BIO OUTLINE Essay Research Paper BIOLOGY 220

BIO OUTLINE Essay, Research Paper BIOLOGY 220 OUTLINE SECTION II Text: Essential Cell Biology I. Opening Comments (Chapter 3) A. Life creates order out of disorder through a never-ending series of chemical reactions

BIO OUTLINE Essay, Research Paper


Text: Essential Cell Biology

I. Opening Comments (Chapter 3)

A. Life creates order out of disorder through a never-ending series of chemical reactions

B. This is Metabolism and the ability to Metabolize

C. Most of the chemical reactions required by the cell would not occur at physiological conditions

D. Control of these reactions is achieved by specialized protein, ENZYMES.

II. Basic Principles of Energy

A. Energy – Basics Principles

1. Define Energy – ability to do work

2. Define Work – the ability to change the way matter is arranged

3. Define Kinetic Energy

4. Define Potential energy – energy of position

5. FIRST LAW of THERMODYNAMICS ?Energy can be transferred or transformed by

never created or destroyed.

6. Explain transferred or transformed ?Different kinds of energy

a. Radiant (solar)

b. Chemical (e.g. gasoline, carbohydrates, fats)

c. Mechanical (involves movement)

d. Atomic.

7. SECOND LAW of THERMODYNAMICS – In any energy transformation or transfer some energy is lost

to the surrounding environment as heat.

a. Define Entropy



B. The Concept of Free Energy

1. Free energy – the portion of a systems energy that can perform work given constant T throughout

system (e.g., living cell)

2. Total free energy of a system (G) is define by this equation G = H – TS

a. H = total energy of system = ENTHALPY

b. T = absolute temp in K (KELVINS) (? C + 273)

c. S = entropy

d. Note that T increases value of S since as Heat increases, molecular motion increases, and

disorder increases.

3. Spontaneous Processes

a.Definition – occur w/o outside help (energy) – energy of system is sufficient to carry out

reaction or process

b.Is not concerned with rate or time, so spontaneous processes will not necessarily occur

in a useful time frame

4. Determining when a system can undergo spontaneous change

a. Stability

b. The change in Free Energy is negative for spontaneous systems .

G = Gfinal state – Ginitial state or .DG = DH – TDS

III. Basics of Chemical Reactions

A.All reactions require an input of energy to get them started


a. Define Activation Energy with overhead

b. For some reactions the activation energy can be provided by the reacting molecules themselves.

c. For others, the activation is very high since the reacting molecules must be brought together in exactly the right orientation in order for the reaction to take place (?effective collision?)..

B.Enzymes reduce activation energy (Chap5. p. 167-69)

1. Define Catalyst

2. Define Substrate

3. Random interactions lead to Enzyme-Substrate Complex formation (effective collision)

4. Enzymes reduce activation energy by

a. Increasing the number of effective collisions between substrates

5. Enzymes are proteins

a. review structure of proteins.

6. Define Active Site

a. Active Site can function by

(1) shape similarities

(2) chemical attraction

(3) both

b. Example: Ribonuclease

c. Review steps of RNAse active site

d.Another example: Lysozyme: pg.170 Figure 5-28

7. Discuss how enzymes are named

a. See Table 5-2 p.169 for list of common enzyme group names and functions..

IV. Factors effecting Reactions (in general, including enzyme-mediated)(Back to Chapter 3)

A. Free energy considerations (as discussed earlier)

1. Free energy change must be negative

B. Concentration of the molecules in the system also determines whether a reaction will occur.

1. As the concentration of one molecule increases the reaction will move toward the production

of the other molecule (Le Chatlier’s Principle).

C. BIG QUESTION – how much of a concentration difference is required to overcome a .G that

might be unfavorable.

1. Rewrite .G to reflect concentration component

2. .G = .G o + 0.616ln[B]/[A]

a. 0.616 is a constant

b. .G o is the Standard Free Energy change (1M @ pH=7) in kcal/mole

c.@37 o C d. Note that when [A] = [B], concentration effects are negated and .G=.G o (ln 1 = 0). D.

For a reversible reaction A ??B (see Figure. 3-20 p.92)

1. One direction is energetically favored (-.G) over the other

2. For example A to B is favored

3. As A converts to B, the concentration effect of greater amounts of B begins to overcome the +??????? G (for B A), to a point where B ??A is equal to A ??B.

