LAB 7: DNA Structure, Replication, and Protein Synthesis
PRE-LAB QUESTIONS
DNA Structure, Replication and Protein Synthesis
1.
What does DNA stand for?
2. What sugar is part of a DNA molecule?
3. What is the function of genes?
4. What are the 4 nitrogenous bases that make up DNA?
5. What happens during transcription, and where does it take place in a eukaryotic cell?
6. What happens during translation, and where does it take place in a eukaryotic cell?
7. List all the cell components that must be present in order for translation to occur:
8. What is a codon?
9. What is a stop codon?
LAB 7: DNA Structure, Replication, and Protein Synthesis
PART A –DNA EXTRACTION
INTRODUCTION
Nucleic Acids:
Nucleic Acids are made up of repeating monomers called nucleotides. They are in charge of heredity
and the production of proteins. Two nucleic acids exist – DNA (deoxyribonucleic acid) and RNA
(ribonucleic acid). DNA is found in all living organisms. DNA in a plant cell is about 100,000 times as
long as the cell itself. However, DNA only takes up about 10% of the cell’s volume. This is because
DNA molecules are full of twists and turns and are packed and compacted into chromosomes in the
cell’s nucleus. The long, thick fibers of DNA store the information for the functioning of the chemistry
of life. DNA is present in every cell of plants and animals. The DNA found in strawberry cells can be
extracted using common, everyday materials. In order to release the DNA contained in the nucleus, it
is necessary to get through three barriers in plant cells: cell wall, plasma membrane, and nuclear
membrane. The toughest barrier is the cell wall. Cell walls are made of cellulose which provide a
tough barrier that protect the cell. Some crushing and mashing will be necessary to burst open the cell
walls and allows the contents to spill out. We will use an extraction buffer containing salt, to break up
protein chains that bind the nucleic acids, and dish soap to dissolve the lipid (fat) part of the
strawberry cell wall and nuclear membrane. This extraction buffer will help provide us access to the
DNA inside the cells.
Materials for the lab group of 2 students:
Heavy duty Zip-lock bag
1 Strawberry
10mL extraction buffer (soapy, salty water)
Cheesecloth
Funnel
50mL vial/ test tube
Glass Rod
20mL Ethanol
Foil- square (1) to fit medicine cup
Medicine Cups (2)
Glass marking pens
Elastic band
LAB 7: DNA Structure, Replication, and Protein Synthesis
Procedure –
1.
2.
3.
4.
5.
Pre-rinse the medicine cup with ethanol. Discard in sink.
Label foil with your name.
Pour 20mL of ethanol in medicine cup, watching for meniscus.
Cover medicine cup with foil and place in freezer.
Place one strawberry in Zip-lock bag.
6. Smash and grind up the strawberry using your fist and fingers for 2 minutes. Be careful not to
break the bag!
7. Pre-rinse the 2nd medicine cup with a drop or two of DNA extraction buffer.
8. Add the provided 10mL of extraction buffer (salt and soapy solution) to the bag.
9. Kneed/ mush the strawberry in the bag for 1 minute.
10.
Assemble the filtration apparatus as shown.
Tubes should fit in the test tube rack.
11. Pour the strawberry slurry into the filtration
apparatus and let it drip directly into the test tube.
Filter for 20 minutes.
12. Slowly pour cold ethanol into the tube from freezer.
Observe.
13. Fix to spooling the glass rod into the test tube where
the strawberry extract and ethanol layers come into
contact with each other. Observe.
LAB 7: DNA Structure, Replication, and Protein Synthesis
POST LAB QUESTIONS PART A -DNA EXTRACTION
1. It is important that you understand the steps in the extraction procedure and why each
step was necessary. Each step in the procedure aided in the isolating the DNA from
other cellular materials. Match the procedure with its function:
PROCEDURE
A.
B.
C.
D.
Filter strawberry slurry through cheesecloth
Mush strawberry with salty/ soapy solution
Initial smashing and grinding of strawberry
Addition of ethanol to filtered ethanol
FUNCTION
______ To precipitate DNA from solution
______ Separate components of the cell
______ Break open the cells
______ Break up proteins and dissolve cell
membranes
2. A person cannot see a single cotton thread 100 feet away, but if you wound thousands of
threads together into a rope, it would be visible much farther away. Is this statement
analogous to our DNA extraction? Explain your answer.
