Today’s Plan: 2/4/11
 Bellwork: Go over test and lab (25
mins)
 Finish AP Lab 3 and work on genetics
problems(45 mins)
 Meiosis Notes (the rest of the period)
Today’s Plan: 2/7/11
 Bellwork: Punnett Squares Quiz(20
mins)
 Go over quiz and questions
 Homework problems (mono and
dihybrids)
 Pack/Wrap-up (last 5 mins)
Today’s Plan: 2/8/11
 Bellwork: Set up lab Part A (20 mins)
 Do the rest of the lab (except for the
overnight)
 Continue with genetics notes (the rest
of class)
Today’s Plan: 2/9/11
 Bellwork: Finish Titrations from
yesterday (40 mins)
 Continue with Notes on Genetics (the
rest of class)
Today’s Plan: 2/10/11
 Bellwork: Pedigree
Discussion/Practice (30 mins)
 Flies!!!!! (30 mins)
 DNA Notes (the rest of class)
 Pack/Wrap-up (last few mins of class)
Today’s Plan: 2/11/10
 Bellwork: Replication Video (10 mins)
 DNA Notes, continued with sticky
board intermittent (the rest of class)
Today’s Plan: 2/14/11
 BW: Q&A (15 mins)
 Genetics Test (as needed)
Sexual Reproduction
 Why sex?




Energetically expensive
Health risks of childbearing
Genetic diversity
Overall heath of the species
 Meiosis makes the gametes
 Spermatogenesis=4 identical sperm cells,
puberty to death
 Oogenesis=1 egg cell per month, initial egg
production before birth then freezing at
Prophase I
Chromosome number and
fertilization
 Diploid (2n)=full set of chromosomes, is the result of
fertilization with gametes
 Homologs for each chromosome, 1 from mom, 1
from dad
 In humans-22 autosome pairs, 1 sex-determing
pair
 Haploid (n)=1/2 set of chromosomes, is the result of
meiosis
 Karyotyping=generation of a picture of one’s
chromosomes for counting/sexing/diagnosing






Kleinfelter’s-Trisomy 23 XXY
Turner’s-monosomy 23 XO
Jacob’s-XYY
Down’s-trisomy 21
Patau’s syndrome=trisomy 13
Crie du Chat=missing a portion a chromosome
Figure 12-1-Table 12-1
Figure 12-1-Table 12-2
Figure 12-6a
Normal human karyotype
Abnormal Chromosome Number
 Failure of chromosome separation during
meiosis is called nondisjunction
 Results in cells with an extra chromosome or
missing a chromosome
 When these cells are fertilized, this causes
trisomies and monosomies (both types of
aneuploidy)
 Polyploidy-occurs when an individual gets
more than two complete chromosome sets
 This is most common in the plant kingdom
Figure 12-15
NONDISJUNCTION
n+1
n+1
n–1
2n = 4
n=2
n–1
1. Meiosis I starts normally.
2. Then one set of
Tetrads line up in middle of cell.
homologs does not
separate (= nondisjunction).
3. Meiosis II occurs normally.
4. All gametes have an abnormal
number of chromosomes—either
one too many or one too few.
Figure 12-15-Table12-4
Figure 12-16
1
12
1
16
1
20
1
36
1
1000
1
300
1
1
100
180
1
60
Alterations to Chromosomes
 These are called chromosome mutations
and occur when a chromosome is damaged
(ex: radiation damage).
