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Bio 97 Midterm Review 1
10/6/10Brett Springer
Brian Dang
DNA polymerase can’t start from scratch: an RNA primer is required
DNA replication requires many different proteins
• Helicase: unwinds/unzips DNA
• SSBs: stabilize single stranded DNA
• Gyrase: relieves torsional stress
• RNA primase: primes DNA synthesis
• DNA polymerase: synthesizes DNA
• DNA ligase: seals ends
2-1
DNA polymerase possesses a proofreading function
Polymerase is more activethan the exonuclease unless
there is a mismatch
2-4
Deoxyribose vs ribose
Deoxyribose Ribose
Characterized by the “H” on carbon 2
Characterized by the “hydroxyl” on carbon 2
One DNA strand is replicated discontinuously
3’
5’
5’3’ 3’5’
5’
3’
5’ 3’
= RNA primer2-6
DNA is generally replicated in both directions at once
5’
5’
3’
3’
origin
5’ 3’
3’ 5’
5’
3’
5’
3’
2-7
Template DNA
• DNA synthesized 5’->3’• Template DNA strand is read 3’->5’
Final Step in Laggin Strand (Okazaki Fragments)
• DNA pol. III extends new DNA fragment to RNA primer
• DNA pol. I comes in, removes RNA primer, leaving a “nick”, a missing phosphodiester bond
• DNA Ligase seals the bond using energy from ATP to generate phosphodiester bond.
• ****Doesn’t use a new nucleotide****
PCR involves repeated cycles of polymerization to amplify DNA
• 95°C→denatures DNA by breaking the hydrogen bonds between strands
• 50-65°C→temperature is reduced to allow primer to bind (anneal) to DNA
• 72°C→temperature is raised to the optimum for the thermostable polymerase (Taq) to extend the DNA strand
2-15
PCR cont.• Requires sequence-specific primer, you must
know part of the DNA sequence to do this• Requires special polymerase (Taq)• No leading or lagging strands• How do you choose temp. for annealing?• A=T 2 Celsius; G=C 4 Celsius• Temp. for Annealing = ((# of A/T) x 2 C) + ((# of
G/C) x 4 C)
The Central Dogma explains how DNA becomes a phenotype
DNA
RNA
protein
transcription
translation
3-1
Proteins give us our phenotypeEnzymes and many structural components of cells are made up of protein. Proteins determine how we function and how we look. DNA controls our phenotype by encoding proteins.
3-2
The conversion of a gene into a protein* through an RNA intermediate is called
GENEEXPRESSION
Gene expression is highly regulated and cell type-specific
3-3*some genes function as RNA and are transcribed but not translated
A liver cell encounters a toxin in the bloodstream. Its genomic DNA is mutated as
a result. The mutation occurs in the rhodopsin gene. Rhodopsin is a protein found
only in the eye.
What is the consequence of this mutation for the
organism?
3-5
DNA is copied into messenger RNA before it becomes protein
RNA
Anti-senseDNA strand
SenseDNA strand
This is called transcription
3-7
RNA polymerase binds to promoter sequences in DNA to initiate transcription
5’ 3’The SENSE strand is shown
3-13
For prokaryotic promoters:
Promoters “make sense” in one direction, they are not palindromic
5’-TTGACA . . . TATAAT-3’
3’-AACTGT . . . ATATTA-5’
PROMOTER
-35 -10
ABLE WAS I ERE I SAW ELBATHIS is a palindrome:
3-15
Promoters determine the start and direction of transcription, it must have consensus
mRNA splicing:
• Occurs at sites determined by consensus sequences
• Requires multiple proteins• Takes place in the nucleus• Most of the RNA is discarded in mammals• Splicing pattern can vary under different
conditions (alternative splicing)
3-21
The primary transcript is processed into mRNA in eukaryotes
5’ methyl-guanosine
cap
PolyA tail
3-20
Alternative splicing of the same transcript can give rise to different proteins
3-22
Alternative splicing can give rise to variant proteins in different tissues
**note that the order of the exons is not altered**3-23
Proteins fold into domains
Structure 13:579 (2005)3-27
•A domain is an independent folding unit
•In general, each domain of the protein is encoded by a different exon
•This allows new proteins to be generated from the different domains/exons/parts
The Exon Shuffle Model explains how new proteins are generated over the
course of evolution
3-25
Intron/Exon organization allows the generation of new proteins: the Exon Shuffle model
Exon 1 Exon 2 Ex 35’ 3’
3-29
Exon 15’ 3’Exon 2
1) 2 distinct genes
Exon 15’ 3’Exon 2
2) DNA breaks occur at the arrows
3) DNA repair accidentally anneals the broken ends of Gene A to Gene B:
4) A new protein will be made—because the recombination occurred in introns, no coding information was destroyed
GENE A:
GENE B:
Ex 3
Exon 1 Exon 25’ 3’Exon 2GENE A/B:
GENE B/A:
The anatomy of a mature eukaryotic mRNA
AUG STOP5’UTR 3’UTR AAAAAA
5’ CAP
ORF:
open reading frame
UTR: untranslated region
4-1
polyA tail
*NO INTRONS*
In eukaryotes, a given mRNA codes for a single protein
In prokaryotes, mRNA can be polycistronic (can code for
multiple proteins)
4-4
Nature of the mRNA:
