welkom op deze vernieuwde site

  


 
 
 

         
 
       
 	 
Buf explaned by Charles Pritzel

Evidence for the inheritance pattern of the Buf gene.


A few of my friends have asked me for clarification on this subject, so I have made this condensed/digested version to hopefully make it clear why the buf gene is listed in the CMG along with other known genes. I think it should also be noted that the goalposts for this gene have been set much higher than other similar genes which were included in recent years without any controversy. Regardless of this double standard, the evidence collected by Jan Notte over the past few years is very compelling.

Following are reasonable hypotheses which could be proposed to explain the "buf" phenomenon, and what the evidence says about each of them.


Buf is recessive Mendelian.
Status: Falsified.
Reason: Crossing Orange X Buf het amel produced a clutch of 8 hatchlings, 2 of which were non-buf. If buf were recessive, both of these parents would have to be homozygous, and thus the entire clutch would have been homozygous for the same. Another cross of the same produced 6 eggs, 4 of which were non-buf.


Buf is dominant (or codominant) Mendelian.
Status: Confirmed.
Reason: Eleven separate clutches producing 44 buf offspring out of 144 total. The 21 which resulted from buf X buf crosses did not show any obvious third (homozygous) phenotype so it is a reasonably safe conclusion that it is not codominant, although further evidence could appear at some future date. If so, we can consider it codominant and carry on.


Buf is not Mendelian.
Status: Falsified.
Reason: There is clear on/off expression of this gene in clutchmates, which occur in predictable ratios of offspring.


Buf is the phenotype of snakes het for caramel.
Status: Falsified.
Reason: First, Buf corns do not resemble normal hatchlings het for caramel, the saddle colors are obviously different. Next, a cross between buf X butter het motley produced 5 orange and 4 amel hatchlings. Every one of these nine hatchlings was het for caramel, so if the phenotype were a result of being het for caramel, the entire clutch would be "orange" corns. Golddust Motley X Orange produced 19 hatchlings between first and second clutches, 8 of which were buf or orange and 11 of which were not buf or orange. The same reasoning applies as did with the butter het motley. Finally, the parents of the 2001 buf matriarch were normal in appearance. If one of them was het caramel to pass this gene down to the 2001 female, they would have looked "buf" too.


Buf is the phenotype of snakes homozygous for caramel.
Status: Falsified.
Reason: See "Buf is recessive" above. If these snakes were homozygous, buf X buf would produce all buf, no non-buf. Results from two clutches exclude this hypothesis by producing non-buf hatchlings.


Buf is an allele to caramel.
Status: Undetermined.
Reason: A few crosses have been performed which include buf on one side and caramel on the other, but so far no conclusive evidence for or against allelism has surfaced. A conclusive test would be het buf het caramel or orange (amel het buf) het caramel x anything homozygous for caramel including butter, amber, golddust, honey, etc. The results of such a breeding would include non-buf, non-caramel hatchlings if buf and caramel do not share a locus. If they share a locus, all hatchlings will express buf or caramel since neither parent would have a wild-type allele to throw to any of its offspring. A significant number of hatchlings would then provide a statistical argument for an allele hypothesis. This is illustrated in the Punnett squares below. It should also be noted that since buf is dominant to wild type, and to caramel if it is an allele (known buf het caramel exhibits the buf phenotype) it is not nearly as important in most cases to know whether or not they share a locus, because it is not necessary to match up buf with caramel, as it is with recessive genes to get them to reappear from hets.
Click image for larger version    Name:	bufhetcarameltest.gif  Views:	148  Size:	9.9 KB  ID:	29818


Buf needs to be crossed to a caramel in order to conclude testing.
Status: False.
Reason: A simple punnett square shows that Buf x caramel (or butter, etc) will produce exactly the same prediction regardless of whether it is assumed to be an allele to caramel or not an allele to caramel. Buf X caramel will not produce data that wasn't already produced by buf x butter and buf x golddust.

Click image for larger version    Name:	bufcarameltest.gif  Views:	153  Size:	7.0 KB  ID:	29817

BTW if anyone wants to discuss the evidence or its meaning please feel free to post. I doubt there's much left, though.

Also, let's try to keep in perspective that we are not mapping a genome, we are determining outcomes of breedings. That is the whole purpose of "proving" a gene: to answer the questions of "what will I get if I breed this to a ____?" We don't need to know the base-pair sequence or which chromosome it's on in order to make reasonably accurate predictions.

In the case of buf, the data is sufficient to describe all breedings except one. We already know what will result from het buf X anything: 50% buf corns, and homozygous buf X anything = 100% bufs. The one single question we do not know at this time is, "what is the result of buf het caramel x caramel?" and we do know that the answer is either going to be 50% buf, 25% caramel, 25% normal, or 50% buf, 50% caramel. The two results are similar enough that in many cases "bad luck" is as much or more of a variable.

There are definitely all kinds of interesting side-questions that we can ask, but the answers to those questions don't affect the outcome in the vast majority of breedings, and the same questions remain unanswered about amel, ultra, hypo, strawberry, sunkissed, lava, dilute, anery, charcoal, caramel, lavender, cinder, kastanie, diffused, masque, piedsided, motley, stripe, terrazzo, tessera, stargazer, and short-tail.