Help understanding Het's

[PythonOutlaw]

BUMBLE BEE X PIEBALD

25% 1/4 Normal Het. Piebald
25% 1/4 Pastel Het. Piebald
25% 1/4 Spider Het. Piebald
25% 1/4 Bumble Bee Het. Piebald

Are the offspring of that pairing %100 Het piebald

[EverEvolvingExotics]

Yes every animal will be 100% het for piebald.

[cmack91]

yes they are

[Maddumpling]

Yes, everything will be 100% het.

[PythonOutlaw]

ok thanks for the replies,

Now where do the %66 and %50 hets come from

[C&H Exotic Morphs]

ok thanks for the replies,

Now where do the %66 and %50 hets come from

100% Het x 100% Het breeding will give you 66% possible Het babies.
100% Het x anything other than a visual or 100% Het will give you 50% possible Het babies.

[JulieInNJ]

I always refer to this website for percentages help.

http://www.ballpython.ca/genetics.html

[rabernet]

ok thanks for the replies,

Now where do the %66 and %50 hets come from

With recessive genes, an animal needs to receive two copies to show the visual mutation.

Hets carry one copy of the gene. In your example, the Piebald is contributing one copy of the pied gene to the pairing, making all the animals be 100% Het Pied.

We'll stick with pieds for this example.

When you breed a Het Pied (1 copy of the pied gene, 1 copy of the normal gene) to a normal (or anything that's not also pied or het pied), the het pied will contribute EITHER a copy of the pied gene, or a copy of the normal gene.

The offspring will be classified as 50% POSSIBLE Het Pied - because each resulting snake had a 50% chance of inheriting the pied gene and a 50% chance of inheriting the normal gene. It can only be changed from the designation of 50% possible het pied once it proves out to indeed carry the gene through future breeding.

Now - when you breed 100% Het Pied to 100% Het Pied, let's say you get four eggs. Statistically, if the statistics played out perfectly - you'd get 1 pied (inherited a copy from both parents of the pied gene, giving it two copies to produce a visual), 1 normal (inherited 2 copies of the normal gene from both parents), and 2 het pieds (received a copy of the het pied gene from one of the parents, and a normal gene from the other).

So that's three visually normal animals. Since you can't tell without breeding, which inherited the gene, and 2 out of 3 (or 66% of the remaining clutch) could carry the gene, they are designated as 66% possible het pieds.

[paulh]

With recessive genes, an animal needs to receive two copies to show the visual mutation.

Hets carry one copy of the gene. In your example, the Piebald is contributing one copy of the pied gene to the pairing, making all the animals be 100% Het Pied.

We'll stick with pieds for this example.

When you breed a Het Pied (1 copy of the pied gene, 1 copy of the normal gene) to a normal (or anything that's not also pied or het pied), the het pied will contribute EITHER a copy of the pied gene, or a copy of the normal gene.

The offspring will be classified as 50% POSSIBLE Het Pied - because each resulting snake had a 50% chance of inheriting the pied gene and a 50% chance of inheriting the normal gene. It can only be changed from the designation of 50% possible het pied once it proves out to indeed carry the gene through future breeding.

Now - when you breed 100% Het Pied to 100% Het Pied, let's say you get four eggs. Statistically, if the statistics played out perfectly - you'd get 1 pied (inherited a copy from both parents of the pied gene, giving it two copies to produce a visual), 1 normal (inherited 2 copies of the normal gene from both parents), and 2 het pieds (received a copy of the het pied gene from one of the parents, and a normal gene from the other).

So that's three visually normal animals. Since you can't tell without breeding, which inherited the gene, and 2 out of 3 (or 66% of the remaining clutch) could carry the gene, they are designated as 66% possible het pieds.

That takes care of possible hets with recessive mutant genes. This is the most common case. But there are also codominant mutant genes (like lesser platinum) and dominant mutant genes (like pinstripe).

A heterozygous snake has a gene pair made up of a mutant gene and a normal gene. A homozygous snake has a gene pair made up of two copies of a mutant gene or two copies of the normal gene.

There are no possible hets with codominant mutant genes. You can tell whether the animal is heterozygous by looking at it. A royal python with a lesser platinum gene paired with a normal gene does not look like a royal python with two normal genes and does not look like a royal python with two lesser platinum mutant genes.

You cannot tell whether a pinstripe royal python is homozygous pinstripe or heterozygous pinstripe by looking at it. You have to use either the pedigree or a breeding test.

When you breed a het pinstripe (1 copy of the pinstripe gene, 1 copy of the normal gene) to a normal (or anything that's not also either homozygous pinstripe or het pinstripe), the het pinstripe will contribute EITHER a copy of the pinstripe gene or a copy of the normal gene. The normal parent contributes a normal gene. Result -- statistically half the babies are pinstripe and half are normals, with two normal genes. All the pinstripes must have a pinstripe and a normal gene making them het pinstripes.

