Lab Activity No. 9
DNA Fingerprinting, Transcription and Translation
Survey of Biology Lab
A. Objectives: Upon completion of this lab activity, you should be able to:
1. Define and correctly use the following terms: restriction enzymes, recognition site/sequence, electrophoresis,
agarose gel, DNA fingerprinting, and DNA fragments.
2. Demonstrate knowledge of the structure of DNA and how DNA sequences can vary between samples.
3. Identify restriction enzyme recognition sequences in DNA fragment sequences and predict the results of the
exposure of the DNA fragment to the restriction enzyme.
4. Interpret DNA fingerprint electrophoresis results.
5. Demonstrate the processes of transcription and translation using the puzzle kits.
6. Explain how mutations in DNA can ultimately affect the proteins in a cell.
B. Introduction: DNA fingerprinting is a technique used to detect differences in the base sequence between two or more
samples of DNA. It has many practical applications including helping crime scene investigators identify possible
suspects who may have left DNA at a crime scene. No two person’s DNA base sequences are identical (except in the
case of identical twins) so samples of DNA can be analyzed and matched to samples from known persons. This
technique, when performed correctly, is extremely accurate.
In this lab, you will apply DNA fingerprinting techniques to different DNA samples to determine if the samples are
from the same individual or from different individuals.
After you set up your DNA fingerprinting apparatus, it must run for 30 minutes. During this wait time, you will use the
puzzle kits to examine the two processes of transcription and translation in which DNA is “read” by the cell as a set of
instructions to ultimately build proteins.
C. DNA Fingerprinting
DNA fingerprinting techniques employ the use of specialized enzymes called restriction enzymes. Restriction enzymes
cut DNA into fragments at specific locations designated by particular base sequences called recognition sites. When
DNA samples from different individuals (so having different base sequences) are cut with the same restriction enzyme
(digested), fragments will be produced that vary in number and/or fragment size. Since it is not possible to directly view
the DNA fragments and observe possible differences in them, it is necessary to separate them using a procedure called
gel electrophoresis. In this procedure, samples of DNA that have been digested by a restriction enzyme are placed into a
space in a gel (sort of like a small piece of jello) and the gel is exposed to an electrical current. Because the DNA
fragments have a negative charge (due to the phosphate groups in the “backbone”), they will be attracted and move
toward the positive pole. As the DNA moves through the gel, which contains lots of internal holes, the smaller DNA
fragments will be able to move faster than the larger fragments. After the fragments have been separated in the gel, a
stain is applied which “sticks” to the DNA and allows you to see the fragments more clearly.
In this activity, you will be given the task of loading a gel with six DNA samples associated with a crime scene. One
sample was collected at the crime scene (CS) and five additional samples were collected from five possible suspects (S1S5). After you complete the fingerprinting process, you will determine which of the five suspects’ DNA matches the
sample left at the crime scene.
Materials:
• Pre-poured agarose gel
• Electrophoresis box and power supply
• TAE buffer solution
• Loading dye
• Micropipettor and tips
• Digested DNA samples (6)
Procedure:
1. Using a micropipettor and a new tip for each sample, add 10 µl of sample loading dye "LD" to each of the following
6 sample tubes (Note: The loading dye functions to cause the DNA samples to sink to the bottom of the spaces in the gel
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and also help you see where you are loading the sample). Close the caps on all the tubes and shake them to mix.
Centrifuge the samples for about 5 seconds to bring the contents to the bottom of the tube.
2. Carefully remove the comb from the gel by pulling it straight up. The indentations/slots that are exposed is where the
DNA samples will be loaded. Also remove the black rubber “bumpers” at each end of the casting try. Your instructor
will demonstrate both procedures.
3. Place the casting tray with the solidified gel in it into the platform in the electrophoresis box. The wells (i.e. the small
spaces where you will load the DNA) should be at the (-) cathode end of the box, where the black lead from the power
supply will connect. Make sure you correctly position the gel in the box!
