GCD Exam 2 Practice Test Comp 86 Questions (chp 9-13) with Verified Answers,100% CORRECT

GCD Exam 2 Practice Test Comp 86 Questions (chp 9-13) with Verified Answers Which of the following could be the components of a single nucleotide found in DNA? a. Ribose, phosphate, and cytosine b. Deoxyribose, phosphate, and uracil c. Deoxyribose, phosphate, and adenine d. Ribose, phosphate, and uracil - CORRECT ANSWER c. Deoxyribose, phosphate, and adenine 2. Which of the following is NOT a feature of the DNA double helix? a. It obeys the AG/TC rule. b. There are 10 nucleotides in each strand per complete 360 degree turn of the helix. c. The DNA strands are anti-parallel. d. The structure is stabilized by base stacking. - CORRECT ANSWER a. It obeys the AG/TC rule 3. The structure of an RNA strand is much like that of a DNA strand. However, RNA molecules are usually ( )-stranded and ( ). a. double, much longer than chromosomal DNA. b. single, much longer than chromosomal DNA. c. double, much shorter than chromosomal DNA. d. single, much shorter than chromosomal DNA. - CORRECT ANSWER d. single, much shorter than chromosomal DNA. An RNA molecule has the following sequence: 5'-CAUCCAUCCAUUCCCCAUCGAUAAGGGGAAUGGAAUCCGAAUGGAUAAC-3' Region 1 Region 2 Region 3 Regions 1-3 can form two different stem-loops. Which stem-loop would you predict more stable? Why? a. Stem-loop involving Regions 1/2, because base paring is longer and contains 6 G:C pairs b. Stem-loop involving Regions 1/3, because base paring is shorter. c. Stem-loop involving with Regions 1/2, because these regions are right next to each other. d. Stem-loop involving Regions 2/3, because these regions are right next to each other. - CORRECT ANSWER a. Stem-loop involving Regions 1/2, because base paring is longer and contains 6 G:C pairs An RNA virus is a virus that has RNA (ribonucleic acid) as its genetic material. This nucleic acid is usually single-stranded RNA but may be double-stranded RNA in viruses . The base compositions of two kinds of RNA viruses are the following: RNA virus (#1) 20.2% A, 20.0% U, 29.8% G, and 30.0% C. RNA virus (#2) 12.2% A, 18.8% U, 40.5% G, and 28.5% C. Which of the following correctly describes the genetic materials of these two RNA viruses? a. #1 virus has double-stranded RNA, #2 virus has single-stranded RNA. b. Both viruses have double-stranded RNA c. Both viruses have single-stranded RNA. d. #1 virus has single-stranded RNA, #2 virus has double-stranded RNA. - CORRECT ANSWER a. #1 virus has double-stranded RNA, #2 virus has single-stranded RNA. 6. ( ) supercoiling can enhance transcription and DNA replication because it ( ). a. Positive, promotes DNA strand separation b. Negative, promotes DNA strand separation c. Positive, makes the DNA more compact d. Negative, makes the DNA more compact - CORRECT ANSWER b. Negative, promotes DNA strand separation 7. Highly repetitive sequences are found in tens of thousands or million times throughout a genome. Which of the following is an example of highly repetitive sequences? a. Genes that encode rRNA b. The Alu sequences c. Introns d. Most protein-encoding genes. - CORRECT ANSWER b. The Alu sequences 8. A segment of DNA had one negative super-coiling. If it was given three complete (360 degree) twists in the right-handed direction (overwinding), it would have a. two negative supercoils. b. two positive supercoils. c. one positive supercoil. d. three positive supercoils. - CORRECT ANSWER b. two positive supercoils. 