Lecture Exam 1

Question 1

Emergent Properties (5)

Levels of Structural Organization:

Answer 1

trait or characteristic possessed by an object due to the arrangement and interaction of its components; property is not found in the components themselves

“eat” Energy utilization
“sleep” Homeostasis
“grow” Growth and development
“repeat” Reproduce (transition of genetic material)
“Evolve”
Possess high level of structural [O]rganization
Life consists of [C]ells

Levels of Structural Organization (big to small): organ systems > organs > tissue > cells > organelles > surpramolecular systems > macromolecules > molecules > atoms

Question 2

Properties of Water (5)

Answer 2

1. Has polar molecules that have a high propensity for forming hydrogen bonds, thus creating a dynamic crystalline lattice
2. High degree of cohesion due to H bonding
3. High specific heat
4. High heat of vaporization
5. Important solvent that can dissolve polar or charged molecules

Question 3

Element Information to Know

Answer 3

92 naturally occurring elements, 25 of which can be found in organisms

4 main players → C, H, O, N ⇒ specifically (by mass) 63% O + 25% C + 9% H + 5% N = 97% total
(( Remaining 3% are the other 21 naturally occurring elements))

Carbon = key player bc can form FOUR covalent bonds and long carbon chains, thus becoming the backbone of biomolecules

Question 4

Functional Groups to Know

• Hydroxyl
• Carbonyl
• Carboxyl
• Carboxylic Acid
• Amino
• Amine
• Sulfhydryl
• Thiol
• Phosphate
• Methyl
• Acetyl

Answer 4

Hydroxyl = R-OH
Carbonyl = R-CO-R
Carboxyl = hydroxyl and carbonyl together → R-CO-OH; also known as “COOH”
Carboxylic acid = molecules w/ carboxyl group
Amino = R-NH2
Amine = compound w/ N, ammonia derivative → eg. N-R3 where R can vary and be different from one another
Sulfhydryl = R-SH
Thiol = compound w/ “sulfhydryl” func group
Phosphate = R-PO4 -2
Methyl = R-CH3
Acetyl = carbonyl + methyl; R-CO-CH3

Question 5

Define the following:

• Isotopes
• Isomers
• Structural Isomer
• Sterioisomer

Answer 5

ISOTOPES: compounds with the same number of protons (thus same chemical identity) but different number of neutrons (thus have different atomic masses); will have keep approximately the same chemical properties

ISOMER: same molecular formula but different structures; result in different properties (eg. melting point)

STRUCTURAL ISOMER: different bonding

STEREOISOMER: retain same boding, diff spatial arrangement → enantiomer (all mirror) and diastereomer (at least one non mirror)

Question 6

Explain how to determine the following:

• Number of Stereoisomers
• D or L conformation
• R or S configuration

Answer 6

How to find number of stereoisomers:
Find asymmetric carbons
Number of stereoisomers: 2n where n is the number of asymmetric carbons
– Number of pairs of enantiomers: (2n) / 2 where n is the number of asymmetric carbons

How to determine D or L conformation?
Look at asymmetric carbon that happens to be farthest from the carbonyl group
– If hydroxyl group is to the LEFT, then will be L → living systems only have L
– If to the RIGHT, then will be D

How to determine R or S configuration?

1. Number the priority groups
2. Look at 1 > 2 > 3
   - - If CLOCKWISE, then S → running out of time [S]ucks on exams
   - - If COUNTER-CLOCKWISE, then R → if only I could [R]ewind the time!

Question 7

Define the following:

• Biomolecules
• Macromolecules
• Disaccharide
• Oligosaccharide
• D-n-Acetylgulcosamine

Answer 7

BIOMOLECULES: carbohydrates, proteins, nucleic acids, lipids → all but lipids are polymers, aka macromolecule with repeating units of monomers

MACROMOLECULES: constructed by forming covalent bonds between building blocks, called linkages which are formed by the removal of H2O (aka dehydration or condensation reaction); can be reversed by hydrolysis (which adds water into the linkage to break it and reform the hydroxyl groups

DISACCHARIDE: 2 monosaccharides joined by covalent bond; used in transport via dehydration reaction

OLIGOSACCHARIDE: contains three to twenty monosacharrides; involved in cell communication and identification; can be found on the surface of the cell membrane (configuration of which results in different blood types)

D-n-acetylglucosamine: located in chitin and made from derived monosaccharides; can be found in skin and cartilage

Question 8

Classification of Monosaccharides (2)

Answer 8

By location of carbonyl: basically, where is the C=O bond?

• • Aldose: on one end of the linear chain → -CHO end
• • Ketone: internal location within the linear chain

By number of carbons: ie. C3 = trioses; C4 = tetroses
– Under physiological conditions, glucose (and other hexoses / pentoses) exist as CLOSED RINGS as equilibrium favors the ring formation more than the linear

Question 9

Functions of Carbohydrates (aka simple sugars) + Formula

Answer 9

monomer as basic building block
Formula: (CH2O)x where x is between 3 and 7 → when numbering, start at the end closest to the carbonyl grou

1. Raw material for biosynthesis + fuel for cellular respiration
2. Monosaccharide derivatives: just replace hydroxyl
   - - Acid sugars: change to carboxyl → C-OOH
   - - Amino sugars: change to amine → C-NH2

Question 10

Types of Polysaccharides (2)

Answer 10

STORAGE polysaccharides: storing carbohydrates / fuel used in respiration; considered as “quick use” → Starch: three sub types

• • Amylose: linear polymer of glucose
• • Amylopectin: branching polymer of glucose
• • Glycogen
• • All three are polymer in glucose ! Have a (loose) spiral form in order to interact with water.

STRUCTURAL polysaccharides are not for quick use (unlike storage); usually form some part of the organism’s structure → Cellulose

Question 11

Alpha and Beta Configuration

• Determination
• Digestion

Answer 11

1. Identify the anomeric carbon.
2. Counting begins from the anomeric carbon if there is not a CHO group attached; otherwise, the anomeric will count as the second carbon while the C in the CHO group will count as the first.
3. Look at the carbon in the ring that is farthest from the anomeric carbon. Determine R or S configuration.
4. IF R CONFIGURATION, return to the anomeric carbon and look at the OH group that is attached. (Invert if S configuration)
   – If below the plane, then ALPHA → alpha looks like a fish and fish live in the sea (below ground)
   – If above the plane, then BETA → beta starts with b and so do birds, live in the sky (above ground)
5. For glycosidic linkages: one of the linkages will come from an anomeric carbon. Use this to orient yourself.
   If two anomeric carbons are involved, start with the one of the lower numbering.
   ** BE AWARE OF FLIPS.

Animals can DIGEST only starch and glycogen BUT not cellulose despite all three being polymers of glucose.

• • Difference lies in the configuration of ring formation (thereby affects the animal’s ability to digest cellulose)
• • Alpha configuration for starch and glycogen
• • Beta configuration for cellulose

Question 12

Define the following:

• ProteinS
• Chaperone proteins
• Conformation

What affects conformation?

Answer 12

PROTEINS: 50% dry weight; amino acids as monomers; function depends on conformation
– Structure: NCC backbone = central carbon atom with four bonds → hydrogen, carboxylic acid (COOH), amino (NH2), and R (side chain)

CHAPERONE PROTEINS: assist in the folding / refolding of a protein into the proper conformation

CONFORMATION: easy to change because of “sliding” motion → contrast to configuration, which is rigid and hard to change

What affects conformation?

