dna replication and repair - bio resource site · 2016-03-21 · step 3 elongation of a new dna...
TRANSCRIPT
DNA Replication
and Repair
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Overview of DNA Replication SWYK CNs – 1, 2, 30 Explain how specific base pairing enables existing DNA strands to accurately replicate themselves.
DNA Replication
• genetic information is passed on to the next generation
• semi-conservative
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Parent molecule with two
complementary molecules
Parental strands
separate
Each parental strand is a template
Each daughter DNA molecule consists of one
parental and one new strand
Overview of replication
• DNA is unwound and stabilized
• Origins of replication: Replication bubble and replication fork
Initiation
• RNA primers bind to sections of the DNA and initiate synthesis
Priming
• Leading strand (5’ 3’) synthesized continuously
• Lagging strand synthesized discontinuously then fragments are joined
• RNA primer replaced by DNA
Elongation
• Mismatch repair by DNA polymerase
• Excision repair by nucleases
Proofreading
Review of DNA structure
• double helix
• each strand has a 5’ phosphate end and a 3’ hydroxyl end
• strands run antiparallel to each other
• A-T pairs (2 H-bonds), G-C pairs (3 H-bonds)
SWYK CNs – 3, 4, 29 1. Sketch and label origins of replication: Replication bubble and replication fork 2. Describe the roles of the following proteins: - helicase - single-strand binding proteins - topoisomerase 3. Give two differences between prokaryotic and eukaryotic DNA replication.
STEP 1 Initiation at origins of replication separation sites on DNA strands
• Depend on a specific AT-rich DNA sequence – Prokaryotes – one site – Eukaryotes – multiple sites
• Replication bubble • Replication fork • Proceeds in two directions from point of
origin
The proteins of initiation and priming
SWYK CNs – 5, 6, 27 4. What is the role of primase? 5. Why do we need RNA primers in replication? 6. What is the leading strand? How is it different from the lagging strand in terms of the priming step?
STEP 2 Priming initiation of DNA synthesis by RNA
RNA primers bind to unwound sections through the action of primase
– leading strand – only 1 primer
– lagging strand – multiple primers
– replaced by DNA later
SWYK CNs – 7, 25, 26 7. Describe the roles of the following proteins: DNA polymerase (I/III) DNA ligase 8. Why is the leading strand synthesized continuously whereas the lagging strand is synthesized discontinuously? 9. Draw parental DNA being replicated. Label the following: parental DNA, leading strand, lagging strand, 3’ and 5’ ends, Okazaki fragments.
STEP 3 Elongation of a new DNA strand lengthening in the 5’ 3’ direction
DNA polymerase III can only add nucleotides to the 3’ hydroxyl end
Leading strand - DNA pol III – adds nucleotides
towards the replication fork; - DNA pol I - replaces RNA with
DNA Lagging strand - DNA pol III - adds Okazaki
fragments to free 3’ end away from replication fork
- DNA pol I - replaces RNA with DNA
- DNA ligase – joins Okazaki fragments to create a continuous strand
SWYK CNs – 10, 11, 24 10. How does DNA polymerase III function in mismatch repair? 11. How do nuclease, DNA pol III, and ligase function in excision repair? 12. What is the problem with telomeres at the 5' end? How is this solved in eukaryotic cells? You may use a diagram and labels for this.
STEP 4 Proofreading correcting errors in replication
Mismatch repair • DNA pol III – proofreads
nucleotides against the template strand
Excision repair • nuclease – cuts damaged
segment • DNA pol III and ligase – fill the
gap left Telomeres at 5’ ends of lagging
strands • no genes, only 100 – 1000
TTAGGG sequences to protect genes
• telomerase catalyzes lengthening of telomeres
Gene Expression From gene to protein
Transcription and Translation
GENE EXPRESSION Genes code for polypeptide chains or for RNA molecules
TRANSCRIPTION
• DNA-directed RNA synthesis
• produces mRNA
TRANSLATION
• mRNA-directed polypeptide synthesis
• occurs on ribosomes
• Prokaryotes – mRNA translated immediately • Eukaryotes – pre-mRNA processed before leaving the
nucleus as mRNA
SWYK CNs – 12, 13, 14, 22, 23 15. What is the role of RNA polymerase II in initiating transcription? 16. Assuming that the DNA template strand has the sequence 3’ A T A T T T T A C G C G C C A 5’, draw a) Nontemplate/coding/sense strand b) RNA strand
Transcription
TERMINATION
ELONGATION
INITIATION
STEP 1 – Initiation Occurs at a promoter
Transcription factors bind to the TATA box on the DNA.
