I need to go back and listen to the whole thing some time but I caught part of a reptile radio interview and it sounds like BHB has proven pinstripe to be the first dominant ball python mutation. Still, almost all pinstripes out there will only be heterozygous for the pinstripe mutation but if both parents of your pinstripe where pinstripes it would be a 33% chance homozygous pinstripe.

But assuming we are talking about the heterozygous pinstripes here it will have one pinstripe mutant copy of the gene at the pinstripe location and one normal for pinstripe version of that gene so each baby would have a 50/50 chance of getting the pinstripe mutant copy. Same for Mojave and Cinnamon. Although most ball python people don't call them hets they are heterozygous for their mutations and knowing that you will see why each of their offspring has a 50/50 chance of getting the mutation from them and since these are all dominant type mutations that one mutant gene will show.

Now ghost is recessive so your visual ghost will be homozygous for the ghost mutation meaning that both copies of the gene at the ghost location will have the ghost mutation so it will pass that mutation to all of its offspring. However, because ghost is recessive that one copy passed to the het ghost offspring will not show and unlike the het pinstripes, mojave, and cinnamon the het ghosts will not show the ghost mutation.

So your steps to predict the offspring are:

1. Figure out the genotypes of the parents realizing that the only ball python hets aren’t recessive.

2. Use the same genotype inheritance rules regardless of the mutation type. If an animal is a het it has two different copies and each has a 50/50 chance of being passed but if it’s homozygous you know 100% which type it will pass because it only has one type available.

3. Figure out what the genotypes look like based on the mutation type. For example apparently now we know a homozygous pinstripe looks like a heterozygous pinstripe which is what defines it as a completely dominant mutation type.

Applying these to your questions:


Pin X Normal

1. Heterozygous for the pinstripe gene with one pinstripe mutant copy and one normal for pinstripe copy X homozygous normal for pinstripe.
2. As a homozygous normal we know the normal will pass the normal for pinstripe version of the gene 100% of the time. As a heterozygous the pinstripe parent has a 50/50 chance of passing the pinstripe mutant copy (otherwise it passes it’s normal for pinstripe copy). So as a het X normal breeding each egg has a 50% chance of being heterozygous pinstripe.
3. Because pinstripe is dominant the babies that hit the 50% chance and are heterozygous pinstripe will be visible pinstripe mutants.


Pin X Ghost

1. This is a little more complicated as a breeding with two genes that you care about but in this case they appear to be unrelated genes so treat them separately. The pin is heterozygous for the pinstripe mutation and homozygous normal for the ghost mutation. The ghost is homozygous normal for the pinstripe mutation and homozygous mutant for the ghost mutation.
2. On the pin side as a het X normal breeding it’s just like above, 50% heterozygous pinstripe offspring. On the ghost side as a homozygous mutant X homozygous normal breeding it’s 100% heterozygous ghost offspring. Overlay the two and you have 50% normal for pinstripe het ghosts and 50% heterozygous for pinstripe het ghosts.
3. As recessive none of the het ghosts show but as dominant you can pick out the ones that are also het pinstripe.


Pin X Mojave

1. Heterozygous pin homozygous normal Mojave X heterozygous Mojave homozygous normal pin.
2. On the pin side a het X normal breeding produces 50/50 hets and normals. Same on the Mojave side. Overlaying them you get 25% normals, 25% het Mojave only, 25% het pinstripe only, and 25% Mojave and pinstripe double hets.
3. As dominant type morphs (i.e. co-dominant is a type of dominant) both the het pinstripes and the het Mojave’s are visible.


Pin X Cinnamon

1. Heterozygous pin homozygous normal for cinnamon X heterozygous cinnamon homozygous normal for pin.

2. On both the cinnamon and pin side we are talking het X normal so each is a 50/50 het and normal split for babies. Overlaying the two mutations we again get 25% normal, 25% het cinnamon only, 25% het pin only, 25% het for both pin and cinnamon.

3. As dominant type morphs both the het cinnamons and het pins are visible.


Ghost X Mojave

1. Homozygous mutant ghost homozygous normal Mojave X heterozygous mutant Mojave homozygous normal ghost.
2. On the ghost side homozygous X normal so babies 100% het ghosts. On the Mojave side het X normal so 50% het ghosts. Overlaying the two mutations you get 50% het Mojave het ghost and 50% het ghost only.
3. As recessive you can’t tell that the are all het ghost but you can pick out the ones that are also het Mojave because it’s a type of dominant mutation.

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