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Chapter 10
Molecular Biology of the Gene
PowerPoint Lectures for
Campbell Biology: Concepts & Connections, Seventh Edition
Reece, Taylor, Simon, and Dickey
© 2012 Pearson Education, Inc.
Lecture by Edward J. Zalisko
Figure 10.0_1
Chapter 10: Big Ideas
The Structure of the
Genetic Material
DNA Replication
The Flow of Genetic
Information from DNA to
RNA to Protein
The Genetics of Viruses
and Bacteria
THE STRUCTURE OF THE
GENETIC MATERIAL
DNA
&
RNA
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10.1 SCIENTIFIC DISCOVERY: Experiments
showed that DNA is the genetic material
 Griffith’s experiment on bacterial transformation
 Hershey-Chase blender experiment
 Skip the experimental details
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10.2 DNA and RNA are polymers of nucleotides
 DNA and RNA are nucleic acids.
 Both are polymers made up of nucleotides.
 DNA is made from 4 DNA nucleotides.
 Nucleotides contain
– Nitrogenous base
– 5-carbon sugar
– Phosphate group
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Figure 10.2A_3
Nitrogenous base
(can be A, G, C, or T)
Thymine (T)
Phosphate
group
Sugar
(deoxyribose)
DNA nucleotide
10.2 DNA and RNA are polymers of nucleotides
 DNA is made up of four DNA nucleotides.
 Each nucleotide has a different nitrogen-containing
base:
– adenine (A),
– cytosine (C),
– thymine (T), and
– guanine (G).
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Figure 10.2B
Thymine (T)
Cytosine (C)
Pyrimidines
Guanine (G)
Adenine (A)
Purines
Figure 10.2A_2
Sugar-phosphate
backbone
A
A
Covalent
bond
joining
nucleotides
C
DNA
nucleotide
T
Nitrogenous
base
Sugar
C
T
G
G
G
G
Two representations
of a DNA polynucleotide
Phosphate
group
Figure 10.5B
3 end
5 end
P
HO
5
4
3
2
1
2
A
T
5
P
C
P
G
C
P
P
T
3 end
P
G
P
OH
3
4
1
A
P
5 end
10.2 DNA and RNA are polymers of nucleotides
 The nucleotides are joined to one another by a sugarphosphate backbone.
 The two DNA strands are anti-parallel. This means
that one strand runs in 5’ to 3’ direction and the other
in 3’ to 5’ direction.
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Figure 10.2A_1
A
G
C
T
T
A
C
G
T
A
C
G
G
A
C
G
T
T
T
A
A
C
G
T
A
A DNA
double helix
10.2 DNA and RNA are Polymers of Nucleotides
 RNA (ribonucleic acid) is made up of 4 RNA
nucleotides
 RNA
– uses the sugar ribose (instead of deoxyribose in DNA)
and
– RNA has the nitrogenous base uracil (U) instead of
thymine.
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Figure 10.2C
Nitrogenous base
(can be A, G, C, or U)
Phosphate
group
Uracil (U)
An RNA nucleotide
Sugar
(ribose)
Figure 10.2D
Cytosine
Uracil
Adenine
Guanine
Ribose
Phosphate
Animation: DNA and RNA Structure
Right click on animation / Click play
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10.3 SCIENTIFIC DISCOVERY: DNA is a
double-stranded helix
 In 1953, James D. Watson and Francis Crick
deduced the secondary structure of DNA, using
– X-ray crystallography data of DNA from the work of
Rosalind Franklin and Maurice Wilkins and
– Chargaff’s observation that in DNA,
– the amount of adenine was equal to the amount of thymine
and
– the amount of guanine was equal to that of cytosine.
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Figure 10.3B
Figure 10.3A
Rosalind Franklin and her
X-ray image
10.3 SCIENTIFIC DISCOVERY: DNA is a
double-stranded helix
 Watson and Crick reported that DNA consisted of
two polynucleotide strands wrapped into a double
helix.
– The sugar-phosphate backbone is on the outside.
– The nitrogenous bases are perpendicular to the
backbone in the interior.
– Specific pairs of bases give the helix a uniform shape.
– A pairs with T, forming two hydrogen bonds, and
– G pairs with C, forming three hydrogen bonds.
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10.3 SCIENTIFIC DISCOVERY: DNA is a
double-stranded helix
 In 1962, the Nobel Prize was awarded to
– James D. Watson, Francis Crick, and Maurice Wilkins.
– Rosalind Franklin probably would have received the
prize as well but for her death from cancer in 1958.