3. Is DNA found is ALL food? Explain your answer.
LAB 7: DNA Structure, Replication, and Protein Synthesis
PART B – DNA Structure
Purpose: The student will be able to demonstrate an understanding of DNA structure by producing a
3-dimensional model containing its components and to simulate the process of replication.
Introduction:
Deoxyribonucleic acid or DNA is the genetic material that is passed down from one generation to
the next. It is the molecule that makes up your genes. Genes contain the instructions that your
body uses to make and assemble all of the different types of proteins that make each individual
unique. Each gene encodes for a particular protein and the traits you inherit through your genes (like
your hair color, eye color, or blood type) are all controlled by the production of proteins. DNA is
present in all living cells and can be extracted to be analyzed and researched further. All cells contain
the entire DNA sequence of the individual. Let’s start at the beginning and learn what makes up a
DNA molecule.
DNA molecules are composed of small monomers or building blocks called nucleotides. Each
nucleotide is composed of three basic parts: a five-carbon sugar (deoxyribose), 1 phosphate, and 1
nitrogenous base (adenine (A), thymine (T), guanine (G), or cytosine (C)). Each DNA
molecule will have 4 different types of nucleotides and each type is characterized by the nitrogenous
base that is associated with it.
DNA Structure
Image 1831001R from Pearson
Library
Figure 7.1
The overall structure of DNA was first discovered by Watson and Crick in 1953. DNA molecules are
made up of two strands and resemble a twisted ladder. The sides of the ladder each represent a
strand of DNA and are composed of alternating sugar and phosphate molecules that are connected
through covalent bonds. Each sugar molecule also has a nitrogenous base that hangs off to the side.
The two strands of DNA or “sides of the ladder” are connected in the middle by hydrogen bonds
between the nitrogenous bases. How these two strands match up with each other is very specific.
Each nucleotide base only pairs with one specific partner. Adenine always bonds with thymine or A
goes with T. Guanine always bonds with cytosine or G goes with C. This is called complementary
base pairing. The order of these nucleotides determines different DNA molecules and different
organisms.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Part A: Constructing a 3-D Model of DNA
Overview: Students will create their own DNA model to enhance their knowledge of the individual
components that make up a DNA double helix.
Materials:
Pipe cleaners (3 inches in length):
Red: sugar
Yellow: phosphate
Black: hydrogen bonds
Beads- red, yellow, blue, green (representing the bases: A, G, C, T)
Index card with nucleotide sequence
Procedure:
1. Obtain an index card from your instructor with a printed DNA sequence. This is a partial
sequence from 1 strand of DNA. It is very short in comparison to a typical chromosome that
would contain around 100 million bases!
2. For each nucleotide choose a bead color to represent it. Make a note of it here, as well as your
selected sequence.
Sequence: ______________________________________________________________
NUCLEOTIDE
A
C
G
T
COLOR
3. Using the various pipe cleaners and beads, build a 3-D double-stranded model of DNA for your
particular sequence. The pipe cleaners will represent the backbone of sugars and phosphates
and the beads will represent the 4 nitrogenous bases. Once you have created your DNA
molecule, notify the instructor to check your work.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Part C: DNA Replication
DNA replication or copying takes place in the nuclei of your cells prior to cell division. During
replication the parental DNA molecule is used as a template for the synthesis of a new daughter
strand. Nucleotides are added following the base pairing rules. The first step of replication is for the
DNA molecule to unwind. Unwinding occurs as the hydrogen bonds between the nitrogenous bases
are broken by an enzyme. Each strand of DNA can then be used as a template for the newly
synthesized daughter strands. DNA polymerase is the enzyme that joins the neighboring nucleotides
together. Figure 7.2 illustrates the process of DNA replication. DNA synthesis can be referred to as
semiconservative replication because each daughter strand actually consists of half of the parental
DNA and half of the newly made daughter strands.
Figure 7.2 showing DNA replication.
Image 1834001R from Pearson Library
Using the molecule you created in Part A, model the process of DNA replication. Do this by first
unwinding the double-stranded DNA and then create 2 newly synthesized daughter DNA strands.
