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Exercise 10 - DNA Fingerprinting
Introduction
Deoxyribonucleic acid (DNA) is a double stranded genetic molecule consisting of many monomers called
nucleotides, hence DNA is a polynucleotide. The two strands of DNA are connected to one another by
hydrogen bonds between the nitrogenous bases of each strand. The DNA base pair sequence and DNA
quantity (base pair total) vary from species to species. There would be less, but still measurable
differences among conspecifics. Indeed, no two organisms of the same species, unless they are clones
or identical twins, share exactly the same base pair sequence. It is the differences in these base pair
sequences that allows for the identification of genetic similarities between DNA from two sources using
a process known as DNA fingerprinting.
This laboratory exercise will investigate the basic concepts in DNA fingerprinting involving techniques
such as polymerase chain reaction (PCR) and the analysis of short tandem repeats (STR’s).
Materials
Equipment
scissors
marking pen or pencils
Part A: Polymerase Chain Reaction (PCR)
The most current form of DNA fingerprinting begins with a technique known as polymerase chain
reaction (PCR). The advantage of PCR is that only a tiny amount of DNA is needed and the sample can
be old, stored under less than ideal conditions or even partially degraded. PCR involves the following
steps:
1. The DNA sample is placed in a small test tube with a solution of deoxyribonucleotides,
small pieces of DNA to act as primers, and the enzyme DNA polymerase. The mixture is
then placed in a thermal cycling device, which will raise and lower the temperature of
the tube at precisely timed intervals.
2. Denaturing – occurs to the DNA when the mixture is raised to 94qC. The hydrogen
bonds between the nitrogenous bases break down from the heat and the two
complementary strands of DNA separate.
3. Annealing – the primers attach themselves to the long pieces of DNA through
complimentary base pairing (A-T; G-C) when the temperature is lowered to 65qC.
4. Extending – DNA polymerase extends the new strands of DNA from the primers as the
temperature is raised to 72qC.
5. The process (denaturing, annealing, extending) is repeated.
The PCR process amplifies the original amount of DNA in a very short period of time (~ 2 minutes per
cycle) so investigators will have much more DNA to use for subsequent analyses with little delay.
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Exercise 10 – DNA Fingerprinting
Procedure
1. On the following page is a “sample” of DNA and some primers
2. Use scissors to cut out all the pieces and lay them on the table
3. Starting with the two strands next to one another and lined up so the complementary base pairs
are aligned
4. Denaturing - slide the two strands of DNA away from one another
5. Annealing - move two of the primers in between the DNA strands, lining them up to form
complementary base pairs
6. Extending - use a marking pen or pencil to write in the complementary bases to the new DNA
strand until both strands are complete
After one round of PCR, one molecule of DNA consisting of two complementary strands yields _____
molecules of DNA for a total of _____ strands.
How many DNA molecules would exist after 2 PCR cycles? _____
5 cycles?
_________________
10 cycles?
_________________
20 cycles?
_________________
30 cycles?
_________________
Write a formula to calculate the number of DNA molecules which will be created for a given number
of PCR cycles.
Are all the DNA molecules created identical? _____
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Exercise 10 – DNA Fingerprinting
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Exercise 10 – DNA Fingerprinting
Part B: Short Tandem Repeat (STR) Analysis
Once the amplification of the original DNA through PCR has occurred, analysis of the DNA fingerprint
can begin. Although estimates of the differences in DNA between individuals are very small (~ 1/10 of
one percent), the sheer volume of DNA an individual possesses results in about 3 million bases pairs of
unique sequence (i.e., each person differs by about 3 million DNA base pairs). The analysis of short
tandem repeats examines some of this individual variation.
Humans, like other eukaryotes, contain interruptions in the protein-coding genes, called introns.
Because introns do not contain protein synthesis information, the base sequences in these regions tend
to be repetitive. The same sequence of four, five, or six bases repeats itself over and over again. For
example, intron 3 of the human α fibrinogen (a blood clotting protein) gene contains a sequence of
bases “TTTC”, which repeats:
TTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTC
Because these repetitive sequences are short (4-6 bases) and occur side-by-side (in tandem) they are
termed short tandem repeats (STR’s). The objective of DNA fingerprinting is to determine how many
times a sequence of an STR is repeated in a DNA sample.
