BIO 150
Spring 2009
LAB PRACTICAL EXAM 2 - KEY
(50 points)
NAME _____________________________
1.
(6 points)
a.
Using the Hispanic allele frequency table provided, calculate the combined Pm
for the green-labeled STR loci in this profile. Show all of your calculations.
Locus
D8S1179
Genotype
Calculation
Pm
16,16
(0.025) + [(0.025)(0.025)(0.01)]
0.000869
D21S11
28,29
2(0.096)(0.200)
0.0384
D18S851
14,19
2(0.139)(0.039)
0.0108
2
3.60 x 10
Combined Pm
b.
-7
Other than genotypic information about STRs, is there any other information
that you can get from this profile? Explain.
Yes. The amelogenin locus provides information about the sex of the
person. In this case, the sample is from a male because the allele on
the X chromosome and the allele on the Y chromosome both amplified.
In a female, only the allele on the X chromosome would amplify.
*2.
(4 points)
The profile in question #1 is that of an alleged father in a paternity test. The profiles
of the mother and child at these same loci are shown in the table below.
Locus
D8S1179
D21S11
D18S531
Mother
14,18
28,28
14,15
Child
14,16
28,28
14,15
a.
Circle each of the obligate paternal alleles in the child’s profile. If an OPA
cannot be determined, circle both alleles carried by the child.
b.
Calculate the Combined Paternity Index generated by the three markers used
in this paternity test. Show all your calculations.
(1/0.025)(0.5/0.096) = 208; The D18S531 locus in non-informative
for paternity calculations but does not exclude the alleged father,
because the father has at least one of the alleles carried by the child
at that locus (14).
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BIO 150
*3.
Spring 2009
(3 points)
Explain how the Quantifiler kit enables a researcher to detect the presence of PCR
inhibitors in a DNA extract.
In addition to the primers and FAM-labeled Taqman probe that target
human DNA, there is a sequence of non-human DNA (IPC = internal positive
control) in the reaction mix that is targeted by an additional set of primers
and a VIC-labeled TaqMan probe. Regardless of how much human DNA is in
the extract being tested, the non-human DNA should amplify efficiently in
the qPCR reaction. If it doesn’t, this is an indication that a PCR inhibitor is
present in the extract.
4.
(3 points)
Examine the CODISmt printout and then answer the following questions:
a.
What is the haplotype of the sample being analyzed? Use the proper
nomenclature!
human mtDNA: hvII 61-241 (195 C)
b.
Calculate the frequency of the haplotype in peoples of Caucasian origin. Show
your calculation and express your answer as a FRACTION.
Pdatabase = X/N = 9/323 = 0.02786
1/2
Ppopulation = Pdatabase + 1.96
(Pdatabase)(1- Pdatabase)
N
=.0458 = 1/22
1 in 22 Caucasians is estimated to have this mtDNA haplotype.
*5.
(2 points)
The repeat unit at STR locus D18S51 is 5’-AGAA-3’. Write out the exact sequence of
the 10.3 allele of this locus in the space below. You only need to write out one strand
of the DNA.
5’ – AGAA AGAA AGAA AGAA AGAA AGAA AGAA AGAA AGAA AGAA AGA – 3’
6.
(2 points)
What are you looking at through the light microscope? Be as precise and accurate as
you can, making no prior assumptions!
Primate sperm. The heads of primate sperm stain differentially, picking up
less stain in the acrosome area. The tails of the sperm can also be seen.
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BIO 150
*7.
Spring 2009
(2 points)
Label the x and y axes of the STR electropherogram below. (HINT: Don't label them
"x" and "y"!)
x = amplicon length (bases) or seconds since injection
y = relative fluorescent units (rfu’s)
*8.
(2 points)
Why is it necessary to use polyacrylamide, instead of agarose, for DNA sequencing?
DNA sequencing requires that fragments differing in length by a single
nucleotide be resolvable by gel electrophoresis. The pore size in an agarose
gel is too large to permit this level of resolution; DNA fragments differing by
as much as 50 bp exhibit essentially the same mobility on agarose gels.
However, the pore size in polyacrylamide gels is much smaller and
fragments that differ in a single nucleotide run through the gel at detectably
different mobilities. Therefore, polyacrylamide gels have sufficient
resolution power for DNA sequencing.
9.