Nobel Prizes are never awarded posthumously.
 The genetic information in a chromosome is
encoded in the nucleotide sequence of DNA.
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Figure 10.3C
Twist
Figure 10.3D_1
C
C
G
G
G
G
C
C
T
Base pair
A
A
T
C
G
A
T
T
C
G
C
A
G
C
G
A
A
T
T
T
Ribbon
model
A
Figure 10.3D_2
Hydrogen bond
G
T
C
A
A
C
T
G
Partial chemical
structure
DNA Problem:
Write the base sequence in the complementary
DNA strand and label the ends
5’ C T A T G C A A G T C A T T __’
__’ ______________________ __’
DNA REPLICATION
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10.4 DNA replication depends on specific base
pairing
 DNA replication refers to making of two strands of
DNA starting with one parental strand
 Takes place in the synthesis (S) phase of the cell
cycle
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10.4 DNA replication depends on specific base
pairing
 DNA replication follows a semiconservative
model.
– The two DNA strands separate.
– Each strand is used as a pattern to produce a
complementary strand, using specific base pairing.
– Each new DNA helix has one old strand with one new
strand.
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Figure 10.4A_s1
A
T
C
G
G
C
A
T
T
A
A parental
molecule
of DNA
Figure 10.4A_s2
T
A
T
A
C
G
C
G
C
G
C
A
T
A
T
T
A
T
A parental
molecule
of DNA
G
A
C
Free
nucleotides
The parental strands
separate and serve
as templates
T
G
A
Figure 10.4A_s3
A
T
A
C
G
C
G
C
A
T
T
A
T
A
T
A
T
G
C
G
C
G
G
C
G
C
G
C
T
A
T
A
T
A
T
A
T
A
T
A
T
A
A parental
molecule
of DNA
G
C
Free
nucleotides
The parental strands
separate and serve
as templates
Two identical
daughter molecules
of DNA are formed
Figure 10.4B
A
T
G
A
A
T
C
T
T
A
Parental DNA
molecule
Daughter
strand
Parental
strand
Daughter DNA
molecules
10.5 DNA replication proceeds in two directions
at many sites simultaneously
 DNA replication begins at the origins of replication
where
– DNA unwinds at the origin to produce a “bubble,”
– replication proceeds in both directions from the origin,
and
– replication ends when products from the bubbles
merge with each other.
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10.5 DNA replication proceeds in two directions
at many sites simultaneously
 DNA replication occurs in the 5 to 3 direction.
– Replication is continuous on the 3 to 5 template. This
strand is called the leading strand
– Replication is discontinuous on the 5 to 3 template,
forming short segments. This DNA strand is called the
lagging strand
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10.5 DNA replication proceeds in two directions
at many sites simultaneously
 Key enzymes involved in DNA replication.
1. DNA helicase unwinds the double helix and
breaks hydrogen bonds
2. DNA polymerase
– adds nucleotides to a growing chain and
– proofreads and corrects improper base
pairings.
3. DNA ligase joins small fragments into a
continuous chain.
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Figure 10.5A
Parental
DNA
molecule
Origin of
replication
“Bubble”
Two
daughter
DNA
molecules
Parental strand
Daughter strand
Figure 10.5C
DNA polymerase
molecule
5
3
Parental DNA
Replication fork
5
3
DNA ligase
Overall direction of replication
3
5
This daughter
strand is
synthesized
continuously
This daughter
strand is
3 synthesized
5 in pieces
THE FLOW OF GENETIC
INFORMATION FROM DNA TO
RNA TO PROTEIN
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10.6 The DNA genotype is expressed as proteins,
which provide the molecular basis for
phenotypic traits
 DNA specifies traits by dictating protein synthesis.
 The molecular chain of command is from
– DNA in the nucleus is transcribed to RNA and
– RNA in the cytoplasm is translated to protein.
 Transcription is the synthesis of RNA under the
direction of DNA.
 Translation is the synthesis of proteins under the
direction of RNA.
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Figure 10.6A_s1
DNA
NUCLEUS
CYTOPLASM
Figure 10.6A_s2
DNA
Transcription
RNA
NUCLEUS
CYTOPLASM
Figure 10.6A_s3
DNA
Transcription
RNA
NUCLEUS
Translation
Protein
CYTOPLASM
10.7 Genetic information written in codons is
translated into amino acid sequences
 The sequence of nucleotides in DNA provides a
code for constructing a protein.
– Protein construction requires a conversion of a
nucleotide sequence to an amino acid sequence.
– Transcription rewrites the DNA code into RNA, using
the same nucleotide “language.”
