King Tut’s Family Secrets
Kuei-Chiu Chen, Premedical Program, Weill Cornell Medical College in Qatar
Their bodies have lain here in the Valley of the Kings near Luxor, Egypt, for more than 3,000
years. The flesh on many of them is still present as a result of embalming. The walls of their
burial chambers are covered with ornate paintings of gods that were said to greet them in the
other world, and the lavishly decorated objects of gold and precious stones buried with them
were intended for their use in the afterlife.
These are the mummies of the royal families of ancient Egypt. One of the members, the
“Boy King” – King Tutankhamun – became pharaoh at the age of nine or ten and died when
he was still a teenager. Other tombs in the Valley of the Kings contain other royal mummies.
Is “King Tut” related to any of these mummies?
2
Discovery of Satellite DNA
• Genomic DNA is fragmented and centrifuged.
• When reaching equilibrium, DNA bands settle based on density.
Uniform
Gradient
DNA fragments
Centrifugation
Satellite bands
of DNA
Main band of DNA
Cell debris
3
DNA from the satellite bands was named
satellite DNA.
Satellite DNA consists of highly repetitive
sequences found mostly in centromeres and
telomeres.
4
• The length of repeat unit varies in satellite DNA.
• DNA with repeat unit from 2 to 6 bases long is
called microsatellite DNA because the repeat
sequence is shorter.
• The term microsatellite DNA is based on the
historical term of satellite DNA. The location of
microsatellite is mostly NOT found on centromeres
or telomeres of the chromosomes.
5
Terms in genetics: example in human chromosomes
Homologous chromosomes
(one from father and one from mother)
Locus is the specific
position on the
chromosome.
Locus (heterozygous: having two
different alleles)
Allele is alternative form
of a gene at a locus.
Homozygous: having the
same allele at a locus.
Heterozygous: having
different alleles at a
locus.
6
Types of Microsatellites
Those with two-base repeat unit
CGACTAGCCACGTGTGTGTGTGTGTGTGTGTGTACAGGACTTAGC
11 repeats (allele 11)
Those with three-base repeat unit
TTCTAGCCGCTTGTTGTTGTTGTTGTTGTTGTTGCCAGTACTCAG
8 repeats (allele 8)
Those with four-base repeat unit
GCGAAGTTCCGAATGAATGAATGAATGAATGAATGAATGCCGCATT
7 repeats (allele 7)
Most microsatellite DNA loci used in human identification have
4-base pair repeats.
Microsatellites are also called Short Tandem Repeats (STRs).
7
CQ#1: A person has the following microsatellite
sequence on one chromosome:
Repeat region
GCGAAGTTCCGGATGGATGGATGGATGGATGGATGCCGCATT
What is the sequence of the repeat unit?
A.
B.
C.
D.
E.
ATG
GATG
ATGGAT
GGATGG
None of the above
8
The genotype of a person for each locus is shown as the
number of repeat units he/she carries. Since a person is
diploid and has a pair of homologous chromosomes, the
genotype is shown as two numbers.
- Genotype 6, 8: For a person heterozygous at a locus
where there are 6 repeats on one chromosome and 8
repeats on the other chromosome within the homologous
pair.
- Genotype 7, 7: For a person homozygous at a locus
where there are 7 repeats on both of his/her homologous
chromosomes.
9
Repeat sequences
Person A
Genotype 7, 7
(7 repeats, 7 repeats)
Person B
Genotype 6, 8
(6 repeats, 8 repeats)
10
CQ#2: If this person’s homologous chromosomes
have the following sequences:
Repeat region
GCGAAGTTCCGGATGGATGGATGGATGGATGCCGCATT
CGCTTCAAGGCCTACCTACCTACCTACCTACGGCGTAA
GCGAAGTTCCGGATGGATGGATGGATGGATGGATGGATGCCGCATT
CGCTTCAAGGCCTACCTACCTACCTACCTACCTACCTACGGCGTAA
What is this person’s genotype for this
locus?
A. 5, 7
B.
C.
D.
E.
