Molecular Bio - Comp Exam Flashcards
Describe the structure of DNA.
DNA Structure
- Consists of 2 long polynucleotide chains with 4 types of nucleotide subunits.
- Nucleotides = 5-carbon sugar and nitrogenous base covalently linked via glycosidic bond.
- Sugar in DNA is deoxyribose.
- Adenine, cytosine, guanine and thymine are bases.
- 3D structure is a double helix.
- 1 turn every 10 bp.
- Antiparallel: each strand’s sequence is complementary to partner.

Describe the universal features of cells on earth.
Universal Features of Cells on Earth
- All Cells Store Their Hereditary Information in the Same Linear Chemical Code: DNA.
- All Cells Replicate Their Hereditary Information by Templated Polymerization.
- All Cells Translate RNA into Protein in the Same Way.
- Each Protein Is Encoded by a Specific Gene.
- All Cells Function as Biochemical Factories Dealing with the Same Basic Molecular Building Blocks.
- All Cells Are Enclosed in a Plasma Membrane Across Which Nutrients and Waste Materials Pass.
- Life Requires Free Energy.
Describe the structural organization of nucleosomes.
Nucleosomes - Structural Organization
- Digestion with nucleases break down DNA by cutting between nucleosomes and degrading the exposed DNA between nucleosome core particles (linker DNA).
- Each individual nucleosome core particle consists of 8 histone proteins (histone octamer):
- 2 molecules each of H2A, H2B, H3, H4.
- And 2x stranded DNA that is 147 nucleotide pairs long.
- Linker DNA can be few to 80nt; nucleosomes repeat every 200 nucleotide pairs or so.

What is a chromatin?
Nuclear DNA + protein.
Describe histone modifications.
Histone Modifications
- Amino acid side chains of histones are subject to a variety of covalent modifications:
- Occurs on the core of the histone as well as the tail.
- Acetylation of lysines (loosens chromatin structure).
- Added by histone acetyl transferases (HATs); removed by histone deacetylase complexes (HDACs).
- Mono, di and tri-methylation of lysines:
- Added by methyl transferases; removed by histone demethylases.
- Phosphorylation of serines.
- Recruitment of these enzymes depends on gene regulatory proteins.
- All are reversible but can persist long after regulatory proteins have disappeared.
- Important consequences for the types of proteins the modified DNA attracts: this determines how/when/if gene expression takes place.
- Example - H3 Tail Modifications to N-terminal Tail:
- Reading histone code involves joint recognition of marks at other sites on nucleosome along with tail recognition.
- Few meanings are known.

Describe DNA packaging.
DNA Packaging
- Eukaryotic DNA is packaged into chromsomes.
- The human genome is 3.2x109 nucleotides distributed over 24 different chromosomes.
- Each chromosome is a single long linear DNA molecule associated with proteins that fold and pack it into compact structure.
- Similar to packing 24 miles of thread into a tennis ball.
- DNA + protein = chromatin.
- Each human cell contains 2 copies of each chromosome (maternal and paternal homologs).
- except germ cells and RBCs.
- 22 pairs of autosomes and 2 sex chromosomes.
- 46 total chromosomes.
Describe the requirements of a chromasome.
Chromosome - Requirements
- A copy must be passed on to each daughter cell at division: requires replication, separation of copies and partitioning to daughter cells.
- DNA replication origin: where duplication of the DNA begins.
- Centromere: Allows one copy of each duplicated and condensed chromosome to be pulled into each daughter cell when the cell divides. The kinetochore protein complex attaches to the centromere.
- Telomeres: at the ends of a chromosome, contain repetitive sequences that enable the ends to be efficiently replicated.

Describe chromatin remodeling.
Chromatin Remodeling
- Nucleosomes are in a constant state of flux.
- DNA is unwrapped in the nucleosome 4 times per second, and remains unwrapped for 10-50 milliseconds before tightening up again.
- Chromatin remodeling complexes allow further loosening of DNA/histone contact.
- Proteins are related to helicases and are ATP dependent; bind to both protein core and DNA.
- Changes structure of nucleosome temporarily, making DNA less tightly bound.
- Repeated cycles catalyzes nucleosome sliding, making DNA available to other proteins in the cell.

Describe nucleosome packing.
Nucleosome Packing
- Forms a dense fibrous structure with diameter of 30 nm.
- Unknown how this fiber is formed.
- Maybe zig-zag model:
- Stacking may be facilitated by histone tails (esp H4).
- Histone H1 “linker histone” is present in 1:1 ratio with nucleosome cores.
- Histone tails help to condense chromatin:
- Histone tails are largely unstructured, suggesting that they are highly flexible.
- Tails can form interactions with adjacent nucleosomes.

