Reptile & Amphibian Forums

I'm just trying to learn and understand genetics, so:

What would you get when you breed a 100% het pied male with a 100% het albino female?

lets say:

NA = het albino
NP = het pied

would you get:

NN = normal
NP = 100% het pied
AN = 100% het albino
AP = albino pied??

This doesn't sound at all right, so can someone please help me understand. I would imagine the AP to be 100% albino and pied. Right? Not an albino pied.

when you breed a het to a normal you get half normals, half hets, but since you cant tell them apart they are 50% hets. so two different hets bred together would all babies are 50% het for both with a small chance one would be a double het
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jake

my addiction:
0.2 normal ball pythons (lazlo and izzy)
0.1 amelenistic corn snake (mazy)
0.1 blizzard corn (blizz)
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hybrid haven

>I'm just trying to learn and understand genetics, so:

>What would you get when you breed a 100% het pied male with a 100% het albino female?

If you are just starting, you'd be better off learning how to work with a single mutation before working on two independent mutant genes. Learn to walk before trying to run.

A = normal gene at the albino locus
a = the albino mutant gene

P = normal gene at the pied locus
p = the pied mutant gene

Heterozygous pied x heterozygous albino is equivalent to combining two single mutant matings:

1. heterozygous pied (Pp) x normal (at the pied locus)(PP) -->
1/2 heterozygous pied (Pp) (looks normal)
1/2 normal (PP) (looks normal)

2. normal (at the albino locus)(AA) x heterozygous albino (Aa) -->
1/2 heterozygous albino (Aa) (looks normal)
1/2 normal (AA) (looks normal)

You will probably want to do the Punnett Squares.

Half of the babies are heterozygous pied and half of the babies are normal at the pied locus. And half of the babies are heterozygous albino and half of the babies are normal at the albino locus. So half of the heterozygous pieds are also heterozygous albino while the other half of the heterozygous pieds are normal at the albino locus. And half of the normals at the pied locus are also heterozygous albino while the other half are normal at the albino locus.

The easiest way to diagram this is to write the pieds in a column and put a forked line after each one, like so:
1/2 Pp <
1/2 PP <

Then at the top of each fork put 1/2 Aa, and at the bottom of each fork put 1/2 AA. This forum will not do partly blank lines. Traveling each fork from left to right produces this:
1/2 Pp - 1/2 Aa
1/2 Pp - 1/2 AA
1/2 PP - 1/2 Aa
1/2 PP - 1/2 AA

Then multiply the fractions and figure out what they look like:
1/2 Pp - 1/2 Aa = 1/4 Pp Aa (heterozygous pied, heterozygous albino) looks normal
1/2 Pp - 1/2 AA = 1/4 Pp AA (heterozygous pied) looks normal
1/2 PP - 1/2 Aa = 1/4 PP Aa (heterozygous albino) looks normal
1/2 PP - 1/2 AA = 1/4 PP AA (normal) looks and is normal

Shorthand for this is normals that are 50% probability heterozygous pied and 50% probability heterozygous albino.

Clear as mud?

By all means check these results with a Punnett Square. What I did was called a forkline or branching system, which takes me half the time as the equivalent Punnett Square. YMMV.

Paul Hollander

Go with the short answer.
All the offspring would be normal in appearance, with each holding a 50% probability of carrying either gene. Therefore, some could carry neither while some COULD carry both.