Fragen Wöhnert

• An sp³ carbon atom forms four sigma bonds and has a tetrahedral geometry (bond angles of 109.5°).

• An sp² carbon atom forms three sigma bonds and one pi bond, with a trigonal planar geometry (bond angles of 120°).

• An sp carbon atom forms two sigma bonds and two pi bonds (linear geometry, bond angle 180°)

Atoms in a covalent bond share electrons. The distribution of electrons depends on electronegativity; the electron density shifts toward the more electronegative atom, resulting in a polar covalent bond if electronegativities differ.

An ionic bond is the electrostatic attraction between oppositely charged ions (formed when an electron is transferred from one atom to another).

Include hydrogen bonds, van der Waals forces (dispersion), dipole-dipole interactions, and ionic interactions.

A Brønsted acid donates a proton (H⁺), while a Brønsted base accepts a proton.

Strong acids dissociate completely in water; weak acids only partially. The pH of a 1M strong acid (e.g., HCl) is approximately 0 (pH = -log[H⁺]). The conjugate base of a strong acid is very weak

A buffer resists pH changes upon addition of acids or bases. The effective pH range is near its pKₐ value. Capacity refers to the amount of acid/base the buffer can neutralize.

Cyclic, planar molecules with delocalized pi electrons fulfilling Huckel's rule (4n+2 π electrons, n=integer). Benzene is classic

Contain a C=O group. Aldehydes, ketones, carboxylic acids, and derivatives are types. Carbonyl carbons are electrophilic; they readily react with nucleophiles.

SN1: two-step, carbocation intermediate

SN2: single-step, backside attack. A leaving group departs with electron pair. Reactivity: stabilized carbocation (SN1) or less hindered substrate (SN2)

Delocalization of electrons across adjacent pi bonds or lone pairs creates resonance structures. Resonance stabilization lowers molecular energy and increases stability.

• nonpolar alkyl

• polar uncharged (hydroxyl, amide, thiol)

• acidic (carboxyl)

• basic (amine)

• aromatic (phenyl, indolyl, imidazolyl) groups

DNA: Adenine, Guanine, Cytosine, Thymine.

RNA: Adenine, Guanine, Cytosine, Uracil.

Right-handed double helix, 10.5 bp/turn. Stabilized by H-bonds (base-pairing) and hydrophobic interactions (base stacking)

Sanger (Chain-Termination) Sequencing

• Principle: The Sanger method uses a DNA polymerase to synthesize new DNA strands from a template. The reaction contains both normal deoxynucleotides (dNTPs) and rare chain-terminating dideoxynucleotides (ddNTPs), each labeled with a different fluorescent dye.

• Steps:

  1. One primer binds specifically to the template strand.

  2. DNA is replicated in the presence of dNTPs and a limiting amount of each ddNTP (A, T, C, G).

  3. Whenever a ddNTP is incorporated, the chain stops because ddNTPs lack the 3′-OH group for further extension.

  4. This creates fragments of varying lengths, each ending at a specific base.

  5. Fragments are separated by size using capillary gel electrophoresis.

  6. The attached dye signals the identity of the terminal base for each fragment; the sequence is read out as a colored chromatogram.

• Result: The DNA sequence is reconstructed base by base, reading the colors corresponding to each base.

Next-Generation Sequencing (NGS)

• Principle: NGS uses massively parallel reactions and sequencing by synthesis, permitting millions of DNA fragments to be sequenced simultaneously.

• Main steps:

  1. Library Preparation: DNA is fragmented and adapters are ligated to both ends.

  2. Template Amplification: Each DNA fragment binds to a solid surface or bead and is amplified (e.g., by bridge PCR).

  3. Sequencing by Synthesis: Fluorescently labeled nucleotides (one at a time) are incorporated. High-resolution imaging records which base was added to every fragment in each cycle.

  4. Data Analysis: The resulting huge number of short sequences (“reads”) are aligned and assembled into the full sequence using bioinformatics tools.

• Advantages: Extremely high throughput, can sequence whole genomes rapidly at relatively low cost, detects SNPs, indels, and structural variants.1.-2.-lecture-DNA-Structure-Replication-2025.pdf​

Key Components and Enzymes

• DNA Polymerase: Extends the template strand during synthesis/chain-termination.

• Primers: Short DNA sequences annealing to the template, defining the start point.

• dNTPs: Standard nucleotides for chain elongation.

• ddNTPs: Chain terminators in Sanger sequencing.

• Fluorescent Dyes: Different labels for each base in modern sequencing.

Further Technologies

• Third-generation sequencing (e.g., PacBio, Nanopore): Sequence single DNA molecules in real-time, allowing very long reads and direct detection of modifications.

• Applications: Research, clinical diagnostics, evolutionary studies, forensic analysis, and more.

DNA sequencing has evolved from labor-intensive, low-throughput methods (Sanger) to massively parallel, automated systems (NGS and single-molecule sequencing), dramatically increasing speed, accuracy, and scale.1.-2.-lecture-DNA-Structure-Replication-2025.pdf​

The emission of light by a substance that has absorbed light or other electromagnetic radiation; emission is at a longer wavelength than excitation (lose some energy as heat).

• The standard reduction potential of a compound describes its tendency to gain electrons (i.e., to be reduced) under standard conditions.

• It plays a key role in redox reactions, determining which direction electrons will flow between two substances. Compounds with higher reduction potentials will accept electrons from those with lower ones, driving the reaction forward in that direction

• The 2nd law states that in any spontaneous process, the total entropy (disorder) of the universe increases.

• For a chemical reaction, this means reactions favor directions that increase overall entropy; systems naturally evolve toward thermodynamically more probable (more disordered) states.

