Mendelian Genetics - Trihybrid Cross?

Never heard of it - too long since I studied biology maybe.

I reckon the chance is 1/16, assuming all these genes are independently assorted.

Ll x ll ?> 50% Ll 50% ll
Ss x Ss ?> 25% SS 50% Ss 25% ss
Dd x dd ?> 50% Dd 50% dd

So that's 50% x 25% x 50%, or 0.5 x 0.25 x 0.5

You can do 3 small Punnett squares to prove those if you like.

Edit: Forked line method is explained halfway down this page.
http://www.emunix.emich.edu/~rwinning/genetics/mendel4.htm
Doesn't sound a million miles from what I've done.

well i can't do the whole forked diagram on here, but it isn't necessary. basically you know that in a test cross (ex. Ll x ll will result in 1/2 Ll and 1/2 ll) and in a monohybrid cross (ex. Ss x Ss will result in 1/4 SS, 1/2 Ss, and 1/4 ss). so determining the chance of each individual gene crossing and multiplying the three together will give you the chance of all three occuring in an individual

ll x Ll . Ss x Ss . dd x Dd
1/2 ll x 1/4 ss x 1/2 dd = 1/16 llssdd

for the other offspring if it does not specifically say dominant homozygous or heterzygous it is assumed that only one dominant allele is necessary, so in the monohybdrid cross there is a 3/4 chance of S_ (1/4 SS and 2/4 Ss) and obviously in a test cross only heterzygotes are made (and the homozygous recessive)

ll x Ll . Ss x Ss . dd x Dd
1/2 Ll x 3/4 S_ x 1/2 dd = 3/16 LlS_dd

feel free to email me if you have questions