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CORE PAPER VI - MICROBIAL GENETICS
UNIT-I
DNA-the genetic material, RNA-the genetic material, characters of a genetic material,
chemistry & molecular structure of DNA, special structure of DNA, structure and types
of RNA.
UNIT-II
Bacterial chromosome, organization of genes in prokaryotes, DNA – replication in
prokaryotes – Meselson and Stahl experiment- mechanism & enzymology of replication –
theta replication & rolling circle replication.
UNIT-III
Transcription in prokaryotes – genetic code – translation of proteins – regulation of gene
expression in prokaryotes – Operon concept – lac & trp Operon.
UNIT-IV
Mutation-spontaneous and induced-mutagen & mutagenesis – DNA repair mechanism.
UNIT-V
Genetic exchange – transduction(specialized & generalized), transformation, conjugation &
Hfr mapping , genetic recombination.
References
1. Gardner, E. J,Simmons, M J& D P Snustard ,1991 , Principles of Genetics, 8th
edition. John Wiley & Sons.NY.
2. Freifelder .S ,1987 Microbial Genetics, Jones & Bartlett, Boston.
3. Robert H .Tamarin. Principles of Genetics, 5th edition, Cm Brown Publishers.
4. Lewin.B, 1990. Genes, 6th edition, Oxford University Press.
5. Klug .W.S. & Cummings,MR, 1996, Essentials of Genetics, Mentics Hail.
NewJersey.
1. DNA as a genetic material - Justify
The most conclusive evidences in support of DNA as the genetic material came from
the following three avenues of approach on microorganisms.
1. THE TRANSFORMATION EXPERIMENTS
2. THE BLENDER EXPERIMENT
3. BACTERIAL CONJUGATION
1. THE TRANSFORMATION EXPERIMENTS
In 1928, Frederic Griffith encountered a phenomenon known as genetic
transformation. Colonies of virulent strain of pneumonia causing bacterium, Diplococcus
pneumoniae grown on nutrient agar, have a smooth (S) glistering appearance due to the
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presence of polysaccharide capsule. The avirulent strains, lack this capsule and they produce
rough dull colonies.
Smooth (S) and rough (R) characters are directly related to the presence of absence of
capsule and this trait is known to be genetically determined.
Griffith injected laboratory mice with live R pneumococci; the mice suffered no
illness because R pneumococci were avirulent. But, when the mice were injected with S
pneumococci; the micr suffered from pneumonia and died. However, when he injected the
heat killed S bacteria, they did not suffer from pneumonia. But, when the mice were injected
with the mixture of living avirulent R and heat killed S virulent, the unexpected symptoms of
pneumonia appeared and high mortality results in them. By postmorteming the dead mice, it
was found that their heart blood had both R and S pneumococci. From these results, Griffith
concluded that the presence of the heat killed S bacteria must have caused a transformation.
Griffith could not understand the cause of transformation. The cause of
transformation was first identified in 1944 by Oswald Avery, Colin MacLeod, and Maclyn
McCarty conducted famous Streptococcus pneumoniae transformation experiments. They
found that the transformation occurred when S cell extract was treated with RNase or
proteinase and mixed with the live R strain as a result; R colonies and a few S colonies
appeared. When the S strain cell extract was treated with DNase and mixed with live R
strain, there were only R strain colonies growing on the agar plates. These experiments
proved fundamentally that DNA is the genetic material that carries genes.
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2. THE BLENDER EXPERIMENT
Alfred Hershey and Martha Chase (1952) used radioactive sulfur (35S) to label
bacteriophage proteins and radioactive phosphate (32 P) to label their DNA. After allowing
viruses to attack the bacterial cells, the bacterial cultures were spun in a blender and
centrifuged. They found that most of the bacteriophage DNA remained with the bacterial
cells while their protein coats were released into the medium. They concluded that the
protein played a role in adsorption to the bacteria and helped inject the viral DNA into the
bacterial cell. Thus, it was the DNA that was involved in the growth and multiplication of
bacteriophage within the infected bacterial cell.
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3. BACTERIAL CONJUGATION
Laderberg and Tatum (1946) found that an F+ (‘male’) E. coli cell conjugated with
an F- (‘female’), unidirectional transfer of F+ factor of ‘male’ cell to F- ‘female’ cell took
place, so that the latter was converted into F+ or ‘male’ strain. The F+ factor was found to be
a fragment of DNA molecule which occurred in the cytoplasm of bacterial cell.
INDIRECT EVIDENCES FOR DNA AS A GENETIC MATERIAL
Indirect evidences came from higher organisms which are not easy to manipulate as
bacteria and viruses.
1. Feulgen techniques have shown that DNA entirely remains restricted to the
chromosomes and it forms one of the major components of chromosomes.
2. Various quantitative measurements of amount of DNA in different cells have
shown that the amount of DNA and the number of chromosomes sets (ploidy).
2. RNA as Genetic Material – Justify
THE TOBACCO MOSAIC VIRUS (TMV)
Some viruses do not contain DNA, being made up instead of protein and RNA (ribonucleic acid). The
tobacco mosaic virus (TMV) is such an RNA virus. H. Fraenkel-Conrat and others were able to dissociate
the TMV into its constituent protein and RNA parts (figure 6.3). When the parts were mixed, they
reformed TMV particles that were normal in every respect. That the RNA contained the genetic
information was demonstrated by isolating protein and RNA from several different types of TMV, with
subsequent combinations of protein and RNA mixed together. These reconstituted viruses, containing
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protein from one type and RNA from another, were then allowed to infect tobacco cells. In every case the
progeny TMVs proved to have the protein coats of the type that had contributed the RNA, and not of the
type that had contributed the protein. Thus, in the tobacco mosaic virus, the RNA, rather than the protein,
must be acting as the genetic material.
– Figure 6.3
–
4. Describe the structure of DNA
In DNA molecule the adjacent deoxyribonucleotides are joined in a chain by
phosphodiester bonds which link the 5’ carbon of the deoxyribose of one mononucleotide
unit with the 3’ carbon of the deoxyribose of the next mononucleotide unit. According to
Watson and Crick DNA molecule consists of two such polynucleotide chains wrapped
helically around each other, with the sugar-phosphate chain on the outside (forming ribbon
like back bone of double helix) and purines and pyrimidines on the inside of the helix. The
two polynucleotide strands are held together by hydrogen bonds between specific pairs of
purines and pyrimidines.
The hydrogen bond between purines and pyrimidines are such that adenine can bond
only to thymine by two hydrogen bonds, and guanine can bone only to cytosine by three
hydrogen bonds and no other alternative possibilities between them. The specificity of the
kind of hydrogen bonds that can be formed assures that for every adenine in one chain there
will be thymine in the other. For every guanine in first chain there will be a cytosine in the
other and so on. Thus, the two chains are complementary to each other; that is, the sequence
of nucleotides in one chain dictates the sequence of nucleotides in the other. The two strands
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run anti-parallely-that is, have opposite directions. One strand has phosphodiester linkage in
3’
5’ direction, while other strand has phosphodiester linkage in just reverse or 5’
3’
direction. Further, both polynucleotides strands remain separated by 20 A° distance. The
coiling of double helix is right handed and a complete turn occurs every 34 A°. Since each
nucleotide occupies 3.4 A° distance along the length of a polynucleotide strand, ten
mononucleotides occur per complete turn (the base pairs are rotated 36° with respect to each
adjacent pair). The helix has two external grooves, a deep wide one, called major groove and
a shallow narrow one, called minor groove: both of these grooves are large enough to allow
protein molecules to come in contact with the bases.
5. Write a note on Components on DNA
Bases : The bases of DNA and RNA are heterocyclic (carbon- and nitrogen-containing)
aromatic rings, with a variety of substituents (Fig. 1). Adenine (A) and guanine (G) are
purines, bicyclic structures (two fused rings), whereas cytosine (C), thymine (T) and uracil
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(U) are monocyclic pyrimidines. In RNA, the thymine base is replaced by uracil. Thymine
differs from uracil only in having a methyl group at the 5-position, that is thymine is 5methyluracil. Nucleosides In nucleic acids, the bases are covalently attached to the 1_position of a pentose sugar ring, to form a nucleoside (Fig. 2). In RNA, the sugar is ribose,
and in DNA, it is 2_-deoxyribose, in which the hydroxyl group at the 2_-position is replaced
by a hydrogen. The point of attachment to the base is the 1-position (N-1) of the pyrimidines
and the 9-position (N-9) of the purines (Fig. 1). The numbers of the atoms in the ribose ring
are designated 1_-, 2_-, etc., merely to distinguish them from the base atoms. The bond
between the bases and the sugars is the glycosylic (or glycosidic) bond. If the sugar is
ribose, the nucleosides (technically ribonucleosides) are adenosine, guanosine, cytidine and
uridine. If the sugar is deoxyribose (as in DNA), the nucleosides (2_-deoxyribonucleosides)
are deoxyadenosine, etc. Thymidine and deoxythymidine may be used interchangeably
Nucleotides A nucleotide is a nucleoside with one or more phosphate groups bound
covalently to the 3_-, 5_- or (in ribonucleotides only) the 2_-position. If the sugar is
deoxyribose, then the compounds are termed deoxynucleotides (Fig. 3). Chemically, the
compounds are phosphate esters. In the case of the 5_-position, up to three phosphates may
be attached, to form, for example, adenosine 5_-triphosphate, or deoxyguanosine 5_triphosphate, commonly abbreviated to ATP and dGTP respectively.In the same way, we
have dCTP, UTP and dTTP (equivalent to TTP). 5_-Mono and -diphosphates are abbreviated
as, for example, AMP and dGDP. Nucleoside 5_-triphosphates (NTPs), or deoxynucleoside
5_-triphosphates (dNTPs) are the building blocks of the polymeric nucleic acids. In the
course of DNA or RNA synthesis, two phosphates are split off as pyrophosphate to leave one
phosphate per nucleotide incorporated into the nucleic acid chain (see Topics E1 and K1).
The repeat unit of a DNA or RNA chain is hence a nucleotide.
Phosphodiester In a DNA or RNA molecule, deoxyribonucleotides or ribonucleotides
respecbonds tively are joined into a polymer by the covalent linkage of a phosphate group
between the 5_-hydroxyl of one ribose and the 3_-hydroxyl of the next (Fig. 4). This kind of
bond or linkage is called a phosphodiester bond, since the phosphate is chemically in the
form of a diester. A nucleic acid chain can hence be seen to have a direction. Any nucleic
acid chain, of whatever length (unless it is circular; see Topic C4), has a free 5_-end, which
may or may not have any attached phosphate groups, and a free 3_-end, which is most likely
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to be a free hydroxyl group. At neutral pH, each phosphate group has a single negative
charge. This is why nucleic acids are termed acids; they are the anions of strong acids.
Nucleic acids are thus highly charged polymers. DNA/RNA Conventionally, the repeating
monomers of DNA or RNA are represented by sequence their single letters A, T, G, C or U.
In addition, there is a convention to write the sequences with the 5_-end at the left. Hence a
stretch of DNA sequence might be written 5_-ATAAGCTC-3_, or even just ATAAGCTC.
An RNA sequence might be 5_-AUAGCUUGA-3_. Note that the directionality of the chain
means that, for example, ATAAG is not the same as GAATA
Levene found that DNA contains three main components: (1) phosphate (PO4)
groups; (2) five-carbon sugars; and (3) nitrogen-containing bases called purines (adenine, A,
and guanine, G) and pyrimidines (thymine, T, and cytosine, C). He concluded that DNA
molecules are made of repeating units of the three components. Each unit, consisting of a
sugar attached to a phosphate group and a base, is called a nucleotide.
In DNA, the sugar, four of the carbon atoms together with an oxygen atom form a
five-membered ring. The carbon atoms are numbered 1′ to 5′, proceeding clockwise from the
oxygen atom; the prime symbol (′) indicates that the number refers to a carbon in a sugar
rather than a base. The phosphate group is attached to the 5′ carbon atom of the sugar, and
the base is attached to the 1′ carbon atom. In addition, a free hydroxyl (—OH) group is
attached to the 3′ carbon atom.
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The 5′ phosphate and 3′ hydroxyl groups allow DNA to form long chains of
nucleotides. The reaction between the phosphate group of one nucleotide and the hydroxyl
group of another is a dehydration synthesis, eliminating a water molecule and forming a
covalent bond that links the two groups. The linkage is called a phosphodiester bond.
Linear strands of DNA will almost always have a free 5′ phosphate group at one end
and a free 3′ hydroxyl group at the other. Therefore, every DNA molecule has an intrinsic
directionality.
5′ pGpTpCpCpApT—OH 3′
Where the phosphates are indicated by “p.”
Levene’s early studies indicated that all four types of DNA nucleotides were present
in roughly equal amounts.