 Deletions-part of a chromosome is missing
 Duplication-a segment of a chromosome is
repeated
 Inversion-two segments of a chromosome
switch places
 Translocation-a piece of a chromosome moves
to a non-homologous chromosome
 In the cases of inversions and translocations,
even though all parts of the chromosome are
present, location seems to affect expression
Mendel and Genetics
 Mendel bred pea plants to determine the rules of
heredity
 Terms:
 P=Parent generation
 F1=First filial generation (offspring from P)
 F2=Second filial generation (offspring from F1)
 Allele=alternate form of a trait
 Dominant=always expressed
 Recessive=only expressed if dominant isn’t present
 Homozygous=2 copies of the same allele
 Heterozygous=2 different alleles
 Genotype=Combination of genes
 Phenotype=Expression of the combination of genes
What Mendel figured out
 Rarely blending of inheritance (purple x
white yields either purple or white, not
lavendar)
 Alternate versions of each gene exist (2
alleles for each gene)
 Each individual inherits 1 allele from each
parent
 Predictable ratios occur when crossing
individuals
 + 2 laws of inheritance
Laws of Genetics
 Law of Segregation-Traits are
separated from one another in the
parents (we now know that this is
due to meiosis)
 Law of Independent Assortment-The
inheritance of one trait doesn’t affect
the inheritance of another (this is
only true if traits are on different
chromosomes, i.e, not “linked”)
Figure 13-7
Rr parent
Dominant allele
for seed shape
Recessive allele
for seed shape
Chromosomes replicate
Meiosis I
Alleles segregate
Meiosis II
Principle of segregation: Each gamete carries only
one allele for seed shape, because the alleles have
segregated during meiosis.
Figure 13-8
R
y
y R
r
Replicated chromosomes
prior to meiosis
r
Y
Y
R
R r
r
R R r
Alleles for seed shape
Alleles for seed color
r
Chromosomes can line up in
two ways during meiosis I
Y Y y y
R
Meiosis I
R
Y
y yY Y
r
r
R
y y
Y
Meiosis I
R
Y
Y
1/4 RY
Y
Meiosis II
r
R
r
Y
y y
Meiosis II
R
r
r
y
y
1/4 ry
r
R
R
y
y
1/4 Ry
r
Y
Y
1/4 rY
Principle of independent assortment: The genes for seed shape and seed color
assort independently, because they are located on different chromosomes.
Types of crosses
 Monohybrids-tracing inheritance of 1
trait at a time
 Dihybrids-tracing inheritance of 2
traits together at a time
 Testcross-crossing individual
expressing dominant gene with an
individual expressing the recessive to
determine the dominant’s genotype
A cross between two homozygotes
Homozygous
mother
Meiosis
Female gametes
Homozygous
father
Meiosis
Offspring genotypes: All Rr (heterozygous)
Offspring phenotypes: All round seeds
A cross between two heterozygotes
Heterozygous
mother
Female gametes
Heterozygous
father
Male gametes
Figure 13-4
Offspring genotypes: 1/4 RR : 1/2 Rr : 1/4 rr
Offspring phenotypes: 3/4 round : 1/4 wrinkled
Figure 13-5a
Hypothesis of independent assortment:
Alleles of different genes don’t stay together when gametes form.
Female parent
F1 PUNNET SQUARE
Female gametes
Male parent
F1 offspring all RrYy
F2 female parent
Alleles at R gene and Y gene
go to gametes independently
of each other
F2 PUNNET SQUARE
Female gametes
F2 male
parent
F2 offspring genotypes: 9/16 R–Y– : 3/16 R–yy : 3/16 rrY– : 1/16 rryy
F2 offspring phenotypes: 9/16
: 3/16
: 3/16
: 1/16
Beyond what Mendel Knew
 There are occasionally traits that
follow different inheritance patterns
 Differing degrees of inheritance
 Codominant alleles-2 dominant alleles,
both are expressed side-by-side
 Incomplete dominance-dominant can’t
completely cover the recessive, so the
recessive is partially expressed (blending
of inheritance
Figure 13-17b
Incomplete dominance in flower color
Parental
generation
F1 generation
Self-fertilization
F2 generation
Purple
Lavender
White
But what is dominant, really?