4. In Table 3-1 some calculations were done to determine when .G=0 (equilibrium),

that is when .G o = -0.616ln[B]/[A] (con’t on next page).

5. It is important to note that it requires significant excess of the favored product (B) to push the reaction back to unfavored product (A).

6. Enzymes do not change the equilibrium point..

V.Factors Affecting Enzyme-Mediated Reactions

A. Physical Parameters affecting Enzyme Activity (use graphs)

1. Temperature

2. pH

B. Concentration effects

1. Unlimited substrate in the presence of limited enzyme

a. Saturation kinetics

b. where did we see this before -answer: membrane transporters

2. Unlimited enzyme in the presence of unlimited substrate.

VI. Regulating Enzyme Reactions

A. Competitive inhibition

1. Reaction rate is [substrate] dependent

B. Non-competitive inhibition

1. reaction rate is [substrate concentration] independent

2. Inhibitor binds at a site other than active site

3. causes conformational change in enzyme – makes active site unavailable

C. Allosteric Control

1. allo = other steric = structure or state

2. Like noncompetitive – Control Molecule

binds at alternate site

3. alternate site = allosteric site

4. Control Molecule called a REGULATORY SUBSTANCE

a. may increase or decrease activity.

5. Allosteric enzymes exist in 2 different states

a. R(elaxed) state = high affinity for substrate

b. T(ense) state = low affinity for substrate

6. Binding of regulatory substance can induce either state.

a. Allosteric Inhibitor – binding causes T state

b. Allosteric activator – binding causes R state.

7. Allosteric enzymes and Reaction rate a. Regulatory substances may have multiple binding sites. Leads to sigmoidal graph of reaction rate

b. For T to R state…enzyme activity is low until sufficient regulator binds to convert enzyme completely to R state

c. For R to T state…enzyme activity is high until sufficient regulator binds to convert enzyme completely to T state

d. Regulator may be substrate or product.

D. Allosteric Feedback Inhibition

1. end product acts as regulator of 1st enzyme in pathway

2. Discuss Threonine to Isoleucine pathway

a. enzyme #1 = threonine deaminase.

E. Regulation by Covalent Modification

1. additions may include

a. Ca


b. PO4 – phosphorylation

(1) Added by protein kinases

(2) Removed by protein phosphotases

c. CH3 – methylation

d. COCH3 – acetylation

3. binding can up or down regulate enzyme.

F. GTP-binding Proteins

1. Binding of GTP or GDP can cause major conformational changes

2. Phosphorylation of bonded GDP and Dephosphorylation of bonded GTP can also cause changes

3. Mode of action

a. Exchange of GTP and GDP

b. Dephosphorylation of bound GTP

4. Exchange and phosphorylation can have different rates

a. Control achieved by different rates for different reactions

b. See Figure 5-37 pg. 176.

G. Ribozymes

1. RNA based catalysts

2. Self splicing RNA molecules

c. also show activity with some proteins

(1) removal of proteins from ribosomes

(2) separation of amino acids from tRNAs.

H. Coenzymes

1. vitamins

2. minerals

3. Carriers

a. Discuss coupled reaction diagrams

b. Electron Carriers

(1) NAD (Figure 13-8) , & NADP

(2) FAD (Figure 4-12)

(3) oxidized and reduced forms

(4) show chemistry

(5) Dehydrogenase ? oxidizing enzymes

(6) Reductase – reducing enzymes

c. Function as cofactors in redox reactions

d. required by enzymes that are involved in oxidations or reductions ?electron donors or receivers.

I. ATP – universal energy currency of the cell

1. Describe molecular structure

a. nucleoside triphosphate

2. Describe cycle ATP ??ADP + P

a. .G 0 = -7.3 kcal/mole

b. Phosphorylation and its relationship to Redox

3. Energy required to make ATP or Energy released from ATP hydrolysis depends on .G 0 and the relative concentrations in the cell

a. For some cells the ATP/ADP ratio approaches 1000

b. Under these conditions, the .G for the hydrolysis of ATP to ADP can approach 11-13 kcal/mole (remember G equation includes a concentration factor).

J. Coupling Reactions to the Hydrolysis of ATP

1. The hydrolysis of ATP can be linked to reactions with +??????? G o Overall reaction: Glu +NH3 ??Gln .G 0 = +3.4kcal/mole

Step 1: Glu + ATP ??Glu-P + ADP .G 0 = -7.3kcal/mole

Step 2: Glu-P + NH3 ??Gln .G 0 = +3.4kcal/mole NET .G 0 = – 3.9 kcal/mole.

