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A back cross is not equal to breeding F1XF1.

back cross is done to create homozigous cattle out of recessive genes, by crossing an F1 with one of its parents. this is equal to inbreeding, while F1XF1 is not necessary inbreeding.

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if you cross an F1XF1 you get an F2 of 25% P1, 25%P2 and 50%F1 (these are the genotypes).

it's just like math, in theory. but if you want to reatain hybrid vigour, you must try to create a, what we call, seed-fast breed, meaning that when breeding from the cross, you always get +- the same genotype out of the f2. but this takes years and years of breeding, trying to retain the qualities that define that particular F1.

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I've always read and heard of back crossing as being breeding back to the breed of one of the parents. If you breed and F1 AngusXHereford back to an angus that is back crossing. If you breed an F1 to an F1 of the same composition, it's in affect backcrossing

dun

> A back cross is not equal to
> breeding F1XF1.

> back cross is done to create
> homozigous cattle out of recessive
> genes, by crossing an F1 with one
> of its parents. this is equal to
> inbreeding, while F1XF1 is not
> necessary inbreeding.
 
I'm not sure of the using of the terms is correct, backcrossing and/or back breeding, but at least what I understand is that a F2 is when the F1 is bred back to one of the parents.

> I've always read and heard of back
> crossing as being breeding back to
> the breed of one of the parents.
> If you breed and F1 AngusXHereford
> back to an angus that is back
> crossing. If you breed an F1 to an
> F1 of the same composition, it's
> in affect backcrossing

> dun

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if you bred F1 X F1 and they were the same breed compositions, then wouldnt that be double backcrossing. and then your F2s would genetically be the same as your F1s..and their heterosis or hybrid vigor would be zero b/c they are the same as their parents, the F1s. correct?

've always read and heard of back
> crossing as being breeding back to
> the breed of one of the parents.
> If you breed and F1 AngusXHereford
> back to an angus that is back
> crossing. If you breed an F1 to an
> F1 of the same composition, it's
> in affect backcrossing

> dun
 
sorry, but you're fundamentally wrong and right there. i will try to illustrate with the next small example.

this is fictive, and only to illustrate the general theory behind the thoughts, it doesn't represent the correct genotype of the cattle.

i will use Angus(P1) and Hereford(P2) as a silly example. both are homozigous for colour(angus is B, herford b) and for hornes (Angus is h, hereford is H)

100% of AngusXhereford is HhBb, therefore black and horned = F1

F1XF1 Black, horned = 9 = 57% red, polled = 1 = 6% Black, polled = 3 = 18.5% Red, Horned = 3 = 18.5% (please allow for rounding off of the decimals) this is also what is understood to be an F2 by most relevant litterature.

if i do a backcross with angus: F1 (BbHh)X BBhh i get the following:

Black, polled = 2 = 50% Black, horned = 2 = 50%

If i do a backcross with the hereford (bbHH), we get:

Red, horned = 2 = 50% Black, Horned = 2 = 50%

hence my statement that it is used to generate or retain recessive qualities from a cross with dominancy over that trait, without loosing all of the heterosis (relative to P1xP2) generated from crossbreeding

Back Cross is absolutely not equal to F1XF1.

i said backcrossing was equal to inbreeding because of the fact that to get a real (ie complete) backcross you must use a genotypically equal parent, ie the orriginal parent.

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I don't mean to be a fly in the ointment here but shouldn't those F1 Angus/Herefords all be POLLED? And going back Angus on them should also all be POLLED.

> sorry, but you're fundamentally
> wrong and right there. i will try
> to illustrate with the next small
> example.

> this is fictive, and only to
> illustrate the general theory
> behind the thoughts, it doesn't
> represent the correct genotype of
> the cattle.

> i will use Angus(P1) and
> Hereford(P2) as a silly example.
> both are homozigous for
> colour(angus is B, herford b) and
> for hornes (Angus is h, hereford
> is H)

> 100% of AngusXhereford is HhBb,
> therefore black and horned = F1

> F1XF1 Black, horned = 9 = 57% red,
> polled = 1 = 6% Black, polled = 3
> = 18.5% Red, Horned = 3 = 18.5%
> (please allow for rounding off of
> the decimals) this is also what is
> understood to be an F2 by most
> relevant litterature.

