Chapter 3ppthorti

CHAPTER -3
The Genetic Material
Plant Biotech By:Kassahun A.1
 Genetics is the scientific study of heredity and the
variation of inherited characteristics.
 It includes the study of:
 Genes,
 How they function,
 Interact, and
 Produce the visible and measurable
characteristics we see in individuals

1. Discovery of DNA
 It became evident that chromosomes were the
organelle of heredity;
 Various attempts were made by early molecular
genetists
 They attempted the physical and chemical nature of
heredity materials.
 Genetic material for most organisms has been
developed by different discoveries.

Evidences to DNA as Genetic
Material
1. Frederick Griffith’s experiments(1928)
2. Avery, MacLeod, and McCarty
experiment(1940)
3. Hershey Chase experiment(1952)
4. Meselson and Stahl experiment (1958)

1.Frederick Griffith’s experiments
 Griffith made experiments on Streptococcus
pneumoniae in 1928.
 Griffith used two strains.
 Smoot type strain identified as S. virulent
type
 The other strain was a mutant non-virulent
type. rough; this strain is called R.
 Griffith killed some virulent cells by boiling
them.
 He injected the heat-killed cells into mice.
The mice survived,

1.Frederick Griffith’s experiments
 Mice injected with a mixture of heat-killed
virulent cells and non-virulent cells did die.
 Live cells could be recovered from the dead
mice;
 These cells gave smooth colonies and were
virulent on subsequent injection.
 The cell debris of the boiled S cells had
converted the live R cells into live S cells.
 This process is called transformation.
 Griffith did not know what the transforming
substance was.

Figure 3.1 Griffith’s experiments on genetic transformation in
pneumococcus.

2.Avery, MacLeod, and McCarty
experiment
 At the time of 1940s, researchers knew that:
 DNA,
 RNA,
 Proteins, and
 Carbohydrates are major constituents of
living cells.
 To determine if any of them was the
genetic material, Avery, MacLeod, and
McCarty used
 Biochemical purification procedures and
 They Prepared bacterial extracts from
type S strains containing each type of

experiment
 They treated samples of the DNA extract with enzymes
that digest:
DNA (called DNase),
 RNA (RNase), and
 Protein (protease).
 When the DNA extracts were treated with RNase or
protease, they still converted type R bacteria into type
S.

experiment
 These results indicated that any remaining RNA or
protein in the extract was not acting as the genetic
material.
 However, when the extract was treated with DNase, it
lost its ability to convert type R into type S bacteria.
 These results indicated that the degradation of the DNA
in the extract by DNase prevented conversion of type R
to type S.

Figure 3.2 Experimental protocol used to identify the transforming
principle.

3.Hershey Chase experiment
 Avery and his colleagues were definitive,
 But scientists were reluctant to accept DNA as genetic
material.
 Additional evidence was provided in 1952 by Alfred
Hershey and Martha Chase.
 They used phage T2, a virus that infects bacteria.
 And reasoned phage must inject into the bacterium the
specific information.

 They find out what material the phage was injecting
into the phage host,
 They decided to label the DNA and protein by using
radioisotopes
 Phosphorus is not found in proteins but in DNA;
 Sulfur is present in proteins but never in DNA.
 Incorporated radioisotope of 32P into phage DNA
 35S into the proteins of a separate phage culture.

 They infected two E. coli cultures with virus particle cell:
E. coli culture received phage labeled with
32P,
 The other received phage labeled with
35S.
 After sufficient time for infection, they sheared the
empty phage carcasses off the bacterial cells by
agitation in a kitchen blender.

 They separated the bacterial cells from the
phage ghosts in a centrifuge and
 Then measured the radioactivity in the two
fractions.
 32P-labeled phages were ended up inside the
bacterial cells, indicating phage DNA entered
cells.
 35S-labeled phages were ended up in the
phage ghosts, indicating the phage protein
never entered the bacterial cell.
 conclusion is : DNA is the hereditary material.

Figure 3.3. The Hershey–Chase experiment demonstrated that the genetic material of
phage is DNA, not protein.

