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Transcript
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)
4. 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'UUUU 5'
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
5. 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 = 41000
6. RNA usually single stranded (can coil back on itself to form hairpin loops)
7. In cells, DNA tends to be supercoiled
C. Replication
1. replication = synthesis of DNA
2. replication begins at the origin of replication
a. one on prokaryotic chromosome
b. several on eukaryotic chromosomes
3. DNA helix unravels and actual replication occurs at the replication fork
a. bidirectional, replicons (portion of genome containing an origin and replicated as
a unit) separate when forks meet opposite the origin
b. replication fork and associated enzymes may be attached to plasma membrane
4. eucaryotic DNA is linear and much longer than procaryotic DNA
a. many replication forks simultaneously
b. several origin sites along the DNA
5. DNA polymerases (3 different types) catalyze DNA synthesis 5' ÷ 3'
a. require dNTP, template
b. pol III main enzyme, some pol I
c. pol I and pol II probably mostly for repair
6. helicases unwind DNA (use ATP for energy)
7. single-stranded DNA binding proteins (SSBs) maintain strand separation
8. topoisomerases relieve supercoiling due to unwinding
9. replication of the leading strand (unwinding 3' ÷ 5') is continuous
10. replication of the lagging strand is discontinuous
a. small (Okazaki) fragments synthesized 5' ÷ 3' (1000-2000 nucleotides in
procaryotes, 100 in eucaryotes)
b. pol I removes RNA primer, synthesizes complementary DNA
c. DNA ligase links the fragments
(1) bacteria use pyrophosphate bond of NAD+ for energy
(2) other ligases use ATP instead
11. pol III has proofreading ability where wrong bases are excised and correct ones
inserted during replication
D. Transcription
1. transcription = synthesis of RNA under direction of DNA
a. RNA has sequence complementary to DNA template
b. uracil replaces thymine
c. mRNA = contains message for protein synthesis
d. tRNA = carries amino acids during protein synthesis
e. rRNA = ribosome components
2. RNA polymerase synthesizes RNA
a. requires NTP, DNA template
b. 5' ÷ 3'
c. transcribes only the sense strand of DNA
(1) different genes may be encoded on opposite strands
(2) gene = DNA segment or sequence that codes for a polypeptide, an rRNA, or a
tRNA
3. RNA polymerase (E. coli) opens double helix and transcribes the sense strand to
produce RNA transcript that is complementary and antiparallel to DNA template
4. synthesis of rRNA and tRNA similar to mRNA except
a. molecules terminated by 5'-monophosphate rather than triphosphate found at end
of all primary transcripts (mRNA)
b. rRNA and tRNA are smaller than primary transcripts
c. tRNA contains bases other than A, G, C, U not in original transcript
d. primary transcripts undergo posttranscriptional modification or RNA
processing
5. 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
1. codon = mRNA code for amino acid = 3 nucleotides
a. 20 amino acids
b. 43 = 64 (42 = 16)
2. code degeneracy = more than one code/amino acid
3. sense codons = direct amino acid incorporation (61)
4. nonsense or stop codons = non-coding codons
a. UGA, UAG, UAA
b. terminate translation
5. 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)
3. Protein synthesis can be divided into initiation, elongation, and termination
4. initiation begins with ribosomes forming over initiation codon (AUG)
5. 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
6. 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
7. protein synthesis stops when ribosome reaches a nonsense or stop codon
G. Regulation is essential for optimizing cell efficiency
1. respond to environment
2. maintain proper levels of cellular constituents
3. goal is to maximize the production and use of energy (cell’s “job” is to make
more cells)
4. 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
1. 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)
2. 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
3. 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
1. protein synthesis requires a lot of energy (ATP), so regulation of protein synthesis is
important to cell's energy economy
2. mRNA transcription is regulated by sigma factors, induction/repression, and
attenuation
3. 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
4. transcription of inducible enzymes can be regulated by induction and repression
5.
6.
7.