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Erwin Chargaff showed that the nucleotide composition of DNA molecules varied in
complex ways, depending on the source of the DNA. This strongly suggested that DNA was
not a simple repeating polymer and might have the information-encoding properties genetic
material must have.
Chargaff observed an important underlying regularity in double stranded DNA: the
amount of adenine present in DNA always equals the amount of thymine, and the amount of
guanine always equals the amount of cytosine. These findings are commonly referred to as
Chargaff’s rules:
1. The proportion of A always equals that of T, and the proportion of G always equals that of
C:
A = T, and G = C.
2. It follows that there is always an equal proportion of purines (A and G) and pyrimidines (C
and T).
6. Explain about three- dimensional structure of DNA
Franklin: X-ray Diffraction Patterns of DNA
British chemist, Rosalind Franklin carried out an X-ray diffraction analysis of DNA.
In X-ray diffraction, a molecule is bombarded with a beam of X rays. When individual rays
encounter atoms, their path is bent or diffracted, and the diffraction pattern is recorded on
photographic film. The patterns resemble the ripples created by tossing a rock into a smooth
lake. When carefully analyzed, they yield information about the three-dimensional structure
of a molecule.
The diffraction patterns she obtained and suggested that the DNA molecule had the
shape of a helix, or corkscrew, with a diameter of about 2 nanometers and a complete helical
turn every 3.4 nanometers.
Watson and Crick: A Model of the Double Helix
In DNA molecule the adjacent deoxyribonucleotides are joined in a chain by
phosphodiester bonds which link the 5’ carbon of the deoxyribose of one mononucleotide
unit with the 3’ carbon of the deoxyribose of the next mononucleotide unit. According to
Watson and Crick DNA molecule consists of two such polynucleotide chains wrapped
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helically around each other, with the sugar-phosphate chain on the outside (forming ribbon
like back bone of double helix) and purines and pyrimidines on the inside of the helix. The
two polynucleotide strands are held together by hydrogen bonds between specific pairs of
purines and pyrimidines.
The hydrogen bond between purines and pyrimidines are such that adenine can bond
only to thymine by two hydrogen bonds, and guanine can bone only to cytosine by three
hydrogen bonds and no other alternative possibilities between them. The specificity of the
kind of hydrogen bonds that can be formed assures that for every adenine in one chain there
will be thymine in the other. For every guanine in first chain there will be a cytosine in the
other and so on. Thus, the two chains are complementary to each other; that is, the sequence
of nucleotides in one chain dictates the sequence of nucleotides in the other. The two strands
run anti-parallely-that is, have opposite directions. One strand has phosphodiester linkage in
3’
5’ direction, while other strand has phosphodiester linkage in just reverse or 5’
3’
direction. Further, both polynucleotides strands remain separated by 20 A° distance. The
coiling of double helix is right handed and a complete turn occurs every 34 A°. Since each
nucleotide occupies 3.4 A° distance along the length of a polynucleotide strand, ten
mononucleotides occur per complete turn (the base pairs are rotated 36° with respect to each
adjacent pair). The helix has two external grooves, a deep wide one, called major groove and
a shallow narrow one, called minor groove: both of these grooves are large enough to allow
protein molecules to come in contact with the bases.
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7. Analyse the different types of DNA structure
Different Forms of DNA
The original Watson–Crick model of DNA is now called the B-form.
In this form, the two strands of DNA form a right-handed helix. If viewed
from either end, it turns in a clockwise direction. B-DNA is the predominant
form in which DNA is found. Our genome, however, also contains several
variations of the B-form double helix. One of these, Z-DNA, so-called because
its backbone has a zig-zag shape, forms a left-handed helix and occurs when
the DNA sequence is made of alternating purines and pyrimidines. Thus the
structure adopted by DNA is a function of its base sequence.
Structural parameters of B-DNA, A-DNA, and Z-DNA
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Circular DNA
The intact DNA molecules of most prokaryotes and viruses are circular. A circular
molecule consists of two unbroken complementary strands called covalently closed circle
where as a circular molecule has one or more nicks in one or both strands are called nicked
circle.
Covalently closed circle
Singly nicked circle
Multiply nicked circle
Superhelical DNA
If covalently closed circles are twisted such circle is said to be a super helix or a super
coil. To form a twisted circle, two ends of linear DNA helix can be brought together and
joined. For formation of twisted circle, one of the two ends is rotated 360° with respect to the
other to produce some unwinding of the double helix. Then the ends are joined as a result
covalent circle, if the hydrogen bonds reform, twist is in opposite sense to form a twisted
circle. Such molecule will look likes figure 8. (i.e., have one cross over point or node).
Instead of 360°, if it is twisted 720° prior to joining, the resulting superhelical molecule will
have two nodes.
The degree of twisting is about same for all molecules; namely, one negative twist is
produced per 200 base pairs. In bacteria, the underwinding of superhelical DNA occurs due
to an enzyme called DNA gyrase.
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SPECIAL STRUCTURE OF DNA
Closed-circular Many DNA molecules in cells consist of closed-circular double-stranded
DNA molecules, for example bacterial plasmids and chromosomes and many viral DNA
molecules. This means that the two complementary single strands are each joined into
circles, 5_ to 3_, and are twisted around one another by the helical path of the DNA. The
molecule has no free ends, and the two single strands are linked together a number of times
corresponding to the number of double-helical turns in the molecule. This number is known
as the linking number (Lk).
Supercoiling A number of properties arise from this circular constraint of a DNA molecule.
A good way to imagine these is to consider the DNA double helix as a piece of rubber ubing,
with a line drawn along its length to enable us to follow its twisting. The tubing may be
joined by a connector into a closed circle. If we imagine a twisting of the DNA helix (tubing)
followed by the joining of the ends, then the deformation so formed is locked into the system
This deformation is known as supercoiling, since it manifests itself as a coiling of the DNA
axis around itself in a higher-order coil, and corresponds to a change in linking number from
the simple circular situation. If the twisting of the DNA is in the same direction as that of the
double helix, that is the helix is twisted up before closure, then the supercoiling formed is
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positive; if the helix is untwisted, then the supercoiling is negative. Almost all DNA
molecules in cells are on average negatively supercoiled. This is true even for linear DNAs
such as eukaryotic chromosomes, which are constrained into large loops by interaction with a
protein scaffold .The level of supercoiling may be quantified in terms of the change in
linking number (_Lk) from that of the unconstrained (relaxed) closed-circular molecule
(Lk°). This corresponds to the number of 360° twists introduced before ring
closure. DNA when isolated from cells is commonly negatively supercoiled by
around six turns per 100 turns of helix (1000 bp), that is _Lk/Lk° = –0.06.
Topoisomer The linking number of a closed-circular DNA is a topological property, that is
one which cannot be changed without breaking one or both of the DNA backbones.A
molecule of a given linking number is known as a topoisomer.Topoisomers differ from each
other only in their linking number.Twist and writhe The conformation (geometry) of the
DNA can be altered while the linking number remains constant. Two extreme conformations
of a supercoiled DNA topoisomer may be envisaged , corresponding to the partition of the
supercoiling (_Lk) completely into writhe or completely into twist . The line on the rubber
tubing model helps to keep track of local twisting of the DNA axis. The equilibrium situation
lies between these two extremes and corresponds to some change in both twist and writhe
induced by supercoiling. This partition may be expressed by the equation:
A, B and Z In fact, a number of different forms of nucleic acid double helix have been
helices observed and studied, all having the basic pattern of two helically-wound antiparallel
strands. The structure identified by Watson and Crick, and described above, is known as BDNA (Fig. 7a), and is believed to be the idealized form of the structure adopted by virtually
all DNA in vivo. It is characterized by a helical repeat of 10 bp/turn (although it is now
Known that ‘real’ B-DNA has a repeat closer to 10.5 bp/turn), by the presence of base pairs
lying on the helix axis and almost perpendicular to it, and by having well-defined, deep
major and minor grooves. DNA can be induced to form an alternative helix, known as the Aform, under conditions of low humidity. The A-form is right-handed, like the Bform,
but has a wider, more compressed structure in which the base pairs are tilted with respect to
the helix axis, and actually lie off the axis (seen end-on,the A-helix has a hole down the
middle). The helical repeat of the A-form is around 11 bp/turn. Although it may be that the
A-form, or something close to it, is adopted by DNA in vivo under unusual circumstances,
the major importance of the A-form is that it is the helix formed by RNA (see below), and by
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DNA-RNA hybrids; it turns out that it is impossible to fit the 2´-OH of RNA into the
theoretically more stable B-form structure.
A further unusual helical structure can be formed by DNA. The left-handed Z-DNA is
stable in synthetic double stranded DNA consisting purely of alternating pyrimidine-purine
sequence (such as 5´-CGCGCG-3´, with the same in the other strand, of course). This is
because in this structure, the pyrimidine and the purine nucleotides adopt very different
conformations, unlike in A- and B-form, where each nucleotide has essentially the same
conformation and immediate environment. In particular, the purine nucleotides in the Zform
adopt the syn conformation, in which the purine base lies directly above the deoxyribose ring
(imagine rotating through 180˚ around the glycosylic bond in Fig. 3; the nucleotides shown
there are in the alternative anti conformation). The pyrimidine nucleotides, and all
nucleotides in the A- and B- forms adopt the anti conformation. The Z-helix has a zig-zag
appearance, with 12 bp/turn, although it probably makes sense to think of it as consisting of 6
‘dimers of base pairs’ per turn; the repeat unit along each strand is really a dinucleotide.
Z-DNA does not easily form in normal DNA, even in regions of repeating CGCGCG, since
the boundaries between the left-handed Z-form and the surrounding B-form would be very
unstable. Although it has its enthusiasts, the Z-form is probably not a significant feature of
DNA (or RNA) in vivo.
8. Discuss On Prokaryotic Chromosome Structure
The Escherichia coli chromosome
Prokaryotic genomes are exemplified by the E. coli chromosome. The bulk of the
DNA in E. coli cells consists of a single closed-circular (see Topic C4) DNA molecule of
length 4.6 million base pairs. The DNA is packaged into a region of the cell known as the
nucleoid. This region has a very high DNA concentration,perhaps 30–50 mg ml–1, as well as
containing all the proteins associated with DNA, such as polymerases, repressors and others
(see below). A fairly high DNA concentration in the test tube would be 1 mg ml–1. In normal
growth, the DNA is being replicated continuously and there may be on average around
two copies of the genome per cell, when growth is at the maximal rate DNA domains
Experiments in which DNA from E. coli is carefully isolated free of most of the attached
proteins and observed under the electron microscope reveal one level of organization of the
nucleoid. The DNA consists of 50–100 domains or loops, the ends of which are constrained
by binding to a structure which probably consists of proteins attached to part of the cell
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membrane .The loops are about 50–100 kb in size. It is not known whether the loops are
static or dynamic, but one model suggests that the DNA may spool through sites of
polymerase or other enzymic action at the base of the loops.
Supercoiling of the Genome
The E. coli chromosome as a whole is negatively supercoiled (_Lk/Lk° = –0.06;
, although there is some evidence that individual domains may be supercoiled independently.
Electron micrographs indicate that some domains may not be supercoiled, perhaps because
the DNA has become broken in one strand (see Topic C4), where other domains clearly do
contain supercoils . The attachment of the DNA to the protein–membrane scaffold may act as
a barrier to rotation of the DNA, such that the domains may be topologically independent.
There is, however, no real biochemical evidence for major differences in the level of
upercoiling in different regions of the chromosome in vivo.
DNA-binding interproteins
The looped DNA domains of the chromosome are constrained further by action with a
number of DNA-binding proteins. The most abundant of these are protein HU, a small basic
(positively charged) dimeric protein, which binds DNA nonspecifically by the wrapping of
the DNA around the protein, and H-NS (formerly known as protein H1), a monomeric
neutral protein, which also binds DNA nonspecifically in terms of sequence, but which
seems to have a preference for regions of DNA which are intrinsically bent. These proteins
are sometimes known as histone-like proteins (see Topic D2), and have the effect of
compacting the DNA, which is essential for the packaging of the DNA into the nucleoid, and
of stabilizing and constraining the supercoiling of the chromosome. This means that, lthough
the chromosome when isolated has a _Lk/Lk° = –0.06, that is approximately one supercoil for
every 17 turns of DNA helix, approximately half of this is constrained as permanent
wrapping of DNA around proteins such as HU (actually a form of writhing; see Topic C4).
Only about half the supercoiling is unconstrained in the sense of being able to adopt the
twisting and writhing conformations described in Topic C4. It has also been uggested that
RNA polymerase and mRNA molecules, as well as site-specific DNA-binding proteins such
as integration host factor (IHF), a homolog of HU, which binds to specific DNA sequences
and bends DNA through 140°, may be important in the organization of the DNA domains. It
may be that the organization of the nucleoid is fairly complex, although highly ordered
DNA–protein complexes such as nucleosomes have not been detected.
