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Dihybrid Inheritance and More Now that you're a Master of monohybrid inheritance, let's move on! So far the genetics covered is pretty basic, taking into account only one gene locus and assuming it's occupied by a single allele, ceteris paribus. However, in reality, things are not too simple. What happens if you were to breed a DT royal blue to a DT turquoise sibling? In this case, 2 different gene loci will have to be considered - that for blue colouration and also tail-split. Such a calculation of genotypes and phenotypes is called dihybrid inheritance. As usual, we assume that the 2 gene loci are unlinked to simplify matters.
As usual, let's consider the above case: a DT royal blue male with a DT turquoise female. Male genotype: dt
dt, Bl bl Hence, by permutation again:
From the above, in the F1 generation, we get 100%
double-tails, among which 50% are turquoise in colour (Bl Bl) and the
remaining 50% are royal blues (Bl bl). Hope you got it!
To better appreciate the phenotypic ratio, let's do a pure-bred cross between 2 parents, both of which are heterozygous for both characteristics. It's also necessary to use completely dominant genes to illustrate this example, so let us use the NR gene. Let: Let's start off with 2 pure-bred parents, one of which is homozygous dominant for both characteristics, and the other which is homozygous recessive. Male
genotype: DT DT NR NR Resulting F1 will produce fry that are all Single tail and Red, but are heterozygous for both characteristics (DT dt NR nr). Then, let's do a F1 x F1 to produce the characteristic 9:3:3:1 ratio of a dihybrid inheritance. F1
Male genotype: DT dt NR nr (Single-tail het. for double-tail, Red het. for
yellow)
The resulting F2 generation will yield: 1/16
DT DT NR NR (Single tail, Red) By grouping the
phenotypes, we have the following ratios: The resulting phenotypic ratio is 9:3:3:1, where 9/16 resemble the original homozygous dominant parent that produced the original F1 generation (ie. the grandparents of these F2 fry), 1/16 resemble the original homozygous recessive "grandparent" and the other 3/16 and 3/16 ratios in the middle are recombinant characteristics of both grandparents. Contrast this to the typical 1:2:1 F2 phenotypic ratio that we have obtained in the typical monohybrid crosses!
Using the basic method outlined above, it is then possible to manually calculate the respective F2 phenotypic ratios of tri-hybrid inheritances, quad-hybrid inheritances and above. However, this manual method is extremely tedious and takes up a fair amount of time. Sometime last year, my friend Aaron and I sat down to work out all the higher order inheritances and attempted to find an easier method to predict the phenotypic ratios of the higher order hybrid inheritances. That we did! =) When I have the time, I shall update this section and share our derived and muuuccch easier method on how to predict the ratios! So ... you know what to do, don't you? Come back again!
So far, we have worked on parents that share exactly the same genes at the same gene loci. However, in the breeding of bettas, we often come across intriguing cases where different gene loci are concerned. Take for example the well-known case of infertile melano females. In order to produce melanos, most breeders breed melano males to steel blue females to produce steel blues that are heterozygous for melano. Also, blues bred to reds often result in multicolour blues having red wash. Theoretically, this is how calculations are done: Let Bl be the dominant allele coding for blue, and NR be the dominant allele coding for red. * denotes a wildcard - which can often be an empty gene locus. Male genotype: * * Bl Bl (Turquoise)
All F1 offspring produced will have a genotype Bl* NR*. OH NO. Here comes the problem. We're told that diploid organisms have 2 sets of alleles each, right? Then what happens when each cell of the F1 fry only have one allele for blue and one for red, and both at different loci? At this point, we call the F1 fry being "Hemizygous" for both blue and red. This means that the genes involved are only present once in the genotype and not in the form of pairs. Hence, at each gene loci, there will be no member whose genotype can mask or affect the expression of the gene. As a result, both the Bl and NR alleles will be expressed. This case is made even more interesting by the fact that both the Bl and NR alleles may be said to be codominant, as they are both expressed as strongly as one another. Hence, both will be expressed to varying degrees, resulting in the fry exhibiting both blue and red. (yes, and this is a pure-colour betta-lover's nightmare!) Of course, in the above example, we are again assuming that the blue and red colours are represented by only one single gene at one single locus. In reality, however, this is not the case. There are many other intrinsic factors that will eventually determine how an F1 fry in the above case will turn out. For example, other inherited properties might determine how much/how fast the various alleles are expressed, and even if the fish has a genotype of Bl* NR*, if the NR allele is expressed much more slowly than the Bl allele, then the copious amounts of blue pigment formed may actually mask the smaller amounts of red pigment formed. This may help to explain why in a cross like that of the above, there might be a small percentage of F1 fry that seem to appear solid blue. These "false solid blues" however, may be easily detected by shining a light through the fins, where the small amounts of red pigment may then show up. Another tell-tale sign of the hypothesis is that when these "false solid blues" are bred to each other, the F2 fry again may exhibit varying amounts of red. However, in this case, why do we not have "false solid reds"?? The answer might lie in the layers of pigmentation in bettas. The NR layer lies right at the bottom, while the Bl layer is right at the top. I then guess that whatever little blue pigment produced will always show up, thus refuting the possibility of producing these "false solid reds". Another example of how this type of hemizygous inheritance is purely theoretical and is subject to the interaction of other genes is the above-mentioned cross of a melano male with a steel blue female. All the F1 offspring will be of the genotype bl* m*. Since the blue layer lies atop the black pigmentation layer, all the F1 offspring should appear blue, or maybe exhibit small amounts of black, especially at the outermost edges of their fins. However, if these F1 heterozygotes were to be bred back to the father whose genotype is mm **, theoretically 50% will have a genotype of **mm and 50% m*bl*. The catch in this is that even the ** mm fish, which should theoretically show no signs of blue, will sometimes show a steel blue sheen on their bodies. This can then be attributed to the presence of either an unconsidered factor that codes for a sheen on the body, or a linked gene that has been inherited from either parent. Phew, this has been a lot to digest in one section, hasn't it? Let my fingers take a break for a while too, before I continue some other time! =) I've been requested to do a section on crowntail (CT) genetics as well .. I'm not really into CTs, but I'll try. =) Just promise you'll be back!
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