GN 311 EXAM 3 QUESTIONS WITH VERIFIED SOLUTIONS LATEST UPDATE 2026

GN 311 EXAM 3 QUESTIONS WITH VERIFIED SOLUTIONS LATEST UPDATE 2026 The distance between base pairs along the length of the DNA helix is - Answers 0.34 nm One turn of the DNA helix is - Answers 3.4 nm The diameter of the DNA helix is - Answers 2 nm The person/people who used X-ray diffraction to obtain the numbers above was/were - Answers Wilkins and Franklin DNA is a - Answers plectonic coil which means that the two strands have to unwind in order to separate from each other. The word anti-parallel - Answers describes the fact that one DNA strand reads 5' to 3' and the other reads 3' to 5' if one reads the sequence from left to right across the molecule. The bases across from each other in the DNA molecule are said to be - Answers complementary to eachother An electrical current passes through a gel to separate DNA molecules based on size in a process called - Answers Electrophoresis In this process, the DNA fragments migrate toward the - Answers positive pole The ___ pieces migrate fastest through the gel. - Answers smallest As DNA is heated... H-bonds are ___ DNA _____ UV absorption _____ - Answers H-bonds are broken DNA denatures UV absorption increases If the nucleosome core occupies 147 bp of DNA and the organism has a linker DNA length of 27 bp, then what is the maximal number of nucleosomes that can occupy a 3062 bp segment of DNA? Your answer must be a whole number. - Answers The nucelosome is the nucleosome core + the linker DNA. The maximal number of nucleosomes indicates that the nucleosome must be complete (no partial nucleosomes are counted). 3062/(147+27)= 17.6 = 17 A DNA molecule that is 350 bp long has 41 complete turns. This DNA molecule is - Answers Relaxed state = 10 bp/turn Fewer turns = Under rotated = negative supercoiling More Turns = Over rotated = positive supercoiling The correct answer is: Positively Supercoiled Which enzyme or protein initiates replication in E. coli by binding to oriC and causing a short segment of DNA to unwind? Pick the best answer - Answers DnaA Which researcher or group of researchers determined that DNA is composed of nucleotides? - Answers Phoebus Levene If the nucleosome core occupies 147 bp of DNA and the organism has a linker DNA length of 60 bp, then what is the maximal number of nucleosomes that can occupy a 8871 bp segment of DNA? Your answer must be a whole number. - Answers (8871)/(147+60)= 42 A DNA molecule that is 350 bp long has 35 complete turns. This DNA molecule is .... - Answers In a Relaxed State Which technique did Taylor, Woods, and Hughes use in their classic experiment regarding DNA replication? - Answers Autoradiography Which researcher or group of researchers is famous for studies involving base composition of DNA in a variety of species? Pick the best answer. - Answers Erwin Chargaff For each stage listed below, select the holoenzyme of RNA polymerase if it is required for proper completion of the stage or select the core enzyme if it can accomplish this step without the rest of the enzyme. Elongation: Initiation: Template Binding: Termination: What component/protein/subunit is present in the holoenzyme but is not present in the core enzyme? What component/protein/subunit is sometimes required for proper termination of transcription? - Answers Elongation Anser: core enzyme Initiation Answer: Holoenzyme Template Binding Answer: Holoenzyme Termination Answer: core enzyme What component/protein/subunit is present in the holoenzyme but is not present in the core enzyme: sigma What component/protein/subunit is sometimes required for proper termination of transcription? Rho Which part of the tRNA does the amino acid bind to? - Answers 3' end What enzyme is responsible for joining the tRNA molecule with its amino acid? - Answers aminoacyl synthetase On which molecule is the Shine-Dalgarno sequence found? - Answers mRNA During translation, the peptide bond formation is catalyzed by - Answers rRNA The sequence of coding strand of a DNA molecule is given below. Assume that it is read from left to right. CCTACCTTATGCCAAGTTGGGGATAAACTC How many amino acids will be in the protein translated from this sequence? What is the name (not abbreviation) of the fourth amino acid in the protein translated from this sequence? The label on the end of the protein that is translated first is the___ end - Answers he left end of this molecule is the Answer: 5' How many amino acids will be in the protein translated from this sequence? 