Biotechnology - Unit 1 Microbiology Part 2

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CONTENTS
Introduction
4
Section 1:
Structure of micro-organisms
5
Section 2:
Microbial metabolism
21
Section 3:
Patterns of growth
31
Section 4:
Copying and translating genes
39
Section 5:
Genetic engineering
56
Section 6:
Infection and immunity
67
Bibliography
74
Appendix: Advice for problem-solving outcomes
76
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SECTION 4
Copying and Translating Genes
In Section 1 of this unit (Structure of Micro-organisms), you were introduced to
the concept that all cells have a genome organised into chromosomes, which
control all the activities of the cell.
The genome itself consists of a series of genes, many of which code for proteins.
These genes are made of a nucleic acid called deoxyribonucleic acid (DNA).
In this section you will find out about the following:



the structure of deoxyribonucleic acid;
how genes control the cell by directing the making of proteins within the
cell;
how genes in prokaryotes are regulated.
The structure of DNA
The genes that make up a chromosome are made of a nucleic acid called
deoxyribonucleic acid (shortened to DNA). This is a long, threadlike molecule
consisting of two strands twisted into a helical (spiral) molecule.
The building blocks of each strand of the DNA molecule are called nucleotides
that are joined together to form a long chain. Each nucleotide consists of three
components: a phosphate group, a sugar molecule called deoxyribose and an
organic base.
These three components are arranged in the following way:
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Copying and Translating Genes
The deoxyribose sugar is given a special numbering system in that each carbon
atom (of which there are 5) is given a number. The carbon atoms of the sugar
that are known as 3´(said 3 prime) and 5´(said 5 prime) are shown in the above
diagram.
There are 4 organic bases found in a DNA molecule, namely ADENINE (A),
GUANINE (G), CYTOSINE (G) and THYMINE (T), and so there are 4 different
types of nucleotides found in DNA, each containing a different organic base.
As mentioned, the nucleotides join together to make a long strand of DNA. The
phosphate group linked to the 5’ end of one nucleotide joins to the 3’ of the sugar
of the neighbouring nucleotide, thus forming a phosphate–sugar backbone. In
this way the nucleotides form a long single strand of DNA, one end with a 5’
phosphate and the other end with a free 3’ group on the sugar.
Two strands of nucleotides link together with weak hydrogen bonds between their
organic bases. The two strands of nucleotides run in opposite directions to each
other, so they are said to be antiparallel. One strand starts with a 5´ end and
finishes with a 3´ end, while the other strand starts with a 3~end and finishes with
a 5´ end:
5´ _________________________________________ 3´
3´ _________________________________________ 5´
Each organic base in a nucleotide from one strand can form a hydrogen bond
with only one other type of organic base in a nucleotide in the other strand:



A bonds only with T
C bonds only with G
A-T and C-G are known as base pairs.
A fragment of double stranded DNA showing only the sequence of organic bases
in each strand is represented below:
5´ AGCTTGCATTAACGTCGC 3´
3´ TCGAACGTAATTGCAGCG 5´
One strand is known as the sense strand, while the other is called the antisense
strand. You will come across these terms again when the making of another type
of nucleic acid (called messenger ribonucleic acid) is discussed.
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Copying and Translating Genes
The double strand of DNA is twisted into a structure called a double helix. This
resembles a spiral staircase with the phosphate-sugar backbone forming the
uprights and the base pairs forming the rungs (Fig. 18).
Chromosomes and genes in eukaryotes
In eukaryotes chromosomes are found in the nucleus. Under a very high
powered microscope the chromosomes appear to be striped. Each stripe
represents one single gene, so eukaryotic chromosomes are made up of lots of
genes.
Many of the genes in a chromosome contain the genetic code (DNA) needed to
make proteins. When a protein is made from a gene (DNA), the gene is said to
be ‘expressed’. It has been found that only some parts of a gene (DNA) are
expressed. These parts are known as exons or coding regions of the gene. The
parts of the gene that do not code for protein are called introns or intervening,
non-coding regions.
Chromosomes and genes in prokaryotes
Prokaryotes have a single circular chromosome. It has been found that genes
which have a related function are grouped together in prokaryotes. This group of
genes is called an operon. You will find out more about operons in prokaryotes
later.
Not all genes needed by a prokaryote are found on the circular chromosome.
Some bacterial genes are found on a plasmid. This is a small circular piece of
double stranded DNA. Plasmids are found naturally in bacteria and generally
carry genes that are advantageous to the bacteria, but are not essential for their
survival. For example, some plasmids have genes that allow the bacteria to grow
on certain antibiotics. Plasmids can be transferred from one bacterium to the
next, so the advantageous genes can be passed on.
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Copying and Translating Genes
Plasmids have been isolated and manipulated by biotechnologists for use in
genetic engineering. It is now possible to use plasmids as cloning vectors to
introduce new genes into bacteria, to grow the bacteria and for the bacteria to
express the new genes and so produce new proteins.
DNA replication
When a cell divides into two, the two new cells are called daughter cells. The
daughter cells have exact copies of the chromosomes that were present in the
original parent cell. Before the cell can divide, the DNA molecules must be
duplicated exactly. The duplication of the DNA molecules is known as DNA
replication.
Several factors are required for the replication of a DNA molecule:




Double stranded DNA (to act as a template for the new DNA)
An enzyme called DNA polymerase
Each of the four nucleotides (A, T, C and G bases);
Energy in the form of ATP.
The steps involved in the replication of DNA are as follows:
1.
The parental double stranded DNA to be replicated (or copied) begins to
untwist from its helical shape.
2.
Hydrogen bonds between complementary bases (A-T and G-C) are
broken. This causes the two strands to separate, forming two single
strands of DNA.