8.
a. constitutive = enzymes that are produced continuously
(1) 60 - 80% of genes
(2) usually code for enzymes required needed in fairly constant amounts for major
life processes (e.g., glycolytic enzymes)
b. transcription of an inducible enzyme increases in the presence of an inducer
c. transcription of repressible enzymes is decreased by corepressors
repression = regulatory mechanism that inhibits gene expression and decreases the
synthesis of enzymes (negative regulation)
a. usually a response to an endproduct of a pathway that acts as corepressor
b. leads to decrease in synthesis of enzymes of that pathway that leads to that
product
c. corepressor activates repressors = regulatory proteins that block RNA polymerase
from initiating transcription
induction = process that turns on the transcription of a gene or genes
a. inducer = substance that induces transcription by binding to and inactivating a
repressor
b. best known example are the genes for lactose metabolism in E. coli
(1) in absence of lactose, $-galactosidase activity is almost absent ($galactosidase hydrolyzes lactose to glucose and galactose)
(2) in presence of lactose, large amounts of $-galactosidase are synthesized
(3) in cell, lactose converted to allolactose, which induces the genes
(4) so presence of lactose indirectly induces the cells to synthesize more enzyme =
enzyme induction
lactose metabolism involves 3 structural gene products (enzymes) located together on
the chromosome along with a control region, all called the lac operon
a. operon = operator + promoter + structural genes
b. structural genes for 3 enzymes are next to each other and regulated together
(1) $-galactosidase
(2) $-galactosidase permease, involved in lactose transport into the cell
(3) thiogalactoside transacetylase, metabolizes certain disaccharides other than
lactose
c. operator = regulatory site that gives a stop or go signal to transcription
d. promoter = region of DNA where RNA polymerase initiates transcription
e. I gene = regulatory gene near the lac operon that codes for a repressor protein
f. in absence of lactose, repressor protein binds to operator site, preventing
transcription (and expression of structural genes)
g. in presence of lactose, some diffuses into cells and is converted into the inducer
allolactose
(1) inducer binds to repressor protein and alters it (allostery) so that it cannot bind
to the operator site
(2) in absence of bound repressor, structural genes are transcribed into mRNA
(3) transcribed mRNA translated into enzymes
(4) lactose induces enzyme synthesis, lac operon is called an inducible operon
catabolite repression = inhibition of metabolism of alternative carbon sources by
glucose
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
9. 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
5. 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
6. 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
1. 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
2. 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
3. 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)
4. 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
c. mutations can have different effects on the phenotype
(1) silent = codon changes, but not amino acid (degeneracy)
(2) missense mutations = amino acid substitution in a protein
(3) nonsense mutation = stop codon in middle of sequence (early termination)
(4) reversion = second mutation which alters phenotype back to wild type
(a) back mutation = mutant nucleotide sequence back to wild type
arrangement
(b) suppressor mutation = overcomes effect of 1st mutation
5. spontaneous mutation = random change in DNA arising from mistakes in replication
or detrimental effects of natural background radiation on DNA
a. rate varies in different species
(1) low = 10-10
(2) high = 10-5
b. high rate of reproduction allow ready observation
6. induced mutations = arise from exposure to mutagens
a. mutagen = physical or chemical agent that interact with DNA is a disruptive
manner
b. valuable laboratory tool
c. poses a health risk
7. Detection and isolation of mutants
a. direct selection
b. replica plating
(1) auxotrophs = mutants with a damaged anabolic pathway
(a) require the end product of that pathway to grow
(b) grow normally when supplied with the end product
(c) typical auxotrophs require an amino acid
(d) prototroph = bacterium with the complete anabolic pathway
(2) temperature sensitive
8. Carcinogen testing (Ames test)
a. mutational reversion assay with specific strains of Salmonella typhimurium
(1) different mutations in histidine biosynthesis operon
(2) cell wall mutations to make cells more permeable to test chemicals
(3) altered repair mechanisms to enhance errors
b. bacteria and test compound mixed in molten top agar with trace histidine
(1) auxotrophs grow slightly (allow DNA replication)
(2) revertants (due to mutagenesis) will continue to grow and produce visible
colonies after a couple of days
(3) colony counts compared to controls
c. the more revertants, the greater the relative mutagenicity of the compound
d. sometimes liver extract added to molten agar to promote transformations that
occur in mammals
L. DNA repair
1. Integrity of DNA important enough to require repair mechanisms in addition to
proofreading by DNA polymerase
2. excision repair is a general repair mechanism that corrects damage that causes
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
3. photoreactivation specific mechanism to split thymine dimers
a. requires light activation
b. does not remove or replace nucleotides
4. 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)
5. 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