EGERTON
UNIVERSITY
COLLEGE OF OPEN AND DISTANCE LEARNING
THE E-CAMPUS
E-LEARNING COURSE
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
By
Prof. M.A. OKIROR
mokiror@egerton.ac.ke
+254722280311
June, 2020
__________________________________________________________
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 1 OF 153
COURSE PRELIMINARIES
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
Is this course for you?
Course Preliminaries
This course is designed for year four Bachelor of Education Science, and
Bachelor of Science with a bias in biology undergraduate students. These are
students who have had a basic understanding of the principles of heredity,
to which they were exposed in the General Genetics course. With such
grounding, these students can now explore the molecular basis of heredity.
Through this course they will understand the basis of how a gene expressed
the trait it is responsible for: the molecular structure of the gene, and
understand how the message of the gene (as nucleic acid) is used to
determine the phenotype. Microbial genetics on the other hand,
demonstrates to the student heredity in microorganisms and the role of
microorganisms in shaping molecular biological events, e.g. gene regulation,
protein synthesis, etc.
You are expected to complete the course in 45 hours within a period of one
semester. There are no pre-requisites for you to study this course.
Introduction to the course
This course comprises two parts (sections): molecular genetics, and
microbial genetics. In the first part, students are introduced the
fundamentals of molecular genetics – history of molecular genetics,
structure and function of the gene, gene regulation, gene mutation, etc. The
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 2 OF 153
gene at molecular level is a length (segment, section, piece, or fragment) of
nucleic acid (DNA or RNA). Therefore, in this part, students are explained
the relationship between the DNA (of which a gene is part of) and the
phenotype! In the second part, micro-organisms and their place in molecular
genetics breakthroughs are examined and appreciated. Mendel, the father of
the modern genetics and contemporaries used larger organisms (macroorganisms) to develop the principles of heredity. However, as newer
investigations were carried out and techniques of inquiry upgraded, it was
realised that the organisms of choice for study of intrinsic details of the gene
were smaller microscopic ones, largely bacteria and bacteriophages.
Course Content
There are NINE (9) topics in this course, namely:
Topic One: Introduction to Molecular and Microbial Genetics
Topic Two: Genes and Chromosomes
Topic Three: Nucleic acids: As repositories of biological information
Topic Four: Nucleic acids synthesis: The Enzymology of replication and
transcription
Topic Five: Translation: The Genetic code concept; Protein synthesis
Topic Six: Mechanisms of gene regulation
Topic Seven: Mutations: Nature and occurrence; DNA repair
Topic Eight: Recombinant DNA technology
Topic Nine: Microbial genetics: genetic exchange; gene mapping
Course Learning Outcomes
Upon successful completion of this course, you should be able to:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 3 OF 153
i.
Explain the evolution of molecular genetics and the place of microorganisms in the advancement and understanding of the molecular
basis of heredity.
ii.
State the molecular structures of gene and chromosome.
iii.
Explain the chemical and physical structure of nucleic acid; dispel
the early notion that protein was the molecule of heredity
iv.
Demonstrate how nucleic acids make copies of themselves, relating
this to chromosome replication.
v.
Show the process of gene expression (i.e. transcription and
translation.
vi.
Demonstrate that whereas organisms have many genes, the
expression of these genes is regulated.
vii.
Explain how phenotype variations is brought about at the molecular
level and show that gene alterations can be corrected.
viii.
Demonstrate that gene manipulation is possible and can be used to
the benefit of organisms and science
ix.
Account for the fundamental place of micro-organisms in the
advancement of molecular genetics
Course Study Skills
As an adult learner your approach to learning will be different to that from
your school days: you will choose what you want to study, you will have
professional and/or personal motivation for doing so and you will most likely
be fitting your study activities around other professional or domestic
responsibilities.
Essentially you will be taking control of your learning environment. As a
consequence, you will need to consider performance issues related to time
management, goal setting, stress management, etc. Perhaps you will also
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 4 OF 153
need to reacquaint yourself in areas such as essay planning, coping with
exams and using the web as a learning resource.
Your most significant considerations will be time and space, that is, the time
you dedicate to your learning and the environment in which you engage in
that learning.
We recommend that you take time now - before starting your self-study - to
familiarize yourself with these issues. There are a number of excellent
resources on the web. A few suggested links are:
http://www.how-to-study.com/
The "How to study” web site is dedicated to study skills resources. You will
find links to study preparation (a list of nine essentials for a good study place),
taking notes, strategies for reading text books, using reference sources, test
anxiety.
http://www.ucc.vt.edu/stdysk/stdyhlp.html
This is the web site of the Virginia Tech, Division of Student Affairs. You will
find links to time scheduling (including a "where does time go?” link), a study
skill
checklist,
basic
concentration
techniques,
control
of
the
study
environment, note taking, how to read essays for analysis, and memory skills
("remembering”).
http://www.howtostudy.org/resources.php
This is another "How to study” web site with useful links to time management,
efficient reading, questioning/listening/observing skills, getting the most out
of doing ("hands-on” learning), memory building, tips for staying motivated,
developing a learning plan.
Need Help?
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 5 OF 153
This course was developed in June 2020 by Michael A. Okiror, Phone:
+254722280311; Email: mokiror@egerton.ac.ke. Prof. Okiror is a Lecturer of
Molecular Genetics in the Department of Biological Sciences at Egerton
University.
This session, the instructor for this course is Prof. Michael A. Okiror. My office
is located in the Department of Biological Sciences Annex (Rm 5), Faculty of
Science. You may consult me during the normal working hours between
Monday and Friday or contact me through: Phone: +254722280311; Email:
mokiror@egerton.ac.ke.
For technical support e.g. lost passwords, broken links etc. please contact
tech-support via e-mail elearning@egerton.ac.ke. You can also reach learner
support through elearnersupport@egerton.ac.ke.
Assignments/Activities
Assignments/Activities are provided at the end of each topic. All
assignments/activities will do not require submission.
Course Learning Requirements

Timely submission of practical schedules (15%)

2 CATs (15%) – each at 7.5 marks.

Final Examination (70% of total score)

Note book, calculator, laptop/computer or ipad

Drawing book, pencil and rubber

Commitment

Discipline
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 6 OF 153
Self-assessment
Self-assessments are provided in order to aid your understanding of the
topic and course content. While they may not be graded, you are strongly
advised to attempt them whenever they are available in a topic.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 7 OF 153
TOPIC ONE: INTRODUCTION TO MOLECULAR AND MICROBIAL
GENETICS
Introduction
Welcome to topic one. This topic is intended to give you a broad preview of
the science of molecular and microbial genetics. You will be guided through
the concept of molecular genetics – definition and its importance. You will also
be inducted into the importance of microbial genetics and its centrality in
molecular discoveries involving the gene. Thus you will be prepared to learn
in detail the two components of this course.
Topic Time

Compulsory online reading, activities, self-assessments and practice
exercises [2 hours]

Optional further reading [1.5 hours]

