Transcript Nucleic Acids

Chapter 21 Nucleic Acids and
Protein Synthesis
21.1
Components of Nucleic Acids
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1
Nucleic Acids
Nucleic acids are
 Molecules that store information for cellular growth
and reproduction.
 Deoxyribonucleic acid (DNA) and ribonucleic acid
(RNA).
 Large molecules consisting of long chains of
monomers called nucleotides.
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Nucleic Acids
The nucleic acids DNA and RNA
consist of monomers called
nucleotides that consist of a
 Pentose sugar.
 Nitrogen-containing base.
 Phosphate.
nucleotide
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Nitrogen Bases
The nitrogen bases in
DNA and RNA are
 Pyrimidines C, T, and U
 Purines A and G.
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Nitrogen-Containing Bases in
DNA and RNA
DNA contains the nitrogen bases
 Cytosine (C)
 Guanine (G)
same in both DNA and RNA
 Adenine (A)
 Thymine (T)
different in DNA than in RNA
RNA contains the nitrogen bases
 Cytosine (C)
 Guanine (G)
same in both DNA and RNA
 Adenine (A)
 Uracil (U)
different in DNA than in RNA
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Pentose Sugars
The pentose (five-carbon) sugar
 In RNA is ribose.
 In DNA is deoxyribose with no O atom on carbon 2′.
 Has carbon atoms numbered with primes to
distinguish them from the atoms in nitrogen bases.
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Nucleosides
A nucleoside
 Has a nitrogen base linked
by a glycosidic bond to C1′
of a sugar (ribose or
deoxyribose).
 Is named by changing the
the nitrogen base ending to
-osine for purines and
-idine for pyrimidines.
HO
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Nucleotides
A nucleotide
 Is a nucleoside that
forms a phosphate ester
with the C5′ –OH group
of a sugar (ribose or
deoxyribose).
 Is named using the
name of the nucleoside
followed by
5′-monophosphate.
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Formation of a Nucleotide
A nucleotide forms when the −OH on C5′ of a sugar
bonds to phosphoric acid.
NH2
NH2
N
N
5’
O
O- P OH
O-
+
O
HO CH2
O
OH
deoxycytidine and phosphate
N
O
5’ O
O- P O CH2
-
N
O
O
OH
deoxycytidine monophosphate (dCMP)
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Nucleosides and Nucleotides with
Purines
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Nucleosides and Nucleotides with
Pyrimidines
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Names of Nucleosides and
Nucleotides
TABLE 21.1
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AMP, ADP, and ATP
 Adding phosphate groups to AMP forms the
diphosphate ADP and the triphosphate ATP.
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Learning Check
Give the name and abbreviation for the following, and
list its nitrogen base and sugar.
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Solution
Guanosine 5′-monophosphate; GMP
nitrogen base: guanine
sugar: ribose
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.2
Primary Structure of Nucleic Acids
21.3
DNA Double Helix
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Primary Structure of Nucleic Acids
In the primary structure of nucleic acids,
 Nucleotides are joined by phosphodiester bonds.
 The 3’-OH group of the sugar in one nucleotide forms
an ester bond to the phosphate group on the
5’-carbon of the sugar of the next nucleotide.
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Primary Structure of Nucleic Acids
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Structure of Nucleic Acids
A nucleic acid polymer
 Has a free 5’-phosphate
group at one end and a free
3’-OH group at the other
end.
 Is read from the free 5’-end
using the letters of the
bases.
 This section is read as:
5’—A—C—G—T—3’.
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Example of RNA
The primary
structure of RNA
 Is a single
strand of
nucleotides.
 Consists of the
bases A, C, G,
and U linked by
3’-5’ ester
bonds between
ribose and
phosphate.
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Example of DNA
In the primary
structure of DNA,
A, C, G, and T
are linked by 3’-5’
ester bonds
between
deoxyribose and
phosphate.
