nirenberg-Leder experiment to assign triplet codons to specific amino acids, radioactively labeled aminoacyl-tRNA with the anticodon of 3'-CUG-5' was retained on the filter at the end of the experiment. Which mRNA was used in this iteration of the experiment? -5'-CAG-3' -5'-GTC-3' -5'-GAC-3' -5'-CUG-3'

-5'-GAC-3'

Both prokaryotes and eukaryotes recognize __________ as a start codon. -GUA -UAC -CAU -AUG

-AUG

Place the following steps in proper order for the translation of a membrane-bound protein. A. GTP binds to SRP. B. Protein synthesis occurs on free ribosome. C. Protein synthesis halts. D. SRP binds to the signal peptide sequence. -A; B; D; C -D; C; B; A -A; B; C; D -B; D; C; A

-B; D; C; A

Which base location in the figure below (A, B, C, D) explains the wobble hypothesis? -A -B -C -D

-D

Which of the following would cause a genetic mutation? -Death of a cell occurs -DNA makes too many copies in the cell -DNA damage is not corrected -DNA synthesis does not occur

-DNA damage is not corrected

The enzyme responsible for generating siRNA from double strand RNA is ____________. -Dicer -RISC -snoRNA particle -Sigma

-Dicer

What explains the observation that a single eukaryotic protein coding gene can give rise to multiple different proteins, i.e., why are there only ~25,000 human genes in the genome, but ~150,000 different proteins in the human "proteome?" Choose the TWO best correct answers. -Genes can contain more than one polyadenylation site, which alters the 3' of the mRNA transcript and the inclusion/exclusion of exons. -mRNA transcripts from the same gene can be differentially spliced to include/exclude exons. Serine phosphorylation of a polypeptide changes its amino acid sequence, and thereby generates multiple proteins from the same gene. Some genes are transcribed from both strands of the DNA and this generates isozymes of the same protein. Polypeptide subunits of the same protein complexes are often differentially cleaved by protease enzymes, resulting in loss of some subunits and duplication of others. There are ~6 times more proteins than genes because each gene makes 6 different proteins.

-Genes can contain more than one polyadenylation site, which alters the 3' of the mRNA transcript and the inclusion/exclusion of exons. -mRNA transcripts from the same gene can be differentially spliced to include/exclude exons.cdc

Which of the following is correct concerning self-splicing introns? -Group I introns are found across all organisms, including vertebrates. -Group I introns form a lariat structure. -Group II introns use an exogenous guanosine as a cofactor. -Group I introns use an exogenous guanosine as a cofactor.

-Group I introns use an exogenous guanosine as a cofactor.

In the figure show below of an E. coli replication fork, which protein is responsible for relieving torsional stress? -Single-stranded DNA binding protein -Gyrase -Helicase -Primase

-Gyrase

Why do helicase and gyrase need to work together to keep the replication fork moving forward? -Helicase unwinds DNA and gyrase relieves the torsional strain created by the unwinding. -Helicase synthesizes DNA and gyrase prevents helicase from dissociating from the DNA at the fork. -Gyrase adds the RNA primer and helicase removes the RNA primer after the fork passes through it. -Helicase unwinds the DNA in front of the fork and gyrase unwinds the DNA behind the fork.

-Helicase unwinds DNA and gyrase relieves the torsional strain created by the unwinding.

Which of the following sequences is most likely a nuclear localization signal? -Lys-Asp-Arg-Glu-Glu -Lys-Lys-Arg-Gly-Arg -Glu-Asp-Asp-Gly-Glu -Ile-Leu-Phe-Leu-Gly

-Lys-Lys-Arg-Gly-Arg

How are RNA tertiary structures different from protein tertiary structures? -RNA structures are more chemically complex than protein structures. -RNA adopts multiple structures, whereas protein structures are restricted. -Protein structures are hydrophobic and only found in membranes, RNA structures are soluble and always only found in the nucleus. -Protein structures have covalent bonds, whereas RNA structures have noncovalent bonds.

-RNA adopts multiple structures, whereas protein structures are restricted.

Identify the two ways in which RNA splicing can give rise to multiple mRNA transcripts from a gene containing 3 exons and 2 introns. (check all that apply) -Splicing between the 3' end of intron 1 and the 3' end of intron 2 -Splicing between the 5' end of exon 1 and the 3' end of exon 3 -Splicing between 3' end of exons 1 and 2, and the 5' end of exons 2 and 3, respectively. -Splicing between the 5' end of exon 2 and the 5' end of exon 3 -Splicing between the 3' end of exon 1 and 5' end of exon 3.

-Splicing between 3' end of exons 1 and 2, and the 5' end of exons 2 and 3, respectively. -Splicing between the 3' end of exon 1 and 5' end of exon 3.