Show your instructor the process of DNA replication and explain how you created this new piece of
DNA.
During replication, there are times when errors or changes in the DNA can occur. These are called
mutations. Mutations are any change in the nucleotide sequence of DNA. This can be a single
nucleotide, like we see in sickle cell anemia, or a larger region of the chromosome.
Part D: Protein Synthesis (DNA > RNA > Protein)
LAB 7: DNA Structure, Replication, and Protein Synthesis
Purpose: To understand the process of protein synthesis in a cell - how a gene is transcribed into
mRNA, and how mRNA is translated into protein.
Introduction:
Deoxyribonucleic acid, or DNA, is the genetic material of all living organisms. Every DNA molecule is
made up of a linear sequence containing four different nucleotides. A DNA nucleotide consists of a
phosphate group, a deoxyribose sugar, and one of four bases: A, G, C or T. The nucleotides are
arranged into segments called genes. Most genes code for proteins, but some code for RNA
molecules instead, for example tRNA, a molecule that plays an essential role in protein synthesis. In
addition to genes, every chromosome contains large amounts of noncoding DNA – DNA that does not
code for any genes. In the humans, for example, only about 1.5% of the genome actually codes for
genes. The remaining 98.5% of the genome contains a large number of repetitive DNA sequences.
Within every cell, at a given moment, a variety of genes are being transcribed into RNA and translated
into protein. These genes are said to be “turned on”, or expressed. Other genes are inactive, or
“turned off”. This is a dynamic process within the cell, called gene regulation. Changes in the
internal and external conditions of a cell can result in new genes being turned on, and currently
expressed genes being turned off.
In multicellular organisms a different subset of genes is expressed in each different cell type. For
example, a gene that codes for the protein insulin is expressed in pancreas cells, but not in muscle
cells, while a gene that codes for the muscle protein myosin is expressed in muscle cells, but not in
pancreas cells.
How can a DNA sequence, made up of only four different types of nucleotides (A, G, C and T), code for
the huge variety of different proteins found in a cell? Each eukaryotic gene is thousands of nucleotides
long, and has a linear arrangement, or sequence of nucleotides, different from that of any other
gene.
Transcription – DNA into RNA
During transcription, the DNA sequence of a gene is transcribed into RNA by the enzyme RNA
polymerase. The strand of the DNA double helix that is transcribed is called the template strand.
Near the beginning of a gene there is a DNA sequence called the promoter. RNA polymerase
recognizes and binds to the promoter, starts to unwind the DNA, and moves along the DNA double
helix until it reaches the transcription initiation site of the gene. At this site RNA polymerase
begins to transcribe the DNA sequence into an RNA copy.
RNA nucleotides are similar, but not identical to DNA nucleotides in structure. They contain a
phosphate group, a ribose sugar (has one more oxygen atom than does deoxyribose), and one of four
bases: A, G, C or U. Note that DNA and RNA nucleotides both contain A, G and C bases, but that T is
found only in DNA, while U is found only in RNA. U and T have similar structures; both U and T form
a complementary base pair with A.
LAB 7: DNA Structure, Replication, and Protein Synthesis
5’
3’
Figure 7.3. Transcription. RNA polymerase inserts an RNA nucleotide complementary to
each DNA nucleotide of the DNA template strand as it moves along the DNA. Courtesy of
Pearson Education, Inc.
Watch the video on transcription shown below:
http://www.youtube.com/watch?v=5MfSYnItYvg
Note that RNA polymerase synthesizes RNA in the 5’ to 3’ direction. Both DNA and RNA are
synthesized only in the 5’ to 3’ direction.
Practice transcribing DNA into RNA in the exercise below.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Procedure:
1. Transcribe the template strand of the DNA sequence below into RNA. Remember that in RNA
the T nucleotide is replaced by a U nucleotide:
3’-CTATAATCGGTACTAGTACTACCGTACTG-5’ template strand
5’-GATATTAGCCATGATCATGATGGCATGAC-3’ coding strand
5’-
-3’ RNA
Part E: Translation –mRNA into protein
In a prokaryotic cell once the RNA transcript is made, it is ready to be translated into protein. In a
eukaryotic cell, the RNA transcript is modified before leaving the nucleus: portions of the transcript,
called introns, are removed from the transcript, the remaining RNA pieces (exons) are spliced back
together, and a cap and tail are added to the RNA. This modified RNA molecule, called mature
messenger RNA, or mRNA, is now ready to leave the nucleus and enter the cytoplasm.