2. Can also be coupled to Dehydration reactions or almost any synthesis reaction that has a +??????? G 0

3. If the desired product has a .G 0 * +7.3 kcal then the reaction is broken down into steps..

K. ATP Production (Some coverage in Chap 13 p.409 – 410)

1. Substrate-level phosphorylation

a. Direct enzymatic transfer of phosphate group & energy to ADP from a high energy substrate

b. low efficiency.

2. Chemiosmotic Phosphorylation ?MITCHELL THEORY

a. Transfer indirectly through proton gradient

(1) electrochemical gradient

(2) stored charge = ENERGY

(3) high efficiency achieved through step-wise transfer = ELECTRON TRANSPORT CHAIN

b. 3 requirements for Chemiosimosis

(1) Selectively Permeable membrane

(2) H+ pumping Enzymes (Active Transport)

(3) ATP Synthase c.Introduce ATP synthase – enzyme that captures energy from proton gradient and transfers it to ATP production – Figure 13-3 & 13-13.

c. Discuss charge separation and release of energy

(1) separate charges across insulator – Battery Analogy: Figure 13-11

(2) Create a charge gradient across an insulator


(4) Release Energy by allowing gradient to dissipate

(5) In living cells, charge separation achieved with different ion concentrations across membranes (ION GRADIENTS)

(6) ex. H+ gradient

(7) Figure 13-15.

3. Jagendorf experiments -a. Knew of existence of pH differences within chloroplast

b. Review experiment with overhead

c. Experiment shows connection between H+ gradient, H+ flow, and phosphorylation of ADP to ATP.

4. Latest information on ATP Synthase

a. Still unknowns

(1) how it works so fast

(2) how it couples proton flow to ATP production

b. Background information on structure

(1) Figure 13-14

(2) 3 parts

(3) F0 – subunit – channel for protons

(4) F1 – subunit – catalytic subunit ?ATP production

(a) called ATPase

(b) uses ATP to pump protons

(5) Stalk – connects F0 to F1.

c. Latest info on structure of F1

(1) made of 6 subunits + ? subunit

(a) 3 ? subunits

(b) 3 ? subunits

(c) arranged in alternating fashion

(d) ? subunit contains catalytic activity

(e) ? subunit extends into stalk region

i) knife shaped protein.

d. Current theory of how it works

(1) H+ moving through F0 causes ? subunit to rotate

(2) knife edge contacts ? subunit

(3) ? subunit is deformed allowing for ADP & P binding,separately, to active site

(4) ? subunit releases contact

(5) ? subunit reformation brings ADP + P in contact and reaction takes place.

VII. PHOTOSYNTHESIS (Chapter 13 – p.430- 438)

A. Overview – Use ?Energy Flow through Living Systems? OH to put things in perspective.

B. Overall reaction

1. Radiant Energy + H2 O + CO2 ??O2 + Glucose

2. Balanced equation:

a. 6CO2 + 6H2 O + Radiant Energy ?? C6 H12 O6 + 6O2

3. Leaf structure

a. Epidermis

b. Spongy and Palisade Mesophyll

(1) Where photosynthesis takes place

c. Stomates.

4. Chloroplast structure ? EMPHASIZE ORGANIZATION

a. Outer membrane

b. Inner membrane

c. Stroma – Glucose production enzymes

d. Thylakoid membrane

(1) Light absorbing molecules or Photosystems

(2) Electron transport chain/Proton pumps

(3) NADP reductase

(4) ATP synthase

e. Thylakoid space or lumen ? proton reservoir.

C. Photosynthesis as a REDOX 2-step



a. Capture energy in the form of ATP and NADPH

b. Use electrons from oxidation of water


a. Take energy from ATP and NADPH and use it to reduce carbon dioxide.

D. Light Dependent Reactions

1. Occur on thylakoid membrane

2. Discuss electromagnetic spectrum

a. Gamma, X, UV, Visible (380nm-750nm), IR, Micro, Radio

b. Violet, Indigo, Blue, Green, Yellow, Orange, Red

c. High energy, Short wavelength, High freq — Low, Long, Low.

