IB Topics DNA HL

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Transcript IB Topics DNA HL

Welcome Back!!
• The SL material you learned last year is
important to review before each unit this
year…
• Today’s Opener:
• Draw and label a simple diagram of the
molecular structure of DNA. (4 marks)
Happy Monday! 8/23/2010
The structure of the DNA double helix was
described by Watson and Crick in 1953.
Explain the structure of the DNA double
helix, including its subunits and the way in
which they are bonded together.
(Total 8 marks)
• subunits are nucleotides;
• one base, one deoxyribose and one phosphate in each
nucleotide;
• description / diagram showing base linked to deoxyribose C1
and phosphate to C5;
• four different bases – adenine, cytosine, guanine and thymine;
• nucleotides linked up with sugar-phosphate bonds;
• covalent / phosphodiester bonds;
• two strands (of nucleotides) linked together;
• base to base;
• A to T and G to C;
• hydrogen bonds between bases;
• antiparallel strands;
• double helix drawn or described;
Accept any of the points above if clearly explained in a diagram.
• [8]
IB Topics: DNA,
Transcription, Translation
3.3, 3.4, 3.5
7.1, 7.2, 7.3, 7.4
7.1
DNA Structure
7.1.1 Describe the structure of DNA, including the
• antiparallel strands
• 3–5 linkages
• H bonding between purines (A, G) and
pyrimidines (C, T)
– A/T 2 H bonds
– G/C 3 H bonds
• Major and minor grooves, direction of the “twist”,
alternative B and Z forms, and details of the
dimensions are not required.
7.1.2 Outline the structure of
nucleosomes.
• Nucleosome = DNA wrapped around 8
histone proteins; held together by
another histone protein
7.1.3 State that nucleosomes help to supercoil
chromosomes and help to regulate transcription.
• A Human chromosome
can be 4 cm long!
• DNA wraps twice
around 8 core histones
• (DNA - , histone +)
• No transcription when
packaged so tightly—
regulation!
7.1.4 Distinguish between unique or singlecopy genes & highly repetitive sequences in
nuclear DNA.
• Highly repetitive sequences constitutes 5–45% of the genome
• sequences typically 5 - 300 base pairs per repeat, and may be
duplicated as many as 105 times per genome
• “satellite DNA” = clustered regions of repeats
– centromeres
• Most is dispersed throughout genome
• Probably don’t code
– TOK: once classified as “junk DNA”, showing a degree of confidence that it
had no role; research has been sparse. This addresses the question: To
what extent do the labels and categories used in the pursuit of knowledge
affect the knowledge we obtain?
• Transposable elements...can move around w/in the genome
(McClintock, 1950)
7.1.5 State that eukaryotic genes can contain
exons and introns.
• Less than 2% of
human genome is
genes that code for
proteins
• HGProject, mid-70s –
2001
• Introns, exons
(expressed)
Happy Thursday!!!! 9/15
Living organisms use DNA as their genetic
material. Explain how DNA is replicated within the
cells of living organisms.
(Total 8 marks)
• helix is unwound;
• two strands are separated;
• helicase (is the enzyme that unwinds the helix separating the
two strands);
• by breaking hydrogen bonds between bases;
• new strands formed on each of the two single strands;
• nucleotides added to form new strands;
• complementary base pairing;
• A to T and G to C;
• DNA polymerase forms the new complementary strands;
• replication is semi-conservative;
• each of the DNA molecules formed has one old and one new
strand;
• [8]
7.2
DNA Replication
7.2.1 State that DNA replication occurs in a 5  3
direction.
• 5 end of free
DNA nucleotide
is added to the
3 end of chain
of nucleotides
that is already
synthesized.
7.2.2 Explain the process of DNA replication in prokaryotes, including
the role of enzymes
• Origin of replication (bubble; prok has one origin,
euk has several)
• Helicase:
• uncoils the double helix
• RNA primase:
• @ replication fork; short sequence of RNA (5-10
n’tides)
• DNA polymerase III:
– adds DNA n’tides in 5’ to 3’ direction
– N’tide is actually a deoxynucleoside triphosphate
(dNTP)
– 2 P groups lost during bonding  energy for
bonding
• DNA polymerase I:
• removes primer from 5’ end and replaces it with
DNA n’tides
7.2.2 Explain the process of DNA replication in prokaryotes, including
the role of enzymes
• ANTI-PARALLEL...OKAZAKI FRAGMENTS!