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9.Briefly describe the DNA replication process
Definition
DNA replication is the process in which a copy (i.e. a replica) of a double-stranded
DNA molecule is synthesized.
Replication of DNA molecule plays a vital role to transfer genetic information to
progeny from parent. When the DNA molecule is replicated, it forms two double helical
molecules (self complementary). Each of the new DNA molecules consists of one old
(conserved) and one newly synthesized complementary strand. This type of replication is
called semiconservative DNA replication.
Experimental evidence for semiconservative DNA Replication
In 1958, M. Meselson and F.W. Stahl proved DNA replicates semiconservatively
through a series of experiments. They grew E. coli cells in a medium containing
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N as the
soul source of nitrogen for many generations. Then the cells are transferred to a medium
containing less dense isotope 14N.
They observed following results after centrifugation in CsCl (density gradient) of
isolated DNA from the cells grown in both in 15N and 14N containing medium.
1. The density of the DNA was about 1.722 g/cc when the DNA isolated from the cells
grown in 15N medium.
2. The density of the DNA was about 1.708 g/cc when the DNA isolated from the cells
grown in 14N medium after many generation.
3. The density of the DNA was about 1.715 g/cc [(1.708 + 1.722) / 2 = 1.715] when the
DNA isolated from the cells grown in 14N medium after one generation.
4. Denaturation of this “hybrid” DNA after one generation yielded two components
having the density of single stranded [15N]DNA and [14N]DNA.
Mechanism of DNA replication in prokaryotes
1. Initiation of DNA replication
Initiation of DNA replication occurs at one or more special sites, called origins
(designated ori ). Dna A (initiator protein) – ATP complex binds at 9bp inverted repeated
regions of ori C of E. coli and promotes opening of the DNA duplex in a region of three
direct repeats of 13-bp sequence (called 13-mers).
Dna B (helicase) is transferred to exposed single stranded DNA and causes
unwinding of the DNA in the presence of ATP, SSB protein and DNA gyrase
(topoisomerase).
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2. Elongation of DNA chain
An enzyme helicase move towards 5’
3’ direction as a result, replication fork has
been formed by opening the duplex DNA. The DNA strand having helicase becomes lagging
strand. DNA primase (dnaG) associates with Dna B helicase, forming the primosome. The
primosome synthesizes multiple copies of short RNA primers for lagging strand and single
copy of short RNA primer for leading strand.
A short RNA primer, synthesized on one of the template strands, is elongated by
DNA polymerase III in the 5’
3’ direction to form the leading strand); synthesis of the
primer may be mediated by RNA polymerase (RNA polymerase is needed for initiation of
DNA replication).
Exposure of the other template strand permits synthesis of the lagging strand. Each
primer is elongated in the 5’
3’ direction to form an Okazaki fragment (~1000–2000 nt
in length) that extends to the 5’ end of the previous primer. The Primers are subsequently
removed and replaced by DNA by using an enzyme DNA polymerase I.
3. Termination of DNA synthesis
Termination of DNA synthesis usually occurs when the entire duplex has been
replicated. It involves dissociation of the replicating machinery and joining of each daughter
and parental strand. Newly synthesized strands usually undergo modification before the next
round of replication can begin.
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10. Write a note on enzymes involved in replication
Gyrase (DNA gyrase) or Type II topoisomerase
Gyrase functions as a tetramer consisting of two of each of two types of subunit, A
and B (i.e. A2B2) The enzyme binds to dsDNA at specific sites and causes conformational
changes in both from supercoiling to relaxed and from relaxed to supercoiling depending
upon the availability of ATP.
Helicases (DNA unwinding proteins)
Helicases can unwind and separate the strands of dsDNA in the presence of ATP.
Escherichia coli has several types of helicase; helicases I, II and III.
Helicase I (or Rep protein) binds to a region of ssDNA in a dsDNA molecule, moving
along (3’ to 5’) into the double-stranded region and unwinding it processively. Two
molecules of ATP are hydrolysed for each base pair separated.
Helicase II (uvrD gene product) is involved in excision repair. This enzyme may act
processively in the 3’to 5’direction
Helicases I and III both act processively and move in the 5’ to 3’ direction along the
strand to which they are bound. This may be involved in unwinding the plasmid DNA for
transfer during conjugation.
primase (DNA primase)
An enzyme which can polymerize molecules of ribonucleoside triphosphate in the 5’
to 3’ direction to form short oligonucleotides on an ssDNA template. The primase enzyme
can bind with helicases to form primosome complex during DNA replication.
21
DNA polymerases
An enzyme which forms a polymer of deoxyribonucleotides by condensing
deoxyribonucleoside triphosphates (dNTPs) with the elimination of pyrophosphate. A DNA
polymerase adds nucleotides to the 3’-OH end of a preexisting strand of DNA or an RNA
primer.
Prokaryotic DNA polymerases
The bacterium Escherichia coli encodes three main DNA polymerases which are
designated I, II and III. All three main polymerases have exonuclease activity as well as
polymerase activity; all can remove nucleotides sequentially from the 3’ end of a strand.
Polymerases I and III can also cleave nucleotides from the 5’ end. (Klenow
fragment).The 3’ to 5’ exonuclease activity of a DNA polymerase is required for proofreading during DNA replication.
Polymerase I (Kornberg enzyme, product of gene polA)
This enzyme is active in DNA repair. It is also involved in the removal of RNA
primers during DNA replication, and is needed for the replication of certain plasmids (e.g.
ColE1). Polymerase I can act as an RNA-dependent DNA polymerase but it exhibits poor
processivity in this capacity and requires longer incubation times compared with a reverse
transcriptase.
Polymerase II (polB product)
This enzume is induced in the SOS DNA repairing system.
Polymerase III
Polymerase III is the main replicative enzyme in E. coli and is responsible for
replication of the chromosome and of phage DNA etc. ‘DNA polymerase III’ refers to a core
enzyme consisting of three different subunits:
(i)
subunit a (product of dnaE/polC ), which is involved in polymerization;
(ii)
subunit e (product of dnaQ/mutD), which has 3’ to 5’ exonuclease activity and
is apparently involved in proof-reading
(iii) subunit q (of unknown function).
Core enzyme
The core enzyme of polymerase III can catalyze limited DNA synthesis on singlestranded gaps in dsDNA, but for full efficiency, accuracy and processivity on longer
templates it requires a number of additional subunits which include:
(i)
the b subunit (dnaN product)
(ii)
the g subunit (dnaZ product)
processivity
(iii) the d subunit (dnaX product)
(iv)
the t subunit (dnaX-Z product - link two core enzymes
Holoenzyme.
The DNA polymerase III refers to the holoenzyme without the b subunit. It functions
as a dimer – the enzymes being physically linked via their t subunits; during replication, one
enzyme forms the leading strand while the other forms the lagging strand, i.e. they work in
opposite directions.
22
DNA ligase
A ligase which can make a phosphodiester bond between the 3’-OH end of one
ssDNA strand and the 5’-phosphate end of the same or another ssDNA strand. Thus, DNA
ligase can repair a nick in one strand of a dsDNA molecule.
Klenow fragment (Klenow enzyme)
The larger of two fragments obtained by proteolytic cleavage (using e.g.
subtilisin) of the DNA polymerase I of Escherichia coli. The Klenow fragment retains
5’ to 3’ polymerase and 3’ to 5’ exonuclease activities but lacks 5’ to 3’ exonuclease
activity. It has various uses in genetic engineering techniques and in DNA sequencing
methods.
Eukaryotic DNA polymerases
Eukaryotes also have a number of DNA polymerases, including (in most
eukaryotic cells) polymerases α, β and γ. Each type of polymerase appears to be
associated with a particular function – for example, the γ polymerase carries out DNA
replication in mitochondria.
23
Rolling circle Replication
A different pattern of DNA replication occurs during E. coli and the reproduction of
viruses, such as phage lambda. In the rolling-circle mechanism, one strand is nicked and the
free 3′- hydroxyl end is extended by replication enzymes. As the 3′ end is lengthened while
the growing point rolls around the circular template, the 5′ end of the strand is displaced and
forms an everlengthening tail. The single-stranded tail may be converted to the doublestranded form by complementary strand synthesis. This mechanism is particularly useful to
viruses because it allows the rapid, continuous production of many genome copies from a
single initiation event.
The Rolling-Circle Pattern of Replication. A singlestranded tail, often composed of more than one genome
copy, is generated and can be converted to the double-stranded form by synthesis of a complementary strand.
The “free end” of the rollingcircle strand is probably bound to the primosome.
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11. Discuss about genetic code
The genetic code is the collection of the base which corresponds amino acids and
involves in stopping and starting the synthesis. These genetic codes are called codons and the
codes which complement to codons are called anticodons.
Since there are 20 amino acids there should be more than 20 codons, that there are
only 4 bases. A single base cannot act as a codon. If the base occurs in pairs there will be
only16 codons (42 = 16). If the occurs in the triplets there will be a 64 codons (43 = 64) and
this is more than enough. So the codons are found in triplets.
Types of codes
There are two types of codes namely; a) Overlapping codes and b) Non-overlapping
codes.
a) Overlapping codes
In this type each base serves as the first base of the codon. The overlapping code is
more economical than non-overlapping because the DNA can carry the information for
amino acid synthesis three times more than non-overlapping codes.
b) Non-overlapping codes
Each base is used only one codon. Single mutagenic base alters only one amino acid.
Overlapping
– ABC
BCD
CDA
DAB
mRNA bases
– ABCD
ABCD
ABCD
ABCD
Non-overlapping
- ABC
DAB
CDA
BCD
ABC
25
Genetic codes are triplet – Experimental evidence
The phenomenon that the genetic codes are triplet can be proved by an experiment
with mutation. The mutation of the DNA bases can be caused by the molecule proflavin
which is a derivative of acridine. The proflavin is mutagenic only in the replication cycle, but
it cannot be seen when added to just phage culture.
The mutation is caused by the interaction of the acridine molecule in between the two
adjacent bases by increasing the thickness of the base. During replication when the
intercalated region meets DNA polymerase an additional base is inserted in that region. So,
that the mutation is caused by the addition of base. Since there is addition there will be some
changes in the reading of the code and there will be some changes in amino acids which
synthesis the varied protein.
The reversion of the mutant can be done by the removal of the addition base. The
reversion is proved by the study of acridine orange induced mutants in the gene called IIIB of
phage T4. This also proves the triplet nature of the codes.
Base composition of the code
When a mixture of ribosomes, tRNA molecules, radioactive amino acids, a self
fraction known to be required for protein synthesis and synthetic polynucleotide, ploy (U)
was incubated. As a result phenyl alanine was synthesized. The codon UUU corresponds to
amino acid phenyl alanine was synthesized. If a single G is added to the terminus of the poly
U chian, it is terminated by leucine. If many guanines (G) is added it will be terminated with
glysine. This type of experiment was repeated with all 64 triplets.
UUU
UUU
UUU
UUU
Phe
Phe
Phe
Phe
UUU
UUU
UUU
UUG
Phe
Phe
Phe
Leu
UUU
UUU
UUU
UGG
Phe
Phe
Phe
Trp
UUU
UUU
UUU
GGG
Phe
Phe
Phe
Trp
Redundancy of the code
Genetic code is highly redundant. The anticodons of some tRNA molecules must be
able to pair with more than one codon.
26
Identification of the stop codons
The stop codons can be identified using polynucleotide having GUAA. This
polynucleotide can be read as
1.
2. G
GUA
AGU
AAG
UAA
GUA
Val
Ser
Lys
???
Val
UAA
GUA
AGU
AAG
???
Val
Ser
Lys
UAA
GUA
AGU
???
Val
Ser
3. GU AAG
Lys
Note that synthesis stop in each case just before a UAA codon which indicates that
UAA must be a stop codon. Similar experiments identified UAA, UAG and UGA are stop
codons. The three stop codons were giving the following names which are now commonly
used.
UAG
-
Amber
UAA
-
Ochre
UGA
-
Opal
Start codons
The codon AUG is more commonly used as start codons. In few instances, codon
GUG is also used as start codon. Initiation of polypeptide chain growth in prokaryotes, the
AUG code are modified amino acid, N-formyl methionine (f-met) and in eukaryotes it is
ordinary methionine.
Universality of the code
The same codon assignment can be made for all organisms that have been examined.
Such organisms are bacteria, yeast, plants and animals except mitochondria of numerous
eukaryotes and Mycoplasma capricolum. Thus the genetic code is said to be universal.
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Wobble hypothesis
In 1965, Francis Crick made a proposal known as Wobble hypothesis. The Wobble
hypothesis is explained that some tRNA molecules must be able to pair with more than one
codon. Experiments with several tRNA molecules showed this is to be the case.