4.Meselson and Stahl experiment
(1958)
 Complementary strands model of DNA,
proposed by Watson and Crick in 1953.
 Three possible mechanisms for DNA replication:
 Semi-conservative,
 Conservative, and
 Dispersive
 Semi-conservative model proposes the two
strands of a DNA molecule separate during
replication.
 Then strand acts as a template for synthesis of a
new, complementary strand.

(1958)
 Conservative model proposes that the entire DNA
duplex
 Acts as a single template for the synthesis of an
entirely new duplex.
 Dispersive model has the two strands of the double
helix
 Which are then replicated and reassembled, with the
new duplexes containing alternating segments from
one strand to the other.

Figure 3.4: The three models of DNA replication possible from the
double helix model of DNA structure

(1958)
 Meselson and Stahl used different isotopes of Nitrogen.
 Nitrogen (14N) is the most abundant natural isotope,
 Nitrogen (15N) is rare, but also denser.
 Neither is radioactive; each can be followed by a
difference in density
 “light” 14 vs “heavy”15 atomic weight in a CsCl density
gradient ultra-centrifugation of DNA.

(1958)
 The experiment starts with E. coli grown for several
generations on medium containing only 15N. It will have
denser DNA.
 When extracted “heavy” DNA will move to bottom of the
tube in the more dense solution of CsCl.
 DNA extracted from E. coli grown on normal, 14 N
containing medium will migrate more towards the less
dense top of the tube.

(1958)
 If these E. coli cells are transferred to a medium
containing only 14N, the “light” isotope, and grown for
one generation, then their DNA will be composed of
one-half 15N and one-half 14N.
 DNA is extracted and applied to a CsCl gradient, the
observed result is that one band appears at the point
midway between the locations predicted for wholly 15N
DNA and wholly 14N DNA .

4.Meselson and Stahl experiment (1958)
 This “single-band” observation is inconsistent with
conservative model of DNA replication,
 But is consistent with both semi-conservative and
dispersive models.
 If the E. coli is permitted to go through another
round of replication in the 14N medium, and
 The DNA extracted and separated on a CsCl
gradient tube, then two bands were seen:
 One at the 14N and 15N intermediate position and
 One at the wholly 14N position .

(1958)
 This result is inconsistent with the dispersive model
and thus disproves this model.
 The two band observation is consistent with the
semi-conservative model which predicts one
wholly 14 N duplex and one 14N-15N duplex.
 Additional rounds of replication also support the
semi-conservative model/hypothesis of DNA
replication.

(1958)
 Thus, the semi-conservative model is the currently
accepted mechanism for DNA replication.
 Note however, that we now also know from more
recent experiments that whole chromosomes, which
can be millions of bases in length, are also semi-
conservatively replicated.

Figure:3.5 Meselson And Stahl Experimental Procedure

3.2. Chemical Subunits in DNA and
RNA
 Nucleic acids play an important role in the storage and
expression of genetic information.
 They are divided into two major classes:
 Deoxyribonucleic acid (DNA)
 Ribonucleic acids (RNAs)
 All nucleic acids are made up from:
 Nitrogenous base,
 A sugar and
 A phosphate residue.
 A sugar and a nitrogenous base without the phosphate
group are called nucleoside.

3.2.1. Deoxyribonucleic acid (DNA)
 DNA is composed of:
 Two strands of deoxyribonucleotides (sugar
deoxyribose)
 Phosphate
 Nitrogenous bases:
Adenine (A),
Guanine (G) (both are purines),
Cytosine (C),
Thymine (T) (both are pyrimidines)

Figure 3.5 Nitrogenous basis of
DNA

 In a nucleotide the base is joined to 1’
carbon of pentose by an N-β- glycosyl
bond.
 Phosphate is esterified to 5’ carbon.
 Phosphate of 5’ carbon reacts with –OH
group attached to 3’ ribose sugar
carbon.
 During this bond formation a water
molecule is removed.

Cont…

 The phosphates of the DNA backbone are negatively
charged, and
 This will allow proteins that have positively charged
domains to bind to the DNA.
 The deoxyribonucleotides of each strand are paired
through specific hydrogen bonding :
 A always pairs with T via two hydrogen bonds,
 G always pairs with C via three hydrogen bonds.

