BIO 1100 Ch.9

Contrast the R and S strains of Streptococcus pneumoniae in Griffith’s 1928 experiment. Explain the "so what?" of the mouse surviving heat-killed S but dying when mixed with live R

• The S (Smooth) strain is virulent (encapsulated) and kills mice; the R (Rough) strain is non-virulent.

• He found that heat-killing the S strain rendered it harmless.

  • However, when heat-killed S was mixed with live R, the mouse died and live S cells were recovered.

• The "So What": This proved bacterial transformation—a "transforming principle" was transferred from the dead S cells to the live R cells, permanently changing their traits.

In the Avery, MacLeod, and McCarty experiment (1944), deduce which enzyme (DNase, RNase, or Protease) halted transformation and explain why this was the definitive proof that DNA is the genetic material

• Only DNase (which digests DNA) halted transformation.

• RNase and Protease had no effect.

• This proved that DNA is the specific molecule required for the "transforming principle," contradicting the then-popular theory that proteins were the primary carriers of genetic information.

Evaluate the use of radioactive sulfur (35S) vs. phosphorus (32P) in the Hershey-Chase experiment (1952) to differentiate between protein and DNA

• Proteins contain sulfur but not phosphorus, while DNA contains phosphorus but not sulfur.

• By labeling viral protein with 35S and viral DNA with 32P, Hershey and Chase observed that only the radioactive phosphorus entered the bacteria to direct viral replication, confirming DNA—not protein—as the hereditary material.

Describe the specific structural features of DNA revealed by the X-ray diffraction patterns produced by Rosalind Franklin

Franklin’s photographs (specifically Photo 51) provided the mathematical evidence that DNA is a helix of uniform diameter and that certain portions of the helix are repeated at regular intervals

Explain the significance of the "ball-and-stick" model used by Watson and Crick in solving the double helix

• Borrowing the methodology of Linus Pauling, Watson and Crick used physical models to ensure their proposed structure was chemically stable and spatially viable.

• This allowed them to integrate Franklin’s X-ray data and Chargaff’s rules into a coherent 3D structure held together by hydrogen bonds.

Identify the three components of a nucleotide and the specific sugar carbon numbers (1’, 3’, 5’) where the base, OH group, and phosphate attach

1. Nitrogenous Base: Attaches to the 1’ carbon.

2. Hydroxyl Group (-OH): Attached to the 3’ carbon (the critical linkage point for the next nucleotide).

3. Phosphate Group: Attaches to the 5’ carbon. (Note: These three components together form the deoxyribonucleotide monomer).

Compare the sugar and base differences between DNA and RNA

• Sugar: DNA uses Deoxyribose; RNA uses Ribose.

• Bases: DNA uses Thymine (T); RNA replaces it with Uracil (U). Both share Adenine, Guanine, and Cytosine.

Distinguish between Purines and Pyrimidines using a mnemonic or structural prompt

• Purines (Double-ring): Adenine (A) and Guanine (G). Mnemonic: "Pure As Gold."

• Pyrimidines (Single-ring): Cytosine (C), Uracil (U), and Thymine (T). Mnemonic: "CUT the PY (Pie)."

Explain how phosphodiester linkages form the "backbone" and the resulting polarity of the strand

• Phosphodiester bonds are covalent bonds that link the 5’ phosphate of one nucleotide to the 3’ hydroxyl of the next.

• This creates a sugar-phosphate backbone with distinct polarity: one end has an exposed 5’ phosphate, and the other has an exposed 3’ hydroxyl.

Apply Chargaff’s Rule: If a genome is 20% Cytosine, calculate the percentage of Adenine

1. If C = 20%, then G = 20% (Total G+C = 40%).

2. The remaining 60% must be split equally between A and T.

3. Therefore, Adenine = 30%.

Contrast the stability of A-T pairs vs. G-C pairs and predict which would require more energy to separate.

• A-T pairs form 2 hydrogen bonds, while G-C pairs form 3 hydrogen bonds.

• Because of this, G-C pairs are more stable; DNA regions with high G-C content require more thermal energy (higher temperatures) to denature.