3.
A free DNA nucleotide finds its complementary base on the single strand
of DNA. For example, if there is a T on the single strand of DNA, a free A
nucleotide lines up with it, similarly if there is a G on the single strand,
then a C nucleotide lines up with it.
4.
A hydrogen bond forms between the free DNA nucleotide and its
complement.
5.
The 5´ phosphate group of this new nucleotide joins to the free 3´ of the
adjacent nucleotide, thus continuing the formation of the new DNA strand.
The enzyme which joins (polymerises) one nucleotide to the next is called
DNA polymerase.
6.
The newly formed daughter DNA molecule rewinds into a double helix.
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Copying and Translating Genes
At the end of DNA replication, two new strands of DNA are formed that are
identical to each other and the parental DNA molecule.
DNA mutations
Sometimes when DNA is being replicated, mistakes happen and the wrong
nucleotide is inserted into or a nucleotide is missed out of the new DNA. This is
known as a mutation in the DNA. Table 6 shows some of the mutations that can
occur in a DNA molecule:
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Copying and Translating Genes
Table 6
Mutation
Description of the mutation
Substitution
This is when a nucleotide is substituted for another
nucleotide. For example, in the DNA sequence
TTGCTAAGCCGT, the 5th T may be substituted for a G.
The new sequence would be TTGCGAAGCCGT.
Insertion
This is when an extra nucleotide is introduced into a
DNA molecule. Taking the above sequence as an
example, the mutated sequence may be
TTGCTAAGACCGT where an extra A nucleotide has
been inserted.
Deletion
This is when a nucleotide is removed from the original
sequence. For example, TTGCTAAGCCGT may become
TTGCTAGCCGT where the 6th A has been deleted from
the sequence.
Inversion
This is when two nucleotides are inverted. For
example, in the sequence TTGCTAAGCCGT the 3rd and
4th nucleotide may change place so that the mutated
sequence becomes TTCGTAAGCCGT.
The structure of protein
As mentioned previously, many genes in the chromosomes of eukaryotes and
prokaryotes code for proteins.
Proteins are large, complex molecules that carry out many functions in the cell as
described below.




Some proteins have a structural role in the cell
Some proteins are enzymes and carry out biochemical reactions in
the cell such as those involved in respiration.
There are proteins that are involved in preventing infection in the
body. These proteins are known as antibodies
Other proteins are involved in the transport of substance around the
body. For example, haemoglobin is involved in the transport of
oxygen in red blood cells.
Proteins are made of building blocks called amino acids that join together by
strong peptide bonds to make large polypeptide (protein) molecules.
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Copying and Translating Genes
The amino acids that join together to form a polypeptide chain are known as the
primary structure of a protein. In a cell this polypeptide chain folds again into a
secondary, then again into a tertiary structure (a three dimensional (3D) shape).
The 3D shape is the most compact, stable structure that the protein can form.
This 3D shape is held in place by weak hydrogen bonds. A protein must be in its
correct 3D shape for it to work (function) properly in the cell. Anything that causes
a change in the 3D shape of the protein (such as a change in temperature or a
change in pH) can affect the function of the protein.
Before proteins can be made by a cell, another type of nucleic acid is needed,
called RNA.
The structure of RNA
Ribonucleic acid (RNA) is the second type of nucleic acid found in the cell. It
consists of nucleotides polymerised together, although the structure of an RNA
nucleotide is slightly different to a DNA nucleotide.
An RNA nucleotide consists of the following:



a phosphate group
a ribose sugar group
an organic base
In RNA, the organic bases are Adenine (A), Cytosine (C), Guanine (G) and Uracil
(U).
Copying and Translating Genes
The ribose sugar group from one RNA nucleotide joins to the phosphate group of
a second RNA nucleotide, forming a single polymerised chain. RNA does not
exist as a double stranded molecule, instead it is single stranded.
The differences between the structures of DNA and RNA are shown in Table 7.
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----------------------------------------------------------------------------------------------------Table 7
Feature
DNA
RNA
Number of nucleotide strands
present in one molecule
Two
One
Bases found in the nucleotides
A, G, C, T
A, G, C, U
Sugar present in the nucleotides
Deoxyribose
Ribose
The synthesis of RNA
RNA is made in the nucleus of the cell, using one of the strands of DNA as a
template. The strand that is used as the template is known as the sense strand.
Thus, the information that is coded for in the DNA molecule is transferred to the
RNA molecule which is then exported from the nucleus to the cytoplasm. The
synthesis of RNA is called transcription.
There are several types of RNA transcribed from DNA. One type of RNA is called
messenger RNA (mRNA) and another type of RNA is called transfer RNA
(tRNA). Both types of RNA are involved in the synthesis of protein.
The synthesis of mRNA
In eukaryotes, the genes from the DNA strand that are used to synthesise mRNA
are not continuous. This means that the DNA contains nucleotide sequences that
do not appear in the mature mRNA. These intervening sequences are called
introns and they are cut out of newly formed mRNA molecules in a process
known as splicing.
This is shown in Fig. 21.
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Copying and Translating Genes
Figure 21
GENE (DNA)
Exon 1
Intron 1
Exon 2
Intron 2
Exon 3
RNA Synthesis
RNA
Exon 1
Intron 1
Exon 2
Intron 2
Exon 3
Introns are removed
Mature mRNA
Exon 1
Exon 2
Exon 3
The mature mRNA molecule is exported from the nucleus consisting only of exon
sequences and is then used to synthesise protein.
Protein synthesis
Each protein in the cell of an organism is coded for by a gene found in the
chromosomes of that organism. The gene (DNA) is used to synthesise a mRNA
molecule which, in turn, is used to direct the synthesis of the protein molecule.
