SBI4U
Grade 12, University Preparation
Biology
Unit 3 – Molecular Genetics
SBI4U – Biology
Unit 3 - Introduction
Introduction
In this unit, you will learn about DNA from the history of its discovery to the methods of
its manipulation in the field of biotechnology. We have come a long way since the early
1950’s when the DNA structure was first determined. In only 60 years we have moved
forward at an exponential pace. We are now using this accumulated knowledge to
discover the origin of disease, solve crimes and heal the sick. But, as our ability to
understand and manipulate this molecule of life increases, it creates its own set of legal,
moral and ethical issues that must be dealt with now and in the future.
Overall Expectations
At the end of this unit, you will be able to:
•
•
•
Analyse some of the social, ethical, and legal issues associated with genetic
research and biotechnology
Investigate, through laboratory activities, the structures of cell components and their
roles in processes that occur within the cell
Demonstrate an understanding of concepts related to molecular genetics, and how
genetic modification is applied in industry and agriculture
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SBI4U
Grade 12, University Preparation
Biology
Lesson 9 – Introduction to DNA
SBI4U – Biology
Lesson 9
Lesson 9: Introduction to DNA
Introduction
It has been philosophically suggested by some that living things are only the vehicle
designed to carry and pass on its cargo of genetic material. This may sound extreme,
but there is some validity to this statement. One of the main goals of all life is to
reproduce itself and in doing so, pass down its genetic information. In many species,
reproduction is the final act before death. In others, parents will sacrifice themselves to
ensure the continuation of their offspring.
What is the molecule of life? How does it create the blueprint for all living things?
These questions have been asked and are still being answered. This lesson will
examine the history of how the molecule of heredity was discovered and how it is able
to reproduce itself. This will be done through the examination of the various
experiments performed by such scientists as Miescher, Chargaff and Waston and Crick,
to name a few. These scientists solidly established the basic knowledge required to
move us forward to where we are today.
What You Will Learn
After completing this lesson, you will be able to
•
•
•
•
Use appropriate terminology related to DNA replication including polymerase, DNA
ligase, helicase, and Okazaki fragment
Conduct an investigation to extract DNA from a specimen of plant or animal protein
Explain the current model of DNA replication, and describe the different repair
mechanisms that can correct mistakes in DNA sequencing
Describe, some of the historical scientific contributions that have advanced our
understanding of molecular genetics
Thought Questions
Use the following questions to begin thinking about DNA and RNA.
A. What is a gene?
B. How can vastly different organisms use the same molecule for passing on genetic
information?
C. What is the difference between DNA and RNA?
D. What is the purpose of having DNA and RNA in the same cell?
E. How can the genetic information be copied time after time without creating errors?
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Discovering the Molecule of Heredity
It has been understood by people for centuries, at least at a basic level, that individuals
of the same family can look like one another. A child may have their mother’s eyes or
their father’s hair. People have even used this idea of inheriting certain traits to breed
better animals such as a cow that provides more milk, or a hen that lays more eggs or
even horses that can run faster. But, the mechanism that caused this was not
understood. The first inkling of understanding came about in the 1860’s when an
Austrian monk, Gregor Mendel, experimented with different traits found in pea plants.
He performed many plant breedings, mixing the various traits and studying the traits
that appeared in the offspring plants. His findings lead him to the conclusion that both
the maternal (female) and paternal (male) sex cells or gametes contributed equally to
the offspring’s development. He found that the information contributed by both parents
was not a blending of the two but rather, each parent passed on discrete bits of
information or factors of inheritance. There were also situations that even though both
parents contributed information, one of them may not be expressed but that this
recessive factor could be passed on to the offspring. It was many years before
scientists accepted Mendel’s ideas.
Within a few years of Mendel publishing his results, another scientist, Friedrich
Miescher, isolated a substance from the nuclei of white blood cells that he called
“nuclein”. Further study showed that this substance had an acidic portion which he
called nucleic acid and an alkaline (base) portion which was later identified to be
protein.
At that point, no connection was made between the work of these two scientists. It
wasn’t until the early 1900’s that scientists began to unravel the mystery. Phoebus
Levene continued the work of Miescher for nearly thirty years. Levene isolated the
nucleic acid substance and found out that there were two types. This acid contained
the five-carbon sugar ribose so he named it “ribose nucleic acid”. The name has been
altered slightly since then and is now referred to as ribonucleic acid or RNA. The other
acid Levene isolated contained a previously unknown sugar. Since this sugar was
similar to ribose in structure, lacking only one oxygen molecule, he named is
deoxyribose and the acid became known as deoxyribonucleic acid or DNA. In his later
years, Levene went on to determine the structure of the nucleic acids. He found that
nucleic acids are made up of long chains of individual units or subunits he called
nucleotides. Both DNA and RNA contain a combination of four different types of
nucleotides that he identified. Each nucleotide is composed of a five-carbon sugar, a
phosphate group and one of four nitrogen-containing (nitrogenous) bases. These bases
are known as adenine (A), guanine (G), cytosine (C) and thymine (T). A fifth base was
later identified as uracil (U) which is only found in RNA and replaces thymine.
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Figure 9-1 The Discovered Nucleic Acid Sugars
Source: Steven M. Carr 2008
Figure 9-2 The Nitrogenous Bases
Source: Blake et al. 219
During the time that Levene was researching the nucleic acids, another scientist
Joachim Hammerling, a Danish biologist, was studying a one-celled green alga called
Acetabularia to see if he could determine the location of the heredity information in the
cell. Although made up of only one cell, this organism grew to be approximately 5 cm
long so it was easy to work with. It also had a very unique structure which consisted of
three parts: a distinct “foot” which contained the nucleus, a stalk and a cap region that
looked something like the cap of a mushroom. In one experiment, Hammerling
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removed the cap from some cells and the foot from others. Those that had lost the cap
were able to regrow another one but those who had had the foot removed were unable
to regenerate it. This result led to the idea that the heredity information being used to
regenerate the missing portion of the cell is found within the foot and possibly the
nucleus.
Figure 9-3 Hammerling’s Experiment with Acetabularia
Source: Di Giuseppe et al. 207
All of this information from the work of Miescher, Levene and Hammerling, gave
scientists the idea that the nucleus held the hereditary information but they were unsure
as to what portion of the nuclear components were responsible. Some suspected that
is was DNA but others thought the molecule was too simple to provide such
complicated information as building an organism. Many thought the proteins were the
molecule everyone was searching for. The answer to this dilemma took a long time and
many experiments to definitively solve.
This quest began with the work of Frederick Griffith who was trying to develop a vaccine
against pneumonia caused by the bacteria Streptococcus pneumoniae. This bacterium
exists in two forms, one has a polysaccharide coating surrounding it called a capsule
while the other form has no coating. The encapsulated form is the deadly or pathogenic
strain while the one with no coating is harmless. In his experiments, Griffith killed the
pathogenic strain using heat and injected it into mice. The mice suffered no effects and
lived but when a live form of the pathogenic strain was injected, they died. This was
expected. He also performed this experiment with the non-pathogenic strain and the
mice lived as well. The surprise came when he mixed heat killed pathogenic bacteria
with live non-pathogenic bacteria and injected. One would expect that the mice would
live because the pathogenic strain was dead and the non-pathogenic strain doesn’t
harm them. Instead, the mice died. Griffith proposed that a chemical in the heat killed
pathogenic strain was able to alter the live non-pathogenic strain, making it deadly. He
called this phenomenon transformation.
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Griffith’s work was continued by another group of scientists, Avery, MacLeod and
McCarty in the 1940’s. They spent a lot of time isolating the agent behind this
transforming principle. The question being asked was whether this transforming
principle was DNA or protein. They reproduced Griffith’s experiments but with one
important difference. In one set of experiments, they treated the bacteria with a proteindestroying enzyme and in another set of experiments they treated the bacteria with a
DNA-destroying enzyme. Their results showed that the transformation still took place
when the protein was destroyed but did not occur when the DNA was destroyed. This
led them to believe that DNA was the transforming agent. Scientists questioned their
results on many levels. Some suggested that not all the proteins were destroyed by the
enzyme and could still have caused the transformation while still others believed that if
DNA was the heredity molecule, it may only work for simple organisms like bacteria.
Many still believed that DNA was too simple to be the molecule of heredity for the most
complex life forms.
Figure 9-4 Griffith’s Experiment with S. pneuminiae
Source: Blake et al. 220
Finally, in the late 1940’s to early 1950’s, two scientists provided enough solid evidence
to convince the scientific community that DNA was the molecule. Erwin Chargaff,
continued Levene’s work and studied the nucleotide composition of DNA. Levene had
mistakenly thought that all the bases were present in the nucleic acid in equal amounts
but Chargaff showed that this was not true. He found that the nucleotide composition of
DNA varied from one species to another but that within a species, the composition was
the same. Most importantly, he discovered that no matter the species, the amount of
adenine present was always exactly the same as the amount of thymine present and
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the amount of guanine always equalled the amount of cytosine. This is now known as
Chargaff’s rule. This result did lend some credence to DNA being the hereditary
molecule.
It was not until 1952, in a very elegant experiment by Alfred Hershey and Martha Chase
that the question of DNA was finally answered to everyone’s satisfaction. Hershey and
Chase used a virus called T2 that infects the bacteria Escherichia coli. This virus will
attach itself to the wall of the bacterial cell and injects its genetic material. The bacterial
cell is then reprogrammed to make virus particles. The scientists made use of the fact
that almost all of the phosphorus within the virus is found within its DNA while sulphur
was only found in the proteins. They then made the phosphorus radioactive in some
samples and the sulphur radioactive in other samples. In their experiments, the
individual radioactive samples were allowed to infect the bacterial cells. Once the
genetic material had been injected, the part of the viral head stuck to the outside of the
bacterial cell was removed by shaking the mixture up in a blender. When the bacterial
cells were analysed, they found that the samples where the DNA had been labelled
showed that the DNA had entered the bacterial cells while the sulphur labelled proteins
were not found to be present. This indicated that no viral protein entered the cells so it
must be the DNA that entered the cells and reprogrammed the bacterial cells using the
DNA. This meant that DNA was the molecule of heredity.
Figure 9-5 Hershey and Chase’s Experiment
Source: Di Giuseppe et al. 209
Once DNA was firmly established as being the genetically inheritable material, the race
was on to discover the actual structure of the molecule. Rosalind Franklin had actually
been working on DNA to try to figure out its configuration. It was already known that
each nucleic acid was a strand of many nucleotides together but its actual arrangement
was unknown. Franklin used X-ray diffraction to study the DNA molecule. This method
involves taking x-ray pictures of the molecule and then studying the resulting light areas
and shadows to figure out where there is empty space and where something solid is
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present. She found two distinct but regularly repeating patterns of shadow which
indicated that DNA was a helical structure. One pattern occurred at intervals of 0.34 nm
and the other at intervals of 3.4 nm. (One nanometre (nm) = 1/1000 of a cm.) Franklin
also observed DNA in water and concluded that the hydrophobic (water-hating)
nitrogenous bases must be located on the inside of the helical structure with the
hydrophilic (water-loving) sugar-phosphate backbone was on the outside, facing the
watery nucleus.
Figure 9-6 Franklin’s X-ray Crystallography of DNA
Source: Blake et al. 224
The collaboration between the American geneticist James Watson and the British
physicist Francis Crick led to the understanding of the structure. Using Franklin’s work,
they tried out a number of models before proposing the double-helix model that soon
became established as the definitive DNA structure. They published a paper in Nature
magazine in 1953 and received the Nobel Prize in 1962.
Support Question
(do not send in for evaluation)
1.
Create a table, briefly describing the contribution of each scientist or group to our
understanding of DNA.
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DNA – The Double Helix
Based on the work of Franklin and Watson and Crick, we now know the structure of
DNA. It is a thread-like molecule that is composed of two strands of nucleotide bases
strung together that wrap around each other. It is often compared to a ladder when it is
unwound with the sugar-phosphate backbone acting as the rails of the ladder while the
nitrogenous bases from both strands, face one another like the ladder rungs. As first
evidenced in Franklin’s X-ray picture, the spaces between the rungs is 0.34 nm and the
strands takes one complete turn every 10 bases or 3.4 nm. The fifth (5′) carbon from
the sugar of one nucleotide is connected to the hydroxyl group attached to the third (3′)
carbon of the next sugar with the phosphate group serving as the bridge between the
two nucleotides. All of the phosphate groups for a strand have the same orientation so
each DNA strand is said to have directionality with a 5′ end and a 3′ end. The two
strands are in opposing directions to each other or said to be antiparallel. By
convention, a strand is “read” in a 5′ to 3′ direction.
