Question for genetic wizzards

I what am I doing wrong here?

Can someone draw me a chart (the way I did it wrong) that will result in the answer of the genetic wizzard?

If s is recessive for stripe, z is recessive for zigzag (which doesn't behave as a recessive trait actually, so that's misleading), and a is recessive for amel, then your grid should look like the following:


Across the top would be: ZSA, ZsA, zSA, zsA.
Down the side would be: ZSA, ZSa

The first row would be: ZZSSAA, ZZSsAA, ZzSSAA, ZzSsAA
The second row would be: ZZSSAa, ZZSsAa, ZzSSAa, ZzSsAa

Therefore you have a 1 in 8 chance of getting any of the above combinations. If you don't put placeholders for the wildtype genes, you wind up forgetting combinations. To correct your diagram, Marcel, you need to have an extra column labeled sz, since it's possible for the offspring to inherit both of the genes the female has. Again, zigzag doesn't act like a regular recessive... but if it did, that's how you would treat it.

-Kathryn

But the male is only het Amel so Vertical should only have A right?.. That's why I donlt get it. I dont' understand the methode.
I just picked three genes. I know ZigZag is not really recessive. But my question is now how do you come up with the horizontal markers and vertical markers (placeholders)?

Like this?

I think you need to make is simpler for me (lol, bare with me). Can you make a Punnett square. with only Het Amel and Het Stripe? Aa, Ss. What will the horizontal and vertical row be?

.......AS.... As....aS....ss
.
.
A AASS AsSS AsSS AsSs
S
.
.
a aAaS aAas aaaS aass
a


Aaarrggg I am lost..

Ok... If your male was het stripe and your female was het amel, this is how they'd look:

Male: AASs
Female: AaSS

Across the top would be: AS As
Down the side would be: AS aS

The first row's results would be: AASS AASs
The second row's results would be: AaSS AaSs

Therefore, you'd have a 1/4 chance of getting a particular combination (AASS - normal, AASs - het stripe, AaSS - het amel, AaSs - het amel stripe).

-Kat

Marcel,

Each parent has two of each letter, and each sperm/egg has only one of each letter.

Each offspring gets one letter from the sperm, and one letter from the egg, so it has two of each letter.

Use the # button on the top of the post, or use the "code" tag, to insert code and you can make tables that line up.

You'll want to use the preview button to double check...

Like this:
[code]
AaSs x AaSs (both normals double het amel stripe)

A S A s a S a s
AS AASS AASs AaSS AaSs
As AAsS AAss AasS Aass
aS aASS aASs aaSS aaSs
as aAsS aAss aasS aass
[/code]


(Easy enough, I just put all of this inside a "code" tag, LOL)
[php]AaSs x AaSs (both normals double het amel stripe)
A S A s a S a s
AS
AASS
AASs
AaSS
AaSs
As
AAsS
AAss
AasS
Aass
aS
aASS
aASs
aaSS
aaSs
as
aAsS
aAss
aasS
aass[/php]

Kat, Serp,

Both of you are great! Thanks a lot.... To proof if I get it can you say if this picture is right?

If the Male amel isn't het stripe and the Female het stripe isn't het amel, then:
(Gotta have both of the genes in each outcome. )

[CODE]aaSS x AASs (Amel x Normal het stripe)
a S a S
AS AaSS AaSS
As AasS AasS[/CODE]

Or in honesty, even more simply since the male can only throw one gene combo:

[CODE]aaSS x AASs (Amel x Normal het stripe)
a S
AS AaSS
As AasS [/CODE]

So 50% Hets for Amel and Stripe, 50% Hets for Amel only.