Question about getting Albinos.

i have a normal male, and was going to try and get a 100% HET albino female. if i bred these 2 is there a chance of getting an albino?

no unless the other is het for albino 2

if i have 2 albino HETS whats the probability of getting an albino?

depends on the percentage of each

1 in 4 i think

i was under the impression either its a het or its not... that the 66%,50%,etc. just ment that was the probability that it WAS a het not that it had 66% het in it. so i am talking about 2 100% HETS.

if you have a 100% male and a 100% female both with the same het albino you have a 25% chance to get an albino out of four eggs now more eges she lays the better your chances are.

thank you.

Quote:

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Originally Posted by gncz73

if you have a 100% male and a 100% female both with the same het albino you have a 25% chance to get an albino out of four eggs now more eges she lays the better your chances are.

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Actually, that's inaccurate. The chance of getting at least one albino with 4 offspring is 1 - (0.75)^4 = 68%

25% is the chance of any given offspring turning out albino. Your odds of getting an albino will be 25% if you only hatch one egg. With more than 1 offspring, your chances will be much better.

i wish i knew what you guys mean!lol.i dont get ball python breeding

Actually i don't think i'm wrong because if your snake lats four eggs and all four hatch then out of those four egg you would have a 25% chance to get an albino. if you go to the Genetics Wizard it will show you that breeding two 100% will give you three that look normal but out of those three two are 66% chance to be hets. while one should be albino but with breeding hets you have no promise of an albino. in fact here is a very good way to look at breeding stright from the nerd site "Something to keep in mind when dealing with punnett squares and simple recessive genetics is that the numbers involved are theoretical. For example, according to theory, breeding a 100% het-albino (Na) ball to an albino (aa) ball should yield half hets and half albinos out of 4 eggs. It doesn't always work this way though...we've bred hets to homozygous animals and gotten all homozygous, we've also bred hets to homozygous animals and gotten all hets. Depending on how the alleles fall, it's possible to breed two hets together and get all possible hets, or do the same and get all homozygous animals as a result. This can be very exciting or somewhat disappointing, depending on what you were expecting out of the clutch. That's a big reason why it's important to keep these numbers in mind as a hypothetical, best-case-scenario according to the laws of genetics. At the same time, that's what makes working with color morphs both extremely fun and also nerve-wracking!"

Basically, (0.75)^4 is the odds of all 4 offspring turning out to be normals or hets, since there is a 75% chance of each one being a normal or het and there are 4 of them.

So, if you subtract it from 1 you get the odds of not all 4 offspring turning out to be normals or hets. In other words, the odds of at least one turning out to be albino.

Quote:

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Originally Posted by gncz73

Actually i don't think i'm wrong because if your snake lats four eggs and all four hatch then out of those four egg you would have a 25% chance to get an albino.

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25% is the odds of 1 turning out to be albino. That is for each and every offspring independent of what the others turn out to be. As I said before, if you hatch out one, you have a 25% chance of getting an albino. For each additional offspring the odds change.

Clutch of 1: 25%
Clutch of 2: 1 - (0.75)^2 = 44%
Clutch of 3: 1 - (0.75)^3 = 58%
Clutch of 4: 1 - (0.75)^4 = 68%

Probabilities change when you have multiple trials. The genetics pages are just to tell you the odds of what any single given offspring turns out to be. It doesn't tell you the odds of getting an albino in a clutch, as they would have to give you percentages for every single clutch size. If you don't believe, go read up on probability and statistics: I was almost a math minor and I took 3 courses on probability & statistics so I'm pretty certain I know what I'm talking about :P

Awesome post hhw!

-adam

what would i get if i bred my normal male with an albino female?

a het?

TokenLs: all the offspring would be 100% het for albino

HHW is correct in his genetic mathmatics.

The only breeding combo's that will equal a CHANCE to produce an albino are:

albino X albino
Het Alb X Het Alb
alb X Het Alb

The Hets may take time to prove out. You may not actually hit an Albino on your first breeding attempt. There are quite a number of variables that must be considered when trying to produce a non-dominate morph. The genetic mathmatics is just one aspect.

Here are some tables I made up with the chance of producing at least one recessive morph with various clutch sizes for various crosses. It's using the same formula HHW showed.

http://snakemorphs.home.comcast.net/odds.htm#table

we shall wait for some1 more experience w/bp breeding suchas KLG to help all of us solve the issues!