 Tay-Sachs-inability to metabolize certain lipids
because of a malfunctioning enzyme
 Only children with 2 copies of the allele have the
disease-at the organismal level, this is recessive
 However, heterozygotes, biochemically appear to be
the result of incomplete dominance-lipid metabolism
levels are intermediate between those who don’t have
Tay-Sachs and those who
 Molecularly, heterozygotes produce equal levels of
functional and non-functional enzymes-appearing to
be co-dominant
Other single-gene locus influences
 Multiple alleles
 Ex: Human Blood type
 IA, IB, i
 Pleiotropy
 When a single gene affects many traits
 Ex: sickle cell disease and its multiple
symptoms
Sex-linked Traits
 X-chromosome inheritance
 b/c males have only 1 copy, they’re most
likely affected by these, but females can
still inherit these
 Pass from mother to children
 Y-chromosome inheritance
 Only males affected
 Pass from father to sons
Two or more genes
 Epistasis
 One gene alters the expression of another at a
separate locus
 Ex: in mice, B=black, b=brown, however, a second
gene determines whether or not pigment will be
deposited in the fur, so if organism is homozygous
recessive for that gene, the mouse is albino
 Polygenic Traits
 Ex: human height, skin color, hair color
 Spectrum of possible phenotypes that exist along a
bell curve
 Multiple genes and therefore lots of combinations of
dominant and recessive alleles influence these traits
Figure 13-19
A phenotype distribution that forms a bell-shaped curve.
Normal distribution—bell-shaped curve
Figure 13-20
Wheat kernel color is a quantitative trait.
Parental
generation
Hypothesis to explain inheritance of kernel color
aa bb cc
(pure-line white)
F1
generation
AA BB CC
(pure-line red)
Aa Bb Cc
(medium red)
Self-fertilization
F2
generation
20
15
6
1
15
6
1
Other Genetic Patterns
 X-Inactivation-In females only, one X chromosome is
inactivated (Barr Body) in each cell.
 Different cells may inactivate a different X
 ex: calico cats (Black, orange), lymph node patterns
in women
 Linkage-Recall that genes found on the same
chromosome are “linked”
 These don’t segregate via meiosis
 The only way to get recombinations is via crossing
over, therefore the further apart these are, the more
chance of a crossover between them
 % recombinants is proportional to the distance linked
traits have between them-see AP Lab 3 Sordaria
Does Genotype determine
Phenotype?
 The short answer is “no!”
 Environmental influences also affect
gene expression, giving us a norm of
reaction for a phenotype (range of
possible phenotypes)
Pedigree Analysis
 Pedigrees are family trees that use
specific notations that geneticists use
to predict the inheritance pattern of a
trait
Figure 13-21
I
Carrier male
Carriers
(heterozygotes)
are indicated
with half-filled
symbols
II
III
Affected
male
IV
Affected
female
Carrier female
Figure 13-23
I
Queen Victoria
Prince Albert
Female carrier of
hemophilia allele
II
Affected
male
III
IV
Commonly Inherited Disorders
 Recessives
 Tay-Sachs, Sickle Cell, Cystic Fibrosis,
albinism
 Dominants
 Marfan syndrome, Huntington’s disease,
Dwarfism
 Sex-linked, X-chromosome
 Color-blindness, several forms of
muscular dystrophy, pattern baldness,
hemophilia
Beyond the Chromosome Theory of
Inheritance
 Genomic Imprinting
 Occurs in about 2-3 dozen autosomally inherited
traits
 Which parent passed the gene matters in the
inheritance pattern
 Ex: Insulin-like growth factor 2 in mice-only
paternally inherited forms are active
 Extranuclear Genes
 Recall that organelles, like mitochondria and
chloroplasts have circular pieces of DNA
 These are capable of replicating and being
passed to daughter organelles during Mitosis
 These are matrolineal
DNA Structure
 Building block=nucleotide
 Double Helix of consisting of
 2 antiparallel strands-one runs 5’ to 3’, the other
in the opposite direction
 A backbone on each strand consisting of a
repeating phosphate-deoxyribose segment
 Complimentary base-paired links between the
strands using hydrogen bonds between a purine
(either A or G) and a pyrimidine (either T or C)
Figure 14-6
Structure of a deoxyribonucleotide
Phosphate group
attached to
5 carbon
of the sugar
Could be
adenine (A),
thymine (T),
guanine (G)
cytosine (C)
5
Hydroxyl (OH)
group on 3 carbon
of the sugar
3
Primary structure of DNA
5 end of strand
Sugar-phosphate
backbone of
DNA strand
Nitrogencontaining bases
project from the
backbone
5
Phosphodiester
bond links
deoxyribonucleotides
3 end of strand
Figure 14-7
The double helix
Complementary base pairing
5
3

5

3

Sugar-phosphate
“backbone” of DNA
Complementary
base pairs joined by
hydrogen bonding
3
5
5
3
Replication
 DNA is copied in preparation for cell division
 Involves a symphony of enzymes
 Is done in a 3’ to 5’ direction
 This means that replication is antiparallel
 One strand is the leading strand and is replicated
continuously, the other is the lagging strand and is
replicated in pieces (called Okazaki fragments),
requiring linking later
 Is semi-conservative
 Has a single point of origin in bacteria, and multiple
points of origin in eukaryotes
 Requires an RNA primer because the enzymes that
synthesize DNA cannot initiate a strand, they can only
add to an existing strand
 Ends with proofreading by enzymes to make sure the
correct bases are in place.