• • Aka denaturation, specifically due to changes in temperature or pH
• • Increased temperatures = increased thermal vibration, which can overcome weak H+ bonds → contrast to low temperatures that lower the kinetic energy so much that no collisions (and therefore no reactions) occur
• • Low pH = excess H+ ions, which can disturb the ionic interactions or neutralize the charge on negatively charged R groups

Question 13

Four Types of Protein Structures

Answer 13

PRIMARY: aka native conformation bc determined by DNA; linear sequence of amino acids in the polypeptide chain

SECONDARY: localized areas of folding stabilized by H bonding btwn double bonded O carbonyl (formerly carboxyl end) of one amino acid and the -H on the amino end of another amino acid

• • Alpha helix: most abundant secondary structure in proteins
• • Beta pleated sheet: also localized but does not cover the entire polypeptide

TERTIARY: global folding with interactions between the R groups (ie. hydrogen bonds, ionic bonds, disulfide bridges), thereby generating a three dimensional structure
– Hydrophobic interactions: water-soluble proteins have a hydrophobic core in which side chains are buried from water, which stabilizes the folded state; charged and polar side chains are situated on the solvent-exposed surface where they interact with surrounding water molecules

QUARTERNARY: more than one polypeptide and their interactions

Question 14

Define the following:

• Lipids
• Fats
• Phospholipids

Answer 14

LIPIDS: once again, not a polymer; Hydrophobic; uses a 3 carbon chain called GLYCEROL as its basic building block
– Each carbon has a hydroxyl; empty spaces are hydrogens.

FATS: aka macromolecule; glycerol and 3 fatty acids that are linked by ester linkages (involves dehydration reaction)

• • Maximum number of fatty acids is 3 (triglyceride) but di/mono also exist.
• • Function: long term storage of energy; cushioning / protection; insulation

PHOSPHOLIPIDS: type of fat; in water, forms as micelles, liposomes, and bilayers (spontaenous formation)
– Replace one fatty acid with a phosphate and a polar molecule → hydrophilic head with long hydrophobic tail

Question 15

Fatty Acids

• Structure
• Naming (2)

Answer 15

STRUCTURE: HOOC - (CH2)x - CH3 → carbonyl - variable amounts of CH2 - methyl
Saturated: full amount of H bc no C=C bonds → tend to be solid at room temperature bc pack together easily
Unsaturated: contains C=C bonds (more than one = polyunsaturated); can be cis or trans → cis introduces a kink, making them oils at room temperature

NAMING: ratio of number of total carbons to number of C=C double bonds

• • Biochemist naming: aka alpha naming system; begins at the CARBOXYL END → attaches Δ #s to denote the location of the double bond STARTING FROM THE CARBOXYL END
• • Dietician naming: aka omega naming system; begins at the METHYL END → attaches omega #’s to denote the location of the FIRST double bond STARTING FROM THE CARBOXYL END

Question 16

Define the following:

• Nucleic acids
• Nucleoside
• Polynucleotide
• other nucleic acid / nucleotide derivatives

Answer 16

NUCLEIC ACIDS: nucleotide as monomer → composed of a phosphate (attached to carbon 5), a 5 carbon sugar, and a nitrogen base (attached to carbon 1) that are linked together by phosphodiester linkages!

• • Nitrogen base: organic ring which can be singular as a pyrimidine (for C, T, U) or doubled as a purine (A, G)
• • DNA: Adenine, Cytosine, Guanine, Thymine → no U for uracil
• • RNA: A, C, G, U → no T for thymine

NUCLEOSIDE: nucleotide minus the phosphate group

POLYNUCLEOTIDE: linear polymer whose molecule is composed of many nucleotides units (usually 13 or more); ie. DNA or RNA

OTHER nucleic acids / nucleotide derivatives:

1. Energy carrier as ATP
2. Coenzymes: entities added to enzymes; vital for enzyme function → NOT A PROTEIN!
   - - Most nucleotide-derived coenzymes function in the energy utilization process → ie. Coenzyme A
3. Signalling molecules → ie. cAMP

Question 17

Define the following:

• Cells
• Cell Theory

Compare electron microscopes with light microscopes.

Answer 17

CELLS: basic unit of life; coined by Robert Hooke in 1665 → ideal for a big surface area (production) to volume (consumption) ratio

CELL THEORY: first two are originally from Schleiden and Schwann in 1839

1. All organisms consist of cells
2. Cells arise from other cells by cell division
3. Chemical reactions occur in cells
4. Cells contain hereditary material (for reproduction)

Electron microscopes > light microscopes bc use smaller wavelength, therefore has better resolution BUT can only see structural artifacts and thing examined must be dead

Question 18

What do all cells have? (4)

Compare the two types of cells.

Answer 18

ALL CELLS HAVE:

1. Plasma membrane (phospholipid bilayer)
2. Genetic material (DNA)
3. Cytoplasm
4. Enzymes (inside cytoplasm)

TWO TYPES OF CELLS: eukaryotic versus prokaryotic

• • PRO = archaea / eubacteria domain; DNA is a single circular molecule with few proteins for packaging; No nucleus with few or no membrane bound organelles; Typically small (1-10um)
• • EU = eukarya domain (including protists, fungi, animals, plants); DNA linear chromosomes; Nucleus present with Many membrane bound organelles; Larger → have internal membranes that partition cells into compartments; have unique lipid and protein compositions; created a localized environment for reactions; isolate incompatible reactions; allow for localized or global concentrations

Question 19

Grams

• Staining Process
• Positive versus Negative

Answer 19

Staining Process:

1. Crystal violet staining, followed by light water wash
2. Iodine → binding agent for violet, thus making it harder to remove
3. Ethanol → dissolves LPS but, if overdone, can overpower the iodine and remove the crystal violet
4. Safranin staining (red)

(+) Positive has a thick layer of peptidoglycan, therefore takes up both dyes but primarily shows up as VIOLET → also has 2 basal rings
(-) Negative has a thin layer of peptidoglycan BUT also has a lipopolysaccharide outer membrane, therefore primarily shows up as RED (bc violet-stained LPS dissolved) → also has four basal rings

Question 20

Endomembrane System (6)

Answer 20

use vesicles to transport things from its lumen to whatever it touches
– Overview: Nucleus creates, ER synthesizes; Golgi modifies and ships out to plasma membrane, lysosomes, or vacuoles

1. Plasma membrane: semipermeable bc of phospholipid bilayer; controls traffic in and out of the cell
2. Nucleus: surrounded by nuclear envelope (inner AND outer phospholipid bilayer membranes); contains nucleolus (produces ribosomes for use in protein synthesis) and nucleolar organizer (contains genes to make ribosomes; specific sections on chromosomes where the nucleolus can form)
3. Endoplasmic reticulum: conglomeration of membranous tubes, the lumen of which is called CISTERNAE
   - - Smooth: no ribosomes; synthesis of lipids; carbohydrate metabolism; detoxifies by adding -OH (more polar / water soluble, therefore easier to get rid of); store Ca++ (calcium; used for muscle contraction and neurotransmitter release)
   - - Rough: synthesis of secretory / membrane / endomembrane proteins; membrane production; products are shipped to the golgia via transport vesicles
4. Golgi: modifies, stores, reroutes products of ER; enzymes can alter phospholipids or target products
   - - Cis side = facing towards ER; receiving side
   - - Trans side = facing away from ER; shipping side
5. Lysosomes: 40+ hydrolytic enzymes (lipases, carbohydrases, proteases, nucleases); used for intracellular digestion, recycling / autophagy (recycle old organelles vs. starving cell will break down functional organelles for cellular respiration)
6. Vacuoles:
   - - Food
   - - Contractile: stores and excretes waste and excess water by merging with the cell membrane to dump its contents outside the cell; found in protists
   - - Central: stores various things; acts like a lysosome sometimes if contains enzymes; can trigger apoptosis (programmed cell death) if contains poison

Question 21

Cytoskeleton + 3 Types of Fibers

Answer 21

network of proteinaceous fibers throughout the cytoplasm to provide framework and support movement

1. Microtubules: largest of the three; originate from the microtubule organizing center; primarily for cilia and flagella
   - - hollow fibers made of alpha and beta tubulin
   - - Dynein (-) motor and kinesin (+) motor
2. Microfilaments: smallest of the three
   - - Solid rods made of Actin (+/ - ends)
   - - Myosin (motor protein)
3. Intermediate fibers: made of keratin subunits, therefore more permanent (harder to depolymerize as opposed to dimers or actin

Question 22

Cilia and Flagella (of eukaryotes)

Answer 22

Cilia is shorter than flagella

Both are hair like structures projecting from the plasma membrane that draw fluid across the cell surface, allowing for movement and replacement of stale material

