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Biology LibreTexts

1: DNA Replication, Transcription and Translation

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  • DNA Replication

    I. Chromosomal DNA

    A. Function: DNA base sequence encodes information for amino acid sequence of proteins. Genetic code: 1 to 1 relationship between a codon (specific sequence of 3 bases) and 1 amino acid. Central Dogma of genetics/info flow in cells -Foundation Figure: Flow of Genetic Info p 1. DNA will be replicated and passed on to “daughter cells”

    B. DNA Structure: figure 8.3 Double stranded (2 strands of DNA), helical “double helix”, antiparallel

    1. Two strands held together by hydrogen bonds between complementary bases inside helix
    2. Strong outer “sugar-phosphate” backbone; covalent phosphodiester bonds link nucleotides
    3. DNA strands: polymers of nucleotides
    4. Nucleotides: 3 components. Sugar=deoxyribose, phosphate, nitrogenous base
    5. Nitrogenous bases of DNA

    a. purines (2 rings)= adenine (A) and guanine (G) pyrimidines (1 ring)= thymine (T) and cytosine (C)
    b. Chagraff’s rules: amount of A=T and amount of C=G; this is because of complementary base-pairing rules

    A=T form 2 hydrogen bonds
    G=C form 3 hydrogen bonds

    *c. complementary base pairing permits the precise replication of DNA

    6. Deoxyribose: pentose 5 carbons. C1' covalently linked to nitrogenous base.

    C3’= free OH (tail)
    C5’ linked to phosphate group (head)

    7. Prokaryotic chromosomes see figure ; Most bacteria have a single circular chromosome. 1 copy of chromosomes=“haploid cells” (most human cells have 2 copies of linear chromosomes and are called “diploid cells” see “eukaryotic chromosomes).

    8. Topoisomerases and Bacterial Gyrase

    -Topoisomerases; Enzymes which “supercoil” DNA or relieve supercoiling different types of toposiomerases in E. coli.
    Type I/III” “relax” DNA supercoils
    Type II= Bacterial Gyrase: introduces negative supercoils
    “Through the action of topoisomerases, the DNA molecule can be alternately coiled and relaxed. Because coiling is necessary for packing DNA into the confines of a cell and relaxing is necessary so DNA can be replicated (and transcribed), these two complementary processes an important role in the behavior of DNA in the cell.“ Brock Biology of Microorganisms 8th edition p 185 )

    -bacterial gyrase is involved in supercoiling/relief of supercoiling of DNA

    -antibiotics quinolones (e.g. nalidixic acid) and fluoroquinolones (such as ciprofloxacin) and novobiocin inhibit bacterial gyrase and interfere with DNA replication/transcription; see p

    C. DNA synthesis by DNA polymerases fig ___; Table _____

    1. DNA polymerase requires template strand (guide), primer strand with free 3’OH group, activated substrates/precursors= nucleoside triphosphates

    *2. DNA replicated in 5’ to 3’ direction (5’->3’). Incoming nucleotides can only be added to 3’OH tail of a growing DNA strand

    3. Oxygen of 3’OH groups makes a nucleophilic attack on inner most phosphorus atom of incoming nucleoside triphosphate. Pyrophosphate split off and will be hydrolyzed by cellular phosphatases with the release of energy to drive synthesis. Nucleotide is linked to primer strand by phosphodiester bond (ester bond= bond between alcohol and acid)

    4. If no 3’OH present , DNA strand cannot be lengthened=DNA chain termination. Use of dideoxynucleoside triphosphates as base analogues and in DNA sequencing reactions.

    II. Replication of Bacterial Chromosome fig ____

    A. Recall bacterial chromosome: singular, circular double stranded DNA in cytoplasm

    B. DNA replication begins at specific site “ori” = origin of replication

    C. DNA replication proceeds bidirectionally from ori, with formation of replication bubble and 2 replication forks. Replication forks= regions where d.s. DNA unwound, form s.s. DNA templates, DNA polymerase makes complementary copy of parent ssDNA template.

    D. DNA replication is semiconservative. 1 parent “old” DNA strand is used as template or guide for synthesis of 1 new daughter DNA strand.-result: 1 parent chromosome -> 2 daughter chromosomes. Each daughter chromosome is a copy of parent chromosome. Each daughter chromosome consists of 1 old parent DNA strand and 1 new daughter DNA strand. 1 parent strand is “conserved” in each new daughter chromosome

    E. Enzymes/proteins involved in DNA synthesis. KNOW FOR EXAM. Fig 8___ Table ___

    1.* Topoisomerases e.g., Bacterial Gyrase; involved in DNA supercoiling/relief of
    supercoiling (target of quinolones e.g., ciprofloxacin “cipro” used to treat/prevent
    inhalation anthrax)

    1. Helicase: unwinds ds DNA, breaks H bonds between bases, forms ss DNA template

    2. Single Strand Binding Proteins SSBP bind, stabilize and protect ssDNA

    3. RNA Primase: an RNA polymerase which does not require a primer strand to start
    primer synthesis. Synthesizes a short complementary RNA primer strand with free 3’OH
    group using ss DNA as template. Creates RNA primer, permitting DNA polymerase to
    start DNA synthesis. (RNA polymerase do not “proof read” and therefore can make
    many mistakes).