> if i do a backcross with angus: F1
> (BbHh)X BBhh i get the following:

> Black, polled = 2 = 50% Black,
> horned = 2 = 50%

> If i do a backcross with the
> hereford (bbHH), we get:

> Red, horned = 2 = 50% Black,
> Horned = 2 = 50%

> hence my statement that it is used
> to generate or retain recessive
> qualities from a cross with
> dominancy over that trait, without
> loosing all of the heterosis
> (relative to P1xP2) generated from
> crossbreeding

> Back Cross is absolutely not equal
> to F1XF1.

> i said backcrossing was equal to
> inbreeding because of the fact
> that to get a real (ie complete)
> backcross you must use a
> genotypically equal parent, ie the
> orriginal parent.
 
no, because i have stated that being polled is a recessive gene, if it isn't, then the outcome is polled, but my example was fictive, and only to point out that backcrossing is not equal to F1XF1, you can aplly this method to any gene-related trait and you will get the correct probability. the thing is you must know which ones are recessive and which ones are dominant. i have taken horned as dominant, because in my experience, it is so. if it is not so in an other breed, then you must make it HH for homozigous instead of hh, which i did, anyhow, the math stays the same, only your outcome will be different.

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I don't know what breeds you deal with, but Polled is dominant in all the breeds of cattle I know of. The African horn gene does throw in some strange curves, but I've never been able to find out how it is different.

dun

> no, because i have stated that
> being polled is a recessive gene,
> if it isn't, then the outcome is
> polled, but my example was
> fictive, and only to point out
> that backcrossing is not equal to
> F1XF1, you can aplly this method
> to any gene-related trait and you
> will get the correct probability.
> the thing is you must know which
> ones are recessive and which ones
> are dominant. i have taken horned
> as dominant, because in my
> experience, it is so. if it is not
> so in an other breed, then you
> must make it HH for homozigous
> instead of hh, which i did,
> anyhow, the math stays the same,
> only your outcome will be
> different.
 
> I remember a question you did ask
> me of what kind of problems I had
> with Gerts (it was posted in
> 11/Dec) And I told you hybrid
> vigor, maybe it is because Santa
> Gertrudis is a 3 breed composite
> so the Santa Cruz will have the
> same problem (hybrid vigor)in the
> future.

Not to be argumentative, but I believe Gerts are a two breed composite (3/8 Brahman and 5/8 Shorthorn), whereas Santa Cruz could be viewed as 3 or 4 breed composite since they are obtained by using 1/4 Gelbvieh, 1/4 Red Angus and 1/2 Gert (Brahman + Shorthorn).

( I'm a little too feeble to readily understand if more F2 heterosis is achieved via the use of 3 or 4 breed composite animals than with a two-breed composite's F2's. )
 
polled was just an example, a poor one, because i am not familiar with polled cattle (had one once and so i presume he was heterozigous for the trait, since all the calves came out horned (only had him breed three times).

OK, let's use the anti-myostatic gene, generating double muscled cattle. this is a recessive gene in real life. now just replace polled with double muscled, and horned with single muscled, and i will hopefully have proven my point.

it's just a sipmle example of how genetics work.

And i believe i have stated in the beginning that it was fictive, and definetely not a representation of real-life genotype.

it was just a way of proving that backcrossing is never ever equal in probable outcome to F1XF1.

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it is a 2 breed composite, sorry for the finger misprint
> Not to be argumentative, but I
> believe Gerts are a two breed
> composite (3/8 Brahman and 5/8
> Shorthorn), whereas Santa Cruz
> could be viewed as 3 or 4 breed
> composite since they are obtained
> by using 1/4 Gelbvieh, 1/4 Red
> Angus and 1/2 Gert (Brahman +
> Shorthorn).

> ( I'm a little too feeble to
> readily understand if more F2
> heterosis is achieved via the use
> of 3 or 4 breed composite animals
> than with a two-breed composite's
> F2's. )

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