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Double Helix of DNA
The DNA structure is a
double helix that
 Consists of two strands of
nucleotides that form a double
helix structure like a spiral
stair case.
 Has hydrogen bonds between
the bases A–T and G–C.
 Has bases along one strand
that complement the bases
along the other.
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Complementary Base Pairs
DNA contains complementary base pairs in which
 Adenine is always linked by two hydrogen bonds
to thymine (A−T).
 Guanine is always linked by three hydrogen bonds
to cytosine (G−C).
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DNA Double Helix Structure
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Learning Check
Write the complementary base sequence for the
matching strand in the following DNA section:
5’—A—G—T—C—C —A—A—T—C—3’
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Solution
Write the complementary base sequence for the
matching strand in the following DNA section:
5’—A—G—T—C—C—A—A—T—C—3’
3’—T—C—A—G—G—T—T—A—G—5’
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.4
DNA Replication
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DNA Replication
DNA replication involves
 Unwinding the DNA.
 Pairing the bases in
each strand with new
bases to form new
complementary strands.
 Producing two new
DNA strands that
exactly duplicate the
original DNA.
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Hydrolysis Energy
 Energy from the
hydrolysis of
each nucleoside
triphosphate
adding to the
complementary
strand is used to
form the
phosphodiester
bond.
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Direction of Replication
During DNA replication,
 An enzyme helicase unwinds the parent DNA at
several sections.
 At each open DNA section called a replication fork,
DNA polymerase catalyzes the formation of
5’-3’ester bonds of the leading strand.
 The lagging strand growing in the 3’-5’ direction is
synthesized in short sections called Okazaki
fragments.
 The Okazaki fragments are joined by DNA ligase to
give a single 3’-5’ DNA strand.
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Direction of Replication
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Learning Check
Match the following:
1) helicase
2) DNA polymerase
3) replication fork
4) Okazaki fragments
A. Short segments formed by the lagging strand.
B. The starting point for synthesis in unwound DNA
sections.
C. The enzyme that unwinds the DNA double helix.
D. The enzyme that catalyzes the formation of
phosphodiester bonds of complementary bases.
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Solution
Match the following:
1) helicase
2) DNA polymerase
3) replication fork
4) Okazaki fragments
A. 4 Short segments formed by the lagging strand.
B. 3 The starting point for synthesis in unwound
DNA sections.
C. 1 The enzyme that unwinds the DNA double helix.
D. 2 The enzyme that catalyzes the formation of
phosphodiester bonds of complementary bases.
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.5
RNA and Transcription
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RNA
RNA
 Transmits information from DNA to make proteins.
 Has several types
Messenger RNA (mRNA) carries genetic
information from DNA to the ribosomes.
Transfer RNA (tRNA) brings amino acids to the
ribosome to make the protein.
Ribosomal RNA (rRNA) makes up 2/3 of
ribosomes where protein synthesis takes
place.
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Types of RNA
TABLE 21.3
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tRNA
Each tRNA
 Has a triplet called an
anticodon that
complements a codon
on mRNA.
 Bonds to a specific
amino acid at the
acceptor stem.
anticodon
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Protein Synthesis
Protein synthesis involves
• Transcription
mRNA is formed from a gene on a DNA strand.
• Translation
tRNA molecules bring amino acids to mRNA to
build a protein.
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Transcription: Synthesis of mRNA
In transcription
 A section of DNA containing the gene unwinds.
 One strand of DNA is copied starting at the initiation
point, which has the sequence TATAAA.
 A mRNA is synthesized using complementary base
pairing with uracil(U) replacing thymine(T).
 The newly formed mRNA moves out of the nucleus
to ribosomes in the cytoplasm.
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RNA Polymerase
During transcription,
 RNA polymerase moves along the DNA template in
the 3’-5’direction to synthesize the corresponding
mRNA.
 The mRNA is released at the termination point.