Why does RNA contain uracil and DNA contain thymine? Choose the ONE best answer to this question. -Spontaneous cytosine deamination generates uracil, which base pairs with adenine during replication, and thereby converts a C-G base pair to a T-A base pair. -Ultraviolet light cause thymine dimers to accumulate in DNA, therefore if uracil were present in DNA, then these uracil dimers would not be recognized by nucleotide excision repair. -Uracil specifies RNA polymerization, and therefore, if uracil is inserted into DNA, then RNA polymerization will occur on DNA strands. -RNA transcription base pairs U-T in the RNA with DNA, whereas this process base pairs A-T in the RNA with DNA. Therefore, if uracil were present in DNA, then U-A and A-U could not be distinguished. -Loss of NH3 from cytosine generates uracil, which base pairs with thymine and cause damage in DNA. If deamination of thymine generated guanine, then this would not be a problem. -DNA damage can alter the nucleotide bases and cause mutations that cannot be repaired if the base is uracil since repair enzymes recognize thymine in DNA, not uracil.

-Spontaneous cytosine deamination generates uracil, which base pairs with adenine during replication, and thereby converts a C-G base pair to a T-A base pair.

The processes of DNA replication and RNA transcription in E. coli are similar in some respects and different in others. Select the THREE statements below that describe BOTH DNA replication and RNA transcription in E. coli. -The mechanism of reaction is attack by the 5'OH group of the pentose on the a-phosphate of an incoming nucleoside triphosphate. -The direction of enzyme movement on the template strand is 3' to 5' -The process includes its own 3' to 5' exonuclease proofreading mechanism. -A specific region of the DNA is recognized and bound by the polymerase. -The direction of polymerization is 5' to 3' -The process requires a primer.

-The direction of enzyme movement on the template strand is 3' to 5' -A specific region of the DNA is recognized and bound by the polymerase -The direction of polymerization is 5' to 3'

In which direction is mRNA synthesized by prokaryotic and eukaryotic RNA polymerases? -It could be 5' to 3' or 3' to 5' depending on the template or coding strands. -The mRNA is synthetized with 5' to 3' phosphodiester linkages. -The mRNA is synthesized with 3' to 5' phosphodiester linkages. -The mRNA is synthesized in the anti-parallel direction.

-The mRNA is synthetized with 5' to 3' phosphodiester linkages.

How do bacterial sigma factor proteins promote RNA synthesis? -They recruit RNA polymerase to specific sequences at the 5' end of genes. -They bind to DNA and induce structural changes that open the DNA helix. -They bind to RNA polymerase and alter its conformation by phosphorylation. -They function as RNA primase proteins that synthesize RNA primers. SUBMIT

-They recruit RNA polymerase to specific sequences at the 5' end of genes.

Which of the following statements best describes how the terms trans-acting and cis-acting apply to gene regulation? -Trans- and cis-acting factors can both bind to specific DNA sequences. -Trans-acting factors are proteins that bind to specific DNA sequences, which are themselves cis-acting sites in DNA. -Cis-acting factors can bind to specific DNA sequences whereas trans-acting sites are DNA sequences. -Trans- and cis-acting factors can both only bind to DNA elements to which they are physically linked.

-Trans-acting factors are proteins that bind to specific DNA sequences, which are themselves cis-acting sites in DNA.

What step in the process of translation of a nascent protein into the lumen of the ER requires hydrolysis of GTP? -Release of SRP from the ribosome. -Binding of SRP to the signal peptide sequence. -Binding of SRP to the SRP receptor. -Transfer of the signal peptide sequence into the translocon.

-Transfer of the signal peptide sequence into the translocon.

Introducing Oct4, Sox2, c-Myc, or Klf4 into a differentiated cell causes -cell death. -a pluripotent state. -no change. -cell meiosis.

-a pluripotent state.

A Holliday junction can best be defined as a region of ________________. -a duplex DNA structure -a pyrimidine dimer -a region of DNA damage -a quadruplex DNA structure

-a quadruplex DNA structure

When a cytosine is deaminated to form uracil and removed by glycosylase enzymes, a __________ site is generated. -abasic -nonbasic -dibasic -deaminated

-abasic

Eukaryotic RNA polymerase II facilitates the -removal of exons. -addition of a poly(A) tail. -addition of 7-methylguanylate cap. -translation of the RNA transcript.

-addition of 7-methylguanylate cap.

When do somatic mutations occur? -G2 phase -G1 phase -S phase -after zygote formation

-after zygote formation

Which component of the Nirenberg-Leder experiment, that assigned triplet codons to specific amino acids, was radioactively labeled? -aminoacyl-tRNA -tRNA -ribosome -mRNA

-aminoacyl-tRNA

Aminoacyl-tRNA synthetases carry out editing _______________. -by binding to a separate editing subunit associated with the enzyme. -before activation of the amino acid with ATP. -at a site that is distinct from the active site. -after the aminoacyl-tRNA has been released by the enzyme.