Total student input [3.5 hours]
Topic Learning Requirements

Participation in one to two chats
Learning Outcomes
By the end of this topic you should be able to:
i.
Define the terms molecular genetics, molecular genetics
ii.
Define the schools that have contributed to the science of molecular
genetics
iii.
State characteristics of genetic material
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 8 OF 153
iv. Explain why protein was considered the molecule of heredity instead of
nucleic acids
Topic Content
Preview
The rapid pace of research in genetics especially after 1900 led to
increasing specialization, jargonisation and fragmentation of this branch of
biology. Thus, today we have many sub disciplines of genetics, e.g.
molecular, microbial, human, medical, ecological, behavioural genetics, etc.
Definitions:
Molecular genetics - is the subdivision of genetics concerned with
understanding the structure and function of genes.
Microbial genetics (also called bacterial and phage genetics) - is the genetic
study of micro-organisms especially their role and usefulness in
investigations on the principles and material basis of heredity and variation.
MOLECULAR GENETICS
1.1 Introduction
Molecular genetics is a recent development in the science of genetics, in
fact of the mid C20th. It is a branch of genetics concerned with the study of
all aspects of the gene, e.g. the identification of the chemical nature of the
gene. It looks at a gene as a unit of biological information. Molecular
geneticists aim to understand the way in which biological information is
stored in genes and how the information is made available to the living cell.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 9 OF 153
However, the earliest reference or study of the chemical composition of
cells is in the early 1860s by the German physician, F. Miescher. His study
based on pus cells led to the isolation and purification of nuclein, the
chemical material of cell nuclei. However, it was not until sometime in the
1940s/1950s that nuclein was demonstrated to contain DNA.
According to Erwin Chargaff (1974), molecular genetics arose with the
discovery of the transforming properties of DNA by Avery and others in 1944
and the introduction of phages as objects of biological research.
By the latter half of C20th, the molecular structure and function of genes
had been determined (ref. Hershey and Chase, Watson and Crick, etc).
Meanwhile J. Monod and F. Jacob through their studies showed how
gene expression is controlled. They proposed that certain genes regulate the
activity of other genes.
Three intellectual currents (also called schools) have helped shape the
advancement of molecular biology. These are:
1. The British school, which was centred at Cambridge with such
personalities as F. H Crick, J. D Watson, etc. This school was responsible for
advancing our understanding of the molecular structure of nucleic acids and
proteins.
2. The American school, based at the California Institute of Technology
(Caltech) through people like Max Delbruck, A. Hershey, Salvador Luria, S.
Benzer, G. Beadle, E. Tatum, J. Lederberg, etc. They are often referred to as
the "Phage group". They contributed tremendously to our knowledge of
DNA- its chemical nature through an experiment by Hershey and Chase, and
replication; finer structure of the gene and certain aspects of phage
genetics. Later, this group was responsible for the birth of bacterial genetics!
Delbruck, Hershey and Luria in 1969 were awarded the Nobel Prize for
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 10 OF 153
Physiology or Medicine for their work concerning the replication mechanism
and genetic makeup of viruses.
3. The French school, which was based at the Louis Pasteur Institute, Paris
was led by Francois Jacob, and had Elie.L. Wollman, Jacques Monod and
Arthur B. Pardee. Their contributions lie mainly in regulation of gene
expression and bacterial sexuality. They are particularly remembered for
their insightful research, designated as the “pajamo” experiment. Jacob and
Monod won the Nobel Prize in Physiology or Medicine in 1965, sharing it with
Andre Lwoff "for their discoveries concerning genetic control of enzyme and
virus synthesis.
The birth of molecular biology not only unified the life sciences and
furnished a firm foothold of these sciences in physics and chemistry but also
marked the commencement of a matchless period of growth of genetics
leading to the establishment of a link between genetics and biochemistry. An
outstanding product of "Genetics-biochemistry" merger was the Cracking of
the genetic code in the 1960s. This and the other major advances during the
1950 - 1980 era can be safely attributed to:
a. the introduction of new methods of analyses;
b. the use of electron microscopic techniques, and
c. the employment of sophisticated biophysical and biochemical instruments
and techniques to the study of micro-organisms and viruses.
In the past three or so decades, genetics has taken a revolutionary and
sometimes controversial turn with the development of recombinant DNA
technology. By this approach, a revolution has slowly but surely taken place
in Science, Agriculture, Medicine and Industry. Recombinant DNA technology
encompasses a variety of techniques for isolating, analysing and
manipulating individual genes. Today, it is now routine to isolate a segment
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 11 OF 153
of DNA, purify it, amplify and sequence it, and use as may be desired, e.g.
introduce it into other organisms where it will be expressed. This technology
is making it possible to study in greater detail the molecular structure and
function of genes, their regulation especially in multicellular organisms
(plants and animals).
Below is a citation of important persons, their studies and the years of
such studies that have made an impact to the development of the science of
molecular genetics:
Year
Discovery / contribution
1928
1934
Discovery of transformation in bacteria (F. Griffith);
Demonstration that certain phages are made of DNA and protein
(M. Schlesinger).
1944
DNA is the hereditary material (O.T. Avery, C.M. MacLeod, and
M. McCarty).
1950
Equivalence between A and T and between G and C in DNA (E.
Chargaff).
1952
When a phage infects a host cell, the phage DNA enters the host
but the protein does not (Hershey/Chase).
1953
The double helix model of DNA structure (J.D. Watson / F.H.C.
Crick).
1955
Elucidation of the structure of a gene and definitions of cistron,
recon and muton (S. Benzer).
1955
Reconstitution of hybrid TMV by mixing protein and RNA
components derived from separate sources (H. Fraenkel Conrat, R.C. Williams).
1956
Enzymatic synthesis of DNA in vitro (A. Kornberg et al.).
1957
Unravelling the molecular basis of the difference between A and
S haemoglobin (V.M. Ingram).
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 12 OF 153
1958
Determination of the distribution of old and new DNA to the
progeny of bacterial cells by means of density-gradient
centrifugation technique (M. Messelson and F.W. Stahl).
1961
mRNA and its significance (S. Brenner, F. Gros, F. Jacob, J.
Monod).
1961
DNA-RNA hybridization (B.D. Hall, S. Spiegelman).
1961
Universality of the genetic code (G. von Ehrenstein, F. Lipmann).
1961
The codons are triplets of the base pairs of DNA (Crick et
1961
1961
Cracking of the genetic code (M. Nirenberg, J.H. Matthaei).
The operon concept in E. coli (Jacob / Monod)
1964
Colinearity between gene and its protein product (A.S. Sarabhai
et al., Yanofsky et al.).
1964
Mechanism of excision repair in bacterial DNA (R.B. Setlow, etc).
1965
Chain-terminating codons AUG, UAA (S. Brenner, A.O.W.
Stretton, S. Kaplan).
1965-68
Restriction (site specific) endonucleases in E. coli (W. Arber et
al.).
1967
Use of polynucleotides with known repeating sequences in the
elucidation and study of the genetic code (H.G. Khorana et al.).
1967
mRNA can originate from both strands of the DNA (K. Taylor, W.
Szybalski et al.).
1970
Reverse transcription: RNA-dependent DNA polymerases occur in
oncogenic viruses (D. Baltimore, H.M. Temin).
1970-80
The age of plasmids, cloning vehicles, restriction endonucleases,
ligases and genetic engineering (Several workers).
al.).
1.2 THE GENETIC MATERIAL
Until 1944, it was not clear what chemical component of the
chromosomes constitutes the genetic material, protein or nucleic acids. Early
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 13 OF 153
molecular geneticists therefore were hard pressed to identify the physical
and chemical nature of the genetic material.
Extensive chemical analyses of chromosomes of different organisms
showed chromosomes to contain proteins and nucleic acids (DNA and RNA).
Because of its abundance, diversity and complexity within the cell, the
role of genetic information carriers was assigned to proteins.
This controversy persisted till about mid-C20th when the first direct
experimental evidence identified DNA as the informational basis for the
process of heredity. DNA is found in most micro-organisms and higher
organisms. Later, RNA was found to be the genetic material of some viruses.
(1) Characteristics of the genetic material
In order to function as an informative family of molecules, the chemical
substances that makeup hereditary determinants must have at least the
following attributes:
(a) Expression. It must be capable of spelling out specific messages just as
the letters of an alphabet make meaningful words. The expression of
alternative traits, e.g. round vs wrinkled, is essential in identifying genes
through the observance of segregation of alleles in mating.
(b) Replication. The material of heredity must be able to replicate itself in
order to be passed on to daughter cells. This it does at interphase.
(c) Variation. The genetic material be able to vary as a result of mutation to
give alternative phenotype.
1.3 Protein as the genetic material
Although proteins and nucleic acids were both considered major
candidates for the role of the genetic material, many geneticists, until the
1940s, favoured proteins. This was for the following reasons:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 14 OF 153
(a) Abundance and diversity. Proteins constitute about 50% dry weight of
the cells and are of a wide variety.
(b) Chemical structure of nucleic acids. It is simple and only repeats of the 4
nucleotide, A, G, C, and T.
(c) Level of research. Before 1940, most geneticists were engaged in the
study of transmission genetics and mutation. The excitement generated in
these areas undoubtedly diluted the concern for finding the precise molecule
that serves as the genetic material.
Topic Summary
In this topic, you have learned how the science of molecular genetics was
incepted and grew to its current status. You have also been demonstrated the
central role microorganisms play in the advancement of molecular biology in
general and molecular genetics in particular. A brief summary has been
presented of individuals/institutions that have incubated this science.
Further Reading
George W. Burns: The Science of Genetics (5/e)
TOPIC ACTIVITIES
Attempt these questions as a test to your understanding of the topic.
1. Which schools and what is the contribution of each to the advancement of
the science of Molecular genetics?
2. For any material to be regarded as heredity, what characteristics should
define it?
3. Why were proteins held for long as the material of heredity?
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 15 OF 153
TOPIC 2. GENES AND CHROMOSOMES
Introduction
Welcome to our topic two of this course. As the title indicates, we are going
to examine in detail genes and chromosomes – organelles considered central
to heredity. We shall give examples of gene and chromosome numbers in
some species. Also to be learned is the relationship between DNA content
and organismal complexity.
Topic Time
.
It is estimated that this topic can be fruitfully covered in one lecture
hour with an additional 1.5 hours for supplementary reading and
assignments.
Topic Learning Requirements
There are no special learning requirements except your Bota 111 notes as
refresher.
Learning Outcomes
After successful completion of this topic, you should be able to:
i.
Give both classical and molecular definitions of the gene
ii.
Explain the “fine structure” of the gene.
iii.
Describe the chemical composition of chromosomes and how this
defines them as carriers of hereditary information
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 16 OF 153
iv.
Describe one study that illustrates gene-protein relationship
v.
Demonstrate variations in C-value with cyclic changes
Topic Content
2.1 GENES
Definition
In year one General Genetics course, we defined a gene as a
“fundamental unit of inheritance” or a “unit of information about a heritable
trait”. However, in at the molecular level, we wish to define a gene in a
corresponding manner, that is as “a stretch of DNA that specifies the
manufacture of a single type of protein (or for some genes, certain RNAs).
Thus, a gene contains a set of instructions (information) about a
heritable trait. Genes are the working subunits of DNA. The DNA in each
chromosome constitutes many genes. Each gene contains a particular set of
instructions, usually coding for a particular protein 0or for a particular
function. Mendel first discovered genes in 1865, but called them factors of
heredity. They remained so until 1909 when Johannsen called them genes.
Their importance was not realized until the start of C20th. For many years
the chemical nature of the gene, the molecular structure of DNA and the
correspondence between genes and proteins were a mystery. These too
were resolved in the 1950s.
Inheritance, distribution and number
Every living thing inherits from its parents a set of genes that determine
its development and appearance. Genes are disposed along the
chromosomes in a fixed linear order characteristic of the spp. The number of
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 17 OF 153
genes for any species is not yet known accurately but probably varies from
several thousand for simple bacteria to many thousands for a mammal. Prior
to the Human Genome Project, a conservative estimate put the number of
genes in a human being at between 50,000 to 140,000 genes distributed
along the 22 pairs of autosomal and sex (X and Y) chromosomes. Thus, each
chromosome contains around 1,000 genes.
2.2 CHROMOSOMES
These are the vehicles of heredity. This is so because they are the
cellular location of genes. This phenomenon was demonstrated by W. Sutton
and T. Boveri in 1902, and confirmed by T.H. Morgan in 1913. He and his
students collected a lot of data on Drosophila. They localised several genes
to specific chromosomes.
They are largely made of nucleoproteins (i.e. a complex of organic acids,
DNA or RNA, and histone proteins). The nucleoprotein complex is called
chromatin. It is chiefly of two types, heterochromatin and euchromatin.
Prokaryotic cells have no nucleus and their ‘chromosome’ is a single circular
molecule of DNA anchored to the cell membrane. The prokaryotic
chromosome is a complex of about 20% proteins and 80% DNA and forms a
compact mass- the nucleoid. The chromosomes of eukaryotic cells on the
other hand are highly organised structures containing about twice as much
proteins as DNA. Chromosomes are characterized on the basis of physical
characteristics, e.g. size, centromere position, etc.
Two cellular processes -mitosis and meiosis- ensure their distribution in a
cell.
Chromosome number in a cell is either haploid (n) or diploid (2n), or greater
(i.e. polyploidy. Single maternal and paternal chromosomes constitute a
homologous pair and are identical in size and shape.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 18 OF 153
2.3 FINE STRUCTURE OF THE GENE
With the acceptance of the chromosome theory of inheritance, genes
came to be thought of as beads on a chromosomal string. Mutant alleles of a
single gene were considered as beads of different colours, with only one
bead of a particular colour on each string. Recombination was believed to
involve breaking and re-joining of the string at positions between beads
while recombination within a gene was thought not to occur. The gene was
regarded as an indivisible entity, and it was even defined as the basic unit of
recombination as well as the unit of function and mutation. Due to lack of
resolving power of the genetic systems, this theory was accepted till about
the 1940s. With the introduction of the microbial genetic systems by
Delbruck and colleagues, the advancement in experimental techniques, the
molecular structure of genes could then be analysed. Seymour Benzer was
able to show that the detailed gene structure would not fit into the definition
of the bead theory. He used a genetic system that detected extremely small
recombination percentages. He demonstrated that whereas a gene can be
defined as a unit of function (Cistron), it could be subdivided into a linear
array of sites that are mutable (Mutons) and that can be recombined
(Recons).
These later studies sharply defined the basic unit of mutation,
recombination and genetic function. These studies destroyed the indivisible
bead theory. In its place came the concept that a gene is sequence of
nucleotide pairs.
2.4 GENE-PROTEIN RELATIONSHIP
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 19 OF 153
The most fruitful endeavors to find correspondence between genes and
proteins examined the ways in which changes in the gene affected kind and
function of proteins in the cell.
Among the pioneers were:
(1) Archibald E. Garrod. Between 1898 and 1902, he examined the human
condition alkaptonuria. This arose due to inability of the individual to
completely metabolise the amino acid tyrosine in the diet producing instead
an intermediate product called homogentisic acid (or phenylpyruvic acid). By
carrying out a pedigree analysis and genetic studies on this trait, Garrod
concluded that this was a heritable abnormality arising out of a mutation in a
single gene. A defective gene led to the production of a non-functional
enzyme, homogentisic acid oxidase! Thus, by the start of C20th Garrod had
demonstrated a relationship between a gene and a protein (or enzyme).
Unfortunately, Garrod's notions, like those of Mendel seem to have been
so far ahead of their time that they had little influence in the market place of
genetic ideas until their rediscovery 30 years later! However, in between
Garrod’s study and that of Beadle and Tatum, (i.e. 1920 to 1940), three
inconclusive studies were made to establish a relationship between genes
and enzymes. These were on plant pigments, eye color in butterflies, and
eye color in Drosophila.
(2) George W. Beadle and Edward L. Tatum. In 1941, they carried out a
classical study and showed that genes controlled enzyme synthesis. They
used a fungus Neurospora that can be grown in a simple, synthetic medium
provided biotin (vitamin) is included. Varying the medium allowed them to
isolate several mutants (1 in 200 spores) that could not grow in the minimal
medium but could grow in the complete medium (enriched medium).
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 20 OF 153
Inference
The unusual fungi are the descendants of mutant spores in whose
hereditary material a gene mutation has blocked an essential metabolic
pathway.
Conclusion
On the basis of such observation, they promulgated the "one gene one enzyme hypothesis". This hypothesis asserts that “each gene has only
one primary function, to direct the formation of one and only one enzyme,
and thus in controlling the single chemical reaction catalysed by that one
enzyme”. Since then, it has been repeatedly demonstrated that a gene can
only control the production of one polypeptide, and therefore this hypothesis
has since been renamed “one gene - one polypeptide”.
In 1958, the two scientists together with Lederberg shared the Nobel
Prize in Physiology or Medicine for their demonstration that a mutation of a
gene would produce a corresponding loss of an enzyme and an alteration of
a cell's metabolism.
2.5 DNA AND MORPHOLOGICAL COMPLEXITY
A genome is an organism’s complete cell of DNA including all the genes.
Each cell has a genome! The genome size does not always correlate with the
complexity of the organism and, in fact, shows great variation in size and
gene number. Genome size is usually measured in base pairs (bps) or bases
in single-stranded DNA or RNA.
The amount of DNA in the nucleus varies between plants (organisms),
both within and between species. Because the amount of DNA also changes
according to the cell cycle, it is common to compare values during the G1
phase, before replication has occurred. This amount is known as the 2C
value, meaning the content of DNA in a somatic nucleus.
Having established the role and function of DNA, speculation logically set
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 21 OF 153
in that the amount of DNA in the nucleus (genome) was related to
the organism’s complexity (Table). For instance, yeasts, Drosophila,
chickens and humans have successively larger amounts of DNA in their
haploid chromosome sets (0.05, 0.15, 0.3, and 3.2 picograms, respectively),
in increasing complexity of these organisms.
The DNA content of various cells
Size of genome:
 bpsa
length (mm)
Organism
Maximum number of:
proteins
encoded
chroms.
b
(n)
1. Prokaryotes:
a) E. coli
4.5 x 106
1.36
3.3 x 103
1
(yeast)
7 x 107
4.60
1.125 x 104
17
D. melanogaster
1.65 x 108
56.00
1.375 x 106
4
990
2.42 x 106
23
2. Eukaryotes:
S. cerevisiae
3. Homo sapiens:
Man
3 x 109
------------------------------------------------------------------------------------------------------a
= bps = base pairs; 1 Kb = 1000 nucleotides,
b
= assuming 1,200 bp/protein
This apparently was the belief of many scientists. Later, experiments
showed that such a relationship was not consistent. The vertebrate species
with the greatest amount of DNA/cell are amphibians/reptiles (e.g. toad =
4.2; lizard 3.5 cf 3.0 for man) which are surely less complex than humans in
both structure and behaviour.
Besides, there is also considerable intra-group variation in DNA content.
For example, all insects or all amphibians would appear to be similarly
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 22 OF 153
complex, but the amount of haploid DNA within each of these phyla varies
by a factor of 100! Also, similar plants can have strikingly different
karyotypes. The broad bean, Vicia faba (2n = 14) has about half the
chromosome number of the Kidney plant (P. vulgaris) (2n = 22). However, it
has about 3-4 times as much DNA/cell as the kidney bean.
Findings such as above have given rise to the “C - value paradox” (i.e.,
failure of C-value to correspond to phylogenetic complexity). The reason is
not clear.
From the above facts it would seem that some of the DNA in certain
organisms is "extra" or expendable; apparently not all DNA in all organisms
is for encoding proteins. For instance, a third of the human genome is
redundant. Of the balance, there are regions of control and introns that
further reduce the effective quantity. In bacteria, however, the entire DNA is
used to code information or control its expression.
2.6 THE C-VALUE CONCEPT
The C-value is another measure of genome size. The C-value refers to
the amount, in picograms, of DNA contained within a haploid nucleus (e.g. a
gamete) or one half the amount in a diploid somatic cell of a eukaryotic
organism.
Since DNA and proteins form the chromatin of which chromosomes are
made of, any changes in chromosome number in the cell affects the amount
of DNA in that cell (Fig. below).
If C represents the amount of DNA in a haploid cell (gamete) before
fertilization, then a gamete will have an amount of 1C and a zygote 2C.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 23 OF 153
During the S period of pre-meiotic mitosis, the DNA content in the cell will
rise to 4C. Mitotic anaphase will bring it back down to 2C. During the S
immediately preceding meiosis, it will rise back to 4C. Anaphase 1 will
reduce the DNA content to 2C and anaphase 2 to the original 1C value.
mitotic divs
amount of DNA/
4C _
S
A
meiotic divs
S
A1
cell (C-value)
3C _
Fertilzn
2C _
zygote
//
1C _gamete
A2
gamete
Sequence of stages
Fig. 1. Changes in the amount of DNA in a cell of a plant or animal, which
undergoes mitosis then meiosis.
Topic Summary
In this topic we have examined the gene and chromosome in molecular
terms and defining their core functions in heredity. At the molecular level
Benzer showed a gene to be a pair of nucleotides each of which can undergo
mutation and/or recombination. Through studies of personalities like Garrod
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 24 OF 153
or Beadle and Tatum we have learnt that there is an intricate relationship
between genes and proteins. As to whether there is a link between the
cellular DNA content and organismal complexity, this has been shown to be
largely so but not universal. Finally, we have considered changes that can
occur to the DNA content in mitotic cells and those that produce gametes.
Further Reading
(1) Suzuki, D.T., Griffiths, Miller, J.H. and Lewontin, R.C. (1986). An
Introduction to Genetic Analysis. (3/e). W.H. Freeman and Co. NY
TOPIC ACTIVITIES
Attempt these questions to evaluate your understanding of the topic
1. Explain what these terms mean: gene, muton, cistron, recon, and c-value
concept.
2. What do you understand by “C-value paradox”?
3. Describe one study to show a relationship between a gene and protein.
4. Explain changes that can occur to DNA content in mitotic and meiotic
cells.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 25 OF 153
TOPIC 3. NUCLEIC ACIDS (DNA AND RNA)
AND THEIR ROLES AS REPOSITORIES OF GENETIC INFORMATION
Introduction
Welcome to our third topic of this course. In this topic it is demonstrated to
us that nucleic acids (DNA and RNA) are molecules of heredity and not
protein. This is followed by examination of the chemical and physical
structures of DNA and elucidation that both these nucleic acids are the
repositories of biological information.
Topic Time
It is envisaged that three lecture hours is adequate for the coverage of this
topic. The student is expected to devote at least two more hours for selfrevision and going through the assignment.
Topic Learning Requirements
No specific requirements are proposed in this topic but accumulated
understanding of what has been covered is an advantage.
Learning Outcomes
After successfully completing this topic, you should be able to:
i.
Describe those experiments that led to the acceptance of DNA and
RNA as the molecules of heredity
ii.
Describe both the chemical and physical structures of DNA and RNA
Topic Content
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 26 OF 153
3.1 INTRODUCTION
In General genetics, we learnt how Mendel and his successors used data
from breeding experiments to deduce the presence and activity of genes,
analyse the effect of alternative alleles on phenotype, etc.
With this approach, it is possible to predict the outcomes of genetic
crosses in the absence of detailed knowledge of the molecules and
biochemical reactions that underlie the events observed. However, it is
impossible to decipher the exact mechanisms by which genes determine
phenotypes, transmit instructions between generations, and evolve new
information without understanding what they are made of. Thus, the
necessity to examine DNA - the molecule that is the genetic material.
3.2 CHEMICAL AND PHYSICAL STRUCTURE OF DNA
1. Chemical structure of DNA
For a long time, scientists did not imagine the physical structure of DNA
to be complex given that it was merely repeats of the 4 (tetra) nucleotides.
However, with the application of novel and more advanced techniques, Erwin
Chargaff, a biochemist (1949-53) was able to measure the amount of DNA in
tissues and cells and determine the relative proportions of the bases present
in the DNA samples of several species. Through his results, Chargaff
convinced the scientific world that DNA had the chemical complexity
necessary of genetic material and thus formulated the following rule:
(1) the amount of DNA extracted from different tissues of the same
organism is the same;
(2) the base composition in a given DNA molecule is constant, i.e. A = T,
and G = C. For instance, in humans, A = 30%; T = 29.4%; G = 19.9%; C =
19.8%.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 27 OF 153
(3) that DNA contains equal amounts of purines and pyrimidines, i.e. A + G:
C + T as a rule is close to 1.00 regardless of the organism used as the DNA
source.
Chargaff's findings, as will be seen, played an important part in the
discovery of the double helix structure of DNA.
DNA like RNA is a polymeric molecule made up of monomeric units called
nucleotides.
A nucleotide has three basic components: a pentose (5-carbon) sugar,
deoxyribose; a phosphate group; and a nitrogenous (nitrogen-containing)
base.
(1) Nitrogenous bases: two kinds - a 9-membered double-ringed purines
(two), Adenine and Guanine; a 6-membered single-ringed pyrimidines
(three), Cytosine, Thymine and Uracil. These bases are commonly
represented by their first letters: A, G, C, T, and U.
(2) Pentose sugar: this 5C-sugar gives the nucleic acid its name,
deoxyribose for DNA, and ribose for RNA. The former has an O group
missing on C2-.
(3) A phosphate group: In a nucleic acid polymer, this group joins two
nucleosides to each other by forming a phosphodiester bridge between the
C5' of one sugar and C3' of another.
Note: A combination of a base and a sugar = nucleoside, while a nucleoside
plus a phosphate group is a nucleotide. A mononucleotide can be described
as a nucleoside monophosphate. When two or three phosphates groups are
on a nucleoside we have nucleoside diphosphate and nucleoside triphosphate
respectively. The triphosphate form serves as a pre-cursor molecule during
nucleic acid synthesis. Also, the triphosphates like ATP and GTP are in the
cell's bioenergetics. Nucleosides and nucleotides are named according to the
specific nitrogenous base that is part of the building block, viz:
(a) RNA
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 28 OF 153
Ribonucleosides
Ribonucleotides
Adenosine (A)
Adenylic acid (AMP)
Cytidine (C)
Cytidylic acid (CMP)
Guanosine (G)
Guanylic acid (GMP)
Uridine (U)
Uridylic acid (UMP)
(b) DNA
Deoxyribonucleosides
Deoxyribonucleotides
Deoxyadenosine (dA)
Deoxyadenylic acid (dAMP)
Deoxycytidine (dC)
Deoxycytidylic acid (dCMP)
Deoxyguanosine (dG)
Deoxyguanylic acid (dGMP)
Deoxythymidine (dT)
Deoxythymidylic acid (TMP)
2. Physical structure of DNA
Maurice Wilkins and Rosalind Franklin in the 1940s used the x-ray
crystallography technique to try and determine the physical structure of a
DNA molecule by capturing diffraction images on a photographic plate. By
analysing the photograph, they obtained information about the molecule’s
atomic structure. They noticed a few repeating distances in the molecule:
0.34 nm (3.4A), 2 nm (20A) and 3.4 nm (34A).
Franklin also saw a pattern indicating that the molecule was helical (corkscrew). She also argued that there were probably two strands in each
molecule not 1 or 3.
3. The DNA model (According to Watson and Crick)
The final model was thus built according to information of Chargaff,
Wilkins and Franklin. Thus, in April 1953 Watson and Crick published their
findings in the scientific journal of Nature – a double helical structure of DNA
held together by cross-connections of paired bases, and resembling a spiral
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 29 OF 153
staircase in which the base to base attachments represent the series of
steps. The two chains are coiled around a common axis called the axis of
symmetry. It is important to note that whereas most DNA is double
stranded, a few viruses contain single stranded DNA!
The measurements:
(i) a residue on each chain every 3.4 A;
(ii) an angle of 360 between adjacent residues in the same chain so that the
structure repeats after ten residues on each chain, i.e. after 34 A.
(iii) the distance of a phosphate atom from the axis is 10A
The backbone of the helix is a chain of S-P alternating. The two chains
(strands) run in a 5' to 3’ direction and are anti-parallel and complementary
(but not identical).
It is noteworthy to mention here that other earlier scientists had given
other models of DNA, e.g. Pauling and Corey; Fraser all of whom had
suggested a 3 chain intertwined model (structure).
A typical eukaryotic chromosome consists of a DNA molecule many
millions of nucleotides long. The length of a DNA molecule is usu. given in
base pairs or kilo base pairs (bps or Kbps).
3.3 DNA AS THE GENETIC MATERIAL
By 1958, Crick had presented the general rules for the unidirectional flow
of genetic information from DNA to the creation of proteins, as follows:
transcription
DNA
translation
RNA
Enzyme
This information transfer did not occur from protein to protein or
"backward" from protein to RNA or to DNA. This one-directional transfer
hypothesis became known as the "Central Dogma of Molecular Biology".
Later however, some scientists have questioned the validity of the
central dogma, especially with regard to the RNA viruses.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 30 OF 153
3.4 DNA AS THE REPOSITORY OF GENETIC INFORMATION
The genetic information is encoded in the sequence of the bases along
the strands of the helix and represents the instructions for making proteins
and RNA types.
In the nucleus, DNA replicates and/or transcription occurs. Special
enzymes are involved. In the cytoplasm, protein synthesis occurs.
3.5 PROOFS OF DNA AS THE GENETIC MATERIAL
1. Chromosome composition: Eukaryotic chromosomes consist of one
long, linear molecule of DNA bound to a complex of proteins to form
chromatin.
2. Experimental Investigations:
(a) First by F. Griffith (1928) through the transformation experiment;
(b) Secondly by Avery, MacLeod and McCarty (1944);
(c) Thirdly by Hershey and Chase (1952) through the Great Kitchen blender
experiment;
pointed to DNA as the genetic material. Their experiments showed that
bacterial cells expressing one phenotype can be transformed into cells with a
different phenotype, and that the transforming agent is DNA;
(a) The transformation experiment
Griffith (1927/8) demonstrated the process of transformation. His
research provided the foundation for the work of Avery et al.
Griffith's experiment:
Heat-killed, virulent S-type cells
mouse
living, non-virulent R-type cells
mouse
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
mouse
PAGE 31 OF 153
healthy mouse
death by pneumonia
healthy mouse
No bacteria
recovered
Living, virulent
s-type bacteria
present
No bacteria
recovered
His results set the stage for the elucidation of the chemical nature of the
"transforming principle".
(b) Identification of the Transforming principle
Oswald T. Avery, C.M. MacLeod and M. McCarty identified the
transforming principle. Published their work in a classical paper in 1944 after
10 years of research.
s
Kill and fraction into:
Polysaccharides
lipids
RNA
protein
DNA
Live=R cells
R
R
R
R
R
S
They found DNA is the only agent that produces smooth (S) colonies
when mixed with the live R cells!
Despite this evidence, many within the scientific community still resisted
the idea that DNA is the molecule of heredity. They argued that perhaps
Avery’s results reflected the activity of contaminants. Unconvinced for the
moment, these scientists remained attached to the idea that proteins are the
prime candidates for the genetic material. This discovery however, did have
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 32 OF 153
the impact of initiating studies in molecular genetics with the use of microorganisms.
(c) The Hershey-Chase bacteriophage experiment
In their now-legendary experiments, Alfred D. Hershey and Martha C.
Chase studied bacteriophage. The phages they used were simple particles
composed of protein and DNA, with the outer structures made of protein and
the inner core consisting of DNA.
Hershey and Chase knew that the phages attached to the surface of a
host bacterial cell and injected some substance (either DNA or protein) into
the host. This substance gave "instructions" that caused the host bacterium
to start making lots and lots of phages—in other words, it was the phage's
genetic material. Before the experiment, Hershey thought that the genetic
material would prove to be protein.
To establish whether the phage injected DNA or protein into host bacteria,
Hershey and Chase prepared two different batches of phage. In each batch,
the phages were produced in the presence of a specific radioactive element,
which was incorporated into the macromolecules (DNA and protein) that
made up the phage.

One sample was produced in the presence of
35S,
a radioactive isotope
of sulfur. Sulfur is found in many proteins and is absent from DNA, so
only phage proteins were radioactively labeled by this treatment.

The other sample was produced in the presence of 32P, a radioactive
isotope of phosphorous. Phosphorous is found in DNA and not in
proteins. So only phage DNA (and not phage proteins) is radioactively
labeled by this treatment.

Each batch of phage was used to infect a different culture of bacteria.
After infection had taken place, each culture was whirled in a blender,
removing any remaining phage and phage parts from the outside of
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 33 OF 153
the bacterial cells. Finally, the cultures were centrifuged, or spun at
high speeds, to separate the bacteria from the phage debris.

Centrifugation causes heavier material, such as bacteria, to move to
the bottom of the tube and form a lump called a pellet. Lighter
material, such as the medium (broth) used to grow the cultures, along
with phage and phage parts, remains near the top of the tube and
forms a liquid layer called the supernatant.