The Properties of DNA
 1. DNA stores and transmits genetic information.
 a. the sequence info can be converted into new DNA, new RNA, and
protein.
 Specific sequences are involved in regulation.
 b. DNA is copied with high-fidelity: only one mistake per every 109 -
1010 bases
 2. DNA is stable
 a. Very resistant to chemical attack
 1. Resistant to alkali treatment (RNA is not)
 2.Resistant to low pH, however it will begin to hydrolyze below pH 2
 b. Its double-strandedness insures that the information is redundant.
 c. Double-strandedness also shelters the bases from chemical
attack. Very hydrophobic internal environment.

The Properties of DNA
 D. One exception is that cytosine is somewhat
unstable; can deaminate and turn into uracil.
 1. 100 cytosines/human genome/day do this!
 2. Thus there is DNA excision/repair system that
always removes uracil from DNA.
 3. DNA does change, it evolves.
 a. This process is called mutation.
 b. It is caused by either a chemical alteration or a
replication error

3.2.2. Ribonucleic Acids (RNA)
 Is the second major nucleic acid in cells,
 Serves as genetic messenger,
 In some retro-viruses RNA is genetic material.
 It is long un branched polymer joined by phosphodiester
bonds in 5’ to 3’ direction.
 RNA differs from DNA in two respects…
 Uracil (U) is found in RNA in place of Thymine (T).
 The sugar units in RNA are ribose rather than
deoxyribose (in DNA).

3.2.2. Ribonucleic Acids (RNA)
 In addition to 3’ - 5’, a 2’ - 5’ linkage is also
possible for RNA.
 2’ – 5’ linkage is important for removal of introns
and joining of exons for the formation of mature
RNA during RNA splicing.
 The presence of OH group at C-2 , RNA is
hydrolysed more rapidly under alkaline
conditions.
 There are four major forms of RNA in cells:
 Messenger RNA (mRNA),
 Transfer RNA (tRNA),
 Ribosomal RNA (rRNA), and

Comparison between DNA and RNA
DNA RNA
Is usual genetic material, in mitochondria
and chloroplast non genetic
Is non genetic material, in some viruses it
is genetic material
Is double stranded except in some phage
viruses
Is single stranded except in some viruses
Pentose sugar is deoxyribose Pentose sugar is ribose
Common basis are AGCT Common basis are AGCU.
DNA is only one type There are different types
DNA molecules can replicate Does not replicate
DNA cant transcribe RNA RNA dose not transcribe except in some
virusis
Contains large number of nucleotides Contains small number of nucleotidesPlant Biotech By:Kassahun A.37

3.3. Structure of DNA
 In the 1950s, Francis Crick and James Watson
determine the structure of DNA.
 Other scientists, such as Linus Pauling and Maurice
Wilkins, were also actively exploring this field.
 Pauling had discovered the secondary structure of
proteins using X-ray crystallography.
 In Wilkins’ lab, researcher Rosalind Franklin was using
X-ray crystallography to understand the structure of
DNA.

 Watson and Crick were able to piece together the
puzzle of the DNA molecule using Franklin’s data.
 Watson and Crick also had key pieces of information
available from other researchers such as Chargaff’s
rules.
 Chargaff had shown that of the four kinds of monomers
(nucleotides) present in a DNA molecule,
 Two types were always present in equal amounts and
the remaining two types were also always present in
equal amounts.

 This was possible only when
 Adenine (A) pairs with thymine (T) and
 Guanine (G) pairs with cytosine (C)
 Adenine forms two hydrogen bonds with thymine
 Guanine forms three hydrogen bonds with cytosine.

 The Watson and Crick structure of DNA (B-DNA)
has following features:
 It consists of two anti-parallel polynucleotide
strands
 Diameter of a double helix will be around 20 Å.
 Each base is hydrogen bonded to a base on
opposite strand (A with T and G with C).
 The ideal B DNA helix has 10 base pairs per turn
and
 The helix rotates 36° per base pair.