The nitrogenous bases also have a specific arrangement. Adenine and guanine belong
to a family of compounds known as purines which have a double-ring structure while
cytosine and thymine are pyrimidines which are single rings. A purine on one strand is
always associated with a pyrimidine on the other strand. This allows for what we know
as the complementary base pairings where adenine will only bond with thymine and
guanine will only bond with cytosine. This set of pairings also corresponds with
Chargaff’s rule that A and T are always found in equal amounts as is C and G. These
configurations allow for hydrogen bonding to occur between the complementary base
pairs. A and T can form two hydrogen bonds between them while C and G can form
three.
This structure allows the DNA molecule to be extremely stable which is a necessary
factor for a molecule that carries the genetic blueprint of an organism. The phosphate
bridges that link the sugars together are strong, maintaining the sugar-phosphate
backbone. The hydrogen bonding between the base pairs keeps the two strands
together. Finally, the hydrophilic and hydrophobic regions of the molecule keeps the
sugar-phosphate backbone on the outside and the nitrogenous bases on the inside.
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Figure 9-7 The Structure of DNA
Lesson 9
Source: Di Giuseppe et al. 214
The RNA molecule has also been studied structurally and found to be very similar. The
differences between the sugar and the bases has already been mentioned. The RNA
molecule also differs in that it is only a single strand rather than a double and as you will
find out later in the unit, serves some very different but equally important roles in the
cell.
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Support Questions
(do not send in for evaluation)
2.
The following is a segment taken from a strand of DNA: 5′-ATGCCTTA-3′. Write
out the complementary strand for this segment. Be sure to show directionality.
3.
A molecule of DNA was analyzed and found to contain 20% thymine. Calculate
the percentage of adenine, guanine and cytosine.
DNA Replication and Repair
Once DNA was found to be the molecule of heredity, this opened the door to an entirely
new series of questions. All organisms, whether single-celled or multi-cellular, have to
reproduce themselves which means all the cellular components, including the genetic
material, need to be capable of duplication. Cells perform this task through the process
of mitosis where the genetic material is replicated and equally divided between two
daughter nuclei. The cell then splits into two new, completely identical cells in a
process called cytokinesis. This leads to the question as to how the DNA molecule
replicates itself during the mitotic process. Three possible models of replication were
suggested. In conservative replication, the original DNA double helix is used as a
template to make an entirely new double-stranded copy. In semi-conservative
replication, the DNA molecule opens up and each of the two original DNA strands acts
as a template to make the opposite strand. This means that every time DNA replicates,
the double-stranded DNA molecule will consist of one new strand (the one just made)
and one original strand (the one used to make the copy). In the dispersive model, the
original DNA molecules are broken into fragments and copied. The DNA molecules
would then be a mix of bits and pieces of original and new DNA.
Meselson and Stahl designed an experiment to determine which model of replication is
used by organisms. They were able to make use of the fact that DNA contains a lot of
nitrogen in its nitrogenous bases and therefore uses nitrogen to make more bases
during replication. They used two isotopes of nitrogen, N-14, which is lighter, and N-15,
which is a little heavier. This slight difference could be detected when molecules
containing these forms of nitrogen were spun at very high speeds in a centrifuge. The
heavier molecules would settle closer to the bottom of the test tube and the lighter
molecules would be closer to the top. They allowed the bacteria, Escherichia coli, (E.
coli) to grow in N-15 until all of the DNA molecules in the organisms carried this heavier
form of nitrogen. The bacteria were then switched to a solution that only contained N14 so that all of the new DNA would use the lighter form of nitrogen. After one round of
replication, the DNA was removed and spun in the centrifuge to check its location in the
test tube. The results indicated that the DNA molecules did not settle at the bottom or
the top, but rather in the middle of the test tube. This suggested that the DNA
molecules contained both the heavier and lighter forms of nitrogen. A second round of
replication showed that the original strands that contained the heavy nitrogen had not
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been broken into pieces as the dispersive model suggested. This meant that the DNA
molecule replicates semi-conservatively. In this process, the fact that the DNA strands
are complementary to one another forms the basis for the synthesis of the new strands.
Complementary means that if one strand reads an adenine nucleotide at position six on
one strand, position six on the complementary strand would contain thymine.
Remember, adenine is complementary to thymine and guanine is complementary to
cytosine. This way, as the sequence of nucleotides is read from the template strand,
the complementary base is placed in position on the strand being synthesized.
Figure 9-8 Meselson and Stahl’s Experiment
Source: © 2008 by Sinauer Associates, Inc, http://www.nature.com/scitable/content/The-Meselson-Stahl-Experiment-18551
The Replication Process
Through the study of organisms such as bacteria, the replication process is now well
understood. The process works much the same in the more complex organisms. The
main difference is that replication in the more complicated cell occurs at many places on
the DNA molecule simultaneously because there is so much DNA to replicate. It takes
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approximately 40 minutes for bacterial DNA to be replicated while in the more complex
organisms such as humans, the process can take anywhere from 5 to 10 hours.
Because the two DNA strands are twisted around one another and held together by
hydrogen bonds, they must be unravelled and kept separate. The enzyme DNA
helicase is responsible for unwinding the helix by temporarily breaking the hydrogen
bonds holding the two strands together. Specialized single-stranded binding proteins
(SSBs) then attach to the individual strands to prevent the hydrogen bonds from
reforming. Since only small areas of DNA are being unravelled at any one time, tension
can form in the twist of the helix further up the line. If the tension were to become too
great, the strands could break apart. To prevent this, DNA gyrase cuts the two DNA
strands and allows them to swivel around each other until the increased tension has
been released and then reseals the strands.
Figure 9-9 The Unwinding of the DNA
Source: Di Giuseppe et al. 219
As soon as DNA helicase opens up the helix and the SSBs stabilize it, replication
begins. The junction where the two strands are yet to be unravelled is called the
replication fork and replication proceeds in the direction of that fork. It is the enzyme
DNA polymerase III that creates the new DNA strands but, certain conditions have to be
met before it can begin. The enzyme is only able to add new nucleotides to the 3′ end
of the strand. This means that the enzyme can only work in a 5′ to 3′ direction. As well,
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the enzyme is incapable of attaching itself to the template strand and simply stringing
the appropriate nucleotides together. Instead, another enzyme, RNA primase, must first
lay down a short string of temporary RNA nucleotides to provide an attachment site for
DNA polymerase III. The polymerase can then start attaching the complementary DNA
nucleotides to the temporary RNA nucleotides.
Figure 9-10 Formation of the New DNA Strand
Source: Di Giuseppe et al. 220
As mentioned earlier, the two DNA strands run antiparallel (in opposite directions) to
one another. This creates an additional issue for DNA polymerase III. It can only work
in a 5′ to 3′ direction. One of the strands, known as the leading strand, runs in that
same direction, so the synthesis of the new strand proceeds smoothly. The opposite
strand, known as the lagging strand, runs in the opposite direction or 3′ to 5′. So the
question is, how do you get an enzyme that only works 5′ to 3′, to work on a strand that
runs 3′ to 5′? The answer is to have DNA polymerase III move past one to two hundred
nucleotides on the template strand in a 3′ to 5′ direction without synthesizing the new
strand. It then stops, reattaches itself to a set of temporary RNA nucleotides set down
by RNA primase and then works in its normal 5′ to 3′ direction away from the replication
fork. Then, just as it does on the leading strand, it reads the order of nucleotides from
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the template strand and creates the complementary copy but this time, in short
segments or fragments named Okazaki fragments after their discoverer. It continues
this action for the entire length of the strand, moving ahead, stopping, reattaching,
making a short segment, moving forward again, stopping, reattaching and so on and so
on.
Figure 9-11 The Formation of the Lagging Strand
Source: Di Giuseppe et al. 221
Once DNA polymerase III has commenced replication and moved forward, the
temporary RNA nucleotides can be removed and replaced by the appropriate DNA
nucleotides. This action is performed by another polymerase called DNA polymerase I.
The only thing left to do is to join the Okazaki fragments together to create one
continuous strand. DNA ligase is an enzyme that forms a phohphodiester bond
between the outer nucleotides on each fragment. As the replication fork continues
forward and more of the DNA strands are synthesized, the two double-stranded DNA
molecules automatically reform their helical twist and the hydrogen bonds holding the
two strands together, form. DNA replication is now complete.
DNA Repair
Under any circumstances, when a copy is being made, there is the chance that an error
could occur during the copying process. If this were to happen regularly, the DNA of
each cell could mutate beyond recognition and no longer function properly. Therefore, it
is extremely important that DNA replication be error free or as close to error free as
possible. The process is actually very accurate. This is due to the proofreading
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mechanisms of both DNA polymerase I and III. As the new DNA strands are being
synthesized, they are re-checked to make sure that the proper base is in place. When
an error is found, both polymerases have exonuclease abilities. This allows them to cut
out an incorrect base, replace it with the correct base and then carry on with the rest of
the strand.
Figure 9-12 An Overview of the Replication Process
Source: Di Giuseppe et al. 222
Please visit the following website(s)
http://www.sinauer.com/cooper/4e/animations0601.html
Support Question
(do not send in for evaluation)
4.
Create and complete a table similar to the one following by describing the
function of each enzyme. (Do not write in this booklet)
Enzyme
DNA gyrase
DNA helicase
DNA polymerase I
DNA polymerase III
DNA ligase
RNA primase
Function
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Lesson 9
Onion DNA Extraction Activity (Support Question #5)
(NOTE: It is highly recommended that you complete this activity. It is not to
be handed in but it will illustrate the concepts explored in this lesson)
Introduction
DNA is found in the cells of all living things. It is the blueprint of life, providing
instructions at a molecular level as to how to create or build the organism itself. It is a
thread-like molecule that is folded up in such a complex pattern that very large amounts
of it are able to occupy the very small space in the nucleus.
Learning about DNA from such a scientific standpoint tends to restrict the idea of it to
specialized laboratories where strange and amazing things are done with it. We forget
that we ourselves as an organism are the product our personal genetic information.
Because DNA is such a stable molecule, it is actually quite easy to remove it from the
safety of the nucleus and see it with our own eyes. It makes it all seem a little less
mysterious.
Objectives
•
•
To understand how science is a part of our everyday lives
To visually see the substance that codes for life itself
Materials
•
•
•
•
•
•
•
•
Part of a white onion
Un-iodized salt
Regular shampoo (not a 2 in 1 just shampoo)
Cheese cloth
Baking soda
95% ethanol (this can be found as rubbing alcohol at the pharmacy)
Tap water
Two small glass containers that can be shaken without spilling the contents or
glassware deep enough to allow for extremely vigorous mixing
Method
1.
In a container, mix 120 mL of lukewarm water with 5 mL of shampoo. Add ¼
teaspoon of un-iodized salt and two teaspoons of baking soda. Set it aside.
2.
Remove the outer skin of a medium sized onion and break off pieces that would
amount to approximately 1/8 of the whole onion.
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3.
Chop and mince the onion adding a small amount of water (1-2 teaspoons) until
the onion has the consistency of applesauce.
4.
Add 5 mL of the minced onion to a glass container and add 10 mL of the
shampoo mixture that was made in step 1.
5.
Either shake or mix vigorously for 3 minutes.
6.
Using the cheese cloth, filter the mixture into a second glass container. The
“solid” portions should remain behind and the liquid portion should move to the
second container.
7.
Slowly and gently, add 15 mL of ice cold ethanol (rubbing alcohol) to the liquid
and let it sit for a couple of minutes.
8.
The DNA sample should become visible and can be spooled around a glass rod
if one is available.
Questions
Using your research skills and knowledge, answer the following questions.
a. The onion had to be minced as finely as possible. What was the purpose of
this step? Hint: What does the mincing help to break down?
b. What was the purpose of the shampoo? Hint: Think of what shampoo is
used for and what cellular structures this may relate to.
c. What was the purpose of the salt?
d. What was the purpose of the alcohol?
e. Describe the appearance of the DNA.
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Key Question #9
1.
Write a short essay explaining the evidence that DNA is the genetic material. Be
sure to use the results of the key experiments to help you with your answer.
(10 marks)
2.
Set up a chart or table compare DNA and RNA in the following ways.
a. Two ways in which they are similar. (2 marks)
b. Three ways in which they are different. (6 marks)
3.
