Chapter 12: Molecular Biology of
the Gene (Outline)
 DNA Structure
 Watson and Crick Model
 DNA Replication
 Semiconservative Replication
 Prokaryotic versus Eukaryotic Replication
 Types of RNA
 Gene Expression
 The Genetic Code
 Transcription
 Translation
Structure of DNA
 DNA contains
 Two Nucleotides with purine bases (double ring)
 Adenine (A)
 Guanine (G)
 Two Nucleotides with pyrimidine bases (single ring)
 Thymine (T)
 Cytosine (C)
 Each nucleotide consists of
 Deoxyribose (5-carbon sugar)
 Phosphate group
 A nitrogen-containing base
Chargaff’s Rules
 In 1947, Erwin Chargaff had developed a series
of rules based on a survey of DNA composition
in organisms
 The amounts of A, T, G, and C in DNA varies
from species to species
 In each species, the amount of A=T and the
amount of G=C
 All this suggests DNA uses complementary
base pairing to store genetic info
Rosalind Franklin’s Work
 Was an expert in X-ray crystallography
 Technique used to examine DNA
fibers (under right conditions form a crystal)
 Concluded that DNA is a double helix
Watson and Crick Model
 Watson and Crick, 1953
 Constructed a model of DNA from Franklin’s
X-ray diffraction
 Double-helix model is similar to a twisted
ladder
 Sugar-phosphate backbones make up the sides
 Hydrogen-bonded bases make up the rungs
‘steps’
 Model agreed with Chargaff’s rule or
complementary base pairing
 Received a Nobel Prize in 1962
Watson/Crick Model of DNA
Watson and Crick Model
(cont.)
 Antiparallel nature: the sugarphosphate backbone of each
strand runs in opposite
directions
 One strand runs 5’ to 3’, while
the other runs 3’ to 5’
 The nucleotides connect at the
hydroxyl group of the 5 carbon
sugar (at the 3’ end)
 DNA strand is made in a 5’ to 3’
direction
Replication of DNA
 DNA replication: the process of copying a DNA
molecule
 Each old DNA strand serves as a template
 Replication involves 3 main steps
 Unwinding – original double helix strands (parental
DNA) are unwound and “unziped” by helicase enzyme
 Complementary base pairing – positioning of new
complementary nucleotides
 Joining – complementary nucleotides join to form
new strands
 Each daughter DNA contains an old & new strand
Semiconservative Replication of
DNA
 DNA polymerase: enzyme
complex that carries out
the last two steps in DNA
synthesis
 DNA replication must occur
before cellular division
 Cancer cells are treated
with chemotherapeutic
drugs “analogs”, which
causes replication to stop
and cells to die off
Replication:
Prokaryotic vs. Eukaryotic
 Prokaryotic Replication
 Bacteria have a single circular loop of DNA
 Replication moves around the circular DNA
molecule in both directions
 The process begins at the origin of replication
and always occur in the 5’ to 3’ direction
 Replication stops when the 2 DNA polymerases
meet at a termination region
 Bacterial cells require about 40 min to replicate
the complete chromosome
Prokaryotic DNA Replication
Replication:
Prokaryotic vs. Eukaryotic
 Eukaryotic Replication
 DNA replication begins at numerous points
(origins of replication) along linear chromosome
 DNA unwinds and unzips into two strands
through the action of helicase enzyme
 Each old strand of DNA serves as a template for
a new strand
 Replication bubbles spread bi-directionally until
they meet
 Replication fork – V shape formed during DNA
replication
Eukaryotic DNA Replication
Eukaryotic DNA Replication (cont.)
 Eukaryotes replicate their DNA at a slower rate
500–5,000 base pair per minute
 Eukaryotic cells, however complete DNA
replication in a matter of hours, how?
 The linear chromosomes also pose a problem:
 DNA polymerase cannot replicate the ends of
chromosomes that contain telomeres (short
segments of DNA repeated over and over)
 Instead, telomerase enzymes add the repeats
after chromosome replication
 In stem cells, this process preserves the ends of
chromosomes and prevents the loss of DNA
Accuracy of Replication
 DNA polymerase is very accurate with approx.
one mistake per 100,000 base pairs
 DNA polymerase is also capable of proof
reading the daughter strand
 It recognizes a mismatched nucleotide and
removes it from a daughter strand, how?
 By reversing direction and removing several
nucleotides
 After removing the mismatched nucleotide, it
changes direction again and continues
 Overall, the error rate for the bacterial DNA
polymerase is only one in 100 million base pairs!
The Genetic Code of Life
 The mechanism of gene expression
 Gene – segment of DNA that specify
information, but information is not structure
and function (i.e. protein)
 Genetic info is expressed into structure and
function through protein synthesis
 DNA in gene controls the sequence of
nucleotides in an RNA molecule
 RNA controls the primary structure of a
protein
RNA Carries the Information
 Like DNA, RNA is a polymer of nucleotides
 RNA nucleotides are of four types: U, A, C & G
 Uracil (U) replaces thymine (T) of DNA
 There are three major classes of RNA
 Messenger RNA (mRNA) - takes a message from
DNA in the nucleus to ribosomes in cytoplasm
 Transfer RNA (tRNA) – transfers amino acids to
the ribosomes
 Ribosomal RNA (rRNA) – and proteins make up
ribosomes which read the message in mRNA
Structure of RNA
The Genetic Code
 The central dogma of molecular biology states
that the flow of genetic information is “DNA to
RNA to protein”
 There must be a genetic code for each of the
20 amino acids found in proteins
 However, can four nucleotides provide enough
combinations to code for 20 amino acids?
 The genetic code is a triplet code, comprised of
three-base code words (e.g. AUG).
 A codon consists of 3 nucleotide bases of DNA,
why?
