# Sticky  Learning Genetics



## imp_Reptile Forums

A good place to start learning about genetics is from the pages provided by NERD. There are excelent discriptions and diagrams of how the genetics work using the punnet square.

Genetics Intro:
http://www.newenglandreptile.com/genetics_intro.html

Recessive Genetics:
http://www.newenglandreptile.com/genetics_simple_recessive.html

Double Recessive Genetics:
http://www.newenglandreptile.com/genetics_double_recessive.html

Co-Dominant / Dominant Genetics:
http://www.newenglandreptile.com/genetics_codom.html

Hope these are of some help to people who didnt know they were there.


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## cornmorphs

yeah thats a good site,that should help a lot.


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## tazzyb

I found the matrix page on Ralph Davis website a god send when trying to understand this genetics malarky.


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## cornmorphs

yeah it is hard work innit, what i did this year was breed almost all the same colours to each other, so i deffo lknow what i'm gonna get


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## SCOTT

I must be stupid reading all that sort of have a ide think i will in time just stick to 2 types the same for breeding,


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## cornmorphs

well that still wont gaurentee that you will get all the same babies lol


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## SCOTT

thankyou my confidence as just flown out the window, lol :? 
long as the little ones turn out well and mother is okay i will be happy


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## Guest

Thats the way to look at it Scott

I just let most of ours get on with it and get a surprise when the eggs hatch out way more fun  

Ryan


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## cornmorphs

if both parents are the same, and they are not het for anything then they will all be the same as the parents.
if the parents are both het for the same thing then you would get 25% of the total that would be the same as whatever the het was. the rest would all be the same as the parents, although they would all be 66% het for whatever morph the other babies turned out to be, lol
i'm not gonna read that again, i just hope t makes sense


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## Guest

This is where the problems come in though isnt it Nigel.You may well have a pair of carolinas but when paired up and the eggs hatch out you can have any number morphs with out knowing that they were even het.Then when you also go into ressesive traits you could end up with almost anything.

Im surprised that i havnt seen royals popping out of corn eggs with all the different genetics hehehehehehehe

Ryan


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## cornmorphs

well i have had a male anery for 15 years, i have mated him plenty of times.
last year i mated him to a normal female for the 1st time not knowing her genetics, she laid 11 eggs and i got 6 morphs, not bad eh


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## t-bo

cornmorphs said:


> well i have had a male anery for 15 years, i have mated him plenty of times.
> last year i mated him to a normal female for the 1st time not knowing her genetics, she laid 11 eggs and i got 6 morphs, not bad eh


Hehe, not bad at all!


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## Guest

See what i mean about the recessive and co-dominant traits.A lot of people seem to forgat all this when looking into genetics  

Great news about the unknown female though i bet you were well pleased

Ryan


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## cornmorphs

yeah i was well pleased. she gave an amel the year before out of a clutch which had just the one live. so i didnt even know she was het amel before that, or him for that matter
yeah, the ghosts they produced looked like lavenders when they hatched


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## Guest

mmmmmmmmmmm lavender ghosts i just love em.One of the prettiest corn snake morphs in my opinion

Ryan


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## cornmorphs

yeah, i was gonna keep him, 2 hatched but 1 died on the way out, so thats why he was gonna stay.
someone turned up a few weeks ago and wanted him a lot, and made me an offer i couldnt refuse, lol, i just hope the same pair produce one this year. she has just had her pre shed, so i'll be able to tell you in just over 2 months if all goes well


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## Guest

got a carolia dropping hers at the moment.She was paired with our creamsicle and the count so far is 23 and still going.  

The creamsicle female had a clutch of 12 but it looks like there are 2 or 3 that are unfertile  

Lets just hope they are ok and the rest hatch out.just keeping an eye on them and counting the days  

Ryan


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## t-bo

Hay congrats greenphase


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## cornmorphs

what else you breeding ryan?
i am half way through, i have 5 clutches and 5 to go


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## Guest

Just those this year as the anthys are to young at the moment but the carolina is a new one for us so we are not sure of her genetics we are also picking up 3 adult females next week.Carolina,amel and miami so they will get fed up for a second breed later in the year.

The final count for the carolina was 25 and all looking good

Ryan


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## cornmorphs

cool, i wont be doing any second breeds this year. katys 9 weeks pregnant so dont wanna have all the hatchlings at that time to be honest, although the money is always nice.


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## Guest

I know what you mean mate.The money is very nice but who wants or needs the trials of hatchlings when you also have to deal with a baby.Hope everything goes really well for you mate and if you do let stuff go give me a shout as and when you are looking to get back into it.You know i will treat you right.You have done me some good deals in the past the least i can do is return the favour  

Ryan


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## Simon

Another good Genetics info base can be found at www.ballpython.com. Certainly worth a look


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## beckys_dad98

Hey all
I had quite a good surprise the other night. I have had 2 males (lol) living together in the same tank for the last 5 years, one is 7, the other is 8... The one is an Everglades ratsnake, the other an albino Blackrat... Well anyway, assuming I don't have gay snakes, it turns out the Blackrat is a female because they have been breeding pretty good for the last couple days. I was sure the Blackrat was male, we probed it at the reptile show about 5 years ago and the probe went in pretty deep, guess was wrong...
Just curious though, anyone wanna hazard a guess as to what the babies will be???
Steve


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## Guest

errrmmmmmm.The best i can come up with is snakes lol


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## beckys_dad98

lol, funny... DUH!!!
Ya know what? I ain't even going to worry bout the genetics and what's het for what, I'm just gonna hope that the babies all turn out healthy...
Steve


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## Guest

Sorry Steve

I didnt mean any offence by that comment.I was just in a really weird mood yesterday and i couldnt for the life of me work out what you would get from that mating.

Whatever it turns out to be i would say you will get some stunning offspring in any case


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## beckys_dad98

My apologies too, Greenphase, I was in a bitchy mood too, had too much shit happening yesterday.... Anyway, figuring that Bubblegums are usually a three-way cross and the Everglades has Yellow blood already crossed into it, I'll probably get more Bubblegums.... Just have to see how the Blackrat part of the Bubblegum figures into the equation...
Thanks, Steve


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## Ssthisto

beckys_dad98 said:


> Just curious though, anyone wanna hazard a guess as to what the babies will be???


"North American Rat Snakes", het for the amelanistic strain found in Black Rat snakes - which may or may not be compatible with Corn Amelanistic.


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## mattb22

Basky it would be really nice if you could get me a male and a female mate but how are you going to be able to get them to me? 

I appreciate it loads mate :smile:


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## SiUK

why when I click on page two of my peta thread does this come up???????????


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## SiUK

I really wanna get the grasp of genetics but its seems so complicated to me I look at the links on this thread and its meant to give a basic understanding of it all in a simple way, and I feel stupid.