Radioactivity was measured in the pellet and liquid (supernatant) for each
experiment.
P was found in the pellet (inside the bacteria), while
32
S was found in
35
the supernatant (outside of the bacteria)
When Hershey and Chase measured radioactivity in the pellet and
supernatant from both of their experiments, they found that a large amount
of
32P,
appeared in the pellet, whereas almost all of the
35S,
appeared in the
supernatant.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 34 OF 153
Conclusion
Based on this experiment, Hershey and Chase concluded that DNA, not
protein, was injected into host cells and made up the genetic material of the
phage.
3.6 RNA AS THE GENETIC MATERIAL
Viruses that infect and parasite plant cells, some animal cells contain
RNA only. In these viruses, RNA acts as genetic material. One plant virus,
Tobacco mosaic virus(TMV), that contains RNA, not DNA was an important
tool for genetic experiments. TMV infects tobacco, causing the infected
regions on leaves to become discolored and bristled. Different strains of TMV
produce clearly different inherited lesions on the infected leaves. The
common virus produces a green mosaic disease, but a variant Holmes rib
grass(TMV-HR), produces ring spot lesions. Moreover, the amino acid
compositions of the proteins of these two strains differ.
The demonstration that RNA is the genetic material came in several
studies:
1. In 1956 A. Gierer and Gerhard Schramm exposed tobacco plant tissue
to purified RNA from tobacco mosaic virus (TMV), and the plants
developed the same types of lesions as if they were exposed to the
virus itself. What would be the results if the RNA was treated with
DNase, RNase, or protease prior to its exposure to the plant tissue?
It was concluded that RNA is the genetic material of this virus.
2. In 1957, Heinz Fraenkel-Conrat and B. Singer reported another type
of experiment with TMV. They use two tobacco viruses – TMV and its
variant, HR virus. From the two strains of TMV they were able to
reconstitute viruses with the RNA from TMV common enclosed in TMVHR protein and TMV-HR RNA with TMV common protein. When these
reassembled viruses were used to infect tobacco leaves, the progeny
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 35 OF 153
viruses produced were always found to be phenotypically and
genotypically identical to the parent strain from which the RNA had
been obtained. The reassembled viruses with the TMV-RNA and TMVHR protein produced a green mosaic disease characteristic of TMVcommon. Recovered virus had protein characteristic of TMV common.
This proved that specificity of virus proteins was determined by RNA
alone and that proteins carried no genetic information. Hence RNA
carries genetic information not proteins.
The genetic RNA is usually found to be single stranded but in some it
is double stranded as in retrovirus, wound tumor virus.
3. In 1965 and 1966 Norman R. Pace and Sol Spiegelman further
demonstrated that RNA from the phage Q-Beta could be isolated and
replicated in vitro. Replication was dependent on an enzyme Q-Beta
RNA replicase, which was isolated from host E. coli cells following
normal infection. When the RNA replicated in vitro was added to E.
coli protoplasts, infection and viral multiplication occurred. Thus, RNA
synthesised in a test tube can amply serve as the genetic material in
the phages.
Topic Summary
In this topic, you have been introduced to the two kinds on nucleic acids,
DNA and RNA, the former having been isolated in pus cells of humans and
the latter from yeast cells. We have examined several aspects of these
acids: their physical and chemical structures; how the physical model was
arrived at by Watson and Crick; and given proofs that they are indeed the
materials of heredity.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 36 OF 153
Further Reading
(1) Burns, G.W. (1983). The Science of Genetics: An introduction to
heredity. (5/e). Macmillan Pub.Co., NY.
(2) Russell, P.J. (1992). Genetics (3/e). HarperCollins Publishers, NY
TOPIC ACTIVITIES
Attempt the questions below as you consolidate your understanding of this
topic.
1. What do you understand by “Chargaff’s rule”?
2. What is a nucleotide? How does it differ from a nucleoside?
3. What is the significance of these values (3.4 A, 34 A, 10 A and 360) in the
DNA physical model?
4. Outline one proof each that DNA and RNA are the genetic material.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 37 OF 153
TOPIC 4. NUCLEIC ACIDS SYNTHESIS: REPLICATION AND
TRANSCRIPTION
Introduction
In topic one, we learned that a molecule that is hereditary must be able to
make copies of itself and also be able to express information therein.
Following the confirmation of DNA as the hereditary molecule, it is necessary
now to demonstrate that in deed replication and transcription can take place.
Topic Time
Consideration of the two fundamental activities – replication and
transcription – require progressive exploration. They can be successfully
covered in five lecture hours with the student availing additional time for
self-study and doing assignments.
Topic learning Requirements
. Notes of BOTA 111 for an overview of this topic
Learning Outcomes
After successfully completing this topic, you should be able to:
i.
List the modes earlier proposed for DNA replication
ii.
Describe how the correct mode was identified and confirmed
iii.
Describe the process of replication of double stranded DNA as well as
single stranded DNA
iv.
Define and demonstrate transcription in prokaryotes and eukaryotes
v.
Define split genes and the need for RNA processing to produce mature
transcripts
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 38 OF 153
Topic Content
Introduction
After proposing the double-helical DNA structure, Watson and Crick
remarked that, "the specific pairing they'd postulated suggested immediately
a possible copying mechanism for the genetic material". Their prediction was
in deed true.
Any molecule that is to serve as a repository of genetic information must be
able to act as a template for the synthesis of an exact copy of itself to pass
on to daughter cells at cell division. Amongst biological molecules, only the
nucleic acids, DNA and RNA, can do this.
Summary of guides to nucleic acids synthesis
(1) Both DNA and RNA chains are produced in cells by copying of a preexisting DNA strand according to the rules of Watson and Crick of base
pairing. In the replication of a duplex, both strands are copied. In some
viruses, the copying of a pre-existing RNA molecule produces RNA
molecules; in the retroviruses, the copying of RNA produces DNA.
(2) Nucleic acid strand growth is in one direction, 5- to 3-. All RNA and DNA
synthesis, both cellular and viral, proceeds in one chemical direction. This
directionality has given rise to the convention that polynucleotide sequences
are read from left to right in the 5- to 3- direction.
The nucleotides that are used in the construction of nucleic acid chains are 5triphosphates of ribo- or deoxy- ribonucleosides.
(3) Special enzymes called polymerases elongate RNA or DNA strands.
These are the enzymes that make the phosphodiester bonds and their
activity results in the production of a polymer of DNA or RNA, viz:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 39 OF 153
(a) the enzymes that make more DNA from DNA are called DNA
polymerases (Dpols);
(b) those that copy RNA from DNA are called RNA polymerases (Rpols).
(4) RNA polymerases can initiate a nucleic acid strand, but DNA polymerases
can’t. A single RNA Polymerase:
(a) can find an appropriate initiation site on duplex DNA,
(b) bind the DNA,
(c) separate the two strands in that region, and
(d) begin generating a new RNA strand.
As for DNA Polymerases:
(a) they cannot initiate synthesis of a DNA chain; they can only elongate a
pre-existing primer strand of DNA or RNA;
(b) all DNA Polymerases catalyse nucleotide addition to the 3- OH end of the
primer and thus direct growth in the 5- to 3- direction.
The terminal 5- end of an RNA strand is chemically distinct from the rest
of the strand. Unlike the nucleotides within the strand, the nucleotide at the
5- end retains all the PO4- groups of the triphosphate.
When each additional nucleotide is added to the 3- end of the growing
strand, only the  PO4- is retained, whereas the  and  PO4- are lost. This
bis-phosphate group is further cleaved to yield inorganic phosphate (Pi),
with release of energy.
4.1 Replication of DNA
Using the ultraviolet microscope, it was found that cells during interphase
undergo cyclic changes in respect of their nucleic acid content. The uv
microscope revealed that just before onset of mitosis, the amount of DNA is
doubled. Furthermore, it was found that the total amount of DNA in the
prophase and metaphase chromosomes is twice the amount found ordinarily.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 40 OF 153
These findings indicated that during interphase (prior to mitosis) not only
do the chromosomes replicate but their genetic material, the DNA, does the
same.
4.1.1 Modes of DNA replication
Two or even three mechanisms (modes) of DNA replication were
envisaged. These are: semi-conservative (according to Watson & Crick);
conservative, and dispersive (according to Delbruck and Stent). Delbruck
and Stent thought that the semi-conservative model was too simple.
(1) Semi-conservative
A daughter duplex contains one parental and one newly synthesised
strand.
(2) Conservative
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 41 OF 153
According to this model, the two original strands remain intact and the
two strands of the daughter molecule are synthesised ex nihilo using
available nucleotides. Thus the parental duplex is conserved.
(3) Dispersive
According to Delbruck and Stent, nothing is conserved, i.e. the parental
DNA molecules were degraded and new DNA molecules were synthesised.
Thus, the daughter duplexes consist of strands containing segments of
parental DNA and newly synthesised DNA.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 42 OF 153
Five years after the physical DNA structure had been given, Matthew
Messelson and Franklin Stahl set about to test which of the three modes of
replication was in agreement with the Crick-Watson model of DNA. They
confirmed the semi-conservative mode.
4.1.2 The experiment
(1) Messelson and Stahl used a heavy isotope of nitrogen,
15N,
to label the
DNA of E.coli by growing cells in a medium in which the only nitrogen source
was
15NH Cl.
4
A dozen generations of growth in such a medium is sufficient to
label uniformly the entire bacterial DNA with
15N
15N.
DNA molecules containing
can be distinguished from DNA molecules containing the lighter, common
isotope,
14N,
on the basis of their densities, because DNA containing
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 43 OF 153
15N
has
a higher mass/nucleotide than the DNA containing
14N.
DNA molecules of
different densities can be separated from one another by centrifugation in a
density gradient of a caesium chloride (Cscl) solution.
(2) The bacterial cells from the
15N
were then transferred to
14N
culture
medium.
(3) Samples of these bacteria were taken immediately, and then after 1, 2,
3 and 4 generations in the new medium, and the DNA extracted.
(4) These DNA extracts were then centrifuged in the Cscl gradient and the
bands of DNA of different densities located by noting their absorption of uv
light (see figure below)
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 44 OF 153
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 45 OF 153
Inference
The formation of the hybrid
15N/14N
DNA molecules after one generation
and their perpetuation are the key to the mode of DNA replication.
Conclusion
On the basis of their observations, Meselson and Stahl concluded that the
DNA of intermediate density, formed after 1 generation of growth in
14N,
is a
hybrid molecule consisting of 1 fully heavy, conserved parental strand and 1
fully, newly synthesised daughter strand, exactly as predicted by the
Watson-Crick model.
4.1.3 Single-Stranded DNA (ssDNA) Replication
Certain small bacteriophages contain a circular, single-stranded DNA (ss
DNA) as genome, e.g. Escherichia coli phages M13, fX174, M13, F1, and
S13. Their replication can be divided into three steps:
(1) conversion of the ss DNA genome to a double-stranded form, called the
replicative form (RF);
(2) multiplication of RF DNA by rolling circle replication; and
(3) generation of ss DNA genome for packaging into phage from the RF
DNA.
Thus, replication of the ss DNA genome depends on the synthesis by DNA
polymerase of a complementary strand! This replication also further confirms
the semi-conservative mode of DNA replication.
4.1.4 Enzymology of replication
The enzymes involved in the DNA replication process are templatedirected polymerases and others.
(1) DNA Topoisomerases, viz:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 46 OF 153
Type I: reversibly nicks one strand of DNA helix allowing the helix to
swivel. These topoisomerases have both nuclease (strand cutting) and ligase
(strand – resealing) activities.
Type II (gyrase) – required to rotate the double helix and thereby relax it.
(2) Helicase - unwinds short segments of the double helix of DNA, being
aided by the single-strand DNA-binding proteins (SSBs). In so doing, it
breaks the base pairing between the two strands of the parent DNA
molecule. To do this it binds to single strand DNA near the replication fork
and then move into it forcing the strands apart, i.e. unwinding the double
helix. When the strands separate, SSB proteins bind, preventing reformation
of the double helix.
(3) Primase: Is a specific RNA polymerase; it synthesises short stretches of
RNA (6-30 nucleotides long). Both lagging and leading strands need RNA
primers, one for the latter and several for the former when replication
commences.
(4) Polymerases: Responsible for copying the DNA templates. They read the
templates only in the 3- to 5- direction. However, they synthesise the new
DNA strands in the 5- to 3- (anti-parallel) direction, one towards the
replication fork and the other away from it. Prokaryotes have 3 (I, II, and
III), and eukaryotes four (, , , and ).
(5) Ligase
Makes the final phosphodiester linkage between the 5- phosphate group
on the DNA chain made by DP III and the 3- OH group made by DP I. The
joining of these two stretches of DNA requires energy, which in humans is
provided by the cleavage of ATP to AMP and Ppi.
4.1.5 The Process
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 47 OF 153
1. Origin of replication
Replication always starts at a special site called replication origin.
Eukaryotic DNA molecule has several origins of replication; prokaryotic DNA
has only one.
Having multiple origins of replication provides a mechanism for rapidly
replicating the great length of the eukaryotic DNA molecules. A DNA that
forms a complete replicating unit is often termed a replicon. In bacteria, the
chromosome functions as a single replicon, whereas eukaryotic
chromosomes contain hundreds of replicons in series.
2. Procedure
DNA synthesis is bi-directional from a single point of origin (in
prokaryotes with circular DNA), i.e. the replication forks move in both
directions away from the origin. The enzyme gyrase (called topoisomerase
II) nicks one strand and thus causes DNA relaxation. Each replicon contains
a segment to which a specific RNA polymerase binds and a replicator locus
at which DNA replication commences.
DNA replication in a eukaryotic chromosome proceeds at a rate of about 50
nucleotides/sec, while in bacteria it is about 500 nucleotides/sec.
The two strands are antiparallel yet both grow in a 5- to 3- direction. How
is this possible? This is accomplished by the discontinuous synthesis of one
of the daughter strand, and continuous synthesis of the other strand. The
discontinuous synthesis occurs through the addition of short segments of
DNA called the Okazaki fragments (100 to 1,000 nucleotides)
complementary to such a parental strand. These short segments are formed
on RNA primers. To link up these short segments, DP I, which has
exonuclease properties chews away these primers and fills the gaps with
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 48 OF 153
relevant nucleotides. When all the ribonucleotides are replaced, DP I stops
and dissociates from the new double helix.
The final reaction needed to complete replication of the lagging strand is
to join up adjacent Okazaki fragments, which are now separated by just a
single gap between neighbouring nucleotides. All that is needed is to
synthesise a phosphodiester bond at this position, a reaction catalysed by
DNA ligase.
The formation of RNA primers is an important key to the dilemma of
replicating anti-parallel strands.
4.2 Synthesis of RNA
1. Introduction
Although the information for the synthesis of all protein is located in the
DNA, the DNA does not serve as a direct template along which the amino
acids are laid down to form protein. Instead, it serves as a template for RNA
formation. The copying of a DNA strand into RNA transcript is called
transcription and is the first step of gene expression, an event that connects
the genotype and phenotype.
The first step in gene expression is the making of a disposable copy of
itself. This expendable copy is either of the RNA types. This occurs by
polymerisation of ribonucleotide subunits according to information on the
DNA strand as follows:
n(NTP)
RNA of length n nucleotides plus n-1(ppi).
Transcription occurs from specific, relatively short lengths of DNA.
2. Enzymes
A transcription enzyme, RNA Polymerase (Rpol) performs the
transcription in a fashion quite similar to replication.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 49 OF 153
Prokaryotes such as bacteria have only one Rpol type for synthesis of all
RNAs (i.e. mRNA, tRNA, and rRNA). Eukaryotes (e.g. humans) on the other
hand have 3 distinct Rpol types, viz: Type I for rRNA;Type II for mRNA; and
Type III for tRNA.
Chemical evidence confirms that transcription takes place on only one
of the two DNA strands at a time (though not necessarily the same strand
throughout the entire chromosome, i.e. it is asymmetrical).
The DNA strand carrying the same base sequence as the RNA is called
the coding strand or antisense strand; the opposite DNA strand, which acts
as a template for transcription, is the anticoding or sense strand.
The RNA is always synthesised in the 5- to 3- direction, i.e. the 5- end of
an RNA is the beginning. This therefore means that the Rpol moves along
the DNA in a 3- to 5- direction, locally separating the DNA strands of the
helix and breaking the weak bonds between the base-paired nucleotides.
The regulatory sequence of DNA nucleotides, i.e., the signal for initiation
and strand selection during transcription is called the promoter site.
Transcription also terminates at specific regulatory sequences (termination
region).
3. RNA types
There are three major ones: Ribosomal RNA (rRNA), Transfer RNA (tRNA),
and Messenger RNA (mRNA).
(1) Transfer RNA (tRNA)
There is no direct interaction between a codon and the amino acid it
represents. Instead, their association is mediated by tRNA molecules.
(a) Structural features:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 50 OF 153
i.
Is clover-leaf shaped due to extensive intra-chain base-pairing. This
figure exposes the anticodon loop at one end;
ii.
At the other end is the aminoacyl attachment site (i.e. the C-C-A-3OH);
iii.
There is at least one specific type for each of the 20 amino acids
found in proteins;
(b) Function:
Functions as adaptor molecule that carries a specific amino acid to the
ribosomal/mRNA complex;
Together, tRNAs make up about 15% of the total RNA in the cell.
(2) Ribosomal RNA (rRNA)
Ribosomal ribonucleic acid (rRNA) is the RNA component of the ribosome, and is
essential for protein synthesis in all living organisms. It constitutes the predominant
material within the ribosome, which is approximately 60% rRNA and 40% protein
by weight. Ribosomes contain two major rRNAs and 50 or more proteins.
Molecules of rRNA are synthesized in a specialized region of the cell nucleus called
the nucleolus, which appears as a dense area within the nucleus and contains the
genes that encode rRNA
Ribosomal RNAs of both prokaryotic and eukaryotic cells are synthesized
from long precursor molecules called pre-ribosomal RNAs. There are three
distinct size spp of rRNAs (23s, 16s, and 5s) in prokaryotic cells and four
(28s, 18s and 5.8s and 5s) in eukaryotic cells. The S in the names of
ribosomal subunits, and other macromolecular particles found in the cell
stands for Svedberg units. This unit is named after Theodor Svedberg, a
Swedish chemist who received the 1926 Nobel Prize in Chemistry for his
research into disperse systems, (in this case, colloids of macromolecules
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 51 OF 153
dispersed in solution). Svedberg pioneered the use of ultracentrifugation to
investigate the properties of macromolecules.
Processing occurs as follows: A cut is made by ribonucleases between the
18s and 5.8s segments. The intron is removed and the two exons linked.
They then associate with several proteins as components of the ribosomes
as follows:
(a) Eukaryotes
(i)
Small ribosomal subunit: 18s rRNA + 30 protein = 40s
(ii)
Large ribosomal subunit: 5.8 s +28s rRNAs + 50 protein = 60s
A functional eukaryotic ribosome (80s) is then made up of a combination
of 40s and 60s ribosomal subunits.
(b) Prokaryotes
(i)
Large sub unit (50s) = 5s rRNA + 34 proteins + 23s rRNA; and
(ii)
Small subunit (30s) = 16s rRNA + 21 proteins.
A functional prokaryotic ribosome (70s) is a combination of 30s and 50s.
Together, rRNAs make up 80% of the total RNA in the cell.
(3) Messenger RNA (mRNA)
It carries genetic information from the nuclear DNA to the cytosol,
where it is used as the template for protein synthesis. The mRNA constitutes
only about 5% of the RNA in the cell. Once formed, it is processed through:
capping, tailing and intron removal before export (for eukaryotes) to the
ribosomes for translation.
4.3 Transcription in prokaryotes
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 52 OF 153
1. The enzyme
(a) Is RNA polymerase (Rpol) for all forms of RNAs,
(b) Has identical requirements as Dpol except that ribonucleoside
triphosphates rather than the deoxyribonucleotides are its raw material.
(c) It comprises five peptide subunits, 2 (alpha), 1 (beta), 1’(beta prime)
and 1 (sigma) and is therefore called a holoenzyme. The 4 subunits (i.e.
minus sigma) are responsible for the 5- to 3- RNA polymerase activity, and
are referred to as the core enzyme. The sigma subunit enables the RP to
recognize the correct point along the DNA template (i.e. the promoter)
where RNA transcription is initiated.
2. The Process
This may be divided into 4 stages, namely template binding, chain
initiation, chain elongation and chain termination.
(1) Template binding
(i) Involves Rpol with DNA,
(ii) The promoter regions along the sense strand are recognized by the
sigma subunit of the holoenzyme.
(iii) The initiation sites consist of < 50 deoxyribonucleotide residues and are
rich in A - T base pairs, a factor that facilitates ease opening of the duplex.
RP binds to any region of the duplex DNA. By binding loosely, releasing
momentarily and then binding again, an RP explores the DNA. With the aid
of the sigma factor, the enzyme recognises a nucleotide sequence (the
promoter region) at the beginning of the DNA at which synthesis can begin.
At this point, the holoenzyme binds tightly causing the duplex to open up
and so allow transcription to begin.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 53 OF 153
(2) Chain initiation
The residues of the promoter region are themselves not transcribed.
Instead, the enzyme moves along the DNA molecule until it encounters the
first 2 nucleotides to be transcribed. The ribonucleoside triphosphate
complements (esp. ATP or GTP) are inserted and linked by a phosphodiester
bond, resulting in chain initiation.
(3) Chain elongation