The information on the DNA is known as the genetic code. The sequence of
bases along a DNA strand represents a code for making proteins.
DNA contains 4 bases (ACGT) yet proteins contain about 20 amino acids. The
relationship cannot be that 1 base represents (codes) for 1 amino acid as this
would allow only 4 amino acids to be coded. Even 2 bases coding for 1 amino
acid is insufficient as this allows for only 16 amino acids.
It has been found that 3 bases in the DNA code for 1 amino acid. The triplet of 3
bases is known as a codon. There are 64 codons and some amino acids have
more than one codon. The codons are arranged in a specific order to code for a
specific protein.
Remember that the DNA is transcribed into mRNA. The mRNA that is produced
contains the complementary sequence of codons to the sense strand of DNA.
Remember also that RNA has uracil (U) instead of thymine (T).
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Copying and Translating Genes
For example:
DNA
{ TTTCTTTAGGGT
{ AAAGAAATCCCA (sense strand)
The sense strand is used as the template
to make mRNA
mRNA
UUUCUUUAGGGU (this sequence is complementary to the
sense strand of DNA)
Table 8 shows some of the codons that specify different amino acids.
Table 8
Codon
Amino acid
Codon
Amino acid
UAG
Tyrosine
GUG
Valine
UUU
Phenylalanine
CCA
Proline
AGU
Serine
UGG
Tryptophan
CUU
Leucine
AGA
Arginine
GGU
Glycine
UCA
Serine
Using this table we can work out the sequence of amino acids that would be
produced using the mRNA (UUUCUUUAGGGU) from the above example:
Phenylalanine-Leucine-Tyrosine-Glycine
This table can also be used to show that a mutation in the gene can cause a
change in the sequence of the protein. If there was a mutation such that the first
U in the sequence was replaced by a C (look back to the previous section to find
out the name of such a mutation), then the sequence of amino acids would
change to:
Leucine-Leucine-Tyrosine-Glycine
Sometimes a change in the amino acid sequence has no effect on the function of
the protein but in some cases, the protein may become inactive.
mRNA is not the only RNA molecule involved in making protein, tRNA is needed
too. tRNA is a small molecule that attaches to an amino acid in the cytoplasm of
the cell. There is a different tRNA molecule for each amino acid. At the opposite
end to where the amino acid is attached to tRNA, there is a triplet of bases called
the anticodon. The anticodon corresponds to a particular amino acid. The tRNA
carries the amino acid to the ribosome, where proteins are made.
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Copying and Translating Genes
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Copying and Translating Genes
The interaction between mRNA, tRNA and the ribosome is shown in Fig. 22.
When two tRNA molecules are present within a ribosome, a peptide bond forms
between the amino acids.
Ribosomes are small spherical structures found in the cytoplasm of the cell. They
are the site of protein synthesis. Each ribosome contains all the components
(proteins and RNA) required to make new proteins in the cell. In prokaryotes,
ribosomes are free in the cytoplasm whereas in eukaryotes they are often found
attached to internal membranes, forming the organelle known as the rough
endoplasmic reticulum.
The rough endoplasmic reticulum is involved in transporting the newly made
protein to another organelle, the Golgi apparatus. Proteins are modified, then
packaged by this organelle before being secreted out of the cell.
Control of gene action
Some proteins are required by a cell only under certain conditions, e.g. E.coli
require the enzyme β-galactosidase only where the bacteria are growing on
lactose. When E.coli are growing on a different medium, such as glucose, the
genes that code for β-galactosidase are switched off.
The advantage of this control is that resources within the bacterial cell are not
wasted.
Three areas of bacterial DNA are involved in the control of β-galactosidase
activity:



The structural gene codes for the enzyme
The regulator gene codes for a protein known as the repressor
The operator is where the repressor binds.
These three areas are found together on the DNA in an area known as the lac
operon as shown in Fig. 23.
Copying and Translating Genes
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Regulator
gene
Operator
Structural
gene
In the absence of lactose
When no lactose is present in the culture medium, E.coli does not need the βgalactosidase enzyme. Therefore the gene coding for this enzyme is switched
off.
The gene is switched off due to the presence of the repressor protein (coded for
by the regulator gene). The repressor protein binds to the operator and switches
off the structural gene. This is shown below:
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Copying and Translating Genes
Figure 24: The lac operon in the absence of lactose
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In the presence of lactose
When E.coli is grown in culture medium containing lactose, β galactosidase is
produced. The enzyme breaks down lactose into glucose and galactose and the
bacteria use the glucose for growth:
Lactose
glucose + galactose
Lactose is called an inducer as it switches the structural gene on, so producing
the enzyme. This is shown in Fig. 25.
Figure 25: The lac operon in the presence of lactose
Regulator
gene
Operator
Structural
gene
Lactose binds to the repressor molecule, which prevents the repressor from binding
to the operator. Therefore RNA is produced from the structural gene. If RNA is
made, then it can be used to synthesise the enzyme.
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Test yourself on DNA structure and protein synthesis
Before you move onto the next part of this unit, spend time reviewing your notes
on the above section. It contains a lot of information.
1.
Give two differences between the structure of DNA and RNA.
2.
The sense strand of a piece of DNA has the following sequence:
5´ AGTGGTACCGAACAC 3´
(a) Write down the sequence of the corresponding antisense strand.
(b) Write down the sequence of mRNA that would be produced if the
sense strand was transcribed.
(c) Use Table 8 to find out the sequence of amino acids that would be
produced using the mRNA from (b).
3.
A DNA molecule consists of 24% cytosine bases. Calculate the percentage
number of thymine bases that would be present in this DNA molecule.
4.
Describe the steps involved in the replication of a DNA molecule.
5.