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10.7 Genetic information written in codons is
translated into amino acid sequences
– The flow of information from gene to protein is based
on a triplet code: the genetic instructions for the
amino acid sequence of a polypeptide chain are written
in DNA and RNA as a series of nonoverlapping threebase “words” called codons.
– Translation involves switching from the nucleotide
“language” to the amino acid “language.”
– Each amino acid is specified by a codon.
– 64 codons are possible.
– Some amino acids have more than one possible codon.
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Figure 10.7
DNA
molecule
Gene 1
Gene 2
Gene 3
DNA
A A A C C G G C A A A A
Transcription
RNA
Translation
U U
U G G C
Codon
Polypeptide
Amino
acid
C G U
U
U U
Figure 10.7_1
DNA
A A
A C
U U
U
C G G
C
A
A
A A
C G U
U
U
Transcription
RNA
Translation
Codon
Polypeptide
Amino
acid
G
G C
U
10.8 The genetic code dictates how codons are
translated into amino acids
 Characteristics of the genetic code
– Three nucleotides specify one amino acid.
– 61 codons correspond to amino acids.
– AUG codes for methionine and signals the start of
transcription.
– 3 “stop” codons signal the end of translation.
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10.8 The genetic code dictates how codons are
translated into amino acids
 The genetic code is
– redundant, with more than one codon for some amino
acids,
– unambiguous in that any codon for one amino acid
does not code for any other amino acid,
– nearly universal—the genetic code is shared by
organisms from the simplest bacteria to the most
complex plants and animals, and
– without punctuation in that codons are adjacent to each
other with no gaps in between.
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Figure 10.8A
Third base
First base
Second base
Figure 10.8B_s1
Strand to be transcribed
T A C T
T
C A A A A T
C
DNA
A T G A A G T
T T
T A G
Figure 10.8B_s2
Strand to be transcribed
T A C T
T
C A A A A T
C
DNA
A T G A A G T
T T
T A G
Transcription
RNA
A U G A A G U U U U A G
Figure 10.8B_s3
Strand to be transcribed
T A C T
T
C A A A A T
C
DNA
A T G A A G T
T T
T A G
Transcription
RNA
A U G A A G U U U U A G
Translation
Start
codon
Polypeptide
Met
Stop
codon
Lys
Phe
DNA Problem:
 Write the mRNA sequence and the tRNA
sequence. Using the genetic code, write the
amino acid sequence
DNA
TAC GCT TAA TCA GGC GTA ATT
mRNA
AUG
tRNA
UAC
Amino acid
Met
UAA
(stop)
10.9 Transcription produces genetic messages in
the form of RNA
 Overview of transcription
– An RNA molecule is transcribed from a DNA template
by a process that resembles the synthesis of a DNA
strand during DNA replication.
– RNA nucleotides are linked by the transcription
enzyme RNA polymerase.
– Specific sequences of nucleotides along the DNA mark
where transcription begins and ends.
– The “start transcribing” signal is a nucleotide sequence
called a promoter.
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10.9 Transcription produces genetic messages in
the form of RNA
– Transcription begins with initiation, as the RNA
polymerase attaches to the promoter.
– During the second phase, elongation, the RNA grows
longer.
– As the RNA peels away, the DNA strands rejoin.
– Finally, in the third phase, termination, the RNA
polymerase reaches a sequence of bases in the DNA
template called a terminator, which signals the end of
the gene.
– The polymerase molecule now detaches from the RNA
molecule and the gene.
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Figure 10.9A
Free RNA
nucleotides
RNA
polymerase
C C A A
A U C C A
T A G G T
Direction of
transcription
Newly made RNA
T
Template
strand of DNA
Figure 10.9B
RNA polymerase
DNA of gene
Terminator
DNA
Promoter
DNA
1
Initiation
2
Elongation
Area shown
in Figure 10.9A
3
Termination
Growing
RNA
Completed
RNA
RNA
polymerase
Figure 10.9B_1
RNA polymerase
DNA of gene
Promoter
DNA
1
Initiation
Terminator
DNA
Figure 10.9B_2
2
Elongation
Area shown
in Figure 10.9A
Growing
RNA
Figure 10.9B_3
3
Completed
RNA
Termination
Growing
RNA
RNA
polymerase
10.10 Eukaryotic RNA is processed before
leaving the nucleus as mRNA
 Messenger RNA (mRNA)
– Has codons that encodes amino acid sequences
– Conveys genetic messages from DNA to the
translation machinery of the cell, which in
– prokaryotes, occurs in the same place that mRNA is made,
but in
– eukaryotes, mRNA must exit the nucleus via nuclear pores to
enter the cytoplasm.
