2. Molecular Biology – 2.7 DNA replication, transcription, & translation
Name:
Understandings, Applications and Skills (This is what you maybe assessed on)
Statement
Guidance
2.7.U1
The replication of DNA is semi-conservative and depends
on complementary base pairing.
2.7.U2
Helicase unwinds the double helix and separates the two
strands by breaking hydrogen bonds.
2.7.U3
DNA polymerase links nucleotides together to form a new The different types of DNA polymerase do
strand, using the pre-existing strand as a template.
not need to be distinguished.
2.7.U4
Transcription is the synthesis of mRNA copied from the
DNA base sequences by RNA polymerase.
2.7.U5
Translation is the synthesis of polypeptides on ribosomes.
2.7.U6
The amino acid sequence of polypeptides is determined by
mRNA according to the genetic code.
2.7.U7
Codons of three bases on mRNA correspond to one amino
acid in a polypeptide.
2.7.U8
Translation depends on complementary base pairing
between codons on mRNA and anticodons on tRNA.
2.7.A1
Use of Taq DNA polymerase to produce multiple copies of
DNA rapidly by the polymerase chain reaction (PCR).
2.7.A2
Production of human insulin in bacteria as an example of
the universality of the genetic code allowing gene transfer
between species.
2.7.S1
Use a table of the genetic code to deduce which codon(s)
corresponds to which amino acid.
2.7.S2
Analysis of Meselson and Stahl’s results to obtain support
for the theory of semi-conservative replication of DNA.
2.7.S3
Use a table of mRNA codons and their corresponding
amino acids to deduce the sequence of amino acids coded
by a short mRNA strand of known base sequence.
2.7.S4
Deducing the DNA base sequence for the mRNA strand.
Recommended resources:
http://bioknowledgy.weebly.com/27-dna-replication-transcription-and-translation.html
Allott, Andrew. Biology: Course Companion. S.l.: Oxford UP, 2014. Print.
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Review questions
1. State during which phase of the cell cycle DNA replication occurs.
- Interphase, more specifically S-phase of Interphase
2. State the function of DNA replication.
- To make new copies of the same DNA
3. Which of the following is the end product of DNA replication in a human somatic cell?
 23 chromosomes
 46 chromosomes
 23 pairs of chromosomes
 23 pairs of sister chromatids
• Remember: A somatic cell is a cell that is not for reproductive purposes
• Please do research this question on your own I’ve found that people either answer C or
D
2.7.U2 Helicase unwinds the double helix and separates the two strands by breaking hydrogen bonds.
4. Outline the molecular structure of Helicase.
- It is an enzyme
- It is formed from multiple polypeptides
- Globular
- Doughnut shape
5. Label and annotate the diagram below to describe the function of helicase.
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2.7.U3 DNA polymerase links nucleotides together to form a new strand, using the pre-existing strand as a
template.
6. Outline the molecular structure of DNA polymerase.
 It is an enzyme, made up of multiple polypeptides
7. Label and annotate the diagram below to describe the function of DNA Polymerase.