9 outer microtubular DOUBLETS (two microtubules that share a common wall) + 2 central microtubules
– Basal body: 9 peripheral TRIPLETS (three microtubules that share a common wall) with nothing in the middle; acts as an anchor by attaching to the cytoskeleton

Movement is due to the motor proteins dynein and kinesin that allow for “walking”

Question 23

Define the following:
- Cell wall
Extracellular matrix

Answer 23

CELL WALL:

• • Found in plants (cellulose), fungi (chitin, specifically n-acetylglucosamine), bacteria, and most algae
• • Basically a rigid, static structure found outside the plasma membrane
• • Used for protection, shape, and preventing the excess uptake of water via osmosis (thus preventing the cell from bursting)

EXTRACELLULAR MATRIX:

• • (in animals) web of glycoproteins (proteins + carbohydrate), extracellular proteins, and polysaccharides that are associated with the OUTER FACE OF THE MEMBRANE
• • Used for support, adhesion, and cell communication / regulation

Question 24

Fluid Mosaic Model of the Membrane Structure

Answer 24

developed by Sanger and Nicolson in 1972

1. Plasma membranes are NOT static.
2. Protein associations are stable due to hydrophobic and hydrophilic interactions.
3. Fluidity rises from the lateral drift of lipids / proteins moving around; faster in phospholipids than in membrane proteins.
4. Flip flopping (movement on one layer of the phospholipid will FLIP onto the other layer) is rare but does happen; only occurs when there is a charge that allows for movement between the hydrophobic tails of the phospholipid bilayer. Catalyzed by phospholipid translocase (aka “flip-flop”-ase)

Question 25

What are factors that affect fluidity?

Answer 25

1. Presence of double bonds, specifically cis (creates kinks)
2. Length of the fatty acid (short = more fluid bc less contact room for reactions)
3. Presence of Sterols: hydrophobic; used for membrane structure (in animals) and hormones; allows for more fluidity at low temperatures by “pushing” apart the packing of phospholipids, and less at high by reducing phospholipid bilayer movement → Cholesterol (animals), sitosterol (plants), ergosterol (fungi)
4. Cellular control: ability to change lipid composition when temperatures vary in order to maintain stable fluidity

Question 26

Membrane Proteins + 2 Types, defined

Answer 26

Functions: passage across membrane; receptor; enzyme; adhesion; cell communication; energy transformation

Bi-facial, therefore can vary in lipids, proteins, carbohydrates, fluidity → therefore, composition will vary.

2. Integral proteins: strong transmembrane association, therefore can operate on both sides; either anchored in hydrophobic regions of the bilayer
   - - Alpha helices: transmembrane; can be single spanning (pass through the membrane once) or multiple spanning (multiple passings; usually common for proteins to be 6 or 7)
   - - Unilaterals: strong association by being covalently bonded to a phospholipid / glycolipid / etc on ONE SIDE of the membrane
3. Peripheral proteins: weak interactions with integral protein / phospholipid / pr carbohydrate therefore only exists on ONE SIDE of the membrane

Question 27

Movement of membranes via vesicles:

• Principle of topological equivalence
• factors that affect permeability

Answer 27

Principle of topological equivalence: The side of the membrane facing the cytosol has always faced the cytosol.

What are factors that affect permeability?
Size of molecule
Polarity
Charge
Consider: Nonpolar molecules (lipid soluble) and small polar uncharged (water, glycerol) can cross the membrane BUT Large polar uncharged (glucose, sucrose) molecules and ions (Na+, Mg++, Cl-) cannot cross.

Question 28

Types of Membrane Movement (3)

Answer 28

1. Simple diffusion: passive transport (no energy cost) powered by concentration gradient; for small polar molecules are lipid soluble molecules
2. Facilitated diffusion: involve transport proteins (channel and carrier) for ions and large polar molecules
   - - Osmosis: movement of water across a membrane from a area of low [solute] to an area of high [solute] bc high [solute] indicates low [water]
   - - channel proteins: gated hydrophilic tunnel, like aquaporins which allow massive amounts of water transport
   - - carrier proteins (permease): physically carry across membrane using conformational changes; Specific for solute transported because the binding site is specific; can be saturated or inhibited by a similar molecule or dissimilar molecule that alters its conformation
3. Active transport: can move material against the concentration gradient WITH AN ENERGY COST; involves only carrier proteins
   - - ATP driven pumps: phosphate group donated by ATP creates a conformational change; can dephosphorylate to return to original shape; creates a concentration gradient for [solute 1]
   - - Co transport: combines features of facilitated diffusion and ATP pumps to allow downhill diffusion of [solute 1] through the carrier protein WHILE ALSO CARRYING [solute 2] AGAINST its concentration gradient → symporter: carry solutes in same direction (H+ / sucrose) VERSUS antiporter: carry solutes in different directions (Na/K pump)

Question 29

Basal and Apical Sides of an Epithelial Cell (ie. glucose from the small intestine)

• defined
• pathway

Answer 29

APICAL: aka the business side; faces towards the lumen of the small intestine; location of NA+ / Glucose co transporter (can also have GLUT-2 but v rare)

BASAL: faces away from the lumen and towards the blood; location of facilitated carrier (specifically GLUT-2 that can transport glucose into the blood) and NA+ / K+ ATPase

PATHWAY:

1. NA+/K+ ATPase: pays ATP to pump 3 NA+ out and 2 K+ in, creating a higher [NA+] outside the cell and a higher [K+] inside the cell → stroma is “positive” and the lumen is “negative”
2. NA+ / Glucose cotransporter: symporter; takes both into the cell bc [glucose] is lower outside than inside and thus is moving AGAINST its gradient
3. GLUT-2: facilitated transport of glucose from cell into the blood; high to low concentration
4. IF [GLUCOSE] IS LOW INSIDE AND HIGH OUTSIDE: the Na+ / Glucose co-transporter shuts down due to the favorable gradient of glucose into the cell and GLUT-2 can be inserted into the apical side and allow for facilitated diffusion

Question 30

Define the following:

• Bulk flow

- Chemical signalling

Answer 30

BULK FLOW: refers to the movement of large polar molecules

• • Requires energy and movement along the cytoskeleton
• • Endocytosis: movement into the cell → ie. phagocytosis
• • Exocytosis: movement out of the cell → ie. transport vesicle fuses with plasma membrane

CHEMICAL SIGNALLING:

• • Respond to surroundings or other cells
• • Multicellularity: coordination of local activities, homeostatic responses within the tissues and organs)
• • Signaling cell secretes a chemical signal (aka ligand; binds to a site on a target cell); remains OUTSIDE the cell membrane
• • Target cell: equipped with a receptor (usually a protein) to accept the ligand, triggering a change in behavior of the cell

Question 31

Types of Membrane Receptors (4; only need to know 2 specifically tho)

Answer 31

1. Ion channel linked receptors: basically ligand gated ion channel; open or closes based on shape
2. Enzyme linked receptors (not tested)
3. G Protein Linked Receptors: type of membrane receptor that interacts with the GTP binding relay protein → G protein: trimer (3 subunits); binding GDP is inactive; GTP is active
   - - One subunit of activated G protein detaches and is transported linearly along the membrane to interact with the effector enzyme (secondary messenger on the cytoplasmic side of the membrane
   - - G protein can activate adenylyl cyclase (membrane enzyme that converts ATP into cyclic AMP); deactivated by phosphodiesterase
   - - Cyclic AMP can activate certain kinases to begin phosphorylation; kinase can turn on unlimited number of proteins, thus creating a phosphorylation cascade
4. Intracellular receptors (not tested)

Question 32

Metabolic Pathways (2) + Regulation of Enzyme catalyzed Metabolic Pathways (4)

Answer 32

METABOLIC PATHWAYS:

• • Anabolic: “building up”, takes reactant from simple to complex, requires energy, endergonic → photosynthesis; creation of macromolecules to monomers
• • Catabolic: degradative, takes reactant from complex to simple, releases energy, exergonic → cellular respiration

REGULATION OF ENZYME CATALYZED METABOLIC PATHWAYS:

• • Post translational modification: takes place after native conformation is set; made inactive but can be later activated with cofactors
• • Allosteric regulation: inhibitors bind to regulatory sites and lock in INACTIVE conformation; activators lock in ACTIVE
• • Cooperativity: subunits of enzymes with quaternary structures can each have an active site; binding of substrate to these sites can enhance binding at other active sites
• • Feedback inhibition: allosteric inhibition of 1st enzyme in pathway by end product of pathway

Question 33

Energy, defined + 2 Laws

Free Energy:

• Equation
• Spontaneity

Answer 33

ENERGY: capacity to do work → kinetic (motion) and potential (stored)

1. First Law of Thermodynamics: aka conservation of energy; energy can change form but is not destroyed
2. Second Law of Thermodynamics: aka entropy; every energy transformation results in greater disorder in the universe → ie. entropy of system may decrease if entropy of system + surroundings increase
   - - Think - Are you a violation of 2nd law? Nope–the 2nd law only relates to a closed system and the only closed system that exists is the universe. Cells, organisms, etc. are open systems meaning energy can be transferred between system and surroundings.

