Chap 10-12 DNA structure and replication

What four key characteristics must all genetic material possess?

1. Storage of complex information\n2. Replication (accurate copying)\n3. Expression of information (determining phenotype)\n4. Capacity for variation (mutation/evolution)

Describe the Griffith (1928) Transformation Experiment and its core hypothesis.

Griffith showed that heat-killed virulent S. pneumoniae (IIIS) passed a heritable "transforming principle" to living avirulent bacteria (IIR), converting them into live, deadly IIIS cells.

How did Avery, MacLeod, and McCarty (1944) identify the chemical nature of the transforming principle?

They treated the transforming factor with protease, RNase, and DNase. Transformation occurred normally in all treatments except the one treated with DNase, proving that DNA is the transforming principle.

Explain how Hershey and Chase (1952) used radioisotopes to confirm DNA as the genetic material.

They labeled T2 phage DNA with ³²P and the protein coat with ³⁵S. After bacterial infection and blending, only the ³²P radioisotope entered the host cell and appeared in progeny phages, proving DNA directs viral reproduction.

What are the three essential components of a nucleotide?

A pentose sugar, a nitrogenous base (attached to the 1′ carbon), and a phosphate group (attached to the 5′ carbon).

What is the structural and functional difference between Ribose and 2-Deoxyribose?

Ribose has a hydroxyl group (-OH) on the 2′ carbon, while 2-deoxyribose has a hydrogen atom (-H). The missing oxygen makes DNA more chemically stable for long-term storage than RNA.

Distinguish between Purines and Pyrimidines in DNA and RNA.

Purines (Double-ring): Adenine (A) and Guanine (G).\nPyrimidines (Single-ring): Cytosine (C), Thymine (T) (DNA only), and Uracil (U) (RNA only).

Define Chargaff’s Rules and how they apply to the DNA double helix.

In double-stranded DNA, the amount of A equals T (A=T) and the amount of G equals C (G=C). Therefore, total purines equal total pyrimidines (A+G = C+T).

Describe the structural properties of the Watson-Crick B-DNA model.

A right-handed double helix consisting of two antiparallel polynucleotide strands. Strands have a strong sugar-phosphate backbone linked by 3′–5′ phosphodiester bonds, with flat nitrogenous bases stacked perpendicular to the axis and held together by weak hydrogen bonds.

Contrast B-DNA, A-DNA, and Z-DNA.

B-DNA: The normal, biologically active right-handed helix under cell conditions.\nA-DNA: A right-handed helix that is slightly more compact, usually found under dehydrated conditions.\nZ-DNA: A rare, left-handed elongated helix configuration.

Why do nucleic acids absorb UV light most strongly at 260 nm?

Due to the resonant ring structures of the purine and pyrimidine bases, which allow scientists to isolate and quantify DNA/RNA.

What is a hyperchromic shift, and how does it relate to Tm?

As double-stranded DNA denatures (melts) into single strands, its UV absorption at 260 nm increases sharply. The midpoint of this curve is the melting temperature (Tm), which increases with higher G-C base-pair content due to their 3 hydrogen bonds.

What dictates DNA reassociation kinetics (C₀t analysis)?

The concentration and sequence complexity (X) of the genome. Highly repetitive sequences find their complementary partners faster and reassociate at lower C₀t values than unique single-copy sequences.

How did Meselson and Stahl (1958) prove the semiconservative model of replication?

They grew E. coli in heavy ¹⁵N media, transferred them to light ¹⁴N media, and analyzed DNA density using gradient centrifugation. Generation 1 showed a single hybrid (¹⁵N/¹⁴N) band, and Generation 2 showed distinct hybrid and light bands, rejecting conservative and dispersive models.

Contrast Theta replication and Linear eukaryotic replication.

Theta replication: Occurs in circular bacterial chromosomes, initiating from a single origin (oriC) and replicating bidirectionally.\nLinear replication: Occurs in eukaryotes, tracking bidirectionally from multiple origins of replication to copy massive chromosomes within a single cell cycle.

Compare the unique enzymatic activities of bacterial DNA Polymerase I and III.

DNA Pol III: Main replicative enzyme; polymerizes 5′ → 3′ and proofreads 3′ → 5′ exonuclease.\nDNA Pol I: Removes RNA primers via unique 5′ → 3′ exonuclease activity and fills the resulting gaps with DNA.

Name the replication role of DnaA, Helicase (DnaB/C), and SSBPs.

DnaA: Binds to the origin (oriC) to initiate local unwinding.\nHelicase: Uses ATP to fully unzip the double-stranded DNA helix.\nSSBPs: Bind to exposed single strands to prevent them from snapping back together.

What problem does DNA Gyrase (Topoisomerase) solve during replication?

Helicase unwinding causes severe positive supercoiling (torsional tension) ahead of the replication fork. DNA Gyrase cuts and reseals the strands to relieve this supercoiling stress. (Fluoroquinolone antibiotics kill bacteria by trapping these enzymes mid-cut.)

Why must DNA replication be discontinuous on the lagging strand?

DNA polymerases can only synthesize in the 5′ → 3′ direction. Because strands are antiparallel, the lagging strand must be built away from the fork in short fragments (Okazaki fragments), each requiring a new RNA primer initiated by primase.

How are Okazaki fragments structurally joined together?

DNA Polymerase I excises the RNA primers and fills the gaps with dNTPs. DNA Ligase then catalyzes the final phosphodiester bond to seal the nicks between adjacent fragments.

Explain the concept of Origin Licensing across the eukaryotic cell cycle.

To prevent duplicating DNA twice, the Origin Recognition Complex (ORC) binds origins in G1 phase and recruits Cdc6 and Cdt1 to load an inactive MCM2-7 helicase complex. In S phase, high kinase activity (CDKs/DDK) fires the origin, activating the helicase and destroying licensing factors to prevent reloading.

What is the "End-Replication Problem" in linear eukaryotic chromosomes?

Once the terminal RNA primer on the extreme 5′ end of the lagging strand is removed, there is no upstream 3′-OH group for a DNA polymerase to extend from. Without a fix, chromosomes would shorten with every cell division.

How does Telomerase fix the end-replication problem?

Telomerase is a ribonucleoprotein. Its intrinsic RNA component (TERC) serves as a template for its reverse transcriptase subunit (TERT) to extend the G-rich 3′ overhang of the telomere. Primase and polymerase can then synthesize a final primer to fill the lagging strand gap.

Outline the major steps of the Double-Strand Break Repair (DSBR) model.

1. A double-strand break is processed by nucleases (resection) to create single-stranded 3′ tails. 2. Strand invasion takes place via RecA/RAD51 filament searching. 3. DNA synthesis extends the strands, leading to second-end capture and ligation. 4. This forms a double Holliday junction which can be resolved into crossover or non-crossover products.

Define Gene Conversion and state its cause.

A nonreciprocal genetic exchange where one allele is physically converted into another. It arises during homologous recombination when mismatch repair enzymes fix base mismatches found within newly formed heteroduplex DNA loops.

Describe the biochemical construction of a eukaryotic nucleosome.

The primary unit of chromatin packing. It consists of roughly 147 bp of DNA wound in a left-handed superhelix around a core histone octamer, which contains two copies each of histones H2A, H2B, H3, and H4. Histone H1 binds externally as a linker between core particles.

Contrast Euchromatin and Heterochromatin.

Euchromatin: Uncoiled, relaxed, less densely stained chromatin that houses actively transcribed genes.\nHeterochromatin: Permanently condensed, darkly stained chromatin that is transcriptionally inactive and abundant at centromeres, telomeres, and repetitive regions.