 The helix has a pitch of 34 Å.
 So per base pair raise in common axis will be
3.4 Å.
 The double helix has major and minor grooves.
 Fibres of DNA assume the so called B-
Conformation,
 when the counter ion is an alkali metal such as
Na+ and the relative humidity is >92%.
 It is the most stable structure for a random
sequence of DNA and is therefore the standard

 The 10 structure of DNA is simple nucleotide sequence.
 20 structure represents regular, stable structure of the
nucleotides in a nucleic acid.
 Further coiling & complex folding of large chromosome
within eukaryotic chromatin & bacterial are 30 structure.
 DNA can exist in 3 forms A, B & Z.
 only B- DNA and Z-DNA have been directly observed
in functional organism.

3.4. DNA Replication
 Three variants for genetic information transfer
occurring in different organisms
1. Replication
2. Transcription.
3. Translation
 All the types of genetic information transfer are based
on the template mechanism.
 During replication, one of the two DNA chains (or RNA
in viruses) serves as a template.

 In transcription, a DNA section (forward transcription),
 RNA section (reverse transcription), and
 In translation, mRNA, that is, only a nucleic acid is
capable of acting as a template.
 By the early 1950’s, it was clear that DNA was a linear
string of deoxyribonucleotides.
 One could postulate three d/t ways to replicate

 First, cell might have DNA-synthesizing "machine“
 Second, replication could break parental DNA into
pieces.
 Third model could be the DNA structure deduced by
Watson and Crick.
 The complementarily b/n base pairs (A=T and G=C) not
only holds the two strands of the double helix together,
 But the sequence of one strand is sufficient to determine
the sequence of the other.

 These three models make different predictions about the
behavior of the two strands of the parental DNA during
replication.
 In the first, programmed machine model, the two
strands of the parental DNA can remain together,
 This model of replication is called conservative:
 The parental DNA molecules are the same in the
progeny as in the parent cell.

 In the second model, each strand of the daughter DNA
molecules would be a combination of old and new DNA.
 This type of replication is referred to as random
(dispersive).
 The third model, strand of the parental DNA serves as
a template directing the order of nucleotides on the new
DNA strand, is semi-conservative mode of replication

Figure 3.7 Possible models of nucleic acid replication.

Enzymes involved in DNA
replication
1. Helicase
2. DNA dependant RNA plymarase(primer)
3. DNA polymeraseIII
4. DNA plymeraseI
5. SBP (single strand binding protein) cofactors
6. DNA ligase

3.4.1. Molecular fundamentals of
replication
 In 1957, Meselson and Stahl established DNA
replication in living organisms.
 proteins (helicase) break hydrogen bonds b/n the
complementary bases of DNA.
 The untwisted portion of DNA is called the replicative
fork.

replication
Figure 18፡ Replication of DNA: Scheme of the replication fork: a: template,
b:leading strand, c: lagging strand, d: replication fork, e: primer, f: Okazaki
fragment

replication
 The initial step of replication is the production of the
RNA primer in the 5’3’ direction,
 Assisted by RNA-polymerase (primase).
 After the synthesis of the short chain of RNA on the DNA
template is completed, the enzyme is detached from
DNA.
 Deoxyribonucleotides are added to the RNA primer
 Through the assistance of in 5’3’ direction. DNA-
polymerase III
 A hybrid chain RNA-DNA is thus formed.

replication

replication
DNA-polymerase III synthesizes
short DNA fragments (Okazaki
fragments) on replicative fork.
If an error in base pairing occurs,
the mismatched nucleotide is
immediately split off by the
enzyme operative.
 Correctly paired new nucleotide is
routinely added.

replication
 RNA primer, after termination of DNA-polymerase III
 is removed from the synthetic chain by specific
ribonuclease H, or DNA-polymerase I.
 At the site of removed RNA primer, the missing fragment
of DNA strand is completed by DNA-polymerase I.
 Colligation of pre-synthesized DNA fragments in the
3’5’ direction is effected through the aid of DNA-ligase.

3.5. The Gene Concept
 Gene is heritable unit of phenotypic variation.
 Gene is the linear DNA sequence.
 Genes can be assigned to one of two broad functional
categories:
 Structural genes and
 Regulatory genes.
 Structural genes code for polypeptides or RNAs
needed for the normal metabolic activities of the cell,
e.g.
enzymes,
structural proteins,
transporters, and
receptors.