Complete Table 9, which is about mutations:
Table 9
Type of mutation
Description of mutation
Substitution
When an extra nucleotide is inserted
into a DNA molecule
Inversion
When a nucleotide is removed from a
DNA sequence
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6.
Describe the process of splicing in the synthesis of mRNA.
7.
What is the role of the following types of RNA in the synthesis of protein:
(a) mRNA (b) tRNA.
8.
Describe the functions of the following organelles in the cell in protein
synthesis:
(a) ribosome
(b) rough endoplasmic reticulum
(c) Golgi apparatus.
9.
Fig. 26 shows the lac operon found in bacteria such as E.coli.
(a) Name protein Y and protein Z.
(b) State whether protein Y and protein Z are produced:
(i) in the absence of lactose
(ii) in the presence of lactose.
Figure 26
Regulator
gene
Protein Y
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Operator
Structural
gene
Protein Z
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SECTION 5
Genetic Engineering
Genetic engineering may be defined as the deliberate change of the genetic
makeup of an organism. This can be achieved by the introduction of genes from
another organism. In this way, organisms with new characteristics are produced
in a way that is not possible using conventional breeding methods.
Genetic engineering is a rapidly growing technology and it is thought that it will
have profound effects on our everyday lives. Some examples of how it may affect
us are given below.




In the field of medicine it may improve the diagnosis and cure of hereditary
defects and disease.
It is being used for the development of new drugs and vaccines for use by
humans and animals.
In agriculture it is being used to improve food production.
It is being used to monitor and reduce environmental pollution.
In Scotland, one of the fastest growing industries is biotechnology. Numerous
biotechnology companies have been set up, many using the techniques of
genetic engineering.
In this section of the unit you will be introduced to some of the techniques used in
genetic engineering. The most basic technique associated with genetic
engineering is gene cloning.
Gene cloning itself involves several techniques including:




the isolation and purification of DNA
cutting DNA into smaller fragments with enzymes
separating fragments of DNA using electrophoresis
introducing fragments of DNA into organisms using cloning vectors.
The end result of gene cloning is the production of an organism that is able to
make many copies of the newly introduced DNA.
Purification of DNA
The first step in many genetic engineering processes is the isolation of DNA from
cells. There are several steps involved in DNA purification and these are outlined
below.
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Firstly, the cells must be disrupted to release the soluble intracellular
components, including the DNA. This can be done mechanically by putting the
cells into a liquidiser/blender – similar to the one in your kitchen!
Alternatively, cells can be disrupted using enzymes. The soluble intracellular
components are separated from insoluble cellular debris by centrifugation, a
technique that separates components using high speed centrifugal forces.
The second step in DNA purification involves separating the DNA from proteins.
This is achieved by extracting the proteins into an organic solvent and/or using
enzymes that degrade the proteins, leaving purified DNA.
Finally, the DNA is precipitated using alcohol and then resuspended in a suitable
buffer.
Restriction endonucleases
After DNA has been purified, it is cut into smaller fragments using restriction
endonucleases. These are enzymes that are found naturally in bacteria. These
enzymes recognise and cut short specific sequences (between 4 and 8 base
pairs) within DNA. Biotechnologists have isolated many of these enzymes and
they are now routinely used in genetic engineering for cutting DNA.
One of the most commonly used restriction enzymes is called EcoR1. It
recognises the following 6-base pair DNA sequence:
5´ GAATTC 3´
3´CTTAAG 5´
EcoR1 then cuts the DNA sequence as follows:
5´G
3´ CTTAA
AATTC 3´
G 5´
When EcoR1 cuts DNA it produces two double stranded fragments, but the cuts
do not occur at the same position. Instead the cut is staggered by four
nucleotides, so that the DNA fragments have single stranded overhangs (known
as sticky ends). If another piece of DNA is cut with the same enzyme and so has
the same sticky ends, the pieces of DNA can be joined together by base pairing
between the sticky ends.
Other restriction endonucleases cut in the middle of their recognition sequence
so producing blunt ends.
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Agarose gel electrophoresis
This is a technique used to separate fragments of DNA according to their size. It
is often used to separate fragments of DNA after digestion with restriction
endonucleases.
A solution of warm agarose is poured into a casting tray. A comb is inserted in
one end of the tray and the gel is allowed to cool causing the agarose to set.
After it has set, the comb is removed forming a number of wells.
Different concentrations of agarose can be used, the higher the concentration of
agarose, the slower the rate of movement of the DNA fragments.
The agarose gel has very small pores that act as a molecular sieve and causes
DNA of different sizes to separate from each other as follows:


Small fragments of DNA move fastest through the gel
Large DNA fragments move slowly through the gel.
The DNA fragments to be separated are mixed with a tracking dye and loaded
into the wells. DNA is negatively charged and, when a voltage is applied to the
gel, the DNA migrates towards the positively charged anode. The power supply is
switched off when the tracking dye reaches the end of the gel.
After electrophoresis the DNA fragments can be visualised by staining the gel
with a dye that binds to the DNA.
Fig. 27 shows an agarose gel with DNA fragments that have been stained and so
can be easily seen.
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Genetic Engineering
Lanes 1 and 7 contain DNA fragments of known size.
Lanes 2–6 contain plasmid DNA of different sizes.
Locating a fragment of DNA separated by electrophoresis
After separating DNA fragments on an agarose gel, one particular DNA fragment
may need to be located. For example, if chromosomal DNA is cut up into smaller
fragments, one of the smaller fragments may contain a gene that a
biotechnologist is interested in. How is this fragment located?
Firstly, the DNA is transferred from the agarose gel to a membrane filter. This
step is needed because the double stranded DNA must be denatured into single
strands. This is almost impossible to do while the DNA is in agarose. The DNA is
transferred to the membrane by a process known as blotting. Then the DNA is
denatured.
