CHapter 16-17 Essay Test

  1. What is the initiation stage of transcription?
    • It begins with a promoter, which determines where transcription starts and acts as a binding site for RNA, which is facilitated by transcription factors.
    • RNA pol attaches to the promoter. It then pries the two strands of DNA apart and joins the RNA nucleotides as they base-pair along the DNA template. (In eukaryotes, RNA pol II is the pol that synthesizes an mRNA transcript. The transcription factors and RNA pol II bound to the promoter create the transcription initiation complex.
    • A promoter DNA sequence called a TATA box is crucial to the formation of the complex.
    • RNA pol is bound to the start point on the template strand, where the codon AUG that codes for the amino acid Met can begin the elongation of the strand.
  2. What is the elongation process of transcription?
    Occurs when RNA pol moves along the DNA, continually untwisting the helix, and adding nucleotides to the 3' end of the growing RNA molecule. In this elongation process, nucleotides are added, which are then grouped into groups of three, called triplet codons. These codons represent a single amino acid molecule. In the wake of RNA synthesis, the new RNA molecule peels away from its DNA template and the DNA double helix reforms. Sometimes many genes can be transcribed simultaneously by several RNA polymerases.
  3. What is the termination stage of transcription?
    • the RNA trancscript is released, and the polymerase detaches from teh DNA. In bacteria, transcription proceeds through a terminator sequence in the DNA. The transcribed terminator functions as the termination signal, causing the polymerase to detach from the DNA and release the transcript, which is available for mRNA.
    • In eukaryotes, RNA pol II transcribes an AAUAAA sequence called the polyadenylation signal sequence, which codes frot this signal in the pre-mRNA. About 10-35 nucleotides downstream from the AAUAAA signal, proteins associated with the growing RNA transcript cut it free from polymerase, releasing the pre-mRNA. Even after the pre-mRNA is released, RNA polymerase continues adding more nucleotides, which are destroyed.
  4. At the replication fork, strand have to unwound, what enzymes assist?
    At the rep fork, the DNA double helix is unwound by ther enzyme helicase, allowing the two DNA strands to be separate, which now become template strands for the synthesis of new DNA. The two strands of DNA are unstable, but single strand binding proteins line up along the unpaired DNA strands, allowing them to remain apart for the synthesis of the daughter strands. Another enzyme called topoisomerase relieves strain caused by the untwisting of the double helix, which causes tighter strain caused by the untwisting of the double helix, which causes tighter twisting and strain ahead of the rep fork. Topoisomerase accomplishes this by breaking, swiveling, and rejoining the DNA strands.
  5. The template strand requires an RNA primer. How is this accomplished?
    The template strand does not just produce a polynucleotide, and do DNA polymerase. BEfore any synthesis of a new DNA strand can occur, an RNA primer, a short chain of about 10-20 RNA nucleotides must be added to the 3' end. An enzyme called primase synthesizes the creation of primer. Once the primer is completed, it can be used as a starting point for making new DNA. RNA primer is an initial nucleotide chain that is base-paired with the template strand. Once the primer is created, DNA pol can do its job.
  6. What's needed? What is the source of energy for these additions?
    Energy is needed for this synthesis. dATP, which is similar to ATP, is the source of the energy. DNA pol catalyzes the addition of a nucleoside triphosphate to the 3' end of a growing DNA strand, with the release of two phosphates into pyrophosphate. Each nucleotide added to the growing DNA strand comes from a nucleoside triphosphate, which contains deoxyribose, is chemically reactive, due to the phosphate tails. Subsequent hydrolysis of pyrophosphate to two molecules of inorganic phosphate helps drive the polymerization reaction- the addition of nucleotides to a daughter strand forming new complimentary DNA.
  7. How does antiparallel elongation occur?
    • Antiparallel elongation occurs when an RNA primer is added to the 3' carbon end of a DNA strand. DNA pol III then adds nucleotides in the 5->3' direction, synthesizing a leading strand in the direction of the replication fork. DNA pol II is located in the rep fork on the template strand and adds nucleotides to the daughter strand as the fork, along with helicases, continues providing more of the DNA strand.
    • A lagging strand, produced away from teh replication fork is synthesized in Okazaki fragments, using more than one RNA primer. THis also is syntehsized in the 5->3 direction, with the RNA primer being syntesized from teh 3' end.
    • DNA ligase, after DNA pol I replaces RNA with DNA nucleotides, connects the Okazaki fragments creating one new DNA daughter strand.
    • While the parental DNA strands are antiparallel, so must the daughter strands be. While one is at the 5'c arbon end, tat same side of the complimentary strand contains 3'. THey run in opposite directions.
  8. How does the issue of adding the 3 carbon end of the deoxyribose sugar get resolved?
    • Adding to the 3' carbon end gets resolved because the DNA pol only adds to the 3' end of a primer, which is lcoated at 3' end. Along one template strand, DNA pol III sythesizes the leading strand in the mandatory 5->3' direciton.
    • The lagging strand, although fragmented is still synthesized in the mandatory 5->3' direction. However, it moves away from teh rep fork. THe DNA replication complex reels DNA in, synthesizing the strands. the 5' carbon end contains a phosphate, while the 3' end contains an -OH group.
  9. How is the DNA strand proofed and repaired?
    The DNA polymerase proofread each nucleotide against its template once added. If a nucleotide is incorrect, the DNA pol will remove the incorrect nucleotide and enters the correct base. If proofreading is dodged, mismatched repair occurs. However, even after replicaiton, errors can occur. Such errors are fixed by nucleotide excision repair, in which a nuclease cuts out (excise) the incorrectly paired base, and DNA polymerase and ligase add and combine it to the strand. This and other DNA repair enzymes are very important for the proper functioning of DNA and correct replication.
  10. What is translation initiation process?
    • Includes ribosomes, mRNA, tRNA, and amino acids. All of these come together to form a translation initiation complex.
    • First, the small ribosomal unit binds to the mRNA at the 5' end site and an initiator tRNA carrying the amino acid, met, the start amino acid.
    • The small subunit, once bound, scans downstream along the mRNA until it reaches the start codon, signaling teh start of translation.
    • Once the large ribosomal subunit binds to the mRNA, initiator tRNA and the small ribosomal subunit, called the translation initiation complex. GTP is the energy source.
    • At the completion, the initiator tRNA sits in the P site of the ribosome and the vacant A site is ready for next aminoacyl tRNA. Synthesis occurs from the N-terminus to the C-terminus end.
  11. What is the elongation process of translation?
    • Codon recognition: anticodon of the incoming aminoacyl-tRNA base pairs with the complementary codon of hte mRNA in the A site. GTP is hydrolyzed.
    • In the next step, an rRNA molecule of the large ribosomal subunit catalyzes the formation of a peptide bond between the new amino acid at the A site and the growing polypeptide chain in the P site. The polypeptide is then removed from tRNA in the P site and attacvhes to the tRNA in the A site, which translocates to the P site.
    • Translocation: third step- the ribosome translcates the tRNA in the A site to the P site. The empty tRNA is moved to the E site, and texits the complex. mRNA moves along with bound tRNAs, bringing the next codon to the A site.
  12. Explain translation termination.
    When a stop codon (UGA, UAA, UAG) enters the A site. A release factor binds directly to the stop codon and causes the hydrolysis of water, breaking the bond between teh completed polypeptide and the tRNA in the P site, releasing the polypeptide throguh the exit tunnel of the ribosome's large subunit. THe translation assembly then comes apart.
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CHapter 16-17 Essay Test