3. Pigments

a. Imbedded in the thylakoid membrane

b. All have hydrophobic tails anchored in thylakoid membrane

c. Chlorophylls (a & b) ? Mg 2+ Center ?Figure 13-30

d. Carotenes – pure hydrocarbons ?aromatics rings linked by polyunsat?d chain

e. Xanthophylls – as above w/ alcohols on rings

f. Draw absorption spectrum on board.

4. Light Energy Absorption

a. achieved by pigment molecules

b. Excitation event & excitation energy (see Figure 13-32)

(1)electrons at ground state (inner orbital) ?excited? by photon of light to

excited state (outer orbital)

(2) can remain in this state for only a billionth of a second

c. Possible fates of excitation energy

(1) If excited electron returns to ground state ? FLUORESCENCE + heat

(2) Excited electron is picked up by stable acceptor molecules. excited electron is transferred to stable orbital of the same energy level (Figure 13-32)

(a) Light energy converted to chemical energy.

(3) Inductive transference

(a) energy is transferred to adjacent pigment molecule

i) vibration of excited e – sets up electromagnetic field.

ii) adjacent e – in equivalent orbitals begin to vibrate in resonance

iii) energy is transferred.

(b) little or no loss of energy.

5. Capturing and Converting Light Energy to Chemical Energy

a. Photosystems (see Figure 13-31 for general diagram)

(1) Photosystem I

(a) 110 chlorophyll a + 16 – bcarotenes = CORE ANTENNA

(b) Reaction Center = special chlorophyll a molecule: P700 (700 refers to light absorbing properties @ 700nm)

(c) Energy is absorbed by Core Antenna and passed by inductive resonance to P700 then to 1? acceptor.

(2) Photosystem II

(a) 40 Chlorophyll a + ** b carotenes = CORE ANTENNA

(b) Reaction Center = special chlorophyll a molecule: P680

(3) Energy is absorbed by Core Antenna and passed by inductive resonance to P680 then to 1? acceptor

b. Light Harvesting Complexes (Not shown in Alberts diagrams)

(1) One associated with each PS

(2) Designated LHC I and LHC II

(3) Collections of pigment molecules imbedded in thylakoid membrane

(4) In close physical proximity if not physically attached to PS?s

(5) Funnel excitation energy to reaction centers via inductive resonance.


6. Non-Cyclic electron Photophosphorylation (Figure13-34 = Z Scheme with electron volt ratings)

a. The players

(1) Mn -center – Water Oxidizing Enzyme

(2) LHC II & PS II

(3) Plastoquinone ? e- carrier ?aromatic ring w/ long chair hydrocarbon ?not attached to PS II

(4) Cytochrome b6 – e- carrier ?Heme (Fe) containing protein Fe2+ ? Fe 3+

(5) Cytochrome f.


(a) b6+f complex = H+ pump

(b) Sets up H+ gradient between Stroma and Thylakoid lumen

(c) H+ pumped from stroma into lumen

(d) Flow out through CF0CF1 ? ATPase imbedded in thylakoid lumen.

(e) Make ATP in Stroma

(7) PC = e- carrier – plastocyanin ?

(a) Cu containing protein: Cu+ ? Cu2+

(8) electrons are passed to PS I

(9) Ferrodoxin – Fe/S center ?mobile – not attached to PS I

(10) NADP+ Reductase ? use electrons to reduce NADP+ to NADPH.

7. Cyclic Photophosphorylation

a. electrons pass from Ferrodoxin to cytochrome b6

b. only produces ATP

c. may be used to produce the additional ATP needed to drive glucose production (~3:2 ATP:NADPH).


1. Occur in stroma

2. Use ATP and NADPH to reduce CO2 to Glucose

3. Review process using an overhead

4. Points to Stress

a. Reducing enzyme = Ribulose bisphosphate carboxylase (rubisco)

b. Ribulose bisphosphate = RuBP

c. PGA = Phosphoglycerate

d. PGAL = Phosphoglyceraldehyde

e. Each turn of the cycle…REFER TO HANDOUT.

f. 1 molecule of glucose requires

(1) 18 ATP ? 7.3 kcal/mole x 18 = 131.4 kcal

(2) 12 NADPH ? 53 kcal/mole x 12 = 636 kcal

(a) Note 53 kcal/mole – ref: Campbell pg. 178 for NADH to O2 ?H2 O

(3) Takes 767.4 kcal to make 1 molecule of glucose (686 kcal)

(a) 686/767.4 = 89% efficiency.