5 What is the name (not abbreviation) of the fourth amino acid in the protein translated from this sequence? Tryptophan The label on the end of the protein that is translated first is the amino end A tRNA molecule has the anticodon 5'-IGA-3'. What does the I in this anticodon stand for? Which amino acid will this tRNA carry? Give the full name of the amino acid, not the abbreviation. What is one codon that this tRNA will bind to? Write only the three letters of the codon in 5' to 3' orientation so that Moodle will grade it correctly. - Answers What does the I in this anticodon stand for? Inosine Which amino acid will this tRNA carry? Give the full name of the amino acid, not the abbreviation. Serine What is one codon that this tRNA will bind to? Write only the three letters of the codon in 5' to 3' orientation so that Moodle will grade it correctly. UCC Which researcher or group of researchers determined that contains the four nitrogenous bases: Adenine, Guanine, Cytosine and Thymine? - Answers Albrecht Kossel Which researcher or group of researchers documented the Ac/Ds transposable element system in maize? Pick the best answer. - Answers Barbara McClintock Meselson and Stahl were able to determine that the semiconservative model for DNA replication was correct (ruling out all others) after _____ round(s) of replication. - Answers 2 The bond catalyzed by DNA polymerase is called a ____ bond. - Answers Phosphodiester Bond The F factor that is involved in bacterial conjugation uses this mode of DNA replication - Answers Rolling Circle Component of ribosomes - Answers rRNA Used in degrading mRNA - Answers siRNA Contains codons - Answers mRNA Processing of mRNA - Answers snRNA Brings amino acid to ribosome - Answers tRNA Processing of rRNA - Answers snoRNA This question refers to the mRNA sequence below: 5' - A G C U G A U G G G C U G G U G C C G A G A A A G U U A G G U A A - 3' As this mRNA is translated, the fourth codon is ____. - Answers UGC If the GC content of a DNA molecule it 56%, what are the percentages of the four bases in this molecule? - Answers - 28%G - 28%C - 22%A - 22%T Imagine you are a student in Alfred Hershey and Martha Chase's lab in the late 1940s. You are given five test tubes containing E. coli bacteria that were infected with T2 bacteriophage that have been labeled with either 32P or 35S. Unfortunately, you forgot to mark the tubes and are now uncertain which were labeled with 32P and which with 35S. You place the contents of the each tube in a blender and turn it on for a few seconds to shear off the protein coats. You then centrifuge the contents to separate the protein coats and the cells. You check for the presence of radioactivity and obtain the following results. Which tubes contained E. coli infected with 32P labeled phage? Tube 1- cells tube 2- protein coats tube 3- protein coats tube 4- cells' tube 5- cells - Answers Tubes 1, 4, and 5. The DNA of the bacteriophage contains phosphorous and the protein contains sulfur. When the bacteriophages infect the cell, they inject their DNA into the cell, but the protein coats stay on the surface of the cell. The protein coats are sheared off in the blender, while the cells with the DNA pellet at the bottom of the tube. Thus, cells infected with 35S -labeled bacteriophage will have radioactivity associated with the protein coats, whereas those cells infected with 32Plabeled bacteriophage will have radioactivity associated with the cells. What results would you expect if the bacteriophage that Hershey and Chase used in their experiment had contained RNA instead of DNA? - Answers The results would be the same. RNA, like DNA, contains phosphorus but not sulfur, so the 32P would have been incorporated into the RNA and entered the cell hosts during the course of the infection. Ultimately the 32P would be incorporated into the RNA of progeny phage during the course of phage reproduction. Protein contains sulfur but not phosphorus, so 35S would have been assimilated into the phage protein coat; it would have remained with the viral coats that do not enter the cells and not transferred to progeny phage. The 35S containing protein coats would still be recovered in the fluid recovered following centrifugation In a typical eukaryotic cell, would you expect to find more molecules of the H1 histone or more molecules of the H2A histone? - Answers Because each nucleosome contains two molecules of histone H2A and only one molecule of histone H1 is associated with each nucleosome, eukaryotic cells will have more H2A than H1. Would you expect to find more molecules of H2A or more molecules of H3? - Answers Because each nucleosome contains two molecules of H2A and two molecules of H3, eukaryotic cells should have equal amounts of these two histones. • Consider the following segment of DNA, which is part of a much longer molecule making a chromosome: • 5'...ATTCGTACGATCGACTGACTGACAGTC... 