(4 points)
Examine the real-time PCR printout. To prepare for quantitation, each “unknown”
extract was diluted 1:30. Then 2 uL of the diluted extract was added to a
Quantifiler reaction.
a.
What is the concentration of human DNA in the diluted sample D0-2? Be sure
to include units!
(.1412 ng)/2 uL) = 0.0706 ng/uL
b.
Calculate the volume of undiluted D0-2 DNA extract you would need to add to
an Identifiler reaction to ensure that the amplified product would fall within
the correct range for detection of the PCR products by capillary gel
electrophoresis. Show your calculation. Would you need to dilute the sample
first? Explain!
(0.0706 ng/uL)(30) = 2.118 ng/uL in the undiluted extract
A total of 1 ng of DNA is required for Identifiler and the total DNA
added should be 10 uL. In order to add 1 ng of the undiluted extract
to the Identifiler reaction, you would have to pipet 1 ng/(2.118 ng/uL)
= 0.47 uL into 9.53 uL of water.
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BIO 150
Spring 2009
However, it is not possible to accurately pipette less than 1 uL, so the
sample must be diluted before adding it to the water. With a dilution
of 1:10, you could add 4.7 uL of DNA to 5.3 uL of water for a total of
volume of 10 uL of DNA extract.
*10.
(4 points)
Draw a diagram showing how capillary electrophoresis is used to detect STR size
polymorphisms. Be sure to label your diagram carefully and thoroughly! What is
being detected and how?
(-)
Identifiler PCR products
are loaded into
capillary
Products run through capillary
according to size; Larger amplicons
run more slowly than small ones
As the products pass the capillary
window, a laser excites the
fluorescent tag and the tag emits a
signal
detector
(+)
*11.
A CCD camera detects the
wavelength, amplitude, and
time since injection of each
amplicon and sends the digital
information to a computer for
analysis; an electropherogram
is generated
(2 points)
Approximately how much larger than the human mitochondrial genome is the human
nuclear genome? Show your calculation.
(3.3 x 109 bp)/16,569 bp) = ~200,000 times larger
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BIO 150
*12.
Spring 2009
(3 points)
In the space below, draw ddATP. You do not need to show the full chemistry of the
nitrogenous base. However, you should clearly indicate where it is attached the rest
of the molecule. Number all of the carbons and show the locations of all charges.
A
*13.
(2 points)
In lab, we extracted DNA from hair, semen, and saliva. Which of these required the
use of DTT and why? In your answer, explain precisely what action DTT performs.
Semen. Sperm DNA is surrounded by proteins called protamines that are
stabilized by strong, disulfide bonds. In order to extract DNA from semen,
DTT is needed because DTT breaks disulfide bonds. Saliva does not contain
proteins stabilized by disulfide bonds, so DTT is not needed. DTT is often
used to isolate DNA from hair because keratin (the major protein in hair)
also contains disulfide bonds. However, we did not use DTT for our hair
extraction in lab because we isolated the mtDNA from the non-keratinized
sheath cells.
14.
(2 points)
The arrows point to artifact peaks in an electropherogram. What are these artifact
peaks called? What causes them?
Stutter peaks. They are caused by strand slippage errors made by taq DNA
polymerase during PCR. Stutter projects are always much lower in
amplitude than “real” peaks and usually occur 4 base-pairs before a “real”
peak. This is because taq polymerase tends to favor slippage in the template,
rather than the growing strand, during replication. This leads to a minor
product in the PCR reaction that is one repeat shorter than the major
product.
*15.
(4 points)
a. For what purpose did we use this item in lab?
A QIAGEN QIAamp spin column. We used it to extract DNA from semen
and saliva samples.
b. Diagram how it works.
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BIO 150
*16.
Spring 2009
(5 points)
In the crime lab, you have received an oral swab from a rape victim who says she
was forced to orally copulate her attacker. Draw a simple diagram showing the steps
you would follow (in the correct order!) to perform your analysis, starting with DNA
extraction and ending in an autosomal STR profile. (You do not need to give all the
details involved in each step, but be sure to clearly show your overall approach.)
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BIO 150
Spring 2009
EQUATIONS:
Pm = p2 + [(p)(1-p)(0.01)
Pm = 2pq
Ppopulation = Pdatabase + 1.96
1/2
(Pdatabase)(1- Pdatabase)
N
Pdatabase = X/N
PI = 0.5/(freq OPA)
PI = 1/(freq OPA)
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