2.7.U1 The replication of DNA is semi-conservative and depends on complementary base pairing.
8. Explain the importance of complementary base pairing in conserving the base-sequence during DNA
Replication.
- Complementary base pairing ensures that the two DNA molecules resulting from DNA replication are
identical in their base sequences to the parent molecule
9. Explain why DNA replication of this kind of referred to as being semi-conservative.
- The products of DNA replication are two strands; Both are composed of one newly synthesised
strand and an original parent strand.
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Notes from Kyla: DNA Replication process
-
-
-
THE FIVE ENZYMES INVOLVED IN DNA REPLICATION
1. Helicase: Using hydrolysis of ATP, helicase breaks hydrogen bonds between complementary bases and unzips
the DNA strands
 SSB proteins (single strand proteins) are used to hold apart the strands while replication occurs
2. Primase: Synthesises RNA primers from a direction of 5’ to 3’ to serve as a starting point for DNA synthesis
3. DNA Polymerase I: Proofreads the DNA and corrects any mismatches and removes the RNA primer and
replaces it with necessary nucleotides;
4. DNA Polymerase III: Synthesises nucleotides in the 5’ to 3’ direction on the leading and lagging strand and does
not have the proofreading function that DNA Polymerase I has
5. Ligase: “Glues” together the DNA strands by catalyzing the formation of the phosphodiester bonds, and also
repairs DNA strands and connects Okazaki fragments together on the lagging strand
KEY INFORMATION THAT I DON’T KNOW WHERE ELSE TO PUT:
1. Okazaki fragments are found on the lagging strand, and are short DNA sequences synthesized discontinuously
and later connected together using ligase
2. Replication is always in the 5’ to 3’ direction
3. Leading vs. Lagging Strand
 Leading is synthesized continuously while Lagging is synthesized discontinuously
 Leading strand is synthesized in the 5’ to 3’ direction
DNA REPLICATION PROCESS:
1. Helicase unwinds the DNA helix by breaking hydrogen bonds between complementary bases
2. Single-Strand binding proteins (SSB) stabilize the unwound parental DNA and hold them apart
3. Leading strand is synthesised continuosly in the 5’ to 3’ direction by DNA polymerase III
4. Lagging strand is synthesised discontinuously . Primase (also used in leading strand) synthesises a short RNA
primer which is extended by DNA polymerase to form an Okazaki fragment
5. After the primer is replaced by DNA, DNA ligase connects the Okazaki Fragment to the growing strand
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2.7.A1 Use of Taq DNA polymerase to produce multiple copies of DNA rapidly by the polymerase
chain reaction (PCR).
10. polymerase chain reaction (PCR) is a key technique in DNA manipulation and analysis.
 State the main uses of PCR.
- Used to make many copies of a selected DNA sequence
- Useful when only a small amount of DNA is available for testing such as a crime scene, samples of
blood
 State the name of the specialised apparatus needed to support this process.
- Thermal Cycler