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Protein Synthesis: Transcription
transcription
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mRNA Processing: Exons and
Introns
 The DNA of eukaryotes contains exons that code
for proteins along with introns that do not.
 The initial mRNA called a pre-RNA includes the
noncoding introns.
 While in the nucleus, the introns are removed from
the pre-RNA.
 The exons that remain are joined to form the
mRNA that leaves the nucleus with the information
for the synthesis of protein.
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Removing Introns from PremRNA
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Regulation of Transcription
Transcription is regulated by
 A specific mRNA synthesized when the cell
requires a particular protein.
 Feedback control, in which the end products
speed up or slow the synthesis of mRNA.
 Enzyme induction, in which high levels of a
reactant induce the transcription process to
provide the necessary enzymes for that reactant.
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Lactose Operon and Repressor
 The lactose operon consists of a control site and
the genes that produce mRNA for lactose
enzymes.
 When there is no lactose in the cell, a regulatory
gene produces a repressor protein that prevents
the synthesis of lactose enzymes.
 The repressor turns off mRNA synthesis.
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Lactose Operon Turned Off
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Lactose Operon and Inducer
 When lactose is present in the cell, some
lactose combines with the repressor, which
removes the repressor from the control site.
 Without the repressor, RNA polymerase
catalyzes the synthesis of the enzymes by the
genes in the operon.
 The level of lactose in the cell induces the
synthesis of the enzymes required for its
metabolism.
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Lactose Operon Turned On
RNA Polymerase
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Learning Check
What is the sequence of bases in mRNA produced
from a section of the template strand of DNA that has
the sequence of bases: 3’–C–T–A–A–G–G–5’?
1. 5’–G–A–T–T–C–C–3’
2. 5’–G–A–U–U–C–C–3’
3. 5’–C–T–A–A–G–G–3’
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Solution
What is the sequence of bases in mRNA produced
from a section of the template strand of DNA that has
the sequence of bases: 3’–C–T–A–A–G–G–5’?
3’–C–T–A–A–G–G–5’?
2. 5’–G–A–U–U–C–C–5’
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.6
The Genetic Code
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Genetic Code
The genetic code
 Is a sequence of amino acids in a mRNA that
determine the amino acid order for the protein.
 Consists of sets of three bases (triplet) along the
mRNA called codons.
 Has a different codon for all 20 amino acids
needed to build a protein.
 Contains certain codons that signal the “start”
and “end” of a polypeptide chain.
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The Genetic Code: mRNA Codons
TABLE 21.4
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Codons and Amino Acids
Suppose we want to determine the amino acids coded
for in the following section of a mRNA.
5’—CCU —AGC—GGA—CUU—3’
According to the genetic code, the amino acids for these
codons are
CCU = Proline
AGC = Serine
GGA = Glycine
CUU = Leucine
The mRNA section codes for the amino acid sequence of
Pro—Ser—Gly—Leu
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Learning Check
Write the order of amino acids coded for by a section
of mRNA with the base sequence
5’—GCC—GUA—GAC—3’.
Some possible codons to use are the following:
GGC = Glycine GAC = Aspartic acid
CUC = Leucine GUA = Valine
GCC = Alanine CGC = Arginine
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Solution
GGC = Glycine
CUC = Leucine
GCC = Alanine
GAC = Aspartic acid
GUA =Valine
CGC = Arginine
5’—GCC—GUA—GAC—3’
Ala——Val——Asp
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.7
Protein Synthesis: Translation
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tRNA Activation
The activation of tRNA
 Occurs when a synthetase
uses energy of ATP
hydrolysis to attach an
amino acid to a specific
tRNA.
 Prepares each tRNA to
use a triplet called an
anticodon to complement a
codon on mRNA.
Anticodon
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Initiation of Protein Synthesis
For the initiation of protein synthesis
 A mRNA attaches to a ribosome.
 The start codon (AUG) binds to a tRNA with
methionine.
 The second codon attaches to a tRNA with the
next amino acid.