The following sentences summarize the process of DNA replication. Fill in the
missing terms to complete the sentences. Remember – DO NOT write in this
booklet. (15 marks)
a. The enzyme _________ removes the tension by unwinding the double helix.
b. The enzyme _________ breaks the hydrogen bonds between the two DNA
strands, resulting in an unzipped helix that terminates at the
_________________.
c. ________________ attach themselves to the individual DNA strands to
prevent them from reforming hydrogen bonds.
d. The enzyme ___________ lays down RNA nucleotides that will be used by
____________ as a starting point to begin synthesis of the new DNA strand.
e. Using the template strand as a guide, __________________ adds DNA
nucleotides to the 3′ end of the growing strand.
f. Replication can proceed continuously with the ___________ strand while the
___________ strand must be made in discontinuous segments called
_____________.
g. _______________ removes the RNA nucleotides and replaces them with
DNA nucleotides. _______________ joins the segments together by forming
a _________________ bond between the nucleotides on the ends of the
segments.
h. _________________ and ______________ proofread the newly formed
strand and remove and replace any incorrectly paired nucleotides.
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SBI4U
Grade 12, University Preparation
Biology
Lesson 10 – Protein Synthesis
SBI4U – Biology
Lesson 10
Lesson 10: Protein Synthesis
Introduction
In the last lesson, you learned about the scientists and the experiments they performed
that led to our current understanding of the DNA molecule. You also examined the
process of DNA replication to understand how the genetic material is able to copy itself
as cells divide. But, DNA is only part of the equation that takes us from the blueprint to
the actual working organism. Lesson 1 focused on the genetic information and how that
information is stored and passed on. The focus of this lesson is how that genetic
information is used to create the proteins that create everything else leading to the
formation of the organism itself.
What You Will Learn
At the end of this lesson, you will be able to
•
•
•
•
Use the appropriate terminology related to protein synthesis including mRNA, tRNA,
rRNA, codon, anticodon, transcription, translation and ribosomes
Analyse a simulated strand of DNA to determine the genetic code and base pairing
by determining the base sequences of DNA for a protein, and analysing the base
sequences in DNA to recognize an anomaly
Explain the steps involved in the process of protein synthesis and how genetic
expression is controlled in prokaryotes and eukaryotes
Explain how mutagens such as radiation and chemicals can cause mutations by
changing the genetic material in cells
Thought Questions
Use the following questions to begin thinking about protein synthesis.
A. What is the role of RNA in protein synthesis?
B. How is the genetic information transformed into a working protein?
C. How do mutations in the DNA code affect the final protein product?
One Gene/One Polypeptide
Beginning with the work of Mendel, it was understood that “factors” played a role in the
transmission of traits from parents to offspring. Each physical trait, whether plant or
animal, has various possibilities or alleles that can be passed on. For example, in
plants, the colour of the flower may be pink or red or a person’s eye colour may be
brown or blue. Mendel showed with his pea plant experiments that each parent in
sexually reproducing species, contribute a possibility for each trait that makes up the
organism. Some of those alleles dominate over others so it is their information that is
expressed in the offspring. The other recessive allele is still present, but is
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overshadowed by the stronger allele. It won’t be seen in that individual but it can be
passed on to the next generation. We know today that these factors are genes and that
they direct the production of proteins whose functions in the cell are expressed as the
physical traits that we see.
In the early 1900’s Garrod first proposed the idea that each gene provided the
information to produce one enzyme. He noticed that some illnesses would appear in
many family members and hypothesized that an error in one of the enzymes could
create the illness. He called them “inborn errors of metabolism”. He proposed that
these enzymes had to be under the control of the genetic material and therefore, an
error in the genetic material led to an error in the enzyme.
Garrod’s ideas were further confirmed by Beadle and Tatum who looked at inborn errors
of metabolism in a bread mould. By studying the mutant strains, they were able to
determine the metabolic pathway responsible for synthesizing one of the amino acids.
They found that an error in one enzyme shut down the pathway at that point since the
product of that particular enzyme was not available for the next step in the pathway.
They concluded that each gene codes for an enzyme.
Vernon Ingram studied this idea very closely in humans, particularly the disease of
sickle cell anemia where the red blood cell takes on a sickle or crescent moon shape,
clogging up blood vessels and creating problems for normal circulation. He discovered
that one of the amino acids that make up the haemoglobin protein was substituted for
another amino acid. This suggested that the one amino acid change from the normal
protein was responsible for the disease. Since haemoglobin is a large protein
composed of two different polypeptide chains that come together to make the protein,
the one gene/one enzyme idea is not entirely accurate. We now know that each gene
codes for one polypeptide chain that may by itself make up a protein or it can combine
with other chains to form larger proteins. Therefore, the current hypothesis is one gene
codes for one polypeptide.
Protein Synthesis
Ribonucleic Acid (RNA)
The basic structure of the RNA molecule has already been discussed in the previous
lesson. RNA, like DNA, is a carrier of genetic information but it carries out its functions
in way that is very different from DNA. The RNA molecules can be divided into three
major classes, each of which performs a different function in protein synthesis. The
mRNA molecule or messenger RNA is responsible for creating a copy of the genetic
information contained in the DNA and moving that information from the nucleus to the
cytoplasm where protein synthesis occurs. Transfer RNA or tRNA is responsible for
supplying the amino acids in the correct order as the polypeptide chain is being
synthesized. The last form, ribosomal RNA (rRNA) plays a structural role and helps
form the ribosome which is the structure that synthesizes the polypeptide chain.
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The Genetic Code
The polypeptide chains that make up proteins are composed of subunits of amino acids
that are joined together. The protein that is formed differs from other proteins in the
number and order of the amino acids that make up its polypeptide chain. This is the
information encoded in the DNA. The number of different proteins that can be created
is almost infinite. There are 20 different amino acids that can make up the chain so a
great deal of variation is possible. Because of this large number of amino acids, three
nucleotides are required to code for each amino acid. This set of three nucleotides is
referred to as a codon. There is some redundancy in the coding where some amino
acids have one then one codon. There is one codon that codes for the start of the
protein. This start codon tells the ribosome that this is the first codon for the amino acid
chain. There are also three stop codons to let the ribosome know that this is the end of
the chain and to stop adding amino acids. A codon chart has been provided so that you
can examine the codons for all of the different amino acids.
Figure 10-1 The Codon Chart
Source: Di Giuseppe et al. 240
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Transcription
Transcription is the first phase of protein synthesis and involves the formation of the
mRNA molecule. This process is divided into four stages.
Stage One: Initiation
The formation of the mRNA is carried out by a polymerase enzyme called RNA
polymerase. When a particular protein is needed, the RNA polymerase binds to a
region on the DNA in front of or upstream of the actual gene. This is called the
promoter and it indicates where the RNA polymerase should start copying or
transcribing the information. It usually consists of a series of a few hundred bases that
are rich in adenine and thymine which means, because this pairing is only held together
by two hydrogen bonds, that it is slightly easier to open the double stranded DNA.
Once the polymerase has bound itself to the promoter, the next step in the process
begins.
Stage Two: Elongation
At this point, the polymerase starts building the single-stranded message in a 5′ to 3′
direction. It does not require a primer like DNA polymerase so as soon as it binds,
transcription occurs. The polymerase only needs one strand to make a copy. The
chosen DNA strand is the template strand and the other is the coding strand. Since the
mRNA being made is complementary to the template strand, it is the same as the
coding strand except that uracil is put in place of thymine.
Stage Three: Termination
Once the end of the gene is reached, RNA polymerase recognizes a terminator
sequence. The RNA message then dissociates from the template strand, and the
polymerase is free to transcribe new messages.
At this point, the process of transcription is completed but the mRNA needs to be
modified before it can leave the nucleus and enter the cytoplasm.
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Figure 10-2 The Process of Transcription
Source: Di Giuseppe et al. 243
Stage Four: Post-transcriptional Modification
The first modification that the primary transcript undergoes is to receive a 5′ cap. The
cap consists of a 7-methyl guanosine, a modified guanine nucleotide. It protects the
mRNA from being digested by nucleases and phosphotases as it leaves the nucleus
and plays a role in the next phase of protein synthesis. A poly-A tail, a string of
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approximately 200 adenine nucleotides, is added to the 3′ end by the enzyme poly-Apolymerase. The next modification occurs only in eukaryotes. The original DNA
consists of coding regions, called exons, that actually contain the needed information
and non-coding regions, called introns, which at this point, are thought to be made up of
unnecessary or “junk” DNA. These non-coding regions need to be removed and the
coding portion of information needs to be strung together. If the intron regions are not
removed, it will change the normal order of amino acids in the message and the protein
will not form properly. These non-coding regions are removed by molecules called
spliceosomes that are composed of RNA and proteins. Once these steps have been
completed, the mRNA transcript is ready to leave the nucleus.
Figure 10-3 Intron Removal from RNA
Source: Di Giuseppe et al. 244
Please visit the following website(s)
http://www.sinauer.com/cooper/4e/animations0701.html
http://www.sinauer.com/cooper/4e/animations0702.html
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Lesson 10
Support Questions
(do not send in for evaluation)
1.
A short fragment of a particular gene includes the following nucleotide sequence:
3′ - GGCATGCACCATAATATCGACCTTCGGCACGG - 5′
a. Identify the promoter region and explain your answer.
b. Explain the purpose of the promoter during transcription.
2.
Typically, DNA is double stranded and RNA is single stranded. Provide reasons
to account for this difference. Consider their roles in protein synthesis.
3.
Explain the ramifications to transcription if the following occurs:
a.
b.
c.
d.
The termination sequence of a gene is removed.
Poly-A polymerase is inactivated.
The enzyme that adds the 5′ cap is non-functional.
Spliceosomes excise exons and join the remaining introns together.
Translation
Translation is the second phase of protein synthesis that takes place in the cytoplasm.
In this phase, the mRNA message is read and the appropriate amino acids are gathered
and strung together in the proper order, forming the polypeptide which folds into the
working protein. There are many components involved in this phase.
The tRNA Molecule
This form of RNA is an 80 base pair sequence that folds itself, due to the interactions of
the nucleotides, into a cloverleaf structure. The bottom arm of the tRNA consists of an
anti-codon triplet that is complementary to the mRNA sequence. The open arm of the
structure bonds to the appropriate amino acid that the tRNA has been designed to
carry. If the mRNA codon is CUU, which codes for the amino acid leucine, the
complementary tRNA anti-codon will read GAA. This means that each tRNA is specific
to only one amino acid. The attachment of the amino acid to the tRNA is performed by
the enzyme aninoacyl-tRNA synthetase.
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Lesson 10
Figure 10-4 The tRNA Molecule
Source: Di Giuseppe et al. 251
The Ribosome
This organelle is composed of two subunits or parts; the small and large ribosomal
subunits. They are both made up of various proteins and the rRNA molecules. There
are a total of three binding sites. One site binds the mRNA transcript and the other two
bind tRNA molecules. The first tRNA binding site is called the peptide site or P site. It
holds the growing amino acid chain. The acceptor site or A site holds the tRNA carrying
the next amino acid to be added to the chain. The two subunits remain separate in the
cytoplasm until an mRNA transcript is available to be translated.
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Lesson 10
Figure 10-5 The Structure of the Ribosome
Source: Kimball
The Process
Just like transcription, translation can be divided up into the three main steps of
initiation, elongation and termination.
Step One: Initiation
Once the modified mRNA transcript enters the cytoplasm, the initiation step begins.
The large ribosomal subunit is attracted to the 5′ cap of the mRNA transcript and binds
the transcript to its binding site. This action prompts the small ribosomal subunit to
bind, sandwiching the transcript between the two subunits. At this point, the fully
formed ribosome moves along the mRNA transcript in a 5′ to 3′ direction. Once the start
codon is reached, the anti-codon of the tRNA carrying the amino acid methionine is
attracted to its complementary codon on the mRNA transcript and binds to the P site.
The amino acid chain is now ready to be built.
Figure 10-6 The Initiation of Protein Synthesis
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Step Two: Elongation
Once the methionine tRNA is present in the P site, the next codon which is level with
the A site is read and the appropriate tRNA binds. The enzyme peptidyl transferase
catalyzes the formation of a peptide bond between the two amino acids. When the
bond forms, the methionine is released from its tRNA and the tRNA dissociates from the
P site. Now both amino acids are attached to the tRNA molecule sitting in the A site.
The ribosome then shifts down one codon. The tRNA carrying the growing chain gets
moved to the P site and the new tRNA carrying the next amino acid moves into the A
site. Again, a peptide bond forms, this time, between amino acids 2 and 3. The tRNA is
the P site is released and the growing chain is attached to the tRNA in the A site. This
process continues until the entire amino acid chain is strung together and the stop
codon is reached.
Figure 10-7 Elongation
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Step Three: Termination
There are no tRNA’s that code for the stop codons. When the ribosome reads the stop
codon, a release factor moves into the A site. When a peptide bond tries to form, it is
obviously unable to do so. This causes the amino acid chain to simply float away from
the ribosome and fold into its proper shape. Once the amino acid chain moves away, it
prompts the dissociation of the ribosomal subunit and the mRNA transcript. The mRNA
transcript is then free to be transcribed again and the ribosomal subunits are free to
bind with another mRNA transcript and translate it.