-
Complete outline of the three keys steps in the process of PCR.

Denaturation: The DNA sample is heated to separate the two strands

Annealing: DNA primers attach to the opposite ends of the target sequence.

Elongation: A heat-tolerant DNA polymerase copies the strands.
If one cycle of PCR yields two identical copies of the DNA sequence. Calculate how many copies
20 cycles would yield.
1,048,576
To figure this out it’s 2^x where x is the number of cycles
So 2^20 = 1,048,576
2.7.S2 Analysis of Meselson and Stahl’s results to obtain support for the theory of semi-conservative
replication of DNA.
The image to the right details the three possible methods of DNA replication.
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https://upload.wikimedia.org/wikipedia/commons/a/a2/DNAreplicationModes.png
Review your understanding of Meselson and
Stahl’s experiments by using using the presentation, the McGraw and Hill animation (clear and accessible)
or the Scitable article (Nature education) (more detailed).
-
Please do research this experiment it is very important
Notes from Kyla: Meselson and Stahl’s experiment
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In the experiment, bacterial cells were grown for several generations on a medium containing a
heavy nitrogen isotope (15N). The DNA in these cells therefore contained 15N
Cells then transferred to a new medium containing the lighter isotope (14N). At various times after
the transfer, samples of the bacteria were collected
DNA then extracted and dissolved in caesium chloride solution
Samples spun rapidly in a centrifuge, establishing a concentration gradient in the tube
DNA molecules in centrifuge are rearranged according to weight
DNA containing 15N sank to a lower position in the gradient than 14N DNA and settled to the
bottom.
After one generation in the 14N medium, the bacteria yielded a single band of DNA with a density
between 14N and 15N, indicating that only one strand of each duplex contained 15N
After two generations in 14N medium, two bands were obtained, one of intermediate density
(containing 15N), and one of low density (only 14N)
Concluded that replication of the DNA duplex involves building new molecules by separating
parent strands and then adding new nucleotides to form the complementary strand on each of
these templates
11. At the start of a Meselson and Stahl experiment (generation 0) a single band of DNA with a density of
1.730 g cm-3 was found. After 4 generations two bands were found, but the main band had a density of
1.700 g cm-3.
a.
-
Explain why the density of the main band changed over four generations. (2)
N15 has a greater mass than N14 due to the extra neutron
Generation 0 contained DNA with exclusively N15, hence the greater density
With each generation, the proportion of N14 increases due to the free nucleotides as the mass of
DNA doubles
After 4 generations, most strands contain only N14 isotope - the dominant band at a density of 1.700
g cm-3.
N15 isotope remains, but is combined in strands with N14 isotope - a second band at a density
between 1.730 and 1.700 g cm-3.
Simple Ver: Generation 0 had DNA with only N15 isotopes, which are heavier than N14. As each
generation occurs, the amount of 14N increases until the 4th generation, when the band contains
mostly N14
b. After one generation only one DNA band appeared, but the density had changed.
i.
Estimate the density of the band. (1)
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- The band would contain equal amounts of N14 isotope and N15 isotope
- Density of an all N15 isotope band is 1.730 g cm-3.
- Density of an all N14 isotope band is 1.700 g cm-3.
- Density of an the mixed isotope band is the average of the two: (1.730 + 1.700) / 2 = 1.715 g cm-3
ii.
-
Which (if any) mechanisms of DNA replication are falsified by this result? (1)
Conservative
iii.
Explain why the identified mechanism(s) are falsified. (1)
Two bands should appear in all generations after generation 0
Additionally, after generation 0 the DNA was half heavy, so half of the DNA contained heavy nitrogen
(N15) and light nitrogen (N14). This therefore falsifies conservative mechanism as it would have
resulted in one strand all heavy and one all light
c. Describe the results after two generations and which mechanisms and explain the identified
mechanism(s) (if any) are falsified as a consequence. (3)
- 2 bands.
-
One band containing a mixture of N15 and N14 isotopes (semi-conservative replication preserves
the DNA strands containing N15 isotopes, but combines them with N14 nucleotides during
replication).
-
Other band containing all N14 isotopes during replication from generation 1 to generation 2. The new
strands consisting of only N14 isotopes are replicated using N14 nucleotides.
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Dispersive replication is falsified as this model would continue to produce a single band,
containing proportionally less N15 isotope.
d. Describe and explain the result found by centrifuging a mixture of DNA from generation 0 and 2. (2)
- 3 bands:
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First band from generation 0 containing all N15 isotopes - no replication has occurred
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Second band from generation 2 containing a mixture of N15 and N14 isotopes (semi-conservative
replication preserves the DNA strands containing N15 isotopes, but combines them with N14
nucleotides during replication).
-
Third band from generation 2 (all replicated DNA) containing all N14 isotopes - during replication
from generation 1 to generation 2. The new strands consisting only of N14 isotopes are replicated
using N14 nucleotides
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Nature of Science: Obtaining evidence for scientific theories—Meselson and Stahl obtained evidence for the
semi-conservative replication of DNA. (1.8)
12. Define the term theory as it is used in science.
- One way is a proposed explanation for a natural phenomenon often based on a general principle
- Another is the body of interconnected concepts supported by scientific reasoning and experimental
evidence, explaining the facts in an area of study
13. If evidence does not exist to support a scientific idea then the idea cannot be described as being a
theory. Without evidence outline the term used to describe a scientific idea.
- Hypothesis
Review questions
14. Transcription and translation is also known as protein synthesis, and is the expression of genes. The
genetic code determines the amino acid sequence of a polypeptide, and the properties of the amino
acids give the final structure and function of the protein. Other than membrane proteins, state four
functions of proteins in the cell.