9. Which of the following is a feature of heterochromatin? a. Less compacted chromosomal regions b. Found in telomeres c. Transcriptionally active d. 30 nm fiber forms loop domains but no further folding or compaction - CORRECT ANSWER b. Found in telomeres 10. Which of the following is a feature of euchromatin? a. Found in telomeres b. Tightly compacted chromosomal regions c. Found in centromeres d. Transcriptionally active - CORRECT ANSWER d. Transcriptionally active 11. DNA replication is a process that produces ( ) and requires the base sequence known as the ( ) to start. a. an RNA copy of a gene, promoter b. a copy of chromosomal DNA, origin of replication c. a polypeptide encoded in a gene, start codon - CORRECT ANSWER b. a copy of chromosomal DNA, origin of replication 12. Which of the following is NOT a property of DNA polymerase III? a. It synthesizes DNA in the 5'-3' direction. b. It can only elongate a strand starting with an RNA primer or an existing DNA strand. c. It can proof-read. d. It removes RNA primers. - CORRECT ANSWER d. It removes RNA primers. 13. Which of the following statements is true for DNA polymerase alpha? a. It synthesizes a short RNA-DNA primer with primase. b. It is responsible leading strand synthesis. c. It is responsible for lagging strand synthesis. d. It is involved in the removal of RNA primers. - CORRECT ANSWER a. It synthesizes a short RNA-DNA primer with primase. 14. Which of the following enzymes is NOT involved in bacterial DNA replication? a. Helicase b. Ligase c. Primase d. Flap endonuclease - CORRECT ANSWER d. Flap endonuclease 15. Transcription is a process that produces ( ) and requires the base sequence known as ( ) to start. a. an RNA copy of a gene, promoter b. a copy of chromosomal DNA, origin of replication c. a polypeptide encoded in a gene, start codon - CORRECT ANSWER a. an RNA copy of a gene, promoter A DNA strand includes the following sequence: 5'-GATCCCCGATCCGAAATTTGACCTTTAA-3' Transcription start site If this strand is used as a template by RNA polymerase, what would be the sequence of the synthesized RNA transcript? In which direction would RNA polymerase slide along this strand (from left to right or from right to left)? a. 3'-CUAGGGGCUAGGCUUUAAACUGGAAA-5' From left to right b. 3'-CUAGGGGCUAGGCUUUAAACUGGAAA-5' From right to left c. 3'-CTAGGGGCTAGGCTTTAAACTGGAAA-5' From left to right d. 3'-CTAGGGGCTAGGCTTTAAACTGGAAA-5' From right to left - CORRECT ANSWER b. 3'-CUAGGGGCUAGGCUUUAAACUGGAAA-5' From right to left 18. If TFIID is missing, how would this affect eukaryotic transcriptional initiation? (See textbook p296-297 and Fig. 12.14) a. No effect on transcription. b. Transcription would be inhibited because the core promoter of a gene would not be recognized. c. Transcription would be inhibited because RNA polymerase II would not be recruited. d. Transcription would be inhibited because the formation of an open complex would not happen. - CORRECT ANSWER b. Transcription would be inhibited because the core promoter of a gene would not be recognized. 19. Which of the following commonly occurs in bacteria and eukaryotes? a. Processing of rRNAs b. Alternative splicing of pre-mRNA. c. 5' Capping of mRNA. d. Splicing of pre-mRNA. - CORRECT ANSWER a. Processing of rRNAs Translation is a process that produces ( ) and requires the base sequence known as ( ) to start. a. a copy of chromosomal DNA, origin of replication b. an RNA copy of a gene, promoter c. a polypeptide encoded in a gene, start codon - CORRECT ANSWER c. a polypeptide encoded in a gene, start codon A codon is 5'-AAU-3'. An anticodon that would recognize this codon would be a. 3'-UUA-5' b. 3'-AAU-5'. c. 5'-UUA-3' d. 3'-TTA-5' - CORRECT ANSWER a. 3'-UUA-5' The genetic code and wobble rules for tRNA-mRNA pairing are shown in Fig. 13.12. If we assume that the tRNAs do not contain modified bases, what is the minimum number of tRNAs that are needed to recognize all the leucine codons? 