When you breed a homozygous pinstripe (2 copies of the pinstripe gene) to a normal royal python (two copies of the normal gene), the homozygous pinstripe must contribute a copy of the pinstripe gene to each baby. The normal parent contributes a normal gene. Result -- all of the babies are pinstripe. All the pinstripes must have a pinstripe and a normal gene making them het pinstripes.

When you breed a homozygous pinstripe (2 copies of the pinstripe gene) to a het pinstripe royal python (a pinstripe mutant gene and a normal gene), the homozygous pinstripe must contribute a copy of the pinstripe gene to each baby. The het pinstripe parent contributes either a pinstripe mutant gene or a normal gene. Result -- all of the babies are pinstripe. Statistically, half of the pinstripes have a pinstripe and a normal gene making them het pinstripes, and half of the pinstripes have 2 pinstripe genes making them homozygous pinstripes. The offspring will be classified as 50% probability het pinstripe. Or the offspring can be classified as 50% probability homozygous pinstripe. It's two sides of the same coin. The classification can only be changed from the designation of 50% probability het pinstripe after a breeding test.

Now - when you breed het pinstripe to het pinstripe, let's say you get four eggs. Statistically, if the statistics played out perfectly - you'd get 1 normal (inherited a copy of the normal gene from both parents, giving it two copies to produce a normal royal python), 1 homozygous pinstripe (inherited 2 copies of the pinstripe gene, one from each parent), and 2 het pinstripe (received a copy of the pinstripe gene from one of the parents and a normal gene from the other).

So that's three pinstripe animals left. Since you can't tell what the genes are without a breeding test, and 2 out of 3 (or 66% of the remaining clutch) are het pinstripes, they are designated as 66% probability het pinstripe or 33% probability homozygous pinstripe.

[rabernet]

That takes care of possible hets with recessive mutant genes. This is the most common case. But there are also codominant mutant genes (like lesser platinum) and dominant mutant genes (like pinstripe).

A heterozygous snake has a gene pair made up of a mutant gene and a normal gene. A homozygous snake has a gene pair made up of two copies of a mutant gene or two copies of the normal gene.

There are no possible hets with codominant mutant genes. You can tell whether the animal is heterozygous by looking at it. A royal python with a lesser platinum gene paired with a normal gene does not look like a royal python with two normal genes and does not look like a royal python with two lesser platinum mutant genes.

You cannot tell whether a pinstripe royal python is homozygous pinstripe or heterozygous pinstripe by looking at it. You have to use either the pedigree or a breeding test.

When you breed a het pinstripe (1 copy of the pinstripe gene, 1 copy of the normal gene) to a normal (or anything that's not also either homozygous pinstripe or het pinstripe), the het pinstripe will contribute EITHER a copy of the pinstripe gene or a copy of the normal gene. The normal parent contributes a normal gene. Result -- statistically half the babies are pinstripe and half are normals, with two normal genes. All the pinstripes must have a pinstripe and a normal gene making them het pinstripes.

When you breed a homozygous pinstripe (2 copies of the pinstripe gene) to a normal royal python (two copies of the normal gene), the homozygous pinstripe must contribute a copy of the pinstripe gene to each baby. The normal parent contributes a normal gene. Result -- all of the babies are pinstripe. All the pinstripes must have a pinstripe and a normal gene making them het pinstripes.

When you breed a homozygous pinstripe (2 copies of the pinstripe gene) to a het pinstripe royal python (a pinstripe mutant gene and a normal gene), the homozygous pinstripe must contribute a copy of the pinstripe gene to each baby. The het pinstripe parent contributes either a pinstripe mutant gene or a normal gene. Result -- all of the babies are pinstripe. Statistically, half of the pinstripes have a pinstripe and a normal gene making them het pinstripes, and half of the pinstripes have 2 pinstripe genes making them homozygous pinstripes. The offspring will be classified as 50% probability het pinstripe. Or the offspring can be classified as 50% probability homozygous pinstripe. It's two sides of the same coin. The classification can only be changed from the designation of 50% probability het pinstripe after a breeding test.

Now - when you breed het pinstripe to het pinstripe, let's say you get four eggs. Statistically, if the statistics played out perfectly - you'd get 1 normal (inherited a copy of the normal gene from both parents, giving it two copies to produce a normal royal python), 1 homozygous pinstripe (inherited 2 copies of the pinstripe gene, one from each parent), and 2 het pinstripe (received a copy of the pinstripe gene from one of the parents and a normal gene from the other).

So that's three pinstripe animals left. Since you can't tell what the genes are without a breeding test, and 2 out of 3 (or 66% of the remaining clutch) are het pinstripes, they are designated as 66% probability het pinstripe or 33% probability homozygous pinstripe.

Wow that muddied the waters, even though you basically just copied my entire post and pasted it word for word as yours, replacing it with pinstripes.

While accurate, it really wasn't what the OP was asking.