Download Chapter Sixteen

Chapter Seventeen
Nucleic Acids
The Third Hour Exam
Will be held on FRIDAY, March 2 nd, 2007,
and will cover Chapters 16 and 17
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16a–2
Nucleic acids comprise the genetic machinery of living cells.
They are polymers composed of nucleotide monomer units.
They were first discovered by F. Miescher in 1869, but their
importance was not recognized for nearly a century.
There are two types of nucleic acids: deoxyribonucleic acid
(DNA) and ribonucleic acid (RNA).
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16a–3
Nucleotides: The Building Blocks
of Nucleic Acids
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16a–4
Nucleotides are three-component molecules containing:
(1) a phosphate group,
(2) a pentose sugar (deoxyribose or ribose),
(3) a heterocyclic nitrogen-containing base ( a purine or
a pyrimidine).
Base
Phosphate
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Sugar
16a–5
The Pentose Sugars
The only difference is that an OH group is bound to carbon 2’
in ribose, and this is replaced by an H in deoxyribose.
RNA has ribose as its sugar, DNA uses deoxyribose.
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16a–6
The Nitrogenous Bases
There are two basic forms: Purines and Pyrimidines:
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16a–7
The Nitrogenous Bases
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16a–8
The Phosphate Group
Phosphoric acid is H3PO4
OOH
|
O==P—OH
|
OH
Phosphoric Acid
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O|
O==P—OH
|
OIonized form at
pH = 7
16a–9
Nucleotide Formation
Nucleotides are formed from their subunits by (you
guessed it!) condensation reactions, splitting out waters.
Both the phosphate-sugar reaction and the sugar-base
reaction split out water molecules.
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16a–10
Nucleotide Nomenclature
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16a–11
Additional corrections to the text
Example 17.1: The original strand is not properly copied
in the illustration at the bottom (there’s an extra T). Thanks to
C. Hager
Question 17.28: Ignore part (d). Thanks to C. Hager
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16a–12
Primary Structure of Nucleic Acids
The backbone of a nucleic acid is composed of sugar-phosphate
bonds. The bases stick off of the sugar units and their sequence
forms the primary structure.
Only four types of bases appear in any nucleic acid.
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16a–13
The Structure of a Nucleic Acid
Note the 3’ and 5’
ends.
The primary structure
has a direction.
The links are 3’,5’
“phosphodiester”
links. Each
phosphate link has a
-1 charge.
By convention, the
strand is read from
the 5’ to the 3’ end.
Here TGCA.
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16a–14
The Double Helix
In 1953 in James D. Watson
and Francis Crick figured
out the structure of DNA—
it turned out to be a double
helix.
They published their results
in the journal Nature.
Watson later wrote a controversial book called The
Double Helix.
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16a–15
Two key papers in 1953
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16a–16
The Key: Hydrogen Bonding Between the Bases
The key discovery was that
hydrogen bonds could form
between the bases on the two
strands. One strand was thus a
sort of alter-ego of the other.
When the two strands separated
they would then form templates
for each other: hydrogen bonds
with the right nucleotides could
then form new helixes just like
the originals.
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16a–17
There Were Clues Earlier
￭ One clue was that in DNA the percentage of the base
adenine was always the same as that of the base thymine,
and the percentage of guanine was always the same as that
of the base cytosine: that is,
%A = %T and %G = %C
￭ Also, experiments indicated that nucleic acids were the
genetic materials in viruses.
￭ But these clues were largely ignored while most workers
focused on proteins as the supposed genetic material.
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16a–18
The Two Strands Of DNA are
Complementary
“A” hydrogen bonds to “T” and “G” hydrogen bonds to “C,”
so one strand is complementary to the other.
Example:
Strand 1:
Strand 2:
AGTCAATGCC
TCAGTTACGG
Remember: We start on strand 1 at the 5’ end and go
from there.
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16a–19
Replication of DNA
In this process DNA molecules produce exact duplicates
of themselves. The strands separate and acquire new
strands. Enzymes direct the process.
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16a–20
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16a–21
Chromosomes and Genes
A chromosome is a DNA molecule bound to a group of
proteins. Typically a chromosome is about 15% DNA and
85% protein.
Humans have 46 chromosomes in each cell; dogs have 78,
frogs 26, mosquitos 6.
Chromosomes come in matched pairs: so humans have 23
pairs, one from the father, the other from the mother, in each
pair.
Each chromosome houses a large number of genes.
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16a–22
Protein Synthesis
The Central Dogma: Information flows from DNA to RNA to
proteins
Information
DNA → RNA → Proteins
The DNA → RNA step is called transcription. In this step DNA
passes on the code for a protein to RNA.