Figure 10-8 Termination
Please visit the following website(s)
http://www.sinauer.com/cooper/4e/animations0801.html
http://learn.genetics.utah.edu/content/begin/dna/transcribe/
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This entire process from DNA to mRNA transcript to amino acid chain to folded protein
is known as the central dogma of molecular genetics.
Figure 10-9 The Central Dogma of Protein Synthesis
Copyright © 2009, Durham Continuing Education
Source: Di Giuseppe et al. 237
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Lesson 10
Support Questions
(do not send in for evaluation)
4.
The following sequence was isolated from a fragment of mRNA:
5′- GGC CCA UAG AUG CCA CCG GGA AAA GAC UGA GCC CCG - 3′
Translate the sequence into protein starting with the start codon.
5.
Explain why the process of translation has been appropriately named.
6.
Describe the consequences to protein synthesis if the following were inactivated:
a.
b.
c.
d.
spliceosomes
RNA polymerase
tRNA
ribosomes
The Regulation of Protein Synthesis
The formation of a protein is a complex process requiring a great deal of cellular energy.
It would be highly inefficient if every single gene was transcribed and translated all the
time unless the presence of its resulting protein is necessary. There are housekeeping
genes which are constantly active, producing proteins whose continuing presence is
necessary for the cell to function properly. But, many of the proteins found in a cell are
only needed in certain cell types or only needed under certain circumstances. In these
cases, it is more efficient for the cell to have mechanisms in place that regulate or
restrict protein synthesis on some level of the process. This gene regulation allows the
organism to control specific genes depending on its requirements.
There are five control points over gene expression in eukaryotic cells.
a. pre-transcriptional control - The cell is able to control how much the DNA is
exposed to the transcription enzymes. Due to the high degree of folding of the DNA,
some areas are more condensed than others. The condensed regions are generally
not accessible to the transcription enzymes and therefore transcription does not
occur while areas that are more loosely folded are accessible.
b. transcriptional control – The cell is able to control whether the exposed DNA is
transcribed. In the case of eukaryotic genes, helper proteins called transcription
factors need to bind to the promoter region of a gene before the RNA polymerase
can bind and begin transcription. Other control elements can affect how well the
transcription factors work and alter the rate at which transcription can occur.
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c. post-transcriptional control – The cell controls the rate at which the unmodified
mRNA transcript is modified into finished mRNA. The addition of the 5′ cap or polyA tail may not occur in which case the
mRNA may be unable to leave the
nucleus or, if it does, it may be digested
without its protective cap and tail. By
this method, the mRNA is made but
never makes it to the ribosome for
translation.
d. translational control - The cell is able
to control how often and how quickly the
mRNA gets translated. Regulatory
proteins in the cytoplasm can bind to the
5′ cap and prevent or at least slow down
the ability of the mRNA to bind to the
ribosome.
e. post-translational control – If the
protein needs additional processing
before it becomes functional, this
processing can be slowed down or
eliminated altogether. Still other control
factors may interfere with the protein’s
ability to leave the cell and move to the
area where the protein is required. (For
these types of proteins, one cell type
makes them but they are active in a
different type of cell.) Lastly, the cell can
control how long the protein is
functional. The addition of different
chemical groups may increase or
decrease the protein’s longevity.
Prokaryotes or bacterial cells, use some of
these same methods but because the
organism is single celled some of the
methods are not effective. The mechanism
most often used is transcriptional control.
Figure 10-10 An Overview of Protein Synthesis Regulation
Source: Blake et al. 277
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Lesson 10
The Lac Operon
An operon is a cluster of genes located on the bacterial chromosome and contains a
series of structural genes, a promoter and, between these two regions, an operator
which acts as a regulatory sequence. The structural genes within this operon code for
proteins involved in lactose metabolism. Lactose is a disaccharide sugar found in milk
consisting of one glucose molecule and one galactose molecule. As this sugar is not
always available to the bacteria, E. coli, control of gene expression is regulated. The
three genes within the operon are called lacZ, lacY and lacA. LacZ codes for the
enzyme beta-galactosidase which is able to break down lactose. LacY codes for the
enzyme beta-galactosidase permease which makes the membrane more permeable to
lactose so it can enter the cell easily. The function of the lacA enzyme is unknown at
this time.
If lactose is not present in the environment, a repressor protein, called Lacl protein,
binds to the lactose operator. Because the promoter and operator regions overlap, the
binding of the Lacl protein covers part of the promoter. This prevents RNA polymerase
from binding to the promoter and transcribing the genes. If, on the other hand, lactose
is present the repressor protein needs to be removed. Lactose itself, acts as an inducer
by binding to the repressor protein. This causes a change in the shape or conformation
of the repressor protein so it can no longer bind to the operator. Once the repressor
protein is removed, the genes can be transcribed and translated into working enzymes.
Figure 10-11 The Lac Operon
Source: Di Giuseppe et al. 256
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Lesson 10
Support Question
(do not send in for evaluation
7.
Describe the state of the lac operon system if the level of lactose is low. What
changes would take place if lactose was made available?
Mutations
When there is a permanent change in the genetic material it is known as a mutation. A
mutation may have harmful or deleterious effects, no effects or positive effects
depending on the nature of the mutation. In many cases, these mutations are induced
or caused by agents in the environment. High-energy radiation from X-rays and gamma
rays are physical mutagens because they can break through the DNA strand causing
random changes in the nucleotide sequence. Ultraviolet radiation from the sun is not as
strong but can still cause damage and is the leading cause of skin cancer. Other
mutagens are chemical mutagens which have the ability to react or interact with the
DNA molecule. Some chemical mutagens have chemical structures similar to that of
the normal nucleotides and can be incorporated into the DNA during replication causing
the polymerase confusion when trying to determine which nucleotide it is supposed to
be. Other chemicals can intercalate or insert themselves into the twists of the double
helix and interact with the nucleotides in such a way that during replication, the
polymerase misses a nucleotide or puts in extras in an effort to read the distorted code.
There are two main categories of mutations that can affect individual genes; point
mutations and frame shift mutations.
Point mutations involve a nucleotide substitution where one nucleotide is replaced by
another. In this type of mutation, only one codon is affected.
In silent mutations, the substituted nucleotide changes the codon but it still codes for
the same amino acid. (As mentioned earlier, some amino acids have more than one
codon). This means that the order of amino acids is unchanged and the protein should
not be affected. A note of caution should be used here. More recent evidence
suggests that there is an effect. When there is more than one codon for an amino acid,
some codons are more commonly used than others. If the silent mutation substitutes a
more commonly used codon for a less common one, the cell may have difficulty finding
it. This substitution does seem to effect the folding of the protein which in turn can
affect protein function but scientists do not yet understand why this effect seems to be
present. Some have suggested that the change alters the protein just enough to affect
its level of function which can have consequences in the cell.
Missence mutations occur when the nucleotide substitution causes a change in the
codon leading to a different amino acid in the protein sequence. Even one change can
have serious effects on the protein. Sickle-cell anemia, a change from normally shaped
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red blood cells and sickle shaped blood cells is caused by a single amino acid change
in the haemoglobin protein.
Nonsense mutations occur when the change in a nucleotide changes the codon
specifying an amino acid into a stop codon. This means that the protein will stop being
translated at this point. Depending on where the change occurs, the protein could be
missing a small portion or the majority of the amino acids that form the sequence.
Frame shift mutations involve either the insertion or deletion of a nucleotide causing
the entire reading frame to shift. This means that all of the codons beyond that point
are affected.
Sometimes, the mutation of the DNA can involve larger groups of nucleotides.
Translocations involve moving a segment of DNA from one area to another while in
inversions; a segment is taken out and put in backwards.
Although these any of these changes can have extreme effects on the organism, it is
believed that mutations are also at least partly responsible for evolutionary changes
which may have led to the formation of new species.
Figure 10-12 An Overview of the DNA Mutations
Source: Di Giuseppe et al. 260
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Support Question
(do not send in for evaluation
8.
For each pair of mutations, identify which type would be the LEAST harmful to
the organism and explain why.
a. substitution versus deletion
b. inversion versus substitution
Key Question #10
1.
Many antibiotics function by inhibiting protein synthesis in bacteria. Research the
antibiotic chloramphenicol. In a paragraph, explain how it inhibits protein
synthesis and why this inhibition kills the bacteria. (8 marks)
2.
You have been provided with a DNA nucleotide sequence that codes for a
hypothetical protein. You will have to transcribe the code into mRNA and then
translate that mRNA into the protein. Remember to pay attention to the start and
stop codons so once the mRNA transcript is written out, scan from the beginning
until you find the AUG start codon and divide the message up into triplet codons
FROM THAT POINT. Then translate the amino acid sequence using the codon
chart provided in the lesson. (10 marks)
DNA sequence
3′ - AACAAACAATTTCCGTATAATCTTAATACATGGTGTAATCGTACCGGGACGGTT
TCGCGAGCATAGTGTACTTATATACAGTACTACGGACATGATAGTGGGTGTAAGA
ATACCCCCGGTGTACCTTGACACGGACTTGCAGGCATGCCTACCGTCG - 5′
3.
Write out the normal sequence the first time and then write out the mutated
sequences to determine whether or not the following mutations would be harmful
to an organism. Explain your reasoning. (10 marks)
a.
AUG UUU UUG CCU UAU CAU CGU
AUG UUU UUG CCU UAC CAU CGU
b.
AUG UUU UUG CCU UAU CAU CGU
AUG UUU UUG CCU UAA CAU CGU
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c.
AUG UUU UUG CCU UAU CAU CGU
AUG UUU CUU GCC UUA UCA UCG U
d.
AUG UUU UUG CCU UAU CAU CGU
AUG UUU UUG CCU AUC AUC GU
e.
AUG UUU UUG CCU UAU CAU CGU
UGC UAC UAU UCC GUU UUU GUA
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Lesson 10
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SBI4U
Grade 12, University Preparation
Biology
Lesson 11 – Biotechnology and Gene Modification
SBI4U – Biology
Lesson 11
Lesson 11: Biotechnology and Gene Modification
Introduction
So far in this unit, the focus has been on the DNA molecule and how information coded
in the DNA is converted into a working protein. Scientific knowledge about this
molecule has advanced rapidly over the last few decades. Now scientists are focusing
on how to manipulate DNA and proteins and this area of study has formed the scientific
field known as biotechnology. Before scientists can manipulate DNA, they have to have
the tools to do so. The tools that are used are living organisms and biological
molecules. Scientists have learned to make use of their natural properties under
controlled conditions. This lesson will examine the various tools of biotechnology and
explain how they work and are controlled in a laboratory setting.
What You Will Learn
At the end of this lesson, you will be able to
•
•
Describe the functions of some of the cell components used in biotechnology such
as the role of plasmids, restriction enzymes, recombinant DNA
Describe some examples of genetic modification such as the processes involved in
the sequencing of DNA bases and the mechanisms of the polymerase chain reaction
Thought Questions
Use the following questions to begin thinking about biotechnology.
A. How can living organisms be used to help scientists manipulate DNA?
B. How can cell components be used in biotechnology?
C. What is the polymerase chain reaction and how is it used?
Biotechnological Tools
The tools of biotechnology allow scientists to remove DNA for closer study, make more
of it if the sample is small, isolate certain DNA fragments and determine the sequence
of bases that make up the code itself, just to name a few.
Restriction Enzymes
When it comes to manipulating DNA, one of the most important abilities is to be able to
remove the DNA segment of interest from the rest of the genetic material. Remember
that the set of human DNA in our cells is vast, approximately 3 billion base pairs of
information. Therefore, the entire human genome or set of genetic material is very
awkward to work with if studying only a very small segment.
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Restriction enzymes or endonucleases act as molecular scissors, cutting double
stranded (ds) DNA at very specific points. These enzymes are found naturally in
bacteria and each species has a specific set of them. Evidence suggests that bacteria
use them as a primitive immune system. Just as there are viruses that invade human
cells, there are viruses that infect bacterial cells. Once bacteria detect the presence of
viral DNA, the restriction enzymes work by cutting up the foreign DNA. Each enzyme
recognizes a specific base pair sequence called a recognition site that is usually four to
eight base pairs long. If the viral DNA carries this sequence, the enzyme will attach
itself at that site and cut the DNA. Once the viral DNA is in pieces, it is unable to disrupt
the cell. Scientists have isolated many different restriction enzymes and determined the
recognition site for each. Now, these enzymes can be used under controlled laboratory
conditions to cut up DNA into fragments which are more manageable to study.