• The change in Gibbs free energy (ΔGΔG) incorporates enthalpy and entropy and must be negative for a process to be spontaneous.

Protein Structure Levels

• Primary structure: Linear sequence of amino acids (polypeptide chain).

• Secondary structure: Local folding patterns such as α-helix and β-sheet stabilized by backbone hydrogen bonds.

• Tertiary structure: Three-dimensional folding of a single polypeptide, involving side chain interactions.

• Quaternary structure: Association of multiple polypeptide chains (subunits) into a functional protein complex.​

• Peptide bonds are covalent links between amino acids, formed by condensation between the carboxyl group of one and the amino group of another.

• They exhibit partial double-bond character (due to resonance), making them planar and rigid, limiting rotation and stabilizing protein structure

• Proteins contain α-helices and β-sheets as major secondary structures, stabilized by hydrogen bonds between backbone amide H and carbonyl O atoms.

• Additional elements include turns and random coils—these help connect α-helices and β-sheets and allow compact protein folding

• Allostery describes the regulation of a protein’s function through binding of an effector molecule at a specific site distinct from the active site, inducing a conformational change.

• This mechanism allows regulation of enzyme activity or protein function, often essential for feedback and signal integration in biological systems.​

• PTMs are chemical modifications to proteins after synthesis, such as phosphorylation, methylation, acetylation, glycosylation, and ubiquitination.

• These modifications regulate protein activity, localization, stability, interactions, and cellular signaling, greatly expanding protein functional diversity.​

• A coenzyme is an organic, non-protein molecule that binds to enzymes and assists in enzyme-catalyzed reactions, often serving as carriers of electrons or functional groups.

• Examples include NAD⁺ (nicotinamide adenine dinucleotide), FAD (flavin adenine dinucleotide), Coenzyme A, biotin, and TPP (thiamine pyrophosphate).

Given:

• Volume after purification: 10 ml

• Dilution: 0.1 ml protein + 0.9 ml water (dilution factor = 10)

• Absorbance (A) at 280 nm: 0.3

• Path length (l): 1 cm

• Absorption coefficient (εε): 15,000 mol−1−1cm−1−1

• Molecular weight: 20 kDa (20,000 g/mol)

A. Find concentration after dilution (cc):

A=ε⋅c⋅l  ⟹  c=Aε⋅l=0.315,000⋅1=2×10−5 mol/lA=ε⋅c⋅l⟹c=ε⋅lA=15,000⋅10.3=2×10−5 mol/l

B. Original solution concentration (before dilution, factor 10):

corig=10×2×10−5=2×10−4 mol/l

C. Total amount in 10 ml:

n=corig×V=2×10−4 mol/l×0.01 l=2×10−6 moln=corig×V=2×10−4 mol/l×0.01 l=2×10−6 mol

D. Convert to mass with MW:

m=n×MW=2×10−6 mol×20,000 g/mol=0.04 g=40 mgm=n×MW=2×10−6 mol×20,000 g/mol=0.04 g=40 mg

So: Protein concentration in original = 2×10−42×10−4 mol/l; total mass in 10 ml = 40 mg.​

Given:

• Enzyme mass: 100 mg (0.10.1 g)

• MW: 20 kDa = 20,000 g/mol

A. Moles:

n=0.1 g20,000 g/mol=5×10−6 moln=20,000 g/mol0.1 g=5×10−6 mol

B. Convert to mmol, μμmol, nmol:

• 5×10−35×10−3 mmol = 55 µmol = 5×1035×103 nmol

C. Molecules using Avogadro’s constant (NA=6.022×1023NA=6.022×1023):

Number=n×NA=5×10−6×6.022×1023≈3.01×1018Number=n×NA=5×10−6×6.022×1023≈3.01×1018

D. If dissolved in 10 ml (0.010.01 l):

• Molar concentration: c=5×10−60.01=5×10−4c=0.015×10−6=5×10−4 mol/l

• Mass concentration: cmg=100 mg10 ml=10 g/lcmg=10 ml100 mg=10 g/l

So: 5×10−65×10−6 mol, 5×10−35×10−3 mmol, 55 µmol, 5×1035×103 nmol, Avogadro’s constant for molecule number, concentration in 10 ml = 5×10−45×10−4 mol/l or 10 g/l.​

Protein sphere radius: 25 Å = 2.5×10−92.5×10−9 m Cube edge: 0.1 mm = 1×10−41×10−4 m

A. Protein sphere volume:

V=43πr3≈43×3.14×(2.5×10−9)3≈6.54×10−26 m3V=34πr3≈34×3.14×(2.5×10−9)3≈6.54×10−26 m3

B. Crystal volume:

Vcube=(1×10−4)3=1×10−12 m3Vcube=(1×10−4)3=1×10−12 m3

C. Number of molecules:

Number=VcubeVprotein≈1×10−126.54×10−26≈1.53×1013Number=VproteinVcube≈6.54×10−261×10−12≈1.53×1013

Convert to mol:

n=1.53×10136×1023≈2.55×10−11 moln=6×10231.53×1013≈2.55×10−11 mol

D. Dissolved in 0.001 ml (1×10−61×10−6 l):

c=2.55×10−111×10−6=2.55×10−5 mol/lc=1×10−62.55×10−11=2.55×10−5 mol/l

E. Mass for 10 kDa molecule:

m=n×MW=2.55×10−11×10,000=2.55×10−7 g=255 ngm=n×MW=2.55×10−11×10,000=2.55×10−7 g=255 ng

So: Number of molecules ≈ 1.5×10131.5×1013; amount ≈ 2.6×10−112.6×10−11 mol; concentration ≈ 2.6×10−52.6×10−5 mol/l; mass ≈ 255 ng if MW = 10 kDa.​