![\"novice-01\"](\"https:\/\/i0.wp.com\/minkhollow.ca\/gene-rummy\/wp-content\/uploads\/2013\/04\/novice-01.png\")
<\/a><\/p>\nAlong the bottom is the notation we use for each of the shades. In this novice game, it doesn’t matter what the other alleles are. Only the B-series and D-series are used.<\/p>\n
This game is a Production game where each player is dealt eight cards. The remaining cards are placed face down on the table. This becomes the “Gene Pool”. On their turn, players select a rabbit from the Gene Pool or Retirement and then try to match it with the rabbits in their hands that could be its legal parents. If they succeed, then all three cards are laid down. The player then chooses a rabbit from their Herd to Retire, and play proceeds to the next player.<\/p>\n
When a trio is located, the offspring may resolve some of the unknown alleles of the parents, and this must be completed before the score is tallied. For example, if a Lilac (bb dd) offspring is matched with 2 Black (B- D-) parents, then the unknown alleles of BOTH parents become known, because the only way for 2 Blacks to produce a Lilac is if they are both heterozygous on both loci. In other words, both parents must be: Bb Dd. That means the genotypes for this trio have to be:<\/p>\n
- \n
- Black Buck: \u00a0 Bb Dd<\/li>\n
- Black Doe:\u00a0 Bb Dd<\/li>\n
- Lilac Kit:\u00a0 bb dd<\/li>\n<\/ul>\n
![\"blk-choc-blk-N\"](\"https:\/\/i0.wp.com\/minkhollow.ca\/gene-rummy\/wp-content\/uploads\/2013\/04\/blk-choc-blk-N.png?resize=203%2C300\")
In other cases it will not be possible to resolve all the unknowns. If a Black buck (B- D-) and a Chocolate doe (bb D-) result in a Black offspring (B- D-), we can fill in one of the unknowns for the kit because of the recessive ‘b’ must be passed on by the doe, but we can’t fill in any of the other unknowns for the parents:<\/p>\n- \n
- Black Buck: \u00a0 \u00a0 B- D-<\/li>\n
- Chocolate Doe:\u00a0 bb D-<\/li>\n
- Black Kit:\u00a0 \u00a0 Bb D-<\/li>\n<\/ul>\n
Some combinations are simply NOT possible. For example,two DILUTE (dd) parents (either Blue or Lilac) simply CAN’T produce a Dense (D-) kit (black or chocolate) because they don’t carry the dense allele ‘D’. Similarly, two browns (Chocolate or Lilac) CAN’T produce a Black (B- D-) or a Blue (B- dd), since they lack the dominant ‘B’.<\/p>\n
If a player ends up with a single card, they don’t need to pick up a new card on their turn; they can simply lay off the one they have left and end the game.<\/p>\n
Scoring<\/h2>\n
Scoring in this game makes use of the dominant\/recessive alleles. The score is tallied using BOTH the known genes of the phenotype and those that can be filled in as a result of the mating. The recessive ones are worth more as it is harder to produce offspring using them (that is to say, the choices are more limited). All ‘resolved’ alleles must be filled in before the points can be tallied.<\/p>\n
- \n
- Dominant = 5 points<\/li>\n
- Recessive = 10 points<\/li>\n
- A homozygous Black: BB DD = 5+5+5+5 = 20<\/li>\n
- A Chocolate that is heterozygous on the D-loci: bb Dd = 10+10+5+10 = 35<\/li>\n<\/ul>\n
The table below shows the 16 possible combinations of these two alleles and their point tallies:<\/p>\n