Describe the regulation of chromatin structure.
Regulation of Chromatin Structure
- Certain types of chromatin structure can be inherited.
- Epigenetics: A form of inheritance that is superimposed on the genetic inheritance based on DNA.
- Examples:
- DNA methylation
- Chromatin structure
- Histone modification

Describe heterochromatin.
Heterochromatin
- Chromatin that is very condensed:
- Stains darkly throughout the cell cycle, even in interphase.
- Thought to be late replicating and genetically inactive.
- Highly concentrated at centromeres and telomeres.
- Contains very few genes; those that are present are resistant to gene expression.
- Position effect: activity a gene depends on position on chromosome:
- Will be silenced if relocated near heterochromatin.
- All the rest is less condensed and known as euchromatin.
How do genomic changes occur?
Genomic Change - How They Occur
- Occur as mistakes in DNA replication and repair:
- Rare occurrence: 1/1000 nucleotide pairs is randomly changed in the germ line every million years
- Movement of transposable elements also play a role.
- Range of changes can occur:
- Base pair substitutions.
- Large scale rearrangements:
- Duplications
- Deletions
- Inversions
- Translocations
Describe human variation in genome evolution.
Genome Evolution - Human Variation
- Human sequences vary 0.1% from one to another:
- Human and chimps differ 1%.
- Majority of mutation are neither harmful or beneficial.
- Neutral mutations can become fixed in a population.
- SNPs - single-nucleotide polymorphisms:
- Points in the genome where one group has one nucleotide and another group as another.
- Variation occurs at a high rate (1% or more).
- CNVs - copy number variants:
- Presence of many duplications and deletions of large blocks of DNA.
- Some blocks are common and others rare; significance of most is unknown.

Describe the importance of high-fidelity DNA replication.
Multicellular Organisms Need High Fidelity Replication
- Germ cells have to have low mutation rates to maintain the species.
- Somatic cells need low mutation rates to avoid uncontrolled proliferation/cancer.

Describe proofreading in DNA replication.
DNA Replication - Proofreading
- DNA polymerase makes 1 mistake out of every 109 nucleotides copied, thanks to proofreading:
- First step is just before a new nucleotide is added: enzyme must tighten its “fingers” around the active site, which is easier if the correct base is in place.
- Exonucleolytic proofreading:
- Takes place immediately after incorrect bases is added.
- DNA polymerase requires a perfectly paired 3’ terminus.
- 3’ to 5’ exonuclease clips off unpaired residues at 3’ primer terminus.
Describe DNA replication.
DNA Replication
- DNA polymerase synthesizes DNA by catalyzing the following reaction: (DNA)n residues + dNTP → (DNA)n+1 residues + P2O74-.
- Template directed—new chain is assembled in a preexisting DNA template that is complementary to the incoming bases.
- Requires separation of the two parental strands.
- Requires dATP, dGTP, dCTP and dTTP.
- DNA polymerase requires a primer with a free 3’ -OH to begin.
- Occurs during DNA synthesis phase (S) which lasts ~8hrs for mammalian cells.
- Chromosomes are replicated to produce two complete copies, joined at centromeres until M phase.
- Replication is activated in clusters (replication units) consisting of 20-80 origins.
- Different regions of each chromosome are replicated in a reproducible order during S phase, depending on chromatin structure.
- Heterochromatin is late-replicating (timing related to packing of DNA in chromatin):
- Example: X chromosomes of females: Almost all of inactive X is condensed into heterochromatin and is replicated late in S phase; the active homolog is less condensed and replicates throughout S phase.
- Regions of genome with less condensed chromatin replicate first.
- Heterochromatin is late-replicating (timing related to packing of DNA in chromatin):

Describe the proteins involved at the replication fork in DNA replication.
Replication Fork - Proteins
-
DNA helicase - unwinds DNA
- Protein with 6 identical subunits that binds and hydrolyzes ATP.
- This causes conformational change that propels it like a rotary engine along single stranded DNA, passing it through a center hole.
- Capable of prying apart the helix at rates of 1000 nucleotide pairs/sec.
- Single-stranded DNA binding proteins: bind tightly and cooperatively to exposed SS DNA:
- Help stabilize unwound DNA.
- Prevent formation of hairpins.
- DNA bases remain exposed.
- Sliding clamp: Keeps DNA polymerase on DNA when moving; releases when double stranded DNA is encountered.
- Assembly requires clamp loader: hydrolyzes ATP as it loads the clamp onto a primer-template junction.
- Leading strand: clamp remains associated with DNA polymerase for long stretches.
- Lagging strand: Clamp loader stays close so it can assemble a new clamp at start of each new Okazaki fragment.