Figure 14-10
Bacterial chromosomes have a single point of origin.
A chromosome being replicated
5
3
5
Replication
proceeds in
both directions
Origin of
replication
Eukaryotic chromosomes have multiple points of origin.
5
3
3
5
Replication
fork
5
5
3
5
3
Old DNA
New DNA
Replication proceeds in both
directions from each starting point
3
Old DNA
New DNA
3
Replication
bubble 3
5
Figure 14-13-Table 14-1-1
Figure 14-13-Table 14-1-2
Figure 14-11
SYNTHESIS OF LEADING STRAND
Primase synthesizes RNA primer
3
5
Topoisomerase relieves twisting forces
5
3
Helicase opens double helix
5
Single-strand DNA-binding proteins (SSBP) stabilize single strands
1. DNA is opened, unwound, and primed.
Sliding clamp holds DNA polymerase in place
3
5
RNA primer
Leading strand
5
2. Synthesis of leading strand begins.
DNA polymerase III works in 5 3 direction, synthesizing leading strand
5
3
Figure 14-13-Table 14-1-3
Figure 14-13-1
SYNTHESIS OF LAGGING STRAND
5
3
RNA
primer
5
3
5
Topoisomerase
SSBPs
Primase
Helicase
1. Primase synthesizes RNA primer.
Figure 14-13-2
SYNTHESIS OF LAGGING STRAND
5
3
Okazaki fragment
3
5
5
3
5
Sliding clamp
DNA polymerase III
2. DNA polymerase III works in 53 direction, synthesizing first Okazaki
fragment of lagging stand.
5
3
Okazaki fragment
3
5
Okazaki fragment
5
5
3
3. Primase and DNA polymerase III synthesize another Okazaki fragment.
Figure 14-13-3
SYNTHESIS OF LAGGING STRAND
5
3
DNA polymerase I
3
5
5
3
5
4. DNA polymerase I removes ribonucleotides of primer, replaces them with
deoxyribonucleotides in 53 direction.
5
3
DNA ligase
3
5
5
5. DNA ligase closes gap in sugar-phosphate backbone.
5
3
Figure 14-15
TELOMERE REPLICATION
5
Missing DNA on
lagging strand
1. When the RNA primer is removed from the
5 end of the lagging strand (see Figure 14.14),
a strand of parent DNA remains unreplicated.
3
Telomerase with its
own RNA template
2. Telomerase binds to the “overhanging” section
5
of single-stranded DNA. Telomerase adds
deoxyribonucleotides to the end of the parent
DNA, extending it.
5
3
3
3. Telomerase moves down the DNA strand and
5
5
3
adds additional repeats.
3
RNA primer
4. Primase, DNA polymerase, and ligase then
5
synthesize the lagging strand in the 53
direction, restoring the original length of the
chromosome.
5
3
DNA polymerase
Sliding clamp
Figure 14-16
Mismatched bases
5
OH 3
3
5
DNA polymerase III can repair mismatches.
5
3
5
Figure 14-17
DNA strand
with adjacent
thymine bases
UV light
Kink
Thymine
dimer
Figure 14-18
NUCLEOTIDE EXCISION REPAIR
1. Enzymes detect an
irregularity in DNA
structure and cut the
damaged strand.