I suppose its taking one bit at a time if I try and learn the meanings for all the words on that first link and just as importantly the understand them so im not just rattling off something I dont understand, then that will be a good start.

You may find me asking a few stupid questions over the next few months so please bear with me.


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## Blazin

my favourite corn morph is Ghost Bloodred corn's. they look amamzing! I really want to get a snake, but wouldnt breed them cause this whole genetics, hets, etc is way to confusing and im tight on cash since im young lol. Cornmorphs, do you sell your baby corns onto Northampton reptile centre? they always have loads of little babies in them tubs. Dan


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## Ssthisto

SiUK said:


> You may find me asking a few stupid questions over the next few months so please bear with me.


The only stupid question is the one you didn't ask despite not knowing the answer.

Some people just take longer to pick things up than others.


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## gargoyle1980

I know the basic of genetics and working out double genetics - but how the hell do you do more than two? I usually use the FOIL technique but can't get my head round if there's more than two. Anybody care to elaborate?


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## Ssthisto

You just have to draw a much, much bigger punnett square.

Or work it out gene pair by gene pair.


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## gargoyle1980

but how do you work out the breakdown of the genes?


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## intravenous

gargoyle1980 said:


> but how do you work out the breakdown of the genes?


What do you mean by "breakdown?"


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## gargoyle1980

When you do a normal punnett you breakdown the genes into pairs - I use FOIL but that doesn't work on anything bigger - at least I can't get my head round it. So if I want to work out something that is het for two or more things for example normal het snow,lavender x lavender, how do I do this as I understand it would look something like this AaBbLl x llll. So how do I fill in the blanks for the lack of amel and anery on the one hand, and how do I break it down to pairs of the same gene?


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## intravenous

gargoyle1980 said:


> When you do a normal punnett you breakdown the genes into pairs - I use FOIL but that doesn't work on anything bigger - at least I can't get my head round it. So if I want to work out something that is het for two or more things for example normal het snow,lavender x lavender, how do I do this as I understand it would look something like this AaBbLl x llll. So how do I fill in the blanks for the lack of amel and anery on the one hand, and how do I break it down to pairs of the same gene?


I don't know what FOIL means but for the example you used:

You need to consider all the possible combinations that a parent can give its offspring:

The normal het amel, het anery (I'm assuming the anery is B?), het lavender (AaBbLl) can give 8 combinations:

ABL
ABl
AbL
Abl
aBL
aBl
abL
abl

The lavender (AABBll) can only give one combination:

ABl


Therefore there is only 8 possible combinations (8x1):

AABBLl
AABBll
AABbLl
AABbll
AaBBLl
AaBBll
AaBbLl
AaBbll


Does that help?


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## intravenous

intravenous said:


> You need to consider all the possible combinations that a parent can give its offspring:
> 
> The normal het amel, het anery (I'm assuming the anery is B?), het lavender (AaBbLl) can give 8 combinations:
> 
> ABL
> ABl
> AbL
> Abl
> aBL
> aBl
> abL
> abl


It's hard to explain how to do this bit exactly. Basically what I do is I pick one trait (in the example amel because I've listed it first) and keep it constant, varying all the other traits. 

So first I picked "A", then I picked "B" and varied lavender giving:

ABL
ABl

then still keeping "A", I picked "b" and varied lavender giving:

AbL
Abl

then you move onto "a".


Does that help? I know I'm not very good at explaining things .


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## toyah

> When you do a normal punnett you breakdown the genes into pairs - I use FOIL but that doesn't work on anything bigger - at least I can't get my head round it. So if I want to work out something that is het for two or more things for example normal het snow,lavender x lavender, how do I do this


I hate doing punnett squares for multiple traits - so I don't bother. There's easier ways.

IMO the easiest way to work it out is just breaking things down into the individual locuses, and then adding them back up.

The cross is: Normal het amel, anery, lavender x lavender. You then break this down into each individual locus.

1st: Het amel x not het amel: 50% het amel, 50% not het amel
2nd: Het anery x not het anery: 50% het anery, 50% not het anery
3rd: Het lavender x lavender: 50% lavender, 50% het lavender.

Now work out the offspring, using the percentages.
Normal het amel, Het anery, het lavender: 50% of 50% of 50%: 12.5%
Normal het anery, het lavender: 50% of 50% of 50%: 12.5%
Normal het amel, het lavender: 12.5%
Normal het lavender: 12.5%
Lavender het amel, het anery: 12.5%
Lavender het anery: 12.5%
Lavender het amel: 12.5%
Lavender: 12.5%

Add the percentages back up. ... if they don't hit 100%, you've missed one!


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## gargoyle1980

intravenous said:


> It's hard to explain how to do this bit exactly. Basically what I do is I pick one trait (in the example amel because I've listed it first) and keep it constant, varying all the other traits.
> 
> So first I picked "A", then I picked "B" and varied lavender giving:
> 
> ABL
> ABl
> 
> then still keeping "A", I picked "b" and varied lavender giving:
> 
> AbL
> Abl
> 
> then you move onto "a".
> 
> 
> Does that help? I know I'm not very good at explaining things .


So I don't need to break it up with sperm cross with egg like a punnett? I can just list all the possibilities and then put them back together again. That's easier.

Sorry Toyah - working with percentages will just get me even more confused lol.


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## intravenous

gargoyle1980 said:


> So I don't need to break it up with sperm cross with egg like a punnett? I can just list all the possibilities and then put them back together again. That's easier.
> 
> Sorry Toyah - working with percentages will just get me even more confused lol.


Well what I've done is sperm cross with egg...just in a different format :razz:. Say the normal het amel, her anery is the dad...then those 8 combinations I've put down are all the different sperm combinations. The single "ABl" would then be the egg combination.

Since the mother only has one possible combination she can offer her offspring then it can be done easily like this but if the mother had multiple hets then you would be better doing it in a punnet/table...thats not much harder though: you just list all the egg combinations down the way and all the sperm combinations along the way and cross them against each other. Its the same principle :smile:.

Toyah's method is probably the best for multiple hets (much less time consuming) but both will get there in the end.


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## GlasgowGecko

Hi guys,

Im currently doing my p.hd in genetics and have written an introduction to breeding genetics on my site. 

Hope it can be useful.


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## Harrison

Something that I think should be made clearer to people new to genetics:

WHen people discuss percentages, they can be misleading. By using percentages near words like "clutch", it can seem that 50% albino means that half of your babies will be albino. This is not true (although could happen of course). The percentage actually applies to every individual baby critter. So being told there is a 25% chance of being BLAH doesn't mean that out of 4 hatchlings, one will be BLAH. This is likely to happen but not for sure... Strictly speaking, the 25% applies to a single individual meaning that there is a 25% chance that the child will be BLAH. Some newbies think that if this first baby actually turns out normal, it increases the chance of the next birth to be BLAH. (You've probably noticed by now that BLAH is a fictional genetic trait). This is not true, as the next baby will also have a 25% chance of being BLAH.