-at a site that is distinct from the active site.

The Tus-Ter complex terminates E. coli DNA synthesis by ____________. -blocking the opening of the DNA helix at the fork by helicase. -blocking DNA methylation on the nascent DNA strand. - addition dNTPs to the lagging DNA strand at the fork. -timulating the dissociation of RNA primase from the lagging strand. SUBMIT

-blocking the opening of the DNA helix at the fork by helicase.

RNA is a highly dynamic biomolecule in that it -can fold and hydrogen bond with itself, DNA, proteins, or small molecules to adopt largely modified tertiary structures. -is less capable of forming stable H-bonds than DNA. -is often shorter than DNA. -is largely composed of a phosphate backbone.

-can fold and hydrogen bond with itself, DNA, proteins, or small molecules to adopt largely modified tertiary structures.

A newly discovered protein is found to be modified by a lipid. Where is this protein most likely located? -proteasome -nucleus -extracellular matrix -cell membrane

-cell membrane

What is a major function of prokaryotic DNA polymerase I besides replication? -very slow polymerization rate (1-2 nucleotides) to prevent mistakes -DNA translation -unlimited processivity -checking for exonuclease activity

-checking for exonuclease activity

In the ER, prenylation can occur. Prenylation is the attachment of an isoprenoid group to a _____________ residue via a(n) _____________. -lysine; amide -cysteine; thioester -serine; ester -threonine; ester

-cysteine; thioester

Inosine is formed through a -deamination of adenosine. -deamination of cytosine. -demethylation of thymine. -deamination of guanosine.

-deamination of adenosine.

Individuals with mutations of BRCA have increased incidence of cancer because of their___________. -increased ability to form pyrimidine dimers -decreased ability to repair single-strand DNA breaks -increased ability to form methylated guanine -decreased ability to repair double-strand DNA breaks

-decreased ability to repair double-strand DNA breaks

The enzyme responsible for making siRNA is -snoRNA. -dicer. -exosome. -RISC.

-dicer.

When trp operon is attenuated, what does that mean for the trp operon? -inhibition of transcriptional initiation by RNA polymerase -disruption of transcriptional elongation by RNA polymerase -initiation by RNA polymerase -enhancement of transcriptional elongation by RNA polymerase

-disruption of transcriptional elongation by RNA polymerase

A nucleophilic amino group of the amino acid bound to the 3' terminus of A-site tRNA attacks the electrophilic carbonyl carbon in the ester bond between the 3' terminus of P-site tRNA and its bound amino acid during the __________ step of translation. -termination -translocation -initiation -elongation

-elongation

Which of the following best describes when a pattern of gene expression is altered without change in the DNA sequence? -epigenetic states -negative autoregulation -positive autoregulation -meiosis

-epigenetic states

A 7-methylguanylate cap and poly(A) tail is added to mRNA to -facilitate binding and translation by the ribosome. -signify the start and end of the gene sequence. -increase mRNA splicing efficiency. -differentiate the mRNA from the tRNA.

-facilitate binding and translation by the ribosome.

Producing an iPS cell is remarkable because the pathway -for conversion is general, not specific. -produces a very unstable cell. -is a highly irreversible reaction. -for conversion is very specific and limited.

-for conversion is general, not specific.

If 15N DNA (heavy) replicated using conservative replication in 14N (light) media, the outcome would be different than semi-conservative replication, and in this case, the two daughter DNA double strands would be ______________after one generation. -heavy only for one daughter DNA and light only for the other daughter DNA -light for both daughter DNA products -heavy for both daughter DNA products -both daughter DNA molecules would have one heavy and one light DNA strand

-heavy only for one daughter DNA and light only for the other daughter DNA

Which protein would initially have been translated on free ribosomes? -collagen -insulin receptor subunit -lysosomal-associated membrane protein 1 (LAMP-1) -histone subunit H2A

-histone subunit H2A

A common feature of protein synthesis in both eukaryotes and prokaryotes is _____________. -hydrolysis of GTP to facilitate binding of the AA-tRNAAA. -hydrolysis of ATP to facilitate translocation of the ribosome. -the P site in the ribosome can bind an uncharged tRNA. -the E site that can bind a tRNA covalently bound to the new protein.

-hydrolysis of GTP to facilitate binding of the AA-tRNAAA.

Lysogeny by bacteriophages causes what result to the host DNA? -integration of virus DNA into host DNA -formation of Holliday junctions -production of new bacteriophage particles -increased transcription of virus genes

-integration of virus DNA into host DNA

Semiconservative, as it relates to DNA replication, can be defined as when the original duplex DNA template -can only be replicated once. -is broken into fragments. -remains intact to make a new DNA duplex. -is separated into single strands before replications.