(i) Once the first two nucleotides have been linked, chain elongation
proceeds rapidly in the 5- to 3- direction, anti-parallel to the DNA template
strand.
(ii) After about 10-12 ribonucleotides have been added to the growing RNA
chain, the  sub-unit is disassociated from the holo-enzyme and elongation
proceeds at the rate of about 50 nucleotides / sec by the addition of NTPs to
its 3- end. As in DNA synthesis, two high-energy bonds are used for the
addition of each nucleotide.
(iii) The elongation process produces a temporary DNA/RNA hybrid with the
DNA unwinding in front of Rpol and closing and reforming behind it. Thus,
we’ve a transcription bubble.
(4) Chain termination
(i) When the enzyme encounters a termination signal (a specific nucleotide
sequence), synthesis is completed. This requires the co-activity of the
termination protein,  (rho) factor.
(ii) The transcribed RNA is released from the DNA template, and the core
enzyme dissociates.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 54 OF 153
(iii) This then marks the end of the 1st step in the information flow within
the cell.
NOTE:
1. Since there is no membrane barrier between the genes that are being
transcribed and the cytoplasmic area where synthesis occurs, the
transcription of RNA can proceed at the same time as its utilization. Thus,
transcription and translation are coupled in prokaryotes.
2. No primer is needed to begin the new chain, only a duplex DNA.
3. Usually the transcript contains information for several polypeptides (i.e. is
polycistronic) that are related.
4.4 Transcription in Eukaryotes
Is similar to that found in prokaryotes and like in prokaryotes, there is no
primer. However, it is far more complicated.
Also,
(i) Eukaryotes do not in general produce polygenic mRNA. Their primary
transcripts are first "processed" into a mature mRNA prior to translation.
(ii) They have 3 Rpols, of which type II is for synthesis of mRNA.
(iii) Eukaryotic DNA is transcribed into rather long, heterogeneous pieces of
RNA of between one to 30 gene lengths. These RNAs together are called
heterogeneous nuclear RNA (hnRNA) or pre-mRNA. However, most of it (i.e.
about 90%) is degraded before it exits from the nucleus. A typical hnRNA
molecule has a 10 - minute life-time.
4.5 Post transcriptional modification of RNA
RNA processing involves at least two nuclear events, modification of one
or both ends of the primary transcript through the addition of nucleotides to
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 55 OF 153
the termini of some RNA chains, and removal of intervening noncoding
sequences (IVs). These modifications are essential to efficient processing
and subsequently, to translation.
(1) Terminal modifications
(i) 5- capping: The 5- end of pre-mRNA is modified by addition of a
structure called a 7-methylguanosine cap. A guanine-containing nucleotide
is added backward to the C5 (5- to 5-) where the triphosphate is found. This
is catalysed by the nuclear enzyme guanylyl transferase. Then the G is
methylated by guanine-7 methyltransferase.
(ii) Poly-A tailing: At the 3- end, a sequence (40-200) rich in A bases, called
poly (A) tail is added. This is thought to occur after RNA has been cleaved.
Capping is mediated by the enzyme polyadenylate polymerase.
(2) Excision of introns
Maturation of eukaryotic mRNA usually involves the removal of RNA
sequences, which do not code for protein from the primary transcript. The
remaining coding sequences, the exons, are spliced together to form the
mature mRNA, a process referred to as RNA splicing.
Genes that contain introns and exons are commonly referred to as split
genes.
The mature mRNA is then shipped to the cytoplasm on its way to the
ribosomes.
4.6 Split Genes
A split gene (also called an interrupted gene) is a gene that contains sections of
DNA called exons, which are expressed as RNA and protein, interrupted by sections
of DNA called introns, which are not expressed. The DNA sequence in the exon
provides instructions for coding proteins.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 56 OF 153
This type of genetic organization is typical of most eukaryotic genes and
some animal viral genomes. Introns are rare in the DNA of prokaryotes.
The existence of split genes was first suggested because of a lack of
correspondence between the genetic map and specific mRNA molecules. For
instance, the ovalbumin gene has 7,700 bps. After splicing, the mature
mRNA has only about 1,872 bases translating to a protein of 674 amino
acids.
4.7 Summary of differences in expression between prokaryotes and
eukaryotes genes
Prokaryotes
Eukaryotes
(i) All RNA species are synthesized
by a single RNA polymerase;
(i) Three different RNA polymerase
are required
(ii) mRNA is translated during
(ii) mRNA processed before exit to
transcription;
cytoplasm by capping, tailing and
RNA splicing
(iii) Genes are contiguous segments
(iii) Genes are often split. They are not
of DNA that are colinear with the
contiguous segments; rather the
mRNA that is translated into
coding sequences are interrupted by
a protein.
intervening sequences.
(iv) mRNA is generally polycistronic;
(iv) mRNA is monocistronic
Topic Summary
Following the acceptance of the two nucleic acids as the genetic material, we
in this topic have learnt how they are synthesized so copies can be availed
to daughter cells following cell divisions. We have learnt that following the
unravelling of the physical structure of a DNA molecule, three modes of
replication were suggested. These were: semi-conservative, conservative,
and dispersive. It took the experiment of Messelson and Stahl to identify the
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 57 OF 153
correct mode by which DNA replicates. However, not all DNA is doublestranded. We have thus also learnt how this occurs if the DNA molecule is
single-stranded. We have also learnt about the enzymes that participate in
this process and how the process occurs at the molecular level. The
formation of RNA from DNA is called transcription. We have studied how this
occurs in eukaryotes and prokaryotes. In the case of eukaryotes, we have
learnt also that the resultant RNA transcripts must be processed in the
formation of the three kinds of RNA- mRNA, rRNA, and tRNA. The roles of
the three types of RNA have also been reviewed.
Further Reading
(1) Russel, P.J. (1992). Genetics (3/e). HarperCollins Publishers, NY
(2) Stine, G.J. (1989). The New Human Genetics. WCB Publishers, Iowa
TOPIC ASSIGNMENT
1. What five enzymes are key to DNA replication?
2. What modes were suggested for DNA replication? Outline them.
3. Describe how Messelson and Stahl confirmed the correct mode.
4. The phage M13 has one strand of DNA molecule. How does it make copies
of itself?
5. Given any three differences of gene expression between prokaryotes and
eukaryotes.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 58 OF 153
TOPIC 5. THE GENETIC CODE, TRANSLATION, AND PROTEIN
SYNTHESIS
Introduction
As stated in the previous topic, gene expression is a two-step process –
transcription and translation. This topic covers the second part of this
process. Therefore, in this part therefore, we consider how the instructions
coded in the mRNA is turned into a chain of amino acids of a protein. At the
conclusion of this topic, it is expected that the student has indeed
understood the relationship between a gene and a phenotype, e.g. the
ability of a gene to define a given phenotype.
Topic Time
To understand this topic well, four lecture hours are estimated to be
sufficient. The student needs additional time to review those aspects than
have not been sufficiently understood and attempt the assignment.
Topic Learning Requirements
There are no specific learning requirements to this topic. However, an
advance study of the topic will make it easy to follow this topic.
Learning Outcomes
After successfully completing this topic, you should be able to:
i.
Define the genetic code and explain how it was broken and size
determined
ii.
Define translation and describe the key steps in this process.
iii.
List and categorise components required for each step of protein
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 59 OF 153
synthesis
iv.
Answer any problem relevant to this topic
Topic Content
5.1 The Genetic Code
Because plants and animals produce an enormous variety of proteins, the
question asked by 1960s researchers was "how does DNA, with only four
bases, provide the instructions that allow for so many different
arrangements of amino acids to form an almost infinite variety of protein?".
This question was answered through the understanding the 4-letter
genetic language of the DNA molecule- a phenomenon popularly called the
"Cracking of the genetic Code" or the “deciphering of the genetic
code”. This was in 1961 by Marshall Nirenberg and J.H. Matthaei.
The discovery of mRNA and how it functions led to the solution of the
genetic code. The genetic code is a dictionary that identifies the
correspondence between a sequence of nucleotide bases and a sequence of
amino acids.
1. Breaking the Code
Cracking the genetic code was done using cell-free extracts from E. coli
and synthetic mRNA (earlier made by Severo Ochoa using polynucleotide
phosphorylase enzyme). This was done at the National Institute of Arthritic
and Metabolic diseases, Washington, USA.
All the necessary components for protein synthesis except mRNA were
present in these extracts.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 60 OF 153
After incubating the mixture for 1 hour at 350C, Matthaei precipitated the
proteins from the mixture using chloroacetic acid, washed it, and placed it in
a radioactivity counter.
Results:
In the presence of “poly-U” nucleic acid but not in its absence, amino
acids were incorporated into a material that could be precipitated in an acid
medium, i.e. into proteins.
Conclusion:
The poly-U had led to the synthesis of a protein, entirely in vitro! This
achievement was on 22nd May 1961. For the next 5 days Matthaei tried to
determine which amino acid(s) was or were incorporated into proteins in the
presence of poly-u. On the 27th May at 6 PM, he got the answer: the poly-u
coded for a monotonous protein consisting of a chain of a single amino acid
– phenylalanine. Thus, in less than a week Matthaei had identified the first
“word” of the genetic code!
Nirenberg, the head of the team finished characterizing the product of
the experiment. In November 1961, in Proceedings of National Academy of
Sciences, they published their work, presenting the results obtained with
different RNAs including the poly-u molecule. These were that:
(i). Bacterial extract plus mRNA: AAA, AAA, AAA = Lys-Lys-Lys protein;
(ii). Bacterial extract plus mRNA: CCC, CCC, CCC = Pro-Pro-Pro protein;
(iii). Bacterial extract plus mRNA: GGG, GGG, GGG = Gly-Gly-Gly protein.
Following this lead, but using mixed-base synthetic mRNA polymers,
Nirenberg and colleagues were able to decipher the complete code by 1966.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 61 OF 153
Since the advent of gene cloning and DNA sequencing, it is now a
straightforward operation to work out the genetic code. It involves
comparison of a protein sequence determined by chemical and enzymatic
techniques with its DNA sequence.
2. Size of word for an amino acid
Having resolved the sequence of the nucleotides, the next challenge was
to determine the number necessary to code for the placement of one amino
acid into the protein chain.
Nirenberg and other Scientists reasoned that because there were 4
different nucleotides in DNA, and the 4 had to be used in some combination
in order to code for all 20 naturally occurring amino acids. Thus they
established that:
i. the genetic code is not singlet,
ii. the genetic code is not doublet, but
iii. the genetic code could be triplet, i.e. a combination of 4 x 4 x4. This
result is 64 triplets comprising both sense and nonsense triplets. The sense
triplets are 61 and code for amino acids, whereas the other three are stop
triplets. These tRNAs do not carry amino acids to the growing peptide chain,
but instead terminate translation causing the release of the newly
synthesised protein and disassembly of the translational machinery. The 3base (triplet) sequence in mRNA was thereafter designated a codon.
In 1968, M. Nirenberg, H.G. Khorana and R. Holley shared the Nobel
Prize for their work in "Cracking the genetic code". They had explained
“how a gene controls the arrangement of amino acids into a specific
protein”.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 62 OF 153
Because the code comes into play only during the translation part of
gene expression, i.e. during the decoding of mRNA to polypeptide,
geneticists usually present the code in the RNA dialect of A, G, C, and U.
5.2 Translation
Translation is the final step on the way from DNA to protein. It is the
synthesis of proteins directed by a mRNA template. The information
contained in the nucleotide sequence of the mRNA is read as three letter
words (triplets), called codons. Each word stands for one amino acid.
Once mRNA makes its way to the cytoplasm, it proceeds through the
three phases of translation – initiation, elongation, and termination.
1. Components
In order to decode the information held within the mRNA and to turn it
into a protein, two other species of RNA are required; tRNA (necessary for
the transport of amino acids to the ribosome, where mRNA is positioned to
produce a protein), and rRNA (an integral part of the ribosome, holds the
entire complex of mRNA, tRNA and ribosomes together).
2. The Process
The synthesis of proteins requires that the instructions in mRNA be
interpreted into an ordered list of amino acids that are assembled into a
particular protein (polypeptide chain). The many components involved are
below):
Component
Function
(i)
Ribosomes
workbench for translation.
(ii)
tRNA
adaptor molecule.
(iii)
Amino acids
building blocks of protein.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 63 OF 153
(iv)
mRNA
template.
(v)
Aminoacyl synthetase
charging tRNA.
(vi)
Initiation factors
mediate the various binding steps of the
initiation process.
(vii) Peptidyl transferase
elongation and translocation of growing
polypeptide.
(viii)
(ix)
Ions: Mg++, K+
ATP
stabilize conformation of ribosomes.
energy source, essential for charging
tRNA.
(x)
GTP
energy source for translocation.
Translation is therefore a chemical reaction that occurs in steps
beginning with charging tRNAs, then chain initiation, chain elongation and
chain termination.
(1) Activation of amino acid or charging a tRNA (also called
aminoacylation)
Each amino acid (AA) is attached to a tRNA molecule that is specific to
that amino acid by a high-energy bond derived from ATP. Catalysis is by a
synthetase. There is a separate synthetase for each amino acid. Then we
can speak of a "charged" tRNA.
synthetase1
aa1 + tRNA1 + ATP
aa1-tRNA1 + ADP.
The energy of the charged tRNA is converted into a peptide bond linking
the amino acid to another on the ribosome.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 64 OF 153
Note: Upon activation, an amino acid is thermodynamically capable of being
efficiently used for protein synthesis.
In all protein syntheses, the first amino acid to be attached to its tRNA is
methionine (Met). The tRNA is tRNAmet and the enzyme for this is methionyltRNA synthetase. Only methionyl-tRNAimet (i.e. methionine attached to
tRNAimet) can directly enter the P site of the ribosome and bind to a small
ribosomal subunit to begin the process of protein synthesis. Whereas all
bacterial polypeptides start with N-formyl methionine, it is just methionine in
eukaryotes.
There are at least two tRNAs for methionine:
i. tRNAimet for initiation of protein synthesis, and
ii. tRNAomet, incorporates methionine to a growing polypeptide chain.
(2) Initiation
AUG is the initiation signal in mRNA. Initiation is a 3-step process:
Step 1: Attachments of: one, IF3 to a free 30s subunit; two, initiator tRNA
(i.e. f-met-tRNA) to an IF2-GTP; and three, the union of the whole complex.
Step 2: The resulting IF3-30s-f.met-tRNA-IF2-GTP complex (now called the
initiation complex) then binds to mRNA, a process that may also require
IF1. This occurs at a site that is located upstream near the AUG codon.
Bacterial mRNA can have several AUG initiation sites, but eukaryotic mRNA
molecules almost always have a single functional AUG near its 5- end.
Step 3: Lastly, the large ribosomal subunit (50 or 60s) then joins the
complex. The bound GTP is hydrolysed to GDP and inorganic P, and the
initiation factors (IFs) detach. The Met-tRNAimet bearing the first amino acid
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 65 OF 153
is now bound to the ribosome at the P site. The initiation complex is ready to
begin synthesis of the peptide chain. In E.coli every protein is initiated by Nformylmethionine. It is non-formylated methionine in higher organisms.
(3) Elongation
The stage in which the mRNA is decoded and the polypeptide is synthesised.
Ribosomes use two tRNA-binding sites, ‘A’ and ‘P’, during protein
elongation. The P-site is the peptidyl site, and the A-site is the aminoacyl
site.
For the peptide chain to begin to grow, a second amino acid that is
correctly bound to its tRNA must be brought into its proper position on the
ribosome.
The ‘A’ site accommodates the incoming aminoacyl-tRNA that is to
contribute a new amino acid to the growing chain.
The attachment of aa-tRNA precursor to ribosome is a complex reaction.
It starts when one of the elongation factors (ef-Tu) reacts with GTP and aatRNA to form an aa-tRNA-GTP-ef-Tu complex. This complex then transfers
its aa-tRNA component to its ‘A’ site with release of a free ef-Tu-GDP
complex and P. The ef-TS is responsible for the dissociation of GDP from efTU to allow another GTP to bind and to allow the cycle to continue.
The P-site contains the peptidyl-tRNA complex, i.e., the tRNA linked to
the amino acids that have sofar been added to the chain.
The general formula for an amino acid is H2N – CHR – COOH, in which
the R group can be anything from a hydrogen atom (as in the amino acid
glycine) to a complex ring (as in the amino acid tryptophan).
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 66 OF 153
A peptide bond is formed through a condensation reaction that involves
the removal of a water molecule, for example between the -COOH group of
amino acid 1 (aa1), methionine and the amino group of the next amino acid
(aa2), e.g. phe-tRNAphe, now in the A-site to form the peptide methionylphenylalanyl-tRNAphe . This is catalysed by an enzyme activity within the
ribosome called peptidyl transferase. The tRNA, which brought the f.met to
P-site needs to be removed. This is done by a 2nd enzyme activity within the
ribosome, tRNA deacylase. This breaks the bond between the amino acid
f.met and the tRNA freeing the tRNA so it may be recharged and recycled.
This is followed by the ribosomal translocation.
In translocation, the ribosome shifts down the mRNA by one codon,
moving the tRNA carrying the growing polypeptide chain to the P site, and
moving the next mRNA codon into the A site.
The average rate of amino acid addition is about 5 amino acids / sec.
It has been found that in E. coli, a protein 300 - 400 amino acids long is
made in 10 - 20 seconds.
(4) Termination
Two conditions are necessary for chain termination:
a. The presence of a codon in the A site (i.e. UAA, UAG or UGA) that
specifically signifies that polypeptide elongation should stop.
b. The presence of a GTP-bound release factor (rf = TF), which reads the
chain-terminating signal.
When any of the stop codons are recognised by the termination factors
(TFs), the peptidyl-tRNA complex is released.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 67 OF 153
Almost simultaneously the complex divides into uncharged tRNA molecule
and a newly completed protein that can either assume its final shape or
combine with additional protein subunits.
After releasing its peptidyl-tRNA, the ribosome disengages from the mRNA
and divides into two subunits, where upon it is ready to start the whole cycle
over again.
Topic Summary
In this topic, you have learnt how a gene product is realized. For a simpler
understanding of this process, we needed to understand the genetic codeboth its size and what ‘letters’ spell which ‘word’. This enabled us to discuss
the process of translation (protein synthesis). In prokaryotes, we saw that
transcription and translation are coupled, that is occur simultaneously but in
eukaryotes translation occurs in the cytoplasm while transcription takes
place in the nucleus. Translation was noted to be a 4-step process involving
aminoacylation, chain initiation, elongation and termination. Thus, we saw
how a gene expression led to a protein which has a bearing on the
phenotype.
Further Reading
1. Russel, P.J. (1992). Genetics (3/e). HarperCollins Publishers, NY
2. Suzuki, D.T., Griffiths, Miller, J.H. and Lewontin, R.C. (1986). An
Introduction to Genetic Analysis. (3/e). W.H. Freeman and Co. NY
TOPIC ACTIVITIES
Use this questions to test your understanding of this topic.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 68 OF 153
1. Explain briefly how the genetic code was ‘cracked’.
2. Give any three characteristics of the genetic code.
3. Give an outline of the process of translation.
4. If a gene size is 720 base pairs. How many amino acids could the
resultant protein comprise assuming a codon each for start and stop signals?
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 69 OF 153
TOPIC 6. MECHANISMS OF CONTROL OF GENE EXPRESSION
Introduction
All organisms have many genes that perform various intricate activities in
them through their gene products. The process of gene expression relates
the gene to is phenotype. The challenge that faced the early molecular
geneticists was to understand whether all the genes in an individual
functioned all the time. Was gene action under any control and if so, how
was this operationalised? The answer was provided through insightful
experiments by some scientists at the French school led by Pardee, Jacob
and Monod in the 1960s.
Topic Time
Due to the newness of this topic to most students a slow incisive
understanding is vital for any grasp to be made. Thus, three lecture hours is
mandatory. However, optional time for a revisit to the content and tackling
the assignment is strongly encouraged.
Topic learning requirements
Prior reading of related content is strongly advised if you are to follow and
easily understand this topic.
Learning Outcomes
After you successfully complete this topic, you should be able to:
i.
Demonstrate that gene expression is indeed a regulated process
ii.
Cite at least four examples of gene regulation in your surrounding
iii.
Define an operon and name several examples thereof
iv.
Demonstrate how Jacob and Monod showed that gene regulation in
organisms occurs
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 70 OF 153
v.
Demonstrate how mutations in an operon may interfere with its
normal function
vi.
Answer any questions relevant to this topic
Topic Content
6.1 Introduction
Control of gene expression, is in essence control over the amount of gene
product present in the Cell.
Q: How can the synthesis rate of a gene product be regulated?
A: By exerting control over anyone, or a combination, of the various steps in
the gene expression pathway.
The possible points are:
1. Transcription,
2. mRNA degradation,
3. mRNA processing,
4. Translation – control could be exerted over the number of ribosomes
that can attach to a single mRNA, or over the rate at which individual
ribosomes translate a message.
However, the best understood mechanisms, in both prokaryotes and
eukaryotes, depend solely on the first possibility, that is, regulatory control
over transcription.
6.2 Gene and protein types in bacteria
These can be categorized into structural and regulatory genes and
proteins.
1. Genes
(1) Structural genes
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 71 OF 153
These encode the largest group of cellular proteins, e.g. enzymes,