Some people don't figure this out it seems because they often _do_ get the results in the way they expected, thinking the percentage applies to a clutch. Due to the statistics involved, you very often will get 3 normals and 1 BLAH if you are told you will get 25% blah. But this may happen sometimes and not other times. You may get all BLAH! Or all normal of course. This is because each individual baby goes through the process, not as a clutch.

Hope I explained that relatively well. Another way to look at it is this: A mother and father go to see a genetics counselor because they fear from their genetics that they will produce a child with a certain disease. They want to know what the chances are that the child will have the disease. The counselor may tell them that because of their genes, there is a 50% chance that the baby will have the disease. Now, many people go away from that, have a baby which unfortunately _does_ have the disease and then they assume that this makes the next baby less likely to have it. Wrong. Assuming we are talking about the same mother and father, the process is starting all over. The next baby also has a 50% chance of having the disease. The parents could have 8 babies all with the the disease or 8 babies without the disease. Because of statistics, it will most likely be relatively even but the point it that each individual child has the percentage applied to them and this does not directly affect any future births.


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## Mrs G

Whoever posted this.....thankyou!!!  I really need to knuckle down on this subject (to learn ANYTHING I have to read, re- read, write, then draw the subject!) It's great having all the links on one page 

I might ask a billion questions as I slowly go through it!: victory:


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## xmashx

the thing with the squares i dont get is that none of them give all the letter genetics. 
like for example a is for albino or aabb is for something else. i think it would be easier if someone puts up or knows the letter combos for colour and pattern morphs. 
: victory:
xsachax


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## eeji

the letters can be anything you want them to be, and everyone seems to use different variations


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## Harrison

xmashx said:


> the thing with the squares i dont get is that none of them give all the letter genetics.
> like for example a is for albino or aabb is for something else. i think it would be easier if someone puts up or knows the letter combos for colour and pattern morphs.
> : victory:
> xsachax


Generally, someone will explain at the start what "P" or "a" or "Jf" means before getting into it. If people just start showing you squares, they may be talking about bananas for all you know!


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## Lisasgeckonursery

really helpful link thnx, although don't know if my normals are het for anything :lol2: so may get some suprises one day


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## Velenon

good links and im sure ill be learnign all about this, but ATM thats making me go cross eyed!

Ill have to get my house mate to explain it to me, although she didnt like genetics very much on her course.


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## GRB

Seems like a good explanation of the basics of mendelian genetics. 

I didn't see it mention the possibility of linkage drag however - this is something that breeders will have to keep an eye out for as more animals are inbred. 

Basically, it is when a trait such as "pastel colour" is joined by another neutral, positive or negative trait that gets carried over during breeding. 

For example, you could breed a particular colour variety but also drag across a gene complex that cannot synthesise a particular protein, or that sensitises the animal to a certain disease. The more animals are inbred, the more they increase in homozygosity - if animals lack variation, it usually tends to increase the build up of negative alleles, so linkage drag of negative alleles could become a real problem. 

I think it would be nice to see some reference to more realistic genetics in the hobby - mendelian ratios are one thing, but they don't really cover all the potential damages that inbreeding can cause.


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## laura1486

Ouch, giving this a read hurts! Back to revision i guess, less testing :lol2:


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## cfcbulldog78

lol yes i think i need to go back to college lol


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## rmb87

SiUK said:


> I really wanna get the grasp of genetics but its seems so complicated to me I look at the links on this thread and its meant to give a basic understanding of it all in a simple way, and I feel stupid.
> 
> I suppose its taking one bit at a time if I try and learn the meanings for all the words on that first link and just as importantly the understand them so im not just rattling off something I dont understand, then that will be a good start.
> 
> You may find me asking a few stupid questions over the next few months so please bear with me.


 
I feel the same! It all seems so complicated, and just when i think ive grasped a bit of it, i find out something else that throws what i thought i had learnt :blush: Not that i have any plans to breed, just the genetic side of things fasinate me.


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## paulh

rmb87 said:


> ... just when i think ive grasped a bit of it, i find out something else that throws what i thought i had learnt ....


I learned genetics in a university course and working in the university genetics lab. Some of the links in this thread look like the blind trying to lead the blind. With the predictable result that they wind up sinking in the bog.  

If you can understand why there are four possible outcomes from flipping two coins (heads, heads; heads, tails; tails, heads; tails, tails), you can understand genetics. This is not a mental exercise; I recommend using a couple of coins. 

Pritzel's Genetics for Herpers is good, if you are willing to spend some money. Wilmer Miller's A Survey of Genetics is available for download at Wilmer Jay Miller's web site. There are other texts available on the web, too.


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## igmillichip

There are some pretty duff explanations of genetics on the web on snake and reptile sites.

I think that it is fine to show the actual results of crosses as that is useful information, but the attempts at explanations using Punnett square and simple mendelian genetics to 'help' futures breeders is a little short-sighted.

The fact is that when you make crosses with real animals, it is not just ONE gene that is a focus; and if you are considering two traits then it is more than 2 unlinked traits that are a focus in the real animal.

Personally, I feel that when one moves into the realms of breeding real animals then one has to move away from simplistic and somewhat confusing genetics.

Reading many posts on genetics, here and elsewhere, people say that they are confused.......if they are confused over the real science then that is one thing, but they seem to be confused over conflicting mis-information on some info on some links.

Empirical observations are good (and they are the basis of where genetics started), but let's move this on towards holistic predicatbility taking into account 'hidden' traits and linkages.

ian


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## Guest

I wrote a guide explaining the terminology for genetics with a few examples to help explain stuff. Planning on adding a calculator (once I've figured out how...). Anyway, thought I'd post a link here: Basic Snake Genetics - explanations of terminology and snake genetics as it might be of use to people.


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## paulh

Moonleh said:


> I wrote a guide explaining the terminology for genetics with a few examples to help explain stuff. Planning on adding a calculator (once I've figured out how...). Anyway, thought I'd post a link here: Basic Snake Genetics - explanations of terminology and snake genetics as it might be of use to people.


It's a start.

What was your source for "homozygotic" and "heterozygotic"? I'd like to check it out as I hardly ever see those words. "Homozygous" and "heterozygous" are much more common.

Some (hopefully constructive) suggestions:
1. Needs some editing. Some of the sentences are long and difficult to understand. 
2. Traits are what you see. They are the effects of the genes. Traits and genes are not the same thing.
3. In a heterozygous pair of codominant genes, the two genes do not interact. Each does its own thing. The combination of effects produces a trait that is different from the trait when either gene is homozygous.

Hope this helps.