4. Pour ~ 275 ml of TAE/TBE electrophoresis buffer into the electrophoresis chamber. Pour buffer in
the electrophoresis box until it just covers the wells of the gel. Once poured, keep movements of the tray to a minimum.
5. Using a separate pipet tip for each sample, load your gel as follows:
Lane 1: CS, 20 µl
Lane 2: S1, 20 µl
Lane 3: S2, 20 µl
Lane 4: S3, 20 µl
Lane 5: S4, 20 µl
Lane 6: S5, 20 µl
Make a diagram and label your lanes so you know where each sample was poured.
6. Secure the lid on the electrophoresis box. The lid will attach to the base in only one orientation: red to red and black to
black. Connect electrical leads to the power supply in the appropriate positions (again red to red and black to black).
7. Turn on the power supply and set it for 100 V (or as close as you can get to that value) and allow electrophoresis of
the samples to occur for 30 minutes.
8. When the electrophoresis is complete, turn off the power and remove the lid from the electrophoresis box. Carefully
remove the gel tray and the gel from the gel box. Be careful, the gel is very slippery! Nudge the gel off the gel tray with
your thumb and carefully slide it into your plastic staining tray.
9. Follow the gel staining procedure your instructor gives you. You can read the results immediately if using a “fast”
staining procedure or you can cover the gel with plastic wrap and let it sit overnight.
10. After the gel has been stained, your group can view it to evaluate the results.
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Questions:
1. Assume that the recognition site of a restriction enzyme is GAATTC and that the cuts/beaks the sugar phosphate
backbone between the G and the A. Therefore, each time the enzyme encounters this recognition site, it will cut the
DNA in the same place as is illustrated below:
How many pieces/fragments of DNA would result from the cut shown above? ___________
2. DNA fragment size can be expressed as the number of base pairs in the fragment. Indicate the size of the fragments
[mention any discrepancy you may detect]. Note that there will be a region of single bases (i.e. not base pairs) at the
region of the cut. Therefore, these would formally not be included in the determination of the fragment size. Normally,
the number of unpaired bases is insignificant compared to the total size of the double stranded region (frequently a
hundred to several thousand bases). However, in such small fragments shown in the example, the effect is exaggerated.
a) The smaller fragment is ___________ base pairs (bp).
b) The larger fragment is __________ bp.
3. Consider the two samples of DNA shown below - single strands are shown for simplicity:
Sample #1: C A G T G A T C T C G A A T T C G C T A G T A A C G T T
Sample #2: T C A T G A A T T C C T G G A A T C A G C A A A T G C A
If both samples are treated with a restriction enzyme [recognition sequence GAATTC], indicate the number of
fragments and the size of each fragment from each sample of DNA that would result.
Sample #1: No. of fragments:________
Sample #2: No. of fragments:_________
List fragment sizes for each sample in order of largest (bp) to smallest (bp):
Sample #1:
Sample #2:
4. If you could not see the original sequences but could only compare the number and sizes of the fragments, would you
say that these samples come from the same person or two different people? Explain your answer:
D. Transcription
Within the cell, DNA serves as a set of instructions that help with the assembly of proteins. As we shall see, the
specific order of the base pairs in the DNA is critical to the proper construction of the huge variety of proteins a cell can
make. DNA resides within the nucleus of eukaryotic cells and does not leave. However, the proteins are assembled
outside the nucleus in the cytoplasm. This means that the information stored in the specific order of the base pairs on the
DNA must be passed from the nucleus to the cytoplasm. This is the job of messenger RNA (mRNA). Messenger RNA
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is assembled in a process know as transcription. During transcription, a region of the DNA double helix unwinds and
unzips and a single strand of mRNA is constructed using the nucleotide sequence of one strand of the DNA as a
template. This mRNA strand then leaves the nucleus via pores in the nuclear membrane and is used to build proteins in
the cytoplasm. The DNA double helix reforms its hydrogen bonds (“re-zips” itself) and can thus be read over and over
again to make more mRNA as it (and the corresponding proteins) are needed by the cell. Two other types of RNA are
also produced by transcription (tRNA and rRNA). While these RNAs are not directly translated, as is mRNA, they do
function in the process of protein synthesis.