1. Rubisco prefers O2 to CO2

2. If rubisco binds O2

a. Process uses 6 additional ATP

b. Regenerates RuBP

c. Produces a 2-C compound (instead of 3-C)

d. This compound is sent to peroxisome and mitochondrion

(1) converted to Glycerate (3C)

(2) transported back to chloroplast

(3) Uses ATP to convert to 3-PGAL


4. Some plants waste as much as 50% of the energy they make on this process

5. For plants under ideal conditions ?photorespiration poses no problems

6. Some plants have evolved structures to reduce photorespiration.

G. C4 plants

1. Use spatial (C4) isolation of Rubisco to prevent Photorespiration

2. Fix Carbon into 4-Carbon organic Acids

3. Specifics (Use Study Sheet)

a. Rubisco sequestered away from O2 in specialized cells ? BUNDLE SHEATH CELLS

b. Capture CO2 using shuttle molecules

c. C4 use PEP (phosphoenol pyruvate) and Pepco (PEP carboxylase) to capture CO2 and funnel it into Calvin Cycle.

(1) C4 comes from intermediates (oxaloacetate and malate) which are 4-C molecules

(2) Other plants called CAM plants.

VIII. Oxidative Respiration – Overview

A. Review cell organization and where reactions are taking place

B. Emphasize role of ATP made in Photosynthesis

IX. Glycolysis – Chapter 4: 110-118

A. Interested only in the big picture (see Figure 4-3, pg. 111)

B. Emphasize 3 major events

1. Energy Investment

2. Cleavage of Glc into 2 3-carbon sugars

3. Energy Generation

a. Stored as NADH

b. ATP


C. Review Figure 4-3 with students

1. If they want to learn the steps, that?s fine.

2. Only be responsible for names in Figure.

a. Glucose

b. Fructose 1,6 bisphosphate

c. Glyceraldehyde 3-Phosphate (PGAL)

d. Pyruvate

D. Other points to emphasize

1. No involvement of molecular Oxygen

2. Direct dependence on the availability of NAD+

a. Use this as a lead into Fermentation.

X. Fermentation – Regeneration of NAD+ in the absence of Oxygen

A. Discuss what is needed to keep Glycolysis going

1. ADP – no problem since cell is using ATP rapidly

2. Glucose

3. NAD+ – must find a way to oxidize NADH to get to ATP generating step.

B. Review 2 Fermentation pathways with study sheet.

XI. Mitochondrial Events – Oxidation of Pyruvate to CO2

A. Review structure of Mitochondrion

B. Transition Reactions – Review with Study Sheet

1. Enzyme – Pyruvate Dehydrogenase Complex

a. 3 enzymes

b. 60 polypeptide chains

2. See Figure 4-8 – p. 118


C. Kreb?s Cycle (Citric Acid Cycle)

1. Good overview – Figure 4-11 pg. 120

2. Review with Study Sheet

a. Students responsible for names and events.

D. Chemiosmotic Phosphorylation – Chap. 13 – p.410 – 429

1. Conversion of stored energy (NADH & FADH2) into ATP

2. Stored Energy used to generate an a H+ gradient

3. Review with Study Sheets

a. Ubiquinone Structure

b. Figure 13-20

c. Figure 13-10 – Summary

d. Figure 13-21 – Shows Redox potentials


a. Doesn?t take into account ATP used for transport out of the Mitochondria

d.See Problem 13-5 pg. 420 ? Yields are 2.5 and 1.5 (for NADH and FADH2 and NADHcytosol.

Uses 6 of 36 for transport.

XII. The Metabolic Pool

A. Use Study Sheet to cover

B. See Figure 4-18 p.127 in text for similar but more detailed treatment.

XIII. DNA as the GENETIC MATERIAL ? A History Lesson (ref: Bio 120 Outline)

XIV. Characteristics of the Genetic Material and the Central Dogma (ref: Bio 120 Outline)


A. Discovery

1. James Watson

2. Francis Crick

3. Maurice Wilkins

4. Rosalind Franklin

B. Basic structure

1. Sugar phosphate backbone

2. Nitrogen bases as rungs

3. Double helix

4. Basic Unit = Nucleotide

a. phosphate

b. sugar

c. nitrogen base.

