Topics Covered:

Carbohydrates • Lipids • Amino Acids & Proteins • Enzymes

Hemoglobin & Myoglobin • DNA & RNA • Thermodynamics

1. Carbohydrates

1.1 Classification of Monosaccharides

Monosaccharides are classified by the number of carbons and the type of carbonyl group:

TermDefinition
AldoseContains an aldehyde group (CHO) at C1. Examples: glucose, galactose, ribose, mannose
KetoseContains a ketone group at C2. Example: fructose
Triose3 carbons (e.g., glyceraldehyde)
Pentose5 carbons (e.g., ribose, deoxyribose)
Hexose6 carbons (e.g., glucose, galactose, fructose)

📝 EXAM ALERT: Galactose is an ALDOHEXOSE (6C aldose). It is NOT a pentose. Fructose is a KETOHEXOSE.

1.2 D vs. L Configuration

Determined by the position of the -OH on the LAST chiral carbon (highest-numbered asymmetric carbon):

  • D-monosaccharide: -OH is on the RIGHT in Fischer projection
  • L-monosaccharide: -OH is on the LEFT in Fischer projection

📝 EXAM ALERT: Most naturally occurring sugars are D-form. To determine D or L, always look at the carbon furthest from the carbonyl.

1.3 Chiral Carbons (Stereocenters)

A chiral carbon is bonded to 4 different groups. For linear sugars:

SugarNumber of Chiral Carbons
Tetrose (4C)2 chiral carbons
Pentose (5C)3 chiral carbons (e.g., ribose)
Hexose (6C)4 chiral carbons (e.g., glucose, galactose, mannose)

📝 EXAM ALERT: Mannose has 4 stereocenters. A hexose shown in Fischer with CHO at top and CH2OH at bottom has 4 chiral carbons (C2, C3, C4, C5).

1.4 Epimers & Isomers

TermExplanation
EpimersDiffer in configuration at only ONE chiral carbon. Glucose & galactose (differ at C4). Glucose & mannose (differ at C2).
EnantiomersMirror images - one is D, the other L
AnomersDiffer at the anomeric carbon (C1 for aldoses, C2 for ketoses). Alpha vs Beta.
Alpha (α) anomer-OH at anomeric carbon is on the SAME side as the ring oxygen (axial in glucose)
Beta (β) anomer-OH at anomeric carbon is on the OPPOSITE side from ring oxygen (equatorial in glucose)

1.5 Haworth Projections & Ring Forms

Cyclic ring forms (pyranose = 6-membered, furanose = 5-membered). Rules for Haworth projections:

  • Groups on the RIGHT in Fischer → point DOWN in Haworth
  • Groups on the LEFT in Fischer → point UP in Haworth
  • In beta-D-glucose: the -OH at C1 is UP (equatorial)
  • In alpha-D-glucose: the -OH at C1 is DOWN (axial)

📝 EXAM ALERT: A structure with -OH at C1 pointing DOWN is alpha-D-glucose. If all -OH groups match glucose except at C4, it is galactose.

1.6 Reducing Sugars

A reducing sugar has a FREE anomeric carbon (not locked in glycosidic bond) that can be oxidized.

  • Reducing sugars: glucose, galactose, fructose, maltose, lactose, ribose
  • NON-reducing sugars: sucrose (anomeric carbons of BOTH units are locked in the glycosidic bond)

📝 EXAM ALERT: Fructose is reducing (free anomeric C2). Maltose and lactose are reducing (one free anomeric carbon). Sucrose is NOT reducing.

1.7 Important Disaccharides

DisaccharideComponentsNotes
MaltoseGlucose + Glucoseα(1→4) bond; reducing sugar
LactoseGalactose + Glucoseβ(1→4) bond; reducing sugar
SucroseGlucose + Fructoseα(1→2)β bond; NON-reducing sugar
CellobioseGlucose + Glucoseβ(1→4) bond; reducing sugar

1.8 Polysaccharides: Starch vs. Glycogen vs. Cellulose

PolysaccharideStructureFunction/Location
StarchAmylose (α(1→4)) + Amylopectin (α(1→4) with α(1→6) branches every 24-30 residues)Plant energy storage
Glycogenα(1→4) with α(1→6) branches every 8-12 residues (more branched than starch)Animal energy storage - liver & muscles
Celluloseβ(1→4) glucose links; linear, no branchesPlant structural polysaccharide; humans cannot digest it

📝 EXAM ALERT: α-amylase hydrolyzes STARCH (α linkages), NOT cellulose (β linkages). Glycogen is MORE branched than starch. Glycogen is found in liver and muscles.