These enzymes can cut the DNA in two different ways. Some enzymes cut the dsDNA
in a staggered fashion between the two strands. This leaves overhangs on each end of
the cut and are referred to as sticky ends. These tend to be the preferred enzymes
because the fragments with sticky ends can be joined to other fragments with sticky
ends, when cut by the same enzyme because of complementary base pairing. Other
enzymes cut straight through both DNA strands creating an even cut. These are called
blunt ends.
Each of these enzymes is named after the bacteria from which they originate. For
example, HindIII is named as follows. H represents the genus Haemophilus, in
represents the species influenzae, d represents the strain Rd and III means that it was
the third enzyme isolated from this strain. Scientists have discovered approximately
2500 restriction enzymes which have about 200 different target sites. This gives
scientists a great deal of choice in the molecular scissors that they use.
Figure 11-1 Restriction Enzymes
Source: ©2006 BarleyWorld.org / http://barleyworld.org/css430_09/lecture%208-
09/figure-11-04.JPG
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Lesson 11
Please visit the following website(s)
http://www.dnalc.org/ddnalc/resources/animations/restriction.swf
Support Question
(do not send for evaluation)
1.
The following DNA sequence was digested with SmaI:
5′- AATTCGCCCGGGATATTACGGATTATGCATTATCCGCCCGGGATATTTTAGCA-3′
3′-TTAAGCGGGCCCTATAATGCCTAATACGTAATAGGCGGGCCCTATAAAATCGT-5′
SmaI recognizes the sequence CCCGGG and cuts between the C and the G.
a. Identify the location of the cuts.
b. How many fragments will be produced if SmaI digests this sequence?
c. What kind of ends does SmaI produce?
Gel Electrophoresis
Once DNA has been cut up into smaller fragments, they can be separated from each
other and ordered according to their size using gel electrophoresis. This process takes
advantage of the chemical and physical properties of DNA. All DNA molecules carry a
negative charge because of the phosphate groups attached to the molecule. The size
of each DNA fragment is consistent with the number of base pairs that form it so
fragments that have the same number of base pairs will be the same size. Gel
electrophoresis acts as a molecular sieve separating the fragments according to their
size using their negative charge as a means of moving them.
The gel itself is a rectangular slab composed of a polysaccharide called agarose which
is extracted from kelp seaweed. When agarose is mixed with a solution of electrolytes
(water full of charged ions) it solidifies into a gel, about 5 times denser than Jell-O. On
a molecular level, the gel is composed of a series of molecules that link up to create tiny
pores throughout.
The cut up DNA can be physically placed within the gel. When an electrical current is
applied, the DNA fragments will move away from the negative electrode and towards
the positive electrode. As the DNA fragments weave their way through the gel, smaller
fragments will be able to move through the pores faster than larger fragments. This will
separate the fragments based on size with the smaller fragments moving further than
the larger fragments in the same amount of time. Picture an obstacle course and think
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of how quickly one person could move through it compared to a group of people holding
hands.
Once separation is complete, the gel is exposed to a dye that will allow the DNA
fragments to become visible. The most common dye is ethidium bromide, a molecule
that attaches itself to the rungs of the DNA ladder. When it is exposed to UV light, the
ethidium bromide will fluoresce or glow, lighting up the fragments.
Figure 11-2 An Agarose Gel
Source: Di Giuseppe et al. 283
Please visit the following website(s)
http://www.dnalc.org/ddnalc/resources/animations/gelelectrophoresis.swf
http://learn.genetics.utah.edu/content/labs/gel/
Support Question
(do not send for evaluation)
2.
The following sized fragments were obtained after digestion o a sequence using
restriction enzymes: 5.3 kb, 2.1 kb, 6.8 kb, 3.6 kb, 1.1 kb, 2.2 kb. Order the
fragments, on a gel with the slowest migration at the top and the fastest migration
at the bottom. Which two fragments would be the hardest to separate? Assume
all the fragments were run in one lane. Label the positions of the positive and
negative electrodes.
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Plasmids
Plasmids are another piece of cellular machinery that can be manipulated by scientists.
They are small, circular pieces of dsDNA that are found in the cytoplasm of various
strains of bacteria. Their natural role is to provide extra genetic information for the
bacterial cell. In many cases they carry the genes that allow the bacteria to be resistant
to antibiotics.
Because they are separate and much smaller than the chromosomal DNA, the plasmids
are easy to remove from the cell. The circular ring of DNA can be cut open and other
genes can be added to the ring. When the plasmid is placed back into the cell, the
cellular machinery will express the foreign gene by synthesizing the protein. This is an
example of recombinant DNA where the genes of two different species have been
mixed and are present in one organism.
The Techniques of Biotechnology
Gene Cloning
This technique involves using bacterial plasmids to express a foreign gene from another
species. Using bacteria is an easy choice because it is easily contained in large vats
under controlled conditions. In this technique, the gene in question is cut from its
original species using a restriction enzyme. Gel electrophoresis can then be used to
isolate it from the rest of the fragments. The plasmid is then removed from the bacterial
cell and cut with the same restriction enzyme so that the sticky ends of the foreign gene
and the plasmid match. Due to the complementary base pairing, the sticky ends will
match up and DNA ligase can be used to seal the foreign gene into the plasmid. The
plasmid can then be reintroduced into the bacterial cell. As the bacteria divide and
grow, the plasmid will be replicated, until the entire bacterial colony carries the
recombinant DNA plasmid and expresses the protein that the gene encodes. The
resulting protein product can then be isolated from the bacteria and purified for other
uses.
The best example of this technique is the production of human insulin for diabetics. In
the past, diabetics injected pig insulin. Insulin forms in the pancreas so these organs
were literally collected, ground up and the insulin was isolated from them. For the last
decade or so, diabetics have been able to use human insulin that was produced in
bacterial cells. The human insulin gene was isolated and recombined into plasmid
DNA. Once it was placed back into the bacterial cells, the bacteria began producing
human insulin which could then be isolated and purified and given to diabetics for their
use.
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Figure 11-3 Gene Cloning in a Plasmid
Source: Di Giuseppe et al. 286
Restriction Fragment Length Polymorphism (RFLP)
This technique makes use of gel electrophoresis and the fact that DNA fragments of
different lengths will move to different positions on the gel based on the size of the
fragment. In this technique, the DNA sample in question is cut with one or more
restriction enzymes to create a series of fragments. The fragments are run on the gel to
separate them out. Because of the amount of DNA that is used, the fragments actually
appear as a large smear on the gel. Then to actually be able to see the fragment
pattern and create a more permanent record of that fragment pattern, a nylon
membrane is placed over the gel. The DNA fragments in the gel will move directly onto
the membrane in the same relative position they held on the gel. Short, complementary
stretches of radioactive DNA are exposed to the membrane. Because of
complementary base pairing, these radioactive probes will bind to the DNA fragments
on the membrane. When the membrane is exposed to an X-ray film, the radioactivity
will mark the film, creating a permanent record of the pattern of the fragments.
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But what makes this technique overly different from basic gel electrophoresis?
Technically, they are relatively the same. But, this technique is used on the regions of
DNA varies greatly from person to person.
In the DNA itself, there are regions of DNA that we know, code for all of the genes that
make up the organism. But, a great deal of the genetic code is essentially “junk DNA”
that, at least at this point in time, is considered to be extra filler that serves no purpose.
When it comes to the coding regions of DNA, there will be little variation between
individuals of the same species. The gene codes for the proper order of amino acids
that is needed to make the protein. As was discussed in the last lesson, any true
variations in the DNA code would change the order of amino acids and the protein
would not function properly. Therefore, mutations cannot accumulate in these regions
of the DNA because it would lead to non-functional proteins and possibly the death of
the organism.
In the non-coding regions of junk DNA, it doesn’t affect the organism if changes occur in
the DNA code because these regions don’t produce a protein. Therefore, generation
after generation of accumulated changes has no effect. In sexually reproducing
organisms, the combining of the maternal and paternal DNA in the offspring leads to a
unique nucleotide sequence in these areas because of the accumulated changes. This
means that if one of these regions was cut up with restriction enzymes, each individual’s
pattern of fragments would be different.
DNA fingerprinting is an excellent example of how this technique can be used. Just as
the fingertip fingerprints are unique to the individual, the DNA fragment pattern is
individual as well. This means that the DNA pattern of fragments can be used to match
a sample of DNA to an individual. Because of the number of people on the planet, it is
possible for more than one person to generate that same fragment pattern when the
DNA is cut with one restriction enzyme just due to chance alone. Therefore, when this
type of analysis is performed, several junk DNA areas are tested with multiple enzymes.
This lowers the chance of two individuals having the same pattern to be one chance in a
few billion or even a trillion.
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Figure 11-4 The RFLP Process
Source: Di Giuseppe et al. 299
Support Question
(do not send for evaluation)
3.
A camper is caught by authorities frying a fish that
is believed to be out of season. Luckily, a piece
of the fish has yet to be subjected to the
barbecue. The authorities want to ensure that
this piece of fish is definitely the type of fish in
question. DNA fingerprinting is conducted on the
fish using RFLP analysis. The gel pattern shown
below is obtained. Determine whether the
investigation should continue, given this
evidence.
Source: Di Giuseppe et al. 321
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Lesson 11
The Polymerase Chain Reaction (PCR)
This technique is used to amplify a small sample of DNA into a large sample. Just as a
teacher will make copies of the original assignment to hand out to the class, the original
DNA can be used to make copies of itself to increase the amount that is present. In this
technique, the natural process of DNA replication and all the biological molecules
involved can be controlled and force replication to occur over and over again until
multiple copies of the DNA have been made.
So, thinking back to DNA replication, what components are needed to replicate DNA?
The double stranded DNA sample must be present to act as the template, a polymerase
to make the new DNA strands, primers for the polymerase to attach to in order to begin,
and a supply of nucleotides to string together to form the new strands.
All of these components are introduced into a test tube. DNA cannot be copied while in
its double stranded form so rather than using the helicase, gyrase, and single stranded
binding proteins, heat is used instead to fully separate the strands. DNA will denature
or separate at 95°C. Once in single stranded form, the primers which are
complementary to the DNA sequence will bind on through complementary base pairing.
The polymerase will then attach and using
the strand as a template, string together
the complementary strand. The DNA is
now replicated and in double stranded
form again. Additional heat is added and
all of those double stranded DNA
molecules will separate, primers will bind
and the polymerase will make another set
of complementary strands. This cycle will
occur over and over again until the original
DNA sample has been amplified to the
desired level. Remember, since each
strand of the double stranded DNA is used
as a template, the amount of DNA will
double with every cycle. After just 20
cycles, a single copy of DNA is amplified
over two million times.
This technique seems relatively
straightforward but there is one small trick.
Proteins, or more specifically enzymes, fall
apart when exposed to high heat levels.
This means that the polymerase should
stop working after the first cycle when heat
is applied to the system. This problem is
solved by using a very thermostable or
heat stable form of polymerase. The bacteria,
Thermus aquaticus, live in hot springs and have a
Copyright © 2009, Durham Continuing Education
Figure 11-5 PCR
Source: Blake et al. 295
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polymerase capable of tolerating high heat. This is the form of the polymerase that is
used for it is able to stay functional despite the temperature.
Please visit the following website(s)
http://learn.genetics.utah.edu/content/labs/pcr/
http://www.dnalc.org/ddnalc/resources/animations/pcr.swf
Dideoxy or Sanger Sequencing
This technique was developed by Frederick Sanger as a method of determining the
base pair sequence of a DNA fragment. This means being able to read the order of
nucleotides that code the genetic information. This is another technique that makes use
of the process of DNA replication. It uses single stranded DNA, radioactive primers,
DNA polymerase, the four nucleotide bases A, C, T and G and four test tubes. All of
these components are put into each test tube. In addition, each tube contains one type
of nucleotide bases that have been altered. This altered base is called a dideoxy
analogue. Therefore, all four test tubes will have all the required components plus one
dideoxy analogue. For example, test tube one will have the dideoxy A analogue, test
tube two will have the dideoxy C analogue, test tube three will have the dideoxy T
analogue and test tube four will have the dideoxy G analogue.
These analogues are altered in that the sugar is missing a hydroxyl group (-OH) on the
3′ carbon where the phosphodiester bond would normally form. This means that during
replication, if one of the analogues is added in, the chain will stop forming because no
other nucleotides can be bonded onto the end of it. If the analogue is present in the test
tube at low concentrations, by chance, it will be added to the growing DNA strand once
in a while. During the replication process, the DNA polymerase will use the single
stranded DNA as a template and produce the complementary strand. In each test tube,
as the bases are required, they will be incorporated into the growing chain and
occasionally the analogue will be added. This means that every test tube will be able to
replicate the entire strand sometimes and in other cases, only fragments of the whole
strand will be produced. Each test tube will contain a series of fragments of different
sizes that can be run out on a gel and separated. Gels can be run that are sensitive
enough to separate fragments that are only one base different in size. If the resulting
fragments of each test tube are run on a gel in different lanes (like lanes on a highway),
the different fragment sizes will spread out and the gel can be read.