Damaged bases
2. An enzyme excises
nucleotides on the
damaged strand.
5
3
5
3. DNA polymerase
fills in the gap in the
53 direction.
3
4. DNA ligase links the
new and old nucleotides.
Repaired damage
From DNA to Chromosomes
 DNA, after replication is bound to
proteins called histones-this is
chromatin
 There are 8 histones per complex, and
DNA winds around these
 In preparation for cell division,
chromatin coils, then supercoils (think
of coiling a phone cord) to become
more compact
Why RNA?
 RNA, like DNA is made of nucleotides,
however:




RNA nucleotides use ribose, not deoxyribose
Uracil, a base, replaces Thymine,
RNA is single-stranded
RNA contains only 1 gene
 RNA is smaller, and can therefore leave the
nucleus through the nuclear pores. DNA
can’t
 RNA code is written as codons (3-base
segments)
Figure 15-5
DNA sequences define the genotype;
proteins create the phenotype.
Information flows from DNA to RNA to
proteins.
3
DNA
(information
storage)
5
Changes in the genotype may
lead to changes in the phenotype.
3
5
TRANSCRIPTION
mRNA
(information
carrier)
5
3
3
5
TRANSLATION
Proteins
(active cell
machinery)
TRANSCRIPTION
TRANSLATION
Figure 15-8
SECOND BASE
Phenylalanine (Phe)
Serine (Ser)
Cysteine (Cys)
Stop codon
Stop codon
Stop codon
Tryptophan (Trp)
Histidine (His)
Leucine (Leu)
Arginine (Arg)
Proline (Pro)
Glutamine (Glu)
Isoleucine (Ile)
Asparagine (Asn)
Serine (Ser)
Lysine (Lys)
Arginine (Arg)
Threonine (Thr)
Methionine (Met)
Start codon
Aspartic acid (Asp)
Valine (Val)
Glycine (Gly)
Alanine (Ala)
Glutamic acid (Glu)
THIRD BASE
FIRST BASE
Leucine (Leu)
Tyrosine (Tyr)
Figure 15-9a
Using the genetic code to predict an amino acid sequence
5
The DNA sequence…
3
…would be transcribed as
…and translated as
3
5
5
Remember that RNA polymerase
works only in the 5 to 3 direction
and that RNA is antiparallel to DNA
3
Also remember that RNA contains
the base uracil (U) instead of
thymine (T), and that uracil forms
a complementary base pair with
adenine (A)
Transcription
 This is the 1st step of protein synthesis, but
also makes any RNA molecule (mRNA,
tRNA, rRNA)
 Begins at the promoter and ends at the
terminator
 Involves:
 Making an RNA copy of a gene from DNA
 Another symphony of enzymes
 Opening DNA at one place only, and when
complete, the DNA closes again
 Making an RNA molecule from 5’ to 3’ (DNA is
still read 3’ to 5’)
Figure 16-1
Non-template
(coding) strand
DNA
3
5
5
RNA
3
5
3
Template strand 3
Phosphodiester bond is
formed by RNA polymerase
after base pairing occurs
5
RNA 5
3
Hydrogen bonds form between
complementary base pairs
DNA template
3
5
Figure 16-3-Table 16-1
Figure 16-3
HOW TRANSCRIPTION BEGINS
Promoter (on non-template strand)
35 box
10 box
Upstream
DNA
Template strand
Upstream
DNA
RNA
+1 site
Sigma
Non-template
strand
+1 site
Active
site
RNA polymerase
Downstream
DNA
RNA
exit
site
Zipper
Rudder
RNA
NTPs
Downstream
DNA
1. Initiation begins
2. Initiation continues
3. Initiation is complete
Sigma binds to promoter
region of DNA.
Sigma opens the DNA helix;
transcription begins.
Sigma releases; mRNA
synthesis continues.
Figure 16-4
HOW TRANSCRIPTION ENDS
Upstream DNA
Hairpin loop
RNA
polymerase
RNA
RNA
Transcription
termination
signal
DNA
Downstream
DNA
1. RNA polymerase reaches a transcription
2. The RNA hairpin causes the RNA strand
termination signal, which codes for RNA
that forms a hairpin.
to separate from the RNA polymerase,
terminating transcription.