2. Lipids

2.1 Fatty Acid Classification

Fatty acids are named by carbon number : number of double bonds, e.g., 18:1 Δ9 = oleic acid.

TypeDetails
SaturatedNo double bonds. Straight chains, pack tightly, HIGH melting point. Example: Palmitic (16:0), Stearic (18:0)
MonounsaturatedOne double bond. Example: Oleic acid (18:1, Δ9)
PolyunsaturatedMultiple double bonds. Examples: Linoleic (18:2 Δ9,12), α-Linolenic (18:3 Δ9,12,15), Arachidonic (20:4 Δ5,8,11,14)
Essential fatty acidsCannot be synthesized by humans - must come from diet. Linoleic & α-Linolenic acids. Palmitic is NOT essential.

2.2 Melting Points & Double Bond Geometry

Key rules for melting points of fatty acids:

  • More double bonds = LOWER melting point (more unsaturation = less packing)
  • CIS double bonds cause kinks in the chain, reducing van der Waals interactions
  • TRANS double bonds behave more like saturated bonds - HIGHER melting point than cis
  • Longer chain = HIGHER melting point
Fatty AcidMelting Point
Stearic (18:0)Highest melting point (~70°C)
Oleic (18:1 cis Δ9)~16°C
Elaidic (18:1 trans Δ9)~44°C - higher than oleic (cis)
Linoleic (18:2)~-5°C
α-Linolenic (18:3)~-11°C - LOWEST of common fatty acids

📝 EXAM ALERT: Converting cis oleic acid to TRANS increases the melting point (trans packs more like saturated). Among 18-carbon acids, more double bonds = lower MP. α-Linolenic (18:3) has the LOWEST MP.

2.3 Lipid Classes

Lipid TypeDescription
Triglycerides (fats)Glycerol + 3 fatty acids. Energy storage. NON-polar lipids.
WaxesLong-chain alcohol + fatty acid. NON-polar lipids. Waterproofing.
GlycerophospholipidsGlycerol backbone + 2 fatty acids + phosphate + head group. POLAR lipids. Membrane components.
SphingolipidsSphingosine backbone + 1 fatty acid. POLAR lipids. Membrane components.
SphingomyelinSphingosine + fatty acid + phosphocholine. A sphingophospholipid.

📝 EXAM ALERT: Waxes are NOT polar lipids. Glycerophospholipids and sphingolipids ARE polar lipids. Membrane phospholipids can have either glycerol or sphingosine as backbone.

2.4 Lipoproteins & Cholesterol

LipoproteinKey Feature
ChylomicronsLargest, lowest density. Highest triglyceride content. Transport dietary fat from intestine.
VLDLVery low density. High triglyceride content. Made in liver.
LDLLow density lipoprotein. HIGHEST cholesterol content (~45%). Transports cholesterol TO cells.
HDLHigh density lipoprotein. Transports cholesterol FROM cells back to liver.

📝 EXAM ALERT: LDL has the HIGHEST cholesterol content. HDL = ‘good’ cholesterol. Chylomicrons have the HIGHEST triglyceride content.

2.5 ABO Blood Group Antigens

The ABO blood group antigens are determined by CARBOHYDRATES (oligosaccharides) attached to glycoproteins and glycolipids on red blood cell surfaces. The genetic difference between A, B, and O blood types is which glycosyltransferase enzyme is expressed, which adds different sugar residues.

📝 EXAM ALERT: Blood group antigens are CARBOHYDRATES - not proteins or lipids.

3. Amino Acids & Proteins

3.1 Amino Acid Basics

All 20 amino acids used in protein synthesis are L-type (L-configuration) and alpha-amino acids (amino group on the alpha carbon adjacent to carboxyl group). Beta-amino acids are NOT used in protein synthesis.

📝 EXAM ALERT: Amino acids in proteins are ALPHA (not beta) type, and are L-configuration (not D).