The RNA → Proteins step is called translation. In this step
(which is actually composed of several steps) the codes in the
RNA are used as blueprints in protein synthesis.
Note that RNA molecules are single-stranded, whereas DNA
is double-stranded. RNA molecules also tend to be much
smaller than DNA molecules.
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16a–23
Ribonucleic Acids
There are several important types of RNA molecules in cells:
Messenger RNA – mRNA molecules carry the codes for
proteins from the DNA (in the nucleus) to the ribosomes (which
are structures in the cell outside the nucleus).
Primary transcript RNA – ptRNA is the “raw material” for
messenger RNA. It will be edited to produce mRNA
Ribosomal RNA – The ribosomes are structures in the cell
where the actual synthesis of proteins occurs. They consist of
RNA and protein. This RNA is designated as rRNA.
Transfer RNA – tRNA molecules transfer specific amino acids to
the ribosomes, where the amino acids are joined into proteins.
Other RNAs – A number of other types of small RNAs play
special genetic roles, such as regulating gene expression.
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16a–24
The Genetic Code
A gene is a segment of DNA that carries the code for the structure of
a protein (or sometimes an RNA molecule). A gene usually consists
of a DNA sequence of about 1000-3500 nucleotides.
The human genome (the entire genetic code on 46 chromosomes)
apparently contains about 25,000 genes (latest number).
The “code” consists of three-nucleotide sequences (codons) that
stand for individual amino acids. For example, the DNA sequence
G-U-C (starting, remember, from the 5’ end) represents the amino
acid valene.
Other coding sequences are:
UUC – phenylalanine
CCA – proline
UCA – serene
AGA – argenine
And so forth: See Table 17.2 for a full list.
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16a–25
Additional Comments on the Code
• The code is almost universal: that is, it used by almost
all organisms: plants, microbes, and humans.
• The code contains start and stop signals.
• With four different “letters” (A,T,G,C), a threeletter sequence has 43 = 4x4x4 = 64 possibilities.

Thus a given amino acid may be specified by several
sequences. For example, the sequences UCU, UCC,
UCA, and UCG all specify the amino acid serine.
(The code is thus said to be “degenerate”, i.e., not
having a single, unique sequence for each amino
acid.)
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16a–26
Transcription of the Code from DNA to RNA
Four steps:
1. A portion of the DNA sequence unwinds and exposes a
sequence of bases (a gene). This is governed by the enzyme
RNA polymerase.
2. Free ribonucleotides align along one of the exposed DNA
strands. They pair up with complementary bases on the DNA.
3. RNA polymerase catalyses the linking of the aligned
ribonuceotides.
3. Transcription ends when the enzyme encounters a “stop”
signal in the DNA sequence.
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16a–27
Editing of the ptRNA: Formation of mRNA
But the RNA sequence (ptRNA) is not quite ready to be used for
protein formation – it must be edited. In the editing some parts
(introns) are spliced out. Exons are kept and joined.
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16a–28
Some Additional Details
• Splicing means that a single gene can yield several protein
codes, depending on how it is spliced. This is called alternative
splicing.
• Both the exons and introns of a gene are initially transcribed
to form ptRNA. Then the ptRNA is spliced to form mRNA for a
given protein. Enzymes direct the splicing.
• The Human Genome Project (which produced a rough
sequence for the human genome) was completed in 2001 with
the sequencing of about 3 billion nucleotides. Two groups were
responsible for it: a group from the NIH led by Francis Collins
and a group from a private company (Celera Genomics) led by
J. Craig Venter.
• Now many additional organisms have had their genomes
sequenced.
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16a–29
Transfer RNA
Transfer RNA (tRNA) molecules pick up amino acids and
transfer them to the ribosomes, where they are aligned and
linked up to form proteins. tRNAs attach amino acids at one
end and have a recognition portion (anticodon) that pairs up
with a complementary sequence on the mRNA that codes for
that amino acid.
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16a–30
Translation: Protein Synthesis
Protein synthesis occurs on the ribosomes, which are cell organelles.
The synthetic process requires a number of ingredients:
Ribosomes are the sites, or “factories”, where proteins are synthesized.
They are comprised of subunits containing about 65% RNA and 35%
protein.
Messenger RNA molecules act like messengers, bringing the code,
or blueprint, for a protein (the “message”) from the DNA gene to
the ribosomes.
Transfer RNA molecules pick up amino acids and carry them to the
ribosomes. There they line them up using their anticodons. Thus they
act like workers who gather and assemble the parts.
Enzymes act like managers and workers who control and finalize the
manufacturing process.
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16a–31
Enzymes direct the process, and the polypeptide continues to
grow as more amino acids are added.