FREE ENERGY: ΔG = ΔH - TΔS

• • G = Free Energy (available to do work)
• • H = total Energy (enthalpy)
• • S = Entropy
• • T = Temperature in Kelvin (0 K = -273 C)
• • TΔS = isothermally unavailable energy

ΔG > 0 reverse reaction is spontaneous
ΔG = 0 equilibrium
ΔG < 0 forward reaction spontaneous
** For forward reaction to be spontaneous, system must give up energy (exothermic) or order

Question 34

Define the following:

• Exergonic
• Endergonic
• Reaction Kinetics

Answer 34

EXERGONIC: negative ΔG; no input of energy bc releases energy; spontaneous → -ΔG is the maximum amount of work that can be done; products have less free energy

ENDERGONIC: positive ΔG; requires energy to occur; free energy stored in molecules; not spontaneous → +ΔG is the minimum amount of energy needed to drive the reaction; products have more free energy
– NOTE: exo/endo refer to ΔH and are NOT THE SAME as exer/endo

REACTION KINETICS: Molecules must collide at correct angle and with sufficient speed

Question 35

Define the following:

• Biological catalysts
• Cofactors
• Coenzymes

Answer 35

BIOLOGICAL CATALYSTS: used to speed up reaction rate; usually enzymes in small amounts to provide an alternate pathway with a lower activation energy for that specific substrate

COFACTORS: non organic; small non protein molecules added to the enzyme and are required for catalysis; may bind to activate sites or substrate → type of post translational modification

COENZYMES: organic; non protein cofactors that are often derived from vitamins → FAD from riboflavin, NAD from niacin, Coenzyme A from pantothenate

Question 36

Enzymes

• Induced fit model, defined
• Kinetics, defined
• Reaction rates + Km
• Inhibition (2)

Answer 36

INDUCED FIT MODEL: Enzyme changes shape in response to substrate (conformational change–original shape restored at end of reaction) → Substrates usually bind by weak forces, occasionally transient covalent bonds

KINETICS: orient reactants for better interaction; distort / weaken chemical bonds; create a favorable microenvironment; temporarily offer side chains for use

REACTION RATES: determined by the substrate UP TO A POINT if [enzyme] is held constant
– Km = Michaelis constant; concentration at which Vmax is halved; does not increase with enzyme concentration; indication of affinity of the enzyme for the substrate (high Km = weak affinity / attraction)

INHIBITION: irreversible if covalent bonds or altering functional groups; reversible if weak bonds

• • Competitive: mimic substrate and block active site; overcome by adding more substrate
• • Non competitive: bind to another part of the enzyme to induce conformational change; can’t be overcome by adding more substrate because substrate is not accepted by the new conformation

Question 37

What can affect enzyme activity?

Answer 37

TEMPERATURE: bell curve; optimum in middle

• • Lower temperatures do not provide enough energy because the molecules are moving too slowly to react.
• • Higher temperatures can initiate denaturation (change or breaking away from the optimum structure) of enzymes OR moves the particles around too much for proper reaction

pH: also bell curve; optimum in middle

• • Most enzymes and biochemical reactions work best at a pH of 7 (bell curve again)
• • R groups can change with the pH which is BAD because now the experiment is not operating under the best / most stable conditions
• • Extreme pH values can result in enzyme denaturation, but smaller changes may affect the reaction rate

Question 38

Textual breakdown of pathway between photosynthesis and cellular respiration

Answer 38

Energy flow in biological systems originate with light going through photosynthesis (chloroplast) to create organic molecules (chemical energy in the form of carbon) that are then taken up by animals through cellular respiration (mitochondria) to create energy in the form of heat and ATP, the latter of which is used to do work (which can give off heat).

ATP changes into ADP after being used and can be regenerated by attaching a phosphate group to the ADP via a process called substrate-level phosphorylation. → Note: ATP needs to be replenished EQUAL or MORE THAN its usage in order for a cell to function

Waste products: Photosynthesis creates oxygen gas; cellular respiration creates CO2 and water.

Question 39

Goal of Cellular Respiration: words and equation

Answer 39

Regenerate ATP through the complete oxidation of glucose.
Not a single step but rather a series of redox reactions

C6H12O6 + 6 O2 + 6 H2O → 6 CO2 + 12 H2O + Energy (ATP, Heat)
==> ΔGo’ = - 686 kcal / mol

ADP + Pi → ATP where Pi denotes an inorganic phosphate ==> ΔGo’ = + 7.3 kcal / mol

Question 40

Define the following:

• Oxidation
• Reduction
• Reducing agent
• Oxidizing agent
• Special electron acceptors

Answer 40

OXIDATION: loss of electrons or H

REDUCTION: gain of electrons or H

REDUCING AGENT: substance that tends to bring about reduction BY BEING OXIDIZED and LOSING ELECTRONS / H

OXIDIZING AGENT: substance that tends to bring about oxidation BY BEING REDUCED and GAINING ELECTRONS / H

SPECIAL ELECTRON ACCEPTORS: specifically, NICOTINAMIDE ADENINE DINUCLEOTIDE (aka NAD+), act as an intermediate to transfer electrons from the substrate (glucose) to the oxygen (creating water) → H is stripped from the transferred when NAD+ is oxidized; this transfer is catalyzed by a dehydrogenase

Question 41

Glycolysis

• Equation, simplified
• Pathway (2)
• Main takeaway

Answer 41

C6 (glucose) → (partial oxidation) → 2 C3 (pyruvate)

ENERGY INVESTMENT PHASE: cost 2 ATP as energy carrier

1. C6 (glucose) + ATP → C6-P (glucose-6-phosphate) + ADP
2. C6-P (glucose-6-phosphate) + ATP → P-C6-P (fructose 1,6 biphosphate) + ADP
   - - This step is also called phosphofructokinase as it has KINASE as the enzyme.
3. P-C6-P (fructose 1,6 biphosphate) splits into 2 C3-P (glyceraldehyde 3 phosphate)

ENERGY YIELDING PHASE: yield 2 NADH and 2 ATP

1. 2 C3-P (glyceraldehyde 3 phosphate) + 2 NAD+ + 2 Pi → 2 P-C3-P (1,3 biphosphoglycerate)
   - - The creation of NADH from NAD+ is EXERGONIC and fuels the endergonic reaction of adding Pi, creating the phenomenon of energy coupling; this step is also catalyzed by triose phosphate dehydrogenase.
2. 2 P-C3-P (1,3 biphosphoglycerate) + 2 ADP → 2 C3-P (phosphoglycerate) + 2 ATP
   - - The creation of ATP from ADP is EXERGONIC.
3. 2 C3-P (phosphoglycerate) + 2 ADP → 2 C3 (pyruvate) + 2 ATP
   - - This step is also powered by the enzyme KINASE.

MAIN TAKEAWAYS:

• • Cost = - 2 ATP (from the energy investment phase)
• • Gain = 4 ATP + 2 NADH (from the energy yielding phase)
• • Net = 2 ATP + 2 NADH

Note: while glycolysis contains both endergonic and exergonic reactions, it is overall exergonic.