The membrane containing the single stranded DNA is incubated with either
single stranded DNA or RNA (known as a probe) that contains some bases
complementary to the fragment of DNA to be located. The complementary bases
in the probe and the desired fragment of DNA join together, forming double
stranded DNA.
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The fragment of DNA located by the probe is visualised because the probe is
labelled either with radioactivity or with a chemiluminescent label, making it easy
to see.
The production of complementary DNA from RNA
Sometimes biotechnologists do not want to work with genes because they
contain introns that are not used to make protein. (Look back at Section 4 to
remind yourself about introns and exons.)
Instead, some biotechnologists work with messenger RNA (mRNA)
which is the expressed form of the gene. However, working with RNA is difficult
because it is single stranded and so it cannot easily be inserted into a cloning
vector such as a plasmid. Also, RNA is degraded very easily and so can be
difficult to use.
However, these problems working with RNA can be overcome by converting
RNA into DNA (known as complementary DNA or cDNA) using an enzyme called
reverse transcriptase.
cDNA is a direct copy of the mRNA but, unlike the original gene, it does not
contain introns. cDNA can be inserted easily into a cloning vector and cloned in
the usual manner.
Fig. 28 shows the steps taken to make cDNA from a mRNA template.
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Firstly the mRNA is used by reverse transcriptase as a template to synthesise the
first strand of DNA. A DNA-RNA hybrid is formed.
The RNA in the DNA-RNA hybrid is removed using alkali.
The remaining single stranded DNA is used as a template by the enzyme DNA
polymerase to make a second complementary strand of DNA.
The cDNA can now be inserted into a cloning vector, such as a plasmid and
cloned to produce many identical copies of the cDNA.
Transformation and cloning
Transformation is the name used to describe the process when a foreign
sequence of DNA (such as a gene or cDNA) is introduced into micro-organisms
such as bacteria and yeast.
Two micro-organisms that are commonly used in transformations are the
bacterium E.coli and the yeast, S. cerevisiae. Both micro-organisms are single
celled (unicellular) organisms that have fast reproduction rates and thus are quick
growing. This makes them ideal for large scale production in industrial fermenters
(bioreactors).
E.coli
This is a prokaryote that is often used as a recipient for foreign DNA. Large
sequences of foreign DNA can be inserted into E.coli using a plasmid. The DNA
is transcribed and translated and it is possible for the protein coded for by the
foreign DNA to account for 60% of the total protein produced by the bacterial cell.
E.coli are relatively easy to transform.
While there are many advantages of using E.coli, there are some disadvantages
– mainly due to the fact that it is a prokaryote and the foreign protein produced
may originally have come from a eukaryote. The disadvantages are outlined
below.
The foreign protein produced is not always secreted easily from E.coli. This may
be due to E.coli not being able to carry out modifications to the protein after it is
made, for example addition of sugar groups. If the protein is not secreted by the
bacterium, it causes problems for the biotechnologist as E.coli must be
harvested, the bacterial cells broken open (lysed), and the protein purified. This
increases the production costs.
E. coli does not always fold the foreign protein into its natural 3D shape. This
causes the protein to be inactive.
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Genetic Engineering
S. cerevisiae
This is a eukaryote (it is a yeast) that can be used instead of E.coli as the
recipient for foreign DNA.
Since it is eukaryotic, it can fold proteins into their 3D shape which allows the
proteins to be active. Foreign proteins made by S. cerevisiae are secreted from
the cell as S. cerevisiae can carry out post-translational modifications (e.g. it can
add sugar groups to proteins) which allows the proteins to cross the cell wall.
Thus proteins secreted by S. cerevisiae can be extracted from the culture
medium.
The disadvantages of using yeast include the following:

It can be difficult to transform, this means that it can be difficult to
introduce the foreign DNA into the yeast
 It produces less protein, so yields of the foreign protein are smaller
 Plasmid vectors may be lost from yeast if there is no advantage to the
yeast in having the plasmid.
Cloning vectors
Cloning vectors are used to introduce foreign DNA into micro-organisms such
as E.coli and S. cerevisiae. Cloning vectors must be able to replicate within these
host cells.
Two types of cloning vectors used to introduce foreign DNA sequences into
micro-organisms are plasmids and bacteriophages. Both of these cloning vectors
have been mentioned previously. Plasmids are discussed in the section on
bacteria and bacteriophages are mentioned in the section on viruses. You might
find it helpful to read these sections again before continuing.
Both occur naturally in bacteria but biotechnologists have genetically engineered
them so that they can be used as cloning vectors.
Cloning vectors have been manipulated so that they have the following
characteristics:
1.
They can be cut with restriction enzymes and foreign DNA sequences
(cut with the same restriction enzymes) can be inserted into them using
an enzyme called DNA ligase.
2.
Antibiotic resistance marker genes have been added to them. These
genes code for proteins that breakdown antibiotics. If a cloning vector is
inserted into a micro-organism, the micro-organism gains the antibiotic
resistance gene and so is able to grow in the presence of this antibiotic.
The micro-organism becomes resistant to the antibiotic.
3.
Some cloning vectors contain part of the lac operon. This is used to
control the expression of the foreign DNA sequences. The foreign DNA
is transcribed and translated only when the lac operon is switched on.
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After a foreign sequence of DNA has been inserted into a cloning vector using
DNA ligase, the cloning vector is mixed with the micro-organism into which it is to
be transformed. Some of the micro-organisms will take up the cloning vector,
some will not. To separate the transformed micro-organism from those that are
not, the micro-organism is grown in media containing the antibiotic to which the
transformed micro-organism has acquired resistance. The transformed microorganism has the cloning vector that has the antibiotic resistance gene, so it is
able to grow in the presence of the antibiotic.