3' • 3'...TAAGCATGCTAGCTGACTGACTGTCAG... 5' • If the DNA polymerase starts replicating this segment from the right, - Which will be the template for the leading strand? - Draw the two complete daughter molecules - Answers - Which will be the template for the leading strand? • The top strand - Draw the two complete daughter molecules • 5' ATTCGTACGATCGACTGACTGACAGTC 3' - 3' TAAGCATGCTAGCTGACTGACTGTCAG 5' • 5' ATTCGTACGATCGACTGACTGACAGTC 3' - 3' TAAGCATGCTAGCTGACTGACTGTCAG 5' • The DNA polymerases are positioned over the DNA segment below (which is part of a much larger molecule) and the fork is moving from left to right. If we assume that an okazaki fragment is made from this segment, what will be its sequence? Label the 5' and 3' ends. 5' ...CCTTAAGACTAACTACTTACTGGGATC... 3' 3' ...GGAATTCTGATTGATGAATGACCCTAG... 5' - Answers - Bottom strand is leading strand template - Top strand is lagging strand template - Okazaki fragment: 3' ...GGAATTCTGATTGATGAATGACCCTAG... 5' • A portion of of DNA has the following sequence: 5' GCT TCC CAA 3' 3' CGA AGG GTT 5' • If the TOP strand is the template strand, a) Write the mRNA transcribed, labeling 5'and 3' b) Assume the first codon for the peptide begins with the first nucleotide of the portion of mRNA and there are no introns (May not start at AUG). Write the amino acid sequence. (Use the one letter abbreviations for the amino acids) - Answers a) Write the mRNA transcribed, labeling 5'and 3' 5' UUG GGA AGC 3' b) Assume the first codon for the peptide begins with the first nucleotide of the portion of mRNA and there are no introns (May not start at AUG). Write the amino acid sequence. (Use the one letter abbreviations for the amino acids) NH2 - L G S - COOH • A portion of of DNA has the following sequence: 5' GCT TCC CAA 3' 3' CGA AGG GTT 5' • If the BOTTOM strand is the template strand, a) Write the mRNA transcribed, labeling 5'and 3' b) Assume the first codon for the peptide begins with the first nucleotide of the portion of mRNA and there are no introns (May not start at AUG). Write the amino acid sequence. (Use the one letter abbreviations for the amino acids) - Answers If the BOTTOM strand is the template strand, a) Write the mRNA transcribed, labeling 5'and 3' 5' GCU UCC CAA 3' Assume the first codon for the peptide begins with the first nucleotide of the portion of mRNA and there are no introns (May not start at AUG). Write the amino acid sequence. (Use the one letter abbreviations for the amino acids) NH2 - A S Q- COOH Which anticodon would you predict for a tRNA species carrying isoleucine (ile)? Give all possible answers. - Answers - Isoleucine has three codons: AUU, AUC, AUA - Anticodons are: UAA, UAG(wobble), UAI(wobble) - Anticodon UAU is complementary, but would wobble pair with AUG, so is not acceptable Where does Replication, Transcription, and Translation occur in a eukaryotic cell? - Answers Replication and transcription in the- nucleus translation-cytoplasm What is the enzyme that carries out Replication, Transcription, and Translation? - Answers Replication: DNA Polymerase Transcription: RNA polymerase Translation: Ribosome What is the template that is read during Replication, Transcription, and Translation? - Answers Replication: DNA Transcription: DNA Translation: RNA In what direction is the template read during Replication, Transcription, and Translation? - Answers Replication: 3' to 5' Transcription: 3' to 5' Translation: 5' to 3' What is the start signal/sequence for this process? - Answers Replication: Origin Transcription: promoter Translation: start codon What is the polymer that is formed during Replication Transcription Translation? - Answers Replication: DNA What monomer is used to form this polymer? - deoxyribonucleoti des Transcription: RNA What monomer is used to form this polymer? - ribonucleotides Translation: polypeptide What monomer is used to form this polymer? - Amino acids What type of bond is formed between monomers? - Answers Replication: Phosphodiester Transcription: phosphodiester Translation: peptide In what direction is the new polymer formed? - Answers Replication: 5' to 3' Transcription: 5' to 3' Translation: Amino to carboxyl What is the stop signal/sequence for Replication Transcription Translation? - Answers Replication: none Transcription: Transcription Terminator Translation: Stop codon Bacterial Transcription - Answers 1) promoter- DNA sequence that transcription apparatus recognizes and binds - specific binding sites include consensus sequences at -35 (closest to 5' end) and -10 (10 and 35 bp upstream of the transcription start site) start site (+1) some promoters have an Upstream element (UP) that takes part in initiation (will be before the -35) 2) RNA coding region- DNA sequence that gets copied into an RNA molecule. 3) Terminator: DNA sequence that signals where transcription ends - may be a rho independent terminator (1 of 2 major types of terminators in bacterial cells.); contains inverted repeats and series of adenine nucleotides that are important for termination (Poly A will be last segment of terminator region, inverted repeat will be near first piece of terminator segment) Initiation of RNA synthesis doesn't require a primer - transcription is initiated when RNA polymerase holoenzyme (composed of core enzyme and sigma subunit) binds to promoter; as holoenzyme unwinds DNA transcription bubble forms creating a single stranded DNA where transcription can occur ; only the template strand gets transcribed.; RNA synthesis is complementary to template strand and is in the 5' to 3' direction (mRNA) RNA polymerase moves downstream and continues adding nucleotides to the RNA molecule according to sequences on template rho independent terminators contain inverted repeats followed by Poly A region. when inverted repeat is transcribed it creates a RNA hair pin which caused RNA polymerase to pause. stretch of A-U base pairs is unstable; causes RNA strand to separate from template during the pause. when transcription has ended; DNA duplex reforms and RNA polymerase dissociates from the DNA. Bacterial translation - Answers amino acids are assembled into proteins initiated in 3 steps - small subunit of ribosome binds to mRNA and P-site is positioned over start codon 2) Initiator tRNA with fMet attached binds to the mRNA through base pairing between codon and anticodon 3) large subunit ribosome joins the complex to complete initiation during elongation: amino acids are joined to growing polypeptide chain - first step is delivery of tRNA and its amino acid to the A-site 2) peptide bond forms between the amino acids 3) translocation: movement of ribosome down the mRNA - afterwards the tRNA formerly in P-site is now in E-site through which it moves to the cytoplasm; now elongation cycle repeats itself- tRNA and amino acid occupy A-site, peptide bond forms, ribosome translocates to next codon - addition of amino acids to the carboxyl end causes polypeptide chain to continue to grow as elongation cycles through Termination - where ribosome reaches termination codon - release factors bind to termination codon causing the release of the polypeptide chain - ribosome translocates one more time - tRNA and release factors exit the ribosome , ribosome dissociates. several unmentioned accessory proteins also assist in translation Lentz and protein coding region - Answers designed mutation on the protein coding regions of a gene What if a mutation occurs that changes C to G in mRNA sequence- when ribosome reaches mutated codon, inserts different amino acid into growing polypeptide chain resulting mutant polypeptide has valine instead of lucine and may or may not be functional. what if we delete a single nucleotide from the sequence? - reading frame gets altered; translation results in non functional shortened polypeptide delete 3 nucleotides (ex: ACU) - may result in original reading frame but results in deletion of one amino acid. may or may not be functional Insertion of 2 nucleotides (ex: GG) - results in a new codon which codes for different amino acid; reading frame disrupted, mutant polypeptide very different from wild type and likely non functional Remember: insertion or deletion of one or two nucleotides will result in a frameshift of all downstream codons insertion or deletion of 3 nucleotides will not cause a frame shift Bidirectional replication of DNA - Answers DNA replication begins at replication origin (specific chromosomal sites)