Materials:
•
1 DNA Puzzle Kit
Before starting this exercise, check that all the pieces in your puzzle kit have the same number written on the back and
that the number matches the one on the box lid. If you have numbered pieces that don’t belong to your puzzle, return
them to the proper box. Then check to make sure your puzzle has the following pieces:
• 24 deoxyribose sugars (dark pink)
• 12 ribose sugars (light pink)
• 24 phosphate groups (yellow)
• 4 adenine bases
• 8 guanine bases
• 8 cytosine bases
• 4 thymine bases
• 2 uracil bases
• 4 transfer RNA (tRNA) (tan and brown)
• 4 amino acids (tan and brown)
• 4 activating units (tan and brown)
• 1 ribosome template sheet (white paper)
Procedure:
1. Using the appropriate puzzle pieces, construct the nucleotides necessary to build one half of a DNA molecule with the
base sequence A-C-C-T-G-C-A-C-C-T-G-C. Remember to include the deoxyribose sugars and the phosphate groups in
the “backbone”. Lay the strand on the table horizontally with the exposed bases pointing away from you with the A on
the left.
2. Now build twelve RNA nucleotides that are complimentary to the exposed DNA bases, which are referred to as the
template strand. Remember to review the differences between DNA and RNA when you build these nucleotides. Build
the RNA strand by pairing the bases between the RNA and DNA and connect the sugar (ribose) and phosphates on the
RNA. Write the base sequences of the two strands below:
mRNA strand sequence:
____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____
DNA template strand sequence:
____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ ____
Separate the DNA template and the mRNA strands. In a cell, the mRNA strand would undergo some “editing” including
the addition of the 5’ CAP, the 3’ poly-A tail, and removal of the introns. Then it would leave the nucleus, and the DNA
strand would rejoin with its complementary half by reforming the hydrogen bonds between complimentary base pairs.
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E. Translation
In the cell, the mRNA leaves the nucleus through a pore in the nuclear membrane. Once in the cytoplasm, it attaches to
a ribosome and is “read” three base pairs at a time to determine the appropriate amino acid that should be inserted in the
growing protein. A block of three consecutive nucleotides in the mRNA is called a codon. The ribosome (which is made
of ribosomal RNA or rRNA) is comprised of two subunits named the 40S subunit and the 60S subunit (You can think of
the “S” as a unit of weight measurement). The transfer RNA (tRNA) function to bring specific amino acids to the
ribosome to be used in building the specific protein. Each tRNA has a three base sequence on one end called the
anticodon and a sequence CCA on the other which serves as an attachment site for the tRNA’s specific activating
enzyme. The activating enzyme is responsible for catalyzing the chemical reaction of joining the amino acid to the
tRNA. The anticodon on the tRNA is complementary to the codon on the mRNA, which explains how the correct amino
acid is brought to the site of translation.
Procedure:
1. Use the mRNA strand that was created in the transcription activity. You will now translate this mRNA into an amino
acid sequence of a protein.
2. Place the large, white sheet of paper from the puzzle kit on the desk in front of you. The black outline on the paper
(template) represents the ribosome. Carefully slide your mRNA strand onto the mRNA binding site of the 40 S subunit
of the ribosome template. The phosphate attached to the G on the right end of the mRNA should line up with the 5’ on
the template. The exposed bases should point toward the P and A sites.
3. Find the four tRNA puzzle pieces and attach their specific activating enzymes to them. Then attach the specific amino
acid to each activating unit.
