1 chapter 17 nucleic acids and protein synthesis 17.1 components of nucleic acids 17.2 primary...
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Chapter 17 Nucleic Acids and Protein Synthesis
17.1 Components of Nucleic Acids
17.2 Primary Structure of Nucleic Acids
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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 RNAconsist of monomers callednucleotides 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 RNA
RNA contains the nitrogen bases • Cytosine (C)• Guanine (G) same in both DNA and RNA• Adenine (A)• Uracil (U) different in DNA than 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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HO
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.
Nucleosides
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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.
Nucleotides
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Formation of a Nucleotide
A nucleotide forms when the −OH on C5’ of a
sugar bonds to phosphoric acid.
deoxycytidine monophosphate (dCMP)
O-
O-
O
P O CH2
O
NH2
N
N
OH
OO-
O-
O
P OH +
deoxycytidine and phosphate
HO CH2
O
NH2
N
N
OH
O
5’ 5’
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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
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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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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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A nucleic acid• 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 example reads —A—C—G—T—.
Structure of Nucleic Acids
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Example of RNA Structure
The primary structure
of RNA,• is a single strand
of nucleotides with bases A, C, G, and U.
• is linked by phosophodiester bonds between ribose and phosphate.
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Chapter 17 Nucleic Acids and Protein Synthesis
17.3 DNA Double Helix
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Example of DNA
In DNA,• nucleotides
containing bases A, C, G, and T are linked by ester bonds between deoxyribose sugars and phosphate groups. Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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DNA Double Helix
A double helix • is the structure of DNA. • has two strands of nucleotides that wind together. • is held in place by of two hydrogen bonds that form
between the base pairs A-T.• is held in place by three hydrogen bonds that form
between the base pairs G-C.
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Complementary Base Pairs
DNA contains complementary base pairs in which• Adenine is always linked by two hydrogen bonds
with thymine (A−T).• Guanine is always linked by three hydrogen with
Cytosine (G−C).
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Double Helix of DNA
In the double helix of DNA
• two strands of nucleotides form a double helix structure like a spiral stair case.
• hydrogen bonds link bases A–T and G–C.
• the bases along one strand complement the bases along the other.
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Write the complementary base sequence for the matching strand in the following DNA section:
—A—G—T—C—C—A—A—T—C—
Learning Check
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Write the complementary base sequence for the matching strand in the following DNA section:
—A—G—T—C—C—A—A—T—C—
Solution
’—T—C—A—G—G—T—T—A—G—
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DNA Replication
In DNA replication • genetic information is
maintained each time a cell divides.
• the DNA strands unwind.• each parent strand bonds
with new complementary bases.
• two new DNA strands form that are exact copies of the original DNA. Copyright © 2005 by
Pearson Education, Inc.Publishing as Benjamin Cummings
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Chapter 17 Nucleic Acids and Protein Synthesis
17.4 RNA and the Genetic Code
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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
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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 tobuild a protein.
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During transcription • a section of DNA containing the gene unwinds.• one strand of DNA bases is used as a template. • 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.
Transcription: Synthesis of mRNA
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RNA Polymerase
During transcription,• RNA polymerase moves along the DNA template to
synthesize the corresponding mRNA.• the mRNA is released at the termination point.
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Protein Synthesis: Transcription
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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: – C –T –A –A –G –G – ?
1. –G –A –T –T –C –C –
2. –G –A –U –U –C –C –
3. –C –T –A –A –G –G –
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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: – C –T –A –A –G –G – ?
–C –T –A –A –G –G –
2. –G –A –U –U –C –C –
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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
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Determine the amino acids from the followingcodons in a section of mRNA.
—CCU —AGC—GGA—CUU—According to the genetic code, the amino acids for thesecodons are
CCU = proline AGC = serine GGA = glycine CUU = leucine
This mRNA section codes for an amino acid sequence of —CCU —AGC—GGA—CUU—
—Pro — Ser — Gly — Leu —
Codons and Amino Acids
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Write the order of amino acids coded for by a section of
mRNA with the base sequence
—GCC—GUA—GAC—
GGC = Glycine GAC = Aspartic acidCUC = Leucine GUA = ValineGCC = Alanine CGC = Arginine
Learning Check
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GGC = Glycine GAC = Aspartic acidCUC = Leucine GUA =ValineGCC = Alanine CGC = Arginine
—GCC—GUA—GAC—
Ala — Val — Asp
Solution
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Chapter 17 Nucleic Acids and Protein Synthesis
17.5 Protein Synthesis: Translation
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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.