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## Guest

paulh said:


> It's a start.
> 
> What was your source for "homozygotic" and "heterozygotic"? I'd like to check it out as I hardly ever see those words. "Homozygous" and "heterozygous" are much more common.
> 
> Some (hopefully constructive) suggestions:
> 1. Needs some editing. Some of the sentences are long and difficult to understand.
> 2. Traits are what you see. They are the effects of the genes. Traits and genes are not the same thing.
> 3. In a heterozygous pair of codominant genes, the two genes do not interact. Each does its own thing. The combination of effects produces a trait that is different from the trait when either gene is homozygous.
> 
> Hope this helps.


Thanks for the input. Don't really have a source for the terms hetero and homozygotic, they were just banded about by my genetics lecturers as an undergrad.

1. I'll take a look in to it thanks
2. I think the two things are essentially interchangable, although I think that is debatable.
3. Yeah I think I worded it poorly, but there is essentially an interaction going on. That said, I do think it could confuse the lay and will therefore edit it.


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## paulh

I wonder if the ****- and heterozygotic terms are a Brit thing. Oh, well, it's not important.

As for "trait". all I know is what I read in the dictionary.

trait = A genetically determined characteristic or condition. Traits may be physical, such as hair color or leaf shape, or they may be behavioral, such as nesting in birds and burrowing in rodents. Traits typically result from the combined action of several genes, though some traits are expressed by a single gene. (American Heritage Science Dictionary at Dictionary.com | Free Online Dictionary for English Definitions)


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## paulh

Whipped this up for another forum, trying to use the minimum of jargon. Thought I'd put it here, too, where it would be easily accessible for use and criticism. Adding a list of the mutants would be nice, but that is available at the JnB boas web site.

NO FRILLS GENETICS GUIDE

Definitions:
Genotype = the actual identity of the genes.
Phenotype = The creature's appearance and any other physical or behavioral manifestation. The phenotype is produced by all the genes and the environment working together.

Normal = wild type = 1. The most common phenotype in the members of a given species in the wild. 2. The most common gene found in a given gene pair in the members of a given species in the wild.
Mutant = 1. NOT the most common phenotype in the members of a given species in the wild. 2. NOT the most common gene found in a given gene pair in the members of a given species in the wild.

Homozygous = the two genes in a gene pair are the same. A heterozygous snake may have either a normal or mutant phenotype. It depends on the genes in the gene pair.
Heterozygous = the two genes in a gene pair are NOT the same. A heterozygous snake may have either a normal or mutant phenotype. It depends on the genes in the gene pair.

Every vertebrate has thousands of gene pairs. For simplicity, we ignore the homozygous normal gene pairs. This is like a repairman. He concentrates on the malfunctioning machine or machines in an assembly line and ignores all the machines that are working as expected.

Each sperm or egg gets one member of each gene pair. If the gene pair is homozygous, then each sperm or egg has the same gene. If the gene pair is heterozygous, then half the sperm or eggs get one gene and the other half of the sperm or eggs get the other gene. When a sperm fertilizes an egg, the gene pairs are reestablished.

Simplest case in genetics: All but one of the gene pairs are homozygous normal, to the best of our knowledge and belief. There is one gene pair of concern. There are two possible genes: A and a. These make three possible gene pairs: AA, Aa, aa. These make six possible matings:
AA x AA --> 1/1 AA
AA x Aa --> 1/2 AA, 1/2 Aa
AA x aa --> 1/1 Aa
Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
Aa x aa --> 1/2 Aa, 1/2 aa
aa x aa --> 1/1 aa
The fractions are the odds of a given outcome for each baby, not for a whole litter.

The appearance of the heterozygous (Aa) snake determines whether the mutant gene is dominant, codominant, or recessive to the normal gene.

If A is the mutant gene and a is the normal gene and A is dominant to a,
AA --> homozygous mutant phenotype
Aa --> same as the AA phenotype
aa --> homozygous normal phenotype

If A is the mutant gene and a is the normal gene and A is codominant to a,
AA --> homozygous mutant phenotype
Aa --> heterozygous mutant phenotype (can be distinguished from both AA and aa phenotypes)
aa --> homozygous normal phenotype

If a is the mutant gene and A is the normal gene and a is recessive to A,
AA --> homozygous normal phenotype
Aa --> same as the AA phenotype
aa --> homozygous mutant phenotype

Lets say there is a different gene pair of interest, the B gene pair. There are two possible genes: B and b. These make three possible gene pairs: BB, Bb, bb. The six possible B gene pair matings are the same as the six possible A gene matings -- simply put in B for A and b for a.

Often a mating has two or more gene pairs of interest. In such cases, determine the parental genotypes for each gene pair. Find the appropriate result for each single gene pair. Then match all the results of the second pair with each result from the first pair and multiply the fractions.

Example: Parental genotypes are Aa Bb mated to Aa Bb.
Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
Bb x Bb --> 1/4 BB, 2/4 Bb, 1/4 bb

Final result:
1/4 AA - 1/4 BB = 1/16 AA BB
1/4 AA - 2/4 Bb = 2/16 AA Bb
1/4 AA - 1/4 bb = 1/16 AA bb

2/4 Aa - 1/4 BB = 2/16 Aa BB
2/4 Aa - 2/4 Bb = 4/16 Aa Bb
2/4 Aa - 1/4 bb = 2/16 Aa bb

1/4 aa - 1/4 BB = 1/16 aa BB
1/4 aa - 2/4 Bb = 2/16 aa Bb
1/4 aa - 1/4 bb = 1/16 aa bb
(The fractions are the odds of a given outcome for each baby, not for a whole litter.)

For a third pair of interest, match all of the results of the third pair of genes with each result from the first two pairs of genes and multiply the fractions. And so on.

When the final genotypes are determined, convert them to the appropriate phenotypes.

From here on, it is mostly learning shortcuts and practice, practice, practice.

Here is one shortcut. Often people are primarily interested in the probability of getting one genotype among the offspring. For example, what is the probability of getting an albino anerythristic (AKA snow) baby boa constrictor from a het albino, het anerythristic (AKA het snow) mated to a het albino, het anerythristic (AKA het snow)?

A = normal, a = albino
B = normal, b = anerythristic

Parental genotypes: Aa Bb x Aa Bb
Snow genotype = aa bb

Aa x Aa --> 1/4 AA, 2/4 Aa, 1/4 aa
Bb x Bb --> 1/4 BB, 2/4 Bb, 1/4 bb

1/4 aa - 1/4 bb = 1/16 aa bb (snow)


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## cree

While trying to learn I found this helpful as I had real problems getting my head around the whole punnett square thing.
Genetics Practice Problems


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## smithdavid

And more details:
Genetic engineering research scientists, or biotechnologists, manipulate and modify the genes, or hereditary makeup, of microorganisms, plants, and animals. They are specialists in the field of genetics and conduct research in a broad range of biological sciences including biochemistry, botany, embryology, and microbiology. They have developed techniques with numerous important applications in the fields of medicine, agriculture, and animal husbandry.

