...
The selected proportion alone is not a very good representation of how much better than average the parents are. It needs to be evaluated in combination with the size of the variation. A way to do that, is expressing the mean of the selected proportion in units of variation: the standard deviation. As is described in the chapter about introduction to statistics, the normal distribution can be divided into standard deviations according to a fixed pattern, such that 68% of the observations lay between plus and minus one standard deviation around the mean, 95% between plus and minus two standard deviations, and 99.7% between plus and minus three standard deviations. Many phenotypes tend to be normally distributed in a population. A phenotypic value can thus be expressed as being so many standard deviations away from the mean. We can use the selected proportion of animals and use properties of the normal distribution to determine the mean of the animals in that selected proportion, expressed in phenotypic standard deviations: the selection intensity.
...
Thus:
| Paneel | ||
|---|---|---|
|
...
Thus: genetic gain is determined by 3 main factors: phenotypic variance, accuracy of selection, and selected proportion. |
The selection intensityis abbreviated as i. In formula:
...
In summary: the selected proportion, in combination with the phenotypic variance, is enough to predict the average performance of selected parents.Thus:
| Paneel | ||
|---|---|---|
| ||
Thus: The selection intensity represents the mean of the selected proportion in phenotypic standard deviations. |
In chapter 9.5.1 you will find a table where you can look up i for any given selected proportion. This table is valid for selection on any trait that is normally distributed, so it is not specific for a trait or a population.
| Onderliggende pagina's (weergave met onderliggende items) |
|---|