Are you kidding me? You've already had breeders and people educated in both genetics and statistics respond on this thread. Not to put Kara down in any way, but she's not going to give you any more qualified response than you've already gotten here.

ooh i didnt see this second page and half of the 1st page<my computer has been actin all weird lately???

i second what marla said and just for the record adam of 8ballpythons that replied also has alot of knowledge in care/breeding. i've purchased from him and asked alot of questions, he's been very helpful before and after the sale.


thanks
vaughn

Wouldn't a normal and an albino produce all guarenteed hets?

Does anyone know the average number of eggs, and how many of those are likely to hatch per breeding?

Oops--I didn't see that a million people had already answered the question I just responded to.

Still--does anyone know the answer to my question about the average number of eggs laid and probability of hatching?

Average clutch sizes for ball pythons is around 6, but it can vary greatly. Some females will lay more smaller eggs and some will lay fewer larger eggs. No matter what though, the heavier the female the better your chances are of producing more eggs.

If you're growing up your own females, chances are they'll be smaller than a mature proven breeder female and therefore likely to lay fewer eggs... however, many females have laid big first clutches, provided they are a good size. I don't think anyone has gathered enough samples of clutch sizes to match them to any probability density function. I would just assume a normal distribution for now, meaning (given an average clutch size of 6) that a clutch will most likely consist of 6 eggs and any clutch size (either less or greater than 6) will decrease in probability the greater the difference is from 6.

Hi hhw,

Being of the mathematically impaired part of our fine planet I, unfortunately, spend a lot of my time in the dark mathematically. My wife on the other hand has a math degree but sadly I'm too ashamed to ask her these types of questions. (insert shame and guilt here)

The question(s):
If you have two critters, one homozygous albino (aa) and one normal/heterozygous for albinism (Aa), that should give you a 50% chance for any given 'individual' egg to pop out an albino, right? If so, does that mean the the increased statistical probability for a larger clutch would be calculated with the remainder of the percentage? (see below) Or shall I tuck my tail between my legs now and go ask my wife to fix me a nice cup of cocoa and give me the simple kindergarten review version of it all? ;)


Clutch of 1: 50%
Clutch of 2: 1 - (0.50)^2 = 75%
Clutch of 3: 1 - (0.50)^3 = 87%
Clutch of 4: 1 - (0.50)^4 = 93%
etc.


In conclusive confusion...

In your earlier example, you showed the math for a 25% chance of throwing a trait using 0.75 in the formula. Does that mean that an 11% chance for some trait would be calculatred with 0.89 in the formula as in: Clutch of 2: 1 - (0.89)^2 ?

Forgive my ignorance. Thanks.

If I may ...

Neumann,

Your formulas are correct for determining the chance of getting one or more albino from homozygous X het clutches of the sizes you have listed.

However I'm not sure I follow your 11% question.

Perhaps explaining what the formula is doing will help us straiten it out.

Going back to the chance from het X het you get an independent 25% chance for each egg to be the desired homozygous. However, there are so many differ rent ways you can get one or more albino (especially after you get into bigger clutches) that it's easier to calculate the chance that you will not get any albinos. If each egg has a 25% chance of being an albino it also has a 75% chance of not being albino. The chance of all the eggs in a clutch of n eggs not being albino is (0.75)^n. This is the one way you can strike out so much easier to figure than all the different ways you can get lucky (at least for clutches bigger than 1). We are just multiplying the 75% chance of each egg missing times it's self once for each egg to find the cumulative chance of all of them not being albinos. We can split the possible outcomes into two groups; success if 1 or more is an albino and failure if none of them are albinos. Since this covers 100% of the possible outcomes we know that the chance of success + chance failure = 1 so chance of success = 1 - chance of failure = 1 - (individual failure)^n. In the case of het X het this is 1 – (0.75)^n

So if there was a genetic combination that would only happen 11% of the time (the 11% is what threw me, it would have to be several odd genes to come out to that chance, most are 25% or 50%) the chance of it not happening for each egg would indeed be 0.89 and the chance of it not happening at all in a clutch would be (0.89)^number of eggs and hence the chance of it happening one or more (two) time in a clutch of 2 would indeed be 1-(0.89)^2.

It's easy to forget what the numbers mean in a homozygous X heterozygous cross because each egg has both a 50% chance of homozygous and a 50% chance of heterozygous so you can forget which 0.5 it is you are using in the formula. In the formula where we are calculating the odds of getting one or more it's the odds of not getting what we are looking for that we plug into the formula.

