Octamer Study guides, Revision notes & Summaries

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MCB EXAM 1 Questions And Answers
  • MCB EXAM 1 Questions And Answers

  • Exam (elaborations) • 11 pages • 2023
  • How large is our genome? - Answer- 3200 million bp How many base pairs does the genome contain? - Answer- 3 × 10^9 base pairs. How long is the genome when fully extended? - Answer- 20,000 phone books long What is Chromatin? - Answer- the fibrous complex of eukaryotic DNA and histone proteins. Which are the main Histones? - Answer- H1, H2A, H2B, H3, and H4. What's the charge (polarity) of Histones? - Answer- Positive What amino acid residues are responsible for that net charge...
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MCB Exam 2 Answers and updated
  • MCB Exam 2 Answers and updated

  • Exam (elaborations) • 17 pages • 2024
  • T/F: DNA looping is important for RNA polymerase activity in the presence of enhancer elements. - answer-True T/F: Genome editing has been greatly facilitated by the adaptation of the bacterial CRISPR-Cas9 system. - answer-True T/F: Transposase is required for non-LTR retrotransposon movement. - answer-false-transposase is required for transposon movement. non-LTR retrotransposon movement requires reverse transcriptase T/F: LTR and non-LTR retrotransposons require reverse transcriptase ...
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USMLE Step 1 First Aid - Biochemistry 2022 (100% Verified)
  • USMLE Step 1 First Aid - Biochemistry 2022 (100% Verified)

  • Exam (elaborations) • 61 pages • 2022
  • Available in package deal
  • Chromatin structure - Negatively charged DNA loops twice around histone octamer (2 each of the positively charged H2A, H2B, H3, and H4) to form nucleosome bead. H1 ties nucleosomes together in a string. (Think of "beads on a string"; H1 is the only histone that is not in the nucleosome core.) In mitosis, DNA condenses to form mitotic chromosomes. <img src="66a - Euchramatin strxr.JPG" /> Heterochromatin - Condensed, transcriptionally inactive ("H eteroC hromatin = H ighly C onde...
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Step 1 First Aid – Biochemistry Questions with answers 2022
  • Step 1 First Aid – Biochemistry Questions with answers 2022

  • Exam (elaborations) • 46 pages • 2022
  • Chromatin structure - Answer - Negatively charged DNA loops twice around histone octamer (2 each of the positively charged H2A, H2B, H3, and H4) to form nucleosome bead. H1 ties nucleosomes together in a string. (Think of "beads on a string"; H1 is the only histone that is not in the nucleosome core.) In mitosis, DNA condenses to form mitotic chromosomes. <img src="66a - Euchramatin strxr.JPG" /> Heterochromatin - Answer - Condensed, transcriptionally inactive ("H eteroC hromatin =...
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ACS BIOCHEMISTRY EXAM Q&A With complete solution
  • ACS BIOCHEMISTRY EXAM Q&A With complete solution

  • Exam (elaborations) • 19 pages • 2023
  • ACS BIOCHEMISTRY EXAM Q&A With complete solution Henderson-Hasselbach Equation - ANSWER pH = pKa + log ([A-] / [HA]) FMOC Chemical Synthesis - ANSWER Used in synthesis of a growing amino acid chain to a polystyrene bead. FMOC is used as a protecting group on the N-terminus. Salting Out (Purification) - ANSWER Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution. Size-Exclusion Chromatography - ANSWER Separates sa...
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BIOS 390 Week 1 Quiz Study Guide (LATEST UPDATE) | Download To Score An A
  • BIOS 390 Week 1 Quiz Study Guide (LATEST UPDATE) | Download To Score An A

  • Exam (elaborations) • 11 pages • 2022
  • BIOS 390 Week 1 Quiz Study Guide BIOS390 – Molecular Diagnostics Week 1 (TCO 1) The basic principles of genetics contributed by Gregor Johann Mendel are: 1. The principle of segregation 2. T he principle of independent assortment (TCO 1) The Augustinian monk, Gregor Johann Mendel, crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F2 generation. The results were as follows: 6018 yellow and 2002 green (8020 total). The allele for gre...
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ACS BIOCHEMISTRY EXAM Q&A
  • ACS BIOCHEMISTRY EXAM Q&A

  • Exam (elaborations) • 19 pages • 2023
  • ACS BIOCHEMISTRY EXAM Q&A With complete solution Henderson-Hasselbach Equation - ANSWER pH = pKa + log ([A-] / [HA]) FMOC Chemical Synthesis - ANSWER Used in synthesis of a growing amino acid chain to a polystyrene bead. FMOC is used as a protecting group on the N-terminus. Salting Out (Purification) - ANSWER Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution. Size-Exclusion Chromatography - ANSWER Separates sa...
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ACS BIOCHEMISTRY EXAM  2023 Q&A With  complete solution
  • ACS BIOCHEMISTRY EXAM 2023 Q&A With complete solution

  • Exam (elaborations) • 19 pages • 2023
  • ACS BIOCHEMISTRY EXAM 2023 Q&A With complete solution Henderson-Hasselbach Equation - ANSWER pH = pKa + log ([A-] / [HA]) FMOC Chemical Synthesis - ANSWER Used in synthesis of a growing amino acid chain to a polystyrene bead. FMOC is used as a protecting group on the N-terminus. Salting Out (Purification) - ANSWER Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution. Size-Exclusion Chromatography - ANSWER Se...
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ACS BIOCHEMISTRY EXAM 2023  Q&A With  complete solution updated
  • ACS BIOCHEMISTRY EXAM 2023 Q&A With complete solution updated

  • Exam (elaborations) • 19 pages • 2023
  • ACS BIOCHEMISTRY EXAM 2023 Q&A With complete solution Henderson-Hasselbach Equation - ANSWER pH = pKa + log ([A-] / [HA]) FMOC Chemical Synthesis - ANSWER Used in synthesis of a growing amino acid chain to a polystyrene bead. FMOC is used as a protecting group on the N-terminus. Salting Out (Purification) - ANSWER Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution. Size-Exclusion Chromatography - ANSWER Se...
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BIOS 390 Week 1 Quiz Study Guid-Latest Update
  • BIOS 390 Week 1 Quiz Study Guid-Latest Update

  • Summary • 11 pages • 2020
  • Available in package deal
  • Question Set 1 (TCO 1) The basic principles of genetics contributed by Gregor Johann Mendel are: (TCO 1) The Augustinian monk, Gregor Johann Mendel, crossed yellow-seeded and green-seeded pea plants and then allowed the offspring to self-pollinate to produce an F2 The results were as follows: 6018 yellow and 2002 green (8020 total). The allele for green seeds has what relationship to the allele for yellow seeds? (TCO 1) In 1953, James Watson and Francis Crick proposed. Question Set 2 (T...
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