Catalysis - Enzymes

Cytoskeletons - Tubulin

Transport of nutrients and gases

Hormones - Insulin
2.7.U4 Transcription is the synthesis of mRNA copied from the DNA base sequences by RNA polymerase.
15. Outline the process of transcription in the nucleus, including the roles of RNA polymerase,
ribonucleoside triphosphates and complementary base pairing.
- Transcription is the synthesis of mRNA, using DNA as a template – Only occurs along one of the 2
DNA strands
1. RNA polymerase binds to a site on the DNA at the start of a gene
2. RNA polymerase moves along the gene separating DNA into single strands and pairing
ribonucleoside triphosphates with complementary bases on the DNA strand
a.
Uracil bonds with adenine since there is no thymine in RNA
b. Ribonucleoside Triphosphates are the RNA bases without the phosphate group, because
RNA polymerase removes the phosphate groups and uses the energy gained from that to
covalently bond
3. RNA polymerase forms covalent bonds between the ribonucleoside triphosphates
4. RNA separates from the DNA and the double helix reforms
5. Transcription stops at the end of the gene and the completed RNA molecule is released]
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The antisense strand is the DNA strand that is transcribed and has a complementary base sequence
to both the mRNA and the DNA strand
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The sense strand is the DNA strand with the same base sequence as mRNA and is not transcribed
the nucleotide to the growing sequence
2.7.U5 Translation is the synthesis of polypeptides on ribosomes.
16. Complete the table to compare and contrast the processes of transcription and translation.
Transcription
Translation
Begins with…
DNA
mRNA
Ends with…
mRNA
Polypeptides
Location
DNA strand
Ribosome
Uses…
RNA polymerase
tRNA
17. Ribosomes are the cell components that carry out the process of translation. Outline the structure of the
ribosome and explain how it is adapted to carry out translation.
 A ribosome is composed of two halves; a large and a small subunit
 During translation, ribosomal subunits assemble together (like a sandwich) on the strand of
mRNA
 Each subunit is composed of RNA molecules and proteins
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
The large subunit contains the binding sites for tRNAs and also the catalysis of peptide bonds
between amino acids
http://www.hartnell.edu/tutorials/biology/translation.html
2.7.U6 The amino acid sequence of polypeptides is determined by mRNA according to the genetic code.
2.7.U7 Codons of three bases on mRNA correspond to one amino acid in a polypeptide.
18. Define mRNA in terms of it’s function
- mRNA is the RNA that carries the information needed to synthesize a polypeptide, or a transcript copy of
a gene used to encode a polypeptide
19. Suggest why the length of mRNA molecules varies.
- Length varies because it depends on the number of amino acids in the polypeptide
20. Describe what is meant by the term ‘genetic code’.
- Genetic code is the set of rules by which information encoded in mRNA sequences is converted into
proteins by living cells
21. Define the term codon.
- A sequence of 3 bases on the mRNA which encodes for 1 amino acid
22. Calculate the number of different codons combinations.
64 (4 *4 *4 = 64)
23. State the number of amino acids that can be translated by ribosomes.
20
24. Explain what is meant by the term degenerate. Refer to the last two questions in your answer.
- Different codons can translate for the same amino acid, so the genetic code is said to be
degenerate, because it has several codons that can code for the same amino acid.
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2.7.U8 Translation depends on complementary base pairing between codons on mRNA and anticodons on
tRNA.
25. State the molecule on which anti-codons, which are complementary to codons, can be found.
tRNA
26. Complete the steps to outline the process of translation.

mRNA binds to the small subunit of the ribosome.

The mRNA contains a series of codons each of which codes for an amino acid.

tRNA molecules contain anticodons which are complementary to the codons on the mRNA.

tRNA molecules bind to a specific amino acid that corresponds to the anticodon

The large subunit binds to the small subunit of the ribosome.

There are three binding sites on the large subunit of the ribosome, but only two can contain tRNA
molecules at a time

The ribosome moves along the mRNA and presents codons in the first two binding sites.

tRNA with anticodons complementary to the codons bind (the bases are linked by the formation
of hydrogen bonds)

A peptide bond is formed between the two amino acids (carried by the tRNAs)

As the ribosome moves along the mRNA a tRNA moves to the third binding site and detaches.

Another tRNA carrying an amino acid binds to the first site and a second peptide bond is formed.

The process (i.e. the last two steps) repeats forming a polypeptide.
Extension: sketch diagrams below
to show translation and use the
steps above as annotations.
2.7.S1 Use a table of the genetic
code to deduce which codon(s)
corresponds to which amino acid.
2.7.S3 Use a table of mRNA codons
and their corresponding amino acids
to deduce the sequence of amino
acids coded by a short mRNA strand
of known base sequence.
2.7.S4 Deducing the DNA base
sequence for the mRNA strand.
The genetic code – how mRNA codons translate to amino acids
Use the genetic code table to help
answer the questions below.
http://www.ib.bioninja.com.au/_Media/genetic_code.jpeg
27. Deduce the codon(s) that translate for Aspartate.
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GAU and GAC
28. If mRNA contains the base sequence CUG ACU AGG UCC GGA
 deduce the amino acid sequence of the polypeptide translated.
Leucine – Threonine – Arginine – Serine - Glycine

-
deduce the base sequence of the DNA antisense strand from which the mRNA was transcribed.
Antisense strand is complementary to the mRNA but has T instead of U
GAC TGA TCC AGG CCT