Initiation of Protein Synthesis
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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 next
codon.
Translocation
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Peptide chain starts to form
anticodons UAC AGA AGA GAG
• • • • • • • • • • • •
AUG UCU CUC UCU CUC UUU
Peptide Formation
Met Ser
Met
Ser LeutRNA
Ribosome ribosome shifts
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Protein Synthesis
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings translation
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In the termination step • all the amino acids are linked. • the ribosome reaches a “stop” codon: UGA, UAA, or
UAG. • there is no tRNA with an anticodon for the “stop”
codons.• the polypeptide detaches from the ribosome.
Termination
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Match the following. 1) Activation 2) Initiation 3) Translocation 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.
Learning Check
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Match the following.
1) Activation 2) Initiation
3) Translocation 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.
Solution
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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 reach the STOP codon.• The polypeptide chain detaches to function as an
active protein.
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The following section of DNA is used to build mRNA for a protein.
—GAA—CCC—TTT—
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?
Learning Check
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A. What is the corresponding mRNA sequence?
—CUU—GGG—AAA— mRNA
B. What are the anticodons for the tRNAs?
GAA for CUU; CCC for GGG; UUU for AAA
C. What is the amino acid order in the peptide?
—CUU—GGG—AAA— mRNA
Leu — Gly — Lys
Solution
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Learning Check
Place the following statements in order of their
occurrence in protein synthesis.
A. mRNA attaches to a ribosome.
B. The ribosome moves along a mRNA to add amino acids to the growing peptide chain.
C. A completed polypeptide is released.
D. A tRNA brings an amino acid to its codon on mRNA.
E. DNA produces mRNA.
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Solution
Place the following statements in order of their
occurrence in protein synthesis.
E. DNA produces mRNA.
A. mRNA attaches to a ribosome.
D. A tRNA brings an amino acid to its codon on mRNA.
B. The ribosome moves along a mRNA to add amino
acids to the growing peptide chain.
C. A completed polypeptide is released.
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17.6 Genetic Mutations
17.7 Viruses
Chapter 17 Nucleic Acids and Protein Synthesis
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A mutation can• alter the nucleotide sequence in DNA.• result from mutagens such as radiation and
chemicals.• produce one or more incorrect codons in mRNA.• produce a protein containing one or more incorrect
amino acids.• produce defective proteins and enzymes.• cause genetic diseases.
Mutations
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Examples of Genetic Diseases
Galactosemia
Cystic fibrosis
Downs syndrome
Muscular dystrophy
Huntington’s disease
Sickle-cell anemia
Hemophilia
Tay-Sachs disease
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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
Substitution• of a base in
DNA changes a codon in the mRNA.
• of a different codon leads to the placement of an incorrect amino acid in the polypeptide. Incorrect order
Wrong amino acid
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Frame Shift Mutation
In frame shiftmutation,• 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. Incorrect amino acids
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Learning Check
Identify each type of mutations as 1) substitution 2) frame shift.
A. Cytidine (C) enters the DNA sequence.B. One adenosine is removed from the DNA sequence.C. A base sequence of TGA in DNA changes to TAA.
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Solution
Identify each type of mutations as 1) substitution 2) frame shift.
A. 2 Cytosine (C) enters the DNA sequence.B. 2 One adenosine is removed from the DNA sequence.C. 1 A base sequence of TGA in DNA changes to TAA.
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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 • is a retrovirus that
infects T4 lymphocyte cells.
• decreases the T4 level and the immune system fails to destroy harmful organisms.
• causes pneumonia and skin cancer associated with AIDS.
HIV virus
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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)
OCH2HO
H H
HN3
N
N
O
O
HH3C
N
N
N
N
O
H
O
H H
H H
CH2HO
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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 Inhibited by AZT, ddI protease inhibitors
reverse transcriptase protease
Viral RNA Viral DNA Viral proteins
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Learning Check
Match the following.
1) Virus 2) Retrovirus
3) Protease inhibitor 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 2) Retrovirus
3) Protease inhibitor 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