The first to bases must form pairs of the usual type but the third pair can be a different
type. This observation was derived from the discovery of the anticodon of yeast tRNAala
contains the nucleoside inosine (I), in the position that pairs with the 3rd base of the codon.
The Wobble hypothesis explains the arrangement of synonyms in the code.
The following points can be made
Third position of the codon base
First position of the anticodon base
A
U, I
G
C, U
U
G, I
C
G, I
1. The codon XYC and XYU are always synonyms.
a) If the anticodon for XYC is GYX, this anticodon can also pair with XYU. Because
G can pair with U in the 3rd position of the codon.
b) If the anticodon is IYX then pairing also occurs with XYU, XYA and XYC.
2. The codon XYA has the anticodons IYX or UYX.
a) If the anticodon id IYX then XYA is synonymous with both XYU and XYC.
28
b) If the anticodon is UYX then XYA is synonymous with XYG. Thus every codon
ending with A is redundant. This point also explains the number of different tRNA
molecules that corresponds to a particular amino acid.
3. The codons AUG (met) and UGG (trp) are non-redundant. The codons XYG can have an
anticodon CYX that responds to no other codon.
4. There is no tRNA having an anticodon complementary to UGA.
Synonyms resolving from Wobble pairing
If
XYG
XYG
Then also
If
two codons are recognized.
XYC
XYU
XYI
XYI
Then also
If
and
XYC
XYU
XYU
XYU
Then also
XYA
If
XYI
three codons are recognized.
XYA
two codons are recognized.
XYG
XYC
The only one codon is recognized.
XYG
11. Illustrate the operon concept
Jacob and Monod proposed the operon model in 1961 for the co-ordinate regulation
of transcription of genes involved in specific metabolic pathways. The operon is a unit of
gene expression and regulation which typically includes:
● The structural genes (any gene other than a regulator) for enzymes involved in a specific
biosynthetic pathway whose expression is co-ordinately controlled.
● Control elements such as an operator sequence, which is a DNA sequence that regulates
transcription of the structural genes.
29
● Regulator gene(s) whose products recognize the control elements, for example a repressor
which binds to and regulates an operator sequence.
The Lactose operon
Escherichia coli can use lactose as a source of carbon. The enzymes required for
operon the use of lactose as a carbon source are only synthesized when lactose is available as
the sole carbon source. The lactose operon (or lac operon) consists of three structural genes:
lacZ, which codes for galactosidase, an enzyme responsible for hydrolysis of lactose to
galactose and glucose; lacY, which encodes a galactoside permease which is responsible for
lactose transport across the bacterial cell wall; and lacA, which encodes a thiogalactoside
transacetylase. The three structural genes are encoded in a single transcription unit, lacZYA,
which has a single promoter, Plac. This organization means that the three lactose operon
structural proteins are expressed together as a polycistronic mRNA containing more than
one coding region under the same regulatory control. The lacZYA transcription unit contains
an operator site Olac which is positioned between bases –5 and +21 at the 5’-end of the
Plac promoter region. This site binds a protein called the lac repressor which is a potent
inhibitor of transcription when it is bound to the operator. The lac repressor is encoded by a
separate regulatory gene lacI which is also a part of the lactose operon; lacI is situated just
upstream from Plac.
The lac repressor
The lacI gene encodes the lac repressor, which is active as a tetramer of identical
subunits. It has a very strong affinity for the lac operator-binding site, Olac, and also has a
generally high affinity for DNA. The lac operator site consists of 28 bp which is
palindromic. (A palindrome has the same DNA sequence when one strand is read left to
right in a 5’ to 3’ direction and the complementary strand is read right to left in a 5’ to 3’
30
direction). This inverted repeat symmetry of the operator matches the inherent symmetry of
the lac repressor which is made up of four identical subunits. In the absence of lactose, the
repressor occupies the operator-binding site. It seems that both the lac repressor and the RNA
polymerase can bind simultaneously to the lac promoter and operator sites. The lac repressor
actually increases the binding of the polymerase to the lac promoter by two orders of
magnitude. This means that when lac repressor is bound to the Olac operator DNA sequence,
polymerase is also likely to be bound to the adjacent Plac promoter sequence.
Induction
In the absence of an inducer, the lac repressor blocks all but a very low level of
transcription of lacZYA. When lactose is added to cells, the low basal level Allolactose acts
as an inducer and binds to the lac repressor. This causes a change in the conformation of the
repressor tetramer, reducing its affinity for the lac operator. The removal of the lac repressor
from the operator site allows the polymerase (which is already sited at the adjacent promoter)
to rapidly begin transcription of the lacZYA genes. Thus, the addition of lactose, or a
synthetic inducer such as isopropyl-_-D-thiogalactopyranoside (IPTG), very rapidly
stimulates transcription of the lactose operon structural genes. The subsequent removal of the
inducer leads to an almost immediate inhibition of this induced transcription, since the free
lac repressor rapidly re-occupies the operator site and the lacZYA RNA transcript is
extremely unstable.
Structures of lactose, allolactose and IPTG
31
Binding of inducer inactivates the lac repressor
cAMP receptor protein
The Plac promoter is not a strong promoter. Plac and related promoters do not have
strong –35 sequences and some even have weak –10 consensus sequences. For high level
transcription, they require the activity of a specific activator protein called cAMP receptor
protein (CRP). CRP may also be called catabolite activator protein or CAP. When
glucose is present, E. coli does not require alternative carbon sources such as lactose.
Therefore, catabolic operons, such as the lactose operon, are not normally activated. This
regulation is mediated by CRP which exists as a dimer which cannot bind to DNA on its
own, nor regulate transcription. Glucose reduces the level of cAMP in the cell. When glucose
is absent, the levels of cAMP in E. coli increase and CRP binds to cAMP. The CRP–cAMP
complex binds to the lactose operon promoter Plac just upstream from the site for RNA
polymerase. CRP binding induces a 90° bend in DNA, and this is believed to enhance RNA
polymerase binding to the promoter, enhancing transcription by 50-fold. The CRP-binding
site is an inverted repeat and may be adjacent to the promoter (as in the lactose operon), may
lie within the promoter itself, or may be much further upstream from the promoter.
Differences in the CRP-binding sites of the promoters of different catabolic operons may
mediate different levels of response of these operons to cAMP in vivo.
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TRP Operon
The trp operon encodes five structural genes whose activity is required for tryptophan
synthesis. The operon encodes a single transcription unit which produces a 7 kb transcript
which is synthesized downstream from the trp promoter and trp operator sites Ptrp and Otrp.
Like many of the operons involved in amino acid biosynthesis, the trp operon has evolved
systems for coordinated expression of these genes when the product of the biosynthetic
pathway, tryptophan, is in short supply in the cell. As with the lac operon, the RNA product
of this transcription unit is very unstable, enabling bacteria to respond rapidly to changing
needs for tryptophan.
The trp repressor
A gene product of the separate trpR operon, the trp repressor, specifically interacts
with the operator site of the trp operon. The symmetrical operator sequence, which forms the
trp repressor-binding site, overlaps with the trp promoter sequence between bases –21 and
+3. The core binding site is a palindrome of 18 bp. The trp repressor binds tryptophan and
can only bind to the operator when it is complexed with tryptophan. When tryptophan is
bound to the repressor as a result, the repressor changes conformation to interact with
successive major grooves of the DNA at the trp operator sequence. Tryptophan, the endproduct of the enzymes encoded by the trp operon, therefore acts as a co-repressor and
inhibits its own synthesis through end-product inhibition. The repressor reduces
transcription initiation by around 70-fold. This is a much smaller transcriptional effect than
that mediated by the binding of the lac repressor.
The attenuator
At first, it was thought that the repressor was responsible for all of the transcriptional
regulation of the trp operon. However, it was observed that the deletion of a sequence
between the operator and the trpE gene coding region resulted in an increase in both the basal
and the activated (derepressed) levels of transcription. This site is termed the attenuator and it
33
lies towards the end of the transcribed leader sequence of 162 nt that precedes the trpE
initiator codon. The attenuator is a rho-independent terminator site which has a short GC-rich
palindrome followed by eight successive U residues. If this sequence is able to form a hairpin
structure in the RNA transcript, then it acts as a highly efficient transcription terminator and
only a 140 bp transcript is synthesized.
Leader RNA structure
The leader sequence of the trp operon RNA contains four regions of complementary
sequence which can form different base-paired RNA structures. These are termed sequences
1, 2, 3 and 4. The attenuator hairpin is the product of the base pairing of sequences 3 and 4
(3:4 structure). Sequences 1 and 2 are also complementary and can form a second 1:2
hairpin. However, sequence 2 is also complementary to sequence 3. If sequences 2 and 3
form a 2:3 hairpin structure, the 3:4 attenuator hairpin cannot be formed and transcription
termination will not occur. Under normal conditions, the formation of the 1:2 and 3:4
hairpins is energetically favorable.
The leader peptide
The leader RNA sequence contains an efficient ribosome-binding site and can form a
14-amino-acid leader peptide encoded by bases 27–68 of the leader RNA. The 10th and 11th
codons of this leader peptide encode successive tryptophan residues, the end-product of the
synthetic enzymes of the trp operon. This leader has no obvious function as a polypeptide,
and tryptophan is a rare amino acid; therefore, the chances of two tryptophan codons in
succession is low and, under conditions of low tryptophan availability, the ribosome would
be expected to pause at this site. The function of this leader peptide is to determine
tryptophan availability and to regulate transcription termination.
34
Attenuation
Attenuation depends on the fact that transcription and translation are tightly coupled
in E. coli; translation can occur as an mRNA is being transcribed. The 3′-end of the trp leader
peptide coding sequence overlaps complementary sequence 1; the two trp codons are within
sequence 1 and the stop codon is between sequences 1 and 2. The availability of tryptophan
(the ultimate product of the enzymes synthesized by the trp operon) is sensed through its
being required in translation, and determines whether or not the terminator (3:4) hairpin
forms in the mRNA.
As transcription of the trp operon proceeds, the RNA polymerase pauses at the end of
sequence 2 until a ribosome begins to translate the leader peptide. Under conditions of high
tryptophan availability, the ribosome rapidly incorporates tryptophan at the two trp codons
and thus translates to the end of the leader message. The ribosome is then occluding sequence
35
2 and, as the RNA polymerase reaches the terminator sequence, the 3:4 hairpin can form, and
transcription may be terminated. This is the process of attenuation.
Alternatively, if tryptophan is in scarce supply, it will not be available as an
aminoacyl tRNA for translation, and the ribosome will tend to pause at the two trp codons,
occluding sequence 1. This leaves sequence 2 free to form a hairpin with sequence 3, known
as the anti-terminator. The terminator (3:4) hairpin cannot form, and transcription continues
into trpE and beyond. Thus the level of the end product, tryptophan, determines the
probability that transcription will terminate early (attenuation), rather than proceeding
through the whole operon.
Importance of attenuation
The presence of tryptophan gives rise to a 10-fold repression of trp operon
transcription through the process of attenuation alone. Combined with control by the trp
repressor (70-fold), this means that tryptophan levels exert a 700-fold regulatory effect on
expression from the trp operon. Attenuation occurs in at least six operons that encode
enzymes concerned with amino acid biosynthesis. For example, the His operon has a leader
which encodes a peptide with seven successive histidine codons. Not all of these other
operons have the same combination of regulatory controls that are found in the trp operon.
The His operon has no repressor–operator regulation, and attenuation forms the only
mechanism of feedback control.
36
12. Write down the steps involved in DNA replication
Definition
Transcription is a process in which a copy of RNA is synthesized from the DNA
sequence encoding the gene. It is the first step in gene expression.
RNA Polymerase
RNA polymerase is best understood in bacteria. Bacterial RNA polymerase is very
large and complex, consisting of five subunits: two α subunits bind regulatory proteins, a β′
subunit binds the DNA template, a β subunit binds RNA nucleoside subunits, and a σ subunit
recognizes the promoter and initiates synthesis.
Only one of the two strands of DNA (template strand) is transcribed. The RNA
transcript’s sequence is complementary to the template strand.
In both bacteria and eukaryotes, the polymerase adds ribonucleotides to the growing
3′ end of an RNA chain. The synthesis of RNA proceeds in the 5′ → 3′ direction. Bacteria
contain only one RNA polymerase enzyme, while eukaryotes have three different RNA
polymerases: RNA polymerase I synthesizes rRNA in the nucleolus; RNA polymerase II
synthesizes mRNA; and RNA polymerase III synthesizes tRNA.