3.5. The Gene Concept
 Regulatory genes code for proteins whose function is
to control the expression of structural genes.
 Genes come in multiple forms called alleles
 Genes encode phenotypes
 Important concepts in genetics is the distinction b/n
traits and genes.
 Traits are not inherited directly.
 Rather, genes are inherited.

3.5.1Characteristics of gene
 Genes are hereditary units that are transmitted
from parent to offspring.
 The genes reside in a long molecule called DNA.
 DNA plus protein make up chromosomes.
 Chromosomes are found in the nucleus of the
cell.
 The behavior of the genes is parallel to that of
the chromosomes.
 specific position on a chromosome called the
gene locus.
 Genes on single chromosome is called linkage
group.

3.6. Gene Expression and
Regulation
 The expression involving two consecutive steps
 Transcription and
 Translation.
Transcription
 The process of copying genetic information from one
strand of the DNA into RNA.
 Transcription units
 Promotr
 Structural gene
 A terminator

Cont…
 Transcription follows the same base pairing rules
as DNA replication
 Transcription does not happen at significant rates
by themselves they are enzyme-catalyzed.
 The enzyme that directs transcription is called
RNA polymerase.

3.6. Gene Expression and Regulation
 A single RNA polymerase is responsible for transcribing
all types of RNA in prokaryotic system.
 However, eukaryotes have three different RNA
polymerases:
 RNA polymerase I (Pol I) - transcribes rRNA
genes.
 RNA polymeraseII (Pol II) - transcribes protein
mRNA.
 RNA polymerase III (Pol III) - transcribes other
functional tRNA.

Regulation
 In eukaryotes, transcription occurs inside the
nucleus.
 In prokaryotes the movement of transcripts
from nucleus to cytoplasm does not take
place,
 Translation take place in cytoplasm, directly
on the growing transcript.
 Transcription has three phases:
 Initiation,
 Elongation, and
Termination.

Regulation
Initiation
 First, the enzyme recognizes a region called a
promoter, “upstream” of the gene.
 The polymerase binds tightly to the promoter and
causes localized melting, or separation, of the two DNA
strands within the promoter.
 Next, the polymerase starts building the RNA chain.
 It uses four ribonucleoside triphosphates: ATP, GTP,
CTP, and UTP.

Regulation
Elongation
 During elongation RNA polymerase directs the
sequential binding of ribonucleotides to the growing RNA
in 5’ →3’.
 It moves along the DNA template, and
 The “bubble” of melted DNA moves with it.
 As soon as the transcription machinery passes, the two
DNA strands wind around each other.
 Fundamental differences b/n transcription and DNA
replication:

Regulation
 A. RNA polymerase makes only one RNA strand during
transcription,
 Transcription is therefore said to be asymmetrical.
 B. In transcription, DNA melting is limited and transient.
 Strand separation occurs to allow the polymerase to
“read” the DNA template strand.

Regulation
Termination
 Just as promoters serve as initiation signals for
transcription, regions at the ends of genes, called
terminators, signal termination.
 These work in conjunction with RNA polymerase to
loosen the association between RNA product and DNA
template.
 The result is that the RNA dissociates from the RNA
polymerase and DNA, thereby stopping transcription.

Figure 3.8 Overview of the early steps of
transcription

Translation
 Genetic code and its characterization
 Genetic code is understood as the relationship b/n the
sequence of bases in a nucleic acid
 The genetic code regarded as a specific dictionary for
translating a text, recorded by means of four
nucleotides, into a protein text,
 The other amino acids found in a protein are modified
forms of one of the 20 amino acids.

Translation
 The genetic code exhibits the following features:
 Triplicity: a triplet of nucleotides corresponds to each
amino acid.
 There are four nucleotides available; it can easily be
seen that, taken as triplets, they can form፡
 43=64 codons.
61 are sense codons, and
 3 are nonsense codons (or termination
codons).
 Non-overlap: The genetic text codons are independent
of each other.
 |CCA| |CGG| |AAC|
Non overlapping

Genetic code

Translation
 Degeneracy or redundancy: certain amino acids may
have more than one codons.
 61 sense codons account for 20 amino acids, i.e. on the
average, more than 3 codons per amino acid.
 Specificity: definite codons correspond to each amino
acid. They cannot be used for another amino acid.
 Colinearity: correspondence b/n the linear sequence of
codons in mRNA and that of amino acids in protein.
 Universality: all the above mentioned features of the
genetic code are characteristic of any living organism

Translation
 Protein biosynthesis divided into three stages:
1. Initiation (start of synthesis),
2. Elongation (polypeptide chain lengthening),
3. Termination (end of synthesis).
Initiation of Translation
 Translation of the mRNA is done in ribosomes
 In eukaryotes, ribosomes are complex and composed
of two subunits, one large and the other small.