Define "antiparallel" geometry and explain the spatial relationship between the two strands

• One strand runs in the 5’ to 3’ direction, while the complementary strand runs 3’ to 5’.

• This "head-to-tail" arrangement allows the nitrogenous bases to face each other in the center to form hydrogen bonds.

Provide the 3’ to 5’ complement for the following sequence: 5’ AGTACTGA 3’

3’ TCATGACT 5’

Define Chromatin and identify its two primary components.

• Chromatin is the material that composes chromosomes.

• Its components are DNA and Histone proteins

Describe the Nucleosome and its role as the fundamental unit of DNA packaging

• A nucleosome is a structural unit consisting of a length of DNA coiled around a core of histone proteins (the "beads on a string" appearance).

• It is the first level of eukaryotic DNA compaction.

Distinguish between Genes and the Genome

• A gene is a specific sequence of DNA that provides instructions for a single protein.

• The genome is the sum total of all genetic information in an organism.

Contrast the physical scale of the human genome with its coding capacity using the "2 meter" and "20 billion kilometer" statistics

• A single human cell contains 2 meters of DNA, housing 20k–30k genes.

• Across the roughly 10 trillion cells in the human body, the total DNA length is over 20 billion kilometers—more than four times the distance from the Sun to Neptune.

The Replication Enzyme Toolkit: Helicase

Unwinds the double helix at the replication fork

The Replication Enzyme Toolkit: Topoisomerase

Relieves torsional strain and prevents supercoiling ahead of the fork.

The Replication Enzyme Toolkit: SSB Proteins

Bind to and stabilize single-stranded DNA to prevent re-annealing.

The Replication Enzyme Toolkit: Primase

Synthesizes a short RNA primer to provide a 3' OH for DNA Polymerase.

The Replication Enzyme Toolkit: DNA Polymerase

Adds dNTPs to the growing strand and proofreads the sequence.

The Replication Enzyme Toolkit: Ligase

"Glues" Okazaki fragments together on the lagging strand

Deduce the necessity of Okazaki fragments based on the 5’ to 3’ synthesis rule

• Because DNA Polymerase can only add nucleotides to a 3' OH (building 5' to 3'), the strand opening in the "wrong" direction relative to the fork cannot be built continuously.

• It must be synthesized in short, backward-moving segments called Okazaki fragments.

Explicitly state the "Directionality Rule": in which direction does DNA Polymerase read the template, and in which direction does it build the new strand?

DNA Polymerase reads the template in the 3’ to 5’ direction and builds the new strand in the 5’ to 3’ direction.

Describe the physical Origin and Fork of replication.

• The Origin is the specific sequence where replication is initiated.

• The Fork is the Y-shaped region where the parent DNA is actively unwound and the new strands are synthesized

 Describe the proofreading mechanism DNA Polymerase uses to correct mismatches

• DNA Polymerase possesses 3’ to 5’ exonuclease activity.

• If it detects an incorrectly paired base, it can reverse, remove the mismatch, and replace it with the correct nucleotide before continuing.

Explain the function of Telomeres and why linear chromosomes shorten over time

• Telomeres are repetitive, non-coding DNA "caps" at the ends of chromosomes.

• Shortening occurs because the RNA primer at the very end of the lagging strand cannot be replaced with DNA, leading to a loss of terminal sequences with each division.

Detail the three steps of PCR and the specific role of temperature in each

1. Denaturation (≈95∘C): Heat breaks hydrogen bonds to separate DNA strands.

2. Annealing (≈55∘C): Cooling allows synthetic primers to bond to the target sequence.

3. Extension (≈72∘C): Polymerase adds dNTPs to the primers to build the new strands.

List the essential "Ingredients" for PCR and explain the necessity of a heat-stable polymerase

• Ingredients: Template DNA, Primers, dNTPs (nucleotides), and Taq Polymerase.

• Necessity: A heat-stable enzyme (like Taq Polymerase) is required because standard human polymerases would denature and become non-functional during the high-heat denaturation step.