So how does this lead to the sequence of bases being identified? In test tube one, if
one of the fragments is only 3 nucleotides long, it means that the last nucleotide in the
fragment was the dideoxy A analogue. This tells researchers that the third base in the
sequence is an A. If test tube two, carrying the dideoxy C analogue, generates a
fragment 4 bases long, then C would be the fourth base. Knowing this, researchers can
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start at the bottom of the gel at the one base pair long fragment, see which test tube
containing which altered base it came from and know that has to be the first base. They
then find the 2 base pair fragment, then the three and so on and so on, walking up the
gel and reading the sequence.
Figure 11-6 DNA Sequencing
Source: Blake et al. 298
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Lesson 11
Please visit the following website(s)
http://www.dnalc.org/ddnalc/resources/animations/sangerseq.swf
Support Question
(do not send for evaluation)
4.
The gel pattern shown below was produced using the
Sanger dideoxy sequencing method. Determine the
sequence of the fragment that produced this gel.
Key Question #11
1.
Source: Di Giuseppe et al. 321
The DNA fragment shown below was digested with HindIII, BamHI and SmaI.
a. What sized fragments would be produced if it is digested only with HindIII?
(3 marks)
b. What sized fragments would be produced if it is digested only with BamHI?
(3 marks)
c. What sized fragments would be produced if it is digested only with SmaI?
(3 marks)
d. Draw the gel electrophoresis pattern that would be obtained from the
digestions in a, b and c. Place each digest in its own lane on the gel.
(3 marks)
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Lesson 11
2.
A genetic engineer wants to isolate gene A and splice it into plasmid B. Write out
the steps that would be used. (8 marks)
3.
PCR is based on the natural process of DNA replication that occurs in the cell. In
a comparative format, list the steps of DNA replication and PCR. Indicate how
the steps of DNA replication are accomplished outside the cell using PCR.
(6 marks)
Copyright © 2009, Durham Continuing Education
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SBI4U
Grade 12, University Preparation
Biology
Lesson 12 – Social, Ethical and Legal Implications of
Biotechnology
SBI4U – Biology
Lesson 12
Lesson 12: Social, Ethical and Legal Implications of
Biotechnology
Introduction
Now that you have been introduced to some of the tools and techniques and hopefully
understand how they work, this lesson will extend your knowledge by examining the
application of these techniques. Many of the techniques you examined have specific
purposes, being used to answer very specific questions. But, many of these techniques
can be applied to the bigger picture of biotechnology. For example, recombinant DNA
technology can be used to place a gene from one species into another so we can make
human insulin in bacteria. But, how can we expand the application of this process or at
least the idea of moving genes from one species to another? We are now using these
techniques to design better crops, adding in genes that provide plants with better
resistance to drought or insects, or making them more nutrient rich. The possibilities
are truly endless. Even though we can do it, the controversy arises with the debate as
to whether we should do it. This lesson will focus on some of the many broader
applications that utilize the techniques that were examined and the implications for
doing so.
What You Will Learn
By the end of this lesson, you will be able to
•
•
•
Describe some examples of genetic modification and explain how it is applied in
industry and agriculture
Analyse, on the basis of research, some of the social, ethical and legal implications
of biotechnology
Analyse, on the basis of research, some key aspects of Canadian regulations
pertaining to biotechnology
Thought Questions
Use these questions to start thinking about the issues surrounding biotechnology.
A. Should scientists be able to move genes from one species to another?
B. If scientists sequence a region of DNA should they be able to patent it?
C. Should the information contained in your DNA be private or should the government
or insurance company have access to it?
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Genetically Modified Organisms (GMO’s)
The term GMO’s is most commonly used to refer to crop plants created for human or
animal consumption using molecular biology techniques. These plants have been
modified in the laboratory to enhance desired traits. In simple language, new genes can
be introduced into the plant’s genetic material to change the plant in some way. The
techniques involve isolating the desired gene from whatever species originally carries it
and transferring it to the new species. It involves using restriction enzymes and various
methods to get the DNA into the plant cells so the gene can be expressed. The ultimate
goal when altering plants and animals is to get this new DNA into the germ line so that
rather than having to physically alter each plant, the new genetic material can be
passed on and expressed in future offspring.
The Advantages of GMO’s
Pest Resistance – Crop losses from insect pests can be staggering, resulting in
devastating financial loss for farmers and starvation in developing countries. Farmers
typically use many tons of chemical pesticides annually. Consumers do not wish to eat
food that has been treated with pesticides because of potential health hazards and runoff of agricultural wastes from excessive use of pesticides and fertilizers can poison the
water supply and cause harm to the environment. Growing foods such as B.t. corn can
help eliminate the application of chemical pesticides and reduce the cost of bringing a
crop to market. B.t. corn is a transgenic (containing genes from a different species)
variety of corn that is engineered to produce Bt toxin. This toxin is naturally produced
by the bacteria Bacillus thuringiensis and is a natural herbicide against the corn borer.
Herbicide Tolerance – For some crops, it is not cost-effective to remove weeds
physically so farmers will often spray large quantities of different herbicides (weed-killer)
to destroy them. These herbicides will end up in the ground water and the drinking
supply. Some crops can be engineered to be resistant to one very powerful herbicide
that could be sprayed so the plants survive but the weeds do not. This would reduce
the amount of spraying needed. For example, the company Monsanto has created a
strain of soybeans that is unaffected by the herbicide Roundup. The weeds can be
killed by the crop is unaffected.
Disease Resistance – There are many viruses, fungi and bacteria that cause plant
diseases. Plant biologists are working to create plants with genetically-engineered
resistance to these diseases.
Cold Tolerance – Unexpected frost can destroy sensitive seedlings. An antifreeze
gene from cold water fish has been introduced into plants such as tobacco and potato.
With this gene, these plants are able to tolerate cold temperatures that would normally
kill the seedlings.
Drought and Salinity Tolerance – As the world population grows and more land is
used for housing instead of crops, farmers will need to grow crops in locations
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previously unsuited for plant cultivation. Creating plants that can withstand long periods
of drought or high salt content in the soil and groundwater will help people to grow crops
in formerly inhospitable places.
Nutrition – Malnutrition is common in third world countries where impoverished people
rely on a single crop such as rice for the main staple in their diet. However, rice does
not contain adequate amounts of all necessary nutrients to prevent malnutrition. If rice
could be genetically engineered to contain additional vitamins and minerals, nutrient
deficiencies could be alleviated. For example, blindness due to vitamin A deficiency is a
common problem in third world countries. Researchers at the Swiss Federal Institute of
Technology- Institute of Plant Sciences have created a strain of “golden” rice containing
an unusually high content of beta-carotene (vitamin A). The institute hopes to offer it to
any third world country that requests it. Plans were also underway to develop a golden
rice that also has increased iron content. However, the grant that funded the creation of
these two rice strains was not renewed and may not come to market at all.
The Criticisms Against GMO’s
Environmental activists, religious organizations, public interest groups, scientists and
government officials have raised concerns about GMO’s. Agribusiness is criticized for
pursuing profit without concern for potential hazards and the government is criticized for
failing to adequately regulate it. Most concerns about GMO’s fall into three categories:
environmental hazards, human health risks and economic concerns.
Environmental Hazards – Some studies have indicated that monarch butterfly
caterpillars that eat the pollen from transgenic Bt corn had high mortality rates of
approximately 56%. They don’t eat the corn pollen but people are concerned that the
pollen may blow onto the milkweed plants that they do eat. It is not possible to develop
a Bt toxin that will only kill the insect pests and remain harmless to all other insects.
Other studies have disproven these claims so the controversy continues.
Some people are concerned that just as mosquito populations developed resistance to
pesticides, other insects may develop resistance to the Bt toxin rendering it useless.
Another concern is that plants containing the transgenic gene developed for resistance
to Roundup may cross pollinate with the weeds and confer Roundup resistance to the
weeds making them almost impossible to kill. There is evidence that the transgenic
corn will cross pollinate. The company Monsanto has actually sued some farmers for
harvesting GM crops after obtaining them from an unknown source and not paying
royalties for their use. The farmers claim that GM corn crops cross pollinated their
unmodified crops from a field or two away. Some farmers have actually been found
guilty and been forced to pay the company.
Human Health Risks – Many children in North America and Europe have developed
life-threatening allergies to peanuts and other foods. It is possible that introducing a
gene into a plant may create a new allergen or cause an allergic reaction in susceptible
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individuals. A proposal to incorporate a gene from Brazil nuts into soybeans was
abandoned because of the fear of causing unexpected allergic reactions. Extensive
testing of GM foods may be required to avoid the possibility of harm to consumers with
food allergies. Labelling of GM foods and food products will acquire new importance.
There is also a growing concern that introducing foreign genes into food plants may
have unexpected and negative impacts on human health. A recent study examined the
effects of GM potatoes on the intestines of rats and found it did cause a change. This
study has been criticized saying its data is flawed. On the whole, scientists believe that
GM foods do not present a human health risk.
Economic Concerns – Bringing a GM food to market is a lengthy and costly process
and the companies that design them want to make a profit. Many new plant genetic
engineering technologies and the GM plants themselves, have been patented and
patent infringement is a big concern to those companies. Consumer advocates are
worried that patenting these new plant varieties will raise the price of seeds so high that
small farmers and third world countries will not be able to afford seeds for GM crops,
thus widening the gap between the wealthy and the poor.
Patent enforcement may also be difficult, as shown with the farmers that involuntarily
grew Monsanto-engineered strains when their crops were cross pollinated. One way to
combat possible patent infringement is to introduce a “suicide gene” into GM plants.
These plants would be viable for only one growing season and would produce sterile
seeds that do not germinate. Farmers would need to buy a fresh supply of seeds each
year. This would be financially disastrous for third world farmers who cannot afford to
buy seed each year.
There are definite benefits and drawbacks to GMO’s with both sides having legitimate
points. Research is going to continue in this area and more GMO’s are going to appear
on the market in the future.
Support Question
(do not send for evaluation)
1.
List 5 advantages and 5 disadvantages of GMO’s.
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Lesson 12
Gene Therapy
Many medical conditions result from a mutation of a gene or genes creating a
malfunctioning protein. When the protein does not function properly, cells that rely on
the protein cannot behave normally which in turn can affect whole tissues or organs.
Gene therapy provides a way to fix the problem at the source; adding a properly
functioning gene to the cells that need to utilize the protein that it makes. But, gene
therapy is not a simple solution and it does not automatically fix the problem forever.
This form of therapy is still in its infancy and scientists still have a lot to learn.
A large number of diseases would benefit from this process but not all diseases are
good candidates for this approach. Before a gene therapy is designed for any disorder,
it must meet a number of certain conditions.
¾ The disorder must be the result of a mutation in one or more genes.
¾ The gene or genes that carry the mutation must be known and a copy of that gene
needs to be available for laboratory study. The best candidates are single mutation
genes because they are much simpler to treat.
¾ It is important to understand the biology behind the disorder. You need to know
what tissues are affected, the role the malfunctioning protein plays in the cell and
how the mutation affects the protein’s function.
¾ There must be a method to deliver the correct copy of the gene to the cells of the
affected tissue.
One of the biggest issues for gene therapy is being able to successfully deliver the good
copy of the gene to the right tissues. If the delivery is successful, the gene will enter the
target cells but this is just the first step. Once the gene enters the cell it must still enter
the cell’s nucleus and be activated. This means that it needs to be transcribed and
translated and the protein must function properly. For the best results, the gene needs
to integrate itself into the cell’s DNA. But, perhaps even more importantly, harmful side
effects need to be avoided. Anytime an unfamiliar substance is introduced into the
body, there is a risk that it will be toxic or that the body will mount an immune response
against it. If the body develops immunity against the gene delivery system, future
rounds of therapy will be ineffective because the delivery system, including the good
gene will be destroyed.