​


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## paulh

chewy86 said:


> Hopefully there are more genetic wizards than useless at genetics people (like me) reading this. (Untill the threads loaded with knowledge atleast, then it needs stickying to help other numptys like me in the future. good referance if unsure on something in the future)
> 
> Long story short, Ive searched the forum and can only find info on the basic genetics, ie recessive genes x recessive genes and their hets, normals to and so on and get all of that. My problem starts when mixing in co-dom, dom, dbl co-dom and so on with other random multi gene morphs.
> If you are one of these gifted genetic wizards and have time to fully explain it start to finish, displaying tables and workings out of all possible genetic out comes, please do below. Of course i dont mean all morphs just how to work out any genetic task that I/others could encounter in the future (snakes is my main interest although im sure they all tie in)
> 
> Massive thanks in advance and hope this thread becomes an absolute gold mine of genetic facts.
> 
> Gav.


This post attempts to summarize a non-stickie thread named "Genetics explained (genetic wizards help needed, to fill in gap)".

A genetics wizard has one attribute. He or she understands that flipping a coin twice produces four possible outcomes -- heads and heads, heads and tails, tails and heads, and tails and tails. Flipping coins multiple times and recording the outcomes will show that to be true. The rest is just learning some jargon and practice.

See post #63 in http://www.reptileforums.co.uk/forums/genetics/2-learning-genetics-7.html for the definitions of genotype vs phenotype, normal (AKA wild type) vs mutant, homozygous vs heterozygous. "Dominant", "codominant" and "recessive" will be defined again here. These words are the basics of the genetics vocabulary. 

Every vertebrate has thousands of gene pairs. For simplicity, we ignore the homozygous normal gene pairs. This is like a repairman. He concentrates on the malfunctioning machine or machines in an assembly line and ignores all the machines that are working as expected.

Each sperm or egg gets one member of each parental gene pair. If the gene pair is homozygous, then every sperm or egg has the same gene. If the gene pair is heterozygous, then half the sperm or eggs get one gene and the other half of the sperm or eggs get the other gene. When a sperm fertilizes an egg, the gene pairs are reestablished.

A gene is not dominant, codominant, or recessive in a vacuum. A gene is dominant, codominant, or recessive in comparison to another gene. When there are two genes, gene1 and gene2, they make three possible gene pairs -- gene1//gene1, gene1//gene2, and gene2//gene2. The phenotype produced by the gene1//gene2 pair determines which gene is dominant, codominant or recessive to the other gene.

Dominance Relations (no standard of reference):
Gene1 is dominant to gene2, and gene2 is recessive to gene1 if
gene1//gene1 genotype produces phenotype1
gene1//gene2 genotype produces phenotype1
gene2//gene2 genotype produces phenotype2

Gene1 is codominant to gene2, and gene2 is codominant to gene1 if
gene1//gene1 genotype produces phenotype1
gene1//gene2 genotype produces phenotype2 (Phenotype2 may be roughly intermediate between phenotype1 and phenotype2, may show both phenotype1 and phenotype2, or may be different from the other two phenotypes in some other way.)
gene2//gene2 genotype produces phenotype3

As more and more genes were discovered, geneticists were getting confused. Different explorers were entering the Unknown Lands and mapping bits of it. Different maps had different scales, different reference points, and different terms. Combining different maps was hard. Geneticists needed a standard of reference to simplify that.

The wild type (or normal) genotype and phenotype became the standard. The wild type or normal phenotype is the most common phenotype in the wild. Normal genes in every gene pair produces the the normal phenotype. Any difference from the normal phenotype is assumed to by produced by one or more mutant genes. This model has served well.

Using the normal gene as standard changes the above dominance relations chart. Now, two genes, the normal gene and a mutant gene, make three possible gene pairs -- mutant//mutant, mutant//normal, and normal//normal. The phenotype produced by the mutant//normal pair determines whether the mutant gene is dominant, codominant or recessive to the normal gene.

Dominance Relations (normal gene is standard of reference):
A mutant gene is dominant to the normal gene if
mutant//mutant genotype produces mutant phenotype1
mutant//normal genotype produces mutant phenotype1
normal//normal genotype produces the normal phenotype

A mutant gene is recessive to the normal gene if
mutant//mutant genotype produces mutant phenotype1
mutant//normal genotype produces the normal phenotype
normal//normal genotype produces the normal phenotype

A mutant gene is codominant to the normal gene if
mutant//mutant genotype produces mutant phenotype1
mutant//normal genotype produces mutant phenotype2 (Mutant phenotype2 may be roughly intermediate between mutant phenotype1 and the normal phenotype, may show both mutant phenotype1 and the normal phenotype, or may be different from the other two phenotypes in some other way.)
normal//normal genotype produces the normal phenotype

If two mutant genes must have their dominance relation determined, then the older chart is used.

Abbreviations and symbols conserve bandwidth. In genetics, they were used as far back as Gregor Mendel's first paper. Any gene symbols can be used, as long as the user understands what they mean. However, uniformity helps communication. There are no standarized rules for all vertebrates, but the number of rat and mouse geneticists means that their guidelines will have a major influence if and when standard rules are adopted. Symbols in this post mostly conform to section 3.1 of the rat/mouse guidelines, MGI-Guidelines for Nomenclature of Genes, Genetic Markers, Alleles, & Mutations in Mouse & Rat. The two major exceptions involve superscripts. As this forum does not support use of superscripts, the character ^ sets off the superscript. In other words, a^p stands for a with the superscript p. And the wild type gene a^+ becomes + all by itself.

Packing the symbols for gene pairs together does not help readability. To minimize confusion, two gene pairs are separated by a space. Singleton genes in a sperm or egg are also separated by a space. The two genes in a gene pair are separated by a double slash, which stand for a pair of chromosomes. The symbol a//a means a gene pair made up of an a gene in one chromosome and a second a gene in the same location in the other chromosome. These conventions convert the run-together genotype T+a^pa into T//+ a^p//a.

All genetics problems are worked out the same way:
1. Identify the parents' genes. Symbolize them to save bandwidth.
2. Determine the gene(s) in the parents' possible sperm and eggs.
3. Determine the genes in all possible fertilized eggs. Use your preferred method -- the Punnett square, branching system, etc.
4. Combine identical genotypes, as necessary.
5. Assign a phenotype to each genotype.
6. Combine identical phenotypes and delete every unnecessary "normal".


Genetics Problem #1: A cross involving one gene pair. The mutant gene is recessive to the normal gene.

The snakes are royal pythons. Each has an albino mutant gene paired with a normal gene.