-is separated into single strands before replications.

What is a difference between group I and group II introns? -linear versus lariat intron products -nucleophilic versus electrophilic hydroxyl attacks -cis versus trans cleaving capabilities -intron cleaving versus exon ligating abilities

-linear versus lariat intron products

Which amino acid residues on histones are acetylated? -serine -histidine -arginine -lysine

-lysine

A common chemical modification found in RNA is __________________. -phosphorylation of the C-terminal domain of rRNA -alkylation of mRNA at the intron-exon border -methylation of nucleotide bases in tRNA -carboxylation of tRNA in the region of the anti-codon

-methylation of nucleotide bases in tRNA

The figure below shows __________ autoregulation and will __________. -negative; be zero in the absence of an activator -positive; reach a steady state -positive; be zero in the absence of an activator -negative; reach a steady state

-negative; reach a steady state

Together, the eight histone molecules are called the histone -octane. -dimer. -tetramer. -octamer.

-octamer.

A protein is targeted to the plasma membrane. Its final functional location is as a subunit of a transmembrane protein that interacts with the central subdomain of the membrane. Analysis of this protein would most likely show modification with a -palmitoylate. -isoprenoid. -myristoylate. -phosphate.

-palmitoylate

AZT is such a good HIV drug because it -prevents new DNA from being replicated. -prevents HIV from binding to the cell. -allows DNA to repair itself after viral DNA is added. -allows DNA to be replicated at a slower rate.

-prevents new DNA from being replicated.

A function of transcriptional activator proteins is to -recruit gene promotors. -recruit other transcription factors. -initiate DNA synthesis. -initiate RNA synthesis.

-recruit other transcription factors.

The initiation of transcription in eukaryotes -requires many more transcription factors than prokaryotic transcription. -requires the complete unfolding of the gene into single-stranded DNA. -occurs in the same manner as prokaryotic transcription. -uses RNA polymerase as well as helicase and primase.

-requires many more transcription factors than prokaryotic transcription.

The spliceosome ___________. -resembles group II introns in its mechanism and product. -performs the same function in prokaryotes and eukaryotes. -is a cis-acting ribozyme that autocleaves mRNA. -is a protein catalyst that functions to ligate exons to introns.

-resembles group II introns in its mechanism and product.

What is the function of telomerase in termination of DNA synthesis? -remove the telomeres -reverse transcription of the telomeres -bind to single-strand DNA to prevent refolding -shorten the DNA strand after each replication

-reverse transcription of the telomeres

The RNA type directly involved in protein synthesis is -short interfering RNA. -small nuclear RNA. -long nc RNA. -ribosomal RNA.

-ribosomal RNA.

The role of Mg2+ in DNA replication is to __________ the incoming deoxynucleotide. -stabilize the positive charges on -act as a nucleophile to attack the alpha-phosphoryl group in -stabilize the negative charges on -protonate

-stabilize the negative charges on

Prokaryotic and eukaryotic DNA polymerases rarely make mismatched base pairs because ____________. -the mismatched pairs make covalent bonds instead of hydrogen bonds. -the active site does not fit mismatches well. -there are no hydrogen bonds that line up between the mismatches. -the DNA primer does not allow for mismatched pairs.

-the active site does not fit mismatches well.

The function of the lac operon is to provide -the enzymes needed to utilize the disaccharide lactose. -the enzymes needed to produce lactose. -lactose. -glucose and galactose to make lactose.

-the enzymes needed to utilize the disaccharide lactose.

The Ames test is designed to identify mutation rates using a strategy based on ___________. -The frequency that a new mutation blocks amino acid biosynthesis. -the frequency by which a DNA damaging agent blocks histidine biosynthesis. -the frequency by which a carcinogen causes uncontrolled bacterial cell growth. -the frequency of new mutations that compensate for an existing mutation.

-the frequency of new mutations that compensate for an existing mutation.

Long noncoding RNA (lncRNA) is generated from -the transcription of genomic DNA. -the splicing of used mRNA. -the tips of chromosomes. -an infection of viral RNA. -excision of intron lariat segments.

-the transcription of genomic DNA.

Long noncoding RNA (lncRNA) is generated from _____________. -telomere sequences at the ends of chromosomes. -viral RNA that did not integrate into host DNA. -intronic sequences spliced out of mRNA. -the transcription of genomic DNA.

-the transcription of genomic DNA.

When there are high levels of tryptophan in the bacterial cell, the effect it has on the Trp repressor is _______________ and the effect on the structure of the Trp operon mRNA is _____________. -to stimulate operon binding; to promote formation of the regions 2-3 stem loop -to inhibit operon binding; to promote formation of the regions 2-3 stem loop -to stimulate operon binding; to promote formation of the regions 3-4 stem loop -to inhibit operon binding; to promote formation of the regions 3-4 stem loop

-to stimulate operon binding; to promote formation of the regions 3-4 stem loop

The control point for most gene regulation occurs at the initiation of -protein synthesis. -transcription. -protein modifications. -RNA processing.