Translation
 The large subunit contains three ribosomal
RNAs:
 28S rRNA,
 5S rRNA, and
 5.8S rRNA, along with 49 proteins.
 small subunit contains the 18S rRNA, and 33
proteins.
 The initiation codon codes for methionine;
brought it via tRNA .
 tRNA is said to be “charged” when it carries an
amino acid.
 The charged tRNA recognizes the codon
through complementary base pairing with a
region of it called an anticodon.

Translation Elongation
 Before elongation can occur, the large ribosomal subunit
joins to create a complete ribosome.
 The ribosome has three sites to accommodate tRNA
molecule:
 Peptidyl (P),
 Aminoacyl (A), and
 Exit (E) site.
 The initiator tRNA occupies the P site of the ribosome,
 Which is positioned over the initiator AUG codon and is
adjacent to the A site,

translation.

Translation Elongation
 Then the appropriately charged tRNA for this next codon
in the A site enters it, and
 its anticodon pairs with the codon.
 A peptide bond then forms between the amino acids that
are attached to the tRNAs in the P and A sites
 The initiator tRNA that no longer is charged is in the E
site and it is then free to leave the ribosome
 This elongation cycle is repeated until the entire
polypeptide chain is made

Translation Termination
 Polypeptide synthesis is over when the ribosome
encounters a stop codon in it’s A site.
 Since no tRNAs can base pair with these stop codons,
proteins called “release factors” bind to the ribosome
instead.
 These release factors allow the polypetide chain to be
released from the P site as well as the mRNA to no
longer bind to the ribosome.
 The ribosome also splits into its two subunits.

3.6.1. Regulation of gene
expression
 Gene regulation is a critical for proper plant growth and
development.
 Regulation entails the “turning on” and “turning off” of genes.
 It is through regulation of:
 Gene expression that cellular adaptation,
 Variation,
 Differentiation, and
 Development occur.
 Some genes are turned on all the time (called constitutive
expression),
 others are turned on only some of the time (called differential
expression).

3.6.1. Regulation of gene
expression
 Gene regulation is regulatory molecules interact with nucleic
acid sequences to control transcription or translation.
 Six potential levels for regulation of gene expression exist in
eukaryotes:
 transcription;
 RNA processing;
 mRNA transport;
 mRNA stability;
 Translation; and protein activity.

There are different expression
regulators
 Promoters: region of a gene hundred nucleotides
long upstream the transcription initiation site.
 They constitutes binding site for enzyme
machinery
 Enhancers: are sequences that increase
transcription initiation
 Unlike promoters, are not dependent on the
distance from the transcription start site.
 Enhancers are short sequences (< 20 to 30 bp)
 Facilitate transcriptional complex (RNA

regulators
 Operators: Operators are nucleotide sequences
positioned b/n promoter & the structural gene.
 They constitute region of DNA to which repressor
proteins bind and prevent transcription.
 Repressor proteins have a very high affinity for
operator sequences.
 Repression of transcription is accomplished by the
repressor protein attaching to the operator.

regulators
 Attenuators: The attenuator sequences are found in
bacterial gene clusters.
 Attenuators are located in leader sequences, a unit of
about 162 bp situated b/n the promoter-operator region
 Attenuation decreases the level of transcription
approximately 10-fold.

Exercise of the chapter
1. What are the differences between DNA and RNA?
2. What are the control points that can regulate gene
expression?
3. Describe the main parts of a gene and their functions.
4. Give evidences for DNA is genetic material.
5. What are DNA, gene, and chromosome? Explain the
difference.
6. Name types of ribosomal RNAs.
7. How DNA replications proceed in prokaryotic and
eukaryotic cell?
8. List the characteristics of gene.
9. What is positive and negative gene regulation?

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