A lesson on probabilites with double recessives

[towelie4365]

So, I saw the thread on the odds of getting 4 out of 5 albinos in a het x het clutch, and figured I'd give a lesson on how to calculate odds.

Let me start first with an example of what I want to do, and look into double recessives:

I have an albino het pied male, and a pied female. When I finally get into breeding them, 50% will be pied het albino, 50% will be double hets. For my purpose, I want a female pied het albino.

The chance I get a female in each egg is 50%, and the chance I get a pied het albino in each egg is 50%, so in order to find out the chance per egg of getting a female pied het albino, I multiple the odds together:
.5 x .5 = .25
Therefore, each egg has a 25% chance of being what I was, and thus, a 75% chance of not being what I want.

In order to figure out the odds of getting at least one female het pied, I need to look at the odds of not getting a female het pied in the clutch. So, if there are "n" eggs in a clutch, the odds of me getting one is:
1 - (.75)ⁿ
To explain that further, if I have 3 eggs, then there is a .75 x .75 x .75 = 42% (or .75³) chance that I don't get one, which means there is a 58% chance that I do get one, which is why you subtract it from one.
With each egg, you have a higher chance of getting what you want, just plug in the number for "n":
3 eggs = 58% chance
4 eggs = 68%
6 eggs = 82%
10 eggs = 94%... etc

So, now that I have the odds of getting the first snake I want, we'll say its an 82% chance with 6 eggs, now I can calculate the odds of getting the albino pied once I raise her up:
Albino het pied x pied het albino: There is a 25% chance I get an albino pied, 25% chance I get an albino het pied, 25% chance I get a pied het albino, 25% chance for a double het
Again, I have a 25% chance at getting what I want, so the odds are the same as before. This means that, if both clutches are 6 eggs, I have an 82% chance at getting what I want each time.

Now I have to calculate the odds of getting what I want both times, so I multiply those together again:
.82 x .82 = .676 = 67.6%

If everything goes right, and each clutch is 6 eggs, I have a 67.6% chance at getting an albino pied in two generations.

Things get messy if you start to look at if you'll get it within 3 or more generations, but if you add the odds of not getting it 3/3 times and not getting it the first and second time, or the first and third time, or the second and third time (we'll just multiply the odds of not getting it 2/3 times by 3), and subtract it from one, you get the odds of getting it at least 2/3 times: 1 - (.18³ + 3*.18²*.82) = 1 - (.00583 + 3*.0265) , or about a 91.5% chance of getting one within 3 generations. You could also do it by adding up the chances of getting what you want 2/3 or 3/3 times: (.82³ + 3*.82²*.18) = .9145, or 91.5% again.

Since some people deal with double hets alone, I'll cover that quickly:
The odds of getting an albino pied from two double hets are 1/16, or 6.25%. So, if you have 6 eggs in a clutch, your odds of getting it are 1 - (1-.0625)⁶, which is a 32% chance. So, the odds of not getting an albino pied two years in a row is .68 x .68, which is 46.2%, making your chances of getting it within two years 53.8% (you can also calculate it based off of number of eggs you get from the two years, so using 12 instead of 6 yields the same results).

Of course this is all assuming you know how many eggs you will have. Now, if you knew the odds of how many eggs you'll get, you could calculate the exact chance you have... but that depends on the snake and other factors I believe we don't understand.

If you want me to explain anything in further detail, I'd be more than happy to.

[towelie4365]

Also, if anyone wants me to help them calculate their odds, I'm willing to help.

This might be better suited for the genetics forum, but also very applicable here

[BHReptiles]

I think this thread is a great idea. Genetics is hard enough...add in the math part and it's a brain breaker for a lot of non-math people.

[Blue Apple Herps]

I think this thread is a great idea. Genetics is hard enough...add in the math part and it's a brain breaker for a lot of non-math people.

I agree. Great thread. However, I don't think genetics is that difficult if people invest the time to learn it. Punnett squares can easily be worked out to see the percentages for those who are not strong at math. I think a big problem is people with no back ground in genetics ask "what will I get If I breed my albino pastel 100% het clown 66% het caramel to my mojave cinnamon 50% het hypo and vpi axanthic together????"

That's like asking how to do differential calculus when you're just learning how to multiply. If more would spend a month starting out with the basics and learn to walk before running, that would help a lot. Just my 2¢.

[towelie4365]

Blue Apple: I agree, people should invest in learning probabilities like that on their own. I think most people understand how to calculate odds with 1 egg, but this is more of odds across a whole clutch. It gets a bit more complicated when you start getting into multiplying odds and having to deal with permutations.

[paulh]

what will I get If I breed my albino pastel 100% het clown 66% het caramel to my mojave cinnamon 50% het hypo and vpi axanthic together????

This is an eight locus problem. It is a logic problem that requires multiplying fractions. A bright 13 year old could solve it if the right method is explained.

1. Break down the problem into 8 single locus problems.
a. albino x normal --> 1/1 het albino
b. pastel x normal --> 1/2 pastel, 1/2 normal
c. het clown x normal --> 1/2 het clown, 1/2 normal. AKA 50% probability het clown.
d. 66% het caramel x normal --> Disregard as het caramel cannot be distinguished from normal. And all may be normals.
e. normal x mojave --> 1/2 mojave, 1/2 normal
f. normal x cinnamon --> 1/2 cinnamon, 1/2 normal
g. normal x 50% het hypo --> Disregard as het hypo cannot be distinguished from normal. And all may be normals.
h. normal x 50% het vpi axanthic --> Disregard as het vpi axanthic cannot be distinguished from normal. And all may be normals.