3.2 Classification of Amino Acid Side Chains

CategoryAmino AcidsNotes
Non-polar (hydrophobic)Glycine, Alanine, Valine, Leucine, Isoleucine, Proline, Phenylalanine, Tryptophan, MethionineNo polar/charged groups on R-chain
Polar, not chargedSerine, Threonine, Cysteine, Tyrosine, Asparagine, GlutaminePolar but uncharged at pH 7
Polar, positively charged (+)Lysine, Arginine, HistidineBasic; protonated at physiological pH
Polar, negatively charged (-)Aspartate, GlutamateAcidic; lose proton at physiological pH

📝 EXAM ALERT: Proline and Alanine are NON-POLAR. Cysteine is POLAR but not charged. Aspartate is POLAR CHARGED (negative). Lysine is POLAR CHARGED (positive). Glutamate IS charged at physiological pH (negatively) - a common false statement says it is not.

3.3 Peptide Bond

The peptide bond forms between the carboxyl group of one amino acid and the amino group of another, with the LOSS OF WATER (condensation reaction). Key properties:

  • Has partial double bond character due to resonance - restricts rotation
  • The peptide bond is PLANAR
  • It is responsible for the PRIMARY structure (sequence) NOT the tertiary structure

📝 EXAM ALERT: Peptide bond formation releases water. The bond has partial double-bond character (it is NOT a single bond). Disulfide bridges (tertiary structure) involve cysteine residues.

3.4 Levels of Protein Structure

LevelDescription
Primary (1°)Sequence of amino acids linked by peptide bonds. Determined by gene sequence.
Secondary (2°)Local folding patterns: α-helix and β-pleated sheets. Maintained by HYDROGEN BONDS between backbone atoms.
Tertiary (3°)Overall 3D folding of a single polypeptide chain. Maintained by: H-bonds, ionic interactions, disulfide bridges, hydrophobic interactions, van der Waals forces.
Quaternary (4°)Association of TWO OR MORE polypeptide chains. Only in multi-subunit proteins. Hemoglobin is a classic example (2α + 2β).

📝 EXAM ALERT: Secondary structure → hydrogen bonds. Tertiary structure → disulfide bridges, hydrophobic interactions, etc. A protein with a SINGLE polypeptide CANNOT have quaternary structure. The PRIMARY structure determines the tertiary structure.

3.5 Important Structural Proteins

ProteinKey Features
α-KeratinFibrous protein. Found in hair, nails. Predominant secondary structure: α-HELIX.
CollagenFibrous protein. Found in connective tissue. Triple helix structure. Requires vitamin C for synthesis (hydroxylation of proline/lysine). Deficiency → Scurvy.
ActinGlobular protein. Component of cytoskeleton and muscle.
MyoglobinGlobular protein. Oxygen storage in muscles. Single polypeptide (NO quaternary structure).
HemoglobinGlobular protein. Oxygen transport in blood. Tetrameric (quaternary structure: 2α + 2β subunits).

📝 EXAM ALERT: α-keratin has α-helix secondary structure. Collagen → scurvy via vitamin C deficiency. Keratin and Collagen are FIBROUS. Hemoglobin, Myoglobin, Actin are GLOBULAR.

3.6 Isoelectric Point (pI)

The pI is the pH at which an amino acid carries NO net charge (zwitterion). Calculation: pI = (pKa1 + pKa2) / 2 for the flanking pKa values of the neutral form.

  • For acidic amino acids (Asp, Glu): pI is below 7 (pI of Glutamate = 3.22)
  • For basic amino acids (Lys, Arg, His): pI is above 7
  • For neutral amino acids: pI is around 5-6

Above pI: protein has net negative charge. Below pI: net positive charge.

4. Hemoglobin & Myoglobin

4.1 Myoglobin

Myoglobin (Mb) is an oxygen-STORAGE protein found in muscle cells.

  • Single polypeptide chain (153 aa), tertiary structure only - NO quaternary structure
  • Contains ONE heme group with one Fe2+ ion
  • Hyperbolic oxygen-binding curve
  • Very high affinity for O2 - holds onto O2 rather than releasing it efficiently
  • Cannot function as an oxygen TRANSPORT protein because it cannot efficiently release O2 at physiological O2 pressures (remains >50% saturated even at venous O2 levels)

📝 EXAM ALERT: Myoglobin cannot carry O2 in blood because its O2 dissociation curve is hyperbolic (too high affinity). It stores O2 in muscles; hemoglobin transports it.