5, 5
7, 7
GATG, GATG
CTAC, CTAC
11
Slippage during DNA replication creates variation in
number of repeat units
Normal
replication
A
Slippage
in new
strand
New strand
GAC GAC GAC GAC GAC GAC
CTG CTG CTG CTG CTG CTG
Template strand
Next
replication
C
G
GAC GAC GAC GAC GAC GAC
CTG CTG CTG CTG CTG CTG
Insertion mutation (n+1)
GAC GAC GAC GAC GAC GAC GAC
CTG CTG CTG CTG CTG CTG CTG
CTG CTG CTG CTG CTG CTG
GAC GAC GAC GAC GAC GAC
Normal length
Next
replication
Slippage
in
template
GAC GAC GAC GAC GAC
CTG CTG CTG CTG CTG
C
G
T
Deletion mutation (n-1)
CTG CTG CTG CTG CTG
GAC GAC GAC GAC GAC
CTG CTG CTG CTG CTG CTG
GAC GAC GAC GAC GAC GAC
Normal length
12
Microsatellite DNA is highly variable between individuals.
Variation forms the basis for individual identification.
Alleles (number of repeats)
Locus Name
TPOX
D3S1358
FGA
CSF1PO
7
14
18
8
8
15
19
9
9
16
20
10
10
17
21
11
11
18
22
12
23
13
24
14
25
15
12
26
27
28
13
Polymerase Chain Reaction (PCR)
 An in vitro reaction (cell-free)
 Required materials:
- template: the DNA sequence to be copied
- dNTPs (dATP, dGTP, dCTP, and dTTP) as building
blocks
- primers: short DNA sequences (19-33 bases in
STRs) complementary to the template to start the
reaction
- DNA polymerase: enzyme to catalyze the reaction
Forward primer
DNA
Reverse
primer
14
Extraction of DNA from Ancient Mummies
Challenges:
1. DNA degradation: by physical,
chemical, and biological causes.
2. Contamination: cells from other
persons who contacted the
mummies from ancient to
modern time.
Solutions?
1. Use deep tissues that were less degraded and less likely
to be contaminated, such as bone.
2. All researchers who handled the mummies should have
their DNA profile analyzed for use as reference samples.
3. Have multiple labs conduct the same PCR and compare
results.
15
Temperature Profile for PCR
94C
55C
72C
16
The number of PCR product increases exponentially following the
number of thermal cycles.
0 cycle
1
molecule
1 cycle
2 molecules
(21)
2 cycles
3 cycles
4 molecules
(22)
8 molecules
(23)
If starting with one molecule, the final number of double stranded molecule is 2n.
17
Polymerase Chain Reaction Animation
http://www.sumanasinc.com/webcontent/animations/content/pcr.html
18
CQ#3: If you start with 1 double-stranded DNA molecule
as a template and conduct a PCR for 4 thermal cycles,
how many DNA molecules of the PCR product will be
there at the end of the reaction?
A.
B.
C.
D.
E.
12
16
24
48
96
19
CQ#4: If you start with 3 double-stranded DNA molecules
as a template and conduct a PCR for 4 thermal cycles,
how many DNA molecules of the PCR product will be
there at the end of the reaction?
A.
B.
C.
D.
E.
12
16
24
48
96
20
Composition of a PCR product from a locus
The length of a PCR product consists of the forward primer, reverse
primer, repeat sequences and flanking sequences.
Forward primer
binding site
Reverse primer
binding site
Repeat sequences
3’
5’
Flanking sequence
Flanking sequence
5’
3’
Length of PCR products
21
Sample from a person homozygous at the locus will
produce PCR fragments of one length
Person A
Genotype 7, 7
Length of PCR products
22
Sample from a person who is heterozygous at a locus will
produce PCR fragments of two different lengths
Length of PCR products
Person B
Genotype 6, 8
Length of PCR products
Allele 6 fragments
Allele 8 fragments
23
CQ#5: In the picture below, if each of the repeat sequences
is 4 base-pair long, what is the total length of the PCR
product?
bp = base pairs
5 repeat
sequences
3’
5’
20 bp
5’
25 bp
35 bp
A.
B.
C.
D.
E.
60 bp
3’
100
115
135
145
160
24
DNA Structure
Phosphate groups give
DNA its negative
charge
25
Gel Electrophoresis
- Gel made of agarose forms a matrix.
- DNA sample is loading into a well on the gel as shown in A.
- Phosphate groups on DNA carry negative charges.
- After applying power, DNA fragments migrate to the positive electrode,
shown in B.