If mRNA contains the base sequence ACU AAC deduce the base sequence of the DNA sense
strand.
The sense strand is the template for the mRNA and identical except for the T
ACT AAC
-
29. Transcribe and translate this DNA sequence.
DNA T
A
C
mRNA A
U
G C
Amino Met
acid
G G G C
Pro
C
C
C
C
G T
G G G C
Gly
His
A
G A
C
C
G U
U
Cyc
A
G C
C
C
A
G G U
Arg
C
T
G A
STOP
30. An mRNA strand has 76 codons. How many amino acids will be in the polypeptide?
75 amino acids, because the last codon is a STOP codon
31. A polypeptide contains 103 amino acids. What is the length of the gene (unit = base pairs)?
309 (312 if you include STOP)
32. A gene is 105kbp (kilobase pairs). How many amino acids are in the polypeptide?
105/3 = 35 – 1 = 34 amino acids
2.7.A2 Production of human insulin in bacteria as an example of the universality of the genetic code
allowing gene transfer between species.
33. Diabetes in some individuals is due to destruction of cells in the pancreas that secrete the hormone
insulin. It can be treated by injecting insulin into the blood. Despite the differences in the amino acid
sequence between animal and human insulin, they all bind to the human insulin receptor and cause
lowering of blood glucose concentration. However, some diabetics develop an allergy to animal insulins,
so it is preferable to use human insulin. In 1982 human insulin became commercially available for the
first time. It was produced using genetically modified E. coli bacteria. Since then methods of production
have been developed using yeast cells and more recently safflower plants.
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 Describe what is meant by the term ‘universality of the genetic code’
- All living things use the same bases and same genetic code, no matter what species. Each codon
produces the same amino acid in transcription and translation, so the sequence of amino acids in a
polypeptide remains unchanged regardless of species. So, we can take genes of one species and
insert them into the genome of another species.

Complete the table to outline the keys steps in the process of gene transfer.
Diagrams
Notes
Restriction enzymes cut the desired gene from the genome
E. coli bacteria contain small circles of DNA known as plasmids,
which can be removed
The same restriction enzyme cuts into the plasmid and because
it is the same restriction enzyme the same bases are left
exposed creating “sticky ends”
Ligase joins the sticky ends together fixing the gene into the E.
coli plasmid
The recombined plasmid is inserted into the host cell, which
now expresses the new gene.
An example of this is human insulin production
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Notes From Kyla: POST-TRANSCRIPTIONAL MODIFICATION
-
-
-
Happens only in eukaryotic cells; mRNA that is transcribed can be modified and you can cut out bits
and pieces not needed for coding and put together only the useful pieces
“Why not just have all of the necessary stuff? Why do we need to remove stuff to make the final
protein?”
o Mutations. If a mutation occurs in the unnecessary part of the strand, you can easily remove
it. It helps to preserve the gene sequence and protect it
o You can also increase the number of proteins an organism can produce by splicing together
(“gluing together”) different combinations of the necessary pieces
Introns are the sequences or parts of the DNA that does not code for the proteins needed and
interrupts the sequence of the genes. You remove them.
Exons are the sequences that contain the needed information for coding a protein. This is what you
keep. The exons are spliced together (gluing together) to form mature mRNA, which is used and kept
and sent to the ribosomes to be translated into the actual protein.
THIS IS ONLY DONE IN EUKARYOTIC CELLS.
Citations:
Allott, Andrew. Biology: Course Companion. S.l.: Oxford UP, 2014. Print.
Taylor, Stephen. "Essential Biology 03.4 DNA Replication.docx." Web. 1 Oct. 2014.
<http://www.slideshare.net/gurustip/essential-biology-34-dna-replication-core>.
Taylor, Stephen. "Essential Biology 03.5 DNA Transcription and Translation.docx." Web. 1 Oct. 2014.
<http://www.slideshare.net/gurustip/essential-biology-35-transcription-translation-core>.
Taylor, Stephen. "4.4 Genetic Engineering and Biotechnology.pptx" Web. 1 Oct. 2014.
<http://www.slideshare.net/gurustip/genetic-engineering-and-biotechnology-presentation>.
http://bioknowledgy.weebly.com/
(Chris Paine)