 A peptide bond forms between the adjacent amino
acids at the first and second codons.
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Translocation
During translocation
 The first tRNA detaches from the ribosome.
 The ribosome shifts to the adjacent codon on the
mRNA.
 A new tRNA/amino acid attaches to the open
binding site.
 A peptide bond forms and that tRNA detaches.
 The ribosome shifts down the mRNA to read the
next codon.
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Protein Synthesis
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translation
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Termination of Protein Synthesis
In the terminiation of protein synthesis
 After a polypeptide with all the amino acids for a
protein is synthesized.
 The ribosome reaches a “stop” codon: UGA, UAA,
or UAG.
 There is no tRNA with an anticodon for the “stop”
codons.
 Protein synthesis ends.
 The polypeptide detaches from the ribosome.
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Learning Check
Match the following:
1) Activation
3) Translocation
2) Initiation
4) Termination
A.
Ribosomes move along mRNA adding amino acids to
a growing peptide chain.
B. A completed peptide chain is released.
C. A tRNA attaches to its specific amino acid.
D. A tRNA binds to the AUG codon of the mRNA on the
ribosome.
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Solution
Match the following:
1) Activation
3) Translocation
2) Initiation
4) Termination
A. 3 Ribosomes move along mRNA adding amino
acids to a growing peptide chain.
B. 4 A completed peptide chain is released.
C. 1 A tRNA attaches to its specific amino acid.
D. 2 A tRNA binds to the AUG codon of the mRNA on the
ribosome.
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Summary of Protein Synthesis
To summarize protein synthesis:
 A mRNA attaches to a ribosome.
 tRNA molecules bonded to specific amino acids
attach to the codons on mRNA.
 Peptide bonds form between an amino acid and the
peptide chain.
 The ribosome shifts to each codon on the mRNA
until it reaches the STOP codon.
 The polypeptide chain detaches to function as an
active protein.
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Learning Check
The following section of DNA is used to build a mRNA
for a protein.
3’—GAA—CCC—TTT—5’
A. What is the corresponding mRNA sequence?
B. What are the anticodons on the tRNAs?
C. What is the amino acid order in the peptide?
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Solution
3’—GAA—CCC—TTT—5’ DNA
A. What is the corresponding mRNA sequence?
5’—CUU—GGG—AAA—3’ mRNA
B. What are the anticodons for the tRNAs?
mRNA codons
CUU GGG AAA
tRNA anticodons
GAA CCC UUU
C. What is the amino acid order in the peptide?
mRNA 5’—CUU—GGG—AAA—3’
Leu — Gly — Lys
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.8
Genetic Mutations
21.9
Recombinant DNA
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Mutations
A mutation
 Alters the nucleotide sequence in DNA.
 Results from mutagens such as radiation and
chemicals.
 Produces one or more incorrect codons in the
corresponding mRNA.
 Produces a protein that incorporates one or more
incorrect amino acids.
 Causes genetic diseases that produce defective
proteins and enzymes.
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Normal DNA Sequence
 The normal DNA sequence produces a mRNA that
provides instructions for the correct series of amino
acids in a protein.
Correct order
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Mutation: Substitution
In a substitution mutation,
 A different base substitutes for the proper base in DNA.
 There is a change in a codon in the mRNA.
 The wrong amino acid may be placed in the polypeptide.
Incorrect order
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Wrong amino acid
Mutation: Frame Shift
In a frame shift mutation,
 An extra base adds to or is deleted from the normal
DNA sequence.
 All the codons in mRNA and amino acids are incorrect
from the base change.
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Incorrect amino acids
Genetic Diseases
TABLE 21.6
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Learning Check
Identify each type of mutation as a
substitution (S)
frame shift (F)
1. Cytidine (C) enters the DNA sequence.
2. One adenosine is removed from the DNA
sequence.