The various cell components needed to convert mRNA into protein are located in the cytoplasm.
Ribosomes are complexes of proteins and ribosomal RNA (rRNA), and consist of a large and a
small subunit. The ribosome binds to the 5’ end of the mRNA, and translation is initiated at the start
codon (AUG) of the mRNA molecule. A codon is three-base unit of mRNA. There are 64 different
codons that can be made from a 3-base sequence with 4 possible bases at each position. Sixty-one of
the codons code for amino acids; the remaining three codons are stop codons – they do not code for
an amino acid, instead they cause the ribosome to stop transcription. The start codon (AUG) has two
functions: it marks the beginning of a protein and it codes for the amino acid methionine (Met).
Shown below (Figure 7.4) is a chart of the genetic code – use this chart as a dictionary to translate
RNA codons into amino acids. This will allow you to translate mRNA sequences into protein.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Figure 7.4. The genetic code: Translating RNA codons into amino acids. Courtesy of
Pearson Education, Inc.
How does an mRNA sequence get translated into the amino acid sequence of a protein? Transfer
RNA molecules (tRNAs) act as translators. At one end of each tRNA molecule is a three-base
sequence called an anticodon, which is complementary to a specific mRNA codon. At the other end
of the tRNA molecule is a site to which a specific amino acid can attach. For each mRNA codon there
is a tRNA with a complementary anticodon that carries the amino acid specified by the mRNA codon.
LAB 7: DNA Structure, Replication, and Protein Synthesis
amino acid
attachment site
anticodon
Figure 7.5. The structure of tRNA. Modified from Pearson Education, Inc.
Translation starts when an mRNA molecule binds to a small ribosomal subunit. The first tRNA binds
to the start codon (AUG) of the mRNA, where translation begins. The large ribosomal subunit then
attaches to the small subunit. A new tRNA enters the ribosome and binds to the second mRNA codon.
The ribosome then shifts both tRNAs and catalyzes the covalent bonding of the first amino acid to the
second amino acid, which is still attached to its tRNA. The first tRNA, which now no longer has an
amino acid attached to it, leaves the ribosome. The ribosome then moves along the mRNA to the next
codon, where the process is repeated. This continues until the ribosome reaches a stop codon. Stop
codons do not code for amino acids – instead they signal the end of translation. The completed
peptide is released from the ribosome, and the translation machinery disassembles. See Figure 7.6
below for an overview of translation.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Polypeptide
Amino
acids
tRNA with
amino acid
attached
tRNA
Anticodon
Codons
5
3
mRNA
Figure 7.6. An overview of translation. Courtesy of Pearson Education, Inc.
Watch the videos on translation shown below:
http://www.youtube.com/watch?NR=1&feature=fvwp&v=Ikq9AcBcohA
http://www.youtube.com/watch?v=TfYf_rPWUdY
In the exercise below you will practice using the codon chart to translate RNA into protein.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Procedure:
1. Transcribe the DNA template below into RNA, and then translate the RNA into protein using the
codon chart in Figure 7.3. Use the three-letter abbreviations shown in the chart for the amino acids
(EX: ARG).
3’-CTATAATCGGTACTAGTACTACCGTACGAGATT-5’ DNA
5’-
_________ -3’ RNA
__________________________________________ protein
Part F: Mutations
A mutation is a change in a nucleotide sequence of the DNA. A mutation can involve one or more
bases. Mutations can occur through errors in DNA replication, or through exposure to chemical or
physical mutagens. UV light and X-rays are examples of physical mutagens. The nitrosamines found
in tobacco are an example of chemical mutagens.
There are several types of mutations that can occur, with different effects on the protein sequence. In
a deletion, one or more bases are lost from the DNA sequence. In an insertion, one or more bases
are added to the DNA. In a substitution, one base is substituted for another.
In the exercise below you will examine the effect of the various types of mutations on a protein
sequence. This exercise will count as the post-lab questions for today’s lab.
