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#92351 - 10/29/10 07:36 AM Re: Mottling and Linkages [Re: Black Feather]
KazJaps Offline
Classroom Professor

Registered: 08/30/02
Posts: 2807
Loc: Australia
Thank you all for the posts, & for the translations smile Helps alot.

So, this is where I got to in the linkages calculations last night, but wasn't sure if they were the correct methodology (still not sure).

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


Put this into a punnett square ...



*Phenotypic % = 74 % P1 & P2 parental phenotypes, 26% crossovers
  • 51.69% black,
  • 23.31% partridge,
  • 23.31% black mottled,
  • 1.69% partridge mottled

(*doesn't quite add up there smile but close)

Of all F2, 75% will be black or black mottled (E/E or E/eb based). Of these black/black mottled:
  • 68.92% black,
  • 31.08% black mottled


-------------
If there were no linkages between E & mo loci:

  • 56.25% black,
  • 18.75% partridge,
  • 18.75% black mottled,
  • 6.25% partridge mottled



Of all F2, 75% will be black or black mottled (E/E or E/eb based). Of these black/black mottled:
  • 75% black,
  • 25% black mottled


----------------
So, I got the same as yourself black_feather, ie in Martin's test breeding, 31% Black Mottled expected if a 26% linkage. If there was no linkage, wouldn't the expected percentage be around 25% Black Mottled in E/E or E/eb F2's? So this is why I was confused with the high 42.26 percentage of Black Mottled F2. In this example, wouldn't this suggest a closer linkage than 26%? Would of helped if they had given the partridge/partridge mottled F2 percentages.

-----------------
I'll give the formula below to work out cM - linkage value:
- how to determine linkage map distance (%)



-----------------
-once again, please correct my errors wink

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#92355 - 10/29/10 09:40 AM Re: Mottling and Linkages [Re: Black Feather]
Sonoran Silkies Offline
Flock Leader

Registered: 10/09/08
Posts: 345
Loc: Arizona
Originally Posted By: Black Feather
I think there is a mistake in this table about Mottling in Linkage Group II. As far as I know, Mottling (Mo) is linked to Extension (E) at 26 cM, as has already been said by Ron Okimoto. As we now know that Extension is MC1R gene, it indicates that Mottling and Extension should both be on Chromosome 11.

Has someone an indication of the linkage between Mottling and other locus from Linkage Group II, as Crest (Cr), Dominant White (I), or Frizzled (F)?


Is there any consolidated list of the gene names as used by breeders with the names used in the scientific literature (example: E = MC1R on Chromosome 11)?

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#92356 - 10/29/10 12:36 PM Re: Mottling and Linkages [Re: Sonoran Silkies]
Black Feather Offline
Bantam

Registered: 09/18/07
Posts: 57
Loc: France
Originally Posted By: Sonoran Silkies

Is there any consolidated list of the gene names as used by breeders with the names used in the scientific literature (example: E = MC1R on Chromosome 11)?


Last year, I have posted a list that have been supplemented by KazJaps :

Chromosome Z:
- Dw : GHR (growth hormone receptor)
- S : SLC45A2 (solute carrier family 45 member 2)

Chromosome 1
- C : TYR (tyrosinase)
- P : SOX5 (SRY-box 5)
- Db : SOX10 (SRY-box 10)

Chromosome 2
- Po : Lmbr1 (limb region 1)

Chromosome 7
- Lav : MLPH (melanophilin)

Microchromosome 11
- E : MC1R (melanocortin-1 receptor)

Microchromosome 24
- W : BCDO2 (beta-carotene dioxygenase 2)

Linkage Group E22C19W28
- I : PMEL17 (Melanocyte protein Pmel 17)


Since that, we can add :

Chromosome Z :
- B : CDKN2A/B (Cyclin-dependent kinas inhibitor 2A/B)


Moreover a publication suggests (but does not prove) some genes for other traits :

- Fm : EDN3 (Endothelin 3, on chromosome 20)

- Po : SHH (Sonic HedgeHog, on chromosome 2), instead of LMBR1

- Id : VCAN (Versican) or B4GALT1 (beta1,4-Galactosyltransferase) on chromosome Z

To be continued...

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#92357 - 10/29/10 01:12 PM Re: Mottling and Linkages [Re: KazJaps]
Black Feather Offline
Bantam

Registered: 09/18/07
Posts: 57
Loc: France
Originally Posted By: KazJaps

So, I got the same as yourself black_feather, ie in Martin's test breeding, 31% Black Mottled expected if a 26% linkage. If there was no linkage, wouldn't the expected percentage be around 25% Black Mottled in E/E or E/eb F2's? So this is why I was confused with the high 42.26 percentage of Black Mottled F2. In this example, wouldn't this suggest a closer linkage than 26%? Would of helped if they had given the partridge/partridge mottled F2 percentages.


