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Biology
Sylvia S. Mader
Michael Windelspecht
Chapter 12
Molecular Biology
of the Gene
Lecture Outline
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1
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12.1 The Genetic Material
• Frederick Griffith investigated virulence of
Streptococcus pneumoniae
 Concluded that virulence could be passed
from a dead strain to a nonvirulent living strain
 Transformation
• Further research by Avery et al.
 Discovered that DNA is the transforming
substance
 DNA from dead cells was being incorporated
into the genome of living cells
2
The Genetic Material
• Griffith’s Transformation Experiment
 Mice were injected with two strains of
pneumococcus: an encapsulated (S) strain
and a non-encapsulated (R) strain.
• The S strain is virulent (the mice died); it has a
mucous capsule and forms “shiny” colonies.
• The R strain is not virulent (the mice lived); it has
no capsule and forms “dull” colonies.
3
Griffith’s Transformation
Experiment
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
capsule
Injected live
R strain has
no capsule
and mice
do not die.
Injected live
S strain has
capsule and
causes mice
to die.
a.
b.
Injected heatkilled S strain
does not cause
mice to die.
c.
Injected heat-killed
S strain plus live
R strain causes
mice to die.
Live S strain is
withdrawn from
dead mice.
d.
4
The Genetic Material
• Transformation of organisms today:
 Result is the so-called genetically modified
organisms (GMOs)
• Invaluable tool in modern biotechnology today
• Commercial products that are currently much used
• Green fluorescent protein (GFP) can be used as a
marker
– A jellyfish gene codes for GFP
– The jellyfish gene is isolated and then transferred to a
bacterium, or the embryo of a plant, pig, or mouse.
– When this gene is transferred to another organism, the
organism glows in the dark
5
Transformation of Organisms
6
The Genetic Material
• DNA contains:
 Two Nucleotides with purine bases
• Adenine (A)
• Guanine (G)
 Two Nucleotides with pyrimidine bases
• Thymine (T)
• Cytosine (C)
7
The Genetic Material
• Chargaff’s Rules:
 The amounts of A, T, G, and C in DNA:
• Are constant among members of the same species
• Vary from species to species
 In each species, there are equal amounts of:
• A and T
• G and C
 All this suggests that DNA uses
complementary base pairing to store genetic
information
 Each human chromosome contains, on
average, about 140 million base pairs
 The number of possible nucleotide sequences
is 4140,000,000
8
Nucleotide Composition of DNA
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NH2
adenine
(A)
C
N
C
HC
C
CH
N
O
HO
P
O
O
5
4
CH2
N
C
thymine
(T)
N
O
HN
nitrogen-containing
base
CH
C
O
CH3
C
N
O
O
C H
H C
3
OH
HO
H C1
C H
2
H
P
O
C
C
HN
4
CH2
O
C H
H C1
C H
2
H
sugar = deoxyribose
NH2
N
O
HO
P
C
O
O
5
4
CH2
O
N
CH
C
CH
N
N
O
O
C H
H C
3
OH
a. Purine nucleotides
C
cytosine
(C)
N
CH
H2N C
phosphate
5
H C
3
OH
O
guanine
(G)
O
HO
H C1
C H
2
H
P
O
b. Pyrimidine nucleotides
O
5
4
CH2
O
C H
H C
3
OH
H C1
C H
2
H
DNA Composition in Various Species (%)
Species
A
T
G
C
Homo sapiens (human)
31.0
31.5
19.1
18.4
Drosophila melanogaster (fruit fly)
Zea mays (corn)
Neurospora crassa (fungus)
27.3
25.6
23.0
27.6
25.3
23.3
22.5
24.5
27.1
22.5
24.6
26.6
Escherichia coli (bacterium)
Bacillus subtilis (bacterium)
24.6
28.4
24.3
29.0
25.5
21.0
25.6
21.6
c. Chargaff’s data
9
The Genetic Material
• X-Ray diffraction:
 Rosalind Franklin studied the structure of DNA using
X-rays.
 She found that if a concentrated, viscous solution of
DNA is made, it can be separated into fibers.