    4-5. DNA polymerase: requires primer strand, template and activated nucleoside
    triphosphates (dATP, dTTP,DCTP,dGTP). Must have DNA template. Synthesizes complementary DNA
    strand using parent strand as template/guide. DNA polymerase have “proofreading abilities”, they “check”
    each nucleotide they add, remove if incorrect and add correct nucleotide. DNA polymerases have high
    fidelity, they make very few mistakes. Original mistake rates 10-4; following proofreading, mistake rate=
    10-9 ie one incorrect base in every 109 bases added E. coli: DNA polymerase III performs most of DNA synthesis
    DNA polymerase I: will remove RNA primer and replace with DNA sequence

    6. Ligase: links short sequences of DNA (called Okazaki fragments) together on “lagging
    strand” homework see inhibition of nucleic acid synthesis. What are nucleotide analogs? What are their uses?

    Compare and contrast bacterial DNA polymerases and RNA polymerases
    Note: ss=single strand ds=double strand P=phosphate

    DNA polymerases synthesize complementary DNA using a DNA template/guide
    e.g., ssDNA template base sequence: A T A G G C
    Complementary DNA sequence T A T C C G dna
    synthesized by DNA polymerase

    RNA polymerases synthesize complementary RNA sequences using DNA as a template/guide
    e.g., ssDNA template base sequence: A T A G G C
    Complementary RNA sequence U A U C C G rna
    synthesized by RNA polymerase

    Synthesis of DNA and RNA require input of energy, both ATP and charged precursors

    Compare and Contrast DNA Polymerase and cellular RNA Polymerase
    DNA Polymerase RNA Polymerase
    Template/guide ss DNA ssDNA
    Synthesize complementary DNA complementary RNA
    Charged precursors deoxyadenosine tri-P= dATP adenosine tri-P= ATP
    deoxythymidine tri-P=dTTP uridine tri-P=UTP
    deoxycytodine tri-P= dCTP cytodine tri-P=CTP
    deoxyguanosine tri-P=dGTP guanosine tri-P=GTP
    primer required? yes no
    proofreading/editing? yes * no

    *DNA polymerase proofreading/editting
    Polymerases have a ”normal” or “intrinsic” mistake rate of approximately 10 -4 – 10 -5 nucleotides (this means the polymerases introduce the incorrect nucleotide every 10,000 to 100, 000 nucleotides). DNA polymerases have the ability to “proofread
    and edit” their mistakes. If they introduce the wrong nucleotide, they can remove or “excise” the wrong nucleotide and try again to make a correct match. This reduces the mistake rate of DNA polymerases to approximately 10-9 – 10 -10 (or only one incorrect
    nucleotide every 1,000,000,000 – 10,000,000,000 nucleotides). RNA polymerase cannot proofread or edit so RNA polymerase make many mistakes (one reason many RNA viruses, for example HIV, mutate so rapidly...more later)

    Transcription Prokaryotic

    Review flow of information in cell
    DNA--------> RNA --------->Protein

    replication transcription translation

    I. Genetic Code: one to one relationship between specific codon (specific 3 base sequence) and an amino acid

    II. Bacterial Transcription: use of DNA as template/guide to synthesize complementary RNA.
    DNA info is rewritten in RNA sequence. Fig ___

    A. First step in gene expression

    B. Products of transcription

    1. messenger RNA=mRNA: will be translated into specific amino acid
    sequence of a protein
    2. transfer RNA=tRNA: actual “translator” molecule, recognizes both a
    specific codon and specific amino acid
    3. ribosomal RNA=rRNA: combined with ribosomal proteins, will form
    the ribosome, the “workbench” at which mRNA is translated into a specific amino acid

    4. additional RNA products

    III. Promoters and Bacterial RNA polymerases

    A. Promoters: specific DNA sequences which signal the “start” points for gene
    transcription. Sigma factor/subunit of RNA polymerase binds to promoters to
    initiate transcription

    B. Bacterial RNA polymerases: enzyme complex which recognizes DNA promoters, binds
    to promoter and synthesizes complementary RNA copy using DNA as

    E. coli RNA Polymerase: 2 subunits, sigma subunit and core

    a. sigma subunit/factor= “brains” of RNA polymerase. Travels
    along DNA until it reaches a promoter, binds promoter

    b. core subunit: binds to sigma attached at promoter. “Workhorse”
    of RNA polymerase, carries out actual RNA synthesis. Requires
    activated precursors and template strand, DOES NOT REQUIRE
    PRIMER (compare to DNA Polymerase). Synthesizes RNA in 5’ -
    to->3’ , similar to DNA polymerase. No proofreading ability
    therefore will make more mistakes than DNA Polymerase

    c. sigma subunit will drop off after the first few ribonucleotides
    have been linked together, core continues alone. Note: core would
    start transcription randomly of DNA without direction of sigma
    subunit. Polycistronic mRNA (prok. only)

    IV. Termination of transcription (over-simplified)

    Terminators: DNA sequences which signal transcription stop signals. RNA
    polymerase releases DNA when transcription terminator sequence encountered
    Homework Describe antimicrobial drugs which bind to and inhibit function of bacterial RNA
    polymerases (answer: rifampin _used to treat which pathogen?)

