EXAM 2 BIOL 213: DNA replication

initiator proteins

helix opening at replication origin

-allow the denaturation of the dsDNA

DNA polymerase

polymerization (5’-3’) proofreading (3’-5’)

1 & 3: prokaryotic

3: does bulk of DNA replication

1: remove RNA primers and replace with DNA

DNA primase

synthesis of short RNA primer

Ribonuclease

enzyme: removes primer

-degradation of RNA primer

Repair polymerase

-form of DNA polymerase

-replace RNA with DNA

DNA ligase

joining of DNA fragments

DNA helicase

unzipping DNA helix prior to replication

single-stranded DNA-binding protein

prevent reannealing

sliding clamp

keeps DNA polymerase attached to template and on lagging strand, releases when Okazaki fragments is completed

Topoisomerase

prevents supercoiling of the DNA double helix by breaking covalent bonds in the backbone of the template strand

replication origins

specific sites DNA replication begins

overall shape DNA forms

double helix

components that make up the backbone of DNA

sugar and phosphate

why the DNA backbone located on the outside of the helix

polar and interacts with the aqueous environment

interactions that help stabilize the DNA double helix

hydrogen bonds and base stacking interactions (hydrophobic + van de waals forces)

end of a DNA strand that has a free phosphate group

5’ end

end of a DNA strand that has a free hydroxyl group

3’ end

meaning of DNA strands are antiparallel

they run in opposite 5’ to 3’ directions

DNA maintain a consistent width throughout the helix

purine always pairs with a pyrimidine

importance of hydrogen bonds for DNA function

allow the strands to separate when needed while maintaining stability

why can DNA polymerase only add nucleotides to one specific end of a growing DNA strand

because new nucleotides are added to the free 3’ OH

genome

complete set of genetic info in an organism

in prokaryotes, where is the genome typically located

single cellular chromosome in the nucleoid region

organization od prokaryotic genomes and eukaryotic genome difference

prokaryotes: one circular chromosome

eukaryotes: many linear chromosomes

why can two organisms with similar genome sizes have different levels of complexity

genome size does not directly correlate with organismal complexity

why does an increase in genome size in eukaryotes not necessarily mean an increase in the number of genes

much of the genome consists of noncoding and repetitive DNA

introns

noncoding sequences within genes that are removed during RNA processing

exons

coding sequences that remain in matura mRNA

how does the presence of introns allow for alternative splicing

different combos of exons can be joined to produce different proteins

alternative splicing increase protein diversity without increasing gene number

a single gene can produce multiple protein variants

mutation changes a G-C base pair to an A-T base pair within a region of DNA

-how could this affect the stability of the DNA double helix and why

DNA region would become less stable because G-C base pairs form three H-bonds, while A-T base pairs form two H-bonds

-fewer H-bonds reduce the overall stability of the helix

histone

proteins that DNA wrap around to form nucleosomes

nucleosome

DNA wrapped around a histone protein core

importance of chromatin packaging in eukaryotic cells

To fit inside the nucleus

chromatin structure influence gene expression

more tightly packed chromatin is generally less accessible for transcription

reason for chromatin structure to change during DNA replication and transcription

DNA must become accessible for enzymes to bind and function

telomeres

prevent chromosome shortening and degradation

chromosome

long, single DNA molecules associated with proteins that fold and pack the DNA into a compact structure

why linear chromosomes require special mechanisms to replicate their ends

DNA polymerase cannot fully replicate the ends of linera DNA

order of chromosome packaging

DNA strand → nucleosome → 30 nm fiber → looped domains → chromatid

problem that would arise if telomeres were not maintained

progressive shortening of chromosomes successive cell divisions

histones in nucleosome core

• 5 histones

• Nucleosome core

  • H2A

  • H2B

  • H3

  • H4

• required for 30 nm fiber packing level

  • H1

amino acids that histones are rich in

lysine and arginine

histones interact with DNA

histones contain many basic amino acids like lysine and arginine that allow them to interact with the negatively charged phosphates in DNA

composition of one nucleosome

146 bp DNA fragment ,2 H2A ,2 H2B, 2 H3, 2 H4 (histone octamer)

determines how a stretch of chromatin is handled by the cell

pattern of modification of histone tails