How many times does the STR “TTTC” repeat itself in the above DNA sequence? _____
For intron 3 in the human α fibrinogen gene, an individual has between a 5 and 20% chance of sharing
the same number of repeats with another individual. Although that seems like a relatively low chance, it
must be remembered that the human population is quite large, so there would still be quite a few
people in this group. Therefore, knowing the number of STR’s for just one particular intron is not
sufficient enough to develop a unique DNA fingerprint. To achieve that, more than one STR must be
examined.
For example, intron 1 of the human tyrosine hydroxylase gene repeats the sequence “AATG”. Like the
human α fibrinogen gene, only about 5-20% of people will repeat the “AATG” sequence the same
number of times. Combining both these STR percentages narrows the field considerably.
As an illustration, assume that 1 out of every 20 people (1/20 or 5%) repeats the intron 3 of the human
α fibrinogen gene STR (“TTTC”) 14 times and that 1 out of every 5 people (1/5 or 20%) repeats the intron
1 of the human tyrosine hydroxylase gene STR (“AATG”) 10 times. The chances that two individuals will
share the same number of repeats for both STR’s would be calculated as such:
Chance of sharing the same number of repeats in both STR’s = chance of sharing the same
number of repeats in the first STR x chance of sharing the same number of repeats in the
second STR
For this example:
Chance of sharing the same number of repeats in both STR’s = 1/20 x 1/20 = 1/400
So, an individual has a 1 in 400 chance (0.25%) of sharing the same number of repeats in both STR’s.
The more STR’s examined, the more that chance decreases until, within statistical certainty, a unique
DNA fingerprint is assembled.
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Exercise 10 – DNA Fingerprinting
Since 1997 the Federal Bureau of Investigation (FBI) has set standards for DNA fingerprinting analysis for
forensic and law enforcement purposes. To meet those standards, 13 specific genes areas (loci; singular
locus) are evaluated. These loci are found on autosomes (non-sex chromosomes). A 14th locus is used
to determine the sex of the individual and measures STR’s on the X and Y chromosomes. Requiring a
match of at least 14 loci virtually guarantees a conclusive identification as the odds of two, unrelated
person matching all 14 loci is approximately 1 in 1,000,000,000,000,000 (1 quadrillion).
After PCR, the scientist or technician must find out how many repeats occur at each loci examined.
There are number of ways this is accomplished, but most methods compare the DNA molecules
produced by PCR to DNA fragments of known lengths. DNA fragments are created by cutting the DNA
sample with special molecules known as restriction enzymes.
Part C: Restriction Enzymes
The preparation of a sample of DNA for fingerprinting involves treating the DNA with specific restriction
enzymes (endonucleases). Restriction enzymes cut both strands of DNA at specific locations known as
recognition sites (Table 10.1). The original intent of these enzymes is to provide protection for bacteria
against some invading bacteriophages. Bacteriophages are viruses that attack bacteria. Cutting DNA
with restriction enzymes results in smaller DNA fragments (restriction fragments) of various base pair
lengths.
Table 10.1 Recognition Sites for Various Restriction Enzymes
Restriction Enzyme
BstEII
EcoRI
HindIII
Recognition Site
---GpGTNACC-----GGTNACC-----CCANTGG-----CCANTGnG---
---GpAATTC-----GAATTC-----CTTAAG-----CTTAAnG---
---ApAGCTT-----AAGCTT-----TTCGAA-----TTCGAnA---
Digested DNA
Fragments
---GXXXXGTNACC-----CCANTGXXXXG--(N = any nucleotide)
---GXXXXAATTC-----CTTAAXXXXG---
---AXXXXAGCTT-----TTCGAXXXXA---
A DNA Restriction Map provides the exact locations of the recognition sites for a particular restriction
enzyme on a DNA sample. The map indicates distances from the origin where the enzyme cut the DNA.
Size of each DNA fragment produced by restriction enzyme treatment can be measured in base pair (bp)
units. Figure 10.2 illustrates a map of a DNA molecule of 50,000 bp in length. This DNA molecule has
been treated with three different restriction enzymes.