C. Backbone

1. Deoxyribose sugar

2. Reason for name

3. Review numbering

4. Where phosphate bonds

5. 5′ to 3′ orientation

6. Antiparallel helix

D. Nitrogen Bases

1. Purines

a. Adenine and Guanine

2. Pyrimidines

a. Thymine and Cytosine

3. AT pair and GC pair

a. Complementary base pairing.

E. Chromosomal organization

1. Purpose

a. packaging

b. organization

c. access and control

2. Genome size

a. E. coli = 4.3 x 106 nucleotide pairs/genome

b. Humans = 2 x 108 nucleotide pairs/chromosome (3 x 109 genome)

(1)stretched out = 6cm.

F. Packaging the Eukaryotic Chromosome (p.250-255)

1. Level 1 (Figure 8-9)

a. utilizes proteins called HISTONES

b. amount of DNA ??amount of histones

c. very basic

(1) high proportion of positively charged amino acids

(a) allows for tight binding to negatively charged DNA

(2) lysine and arginine

d. DNA + Histone = Chromatin

e. Five types of histones

(1) very similar from species to species

(a) ex. some cow and pea histones differ by 2 aa)

(2) highly conserved genes.

f. DNA + Histone core form NUCLEOSOMES (Figure 8-9)

(1) Beads on a String appearance

(2) basic unit of DNA packing

(3) Histone core = 8 “nucleosomal” histone molecules

(a)nucleosomal histones = H2A, H2B, H3, H4

i) 2 molecules each compose core

(b)small proteins (102-135 aa)


2. Level 2 – 30nm fiber = SOLENOID (Figure 8-10)

a. appear to be mediated by 5th histone = histone H1

3. Higher levels of packaging

a. refer to overhead of Figure 8-10

b. not clearly understood.

4. Heterochromatin vs. Euchromatin

a. hetero = chromosomes in condensed state during interphase

b. eu = chromosomes in less condensed state

c. only euchromatin is actively transcribed

d. may be a coarse form of gene control

e. most widely known example

(1)Barr Body in females

(a) one of two X chromosomes is always in most condensed form (during interphase)

(b) Only genes on other chromosome are expressed

(c) females are a mosaic since different X-chromosomes can be condensed in different cells..


A. Replication Models

1. Watson-Crick model implied Semi-Conservative

2. Conservative – possible model

3. Meselsohn & Stahl experiments

a. Use 15 N labeled DNA (14 N = normal)

B. Basic Steps

1. Must take into account double helical structure

2. Step 1 – separate strands to access information

3. Step 2 – Make copies using old as model

4. Step 3 – Reform old and new as double helix.

C. Step 1 – Separating Strands

1. unwind helix at specific starting point(s)

a. ORI = origins of replication


2. stabilize unwound helix so it doesn?t reanneal


3. prevent supercoiling


D. Step 2 – Making copies

1. Use template synthesis

2. Complementary base pairing

3. Need enzyme that can make new DNA polymer.

4. DNA POLYMERASE (Figure 6-21 & 22)

a. 5′ to 3′ polymerase

b. Sliding clamp protein ? moderates attachment of DNA pol to template.

c. uses 5′-nucleotide triphosphates (ATP, GTP, CTP, TTP)

(1)provide energy for bond formation

d. Review antiparallel structure

(1)requires bidirectional synthesis

(2)DNA pol can only synthesis unidirectionally (eg. 5′ to 3′)

(3)Synthesis occurs continuously on 3′ to 5′ strand = leading strand

(4)Synthesis occurs discontinuously on 5′ to 3′ strand = lagging strand

(a)DNA LIGASE connects pieces.

e. Requires a primer

(1)de novo synthesis not possible

(2)Primase enzyme lays down RNA primer

(3)Primer must be removed

(a)see below

f. Replicating the ends of lagging strands (p.249-250)

(1)requires special enzyme to add tails onto template strand

(a)tails are called – TELOMERES

(2)Enzyme that duplicates them ?TELOMERASE (Figure 8-6)

(a)Contains a short piece of RNA

(b)in humans = CCCCAAU

(c) Creates tandem repeats on ends of lagging strand (GGGGTTA)

(d)allows the end of the chromosome to be replicated

(3)leaves a 3′ tail on template strand.

g. Accuracy

(1)3′ to 5′ exonuclease activity acts as proofreader

(2)?senses? mismatch, backs up, removes mismatch, and corrects

(3)methylation of parent strand.

