Forensic evidence in a criminal court case. In our legal system,the use of DNA as forensic evidence is often regarded as the most reliable type of evidence. However, most of the DNA code is the same for all humans. Consequently, assessing the probability of the DNA code that varies among individuals is the key to a successful case. Chance (Vol. 28, 2015) published an article on the use of DNA in a criminal case. The evidence found at the crime scene consisted of two alleles (sequences of DNA code), denoted {6/9}. One of these alleles comes from the individual’s mother and one from the individual’s father, but it is not known which allele-6 or 9-is from which parent. In forensic science, it is assumed that the two outcomes (alleles) are independent.

1. DNA taken from the suspect resulted in a sequence of {6/9}. Given the evidence (E) comes from the suspect, find the probability of a DNA sequence of {6/9}. This probability-denoted$\mathbf{P}\left(\frac{E}{{\mathbf{H}}_p}\right)$-is used by the prosecution to support a claim of guilt.
2. In the general population, the probability of observing an allele of 6 is 0.21 and the probability of an allele 9 is 0.14. Given the evidence (E) comes from a randomly selected person in the general population, find the probability of a DNA sequence of {6/9}. This probability-denoted$\mathbf{P}\left(\frac{E}{{\mathbf{H}}_d}\right)$-is used by the defense to support the suspect’s claim of not guilty.
3. In a court of law, the likelihood ratio$\mathbf{P}\left(\frac{E}{{\mathbf{H}}_p}\right)\mathbf{/}\mathbf{P}\left(\frac{E}{{\mathbf{H}}_d}\right)$is used to help decide the case. A ratio greater than 1 supports the prosecution, while a ratio less than 1 supports the defendant. Compute this likelihood ratio from the results in parts a and b and use it to make an inference.

Step by step solution

01

Given information

Two alleles (sequences of DNA code), denoted {6/9}. One of these alleles, either allele-6 or allele-9, is inherited from the mother and the other from the father.

02

The probability of a DNA sequence of {6/9}

The required formula is $\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{p}}}\right)$.

The probability is

$\begin{array}{c}P\left(\frac{E}{H_p}\right)=\frac{\frac{6}{9}}{\frac{6}{9}}\\ P\left(\frac{E}{H_p}\right)=1\end{array}$

03

find the probability of a DNA sequence of {6/9} for part b

The formula used$\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{d}}}\right)$

The probability Is$\begin{array}{c}\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{d}}}\right)=\mathrm{P}\text{(}6.9\text{)}+\mathrm{P}\text{(}9.6\text{)}\\ =\text{(}0.21\times 0.14\text{)(}0.14\times 0.21\text{)}\\ =0.0588\end{array}$

04

Find the solution for part c

The require formula is $\frac{\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{p}}}\right)}{\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{d}}}\right)}$.

The probability is $=\frac{\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{p}}}\right)}{\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{H}}_{\mathrm{d}}}\right)}$

$\begin{array}{l}=\frac{1}{0.0588}\\ =17\end{array}$

Therefore, the probability is 17.

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