So how is a good copy of the gene delivered to the cells that need it? These delivery
systems or transports are called vectors. Again, as in any molecular biology technique,
scientists use biological organisms and molecules that perform the task naturally and
simply manipulate and control how the system works. Many of the vectors used in gene
therapy are viruses that have been altered. Viruses are very good at entering cells in
the human body and causing illness. Scientists have removed the parts of these
viruses that make us sick but have maintained the part of the virus that knows how to
get into human tissues. The good copy of the gene is inserted into the virus and
hopefully the virus delivers it to the cells. There are many viral vectors that can be
used, depending on what cells in the body are targeted for therapy and the size of the
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Lesson 12
gene that needs to be delivered. Scientists are also working with some non-viral
vectors but so far, they are not as effective in delivering their genes into the target cells.
There are six main vectors that are used in gene therapy today.
Retrovirus – This is an RNA virus that uses an enzyme called reverse transcriptase to
produce a DNA copy of its genetic information which can integrate into the host’s cells.
The most well known example of this type of virus is HIV. It can carry up to 8 000 base
pairs (bp) of genetic information. One advantage is that it only infects dividing cells but
it tends to integrate into the host cell’s genome at random. This could cause disruption
of another gene which could negatively affect the cell.
Adenovirus – This is the common cold virus. It can carry up to 7 500 bp of information.
There are definite advantages to using this virus. It infects both dividing and nondividing cells very effectively and it is possible to target specific cell types by
engineering proteins on the virus surface that recognize special proteins on the surface
of the target cell. The disadvantages are that it will not integrate into the host’s genome
so after a period of time, the cell will discard the virus and gene activation will be lost. It
can also cause an immune response.
Adeno-associated virus – This is an offshoot of the cold virus but it lacks some of the
negative effects. It can carry only a maximum of 5 000 bp. It does not cause illness in
humans and infects a wide range of cell types very effectively. Like the adenovirus, it is
possible to target certain cell types and it will integrate into the host cell’s genome 95%
of the time in a specific region on chromosome 19, greatly reducing the chance that it
will disrupt the function of other genes. But, there is a 5% chance that it will integrate
and disrupt another gene.
Herpes-simplex virus – This virus will carry up to 20 000 bp of genetic information. It
infects the cells of the nervous system. It will not integrate into the host’s genome but
acts as a circular piece of DNA that replicates when the cell divides and stays in the
nucleus for a long time. Unfortunately, it can cause an immune response.
Liposomes – These are plasmid DNA surrounded by a mini membrane that can fuse
with the cell membrane. The DNA is released and transported into the nucleus. Any
reasonable amount of DNA can be used. It will not generate an immune response but
will enter cells less effectively than a virus and cannot integrate with the host’s genome.
Naked DNA – This is simply plasmid DNA that can be taken up by some cells and
transported to the nucleus. It can handle any size DNA, will not generate an immune
response and is generally not toxic. But it is not taken up by cells very effectively, will
not integrate into the host’s genome and is unstable in most body tissues.
Once scientists have studied the condition and understand the mutation, the most
appropriate of these vectors will be chosen.
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Successes and Failures
So far, only a small number of disorders have been treated with gene therapy. One of
the more successful treatments has been with Severe Combined Immune Deficiency or
SCID where individuals have virtually no immune system. This therapy provided the
sufferers with a proper form of the gene and they were cured. Unfortunately, two
patients had the gene integrate into the middle of a gene that helps regulate the rate at
which cells divide and they developed leukemia which has since been treated.
In 1999, Jesse Gelsinger who had a rare liver disorder participated in a gene therapy
treatment. He died because of an immune response to the engineered adenovirus
vector. His death caused gene therapy trials to halt for a period while scientists learned
more about the immune system and what it does.
One new therapy that has emerged with great success is the treatment of Leber
congenital amaurosis, an eye disorder that causes progressive loss of vision leading to
complete blindness around age 40. This condition can be caused by mutations in
several genes including RPE65. This gene manufactures an enzyme needed by the
light sensing rods and cones in the eyes. A tiny dose of the proper RPE65 gene was
inserted into an adeno-associated virus and injected into the retina of the eye. The
gene stayed in place like an extra chromosome and produced the working enzyme.
The retinal cells do not divide so the new gene will stay for a long time. So far, all the
results have been positive with all patients reporting an improvement in their vision.
Scientists have recently discovered a large number of variants of adeno-associated
virus. Some of the variants seem to have an affinity for certain tissues which would help
in delivering genes to specific tissues. The range of disorders that benefit from gene
therapy will no doubt expand in the future and hopefully the safety issues will become a
thing of the past.
Please visit the following website(s)
http://learn.genetics.utah.edu/content/tech/genetherapy/
Support Question
(do not send for evaluation)
2.
Describe some of the benefits and some of the risks of gene therapy.
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Lesson 12
Animal Cloning
There has been a great deal of controversy surrounding the idea of cloning animals and
one day even humans. Cloning, at its most basic, involves creating a genetically
identical organism. This occurs naturally when identical twins or triplets are born when
a single fertilized egg divides in the early stages of development. Embryo twinning has
been used for many years in the agricultural industry as the cells of a growing embryo
are manually divided and placed in the uterus. This has been used for prized cattle,
sheep and horses.
The cloning of today is a little more high tech, creating a genetically identical organism
from regular body cells rather than sex cells. The process used is called somatic cell
nuclear transfer. There are five basic steps involved in the process:
1.
2.
3.
4.
5.
the donor’s somatic (body) cells and egg cells are isolated
the nucleus is removed from the egg cell
the nucleus from the somatic cell is transferred into the egg cell
the egg cell is chemically stimulated to start dividing
the growing embryo is transplanted into a surrogate mother
The first organism to be cloned in this manner was a frog. The most famous cloned
animal was Dolly, the sheep, born in 1997. She was the only one of 277 cloned
embryos to survive. She was put down in 2003 suffering from a virus-born form of lung
cancer common in sheep. During her lifetime, she had six lambs under natural
conditions. Since then, a range of different animals have been cloned in various places
around the world. The following is a list of some of those animals.
Goat- June 16, 2000 in China
Housecats- Dec 21, 2001 Texas, USA
White-tailed deer- May 23, 2003 Texas, USA
Horse- May 28, 2003 Italy
Dog- April 24, 2005 Korea
Camel – April 8, 2009 Dubai
Reasons for Cloning
Scientists have given a number of reasons for the importance of continuing research
into cloning.
Cloning Animal Models of Disease – Most research about human diseases uses
animal models. It takes a long time to genetically engineer a transgenic animal to carry
and express a human disease. Cloning may make the time frame shorter and produce
a large number of animals for study
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Lesson 12
Cloning Stem Cells – Research in this area may lead to using human stem cells to
repair damaged tissues and organs. Cloning is a way to create human stem cells for
research.
Pharming for Drug Production – Some farm animals are being genetically engineered
to produce drugs and proteins that are useful in research. Cloning may provide a faster
way to create large herds of these animals.
Reviving Endangered or Extinct Species – This would provide a means to increase
the number individuals of an endangered species or even recreate an extinct species if
cell samples are available.
On less of a medical front, it could be possible in the future to clone a pet that has died
or even a family member. This idea was explored and taken to the extreme in the
Arnold Schwarzenegger movie, “The Sixth Day” and the more recent movie, “The
Island”.
In reality, any clone that is created will not be identical to the original. Clones are not
carbon copies in much the same way that identical twins are not exact copies. A clone
is born as a baby and has to go through the regular growth phase of its species.
Different environmental factors lead to different personalities, behaviours and interests.
Problems with Cloning
Despite all of the media attention over the miracle of cloning, there are a number of
problems that scientists have yet to fully understand and overcome. There is a very
high failure rate with cloning. For every success story, there are hundreds of failed
attempts. More recently, study of the cloned animals has revealed that many have
severe biological problems ranging from developmental delays to heart and lung
problems and malfunctioning immune systems. It is believed that the cloning process
creates random errors of gene expression. The egg cell only has a matter or hours to
reprogram the donor cell genes and scientists think this may be where the genetic
problems are created. The reprogramming many introduce random errors into the
clone’s DNA subtly altering the genes. This may be one of the reasons why so many of
the cloning attempts fail, the genetic damage is too great. The clones that are born,
may have just been lucky that the errors generated did not affect them in the embryo
stage.
Despite these problems, scientists say it may only be a matter of time before human
cloning becomes a reality.
Please visit the following website(s)
http://learn.genetics.utah.edu/content/tech/cloning/clickandclone/
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Lesson 12
Figure 12-1 Cloning by Somatic Nuclear Transfer
Source: Blake et al. 314
Support Question
(do not send for evaluation)
3.
From a scientific standpoint, are we ready to clone humans?
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SBI4U – Biology
Lesson 12
The Human Genome Project
This massive initiative began in 1990 in research labs all over the world. The idea was
to sequence the entire human genome, all 3 billion base pairs. Various research
laboratories in North America, Europe and Asia were supplied with parts of
chromosomes to sequence. The data would then be brought together and the
information would be available to one and all. Many expected that it would take
decades to sequence all the DNA and work went on slowly for years. As sections of
DNA were sequenced, page after page of A, C, T and G bases, in their proper order,
were made available on line. About five or six years into the project, a scientist named
Craig Ventor and his company Celera announced that they were getting into the
sequencing game and actually announced publically that they would complete the
project before the government sponsored scientists. Up until this point, sequencing was
done by hand just as it was explained in Lesson 3. This method was very slow and
prone to human errors of reading the gel by hand. Ventor and his colleagues had
created high-speed computers that were able to read in a few minutes what would take
scientists days or weeks. The race for the genome was on, publically-funded labs
versus the private sector. Finally, in April of 2003, both groups were hailed for finishing
at the same time. Thus, a high-quality reference sequence was completed, showing the
order of all 3 billion bases that make up the human genome. The question then
becomes, what do we do with it or better yet, what can be done with it?
Insights from the Human DNA Sequence
One of the initial steps, since the sequence was completed, was to examine the
genome in general and learn more about the arrangement of the genome. The human
genome actually contains 3.2 billion base pairs. There are approximately 25 000 genes
with the average gene consisting of 3000 base pairs. The largest gene codes for the
protein dystrophin at 2.4 million base pairs. The function of over 50% of the discovered
genes is unknown. The human genome is almost exactly same for all people, 99.9%
identical. The genes seem to be concentrated in random areas along the genome with
vast regions of non-coding DNA between. Scientists have identified millions of locations
where single-base DNA differences occur in humans. It is being discovered that these
slight changes can drastically change how the body reacts to certain drugs and other
therapies.
Using the Genome in Medicine
DNA underlies almost every aspect of human health directly or indirectly. Obtaining a
detailed picture of how genes and other regulatory sequences interact with each other
and the influence of environmental factors, will uncover the pathways of normal body
functioning and the disease process. This will impact how disease is diagnosed, treated
and prevented.
One of the first commercial medical application of the new genetic discoveries is DNAbased tests. Gene testing can be used to diagnose or confirm disease. There are
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Lesson 12
currently several hundred genetic tests in clinical use and their number is expected to
increase. In the future, upon birth, babies can be tested for hundreds of genetic
diseases. Some of the testing used today can help predict the risk to individuals with a
family history of a disorder, Huntington’s disease being a prime example. One benefit
from this could help doctors and patients manage the disorder more effectively and from
an earlier stage.
Pharmacogenomics is another area that has sprung to life with the sequencing of the
genome. Single base variations in some genes can affect how people respond to
medications. In the future, drugs could be tailored to the individual based on their
genomic sequences, maximizing the effectiveness of drugs and reducing the chances of
negative side effects.
Into the Future
So much rapid progress has been made in the area of genetics and the study of the
genome that scientists are predicting that biology will be the foremost science of the 21st
century. It is believed that the initial sequencing and further analysis of the genome will
open vast areas of biological research and development. Listed below are just a few of
the applications from the study of this field of biology.
Molecular Medicine
¾
¾
¾
¾
Improve disease diagnosis
Detect genetic predispositions to disease
Create drugs based on molecular information
Use gene therapy as treatment for disease
Microbial Genomics
¾
¾
¾
¾
Rapidly detect and treat pathogens (disease-causing microbes)
Develop new biofuel energy sources
Monitor the environment to detect pollutants
Clean up toxic waste safely and effectively
DNA Identification
¾ Identify potential suspects whose DNA may match evidence from crime
scenes
¾ Exonerate people wrongly accused of crimes
¾ Identify crime, catastrophe and other victims
¾ Establish paternity and other family relationships
¾ Match organ donors with recipients in transplant programs
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Lesson 12
Societal Concerns
Although the sequencing of the genome has led to a deeper understanding of the
human system which can be applied to improve the quality of life for many, there are
concerns as well. This area is moving forward at such a fast pace it is difficult for the
general public to even understand all of the implications that we face.
Who owns our personal genetic information and is genetic privacy the same as medical
privacy?
Who should have access to our personal genetic information and how will it be used?
Will an individual’s genetic makeup cause discrimination and how will that affect the
individual psychologically?
