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 X 10^24  34,599,800,000 = 3.45998 X 10^10 12.56 = 1.256 X 10^1 0.3456789 = 3.456789 X 10^-1  0.0000000332 = 3.32 X 10^-8 0.000000000000000000000000000000123456789 = 1.23456789 X 10^-31  6 trillion = 6 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 = .000000000000000000000000000000000662 1.5 x 10^11 = 150,000,000,000  1.44 x 10^-1 = .144


 * Exercise 2: DNA Probability Vs. Frequency]**

ABO genotype probability <span style="font-family: Arial,Helvetica;">AA = 11.1% <span style="font-family: Arial,Helvetica;">AB = 22.2% <span style="font-family: Arial,Helvetica;">Ai = 22.2% <span style="font-family: Arial,Helvetica;">BB = 11.1% <span style="font-family: Arial,Helvetica;">Bi = 22.2% <span style="font-family: Arial,Helvetica; font-size: medium; line-height: normal;"> ii = 11.1%
 * || A || B || i ||
 * A || AA || AB || Ai ||
 * B || AB || BB || Bi ||
 * i || Ai || Bi || ii ||

<span style="font-family: Arial,Helvetica; font-size: medium; line-height: normal;">phenotype probability <span style="font-family: Arial,Helvetica;">A = 11.1% <span style="font-family: Arial,Helvetica;">AB = 22.2% <span style="font-family: Arial,Helvetica;">B = 11.1% <span style="font-family: Arial,Helvetica; font-size: medium; line-height: normal;"> O = 55.5% (%AI + %BI + %ii)

<span style="font-family: Arial,Helvetica; font-size: medium; line-height: normal;">ACTUAL PHENOTYPE FREQUENCY <span style="font-family: Arial,Helvetica;">A = 40% of white non-Hispanics, 31% of Hispanics, 26% Black Non-Hispanics, 28% Asians, 35% Native Americans

<span style="font-family: Arial,Helvetica;">AB = 4% of white non-Hispanics, 2% of Hispanics, 4% of Black non-Hispanics, 7% of Asians, 2% of North American Indians

<span style="font-family: Arial,Helvetica;">B = 11% of white non-Hispanics, 10% of Hispanics, 20% of Black non-Hispanics, 25% of Asians, 9% of North American Indians

<span style="font-family: Arial,Helvetica; font-size: medium; line-height: normal;">O = 45% of white non-Hispanics, 57% of Hispanics, <span style="font-family: Arial,Helvetica;">50% of Black non-Hispanics, 40% Asians, 54% Native Americans

http://weloveteaching.com/2011/labs/forensics/aboethnic.jpg ABO

A AB B O (Blue is actual frequency, Red is the projected probability)

The B allele is most common in Central Asia and in many parts of Africa, where it comprises about 20 - 25% of the population. In most of the world outside of these regions, however, the amount of the B allele is typically between 0 and 5%. It only makes up 16% of the world's population. About 21% of the world shares the A allele and its prevalence is quite random: pockets of Europe, especially Scandinavia, much of Australia, and part of the northern United States have preponderant amounts of the A allele, between 30 and 40%. Type O allele is the most common in the world, equating to about 63% of the world's population. Practically all of Central and South America share the O - type and the only area not to have a plethora of O-type is in Central Europe. This could be because typically people marry and breed with people of a similar background and because certain regions usually have the same blood types, the homogeny of blood types in certain regions perpetuates.

http://anthro.palomar.edu/vary/vary_3.htm


 * Exercise 3: Forensic DNA Fingerprinting**

Paternity Test 1 Pop 1- included, no B in GYPA stage Pop 2 - excluded
 * (.43)^2 * 2(.55 * .46) * 2(.48 * .52) * (.40)^2 * (.56)^2 = 2.3 * 10^-3 **

Paternity Test 2 Pop 1- excluded, no A in LDLR Pop 2- included (.43)^2 * (.46)^2 * 2(.48 * .52) * (.40)^2 * 2(.29 * .56) = 1.01 x 10^-3

Paternity Test 3 Pop 1- excluded because of HBGG Pop 2- excluded because of GC

Paternity Test 4 Pop 1- included Pop 2- excluded, no B in the GYPA stage. Probability = (.43)^2 * 2(.55 * .46) * 2(.48 * .52) * (.40)^2 * (.56)^2 = 2.3 * 10^-3

Criminal Test 1 It is suspect's blood, suspect cannot be excluded. 2(.43 * .57) * (.46)^2 * 2(.48 * .006) x (.6)^2 * 2(.29 * .15) = **1.9 * 10^-5**

Criminal Test 2 There's none of the suspect's blood; there is victim blood on suspect's jacket and at scene of the crime. Frequency of blood = 2(.43 * .57) * (.46)^2 * 2(.48 * .006) x (.6)^2 * 2(.29 * .15) = **1.9 * 10^-5**

Criminal Test 3 The suspect is excluded because the blood on suspect's jacket is not the victim's and only the victim's blood was at crime scene.

Criminal test 4 There are 2 bloods at crime scene so suspect cannot be excluded, because it is victim's and someone else's. The victim's blood is on the suspect's jacket. Frequency = 2(.43 * .57) * (.46)^2 * (.006)^2 * (.6)^2 * 2(.29 * .15) = 1.2 * 10^-7

__Probability__ LDLR = 2(.43 * .57) = .49 GYPA = .46^2 = .21 HBGG = 3(.48 * .006) = .0058 D7S8 = .6^2 = .36 GC = 2(.29 *.15) = .087 Total = (.49) (.21) (.0058) (.36) (.087) = .000019

Formulas employed from http://weloveteaching.com/2011/labs/forensics/calculatefreq.html