- Fragments are separated based on their sizes, smaller fragments moving
faster while larger fragments moving more slowly.
- Fragments of the same size will settle at the same location on the gel.
26
• DNA bands do not have colors.
• Ethidium bromide (a chemical) inserts itself between DNA bases.
• When a gel stained with ethidium bromide is put on a UV table, DNA
bands glow orange light.
• Photography of the gel produces pictures in which DNA bands are
white on a dark background.
• The brighter the DNA bands, the more molecules there are in the
bands.
Sizing
ladders
(standard)
27
Ethidium Bromide
A very planar molecule that can insert itself between DNA bases.
28
CQ6. A person has the genotype of 9, 9 on one of his
microsatellite loci. An electrophoresis was conducted
on the PCR products of this locus. Which lane on the
gel below is most likely the result of the PCR fragments?
(L: size ladder)
L
A
B
C
D
29
Analysis of one locus is often insufficient to definitively
identify a person, so multiple loci are required.
Eight loci were amplified in the seven Egyptian mummies,
each locus requiring one specific pair (two unique sequences)
of primers.
D2S1338
FGA
CSF1PO
D7S820
D13S317
D16S539
D18S51
D21S11
30
Conventional agarose gels have difficulty detecting
small differences in fragment size.
Example:
Genotype 7, 8
Using other gel medium in conjunction with
automated PCR, the person’s genotype may be
analyzed accurately and quickly.
31
Capillary tube
(-)
Larger fragment
DNA Analyzers
Detector
Laser
Smaller fragment
(+)
32
Results from DNA Analyzer
If rotate counterclockwise…
Smaller
fragments
Locus 1
181
193
90°
Larger fragment
Locus 2
225
Locus 3
278
286
Heterozygous
(two peaks)
Homozygous
(one peak)
Heterozygous
(two peaks)
33
286
Locus 3
278
225
193
Height of the peak is equivalent
to the intensity of the DNA band
and the quantity of DNA
molecules on a gel.
181
Smaller
fragments
Locus 1
Locus 2
Larger
fragment
Location of hypothetical
well in a gel
34
The Mystery of
Tutankhamun’s
Family Tree
35
The Seven Mummies
Amenhotep III
KV35YL
Tiye
Thuya
Akhenaten
Tutankhamun
Yuya
36
In a known parentage relationship, you can trace which allele
is passed down from each of the parents to the child.
13, 15
14, 17
13, 17
37
CQ7. The following are the results of an analysis on two
microsatellite loci from a mother, a father and their child. Which one
is the child of the other two? (Number below each peak is the size of
that fragment in base pair.)
Locus 1
Locus 2
A
190
194
282
286
B
290
190
198
C
194
198
286
290
38
Seven strips of paper showing genotypic profiles
similar to the following will be distributed to each
group.
Locus
Female
D13S317
Tiye
11
Name
12
D7S820
10
15
D2S1338
22
26
D21S11
26
29
D16S539
6
11
D18S51
19
22
CSF1PO
9
12
FGA
20
26
Allele
39
D13S317
Tutankhamun
Yuya
10
12
Amenhotep III
KV35YL
16
26
D2S1338
11
6
22
13
9
12
11
12
15
D7S820
10
13
D7S820
10
15
27
D2S1338
19
26
D2S1338
22
26
D13S317
D7S820
D2S1338
10
6
16
16
D13S317
Akhenaten
15
D7S820
D13S317
Tiye
10
D2S1338
D13S317
D13S317
Thuya
D7S820
10
12
15
D7S820
15
15
27
D2S1338
16
26
D13S317
D7S820
D2S1338
10
6
16
12
10
26
D21S11
29
34
D21S11
29
34
D21S11
26
35
D21S11
26
29
D21S11
25
34
D21S11
29
34
D21S11
25
29
D16S539
8
13
D16S539
6
10
D18S51
19
19
D18S51
12
22
CSF1PO
6
12
CSF1PO
9
12
D16S539
D18S51
CSF1PO
11
8
7
13
D16S539
6
11
D16S539
8
13
D16S539
11
13
D16S539
8
11
19
D18S51
19
22
D18S51
16
22
D18S51
16
19
D18S51
16
19
12
CSF1PO
9
12
CSF1PO
6
9
CSF1PO
9
12
CSF1PO
6
12
FGA
23
23
FGA
20
Genotypic profiles
of King Tut and six
family members.