2. Make a branching system. All babies are either pastel or normal (not pastel). Each of those 2 branches is either mojave or normal (not mojave), making 4 branches. Each of those four branches is either cinnamon or normal (not cinnamon), making 8 branches.

---------------------------1/2 cinnamon
--------------1/2 mojave <
---------------------------1/2 normal
1/2 pastel <
---------------------------1/2 cinnamon
--------------1/2 normal <
---------------------------1/2 normal
---------------------------1/2 cinnamon
--------------1/2 mojave <
---------------------------1/2 normal
1/2 normal <
---------------------------1/2 cinnamon
--------------1/2 normal <
---------------------------1/2 normal

3. Follow each branch from left to right and multiply the fractions to get the final probability.

1/2 pastel - 1/2 mojave - 1/2 cinnamon = 1/8 pastel mojave cinnamon
1/2 pastel - 1/2 mojave - 1/2 normal = 1/8 pastel mojave
1/2 pastel - 1/2 normal - 1/2 cinnamon = 1/8 pastel cinnamon
1/2 pastel - 1/2 normal - 1/2 normal = 1/8 pastel
1/2 normal - 1/2 mojave - 1/2 cinnamon = 1/8 mojave cinnamon
1/2 normal - 1/2 mojave - 1/2 normal = 1/8 mojave
1/2 normal - 1/2 normal - 1/2 cinnamon = 1/8 cinnamon
1/2 normal - 1/2 normal - 1/2 normal = 1/8 normal
And all babies are het albino and 50% probability het clown
Fractions are the probability of a given result for each egg, not per clutch.

[grcforce327]

So, I saw the thread on the odds of getting 4 out of 5 albinos in a het x het clutch, and figured I'd give a lesson on how to calculate odds.
Let me start first with an example of what I want to do, and look into double recessives:
I have an albino het pied male, and a pied female. When I finally get into breeding them,it is possible for 50% to be pied het albino,and 50% to be double hets. For my purpose, I want a female pied het albino.

I figured I'd give a lesson on the word "possible".

[scooter11]

I was wondering something and I assume this is as good a place as any to get an answer. First, does anyone know of a book or article they can recommend on punnett squares and probabilities? Preferably one that isn't to heavy or overly technical. Second, I have been trying to figure out the probabilities of a double hey breeding and what the hets will be percentage wise. So its a double het caramel/pied to double het caramel pied. There are for normal looking snakes. 1/16 normal, 2/16 het pied, 2/16 het caramel, 4/16 double het pied/caramel. So if I break these down for each normal in the litter, what do I have as a percentage for each. Again if anyone can recommend a good study guide to learn this I'd appreciate it. Thanks

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[BHReptiles]

I was wondering something and I assume this is as good a place as any to get an answer. First, does anyone know of a book or article they can recommend on punnett squares and probabilities? Preferably one that isn't to heavy or overly technical. Second, I have been trying to figure out the probabilities of a double hey breeding and what the hets will be percentage wise. So its a double het caramel/pied to double het caramel pied. There are for normal looking snakes. 1/16 normal, 2/16 het pied, 2/16 het caramel, 4/16 double het pied/caramel. So if I break these down for each normal in the litter, what do I have as a percentage for each. Again if anyone can recommend a good study guide to learn this I'd appreciate it. Thanks

Sent from my EVO using Tapatalk 2

I don't own this book, per say, but it looks like a great reference for genetics that's geared towards herps! It's got explanations and pictures that are designed for reptile breeders. I've heard good things and will probably invest in this book in the near future.

http://geneticsforherpers.com/

[paulh]

I don't own this book, per say, but it looks like a great reference for genetics that's geared towards herps! It's got explanations and pictures that are designed for reptile breeders. I've heard good things and will probably invest in this book in the near future.

http://geneticsforherpers.com/

I do own the book. It is worth getting.

Second, I have been trying to figure out the probabilities of a double hey breeding and what the hets will be percentage wise. So its a double het caramel/pied to double het caramel pied.

I ad to memorize this in my univ genetics course.

Both pied and caramel are recessive to the respective normal gene. The results would be different if one or both mutant genes is dominant or codominant to the alternative normal gene.

+ = normal alternative to the caramel gene
ca = caramel mutant gene.
+//+ = two normal genes (normal)
+//ca = normal gene and caramel gene = het pied (looks normal)
ca//ca = two caramel genes (caramel appearance

+ = normal alternative to the pied gene
ca = pied mutant gene.
+//+ = two normal genes (normal)
+//pi = normal gene and pied gene = het pied (looks normal)
pi//pi = two pied genes (pied appearance

double het caramel/pied (+//ca +//pi) x double het caramel/pied (+//ca +//pi)
1/16 +//+ +//+ (normal)
2/16 +//+ +//pi (normal looking, het pied)
1/16 +//+ pi//pi (pied)
2/16 +//ca +//+ (normal looking, het caramel)
4/16 +//ca +//pi (normal looking, double het caramel/pied)
2/16 +//ca pi//pi (pied, het caramel)
1/16 ca//ca +//+ (caramel)
2/16 ca//ca +//pi (caramel, het pied)
1/16 ca//ca pi//pi (caramel pied)

Or:
916 normal (66% probability het caramel, 66% probability het pied)
3/16 caramel (66% probability het pied)
3/16 pied (66% probability het caramel)
1/16 caramel pied
The fractions are the probability of an outcome per baby. They are not the probability per clutch.