 Under the right conditions, the fibers can produce an
X-ray diffraction pattern
• She produced X-ray diffraction photographs.
• This provided evidence that DNA had the following features:
– DNA is a helix.
– Some portion of the helix is repeated.
10
X-Ray Diffraction of DNA
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Rosalind Franklin
diffraction pattern
diffracted
X-rays
a.
X-ray beam
Crystalline
DNA
b.
c.
© Photo Researchers, Inc.; c: © Science Source/Photo Researchers, Inc.
11
The Genetic Material
• The Watson and Crick Model (1953)
 Double helix model is similar to a twisted ladder
• Sugar-phosphate backbones make up the sides
• Hydrogen-bonded bases make up the rungs
 Complementary base pairing ensures that a purine is
always bonded to a pyrimidine (A with T, G with C)
 Received a Nobel Prize in 1962
12
Watson and Crick Model of DNA
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
3.4 nm
0.34 nm
2 nm
b.
d.
C
a.
G
5′ end
sugar-phosphate
backbone
T
3′ end
P
A
C
G
S
P
S
A
T
P
P
T
S
3′ end
A
P
S
G
P
5′ end
C
P
complementary
base pairing
c.
C
G
P
sugar
hydrogen
bonds
a: © Photodisk Red/Getty RF; d: © A. Barrington Brown/Photo Researchers
13
12.2 Replication of DNA
• DNA replication is the process of copying
a DNA molecule.
• Semiconservative replication - each
strand of the original double helix (parental
molecule) serves as a template (mold or
model) for a new strand in a daughter
molecule.
14
Semiconservative Replication
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5′
3′
G
G
C
C
A
A
region of parental
DNA double helix
T
C
G
T
A
G
A
A
DNA
polymerase
enzyme
G
G
C
C
T
G
A
region of
replication:
new nucleotides
are pairing
with those of
parental strands
region of
completed
replication
new
strand
old
strand
daughter DNA double helix
5′
old
strand
3′
new
strand
daughter DNA double helix
15
Replication of DNA
• Replication requires the following steps:
 Unwinding, or separation of the two strands of the
parental DNA molecule
 Complementary base pairing between a new
nucleotide and a nucleotide on the template strand
 Joining of nucleotides to form the new strand
• Each daughter DNA molecule contains one old strand and one
new strand
16
Replication of DNA
• Prokaryotic Replication
 Bacteria have a single circular loop of DNA
 Replication moves around the circular DNA molecule
in both directions
 Produces two identical circles
 The process begins at the origin of replication
 Replication takes about 40 minutes, but the cell
divides every 20 minutes
• A new round of replication can begin before the previous
round is completed
17
Replication of DNA
• Eukaryotic Replication
 DNA replication begins at numerous points
along each linear chromosome
 DNA unwinds and unzips into two strands
 Each old strand of DNA serves as a template
for a new strand
 Complementary base-pairing forms a new
strand paired with each old strand
• Requires enzyme DNA polymerase
18
Replication of DNA
• Eukaryotic Replication
 Replication bubbles spread bidirectionally until
they meet
 The complementary nucleotides are joined to form
new strands. Each daughter DNA molecule
contains an old strand and a new strand.
 Replication is semiconservative:
• One original strand is conserved in each daughter
molecule, i.e., each daughter double helix has one
parental strand and one new strand.
19
Replication of DNA
• Accuracy of Replication
 DNA polymerase is very accurate, yet makes
a mistake about once per 100,000 base pairs.