25
FGA
24
26
FGA
20
26
FGA
23
31
FGA
20
23
FGA
20
23
40
Exercise 1: Determine the members in King Tut’s
family tree
Male
Female
Marriage
Giving rise to
descendant
Giving rise to two
descendants
Sibling marriage
41
The family tree has been determined
But how confident are you in this family tree?
Paternity index
(or maternity index if the paternity is certain)
42
If only one of the parents is known, the allele in the child not
found in one of the parents must have come from the other
parent. This allele is called the obligate allele.
14, 17
13, 17
43
SAQ 1. Determine the obligate allele of each of the
loci. Determine the partial genotype of the father.
sample
locus
Mother
Child
CSF1PO
10, 11
10, 13
TPOX
5, 5
5, 11
D7S820
12, 15
12, 12
D13S317
8, 10
10, 11
44
CQ8. Which of the accused men is more likely to be the
father of the child? A or B?
sample
locus
CSF1PO
Mother
Accused Accused
Child Man A
Man B
10, 11
10, 13
10, 17
13, 21
5, 5
5, 11
6, 7
6, 11
D7S820
12, 15
12, 12
7, 15
8, 12
D13S317
8, 10
10, 11
9, 15
11, 11
TPOX
45
You have determined that one of the men in the
previous slide is a possible father to the child.
But how certain are you?
In legal situations, Paternity Index (or Parentage
Index, PI) is often used to provide mathematical
certainty to parentage.
PI is applied only when an accused man is a
match by having the set of obligate alleles.
46
Alleged
father
Mother
AB
CD
BC
Child
Paternity Index (PI) =
Probability of passing the obligate
allele given the alleged man is the
father (X)
Probability of a random man in the
population passing the obligate allele to
47
the child (Y)
PI =
X
Y
X = 0.5 if the alleged father is heterozygous for the obligate allele;
1.0 if the alleged father is homozygous for the obligate allele
Y = the frequency of the obligate allele found in the population
X and Y are two competing hypotheses; the higher the X is
and the lower the Y is, the greater the PI will be, and
stronger the implication that the alleged man is the father of
the child.
48
If the alleged father is heterozygous for the obligate allele…..
Alleged
father
Mother
AB
CD
0.5
BC
Child
0.5
Paternity Index =
pC
pC – probability of allele C, which is the frequency of allele C in the
population
49
If the alleged father is homozygous for the obligate allele…..
Alleged
father
Mother
AB
CC
1.0
BC
Child
1.0
Paternity Index =
pC
50
SAQ2. The tables below list the genotype of a locus from
three people involved in a paternity case and the
frequency of each of the alleles of this locus in the entire
population. (Frequency here is a number ranging from 0
to 1.) What is the paternity index?
Allele
Frequency
6
0.10
Individual
Genotype
7
0.01
Mother
8, 9
8
0.37
Child
8, 11
9
0.18
Accused man
6, 11
10
0.09
11
0.21
12
0.02
Total
1.00
51
If the mother and the child are heterozygous and share
both alleles, we don’t know which of the two alleles is
actually the allele that has been passed on by the mother.
Because of this uncertainty, either of the alleles could be
from the mother, and the other allele would be from the
father.
In probability, if either of the two events could be true,
the probability for either of the events to take place is
the sum of the individual probability of the event.
52
If the mother and the child share both heterozygous alleles….
Mother
AB
AB
Child
If the mother passed allele A to the child, then the probability of a random man
passing allele B to the child is the frequency of allele B in the population, or pB.
If the mother passed allele B to the child, then the probability of a random man
passing allele A to the child is the frequency of allele B in the population, or pA.
53
If the mother and the child share both heterozygous
alleles….
Alleged
father
Mother
AB
BC
0.5
AB
Child
0.5
Paternity Index =
Probability of a random
man passing allele A
pA + pB
or
Probability of a random
man passed allele B
54
• Multiple markers are often used simultaneously to increase the
certainty of parentage.
• If these markers are not linked genetically, multiplication rule may be
applied to all the paternity indices.
• The mathematical product of paternity indices is called combined
paternity index.
D2S1338
FGA
CSF1PO
D7S820
D13S317
D16S539
D18S51
D21S11
55
CQ9. If you want to use PCR technique to amplify markers
located on six unlinked locations on the chromosomes,
how many total unique primer sequences do you need?
