RNA Polymerase & Transcription: RNA & Protein Synthesis in Prokaryotes & Eukaryotes - Prof, Study notes of Biology

An overview of rna polymerase, the process of transcription, and the synthesis of rna and proteins in both prokaryotes and eukaryotes. Topics include the role of ntp and dna templates, the sense strand of dna, the synthesis of rrna, trna, and mrna, codons and anticodons, translation, and regulatory mechanisms such as allosteric regulation and transcriptional control. The document also covers the regulation of enzyme activity and the control of transcription through sigma factors, induction and repression, and the lac operon.

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I. Microbial Genetics (Chapter 7)
A. Overview
1. all of the information necessary for cellular function and reproduction is stored in an
organism’s genetic material
a. genetic material can be transferred to progeny (heredity)
b. genetic material is typically DNA although some viruses use RNA
c. genetic material is contained within the chromosome
(1) circular molecule in prokaryotes
(2) linear molecules in eukaryotes
2. DNA is a large polymer made up of nucleotides
a. nucleotides are monomer composed of a nitrogen base (adenine, thymine,
guanosine, cytosine, uracil), a five carbon sugar (ribose or deoxyribose), and
phosphate
b. DNA exists as two strands, held together in a helix by hydrogen bonds
(1) complementary
(2) antiparallel
3. A gene is a sequence of DNA that carries information for a functional product
a. basic unit of heredity
b. a mutation is an inheritable change in DNA that may code for an altered
phenotype
4. The Central Dogma of Biology describes the transfer of information from DNA to
proteins
a. DNA serves as the template for its own synthesis (replication) as well as the
synthesis of RNA
b. transcription” describes the synthesis of RNA from DNA (ribonucleotides from
deoxyribonucleotides)
(1) same “language” (nucleotides) with slight variation
(2) three species of RNA
(a) messenger RNA (mRNA)
(b) transfer RNA (tRNA)
(c) ribosomal RNA (rRNA)
B. Structure of DNA
1. purine or pyrimidine base + ribose or deoxyribose = nucleoside or deoxynucleoside
2. nuclecoside or deoxynucleoside + phosphoric acid = nucleotide or deoxynucleotide
3. gigantic nucleic acid molecule with two polynucleotide strands arranged in a double
helix
pf3
pf4
pf5
pf8
pf9
pfa

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I. Microbial Genetics (Chapter 7) A. Overview

  1. all of the information necessary for cellular function and reproduction is stored in an organism’s genetic material a. genetic material can be transferred to progeny ( heredity ) b. genetic material is typically DNA although some viruses use RNA c. genetic material is contained within the chromosome (1) circular molecule in prokaryotes (2) linear molecules in eukaryotes
  2. DNA is a large polymer made up of nucleotides a. nucleotides are monomer composed of a nitrogen base (adenine, thymine, guanosine, cytosine, uracil), a five carbon sugar ( ribose or deoxyribose ), and phosphate b. DNA exists as two strands, held together in a helix by hydrogen bonds (1) complementary (2) antiparallel
  3. A gene is a sequence of DNA that carries information for a functional product a. basic unit of heredity b. a mutation is an inheritable change in DNA that may code for an altered phenotype
  4. The Central Dogma of Biology describes the transfer of information from DNA to proteins

a. DNA serves as the template for its own synthesis ( replication ) as well as the synthesis of RNA b. “ transcription” describes the synthesis of RNA from DNA (ribonucleotides from deoxyribonucleotides) (1) same “language” (nucleotides) with slight variation

(2) three species of RNA (a) messenger RNA (mRNA) (b) transfer RNA (tRNA) (c) ribosomal RNA (rRNA) B. Structure of DNA

  1. purine or pyrimidine base + ribose or deoxyribose = nucleoside or deoxynucleoside
  2. nuclecoside or deoxynucleoside + phosphoric acid = nucleotide or deoxynucleotide
  3. gigantic nucleic acid molecule with two polynucleotide strands arranged in a double helix

a. nucleic acid (RNA or DNA) = polymers of nucleotides b. nucleotides = nitrogen base + pentose sugar + phosphate c. nitrogen base = purines (2 rings) or pyrimidines (one ring) (1) purines = adenine (A) or guanine (G) (2) pyrimidines = thymine (T), cytosine (C) or uracil (U)

  1. The DNA molecule consists of a deoxyribose sugar -phosphate backbone attached to nitrogenous base crosspieces a. phosphate substitution on 5' carbon on deoxyribose joins to 3' hydroxyl group on

neighbor 3'UUUU5'

b. bases attached to 1' C of deoxyribose c. hydrogen bonds between anti-parallel strands join the two helices (1) G-C, 3 bonds (2) A-T, 2 bonds