membrane proteins and ribosomal components. They do not regulate
transcription but are themselves regulated.
(2) Regulatory genes
Code for proteins that help the cell sense its environment and regulate
transcription of structural genes by binding to DNA. Regulatory proteins can
be divided into negatively acting and positively acting.
a. Negatively acting proteins (also called repressors)
Repress a gene or an operon by binding to DNA at or near a promoter site
(usu. Operator) and physically preventing mRNA synthesis by denying RNA
Polymerase access to that gene or operon.
They combine with effectors thereby preventing them from the operator
sites.
Effectors are either inducers like lactose or co-repressors like tryptophan.
(i)
Inducers: The binding with an inducer produces a change in the
shape of the repressor protein, and in this way decrease its binding
affinity for the operator locus;
(ii)
Co-repressors: Co-repressors combine with repressors (that usu. are
not functional without the co-repressor) to become functional. For
example, the amino acid tryptophan in the tryptophan operon)
combines with trp repressor protein and only then can the repressor
protein bind to the operator and inhibit transcription of the genes for
the enzymes that make tryptophan. Thus, the amino acid is acting as
a co-repressor, and the addition of tryptophan to an E. coli culture
stops the synthesis of the tryptophan-producing enzymes.
b. Positively acting proteins (also called activators)
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 72 OF 153
These bind to DNA at or near a promoter site and increase the efficiency
with which RNA Polymerase binds to the promoter. The sites to which they
bind are called activator sites. Example is the Ara C protein. Ara C is
necessary in the transcription of genes responsible for arabinose utilization.
6.3 Purpose of Gene Regulation
Gene regulation prevents a cell from wasting energy by making
unnecessary components. For instance:
1. Bacteria
By regulating expression of their genes, E.coli and other bacteria are
able to respond quickly to changes in their environments without
wasting energy by maintaining in an active state, genes whose products
are of no immediate use.
2. Eukaryotes
Gene regulation in eukaryotes is similar to that in prokaryotes, but
is more sophiscated to both the signals and the impact gene
regulation has on the organism, viz:
a. Eukaryotic cells respond to a greater range of regulatory stimuli,
e.g. plant cells switch on genes for photosynthetic proteins in
response to light.
b. Some eukaryotic genes are developmentally regulated, e.g.
human globin genes – different members of the  and  gene
families being expressed in the embryo, foetus and adult.
c. Gene regulation in multicellular organisms results in cell
specialization. In humans there are about 250 different cell
types, each with a different morphology, biochemistry and role
to play in the organism.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 73 OF 153
6.4 Understanding Gene Regulation
Studies using bacterial cells provide the foundations of our
understanding of gene control. In bacteria, regulatory proteins that bind to
specific sites on DNA carry out most gene control. The binder either prevents
or increases the synthesis of specific mRNA.
6.5 Prokaryotic Gene Regulation
It is common in bacteria for one promoter to serve a series of clustered
genes and such genes often produce enzymes that are active in a single
metabolic pathway. Such a gene cluster constitutes the operon and all
transcribe into a single polycistronic mRNA (i.e. the mRNA encodes more
than one polypeptide).
Each gene of the operon is represented in the mRNA, and each section
of the mRNA is independently translated as shown below.
DNA strand
P
Gene A
Gene B
gene C
transcription
5’ AUG
UGA AUG
UGA AUG
Ribosomal subunits attach to each
initiation site
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 74 OF 153
UGA
translation
Peptide A
peptide B
peptide C
6.6 Operons
Through their investigations on transcription, translation and protein
synthesis of E. coli, Francois Jacob and Jacques Monod (1950s to 1960s)
provided a definite answer to the question of gene regulation in the living
system. This concept came to be called the Lactose operon and was the first
documented system of gene regulation. Jacob and Monod received the Nobel
Prize for their operon theory in 1965.
An operon is a unit consisting of adjacent cistrons that function coordinately under the control of an operator gene.
1. The lactose Operon
Jacob and Monod predicted, based on genetics studies on the lactose genes,
that the transcription of the lactose gene DNA into RNA was controlled by 3
classes of genes:
*
A regulator gene (I),
*
An operator gene (O), and
*
One or more structural genes
In this operon, gene Z is responsible for the production of (beta) βgalactosidase, an enzyme that cleaves lactose into glucose and galactose.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 75 OF 153
I
P
O
Z
Y
A
mRNA
transacetylase
repressor protein
permease
β-galactosidase
Key:
Gene I = produces a repressor protein that binds to the operator;
Gene O = the operator and is the one repressed by the protein from
gene I;
Genes Z, Y and A = produce the enzymes galactosidase, permease
and transacetylase respectively.
In the wild E. coli, these genes are in a repressed state, i.e. no
enzymes are formed.
If glucose and lactose are present, glucose is preferentially used
while the operator remains shut down. However, when glucose is used
up, the lactose molecules interact with the repressor protein, stopping
this protein from repressing the operator gene function. The
unrepressed O gene allows the attachment of RNA Polymerase, and
transcription begins throughout the length of the O and structural
genes.
In this system therefore, we have an active repressor plus inducer
= inactive repressor which then permits structural genes to produce
mRNA.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 76 OF 153
2. Faults in the Lactose Operon
Mutations in the I and O genes have been found that alter the control
of the operon.
Mutations in Z can make normal I and O genes useless with
respect to control of the production of a normal protein because the
transcription of the structural genes to produce the protein is stopped
at the mutant Z gene.
Possible mutations that may alter the suppression and transcription of
this operon:
I O Z designate normal operon: inducible enzyme; no transcription;
repressor Protein made.
I- O Z, an operon with a mutant regulator gene and therefore no
repressor protein made: constitutive enzyme continuous transcription
occur.
I Oc Z, an operon with a mutant O gene and therefore whereas the
repressor protein made but will not attach to operator site: Thus,
constitutive enzyme transcription (i.e. continuous transcription)
occurs.
I O Z-, an operon in which a mutant structural gene for galactosidase has occurred. Therefore, no transcription due to
structural defect will occur.
The other example of negative control is the tryptophan operon.
This system as already stated operates on the basis of a repressed
system, i.e. an inactive repressor plus co-repressor results in an
active repressor which then prevents structural genes from producing
mRNA.
6.7 Negative and Positive Controls of Transcription
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 77 OF 153
1. Negative Control
The lactose operon is an example of this kind of control in the sense that this
operon is potentially operative and active but is actually switched off by the
regulator gene.
Thus, this operon is not permitted to express itself unless
actually required. The product of the regulator gene represses the enzyme
synthesis. When the repressor is absent, protein synthesis occurs even
without the aid of the inducer
2. Positive control (Activation)
A case where the product of the regulator gene is needed for the initiation of
transcription. An example is the Ara C operon.
When E. coli cells are grown on arabinose as an energy source, they produce
the three enzymes needed to convert arabinose into xylulose (which then
enters the glycolytic cycle).
These three enzymes, an isomerase, a kinase and an epimerase, are the
products of three genes of this operon. They are produced to metabolise
arabinose. These three genes (B, A and D) are regulated by the ara
regulatory protein encoded by the ara C locus.
In the absence of arabinose, the protein behaves as a repressor, binding to
the "O" DNA preceding the ara B, A and D genes.
Addition of arabinose to the medium results in the release from the operon
of the ara repressor protein, which binds to arabinose. This newly formed
complex then becomes an activator, attaching to a separate initiator site on
the operon, stimulating RNA Polymerase binding and transcription of the B,
A and D genes. Thus the ara regulatory protein can alternatively behave as a
repressor or an activator, depending on the absence or presence of
arabinose, respectively.Extension of the operon-regulator gene theory may
account for some aspects of cellular differentiation. It is believed that all
cells of eukaryotes contain the entire genome of the organism, but much of
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 78 OF 153
it is repressed. Differentiation is thought to be the result of the programmed
de-repression of different operons.
6.8 Other bacterial operons:
1. Galactose operon.
This is a set of genes, E, T, and K producing enzymes epimerase,
transacetylase and kinase respectively needed to convert galactose into
glucose.
2. Arabinose operon
This is a set of genes ara B, ara A, and ara D producing enzymes isomerase,
kinase and epimerase respectively needed to convert arabinose to xylulose.
3. Tryptophan operon
This is an operon comprising five genes required for the synthesis of
the
amino acid tryptophan. This operon is an example of a repressed system.
The induction and repression of bacterial enzymes occur mainly through the
control of the transcription of the appropriate genes.
Topic Summary
In the previous topic, we learnt that genes express themselves in a two-step
process leading to formation of gene products. It is these gene products that
influence the function and phenotypes of organisms. The question that arose
subsequently was whether all genes of an organisms express themselves or
do so only when required. In this topic, we have learnt that gene expression
is a highly regulated process, thanks to the work of Francois Jacob and
Jacques Monod. This breakthrough was achieved in bacteria by studying how
the Lactose operon. We also considered the levels at which gene regulation
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 79 OF 153
can be instituted, e.g. at transcription, translation, etc. Mutations in this
bacterial system alters the function of gene regulation.
Further Reading
(1) Russell, P.J. (1992). Genetics (3/e).
(20 Stine, G.J. (1989). The New Human Genetics. WCB Publishers, USA.
Topic Activities
1. What is the purpose of gene regulation in organisms?
2. At what points can gene activity possibly be regulated?
3. If you were to classify genes and proteins, how would you do so?
4. Define ‘operon’ and give possible examples in bacteria.
5. name any familiar examples of gene regulation that you are familiar with.
6. Discuss the operation of the ‘Lactose operon’.
7. What are the possible points of mutation in lactose operation and their
effect.
TOPIC 7. GENE MUTATIONS AND REPAIR
In the foregoing topics we have examined or considered the gene or DNA as
a normal entity and therefore there was great accuracy in their expression.
The existence of alternative phenotypes indicates changes do occur to the
gene and these changes have significant consequences to the organism both
in its function and phenotype. This variation can occur through either
recombination or mutation. In the latter case, how does this occur and when
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 80 OF 153
it does occur, is there provision for restoration? This is the gist of this topic.
Learn and appreciate its significance.
Topic Time
It is envisaged that four lecture hours are sufficient to slowly and effectively
learn this topic. Any additional time goes to further review and assignments.
Topic Learning Requirements
In BOTA 111, you covered preliminaries of this topic. So, a revisit to the
relevant notes to this topic is strongly encouraged.
Learning Outcomes
After you have successfully completed this topic, you should be able to:
i.
Define the two possible sources of heritable variations in an organism
ii.
Identify the possible types of gene mutations
iii.
Describe the sources of spontaneous and induced mutations
iv.
Describe at least one DNA repair mechanism available to the organism
v.
Answer any question relevant to this topic
Topic Content
General Introduction
Types of variation in a species
These are: environmental, genetic, or a combination of both
environmental and genetic.
Changes, which are independent of environment and are heritable,
provide a permanent step on which selection can be made. Such changes
occur in the genome of an individual and may be caused by recombination
and mutation. Recombination usually causes no marked variations because it
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 81 OF 153
merely redistributes existing genetic material among different individuals.
Mutation, on the other hand causes a marked variation as it affects changes
in the alleles (genes) and/or chromosomes. Usually they can be heritable or
non-heritable.
7.1 MUTATION
Introduction
Biological evolution occurs because the hereditary material, DNA, can
change from generation to generation. These changes affect nucleotide
sequence or amount of DNA and can occur spontaneously or by induction.
Hereditary information is usually maintained intact by a complex metabolism
involving both replication and repair functions. Thus, mutations may be the
result of errors in these processes.
Mutations provide species with different forms of the genetic material,
and also they are the working tools of the geneticist. The resulting
phenotypic variability allows the geneticist to identify and study the genes
that control the modified traits.
Even though mutations are known to be harmful, there is little that can
be done to stop their occurrence. As mutations accumulate in a species,
their burden, called the genetic load increases.
One can classify mutations either as point or gross. In this topic we
consider point or gene mutations.
7.1.1 Point mutations
Also called gene mutations, affect one or a few nucleotides within a
gene. They show a tendency to back mutate, i.e. revert to the wild-type.
TYPES
They can be divided into two general classes: base-pair substitutions
(<20% of species mutations) and frameshift mutations.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 82 OF 153
1. Substitutions. These are mutations that involve alteration in the
sequence but not the number of nucleotides in a gene.
a. Transitions – Caused mainly by tautomerization, deaminations and base
analogs. They involve replacement of a purine by a purine (G-A or A-G)
or a pyrimidine by a pyrimidine (C-T or T-C).
b. Transversions. Here, a purine replaces a pyrimidine (T or C changes to A
or G; A or G replaces T or C).
Obviously these mutations cannot be seen but their effects occur in the
proteins formed. At protein level these mutations are manifested as:
a. Silent mutation: where the triplet codes for same amino acid, e.g. AGG
CGG, both code for Arg.
b. Neutral mutation: triplet codes for different but functionally equivalent
amino acid, e.g. AAA
AGA, changing basic Lys to basic Arg at many
places will not alter protein function.
c. Sense mutation: this leads to proteins that are longer than normal by
changing a termination codon to one that codes for an amino acid.
d. Missense mutation: triplet codes for different and non-functional amino
acid. E.g. -globin:
Hb A
Hb S
Codin:
CTC
CAC
Codon:
GAG
GUG
Amino acid:
glu
val
e. Nonsense mutation: this codon does not code for the incorporation of
any amino acid, but behaves as amino acid chain terminator. It may
arise within a sense codin, e.g. ATG (UAC) (Tyr) ATT (UAA) (nonsense
triplet). A mutation that affects a nonsense codon serves to elongate the
protein being formed.
While those mutations imply a forward occurrence, reverse mutations
can also occur as follows:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 83 OF 153
a. Exact reversion.
forward
reverse
AAA (lys)
GAA (glu)
Wild type
AAA (lys)
mutant
wild type
b. Equivalent reversion.
Forward
reverse
UCC (Ser)
UGC (cys)
Wild type
AGC (ser)
mutant
wild type
Note:
a. Not all point mutations are observable,
b. All point mutations do not occur only in coding regions but can
occur in other areas of the gene, e.g. regulatory sites and are
therefore called regulatory mutations.
2. Frameshifts. These involve insertion (addition) or deletion of nucleotides
from the DNA that are not a multiple of three. Insertions are induced by
treating cells with acridine derivatives while deletions result from hydrolytic
loss of a purine base because of low pH or temperature by alkylating or
deaminating agents.
Frameshifts cause all DNA beyond the point of mutation to be misread.
They are also usually lethal.
Whereas in nucleotide substitutions only one amino acid in the protein is
altered, in frameshifts, the addition or deletion of a single base can cause
large-scale changes in amino acid composition of the polypeptide chain
leading to a non-functional gene product.
Example: Haemoglobin :
Position 
138
139
140
141
Termination
codon
UCC
AAA
UAC
CGU
UAA
amino acid
ser
lys
tyr
arg
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 84 OF 153
A deletion in last base of codon 139 produces a frameshift thus:
codon
UCC
AAU
ACC
GUU
AA
amino acid
ser
asn
thr
val
-
7.1.2 Spontaneous vs Induced Mutations
1. Spontaneous mutations
These arise in nature by chance due to natural forces such as radiation
from cosmic rays and radioactive mineral. They include:
a. Errors in DNA duplication,
b. Spontaneous lesions, and
c. Transposable genetic elements, e.g. some phages, transposons, etc.
a. Errors in DNA replication
These result from an illegitimate nucleotide pair (e.g. A – C) being
incorporated into a newly synthesised strand. This is however of a very rare
occurrence, i.e. at a frequency of between 108 and 1012 bases. Where it
occurs, the DNA polymerases (DPs) correct this anomaly. All prokaryotic DPs
have exonuclease activity.
Spontaneous mutations may also occur during DNA replication because
of tautomerism, i.e. a shift in the position of a proton that changes the
chemical properties of a molecule. The nitrogen bases exist in normal forms:
keto (C=O) for G and T; and amino (NH2) for A and C. The rarer forms are
enol (COH) and imino (NH) respectively as below.
Base
Normal
imino (NH)
Adenine
amino = NH2
A*
Cytosine
amino = NH2
C*
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
enol(COH)
PAGE 85 OF 153
Guanine
keto
= C=O
G*
Thymine
keto
= C=O
T*
Tautomeric shifts in the bases change the hydrogen bonding properties
such that Adenine assumes those of Guanine, Guanine those of Adenine,
Cytosine those of Thymine, and Thymine those of Cytosine. Tautomers
cause transitions, e.g. A-T may become G-T leading to G-C mutant (below).
The base analog mutagen 5-Bu, an analogue of Thymine exerts its
mutagenic activity through tautomerism.
A:T
5-Bu
A : 5 Bu (k)
1st replication
A:T
G : 5 Bu (e)
2nd repl.
A:T
A:T
G : C (mutant)
A : 5 Bu (k)
b. Spontaneous lesions
Types
These are largely two, depurination and deamination.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 86 OF 153
Depurination: Is the more common of the two. It involves the interruption
of the glycosidic bond between base and sugar and subsequent loss of G or
A from the backbone. A mammalian cell spontaneously losses about 10 3
purines from its DNA during a 20-hr generation period at 370C. If these
lesions were to persist, they’d result in significant genetic damage, since
during replication the resulting apurinic sites cannot specify a base
complementary to the original purine.
Deamination: Is the loss of an amino group (-NH2) due to oxidative
deamination of the bases. The bases A, G and C have NH2 groups. The –NH2
group is replaced by a hydroxyl (-OH) group, viz:
(i) A is deaminated to Hypoxanthine (H), which pairs with C, as below
A:T
Rare tautomer
H:T
H:C
A:T
(normal replication)
keto tautomer
H:C
G:C
mutant
Thus, due to this act a base pair such as A:T changes to G:C, a
transition.
(ii)
G to Xanthine, which pairs with C (no change);
(iii)
C to uracil, which pairs with A during replication.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 87 OF 153
Luckily, enzyme systems exist within the cell to restore such damage.
For instance, the enzyme U-DNA glycosylase, recognises U residues in the
DNA and excises them, leaving a gap that is subsequently filled in.
2. Induced mutations
A mutation is induced when it is caused by the application of a mutagen.
The mutagens used are either physical or chemical. Chemicals include:
Industrial chemicals used in paints, solvents, cleaning agents, etc.;
derivatives of petroleum compounds; reagents in pesticides; food
derivatives, and various drugs. Radiation and its sources such as nuclear
armament testing (fallout), and nuclear power plants, naturally from
40K,
14C.
Action:
Mutagens induce mutations by at least three different mechanisms:
a. Replacement of a base in DNA (base analogs),
b. Alteration of a base so that it specifically mispairs with another
base (alkylating agents), and
c. Damaging a base so that it can no longer pair with any base under
normal conditions (deaminating agents).
Incorporation of base analogs
Base analogs are mutagenic chemicals whose structure so mimics that
of a naturally occurring base that they’re incorporated by a cell into DNA.
Examples are 5-Bu (which is similar to thymine), and 2-Ap, similar to
adenine. Once in place, these bases have pairing properties unlike those of
the bases they replace and thus can produce mutations by causing insertions
of incorrect nucleotides opposite them during replication.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 88 OF 153
Specific mispairing
Both alkylating and deaminating agents can cause specific mispairing.
Both modify side groups of bases in the DNA in situ.
a. Alkylating agents: Include nitrogen and sulphur mustards, methyl
methane sulfonate (MMS), ethyl ethane sulfonate (EES), ethyl methane
sulfonate (EMS). They cause the addition of methyl, ethyl or (CH3, C2H5,
C3H7) groups. This results in a modification of the bases’s pairing potential
and thus mispairing. They generally cause transversions. Once they are
modified the bases now will pair with non-traditional partners.
For example, EMS adds an ethyl group to 6th position of G to result in 6ethylguanine. This then acts as a base analog of A causing it to pair with T.
This leads to direct mispairing with T and results in a G: C to A: T transition
at the next replication. EMS can also alkylate the keto group in the 4th
position of T.
b. Deaminating agents include nitrous acid and formaldehyde.
Three of the four natural bases in DNA (A, G, C) have NH2 groups that
can be removed by nitrous acid. The products of the reaction and their
pairing properties are: A to H, which pairs with C; G to X, which pairs with
C; C to U, which pairs with A, and 5-m C to T, which pairs with A.
Another agent, hydroxylamine (HA or NH2OH) causes G-C to A-T
transitions only. It acts by adding a hydroxyl group to its existing amino
group. The new product, hydroxylaminocytosine, may undergo a tautomeric
shift allowing it to pair with A. Following two replications, a G-C pair will be
converted to an A-T pair.
In conclusion, we may now summarise the kinds of damage to DNA and
possible causes as below.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 89 OF 153
Summary of the general damages to DNA and their causes
DNA lesion
cause
a. Missing base
acid and heat remove purines;
b. Altered base
ionizing radiation, alkylating
