# BME100 f2018:Group9 T1030 L6

BME 100 Fall 2018 | Home People Lab Write-Up 1 | Lab Write-Up 2 | Lab Write-Up 3 Lab Write-Up 4 | Lab Write-Up 5 | Lab Write-Up 6 Course Logistics For Instructors Photos Wiki Editing Help
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## Contents## OUR TEAM
## LAB 6 WRITE-UP## Bayesian Statistics
In the lab 17 groups were assigned two different patients, so a total of 34 patients, and our job was to determine if the patients genes contained the SNP of a disease associated with Parkinson's. We used PCR (Polymerase Chain Reaction) to replicate the copy of the SNP in order to be able to detect if it was present in the patient’s DNA sample. Every group was given a positive and negative control to help prevent error and have a comparable result to definite controls. Three samples from each patient was also given to help prevent error with false results or contamination. Three pictures were taken for each sample to reduce error and increase reliable. The images were analyzed through ImageJ . The droplet was outlined in order to have better results from the analysis on ImageJ. From the data collected, the class had 14 patients test positive, 17 patients test negative, one patient’s test was inconclusive, and one group did not provide data for their tests. There could have also been discrepancies in the micropipetting resulting in various measurements and inconsistencies in the amount of DNA.
Calculation 1 was “What is the probability that a patient will get a positive final test conclusion, given a positive PCR reaction?” and the number was 0.94. The equation to find this was (0.90 x 0.44)/(0.42). 0.90 is the frequency of total pos with POS, 0.44 is the frequency of total POS conclusions, and 0.42 is the frequency of the total pos PCRs. Calculation 1 shows the sensitivity of the system regarding the ability to detect the disease SNP. Calculation 2 was “What is the probability that a patient will get a negative final test conclusion, given a negative diagnostic signal?” and the number was 0.98. The equation to find this was (0.94 x 0.53)/ (0.51). 0.94 is the frequency of toal neg with NEG, 0.53 is the frequency of total NEG conclusions, and 0.51 is the frequency of total neg PCRs. Calculation 2 describes the specificity of the system regarding the ability to detect the disease SNP. Calculation 3 was “What is the probability that a patient will develop the disease, given a positive final test conclusion?” and the number was 0.33. The equation to find this was (0.43 x 0.34)/(0.44). 0.43 is the frequency of total “yes” diagnoses with POS conclusion, 0.34 is the frequency of total “yes” diagnoses, and 0.44 is the frequency of total POS conclusions. Calculation 3 describes the sensitivity of the system regarding the ability to predict the disease. Calculation 4 was “What is the probability that a patient will not develop the disease, given a negative final test conclusion? and the number was 0.96. The equation to find this was (0.77 x 0.66)/(0.53). 0.77 is the frequency of total “no” diagnoses with NEG conclusions, 0.66 is the frequency of total “no” diagnoses, and 0.53 is the frequency of total NEG conclusions. Calculation 4 describes the specificity of the system regarding the ability to predict the disease. ## Intro to Computer-Aided Design
## Feature 1: Consumables## Feature 2: Hardware - PCR Machine & Fluorimeter |