Black or brindle?

This is The Masked Wonder. He is definitely c^ch/c^e. Is he A^vy/* or a/a? Anybody familiar with what c^ch/c^e looks like on either of these colors?





What is funny is that the dark side of his face looks browner than black, but I don't see any ticking.

Wow! What a beauty! I don't really know the answer to your question, but I'd guess he was a/a. My a/a tri-colors (of varying combos of c-dilutes) are usually shades of black, brown and grayish-beige.

Thanks for weighing in!

If it doesn't look ticked, I'd say he's probably a/a. The slight discoloration of the black spots is because even with splashed, it's difficult for the color to come through quite as brightly as it would undiluted. It means black is a little off, agouti is a little muted, and red or red-based colors are a little pale.

OK. I think I understand. I am happy to know you think he's probably a/a. What color would a c^ch/c^ch dilution of A^vy/a be then?

Chinchillated brindle would probably wipe out your orange coloring and give your agouti bits the silver agouti color. Your splashing would have very faint orange on the orange bits and brown in your chinchilla bits. On the other hand, chinchilla doesn't dilute black but hardly at all. If he's a/a, he's not cch/cch.

Yes, but what color would c^ch/c^ch be on an A^vy/a (black-based brindle) mouse? I want to wipe out the orange in the diluted parts, so I'm hoping it does.

There's not way the Masked Wonder could be c^ch/c^ch. He's c^ch/c^e. At any rate, he's in with his mate, an A^vy/* c^e/c^e s/s Spl/* doe (his sister!), busily making babies with her. I'll get to see what c^ch/c^e brindles look like, hopefully.