• Can only go 5’ to 3’ b/c DNA Pol III
• Leading strand
– Fast; toward rep fork in 5’  3’
– Needs primase, primer, DNA pol III only once
• Lagging strands
– Slower; Fragments away from rep fork, 5’  3’
– Each fragment needs primase, primer, DNA pol III
– DNA ligase attaches S-P backbones of Okazaki
fragments to make 1 strand
IB Book, fig 7.7 page 200
• Examiner’s Hint
– Draw the figure from memory
– Annotate what’s happening at specific
locations
7.2.3 State that DNA replication is initiated at
many points in eukaryotic chromosomes.
1. replication begins at origin, strands separate b/c
helicase breaks H bonds
2. Replication fork at each end of bubble (DBL strand
opens to expose 2 template strands)
3. Bubble enlarges in both directions (bidirectional) &
eventually fuse together
Multiple bubbles  replication faster for the BIG
eukaryotic genomes
Primer
Primer
LIFE book
Happy ….
• Explain the process of transcription in
eukaryotes.
• (8)
• RNA polymerase controls transcription / is the enzyme used in
transcription;
• DNA is unwound by RNA polymerase;
• DNA is split into two strands;
• mRNA is made by transcription;
• promoter region (by start of gene) causes RNA polymerase to bind;
• anti-sense / template strand of DNA is transcribed;
• direction of transcription is ;
• free nucleotide triphosphates used;
• complementary base pairing between template strand and RNA
nucleotides / bases;
• Accept this marking point if illustrated using a diagram
• RNA contains uracil instead of thymine;
• terminator (sequence) stops RNA polymerase / transcription;
• mRNA is released / RNA polymerase released;
8 max
TUUUUUUUESDAY, 8/31!!! 
• List three of the other molecules, apart from
mRNA, required for transcription.
• 3 mks
• DNA;
• RNA polymerase;
• (ribose) nucleotides / ribonucleotides / RNA
nucleotides;
• transcription factors;
• nucleoside / ribonucleoside triphosphates;
3 max
• Any two of the following: A / C / G / U;
7.3
Transcription
7.3.1 State that transcription is
carried out in a 5  3 direction.
• SIMILAR to replication
– No helicase!
– RNA polymerase separates DNA strands
– binds to promoter region of DNA
– 5’  3’
• The 5 end of the free RNA nucleotide is added
to the 3 end of the RNA molecule that is
already synthesized.
7.3.2 Distinguish between the
sense & antisense strands of
DNA.
• sense strand = coding strand, carries the
genetic code
– has the same base sequence as mRNA with uracil
instead of thymine
• antisense strand = template strand
– is transcribed, has the same base sequence as
the tRNA
• Promoter region determines which is
antisense
– Is always the same strand for a particular gene
– Can differ for different genes
7.3.3 Explain the process of transcription in prokaryotes, including the
role of the promoter region, RNA polymerase, nucleoside triphosphates
terminator.
• Front of RNA polym unwinds the DNA helix
• Adds nucleoside triphosphates to produce
mRNA (2 P groups released), nucleotides added
to 3’ end of growing strand
• Anti-sense is template
• Back of RNA polym rewinds DNA strands
• Terminator: DNA sequence, transcribed, causes
RNA polym to detach & transcription stops
7.3.4 State that eukaryotic RNA needs the removal
of introns to form mature mRNA.
No introns in prokaryotic mRNA
Eukaryotes—must be removed to get functional mRNA
THURSDAY, 9/2
• The information needed to
make polypeptides is carried in
the mRNA from the nucleus to
the ribosomes of eukaryotic
cells. This information is
decoded during translation. The
diagram below represents the
process of translation.
• State the name of the next
amino acid which will attach to
the polypeptide. (1)
• Explain how the amino acid was
attached to the tRNA. (3)
• Alanine / Ala 1
• an activating enzyme attaches amino acid to
the tRNA;
• specific enzyme for specific tRNA;
• recognizes tRNA by its shape / chemical
properties;
• energy (ATP) is needed;
• amino acid attached at end;
• amino acid attached at CCA; 3 max
Happy Thursday! 9/2/2010
Explain the process of translation.