• Capable of identifying and correcting errors
20
12.3 The Genetic Code of Life
• Genes Specify Enzymes
 Beadle and Tatum:
• Experiments on the fungus Neurospora
crassa
• Proposed that each gene specifies the
synthesis of one enzyme
• One-gene-one-enzyme hypothesis
21
The Genetic Code of Life
• The mechanism of gene expression
 DNA in genes specify information, but
information is not structure and function
 Genetic information is expressed into
structure and function through protein
synthesis
• The expression of genetic information into
structure and function:
 DNA in a gene determines the sequence of
nucleotides in an RNA molecule
 RNA controls the primary structure of a
protein
22
The Central Dogma of Molecular Biology
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
nontemplate strand
5
DNA
3
A
G
C
G
A
C
C
C
C
T
C
G
C
T
G
G
G
G
3
5
template strand
transcription
in nucleus
5
mRN A
translation
at ribosome
3
A
G
C
codon 1
G
A
N
C
codon 2
O
polypeptide
C
C
C
C
codon 3
O
N
C
C
O
N
C
C
R1
R2
R3
Serine
Aspartate
Proline
23
The Genetic Code of Life
• RNA is a polymer of RNA nucleotides
 RNA nucleotides contain the sugar ribose instead of
deoxyribose
 RNA nucleotides are of four types: uracil (U),
adenine (A), cytosine (C), and guanine(G)
 Uracil (U) replaces thymine (T) of DNA
• Types of RNA
• Messenger (mRNA) - Takes a message from DNA in the
nucleus to ribosomes in the cytoplasm
• Ribosomal (rRNA) - Makes up ribosomes, which read the
message in mRNA
• Transfer (tRNA) - Transfers the appropriate amino acid to
the ribosomes for protein synthesis
24
Structure of RNA
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5′ end
G
P
U
S
A
base is
uracil instead
of thymine
P
C
S
P
S
P
S
ribose
3′ end
one nucleotide
25
RNA Structure Compared to
DNA structure
26
The Genetic Code of Life
• The unit of the genetic code consists of codons,
each of which is a unique arrangement of symbols
• Each of the 20 amino acids found in proteins is
uniquely specified by one or more codons
 The symbols used by the genetic code are the mRNA bases
• Function as “letters” of the genetic alphabet
• Genetic alphabet has only four “letters” (U, A, C, G)
 Codons in the genetic code are all three bases (symbols)
long
• Function as “words” of genetic information
• Permutations:
– There are 64 possible arrangements of four symbols taken
three at a time
– Often referred to as triplets
• Genetic language only has 64 “words”
27
Messenger RNA Codons
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
First
Base
U
C
A
G
Second Base
Third
Base
U
C
A
G
UUU
phenylalanine
UCU
serine
UAU
tyrosine
UGU
cysteine
U
UUC
phenylalanine
UCC
serine
UAC
tyrosine
UGU
cysteine
C
UUA
leucine
UCA
serine
UAA
stop
UGA
stop
A
UUG
leucine
CUU
leucine
UCG
serine
UAG
stop
UGG
tryptophan
G
CCU
proline
CAU
histidine
CGU
arginine
U
CUC
leucine
CCC
proline
CAC
histidine
CGC
arginine
C
CUA
leucine
CCA
proline
CAA
glutamine
CGA
arginine
A
CUG
leucine
CCG
proline
CAG
glutamine
CGG
arginine
G
AUU
isoleucine
AUC
isoleucine
ACU
threonine
ACC
threonine
AAU
asparagine
AAC
asparagine
AGU
serine
AGC
serine
U
AUA
isoleucine
ACA
threonine
AAA
lysine
AGA
arginine
A
AUG (start)
methionine
ACG
threonine
AAG
lysine
AGG
arginine
G
GUU
valine
GCU
alanine
GAU
aspartate
GGU
glycine
U
GUC
valine
GCC
alanine
GAC
aspartate
GGC
glycine
C
GUA
valine
GUG
valine
GCA
alanine
GCG
alanine
GAA
glutamate
GAG
glutamate
GGA
glycine
GGG
glycine
A
C
28
G
The Genetic Code of Life
• Properties of the genetic code:
 Universal
• With few exceptions, all organisms use the code the same
way
• Encode the same 20 amino acids with the same 64 triplets
 Degenerate (redundant)
• There are 64 codons available for 20 amino acids
• Most amino acids encoded by two or more codons
 Unambiguous (codons are exclusive)
• None of the codons code for two or more amino acids
• Each codon specifies only one of the 20 amino acids
 Contains start and stop signals
• Punctuation codons
• Like the capital letter we use to signify the beginning of a
sentence, and the period to signify the end
29
12.4 First Step: Transcription
• Transcription
 A segment of DNA serves as a template for
the production of an RNA molecule
 The gene unzips and exposes unpaired bases
 Serves as template for mRNA formation
 Loose RNA nucleotides bind to exposed DNA
bases using the C=G and A=U rule
 When entire gene is transcribed into mRNA,
the result is a pre-mRNA transcript of the
gene
 The base sequence in the pre-mRNA is
complementary to the base sequence in DNA
30
First Step: Transcription
• A single chromosome consists of one very long molecule encoding
