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Post by TJ's Rodent Ranch on Jan 30, 2024 12:16:15 GMT -8
Sorry, I've been away from the forum for a while now. Yesterday I had an intensive violin retreat going on, which was lots of fun but had absolutely no internet, and therefore, would not let me check the forum! My brain has gone rather numb recently... however, I did read your posts, and I think I understand them for the most part. I'll come read them again as soon as I'm feeling a bit better and can form a coherent thought about it. I just wanted to reply so that you know I wasn't ignoring this thread
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Post by betty on Jan 31, 2024 10:50:24 GMT -8
Excellent update.
Genetics (unfortunately) is something that always requires reading and re-reading - because it builds on triggers and steps. Once you click on to a new step - you'll realise that more of the last thing you read about it could have made sense - and want to re-read it, etc.
And that would be true even if you only had genetics to think about - but looks like you have a few other things competing for headspace right now.
Plus: that's the great thing about a forum though I suppose - it isn't time sensitive and it is widely accessible.
No rush to some back and read it right now - come when you are ready...
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Post by TJ's Rodent Ranch on Feb 8, 2024 12:21:32 GMT -8
Hi again!
So I went back and read your past posts. I don't think I have any specific questions regarding them, only ones that had really sprung up in the past few weeks.
Beggining with the Agouti and Non-agouti side. So, on the subject of being able to get non-agouti pups from Agouti side parents, and the like. So if you had two Agouti side parents... (just writing down these nomenclatures rather randomly).
The Father, Agouti: AA C* D* E* P* Uw* The Mother, Agouti: AA C* D* E* P* Uw*
So you could only ever get Agouti from this pairing, right?
Where as, if you had one little a in the mix, which would look like: (I'm not actually sure if this nomenclature would work...)
The Father, Polar Fox: Aa C* D* ee Uw* The Mother, Agouti: AA C* D* E* Uw*
Then roughly 25% of the pups would be Non-agouti side?
Or....
the father, Agouti: AA C* D* E* Uw* The mother, Black: aa C* D* E* Uw*
Would produce 50% percent Agouti side?
Etc...
My other question...
I was wondering if we could build a list of Agouti and Non-agouti twins... There are a few colors (well, more than a few, haha) that still are a little odd to me, and I think knowing their twins might be helpful...
So, I maybe a few to the list?...
AA - aa
Agouti - Black Nutmeg - Honeycream Argente - Schimmel...?
some of these are complete shots in the dark, so I'm really not sure. I was wondering if you could help me expand on this tiny list...
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Post by betty on Feb 9, 2024 9:32:33 GMT -8
Hi again! Yay - you're back in the genetics game!!!
So I went back and read your past posts. I don't think I have any specific questions regarding them, only ones that had really sprung up in the past few weeks.
Beggining with the Agouti and Non-agouti side. So, on the subject of being able to get non-agouti pups from Agouti side parents, and the like.
So if you had two Agouti side parents... (just writing down these nomenclatures rather randomly - nicely done).
The Father, Agouti: AA C* D* E* P* Uw* The Mother, Agouti: AA C* D* E* P* Uw*
So you could only ever get agouti from this pairing, right?
Yes - because of the double-dominant all the offspring would be agouti-colored (not all the pups would be the color Agouti of course, because you don't know what all those *s are). If any parent has double-dominant at the gene you are investigating (in this case AA) - then 100% of their pups will have to display that. 100% of the pups in this instance will be an agouti color. However, as BOTH the parents in this example are double-dominant then 100% of their pups will also be double-dominant as well as agouti-colored.
The reason is because the single A will ALWAYS produce an agouti-side pup - and these parents have 2 As. Whichever of their 2 genes the parents give to their offspring - it will always be the dominant A.
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Post by betty on Feb 9, 2024 10:15:36 GMT -8
I have put this second part separately because although it follows the exact same logic pattern as the above example - and although your stats breakdowns of 25% and 50% are often quoted for punnet squares - the examples you use are causing the sums to go wrong. This can happen when the genes are counted individually - rather than added together into a single creature.
Always remember that you need 2 of the recessive versions of the gene present INSIDE EACH PUP to make it be seen outside of each pup. The genes have to have paired up from both parents - 1 recessive gene from each parent. ___________________
So, in your first example here - there is only 1 single recessive a gene in the whole pairing: you have 3 large As and only 1 little a. However, as we know we need 2 little a genes to make a non-agouti pup - it looks like that would be impossible here.
The Father, a DEH: Aa C* D* ee P* Uw* The Mother, Agouti: AA C* D* E* P* Uw*
Then roughly 25% of the pups would be Non-agouti side? Sadly, in this instance: 0% of the pups would be a non-agouti color. Every pup would inherit a dominant A from their mother because she is double-dominant - and the double-dominants hide anything else. The recessive a from the dad would still get passed on to half the pups - but just like with him - we can't see (the actions of) it.
