Lab 6: Forensic DNA

Exercise 1: Scientific Notation

What are the following in scientific notation?

6,000,000,000,000,000,000,000,000
= 6.0 X 10^24
34,599,800,000
=3.45998 X 10^10
12.56
=1.256 X 10
0.3456789
=3.456789 X 10^-1
0.0000000332
=3.32 X 10^-8
0.000000000000000000000000000000123456789
=1.23456789 X 10^-31
6 trillion
=6.0 X 10^12
What are the following in standard notation?


3 x 10^5
=300,000
2.44 x 10^-7
=.000000244
6.02 x 10^23
602,000,000,000,000,000,000,000
6.62 x 10^-34
=.00000000000000000000000000000000662
1.5 x 10^11
=150,000,000,000
1.44 x 10^-1
=.144

Exercise 2: DNA Probability vs. Frequency

ABO

ARSC_graph_lab_6.png

A
B
i
A
AA
AB
Ai
B
AB
BB
Bi
i
Ai
Bi
ii

Genotype Probability
AA = 1/9
AB = 2/9
Ai = 2/9
BB = 1/9
Bi = 2/9
ii = 1/9

Phenotype Probability
A = 3/9
AB = 2/9
B = 3/9
O = 1/9

Actual Phenotype Frequency
A = 40% (36% A+ and 4% A-)
AB = 5% (4% AB+ and 1% AB-)
B = 12% (10% B+ and 2% B-)
O = 43% (37% O+ and 6% O-)
http://www.infoplease.com/ipa/A0877658.html

The A blood allele is found scattered about the world. The areas it is most concentrated in are North America, Australia, and especially Norther Scandinavia while it is essentially non-existent in Central and South America.
The B blood allele is less common than the A blood allele and heavier concentrations can be found primarily in Central Asia and parts of Africa. There are very low concentrations in Australia and the Americas. Only about 16% of humans in the world today have it.
The O blood allele is the most common weighing in at about 63% of today's population. Central and South America have concentrations nearing 100% of O blood type people while it is much less common in Eastern Europe and Central Asia.
Evolution, immigration, and random mating are all different ways that allele populations in different areas have changed. Europe's eventual immigration to the Americas probably had a pretty significant effect on the distribution of the frequency of alleles especially when Europeans began to breed with Native Americans.

Exercise 3: Forensic DNA Fingerprinting

Paternity Case 1
Pop 1:
A^2 X 2(AB) X 2(AB) X B^2 X C^2 = (.43)^2 X 2(.55)(.46) X 2(.48)(.52) X (.40)^2 X (.56)^2 = .002343
Paternity Case 2
Neither
Paternity Case 3
Neither
Paternity Case 4
Neither

Criminal Case 1
Neither
Criminal Case 2
Both Match
Suspect:
2(AB) X B^2 X 2(AC) X A^2 X 2(AB) = 2(.43)(.57) X .46^2 X 2(.48)(.006) X .29^2 X 2(.29)(.15) = 4.371 X 10^-6
Criminal Case 3
Only Blood at Crime Scene Matched
Criminal Case 4
Both Match
Suspect:
2(AB) X B^2 X C^2 X A^2 X 2(AB) = 2(.43)(.57) X .46^2 X .006^2 X .29^2 X 2(.29)(.15) = 2.7 X 10^-8



Probability
external image dna6.gif

2(AB) X B^2 X 2(AC) X A^2 X 2(AB) = 2(.43)(.57) X (.46)^2 X 2(.48)(.006) X (.6)^2 X 2(.29)(.15) = 1.871 X 10^-5