hundreds or thousands of genes
• The genetic information in a gene describes the amino acid
sequence of a protein
 The information is in the base sequence of one side (the “sense”
strand) of the DNA molecule
 The gene is the functional equivalent of a “sentence”
• The segment of DNA corresponding to a gene is unzipped to expose
the bases of the sense strand
 The genetic information in the gene is transcribed (rewritten) into an
mRNA molecule
 The exposed bases in the DNA determine the sequence in which the
RNA bases will be connected together
 RNA polymerase connects the loose RNA nucleotides together
• The completed transcript contains the information from the gene, but
in a mirror image, or complementary form
31
Transcription
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3′
5′
terminator
template
strand
gene strand
C
3′
direction of
polymerase
movement
T
RNA
polymerase
DNA
template
strand
mRNA
transcript
promoter
5′
32
5′
3′
to RNA processing
RNA Polymerase
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
200m
a.
spliceosome
DNA
RNA
polymerase
RNA
transcripts
b.
© Oscar L. Miller/Photo Researchers, Inc.
33
First Step: Transcription
• Pre-mRNA is modified before leaving the
eukaryotic nucleus.
 Modifications to the ends of the primary
transcript:
• Cap on the 5′ end
– The cap is a modified guanine (G) nucleotide
– Helps a ribosome determine where to attach when
translation begins
• Poly-A tail of 150-200 adenines on the 3′ end
– Facilitates the transport of mRNA out of the nucleus
– Inhibits degradation of mRNA by hydrolytic enzymes.
34
First Step: Transcription
• Pre-mRNA, is composed of exons and introns.
 The exons will be expressed,
 The introns, occur in between the exons.
• Allows a cell to pick and choose which exons will go into a
particular mRNA
 RNA splicing:
• Primary transcript consists of:
– Some segments that will not be expressed (introns)
– Segments that will be expressed (exons)
• Performed by spliceosome complexes in nucleoplasm
– Introns are excised
– Remaining exons are spliced back together
• Result is a mature mRNA transcript
35
First Step: Transcription
• In prokaryotes, introns are removed by
“self-splicing”—that is, the intron itself has
the capability of enzymatically splicing
itself out of a pre-mRNA
36
Messenger RNA Processing in Eukaryotes
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
exon
exon
exon
DNA
intron
intron
transcription
exon
exon
exon
pre-mRNA
5
intron
exon
3
intron
exon
exon
3
poly-A tail
5
intron
cap
intron
spliceosome
exon
exon
exon
3
poly-A tail
5
cap
pre-mRNA
splicing
intron RNA
mRNA
5
cap
3
poly-A tail
nuclear pore
in nuclear envelope
nucleus
cytoplasm
37
First Step: Transcription
• Functions of introns:
 As organismal complexity increases;
• The number of protein-coding genes does not keep pace
• But the proportion of the genome that is introns increases
 Possible functions of introns:
• Exons might combine in various combinations
– Would allow different mRNAs to result from one segment of DNA
• Introns might regulate gene expression
• Introns may encourage crossing-over during meiosis
 Exciting new picture of the genome is
emerging
38
12.5 Second Step: Translation
•
Translation
 The sequence of codons in the mRNA at a ribosome directs the sequence of
amino acids into a polypeptide
 A nucleic acid sequence is translated into a protein sequence
•
tRNA molecules have two binding sites:
 One associates with the mRNA transcript
 The other associates with a specific amino acid
 Each of the 20 amino acids in proteins associates with one or more of 64
types of tRNA
•
Translation
 An mRNA transcript associates with the rRNA of a ribosome in the
cytoplasm or a ribosome associated with the rough endoplasmic reticulum
 The ribosome “reads” the information in the transcript
 Ribosome directs various types of tRNA to bring in their specific amino acid
“fares”
 The tRNA specified is determined by the code being translated in the mRNA
transcript
39
Second Step: Translation
• tRNA molecules come in 64 different kinds
• All are very similar except that
 One end bears a specific triplet (of the 64
possible) called the anticodon
 The other end binds with a specific amino acid
type
 tRNA synthetases attach the correct amino acid
to the correct tRNA molecule
• All tRNA molecules with a specific anticodon
will always bind with the same amino acid
40
Structure of a tRNA Molecule
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
amino
acid
leucine
3’
5’
hydrogen
bonding
amino acid end
anticodon
G
C
A
G
U
C
A
C U
anticodon end
A
U C
C
U C
mRNA
5’
3’
codon
a.