– Eukaryotic mRNA has
– introns, interrupting sequences that separate
– exons, the coding regions.
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10.10 Eukaryotic RNA is processed before
leaving the nucleus as mRNA
 Three ways Eukaryotic RNA is processed
– RNA splicing removes introns and joins exons to
produce a continuous coding sequence.
– Addition of a 5’ cap
– Addition of a 3’ tail
 The cap and tail protects mRNA
– facilitate the export of the mRNA from the nucleus,
– protect the mRNA from attack by cellular enzymes, and
– help ribosomes bind to the mRNA.
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Figure 10.10
Exon Intron
Exon
Intron
Exon
DNA
Cap
RNA
transcript
with cap
and tail
Transcription
Addition of cap and tail
Introns removed
Tail
Exons spliced together
mRNA
Coding sequence
NUCLEUS
CYTOPLASM
Translation
 Four requirements for translation to occur
– mRNA
– tRNA
– Ribosome
– Amino acids
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10.11 Transfer RNA molecules serve as
interpreters during translation
 Transfer RNA (tRNA) molecules function as a
language interpreter,
– converting the genetic message of mRNA
– into the language of proteins.
 Transfer RNA molecules perform this interpreter
task by
– picking up the appropriate amino acid and
– using a special triplet of bases, called an anticodon, to
recognize the appropriate codons in the mRNA.
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Figure 10.11A
Amino acid
attachment site
Hydrogen bond
RNA polynucleotide
chain
Anticodon
A tRNA molecule, showing
its polynucleotide strand
and hydrogen bonding
A simplified
schematic of a tRNA
10.12 Ribosomes build polypeptides
 Ribsome: Translation occurs on the surface of the
ribosome.
– Ribosomes coordinate the functioning of mRNA and
tRNA and, ultimately, the synthesis of polypeptides.
– Ribosomes have two subunits: small and large.
– Each subunit is composed of ribosomal RNAs and
proteins.
– Ribosomal subunits come together during translation.
– Ribosomes have binding sites for mRNA and tRNAs.
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Figure 10.12B
tRNA binding sites
Large
subunit
P A
site site
Small
subunit
mRNA binding site
Figure 10.12C
The next amino
acid to be added
to the polypeptide
Growing
polypeptide
mRNA
tRNA
Codons
10.13 An initiation codon marks the start of an
mRNA message
 Translation can be divided into the same three
phases as transcription:
1. initiation,
2. elongation, and
3. termination.
 Initiation brings together
– mRNA,
– a tRNA bearing the first amino acid, and
– the two subunits of a ribosome.
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10.13 An initiation codon marks the start of an
mRNA message
 Initiation establishes where translation will begin.
 Initiation occurs in two steps.
1. An mRNA molecule binds to a small ribosomal subunit and
the first tRNA binds to mRNA at the start codon.
– The start codon reads AUG and codes for methionine.
– The first tRNA has the anticodon UAC.
2. A large ribosomal subunit joins the small subunit, allowing
the ribosome to function.
– The first tRNA occupies the P site, which will hold the growing
peptide chain.
– The A site is available to receive the next tRNA.
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Figure 10.13B
The initiation of translation
Large
ribosomal
subunit
Initiator
tRNA
P
site
mRNA
U A C
A U G
Start codon
1
Small
ribosomal
subunit
2
A
site
U A C
A U G
10.14 Elongation adds amino acids to the
polypeptide chain until a stop codon
terminates translation
 Once initiation is complete, amino acids are
added one by one to the first amino acid.
 Elongation is the addition of amino acids to the
polypeptide chain.
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10.14 Elongation adds amino acids to the
polypeptide chain until a stop codon
terminates translation
 Elongation: Each cycle of elongation has three
steps.
1. Codon recognition: The anticodon of an incoming
tRNA molecule, carrying its amino acid, pairs with the
mRNA codon in the A site of the ribosome.
2. Peptide bond formation: The new amino acid is
joined to the chain.
3. Translocation: tRNA is released from the P site and
the ribosome moves tRNA from the A site into the P
site.
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10.14 Elongation adds amino acids to the
polypeptide chain until a stop codon
terminates translation
 Termination:
– When the ribosome reaches a stop codon,
– the completed polypeptide is freed from the last tRNA,
and
– the ribosome splits back into its separate subunits.
© 2012 Pearson Education, Inc.