Any micro-organism that does not possess the cloning vector is unable to grow in
this medium.
The transformed micro-organism is isolated from the medium and transferred to
another medium where it is allowed to reproduce and grow in large quantities.
Each new micro-organism that is produced is genetically identical to the original
transformed micro-organism. Each genetically identical micro-organism is called
a clone. The process of producing lots of genetically identical micro-organisms is
known as cloning. This is shown in Fig. 29.
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Genetic Engineering
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Genetic Engineering
Test yourself on genetic engineering
Before you move onto the next part of this unit, spend a little time reviewing your
notes on genetic engineering, then see if you can answer the questions below.
1.
Fig. 30 represents a human chromosome showing the possible position
of the human insulin gene.
(a) Name the type of enzyme that can be used to break the chromosome into
smaller fragments.
(b) The above chromosome is broken into smaller fragments with the following
sizes:
Table 10
Fragment
Size of fragment (base pairs)
W
250
X
345
Y
400
Z
750
Fragments were separated by agarose gel electrophoresis.
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Genetic Engineering
Complete the diagram to identify which band corresponds to which fragment.
(c) One of the bands is known to contain the gene for insulin.
Describe how you might use a probe to find out which band contains the
gene.
2.
Given the following components, describe how you could obtain clones of
an insulin gene:
Components available:
Insulin gene
plasmid vector with ampicillin resistant gene
Bacterial cells
restriction enzymes
Ligase
nutrient medium containing ampicillin
(Note: ampicillin is an antibiotic)
3.
State 2 advantages of using yeast rather than bacteria in producing clones
of a gene.
4.
The following questions refer to the making of cDNA.
(a)
Name the enzyme used to convert RNA into DNA.
(b)
What is the purpose of incubating the RNA/DNA hybrid with alkali?
(c)
Name the enzyme that is used to make the second complementary strand
of DNA.
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SECTION 6
Infection and Immunity
Micro-organisms as pathogens
Micro-organisms such as bacteria and fungi can be advantageous to man in that
they can be used to produce useful substances such as yoghurt, cheese, beer,
wine and antibiotics, to name but a few.
However, it must be remembered that not all micro-organisms are beneficial,
some are harmful and cause disease. Micro-organisms that cause disease are
known as pathogens. Many species of bacteria, fungi and viruses are
pathogenic.
However, your body has developed an immune system that removes pathogens
and provides you with natural immunity if the pathogen should enter your body
again.
Production of antibodies and the role of blood cells
When a pathogen enters your body, your immune system responds by producing
antibodies. Any substance that causes your immune system to produce
antibodies is known as an antigen. So a pathogen is also an antigen. An antigen
is generally anything that is foreign to (or not normally part of) your body.
Antibodies are protein molecules that have the following basic structure:
Figure 31
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Infection and Immunity
There are generally two sites on each antibody molecule that bind specifically to
a particular antigen.
The production of antibody molecules is part of your natural immunity. You are
constantly being exposed to pathogens (and other antigens) and so you produce
antibodies to build up a natural immunity to them.
There are two main cells that are involved in natural immunity: B-lymphocytes
and T-lymphocytes. Each of these different cell types is discussed below:
B-lymphocytes and the humoral response
When a pathogen enters your body, a group of cells known as B-lymphocytes
bind to the pathogen. This causes the B-lymphocytes to multiply into two different
types of B-lymphocytes.
The first type of B-lymphocyte produces antibodies that then bind to the pathogen
and help to remove it from your body. The production of antibodies by this type of
B-lymphocyte is known as the humoral response. It takes about two weeks for
antibodies to be produced and a pathogen cleared from your body.
The second type of B-lymphocyte circulates in your blood for many years after
the pathogen has first entered your body and been destroyed. If the pathogen
enters your body again at a later date, these B-lymphocytes produce and secrete
many antibodies very quickly and these help to destroy the pathogen before it
can do harm to your body and before any symptoms of the disease appear.
T-lymphocytes and the cell-mediated response
When a T-lymphocyte is involved in immunity, it is known as the cell-mediated
response.
There are several different types of T-lymphocytes.
The first type of T-lymphocyte is one of the most important cells in the immune
system because it has a regulatory role. It activates and controls B-lymphocytes,
other T-lymphocytes and other cells of the immune system.
The second type of T-lymphocyte destroys any body cell that has been infected
by a pathogen.
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Infection and Immunity
The function of macrophage
B- and T-lymphocytes are not the only cells involved in pathogen removal.
The antibodies produced by B-lymphocytes bind to the pathogen but the antibody
does not directly remove the pathogen. Instead, the antibody acts as a chemical
tag informing other cells in the immune system that the pathogen is foreign and
must be removed from your body.
One of the cells of the immune system that is involved in removing the pathogen
is called a macrophage and the process by which it removes the pathogen from
your body is known as phagocytosis. This process uses the organelle called the
lysosome. Lysosomes are sacs that contain digestive enzymes.
The process of phagocytosis is shown in Fig. 32 and the steps are outlined
below:




Firstly the macrophage recognises and binds to the pathogen
A vacuole then forms around the pathogen and it is engulfed within the
macrophage
Lysosomes within the macrophage move towards the engulfed
pathogen and fuse with the vacuole surrounding the pathogen
Enzymes are released into the vacuole from the lysosomes and the
pathogen is digested.
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Infection and Immunity
Immunity
The action of macrophage is considered to be part of the innate immune
response.
Innate immunity is a non-specific response to a pathogen. This means that a
macrophage will digest any pathogen that it encounters.
Other examples of innate immunity include:




Skin which acts as a physical barrier against infection
Acid in the stomach and in sweat. Pathogens are less likely to grow in these
acidic environments
Lysozyme which is an enzyme found in tears that kills bacteria
Interferon which is a molecule that stops viruses from replicating in your body
cells.
