One problem with reduction in numbers of a species is that if you do it too fast, the resulting inbreeding causes deadly recessive mutations to come together to wipe out the species. The most general question is: given a load of recessive mutations in a general populace, how rapidly can the number of individuals safely be reduced.

Selective inbreeding, eg. in horse racing, is used to weed out recessive mutations, leaving those that have fewest recessive mutations in the gene pool. ie. artificial selection.

The mathematical question is as follows. In a large population of male and female animals, each with 3 deadly recessive mutations (DRM), assumed independent, design an inbreeding scheme that produces a male female pair, neither of which has a deadly recessive mutation.

One way of doing it is as follows.

After one breeding of a M/F pair, the number of DRMs in the children is given by the rolling of two three-sided dice.
0 DRMs – probability 1/16
1 DRM – probability 1/8
2 DRMs – probability 3/16
3 DRMs – probability 1/4
4 DRMs – probability 5/16
5 DRMs – probability 1/8
6 DRMs – probability 1/16

eg. 16 children, 8 of each sex.
Breed back to parents
If that child has 4, 5 or 6 DRMs then there’s a 100% chance of elimination (a pair of identical DRMs)
If that child has 3 DRMs then there’s a 7/8 chance of elimination
If that child has 2 DRMs then there’s a 3/4 chance of elimination
If that child has 1 DRM then there’s a 1/2 chance of elimination
If that child has 0 DRM then there’s no chance of elimination

So if a child of the first generation has no children eliminated in the second generation then it is highly likely that that child of the first generation has no deadly recessive mutations.

mollwollfumble said:

One problem with reduction in numbers of a species is that if you do it too fast, the resulting inbreeding causes deadly recessive mutations to come together to wipe out the species. The most general question is: given a load of recessive mutations in a general populace, how rapidly can the number of individuals safely be reduced.

Selective inbreeding, eg. in horse racing, is used to weed out recessive mutations, leaving those that have fewest recessive mutations in the gene pool. ie. artificial selection.

The mathematical question is as follows. In a large population of male and female animals, each with 3 deadly recessive mutations (DRM), assumed independent, design an inbreeding scheme that produces a male female pair, neither of which has a deadly recessive mutation.

One way of doing it is as follows.

After one breeding of a M/F pair, the number of DRMs in the children is given by the rolling of two three-sided dice.
0 DRMs – probability 1/16
1 DRM – probability 1/8
2 DRMs – probability 3/16
3 DRMs – probability 1/4
4 DRMs – probability 5/16
5 DRMs – probability 1/8
6 DRMs – probability 1/16

eg. 16 children, 8 of each sex.
Breed back to parents
If that child has 4, 5 or 6 DRMs then there’s a 100% chance of elimination (a pair of identical DRMs)
If that child has 3 DRMs then there’s a 7/8 chance of elimination
If that child has 2 DRMs then there’s a 3/4 chance of elimination
If that child has 1 DRM then there’s a 1/2 chance of elimination
If that child has 0 DRM then there’s no chance of elimination

So if a child of the first generation has no children eliminated in the second generation then it is highly likely that that child of the first generation has no deadly recessive mutations.

Do you see a problem?
One problem is that I’m assuming that each parent can have lots and lots of children. Of order 150 children.
Possible for males but not for females.
What’s the best strategy if the number of children each animal can have is limited?

• Do you see a problem?

three-sided dice…

furious said:

• Do you see a problem?

three-sided dice…

What about genetic manipulation to try and alleviate inbreeding genetic problems if you have limited numbers to breed from

furious said:

• Do you see a problem?

three-sided dice…

You’re right. It was actually a pair of four sided dice I needed.

Sides labelled 0,1,2,3.

Cymek said:

What about genetic manipulation to try and alleviate inbreeding genetic problems if you have limited numbers to breed from

Yes.

Wait, that could work. A simple genetic test to determine if any particular animal has any deadly recessive mutations.

Use inbreeding to reduce the average number of DRMs,

Use genetic engineering to eliminate the last ones.

I have to be careful because I carry a deadly mutation that does not show up on gene by gene scans.

mollwollfumble said:

One problem with reduction in numbers of a species is that if you do it too fast, the resulting inbreeding causes deadly recessive mutations to come together to wipe out the species. The most general question is: given a load of recessive mutations in a general populace, how rapidly can the number of individuals safely be reduced.

Selective inbreeding, eg. in horse racing, is used to weed out recessive mutations, leaving those that have fewest recessive mutations in the gene pool. ie. artificial selection.