5'-UUA-3' 5'-UUG-3' 5'-CUU-3' 5'-CUC-3' 5'-CUA-3' 5'-CUG-3' A: U U: A G U I G: C A U C: G A I a. 2 b. 3 c. 4 d. 6 - CORRECT ANSWER (Comprehension Q) The total amount of G plus C in an organism's DNA is 46% of the total base content of that DNA. What are the percentage of A, T, G, and C in the DNA? A (27%), T (27%), G (23%) , C (23%). A (18%), T (18%), G (32%) , C (32%). A (27%), T (23%), G (27%) , C (23%). A (23%), T (23%), G (27%) , C (27%). - CORRECT ANSWER A (27%), T (27%), G (23%) , C (23%). (Comprehension Q) Which of the following DNA double helices would be more difficult to separate into single-stranded molecules by treatment with heat, which breaks hydrogen bonds? 5'-CCCCTACCAGCCCA-3' 3'-GGGGATGGTCGGGT-5' 5'-CCAATACCAGCCCA-3' 3'-GGTTATGGTCGGGT-5' 5'-AAAATTTTAAGGAT-3' 3'-TTTTAAAATTCCTA-5' 5'-AAAATTTTCCGGAT-3' 3'-TTTTAAAAGGCCTA-5' - CORRECT ANSWER 5'-CCCCTACCAGCCCA-3' 3'-GGGGATGGTCGGGT-5' (Comprehension Q) A double-stranded DNA molecule contains 720 nucleotides. How many complete turns occur in this double helix? 36 72 30 60 - CORRECT ANSWER 36 This double-stranded DNA molecule contains 360 base pairs (720/2). There are 10 base pairs per complete 360 degree turn of the helix. 360/10=36 turns (Comprehension Q) The base composition of a DNA virus was analyzed and found to be 15% A, 35% T, 15% G, and 35% C. Which of the following statements best describes the genetic material of this virus? It is mostly likely a single-stranded DNA, because it does not follow Chargaff's rule. e.g., the amount of A does NOT equal that of T. It is mostly likely a double-stranded DNA, because the amount of A equals that of G. It is mostly likely a double-stranded DNA, because the amount of G equals that of C. It is mostly likely a single-stranded DNA, because the amount of A equals that of G. - CORRECT ANSWER It is mostly likely a single-stranded DNA, because it does not follow Chargaff's rule. e.g., the amount of A does NOT equal that of T. (Comprehension Q) A segment of DNA had one positive supercoiling. If it was given two complete (360 degree) twists in the left-handed direction (underwinding), it would have (See textbook p233 and Fig. 10.4) • three positive supercoils. • two negative supercoils. • one negative supercoil. • three negative supercoils. - CORRECT ANSWER • one negative supercoil. (Comprehension Q) If you assume that the average length of a DNA linker region is 54 bp, approximately how many nucleosomes can be found in the haploid genome, which contains 3 billion (3,000,000,000) bp? Note: Nucleosome:~146 bp of DNA wrapped around a histone octamer. • 10 million • 15 million • 30 million • 20 million - CORRECT ANSWER • 15 million Linker (54bp) +Nucleosome (146bp) = 200bp 3,000,000,000/200bp = 15,000,000 (Comprehension Q) Let's assume that the linker DNA averages 154 bp in length (Nucleosome:~146 bp of DNA wrapped around a histone octamer). How many molecules of H2A would you expect to find a DNA molecule that is 75,000 bp in length? • 500 • 750 • 125 • 250 - CORRECT ANSWER • 500 Linker (154bp) +Nucleosome (146bp) = 300bp 75,000/300bp = 250 nucleosomes 250 x 2 (2 H2A molecules) = 500 (Comprehension Q) What would be the expected results for the Meselson and Stahl experiment after 3 generations if the correct mechanism was conservative? Note: Generation zero has all heavy DNA, and all subsequent generations use light nitrogen to make new DNA strands. - 1/8 heavy, 7/8 light • 1/16 heavy, 15/16 light • 1/8 half-heavy, 7/8 light • 2/8 half-heavy, 6/8 light - CORRECT ANSWER 1/8 heavy, 7/8 light (Comprehension Q) In order to replicate the entire circular chromosome of E. coli, how many replication forks are used? • One • Two • Three • Four - CORRECT ANSWER • Two (Comprehension Q) A bacterial chromosome contains 4.6 million base pairs. How long does it take to replicate this chromosome? DNA replication proceeds bi-directionally, and DNA synthesis occurs at a rate of 750 nucleotides per second at each strand. • 10 min • 25 min • 50 min • 100 min - CORRECT ANSWER • 50 min 750 nucleotides x 2 (two replication forks) = 1500 nucleotides per second. 