Example
Pg 170, Statistical Genomics: Linkage, Mapping and QTL Analysis.
AaBb x aabb testcross data.
| class |
AaBb |
Aabb |
aaBb |
aabb |
Total |
| observed |
310 |
287 |
288 |
315 |
1200 |
| expected |
300 |
300 |
300 |
300 |
1200 |
| z = Log10 |
[ |
L( ) |
] |
| L(0.5) |
z = Log10[L()] - Log10[L0.5)]To find L() we have to estimate ().
() = 
= Nr/N = 575/1200 = 0.479
Then L() = (0.479)575 (1-0.479)625
and L(0.5)
= (0.5)575(0.5)625 = (0.5)1200
| z = Log10 |
[ |
(0.479)575 (1-0.479)625 |
]
|
| (.05)1200 |
= 575 Log10(0.479) + 625 Log10(1-0.479)
- 1200 Log10(0.5) = 0.4525 = Lod Score
A Lod score of 3 means that the probability that = 0.479 is (10)3 more likely than the probability that = 0.5. In our example, the lod score for = 0.479 was 0.4525 which means that the likelihood that = 0.479 is 100.4525 = 2.83 times more likely than the probability that = 0.5.
*As changes, the lod score changes. When the value of that maximizes the lod score is found, this is the most probable position of the QTL relative to the marker position - see pg 401. Statistical Genomics.