4.2 Hemoglobin

Hemoglobin (Hb) is an oxygen-TRANSPORT protein in red blood cells.

  • Tetrameric: 2α + 2β subunits (adult HbA); QUATERNARY structure
  • Contains FOUR heme groups (one per subunit), binds 4 O2 molecules
  • Sigmoid (S-shaped) oxygen-binding curve - due to COOPERATIVITY
  • Sickle cell anemia: Glutamate (Glu) at position 6 of β-chain is replaced by VALINE (Val) → HbS

4.3 Cooperative Oxygen Binding

Hemoglobin shows cooperative (allosteric) binding:

  • When the FIRST O2 binds, it causes a conformational change in the subunit
  • This change is transmitted to other subunits, increasing their affinity for O2
  • The T-state (tense, deoxy) has LOW affinity; the R-state (relaxed, oxy) has HIGH affinity
  • Binding of each O2 shifts Hb toward R-state, making subsequent O2 binding easier
  • This creates the SIGMOIDAL curve - allows efficient loading at high pO2 (lungs) and unloading at low pO2 (tissues)

4.4 Modifiers of Hemoglobin Affinity

ModifierEffect
2,3-BPG (bisphosphoglycerate)Binds in the central cavity at the β-subunit INTERFACE (between the two β chains). Stabilizes T-state, DECREASES O2 affinity, promotes O2 release.
CO2 & H+ (Bohr effect)Decrease O2 affinity; promote O2 release in metabolically active tissues
Fetal Hemoglobin (HbF)Has γ-chains instead of β-chains. Lower affinity for 2,3-BPG → HIGHER O2 affinity than maternal Hb → fetus can extract O2 from mother’s blood

📝 EXAM ALERT: 2,3-BPG binds the β-subunit interface, NOT the heme. Fetal Hb has HIGHER O2 affinity than adult Hb (crucial for placental O2 transfer). Among Hb(O2), Hb(O2)2, Hb(O2)3 - Hb(O2)3 has HIGHEST affinity due to cooperativity.

5. Enzymes

5.1 Enzyme Basics

Enzymes are biological catalysts (proteins, mostly) that:

  • Increase the RATE of biochemical reactions
  • Do NOT change the equilibrium constant (Keq) or the ΔG of the reaction
  • Do NOT get consumed - they are regenerated
  • Lower the activation energy (ΔG‡)

📝 EXAM ALERT: Enzymes do NOT change Keq. They DECREASE activation energy, not overall ΔG of the reaction.

5.2 Michaelis-Menten Kinetics

The Michaelis-Menten equation: v = Vmax[S] / (Km + [S])

ParameterMeaning
VmaxMaximum reaction velocity when enzyme is saturated with substrate
Km (Michaelis constant)Substrate concentration at which v = Vmax/2. It is a measure of enzyme-substrate affinity.
High KmLOW affinity (more substrate needed to reach half-max velocity)
Low KmHIGH affinity (enzyme reaches half-max velocity at low substrate concentration)

📝 EXAM ALERT: Km is NOT the substrate concentration at Vmax. At [S] = Km, v = Vmax/2. Lower Km = higher affinity. Among hexokinase substrates: D-glucose (Km = 0.05 mM) has HIGHEST affinity; D-fructose (Km = 1.5 mM) has LOWEST.

5.3 Enzyme Inhibition

TypeMechanismKinetic Effect
CompetitiveInhibitor resembles substrate; binds to ACTIVE SITE. Km INCREASES; Vmax UNCHANGED. Can be overcome by increasing [S].
Non-competitiveInhibitor binds ALLOSTERIC site (not active site); can bind with or without substrate present. Km UNCHANGED; Vmax DECREASES. Cannot be overcome by increasing [S].
IrreversibleCovalently modifies essential amino acid in enzyme. Permanently inactivates the enzyme. Often called ‘suicide inhibitors’ when they require activation by the enzyme.

📝 EXAM ALERT: Competitive inhibitor → Km increases, Vmax unchanged. Non-competitive → Vmax decreases, Km unchanged. A competitive inhibitor does NOT lower Vmax. Reversible competitive inhibitors can be overcome by excess substrate.