A. 1
B. 3
C. 6
D. 8
E. 12
56
Combined Paternity Index = PI1 x PI2 x …x PIn
Combined paternity index in theory can reach infinity
if number of loci used in the analysis is large.
In forensic community, paternity is accepted if CPI >
100
It essentially means that the alleged man is100 times
or more likely than a randomly selected man to be
the father of the child.
57
SAQ3. If the three people involved in a paternity dispute
have the genotypic profiles shown in the following table,
determine the formula for the combined paternity index
(CPI). Note, use the appropriate symbols, e.g., P8 to
indicate the frequency of allele 8.)
TPOX
D3S1358
CSF1PO
Mother
8, 9
16, 17
10, 12
Child
9, 10
16, 16
10, 11
Accused
man
10, 10
15, 16
11, 11
58
SAQ4. If the CPI formula is the one listed below, what is
the CPI value based on the frequency of each allele in the
table?
Locus and frequency of each
allele
1.0 0.5 1.0
CPI =
´
´
P10 P16 P11
Allele
TPOX
D3S1358
CSF1PO
6
0.10
-
-
7
0.01
-
-
8
0.37
-
0.01
9
0.18
-
0.01
10
0.09
-
0.21
11
0.21
-
0.30
12
0.02
-
0.36
13
-
-
0.10
14
-
0.04
0.01
15
-
0.37
-
16
-
0.37
-
17
-
0.16
-
18
-
0.06
-
Total
1.00
1.00
1.00
59
SAQ5. Again, based on the same CPI formula below,
which of the three loci produces the highest PI value so
that it makes the greatest impact on the CPI value?
Why?
Locus and frequency of each allele
1.0 0.5 1.0
CPI =
´
´
P10 P16 P11
Allele
TPOX
D3S1358
CSF1PO
6
0.10
-
-
7
0.01
-
-
8
0.37
-
0.01
9
0.18
-
0.01
10
0.09
-
0.21
11
0.21
-
0.30
12
0.02
-
0.36
13
-
-
0.10
14
-
0.04
0.01
15
-
0.37
-
16
-
0.37
-
17
-
0.16
-
18
-
0.06
-
Total
1.00
1.00
1.00
60
SAQ6. If the CPI is < 100 from the analysis of
three loci, what is the best course of action or
conclusion?
61
SAQ7. Now calculate the combined paternity index
on the first three members in the family tree using
the provided Egyptian allele frequency handout.
You may choose either one of the parents as the
unknown.
If the obligate allele is
not found in the table,
skip that locus. We will
discuss this later.
62
DISCUSSION: What are the possible reasons
that some alleles in the members are not found
in the Egyptian allele frequency table provided?
63
Image Credits
Unless specified differently below, images were created by the case author.
Slide 1
Description: King Tut Ankh Amun Golden Mask.
Source: Steve Evans, accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:King_Tut_Ankh_Amun_Golden_Mask.jpg
Clearance: Used in accordance with the Creative Commons Attribution 2.0 Generic license.
Slide 2
Description: Mural from KV62 in Valley of the Kings.
Source: Kenneth Atherton and family, http://www.theathertons.info/2010/06/the-valley-of-the-kings/
Clearance: Used in accordance with the Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License.
Slides 4, 6, 30
Description: Normal male human karyotype.
Source: National Cancer Institute, accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Karyotype_%28normal%29.jpg
Clearance: This image is a work of the National Institutes of Health, part of the United States Department of Health and Human Services. As a work
of the U.S. federal government, the image is in the public domain.
Slide 15
Description: Mummy on display at the Philadelphia Academy of Natural Sciences.
Source: By “Smallbones,” accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Mummy_PANS.JPG
Clearance: Released into public domain by the photographer.
Slide 25
Description: DNA strand showing Chemical structure of nucleic Acid Bases.
Source: By “Boumphreyfr,” accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Dna_strand3.png
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 3.0 Unported license.
Slide 26
Description: Diagram illustrating gel electrophoresis procedure.
Source: By “Jjw,” accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Gel_electrophoresis_procedure.png
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 3.0 Unported license.
Slide 27
Description: Gel electrophoresis image.
Source: By Kuei-Chiu Chen, case author.