RNA polymerase II and other transcription factors
bind to promoter.
DNA strands unwind.
Polymerase initiates RNA synthesis at the start point.
STEP 2 – Elongation mRNA transcript lengthens
RNA polymerase
unwinds DNA.
Polymerase adds
nucleotides to 3’ end.
New RNA strand
peels away from
template.
DNA strands re-
form a double helix.
Nontemplate
• Coding
• Sense
Template
• Noncoding
• Antisense
STEP 3 – Termination mRNA transcript released
Polymerase transcribes
polyadenylation signal sequence
(AAUAAA)
pre-mRNA is released
Polymerase detaches from
DNA.
SWYK CNs – 15, 21 17. mRNA in eukaryotes undergoes RNA processing. Draw, label, and explain the significance of the following structures a. 5’ cap b. poly-A tail c. spliceosome
RNA processing alteration of pre-mRNA
5’ cap
Modified G added to 5’ end
Poly-A tail
50-250 A added to 3’ end
RNA splicing
Facilitate export of mRNA from nucleus
Protect mRNA from degradation by
enzymes
Help ribosomes attach to 5’ end of RNA
RNA processing RNA splicing
• Introns – intervening sequences – noncoding segments on pre-mRNA – May regulate gene activity
– Enable genes to give rise to two or more different polypeptides
– Facilitate evolution through exon shuffling
• Exons – expressed sequences on pre-mRNA
• Signal for splicing is a short sequence at the ends of introns
• small nuclear ribonucleoproteins (snRNPs) with snRNA recognize splice sites
• snRNPs + proteins spliceosome
– release introns
– join together exons that flank introns
SWYK CNs – 16, 17, 18, 19, 20 Initiator tRNA base with the anticodon (17. _________) pairs with the start codon (18. _________ ) 19. Draw and label the initiator tRNA, mRNA, start codon, and large ribosomal subunits with EPA sites during initiation of translation. Ribosome reaches STOP codons (20.) ___________, _________, _________)
Translation Overview • mRNA moves
through ribosome • codons are
translated into amino acids
• tRNA molecules: anticodon and amino acid ends
• amino acids added to a growing polypeptide chain
• rRNA molecules + proteins ribosomes
tRNA structure
SWYK CNs – 8, 9 21. What is wobble? What is its significance? 22. What is the genetic code? Why is it described as redundant but not ambiguous? 23. Given a prokaryotic DNA template with the following sequence, write the sequence of the mRNA transcript formed and the polypeptide synthesized: 3’ T A T A A T C T A C A C A T T G C C G T A C T A A A T A 5’,
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Inosine
Ribosome structure
Building a polypeptide
Initiation
Translation
Termination
Step 1 - Initiation
Small R subunit binds to mRNA
Initiator tRNA base pairs with
start codon (AUG)
Large R subunit binds to complete the initiation
complex
Initiator tRNA in the
P site, A site is empty
Step 2 Elongation
Step 3 – Termination
Ribosome reaches STOP codon (UAG, UAA, UGA)
A site receives a
release factor
Release factor cleaves bond
between tRNA and the
last amino acid
Two ribosomal subunits
disassemble
Given the following sequence on a template DNA strand 3’ AAA TAT TTT CCG TAC GGA TAG ACA CCG AAA ATC CGG GCA 5’
• What is the sequence on the non-template strand? 5’ TTT ATA AAA GGC ATG CCT ATC TGT GGC TTT TAG GCC CGT 3’ • What is the mRNA sequence transcribed (assuming
transcription right occurs after the TATA sequence)? 5’ AAA GGC AUG CCU AUC UGU GGC UUU UAG GCC CGU 3’
• What is the STOP codon? UAG
• What is the anticodon attached to the tRNA that corresponds to the STOP codon?
There is no tRNA that corresponds to the STOP codon. Release factors take their place.
• What is the amino acid sequence in the polypeptide product?
Met – Pro – Ile – Cys – Gly - Phe