  1. Significance of DNA structure a. maintenance of code during replication (1) base pairing helps retain base sequence (2) each strand provides a template b. provides variety, since order of bases constitutes the genetic code (1) for DNA 1000 bases long, combinations = 4^1000
  2. RNA usually single stranded (can coil back on itself to form hairpin loops)
  3. In cells, DNA tends to be supercoiled C. Replication
  4. replication = synthesis of DNA
  5. replication begins at the origin of replication a. one on prokaryotic chromosome b. several on eukaryotic chromosomes

d. primary transcripts undergo posttranscriptional modification or RNA processing

  1. In eucaryotes 3 RNA polymerases instead of 1 a. different promoters b. heterogeneous nuclear RNA (hnRNA) = large RNA precursors, about 5000 - 50000 nucleotides long c. posttranscriptional modification = hnRNA cleaved to final mRNA d. RNA splicing removes introns from initial RNA transcript (1) small nuclear RNA (snRNA) binds to splice junctions (2) splicing of pre-mRNA occurs in large complex called a splicosome e. ribozyme = self-splicing pre-rRNA molecules (1) medical applications (2) origin of life hypotheses f. rRNA and tRNA result from posttranscriptional processing E. The genetic code
  2. codon = mRNA code for amino acid = 3 nucleotides a. 20 amino acids b. 43 = 64 (4^2 = 16)
  3. code degeneracy = more than one code/amino acid
  4. sense codons = direct amino acid incorporation (61)
  5. nonsense or stop codons = non-coding codons a. UGA, UAG, UAA b. terminate translation
  6. codons match anticodons on tRNA a. generally 2nd and 3rd nucleotides in anticodon have more weight (or 1st and 2nd

in codon, reading 5' ˜ 3')

b. wobble = "sloppy" base pairing c. cells don't have to synthesize 61 different tRNAs F. Translation

  1. mRNA nucleotide sequence is translated into an amino acid sequence a. amino acids added to growing chains at the carboxyl end (C-terminal) b. mRNA often complexed with several ribosomes at once, synthesizing several copies at once ( polyribosome ) (1) only occurs in prokaryotes (2) in eukaryyotes, transcription occurs in nucleus and translation occurs in cytoplasm c. ribosomes can attach to mRNA being synthesized, so transcription and translation can occur simultaneously (in eucaryotes, transcription and translation machinery separated by nuclear membrane)
  2. Ribosomes are 70S organelles composed of a 50S and 30S subunits a. they have three important sites (1) A site = where tRNA carrying the newest amino acid binds (2) P site = where the tRNA with the growing peptide chain binds (3) E site = where the “empty” tRNA is ejected b. eucaryote ribosomes are 80S (60S + 40S)
  1. Protein synthesis can be divided into initiation, elongation, and termination
  2. initiation begins with ribosomes forming over initiation codon (AUG)
  3. elongation divided into binding of tRNA with the next amino acid, linking the growing peptide chain to the new amino acid, and movement of the ribosome to the next codon in the sequence
  4. in translocation: a. peptidyl-tRNA moves from A site to P site b. ribosome moves one codon along mRNA so that new codon is positioned at the A site c. empty tRNA leaves the E site
  5. protein synthesis stops when ribosome reaches a nonsense or stop codon G. Regulation is essential for optimizing cell efficiency
  6. respond to environment
  7. maintain proper levels of cellular constituents
  8. goal is to maximize the production and use of energy (cell’s “job” is to make more cells)
  9. two major types of regulatory mechanisms a. allosteric regulation of specific enzymes (effector molecule influences enzyme activity) b. transcriptional control (amount of enzyme synthesized) H. Control of enzyme activity
  10. activity of allosteric enzymes can be regulated by noncovalent binding of effectors or modulators to a regulatory site a. binding of effector alters affinity of catalytic site for substrate b. regulation can be positive (enhanced activity) or negative (inhibition)
  11. reversible covalent modification involves the addition or removal of a specific group to an enzyme a. does not involve a regulatory site b. covalent rather than noncovalent bonds c. often involves phosphate or AMP
  12. Feedback (endproduct) inhibition occurs with many biosynthetic pathways a. an endproduct inhibits activity of a pacemaker (rate limiting) enzyme b. frequently found in branched pathways (helps balance carbon flow) c. branched chain pathways can also use isozymes (different enzymes that catalyze the same reaction) I. Transcriptional control
  13. protein synthesis requires a lot of energy (ATP), so regulation of protein synthesis is important to cell's energy economy
  14. mRNA transcription is regulated by sigma factors, induction/repression, and attenuation
  15. the use of different sigma factors regulates expression of different genes by controlling the way RNA polymerase binds to the promoter a. different sigma factors allow RNA polymerase to recognize specific promoters b. common in G+, G-, and bacteriophages
  16. transcription of inducible enzymes can be regulated by induction and repression

a. not really repression since it is slow metabolism that stimulates synthesis of the enzymes b. also called the glucose effect c. results in diauxic growth (1) growth curve has two exponential phases (2) glucose is depleted (first exponential phase) before other substrate is used (3) use of second substrate produces the second exponential phase