Naturally acquired immunity
When a pathogen enters your body naturally (for example, if you sit beside
someone who has chickenpox and is coughing and you breathe in their
chickenpox virus) your B-lymphocytes produce antibodies that help you to
remove this virus from your body. Unfortunately, this takes about two weeks, so
you get the symptoms of chickenpox too!
However, remember when the humoral response was discussed previously, a
second type of B-lymphocyte was mentioned. This other B-lymphocyte circulates
in your blood for many years after you have first had chickenpox and if the
chickenpox virus enters your body again, this other type of B-lymphocyte quickly
produces many antibodies and the virus is removed before you get the symptoms
of chickenpox again.
It is because of this natural acquired immunity that someone who has had
chickenpox as a child rarely gets chickenpox again. As you get older, your
naturally acquired immunity to many pathogens increases.
Artificially acquired immunity
Immunity can also be acquired artificially by the process of vaccination.
In the case of a vaccine, the pathogen (which has been weakened or killed in
some way) is injected into a person. This means that the person has been
artificially exposed to the pathogen. When the weakened or killed pathogen
enters their body, the immune system sets to work. Antibodies are produced
which help to remove the pathogen from their body. (Remember that the
pathogen has been weakened or killed and so does not cause any symptoms in
their body). Also, B-lymphocytes are produced that circulate in the blood and will
produce many antibodies quickly if the natural pathogen enters their body at a
later date. Thus the person has artificially acquired immunity to the pathogen.
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Infection and Immunity
An example of a vaccine is the tetanus vaccine. Tetanus is the uncontrolled
contraction of muscles and can cause death in an individual.
Tetanus is caused by a toxin produced by a bacterium. The tetanus vaccine is
made by purifying the toxin and then inactivating it to produce a toxoid.
The toxoid is injected into an individual who then makes antibodies against the
toxoid to remove it from their body. The individual also produces B-lymphocytes
that circulate in the blood and which will secrete antibodies if the naturally
occurring toxin enters their body. The antibodies that are produced are called
antitoxins. These antibodies are able to bind to and neutralise the naturally
occurring toxin produced by the bacterium.
Thus, if the individual is infected by the bacterium that causes tetanus, they can
quickly produce antitoxins that prevent the effects of the toxin.
Active immunity
This refers to the production of antibodies by an individual. The antibodies can be
made by the individual in response to a naturally occurring infection or to the
artificial injection (vaccination) of a pathogen or toxoid.
Passive immunity
This refers to an individual receiving ready-made antibodies. These ready-made
antibodies can be gained either by natural or by artificial means.
Natural passive immunity
This refers to someone receiving ready-made antibodies naturally. A baby
receives antibodies from its mother through the placenta and through breast milk.
Artificial passive immunity
This refers to someone receiving ready-made antibodies through a vaccine. For
example, if someone cuts themselves badly and if they do not have any natural
antitoxins against tetanus in their blood (they may not have kept up to date with
their tetanus vaccines), then they can be given ready-made antitoxins in a
vaccine that allows them to fight the bacteria that causes tetanus, if it has entered
their body through the cut.
Generally, natural and artificial passive immunity do not last long as the readymade antibodies are removed from the body within a few months.
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Infection and Immunity
Test yourself on infection and immunity
Spend time reviewing your notes on infection and immunity, then see if you can
answer the questions below.
1.
What do the following terms mean?
(a) pathogen
(b) antigen
(c) antibody
(d) humoral response
(e) cell-mediated response
2.
White blood cells (wbc) are involved in the immune response. Some of
these wbc are listed below:
B-lymphocytes
T-lymphocytes macrophage
Use the list to complete the following sentences:
(a) The wbc involved in humoral immunity is______________________
(b) The wbc involved in regulating the immune response is
______________________
(c) The wbc involved in phagocytosis is __________________________
3.
Put the following statements into the correct order to describe
phagocytosis:
(a) Digestive enzymes are released from the lysosomes into the vacuole.
(b) The pathogen is digested.
(c) The phagocyte recognises and binds to the pathogen.
(d) Lysosomes within the phagocyte move towards the engulfed
pathogen.
(e) A vacuole forms around the pathogen and it is engulfed within the
phagocyte.
(f) Lysosomes fuse with the vacuole surrounding the pathogen.
4.
Describe what is meant by the terms ‘active’ and ‘passive’ with reference
to immunity.
5.
Describe two ways that a person may acquire natural passive immunity.
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BIBLIOGRAPHY
Some suggested staff reading materials
The following is a commentary on some published reading materials that may be
useful when delivering Higher Biotechnology. This list is in no way exhaustive
and is meant only as a starting point for any tutor delivering the units for Higher
Biotechnology for the first time.
Foundations in Microbiology (3rd edition)
by Kathleen Park Talaro and Arthur Talaro
Published by WCB/McGraw-Hill
ISBN: 0-697-35452-0
This is a general introductory microbiology book that is a good teacher’s
resource, especially if you do not have a microbiology background. The book is
aimed at undergraduates, so it is too detailed and advanced to be used as a
student resource. But it is easy to read and has lots of good illustrations and
diagrams. There is an interactive CD-ROM that can be purchased to accompany
the book. It provides lots of detailed background knowledge on many of the
topics in all of the three units that comprise Higher Biotechnology.
Fundamentals of Microbiology (5th edition)
by I Edward Alcamo
Published by Benjamin/Cummings Publishing Company
ISBN: 0-8053-0532-7
This is another general microbiology book that is a good teacher’s resource.
Again, it is easy to read with lots of diagrams and anecdotes (although they are
all American). This book is a good source of graphs that could be the basis for
problem-solving questions. It also provides lots of detailed background
information for all three units of Higher Biotechnology.
