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Discussion Starter · #1 · (Edited)
Would someone please confirm that normal split for recessive pied **** bird x dilute hen would produce:

50% visual normals and 50% visual recessive pieds?

and

All would be split for dilute?

Thanks in advance. :D

(By the way, Mr. Blue (n/rp) x Nibbles (dilute) produced 4 visual normals and 4 visual recessive pieds: http://talkbudgies.com/showthread.php?t=33077. One of the normals has a pied thumbprint indicating he's split for recessive pied.)
 

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Don't know about the dilute but to obtain visual recessive pieds the gene must come from both parents meaning that your dilute hen is split for it as well.
 

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Actually I believe it would be 50% visual normal, and 50% visual normal split with recessive pied.

And yes, all would be split for dilute.
 

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Discussion Starter · #5 · (Edited)
Don't know about the dilute but to obtain visual recessive pieds the gene must come from both parents meaning that your dilute hen is split for it as well.
I've read the same thing frequently here, i.e., both parents must have recessive pied genes to produce a recessive pied baby. However, according to this site http://www.***************/gen_dilution.html, dilutes have two dilute genes:
When a budgie has two of the recessive dilute genes it shows the traits of dilute with about 70% washed out markings/color all over.
I think this means a dilute cannot be split for anything. If so, then I think the recessive pied gene (contributed by N/rp ****) dominates the dilute gene (contributed by d/d hen) to produce a visual recessive pied split for dilute. Make sense? :p
 

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Discussion Starter · #6 · (Edited)
Actually I believe it would be 50% visual normal, and 50% visual normal split with recessive pied.
If that were true, then all of Mr. Blue and Nibbles' babies would be visual normals. But 1/2 their babies were visual normals and 1/2 were visual recessive pieds.

And yes, all would be split for dilute.
I'm glad you agree. :)
 

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Would someone please confirm that normal split for recessive pied **** bird x dilute hen would produce:

50% visual normals and 50% visual recessive pieds?

and

All would be split for dilute?

Thanks in advance. :D

(By the way, Mr. Blue (n/rp) x Nibbles (dilute) produced 4 visual normals and 4 visual recessive pieds: http://talkbudgies.com/showthread.php?t=33077. One of the normals has a pied thumbprint indicating he's split for recessive pied.)
As has already been said: The dilute hen must be split for recessive pied as well because to get visual recessive pieds the gene must come from both sides. If both parents are split you'd only expect 25% recessive pied so you were lucky to get 50%. All the chicks will be split for dilute but only half of the non-resessive pieds will be split for recessive pied

Incidentally only a small percentage of birds that are split for recessive pied have the thumb print

Dilutes can be split for any other mutations that are not on the same allele. This includes all the sex-linked mutations and most of the recessive mutations except greywing & clearwing
 

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Discussion Starter · #9 ·
As has already been said: The dilute hen must be split for recessive pied as well because to get visual recessive pieds the gene must come from both sides. If both parents are split you'd only expect 25% recessive pied so you were lucky to get 50%. All the chicks will be split for dilute but only half of the non-resessive pieds will be split for recessive pied

Incidentally only a small percentage of birds that are split for recessive pied have the thumb print

Dilutes can be split for any other mutations that are not on the same allele. This includes all the sex-linked mutations and most of the recessive mutations except greywing & clearwing
Apparently, I'm confused about genetics. :hammer: I thought every bird has two genes. If those two genes are homozygous dilute, the bird is visually dilute. But if those two genes are heterozygous dilute and recessive pied, the bird is visually recessive pied split for dilute because dilute is recessive to recessive pied. Where am I wrong? :p
 

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You are wrong in that birds have millions of genes :p

Recessive pied and the three dilute genes (greywing, clearwing, dilute) are on different genes entirely.

However the three dilute genes are in different parts of the same gene, this is why they are semi-dominant and co-dominant to eachother with dilute being the least dominant of the three.
Normal > clearwing / greywing > dilute.

I think you are getting confused because most mutations state 'there is a normal gene, and a mutated gene'. What they are referring to is the normal variation of EACH mutated gene we are talking about.

Hope that doesnt just confuse you more lol i am typing in a rush!
 

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Apparently, I'm confused about genetics. :hammer: I thought every bird has two genes. If those two genes are homozygous dilute, the bird is visually dilute. But if those two genes are heterozygous dilute and recessive pied, the bird is visually recessive pied split for dilute because dilute is recessive to recessive pied. Where am I wrong? :p
Budgies, like other birds and animals have multiple genes. The pair of genes that makes a bird a recessive pied is not connected to the genes that makes it a dilute or a cinnamon or some other mutation. So it is possible for the bird to be all these mutations at the same time or be one of the mutations and be split for some or all of the others. Sometimes it may be difficult to see a second or third mutation because it is masked. For example it might be very hard to tell if a very pale dilute was also opaline because the markings would be very light but if the bird was bred the results would prove the presence of the mutation

Mutations like dilute and recessive pied are not dominant or recessive to each other, they are separate. When you calculate expected mutation outcomes each mutation is worked out individually then the results are combined.

