homozygous form

a homo-spider x normal = all spiders?
homo-pinestripe x normal = all pinestripe?
spider x spider = 1/4 homo-spiders? does this also mean 66% possible homo-spiders?
how is this proved? :confused:

There is not thought to be any homozygous form. Apparently the hatchlings only get one copy of the gene, not two.

Quote:

---

Originally Posted by ballpython911

a homo-spider x normal = all spiders?
homo-pinestripe x normal = all pinestripe?

---

Strictly speaking yes. However, to my knowledge no homozygous spider has ever been produced. Many believe the homozygous form is lethal. Since spider is dominant to wild type (WT), there is a single WT copy and a single spider copy. So spider x normal = 50/50 spider/normal.

IIRC a homozygous pinstripe has been produced by BHB, but don't hold me to that. But in that case there wasn't a super form that I remember. So phenotypically it just looked like a pinstripe. And if there was actually a homozygous pinstripe, then all of the F1 would be pins since it, too, is dominant to WT.

Quote:

---

spider x spider = 1/4 homo-spiders? does this also mean 66% possible homo-spiders?
how is this proved? :confused:

---

Not really. Lets pretend just for a sec that a homozygous spider can be made. In that case, 1/4 would be homozygous spider, 1/4 would be WT, and 2/4 (or 1/2) would have a single copy of each allele. Since there is no super form of the spider, 3/4 would visually appear as spiders, 1/4 would appear normal and only breeding trials could distinguish the hets from the homozygous animals.

This of course assumes that a homozygous spider can be made and that there is NOT a super form, i.e. that the homozygous animal is visually the same as the heterozygous animal.

Good reply blue.

To the OP: For the sake of the hypothetical, lets assume the spider gene is not homozygous-lethal.

If you bred spider x spider, any hatchling that hatches out that appears to be a spider has a 33% chance of being a homozygote.