1. Identify the parents' genes. Symbolize them to save bandwidth.

Symbols:
a = albino mutant gene. Recessive to normal gene
a^+ = + = normal gene

+//+ genotype produces normal phenotype
+//a genotype produces normal phenotype
a//a genotype produces albino phenotype

Mating: +//a x +//a

2. Determine the gene(s) in the parents' possible sperm and eggs.

Half of the sperm have a + gene and the rest have an a gene. Half of the eggs have a + gene and the rest have an a gene.

3. Determine the genes in all possible fertilized eggs. Use your preferred method -- the Punnett square, branching system, etc.

Punnett square:

........+ ........a
+........ +//+.....+//a
a......... a//+......a//a

Branching system. The Punnett square's left column is the branching system's first (leftmost) column. The Punnett square's top row is the branching system's second column. 

..... + = +//+
+ <
..... a = +//a

..... + = a//+
a <
..... a = a//a

The results of the Punnett square and branching system are the same.

4. Combine identical genotypes, as necessary.

The genotype +//a and the genotype a//+ are the same. Here are the combined genotypes:
1/4 +//+ 
2/4 +//a 
1/4 a//a 

5. Assign a phenotype to each genotype.

1/4 +//+ genotype produces normal phenotype
2/4 +//a genotype produces normal phenotype
1/4 a//a genotype produces albino phenotype

6. Combine identical phenotypes and delete every unnecessary "normal".

The +//+ phenotype and the +//a phenotype are the same. Here are the combined phenotypes:
3/4 normal
1/4 albino

Exercize: Using paper and pencil, check the following matings for accuracy and determine the phenotypes:
+//+ x +//+ --> 1/1 +//+ 
+//+ x +//a --> 1/2 +//+, 1/2 +//a 
+//+ x a//a --> 1/1 +//a 
+//a x a//a --> 1/2 +//a, 1/2 a//a
a//a x a//a --> 1/1 a//a 


Genetics Problem #2: A cross involving one gene pair. The mutant gene is dominant to the normal gene.

The snakes are royal pythons. Each has pinstripe mutant gene paired with a normal gene.

1. Identify the parents' genes. Symbolize them to save bandwidth.

Symbols:
Pn = pinstripe mutant gene. Dominant to normal gene
Pn^+ = + = normal gene

+//+ genotype produces normal phenotype
Pn//+ genotype produces pinstripe phenotype
Pn//Pn genotype produces pinstripe phenotype

Mating: Pn//+ x Pn//+

2. Determine the gene(s) in the parents' possible sperm and eggs.

Half of the sperm have a + gene and the rest have a Pn gene. Half of the eggs have a + gene and the rest have a Pn gene.

3. Determine the genes in all possible fertilized eggs. Use your preferred method -- the Punnett square, branching system, etc.

Punnett square:

..............Pn.............+
Pn........Pn//Pn........Pn//+
+ .........+//Pn.........+//+

Branching system:
........Pn = Pn//Pn
Pn <
........+ = Pn//+

........Pn = +//Pn
+ <
........+ = +//+[/font]

The results of the Punnett square and branching system are the same.

4. Combine identical genotypes, as necessary.

The genotype +//Pn and the genotype Pn//+ are the same. Here are the combined genotypes:
1/4 Pn//Pn 
2/4 Pn//+ 
1/4 +//+ 

5. Assign a phenotype to each genotype.

1/4 Pn//Pn genotype produces pinstripe phenotype
2/4 Pn//+ genotype produces pinstripe phenotype
1/4 +//+ genotype produces normal phenotype

6. Combine identical phenotypes and delete every unnecessary "normal".

The Pn//Pn phenotype and the Pn//+ phenotype are the same. Here are the combined phenotypes:
3/4 pinstripe
1/4 normal

Exercize: Using paper and pencil, check the following matings for accuracy and determine the phenotypes:
+//+ x +//+ --> 1/1 +//+ 
+//+ x Pn//+ --> 1/2 +//+, 1/2 Pn//+ 
+//+ x Pn//Pn --> 1/1 Pn//+ 
Pn//+ x Pn//Pn --> 1/2 Pn//+, 1/2 Pn//Pn
Pn//Pn x Pn//Pn --> 1/1 Pn//Pn 


Genetics Problem #3: A cross involving one gene pair. The mutant gene is codominant to the normal gene.

The snakes are reticulated pythons. Each has tiger mutant gene paired with a normal gene.

1. Identify the parents' genes. Symbolize them to save bandwidth.

Symbols: 
T = tiger mutant gene. Codominant to normal gene
T^+ = + = normal gene

+//+ genotype produces normal phenotype
T//+ genotype produces tiger phenotype
T//T genotype produces super tiger phenotype

Mating: T//+ x T//+

2. Determine the gene(s) in the parents' possible sperm and eggs.

Half of the sperm have a + gene and the rest have a T gene. Half of the eggs have a + gene and the rest have a T gene.

3. Determine the genes in all possible fertilized eggs. Use your preferred method -- the Punnett square, branching system, etc.

Punnett square:

....,,...T .........+
T.......... T//T.......T//+
+......... +//T.......+//+

Branching system:

..... T = T//T
T <
..... + = T//+

..... T = +//T
+ <
..... + = +//+

The results of the Punnett square and branching system are the same.

4. Combine identical genotypes, as necessary.

The genotype +//T and the genotype T//+ are the same. Here are the combined genotypes:
1/4 T//T 
2/4 T//+ 
1/4 +//+ 

5. Assign a phenotype to each genotype.

1/4 T//T genotype produces super tiger phenotype
2/4 T//+ genotype produces tiger phenotype
1/4 +//+ genotype produces normal phenotype

6. Combine identical phenotypes and delete every unnecessary "normal".

Each genotype has a unique phenotype. No combining is necessary.

Exercize: Using paper and pencil, check the following matings for accuracy and determine the phenotypes:
+//+ x +//+ --> 1/1 +//+ 
+//+ x T//+ --> 1/2 +//+, 1/2 T//+ 
+//+ x T//T --> 1/1 T//+ 
T//+ x T//T --> 1/2 T//+, 1/2 T//T
T//T x T//T --> 1/1 T//T 


Genetics Problem #4: A cross involving two gene pairs.

The snakes are reticulated pythons. In the first gene pair, both parents have a tiger mutant gene paired with a normal gene. In the second gene pair, the male has a normal gene paired with a white albino mutant gene, and the female has a purple albino mutant gene paired with a white albino mutant gene.