5.4 Allosteric Enzymes

Allosteric enzymes have regulatory sites separate from the active site:

  • Do NOT follow standard Michaelis-Menten kinetics - they show SIGMOIDAL curves
  • Allosteric inhibitors bind the regulatory site and decrease activity
  • Often involved in FEEDBACK REGULATION - product of a pathway inhibits an early enzyme

📝 EXAM ALERT: Allosteric enzymes do NOT follow Michaelis-Menten kinetics. Allosteric inhibitors participate in feedback regulation. They do NOT denature the enzyme.

5.5 Induced-Fit Model

In the induced-fit model of enzyme action, the active site does not have a rigid, pre-formed shape. Instead, when the substrate binds, it INDUCES A CONFORMATIONAL CHANGE in the enzyme’s active site to better accommodate the substrate. This is different from the ‘lock-and-key’ model where the shape is static.

📝 EXAM ALERT: Induced fit = the ACTIVE SITE changes shape upon substrate binding (not the substrate changing shape).

6. DNA, RNA & Replication

6.1 DNA Structure

Form/FeatureDetails
B-form DNAMost common form under physiological conditions. Right-handed double helix. ~10 bp per turn.
A-form DNAShorter, wider right-handed helix. Seen in dehydrated conditions and RNA:DNA hybrids.
Z-form DNALeft-handed helix. Found in GC-rich regions under high salt.
Base methylationCytosine and Adenine are the most commonly methylated bases in DNA (epigenetic regulation).

6.2 DNA Replication Key Concepts

  • DNA replication is SEMICONSERVATIVE - each new double helix contains one old and one new strand
  • Requires RNA PRIMERS (made by PRIMASE) to initiate synthesis - DNA polymerase cannot start de novo
  • DNA synthesis always proceeds 5’ → 3’
  • Leading strand: synthesized CONTINUOUSLY in 5’→3’ direction
  • Lagging strand: synthesized DISCONTINUOUSLY (Okazaki fragments) in 5’→3’ direction
  • Primase makes RNA primers; side products are PYROPHOSPHATE groups (PPi)
  • DNA Polymerase III (E. coli): responsible for the ELONGATION of the new strand

📝 EXAM ALERT: RNA primers ARE required for replication. Leading strand IS synthesized continuously in 5’→3’. Primase side products are PYROPHOSPHATE. DNA Pol III does elongation; Pol I does nick translation/primer removal.

6.3 RNA & Transcription

RNA differs from DNA: uses uracil instead of thymine, ribose instead of deoxyribose, usually single-stranded.

  • RNA backbone (2’-OH on ribose) is MORE susceptible to hydrolysis than DNA - especially in alkaline conditions
  • mRNA maturation (eukaryotes) involves: 5’ capping, 3’ polyadenylation (poly-A tail), splicing (removal of introns)
  • Shine-Dalgarno sequence: Purine-rich sequence in bacterial mRNA that pairs with a purine-rich sequence near the 3’-END of 16S rRNA of the 30S subunit to initiate translation

📝 EXAM ALERT: RNA backbone is subject to hydrolysis in ALKALINE conditions (the 2’-OH acts as nucleophile). Shine-Dalgarno pairs with 3’-end of 16S rRNA (NOT 5’-end).

7. Thermodynamics & Bioenergetics

7.1 Gibbs Free Energy (ΔG)

ΔG = ΔH - TΔS

ConditionSpontaneity
Exergonic (spontaneous)ΔG < 0. Releases free energy. Proceeds spontaneously.
Endergonic (non-spontaneous)ΔG > 0. Requires energy input.
ΔH < 0, ΔS > 0Always spontaneous at any temperature
ΔH > 0, ΔS < 0Always NON-spontaneous at any temperature
ΔH < 0, ΔS < 0Spontaneous at LOW temperature; depends on T
ΔH > 0, ΔS > 0Spontaneous at HIGH temperature; depends on T

📝 EXAM ALERT: ΔH<0 and ΔS<0 → spontaneity DEPENDS ON TEMPERATURE (not always spontaneous). ΔH>0 and ΔS<0 → NEVER spontaneous.

