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#91023 - 08/13/10 03:44 AM Re: Mottling and Linkages [Re: Black Feather]
Wieslaw Offline
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Classroom Professor

Registered: 09/18/09
Posts: 3824
Loc: Denmark
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.

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#91028 - 08/13/10 05:07 AM Re: Mottling and Linkages [Re: Wieslaw]
Black Feather Offline
Bantam

Registered: 09/18/07
Posts: 57
Loc: France

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%).

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#91058 - 08/14/10 09:06 AM Re: Mottling and Linkages [Re: Black Feather]
Wieslaw Offline
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Registered: 09/18/09
Posts: 3824
Loc: Denmark
Black Feather, you are priceless smile

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#92332 - 10/28/10 11:55 AM Re: Mottling and Linkages [Re: Wieslaw]
KazJaps Offline
Classroom Professor

Registered: 08/30/02
Posts: 2864
Loc: Australia
I'm hoping for some help with French wink

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

Quote:
J.C.Martin (communication personalle), en croisant des animaux Bantam de Pekin cailloute noir avec des Bantam de Pekin perdrix, obtient une F1 noire; en F2 il denombre 131 noirs cailloutes sur 310 noirs totaux. Cette proportion de 42,26%, proche des 43,84% attendus avec 265 de recombinants, est egalement proche de 7/16c soit 43,75% que donnerait une F2 avec deux alleles recessifs appartenant a deux loci independants.


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:

Quote:
in F2 it denombre(?) 131 Black Mottled on 310 black totals. This proportion of 42.26%, close of the 43.84% awaited with 265 of recombinants, is egalement(?) close of 7/16 is 43.75% that would give one F2 with two alleles recessives belonging has two loci independants.


"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 smile

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).

---------------
Unfortunately, personally although I had many mottled & non-mottled (mostly mo/Mo+) phenotype birds with E locus alleles eWh, 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.

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#92333 - 10/28/10 12:40 PM Re: Mottling and Linkages [Re: KazJaps]
KazJaps Offline
Classroom Professor

Registered: 08/30/02
Posts: 2864
Loc: Australia
Just for the fun of it....

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



(* Henk's punnett square image generator)
-------------
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+ eb, 1/4 (25%) mo eb, 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 -eb),
1/16 (6.25%) partridge mottled (brown eb/eb mo/mo))

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

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

Any errors in the above?

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#92335 - 10/28/10 03:04 PM Re: Mottling and Linkages [Re: KazJaps]
RuffEnuff Offline
Ruler of the Roost

Registered: 01/27/06
Posts: 1154
Loc: Australia
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.

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#92345 - 10/29/10 04:17 AM Re: Mottling and Linkages [Re: RuffEnuff]
Wieslaw Offline
Moderator
Classroom Professor

Registered: 09/18/09
Posts: 3824
Loc: Denmark
Kazjaps, I was rather good at math, but rather long time ago. I calculated the number of eb-mottleds this way(provided there is a crossover in 1 of 4 cases(25%): you need 1 out of 4 chicks to allow 2 eb-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 eb/eb mo/mo. Do your gametic ratios lead to the same (or close) result? (I sincerely hope my logic holds)

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#92346 - 10/29/10 05:06 AM Re: Mottling and Linkages [Re: Wieslaw]
Henk69 Offline
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Classroom Professor

Registered: 02/13/06
Posts: 3228
Loc: Netherlands
Glad to see that the green square is working on this board.
Also glad to see you post again, Kaz smile
(Degree of gladness not in that order)

Originally Posted By: Henk69

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#92348 - 10/29/10 05:30 AM Re: Mottling and Linkages [Re: Henk69]
Sigi Offline
Ruler of the Roost

Registered: 11/23/06
Posts: 1150
Loc: Holland
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 eb/eb 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.


Edited by Sigi (10/29/10 05:37 AM)

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#92349 - 10/29/10 05:45 AM Re: Mottling and Linkages [Re: KazJaps]
Black Feather Offline
Bantam

Registered: 09/18/07
Posts: 57
Loc: France
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.

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