Promoter
Transcription starts at RNA polymerase binding sites called promoters on the DNA
template strand. A promoter is a short sequence that is not itself transcribed by the
polymerase that binds to it. Striking similarities are evident in the sequences of different
promoters. For example, two six-base sequences are common to many bacterial promoters, a
TTGACA sequence called the –35 sequence, located 35 nucleotides upstream of the position
where transcription actually starts, and a TATAAT sequence called the –10 sequence,
located 10 nucleotides upstream of the start site. In eukaryotic DNA, the sequence TATAAA,
called the TATA box, is located at –25 and is very similar to the prokaryotic –10 sequence
but is farther from the start site.
Promoters differ widely in efficiency. Strong promoters cause frequent initiations of
transcription, as often as every 2 seconds in some bacteria. Weak promoters may transcribe
only once every 10 minutes. Most strong promoters have unaltered –35 and –10 sequences,
while weak promoters often have substitutions within these sites.
Initiation
The binding of RNA polymerase to the promoter is the first step in gene transcription.
In bacteria, a subunit of RNA polymerase called σ (sigma) recognizes the –10 sequence in
37
the promoter and binds RNA polymerase there. Importantly, this subunit can detect the –10
sequence without unwinding the DNA double helix. In eukaryotes, the –25 sequence plays a
similar role in initiating transcription, as it is the binding site for a key protein factor. Other
eukaryotic factors then bind one after another, assembling a large and complicated
transcription complex.
Once bound to the promoter, the RNA polymerase begins to unwind the DNA helix.
Measurements indicate that bacterial RNA polymerase unwinds a segment approximately 17
base-pairs long, nearly two turns of the DNA double helix. This sets the stage for the
assembly of the RNA chain.
Elongation
The transcription of the RNA chain usually starts with ATP or GTP. One of these
forms the 5′ end of the chain, which grows in the 5′ → 3′ direction as ribonucleotides are
added. Unlike DNA synthesis, a primer is not required. The region containing the RNA
polymerase, DNA, and growing RNA transcript is called the transcription bubble because it
contains a locally unwound “bubble” of DNA. Within the bubble, the first 12 bases of the
newly synthesized RNA strand temporarily form a helix with the template DNA strand.
Corresponding to not quite one turn of the helix, this stabilizes the positioning of the 3′ end
of the RNA so it can interact with an incoming ribonucleotide.
The transcription bubble moves down the DNA at a constant rate, about 50
nucleotides per second, leaving the growing RNA strand protruding from the bubble. After
the transcription bubble passes, the now transcribed DNA is rewound as it leaves the bubble.
Unlike DNA polymerase, RNA polymerase has no proofreading capability.
Transcription thus produces many more copying errors than replication. These mistakes,
however, are not transmitted to progeny. Most genes are transcribed many times, so a few
faulty copies are not harmful.
38
Termination
At the end of a gene are “stop” sequences that cause the formation of phosphodiester
bonds to cease, the RNA-DNA hybrid within the transcription bubble to dissociate, the RNA
polymerase to release the DNA, and the DNA within the transcription bubble to rewind.
The simplest stop signal is a series of GC base-pairs followed by a series of AT basepairs. The RNA transcript of this stop region forms a GC hairpin, followed by four or more U
ribonucleotides. The hairpin causes the RNA polymerase to pause immediately after the
polymerase has synthesized it, placing the polymerase directly over the run of four uracils.
The pairing of U with DNA’s A is the weakest of the four hybrid base-pairs and is not strong
enough to hold the hybrid strands together during the long pause. Instead, the RNA strand
dissociates from the DNA within the transcription bubble, and transcription stops. A variety
of protein factors aid hairpin loops in terminating transcription of particular genes.
Posttranscriptional Modifications [RNA Processing]
In eukaryotes, every mRNA transcript must travel a long journey out from the
nucleus into the cytoplasm before it can be translated. Eukaryotic mRNA transcripts are
modified in several ways to aid this journey:
5′ caps. Transcripts usually begin with A or G, and, in eukaryotes, the terminal
phosphate of the 5′ A or G is removed, and then a very unusual 5′-5′ linkage forms with GTP.
Called a 5′ cap, this structure protects the 5′ end of the RNA template from nucleases and
phosphatases during its long journey through the cytoplasm. Without these caps, RNA
transcripts are rapidly degraded.
3′ poly-A tails. The 3′ end of eukaryotic transcript is cleaved off at a specific site,
often containing the sequence AAUAAA. A special poly-A polymerase enzyme then adds
about 250 A ribonucleotides to the 3′ end of the transcript. Called a 3′ poly-A tail, this long
string of A is protects the transcript from degradation by nucleases. It also appears to make
the transcript a better template for protein synthesis.
39
13.Write a note on translation
Overview The actual mechanism of protein synthesis can be divided into three stages:
● Initiation – the assembly of a ribosome on an mRNA molecule;
● Elongation – repeated cycles of amino acid addition;
● Termination – the release of the new protein chain.
In prokaryotes, the factors are abbreviated as IF or EF for initiation and elongation
factors respectively, whereas in eukaryotes they are called eIF and eEF. There are distinct
differences of detail between the mechanism in prokaryotes and eukaryotes, and most of
these occur in the initiation stage.
Initiation
The purpose of the initiation step is to assemble a complete ribosome on to an mRNA
molecule at the correct start point, the initiation codon. The components involved are the
large and small ribosome subunits, the mRNA, the initiator tRNA in its charged form,
three initiation factors and GTP.
The initiation factors IF1, IF2 and IF3 are all just over one-tenth as abundant as
ribosomes, and have masses of 9, 120 and 22 kDa respectively. Only IF2 binds GTP.
Although the finer details have yet to be worked out, the overall sequence of events is as
follows:
● IF1 and IF3 bind to a free 30S subunit. This helps to prevent a large subunit binding to it
without an mRNA molecule and forming an inactive ribosome.
● IF2 complexed with GTP then binds to the small subunit. It will assist the charged initiator
tRNA to bind.
● The 30S subunit attaches to an mRNA molecule making use of the ribosome binding site
(RBS) on the mRNA.
● The initiator tRNA can then bind to the complex by base pairing of its anticodon with the
AUG codon on the mRNA. At this point, IF3 can be released, as its roles in keeping the
subunits apart and helping the mRNA to bind are complete. This complex is called the 30S
initiation complex.
● The 50S subunit can now bind, which displaces IF1 and IF2, and the GTP is hydrolyzed in
this energy-consuming step. The complex formed at the end of the initiation phase is called
40
the 70S initiation complex. The assembled ribosome has two tRNA-binding sites. These are
called the A- and P-sites, for aminoacyl and peptidyl sites. The Asite is where incoming
aminoacyl-tRNA molecules bind, and the P-site is where the growing polypeptide chain is
usually found. These sites are in the cleft of the small subunit and contain adjacent codons
that are being translated. One major outcome of initiation is the placement of the initiator
tRNA in the P-site. It is the only tRNA that does this, as all others must enter the A-site.
Initiation of protein synthesis in E. coli.
Elongation
With the formation of the 70S initiation complex, the elongation cycle can begin. It
can be subdivided into three steps as follows:
(i) aminoacyl-tRNA delivery,
(ii) peptide bond formation
(iii) translocation (movement).
41
Beginning where the P-site is occupied and the A-site is empty. It involves three
elongation factors EF-Tu, EF-Ts and EF-G which all bind GTP or GDP and have masses
of 45, 30 and 80 kDa respectively. EF-Ts and EF-G are about as abundant as ribosomes, but
EF-Tu is nearly 10 times more abundant.
(i) Aminoacyl-tRNA delivery. EF-Tu is required to deliver the aminoacyltRNA to the A-site
and energy is consumed in this step by the hydrolysis of GTP. The released EF-Tu_GDP
complex is regenerated with the help of EF-Ts. In the EF-Tu–EF-Ts exchange cycle, EF-Ts
displaces the GDP and subsequently is displaced itself by GTP. The resultant EF-Tu_GTP
complex is now able to bind another aminoacyl-tRNA and deliver it to the ribosome. All
aminoacyl-tRNAs can form this complex with EF-Tu, except the initiator tRNA.
(ii) Peptide bond formation. After aminoacyl-tRNA delivery, the A- and P-sites are both
occupied and the two amino acids that are to be joined are in close proximity. The peptidyl
transferase activity of the 50S subunit can now form a peptide bond between these two
amino acids without the input of any more energy, since energy in the form of ATP was used
to charge the tRNA.
(iii) Translocation. A complex of EF-G (translocase) and GTP binds to the ribosome and,
in an energy-consuming step, the discharged tRNA is ejected from the P-site, the peptidyltRNA is moved from the A-site to the P-site and the mRNA moves by one codon relative to
the ribosome. GDP and EF-G are released, the latter being re-usable. A new codon is now
present in the vacant A-site. Recent evidence suggests that in prokaryotes the discharged
tRNA is first moved to an E-site (exit site) and is ejected when the next aminoacyl-tRNA
binds. In this way the ribosome maintains contact with the mRNA via 6 base pairs which
may well reduce the chances of frameshifting.
One cycle of the three-step elongation cycle has been completed, and the cycle is
repeated until one of the three termination codons (stop codons) appears in the A-site.
42
Elongation stage of protein synthesis.
Termination
There are no tRNA species that normally recognize stop codons. Instead, protein
factors called release factors interact with these codons and cause release of the completed
polypeptide chain. RF1 recognizes the codons UAA and UAG, and RF2 recognizes UAA
and UGA. RF3 helps either RF1 or RF2 to carry out the reaction. The release factors make
peptidyl transferase transfer the polypeptide to water rather than to the usual aminoacyltRNA, and thus the new protein is released. To remove the uncharged tRNA from the P-site
and release the mRNA, EF-G together with ribosome release factor are needed for the
complete dissociation of the subunits. IF3 can now bind the small subunit to prevent inactive
43
70S ribosomes forming.
There are mechanisms that deal with the problems of mutant or truncated mRNAs
being translated into defective proteins. In the case of truncated mRNAs in prokaryotes, a
special RNA called tmRNA (transfer-messenger RNA), that has properties of both tRNA
and mRNA, is used to free the stalled ribosome and ensure degradation of the defective
protein. The ribosome becomes stalled when there is no complete codon in the A-site for
either a tRNA or a release factor to recognize. The tmRNA behaves firstly like a tRNA in
delivering an alanine residue to the A-site and allowing peptide bond formation to take place.
Then translocation occurs which places part of the tmRNA in the A-site where it behaves as
an mRNA and allows translation of 10 codons in total and then normal termination at a stop
codon. The released protein has a tag of 10 amino acids at its carboxy terminus which target
it for rapid degradation. A different mechanism exists in eukaryotes.
44
UNIT-IV
DNA DAMAGE
DNA lesions
The chemical reactivity of DNA with exogenous chemicals or radiation can give rise
to changes in its chemical or physical structure. These may block replication or transcription
and so be lethal, or they may generate mutations through direct or indirect mutagenesis. The
chemical instability of DNA can generate spontaneous lesions such as deamination and
depurination.
Oxidative damage
Reactive oxygen species such as superoxide and hydroxyl radicals produce a variety
of lesions including 8-oxoguanine and 5-formyluracil. Such damage occurs spontaneously
but is increased by some exogenous agents including _-rays.
Alkylation
Electrophilic alkylating agents such as methylmethane sulfonate and ethylnitrosourea
can modify nucleotides in a variety of positions. Most lesions are indirectly mutagenic, but
O6-alkylguanine is directly mutagenic.
Bulky adducts
Bulky lesions such as pyrimidine dimers and arylating agent adducts distort the
double helix and cause localized denaturation. This disrupts the normal functioning of the
DNA.
DNA lesions
A lesion is an alteration to the normal chemical or physical structure of the DNA.
Some of the nitrogen and carbon atoms in the heterocyclic ring systems of the bases and
some of the exocyclic functional groups (i.e. the keto and amino groups of the bases) are
chemically quite reactive. Many exogenous agents, such as chemicals and radiation, can
cause changes to these positions. The altered chemistry of the bases may lead to loss of base
pairing or altered base pairing (e.g. an altered A may base-pair with C instead of T). If such a
lesion was allowed to remain in the DNA, a mutation could become fixed in the DNA by
direct or indirect mutagenesis. Alternatively, the chemical change may produce a physical
distortion in the DNA which blocks replication and/or transcription, causing cell death. Thus,
DNA lesions may be mutagenic and/or lethal. Some lesions are spontaneous and occur
because of the inherent chemical reactivity of the DNA and the presence of normal, reactive
chemical species within the cell. For example, the base cytosine undergoes spontaneous
hydrolytic deamination to give uracil. If left unrepaired, the resulting uracil would form a
base pair with adenine during subsequent replication, giving rise to a point mutation (see
45
Topic F1). In fact, the generation of uracil in DNA in this way is the probable reason why
DNA contains thymine instead of uracil. Any uracil found in DNA is removed by an enzyme
called uracil DNA glycosylase and is replaced by cytosine (see Topic F3). 5- Methylcytosine,
a modified base found in small amounts in DNA, deaminates to thymine, a normal base. This
is much more difficult to detect.