Exons contain non-overlapping triplets of nucleotides called codons that indicate the order of the amino
acids in the protein
4-6
The genetic code
4-7
Aminoacyl tRNA synthetases translate the genetic code
AMINOACID
PAIRS WITH CODON IN mRNA
4-8
The “start” codon is AUG and codes for methionine
Met
mRNAAUG
UAC
4-9
Amino acid structure
4-11
Proteins have polarity
amino (N)terminus
carboxyl (C)terminus
4-12
Stage 1: Translation initiation• Initiation factors bind to the 5’ cap• The initiation complex forms (includes eIFs,
initiator tRNA, and small ribosomal subunit)• The initiation complex scans along the mRNA
5’ to 3’ until it finds an AUG• eIFs leave and large ribosomal subunit is
recruited• The initiating methionine tRNA is in the P
(peptidyl) site
4-15
Stage 2: Elongation
• Ribosome moves one codon down on mRNA• New tRNA enters the A (acceptor) site• Peptide bond is formed with amino acid on tRNA in
P site• Repeat: Ribosome moves down one codon, tRNA in
the P site moved to E (exit) site and exits• Energy for amino acid addition and ribosomal
translocation comes from GTP hydrolysis by elongation factors
4-18
Stage 3: Termination
No tRNA for a STOP codon, release factor
binds instead
The portion of the mRNA after the stop codon is the 3’UTR
4-19
What is “wobble”?
• All amino acids except Trp and Met are specified by multiple codons
• Synonomous codons generally differ only in the third base
• Generally the third base is either a purine or pyrimidine in both codons
4-23
mRNA sequences have 3 possible reading frames
4-25
The Exon Shuffle Model works because introns do not code for protein and do
not have a reading frame
4-28
DNA in mammalian cells exists as chromatin
• Chromatin is a complex of DNA and proteins
• Chromatin is the material of which chromosomes are composed
• Chromatin in non-dividing cells is not condensed enough to see chromosomes by light microscopy
5-2
The nature of chromatin:
• All eukaryotic DNA is associated with numerous protein molecules
• Histones are the major class of proteins associated with DNA
• 75-90% of genome is wrapped around histones
• There are 5 major types of histones in all eukaryotes H1, H2A, H2B, H3 and H4
• 20-30% of the amino acids in histones are lysine and arginine—DNA has a negative charge
5-5
The core histone particle contains two molecules of H2A, H2B, H3, and H4
The histone “octamer” or core particle
5-6
Histone placement affects gene expression: The Histone Code
• Position of histones affect ability of other proteins to access the DNA
• Histone placement is determined by DNA sequence and by accessory proteins
• Histones are removed from areas of active transcription
• Transcriptionally active sites tend to have low levels of histones
5-8
Nucleosomes are compacted into a 30 nm fiber
Molec. And Cellular Biology Fig. 9-31
5-10
Histone H1 is not a part of the octamer core
Molec. Biol. of the Cell, Fig. 4-31
Developmental Biology, Fig. 5-1
5-11
Chromatin is “decondensed” in areas of active gene expression
5-15
Things to remember about centromeres
• Centromeres are visible only in condensed chromosomes
• The centromere is composed of DNA repeats and associated with kinetochore proteins
• Spindle fibers attach to the centromere to pull the chromosomes to opposite poles during mitosis
5-20
3’
5’
5’3’ 3’5’
5’
3’
5’ 3’
= RNA primer5-22
The lagging strand is synthesized discontinuously and cannot be copied all the
way to the end
Eukaryotes deal with this problem by placing repetitive sequences at the ends of chromosomes
TELOMERES
Human telomeres are composed of many repeats of
this sequence
5-24
Telomerase adds these repeats to the chromosome ends
5-25
Without active telomerase, mammalian chromosomes shorted by 100-200 nucleotides per cell division
Telomerase and DNA polymerase work together to build telomeres
5-26
t-loops (telomeric loops) form at the ends of mammalian chromosomes
• The single stranded 3’ end of the chromosome folds is tucked away in a loop structure
• This loop is stabilized by proteins
• The loop structure protects the single stranded DNA end from nucleases
• The loop prevents the end of the chromosome from being recognized as damaged DNA that needs to be “repaired”
5-29
Loss of telomeres can produce end-to-end chromosome fusions
http://www.med.nus.edu.sg/phys/Projects_Telomere_Prakash.htm 5-30
Telomere length limits how many times human cells can divide in culture
5-31
Things to remember about telomeres
• Telomeres are found at the ends of chromo- somes and are important for chromosome integrity
• Because DNA polymerase cannot replicate the very 3’ end of the template DNA strand, chromosomes would shorten with each cell division
• Telomerase is a complex of proteins and an RNA template that adds DNA repeats to chromosomes and prevents shortening
• Telomeres can serve as a cellular “clock”
5-35