agents;
c. Incorrect base
spontaneous deaminations: C to U;
A to H, etc;
d. Deletions, insertions
intercalating agents, e.g. acridine
dyes+;
e. Cyclobutyl dimers
uv radiation;
f. Strand breaks
ionising radiation, chemicals;
g. Cross linking of strands
light induced or due to
antibiotics, e.g. mitomycin c;
+ Acridines have a structure that allows insertion or intercalation between
bps along the DNA strand. It leads to outstretching of the DNA backbone
thereby causing a deletion or an addition of a nucleotide pair.
7.2 DNA REPAIR
From the foregoing, it is evident that DNA molecules may be damaged
by a variety of agents and this damage must be repaired to prevent
accumulation of harmful mutations. If left uncorrected, both growing and
non-growing somatic cells will suffer genetic damage and no longer function.
Moreover, DNA in germ cells with too many mutations will lead to unviable
offspring. Repair is done by a number of genetically controlled systems.
1. Mutation rates
The fact that the level of mutation is kept low is due to the efficiency of
DNA polymerase and the repair mechanisms. For instance, single breaks of
DNA occur with a frequency of 2.3 x 102 hr-1. Repair occurs at 2.0 x 105 hr-1.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 90 OF 153
Estimates put mutation rates per cell in bacteria and viruses at 10-3 to 10-4
and in humans at 10-5 (i.e. 1:100,000 gametes). However, the determined
value for humans is 4: 100,000 gametes. It appears very low. But when we
consider that male ejaculates 300m sperm, we see that amongst them are
12,000 mutants!
2. Repair mechanisms
Besides enzyme-based repair mechanisms, DNA by virtue of its double
stranded nature repairs itself. If mutations are left uncorrected in both
growing and non-growing somatic cells, the damage will be such that they
are rendered unable to survive. Thus for survival, all such mutations must
be corrected.
DNA can therefore be restored by any of the aforementioned
mechanisms as follows:
a. Natural DNA repair
By virtue of its double helical structure, nature has taken this advantage
of DNA to evolve processes that use the redundancy of the two
polynucleotide chains to raise enormously their stability as information
carriers. The principle of error detection and correction operates very
efficiently.
b. Proof reading
This is done by the polymerase enzymes. Proof reading helps to nullify
errors of DNA copying. They demand accurate base pairing in the
copying of a DNA strand. All prokaryotic polymerases (I, II, III) have a 3to 5- exonuclease activity (but only I and III also have a 5- to 3- activity
inbuilt in them). This gives them the ability to excise any incorrectly
placed base. Therefore, this proof reading property of DNA Polymerases
results in low mutation frequencies.
c. When DNA is damaged by ultra-violet (uv) light. This mostly
affects the DNA of lower organisms such as bacteria (e.g. E. coli).
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 91 OF 153
Several ways of restoring such damage exist: photo reactivation,
excision, recombination, and systems of survival (sos).
When DNA absorbs UV light, neighbouring pyrimidine bases can become
chemically rearranged and covalently linked into a pyrimidine dimmer.
Because they stop DNA replication, these uv damages can be lethal. But
cells contain repair enzymes that cut out the damaged region, replace it with
a new segment of polynucleotide using the intact strand as a template, and
reseal the molecule with covalent bonds (detailed below).
Excision (or dark) repair
Absorption of uv was demonstrated by R. Setlow et al to cause
cyclobutane rings (i.e. pyrimidine dimers between T-T, T-C, and C-C). They
also demonstrated that repair through excision of such damage can occur in
cells kept in the dark. This kind of DNA damage usu. distorts the DNA
backbone and its genetic consequence is hindered replication or
transcription. It has been shown that transcription is stopped at such points
resulting in a shortened species of mRNA.
Excision repair is also called the cut, patch, and seal repair process and
may be undertaken as follows:
Step 1: Cutting or incision. Done by a specific endonuclease that recognises
the distorted area and cuts that strand on either sides of the damage.
Step 2: Excision. This involves the 5- to 3- exonuclease activity of DNA
Polymerase I, which digests away abnormal or chemically modified bases.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 92 OF 153
Step 3: Patching. Complementary base pairing in a 5- to 3- direction by DNA
Polymerase I fills the gap created by excision.
Step 4: Sealing or bridging. This is done by DNA ligase. This then completes
the process.
Photoreactivation
The effect of uv radiation is to cause linkage among adjacent pyrimidine
bases to form dimers, for instance adjacent thymines through C5 and C6.
Exposing the irradiated cells to visible (or white) light can restore such
damaged DNA. Photoreactivation is light-dependent and involves an enzyme
photolyase (or PRE) that cleaves such links and so restores to normal the
molecule.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 93 OF 153
c. SoS
(i) An enzyme system involving DNA methyl transferase reverses the
action of alkylating agents by transferring methyl or ethyl groups from
a G or T to an internal cysteine residue;
(ii) Also glycosylase enzymes remove such alkylated bases to leave sites,
which need to be filled by an apurinic endonuclease repair system; for
instance enzyme u-DNA glycosylase recognises u residues in the DNA
and excises them leaving a gap.
Topic Summary
In this topic, we examined mutation as both source of heritable variation
and source of harmful effects on the individual. This variation was internal
and external to the gene and was caused in several ways. The reprieve
however, was that these changes in the gene could be restored – thus the
various repair mechanisms available to the cell. These repair mechanisms
include: (1) repair by Dpol proofreading; (2) photoreactivation; (3) excision
repair.
Further reading
(1) Suzuki, D.T., Griffiths, A.J.F., Miller, J.H., and Lewontin, R.C. (1986). An
Introduction to Genetic Analysis (3/e). W.H. Freeman & Co./NY
(2) Russell, P.J. (1992). Genetics (3/e). HarperCollins Publishers, NY
Topic Activities
Use the questions to gauge your understanding of this topic.
1. How would you classify gene mutations.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 94 OF 153
2. Qualify these statements as either true or false:
(a) Mutations are caused by genetic recombination.
(b) Mutations occur more frequently if there is a need for them.
(c) Frameshifts are not gene mutations.
(d) Ultraviolet light usually causes deletion of DNA segments.
(e) Sense mutations lead to a longer polypeptide.
3. Mutagens induce gene mutations by at least three mechanisms. Which are
these?
4. Demonstrate how a base analog may cause mutation.
5. Describe anyone repair mechanism familiar to you.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 95 OF 153
TOPIC 8. RECOMBINANT DNA TECHNOLOGY (RDT)
Introduction
Welcome to topic 8 of this course. Now you will learn and appreciate the
revolution that followed the understanding the function of nucleic acids
particularly DNA as the hereditary material. Scientists were challenged to
extend their studies and research into the potential of developing techniques
to cause DNA from unrelated organisms to recombine. The success of this
approach has revolutionized agriculture, medicine, conservation, forensic
science, and many other areas of life.
Topic Time
As a new and fairly recent topic, a slow presentation is necessary. Thus four
lecture hours are projected for its coverage. The student needs more time
for literature supplementation and/or do assignments that may arise.
Topic Learning Requirements
At the conclusion of this topic, there is strong need for the students to be
taken to a molecular research laboratory for a hands-on exercise on DNA
work – DNA extraction, manipulation and related studies.
Learning Outcomes
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 96 OF 153
After successful completion of this topic, you should be able to:
i.
Define and explain recDNA technology (RDT).
ii.
Explain the basic steps in recombinant RNA work
iii.
Define both the promises and fears of this technology
iv.
Name some products that have arisen through molecular work
Topic Content
8.1 Introduction
From about the third quarter of C20th, public interest in genetics was
heightened due to the development and application of particular molecular
techniques collectively called recombinant DNA technology. Research in this
area has led to the development of industries hinged on biotechnology, or
genetic engineering. Benefits of applied genetic engineering have been
realized and are growing particularly in Agriculture, medicine, forensic
science, conservation, etc.
Definition
Recombinant DNA technology (or Gene cloning) is one aspect of genetic
engineering and refers to a collection of experimental techniques that
enables a scientist to identify, isolate and propagate fragments of the
genetic material (DNA) in pure form. By this approach a scientist is then able
to modify a DNA fragment and/or transfer them from one organism to
another.
The term RDT refers to the creation of a new association between DNA
molecules or parts of DNA molecules which are not found together naturally.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 97 OF 153
The idea of deliberately changing the genetic makeup of an organism is
as old as genetics itself, e.g. in 1927 H Muller showed that it was possible to
induce mutations by x-rays. In fact since the 1970s genetic analyses have
taken a biochemical approach and DNA sequence analysis is now a routine
procedure. RDT techniques are now being used to understand heredity,
evolution, gene control and human diseases
Populations of organisms are able to adapt to changes in the
environment because the populations are heterogeneous due to acquisition
of new genes or the alteration of the existing genes.
These natural processes of genetic restructuring form the essence of RDT
with the important difference that DNA results are by man in the laboratory
and natural processes are in the environment and are due to pressure of
natural selection.
Justification
Early breakthroughs on DNA were made without actually being able to
determine the nucleotide sequence of the DNA involved. These
breakthroughs were on:
1) The genetic code, its nature;
2) The mRNA, its existence and role;
3) Translation, its mechanism; and
4) Gene regulation, its basic principles.
That the breakthrough was indirect was because of:
1) The great size of the DNA molecule,
2) Chemical similarity of DNA molecules, and
3) Insufficient quantities of DNA of a given protein-coding gene.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 98 OF 153
Therefore, ways had to be found for isolating identifiable fragments of DNA
of manageable length in sufficient quantity for detailed molecular
analysis.
This became possible in the 1970s with the identification in bacteria of
enzymes called restriction endonucleases. These enzymes cut particular
pieces from heterogeneous mixtures of DNA molecules. With these enzymes
now on hand, there was an explosion of scientific activity as they now
became manipulators instead of observers. They now had the means of
reproducibly creating and combining fragments from different sources.
8.2 Legitimate vs illegitimate DNA combination
Legitimate = recombination of DNA free of human intervention and within
species. DNA recombination formerly was exclusively by meiosis in higher
organisms and by conjugation, transduction and transformation in such
organisms as bacteria. Where DNA was introduced across species its success
depended on the existence of regions of genetic homology. When humans
tried to take DNA across species, e.g. in interspecific and/or intergeneric
hybridization (through cross pollination in plants and artificial insemination in
animals) this was usually unsuccessful. Perhaps these failures were due to
lack of genetic homology.
Illegitimate = recombination of DNA molecules from different species with
the help of restriction endonucleases. These enzymes therefore eliminated
the need for regional genetic homology. All that was required exposed
unpaired bases (Fig. 1).
Illegitimate rDNA technology therefore involves:
1) Obtaining a DNA fragment of interest (insert DNA) and placing it into
a vector (a biological carrier, e.g. plasmid, virus, etc),
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 99 OF 153
2) Introduction of such a molecule (now called rDNA molecule or clone)
into a compatible host cell (a process called transformation). The
host cell, as it divides, amplifies this DNA fragment to several copies
8.3 Promises of rDNA technology
This technology swiftly, dramatically and forever changed the evolutionary
face of the earth.
In Scientific research, the last century can be remembered by at least two
significant eras:
1) The Atomic age era (1945), and
2) The age of Molecular genetic engineering (1970).
Both have undeniably changed the image, enlarged the ears, and promised
unparalleled benefits to humankind. For instance, the Atomic age promised
endless, cheap, clean energy. So far not fully realized and instead there is
more worry about a radioactive polluted planet.
Molecular genetic engineering (rDNA) promised:
1) Mass production of the then rare and expensive pharmaceuticals;
2) Inexpensive source of bioenergy;
3) Genetic therapy (the transfer of one or more normal or modified genes
into an individual’s body cells to correct a genetic defect or boost
resistance to disease);
4) New vaccines;
5) Super plants and animals; and
6) Infants largely free of defects due to prenatal diagnosis of genetic
diseases
Have any of the above been achieved/realized? This topic will endeavor to
answer.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 100 OF 153
8.4 Techniques of recombinant DNA Technology
The central technique in rDNA work is gene cloning (i.e. the production of
many copies of an identical gene).
Stages of gene cloning
1) Splicing DNA of interest to a cloning vector;
2) Introducing the rDNA into a host cell for amplification.
1. Producing rDNA : Splicing DNA
In order to produce rDNA via DNA splicing, six requirements must be
satisfied:
1) A restriction enzyme must be present to cut the cloning vehicle
(vector);
2) The foreign DNA (insert DNA) must contain restriction sites similar to
the restriction of the cloning vehicle;
3) A DNA ligase enzyme is necessary to seal the foreign DNA into the
cloning vehicle;
4) A means of inserting the recombinant cloning vehicle into a host cell
for multiplication be available;
5) The cloning vector must be able to replicate in a host cell;
6) A means of selecting those cells carrying the cloning vehicle and
cloned DNA is necessary
(1) Restriction Enzymes
Some scientists argue that the discovery, characterization, and use of
restriction enzymes marked the origin of genetic engineering.
Discovery
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 101 OF 153
Occur only in micro-organisms where they act as an immune system by
destroying infecting (invading) DNA molecules. However, host-cell DNA is
not cut by its own restriction enzymes because the enzyme methylase
has modified its sites with a methyl group (-CH3). In 1970, three
scientists – Werner Arber, Daniel Nathans, and Hamilton O. Smith –
discovered them. For this achievement, they shared the 1978 Nobel Prize
in Medicine & Physiology- the first award on work related to genetic
engineering.
Types
There are two or even three types: type I, type II, and type III. However,
due to its specificity, only type II is widely used. The vast majority of type
II recognize short, specific sequences that are 4, 5, 6 or 8 nucleotides
(bps) in length and display two-fold symmetry (palindromes):
i)
EcoR1 (from E coli) cleaves
ii)
Pst1 (from Providencia stuartii) cleaves
iii)
Taq 1 (from Thermus aquaticus) cleaves
iv)
Not 1 cleaves G C
Note:
V
v
5’
G
V
AATTC
5’
3’
C
V
5-T v
T G C A G3’
CGA
3-
GGCCGC
= point of incision
These molecular scalpels provide a powerful tool for generating unique
fragments from very large DNA molecules. They are sequence specific
irrespective of the source of such DNA. They also facilitate the subsequent
joining of DNA.
Number
Today thousands of these enzymes having cleavage specificities (varying in
both nucleotide sequences and length of the recognition sites) are
commercially available as analytical reagents.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 102 OF 153
Nomenclature
A restriction enzyme is named according to the organism from which it was
isolated. For instance the enzyme EcoR1, the 1st letter is from the genus of
the bacterium (i.e. Escherichia). The next two letters are from the name of
the species (i.e. coli). An additional subscript letter indicates the type or
strain, and a final number is appended to indicate the order in which the
enzyme was discovered in that particular organism.
(2) Restriction sites and DNA cleavage
Restriction sites
A restriction site is a sequence of DNA that is recognized by a restriction
enzyme. DNA molecules are extremely long and featureless. Thus, it is
hard to isolate a particular fragment of DNA or determine its sequence.
The restriction sites that occur by chance in all DNAs provide natural
chemical guideposts along the DNA. Because most restriction sites are only a
few bps long, they will occur in any DNA of sufficient length whatever its
origin.
A particular restriction enzyme will therefore always cut a given DNA at the
same point (s) generating a set of specific DNA fragments (restriction
fragments) that can be sized and separated by starch gel electrophoresis.
For each restriction enzyme, it is possible to calculate the average length of
the fragments the enzyme will generate and then use that information to
estimate the approximate number and distribution of recognition sites in the
genomes. The estimate depends on two simplifying assumptions:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 103 OF 153
(1) that each of the 4 bases occurs in equal proportions such that a genome
is composed of 25% A, 25% G, 25% C and 25% T.
(2) that the bases are randomly distributed in the DNA sequence.
NOTE:
These assumptions are never precisely valid, but they enable us to
determine the average distance between recognition sites of any length by
the general formula, 4n , (n = number of bases in the sites), viz:
(a) The probability that a 4-base recognition site will be found in a genome
is (¼)4 or ¼ x ¼ x ¼ x ¼ = 1/256.
(b) The probability that a 6-base recognition site will be found in a genome
is (¼)6 = 1/4,096.
Based on this premise, an enzyme that:
(a) recognizes 4-base sequence (e.g. GTAC for Rsa 1) will cut on average,
once every 44, or every 256 bps creating fragments averaging 256 bps in
length.
(b) EcoR1 that recognizes the 6-base sequence (GAATTC) will cut on
average once every 46, or 4,096 bps (or 4.1 kb).
(c) Not1 which recognizes the 8-bases GCGGCCGC will cut on the average
every 48 bp (or every 65.5 kb).
However, because the actual distances between restriction sites for any
enzyme vary considerably, very few of the fragments produced by the three
enzymes above will be precisely 65.5 kb, 4.1 kb, or 256 bp in length.
The duration of contact between the restriction enzyme and the DNA
determines fragment size. If DNA is exposed long enough to any restriction
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 104 OF 153
enzyme, the result is a complete digest, i.e. DNA cut at every one of the
recognition sites it contains. Partial digest leads to larger fragments and
follows controlled enzyme quantity or duration of digestion.
Because the exact cut sites in the DNA made by each restriction enzyme are
known, computers can be used to predict if and where a gene may be cut by
particular enzymes and specific enzymes chosen accordingly.
DNA cleavage
Restriction enzymes cleave DNA so as to produce a 3- OH group on one end
and a 5- PO4 group on the other. In so doing:
(i)
Some restriction enzymes, such as Taq 1, cleave each strand at
similar locations, on opposite sides of the axis of symmetry,
generating fragments of DNA that carry protruding single strand
termini (i.e. “sticky” or cohesive ends).
cutting point
G A A T T C
G G A T C C
C T T A A G
C C T A G G
EcoR1
(E. coli)
(ii)
Bam1
(B. amyloliquefaciens)
Other restriction enzymes, such as Hpa 1, Hae III, cleave in the
middle of their recognition sequence to produce fragments that
have blunt ends that do not hydrogen-bond.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 105 OF 153
Hae III ---
G G
C C ---
---
C C
G G ---
=
----
G G
----
C C
+
C C --G G ---
Cloning vector (vehicle)
Definition:
A vector is a DNA molecule which can replicate in a suitable host
organism and into which a fragment of DNA may be introduced.
Characteristics:
(i)
Be easily introduced into the host organism by transformation
(plasmids) or transfection (phages);
(ii)
Have an origin of replication. This is a sequence of DNA that
allows the vector DNA to be replicated in the host cell;
(iii)
Have at least one cleavage site for a restriction enzyme. This site
called the cloning site is where the insert DNA is incorporated into
the cloning vector
(iv)
Must encode a gene whose product distinguishes transformed host
cells from untransformed cells. For example, many cloning vectors
carry a gene that confers resistance to an antibiotic; cells
transformed with such vectors will grow in media containing the
antibiotic, whereas untransformed cells will be killed.
Examples
Three types of vectors are commonly used for cloning DNA in
bacterial cells: plasmids, bacteriophages, and genetic elements
called cosmids. Each differs in the way it must be manipulated to
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 106 OF 153
clone pieces of DNA and in the maximum amount of DNA that can
be cloned using it.
(i)
Bacterial plasmids are extrachromosomal genetic elements that
replicate autonomously within bacterial cells, e.g. Ti plasmids in
Agrobacterium tumefaciens;
(ii)
Cosmids: constructed DNA carriers similar to naturally occurring
plasmids. Can carry up to 49 kbps. Thus are suitable to carry
large eukaryotic genes DNA.
(iii)
Viruses, examples are sv 40 and lambda phages.
2. Host Systems
Recombinant DNA is introduced into a compatible host in a
process called transformation. Host cells that pick up a DNA
molecule are called transformants or transformed cells. A single
transformant then undergoes many cycles of cell division to yield
a colony of cells with the recombinant DNA molecule. This DNA
can then be isolated, purified, and analyzed. Potentially, the
cloned DNA may be transcribed, its mRNA translated, and the
gene product isolated and studied.
The common hosts are E. coli or S. cerevisiae. After
amplification, hosts carrying the hybrid vector are selected, and
the DNA therein isolated.
Topic Summary
In this topic you have been introduced to a more recent is
approach for DNA isolation, manipulation, analysis and potential
of this technology in biological, medical, agricultural and forensic
research. This approach is the ‘recombinant DNA technology’
(recDNA or RDT). The key element in this technology are the
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 107 OF 153
restriction enzymes popularly called restriction endonucleases.
The technology revolves around enzymes, vectors and host
systems. Through them we are able to amplify an identified DNA
piece that can then be used as needed. The emergence of the
polymerase chain reaction (PCR) has reshaped this procedure.
This you will cover at the next level.
Further Reading
(1) Russel, P.J. (1992). Genetics (3/e). HarperCollins Publishers,
NY.
(2) Jones, M., and Jones, G. (1997). Advanced Biology.
Cambridge Univ. Press, UK
Topic Activities
Use these questions to test your understanding of this topic.
1. Define RDT and give a brief background to its origin and
growth.
2. What were the promises and promises of this technology?