Post-transcription RNA modification


After transcription, the new RNA molecules is called preRNA because it’s not yet translation-ready
End-modification



5’ end receives a cap with modified Guanine when transcription
has gone about 20-40 nucleotides
3’ end receives a poly-A tail (30-250 Adenines) that help the
RNA leave the nucleus and attach to the ribosome for
translation
RNA splicing





Often, RNA has large, noncoding sections tha must be
removed
Introns are noncoding regions between coding regions
Exons are regions that actually get translated (the exit the
nucleus)
Small nuclear ribonucleoproteins (snRNPs-”snurps”) join with
other proteins to form a spliceosome.
Spliceosomes cut out the introns and join their flanking exons
together
Figure 16-8
5 cap
Poly(A) tail
5
3
5 untranslated
region
Coding region
3 untranslated
region
Figure 16-7a
Introns must be removed from RNA transcripts.
Intron 1 Intron 2
DNA
3
Promoter
5
Exon 1
Exon 2
Exon 3
Primary RNA transcript 5
Spliced transcript
3
5
3
Figure 16-7b
snRNPs ARE THE EDITORS.
Primary
RNA
5
snRNPs
3
A
Exon 1
Intron
Exon 2
1. Several snRNPs and
A
5
3
Spliceosome
proteins assemble to
form a spliceosome.
The 2 hydroxyl group
on an adenine
nucleotide (A) reacts
with the 5 end of the
intron, breaking RNA.
2. The 5 end of the
5
A
3
5
5
3
Excised
intron
A
5
Exon 1 Exon 2
intron becomes attached
to the A nucleotide,
forming a loop of RNA.
The free 3 end of one
exon reacts with the 5
end of the other.
3. The 3 and 5 ends
of adjacent exons bond
covalently, releasing the
intron (which is then
degraded).
3 Mature mRNA
Other transcripts
 A spliceosome is not always used for other transcripts
(rRNA, tRNA)
 Ribozymes are RNA molecules that act as enzymes
and actually catalyze their own splicing reactions
 For example: the intron can splice itself from a
protozoan, Tetrahymena
 Why introns?
 Some introns may actually help regulate the cell
 Proteins are actually synthesized in Domains
(segments). Each exon codes for a different domain,
and if each domain was continuously coded, there is
a smaller chance that a crossover could alter a
domain. Introns elongate the gene, allowing for a
greater chance of crossovers, and possibly more
evolutionary variety in proteins
Translation
 mRNA goes to the ribosome to be “read” 5’ to 3’
 tRNA carries amino acids to the ribosome to be
assembled when it’s anticodon matches with the
mRNA codon that’s in place in the ribosome
 Aminoacyl-tRNA Synthase ensures that tRNA picks up
an amino acid
 Ribosomes have 2 parts and are made of rRna and
protein
 Small subunit of the ribosome clamps onto the mRNA
 Initiators help this process by attaching at the start
codon sequence (Met)
 Large subunit of the ribosome:
 E site-exit site holds the tRNA that’s about to leave
 P site-pepdityl-tRNA site holds the growing
polypeptide
 A site-aminoacyl-tRNA binding site is where the new
tRNA and amino acid enter the ribosome
Figure 16-12
HOW AMINO ACIDS ARE LOADED ONTO tRNAs
ATP
1. Active site on aminoacyl
Aminoacyl tRNA
synthetase specific
to leucine
tRNA synthetase binds ATP
and amino acid. Each
aminoacyl tRNA synthetase
is specific to one amino acid.
2. Reaction leaves AMP and
Activated
enzyme
complex
amino acid bound to enzyme;
two phosphate groups
released. “Activated” amino
acid has high potential energy.
tRNA
specific to
leucine
3. The activated amino acid is
transferred from tRNA
synthetase to the tRNA
specific to that amino acid;
AMP leaves.
4. The finished aminoacyl
tRNA is ready to participate
in translation.