1. Identify the parents' genes. Symbolize them to save bandwidth.

Symbols: 
T = tiger mutant gene. Codominant to normal gene
T^+ = + = normal gene

+//+ genotype produces normal phenotype
T//+ genotype produces tiger phenotype
T//T genotype produces super tiger phenotype

a = white albino mutant gene. Recessive to normal gene, codominant to a^p gene
a^p = purple albino mutant gene. Recessive to normal gene, codominant to a gene
a^+ = + = normal gene

+//+ genotype produces normal phenotype
+//a genotype produces normal phenotype
+//a^p genotype produces normal phenotype
a//a genotype produces white albino phenotype
a^p//a^p genotype produces purple albino phenotype
a^p//a genotype produces lavender albino phenotype

Mating: T//T^+ a^+//a (= T//+ +//a) x T//T^+ a^p//a (= T//+ a^p//a)

2. Determine the gene(s) in the parents' possible sperm and eggs.

Half of the sperm have a T gene and the rest have a T^+ gene. Half of the sperm have an a^+ gene and the rest have an a gene. This gives four types of sperm.


.......... a^+ = T a^+ = T +
T <
.......... a = T a

.......... a^+ = T^+ a^+ = + +
T^+ <
.......... a = T^+ a = + a

Half of the eggs have a T gene and the rest have a T^+ gene. Half of the eggs have an a^p gene and the rest have an a gene. This gives four types of eggs.

.......... a^p = T a^p
T <
.......... a = T a

.......... a^p = T^+ a^p = + a^p
T^+ <
........... a = T^+ a = + a

3. Determine the genes in all possible fertilized eggs. Use your preferred method -- the Punnett square, branching system, etc.

Punnett square. Row 1 corresponds to the four types of eggs. Column A corresponds to the four types of sperm.

................. Column A ........... Column B ........... Column C ........... Column D ........... Column E
Row 1..................................... T a^p................... T a.................... + a^p..................... + a
Row 2............. T +............... T//T +//a^p........... T//T +//a........... T//+ +//a^p...........T//+ +//a
Row 3............. T a............... T//T a^p//a............ T//T a//a........... T//+ a^p//a........... T//+ a//a
Row 4............. + +.............. +//T +//a^p.......... +//T +//a........... +//+ +//a^p...........+//+ +//a
Row 5............. + a............... +//T a^p//a........... +//T a//a........... +//+ a^p//a.......... +//+ a//a

A branching system can be drawn using the four types of sperm and four types of eggs. But it takes as much time and space as the equivalent Punnett square.

4. Combine identical genotypes, as necessary.

The genotype +//T and the genotype T//+ are the same. Cells B4 and D2 are the same. Cells C4 and E2 are the same. Cells B5 and D3 are the same. Cells C5 and E3 are the same. That makes the Punnett square's top right quarter identical to the bottom left quarter. Here are the combined genotypes and probabilities:
1/16 T//T +//a^p
1/16 T//T +//a
1/16 T//T a^p//a
1/16 T//T a//a
2/16 T//+ +//a^p
2/16 T//+ +//a
2/16 T//+ a^p//a
2/16 T//+ a//a
1/16 +//+ +//a^p
1/16 +//+ +//a
1/16 +//+ a^p//a
1/16 +//+ a//a

5. Assign a phenotype to each genotype.

1/16 T//T +//a^p produces super tiger, normal phenotype
1/16 T//T +//a produces super tiger, normal phenotype
1/16 T//T a^p//a produces super tiger, lavender albino phenotype
1/16 T//T a//a produces super tiger, white albino phenotype
2/16 T//+ +//a^p produces tiger, normal phenotype
2/16 T//+ +//a produces tiger, normal phenotype
2/16 T//+ a^p//a produces tiger, lavender albino phenotype
2/16 T//+ a//a produces tiger, white albino phenotype
1/16 +//+ +//a^p produces normal, normal phenotype
1/16 +//+ +//a produces normal, normal phenotype
1/16 +//+ a^p//a produces normal, lavender albino phenotype
1/16 +//+ a//a produces normal, white albino phenotype

6. Combine identical phenotypes and delete every unnecessary "normal".

The first and second genotypes have the same phenotypes. The fourth and fifth genotypes have the same phenotypes. And the eigth and nineth genotypes have the same phenotypes. Here are the combined phenotypes:

2/16 super tiger 
1/16 super tiger lavender albino 
1/16 super tiger white albino 
4/16 tiger 
2/16 tiger lavender albino 
2/16 tiger white albino 
2/16 normal 
1/16 lavender albino 
1/16 white albino


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## paulh

Shortcuts

A Punnett square for multiple gene pairs usually requires identification and adding of identical genotypes. That takes a lot of time. To minimize time and reduce mistakes, we change the 6-step method in the previous post to a 9-step method. There are more steps, but those steps go much faster.

1. Identify the parents' genes. Symbolize them to save bandwidth.
2. Determine the gene(s) in the parents' possible sperm and eggs for the first gene pair.
3. Determine the genes in all possible fertilized eggs for the first gene pair. Use your preferred method -- the Punnett square, branching system, etc. 
4. Combine identical genotypes, as necessary for the first gene pair. 
5. Assign a phenotype to each genotype for the first gene pair.
6. Combine identical phenotypes for the first gene pair. 
7. Repeat steps 2-6 for the second and each additional gene pair.
8. Combine all the genotypes, multiplying the fractions.
9. Combine all the phenotypes, multiplying the fractions, and delete every unnecessary "normal".


Genetics Problem #4 (9-step method): A cross involving two gene pairs.

The snakes are reticulated pythons. In the first gene pair, both parents have a tiger mutant gene paired with a normal gene. In the second gene pair, the male has a normal gene paired with a white albino mutant gene, and the female has a purple albino mutant gene paired with a white albino mutant gene. This repeats the last problem in the previous post in this thread.

1. Identify the parents' genes. Symbolize them to save bandwidth.

The gene symbols have already been assigned in the previous post.

Mating: T//T^+ a^+//a (= T//+ +//a) x T//T^+ a^p//a (= T//+ a^p//a) 

2. Determine the gene(s) in the parents' possible sperm and eggs for the first gene pair. 

Sperm = 1/2 T, 1/2 T^+ (= +) 
Eggs = 1/2 T, 1/2 T^+ (= +) 

3. Determine the genes in all possible fertilized eggs for the first gene pair.

Punnett square:

............T.........+
T........T//T.....T//+
+.......+//T.....+//+

Branching system:

..... T = T//T
T <
..... + = T//+

..... T = +//T
+ <
..... + = +//+

The results of the Punnett square and branching system are the same.

4. Combine identical genotypes, as necessary for the first gene pair. 

The genotype +//T and the genotype T//+ are the same. Here are the combined genotypes:
1/4 T//T 
2/4 T//+ 
1/4 +//+ 

5. Assign a phenotype to each genotype for the first gene pair.

1/4 T//T genotype produces super tiger phenotype
2/4 T//+ genotype produces tiger phenotype
1/4 +//+ genotype produces normal phenotype


6. Combine identical phenotypes for the first gene pair.

Each genotype has a unique phenotype. No combining is necessary.