Procedure
1. Examine Fig. 10.2 and notice that each enzyme (A, B, and C) cuts the DNA sample at different
recognition sites (for example, enzyme A cuts the DNA at 15,010 bp, again at 24,650 bp, and a
third time at 30,003 bp)
2. Calculate the base pair (bp) size for each DNA fragment and write that number above the
fragment
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Exercise 10 – DNA Fingerprinting
Fig. 10.2 Restriction Map of 50,000 Base Pair DNA Sample
uncut DNA
enzyme A
enzyme
y
B
enzyme
y
C
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Exercise 10 – DNA Fingerprinting
Part D: Gel Electrophoresis and Fingerprint Analysis
Electrophoresis means to “carry with an electric current.” The digested DNA fragments created by the
restriction enzymes are loaded into an agarose gel in which an electric current will flow through.
Because DNA is a negatively charged molecule (due to the presence of large numbers of phosphate
groups in its backbone), it will migrate through the gel towards the positive pole of the electrophoresis
chamber. This procedure will separate the digested DNA fragments based upon molecular size. Smaller
fragments will migrate further through the gel than larger ones. The different fragments can then be
stained to make them visible. The resultant pattern of stained DNA “bands” or fragments constitutes a
DNA fingerprint (see Fig. 10.3 as example).
Fig. 10.3 Sample DNA Fingerprint
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Exercise 10 – DNA Fingerprinting
Part D1: Restriction Maps
Procedure
1. Compare the Restriction Map in Fig. 10.2 to the DNA fingerprint in Fig. 10.4. Does the number
of bands or fragment created by each restriction enzyme in the fingerprint coincide with the
number of fragments illustrated in the Restriction Map? __________
2. Label each of the DNA bands in Fig. 10.4 with its actual size in base pairs. Remember, in
electrophoresis fragments are separated upon size, with smaller fragments moving farther from
the wells at the negative end of the chamber than larger ones
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Exercise 10 – DNA Fingerprinting
Fig. 10.4 DNA Fingerprint of 50,000 Base Pair DNA Sample
uncut
DNA
enzyme
A
-
enzyme
B
enzyme
C
+
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Exercise 10 – DNA Fingerprinting
Part D2: Using a DNA Fingerprint to Determine the Number of STR’s
The examination of banding patterns in gel electrophoresis can be used to determine the number of
STR’s in a given DNA sample. Intron 3 of human α fibrinogen and intron 1 of the human tyrosine
hydroxylase genes may result in a banding pattern such as Fig. 10.5.
Fig 10.5 STR Banding Patterns for Two Human Genes
α fibrinogen locus
-
tyrosine hydroxylase locus
size markers
72 bp
68 bp
64 bp
60 bp
56 bp
52 bp
48 bp
44 bp
40 bp
+
Why is there only one band for the human α fibrinogen loci? (recall an organism can be homozygous or
heterozygous for a particular gene)
To determine the number of STR’s for each band, divide the size of the DNA fragment by the number of
bases that repeat (usually 4-6) in that particular STR. For example, a DNA fragment of an STR which
repeats the 5 bases “AGGTA” that is 55 base pairs long would repeats that STR 55/5 or 11 times.
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Exercise 10 – DNA Fingerprinting
For each allele in Fig. 10.5, how many times does the STR in intron 3 of the human α fibrinogen gene
repeat?
_____ _____
For each allele in Fig. 10.5, how many times does the STR in intron 1 of the human tyrosine hydroxylase
gene repeat? _____ _____
An example of a DNA fingerprint from 14 different loci is presented in Table 10.2.
Table 10.2 Sample DNA Fingerprint from 14 Loci
Locus
D3S1358
vWA
FGA
Genotype
15, 18
16, 16
19, 24
Locus
Genotype
D13S17
10, 11
D7S820
10, 10
D16S539
11, 11
D8S1179
12, 12
D21S11
29, 31
D18S51
12, 13
D5S818
11, 11
THO1
9, 10
TPOX
8, 8
CSF1PO
11, 12
AMEL
X, Y
Is the person with this DNA fingerprint male or female? __________ {10.9}
Which loci are heterozygous?
Which loci are homozygous?