-transcription.

Where would an amino acid be attached to the tRNA below? A B C D

A

Where might the molecule shown below be located in both the prokaryotic and eukaryotic ribosome? -A site -E site -40S complex -70S complex

A site

Briefly describe the structure and function of the 1) one key E. coli protein that are each needed for high fidelity termination of DNA replication and 2) one key human protein and the one key. A. 1) Tus protein prevents fork replication past termination, 2) Telomerase prevents lagging strand deletion. B. 1) Rho protein functions in E. coli., 2) HDAC/HAT proteins function in humans to stimulate termination. C. 1) Ter protein prevents fork replication past termination, 2) Telomerase prevents leading strand deletion. D. 1) Telomerase binds to E. coli Tus sequence, 2) Ter protein works in humans to prevent over-replication. E. 1) Primase functions in E. coli, 2) Gyrase functions in humans to prevent early termination.

A. 1) Tus protein prevents fork replication past termination, 2) Telomerase prevents lagging strand deletion

Francis Crick hypothesized that the Genetic Code must require 3 nucleotides in the mRNA to specify each amino acid based on a simple calculation. What was the logic behind this calculation? A. Based on the number of known amino acids and the number of known nucleotide bases. B. The square root of 16 equals 4 and the square root of 9 equals 3. C. 4 x 4 = 16, whereas 4 x 4 x 3 = 48; since there are 20 amino acids, it must require 3 codons (48 > 16). D. Based on the number of known amino acids and the number of known tRNA molecules. E. Based on the fact that thymine is present in DNA and uracil in RNA; both base pair with adenine.

A. Based on the number of known amino acids and the number of known nucleotide bases

Section of DNA sequence shown is located somewhere in the middle of the gene not at the start site. Promoter 5'...CCA TAC CGG...3' Termination 3'...GGT ATG GCC...5' The correct order for codons written 5' to 3' on the mRNA strand is: A. CCA, UAC, CGG B. GGU, AUG, GCC C. GGC, CAU, ACC D. CCG, GUA, UGG E. GGG, CAU, CCC

A. CCA, UAC, CGG

In the wobble position, inosine binds to adenine, cytosine, and uracil, but not guanine. Why not? A. Inosine and guanine are chemically incompatible to form hydrogen bonds. B. Cytosine is created by chemical modification of adenine. C. Guanine is a pyrimidine, and therefore too large to bind to inosine. D. Guanine is a pyrimidine and inosine is a purine, whereas cytosine is a purine. E. Inosine is created by chemical modification of guanine.

A. Inosine and guanine are chemically incompatible to form hydrogen bonds.

Why does the initiator fMet-tRNAfMet bind to the P site rather than the A site of the ribosome? A. It needs to bind to the P site so AA2-tRNAAA2 can bind to the A site and a peptide bond can be formed. B. It needs to bind to the P site because the fMet chemical structure only fits the P site ribosomal active site. C. It binds to the A site as do all incoming AA-tRNAAA molecules so that a peptide bond can be formed. D. It needs to bind to the P site first and then it moves to the A site so that a peptide bond can be formed E. It needs to bind to the P site first so that the E site is open for the incoming AA-tRNAAA molecules.

A. It needs to bind to the P site so AA2-tRNAAA2 can bind to the A site and a peptide bond can be formed.

What accounts for the accuracy of DNA synthesis at the replication fork? A. Proofreading function of DNA polymerase that removes incorrect nucleotides in 3' to 5' direction. B. The ability to replicate one DNA strand twice (lagging + lagging) in one round of replication. C. Sieve function in the editing site that removes incorrect amino acids from the replication fork. D. DNA polymerase fidelity is ensured by the G-A and C-T base pairs forming between two DNA strands. E. Mg2+ in the active site provides proofreading functions through ionic bonds with only correct nucleotides.

A. Proofreading function of DNA polymerase that removes incorrect nucleotides in 3' to 5' direction.

A ribozyme is an RNA enzyme that functions as a catalyst to mediate substrate reactions. Which of the following is true regarding ribozyme-mediated cleavage? A. The ribozyme reaction involves a phosphodiester cleavage through acid-base and metal ion catalysis B. Ribozymes decrease the standard change in free energy of the reaction but not the reaction rate. C. The ribozyme reaction involves a transacetylation reaction which removes exons from mRNA. D. Ribozymes protect cells against gene silencing by blocking the ability of Dicer to cleave dsRNA E. Ribozymes are required for use of alternate promoters, which gives rise to multiple proteins

A. The ribozyme reaction involves a phosphodiester cleavage through acid-base and metal ion catalysis

What is the main purpose of the Ames test? A. To determine if chemicals mutate DNA by quantifying their effect on reverting a His- phenotype. B. It allows researchers to study salmonella digestion of liver cell extracts as a function of mutations. C. To determine if chemicals mutate DNA by quantifying their effect on causing a His- phenotype. D. It allows researchers to find mutations that cause bacteria to die when grown on Histidine plates. E. It allows researchers to study antibiotic resistance as a function of DNA mutations.