Growth continues until a “stop” codon is encountered.
Finally, the polypeptide is cleaved from the tRNAs by
hydrolysis.
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16a–32
Protein synthesis involves
a number of different
stages.
Normally,some additional
processing of the protein
takes place after it
separates
from the ribosome.
A single mRNA molecule
can code for the production
of a large number of
Identical protein molecules.
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16a–33
Genetic Engineering
• In general genetic engineering involves placing a gene
from one organism into another organism.
• Example: Bacteria have been genetically altered to
produce human insulin, and this can be used to treat
diabetics.
• The bacteria can be grown in large numbers and the
insulin harvested for medical use.
• Special enzymes are used to extract and insert the
desired genes.
• In some cases genes can be artificially constructed for
such purposes.
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16a–34
Misfolded Proteins
• Just as misguided people can cause problems, so also can
misfolded proteins.
• Disease-causing misfolded proteins are called prions
(pronounced “PREE-ons’). They are believed responsible for the
following diseases:
Sheep – scrapies
Cattle – mad cow disease
Humans – human variant Creutzfeld-Jacop disease
(and perhaps Alzheimer’s disease, Parkinson’s disease)
• Stanley Prusiner (UCSF) won the 1997 Nobel Prize for first
suggesting that misfolded proteins were responsible for disease.
This was, and is, a very controversial idea.
• A tribe of cannibals in New Guinea honored their dead by
eating their brains—and developed a neurological disease (kuru).
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16a–35
What you absolutely must understand from Chapter 17
Understand that nucleic acids form the genetic machinery of cells, and
that they are polymers of nucleotide monomers.
Understand that there are two types of nucleic acids: RNAs and DNAs.
Understand that nucleotides have three parts: a pentose sugar, a
phosphate group, and a nitrogen-containing base.
Be able to distinguish ribose from deoxyribose.
Be able to distinguish a purine from a pyrimidine.
Know that T, C, and U are pyrimidines, and G and C are purines.
Know that DNA contains deoxyribose with A, T, G, and C, whereas
RNA contains ribose with A, U, G,and C.
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16a–36
What you must know (cont.)
Know that the nucleotides are called adenosine, guanosine, etc.
Understand that the pentose sugars of nucleotides are joined to both
phosphate and base units by means of condensation reactions. Know
what a condensation reaction is and what its reverse reaction is called.
Appreciate what the “primary structure” of a nucleic acid refers to,
and that it has a 5’ and a 3’ end. Be able to contrast this primary
structure with that of a protein. (What types of links are involved in
each? What types of side groups?)
Understand that the secondary structure of DNA is a double helix,
with two strands wrapped around each other in a spiral, with the bases
inside, and held together by hydrogen bonds between the bases.
Know what bases bond to each other in this way.
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16a–37
What you must know (cont.)
Know who first determined the DNA structure and when.
Appreciate that they relied on X-ray diffraction studies to find the
structure.
Appreciate why the fact that the % G = % C in DNA was an
important clue for the structure determination.
Appreciate that there are three hydrogen bonds between G and C,
and just two between A and T.
Understand the process by which DNA replicates itself so that two
identical copies are made.
Understand the terms gene, genome, and chromosome. Know how
many chromosomes humans have.
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16a–38
What you must know (cont.)
Understand the general idea of the “Central Dogma” (even though
some exceptions are now known).
Understand the different types of RNA (rRNA, ptRNA, mRNA,
tRNA) present in a cell and what their roles in protein synthesis are.
Understand the general process by which proteins are made in a cell:
where it happens and how it happens.
Understand the basic idea of the genetic code —that each amino acid is
coded for by a sequence of three nucleotides (a codon). Appreciate that
the human genome has about 3 billion nucleotides and 25,000 genes.
Understand the idea of splicing of ptRNA, and how it allows a single
gene to yield several different mRNAs that code for several different
proteins.
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16a–39
What you must know (cont.)
Appreciate that an amino acid may be represented in the code by
several different codons.
Understand the basic structure of a tRNA molecule and how its
attachment site and anticodon region contribute to its action.
Appreciate that it forms an ester link to its amino acid, and that this
link is hydrolyzed when the protein chain is formed.
Appreciate that protein formation on the ribosome is ended when a
stop condon is encountered. Also understand that the whole process
is controlled by several enzymes.
Understand that a single messenger RNA can be used repeatedly as
the blueprint for a number of identical proteins.
Understand that genetic engineering generally involve the
placement of a gene from one organism into the genome of another
organism.
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16a–40
To Do List
• Read chapter 17!!
• Do additional problems
• Do practice test T/F
• Do practice test MC
• Review Lecture notes for
Chapter Seventeen
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16a–41