3. A base sequence of TGA in DNA changes to TAA.
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Solution
Identify each type of mutation as a
substitution (S)
frame shift (F)
1. F Cytosine (C) enters the DNA sequence.
2. F One adenosine is removed from the DNA
sequence.
3. S A base sequence of TGA in DNA changes to TAA.
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Recombinant DNA
In recombinant DNA,
 A DNA fragment from one organism is combined
with DNA in another.
 Restriction enzymes are used to cleave a gene from
a foreign DNA and open DNA plasmids in E. coli.
 DNA fragments are mixed with the plasmids in E.
coli and the ends are joined by ligase.
 The new gene in the altered DNA produces protein.
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Recombinant DNA
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Products of Recombinant DNA
TABLE 21.7
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DNA Fingerprinting
In DNA fingerprinting (Southern transfer):
 Restriction enzymes cut a DNA sample into smaller
fragments (RFLPs).
 The fragments are sorted by size.
 A radioactive isotope that adheres to certain base
sequences in the fragments produces a pattern on Xray film, which is the “fingerprint”.
 The “fingerprint” is unique to each individual DNA.
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Polymerase Chain Reaction
A polymerase chain reaction (PCR):
 Produces multiple copies of a DNA in a short
time.
 Separates the sample DNA strands by heating.
 Mixes the separated strands with enzymes and
nucleotides to form complementary strands.
 Is repeated many times to produce a large
sample of the DNA.
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Polymerase Chain Reaction
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Chapter 21 Nucleic Acids and
Protein Synthesis
21.10
Viruses
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Viruses
Viruses
 Are small particles of DNA or RNA that require a
host cell to replicate.
 Cause a viral infection when the DNA or RNA
enters a host cell.
 Are synthesized in the host cell from the viral
RNA produced by viral DNA.
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Viruses
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Reverse Transcription
In reverse transcription
 A retrovirus, which contains viral RNA, but no
viral DNA, enters a cell.
 The viral RNA uses reverse transcriptase to
produce a viral DNA strand.
 The viral DNA strand forms a complementary
DNA strand.
 The new DNA uses the nucleotides and enzymes
in the host cell to synthesize new virus particles.
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Reverse Transcription
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HIV Virus and AIDS
The HIV-1 virus
HIV virus
 Is a retrovirus that
infects T4
lymphocyte cells.
 Decreases the T4
level making the
immune system
unable to destroy
harmful organisms.
 Causes pneumonia
and skin cancer
associated with
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AIDS.
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AIDS Treatment
 One type of AIDS treatment prevents reverse
transcription of the viral DNA.
 When altered nucleosides such as AZT and ddI are
incorporated into viral DNA, the virus is unable to
replicate.
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AIDS Treatment
Azidothymine (AZT)
Dideoxyinosine (ddI)
O
H
H 3C
HO CH2
O
N
N
O
H
N
O
HO CH2
N
O
H
H
H
H
N3
H
H
H
N
N
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AIDS Treatment
 Another type of AIDS treatment involves protease
inhibitors such as saquinavir, indinavir, and ritonavir.
 Protease inhibitors modify the active site of the
protease enzyme, which prevents the synthesis of
viral proteins.
Inhibited by
AZT, ddI
reverse
transcriptase
Viral RNA
Viral DNA
Inhibited by
protease inhibitors
protease
Viral proteins
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Learning Check
Match the following:
1) Virus
3) Protease inhibitor
2) Retrovirus
4) Reverse transcription
A. A virus containing RNA.
B. Small particles requiring host cells to replicate.
C. A substance that prevents the synthesis of viral
proteins.
D. Using viral RNA to synthesize viral DNA.
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Solution
Match the following:
1) Virus
3) Protease inhibitor
2) Retrovirus
4) Reverse transcription
A. 2 A virus containing RNA.
B. 1 Small particles requiring host cells to replicate.
C. 3 A substance that prevents the synthesis of viral
proteins.
D. 4 Using viral RNA to synthesize viral DNA.
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