A. To determine if chemicals mutate DNA by quantifying their effect on reverting a His- phenotype.

Name two reasons why the regulation mechanisms of trp biosynthetic enzymes in bacterial cells are not found in yeast cells, i.e., how are yeast different? A. 1) hairpin loops do not form in eukaryotic mRNA; 2) bacterial ribosomes are 70S eukaryotic are 80S B. 1) trp is an essential amino acid:no enzymes 2) transcription/translation are uncoupled:no attenuation; C. 1) chromatin packaging of mRNA blocks hairpin loops; 2) trp synthesis in yeast is blocked by glyphosate D. 1) bacterial genes are in operons: not so in yeast; 2) yeast have more than one trp biosynthetic pathway E. 1) trp is a nonessential amino acid:no enzymes; 2) attenuation would not work because of RNA splicing

B. 1) trp is an essential amino acid:no enzymes 2) transcription/translation are uncoupled:no attenuation;

Number the steps from start of RNA synthesis to start of protein synthesis in eukaryotes (a->g). (a)____ U1/U2 form a complex at the 5' end of introns (b) ____Poly(A) polymerase adds ~200 adenine residues (c)____Guanine-N7-methyltransferase adds the m7 G cap (d)____ CStF protein binds to AAUAAA sequence at 3' end (e)____ RNA Pol II CTD is hyperphosphorylated by TFIIH (f)_____U6/U5/U2 complex cleaves the 3' end of introns (g)____mRNA is exported through the nuclear pore complex A. 2, 6, 7, 3, 5, 1, 4 B. 3, 6, 2, 5, 1, 4, 7 C. 7, 3, 6, 1, 5, 2, 4 D. 3, 5, 2, 6, 1, 4, 7 E. 4, 6, 2, 5, 1, 3, 7

B. 3, 6, 2, 5, 1, 4, 7 (e)____ RNA Pol II CTD is hyperphosphorylated by TFIIH (c)____Guanine-N7-methyltransferase adds the m7 G cap (a)____ U1/U2 form a complex at the 5' end of introns (f)_____U6/U5/U2 complex cleaves the 3' end of introns (d)____ CStF protein binds to AAUAAA sequence at 3' end (b) ____Poly(A) polymerase adds ~200 adenine residues (g)____mRNA is exported through the nuclear pore complex

When the signal recognition particle (SRP) binds to the endoplasmic reticulum (ER) signal peptide sequence on the amino terminal end of the polypeptide in the early stages of translation, the ribosome pauses on the mRNA and stops translation. What is the purpose of this ribosome pausing? A. Permits SRP dephosphorylation and ribosome disassembly. B. Allows time for the SRP-Ribosome complex to bind the SRP receptor C. Permits the translocon protein to be exported to the nucleus. D. The signal sequence gets hyperacetylated and bound by histones. E. The mRNA gets cleaved by an endonuclease and releases the m7 G

B. Allows time for the SRP-Ribosome complex to bind the SRP receptor

What is the function of the resolvase enzyme in DNA recombination? A. Base excision repair of DNA mutations. B. Repair of Holliday junctions. C. Adenylation of DNA parental strands. D. Methylation of DNA daughter strands. E. Mismatch repair of DNA mutations.

B. Repair of Holliday junctions.

Which of the following components of eukaryotic mRNA are removed by RNA processing? A. poly (A) sequences B. intronic sequences C. the m7 G cap D. exonic sequences E. all hairpin loops

B. intronic sequences

Number the steps from start to finish for prokaryotic mRNA translation (a->e). (a)____Translational elongation proceeds in the 3' direction along the mRNA strand (b) ___ Binding of 50S subunit, GTP hydrolysis, release of IFs, 70S ribosome forms (c)____Initiation factors (IFs) bind to the 30S ribosome along with GTP (d)____Stop codon enters A site, binding of RF protein, disassembly of ribosome. (e)___ mRNA Shine-Dalgarno sequence base-pairs with 16S rRNA in 30S subunit A. 3, 4, 1, 5, 2 B. 4, 1, 3, 5, 2 C. 4, 3, 1, 5, 2 D. 4, 3, 5, 1, 2 E. 4, 3, 1, 2, 5

C. 4, 3, 1, 5, 2 (c)____Initiation factors (IFs) bind to the 30S ribosome along with GTP (e)___ mRNA Shine-Dalgarno sequence base-pairs with 16S rRNA in 30S subunit (b) ___ Binding of 50S subunit, GTP hydrolysis, release of IFs, 70S ribosome forms (a)____Translational elongation proceeds in the 3' direction along the mRNA strand (d)____Stop codon enters A site, binding of RF protein, disassembly of ribosome.

