Biol280 Final Exam Unit 4 Based on TA Lecture Review

a form dna

dna-rna, rna-rna helix, 11 bp per turn, right handed

b form dna

most stable, 10.5 bp per turn, right-handed helix

Genes

sequences of nucleotides that specify protein sequences

Nucleotides are

nucleic acid monomers

nucleotide is made up of

phosphate group, sugar (pentose), nitrogenous base

nucleoside is made up of

sugar (pentose) and nitrogenous base

nucleosides are similar to nucleotides but lack a ____ ____

phosphate group

Purines how many rings

two

Pyrimidines how many rings

one

purines include what bases

adenine, guanine

pyrimidines include what bases

cytosine, thymine, uracil

what nitrogenous base is different in DNA and RNA? ______ in DNA, and _____ in RNA

thymine in DNA, uracil in RNA

what links nucleotides

phosphodiester linkage

How are phosphodiester linkages formed

3’ end 3rd carbon pentose ring attacks phosphate group

5’ of DNA characterized by

free phosphate group on C5’ of pentose sugar

3’ of DNA characterized by

-OH on C3’ of pentose sugar

Polarity of DNA

5’ → 3’

Histone

DNA packaging protein

Histone residues

Lys and Arg rich

Histone tails

modified for regulation

Nucleosome is made of

DNA + histone

heterochromatin

tightly packed, no gene expression

euchromatin

loosely packed, gene expression

Supercoiling

relieves strain

What introduces strain and how

polymerase by seperating the strands

Linking number specifies the number of…

helical turns in closed circular DNA

Overwound DNA

positive Lk, + supercoiling

Positive supercoiling

overwound DNA

Underwound DNA

negative Lk, - supercoiling

Negative supercoiling

underwound DNA

Topoisomers

differ in Lk only

If there is a broken DNA strand, what is lk

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Topoisomerase I

changes Lk by 1, cleaves one strand, can relax positive and ngetaive supercoils

Topoisomerase II

changes Lk by 2, cleaves both strands, can relax positive and negative supercoils, can introduce negative supercoils (prokaryotes only), hydrolyzes ATP

Topoisomerase I steps

binds DNA, cleaves on strand, passes single strand through the break, DNA is religated

Topoisomerase II steps

1) multi-subunit enzyme binds at a segemnt of a DNA molecule 2) a second segement of the same DNA molecule is bound at the N gate 3) the second segement of DNA is trapped. the light blue segement is cleaved on both strands to form two 5’ phosphotyrosyl linkages to the enzyme 4) the second DNA segment is passed through the break 5) the broken dna is religated, and the second DNA segement is released through C gate

What decatenates DNA circles

topoisomerase II

Prokaryotes use a specific type of topoisomerase called

toposiomerase II (DNA gyrase)

DNA replication initiation place

origins of replication (ori)

DNA replication elongation

prime, polymerize, ligate

DNA replication termination location in prokaryotes

terminator region

SSB

binding to single stranded DNA, stabilize, prevent reannealing

Helicase (DNA B) protein

DNA unwinding

primase (DNA G) protein

RNA primer synthesis

DNA polymerase III

synthesizing new DNA strands

DNA Pol I

fill in gaps, remove primers

DNA ligase

ligation, connect DNA strands

DNA gyrase

introduces negative supercoils into DNA, relieve torsional strain via supercoiling

Leading strand

continuous synthesis

Lagging strand

fragment synthesis

Okazaki fragment

lagging fragment

What removes RNA primers in DNA replication

DNA Pol I

DNA polymerase

synthesizes DNA

Parent DNA

original strand

3’→5’ exonuclease activity

proofreading

Most dna polymerase have what exonuclease activity

3’ → 5’

5’→3’ exonuclease that removes RNA activity is only found in what

Pol I

Pol I has ___ —> ____ exonuclease

5’→3’

difference in pro and eukaryotice replication… eukaryotic is

slower and chromosomes are longer

Processive enzymes

catalyze repeatedly without releasing substrate

Mismatch repair

fixes replication errors

_____ distinguishes between template and newly synthesized strands

methylation

In mismatch, what strand is methylated

parent/template strand

Mismatch repair steps

1) exonuclease activity degrades DNA from methyl past mismatch 2) DNA polymerase III replaces DNA 3) result: DNA containing mismatch is resynthesized

Base excision repair

fixes damaged bases

Base excision repair steps

1) cleaves N-glycosyl bond with DNA glycolase 2) AP endonuclease cleaves phosphate backbone 3) DNA pol I synthesize new DNA 4) DNA ligase seals the nick

______ glycosylase for each base lesion

different

Nucleotide-excision repair fixes

bulky lesions (pyrimidine dimers)

excinuclease

excision endonuclease, makes 2 cuts, excises the damgaed DNA

RNA structure always single stranded

NO

Central dogma

DNA → (transcription) → RNA → (translation) → protein

Core rule of transcription

RNA polymerase is processive

transcription begins at

promoter

RNA synthesized

5’→3’

RNA polymerase reads

3’→5’

DNA nontemplate/coding strand

matches RNA except T/U

transcription in E coli

1) RNA polymerase core and the sigma 70 subunit bind to the DNA promoter 2) transcription bubble forms (open complex) 3) transcription is initiated 4) promoter clearance followed by elongation 5) elongation continues, sigma 70 dissociates, and it is replaced by NusA 6) transcription is terminated NusA dissociates and the RNA polymerase is recycled

____ ___ and ___ assemble at promoter

transcription factors (like sigma factors) and RNA polymerase

Supercoils in elongation

positive (direction of transcription) negative supercoils (opposite direction of transcription)

Rho-dependent termination

helicase binds rut element, p helicase separates the mRNA from the DNA template

Rho-independent termination

An RNA hairpin forms at a palindromic sequence and disrupts the interaction between the RNA and DNA template within the polymerase, mRNA is released

prokaryotice genes are arranged in ____ producing ____ mRNA

operons, polycistronic

eukaryotes have ___ polymerase

III

5’ cap function

enhance stability, only in eukaryotes, roles in processing, roles in translation

5’ cap what linkage and what is added

5’5 condensation, 7-methylguanosine

polyA tails function

contribute to mRNA stability and translation (only in eukaryotes)

polyA tail steps

polyadenylation signal, cleavage with endonuclease, polyadenylation with polyadenylate polymerase

mRNA splicing only in

eukaryotes

exons

coding (expressed) sequences are spliced together

introns

noncoding (intervening) sequences removed during splicing

Translation direction

5’ → 3’

Where does translation start

start codon, AUG

Where does translation end

stop codon

tRNA structure

4 leaf clover

Wobble hypothesis

the third base of an mRNA codon (3’ end) can form non-standard, flexible base pairs (a "wobble") with the first base of a tRNA anticodon (5’ end)

one codon recognized

C, A

Two codons recognized

A and G

3 codons recognized

I

I pairs with

AUC

Differences between prokaryotic and eukaryotic transcription

Prokaryotes = cytoplasm, coupled to translation, little/no mRNA processing, one RNA polymerase, simpler promoters; Eukaryotes = nucleus, separated from translation, capped/spliced/poly-A mRNA, multiple RNA polymerases, more transcription factors