4.600,000/1500 = 3,067 seconds. 3,067/60 = 51 min. (Comprehension Q) DNA replication in early Drosophila embryos occurs every 5 mins. The Drosophila genome contains ~1.8 x 108 base pairs. Eukaryotic DNA polymerases synthesize DNA at a rate of approximately 40 nucleotides per second in each strand. Approximately, how many origins of replication are found for this rate of replication? • 3000 • 6000 • 7500 • 15000 - CORRECT ANSWER • 7500 From each origin of replication, two replication forks would be generated. 5 min = 300 seconds 40 nucleotides x 300 seconds x 2 replication forks = 24,000 nucleotides. • 180,000,000/24,000 = 7500 (Comprehension Q) A mutation inhibits the kinase function of TFIIH. This mutant TFIIH no longer phosphorylates CTD of RNA polymerase II. How would this mutation affect transcription? • It would inhibit the binding of RNA polymerase to the core promoter • It would inhibit the formation of preinitiation complex. • It would inhibit the formation of the open complex • It would inhibit the transition between initiation and elongation. - CORRECT ANSWER • It would inhibit the transition between initiation and elongation. (Comprehension Q) A gene encodes a pre-mRNA with 5 exons and 4 introns. A mutation eliminates the 3' splice site from intron 3. What exons would be found in the mRNA after splicing was completed? Exon 1, 2 Exon 1, 2, 3 Exon 1, 2, 3, 5 Exon 1, 2, 4, 5 - CORRECT ANSWER Exon 1, 2, 3, 5 (Comprehension Q) Let's suppose a mutation changed the 3rd codon in the coding sequence of a gene into a stop codon. This would cause • transcription to end too early. • transcription to end too late. • translation to end too early. • transcription AND translation to end too early. - CORRECT ANSWER • translation to end too early. (Comprehension Q) A codon is 5'-GGU-3'. An anticodon that would recognize this codon would be • 5'-GGU-3' • 3'-GGU-5' • 5'-CCA-3' • 3'-CCA-5' - CORRECT ANSWER • 3'-CCA-5' Codon: anti-codon interaction • Antiparallel • The AU/GC rule An RNA virus is a virus that has RNA as its genetic material. The base compositions of two kinds of RNA viruses are the following:RNA virus (#1) 20.2% A, 20.0% U, 29.8% G, and 30.0% C.RNA virus (#2) 12.2% A, 37.8% U, 12.2.% G, and 37.8% C. Which of the following correctly describes the genetic materials of these two DNA viruses? #1 virus has single-stranded RNA, #2 virus has double-stranded RNA. Both viruses have single-stranded RNA. #1 virus has double-stranded RNA, #2 virus has single-stranded RNA. Both viruses have double-stranded RNA. - CORRECT ANSWER #1 virus has double-stranded RNA, #2 virus has single-stranded RNA. A segment of DNA had one negative super-coiling. If it was given two complete (360 degree) twists in the left-handed direction (under-winding), it would have - two negative supercoils. - three positive supercoils. - three negative supercoils. - one positive supercoil. - CORRECT ANSWER - three negative supercoils. Let's assume that the linker DNA averages 254 bp in length (Nucleosome:~146 bp of DNA wrapped around a histone octamer). How many molecules of H2B would you expect to find a DNA molecule that is 120,000 bp in length? 