3. What do you consider are the ethical, moral and legal issues
in this technology?
4. The restriction enzymes identified cut DNA in different sizes
generating various numbers of fragments. Suppose the DNA
molecule is 3billion bps and you are given 4-base, 6-base and 8base cutters, how many fragments will each generate?
5. Outline the process of cloning a DNA fragment.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 108 OF 153
TOPIC 9: GENETICS OF MICRO-ORGANISMS (BACTERIA AND
PHAGES)
Introduction to the Topic
This topic forms the second part of this course, that is microbial genetics.
In it we review the decision by researchers of genetic events to revert to
microorganisms as organisms of choice in understanding molecular heredity.
Microorganisms, like macroorganisms have nucleic acids in particular DNA as
the blue print of life. Scientists involved in molecular heredity noted that
these organisms offered numerous advantages compared to large organisms
in studying molecular hereditary events and thus the shift in using them as
tools for molecular research. Whereas most microorganisms reproduce
asexually, thus limiting variation which is fundamental in phenotypic
comparisons, variation nonetheless was possible. Scientists like Benzer
analysed the fine structure of a gene through a bacterium and so did many
others in specific aspects of molecular heredity.
Topic Time
A total of 10 hours is allocated to this topic given both its importance and
the aspects to be covered.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 109 OF 153
Learning Requirements
No specific tools or items are required. All you need to do is supplement
these notes with relevant references (both online and hardcopy textbooks).
Learning Outcomes
After successful completion of this topic, you should be able to:
i.
List the advantages and disadvantages of use of microorganisms in
molecular studies/research.
ii.
List some of the main microorganisms used in molecular studies
and the areas of studies they have been used.
iii.
Demonstrate how variation can be generated in bacteria, organisms
that largely reproduce asexually.
iv.
Explain the possible mechanisms of genetic exchange in bacteria
and phages.
v.
Illustrate how gene mapping can be done using the three
mechanisms of genetic exchange in microorganisms.
Topic Content
9.1 Introduction
A very large part of the history of genetics and of current genetics analysis
(particularly molecular genetics) is concerned with microbes and viruses.
Microbes include bacteria, fungi, algae, etc. All share a common property
with the eukaryotes - they have DNA, the material of heredity.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 110 OF 153
Microorganisms are a useful biological material. They are chosen because
they provide good illustrations of important molecular events, e.g. the
replication, recombination and rearrangement of DNA; the control of genes,
etc.
Those scientists, who studied heredity beginning in the 1940s, utilized
mainly microorganisms for their genetic studies. Until the 1930s, virtually all
genetic research had been done with animals and plants. Multicellular,
macroscopic organisms offer the convenience that the characters whose
hereditary transmission is to be studied can generally be recognized at sight.
However, plants and animals have the disadvantage that the number of
individuals that can be examined in any one experiment is limited to a few
hundreds or at most a few thousands. Also their lifecycles are long.
To overcome these limitations, the founders of molecular genetics turned to
bacteria- microscopic organisms whose lifecycles last less than one hour,
and of which billions of individuals can be raised overnight in a Petridis of
nutrient growth medium.
Although Anton van Leeuwenhoek discovered bacteria in the latter part
of the C17th, little else was done on bacteria for a further 200 years. Then
microbiology took its upswing in C19th. This was in response to the
spontaneous generation theory widely acclaimed at the time.
In late C19th a vast ensemble of bacteria of the most diverse shapes,
sizes, and functions were discovered.
These discoveries were soon to alter the human condition, in that they
led to the rationalization of the making of good wine, beer and cheese, and
to the conquest of infectious diseases.
Other than being man's little friends and enemies, bacteria were soon
acknowledged as a major component in the balance of the biosphere.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 111 OF 153
Indeed, macroscopic life as we know it would not be long for this world if
it were not for bacterial intervention in the constant recycling of Nitrogen
and Carbon between the atmosphere and organic matter!
The use of viruses and bacteria as genetic research tools has led to
major breakthroughs in our understanding of genetic processes.
9.2 Advantages of micro-organisms in Genetic Studies
To verify traits or properties of genes can be complex especially in higher
organisms. Microorganisms have been found less complex to study such
traits for the following reasons:
(1). Simplicity of the growth medium and the rapid growth rate. Bacteria can
grow on minimal medium which often comprises: a carbon source e.g.
glucose or glycerol; a nitrogen source (as NH4+ or an organic compound)
such as histidine; salts, e.g. Na+, K+, Ca2+, Mg2+, S042- and PO43-, and trace
elements.
(2) Short life cycles: Micro-organisms have extremely short reproductive
lifecycles of between 20 minutes to 1 hour. This together with their very
simple growth requirements, makes them ideal research organisms.
(3). Ease with which genetically distinct populations of bacterial cells can be
obtained and thereby ensure pure cultures.
(4). Haploid natures, i.e. microorganisms have only a single copy of its
hereditary information. Dominance effects hide none of its genes.
(5). Save for interaction, every gene, which the individual possesses, can be
expressed in its phenotype. Thus their study is simple.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 112 OF 153
(6). Large populations of them, which are generated rapidly and are
economical to keep. Such large populations allow for evaluation of such rare
events as mutations.
9.3 Disadvantages
(1). Scarcity of recombination events.
(2). Reproduction almost wholly asexual, thus generating little genetic
variability.
(3). Lack of morphological variation at individual level. Variations could occur
at clone level affecting such traits (phenotypes) as size/unit time, shape of
perimeter or growth habit, texture, color, etc.; response to nutrients,
resistance to antibiotics and bacteriophages, etc.
9.4 Important microbial and phage strains
In the variety of micro-organisms, several strains of each are studied,
viz:
Bacteria
(1). Escherichia coli - this is the most widely studied species of bacteria. It is
usually prototrophic (i.e. capable of growing on a defined minimal medium)
but occasional auxotrophs (mutant strains) exist. E. coli has an estimated
2,000 genes with about 1/2 already mapped;
(2). Salmonella species e.g. S. typhi, the cause of typhoid, and S.
typhimurium, which is popular among molecular biologists because one of
the most important processes of genetic transfer between bacterial cells
happened to be discovered with it.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 113 OF 153
(3). Streptococcus pneumoniae and Bacillus subtilis; B. thuringiensis; B.
anthracis
(4). Agrobacterium tumefaciens- carries the Ti plasmid
Phages
Work on phage genetics was initiated in the 1939s by Max Delbruck and
colleagues. In the 1940s, he, along with Salvatore Luria and A. D. Hershey
discovered genetic recombination in phages. Thereafter, phages have been
extensively used as tools for the study of gene structure and function. The
much investigated phages are:
(1) Phage lambda (). It was studied exclusively as a model of phage gene
expression.
(2) The T- phages that infect E. coli, both T-even and T-odd. The T-even
include T2, T4. The T4 (whose size is 166 Kbps; with 150 genes); T2 was
used by A. Hershey and M. Chase to prove that DNA is the genetic material;
The T-odd include T-1, T-3, T-5, and T-7.
(3) Phage M13 has single stranded circular DNA with mol. Wt. 1
9.5 Bacterial Genetics
Bacteria affect every aspect of our daily lives. They cause disease, provide
us with food and medicines, and dispose our wastes. Bacteria are essential
to the genetic engineer and are crucial research tools in biochemical
genetics.
9.5.1 Bacterial phenotypes
Bacteria show a variety of phenotypes that are expressed mainly through
mutations in the wild type through:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 114 OF 153
a) Colony morphology;
b) Inability to utilise a given sugar or synthesise an essential growth
substance,
c) Resistance to drugs, antibiotics and bacteriophages,
9.5.2 Genetic Recombination
Although bacteria and viruses are ideal subjects for biochemical analysis,
they'd not be useful for genetic study if they did not have sexual processes.
Such exchange processes create genetic variation in the various species and
strains of bacteria and can be used to map bacterial and viral chromosomes.
In the 1940s and the 1950s, the availability of genetically different bacterial
strains led to the study of a number of means by which genes could be
exchanged between bacteria. These mechanisms or processes are:
conjugation, transduction, transformation and to a less extent sex-duction.
a) Conjugation is the unidirectional transfer of genetic information through
the formation of a cytoplasmic bridge from one cell to another.
b) Transduction - gene transfer from one bacterium to another mediated by
a phage.
c) Transformation - direct uptake of free DNA by bacteria.
d) Sexduction - bacterial genes carried from one cell to another on a
molecule of DNA which acts as a sex factor called the F. This sex factor can
reside in a bacterial chromosome or it may exist as an autonomous unit
extrachromosomally. Genetic elements with such dual capacities are called
episomes.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 115 OF 153
These mechanisms, in conjunction with recombination processes, are largely
responsible for the rapid evolution of bacteria. The exploitation by
geneticists of all three systems of gene transfer has been of crucial
importance for the construction of new strains and the analysis and
manipulation of a wide range of biological processes.
1. Transformation
Griffith discovered this mechanism of gene transfer in 1928 with
Streptococcus pneumoniae. Thereafter, DNA was identified as the active
principle in transformation (Avery et al., 1944; Hershey and Chase, 1952)
(Fig. 1). This was thus the earliest known means of gene transfer between
bacterial cells. The frequency of occurrence of transformation is about 25%
or less.
Bacteria that are able to take up and incorporate free DNA into their
genomes are said to be competent. Some bacterial strains are highly
competent during one or more phases of growth under normal laboratory
conditions, whereas others require special treatments to be rendered
competent.
In E. coli, transformation is only possible after inducing competence through
Ca++ treatment and subjection to heat shock.
The process of transformation can be a modification of the following:
1. Development of competence;
2. Binding of ds DNA molecules to bacteria;
3. Uptake of ds into the bacterium followed by its degradation into single
strand;
4. Coating of ss molecules with a specific protein that protects DNA from
nucleases.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 116 OF 153
5. Integration by recombination of the transformed DNA strand into recipient
chromosome at the point at which it is homologous.
6. Replication of integrated DNA segment and segregation of recipient and
donor alleles in progeny bacteria to yield transformed DNA clones with new
phenotypes.
2. Conjugation
1) The discovery of bacterial gene transfer
Can bacteria exchange hereditary information?
Joshua Lederberg, a 21-year graduate student supervised by Edward Tatum
provided the proof in 1946. They observed that two nutritional mutant
strains of E. coli K-12 have the ability to complement each other by
transferring genetic material during cellular contacts.
Illustration:
Strain A could only grow on a medium supplemented with the amino acid
methionine (met) and the vitamin biotin (bio) (genotype: met- bio- thr+ leu+
thi+);
Strain B could only grow on a medium supplemented with acids threonine
(thr) and leucine (leu), and vitamin thiamine (thi) (genotype: met+ bio+ thrleu- thi-).
Experimental set-up:
Cultures of the two strains growing on supplemented minimal media were
set on plates of minimal media as follows:
Plate 1 - strain A,
Plate 2 - strain B,
Plate 3 - a mixture of the two strains.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 117 OF 153
Observations were made for any colony formations.
Results:
Plate 1- No colonies (i.e. no growth),
Plate 2- No colonies (i.e. no growth,
Plate 3 - A few colonies were observed, and they arose at a frequency of
about 1 in 10 million cells (i.e.1 x 10-7).
Interpretation:
Because only prototrophic colonies, met+ bio+ thr+ leu+ thi+, are able to
grow on minimal medium, the appearance of some colonies suggested that
is recombination of genes had occurred.
The question then arose on the origin of prototroph colonies whether from
mutation, transformation or recombination by some form of sexual union.
Mutation was ruled out as a source of recombination because it should have
occurred in all plates.
Conclusion:
The wild-type colonies were produced by an exchange of genetic material
between the two strains i.e. E. coli has a sexual phase that can bring
together the chromosomes of the two different cells. Crossing over could
then place in one chromosome good copies of all its necessary genes.
2) The discovery of the fertility factor, F.
In 1953, William Hayes, a British geneticist, and Jacob and Wollman, two
French geneticists obtained evidence that the genetic exchange in E. coli is
unidirectional, not reciprocal. This was through the following study:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 118 OF 153
Materials
Used two nutritional mutant strains, A and B.
Strain A: met- thr+ leu+ thi+ (auxotrophic for met allele),
Strain B: met+ thr- leu- thi- (auxotrophic for thr, leu and thi alleles).
Experimental procedure:
1. Strain A culture treated with streptomycin (an antibiotic that does not kill
but prevents cells division), then mixed with strain B and observed for
growth;
2. Strain B culture treated with streptomycin and then strain A added to it
and observed for growth.
3. The 2 strains mixed together without antibiotic, then plated on minimal
medium.
Results
There was growth in experiment 1 with colony frequency similar to that in
experiment 3 but none in experiment 2.
Explanation
The division of strain B and not of strain A was necessary for production of
colonies. The donor could still transfer genes even after exposure to the
antibiotic but the recipient could not divide to produce colonies if it had been
exposed to the antibiotic.
Conclusion
Genetic transfer in bacterial cells is non-reciprocal; cells of A donate genetic
material to cells of B, which then can divide to produce colonies containing
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 119 OF 153
genes from both strains. Therefore, A is the donor and B the recipient. They
termed the process conjugation.
Bacterial cells carry besides the main chromosome, one or more small DNA
molecules in the cytoplasm called plasmids. Of the various kinds of plasmids,
a few are involved in conjugation and are called conjugative plasmids. One
such conjugative plasmid is the sex element or fertility factor or F factor.
Therefore, in a subsequent study, Hayes discovered this factor (i.e. F). The
donor cells have it (and are designated F+) while the recipients lack it (and
are F-). The F+ cells are potentially male since they cannot transfer their
genes to the F- cells. This F factor can be autonomous or be integrated. In
the latter case it is referred to as Hfr (High frequency of recombination) and
in this instance can donate its genes. The F is seldom transferred. Thus
unless the entire chromosome is transferred, the ex-conjugant (the recipient
cell after conjugation is completed) of Hfr x F- is always F-.
The newly introduced DNA strand recombines with the DNA of the recipient
cell. Subsequent DNA replication and cell division give rise to a new
recombinant cell that has characteristics derived from each of the parental
cells (Fig. 2). The partial diploid cells after conjugation are called
merozygotes. The relative frequency with which male gametes are
incorporated into the recombinant chromosome is a measure of how close
they are to the entering end.
3. Transduction
Most phages contain DNA. During growth and reproduction, some phages
incorporate host cell DNA into their own DNA. If the phage then infects
another cell, the DNA of the first host cell may be transferred to the second.
This is transduction and was demonstrated by J. Lederberg and Z. Norton,
his postgraduate student in 1951 (Fig. 3).
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 120 OF 153
They used two auxotrophic strains of Salmonella typhimurium.
Strain 1 (LA 22): tyrStrain 2 (LA 2):
tyr+
phe-
trp-
met+
his+,
phe+ trp+ met-
his-.
When they were separately plated on minimal medium, there was no
growth.
Then using a Davis U-tube each arm had one of the two strains (diagram).
After a time, cultures from each arm were removed and examined for
growth in minimal media.
Results
The culture of LA22 grew while that of LA2 did not. The transducing agent
was a phage P22, that is transient in LA22 but lytic to LA2. When it attaches
to a new host cell the fragment of bacterial chromosome in injected into the
cell. It then can engage in crossing over with the host chromosome
ultimately leading to the production of a genetically altered bacterium. For
example a P1 particles grown on E. coli able to grow on lactose has a
number of lactose genes. Addition of such phages to an E. coli unable to use
lactose (Lac-) transforms some Lac- bacteria to the Lac+ form by means of
genetic recombination. However, it is generally a very rare process.
Phages are categorised into special (restricted) and general types. The
special types pick up specific regions of the host bacterial chromosome,
e.g. in E. coli on a region that includes a cluster of genes controlling
galactose fermentation. General transducing phages acquire DNA randomly
from the bacterial chromosome, e.g.22 on Salmonella.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 121 OF 153
9.5.3 Mapping the bacterial chromosome
Bacterial gene mapping is done somewhat differently from higher organisms.
Methods include interrupted conjugation, uninterrupted conjugation,
complementation mapping, mapping by deletion mutants.
1. Interrupted conjugation
Interrupted mating is a technique used to map bacterial genes by
determining the sequence in which donor genes enter recipient cells.
Conjugation is disrupted after specified time intervals. It was pioneered by
Francois Jacob and Ellie Wollman (1952). They used the Hfr strain- a strain
in which the F factor was integrated in the chromosome. This strain
therefore acted as the donor. They noted that when the Hfr strain
conjugated, the genes closely behind the leading part of the F plasmid were
transferred efficiently but the trailing genes made it less frequently since the
conjugation tube was more likely to break before they made it though. They
devised a method to control it. The strains used were:
Hfr: Strs a- b+ c+ d+ ,and
F-: Strr a- b- c- d-.
These were put together for some time. At specific time intervals (e.g. 5, 10
minutes), they removed samples. Each such sample was then put into a
warring blender for a few seconds to disrupt the mating cell pairs and then
plated on minimal medium containing streptomycin. On such medium all
unmated Hfr cells were killed, unmated F- cells were unable to grow. This
treatment therefore allowed only F- cells with recombinant genotype to form
colonies. The strr cells then were tested using selective media for the
presence of marker alleles from the donor. Transfer of donor allele of each of
these steps is obviously dependent on the time that conjugation is allowed
to continue.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 122 OF 153
Results
i) They found that when the genetic transfer was interrupted after a few
minutes, some of the genetic markers would make it across, while others
always remained behind.
ii) As the periods of mating were extended, more and more markers were
transferred and could be recovered in the progeny.
iii) The markers that appeared with greater frequency obviously were nearer
the leading end of the length of the DNA. The frequency of appearance fell
off for those that were more distant. In this way, the gene sequence could
be mapped. This experiment also showed that bacterial genes occur in a
definite sequence, the sequence in which they are transferred into the F- cell.
This information was useful in the construction of a gene map using as a
measure of "distance" the time at which the donor alleles first appear after
mating. The units of distance in this case are minutes. Thus if marker b+
begins to enter the F- cell 10 minutes after a+ had entered, then a+ and b+
markers are 10 minutes (units) apart on the genetic map. Their study
involved the following cross:
Donor:
Hfr H
strs
thr+
leu+ azir
tonr
lac+
gal+
strr
thr
leu
tons
lac
gal
Recipient:
F-
azis
The Hfr H strain is prototrophic and is sensitive to the antibiotic
streptomycin. The F- (strain) cell carries a streptomycin resistance gene and
also a number of mutant genes. Thus it was auxotrophic for threonine (thr)
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 123 OF 153
and leucine (leu), to be sensitive to sodium azide (azis) and infection by the
phage T1 (tons), and to be unable to ferment lactose (lac) or galactose (gal)
as a carbon source. Following the above procedure, they produced the gene
map below:
Minutes
0
5
10
azi
Origin
15
ton
20
lac
25
gal
The genetic map obtained through analysis of interrupted matings is the
same as that arrived at by analysis of frequencies of various recombinant
classes.
Note: Interrupted mating is however useful for physically mapping genes
separate by large distances but is not at all useful for mapping markers
separated by only 1 or 2 minutes on the chromosome. These can be ordered
and mapped by recombinational analysis.
Example
An Hfr strain carrying the prototrophic markers a+ b+ c+ is mixed with a Fstrain carrying the auxotrophic alleles a, b, c. Conjugation was interrupted at
5 minutes intervals and plated on media which revealed the presence of
recombinants.
Time (minutes)
Recombinants
5
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
a b+ c
PAGE 124 OF 153
10
a b+ c+
15
a+ b+ c+
Determine the gene order.
Solution:
The order of the genes in the Hfr donor strain is b+ c+ a+
b is < 5 time units from the origin;
c is < 10 time units from b,
a is < 10 time units from c.
9.6 Phage Genetics
9.6.1 Introduction
Depending upon the host which a virus infects, they can be broadly placed in
three categories: (i) animal viruses, (ii) plant viruses and (iii) bacterial
viruses or bacteriophages. The simplest of these three classes are
bacteriophages.
Bacteriophages are also a class of viruses, which have been most
extensively utilised for study of genetics of viruses. Contemporary with the
discovery of sexuality in bacteria in 1946 by Lederberg and Tatum, genetic
recombination was also demonstrated between different strains of
bacteriophages by M. Delbruck, W.T. Bailey and A.D. Hershey. These
workers were also able to detect and artificially induce mutations in these
organisms. Considerable genetic work has since been undertaken by these
and other workers using bacteriophages. Drs. A.D. Hershey, M. Delbruck
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 125 OF 153
and S.E. Luria shared the 1969 Nobel Prize in medicine for their contribution