Procedure
1. Assume the DNA fingerprint in Table 10.2 above comes from a blood sample collected at a crime
scene. Law enforcement authorities currently hold three suspects. A sample of DNA has been
collected from each suspect and their DNA fingerprints for the loci in Table 10.2 have been
determined (Table 10.3)
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Exercise 10 – DNA Fingerprinting
Table 10.3 DNA Fingerprints for Three Suspects
Suspect “A”
Locus
D3S1358
vWA
FGA
Genotype
15, 18
16, 17
20, 24
D8S1179
12, 14
D21S11
29, 31
D18S51
10, 12
D5S818
11, 13
Locus
Genotype
D13S17
12, 12
D7S820
13, 13
D16S539
11, 12
THO1
9, 9
TPOX
7, 8
CSF1PO
12, 12
AMEL
X, Y
Suspect “B”
Locus
Genotype
D3S1358
15, 15
vWA
16, 16
FGA
19, 25
D8S1179
12, 12
D21S11
30, 31
D18S51
10, 11
D5S818
11, 13
Locus
Genotype
D13S17
12, 13
D7S820
9, 10
D16S539
11, 11
THO1
9, 10
TPOX
9, 9
CSF1PO
11, 14
AMEL
X, Y
Suspect “C”
Locus
Genotype
D3S1358
15, 18
vWA
16, 16
FGA
19, 24
D8S1179
12, 12
D21S11
29, 31
D18S51
12, 13
D5S818
11, 11
Locus
Genotype
D13S17
10, 11
D7S820
10, 10
D16S539
11, 11
THO1
9, 10
TPOX
8, 8
CSF1PO
11, 12
AMEL
X, Y
Based on these data, which of these suspects is the most likely culprit? __________
How would the answer change if it was found out Suspect “C” had recently been the recipient of a bone
marrow transplant? (hint: what are the functions of bone marrow?)
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Exercise 10 – DNA Fingerprinting
Practice Problems and Review Questions
1. Define or describe the following:
DNA
STR
PCR
locus (loci)
annealing
2. In DNA, __________ binds to thymine, guanine binds to __________ through __________
bonds.
3. How many loci are required by the FBI to positively identify an individual? Is it reasonably
possible for two, non-related persons to share the same genotypes at all these loci? Why or
why not?
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Exercise 10 – DNA Fingerprinting
Lake-Sumter State College, Leesburg Laboratory Manual for BSC 1010C
blood stain on
victim’s clothes
victim
size markers
suspect 2
suspect 1
4. The following set of DNA fingerprints is from an actual crime scene. It contains a set of size
markers and banding patterns from blood samples collected from the victim, two suspects, and
blood found on the victim’s clothing. From what is shown here, determine which suspect’s
blood more likely matches the blood found on the victim’s clothes. Justify your decision.
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Exercise 10 – DNA Fingerprinting
5. DNA fingerprinting can also be used in issues of family relationships. Since each person receives
half their chromosomes from each parent, half of a person’s DNA fingerprint should match the
mother and half should match the father.
Consider the following DNA fingerprints as evidence in a paternity case. Is it reasonably possible
the male in question is the baby’s biological father? __________
How would the answer change if the male in question had an identical twin?
Mother
Locus
Genotype
D3S1358
13, 18
vWA
15, 15
FGA
18, 24
D8S1179
12, 12
D21S11
28, 31
D18S51
13, 13
D5S818
11, 11
Locus
Genotype
D13S17
10, 10
D7S820
11, 11
D16S539
10, 11
THO1
9, 9
TPOX
10, 11
CSF1PO
11, 11
AMEL
X, X
Baby
Locus
Genotype
D3S1358
15, 18
vWA
15, 16
FGA
19, 24
D8S1179
12, 12
D21S11
29, 31
D18S51
12, 13
D5S818
11, 11
Locus
Genotype
D13S17
10, 11
D7S820
10, 11
D16S539
11, 11
THO1
9, 10
TPOX
8, 10
CSF1PO
11, 12
AMEL
X, X
Alleged Father
Locus
D3S1358
Genotype
13, 15
vWA
16, 16
FGA
18, 19
D8S1179
12, 14
D21S11
29, 30
D18S51
12, 12
D5S818
10, 11
D7S820
10, 10
D16S539
11, 11
THO1
10, 10
TPOX
8, 8
CSF1PO
11, 12
AMEL
X, Y
Locus
Genotype
D13S17
11, 12
6. Calculate the number of repeats in the DNA fragments for the STR’s given.
a.
STR: “AAGCTA”
DNA fragment length: 72 bp
# of repeats: _____
b.
STR: “TTAT”
DNA fragment length: 52 bp
# of repeats: _____
c.
STR: “TAGGG”
DNA fragment length: 105 bp
# of repeats: _____
d.
STR: “AAGCT”
DNA fragment length: 60 bp
# of repeats: _____
e.
STR: “GAGGCT”
DNA fragment length: 144 bp
# of repeats: _____
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