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Essential Genetics for the Terrified
by Chris Rutt

Part 5: Multiple Genes (2)

You were asked at the conclusion of Part 4 to consider what phenotype is produced by the two gene pairs, "Marine:Green" "Dark:Not dark". I do hope you took the trouble to do that and are not reading straight through! We have established that "Green" is dominant to "Marine", so all chicks will be of the green series. We know that "Dark" is an incompletely dominant gene, and that the mixture "Dark:Not dark" will produce an intermediate shade dark birds, which when combined with Green are known as "Jades" (and combined with Marine are known as "Cobalts"). I hope you did not find that difficult.

Now we will mate two birds, each of the "Marine:Green" "Dark:Not dark" make up. Before considering the results, I will introduce you to another piece of shorthand which will simplify things now that we are dealing with the inheritance of more than one character.

Each GENE can be represented by a single letter. A capital letter indicates a dominant, or partially dominant, gene, and a small letter a recessive gene. There is a convention for the letters used for each characteristic, which I will deal with later in these articles, but for the moment I will use letters which I hope will be more easily understood by the beginners for whom these articles are intended. Going back then to the example of Olive x Marine, we can write out their genetic make up as DDMM and ddmm respectively.

One gene from each pair will therefore always produce DM from the Olive and dm from the Marine. Each chick will therefore be of the composition DdMm. We know from our previous examples that Dd produces a bird of intermediate "darkness" and Mm produces a Green, so our chicks from this pairing are all "Jades" split for "Marine". Now to consider the next generation, where we mate DdMm to DdMm.

What are the possible results of randomly choosing one "D" gene and one "M" gene from each pair in each bird? You may get:- DM or Dm or dM or dm, so this is the range of gene combinations that can be passed to the offspring. In this case the results for the two parent F1 birds will be the same. The simplest way to obtain the results of this mating is to set out a grid, with the possible gene contributions for one parent across the top as column headings, and from the other parent, at the start of each line

Now fill in each box of the grid by taking the Gene set from the left end of that row and the set from the top of that column. Do it like this....

Now we have to count how many different combinations we have in the chicks. We have 16 seemingly different combinations, but note carefully that, IN THESE CASES, the ORDER of the genes in each pair does NOT matter. That is to say that Dd and dD, or Mm and mM produce the same effect in the phenotyope bird's colour. We want to work out what colours are being produced - that is after all the object of learning genetics!

There are various ways to go about this, but try this one. Count how many of the boxes (each of which of course represents a possible chick) have AT LEAST ONE "M". All the boxes in the first and third ROWS and the first and third COLUMNS. That makes 12 of the sixteen. Don't count boxes twice; remember that some boxes are in the first row AND the first column, and so on.

Now remember back; "M" means Green, and "M" is dominant so all these birds AND ONLY THESE birds, will be green. The only birds which are Marine are those with "mm" and you could go back and look for those on the grid - though simple mathematics tells us that if 12 of the 16 are green, and the only other colour we are examining is Marine, then there are 4 with the "mm" combination. Now to look at the Dark factor. It will be simpler to look at those four Marine birds first. To help you they occur in the second and fourth rows and the second and fourth columns.

You will find that of these, one has NO "D"s, two have ONE "D" and one has 2 "D"s. Translating these into phenotypes (a quick last reminder; "phenotype" means actual appearance to the eye) then, we have one Slate (2 dark factors), two cobalts (1 dark factor) and a Marine (NO dark factors). Among the green birds, how many will be Olive? Try to work this out for yourself before proceeding, while I nip off for a cuppa³!

Now I hope you have remembered that only birds with two "D" genes can be Olive. Therefore you will NOT expect to find them in the third or fourth row, nor the third or fourth column, as these represent cases where one or other parent has contributed a "d". There are in fact three such birds - in row 1, columns 1 & 2, and row 2, column 1. (Row 2, column 2 has two "D"s, but also 2 "m"s so this is the Slate we already found.) How many Light Greens are there? They must not have any "D" in them of course, and AT LEAST one "M". They are all to be found in the block of four on the bottom right of the grid, and there are THREE such. Note that the fourth in that block is the single Marine bird we have already counted. Of the 12 green birds we found, we now know that 3 are Olive and three are Light Green, so the other six are Jade, having just one "D" and either one or two "M"s. Go back and check!

There is another aspect which will be of interest to those breeding for colours. How many birds are capable of passing "Marine" to some of their offspring? They must have the composition "Mm" (or "mM") or "mm". The answer is 12. To round off this section, we could consider a more realistic way of expressing the numbers as proportions. In everything we have dealt with thus far, it all boils down to the two ratios, 3:1 or 1:2:1. In the complex situation we have just examined we found 3:1 Green:Marine. We can also say we have 3:1 ratio of birds which are capable of passing on Marine to those which cannot (i.e are pure green). We also found 1:2:1 in the Olive:Jade:Light Green AND in the Slate:Cobalt:Marine proportions.

So can we tie up these ratios to specific types of mating and inheritance? Think about this one for a while and review the earlier parts of the series, while I attempt to write the next part!

³"cuppa" = "a cup of tea"

Part 6: Alleles

All rights reserved © 1997-2000 Chris Rutt