  1. lac operon is also regulated by catabolite repression a. catabolite activator protein (CAP) or cyclic AMP receptor protein (CRP) and the small cyclic nucleotide 3',5'-cyclic adensosine monophosphate (cAMP) ( positive regulation ) b. CAP must bind before to CAP site on promoter before RNA polymerase can attach and begin transcription (= activator ) c. CAP can only bind when complexed with cAMP d. cell prefers glucose over lactose because it's more directly metabolized as carbon and energy source (1) glycolytic enzymes constitutive (2) at low levels of glucose, cell makes cAMP

e. cAMP = alarmone an alarm signal in response to environmental or nutritional stress (i.e., lack of glucose) J. attenuation

  1. control mechanism that allows transcription to start, but then prematurely terminates due to secondary structures (hairpin loops) in mRNA
  2. common with amino acid synthesis, trp operon best studied
  3. mechanism requires simultaneous transcription and translation (so it only occurs in prokaryotes)
  4. the leader sequence contains four regions that have complementary sequences that can form secondary structures (hairpin loops) a. if region 1 binds to region 2, than region 3 binds to region 4 and forms a termination loop b. if region 1 is prevented from binding to region 2, than region 2 will bind to region 3 and a termination loop does not form
  1. there are several tryptophan codons in region 1 a. if the cell has low concentrations of tryptophan, the ribosomes pause in region 1 and region 2 binds to region 3 (1) the termination loop (region 3 binding to region 4) does not form (2) RNA polymerase continues to transcribe the gene b. if the cell has high levels of tryptophan, the ribosomes follow RNA polymerase, not pausing over region 1 (1) region 1 and region 2 form a loop (2) region 3 and region 4 form a termination loop (3) RNA polymerase is released from transcription prematurely
  2. so, termination is determined by RNA secondary structures a. two distinctive, mutually exclusive structures b. structure formed based on rate of translation of the previous region of the leader sequence into the leader peptide (1) leader peptide has no cellular function other than regulation (2) contains two sequential trp codons (3) in presence of trp, translation continues until a stop codon (UGA) (4) in absence of trp, translation stops earlier and terminator hairpin doesn't form, so transcription continues through structural genes K. Mutations
  3. mutation = permanent, inheritable change in the genetic information a. loss, addition or rearrangement in base sequence b. wild type = strain with non-mutated characteristic c. mutant strain = strain with a mutated characteristic
  4. mutant strains can show difference in morphology, nutritional characteristics, genetic control mechanisms, resistance to chemicals, temperature preference, nearly any type of enzymatic function a. useful for tracking genetic events, determining genetic organization, mapping genes b. detected by replica plating or biochemical indicators c. many mutations are neutral (no phenotypic change) (1) usually single nucleotide substitution, "corrected" by degeneracy of the genetic code (2) mutation could also effect nonvital portion of protein
  5. mutations can be useful tools for studying cell function a. conditional mutations = expressed only under specific environmental conditions b. auxotroph = mutant missing an enzyme in a key anabolic pathway (prototroph = strain able to grow on minimal medium)
  6. different types of mutations a. point mutation = involves changing single base pair (affects only a single gene)

b. frameshift mutations = insertion or deletion so that the natural order of message is shifted (1) loss or change in entire or large portion of chromosome (deletion or insertion) (2) could occur as a point mutation or involve several nucleotides

distortion in the double helix a. mismatch correction enzyme recognizes parental strand (methylated to protect from restriction enzymes) b. wrong nucleotide excised and DNA polymerases fill the gap c. repairs thymine dimers

  1. photoreactivation specific mechanism to split thymine dimers a. requires light activation b. does not remove or replace nucleotides
  2. recombination repair restores damaged DNA for which there is no longer a template (entire pair is missing; gap opposite a lesion) a. recA protein cuts a piece of template DNA from a sister molecule and uses it to replace a damaged strand b. pol I repairs gap in sister strand left by excision (since template exists)
  3. SOS repair = inducible repair system that operates with great DNA damage a. so many gaps, DNA synthesis stops completely b. rec A initiates strand exchange, destroys lex A repressor protein, increasing repair rate c. highly error prone (last ditch effort)General Genetics