Aminoacyl tRNA
Figure 16-14
Secondary structure of tRNA
Early model of aminoacyl tRNA function
3
Amino acid
3
Binding site for
amino acid
5
5
Binding site for
mRNA codon
Serine anticodon
5
3
mRNA
Serine codon
Revised model incorporating tertiary structure of tRNA
5
3
Anticodon
Codon
5
mRNA
3
Figure 16-15
Ribbon model of ribosome during translation
Diagram of ribosome during translation
Polypeptide grows in amino
to carboxyl direction
(amino acids in green)
Large
subunit
Peptide bond
formation occurs
here
Aminoacyl
tRNA
Large
subunit
Anticodon
mRNA
3
5
Small
subunit
tRNA in E site
(blue)
Small
subunit
tRNA in P site tRNA in A site
(green)
(red)
The E site
holds a tRNA
that will exit
Codon
The P site holds
the tRNA with
growing polypeptide
attached
The A site
holds an
aminoacyl
tRNA
Figure 16-17l
ELONGATION OF POLYPEPTIDES DURING TRANSLATION
Ribosome
tRNA
Peptidyl site
Exit site
Aminoacyl site
mRNA
5
3
5
3
5
3
Start
codon
1. Incoming aminoacyl tRNA
New tRNA moves into A site, where
its anticodon base pairs with the
mRNA codon.
2. Peptide bond formation
3. Translocation
The amino acid attached to the tRNA
in the P site is transferred to the
tRNA in the A site.
Ribosome moves down mRNA. The
tRNA attached to polypeptide chain
moves into P site. The A site is empty.
Figure 16-17r
ELONGATION OF POLYPEPTIDES DURING TRANSLATION
Exit tunnel
5
3
3
5
Elongation cycle
continues
3
5
4. Incoming aminoacyl tRNA
5. Peptide bond formation
6. Translocation
New tRNA moves into A site, where
its anticodon base pairs with the
mRNA codon.
The polypeptide chain attached to
the tRNA in the P site is transferred
to the aminoacyl tRNA in the A site.
Ribosome moves down mRNA. The
tRNA attached to polypeptide chain
moves into P site. Empty tRNA from
P site moves to E site, where tRNA is
ejected. The A site is empty again.
Figure 16-19
TERMINATION OF TRANSLATION
Large
subunit
Hydrolysis of
bond linking
tRNA and
polypeptide
5
Release
factor
tRNA
mRNA
5
3
mRNA
5
mRNA
3
3
STOP
codon
Small
subunit
1. When translocation opens the A site
2. The hydrolysis reaction frees the
and exposes one of the stop codons, a
protein called a release factor fills the A
site. The release factor catalyzes the
hydrolysis of the bond linking the tRNA
in the P site with the polypeptide chain.
polypeptide, which is released from the
ribosome. The empty tRNAs are released
either along with the polypeptide or…
3. …when the ribosome separates from the
mRNA, and the two ribosomal subunits
dissociate. The subunits are ready to
attach to the start codon of another
message and start translation anew.
What about prokaryotes?
 In bacteria, transcription and
translation are closely coupled
processes, since mRNA doesn’t need
to leave a nucleus to occur.
 Transcription and translation are
simultaneous
Mutations
 Point Mutations are changes to the
sequence of DNA that occur at one point
 Substitutions-one base pair is substituted with
another
 Cause missense mutations-the new base pair
codes for an amino acid, but the amino acid is
not the correct one
 Insertions and deletions-also called frameshift
 Mutagens are physical or chemical agents
that cause mutations by interacting with
the DNA molecule
 Ex: UV radiation
 Base analogues are similar to the DNA bases,
but pair incorrectly during replication
Figure 16-21-Table 16-3
Figure 16-21
DNA point mutation can lead to a different amino acid sequence.
DNA sequence 5
of non-template
(coding) strand
Phenotype
3
Amino acid
sequence
Normal
Normal red blood cells
DNA sequence 5
of non-template
(coding) strand
3
Amino acid
sequence
Mutant
Sickled red blood cells
Figure 16-21-Table 16-4