• (Total 9 marks)
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consists of initiation, elongation and termination;
mRNA translated in a 5' to 3' direction;
binding of ribosome to mRNA;
small sub-unit then large;
first / initiator tRNA binds to start codon / to small subunit of ribosome;
AUG is the start codon;
second tRNA binds to ribosome;
large subunit moves down mRNA after a second tRNA binds;
amino acid / polypeptide on first tRNA is transferred / bonded to amino acid
on second tRNA;
peptide bonds between amino acids / peptidyl transferase;
requires GTP;
movement of ribosome / small subunit of ribosome down the mRNA;
loss of tRNA and new tRNA binds;
reach a stop codon / termination;
polypeptide released;
tRNA activating enzymes link correct amino acid to each tRNA;
(activated) tRNA has an anticodon and the corresponding amino acid
attached;
[9]
7.4
Translation
7.4.1 Explain that each tRNA molecule is recognized by a
tRNA-activating enzyme that binds a specific amino acid to
the tRNA, using ATP for energy.
• 3’ end is free, has CCA
– Site of a.a. attachment
• H bonds form in 4 areas, create “clover”
• 1 loop of clover has anticodon
– (unique to each tRNA)
• Each a.a. has a specific tRNA-activating enzyme
(aminoacyl-tRNA synthetase; 20 of them)
– Active site fits only 1 a.a. & its tRNA, rxn requires ATP
– “activated” amino acid, tRNA takes it to ribosome
• The shape of tRNA and CCA at the 3’ end should be
included.
7.4.2 Outline the structure of ribosomes,
including protein and RNA composition, large
and small subunits, three tRNA binding sites,
mRNA binding sites.
• Lg & sm subunits
– rRNA (2/3 of its mass) and many RNA proteins
– Prokaryotic – smaller than eukaryotic
• 3 tRNA binding sites
– A site: holds tRNA carrying the next amino acid
to be added to the polypeptide chain
– P site: holds the tRNA carrying the growing
polypeptide chain
– E site: site from which tRNA that has lost its
amino acid is discharged
• mRNA binding site – in cavity b/w 2 subunits
7.4.3 State that translation consists of initiation,
elongation, translocation, termination.
• Initiation
– Start codon AUG on 5’ end of all mRNA
– Activated amino acid (methionine + tRNA w/anticodon UAC)
attaches to mRNA & small ribosomal subunit
– Small subunit travels down mRNA to start codon (AUG)
– Starts translation
– H bonds b/w initiator tRNA & start codon
– Large ribosomal subunit attaches
• Elongation
• Translocation
• Termination
7.4.3 State that translation consists of initiation,
elongation, translocation, termination.
• Initiation
• Elongation
– tRNAs bring amino acids to mRNA-ribosomal
complex in order specified by codons on mRNA
– Elongation factors (ptns) bind tRNAs to exposed
codons @ A site
– tRNA  P site next
– Ribosomes catalyze peptide bonds forming b/w
amino acids (condensation)
• Translocation
• Termination
7.4.3 State that translation consists of initiation,
elongation, translocation, termination.
• Initiation
• Elongation
• Translocation
– During elongation phase
– tRNAs move down mRNA
• Binds with A site
• Its amino acid bonds to polypeptide (attached to tRNA @ A
site)
• Moves to P site
• Transfers polypeptide to new tRNA in A site
• Empty tRNA to E site, released
• 5’ to 3’ direction (ribosome moves along mRNA TOWARD 3’
end—start codon was on 5’ end)
• Termination
7.4.3 State that translation consists of initiation,
elongation, translocation, termination.
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Initiation
Elongation
Translocation
Termination
– 1 of 3 stop codons @ A site
– Release factor (protein) fills A site (doesn’t have
amino acid)
– Catalyzes hydrolysis of bond linking P site’s tRNA
to polypeptide
– All released
7.4.4 State that translation occurs in a 5  3
direction.
• During translation, the ribosome moves
along the mRNA towards the 3 end. The
start codon is nearer to the 5 end.
7.4.5 Draw and label the structure of a peptide
bond between two amino acids.
7.4.6 Explain the process of translation, including
ribosomes, polysomes, start codons and stop codons.
• Polysomes: string of ribosomes all translating the
same mRNA
– common
7.4.7 State that free ribosomes synthesize
proteins for use primarily within the cell, and that
bound ribosomes synthesize proteins primarily for
secretion or for lysosomes.
• “bound” to what???
• ER!