The mathematical question is as follows. In a large population of male and female animals, each with 3 deadly recessive mutations (DRM), assumed independent, design an inbreeding scheme that produces a male female pair, neither of which has a deadly recessive mutation.

One way of doing it is as follows.

After one breeding of a M/F pair, the number of DRMs in the children is given by the rolling of two three-sided dice.
0 DRMs – probability 1/16
1 DRM – probability 1/8
2 DRMs – probability 3/16
3 DRMs – probability 1/4
4 DRMs – probability 5/16
5 DRMs – probability 1/8
6 DRMs – probability 1/16

eg. 16 children, 8 of each sex.
Breed back to parents
If that child has 4, 5 or 6 DRMs then there’s a 100% chance of elimination (a pair of identical DRMs)
If that child has 3 DRMs then there’s a 7/8 chance of elimination
If that child has 2 DRMs then there’s a 3/4 chance of elimination
If that child has 1 DRM then there’s a 1/2 chance of elimination
If that child has 0 DRM then there’s no chance of elimination

So if a child of the first generation has no children eliminated in the second generation then it is highly likely that that child of the first generation has no deadly recessive mutations.

No!

Breeding two organisms with the same recessive gene does not guarantee elimination.

If that child has 6 DRMs then there’s a 7/8 chance of elimination
If that child has 5 DRMs then there’s a 13/16 chance of elimination
…

Complete elimination of deadly recessive genes by line breeding is a slow process involving very many generations.

mollwollfumble said:

One problem with reduction in numbers of a species is that if you do it too fast, the resulting inbreeding causes deadly recessive mutations to come together to wipe out the species. The most general question is: given a load of recessive mutations in a general populace, how rapidly can the number of individuals safely be reduced.

Selective inbreeding, eg. in horse racing, is used to weed out recessive mutations, leaving those that have fewest recessive mutations in the gene pool. ie. artificial selection.

The mathematical question is as follows. In a large population of male and female animals, each with 3 deadly recessive mutations (DRM), assumed independent, design an inbreeding scheme that produces a male female pair, neither of which has a deadly recessive mutation.

> 3 deadly recessive mutations (DRM), assumed independent

I’ve been working on this.

If a couple have 64 children then there is a 50% chance of there being one child with no deadly recessive mutations.

To improve that to a 95% chance of there being one child with no deadly recessive mutations then the couple would have to have 277 children.
Yuk!

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 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 descendants 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 descendants 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 descendants overall.

mollwollfumble said:

mollwollfumble said:

One problem with reduction in numbers of a species is that if you do it too fast, the resulting inbreeding causes deadly recessive mutations to come together to wipe out the species. The most general question is: given a load of recessive mutations in a general populace, how rapidly can the number of individuals safely be reduced.

Selective inbreeding, eg. in horse racing, is used to weed out recessive mutations, leaving those that have fewest recessive mutations in the gene pool. ie. artificial selection.

The mathematical question is as follows. In a large population of male and female animals, each with 3 deadly recessive mutations (DRM), assumed independent, design an inbreeding scheme that produces a male female pair, neither of which has a deadly recessive mutation.

> 3 deadly recessive mutations (DRM), assumed independent

I’ve been working on this.

If a couple have 64 children then there is a 50% chance of there being one child with no deadly recessive mutations.

To improve that to a 95% chance of there being one child with no deadly recessive mutations then the couple would have to have 277 children.
Yuk!

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 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 descendants 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 descendants 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 descendants overall.

Now, if Noah had access to a lab on a chip, then he could do very much better.
Instead of needing 200 or more interbred children, he would need more like 20.
Yes, even less than 64, I think. Will calculate.

mollwollfumble said:

Now, if Noah had access to a lab on a chip, then he could do very much better.
Instead of needing 200 or more interbred children, he would need more like 20.
Yes, even less than 64, I think. Will calculate.

Assuming that Noah had access to a lab on a chip that could count the number of dangerous recessive mutations, I ran a simulation 3 times.

All I did was breed from the male and female with the lowest number of recessive genes.

The three simulations required only 7, 4 and 12 children to find the first individual with no recessive genes.

What an enormous difference that makes! Compared to 200+ children without lab on a chip.

But making that lab on a chip in the first place would not be easy. It would need a very big number of wild individuals, first eliminating the need to test all gene variants found in pairs.

However, for making an interstellar eco-ark in the future it looks very promising.