600 200 400 300 - CORRECT ANSWER 600 What would be the expected results for the Meselson and Stahl experiment after 4 generations? (see Figure 11.2) 2/16 heavy, 14/16 light 1/16 heavy, 15/16 light 2/16 half-heavy, 14/16 light 1/16 half-heavy, 15/16 light - CORRECT ANSWER 2/16 half-heavy, 14/16 light As shown in Figure 11.24, telomerase adds additional nucleotides, six at a time, to the ends of eukaryotic chromosomes. However, it makes only one DNA strand. How is the opposite (complementary) strand replicated? a. DNA polymerase delta or epsilon synthesizes it using an RNA primer made by primase, and ligase connects the newly made strand with the preexisting strand. b. DNA polymerase III synthesizes it using an RNA primer made by primase, and ligase connects the newly made strand with the preexisting strand. c. DNA polymerase alpha synthesizes it all. d. DNA polymerase I synthesizes it all. - CORRECT ANSWER a. DNA polymerase delta or epsilon synthesizes it using an RNA primer made by primase, and ligase connects the newly made strand with the preexisting strand. One strand of a double-stranded DNA molecule includes the following sequence: +1 (transcriptional start site) 5'-......GATCCCGATCCCGGCCGCATTTTTACCAGTTAAATACCACC.......-3' A promoter is located 5' upstream from the transcriptional start site (as indicated in the BOLD A). Which of the following sequences is the correct transcript from this DNA molecule? c. 3'-AUCCCGGCCGCAUUUUUACCAGUUAAAUACCACC......-5' d. 3'-UAGGGCCGGCUAAAAAUGGUCAAUUUAUGGUGG......-5' a. 3'-....CTAGGGCTAGGGCCGGCGTAAAAATGGTCAATTTATGGTCC.....-5' b. 5'-AUCCCGGCCGCAUUUUUACCAGUUAAAUACCACC......-3' - CORRECT ANSWER b. 5'-AUCCCGGCCGCAUUUUUACCAGUUAAAUACCACC......-3' A tRNA molecule carries a glutamic acid, what are the two possible anticodon sequences that it could contain? (See Table 13.1. the Genetic code) c. 5'-CUU-3' and 5'-CUC-3' b. 3'-GAA-5' and 3'-GAG-5' a. 5'-GAA-3' and 5'-GAG-3' d. 3'-CUU-5' and 3'-CUC-5' - CORRECT ANSWER d. 3'-CUU-5' and 3'-CUC-5' The genetic code and wobble rules for tRNA-mRNA pairing are shown in Fig. 13.12. If we assume that the tRNAs do not contain modified bases, what is the minimum number of tRNAs that are needed to recognize all of the isoleucine codons? (Also see Table 13.1. the Genetic code) 4 2 3 1 - CORRECT ANSWER 2 In the experiment of Avery, McLeod, and McCarthy, the addition of protease to a DNA extract - prevented the conversion of type R bacteria into type S bacteria. - prevented the conversion of type S bacteria into type R bacteria. - allowed the conversion of type S bacteria into type R bacteria. - allowed the conversion of type R bacteria into type S bacteria - CORRECT ANSWER - allowed the conversion of type R bacteria into type S bacteria What occurred in the transformation observed by Griffith? A smooth strain underwent mutation to convert it to a rough strain. A rough strain passed genetic information to a smooth strain. A rough strain underwent mutation to convert it to a smooth strain. A smooth strain passed genetic information to a rough strain. - CORRECT ANSWER A smooth strain passed genetic information to a rough strain. A key difference between the nucleotides found in DNA and those in RNA is that Those in DNA have phosphates, and those in RNA do not Those in DNA have deoxyribose, and those in RNA have ribose Those in RNA have phosphates, and those in DNA do not Those in DNA have ribose, and those in RNA have deoxyribose - CORRECT ANSWER Those in DNA have deoxyribose, and those in RNA have ribose Which of the following is NOT a feature of the DNA double helix? There are 10 nucleotides in each strand per complete 360 degree turn of the helix The structure is stabilized by base stacking. It obeys the AT/GC rule. The DNA strands are parallel. - CORRECT ANSWER The DNA strands are parallel. Which of the following DNA double helices would be easier to separate into single-stranded molecules by treatment with heat, which breaks hydrogen bonds? 