XVII. DNA REPAIR (pg. 198-205)

A. Define Mutation

1. Permanent change in DNA code

B. Mismatch Repair system catches errors Replication Machinery misses

1. Rpn machinery error rate 1 in 10 7 nucleotides copies

2. Roughly 10 mistakes/chromosome/rpn cycle

3. Mismatch = mispaired nucleotide (Figure 6-25A p.201)

4. Mismatch repair enzymes recognize the mismatches

a. Excise incorrect nucleotide

b. Repair

c. Must be able to recognize the newly synthesized strand

(1)Nick system – new strands have transient nicks

(2)Methylation system – Parent is methylated

5. Reduces error rate to 1 in 10 9.

C. DNA Damage outside of Rpn

1. Types of Damage (Figure 6-27 p.202)

a. Depurination – spontaneous loss of A or G

(1)Leaves a depurinated sugar

b. Deamination – loss of amine group on Cytosine

(1)Converted to Uracil

c. Thymine Dimer formation due to UV light exposure

d. Many other types caused by reactive metabolic by-products

2. Effects can cause

a. Single base pair changes (deamination) (Figure 6-29A)

b. Single-base pair deletions (depurination) (Figure 6-29B)

c. Stalled or incomplete rpn (thymine dimers).

D. DNA Repair Mechanism (Figure 6-30 p204)

1. Excision of Defect

a. Requires specialized nucleases for each type of damage

2. Repair

a. Uses DNA pol other than rpn DNA pol

3. Ligation

a. Uses DNA Ligase.


A. Central Dogma – From DNA to Protein ?Figure 7-1

B.Discuss this as the first step in Gene Expression


? Using Genetic Information to make the molecules necessary for cellular functions.

? Ultimately, every process within a living organism is controlled by theavailability of specific gene products

C. Definition of a Gene – Stage #1

1. Region of DNA contains some information that needs to be accessed.

D. The Players

1. The information – DNA

a. double helix

b. 5′ to 3′ and 3′ to 5′ strands

2. The enzyme – RNA Polymerase

a. 5′ to 3′ polymerase activity

b. substrate

(1)ribonucleoside triphosphates

c. local unwinding capabilities

d. DNA binding properties

e. 3 in Eukaryotes

(1)I – rRNA

(2)II – mRNA + others

(3)III – tRNA + rRNA.

3. The messenger – mRNA

a. Structure – Use Figure 7-3 to compare and contrast with DNA

(1)single strand

(2)uracil for thymine

(3)ribose for deoxyribose

E. The Basic Steps (Figure 7-9)

1. Find the region to be copied

a. which strand

b. where to begin

2. Attach enzyme

3. Copy

4. Stop.

F. Finding the region to be copied

1. Which strand

a. Discuss sense vs. non-sense

(1)gene is always read 5′ to 3′ regardless of which strand its on

(2)template is always 3′ to 5′ regardless of which strand its on

(3)Genes on different strands: Fig 7-10

b. upstream vs. downstream

(1)up = toward 5′

(2)down – toward 3′

2. Where to Begin

a. Promoters – consensus sequences

(1) TATA box: @ ~ -25

(a) RNA pol binding site

(2)CAAT box: @ ~ -80

(a)bind regulatory proteins

(b)effect rate of initiation

b. More on this when we cover gene regulation.

G. Attaching the enzyme

1. Transcription factors & RNA Pol bind at promoter region

a. More details when we cover gene regulation

2. Transcription begins

H. Elongation

1. Uses anti-sense as template

2. copies in 3′ to 5′ direction producing 5′ to

3′ transcript (mRNA)

I. Termination

1. Probably requires termination factors

2. Specific DNA sequence signals termination

a. in eukaryotes – most common = AATAAA.

J. Compare Eukaryotic and Prokaryotic Transcripts

1. Use Figure 7-13.

K. The Eukayotic mRNA – use Study Sheet

1. Structure

a. Use Figure 7-12

b. 5′ cap

(1) 5′ to 5′ linkage to…


(3) fxn -

(a) in translation (later)

(b) transport out of nucleus

i) pores recognize cap

(c) prevent RNAse degradation

c. 5′ UTR (10-200 nucleotides)

d. coding sequence

e. 3′ UTR

(1) highly conserved

(2) fixes length of UTR and site for 3′tail attachment

f. 3′ poly A (30-200+)

(1) probable fxn – stability

(2) mRNA w/o poly A degraded quickly.