Who will benefit from genetic technologies? Will they be provided to everyone?
Should testing be available for the presence of genetic mutations leading to disease if
there is no cure or treatment?
Should parts of the genome be subject to patents? Will patenting DNA sequences limit
the potential development of useful products?
These questions will all have to be answered and laws will need to be put in place to
ensure that there are not abuses. In the United States, the Genetic Information
Nondiscrimination Act (GINA) became law on May 21, 2008. This law inhibits health
insurance companies and employers from discrimination based on information derived
from genetic tests. As well, insurers and employers are not allowed to request or
demand genetic tests.
Regulation of Biotechnology in Canada
As of 2009, there is no direct legislation that addresses genetic information and privacy.
For the moment, the basic privacy rights cover these issues to a very limited extent
through interpretation of the privacy laws. A research study of Canadian attitudes and
concerns about biotechnology was recently submitted to the Department of Justice
along with recommendations of changes to the existing laws.
The Canadian Food Inspection Agency is responsible for the regulation of all food and
food related products derived through biotechnology. Canada has very stringent
guidelines concerning the study, testing and public consumption of genetically modified
substances. This includes genetically modified plants, animal feeds and animal feed
ingredients, fertilizers and veterinary medicines.
Health Canada is responsible for overseeing the human health aspects of all products
created through biotechnology. This includes food, drugs, cosmetics, medical devices
and pest control products.
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Lesson 12
Support Question
(do not send for evaluation)
4.
List five facts about the human genome that scientists did not know prior to
undertaking the Human Genome Project.
Key Question #12
The first three questions deal with issues that could very easily arise in the future and
will have to be dealt with by society. These are meant to be opinion questions to get
you to think about some of these issues and what should be done. As of this point in
time, these scenarios are not real.
1.
Dr. Sharon Martin and her company have spent years working to identify how
the gene for albinism works. The mutation in this gene causes no pigment to be
produced in the hair, skin or eyes. Identifying the gene would open the door to
curing the condition. Finally, her team succeeds. But, the years of research
have been expensive. One way to make back the money is to patent the gene
that the team members have identified. Then, anyone who wanted to develop
either treatments or tests would have to pay a fee to use the gene.
a. Do you think the government should allow this gene to be patented? Explain.
(3 marks)
b. If Dr. Martin’s company develops a test for this gene, should they be allowed
to patent the test? Explain. (3 marks)
2.
John Smith is 30 years old when his father dies of complications of Huntington’s
disease, a genetic condition that usually does not show up until a person is 35-40
years old. Huntington’s is characterized by a slow progression of physical and
mental deterioration leading to death. There is now a test available to see if the
Huntington’s gene is present. John decides to take the test to see if he will suffer
from the disease in the future. Somehow, John’s health and life insurance
company learn about the results of the test and cancel his insurance. Then, he is
released from his job of more than eight years. Company officials are afraid that
the medical costs of caring for John will increase the group insurance rate.
a. Should the company be allowed to make decisions based on medical
information from John’s DNA? Explain. (3 marks)
b. Should the company be able to make John’s information available to other
companies that John is interviewing with? Why or why not? (3 marks)
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3.
Lesson 12
Scientists have recently inserted a gene to create a mouse with an increased
capacity for learning and memory- basically increasing the mouse’s IQ. Normal,
average human IQ is around 100, above that is considered highly intelligent while
an IQ below 70 is considered to indicate mental disability. In the future, if this
treatment could be applied to humans, it could increase IQ by 30 points.
A couple has a son with Down’s syndrome giving him an IQ of 70. They want to
use the therapy to increase his IQ to 100 in order for him to function normally.
This is considered to be gene therapy, where technology is used to help a person
function better.
A second couple has a son with an IQ of 120. They want to use the technology
to bring his IQ up to 150. They feel he would have a better chance of getting into
a prestigious university. This is called gene enhancement where technology is
used to help a person who is already functioning normally enhance a particular
characteristic.
a. Should this technology be used for gene therapy as in the first couple’s
situation? Why or why not? (3 marks)
b. Should this technology be used for gene enhancement as in the second
couple’s situation? Why or why not? (3 marks)
c. Who should get to decide the way in which this technology is used? (3
marks)
4.
In the past few years, there has been controversy surrounding the use of bovine
somatotropin in cows to increase milk production. The United States has granted
approval but Canada has not.
Using your research skills, create a small report based on the following: (10
marks)
a. the health and safety issues. (4 marks)
b. the economic issues. (2 marks)
c. the Canadian regulations and policies. (4 marks)
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SBI4U
Grade 12, University Preparation
Biology
Support Question Answers
SBI4U – Biology
Support Question Answers
Support Question Answers
Lesson 9
1.
Create a chart;
Scientist
Miescher
Levene
Hammerling
Griffith and Avery,
MacLeod, McCarty
Chargaff
Hershey and Chase
Franklin
Watson and Crick
Contribution
Discovered “nuclein” in cells and called it nucleic acid
Found nucleic acids were of two types, one with ribose sugar
(RNA) and the other deoxyribose sugar (DNA) and found they
were made up of nucleotides
Provided evidence that the heredity material was located in the
nucleus
Worked with a strain of pnemoniae bacteria in mice that
provided evidence that DNA was the heredity material
Continued Levene’s work by studying the nucleic acids and
discovered that A bases were found in equal quantities with T
bases and C bases with G bases-Chargaff’s Rule
Used radioactively labelled phosphorus in DNA and sulphur in
proteins to concisely show that a virus injects its DNA into
bacteria to control the cell showing DNA is the genetic material
Used X-ray crystallography to study the structure of DNA and
showed it to be helical in nature
Developed the model of the DNA double helix that is known
today
2.
The sequence should read, 3′-TACGGAAT-5′
3.
If the DNA contains 20% thymine it will have to contain 20% adenine. This
leaves 60% to divide between the other two bases so it will contain 30% guanine
and 30% cytosine. (Chargaff’s Rule)
4.
Create a chart;
Enzyme
DNA gyrase
DNA helicase
DNA polymerase I
DNA polymerase III
DNA ligase
RNA primase
Function
Relieves tension produced by the unwinding of DNA during
replication
Unwinds the double helix by disrupting the hydrogen bonds
Removes the RNA primers and replaces them with the
appropriate DNA bases
Responsible for synthesizing complementary strands of DNA
during replication
Joins the Okazaki fragments together by creating a
phosphodiester bond between the nucleotides
Adds in the RNA primers so DNA polymerase III can bind
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5.
Support Question Answers
“Onion DNA Extraction” Activity
a. The physical mincing of the onion helped to break open the cells to allow the
DNA to get out. The onion is a plant cell so it had a tough cell wall.
b. Shampoo’s basic purpose in hair washing is to break down and surround oil
droplets so they can be removed from the hair when rinsed with water. The
shampoo in the activity helped to break down the cell membrane and nuclear
membrane which are composed of lipids.
c. The salt is able to bind to the negative portions of the DNA molecule allowing
the strands move more closer together. The normally negative charge would
cause the strands to repel one another.
d. DNA does not dissolve in alcohol so when exposed to it, the alcohol causes
the DNA to precipitate out of the solution and become visible.
e. The DNA generally look like white clumps with some areas where wisps of
threads are visible.
Lesson 10
1.
3′ - GGCATGCACCATAATATCGACCTTCGGCACGG - 5′
a. This underlined section is the promoter because it has a high concentration of
A’s and T’s. Since these bases only hold together with double bonds it takes
less energy to open the double helix at this point.
b. Because the promoter lies upstream of the gene, it acts as a signal for RNA
polymerase to bind and transcribe the gene found downstream. As well,
mentioned above, less energy is required to start the unravelling of the DNA
at this point.
2.
Some of the possible reasons for RNA being single stranded are: it is more
efficient, only one strand is necessary to copy and carry the message so to make
a double stranded copy would waste energy and it is easier for the ribosome to
read the single copy rather than having to expend the energy to separate a
double stranded copy.
3.
a. If the termination sequence of a gene is removed, RNA polymerase will
continue to transcribe and build a primary mRNA transcript beyond the gene.
b. If the poly-A polymerase is inactivated, the protective tail would not be added
leaving the mRNA susceptible to degradation upon leaving the nucleus.
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Support Question Answers
c. If the enzyme that adds the 5′ cap is dysfunctional, then the 7-methyl
guanosine molecule will not be added to the mRNA transcript, again leaving
the mRNA susceptible to degradation upon leaving the nucleus.
d. If splicesomes excise exons and join introns together, then the mRNA
transcript will consist of non-coding sequences and will produce a nonfunctional protein.
4.
The first codon in any protein is always methionine, the start codon so the amino
acid sequence would read:
Met-Pro-Pro-Gly-Lys-Asp-Stop
5.
Translation basically means to express in another language which is what
happens. The information written as a series of bases is then expressed in the
language of amino acids.
6.
a. If splicesomes were inactivated, introns would not be removed from the
mRNA. This would result in dysfunctional proteins.
b. If poly-A polymerase were inactivated, the mRNA transcript would be
degraded in the cytoplasm since the poly A tail would not be added.
c. If the tRNA were inactivated, translation would not take place because the
amino acids would not be delivered to the ribosome.
d. If the ribosomes were inactivated, proteins could not be synthesized.
7.
If the level of lactose is low, none of the enzymes in the operon would be active.
The LacI repressor protein would be bound to the operator, blocking the RNA
polymerase from binding to the promoter. This prevents the enzymes from being
transcribed and translated. If the lactose levels rise, the lactose molecule will
bind to the LacI repressor protein changing its shape and causing it to move
away from the operator. This allows room for the RNA polymerase to bind to the
promoter region and begin transcription of the enzymes needed to break down
lactose.
8.
a. A substitution would be less harmful than a deletion because it does not
result in a shift of the reading frame. The substitution may cause one amino
acid to change but a deletion would create many incorrect amino acids in the
chain.
b. An inversion is more harmful than a substitution because, just as above, it will
cause a change in more amino acids than in a substitution.
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Support Question Answers
Lesson 11
1.
a. The location of the cuts created by SmaI are in between the C and the G in
the bolded sequences.
5′-AATTCGCCCGGGATATTACGGATTATGCATTATCCGCCCGGGATATTTTAGCA-3′
3′-TTAAGCGGGCCCTATAATGCCTAATACGTAATAGGCGGGCCCTATAAAATCGT-5′
b. Three fragments will be produced if SmaI cuts this sequence.
c. SmaI will produce blunt ends on its generated fragments.
2.
The order of fragments from slowest to fastest would be: 6.8 kb, 5.3 kb, 3.6 kb,
2.2 kb, 2.1 kb, and 1.1 kb. The 2.1 kb and 2.2 kb fragments will be the most
difficult to isolate on the gel because they are very similar in size and will remain
close together on the gel.
Source: Di Giuseppe et al. Solutions p. 105
3.
No. The investigation does not need to continue because enzyme 2 in the RFLP
analysis shows that the out of season fish is not the same species of fish that the
camper caught and was cooking. The bands are not identical between the two
fish.
4.
The sequence of the fragment that is produced in the gel is:
5′- TTCGATCCGCGA -3′
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Lesson 12
1.
Five advantages of GMO’s are: creating pest resistance in plants, creating
herbicide tolerance, creating disease resistance in plants, creating cold tolerance
in plants and creating drought resistance in plants.
Five disadvantages of GMO’s are: they may hurt other organisms, insects may
develop resistance, weeds may develop resistance, people may develop life
threatening allergies, and farmers may not be able to afford the genetically
altered seeds.
2.
The benefits to gene therapy are: it may alleviate the symptoms of various
diseases that cannot be controlled with medication, it may save the lives of
people with diseases where no medication is available and it may one day be
used to enhance positive traits in individuals.
The risks are: the vector that delivers the good gene may cause an immune
response that could be deadly, the new genetic material may insert itself into the
genome in a bad place, disrupting other genes and causing new problems and
the therapy only lasts as long as the good copy of the gene stays in the nucleus
so the therapy may have to be repeated many times.
3.
According to scientists, we are not ready to clone humans. There is still too
much about the process of cloning using somatic cells that is not known.
Evidence is beginning to suggest that the cloned animals have improper
expression of some of their genes which can create health problems. Only a
fraction of the number of clones that are created actually survive embryonic
development so process of cloning itself has a very low success rate and many
human embryos would be lost during the many attempts that would be needed.
4.
Five facts that scientists did not known before analyzing the results of the human
genome sequence are: there are 3.2 billion bases that make up the human
genome, there are about 25 000 genes that code for humans, the genes are
clustered in random areas around the chromosomes, humans are 99.9%
identical and there are one base variations in human genes which can affect how
people react to medication.
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