Copyright information:  owned by Cat World expiry date 70 years from July 1973 at which time this information may be placed in the public domain minus the research notes.
Research note:  Here color is used to emphasize information presentation, bars are used to break the information into chunks and pictures have been added to the technical explanations to enhance the information presentation and make the point clearer.  Question:  does using colored cats to express the concept aid in comprehension.

College of Cat Genetics: Part III
by Patricia Turner
 

Study Unit 3
 


Mendel chose the garden pea in his experiments probably because plant breeding was his hobby as much as cat breeding in the hobby of many readers.  The experiments he made were all with pure breeding pea plants, but cat breeders can probably understand them more easily if they are reconstructed with cats as the subjects under study.  So if two cats known to be homozygous, one for black and one for blue  are used to reconstruct Mendel's first experiment the processes and results are as follows:

The 3:1 Ratio

In the first cross of a homozygous black cat with a homozygous blue cat the first generation progeny, known as the first filial generation and symbolized F1, will be found to be all black.  Black is dominant to blue and blue is recessive to black.  The next part of the experiment crosses two of these identical black cats and produces the next filial generation, F2. The kittens of the F2 generation will be found to be black and blue in the ration of 3 black to 1 blue.  However, this is one point where Mendel's experience cannot be reproduced as he obtained 1064 pea plants in his F2 generation and the likelihood of 1064 kittens from matings between littermates of F1 litters are rather remote!  However, the ratios remain the same.  Many cat breeders would feel hesitant about mating littermates even in the course of an experiment, but of course the ratio of three black to one blue from heterozygotes applies even when the mates are not related.  A mating between F1 progeny can be taken to mean either between littermates or between cats of the same genotype as the littermates.
 
 
Thus it can be seen that in this experiment the recessive characteristic (blue) which disappeared in the F1 generation, reappears in the F2 generation intact and not blended in any way with the black.  The experiment suggest that the genes responsible for the production of black and blue are transmitted together to the F1 kittens and are then separated and segregated in the F2 kittens.  The experiment also suggests that each gamete of the original parents contains two genes (either two black genes or two blue genes) and that the two genes then separate or segregate in the formation of the gametes, each gamete receiving only one of the genes.

If the gene producing black is described as D and that producing blue as d the original parents are DD and dd respectively.  The gametes of the DD cat must all be D and the gametes of the dd cat must all be d. So all the F1 kittens must inherit alternative genes Dd. Since D is dominant to d, Dd would appear as D or black. In the second generation (F2) the gametes of one cat must be either D or d as must be those of its mate.

Mendel proposed that hereditary units (in this case D and d) segregated in the gametes and that the gametes themselves united at random.  Thus the F2 kittens must be either Dd like the F1 parents or DD or dd like the original parents which are known as the first parental generation (P1), and the ratio of one genotype to another in the F2 must be 1 DD:2Dd:1dd.  Since both DD and Dd kittens appear black, this results in F2 progeny that are in the ratio of 3 black:1 blue.

These test matings reconstruct the experiments which led to Mendel's first law or principle of segregation, and help to explain that while cats may appear alike they may have a different coding.  This distinction between appearance (phenotype) and coding (genotype) must often be made.

A reconstruction of Mendel's experiment is shown diagrammatically in the next column.

The chequerboard, a Punnett's square can be rewritten as follows:

  1. Cat No. 1 = DD - homozygous black
  2. Cat No. 2 = dd - homozygous blue
  3. The DD cat will produce gametes carrying D.
  4. The dd cat will produce gametes carrying d.
  5. Therefore fertilization of any one gamete by another will result in the genotype Dd and since D is dominant to d all the F1 kittens will be black.
P1 phenotypes
P1 genotypes
DD                                  dd
Segregation into gametes D             D                      d          d
F1 genotypes
Dd Dd
Dd Dd
F1 phenotypes Dd                                Dd
Segregation into gametes D           d                      D          d
F2 genotypes
DD D
D dd 
F2 phenotypes

To obtain the F2 as a mating is made between the F1 littermates or to unrelated cats of the same genotype as the F1 littermates--each of these cats producing two types of gametes in equal proportions--one carrying D and the other carrying d.  The chances for any male gamete to fertilize a female gamete similar to itself are equal so that four types of fertilization are possible.
 
 
Genotype Phenotype
1 DD Black 
2 Dd Black 
1 dd Blue 
Ratio 3:1
M (Black)

F (Black)
D d
D DD
Kitten 1
Dd
Kitten 2
d Dd
Kitten 3
dd
Kitten 4

The experiment between black and blue cats shows the method by which a single dominant or recessive characteristic is inherited and the ratios will apply to any matings between blacn and blue breeds and similarly to any matings between cats with dominant and recessive characteristics.

Since any recessive coloured cat must be homozygous in order to express that colour, the only doubt may be concerning the genotype of the dominant mate.  If, by its pedigree, homozygosity is uncertain then its previous breeding records could give the answer.

So the kittens of any blue cat x homozygous black cat are black and the kittens of any cat of recessive characteristic x cat of homozygous dominant characteristic are of the dominant characteristic.

Keeping to the blue x black matings a few examples follow:

  1. Blue Persian x homozygous Black Persian = all Black
  2. Blue Smoke x homozygous Black Smoke = all Black Smoke
  3. Blue Chinchilla x homozygous Normal Chinchilla = all Normal Chinchilla
  4. Blue British x homozygous Black British = all Black
  5. Blue Burmese x homozygous Brown Burmese = all Brown
  6. Blue Point Siamese x homozygous Seal Point Siamese = all Seal Point
  7. Blue Tabby x homozygous Brown Tabby = all Brown Tabby.
The above matings are all between cats with either D or d in their genotype as Brown Burmese is black amended by a gene for burmese; Seal Point Siamese is black amended by a gene for Siamese; Brown Tabby is black amended by a gene for agouti; Chinchilla is black amended by another, etc.

All the Black, Black Smoke, Normal Chinchilla, Brown Burmese, Seal Point Siamese and Brown Tabby kittens will, of course, be heterozygous for blue.

The kittens of any mating between these cats heterozygous for blue or any one other recessive characteristic will assort in a ratio of three dominant to one recessive and taking the same examples again the results would be as follows:
 

  1. Heterozygous Black Persian x heterozygous Black Persian =  3 Black:1Blue
  2. Heterozygous Black Smoke x heterozygous Black Smoke =  3 Black:1Blue
  3. Heterozygous Normal Chinchilla x heterozygous Normal Chinchilla =  3 Normal Chinchilla:1Blue Chinchilla
  4. Heterozygous Black British x heterozygous Black British =  3 Black:1Blue
  5. Heterozygous Brown Burmese x heterozygous Brown Burmese =  3 Brown:1Blue
  6. Heterozygous Seal Point Siamese x heterozygous Seal Point Siamese =  3 Seal Point:1Blue Point
  7. Heterozygous Brown Tabby x heterozygous Brown Tabby =  3 Brown Tabby:1Blue Tabby


Study Problems.