Question 42

Conversion of Pyruvate to Acetyl-CoA:

• Pathway
• Main Takeaways

Answer 42

1. 2 Pyruvate are transported to the mitochondrion.
2. As they passes through the membrane, they are catalyzed by a carboxylase and create 2 CO2
3. They are also catalyzed by pyruvate dehydrogenase, which is involved in H transfer, and regenerate 2 NADH from 2 NAD+
4. 2 Co-A come in and attach themselves to the free carbon, thus creating 2 Acetyl-CoA

MAIN TAKEAWAYS: Gain = 2 NADH

Question 43

Krebs Cycle

• Pathway
• Main takeaways

Answer 43

aka citric acid cycle, aka tricarboxylic acid (TCA) cycle

1. 2 C4 (oxaloacetate) + 2 Acetyl-CoA → 2 C6 (citrate) + 2 Co-A
2. 2 C6 (citrate) + 2 NAD+ → 2 C5 + 2 CO2 + 2 NADH
3. 2 C5 + 2 NAD+ → 2 C4 + 2 CO2 + 2 NADH
4. 2 C5 + 2 Pi + 2 GDP → 2 C4 + 2 GTP
   - - GTP is the energy equivalent of ATP as it can go through substrate phosphorylation to become ATP.
5. 2 C4 + 2 FAD + 2 NAD+ → 2 C4 (oxaloacetate) + 2 FADH2 + 2 NADH
   - - Regenerates oxaloacetate so the cycle can begin again.

MAIN TAKEAWAYS: Gain = +2 ATP +6 NADH +2 FADH2

Note: 2 H2O is fed into the equation to supplement the H+ ions.

Question 44

Electron Transport Chain

• co factors involved
• 4 complexes + bacteria?
• ubiquinone
• complex c oxidase

Answer 44

series of electron carrier molecules in the inner mitochondrial membrane; also present in the thylakoid membrane in between photosystems; electrons passed from NADH and FADH2 to O2 to form H2O (ie. one glucose creates 24 H+, which can be used to create 12 H2O) – organized in a very specific way so that the energy lost during the transfers allow for ATP synthesis to occur

COFACTORS involved: flavoproteins, iron-sulfur proteins, cytochromes

4 PROTEIN COMPLEXES in the mitochondria can be found in the plasma membrane.
– Mitochondrion are related to the ALPHAPROTEOBACTERIA, both of which are GRAM NEGATIVE.

UBIQUINONE: found in the plasma membrane of mitochondria; hydrophobic; shuttles electrons from complex I (which can be bypassed) and II into complex III

COMPLEX C OXIDASE: hydrophilic; takes electrons from complex III and shuttles them into complex IV

Question 45

Oxidative Phosphorylation

Answer 45

proposed in 1960 by Peter Mitchell

Takes place in the inner mitochondrial membrane
ETC generates H+ as the electrons pass through it, thus lowering the pH in the intermembrane space.

The proton gradient favors coming into the mitochondria, which it does through the ATP synthase.
– The chemiosmosis torque of H entering turns the rotor, creating energy that allows for the addition of Pi to ADP to regenerate ATP.

Question 46

MATH:

• Conversions
• Total Yield
• Other cells tho
• 3 Sources of ATP

Answer 46

4 H+ ions are required by ATP synthase to create 1 ATP.
– Best experimental data indicates that 10 H+ ions are transferred per pair of electrons from NADH → 10/4
THUS ⇒ 1 NADH = 2.5 ATP
– Best experimental data indicates that 6 H+ ions are transferred per pair of electrons from FADH2 → 6/4
THUS ⇒ 1 FADH2 = 1.5 ATP

TOTAL YIELD (including oxidative phosphorylation to convert NADH and FADH2 into ATP)
-- Remember summary yield: 4 ATP + 10 NADH + 2 FADH2
10 NADH * 2.5 = 25 ATP
2 FADH2 * 1.5 = 3 ATP
-- Therefore, 4 ATP + 25 ATP + 3 ATP = 32 ATP net (in animal heart, liver, and kidney) thanks to the malate-aspartate shuttle.

OKAY BUT THAT DOESN’T EXPLAIN WHY IN OTHER CELLS, THE TOTAL YIELD IS 30 ATP

• • In other cells, only 30 ATP net because the two NADH are made in glycolysis, thus are in the cytosol.
• • Instead of the malate-aspartate shuttle, all other cells have a glycerophosphate shuttle that uses G3P (glycerol 3 phosphate) dehydrogenase.
• • Rough pathway: electrons from NADH > G3P shuttle > FADH2 > ubiquinone > complex III
• • COST OF TRANSPORT: 1 ATP per NADH → therefore, while the 2 NADH can create 5 ATP via oxidative phosphorylation, other cells must be 2 ATP to the shuttle to get them in.

THREE SOURCES OF ATP: glycolysis, krebs cycle, oxidative phosphorylation
Vast majority of ATP is created from oxidative phosphorylation.

Question 47

Define the following:

• Oxidative phosphorylation
• Substrate level phosphorylation
• Malate-aspartate shuttle
• Glycerol-3-phosphate shuttle

Answer 47

OXIDATIVE PHOSPHORYLATION: production of ATP using the energy released by the redox reactions of the electron transport chain; proton gradient is coupled to the ATP synthesis

SUBSTRATE LEVEL PHOSPHORYLATION: metabolic reaction that results in the formation of ATP or GTP by the direct transfer of a phosphoryl (PO3) group to ADP or GDP; uses the enzyme KINASE

MALATE-ASPARTATE SHUTTLE: allows for transport of NADH (with no cost!) from the cytosol to matrix in animal hearts, liver, and kidneys which have an inner membrane that is not permeable to NADH; allows for the additional creation of 2 ATP in those organs

GLYCEROL 3 PHOSPHATE SHUTTLE: allows NADH created from glycolysis to pass from the cytosol into the mitochondrion to contribute to the oxidative phosphorylation at the cost of 1 ATP per NADH

Question 48

Beta Oxidation of Fatty Acids:

• Process
• Helpful Equations

Answer 48

Uses the organelle peroxisome to break down longer fatty acids into shorter ones, which can move into the mitochondrion freely.

1. Cost 2 ATP to attach a Co-A-S to the fatty acid.
2. Thiolysis cuts off 2 C of the main fatty acid to create Acetyl-S-CoA → each cut creates 1 NADH + 1 FADH2
3. Repeats for free until you only have units of Acetyl-S-CoA → these units can then proceed thru Krebs

Given fatty acid with n carbons (assumed unsaturated)

• • Initial cost of 2 ATP
• • Can create [ (n / 2) - 1 ] amounts of NADH and FADH2 each
• • Can create (n / 2) amounts of Acetyle-S-CoA

Can therefore calculate the amount of ATP generated → REMEMBER TO SUBTRACT 2 ATP bc of cost.

Question 49

Fermentation (2)

Answer 49

basically glycolysis with a few extra steps to regenerate NAD+ → got two types!

• • under anaerobic conditions, glycolysis can continue even when krebs and ETC stop BUT there is a limited amount of NAD+
• • Limited quantities arise because ETC is paused, which is what converts the NAD+ into NADH, and the Krebs cycle, which converts the NADH back into NAD+ for USE BY GLYCOLYSIS, also is paused.

LACTIC ACID: one step

• • Usually found in heavily exercised muscles
• • Takes pyruvate and reduces it into lactate by reducing NADH into NAD+

ALCOHOL: two step

• • Takes pyruvate and decarboxylates it into acetaldehyde
• • Acetaldehyde is then turned into ethanol by reducing NADH into NAD+

Question 50

Anatomy of a Leaf (4)

Answer 50

bc autotrophs need light, H2O, and minerals

UPPER EPIDERMIS: secretes waxy cuticle to prevent water loss and osmotic exchange

MESOPHYLL: photosynthetic cells; contains chloroplasts

• • Palisade: near upper; cuboidal and neatly packed; no room for gas exchange; absorb radiant energy via chlorophylls
• • Spongy: near lower; round with lots of room; allows for gas exchange with CO2

VEIN: carries water, minerals, and other soil-based nutrients

LOWER EPIDERMIS: also has waxy cuticle BUT allows osmotic exchange thru presence of stomata and guard cells (which open and close in response to osmotic pressure)

STOMATA: allows for entry of CO2 and exit of O2

Question 51

Anatomy of a Chloroplast (4)

Answer 51

can have up to 8 mono layers

CELL MEMBRANE: phospholipid bilayer

ENVELOPE MEMBRANE:

• • Outer membrane: phospholipid bilayer
• • Inner membrane: phospholipid bilayer

STROMA: space between inner membrane and the thylakoid lumen; location of calvin cycle

THYLAKOID: phospholipid bilayer as membrane; contains the chlorophyll; location of light reactions

• • Granum: stack of thylakoids
• • Lumen: inside of thylakoid membrane

Question 52

How does photosynthesis happen?