b.
41
Second Step: Translation
• Ribosomes
 Ribosomal RNA (rRNA):
• Produced from a DNA template in the nucleolus of a nucleus
• Combined with proteins into large and small ribosomal
subunits
 A completed ribosome has three binding sites to
facilitate pairing between tRNA and mRNA
• The E (for exit) site
• The P (for peptide) site, and
• The A (for amino acid) site
42
Ribosome Structure and Function
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
large subunit
3
5
mRNA
tRNA binding
sites
small subunit
b. Binding sites of ribosome
a. Structure of a ribosome
outgoing
tRNA
polypeptide
incoming
tRNA
mRNA
c. Function of ribosomes
d. Polyribosome
Courtesy Alexander Rich
43
Second Step: Translation
• 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
44
Second Step: Translation
• Small ribosomal subunit attaches to
mRNA transcript
 Beginning of transcript always has the START
codon (AUG)
• Initiator tRNA (UAC) attaches to P site
• Large ribosomal subunit joins the small
subunit
45
Initiation
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amino acid methionine
Met
initiator tRNA
5
E site
mRNA
3
P site A site
Met
small ribosomal subunit
large ribosomal subunit
U AC
AUG
5
A small ribosomal subunit
binds to mRNA; an initiator
tRNA pairs with the mRNA
start codon AUG.
start codon
3
The large ribosomal subunit
completes the ribosome.
Initiator tRNA occupies the
P site. The A site is ready
for the next tRNA.
46
Initiation
Second Step: Elongation
• “Elongation” refers to the growth in
length of the polypeptide
• RNA molecules bring their amino acid
fares to the ribosome
 Ribosome reads a codon in the mRNA
• Allows only one type of tRNA to bring its amino
acid
• Must have the anticodon complementary to the
mRNA codon being read
• The incoming tRNA joins the ribosome at its A site
 Methionine of the initiator tRNA is connected
to the amino acid of the 2nd tRNA by a peptide47
bond
Second Step: Elongation
• The second tRNA moves to P site
(translocation)
• The spent initiator tRNA moves to the E site and
exits
• The ribosome reads the next codon in the
mRNA
 Allows only one type of tRNA to bring its amino acid
• Must have the anticodon complementary to the mRNA codon
being read
• Joins the ribosome at its A site
 The dipeptide on the 2nd amino acid is connected to
the amino acid of the 3rd tRNA by a peptide bond
48
Elongation
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Met
peptide
Ser bond
tRNA
Trp
3
A tRNA–amino acid
approaches the
ribosome and binds
at the A site.
C A U
G U A
3
2
Two tRNAs can be at a
ribosome at one time;
the anticodons are
paired to the codons.
Val
Asp
Asp
C U G
G A C
5
Trp
peptide
bond
Val
Asp
C A U
G U A
G A C
5
1
Trp
Val
Val
Ala
Ala
Trp
C U G
C U G
G A C
G U A
3
5
3
Thr
Ser
Ser
Ala
anticodon
C A U
G U A
Met
Ser
C U G
Ala
Met
Met
asp
Peptide bond formation
attaches the peptide
chain to the newly
arrived amino acid.
