Table 1 Estimation from two-point analysis of the recombination fraction (
± SD) and the parental diplotype probability of parent P (
) and Q (
) for five markers in a full-sib family of n = 100
± SD) and the parental diplotype probability of parent P (
) and Q (
) for five markers in a full-sib family of n = 100Parental diplotype | r = 0.05 | r = 0.20 | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
Marker | Pa | × | Qa |
|
|
|
|
|
| ||
| | | | | | | | ||||||||
| a | b | c | d | |||||||
| | | | | | | | 0.530 ± 0.0183 | 0.2097 ± 0.0328 | ||||||
| a | b | a | b | 0.9960 | 0.9972 | 0.9882 | 0.9878 | |||
| | | | | | | | 0.0464 ± 0.0303 | 0.2103 ± 0.0848 | ||||||
| a | o | × | o | a | 1 (0b) | 0(1b) | 1 (0b) | 0(1b) | ||
| | | | | | | | 0.0463 ± 0.0371 | 0.1952 ± 0.0777 | ||||||
| a | b | b | b | 1 | 1/0c | 1 | 1/0c | |||
| | | | | | | | 0.0503 ± 0.0231 | 0.2002 ± 0.0414 | ||||||
| a | b | c | d | 1 | 1/0c | 1 | 1/0c | |||
-
aShown is the parental diplotype of each parent for the five markers hypothesized, where the vertical lines denote the two homologous chromosomes. bThe values in the parentheses present a second possible solution. For any two symmetrical markers (2 and 3), = 1, = 0 and = 0, = 1 give an identical likelihood ratio test statistic (Wu et al. 2002a). Thus, when the two parents have different diplotypes for symmetrical markers, their parental diplotypes cannot be correctly determined from two-point analysis. cThe parental diplotype of parent P
2 cannot be estimated in these two cases because marker 4 is homozygous in this parent. The MLE of r is given between two markers under comparison, whereas the MLEs of p and q given at the second marker.