Micro-organisms and Biotechnology (1st and 2nd editions)
by Jane Taylor
Published by Nelson Thornes
ISBN: 0-17-448255-8 (second edition)
This book is now into its second edition and may be used as a teacher and
student resource. Both the first and second edition provide background
knowledge for all three units comprising Higher Biotechnology and the book is
especially good for the enumerating micro-organisms section in Unit 2
(Microbiological Techniques). The second edition also covers some ethical
issues surrounding some biotechnology processes.
Basic Biotechnology (2nd edition)
Edited by Colin Ratledge and Bjorn Kristiansen
Published by Cambridge University Press
ISBN: 0-521-77917-0
This is a book for teachers who are enthusiasts and want to have a detailed
knowledge of biotechnology. It provides all the background knowledge (and
more!) required for delivering Unit 3 (Biotechnology).
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Bibliography
Some suggested websites
www.Biotechinstitute.org
This is an American website that has lots of biotechnology information. It has
links to biotechnology-related news stories from a range of sources e.g. ‘Nature’,
Yahoo and the BBC. There are teachers’ resources and links to other websites.
Also, you can download back copies of the magazine Your World; this is aimed at
post-16 students. Each issue covers one particular biotechnology topic and so
can be used as a classroom resource.
www.biowise.org.uk
This website provides downloadable case studies on industrial biotechnology that
may be useful for Unit 3 (Biotechnology). The case studies highlight companies
in the UK that actively use biotechnology; so they are a good introduction to
students to show the practical relevance of what they are studying.
www.sgm.ac.uk
This is the Society for General Microbiology website which has links to current
‘hot’ topics and news items, so it is a good way of keeping up to date with issues
in microbiology. It also has educational resources and links to online microbiology
resources.
www.ncbe.reading.ac.uk
This website provides downloadable protocols for practical exercises, as well as
online learning materials. It has a good section on safety issues to be taken into
consideration when carrying out biotechnology practical exercises. It also
provides information about the Scottish Centre for Biotechnology Education.
http://www-saps.plantsci.cam.ac.uk
This website has protocol information, details on how to purchase kits that can be
used as learning activities, and details of biotechnology workshops for teachers
and the annual biotechnology summer school.
www.scottishbiotech.org
This is the website of the Scottish Colleges Biotechnology Consortium who
deliver technical training to industry and schools. Online courses are available.
www.sserc.org.uk
This website provides information about the Scottish Institute of Biotechnology
Education (SIBE) who run workshops for teachers and pupils.
www.sebiotech.org.uk
This is the website of Scottish Enterprise that is dedicated to the Scottish
biotechnology industry. It is very useful for keeping up to date with the
biotechnology companies in Scotland.
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APPENDIX
Advice For Problem-Solving Outcomes
Advice for problem-solving outcomes
Unit 1: Microbiology, Outcome 3 and
Unit 3: Biotechnology, Outcome 2
Candidates are required to produce one report on a problem-solving activity as
part of the evidence for the achievement of Higher Biotechnology. The report can
be used as evidence for Outcome 3 to achieve the unit ‘Microbiology’ and for
Outcome 2 in the unit ‘Biotechnology’. The report must be the individual work of
the candidate.
One way that a problem can be solved is to carry out a practical investigation,
either as an individual or as part of a group. This enables candidates to fulfil the
required performance criteria (PC):
(a) The problem to be solved is identified.
(b) Resources required to solve the problem are identified and obtained.
(c) Procedures appropriate to solving the problem are planned and designed.
(d) The planned procedures are carried out.
(e) The problem-solving procedure is evaluated.
Alternatively, candidates can undertake a paper-based investigation by
identifying a particular problem, obtaining data from other sources (for example
biotechnology journals or the internet), then analysing, presenting and evaluating
this data.
Whichever method is used to solve the problem, it is essential to ensure that
candidates produce sufficient evidence to fulfil all the required performance
criteria. Suggestions to aid professional judgement in ensuring that performance
criteria are covered are given in the support notes of both unit specifications.
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A case study of practical investigation that was used to solve problems by
candidates in a presenting centre is describe below:
Title
‘Immobilisation of enzymes’
Introduction
As a learning activity to demonstrate immobilisation, candidates entrapped yeast
invertase within alginate beads, then assayed the immobilised enzyme by
quantitatively measuring product formed using a standard curve. (Many
experiments used as learning activities can form the basis of problem-solving
exercises.)
The problem
Following this activity, several candidates started to identify potential problems
associated with immobilisation. Some wanted to know if immobilisation changed
the pH and temperature optima of the enzyme; others wanted to know how often
the immobilised enzyme could be used before it stopped making product. Both
groups realised that these problems may be genuine in the biotechnology
industry if an enzyme is to be immobilised for commercial purposes. (Note that
these problems have a real practical application that can help in the evaluation of
the exercise.)
The procedure
These candidates used the knowledge and practical skills they had previously
gained from immobilising enzymes to identify the resources and to plan and
design their problem-solving activities.
The evaluation
The candidates found out that the pH optimum changed, the temperature
optimum stayed the same and the immobilised enzyme could be used three
times before the quantity of product decreased.
Other learning activities that can be used as the basis of problem-solving
activities are given in the support notes of each unit specification. They are as
outlined below:






Set up a small-scale laboratory fermenter and monitor and control
various conditions such as pH and temperature
Autolyse yeast and test viability at different stages in a downstream
process
Investigate the effect of pectinase, amylase, cellulose and RGase on the
production and clarity of fruit juice
Investigate the action of cellulase on cellulose
Investigate methods of removing immobilised enzyme beads from the
substrate
Analyse data on DNA profiling.
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