• Photosystem 1
• Photosystem 2
• ETC
• Proton Gradient
• Creating NADPH
• Notes

Answer 52

Let’s start with PHOTOSYSTEM I:
1. Light passes through in the form of radiant energy / photons which are absorbed by pigments (chlorophyll A/B + carotenoids, which all function at different wavelengths) inside the antenna complex.
2. Pigment molecules located close to one another so that light energy is transferred from one to the other → Resonance energy transfer: give to a neighboring chlorophyll molecule
BUT the energy they pass on is always less (shorter wavelengths) because of the decay by successive electron transfers that emit energy and longer wavelengths (closer to red)
3. Eventually, this energy is brought to a special set of chlorophyll A in the reaction center which can absorb P700 wavelengths (red)

Filling the hole with PHOTOSYSTEM II:

1. Hole created when P700 depletes itself of electrons and new ones are provided by Photosystem II.
2. Similar procedure as in PS I but instead uses specialized chlorophyll a molecule of P680, which has an electron acceptor that initiates the ETC from PS I to PS II
   - - Creates a high energy electron

Explaining the ELECTRON TRANSFER CHAIN:

1. Each high energy electron goes through the ETC.
   - - ETC includes the hydrophobic organic compound plastoquinone (Pq) and cytochrome protein complex plastocyanin (Pc).
   - - Both of these are reduced and then oxidized as the electrons pass along this chain.
2. Pq and Pc use the energy from these electrons to transport H+ from the chloroplast stroma to the thylakoid lumen (where [H+] increases and thus, pH decreases) via the B6-F complex. Electron transport and H+ transport are coupled in the ETC.

Creating the PROTON GRADIENT:

1. Water is brought via veins.
2. Cascade of ETC (releases energy)+ enzyme + water = split water into oxygen, hydrogen ions, and electrons
   - - Oxygen considered waste product and goes back into atmosphere.
   - - Electrons (high energy) replenish hole created in PS II.
   - - Hydrogen ions are released into the lumen.
3. The energy provided by the H+ concentration gradient powers the production of ATP (phosphorylation of ADP by ATP synthase in the thylakoid membrane).
   - - Excess hydrogen ions are formed by 2 sources: the energy used in ETC by Pq and Pc, and the splitting of water

Alternatively, creating NADPH:
1. Light energy has been used to boost energy of / excite electron in P700 chlorophyll a dimer. This electron is transferred to electron acceptor in reaction center of PSI.
2. Electrons passed to second ETC (including ferredoxin (Fd)).
Fd passes two e- to NADP+ to form NADPH + H+. This is used in the Calvin Cycle to form sugar. The 2H+ are located in the stroma.

Note:

• • Antenna complex can also refer to chlorophyll proteins.
• • Cyclic: only produces ATP via the proton gradient and B6-F complex; uses only Photosystem I
• • Non cyclic: produces both ATP and NADPH; uses both Photosystem I and II → 8 photons of light are use to produce approximately 3 ATP and 2 NADPH; 4 from each PS.

Question 53

Calvin Cycle:

• Phase I
• Phase II
• Phase III
• Main takeaways

Answer 53

located in stroma; responsible for the creation of RuBisCO via the reduction of CO2 into sugars

PHASE I - Carbon Fixation :

1. 3 Ribulose bisphosphate (RuBP, 5C each)
2. Attach 3 CO2 → Rubisco (6C each)
3. Split to become 6 3-Phosphoglycerate (3C each)

PHASE II - Reduction

1. 6 3-Phosphoglycerate phosphorylated (3C each) by 6 ATP → 6 ADP
2. 6 1,3-bisphosphoglycerate (3C each) reduced by 6 NADPH → 6 NADPH+ + 6 Pi
3. 6 Glyceraldehyde 3-phosphate (3C each)
   - - Output: 1 G3P (3C)

PHASE III - Regeneration of CO2 Acceptor (RuBP)

1. 5 G3P (3C each) phosphorylated by 3 ATP → ADP + Pi
2. Back to 3 RuBP (5C each)

MAIN TAKEAWAYS:

• • Cost = 9 ATP + 6 NADPH
• • Gain = G3P (Glyceraldehyde 3-phosphate) → two cycles yield 2 G3P which can be joined to make glucose
• • THEREFORE, to make glucose, cost 18 ATP + 12 NADPH
• • Remember that 8 photons of light make 3 ATP + 2 NADPH
• • THEREFORE, 48 photons of light are needed to make 1 glucose

Question 54

Relate the following to each other:

• Photorespiration
• C3 Plants
• C4 Plants
• CAM Plants

Answer 54

PHOTORESPIRATION: taking up of oxygen in the presence of light to produce CO2; contrary to photosynthesis – found mostly in C3 plants on hot days

• • High temperatures close the stomata
• • As [CO2] decreases, O2 favored as substrate
• • Chloroplast → C2 (glycolate) → 2 O2 in peroxisome → Mitochondria → CO2
• • CO2 used for photosynthesis when concentrations have returned to favorability.

WHAT ABOUT C4 PLANTS?

• • Focus on spatial differentiation bc of special leaf anatomy
• • Circumvents photorespiration by using PEP carboxylase to fix CO2 into the mesophyll cell, therefore CO2 concentration in bundle sheath cell (location of Calvin cycle) remains high
• • Pathway: CO2 attached to PEP → PEP carboxylase as organic acid → Oxaloacetate (4C) → Malate (4C) → Pyruvate (3C) + CO2 released into bundle sheath → Phosphorylation of pyruvate to regenerate PEP
• • COST – 1 ATP to reconstitute PEP from pyruvate, therefore 6 more ATP needed to make a glucose (24 total because 18 + 6)

WHAT ABOUT CAM PLANTS?

• • Focus on temporal division → stomata closed during day to prevent water loss but are open at night, aka Crassulacean acid metabolism
• • Also contain PEP carboxylase but only happens at night, allowing for accumulation of the organic acid (pH goes down) → CO2 released from organic acid during the day, creating sugars (pH goes up)

Question 55

Cell Division and its Problems

Answer 55

CELL DIVISION: reproduction as an emergent property of life; different for unicellular and multicellular

• • Unicellular organisms: reproduce the ENTIRE ORGANISM → ie. bacteria, yeast, amoebae
• • Multicellular organisms: go through growth and development (look, another emergent property of life), replace dead or damaged cells, and have reproduce via fertilizing an egg that goes through mitotic cell division to produce another multicellular organism.
• • Sexual life cycles of both unicellular and multicellular eukaryotic organisms go through a special type of cell division called MEIOSIS involved in sexual cycles → use gametes (in animals) and spores (in plants).

PROBLEMS:

• • Consider this: there needs to be a faithful transmission of DNA from one generation to the next with minimum errors; the organisms is what it is by virtue of the interactions of its proteins.
• • Okay but DNA can code for many many MANY different genes → eukaryotes (~20k) and prokaryotes (~1k)
• • Brings up the concern about HOW all of those genes can be replicated exactly.

Question 56

Prokaryotes and Binary Fission:

• Definition
• Basic Steps
• Yield

Answer 56

The whole genome happens to be on ONE circular DNA molecule associated with a few proteins BUT it is highly folded to fit inside the cell.