Figure 10.14_s1
Polypeptide
P
site
mRNA
Amino
acid
A
site
Anticodon
Codons
1
Codon recognition
Figure 10.14_s2
Polypeptide
P
site
mRNA
Amino
acid
A
site
Anticodon
Codons
1
Codon recognition
2
Peptide bond
formation
Figure 10.14_s3
Polypeptide
P
site
mRNA
Amino
acid
A
site
Anticodon
Codons
1
Codon recognition
2
New
peptide
bond
3
Translocation
Peptide bond
formation
Figure 10.14_s4
Polypeptide
P
site
mRNA
Amino
acid
A
site
Anticodon
Codons
1
Codon recognition
mRNA
movement
Stop
codon
2
New
peptide
bond
3
Translocation
Peptide bond
formation
10.15 Review: The flow of genetic information in
the cell is DNA  RNA  protein
 Transcription is the synthesis of RNA from a DNA
template. In eukaryotic cells,
– transcription occurs in the nucleus and
– the mRNA must travel from the nucleus to the cytoplasm.
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10.16 Mutations can change the meaning of genes
 A mutation is any change in the nucleotide
sequence of DNA.
 Types of Mutations
– Single base substitutions
– Frame-Shift mutations (through deletions and insertions)
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10.16 Types of Mutations
1. Base substitutions involve the replacement of one
nucleotide with another. Base substitutions may
– have no effect at all, producing a silent mutation,
– change the amino acid coding, producing a missense
mutation, which produces a different amino acid,
– lead to a base substitution that produces an improved
protein that enhances the success of the mutant organism
and its descendant, or
– change an amino acid into a stop codon, producing a
nonsense mutation.
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Figure 10.16A
The molecular basis of sickle-cell disease
Normal hemoglobin DNA
C T
Mutant hemoglobin DNA
C A T
T
mRNA
mRNA
G A A
G U A
Normal hemoglobin
Sickle-cell hemoglobin
Val
Glu
Figure 10.16B
Normal
gene
mRNA
Protein
Nucleotide
substitution
A
U G A A G U
Met
A U G A
Met
Lys
U
U G G C G
C
Phe
Gly
Ala
U A G C
A G U U
Lys
Phe
Ser
G C
A
A
Ala
U Deleted
Nucleotide
deletion
A U G A A G
Met
U
U G G C G
Ala
Leu
Lys
C A
U
His
Inserted
Nucleotide
insertion
A U G A A G
Met
Lys
U
U G
Leu
U G G
C G C
Ala
His
Figure 10.8A
Third base
First base
Second base
10.16 Types of Mutations
2. Frame-Shift mutations
– Mutations that result in deletions or insertions that may
alter the reading frame (triplet grouping) of the mRNA, so
that nucleotides are grouped into different codons,
– lead to significant changes in amino acid sequence
downstream of the mutation, and
– produce a nonfunctional polypeptide.
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DNA Problem: Normal DNA
 Write the mRNA sequence and the tRNA
sequence. Using the genetic code, write the amino
acid sequence
DNA
TAC GCT TAA TCA GGC GTA ATT
mRNA
AUG …… ….. AGU ……. …… UAA
tRNA
UAC
Amino acid
Met ……. ……. Ser ……. ……. (stop)
DNA Problem: Mutation #1
 Write the mRNA sequence and the tRNA
sequence. Using the genetic code, write the amino
acid sequence for the mutated DNA
DNA
TAC GCT TAA GCA GGC GTA ATT
mRNA
AUG …… ……. CGU ……. …… UAA
tRNA
UAC
Amino acid
Met …… ……. Ala …….. ……. (stop)
What kind of mutation is this?
DNA Problem: Mutation #2
 Write the mRNA sequence and the tRNA
sequence. Using the genetic code, write the
amino acid sequence for this mutated DNA.
DNA
TAC ACT TAA TCA GGC GTA ATT
mRNA
AUG …...
tRNA
UAC
Amino acid
Met
What kind of mutation is this?
DNA Problem: Mutation #3
 Write the mRNA sequence and the tRNA
sequence. Using the genetic code, write the
amino acid sequence, if the T base is deleted
DNA
TAC GCT TAA TCA GGC GTA ATT
mRNA
AUG CGA
tRNA
UAC
Amino acid
Met
10.16 Mutations can change the meaning of genes
 Mutagenesis is the production of mutations.
 Mutations can be caused by
– spontaneous errors that occur during DNA replication
or recombination or
– mutagens, which include
– high-energy radiation such as X-rays and
ultraviolet light and
– chemicals.
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