If you had chosen the example below - then you would expect 25% to be non-agouti (because both the mum AND dad have a small a to pass on to half their pups - but half of these would be paired with a A and so it leaves only a quarter (25%)):
The father, Agouti: Aa C* D* E* Uw* The mother, Agouti: Aa C* D* E* Uw*
____________________
Your second example also falls foul of the '2 recessive genes' problem AND the double-dominant problem. Although in this example there are definitely 2 recesive genes present - they are both in the same parent (the Black mother). As the mother can only give 1 of her 2 genes to each pup - the pups will only ever receive one each - which is 1 short of what they need. The father however, will only need to give one of his dominant As to each pup to make them agouti-colored, and as he only has dominant As - all his pups will get one.
the father, Agouti: AA C* D* E* Uw* The mother, Black: aa C* D* E* Uw*
Would produce 50% percent agouti side colors? No, sadly. In this instance, 100% of the pups would be an agouti color. There would be no non-agoutis.
If you had chosen the example below - then you would expect 50% to be non-agouti (because the dad can only pass on the dominant A to half his offspring - the other half would be free to be aa:
The father, Agouti: Aa C* D* E* Uw* The mother, Black: aa C* D* E* Uw*
Has that just totally thrown a spanner in the works? Hopefuly not...
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Post by betty on Feb 9, 2024 10:42:41 GMT -8
"My other question...
I was wondering if we could build a list of Agouti and Non-agouti twins... There are a few colors (well, more than a few, haha) that still are a little odd to me, and I think knowing their twins might be helpful...
Absolutely - I think that you compiling the twins list yourself as a new thread (that you keep going into and out of to edit as you find more) will be super helpful.
If you brain works a bit like mine does - than I found that writing all the codes down and making charts in a notebook over and over was how I remembered them all.
You need to follow the correct patterns though - otherwise it won't make sense as it gets bigger. You will also learn at the same time what to ignore and what is essential; so for example - your AA only needs to be A* - because you only need the one A for the colour to show:
A* (agouti-based color) vs aa (non-agouti color)
Additionally - you always need to make sure that the A* one is always on the same side - otherwise you will get your own head muddled up as to who is agouti-based and who is black-based (or self) Agouti - Black (spot on - the two basics)
Nutmeg - Dark-Eyed Honey (these are the wrong way around - Nutmeg is on the black-based side)
Argente - Schimmel...? (Schimmel isn't A* or aa - it is an e gene that lays on top of either)
You will have so much fun (and exasperation) as you build your twins list!
Remember some simple rules for your twins list:
1) You are ONLY changing the A* to aa each time - all the other genes should be identical in each twin:
A* C* D* ee pp Uw* - Red-Eyed Honey aa C* D* ee pp Uw* - Saffron
2) Use * for all the genes that don't matter - only those that actually affect the colour (and change the color name). This makes it much simpler - and shorter:
Yes: aa C* D* ee pp Uw* - Saffron No: aa Ccch Dd eef pp Uwuw - still just a Saffron on the outside (but not all saffrons will carry those recessives)
3) Make sure you leave space for all the different combinations that change colour in your chart - there are actually more than you think. Several of these gaps will be filled with the same colors - even though the genes are different - but at least you know you have everything covered in there.
4) Enjoy it - if it's not working out or not fun - take a break. You can't learn something that is annoying you!
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Post by TJ's Rodent Ranch on Feb 9, 2024 14:16:05 GMT -8
Thank you so much! No it definitely hasn't thrown a spanner in... it makes lots of sense now. So, for the E gene, for example, would this be true?
the parents pass one of their E genes on to the pups, and it would split evenly. So if you had a parent with EE and a parent with ee, then you'd get all Ee pups? There might be things about the E gene specifically that I'm not thinking of... but would this theory of mine work with the A gene? Or is it incorrect all together?
Okay, I think I'll do that. I think it might be rather interesting to focus in on a color, then change the aa to A* or A* to aa, and figure out what that color is. Thank you again!
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Post by betty on Feb 10, 2024 11:26:01 GMT -8
It is 100% the same for any gene - the gene itself doesn't matter per se - the logic and pattern in inheritance are always the same for these simple genes: you only get one from each parent.
If I take your above sentence - it would be true for everything:
if you had a parent with EE and a parent with ee, then you'd get all Ee pups if you had a parent with AA and a parent with aa, then you'd get all Aa pups if you had a parent with PP and a parent with pp, then you'd get all Pp pups if you had a parent with DD and a parent with dd, then you'd get all Dd pups etc
Don't forget though, we are only working here on the theory of gene inheritance for single genes - learning how they all add up together to create different (named) colors in actual gerbils is a combination of these single genes. Creating gerbil colors uses these simple rules - but layers up in real life.
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Post by TJ's Rodent Ranch on Feb 10, 2024 22:19:55 GMT -8
Okay, it's starting to make more sense!
I'm re-reading Shooting Star's page again, so I'll probably have more questions in a few days...
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Post by betty on Feb 11, 2024 12:15:17 GMT -8
Great news.
It is definitely like that - it all builds on what you read before - even when what you read before didn't seem to make any sense at first. Or you thought of it in one way - only to learn the next part and then the first part now seems to make even more sense!
I loved it all - certainly was a little project once it got started. Hit a few hurdles, and found out why they didn't make sense - then everything came together.
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