Clear as mudd?

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Originally Posted by RandyRemington

If I may ...

Neumann,

Your formulas are correct for determining the chance of getting one or more albino from het X het clutches of the sizes you have listed.

Clear as mudd?

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Hi Randy,

Thank you for the reply. It's cystal clear- like the eyes of a freshly shed royal.

Sorry for the "11%" confusion. It was a totally random percentage that I selected from my, ahem, feeble mind for the purpose of clarifying the relationship between the chance of a success or a failure, and it had nothing to do with any genetic trait specific to man or beast. I thought my example using a 50/50 chance made things a little confusing (err, in my mind anyway) but it was a simple genetic example pertaining to snakes that I could use, so I grabbed 11% for the heck of it to further clarify the relationship between the numbers.

Thanks again.

Basically, the equation works by calculating the odds of all eggs being normal, then subtracting this number from one to get the odds of getting at least one morph in the clutch. One, being the sum of the probabilities of all possibilities (i.e. 100%).

So in the examples, if your odds are 50% of hitting, there is also a 50% chance of not hitting. You then calculate the odds of all eggs not hitting, which is... (0.50) x (0.50) x ....
To simplify this formula, you then just use an exponent for the number of eggs i.e. (0.50)^#eggs

You can also calculate the odds of getting exactly one, exactly two, etc... but that becomes a lot more difficult because order matters. For instance, to get one morph in a clutch of four, you need to calculate the odds of egg#1 being a morph and all other eggs normal + the odds of egg#2 being a morph and all other eggs normal, etc... There are 4 ways of ordering this, so your total odds are:
4 x (0.50)^1 x (0.50)^3 = 0.25

where 0.50^1 is the odds of getting the morph and 0.50^3 is the odds of the other 3 eggs being normal.

This gets a lot more complex when you have 2 eggs in a clutch that turn out to be a morph, because the number of ways you can order it are more complex. For instance, the 2 eggs being the morph in a clutch of 4 can be: 1&2, 1&3, 1&4, 2&3, 2&4, and 3&4, a total of 6 combinations. You also need to be careful not to count any combination twice. For a general formula to calculate the number of orderings, you can use the formula for combinatorics:
For nCk (read n choose k):

Code:

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      n!
--------------
k! x (n - k)!

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where ! means factorial. e.g. 3! = 3 x 2 x 1

In our example, this would be:

Code:

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      4!              4!        4 x 3 x 2 x 1
--------------  =  ----------  = ---------------  = 6, as we counted above
2! x (4 - 2)!        2! x (2)!      2 x 1 x 2 x 1

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So, in other words, the general formula for getting a number of mutations k of probability p in a clutch of size n is:

Code:

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(# of combinations) x (probability of the combination)

      n!             
--------------  x  [ (p)^k x (1 - p)^(n-k) ]
k! x (n - k)!

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Ack, lack of whitespace makes that hard to read...

[edit] Alright, added in code tags to preserve whitespace and fixed a bunch of mistakes just now... you may want to read it again now that I've corrected them.[/edit]

Hey all, fellow math nerd here. Hhw is totally right on in using the binomial distribution to figure the exact percentages of getting a certain number of albinos from a certain number of eggs. However, only nerdy math teachers like me have the patience or desire to crank out that formula.

IMO, I think the simple expected number formula would be much easier and practical in answering this type of question. The expected number tells you the successes you can expect out of a given number of trials. Or this case the number of albinos you can expect from a given number of eggs. The expected number can be figured out as follows:

Expected Number = (Number of trials) * (% chance of success)

or in this case:

Expected # of Albinos = (# of Eggs) * (% Chance of an individual egg being albino)

Lets take the example of breeding two Het Albinos together. As mentioned earlier the chance of an individual egg in this case would be 25%. So lets look at the number of albinos you could expect from some different numbers of eggs:
4 Eggs:4 * .25 = 1 Albino
6 Eggs:6 * .25 = 1.5 Albinos
8 Eggs:8 * .25 = 2 Albinos
10 Eggs: 10 * .25 = 2.5 Albinos
12 Eggs: 12 * .25 = 3 Albinos
etc.

Now obviously you can't have exactly 1.5 snakes, but you if you have six eggs you can reasonably expect 1 or 2 albinos. The expected number isn't as exact as the binomial distribution, but is much easier in figuring out how many morphs you can expect out of a clutch.

Lee