7.2 Identifying Exo/Endergonic Reactions

The MOST EXERGONIC reaction has the MOST NEGATIVE ΔG (largest magnitude negative value). The MOST ENDERGONIC has the MOST POSITIVE ΔG.

ReactionΔG & Classification
Maltose + H2O → 2 glucoseΔG°’ = -15.5 kJ/mol → MOST EXERGONIC
Fructose-6-P → Glucose-6-PΔG°’ = -1.7 kJ/mol → mildly exergonic
Malate → fumarate + H2OΔG°’ = +3.1 kJ/mol → MOST ENDERGONIC

7.3 Coupled Reactions

A coupled reaction uses the energy released from a spontaneous (exergonic) reaction to drive a non-spontaneous (endergonic) reaction. Both reactions occur simultaneously. The classic example is ATP hydrolysis (highly exergonic, ΔG°’ = -30.5 kJ/mol) coupled to biosynthetic reactions.

8. Quick-Reference: Common Exam Traps & Key Answers

True/False Statements to Know

StatementT/FExplanation
Galactose is an aldoseTRUEGalactose has an aldehyde group; it’s an aldohexose
Galactose is an aldoPENTOSEFALSEGalactose is an ALDOHEXOSE (6 carbons)
Amino acids in protein synthesis are beta typeFALSEThey are ALPHA-amino acids, L-configuration
α-amylase hydrolyzes celluloseFALSEα-amylase hydrolyzes α-linkages (starch/glycogen); cellulose has β-linkages
Glycogen is synthesized in liver and musclesTRUEBoth liver and skeletal muscle synthesize glycogen
Palmitic acid is an essential fatty acidFALSEPalmitic (16:0) can be synthesized de novo; linoleic & α-linolenic are essential
Peptide bond formation releases waterTRUEIt is a condensation (dehydration) reaction
Lactose is a reducing sugarTRUELactose has a free anomeric carbon on the glucose unit
Allosteric enzyme follows Michaelis-MentenFALSEAllosteric enzymes show sigmoidal kinetics
Competitive inhibitor lowers VmaxFALSECompetitive inhibitor increases Km; Vmax is UNCHANGED
Glutamate is uncharged at physiological pHFALSEGlutamate is NEGATIVELY CHARGED at pH 7.4
Galactose is an epimer of glucoseTRUEThey differ only at C4
Competitive inhibitor reduces KmFALSEIt INCREASES Km (apparent lower affinity)
KM corresponds to [S] required to reach VmaxFALSEKM = [S] at which v = Vmax/2
Leading strand synthesized continuously in 5’→3’TRUE
Replication of DNA requires RNA primersTRUEPrimase makes RNA primers

Key Multiple Choice Answers

QuestionAnswerNotes
Reducing sugar (of fructose, maltose, sucrose)MaltoseBoth fructose and maltose reduce, but sucrose does not. Fructose is also reducing. In exams listing these three, maltose is the most classic answer.
Fatty acid with LOWEST melting point (18:1, 18:2, 18:3)α-Linolenic (18:3)More double bonds = lower melting point
Lipoprotein with HIGHEST cholesterolLDL~45% cholesterol content
Blood group antigens natureCarbohydratesOligosaccharides on cell surface
Most exergonic of given reactionsMaltose + H2O → 2 glucose (ΔG = -15.5)Most negative ΔG
Vmax/2 occurs at what [S]?At [S] = KmDefinition of Km
Non-competitive inhibitor effectDecreases VmaxKm unchanged
Amino acid replacing Glu in HbSValine (Val)Sickle cell anemia - Glu6Val in β-chain
Allosteric inhibitor roleFeedback regulationDoes NOT denature enzyme
2,3-BPG binds where in Hb?β-subunit interfaceCentral cavity between the 2 β chains
Highest O2 affinity Hb speciesHb(O2)3Cooperativity → each O2 bound increases affinity
Induced-fit: active site does what?Changes shape upon substrate bindingActive site (NOT substrate) changes shape
Most stable DNA form at physiological conditionsB-form
Which is NOT a globular protein?CollagenCollagen is fibrous (triple helix)
Shine-Dalgarno pairs with:3’-end of 16S rRNA30S ribosomal subunit
Which amino acid is aliphatic non-polar?LeucineThreonine is polar; Phe is aromatic

Good luck on your exam! Focus on understanding the mechanisms - not just memorizing answers.