Farbvererbung-e

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The teacher cannot do without it!
Heredity of Colors at Warmblood Horses
*How colours on horses do pass on to the offsprings?* A horse has 64 chromosomes, on which, as like as it is in human beings, genes are located. They all exist twice, one by the father and one by the mother. The inheritance is dominant-recessive, so I write in the following sentences capitals for the dominant features and for recessive features I will use small letters. (Geneticists speak about alleles) The horses colour is the result of the interaction of 7 of that genes. In the following text, I will include only three of them, because the other 4 genes are responsible for the paint drawing or for the development of palominos. For our basic colours Gray, bay, chestnut and black it is enough to look at the following three genes: 1. Gene G (gray): It is responsible for the gray colour (or better, if there exist pigments in the hair at all) and will be passed dominant on to the offsprings over all the other colours. A gray horse mostly has the genes Gg (heterocygous = mixed heredity) and so it transmits the G to a percentage of 50% (i.e. 50% of its offsprings are gray). The offsprings of a homocygous gray horse (GG) will become all gray (very seldom, f.e.Godehard). To get a gray horse, one of the parents must be gray, what means, that each gray horse must have one gray parent at least. 2. Gene E: This gene E determines, if a black pigmentation of the hairs exists at all. If it exists, the horse either is bay or black (for both it is necessary Ee or EE). Does not exist any black pigmentation, the horse will be chestnut (ee). 3. Gene A (agouti=wildcoloured): This gene must be seen in connection with gene E. Is the horse ee (what means, it is a chestnut) then the A does not play any role. Is there indeed Ee or EE, then gene A determins, wether the horse is bay or black. At least one big A (Aa or AA) effectuates a bay colour in the different graduations (bay-blackbrown). aa effectuates the black colour.
	a a ---> wild colour factor ---> A _ ---> wild colour factor ---> E _ ---> Black factor ---> E E ---> Black factor ---> e e ---> Black factor absent ---> --------------------------------------- G g ---> Grey factor ---> G G ---> Grey factor --->	black, if bay, if bay, if bay, homocygous, if chestnut -------------------------- gray, heterocygous gray, homocygous	E _ E _ A _ A _ AA or Aa or aa ---------------------

	So have........ chestnuts	A a e e or A A e e or a a e e	- throws all colours - throws no black - throws with black no bay
	black	a a E e a a E E	- throws also chestnuts, throws with black no bay - homocygous dark, throws with black only black
	bay	A a E e or A A E e or A a E E or A A E E	- throws all colours - throws no black - throws no chestnuts - throws no chestnuts and no black
	gray	_ _ _ _ G g _ _ _ _ G G	- heterocygous gray, throws 50 % gray - homocygous gray, throws 100% gray
nach Dr. Nebe in "Der Hannoveraner", um 1990

Die Farbformeln unserer Zuchtstuten bzw. der Väter unserer Pferde
	PHorse	has the formula	you can conclude ....
	SPS Auguste heterocyg. bay	A a E e or A A E e	A must be, otherwise she would be black e must be, because she throws chestnuts Question: Is she able to throw black, then Aa?
	SPS Fabienne homocygous bay	A a E E	A must be, otherwise she would be black; a must be, because she threw black; E must be, because she is bay, 2. E is here because of gene test, so she is homocygous bay, throws no chestnuts, also when crossed with chestnuts.
	SPS Watt Nuh chestnut	A a e e or a a e e	a must be, because she threw black
	SPS Shari homocygous black	a a E E	homocygous dark, because Sherlock Holmes has EE and Fabienne has E E, additional gene test, throws allways black when crossed with black, throws no chestnuts
	SPS Floria tosca chestnut	A a e e or a a e e	ee must be, because all chestnuts have this part of formula a must be, because she threw black
	SPS Radunja heterocygous black	a a E e	aa must be, because all black have this part of formula, e must be, because her dam is chestnut, E must be , otherwise she would be chestnut, throws with black 75 % black and 25 % chestnuts
	SPS Ananova Fuchs	A a e e oder a a e e	a must be, because she threw black
	SPS Donnanova Rappe	a a E e	aa must be, because all black have this part of formula, e must be, because her dam is chestnut, E must be , otherwise she would be chestnut, throws with black 75 % black and 25 % chestnuts
	SPS Richenza Rappe	a a E E oder a a E e	E must be, because she is black, if shewould have the second E, she is homocygous black

	Rosentau Black	a a E e	e must be, because he throws chestnuts
	Fabriano brown	A a E e	a and e must be, because he throws black and chestnuts
	Sherl. Holmes bay	A a E E	a must be, because he throws black; A must be, otherwise he would be black; EE must be, because he throws no chestnuts.
	Augustinus xx brown	A a E e	a and e must be, because he throws black and chestnuts
	Glückspilz bay	A A E E	throws no chestnuts, throws no black delivers only bay, with gray self.evident also gray
	Graf Grannus homocygous black	a a E E	homocygous dark, throws with black allways black, throws no chestnuts
	Likoto xx blbr, homocyg. bay	A a E E	a must be, because he throws black; A must be, otherwise he would be black; EE must be, because he throws no chestnuts.
	Weltmeyer chestnut	A a e e a a e e	a must be, because he throws black; ee must be, because all chestnuts have this part of formula
	Weltruhm Black	a a E e	e must be, because he throws chestnut and because one parent is chestnut
	Londonderry Chestnut	A a e e a a e e	a must be, because he throws black; ee must be, because all chestnuts have this part of formula
	Alabaster Chestnut	A a e e a a e e	a must be, because he throws black; ee must be, because all chestnuts have this part of formula
	Metternich blbr, homocyg. bay	A a E E	a must be, because he throws black; A must be, otherwise he would be black; EE must be, because he throws no chestnuts.
	Sir Donnerhall heterocyg. bay	A a E e	A must be, otherwise he would be black a and e must be, because he throws black and chestnuts