Lynch syndrome is marked by which of the following characteristics? A. Cancer resistance due to autosomal dominant mutations in the enzyme telomerase that prevents the shortening of telomere fibers. B. Higher cancer rates due to X-linked inherited mutations in DNA ligase, which prevents the repair of single strand breaks in DNA. C. Higher cancer rates due to autosomal dominant mutations in genes encoding mismatch repair enzymes hMLH1 and hMSH2. D. Cancer resistance due to autosomal recessive mutations in MutS/MutL, which prevent the deamination of cytosine to uracil. E. The inability to excrete uric acid due to a deficiency in the HGPRT enzyme as a result of X-linked mutations, resulting in severe gout and colon cancer.

C. Higher cancer rates due to autosomal dominant mutations in genes encoding mismatch repair enzymes hMLH1 and hMSH2.

What is the type of molecular defect that has been linked to the disease retinitis pigmentosa? A. Mutation in the mRNA sequence causing an alternate splicing reaction to occur. B. Mutation in the mRNA pathway owing to deadenylation of poly(A) tail. C. Mutation in a splicing protein causing defects in the spliceosome machinery D. Mutation causing mRNA redirection during nuclear export owing to loss of m7 G cap E. Mutation causing insertion of a lariat structure into the middle of an exon sequence.

C. Mutation in a splicing protein causing defects in the spliceosome machinery

Record T (true) or F (false) for statements below regarding protein synthesis. Record as a->e a. Prokaryotic and eukaryotic ribosomes both contain RNA that functions as a catalytic enzyme. b. Accurate protein synthesis requires tRNA synthetases to link GTP to 3' end of tRNA molecules. c. Ribosomal E, P, and A sites align tRNA anti-codons with the corresponding mRNA codons. d. Cells contain only ~30 tRNA molecules, of which 3 correspond to termination tRNA molecules. e. The wobble position in tRNA is 5' nucleotide in the anti-codon and 3' nucleotide in mRNA codon. A. T, T, T, F, T B. T, F, F, F, T C. T, F, T, F, T D. T, F, T, T, T E. T, F, T, F, F

C. T, F, T, F, T

What is the BEST explanation for why there is redundancy in the genetic code? A. tRNA wobble position at 3'-position of tRNA anticodon and 5'-position of mRNA codon allows non canonical base pairing to occur B. Aminoacyl tRNA synthetase charges tRNAs with amino acids with low specificity C. tRNA wobble position at 5'-position of tRNA anticodon and 3'-position of mRNA codon allows non canonical base pairing to occur D. tRNA lacks the high-fidelity 3'→5' exonuclease activity, so reading errors are common. E. Cytosine deamination to uracil is more common at the 3' position on the mRNA codon.

C. tRNA wobble position at 5'-position of tRNA anticodon and 3'-position of mRNA codon allows non canonical base pairing to occur

DNA replication is conservative (1) / semiconservative (2), which requires that helicase (3) / gyrase (4) unwinds the DNA double helix. Next, DNA Pol III (5) / DNA Pol II (6) binds and forms the replisome and moves along the template strand in a 5'--3'direction (7) / 3'--5'direction (8). If a mistake is made on the newly synthesized strand of DNA, DNA Pol III corrects the error in a 5'--3'direction (9) / 3'--5'direction (10). After DNA replication is complete, the double stranded DNA is a left (11) / right (12) -handed DNA helix.

D. 2, 3, 5, 7, 10, 12 DNA replication is semiconservative which requires that helicase unwinds the DNA double helix. Next, DNA Pol III binds and forms the replisome and moves along the template strand in a 5'--3'direction . If a mistake is made on the newly synthesized strand of DNA, DNA Pol III corrects the error in a 3'--5'direction.After DNA replication is complete, the double stranded DNA is a right-handed DNA helix.