2. Processing – Use Study Sheet

a. Intron/Exon structure

b. mediated by a group of snRNA?s and proteins

c. snRNA + Proteins = snRNP’s

d. Several snRNP’s take part in each splicing event

e. A complex of functioning snRNP;’s is sometimes referred to as a Spliceosome

f. Can be cis or trans

(1) cis = connecting exons in same mRNA

(2) trans = connectiong exons from different mRNA?s

g. Same mRNA can be spliced into different genes = ALTERNATIVE SPLICING

L. Summarize – Use Figure 7-19.

M. Revision of Gene Definition

1. Includes promoter and termination regions

2. Possibly other control sequences


A. Machinery

1. The code

a. minimum number to cover all AA = 64

b. degeneracy

c. Advantage – can absorb some amount of mutation

2. Ribosomes – Use Figure 7-25

a.rRNA – serves to align ribosome with message and *new evidence* showsit carries out the

enzymatic reactionneeded for peptide bond formation (ref: Science, 11Aug00, p.878).

b. Protein – Structural (Figure 7-26)

c. Small and large subunits

(1)small – recognition and alignment

(a)A & P binding sites

i) lower portion of these sites

ii) involved in binding tRNAs to Codon (mRNA)

(2)large – binding tRNA and making peptide bond

(a)peptidyl-transferase activity (rRNA)

(b)GTP hydrolysis activity (proofreading)

(c) A & P binding sites

i) Major portion of these sites

ii) binds majority of tRNA with AA attached

(d)E (Exit) Site.

3. tRNA

a. transfer RNA

b. structure

(1)anticodon region

c. Amino Acyl-tRNA synthetase

(1)one for each amino acid (20)

(2)attach AA to correct tRNA in 2 step process



(3)proofreading step

(a)accuracy 1,2 per 40,000

(b)done at 2nd step

(4)only process that ensures the correct codon/a.a. pairing.

(5)Active site of enzyme screens Amino Acids based on size.

(a)Coarse sieve removes AA too large for active site.

(b)Fine sieve removes those small enough to fit but not correct

(c)See article: Sieves in Sequence

d. Joins 3′-OH of tRNA to carboxyl group of Amino Acid.

B. Basic steps

1. Start – connect message with ribosome


2. Build protein



C. INITIATION – refer to Figure 7-28

1.Binding of small ribosomal subunit + initiator tRNA ( tRNA met ) + initiation factors (not shown in


a. Initiator tRNA is only tRNA that can bind to small subunit alone

b. Binds at P site

2. Complex binds to 5′ end of mRNA

a. 5′ cap is critical

3. Complex ?scans? mRNA 5′ to 3′ for start codon

a. When found, some IFs dissociate to allow for subsequent steps.

4. Large Ribosomal subuint binds ?Translation begins

D. Elongation

1. assisted by elongation factors

2. EF-tRNANEXT-GTP binds at A site ? Use overhead in binder

a. Proofreading step

b. involves GTP hydrolysis

c. delays peptide bond formation

d. permits incorrect tRNA to diffuse out of ribosome

3. Peptide Bond Formation – use overhead in binder

a. Catalyzed by peptidyl transferase

b. Aminoacyl (3′-OH–carboxyl) bond between tRNAP-AAP

c. AAP-Carboxyl attaches to AAA-NH2

(1)transfers chain from tRNAP to tRNAA


d. empty tRNAP is released.

4. Small Ribosomal Subunit shifts down one codon (use Figure 7-27)


b. Shifts tRNA attached to nascent chain to from A to P site

c. Empty tRNA shifts to E site -dissociated

d. Small subunit shifts back 1 codon to realign with Large subunit

e. Next tRNA binds at A site ? process continues

f. Specific elongation factors (EF) have been identified for this process

5. A -site is now empty – next tRNA binds ?cycle repeats.

E. Termination (refer to Figure 7-30)

1. A-site is occupied by on of the termination codons

2. Release factor protein binds at A-site

3. Peptidyl transferase hydrolyses last amino-acyl bond

4. New protein is released -Ribosome/ mRNA complex dissociates

F. Final Gene Definition

1. A region of DNA containing the code for

a specific protein or RNA (e.g. tRNA & rRNA, snRNA) plus all the adjoining DNA sequences that act as controllers.

G. Final Review of Process (use Figure 7- 33)