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Here is an APPEAL TO ALL AMERICANS: please check your family trees(or family trees of your friends), if there are some 'roots' reaching Iowa State Journal. It would be very lovely to have SUCH AN IMPORTANT GENE LINKAGE finally confirmed.

I have another reference about this linkage :

Brumbaugh & Hollander, Genetics 48:884, 1963

It's an abstract of a presentation in congress, where they say that pourcentage of recombinants between E and Mo is 26% (+/- 4,5%).

Black Feather, you are priceless

I'm hoping for some help with French

The following from the French book: Les Poules. Diversite Genetique Visible - page 87



What I've gathered is that Black Mottled Pekin were crossed with Partridge Pekin. All F1 were black. But then it gets confusing, eg - poor translation:



"131" is 42.26% of 310. So, are they saying that in the F2:

* 131 black mottled
* 179 solid black ?
= 310 E phenotype total (ie both solid black & black mottled).

Then I'm lost in the maths

I "think" they are saying (by the last sentence)" that the two loci (mo & E) segregated independently?

P.s. they also mention on another page (p63) the 26% linkage of mo - E (Brumbaugh & Hollander, 1965).

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Unfortunately, personally although I had many mottled & non-mottled (mostly mo/Mo+) phenotype birds with E locus alleles e^Wh, e+, E (or ER), but they were, eg: e+/e+ mo/Mo+, E/E mo/Mo+, ER/e+ mo/Mo+ etc in foundation birds, so a nightmare to work out if any linkages between mo & E loci.

Just for the fun of it....

J.C.Martin's F1 (E-mo/e^b-Mo+) x F1 (E-mo/e^b-Mo+)= F2 result:



(* Henk's punnett square image generator)
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If the E locus & mo locus had no linkage (segregated independently):

Gametic Ratio = 1.1.1.1 (ie 1/4 (25%) Mo+ E, 1/4 (25%) Mo+ e^b, 1/4 (25%) mo e^b, 1/4 (25%) mo E)
Genotypic Ratio = 1:2:1: 2:4:2: 1:2:1

Phenotypic Ratio = 9:3:3:1
9/16 (56.25%) black,
3/16 (18.75%) black mottled,
3/16 (18.75%) partridge (brown -e^b),
1/16 (6.25%) partridge mottled (brown –e^b/e^b mo/mo))

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If the E locus & mottled locus had 26% linkage, & parental birds were P1: e^b-Mo+ plus P2: E-mo linkage, the F1 X F1 (F2) dihybrid crosses gametic ratios would be:
37% e^b Mo+,
13% E Mo+,
37% E mo,
13% e^b mo

*Phenotypic % = 74% P1 & P2 parental phenotypes (e^b-Mo+ or E-mo), 26% crossovers (E-Mo+ or e^b-mo).

Any errors in the above?

no wander i am having such a hard time getting mottled partridge. i am presuming the same goes from moving the mottling from wheaten to partridge.

nice to see you still around the fringes kazjaps.

Kazjaps, I was rather good at math, but rather long time ago. I calculated the number of e^b-mottleds this way(provided there is a crossover in 1 of 4 cases(25%): you need 1 out of 4 chicks to allow 2 e^b-loci meet. From them 1 out of 4 has had a crossover on the first chromosome , which gives 1 out of 16 chicks. From them 1 out of 4 has had a crossover on the second chromosome. It gives : 1 out of 56 chicks would be e^b/e^b mo/mo. Do your gametic ratios lead to the same (or close) result? (I sincerely hope my logic holds)

Glad to see that the green square is working on this board.
Also glad to see you post again, Kaz
(Degree of gladness not in that order)

Translation:
JCMartin (personal communication), crossed black mottled Pekin bantams to partridge Pekins, and got black F1; in F2 he countes 131 black mottled and 310 blacks in total. This proportion of 42,26%, comes near to 43,83% of the expected 265 recombinants, and it equal to almose the 7/16c which is 43,75% which were given in the F2 on two recessive alleles which appear to be on two independent loci.

I've crossed E Mo+/mo po+/po+ Pti1/pti1+(incl. brachydactyly) x e^b/e^b Mo+/mo Po/Po Pti-2/Pti2.
Of 26 chicks, all mottleds (6) had 4 toes and Pti1.
The others were self choc, had all 5 toes, all Pti1 pure or Pti2 impure (feathered shanks but clean middle toe), and app. 1/3 brachydactyly.

I wondered about why all mottleds had 4 toes instead of 5.
Of the self chocs only one had 4 + 5 toes.
It looked as if the mottleds did not want to have 5 toes.

Owh, PS: of those 6 mottleds, 4 died of Mareks, none of the others.

Hello,

Maybe I can help for translation :

"J.C. Martin (personal communication), by crossing black mottled Pekin Bantam with partridge Pekin Bantam, obtains a black F1; in F2, he counts 131 black mottled out of 310 black in total. This proportion of 42.6%, close to the expected 43.84% with 26% of recombinants, is also close to 7/16th, so 43.75%, that would give a F2 with two recessive alleles belonging to two independant loci."

I agree with the expected values calculated by KazJaps. Then, I'm not so sure to understand how the values of 43.84% or 43.75% have been obtained.

If I follow the gametic ratios provided by KazJaps if E and Mo are linked (26 cM), I find a proportion of 31% of mottled among blacks.

Best regards.