Depurination
Depurination is another spontaneous hydrolytic reaction that involves cleavage of
the N-glycosylic bond between N-9 of the purine bases A and G and C-1_ of the deoxyribose
sugar and hence loss of purine bases from the DNA. The sugar–phosphate backbone of the
DNA remains intact. The resulting apurinic site is a noncoding lesion, as information
encoded in the purine bases is lost. Depurination occurs at the rate of 10 000 purines lost per
human cell per hour at 37°C. Though less frequent, depyrimidination can also occur.
Oxidative damage This occurs under normal conditions due to the presence of reactive
oxygen species (ROS) in all aerobic cells, for example superoxide, hydrogen peroxide and,
most importantly, the hydroxyl radical (-OH). This radical can attack DNA at a number of
points, producing a range of oxidation products with altered properties, for example 8oxoguanine, 2-oxoadenine and 5-formyluracil. The levels of these can be increased by
hydroxyl radicals from the radiolysis of water caused by ionizing radiation.
Examples of oxidized bases produced in DNA by reactive oxygen species.
Alkylation
Alkylating agents are electrophilic chemicals which readily add alkyl (e.g. methyl)
groups to various positions on nucleic acids distinct from those methylated by normal
methylating enzymes (see Topics C1 and G3). Common examples are methylmethane
sulfonate (MMS) and ethylnitrosourea (ENU). Typical examples of methylated bases are
7-methylguanine, 3-methyladenine, 3-methylguanine and O6-methylguanine. Some of these
lesions are potentially lethal as they can interfere with the unwinding of DNA during
replication and transcription. Most are also indirectly mutagenic; however, O6methylguanine is a directly mutagenic lesion as it can base-pair with thymine during
replication.
46
Examples of (a) alkylating agents; (b) alkylated bases.
Bulky adducts
Cyclobutane pyrimidine dimers are formed by ultraviolet light from adjacent
pyrimidines on one strand by cyclization of the double-bonded C5 and C6 carbon atoms of
each base to give a cyclobutane ring. The resulting loss of base pairing with the opposite
strand causes localized denaturation of the DNA producing a bulky lesion which would
disrupt replication and transcription. Another type of pyrimidine dimer, the 6,4photoproduct, results from the formation of a bond between C6 of one pyrimidine base and
C4 of the adjacent base. When the coal tar carcinogen benzo[a]pyrene is metabolized by
cytochrome P-450 in the liver one of its metabolites (a diol epoxide) can covalently attach to
the 2-amino group of guanine residues. Many other aromatic arylating agents form covalent
adducts with DNA. The liver carcinogen aflatoxin B1 also covalently binds to DNA.
47
Formation of (a) cyclobutane thymine dimer from adjacent thymine residues. S = sugar, P = phosphate;
(b) guanine adduct of benzo[a]pyrene diol epoxide.
48
DNA REPAIR
Photoreactivation
The best defence that is mounted against damage by UV irradiation is known as
photoreactivation. This is catalysed by an enzyme (photolyase) within the cells that in the
presence of visible light can break the covalent bonds linking the two pyrimidine residues,
thus re-establishing the original nature of the base sequence at that point. This method is very
efficient and clearly does not lead to the establishment of mutations. For this reason, when
UV mutagenesis is being performed in the laboratory, it is necessary to exclude light from
the cultures (e.g. by wrapping the bottles in foil) while the cells are recovering from the UV
treatment.
Excision repair
Other types of DNA damage, in particular the formation of pyrimidine dimers by
ultraviolet irradiation, give rise to distortion of the double helix, which can activate a repair
mechanism known as excision repair. The process is initiated by an endonuclease (a complex
enzyme, coded for by genes known as uvrA, uvrB and uvrC, since mutations in these genes
cause reduced Ultra Violet Resistance). This enzyme cuts the DNA strand on either side of
the damage, which exposes a 30 OHgroup; this can be used as a primer byDNApolymerase I
to replace the short region of DNA between the nicked sites (15–20 bases long). The final
step is the joining of the newly repaired strand to the existing DNA by DNA ligase.
Mechanism of excision repair. Endonuclease cleavage removes a portion of the damaged strand. The gap is
filled in by DNA polymerase I; the 5’ – 3’ exonuclease action of DNA polymerase I allows it to remove more
DNA and replace it. Finally, the sugar-phosphate backbone is resealed by DNA ligase
Mismatch repair
The simplest of replication errors is one that leads to the wrong base being
incorporated into the new strand. If this occurs, and is not dealt with by the proof-reading
mechanism, it would lead to mutation. However, the cell has an effective mechanism for
removing such mismatches and replacing them with the correct nucleotide. In order to do
this, it has to know which of the two strands contains the correct information. The
methylation of the DNA, as referred to above, identifies the new and old strands, at least until
the new strand becomes methylated – hence the mechanism is referred to as methyl-directed
49
mismatch repair. The system recognizes the mismatched bases, removes a short region of the
non-methylated strand and fills in the gap.
Recombination (post-replication) repair
There is another type of mutant that is abnormally sensitive to UV, although
possessing a functional excision repair system. These bacteria are defective in a gene (recA)
that is amongst other things responsible for general recombination. This indicates that
excision repair is not the only mechanism for dealing with UV damage, but that there is a
further repair mechanism involving general recombination. The double mutant (uvrA recA)
is even more sensitive than either of the single mutants.
Forms of DNA damage that interfere with the base pairing between the strands will
normally prevent replication, due in part to the requirement of the DNA polymerase for an
accurately paired 3’ end. The replication fork will therefore pause. It is possible however for
replication to restart beyond the lesion, thus leaving a gap in the newly synthesized strand.
This portion of DNA cannot be repaired by excision repair, which requires one intact strand.
The gap can however be filled using a portion of DNA from the other pair of strands by a
recombination process (i.e. cutting and rejoining the DNA). Although this merely re-assorts
the damage rather than directly repairing it, it does achieve a situation where the damage is
repairable. The original damage can now be repaired by excision repair while the gap in the
other DNA molecule can be filled in by DNA polymerase I and DNA ligase.
Post-replication repair. After replication stops at a damaged site (i), subsequent re-initiation leaves a gap in the
new strand (ii). This can be repaired by exchange of DNA (iii), allowing the original lesion to be repaired by
excision repair (iv)
SOS repair
An alternative strategy, when faced with overwhelming levels of DNA damage
preventing normal replication, is to temporarily modify or abolish the specificity of the DNA
polymerase. This enables it to continue making a new DNA strand, despite the absence of an
50
accurately paired 3’ end. Although the new strand is produced, it is obviously likely to
contain many mistakes and the process is therefore described as ‘error-prone’.
The mechanism of induction involves the products of two genes: recA and lexA. The
LexA protein acts as a repressor of the genes of the SOS response, including both recA and
lexA itself. The LexA protein also has the ability to cleave itself but only after the RecA
protein binds to it. The RecA protein has a co-protease function in stimulating the selfproteolysis of LexA. This activity of RecA arises after it binds to single-stranded DNA which
arises as a consequence of DNA damage. This causes a conformational change in the protein
that enables it to bind to LexA, resulting in cleavage of LexA and expression of the SOS
genes.
Two of the SOS genes in particular, umuC and umuD, are involved in mutagenesis,
since strains that are defective in these genes not only have increased UV sensitivity but also
are not susceptible to UV-induced mutagenesis. Certain plasmids carry analogous genes
(mucA and mucB); the presence of these genes increases resistance to UV and also results in
an increased level of mutagenesis by UV and many other mutagenic agents. The UmuCD
complex is able to act as a DNA polymerase, taking over from the normal polymerase (DNA
polymerase III) when that enzyme stalls due to the presence of DNA damage. The low
specificity of the UmuCD polymerase enables it to continue synthesis of DNA past the
lesion, but at the expense of producing errors in the new DNA strand.
51
MUTATION
Mutation
Mutations are permanent, heritable alterations in the base sequence of the DNA. They
arise either through spontaneous errors in DNA replication or as a consequence of the
damaging effects of physical or chemical agents on the DNA.
The simplest mutation is a point mutation – a single base change. This can be either
a transition, in which one purine (or pyrimidine) is replaced by the other, or a transversion,
where a purine replaces a pyrimidine or vice versa.
The phenotypic effects of such a mutation can be various.
 If it is in a non coding piece of DNA or in the third position of a codon, which
often has no effect on the amino acid incorporated into a protein, then it may
be silent.
 If it results in an altered amino acid in a gene product then it is a missense
mutation whose effect can vary from none to lethality, depending on the
amino acid affected.
 Mutations which generate new stop codons are nonsense mutations and give
rise to truncated protein products.
 Insertions or deletions involve the addition or loss of one or more bases. These
can produce frameshift mutations in genes, where the translated protein
sequence to the C-terminal side of the mutation is completely changed.
Mechanisms of mutation
Spontaneous mutation
Spontaneous mutation occurs through errors in the replication of DNA. Within the
normal structure of the double helix of the DNA molecule, the only base-pairing
combinations that are allowed are A-T and G-C; any other combinations would result in a
distortion of the helix. Each of the bases can exist in alternative tautomeric forms with
different hydrogen bonding capabilities. Two examples of tautomerism, the change from an
amino to an imino form and a keto-enol tautomerism.
The normal pairing of adenine and thymine occurs with adenine in the amino form
and the keto form of thymine. The imino form of adenine however, will base pair with
cytosine rather than thymine, while the enol form of thymine will hydrogen bond to guanine.
Induced Mutation
Chemical mutagens
Many different chemical agents interact with DNA or the replication machinery so as
to produce alterations in the DNA sequence. Of these, the simplest to understand are those
agents that act by chemically modifying a base on the DNA so that it resembles a different
base.
For example,
 Nitrous acid causes an oxidative deamination in which amino groups are
converted to keto groups and thus cytosine residues for example will be
converted to uracil. Uracil is not a normal base in DNA and the cell contains
enzymes that will remove it. However, if it persists through to replication it
will be capable of pairing with adenine, thus causing a change from a C-G
52
pair to U-A and ultimately T-A. Similarly deamination of adenine creates the
base hypoxanthine which will base-pair with cytosine.
 Alkylating agents such as ethyl methane sulphonate (EMS) and 1-methyl-3nitro-1-nitroso-guanidine (MNNG) are extremely powerful mutagens. They
act by introducing alkyl groups onto the nucleotides at various positions,
especially the O6 position of guanine, and tend to cause multiple closely
linked mutations in the vicinity of the replication fork.
 The intercalating agents, such as acridine orange and ethidium bromide,
have a different mechanism of action. These molecules contain a flat ring
structure which is capable of inserting (intercalating) into the core of the
double helix between adjacent bases. The consequences of this are the
addition.
53
 The base analogues such as 5-bromouracil, is an analogue of thymine in
which the methyl group is replaced by a bromine atom which is a similar size.
5BU can be incorporated into DNA in place of thymine, since it will form
base pairs with adenine residues on the template strand. However, the
tautomerism is much more pronounced with 5BU. Therefore, in subsequent
rounds of replication, it may pair with guanine rather than with adenine, thus
giving rise to an A-T to G-C mutation.
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Ultraviolet irradiation
Many types of irradiation have been used to generate mutations. The higher energy
rays such as X-rays and gamma rays. They produce an excessive amount of gross
chromosomal damage that is not easily repaired by the micro-organism. Ultraviolet
irradiation on the other hand is easily controlled.
The principal effect of UV irradiation with which is the production of pyrimidine
dimers (commonly referred to as thymine dimers). Where two pyrimidine residues are
adjacent on the same DNA strand the result of UV irradiation is the creation of covalent links
between them. These pyrimidine dimers cannot be replicated and are therefore lethal to the
cell unless it is able to repair the damage.
55
MUTANT SELECTION
An effective mutation selection technique is uses of incubation conditions under
which the mutant will grow whereas the wild type will not. Mutation selection methods
involve reversion mutations or the development of resistance to an environmental stress.
For example, if the intent is to isolate revertants from a lysine auxotroph (Lys-), the
approach is quite easy. A large population of lysine auxotrophs is plated on minimal medium
lacking lysine, incubated, and examined for colony formation. Only cells that have mutated
to restore the ability to manufacture lysine will grow on minimal medium. This method has
proven very useful in determining the relative mutagenicity of many substances.
Resistance selection methods follow a similar approach. Often wild-type cells are not
resistant to virus attack or antibiotic treatment, so it is possible to grow the bacterium in the
presence of the agent and look for surviving organisms. Consider the example of a phagesensitive wild-type bacterium. When the organism is cultured in medium lacking the virus
and then plated out on selective medium containing phages, any colonies that form will be
resistant to phage attack and very likely will be mutants in this regard. Resistance selection
can be used together with virtually any environmental parameter; resistance to
bacteriophages, antibiotics, or temperature are most commonly employed.