5'-AAAATTTTAAGGAT-3' 3'-TTTTAAAATTCCTA-5' 5'-CCAATACCAGCCCA-3' 3'-GGTTATGGTCGGGT-5' 5'-AAAATTTTCCGGAT-3' 3'-TTTTAAAAGGCCTA-5' 5'-CCCCTACCAGCCCA-3' 3'-GGGGATGGTCGGGT-5' - CORRECT ANSWER 5'-AAAATTTTAAGGAT-3' 3'-TTTTAAAATTCCTA-5' A bacterial chromosome typically contains telomeres. a few million genes. one origin of replication. a centromere. - CORRECT ANSWER one origin of replication. A human chromosome contains a million genes. telomeres at both ends. one origin of replication. very few repetitive sequences. - CORRECT ANSWER telomeres at both ends. Highly repetitive sequences are found in tens of thousands or million times throughout a genome. Which of the following is highly repetitive sequences? genes that encode rRNA the Alu sequences most protein-encoding genes introns - CORRECT ANSWER the Alu sequences What are component of a single nucelosome? About 146 bp of DNA and eight core histone proteins. About 146 bp of DNA and four core histone proteins. About 200 bp of DNA and eight core histone proteins. About 200 bp of DNA and eight core histone proteins. - CORRECT ANSWER About 146 bp of DNA and eight core histone proteins. Which of the following is NOT a feature of heterochromatin? tightly compacted chromosomal regions. transcriptionally active found in cetromeres found in telomeres - CORRECT ANSWER transcriptionally active What would be the expected results for the Meselson and Stahl experiment after 4 generations if the correct mechanism was conservative? Note: Generation zero has all heavy DNA, and all subsequent generations use light nitrogen to make new DNA strands. 2/16 heavy, 14/16 light 1/16 half-heavy, 15/16 light See Fig. 11.2. 2/16 half-heavy, 14/16 light 1/16 heavy, 15/16 light - CORRECT ANSWER 1/16 heavy, 15/16 light In the leading strand, DNA is made in the direction ( ) the replication fork and is made as ( ). toward, Okazaki fragments. away from, Okazaki fragments. away from, continueous strand. toward, one continuous strand. - CORRECT ANSWER toward, one continuous strand. The enzyme known as ( ) covalently attaches adjacent Okazaki fragments. DnaA protein DNA ligase DNA polymerase III helicase - CORRECT ANSWER DNA ligase In bacterial, the enzyme that removes short RNA primers and fills in gap with DNA is DNA polymerase I helicase DNA polymerase III DNA ligase - CORRECT ANSWER DNA polymerase I In eukaryotes, RNA primers are primarily removed by DNA polymerase α DNA polymerase III DNA polymerase I flap endonuclease - CORRECT ANSWER flap endonuclease Which of the following base sequences is used during transcription? Origins of replication Start and stop codons Promoter and terminator Ribosome-binding site - CORRECT ANSWER Promoter and terminator During bacterial transcription, the function of sigma (σ) factor is to synthesize mRNA. recognize the transcription start site. terminate transcription. recognize a promoter. - CORRECT ANSWER recognize a promoter. What is the consensus sequence of the following six DNA molecules? 5'-GGCGCATT-3', 5'-GCCGGATT-3'‚ 5'-CGCCGATA-3' 5'-GGGGGATT-3' 5'-GGCGGTTT-3' 5'-GGCGCATA-3' 3'-GGCGGATT-5' 5'-GCCGGATA-3' 5'-GGCGGATT-3' 5'-GGCGGATA-3' - CORRECT ANSWER 5'-GGCGGATT-3' What do both the rho-dependent and rho-independent mechanisms of termination have in common? A uracil-rich sequence in mRNA Formation of a stem-loop structure Require a helicase to separate the DNA-RNA complex Terminate transcription immediately after the stop codon - CORRECT ANSWER Formation of a stem-loop structure Which of the following binds to the TATA box of a core promoter in eukaryotes? TFIIF TFIIE TFIID TFIIB - CORRECT ANSWER TFIID (Comprehension Q) A mutation inhibits the helicase function of TFIIH. How would this mutation affect transcription? It would inhibit the formation of the open complex It would inhibit the binding of RNA polymerase to the core promoter It would inhibit the formation of preinitiation complex. It would inhibit the transition between initiation and elongation. - CORRECT ANSWER It would inhibit the formation of the open complex Which of the following statements is incorrect about the function of polyA tail of eukaryotic mRNA? A polyA tail facilitates translocation of