For example if you mate a spangle to a dominant pied. Look first at the spangle gene and find that your pair should half spangle & half normal chicks:

Then you look at the dominant pied gene and find that they will produce half dominant pied and half normals

When the two are combined you find that half of the normals will dominant pied and half the spangles will be dominant pied as well as spangle

So the result is 25% spangle, 25% dominant pied, 25% normal & 25% spangle dominant pied combined
 

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Discussion Starter · #12 ·
I think you are getting confused because most mutations state 'there is a normal gene, and a mutated gene'. What they are referring to is the normal variation of EACH mutated gene we are talking about.
I'm confused because this site http://www.***************/gen_dilution.html shows punnet squares for homozygous and heterozygous genes, and when I replace "normal" with "recessive pied," I get 100% recessive pied babies. :p

How can a dilute split for recessive pied appear dilute when recessive pied is dominant to dilute???

(I need a "Budgie Genetics for Dummies" book.)
 

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Thats what i thought you were doing wrong. You cant replace 'normal' with recessive pied. What that is referring to is SOLELY about the DILUTE gene and its NORMAL original. The recessive pied gene is completely seperate, and there is another NORMAL varient of that gene.

You dont replace normal with another mutation, you do seperate punnet sqaures.

To find out chances of a normal split recessive pied X a dilute split recessive pied you seperate the different genes and do punnet squares for each.

For recessive pied split parents, each side would have (normal / recessive) X (normal x recessive). The punnet squares would then be 25% normal, 50% normal/recessive, 25% recessive.

THEN you do the dilute gene: (normal/normal) x (dilute/dilute) which = 100% normal/dilute.

So you can say from such a pairing that all chicks will be split dilute, but 25% will be normal (split for dilute), 50% will be normal split for recessive AND dilute, and 25% will be recessive split dilute.
 

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so basically i think what they are saying is that the dilute and the r pied are on completely seperate alleles. so you need two punnet squares one for each mutation.
 

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Discussion Starter · #15 · (Edited)
Thank you all, especially Nev and Dean, who've been very patient with me as I've asked the same questions several times. :p

I think I understand now: Each mutation gets its own punnet square. If I study the punnet squares on this site http://www.***************/genetics.html I hope I will fully understand.

I would also like to get firm definitions for the terms "split" and "masking" and how each term applies to genes/punnet squares.
 

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Optimus Prime
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well split means that it has one of each gene for a specific mutation. like a budgie that has one gene for green and one for blue would be green split for blue. green because the green is dominant over blue, and split for blue because it still has the gene for blue. so if it had babies it could pass on either the blue gene or the green gene, so potentially it could have blue babies. it is the same for dilute i think. a budgie can have a gene for dilute but if the other gene is for normal then it will look normal. but it would be split for dilute because there is one of each.

im not so sure about masking but i think it means that one gene is dominant over the other variation of the same gene and therefore masks it?
 

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Masking is when the bird has two mutations but the second mutation can't be seen because it is covered by the first one. For example if a bird was both lutino and dominant pied you wouldn't be able to see the dominant pied because the lutino gene had removed all the markings. So we say that the dominant pied is masked. But if a bird is both spangle & dominant pied there are enough markings to see both mutations and neither is masked

Split is when a bird has one copy of a gene that requires two copies to be visable. That is why if two birds with one copy of the same gene are mated together some of the chicks can get two copies of the gene and the mutation will be be visible
 

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Discussion Starter · #18 · (Edited)
Masking is when the bird has two mutations but the second mutation can't be seen because it is covered by the first one. For example if a bird was both lutino and dominant pied you wouldn't be able to see the dominant pied because the lutino gene had removed all the markings. So we say that the dominant pied is masked. But if a bird is both spangle & dominant pied there are enough markings to see both mutations and neither is masked
Thank you! :thumbsup: Sounds like "a polar bear in a snowstorm"; even though the polar bear (the mutation) is fully there, you can't see him (the mutation) behind the snow.
Split is when a bird has one copy of a gene that requires two copies to be visable. That is why if two birds with one copy of the same gene are mated together some of the chicks can get two copies of the gene and the mutation will be be visible
Fantastic! And the example here are my two budgies (normal and dilute), both of whom were split for recessive pied, thus producing some recessive pied babies. :D
 
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