Basic steps:

1. Chromosome is replicated; each copy is attached to the cell membrane
2. Membrane grows and separates the two copies
3. Plasma membrane pinches inwards
4. Cell wall develops between two chromosomes, dividing cell in two

YIELD: two daughter cells that are genetically identical → aka clones

Question 57

Eukaryotic DNA:

• Difference from pro
• Chromosomes
• Uniqueness

Answer 57

Differs from prokaryotes by being LINEAR and not circular → specifically, one linear DNA per chromosome
– Also distributed into 2 or more chromosomes, made of chromatin

Each chromosome consists of:

• • Single, long, double-stranded molecule of DNA
• • Proteins involved in the folding of DNA; make up 50% of the chromosome’s mass → ie. histones, etc.
• • Proteins involved in the expression of genes (considered the functional segments of DNA)

Unique Chromosome count: DOES NOT FLUCTUATE and will exist in different states, depending on the stage of the cell cycle → humans have 46
Note: the number of chromosomes DOES NOT reflect the complexity of an organism → ie. carp has 104 but obv they’re not as smart as us

Question 58

Discuss chromatin packing and chromosome pairs.

Answer 58

CHROMATIN PACKING: several levels; generally need 8 histones to wrap the DNA twice in order to form nucleosomes

• • Histones are used because they are somewhat positively charged and the DNA has a somewhat negative charge (due to its negatively charged phosphate groups).
• • They also have a tail that can tighten the nucleosomes, causing the scaffolding protein to condense even more.
• • looseness of chromatin is proportional to how much of a gene is expressed → more loose, more expression

CHROMOSOMES ARE FOUND IN PAIRS: one from each parent, therefore yielding two copies of every chromosome.
– Humans have 22 pairs of autosomes (paired) and 2 sex chromosomes (not paired) as they are DIPLOID → human haploid set refers to having 22 autosomes and one sex chromosome (NO PAIRS)
– Can name chromosomes based on where their centromere is
A. Metacentric = centromere in middle
B. Sub-metacentric = on the side / slightly skewed
C. Acrocentric = on extremes

Question 59

Interphase

• 3 sub phases
• end result
• note

Answer 59

consists of three subphases; basically all synthesis but different things for different goals; takes a lot of time

1. G1 phase: “main” growth stage for the cell; biosynthesis of cell components
2. S phase: only happens when cell division is for sure gonna happen; synthesis of DNA; chromosomes duplicate; single becomes double; sister chromatids emerge
3. G2 phase: synthesis of components needed for cell division

At the end of the G2 phase, the nucleus should be intact with one or two nucleoli present / visible (not able to see chromosomes yet though). Two centrosomes (found in both plants and animals) can be found adjacent to the nucleus.

Note: in animals, the centrosomes contain a pair of centrioles, which is the location that microtubular asters are formed. Plants have neither of these. → think: centrosomes > animals > centrioles > asters

Question 60

Mitosis:

• some notes
• 5 phases

Answer 60

five phases in continuum with the goal of dividing the nucleus; happens very quickly

Prophase > Prometaphase > Metaphase > Anaphase > Telophase = PPMAT = “pee-pee mat” lol

Note: telophase is usually accompanied by cytokinesis BUT, if not, then yields a multinucleated (2+ nuclei) cell → ie. muscle is a multinucleated cell

1. PROPHASE: nucleoli disappear; chromatin condenses into visible chromosomes; centromeres move apart to opposite sides; mitotic spindle forms
2. PROMETAPHASE: nuclear envelope fragments and “disappears”; spindle fibers extend from the spindle pole towards the cell’s’ equator
3. METAPHASE: chromosomes align on the metaphase plate; centromeres align as well; kinetochores of sister chromatids face opposite poles
4. ANAPHASE: centromeres divide; sister chromatids split apart to become daughter chromosomes, moving towards separate poles; cell begins to elongate
5. TELOPHASE: elongation continues as the daughter chromosomes (there’s 2; diploid) arrive at their respective spindle poles; chromosomes start to relax and become less visible; nuclear envelope redevelops; cytokinesis usually occurs

Question 61

Mechanism by which the chromosome move

also: mechanism by which the cells are officially divided

Answer 61

MOVEMENT: aka the illusion of being “pulled away”

• • Microtubules (referring to spindle) consist of ALPHA and BETA tubulins → can shorten by depolymerizing or lengthen by polymerizing at the kinetochore end (see right)
• • Motor proteins (specifically kinesin) allow separated chromosomes to “walk” towards the opposite poles by pulling the chromosome along as the motor proteins move along the kinetochore.
• • NON kinetochore microtubules overlap in the middle and “walk” by sliding the fibers along each other and are responsible for elongating the cell.

DIVISION:
Animals cells are pinched in two by the cleavage furrow.
Plants have a cell wall which complicates things a bit → golgi derived vesicles collect at the cell center and fuse together to form a cell plate; the new cell wall will develop a the location of the cell plate, yielding two genetically identical daughter cells (once again, called clones).

Question 62

Asexual mitosis versus Sexual reproduction

Answer 62

ASEXUAL:

• • Unicellular: single parent give all genes to offspring, creating a clone
• • Multicellular: can reproduce asexually via fission or fragmentation → ie. starfish
• • Most eukaryotic organisms can do both asexual and sexual reproduction.

SEXUAL: a more in depth look ⇒ sperm + egg → zygote → new individual created via mitosis

• • Two parents donate half their gene each to the offspring
• • Problem: remember that humans have somatic cells that total up to 46 chromosomes → if gametes, had that same number, then the offspring would total to 92 (and that offspring’s offspring would double again!) ⇒ BAD
• • Therefore, undergo meiosis to compensate and also create genetic diversity

Question 63

Animal vs Plant Life Cycle

Answer 63

ANIMAL life cycle: mostly diploid; only haploid in gametes
– Alternation of gametes → 2n multicellular with N gametes
– Meiotic cell division allows germ cells to form gametes.
Egg (n) + sperm (n) = zygote (2n) → fertilization brings back diploid state

PLANT life cycle:

• • Alternation of generations: do not look like each other
• • Sporophyte (2n) produces spores (n) in the sporangium via meiosis.
• • Gametophyte (n) produces gametes in the gametangium via mitosis.

Question 64

Meiosis

• Interphase 1
• PMAT I
• Yield
• Interphase II
• PMAT II

Be sure to note all the things that happen here that DO NOT HAPPEN in mitosis.

Answer 64

Interphase I: chromosome replicates in S phase; centrioles in animals also replicate.

Prophase I: chromosomes condense; centrosomes move apart; tetrads form via synapsis; crossing over at chiasmata occurs; nuclear membrane and nucleoli disappear
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: synapsis and crossing over

Metaphase I: tetrads align on metaphase plate; each homologue is attached to a kinetochore microtubule from the pole it faces
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: kinetochore microtubules are no longer connected to both homologue; instead now only one per pole

Anaphase I: homologues move toward opposite poles by motor proteins; chromosomes are still doubled and the sister chromatids remain attached but are no longer identical
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: sister chromatids are no longer alike due to crossing over

Telophase I: chromosomes arrive at spindle poles; (sometimes) the nuclear membrane will reappear; cytokinesis
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: no reappearance of the nuclear membrane (usually)

At the end of Meiosis I, each daughter cell (there’s 4) is HAPLOID; chromosomes are still duplicated (therefore, still exist in the form of paired chromatids).

Interphase II: short; no DNA replication before Meiosis II
– no S phase as all the DNA was already duplicated in Interphase I.

Meiosis II: just like mitosis

Question 65

Benefits of Meiosis

Answer 65

Meiosis REDUCES the chromosome SET number (and, therefore, the chromosome number).

GENETIC DIVERSITY

• • Independent assortment (occurs in Metaphase 1), yielding unique combinations of chromosomes. There variations are the set up for Darwin’s “survival of the fittest” theory.
• • Number of possible combos = 2n where n is the number of haploids → ie. for humans, n = 23 yields ~8 million
• • Random fertilization in sexual reproduction → Therefore, number of possible combos is now SQUARED !! ~64 million combinations !!!
• • Crossing over produces unique chromosomes that contain the genes from both parents. → There can be multiple crossovers per chromosome → → to infinity and beyond with genetic variation!