Clearance: Used with permission.
64
Slide 28
Description: Chemical structure of ethidium bromide.
Source: By “Calvero,” accessed on Wikimedia Commons at http://en.wikipedia.org/wiki/File:Ethidium_bromide.svg
Clearance: Released into the public domain by the image creator.
Slide 32
Description: A row of DNA sequencing machines (3730xl DNA Analyzer machines from Applied Biosystems).
Source:
This image was originally posted to Flickr by jurvetson at http://www.flickr.com/photos/jurvetson/57080968/, accessed on Wikimedia
Commons at http://commons.wikimedia.org/wiki/File:DNA-Sequencers_from_Flickr_57080968.jpg.
Clearance: Used in accordance with the Creative Commons Attribution 2.0 Generic license.
Slides 35
Description: Tutankhamun.
Source: Jerzy Strzelecki, http://commons.wikimedia.org/wiki/File:Tutanchamon_(js)_1.jpg
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 3.0 Unported license.
Slide 36
Description: Colossal statue of Amenhotep III in the British Museum.
Source: Photograph by A. Parrot, accessed on Wikimedia Commons at
http://commons.wikimedia.org/wiki/File:Colossal_Amenhotep_III_British_Museum.jpg
Clearance: Released into public domain by photographer.
Slide 36
Description: Tiye, Ägyptisches Museum (Egyptian Museum), Berlin.
Source: Einsamer Schütze, accessed on Wikimedia Commons at http://en.wikipedia.org/wiki/File:%C3%84gyptisches_Museum_Berlin_027.jpg
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 3.0 Unported license.
Slide 36
Description: Akhenaten, Pharaoh of Egypt. Egyptian Museum, Cairo.
Source: User:Hajor, accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Pharaoh_Akhenaten.jpg
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 1.0 Generic license.
Slide 36
Description: Gilded cartonnage mummy mask of Thuya, wife of Yuya and mother of Queen Tiye
Source: Jon Bodsworth, accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Mummy_mask_of_Thuya.jpg
Clearance: Copyright free.
65
Slide 36
Description: KV35YL--Front view of the "Younger Lady Mummy" found in tomb KV35.
Source: Image derived from Grafton Elliot Smith’s The Royal Mummies, Plate XCIX, first published in 1912, accessed on Wikimedia Commons at
http://commons.wikimedia.org/wiki/File:TheYoungerLady-61072-FrontView-PlateXCIX-TheRoyalMummies-1912.gif
Clearance: Public domain due to expired copyright.
Slides 36
Description: Tutankhamun.
Source: Jerzy Strzelecki, http://commons.wikimedia.org/wiki/File:Tutanchamon_(js)_1.jpg
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 3.0 Unported license.
Slide 36
Description: The gilded cartonnage mummy mask of Yuya, the father of Queen Tiye. Yuya was the father-in-law of pharaoh Amenhotep III, one of
the most powerful kings of Egypt’s 18th dynasty.
Source: Jon Bodsworth, accessed on Wikimedia Commons at http://commons.wikimedia.org/wiki/File:Mummy_mask_of_Yuya.jpg
Clearance: Copyright free.
Slides 37 and 43
Description: Cropped screenshot of Richard Burton and Elizabeth Taylor from the trailer for the film Cleopatra.
Source: Jon Bodsworth, accessed on Wikimedia Commons at
http://commons.wikimedia.org/wiki/File:1963_Cleopatra_trailer_screenshot_%2823%29.jpg
Clearance: This work is in the public domain because it was published in the United States between 1923 and 1963 and although there may or may
not have been a copyright notice, the copyright was not renewed. Unless its author has been dead for the required period, it is copyrighted in the
countries or areas that do not apply the rule of the shorter term for US works, such as Canada (50 pma), Mainland China (50 pma, not Hong Kong
or Macao), Germany (70 pma), Mexico (100 pma), Switzerland (70 pma), and other countries with individual treaties.
Slides 37 and 43
Description: Sleeping infant.
Source: Andrés Nieto Porras from Palma de Mallorca, España, (cropped and tinted), accessed on Wikimedia Commons at
http://commons.wikimedia.org/wiki/File:Sleeping_newborn_infant.jpg
Clearance: Used in accordance with the Creative Commons Attribution-Share Alike 2.0 Generic license.
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