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Cross Breeding with Nut's Essence
Determining the length of genome of a child plant
1) Add genome length of Left Splint plant to genome length of Right Splint plant.
2) Divide this *sum by 2.
3) Round up if the *quotient is not an *integer. This is the length of genome for the child plant except...
4) There is a 25% chance of a duplicated gene at the splint point. There is a 25% chance of a destroyed gene at the splint point. In half the crosses the length of the child genome length might be +1 more or -1 less.
Left Splint genome length . 37
Right Splint genome length +26
2) Divide by 2. . . . . . . 63 / 2 = 31.5
3) Round up 31.5 ----> 32 <---- child genome length (see note 4 above)
Two numbers added together equals a sum.
A dividend is divided by a divisor and that equals a quotient.
An integer is a whole number, not a fraction.
Note that you should use Notepad or another 'scratch pad' that uses a non true type font so the letters line up vertically for the calculations below.
Knowing the genome of your child plant
1) Put the Left Splint genome on top and the Right Splint genome underneath.
Example: Left Splint Crown/ Right Splint Blush
2) Center the shortest genome.
If the child genome length had to be rounded up, shift the shortest parent genome by 1 gene in the direction of the Splint the shortest parent genome was in.
......GRORVGOOOVUROOVIOVIOOVIOOO..... Had to shift this shortest Right Splint genome to the right by 1
3) The splice point will be as if you draw a vertical line straight down between the two sets of genomes (denoted by vertical bars as an example). In reality usually you will only know a range of genes where the splice point might be so there are two sets of vertical lines to show that range. Refer to one of the Plant Genome Theories pages links below to interpret gene sets to know approximately where the splice point happens. In this example we would have noticed the child lily included a shade 3 magenta inner west petal gene set instead of Crown's shade 4 thus the splice point must intersect a UROO from Crown's genome. We would also notice the child lily included the rightmost gene sets but was missing a magenta outer south petal gene set GOOO and so the splice point must have at least cut off the 'G' of OOO in Blush.
4) Your child genome will start with the leftmost gene of the Left Splint, go to the splice point, then include the genes from the splice point to the rightmost gene of the Right Splint.
As a general rule there are (shortest genome length + 1) genome possibilities for any given child plant genome (excluding a spontaneous gene subtraction or addition as mentioned above).
By looking at the colors of the hybrid, we narrow the splice possibilities down to 4 for this example.
5) Four basic possible child genomes fit this child lily (not taking spontaneous gene subtractions and additions into account):
The other 8 possiblities are a deleted or a doubled gene at the splice point for each of the basic 4 possible genomes in our example.
Moving gene sets to desired locations
By using parent plants of longer or shorter genome lengths you can 'scootch' desired gene sets left or right in a child plant genome.
We will use wanting a shade two cyan petal set as our goal.
In sea lilies there are few cyan gene sets (GRRR is one). Morning has GRRR in the center. We want to move as far to the right of a child genome as possible. Since Morning has 44 genes and I want GRRR included, we see that there are 12 genes after that in the Morning genome. Those can be cut out. From 'Knowing the genome of your child plant' section above, we will double the 12 gene amount we can cut out from Morning, then subtract that 24 from Morning genome length of 44. We need another parent genome length of at least 20 to keep the GRRR from Morning.
The larger the difference in plant genome lengths, the farther gene sets can be 'scootched'.
Blush is the best possibility keeping the above fact in mind.
Morning Left Splint / Blush Right Splint
Child plant genome length will be ((44 + 26)/2) or 35
We will want to have the genome of IRRRIYYYIYYYUYYYUYYYURORURORGRRRxxx (where x equals genes we do not care which plant genome they are from). Those last three genes in the child genome can be either from Blush or Morning as long as the GRRR cyan gene set is retained. The total number of possible genomes is the shortest parent genome length (Blush at 26) plus 1 or 27 possibilities. The number of desired genomes is the number of 'junk' genes in the child genome + 1 or 4 in this case.
To calculate the chances this will happen...number of desired genomes divided by total number of possible child genomes. In this case 4 out 27 crosses will statistically have the genome you hope for.
Use the various genome theories pages to visually intepret child flowers or harvest results for flax, wheat and vines.
Now you have the desired gene set at the end of your genome. Multiply your precious baby then you want to breed for a second shade of cyan in that petal set.
Morning Left Splint / Child Right Splint
Second Child genome length will be ((44 + 35) / 2) or Round up(39.5) or 40
Remember to shift the shortest parent genome by 1 in the direction of the Splint the shortest parent genome is in because the second child genome had to be rounded up. In this case shift the Child plant to the right by 1.
We see above that there are only two splice possibilities that will work to have a shade two cyan petal set. The chances to attain this are number of desired genomes (1 junk gene + 1) divided by total number of possible child genomes (35 + 1). Or 2 out of 36. Not a good ratio, but doable with patience. So you could do another step before this one and elongate the child genome with Energy (because Energy has a longer genome) so that the GRRR will shift to the right compared to Morning and make a child genome length of 39. Plus you will gain 'V' genes which will speed up bulb split rate. The ratio of success would change to 4 chances in 40 of success.