to replication and recombination in viruses.
9.6.2 Reasons for suitability of phages as genetics research tools
(1)
They have a very rapid rate of multiplication. After infecting a
host cell, it takes about 20 minutes to lyse it releasing hundreds of
phages into the environment.
(2)
For the reasons above, phages can be used to observe and score
rare genetic events, such as mutation and recombination. For
example, if a cell is double infected with two genetically different
strains of a particular phage, then the two parental types, together
with both types of recombinant, can be recovered only after 20
minutes. By plating these progeny phages on a suitable host strain of
bacteria, the parental and recombinant types may be recognised after
24 hours or less. A similar experiment with Drosophila would take at
least 3 – 4 weeks, and over a year with peas!
(3)
Simple size and simplicity of genetic organization. The average
phage genome is very small. For example, the genome of phage X174
has only got 9 genes whilst that of lambda (λ) has less than 60 genes.
This fact means that it is possible to identify practically all the genes
within a viral genome. Therefore, one can investigate and understand
the organization and regulation of an entire genome.
Aside from their utility in the study of gene expression, phage
genetics has been put to practical use as well. Cloning of the human
insulin gene in bacteria was accomplished using a bacteriophage as a
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 126 OF 153
vector. The phage delivered to the bacterium a recombinant plasmid
containing the insulin gene.
Viruses consist of a genome (single stranded or double stranded RNA or
DNA) enclosed in a capsid. They replicate using the metabolic machinery of
their specific host cell.
Phages may be virulent and follow the lytic cycle or temperate with a
lysogenic replication cycle.
9.6.3 Phage Crosses
Viruses can also transfer some of genetic material to one another, a kind of
sexual reproduction.
That recombination can occur in some phages was first demonstrated in
1946 by Max Delbruck and Bailey. They used two virus types: T2 and its
mutant. The phage T2 forms small colonies (plaques) due to slow lysis of the
host, and thus designated T2r+; Its mutant (T4r) forms large, circular
plaques due to rapid lysis of host bacterium. By simultaneously infecting E.
coli strain B and examining the lysate, progeny T2r and T4r+ was found.
These are new types (recombinants). Their frequency ruled out back
mutation. A transfer of genes controlling plaque type had taken place
between these strains.
Since recombination is possible even in viruses, their genomes can as well
be mapped.
1. Recombination in phages
After Delbruck and Bailey's studies, Seymour Benzer also studied genetic
exchange in viruses and made his findings in 1953. He worked with the
bacterium E. coli and the phage T4 and is class of mutants that produces a
phenotype called rapid lysis, r. There are 3 genes, I, II and III in the фT4
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 127 OF 153
genome that can mutate to rapid lyser forms symbolised as rI, rII and rIII.
Benzer chose and worked with the rII locus basically because all rII alleles
are conditional lethals, i.e. an rII mutant will grow and form large plaques on
E. coli strain B but it will not grow at all on strain K12. In contrast, rII+
(normal or wild-type) phages will grow and form small, ragged plaques on
both strains (below).
This conditional growth was to prove essential for identifying recombinants.
Plaque phenotypes produced by different combinations of E. coli and phage
strains
T4 phage strain
E. coli str. B
E. coli str. K 12
rII
Large, round
No plaque
rII+
Small, ragged
Small, ragged
Benzer started with a sample of 8 independently derived rII mutant strains
and set about crossing them in all possible combinations by double infection
(mixed infection) of the B strain, as below:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 128 OF 153
rII-- allele
rII- allele
E.coli str B
Plate on E. coli K12
Results:
(i) rII- won't grow, i.e. is parental
(ii) rII-- won't grow, i.e. parental
(iii) rII+ will grow, i.e. forms plaques*
* The formation of plaques indicates that recombination has occurred within a gene and not
just between genes.
The frequency of plaques was too large for them to be due to back
mutations. Furthermore, the alleles could be mapped unambiguously to the
right or left of each other to give a gene map as below with the map units
being the frequency of rII+ plaques, i.e.
Map distance =
2 𝑥 𝑛𝑜.𝑜𝑓 𝑟𝐼𝐼 𝑝𝑟𝑜𝑔𝑒𝑛𝑦
𝑡𝑜𝑡𝑎𝑙 𝑛𝑜.𝑜𝑓 𝑝𝑟𝑜𝑔𝑒𝑛𝑦
x 100
The result was an intragenic map below:
rII1
I
rII7
I
rII4
I
rII5
rII2
I
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
rII8
I
rII3
I
I
PAGE 129 OF 153
rII6
I
Mixed infection therefore provides a technique for testing functional and
structural allelism between different mutants of a phage in pairwise
combinations.
If there was no growth this suggested functional allelism, and if there was
growth, complementation in trans arrangement.
Benzer then isolated more rII mutants and carried out double infection on E.
coli K12. Benzer was able to categorise these mutants into complementary
groups, he arbitrarily called A and B. Any member of group A will
complement any member of group B, but no complementation is possible
between any pairs of alleles in the same group. Mutations that fail to
complement must be affecting the same unit of function.
2. Mapping
This requires consideration of deletions (mutants) and therefore an
understanding of the technique of complementation or helping.
(a) Complementation
Two mutants complement each other if each one can supply a function the
other lacks. A complementation test therefore defines genes in any organism
or virus. Recessive mutations with similar phenotypic effects can be tested
for functional allelism by determining the phenotypes of the cis and trans
heterozygotes (Fig. below).
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 130 OF 153
cis test
Mutant sites in the same cistron
Ix
x
I
I
I
I
I
A
Mutant sites in different cistrons
x
I
I
I x
I
B
I
x
I
A
trans test
I
I
xI
I
A
B
I
I x
I
I
I
I
B
A
I
x
I
B
x = mutant sites
A, B = parts of rII locus (cistrons).
A cistron is a genetic region within which there is no complementation
between mutations.
If two different rII mutants of the phage T4 infect the same bacterium (trans
configuration) and no growth of phage occurs, the mutants are said to be in
the same cistron. But if complementation occurs (i.e. progeny phage grow),
the mutants are in different functional units (lower right box above). In the
upper right box, only the B function can be expressed normally.
If the trans heterozygote shows a wild-type phenotype, the mutants are
supposed to have occurred in different alleles of the same gene. Cis
heterozygotes always show a wild phenotype. Thus mutations are allelic if
cis and trans arrangement result into different phenotypes. They are nonallelic if both these arrangements result into the same (wild-type)
phenotype. When two mutations in the trans position restore the wild
phenotype, they are said to complement with each other.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 131 OF 153
Using complementation data, complementation maps can be drawn in which
the extent of non-complementation is represented by a straight line and
complementation by the absence of an overlap between these lines. In the
case of Benzer, he confirmed complementation among various rII mutants of
T4.
Worked out Examples
(1) Two independent mutations of the T4 have been isolated. Both lack the
ability to form plaques in bacterial cultures. In an attempt to determine
whether these mutants were allelic or simply different structural genes
responsible for similar phenotypes, cross infections of bacterial cells were
initiated. The cultures were examined for plaques, and were noticed. Are
these two mutations alleles of the same locus?
Answer
Since neither mutation by itself can produce plaques in a bacterial
culture, their phenotype is indeed similar. Simultaneous infection leads to a
diploid state of phage chromosome within the same cell.
If mutations are alleles of one another at the same locus, neither
chromosome will have the ability to produce plaques. If the mutations are
not alleles, but each represents a mutation of different structural genes, loci
A and B, then when the diploid state is present, there will be a wild type
allele present for both locus A and locus B, each complementing one another
to produce plaques. The presence of plaques identifies that these two
independent mutations are not alleles of one another.
(2) Seven new mutations were identified in the bacterium E. coli during
some sexual conjugation tests for recombination. Each of the mutations
prevented normal metabolism of galactose and were identified as galA, galB,
galC, galC1, galC2, galD and galD1. When partial diploids in all pairwise
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 132 OF 153
combinations were examined by complementation tests, the following results
were obtained (+ = normal complementation; 0 = no complementation).
Mutant
A
B
C
C1
C2
D
D1
A
0
+
0
0
+
0
+
0
+
+
0
+
+
0
0
+
0
+
0
+
0
+
0
+
+
0
+
B
C
C1
C2
D
D1
0
Questions:
a) How many loci are there?
b) Which mutations are alleles of one another?
Solution
i. Mutant A complements B, C2 and D1 but not C, C1 and D.
Thus
A, C, C1, D;
B, C2, D1
ii. Mutant B complements A, C, C1, D and D1 but not C2.
Thus
A, C, C1, D; B, C2; D1
iii. Mutant C complements B, C2 and D1 but not A, C1 and D
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 133 OF 153
Thus the loci are as above.
iv. Mutant C1 complements B, C2 and D1 but not A, C, and D.
Thus loci are as above.
v. Mutant D1 complements all the other mutants.
Thus loci are as above.
Therefore:
a) There are 3 loci
b) A, C, C1 and D, are alleles of one locus; B and C2 are alleles of a second
locus and D1 is the only gene for the third locus.
(3) Six mutations are known to belong to three cistrons. From the results of
the complementation tests, determine which mutants are in the same
cistron.
1
2
3
4
5
6
1
2
3
0
+
+
0
0
4
5
+
+
+
0
0
6
+
0
+
0
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 134 OF 153
Key: + = complementation; 0 = non-complementation; blank = not tested
Solution
i. Obviously mutations 3 and 5 are in the same cistron - they fail to
complement each other.
ii. Mutations 1 and 3 are in different cistrons - they complement each other.
Thus are assigned as below to two cistrons, A and B.
Cistron A
Cistron B
Cistron C
iii. 1 and 2 are in different cistrons, but we do not know whether 2 is in A or
C. However, 5 and 2 complement and therefore cannot be either in cistron A
or B and thus must be in C.
Cistron A
Cistron B
Cistron C
iv. 3 and 4 complement, thus 4 must be in either B or V. But 2 and 4 also
complement, thus 4 cannot be in C and must reside in B.
Cistron A
Cistron B
Cistron C
v. Mutant 6 cannot be in A since it complements with 5. Thus 6 is either in B
or C. Since 6 and 4 complement, they are in different cistrons. If 6 cannot
be in A or B, it must be in C. The mutants are thus grouped:
Cistron A
Cistron B
Cistron C
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 135 OF 153
3. Six point mutants are known to reside in 3 cistrons. Complete the
following table (with + for complementation; 0 = non-complementation)
1
1
2
0
+
2
0
3
3
4
5
6
+
+
+
+
0
4
+
0
0
5
0
6
+
0
Answer:
Cistron A
Cistron B
1
Cistron C
2
3
4
5
6
+
+
+
+
+
0
+
+
0
+
0
0
+
0
0
+
0
5
0
+
6
0
1
2
3
4
0
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 136 OF 153
Deletion mapping
Besides using the properties of mutational sites for intragenic mapping,
Benzer introduced deletion mapping to enhance understanding of the finer
gene structure.
Deletions are mutations that result from the elimination of segments of DNA.
Mutations cannot be converted into wild-types in recombination, because the
DNA corresponding to the wild-type region for that particular mutation is no
longer present.
To understand deletions, take an analogy with record tapes. Given 3 tapes,
each with a blemish or deletion in a different place labelled A, B and C.
Imagine these deletions are so located that deletion B overlaps deletions A
and C but that A and C do not overlap each other. In this case only
recombining A and C can recreate a good performance.
A deletion is shown as a bar because it represents the loss of several sites. If
the nucleotides of a gene are: a, b. c, etc., then each point mutation is a
change in one nucleotide but a deletion is the loss of a whole series of
nucleotides, viz: a b c d e f g h; a b c d e f g h represent point mutations,
whereas a b c f g h represent a deletion. A number of rII mutants are
deletion mutants, which are missing small segments of the genome. Point
mutants can back-mutate to wild-type, but deletion mutants never do
because it would be nearly impossible to add back the right sequence of
nucleotides by accident.
Where comparable deletions are contained in mutants, recombination
experiments do not yield wild types. Mutants that do not revert to the
standard type when they reproduce contain same or overlapping deletions.
To test thousands of mutants through cross infection would be cumbersome.
The way-out would be to use mutants of the non-reverting type - whose
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 137 OF 153
deletions divide up the rII region into segments. Each point mutation is
tested against these reference deletions. The recombination test gives a
negative result if the deletion overlaps the point mutation and a positive
result if it does not overlap. In this way a mutation is quickly located within
a particular segment of the map.
a. Test mutants
rII region
Test mutant A:
Deleted
Test mutant B:
Deleted
Test mutant C:
Deleted
Test mutant D:
Del
b. Point mutations testing, e.g. using mutant X.
Test mutant A:
Mutant X
*
No standard recombinants
Test mutant B:
Mutant X
*
No standard recombinants
Test mutant C:
Mutant X
*
Standard recombinants possible
Test mutant D:
Mutant X
*
Standard recombinants possible
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 138 OF 153
Benzer realized that deletions could be used for the rapid location of
mutational sites in newly obtained mutants.
Illustration
Consider the following gene map showing 12 identifiable mutational sites:
1
2
3
4
5
6
7
8
9
10
11
12
I
I
I
I
I
I
I
I
I
I
I
I
i. One special mutant D1 fails to give rII+ recombinants when crossed with
mutants carrying altered sites 1, 2, 3, 4, 5, 6, 7, or 8. Therefore, D1 behaves
as if carries deletions of sites 1 to 8.
9
10
11
12
I
I
I
I
ii. Another special mutant D2 fails to give rII+ recombinants when crossed
with mutants carrying altered sites 5,6,7,8,9,10,11 or 12. Therefore D2
behaves as if it involves a deletion of sites 5 to 12.
1
I
2
3
I
4
I
I
These overlapping deletions now define 3 areas of the gene, designated i, ii
and iii.
i
ii
iii
D1 =====================---------------------------BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 139 OF 153
D2 ---------------------=========================
Thus,
(i)
A new mutant that gives rII+ recombinants when crossed with D1
but not when crossed with D2 must have its mutational site in
area iii;
(ii)
One that gives rII+ recombinants when crossed with D2 but not
with D1 must have its mutational site in area I;and
(iii)
A mutant that does not give rII+ recombinants with either D1 or
D2 must have its mutational site in area ii.
Deletions themselves can be intercrossed and mapped just like point
mutations. A bar represents the deleted region. If no wild-type recombinants
are produced in a cross between different deletions, then the bars are shown
as overlapping.
Worked out Examples
(1) The following deletion map shows four deletions (1-4) involving the rIIA
cistron of T4:
1 -------------2 -----------------3 ----------------------4 --------------------------Five point mutations (a-e) are tested against these 4 deletion mutants for
their ability to give wild-type (r+) recombinants; the results are:
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 140 OF 153
a
b
c
d
e
1
+
+
+
+
+
2
+
+
+
-
-
3
+
-
+
-
-
4
-
-
+
-
-
What is the order of the point mutations?
Answer
The key principle here is that point mutations can recombine with deletions
that do not extend past the mutation but cannot recombine to yield wildtype phages with deletions that do extend past the mutation.
Looking at the test results given in the problem, any mutation that
recombines with del. 1 must be to the right of the deletion, any mutation
that recombines with deletion 2 must be to the right of deletion 2, etc.
Therefore,
(i) Consider point mutation “a”. It recombines with del. 1, 2 and 3 but not
with del. 4. Therefore it is to the right of deletions 1, 2 and 3 but not to the
right of del. 4. We can then easily place point mutation “a” in the interval
between deletions 3 and 4.
(ii)Point mutation “b” recombines with deletions 1 and 2 and therefore must
be to the right of them. It does not recombine with deletions 3 and 4, so it is
in the interval between deletions 2 and 3.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 141 OF 153
iii. Point mutation “c” recombines with all the deletions and thus is to the
right of all deletions.
iv. Finally both point mutations “d” and “e” recombine only with deletion 1
and must therefore be in the interval between deletions 1 and 2.
The results therefore are as follows:
1. --------------- e, d
2. ---------------------- b
3. ---------------------------- a
4. ----------------------------------- c
(2) In a phage, a set of deletions is intercrossed in pairwise combinations.
The following results are obtained (a + = wild-type):
1
2
3
4
5
1
-
+
-
+
-
2
+
-
+
+
-
3
-
+
-
-
-
4
+
+
-
-
+
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 142 OF 153
5
-
-
-
+
-
Using this table construct a deletion map
Solution
If a "+" implies non-overlap and a "-" an overlap,
Then: i. 1 overlaps 3 and 5;
ii. 2 overlaps 5 only;
iii. 3 overlaps all but 2;
iv. 4 overlaps 3 only; and
v. 5 overlaps all but 4.
Results
I
II
III
4
-----------
1
-----------
2
-----------
3
------------------------
5
----------------------(3) Seven deletion mutants within the A cistron of the rII region of phage T4
were tested in all pair-wise combinations for wild-type recombinants. In the
table of results below, + = recombination; 0 = no recombination. Construct
a topological map for these deletions.
1
1
2
3
4
5
6
7
0
+
0
0
+
0
0
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 143 OF 153
2
0
0
0
+
+
0
0
0
+
+
0
4
0
+
0
0
5
0
0
0
0
0
3
6
0
7
Solution
Note: if any two deletions overlap to any extent, no recombinants (+). Thus,
(i) Deletion 1 overlaps with 3,4,6,7 but not with 2 or 5. Thus:
1
2, 5
----------
------------
3,4,6,7
-----------
(ii) Deletion 2 overlaps with 3, 4 and 7 but not with 1, 5 or 6.
1
2
---------- ----------
5
---------
3,4,7
----------------------6
--------BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 144 OF 153
(iii) Deletion 3 overlaps with 1, 2, 4 and 7 but not with 5 or 6.
1
-----------
2
-----------
6
3
---------
5
------------
---------------4, 7
----------------
(iv) Deletion 4 overlaps with 1, 2, 3, 6 and 7 but not with 5.
1
----------
2
------------
6
5
----------
3
----------
------------------4
--------------------------7
------------------------------
(v) Deletion 5 overlaps with 6 and 7 but not with 1,2,3, or 4.
1
2
-----------
---------
6
3
--------------5
----------
---------------------4
-----------------------------
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 145 OF 153
7
-----------------------------------------
(4) Six deletion mutants in rIIA of T4 were tested in pairwise combinations
for wild-type recombinants. The results are as follows:
1
2
1
2
3
4
5
6
0
0
0
0
0
0
0
0
0
0
+
0
+
0
0
0
+
+
0
+
3
4
5
6
0
Key: + = recombination; 0 = no recombination
Question: Construct a topological map for these deletions.
Solution
1
-------------------------------2
--------------------3
----------------BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 146 OF 153
4
5
-----------
6
--------
--------
(5) The following map shows 4 deletions (1 -4) involving the rIIA cistron of
T4:
1 ----------c
e
2 ------------------d
3
a
-------------b
4
------
Five point mutations (a-e) in rIIA are tested against these 4 deletion
mutants for their ability to give r+ recombinants, with the following results:
a
b
c
d
e
1
+
+
-
+
+
2
+
+
-
-
-
3
-
-
+
-
+
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 147 OF 153
4
+
-
+
+
+
What is the order of the point mutants?
(See above, bold letters; i.e. order is c e d a b).
(6) The following 5 histidine-requiring mutations of Neurospora were studied
and allelic relationship determined thus:
a. No complementation occurs between CD-16 and the other mutations.
b. Mutants 245 and 261 complement and also complement with D-556 and
1438 but not with CD-16.
c. Mutants D-556 and 1438 do not complement with each other but
complement with 245 and 261.
Complementation matrix:
CD-16
245
CD-16
245
261
D-556
1438
0
0
0
0
0
0
+
+
+
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 148 OF 153
261
0
D-556
+
+
0
0
1438
0
Key: 0 = no complementation; + = complementation
Construct a gene map of this histidine locus
Answer
From the various deletion tests, 3 cistrons can be categorised. Orders and
relative positions of various sections can be precisely determined by isolating
mutations, which overlap two adjacent sections.
(i) Complementation map
I
CD -16
245
II
III
=====================================
============
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 149 OF 153
261
===============
D-556
=========
1438
=========
Genetic map
CD-16
245
261
D-556
1438
4. Abortive transductants are relatively stable merozygotes that can be used
for complementation tests. Six mutants were tested in all pairwise
combinations, yielding the results shown below (+ = complementation, 0 =
non-complementation). Construct a complementation map consistent with
the data.
1
2
3
4
5
1
2
3
4
5
6
0
+
0
+
+
+
0
0
+
+
+
0
+
+
+
0
0
+
0
0
6
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
0
PAGE 150 OF 153
Answer:
1
2
______
_______
3
_______________
4
______
6
________
5
__________________
Topic Summary
In this topic you have learnt all fundamental aspects of bacteria and viruses
that make them essential tools in molecular genetics, especially in our
understanding of the structure and function of genes. Though largely
reproducing asexually, it was shown that genetic exchange could occur in
bacteria through conjugation, transduction and transformation. It was also
shown that viruses too could exchange genetic materials leading to creation
of recombinants. Such was responsible for genetic diversity in the two. Using
genetic systems, Benzer was able to map the details of the gene and thus
the fine structure of the gene studies. From the recombinants generated in
both bacteria and viruses, gene making in prokaryotes was demonstrated.
Further Reading
(1) Russell, P.J. (1992). Genetics (3/e). HarperCollins Publishers, NY
(2) Suzuki, D.T., Griffiths, A.J.F., Miller, J.H., Lewontin, R.C. (1986). An
Introduction to Genetic Analysis (3/e). W.H. Freeman & Co., NY
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 151 OF 153
Topic Activities
1. What reasons make viruses and bacteria suitable tools for molecular
work?
2. What are the various modes of genetic exchange available to a
bacterium?
3. Explain how it was shown that genetic exchange occurs in bacteria and
viruses.
4. How was genetic exchange in bacteria shown to be unidirectional?
5. Benzer infected strain K of E. coli with rII mutants two at a time and
plaque-assayed the lysates obtained. The following results were obtained.
47
51
101
102
104
47
51
101
102
104
106
0
+
0
+
0
0
0
+
0
+
+
0
+
0
0
0
+
+
0
0
106
0
Question: Does the rII region consist of one, two or more cistrons? Explain.
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 152 OF 153
BOTA 413: MOLECULAR AND MICROBIAL GENETICS
PAGE 153 OF 153