Describe the reassembly of chromatin after replication.
Reassembly of Chromatin After Replication
- Replication requires not only DNA replication but synthesis and assembly of new proteins.
- Eucaryotes have multiple copies of genes for each histone.
- Histone proteins are synthesized mainly in S phase; amount made is highly regulated to meet requirements.
- For efficient replication, chromatin-remodeling proteins are needed to destabilize DNA-histone interface.
- As replication fork passes through chromatin, histone octamer breaks into:
- an H3-H4 tetramer, distributed randomly to daughter duplexes.
- 2 H2A-H2B dimers which are released from the DNA.
- Freshly made H3-H4 fills in spaces; H2A/H2B dimers are 1⁄2 old and 1⁄2 new; they are added at random to complete complex.
- This orderly addition requires histone chaperones (chromatin assembly factors).
- Directed to DNA with sliding clamp called PCNA.

Describe telomeres.
Telomeres
- End replication problem on lagging strand: no place for RNA primer.
- Bacteria have circular genomes; eukaryotes have telomeres.
- Special sequence GGGTTA at the end of each chromosome repeated ~1000x.
- Enzyme called telomerase replenishes these sequences by elongating parental strand in 5’ to 3’ direction using an RNA template on the enzyme.
- After extension of parental strand by telomerase, replication of lagging strand can be completed by DNA polymerase, using extension as template.
- This mechanism (plus a 5’ nuclease) ensures 3’ end is longer, leaving a protruding SS end that loops back and tucks into the repeat.
- T-loops:
- Structures protect ends and distinguishes them from broken ones that need to be repaired.
- Shelterin - protective chromosome cap made up of proteins.

Describe the types of DNA damage.
DNA Damage
- Mutations are not only caused by mistakes in replication.
- 5000 purine bases are lost every day due to a spontaneous reaction called depurination.
- Spontaneous deamination of C to U occurs at 100 bases/day.
- Can also occur from exposure to reactive forms of O2 in the cell or chemicals in environment.
- UV radiation from sun can produce a covalent linkage between two adjacent pyrimidines (T-T or C-T):
- Pyrimidine dimers.
- If unrepaired when DNA replicated, these changes lead to either a deletion or a base pair substitution in the daughter strand.

Describe the DNA repair pathways.

Describe the regulation of homologous recombination.
Homologous Recombination - Regulation
- Accurate repair process can still cause problems for a cell:
- Use of a non-functioning homolog to “repair” the other homolog.
- Loss of heterozygosity:
- Critical first step in cancer development.
- Rare occurrence.
- Loss of heterozygosity:
- Use of a non-functioning homolog to “repair” the other homolog.
- Processing of broken ends is coordinated with the cell cycle:
- Nucleases for generating 3’ invading strand are only active in S and G2 phase.
- Ensures a replicated chromosome or sister chromatid will be the most likely template for repair.
- Prevention of repair in the absence of damage:
- Loading of Rec A on DNA is tightly controlled.
- Repair proteins dispersed throughout the cell:
- After damage, repair occurs in “factories” or “foci” at the sites of damage.
- Nucleases for generating 3’ invading strand are only active in S and G2 phase.
- Mutations in proteins involved in recombination can cause cancer:
- Brca1 and Brca2 lead to increased rates of breast cancer.
- Brca1 regulates the processing of broken ends of chromosomes.
- Mutations lead to use of non-homologous end-joining process.
- Brca2 maintains Rad51 (RecA) inactive until it is at site of damage.
- Does not bind to DNA to form invading strand.
- Brca1 regulates the processing of broken ends of chromosomes.
- Brca1 and Brca2 lead to increased rates of breast cancer.
Describe the repair of double-strand breaks.
DS-Break Repair
- Non-homologous end joining (NHEJ) and homologous recombination (HR) in mammals repair DNA double-strand breaks.
- The process involves:
- End binding and tethering.
- End processing (removing mismatched or damaged nucleotides and replacing them).
- Ligation.
- Several human syndromes are associated with dysfunctional NHEJ, including SCID (Severe combined immunodefiency).
- HR requires a homologous section of DNA to act as a template for repair of the damaged/broken fragment.
- HR is more accurate than NHEJ because of the template. The importance of HR is derived from the fact that the mechanism is conserved throughout evolution.

Describe Holliday junctions and how they are resolved.
Holliday Junctions
- Structures are present only transiently.
- Resolution - strands of the helices are cleaved by endonuclease (RuvC).
- Resolution has two outcomes:
- Crossing over:
- Rare event.
- Only 2 cross over events/ chromosome.
- Gene conversion:
- 90% of Holliday junctions in humans resolve this way.
- Crossing over:








































































































































































