mRNA from the nucleus. PolyA-polymerase adds adenine nucleotides at the 3' end of mRNA. A polyA tail promotes mRNA stability. A PolyA tail promotes degradation of mRNA. - CORRECT ANSWER A PolyA tail promotes degradation of mRNA. ( ) are parts of ( ) that do not code for amino acids. Exons, pre-mRNA Introns, rRNA Introns, pre-mRNA Exons, rRNA - CORRECT ANSWER Introns, pre-mRNA Which of the following does NOT occur in bacteria? Poly-A tailing at the 3' end of mRNA to promote mRNA degradation Processing of rRNAs Processing of tRNAs 5' capping of mRNA - CORRECT ANSWER 5' capping of mRNA An mRNA encodes a polypeptide that is 312 amino acids in length. The fifty-third codon in this polypeptide is a cysteine codon. A mutation changes this cysteine codon into a stop codon. How many amino acids would be in the resulting polypeptide? 52 311 53 17 - CORRECT ANSWER 52 The function of aminoacyl-tRNA-synthetase is to cause the initiator tRNA to bind to the small ribosomal subunit. attach an amino acid to a tRNA. catalyze peptide bond formation between an incoming amino acid and a polypeptide. promote the binding of a tRNA to an mRNA. - CORRECT ANSWER attach an amino acid to a tRNA. Codons that are closer to the 5' end of a mature mRNA specify amino acids that are duplicated in the polypeptide. closer to the carboxyl-terminal end of the polypeptide. spliced out and not found in the polypeptide. closer to the amino-terminal end of the polypeptide. - CORRECT ANSWER closer to the amino-terminal end of the polypeptide. A codon is 5'-UUC-3'. An anticodon that would recognize this codon would be 5'-AAG-3' 3'-UUC-5' 3'-AAG-5' 5'-UUC-3' - CORRECT ANSWER 3'-AAG-5' During the elongation stage of translation, a tRNA without an amino acid attached could be found at the P or A site of a ribosome. the E or P site of a ribosome. only the E site of a ribosome. See Fig. 13.17. only the A site of a ribosome. - CORRECT ANSWER the E or P site of a ribosome. heterochromatin vs. euchromatin - CORRECT ANSWER HeteroChromatin = Highly Condensed (transcriptionally inactive), loop domains condense even further euchromatin = less condensed, transcriptionally active ("truly transcribed"), 30nm fiber forms loops Types of Heterochromatin - CORRECT ANSWER constitutive and facultative Conservation DNA replication - CORRECT ANSWER Both parental strands stay together after DNA replication Semiconservative model - CORRECT ANSWER The double-stranded DNA contains one parental and one daughter strand following replication • This is the correct mode Dispersive model - CORRECT ANSWER Parental and daughter DNA segments are interspersed in both strands following replication Eukaryotic DNA polymerases - CORRECT ANSWER alpha = Function together w/ primase to make primers delta = Synthesizes lagging strand Epsiolon = Synthesizes leading strand Central dogma of genetics - CORRECT ANSWER DNA transcribed to RNA RNA translated to form polypeptides TFIID - CORRECT ANSWER the first general transcription factor to bind the promoter, binds to the TATA box through the TATA binding protein (TBP) TFIIH - CORRECT ANSWER unwinds DNA at the transcription start point, phosphorylates Ser5 of the RNA polymerase CTD; releases RNA polymerase from the promoter stages of translation - CORRECT ANSWER 1. Initiation 2. Elongation 3. Termination Prokaryotes vs. Eukaryotes - CORRECT ANSWER Prokaryotes: No nucleus - Transcription - Cytoplasm - Translation - Cytoplasm Eukaryotes: Have nucleus -Transcription - Nucleus -Translation - Cytoplasm DNA - CORRECT ANSWER ○ Hydrogen bond with AT/GC rule ○ 2 strands Antiparallel with 5' to 3' directionality ○ 10 nucleotides per 360 turn RNA - CORRECT ANSWER ○ U as base instead of T; H-bond with same rule (A to U though) ○ Antiparallel with 5' to 3' directionality ○ RNA shorter in length than DNA