Central Dogma in Molecular Biology
 Transcription:
DNA serves as a
template for RNA
formation
 Translation:
mRNA transcript
directs the amino
acid sequence in a
polypeptide
Finding the Genetic Code
 Nirenberg and Matthei (1961) found that an
enzyme that could be used to construct a
synthetic RNA in a cell-free system; they
showed the codon UUU coded for
phenylalanine
 By translating just three nucleotides at a
time, they assigned an amino acid to each of
the RNA codons and discovered important
properties of the genetic code
Properties of the Genetic Code
 The code is degenerate
 There are 64 codons available for 20 amino
acids
 Most amino acids encoded by two or more
codons (e.g. luecine and serine), why?
 The genetic code is unambiguous
 Each triplet codon specifies one and only
one amino acid
 The code has start and stop signals
 There are one start codon and three stop
codons (sequences)
The Code is Universal
 With few exceptions, all organisms use the
code the same way
 Genetic code used by mammalian mitochondria
and chloroplasts differs slightly
 The universal nature of the genetic code
suggests the code dates back to the very first
organisms and that all organisms are related
 It is possible to transfer genes from one
organism to another – genetic engineering
 Example: Glowing mice
mRNA Codons
Steps in Gene Expression:
(1) Transcription
 Messenger RNA is formed
 A DNA segment helix unwinds and unzips,
thus serving as a template for mRNA
formation
 Loose RNA nucleotides bind to exposed DNA
bases using the C=G AND A=U rule
 The information is in the base sequence of
the “template” strand of the DNA molecule
 RNA polymerase connects the loose RNA
nucleotides together in the 5’ → 3’ direction
Transcription of mRNA
 Transcription (initiation) begins when RNA
polymerase attaches to a promoter on DNA

Promoter – region of DNA which defines the
start of the gene, the direction of transcription,
and the strand to be transcribed
 The RNA-DNA association is not as stable as
the DNA double helix
 Only the newest portion of the RNA molecule
with RNA polymerase is bound to DNA; the rest
dangles off to the side
Transcription of mRNA (cont.)
 Elongation of mRNA continues until RNA
polymerase comes to a DNA stop sequence
 Results in the release the mRNA transcript
 Many RNA polymerase molecules work to
produce mRNA from the same DNA region at
the same time
 Either strand of DNA can be a template strand
but for a different gene
Transcription
Steps in Gene Expression:
(2) Translation
 Translation takes place in the cytoplasm of
eukaryotic cells
 Translation is the second step by which gene
expression leads to protein synthesis
 The sequence of codons in the mRNA at a
ribosome directs the sequence of amino acids
in a polypeptide
 One language (nucleic acids) is translated into
another language (protein)
The Role of Transfer RNA
 The tRNA molecule transfers amino acids to the
ribosomes
 The amino acid binds to the 3’ end; the
opposite end of the molecule contains an
anticodon that binds to the mRNA codon in a
complementary fashion
 There is at least one tRNA molecule for each of
the 20 amino acids found in proteins
 There are fewer tRNAs (40) than codons (61)
as some tRNAs pair with more than one codon
Structure of tRNA
Translation Requires Three Steps
 During translation, mRNA codons base-pair
with tRNA anticodons carrying specific amino
acids
 Codon order determines the order of tRNA
molecules and the sequence of amino acids in
polypeptides
 Protein synthesis involves 3 steps: initiation,
elongation, and termination
 Enzymes are required for all three steps;
energy (ATP) is needed for the first two steps
Steps in Translation:
1. Initiation
 Components necessary for initiation are





Small ribosomal subunit
mRNA transcript
Initiator tRNA, and
Large ribosomal subunit
Initiation factors - special proteins that bring the
above together
 Initiator tRNA
 Always has the UAC anticodon
 Always carries the amino acid methionine
 Capable of binding to the P site of ribosome
Steps in Translation:
1. Initiation (cont.)
 Chain Initiation
 In prokaryotes, a small ribosomal subunit attaches to
mRNA at the start codon (AUG)
 Initiator tRNA (UAC) pairs with this codon; then the
large ribosomal subunit joins to the small subunit
 Each ribosome contains three binding sites – the P
site (for peptide), the A site (for amino acid), and the
E site (for exit)
 The initiator tRNA binds to the P site
 The A site is for the next tRNA carrying the next aa
 The E site is to discharge tRNAs from the ribosome
Initiation
Steps in Translation:
2. Elongation
 Elongation – refers to the growth in length of
the polypeptide one amino acid at a time
 The tRNA with attached polypeptide is at the P
site; a tRNA-amino acid complex arrives at the
A site
 Elongation factors – proteins that facilitate
complementary base pairing between the tRNA
anticodon and the mRNA codon at the ribosome
 The polypeptide is transferred and attached by
a peptide bond to the newly arrived amino acid
in the A site via a ribozyme and energy (ATP)
Steps in Translation:
2. Elongation (cont.)
 The tRNA molecule in the P site is now empty
 Translocation occurs with mRNA, along with
peptide-bearing tRNA, moving to the P site and
the spent tRNA moves from the P site to the
E site → exits the ribosome
 As the ribosome moves forward three
nucleotides, there is a new codon now located
at the empty A site
 The complete cycle is rapidly repeated, about
15 times per second in the bacterium E. coli
Elongation
Steps in Translation:
3. Termination
 Termination of polypeptide synthesis occurs at
a stop codon
 UAA, UAG, or UGA
 Does not code for an amino acid
 The polypeptide is enzymatically cleaved from
the last tRNA by a release factor
 The tRNA and polypeptide leave the ribosome,
which dissociates into its two subunits
 The released polypeptide begins to take on its
3D shape
Termination