4. The order in which the amino acids are connected to build the protein is determined by the sequence of the bases on
the mRNA. Recall that this sequence was ultimately determined by the sequence of the bases on the DNA strand from
which it was transcribed. Pick up the t-RNA/amino acid unit whose anticodon complements the mRNA codon above the
P site of the 60S subunit. Place it on the P site with the codon and anticodon paired. Do the same for the A site.
5. The amino acids form their peptide bonds with each other when their t-RNAs are base paired with the mRNA in the A
and P sites. Detach the P site amino acid from its tRNA activating unit and slide it over the tRNA, attaching it to the A
site amino acid. Move the mRNA to the right until the tRNA activating unit amino acid chain occupies the P site (the
tRNA-activating unit amino acid chain is moved from the A site to the P site). Carefully lift the empty tRNA-activating
unit complex, slide it out from under the amino acid chain, and place it to one side. Fill the A site with another tRNAamino acid unit and repeat the process until you have produced a chain of four amino acid units. Then complete the table
below and on your group lab report form:
Template DNA base sequence: A C C
T G C
A C C
T G C
mRNA codons:
______
______
______
______
tRNA anticodons:
______
______
______
______
Amino acid sequence:
______
______
______
______
6. In the previous lab on DNA replication, we discussed that errors can occur during the replication process. Typically,
these errors are corrected by proofreading enzymes, but sometimes mistakes persist. A mistake that results in a single
base change is called a point mutation and may affect the amino acid composition of the resulting protein.
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a) How would the protein have been different if the DNA template sequence was G-C-C-T-C-C-A-G-C-T-G-A?
Note: Use the table of the genetic code included at the end of the lab:
Template DNA base sequence: G C C
T C C
A G C
T G A
mRNA codons:
______
______
______
______
tRNA anticodons:
______
______
______
______
Amino acid sequence:
______
______
______
______
b) Do all point mutations cause a different amino acid to be inserted?
c) Notice from the table of the genetic code (included at the end of this lab) that 61 codons represent the 20 different
amino acids. Why do you think it is advantageous, from a genetic perspective, to have this redundancy (i.e. the
same amino acid is represented by more than one codon)?
7. Once translation is complete, the protein, the mRNA, and the two subunits of the ribosome separate. In a cell, the
resulting protein would be passed into the endoplasmic reticulum where it would undergo further modifications before
being used within the cell or exported to other locations within an organism.
Questions:
1. If the base sequence on a strand of DNA is TACCCGTATACT, what would be the sequence of the codons on the
mRNA transcribed from this DNA?
_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____
2. What would the sequence of bases of the anticodons on the tRNA be?
_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____
3. What would be the sequence of amino acids that would be translated from this DNA?
_________ _________ _________ _________
4. If a mistake (point mutation) is made during replication of this strand so that the new DNA strand had the base
sequence TACACGTATACT, what would be the resulting change in the protein built from this strand?
_________ _________ _________ _________
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5. Complete the following table using the information provided:
Template
DNA Strand
NonTemplate
DNA Strand
GGG
TAC
mRNA
tRNA
anticodon
Amino Acid
CCU
UCG
Leucine
F. Review Questions
1. The electrophoresis apparatus creates an electrical field with positive and negative poles at the ends of the gel.
DNA molecules have a ______________ charge. To which electrode pole of the electrophoresis field would you
expect DNA to migrate? (+ or -)?
2. After DNA samples are loaded into the sample wells, they are “forced” to move through the gel matrix. What
size fragments (large vs. small) would you expect to move toward the opposite end of the gel most quickly?
3. What is the difference, with respect to transcription, between the template and non-template DNA strands?
4. What are the DNA-DNA rules of complementary base pairing?
5. What are the DNA-RNA rules of complementary base pairing?
6. What are the RNA-RNA rules of complementary base pairing?
7. Distinguish between a codon and an anticodon. How are they similar? How are they different?
8. What are the functions of the mRNA, tRNA, and rRNA in translation.
9. In the genetic code table, three codons (UAA, UAG, and UGA) are associated with “STOP”. What does this
mean with regards to translation?
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