4
G A C
A C C
3
5
The ribosome moves forward; the
“empty” tRNA exits from the E site;
the next amino acid–tRNA complex
is approaching the ribosome.
Elongation
49
Second Step: Translation
• Termination:
 Previous tRNA moves to the P site
 Spent tRNA moves to the E site and exits
 Ribosome reads the STOP codon at the end of the
mRNA
• UAA, UAG, or UGA
• Does not code for an amino acid
 A protein called a release factor binds to the stop
codon and cleaves the polypeptide from the last tRNA
 The ribosome releases the mRNA and dissociates
into subunits
 The same mRNA may be read by another ribosome
50
Termination
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Trp
release factor
Val
Trp
Val
U U
A A
U G A
stop codon
5′
3′
The ribosome comes to a stop
codon on the mRNA. A release
factor binds to the site.
3′
5′
The release factor hydrolyzes the bond
between the last tRNA at the P site and
the polypeptide, releasing them. The
ribosomal subunits dissociate.
Termination
51
Summary of Protein Synthesis in Eukaryotes
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
TRANSCRIPTION
TRANSLATION
1. DNA in nucleus serves
as a template for mRNA.
DNA
2. mRNA is processed
before leaving the nucleus.
introns
large and small
ribosomal subunits
mRNA
5
3. mRNA moves into
cytoplasm and
becomes associated
with ribosomes.
pre-mRNA
3
mRNA
amino
4. tRNAs with
anticodons carry
amino acids
to mRNA.
acids
nuclear pore
peptide
ribosome
5
tRNA
U A C
A UG
3
anticodon
codon
5. During initiation, anticodon-codon
complementary base pairing begins
as the ribosomal subunits come
together at a start codon.
5
CCC UGG UU U
GGG AC C AA A G
3
6. During elongation, polypeptide
synthesis takes place one
amino acid at a time.
8. During termination, a
ribosome reaches a stop
codon; mRNA and
ribosomal subunits
disband.
7. Ribosome attaches to rough
ER. Polypeptide enters lumen,
where it folds and is
modified.
52
12.6 Structure of the
Eukaryotic Chromosome
• Each chromosome contains a single linear DNA
molecule, but is composed of more than 50%
protein.
• Some of these proteins are concerned with DNA
and RNA synthesis
• Histones play primarily a structural role
 The most abundant proteins in chromosomes
 Five primary types of histone molecules
 Responsible for packaging the DNA
• The DNA double helix is wound at intervals around a core of
eight histone molecules (called a nucleosome)
• Nucleosomes are joined by “linker” DNA.
53
Structure of Eukaryotic Chromosomes
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2nm
DNA
double helix
11 nm
1. Wrapping of DNA
around histone proteins.
a. Nucleosomes (“beads on a string”)
histones
nucleosome
2. Formation of a three-dimensional
zigzag structure via histone H1
and other DNA-binding proteins.
histone H1
b. 30-nm fiber
30 nm
300 nm
3. Loose coiling into radial loops .
c. Radial loop domains
euchromatin
700 nm
4. Tight compaction of radial
loops to form heterochromatin.
d. Heterochromatin
5. Metaphase chromosome forms
with the help of a protein
scaffold.
1,400 nm
e. Metaphase chromosome
a: © Ada L. Olins and Donald E. Olins/Biological Photo Service; b: Courtesy Dr. Jerome Rattner, Cell Biology and Anatomy, University of Calgary; c: Courtesy of Ulrich Laemmli and J.R. Paulson, Dept. of
Molecular Biology, University of Geneva, Switzerland; d: © Peter Engelhardt/Department of Pathology and Virology, Haartman Institute/Centre of Excellence in Computational Complex Systems Research,
Biomedical Engineering and Computational Science, Faculty of Information and Natural Sciences, Helsinki University of Technology, Helsinki, Finland
54