LAB 7: DNA Structure, Replication, and Protein Synthesis
Name: _________________________
Date: ________________________
POST-LAB QUESTIONS
1. What is the genetic material that is passed from generation to generation? Through which
process do we make more?
2. If one strand of a DNA molecule has the sequence GCTTTATCA, what would the other strand’s
sequence be? What rule does this follow?
3. List the components of a nucleotide. Which part is different in the 4 potential nucleotides?
4. What is a mutation? Are all mutations harmful?
5. Considering that we are all made up of the same 4 nucleotides in our DNA, the same 4
nucleotides in our RNA, and the same 20 amino acids in our proteins, why are we so different
from each other?
LAB 7: DNA Structure, Replication, and Protein Synthesis
Mutations Worksheet
(adapted from: mrsburse.cmswiki.wikispaces.net/file/view/MutationsWS.doc)
There are several types of mutation:
DELETION (a base is lost)
INSERTION (an extra base is inserted)
Deletion and insertion may cause what’s called a FRAMESHIFT, meaning the reading
“frame” changes, changing the amino acid sequence.
SUBSTITUTION (one base is substituted for another)
If a substitution changes the amino acid, it’s called a MISSENSE mutation.
If a substitution does not change the amino acid, it’s called a SILENT mutation.
If a substitution changes the amino acid to a “stop,” it’s called a NONSENSE mutation.
Complete the boxes below. Classify each as a deletion, insertion, or substitution AND as either
frameshift, missense, silent or nonsense (hint: deletion or insertion will always be frameshift).
Use the codon chart in Figure 7.4 to determine amino acid sequences. Use the three letter
abbreviations for the amino acids (ex: val for valine, see chart)
Original DNA Sequence: T A C A C C T T G G C G A C G A C T
mRNA Sequence:
Amino Acid Sequence: _________________________________________________________
Mutated DNA Sequence #1: T A C A T C T T G G C G A C G A C T
mRNA sequence:
(Circle the change)
amino acid sequence:
Will protein be affected?
Type of mutation:
LAB 7: DNA Structure, Replication, and Protein Synthesis
Mutated DNA Sequence #2: T A C G A C C T T G G C G A C G A C T
mRNA sequence:
(Circle the change)
amino acid sequence:
Will protein be affected?
Type of mutation:
LAB 7: DNA Structure, Replication, and Protein Synthesis
Original DNA Sequence: T A C A C C T T G G C G A C G A C T
mRNA Sequence:
Amino Acid Sequence: _____________________________
Mutated DNA Sequence #3: T A C A C C T T A G C G A C G A C T
mRNA sequence:
(Circle the change)
amino acid sequence:
Will protein be affected?
Type of mutation:
Mutated DNA Sequence #4: T A C A C C T T G G C G A C T A C T
mRNA sequence:
(Circle the change)
amino acid sequence:
Will protein be affected?
Type of mutation:
Mutated DNA Sequence #5: T A C A C C T T G G G A C G A C T
mRNA sequence:
amino acid sequence:
Will protein be affected?
Type of mutation:
7. Which type(s) of mutation resulted in a new variation of a trait?
8. Which type(s) of mutation resulted in an abnormal amino acid sequence?
LAB 7: DNA Structure, Replication, and Protein Synthesis
9. Which type of mutation stops the translation of the mRNA?
Sickle Cell Anemia
Sickle cell anemia is the result of a mutation in the gene that codes for hemoglobin, a molecule
found in your red blood cells which binds oxygen. The mutation causes the red blood cells to
become sickle-shaped when they release their oxygen. The sickled cells tend to get stuck in
blood vessels, causing pain and increased risk of stroke, blindness, damage to the heart and
lungs, and other conditions.
Analyze the DNA strands below to determine which amino acid is changed and what type of
mutation occurred.
Normal hemoglobin DNA
CACGTGGACTGAGGACTCCTC
Normal hemoglobin mRNA
Normal hemoglobin amino acid sequence
Sickle cell hemoglobin DNA
CACGTGGACTGAGGACACCTC
Sickle cell hemoglobin mRNA
Sickle cell hemoglobin
amino acid sequence
10.
Which amino acid is different in the sickle cell hemoglobin? What type of mutation is
this?