    Bacterial Translation fig

    Translation: RNA base sequence translated into amino acid sequence of protein. mRNA is template for
    polypeptide synthesis. Second step in gene expression.

    A.Translation of mRNA into a polypeptide chain is possible because of the genetic code:

    1. genetic code: One to one relationship between a codon (specific sequence of 3 bases)
    and a specific amino acid. Figure __ Genetic code table

    mRNAcodon=amino acid
    Genetic code: Redundant (more than one codon for each amino acid) yet specific (each codon
    encodes info for 1 amino acid only). Universal; most cellular organisms use same genetic code;
    some exceptions

    B. Translation requires tRNA, amino acids, ATP/GTP, ribosomes and mRNA

    C. tRNA =transfer RNA. Adaptor/translator molecule. Only molecule which can "recognize" correct amino acid AND correct codon

    1. structure: ss RNA, stem and loop

    a. amino acid attachment site at one end
    b. anticodon which "recognizes"(forms H bonds with) codon of mRNA

    2. *45 different tRNA’s for 20 different amino acids; “wobble” permits some tRNA’s to bind to more than one codon (“relaxed”/improper base painring between 3 base of codon and anticodon)

    D. amino acyl tRNA synthetases* : “load” proper amino acid on proper tRNA= amino acid activation. 20 different transferases for 20 different amino acids/tRNA’s amino acid x+ ATP + tRNAx--> tRNAx:amino acid x + AMP + 2 P* “charged tRNA” or “activated amino acid”

    E. Ribosomes: 70S in prokaryotes. 2 subunits 30S (small subunit) + 50S (large subunit) S=Svedberg Unit, use to express sedimentation rates, ultracentrifugation

    made of rRNA and ribosomal proteins. “Workbench” at which mRNA will be translated into a polypeptide. 16s rRNA binds RBS (Ribosomal Binding Site on mRNA). 23s rRNA acts a ribozyme, forms peptide bonds between amino acids E, P and A sites.

    F. Mechanics of translation: text. GTP is hydrolyzed during translation

    Translation Initiation (note: tRNA-f met may first bind 30S subunit before 30S subunit binds RBS)

    1. 30S subunit recognizes ribosomal binding site RBS/Shine-Dalgarno sequence. Complementary to 16s rRNA sequence of ribosome.
    2. Translation begins at start codon AUG closest to ribosomal binding site
    3. An initiator tRNA:methionine ( more precisely a formyl methionine in bacteria) enters the “P” or peptidyl binding site of the ribosome. A tRNA fits into the binding site when its anticodon base-pairs with the mRNA codon
    4. The larger ribosomal 50S subunit then binds the complex
    5. Additional proteins called initiation factors are required to bring all components together

    Translation Elongation: amino acids are added one by one to first amino acid. Additional protein
    elongation factors required

    1. A second appropriately charged tRNA enters the “A” or aminoacyl binding site of the ribosome, bearing the next amino acid.
    2. Peptide bond formation. 23s rRNA of large subunit catalyzes formation of peptide bond between amino acid at P site and amino acid at A site (rRNA acts as a “ribozyme”, RNA catalyst) -amino acid of tRNA at P site is transferred to amino acid bond of tRNA at A site
    3. Now ribosome moves “downstream” by one codon. tRNA carrying dipeptide is now in P site, A site is empty.
    4. New appropriately amino acid charged tRNA enters A site
    5. Ribosome catalyzes peptide bond formation between dipeptide and new incoming amino acid. Tripeptide is carried by tRNA at A site
    6. Translocation:
    7. Requires energy (GTP )

    Translation Termination

    1. Ribosome reaches one of 3 nonsense codons/stop codons: UAA, UGA, UAG
    2. Release factor binds A site, causes polypeptide and ribosome to be released from mRNA (by activation of ribozyme)

    G. Polycistronic mRNA in prokaryotes permit coordinated gene expression in prokaryotes

    mRNA encodes more than one gene so ribosomes can coordinately produce several
    different proteins. For example 3 genes for proteins involved in lactose transport/metabolism in E. coli are
    transcribed into a single mRNA molecule. Ribosomes translate all 3 into proteins at same time

    H. Simultaneous transcription and translation in prokaryotes only. Ribosomes can bind mRNA and begin translation before transcription is finished. Very efficient. Fig ____