E. LIGHT-INDEPENDENT REACTIONS (Calvin Cycle)
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.
F. PHOTORESPIRATION (Use Study Sheet)
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
3. NET LOSS OF ENERGY
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. FILL OUT ATP SUMMARY TABLE.
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
3. FILL OUT ATP SUMMARY TABLE
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
4. COMPLETE ATP SUMMARY TABLE
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)
XV. DNA STRUCTURE
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)
(c) Core = HISTONE OCTAMER.
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..
XVI.DNA REPLICATION
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
b. DNA HELICASE
2. stabilize unwound helix so it doesn?t reanneal
a. SINGLE STRAND BINDING PROTEINS (SSB)
3. prevent supercoiling
a. DNA TOPOISOMERASE.
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.
XVIII. TRANSCRIPTION – ACCESSING THE CODE
A. Central Dogma – From DNA to Protein ?Figure 7-1
B.Discuss this as the first step in Gene Expression
PROCESSES INVOLVED 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…
(2)..GTP
(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
XIX.TRANSLATION
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
(a)AA + ATP AA-AMP + PP
(b)AA-AMP + tRNA AA-tRNA + AMP
(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
a. INTIATION
2. Build protein
a. ELONGATION
3. TERMINATION
C. INITIATION – refer to Figure 7-28
1.Binding of small ribosomal subunit + initiator tRNA ( tRNA met ) + initiation factors (not shown in
diagram)
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
(2)PEPTIDE BOND FORMATION
d. empty tRNAP is released.
4. Small Ribosomal Subunit shifts down one codon (use Figure 7-27)
a. TRANSLOCATION
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)
H. READING ASSIGNMENT – PG. 234-240 ?RNA AND THE ORIGINS OF LIFE