Question 66

Blending Theory of Heredity

Answer 66

basically says that all the mixing of genes will result in a homogenous population

Assumes no new input of genetic material, therefore offsprings of subsequent generations will slowly become genetically similar

Once traits are blended, they cannot be separated again → think: mixed paint

Question 67

Why use Pea Plants? (3)

Answer 67

Mendel

Easy to ensure true-breeding (specializes genetic “varieties” producing offspring with the same traits as the parents, generation after generation; yields all or nothing traits)

Self-fertilizing ability allowed Mendel to strictly control the matings to ensure the parentage of offspring (as opposed to natural fertilization taking place by insects)

• • Attained this by cutting off the pollinated stamens (male parts) of the flower, leaving only the carpels (female parts) intact
• • Manually transferred the pollen (via paintbrush) when fertilization was desired

All or nothing traits refers to the lack of gradation /blending present in the phenotypes → ie. either purple or white, no lavender

Question 68

Mendel’s First Cross

Answer 68

P = Purple flowers x White flowers
F1 = all purple
F1 x F1 → self fertilization within F1 population, yields F2
F2 = 705 purple : 224 white
Not a perfect 3:1 ratio but close enough

OBSERVATIONS:

• • No blending BUT white did not disappear as it reappeared in F2
• • Quantification of the 3:1 ratio is approximated via repeated experiments

Question 69

Types of Crossing (3) + How

Answer 69

TEST CROSSING: lets you know what trait is dominant since you’re crossing the partially unknown (which requires at least ONE dominant allele) with a homo-rec (which can only pass on recessive alleles). THEREFORE, the phenotype of the offspring is dependent on the P/? (aka the partially unknown)

• • Ie. If P/P then offspring will be P/p and all purple == OR == If P/p then offspring will be 1:1 ratio of purple to white.
• • Can also use the Punnett square as a better visualization.

DIHYBRID CROSS: lets you hybridize for two traits.
– Consider:
P = Yellow; Round seeds x Green, Wrinkled seeds
F1 = All Yellow; Round
– Therefore, from the F1 generation, you know that Yellow and Round are the dominant alleles and that the offspring of this generation are all heterozygous.
– F2 generation is as follows → 9:3:3:1 ratio BUT still follows 3:1 ratio if genes are considered individually
== 9/16 Yellow; Round
== 1/16 Green; Wrinkled
== 3/16 Yellow; Wrinkled Novel combinations
== 3/16 Green; Round
** first two are the same combination as in the P generation; last two are novel combinations, with one gene dominant and the other recessive

TRIHYBRID CROSS:
Consider this:
– P = Yellow seed ; Round seed ; Long stem x Green seed ; Wrinkled seed ; Short stem
– F1 = Y/y ; R/r ; L/l
– F2 = UUUUUH → actually, don’t freak out. Just do your Punnett squares (separately, then multiply the ratios of what you want together).

Question 70

Mendel’s Conclusion; aka Particulate Theory of Inheritance (5)

Answer 70

1. Alternative versions of genes (alleles) account for variations in inherited characters
2. For each character, an organism inherits two alleles, one from each parent; paired condition (of alleles) are restored by random fusion of gametes at fertilization
3. If two alleles differ, then one (dominant) is fully expressed in the organism’s appearance; the other (recessive) has no noticeable effect on the organism’s appearance
4. Law of Segregation: two alleles for each character segregate during gamete production
   predicts that in a monohybrid (one factor) cross, the genotype of the F2 generation will approximate a 3:1 ratio (dominant to recessive) or a 1:2:1 ratio (homo-dom, hetero, homo-rec)
5. Law of Independent Assortment: each allele pair segregates independently during gamete formation via separate processes

Question 71

Particulate Behavior of Genes

Answer 71

Inheritance is due to discrete factors (genes) that are passed on from generation to generation

Segregation and Assortment are RANDOM events that obey simple laws of probability

If F2 seed is planted, we will not be able to apriori predict the phenotype (basically, can’t theorize / calculate the outcome of the offspring) BUT can state that there will be a ¼ chance that it will have white flowers.
– Among a large sample size, about ¼ or 25% will have white flowers based on the 3:1 ratio. (Because small systems will have microcosms that will deviate the result from the ratio.)

** The larger the sample size, the closer results will fit the prediction.

Question 72

Multiple Alleles + ABO Blood Example

Answer 72

remember that any one individual can only have TWO alleles even if there exist more than two alleles for that gene.

BIG EXAMPLE: ABO Blood Groups arise from a gene that code for enzymes that attach to a particular sugar molecule to the lipid on the surface of a red blood cell. → 3 Alleles: IA and IB (codominant) and iO (recessive to IA and IB)

• • 8 different genotypes with 4 different phenotypes
• • Antigens created match the blood type; antibodies match the opposite of the blood type → ie. type A would have A antigens and Anti-B antibodies; Type O would have no antigens and both Anti-A and Anti-B oxygens
• • Antigens are located on the RBC while antibodies are in the serum → individual produces antibodies against foreign antigen; antibodies react with the foreign antigen by agglutination (or clumping) which make it easier for macrophages to remove them
• • Problematic for blood transfusions because the clumps can clog / rupture capillaries, which can be fatal. – Type AB as universal recipient; type O as universal donor (but can only accept O blood)

Question 73

Define the following:

• Rhesus Factor

- MN Blood Groups

Answer 73

Both are genes related to blood.

RHESUS FACTOR:

• • Refers to a protein found on the surface of RBC; alters a second type of oligosaccharide
• • If present, then positive (dominant); if absent, then negative.
• • More of an issue for pregnancies → ie. Rh- mother may develop antibodies against her Rh+ baby (gained positive allele from father), which can lead to miscarriage or anemia in the child.

MN BLOOD GROUPS:

• • Refers to a glycoprotein found on the surface of RBC as a native antigen; involves a enzyme that alters a third type of oligosaccharide
• • M and N are codominant and act similarly to AB blood interactions.

Question 74

Polygenic:

• Examples
• Multifactorial
• Epistasis

Answer 74

eg. intelligence = genetics + environment == OR == stature (height potential) = genetics + diet

MULTIFACTORIAL: determined by both genes and environment; emphasis on outside forces playing a role in the individual

EPISTASIS with SWEET PEAS as an example
– Defined: One gene can inhibit another when a characteristic requires 2 or more genes since, if one of the genes fails to produce a dominant allele, the pathway is interrupted (as only dominant alleles can call upon the appropriate enzyme to continue the pathway) and the expression is “inhibited”
– P = White x White → true breeding
– F1 = All Purple → hybrids
– F2 = 9/16 Purple : 7/16 White → NOT a 3:1 ratio
== Why not? 2 genes determine flower color, each with a dominant and recessive allele → therefore, 2 alleles exist at the same locus (refers to alleles that belong to the same gene at the same location; very similar to centromeres in chromosomes!)
– To produce flowers, there must be at least ONE DOMINANT allele at BOTH gene loci (doesn’t matter what the second allele of each is, as long as one of them is dominant)

Question 75

Relate the following to each other:

• Dominance
• Recessive
• Incomplete Dominance

Answer 75

Dominance / Recessive pertains only to the expression of alleles in the heterozygous state. → dominant is expressed if present; recessive is only expressed when the dominant is absent (basically: complete dominance in pea plants)

The dominant allele is NOT:
More common or more widespread
Better or allow for greater survivability
Eg. Type O blood is recessive but very common == OR == Polydactyl (having 6 or more digits) is dominant but very rare.

WHAT ABOUT INCOMPLETE DOMINANCE?? emerges with SNAPDRAGONS → intermediate phenotype (halfway between the parents) emerges as PINK, therefore now have three different phenotypes as a result of three different genotypes. Also, new denotation.

Question 76

Equations and Denotations to Know (from “Mendelian Genetics”)

Answer 76

EQUATIONS !!

• • Number of gametes = 2n where n = number of genes being considered
• • Number of boxes in a Punnett square = (number of gametes) ^2

DENOTATION !!!

• • Uppercase = dominant allele
• • Lowercase = recessive allele
• • Semicolon = separation of genes
• • Apostrophe (eg. P’) = intermediate phenotype (from “incomplete dominance” section)
• • [Uppercase] ^letter = dominant with more than one allele possible (from “multiple alleles” section)
• • [lowercase] ^letter = recessive with more than one allele possible (from “multiple alleles” section)