Which statement below is most correct regarding the function of a) mRNA and b) rRNA? A. a) mRNA is the major RNA constituent of ribosomes, b) rRNA is an adaptor molecule in protein synthesis B. a) mRNA is the carrier of genetic information, b) rRNA is an adaptor molecule in protein synthesis. C. a) mRNA is the major RNA constituent of ribosomes, b) rRNA is the carrier of genetic information D. a) mRNA is the carrier of genetic information, b) rRNA is the major RNA constituent of ribosomes. E. a) mRNA is the carrier of genetic information, b) rRNA is the major RNA constituent of RNA polymerase

D. a) mRNA is the carrier of genetic information, b) rRNA is the major RNA constituent of ribosomes.

Which of statement is most correct regarding characteristics of codons in the Genetic Code? A. Genetic code is 54 codons with 51 corresponding to the 20 amino acids and 3 that are stop codons B. Genetic code is 64 codons with 60 corresponding to the 20 amino acids and 4 that are stop codons C. Genetic code is 54 codons with 50 corresponding to the 20 amino acids and 4 that are stop codons D. Genetic code is 64 codons with 62 corresponding to the 20 amino acids and 2 that are stop codons E. Genetic code is 64 codons with 61 corresponding to the 20 amino acids and 3 that are stop codons

E. Genetic code is 64 codons with 61 corresponding to the 20 amino acids and 3 that are stop codons

What function do HAT and HDAC enzymes perform in the chromatin-modifying process? A. HAT acetylates and represses the gene. HDAC deacetylates and activates the gene. B. HAT acetylates the gene and HDAC deacetylates the gene. Both repress the gene. C. HAT enzymes methylate DNA, HDAC enzymes do not. D. HAT acetylates the gene and HDAC deacetylates the gene. Both activate the gene. E. HAT acetylates and activates the gene. HDAC deacetylates and represses the gene.

E. HAT acetylates and activates the gene. HDAC deacetylates and represses the gene.

What is the amino acid sequence (start at N-term) encoded by these three mRNA codons? Promoter 5'...CCA TAC CGG...3' Termination 3'...GGT ATG GCC...5' A. Pro-Val-Gln B. Gly-Met-Glu C. Pro-Val-Trp D. Gly-His-Ile E. Pro-Tyr-Arg

E. Pro-Tyr-Arg

What is the difference between models for semiconservative and dispersive DNA replication? A. Semiconservative replication yields two DNA molecules: one that is all template DNA and one that is all new DNA; dispersive replication yields two DNA molecules that are mixed DNA. B. Semiconservative replication yields two DNA molecules; where one is all new DNA strands and the other is mixed DNA strands; dispersive replication yields two DNA molecules that are both new DNA. C. Semiconservative replication yields two DNA molecules where each contains one template and one new DNA strand; dispersive replication yields two DNA molecules with one mixed strand and one new strand. D. Semiconservative replication yields two DNA molecules that are both made of a mixed DNA; dispersive replication yields two new DNA molecules and two old DNA molecules, so four total DNA molecules. E. Semiconservative replication yields two DNA molecules where each contains one template strand and one new DNA strand; dispersive replication yields two DNA molecules that a both mixed DNA.

E. Semiconservative replication yields two DNA molecules where each contains one template strand and one new DNA strand; dispersive replication yields two DNA molecules that a both mixed DNA.

Choose the one correct statement regarding tRNA synthetases below. A. Hydrolysis of GTP drives conformational changes resulting in formation of the tRNA-amino acid complex. B. tRNA synthetase is a ribozyme that uses RNA to catalyze the formation of the tRNA-amino acid complex. C. tRNA synthetases use small RNA molecules to correctly recognize the anticodon of the tRNA molecules. D. The amino acid is linked to the 3'OH of the ATP before formation of the tRNA-amino acid complex. E. The editing site hydrolyzes an amino acid-AMP or a tRNA-amino acid if the amino acid is not correct

E. The editing site hydrolyzes an amino acid-AMP or a tRNA-amino acid if the amino acid is not correct

Which of statement is most correct regarding the "trombone model" of DNA synthesis? A. The Pol III core on the lagging strand stays bound to the DNA and only releases at termination. B. The Pol III core on the lagging strand synthesizes Okazaki fragments in the 3'→5' direction. C. The lagging strand template is released by Pol III when it reaches the 5' end of the RNA primer. D. DNA Pol II mediates leading strand synthesis in the 5'→3' direction and then reverses to 3' to 5'. E. The leading and lagging strands are synthesized by different types of DNA polymerase enzymes.

E. The leading and lagging strands are synthesized by different types of DNA polymerase enzymes.

What happens during protein synthesis if a tRNA is charged with the wrong amino acid? A. The protein will not be synthesized because the incorrectly charged tRNA will be stuck in the P site. B. Protein synthesis will continue, however the RNA splicing reaction will be blocked by the amino acid. C. The tRNA will be excluded from the A site of the ribosome but not from the P site. D. The amino acid will be removed by a ribozyme that cleaves tRNA molecules at the CCA end. E. The protein will contain the wrong amino acid in the corresponding codon position.

E. The protein will contain the wrong amino acid in the corresponding codon position.

Individuals with mutations of BRCA have increased incidence of cancer because of -increased ability to form pyrimidine dimers. -decreased ability to repair single-strand DNA breaks. -decreased ability to repair double-strand DNA breaks. -increased ability to form methylated guanine.

decreased ability to repair double-strand DNA breaks.