mollwollfumble
New member
- Joined
- May 4, 2019
- Messages
- 12
Noah wants to eliminate deadly recessive genes in the animals before he puts them on his ark. He can only do this by inbreeding. What is his best inbreeding strategy?
This question has three parts: producing an animal with no deadly recessive genes, selecting animals that may have no deadly recessive genes, and confirming (to prespecified probability) that they have no deadly recessive genes.
Case 1, a possible strategy.
Suppose the wild population has 3 deadly recessive genes (DRG) each and is large enough that two individuals from the wild population will have different recessive genes. So their children will have up to 6 DRG, but none die.
If male and female of generation zero (m0 & f0) have 64=2^6 children (generation 1) then we expect, on average, 1 of those generation 1 children to have no deadly recessive genes. More children (how many?) will give a 95% chance of one having no DRG.
Breed every male child of generation 1 with every female child of generation 1 to get generation 2. Those with no deaths in generation 2 are highly likely to have no DRGs.
An improved strategy.
Two problems with the first strategy are that 64 children is excessive for most female mammals, and that it requires about 1,000 descendents of m0 & f0 to find an animal with no deadly recessive genes.
Let the male and female of generation zero (m0 & f0) have 16 children in generation 1. Breed every male child of generation 1 with every female child of generation 1 to get generation 2, which has up to 64 members. Roughly 27 members of generation 2 will die because of doubling up of recessive genes.
For each male and female of generation 1, count how many of their children have died, and select the best 8 and worst 8 for both males and females. Breed the best 8 females to the worst 8 males, and breed the best 8 males to the worst 8 females. If any of the best 8 females or best 8 males have no dead children, then accept one of them for the ark.
This is better than the first strategy because it only requires about 16 children for each female and about 200 descendents overall.
If none of the best 8 females or best 8 males have no dead children, then breed the best to the best to get 64 more children in generation 2. Then for each male and female of generation 2, count how many of their children have died, and select the best 8 and worst 8 for both males and females. Breed the best 8 females to the worst 8 males, and breed the best 8 males to the worst 8 females. If any of the best 8 females or best 8 males have no dead children, then accept one of them for the ark. That makes about 400 descendents overall.
This question has three parts: producing an animal with no deadly recessive genes, selecting animals that may have no deadly recessive genes, and confirming (to prespecified probability) that they have no deadly recessive genes.
Case 1, a possible strategy.
Suppose the wild population has 3 deadly recessive genes (DRG) each and is large enough that two individuals from the wild population will have different recessive genes. So their children will have up to 6 DRG, but none die.
If male and female of generation zero (m0 & f0) have 64=2^6 children (generation 1) then we expect, on average, 1 of those generation 1 children to have no deadly recessive genes. More children (how many?) will give a 95% chance of one having no DRG.
Breed every male child of generation 1 with every female child of generation 1 to get generation 2. Those with no deaths in generation 2 are highly likely to have no DRGs.
An improved strategy.
Two problems with the first strategy are that 64 children is excessive for most female mammals, and that it requires about 1,000 descendents of m0 & f0 to find an animal with no deadly recessive genes.
Let the male and female of generation zero (m0 & f0) have 16 children in generation 1. Breed every male child of generation 1 with every female child of generation 1 to get generation 2, which has up to 64 members. Roughly 27 members of generation 2 will die because of doubling up of recessive genes.
For each male and female of generation 1, count how many of their children have died, and select the best 8 and worst 8 for both males and females. Breed the best 8 females to the worst 8 males, and breed the best 8 males to the worst 8 females. If any of the best 8 females or best 8 males have no dead children, then accept one of them for the ark.
This is better than the first strategy because it only requires about 16 children for each female and about 200 descendents overall.
If none of the best 8 females or best 8 males have no dead children, then breed the best to the best to get 64 more children in generation 2. Then for each male and female of generation 2, count how many of their children have died, and select the best 8 and worst 8 for both males and females. Breed the best 8 females to the worst 8 males, and breed the best 8 males to the worst 8 females. If any of the best 8 females or best 8 males have no dead children, then accept one of them for the ark. That makes about 400 descendents overall.