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