1/4 super tiger 
2/4 tiger 
1/4 normal 

7. Repeat steps 2-6 for the second and each additional gene pair.

Sperm = 1/2 a^+(= +), 1/2 a
Eggs = 1/2 a^p, 1/2 a 

Fertilized eggs: 

Punnett square:

............a^p ............a
+........+//a^p .......+//a
a.........a^p//a .........a//a

Branching system:

........ a^p = +//a^p
+ <
........ a = +//a

........ a^p = a^p//a
a <
........ a = a//a

The results of the Punnett square and branching system are the same.

Combine identical genotypes for the second gene pair. There are no combinations. Here are the final genotypes and their probabilities:
1/4 +//a^p
1/4 +//a 
1/4 a^p//a 
1/4 a//a 

Assign a phenotype to each genotype for the second gene pair.

1/4 +//a^p produces normal phenotype
1/4 +//a produces normal phenotype 
1/4 a^p//a produces lavender albino phenotype
1/4 a//a produces white albino phenotype

Combine identical phenotypes for the second gene pair.

2/4 normal 
1/4 lavender albino 
1/4 white albino

If the problem has more than two gene pairs, steps 2-6 would be repeated here for the third gene pair. Then steps 2-6 for the fourth gene pair, and so on. 

8. Combine all the genotypes, multiplying the fractions.

Here are the results of the original two gene pair Punnett square:
T//T +//a^p........T//T +//a........T//+ +//a^p........T//+ +//a
T//T a^p//a.........T//T a//a........T//+ a^p//a.........T//+ a//a
+//T +//a^p.......+//T +//a........+//+ +//a^p.......+//+ +//a
+//T a^p//a........+//T a//a........+//+ a^p//a........+//+ a//a

Those results can be rewritten as follows:
+//a^p .......+//a ........| ........+//a^p........+//a
..........\ ...../ ..............| ..................\...... /
...........T//T ................| ...................T//+
........../ .....\ ..............| .................../ ....\
a^p//a ........a//a ........| ........a^p//a .......a//a
-------------------------------------------------------------
+//a^p .......+//a ........| ........+//a^p ......+//a
..........\ ...../ ..............| ...................\ ...../
...........+//T ...............| ....................+//+
........../ .....\ ..............| ..................../ ....\
a^p//a ........a//a ........| ..........a^p//a ......a//a

Each quarter of the original two gene pair Punnett square corresponds to a quarter of the T//+ x T//+ Punnett square. The +//a x a^p//a Punnett square is superimposed on each quarter of the T//+ x T//+ Punnett square.

That can be rewritten as a branching system. The results of the T//+ x T//+ Punnett square go in the branching system's first column. The results of the +//a x a^p//a Punnett square fork from each entry in the first column to make the second column.
.............+//a^p
............ +//a
T//T <<
............ a^p//a
............ a//a

............ +//a^p
............ +//a
T//+ <<
............ a^p//a
............ a//a

............ +//a^p
............ +//a
+//T <<
............ a^p//a
............ a//a

............ +//a^p
............ +//a
+//+ <<
............ a^p//a
............ a//a

This branching system is no advance on the equivalent Punnett square because identical genotypes must still be added. However, the final genotype results of the T//+ x T//+ Punnett square reduce from four to three entries: 
1/4 T//T 
2/4 T//+ 
1/4 +//+ 

To make the genotype combinations, use the final results of the T//+ x T//+ Punnett square as the branching system's first column. The final results of the +//a x a^p//a Punnett square fork from each entry in the first column to make the second column. If there is a third gene pair, that set of results forks from each entry in the table's second column, making a third column. If there is a fourth gene pair, that set of results forks from each entry in the third column of the table, making a fourth column. And so on. The final probabilities are determined by multiplying the fractions as we move from the base of the branch to the end of each twig. No further additions are needed.

................... 1/4 +//a^p = 1/4 x 1/4 T//T +//a^p = 1/16 T//T +//a^p
................... 1/4 +//a = 1/4 x 1/4 T//T +//a ... = 1/16 T//T +//a
1/4 T//T <<
................... 1/4 a^p//a = 1/4 x 1/4 T//T a^p//a = 1/16 T//T a^p//a
................... 1/4 a//a = 1/4 x 1/4 T//T a//a ... = 1/16 T//T a//a

................... 1/4 +//a^p = 2/4 x 1/4 T//+ +//a^p = 2/16 T//+ +//a^p
................... 1/4 +//a = 2/4 x 1/4 T//+ +//a ... = 2/16 T//+ +//a
2/4 T//+ << 
................... 1/4 a^p//a = 2/4 x 1/4 T//+ a^p//a = 2/16 T//+ a^p//a
................... 1/4 a//a = 2/4 x 1/4 T//+ a//a ... = 2/16 T//+ a//a

................... 1/4 +//a^p = 1/4 x 1/4 +//+ +//a^p = 1/16 +//+ +//a^p
................... 1/4 +//a = 1/4 x 1/4 +//+ +//a ... = 1/16 +//+ +//a
1/4 +//+ << 
................... 1/4 a^p//a = 1/4 x 1/4 +//+ a^p//a = 1/16 +//+ a^p//a
................... 1/4 a//a = 1/4 x 1/4 +//+ a//a ... = 1/16 +//+ a//a

This chart is the same as the final genotype chart in the original version of the two gene pair mating.

9. Combine all the phenotypes, multiplying the fractions, and delete every unnecessary "normal".

The final phenotype results from the T//+ x T//+ Punnett square become the branching system's first column. The final phenotype results from the +//a x a^p//a Punnett square fork from each entry in the first column to make the second column. If there is a third gene pair, that set of results forks from each entry in the table's second column, making a third column. If there is a fourth gene pair, that set of results forks from each entry in the third column of the table, making a fourth column. And so on. The final probabilities are determined by multiplying the fractions as we move from the base of the branch to the end of each twig. No further additions are needed.

........................... 2/4 normal .......... = 2/16 super tiger
1/4 super tiger <- 1/4 lavender albino = 1/16 super tiger, lavender albino
........................... 1/4 white albino ... = 1/16 super tiger, white albino

........................... 2/4 normal ........... = 4/16 tiger
2/4 tiger <-.......... 1/4 lavender albino = 2/16 tiger, lavender albino
........................... 1/4 white albino .... = 2/16 tiger, white albino

............................ 2/4 normal ........... = 2/16 normal
1/4 normal <- ...... 1/4 lavender albino = 1/16 lavender albino
............................ 1/4 white albino ... = 1/16 white albino

This chart is the same as the final phenotype chart in the original version of the two gene pair mating.

One of the nicest things about this method is that many questions can be answered without going through the full procedure.

There are an assortment of problems in posts 173-178 in http://www.reptileforums.co.uk/forums/genetics/258989-boa-genetics-outcomes-albino-anery-18.html that can be used for practice.


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