Substrate utilization mutations also are employed in bacterial selection. Many
bacteria use only a few primary carbon sources.
56
Mutant Selection. The production and direct selection of auxotroph revertants. In this example, lysine
revertants will be selected after treatment of a lysine auxotroph culture because the agar contains minimal
medium that will not support auxotroph growth.
Replica plating technique for auxotropic isolation
The replica plating technique is used to detect auxotrophic mutants. It distinguishes
between mutants and the wild-type strain based on their ability to grow in the absence of a
particular biosynthetic end product. A lysine auxotroph, for instance, will grow on lysinesupplemented media but not on a medium lacking an adequate supply of lysine because it
cannot synthesize this amino acid.
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Replica Plating. The use of replica plating in isolating a lysine auxotroph. Mutants are generated by treating a
culture with a mutagen. The culture containing wild type and auxotrophs is plated on complete medium. After
the colonies have developed, a piece of sterile velveteen is pressed on the plate surface to pick up bacteria from
each colony. Then the velvet is pressed to the surface of other plates and organisms are transferred to the same
position as on the master plate. After determining the location of Lys- colonies growing on the replica with
complete medium, the auxotrophs can be isolated and cultured.
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CARCINOGENICITY TESTING
Ames test (Mutatest; Salmonella/microsome assay)
A test for detecting whether or not a particular agent is mutagenic. The association of
particular chemicals with cancer, particularly chemicals that are potent mutagens, led
researchers early on to the suspicion that cancer might be caused, at least in part, by
chemicals, the so-called chemical carcinogenesis theory.
Agents thought to cause cancer are called carcinogens. A simple and effective way to
test if a chemical is mutagenic is the Ames test, named for its developer, Bruce Ames. The
test uses a strain of Salmonella bacteria that has a defective histidine-synthesizing gene.
Because these bacteria cannot make histidine, they cannot grow on media without it. Only a
back-mutation that restores the ability to manufacture histidine will permit growth. Thus the
number of colonies of these bacteria that grow on histidine-free medium is a measure of the
frequency of back-mutation. A majority of chemicals that cause back-mutations in this test
are carcinogenic, and vice versa. To increase the sensitivity of the test, the strains of bacteria
are altered to disable their DNA repair machinery. The search for the cause of cancer has
focused in part on chemical carcinogens and other environmental factors, including ionizing
radiation such as X rays.
The Ames test. This test is uses a strain of Salmonella bacteria that requires histidine in the growth medium due
to a mutated gene. If a suspected carcinogen is mutagenic, it can reverse this mutation. Rat liver extract is added
because it contains enzymes that can convert carcinogens into mutagens. The mutagenicity of the carcinogen
can be quantified by counting the number of bacterial colonies that grow on a medium lacking histidine.
59
UNIT –V
BACTERIAL TRANSFORMATION
Bacterial transformation is the process by which bacterial cells take up naked DNA
molecules. If the foreign DNA has an origin of replication recognized by the host cell DNA
polymerases, the bacteria will replicate the foreign DNA along with their own DNA.
Competent
Bacteria which are able to uptake DNA are called "competent" and are made so by
treatment with calcium chloride in the early log phase of growth. The bacterial cell
membrane is permeable to chloride ions, but is non-permeable to calcium ions. As the
chloride ions enter the cell, water molecules accompany the charged particle. This influx of
water causes the cells to swell and is necessary for the uptake of DNA. The exact mechanism
of this uptake is unknown.
Bacterial strains such as those used by Griffiths and Avery have natural
competence, i.e. they have the ability to take up DNA from the medium. Natural competence
is a genetically programmed physiological state. Natural transformation is distinct from
artificial transformation by techniques such as electroporation, protoplast formation, and
microprojectiles. In addition, some bacterial strains, such as E. coli, can be made artificially
competent using CaCl2 and heat shock treatment.
DNA uptake
DNA uptake may be facilitated by DNA interaction in two ways- reversible binding
and irreversible interaction.
Reversible binding – it can be washed off / degraded by DNase.
Irreversible interaction – passage of DNA through cell membrane which requires energy.
Uptake is non specific except in case of H.influenzae.
Two experiments were performed.
1. Strs + DNA of Strr incubated for 15mins and washed free of unincorporated DNA. Then
the cells were taken – one part plated give rise to Strr cells- from other part DNA extracted
and used to transform fresh Strs gave rise to Strr cells.
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2. Strs + DNA of Strr incubated for two min and washed free of unincorporated DNA. The
cells were taken from which one part plated give rise to Strr cells and from other part DNA
was extracted and used to transform fresh Strs. The DNA did not transform.
These experiments prove that a period during which no transforming DNA can be
isolated from potentially transformed cells (about first ten minutes) called eclipse exists.
During uptake double strand binds wherein one strand is degraded and only one strand enters
the host (Gram positive and Gram negative except H. influenzae).
Molecular mechanism of transformation
Two possible mechanisms – (a) incoming donor allele replaces the allele of the
recipient, (b) donor allele is added to the genome.
An experimental evidence to confirm the molecular mechanism was performed.
When x - strain and DNA from x + strain were mixed x + strain was formed. The DNA from
this transformed cell is mixed with x - strain. After transformation both x
+
and x - strains
appeared. If (b) mechanism is true then after transformation no x- strain should be there
because all have x + and x - genes. Hence (a) mechanism proved to be true.
It is not known if transformation is a natural phenomenon in all bacteria -- it may be
widespread in nature but it can be difficult to observe in the laboratory. As mentioned before,
it was discovered first in the gram positive Streptococcus pneumoniae. It has since been
observed and characterized in the gram negative Haemophilus influenzae and in the gram
positive Bacillus subtilis.
Cells need to be competent to take up DNA from the external milieu. In gram positive
bacteria this requires the presence of a DNA-binding protein on the surface of the cell. The
presence of this protein is correlated with nutritional shift-down -- i.e. when the cells start to
run out of nutrients.
The existence of a ssDNA intermediate within the cell has been inferred from the fact
that the cell enters an eclipse period after it has been transformed. In theory, once a cell has
taken up DNA -- containing a specific marker -- from the medium then it should be possible
61
to isolate total DNA immediately from the newly transformed cells and then use that DNA in
a second transformation -- selecting for the same marker.
However, in naturally transformed B. subtilis cells, this is not possible -- no
successful transformants will be found. They will only be found if one waits a period of time
before isolating total DNA to carry out the second transformation.
During transformation of B. subtilis, DNA from the medium is taken up as ssDNA
molecules -- but B. subtilis cannot be transformed with ssDNA. So, until the ssDNA is
converted into dsDNA as a result of recombination with the host chromosome, it will not be
possible to obtain any transformants for the selected marker. The eclipse period is the time
required to convert ssDNA into a stable dsDNA form.
Uses of transformation
 Mapping: based on recombination frequencies between markers. But frequency
varies with size hence not reliable. Based on the principle that two markers transform
together if they are near enough – but varies with competence and DNA
concentration.
 Determine effects of chemical agents on DNA. When a bacterium or an animal is
exposed to a particular chemical and mutagenesis occurs, one cannot be sure that the
chemical acts directly on DNA. To ensure that pure DNA with a marker is exposed to
the chemical and used for transformation. If the marker is inactive it may be sure that
the chemical is active on DNA.
 Identifying DNA can be done by transforming the isolated DNA and testing the
transformants.
62
TRANDUCTION
Transduction is the process by which bacterial DNA is moved from one bacterium
to another by a virus.
Bacteriophages are viruses that infect bacteria. In the process of assembling new
virus particles, some host DNA may be incorporated in them. The virion head can hold only
so much DNA so these viruses
 while still able to infect new host cells
 may be unable to lyze them.
The hitchhiker bacterial gene (or genes) may be inserted into the DNA of the new host,
replacing those already there and giving the host an altered phenotype. This phenomenon is
called transduction.
Transduction are of two types. They are;
1. Generalized: produce particles with only bacterial DNA – bacterial DNA fragment
got from any part.
2. Specialized: produce aprticles with both phage and bacterial DNA in a single DNA
molecule – bacterial genes got from particular region of the bacterial chromosome.
Generalized transduction
If bacteriophages undertake the lytic cycle of infection upon entering a bacteria, the
virus will take control of the cell’s machinery for use in replicating its own viral DNA. If by
chance bacterial chromosomal DNA (instead of viral DNA) is inserted into the viral capsid
used to contain the viral DNA, while this lytic pathway is proceeding, the mistake will lead
to generalized transduction. The new virus capsule now loaded with part bacterial DNA is no
longer infectious, but will still attempt to infect another bacterial cell. When the new DNA is
inserted into this recipient cell it can fall to one of three fates:
1. The DNA will be absorbed by the cell and be recycled for spare parts.
2. If the DNA was originally a plasmid, it will re-circularize inside the new cell and
become a plasmid again.
3. If the new DNA matches with a homologous region of the recipient cell’s
chromosome, it will exchange DNA material similar to the actions in conjugation.
This type of recombination is random and the amount recombined depends on the
size of the virus being used. E.coli phage p1 and S.typhimurium phage p22 are the viruses
that cause transduction. p22 is most studied.
63
p22 infects S. typhimurium by binding to O antigen part of lipopolysaccharide on
outer membrane. After infection recircularizes by binding of cohesive ends. Initially
undergoes θ replication then rolling circle replication. Long concatamers of double strand
DNA are produced. Nuclease cuts the DNA at pac site and DNA is packed into the head.
In S. typhimurium ‘pseudo pac’ sites – phage nuclease cuts ‘pseudo pac’ site – packs
into head – phage contains only bacterial DNA. It can transfer any region chromosomal DNA
and hence called generalized transduction. After injection of bacterial DNA by phage into
another host it transduces host. Recipient should be rec+ and absence of rec+ no integration
can occur. Hence no continuous replication occurs and they form – only tiny micro colonies
called abortive transductants.
Proof for the presence of only bacterial DNA has been observed when bacteria was
grown in 15N medium. p22 DNA was labeled with 14N labeled DNA. Infected the bacteria
by p22 and allowed to grow. Phage progeny centrifuged in CsCl2 and I DNA isolated by
density based separation. Viable phage had only
15
N labeled DNA. This proved that the
phage had only bacterial DNA. Transduction can be quantified by EOP and MOI units.
 EOP – efficiency of plating – fraction of phage particles that can form a plaque.
 MOI – multiplicity of infection – average number of adsorbed phage per bacterium.
Cotransduction ie. simultaneous transduction of two gene. Mapping by co transduction
is possible. Closer the genes are to each other greater the probability they will be co
transduced. Frequency of co transduction is inversely proportional to distance between the
two genes.
64
Specialized transduction
The second type of mistake is called specialized transduction. If a virus removes itself
from the chromosome incorrectly, it can leave part of the viral DNA in the
chromosome.Some of the bacterial DNA can be packaged into the virion. Mistakes in this
process of viral DNA going from the lysogenic to the lytic cycle lead to specialized
transduction. There are three possible results from specialized transduction:
1. DNA can be absorbed and recycled for spare parts.
2. The bacterial DNA can match up with a homologous DNA in the recipient cell and
exchange it. The recipient cell now has DNA from both itself and the other bacterial cell.
3. DNA can insert itself into the genome of the recipient cell as if still acting like a
virus resulting in a double copy of the bacterial genes.
This type of recombination is not random and only small portions of genes are
recombined. Example of specialized transduction is λ phages in Escherichia coli.
Specialized transduction by non lysogen
When Gal - cell is infected with mutant λ and mixed with lysate of Gal + lysogenic for
λ, lysogeny attained will contain almost entire normal λ phage. When plated on galactose
only Gal + grows. Based on this observation there are two types of transduction.
λ Type I: consists of nonlysogenic Gal+. All colonies are Gal + if again resuspended.
As they have arisen as a result of two cross overs they replace gal
-
gene with gal
+
and
contain one gal + gene – haploids.
λ Type II: λ d gal+. About 1% is Gal - and lost prophage. As they have arisen as a
result of single cross over both Gal - and Gal
+
are present which are heterogenotes/partial
diploids. They are incapable of tail producing and are unstable.
65
Specialized transduction by lysogen
This produce a heterogenate which occurs at high frequency by single cross over. But
are highly unstable than type II and gal- production is greater.
High frequency – transducing lysates (HFT)
Dilysogen is formed by sequential insertion of two phage - λ dgal- λ+ - useful λ+
substitutes the lacking gene in λ dgal – yield lysates half of whose phage are transducing
particle – called HFT; lysate from single lysogen LFT.
Specialized transducing phage as cloning vehicle
Proved by φ80 specialized transducing phage to clone supF gene from E.coli. used as
hybridization probes. lacDNA in φ80 as probe, Lac operon induced – mRNA produced
radiolabeled – hybridization – determine amount of radioactive mRNA that are hybridized.
At present specific cloning vectors are being used.
66
CONJUGATION
Definition
Conjugation is a process in which gene transfer (or, exceptionally, complete fusion)
follows the establishment of direct contact between two or more microbial cells.
In bacterial conjugation, DNA is transferred from a ‘male’ (donor) bacterium to a
‘female’ (recipient) bacterium while the cells are in physical contact; a recipient which has
received DNA from a donor is called a transconjugant.
Discovery
The evidence for bacterial conjugation came from an experiment performed by
Joshua Lederberg and Edward L. Tatum in 1946. They mixed two auxotrophic strains,
incubated the culture for several hours in nutrient medium, and then plated it on minimal
medium. For example, one strain required biotin (Bio -), phenylalanine (Phe -), and cysteine
(Cys -) for growth, and another needed threonine (Thr -), leucine (Leu -), and thiamine (Thi -).
Recombinant prototrophic colonies appeared on the minimal medium after incubation. Thus
the chromosomes of the two auxotrophs were able to associate and undergo recombination.
67
Lederberg and Tatum did not directly prove that physical contact of the cells was
necessary for gene transfer. This evidence was provided by Bernard Davis (1950), who
constructed a U tube consisting of two pieces of curved glass tubing fused at the base to form
a U shape with a fritted glass filter between the halves. The filter allows the passage of media
but not bacteria. The U tube was filled with nutrient medium and each side inoculated with a
different auxotrophic strain of E. coli. During incubation, the medium was pumped back and
forth through the filter to ensure medium exchange between the halves. After a 4 hour
incubation, the bacteria were plated on minimal medium. Davis discovered that when the two
auxotrophic strains were separated from each other by the fine filter, gene transfer could not
take place. Therefore direct contact was required for the recombination that Lederberg and
Tatum had observed.
F + X F - Mating
In 1952 William Hayes demonstrated that the gene transfer observed by Lederberg
and Tatum was polar. That is, there were definite donor (F +) and recipient (F -) strains, and
gene transfer was nonreciprocal. He also found that in F
+
X F - mating the progeny were
only rarely changed with regard to auxotrophy (that is, bacterial genes were not often
transferred), but F - strains frequently became F +.
These results are readily explained in terms of the F factor. The F + strain contains an
extrachromosomal F factor carrying the genes for pilus formation and plasmid transfer.
During F
+
X F
-
mating or conjugation, the F factor replicates by the rolling-circle
68
mechanism, and a copy moves to the recipient. The entering strand is copied to produce
double-stranded DNA. Because bacterial chromosome genes are rarely transferred with the
independent F factor, the recombination frequency is low. It is still not completely clear how
the plasmid moves between bacteria. The sex pilus or F pilus joins the donor and recipient
and may contract to draw them together. The channel for DNA transfer could be either the
hollow F pilus or a special conjugation bridge formed upon contact.
Although most research on plasmids and conjugation has been done using E. coli and
other gram-negative bacteria, selftransmissible plasmids are present in gram-positive
bacterial genera such as Bacillus, Streptococcus, Enterococcus, Staphylococcus, and
Streptomyces. Much less is known about these systems. It appears that fewer transfer genes
are involved, possibly because a sex pilus does not seem to be required for plasmid transfer.
For example, Enterococcus faecalis recipient cells release short peptide chemical signals that
activate transfer genes in donor cells containing the proper plasmid. Donor and recipient cells
directly adhere to one another through special plasmid-encoded proteins released by the
activated donor cell. Plasmid transfer then occurs.
69
Hfr Conjugation
Because certain donor strains transfer bacterial genes with great efficiency and do not usually
change recipient bacteria to donors, a second type of conjugation must exist. The F factor is
an episome and can integrate into the bacterial chromosome at several different locations by
recombination between homologous insertion sequences present on both the plasmid and
host chromosomes. When integrated, the F plasmid’s tra operon is still functional; the
plasmid can direct the synthesis of pili, carry out rolling-circle replication, and transfer
genetic material to an F
-
recipient cell. Such a donor is called an Hfr strain (for high
frequency of recombination) because it exhibits a very high efficiency of chromosomal gene
transfer in comparison with F
+
cells. DNA transfer begins when the integrated F factor is
nicked at its site of transfer origin. As it is replicated, the chromosome moves through the
pilus or conjugation bridge connecting the donor and recipient. Because only part of the F
factor is transferred at the start (the initial break is within the F plasmid), the F - recipient
does not become F + unless the whole chromosome is transferred. Transfer is standardized at
100 minutes in E. coli, and the connection usually breaks before this process is finished. Thus
a complete F factor usually is not transferred, and the recipient remains F -.
When an Hfr strain participates in conjugation, bacterial genes are frequently
transferred to the recipient. Gene transfer can be in either a clockwise or counterclockwise
direction around the circular chromosome, depending on the orientation of the integrated F
factor. After the replicated donor chromosome enters the recipient cell, it may be degraded or
incorporated into the F - genome by recombination.
70
F′ Conjugation
Because the F plasmid is an episome, it can leave the bacterial chromosome.
Sometimes during this process the plasmid makes an error in excision and picks up a portion
of the chromosomal material to form an F′ plasmid. It is not unusual to observe the inclusion
of one or more genes in excised F plasmids. The F′ cell retains all of its genes, although some
of them are on the plasmid, and still mates only with an F - recipient. F′ X F - conjugation is
virtually identical with F
+
X F - mating. Once again, the plasmid is transferred, but usually
bacterial genes on the chromosome are not. Bacterial genes on the F′ plasmid are transferred
with it and need not be incorporated into the recipient chromosome to be expressed. The
recipient becomes F′ and is a partially diploid merozygote since it has two sets of the genes
carried by the plasmid. In this way specific bacterial genes may spread rapidly throughout a
bacterial population. Such transfer of bacterial genes is often called sexduction.
F′ conjugation is very important to the microbial geneticist. A partial diploid’s
behavior shows whether the allele carried by an F′ plasmid is dominant or recessive to the
chromosomal gene. The formation of F′ plasmids also is useful in mapping the chromosome
since if two genes are picked up by an F factor they must be neighbors.
71
LINKAGE AND GENETIC MAPS
Finding the location of genes in any organism’s genome is a very complex task.
Following below mentioned approaches can be done to mapping the bacterial genome, using
E. coli as an example. All three modes of gene transfer and recombination have been used in
mapping.
Hfr conjugation is frequently used to map the relative location of bacterial genes. This
technique rests on the observation that during conjugation the linear chromosome moves
from donor to recipient at a constant rate. In an interrupted mating experiment the
conjugation bridge is broken and Hfr X F - mating is stopped at various intervals after the
start of conjugation by mixing the culture vigorously in a blender. The order and timing of
gene transfer can be determined because they are a direct reflection of the order of genes on
the bacterial chromosome. For example, extrapolation of the curves back to the x-axis will
give the time at which each gene just began to enter the recipient. The result is a circular
chromosome map with distances expressed in terms of the minutes elapsed until a gene is
transferred. This technique can fairly precisely locate genes 3 minutes or more apart. The
heights of the plateaus are lower for genes that are more distant from the F factor (the origin
of transfer) because there is an ever-greater chance that the conjugation bridge will
spontaneously break as transfer continues. Because of the relatively large size of the E. coli
genome, it is not possible to generate a map from one Hfr strain. Therefore several Hfr
strains with the F plasmid integrated at different locations must be used and their maps
superimposed on one another.
72
Gene linkage, or the proximity of two genes on a chromosome, also can be
determined from transformation by measuring the frequency with which two or more genes
simultaneously transform a recipient cell. Consider the case for cotransformation by two
genes. In theory, a bacterium could simultaneously receive two genes, each carried on a
separate DNA fragment. However, it is much more likely that the genes reside on the same
fragment. If two genes are closely linked on the chromosome, then they should be able to
cotransform. The closer the genes are together, the more often they will be carried on the
same fragment and the higher will be the frequency of cotransformation. If genes are spaced
a great distance apart, they will be carried on separate DNA fragments and the frequency of
double transformants will equal the product of the individual transformation frequencies.
Generalized transduction can be used to obtain linkage information in much the
same way as transformation. Linkages usually are expressed as cotransduction frequencies,
using the argument that the closer two genes are to each other, the more likely they both will
reside on the DNA fragment incorporated into a single phage capsid. The E. coli phage P1 is
often used in such mapping because it can randomly transduce up to 1 to 2% of the genome.
Specialized transduction is used to find which phage attachment site is close to a
specific gene. The relative locations of specific phage att sites are known from conjugational
mapping, and the genes linked to each att site can be determined by means of specialized
transduction. These data allow precise placement of genes on the chromosome.
73
Mapping by Recombination Frequencies
E. coli strain B can be infected by both h+ and h strains of T4. In fact, a single
bacterial cell can be infected simultaneously by both.
Let infect a liquid culture of E. coli B with two different mutant T4 viruses


h r+ and
h+ r
When this is done in liquid culture, and then plated on a mixed lawn of E. coli B and B/2,
four different kinds of plaques appear.
Phenotype
Number
Plaques
hr+
clear, small
460
h+r
turbid, large 460
of
Genotype
+ +
hr
turbid, small 40
hr
clear, large
40
Total =
1000
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The most abundant (460 each) are those representing the parental types; that is, the
phenotypes are those expected from the two infecting strains. However, small numbers (40
each) of two new phenotypes appear. These can be explained by genetic recombination
having occasionally occurred between the DNA of each parental type within the bacterial
cell.
Just as in higher organisms, assumes that the the frequency of recombinants is
proportional to the distance between the gene loci. In this case, 80 out of 1000 plaques were
recombinant, so the distance between the h and r loci is assigned a value of 8 map units or
centimorgans.
Now repeat coinfecting E. coli B with two other strains of T4:


hm+ and
h+ m
hm+
470
h+m
+
+
470
hm
30
hm
30
Total =
1000
Again, 4 kinds of plaques are produced: parental (470 each) and recombinant (30 each).
The smaller number of recombinants indicates that these two gene loci (h and m) are
closer together (6 cM) than h and r (8 cM).
But the order of the three loci could be either


m–6–h—8—r
or
h–6–m-2-r
To find out which is the correct order, perform a third mating using


mr+ and
m+r
75
440
mr+
m+r
440
m+r+
60
mr
60
Total =
1000
This makes it clear that the order is m—h—r, not h—m—r.
But why only 12cM between the outside loci (m and r) instead of the 14cM produced
by adding the map distances found in the first two matings?
A Three-Point Cross
The answer comes from performing a mating between T2 viruses differing at all three loci:


hmr
and
h+m+r+
(Note: this time one parent has all mutant; the other all wild-type alleles)
Group 1
hmr
435
Group 2
+
hmr
435
Group 3
h+mr+
25
+ +
+
Group 4
hm r
25
Group 5
hmr+
35
+
+
Group 6
hmr
Group 7
+ +
35
hm r
5
Group 8
h+mr
5
Total =
1000
The result: 8 different types of plaques are formed.
76


parentals; that is, nonrecombinants in Groups 1 and 2;
recombinants - all the others
Analyzing these data shows how the two-point cross between m and r understated the true
distance between them.
Let's first look at at single pairs of recombinants as we did before (thus ignoring the third
locus).




If we look at all the recombinants between h and r but ignore m (as in the first
experiment), we find that they are contained in Groups 5, 6, 7, and 8 — giving the
total of 80 that we found originally.
If we look at recombinants between h and m but ignore r (as in the second
experiment), we find that they are contained in Groups 3, 4,7, and 8 — giving the
same total of 60 that we found before.
But if we focus only on m and r (as we did in the third experiment), we find that the
recombinants are contained in Groups 3, 4, 5, and 6 — giving the same total of 120 as
before while the non-recombinants are not only in Groups 1 and 2 but also in Groups
7 and 8. The reason: a double-crossover occurred in these cases, restoring the
parental configuration of the m and r alleles.
Because these double crossovers were hidden in the third experiment, the map
distance (12 cM) was understated. To get the true map distance, we add their number
to each of the other recombinant groups (Groups 3, 4, 5, and 6) so 25 + 5 +25 +5 +35
+ 5 + 35 + 5 = 140, and the true map distance between m and r is the 14 cM that we
found by adding the map distances between h and r (8 cM) and h and m (6 cM).
The three-point cross is also useful because it gives the gene order simply by inspection:

Find the rarest genotypes (here Groups 7 and 8), and the gene not in the parental
configuration (here h) is always the middle one.
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78