I have extended the calculus.

If r is the recombination rate between E and Mo, the expected frequency F of mottled among black for this cross should be :

F = (1-rē)/3

With r=0,26, F=0,3108 (31,08%, it's OK)

The problem is that with this formula, I have :

rē = 1-3F

With F=0,4226 (42,26%), rē is negative, there is no solution !

There is probably a problem somewhere...

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#92368 - 10/30/10 04:22 AM Re: Mottling and Linkages [Re: KazJaps]
Wieslaw Offline
Moderator
Classroom Professor

Registered: 09/18/09
Posts: 3770
Loc: Denmark
Originally Posted By: Kazjaps
Would of helped if they had given the partridge/partridge mottled F2 percentages.

My thought exactly. The most important part of the experiment is missing in my opinion. I can see that you've reached the same result as I, as far as the number of mottled eb birds is concerned.

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#92370 - 10/30/10 03:02 PM Re: Mottling and Linkages [Re: Wieslaw]
RuffEnuff Offline
Ruler of the Roost

Registered: 01/27/06
Posts: 1148
Loc: Australia
it would be also interesting to know how the data was gathered because on new hatched chicken down homo mottling can be hard to tell and so can normal no mottling compared to hetero mottling. possible the reason no percentages of the partridge were given was because you cannot see it on chicken down.

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#92379 - 10/31/10 07:50 AM Re: Mottling and Linkages [Re: RuffEnuff]
KazJaps Offline
Classroom Professor

Registered: 08/30/02
Posts: 2807
Loc: Australia
Thank you black_feather for the new formulas. Much easier than my long hand way wink. Yes, it just doesn't add up.

It would make more sense the breeding results if the total E/E or E/eb segregates (black or black mottled) were 441, ie add 131 black mottled + 310 solid blacks = 441 total. This would make black mottled segregates around 29.7% & solid blacks around 70.3%. The recombination rate / linkage percentage would be approx. 33.01%. But, that is not what was stated (ie 42.7% black mottled segregates, not 29.7%). It's a shame about the discrepancies, as this test breeding result would be very handy to know.

Thanks again for your help black_feather. Much appreciated.

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#92380 - 10/31/10 07:55 AM Re: Mottling and Linkages [Re: KazJaps]
KazJaps Offline
Classroom Professor

Registered: 08/30/02
Posts: 2807
Loc: Australia
Yes Ruff - hopefully they waited till birds reached adulthood, as it's very unreliable relying on chick down or juvenile feathers alone. Eg, one of my family lines of Black Mottled d'Uccles were nearly solid black as day-olds, & mottled up very late (& didn't go through the mottled 'penguin' juvenile pattern).
Some lines of solid black or birchen grey Japs had much more cream/white on the down, yet none were carriers of the mottled gene. You have to compare homozygous family lines of mo/mo & Mo+/Mo+ before crossing. Make sure no other modifiers of chick down & adult plumage are segregating randomly (eg modifiers that produce white undercolour, white flights, sickles/base of tails, neck, etc of adults - with no mo gene present).

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#92391 - 11/01/10 12:38 PM Re: Mottling and Linkages [Re: KazJaps]
Chook-in-Eire Offline
Coop Keeper

Registered: 05/20/05
Posts: 733
Loc: Ireland
Hi guys,
I have also struggled through this, longhand like Kazjaps, using Excel and concur with her results. So going back to the sentence, I was wondering why the author mentions 7/16 or 43.75%. I reckon that none of us (myself included) understands the last part of the sentence correctly "que donnerait une F2 avec deux alleles recessifs appartenant a deux loci independants."

The 7/16 can only be the Black mottled (18.75%) plus Partridge (18.75%) plus Partridge mottled (6.25%) with independent assortment. 18.75+18.75+6.25=43.75%.
These are the phenotypes carrying at least 1 pair of recessive alleles (not quite the same as "deux alleles recessifs"), namely E E mo mo, E eb mo mo, eb eb Mo+ Mo+, eb eb mo Mo+, eb eb mo mo.

Why he throws these together I do not understand.

And when you do the same exercise with the 26% linkage, (23.31% black mottled, 23.31% partridge, 1.69% partridge mottled), the total comes to 48.31%, not 43.84%.

Google translation:
Quote:
This proportion of 42.26%, close to the expected 43.84% with 26% recombinants, is also close to 7/16 or 43.75%, which would give an F2 with two recessive alleles belonging to two independent loci.


I work as a translator so I know that machine translations are VERY ROUGH and usually miss and warp important points. So either something is lost in translation or the author made a mistake, or both.

I have his e-mail address. Anyone brave enough to ask?

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#92392 - 11/01/10 12:55 PM Re: Mottling and Linkages [Re: Chook-in-Eire]
Wieslaw Offline
Moderator
Classroom Professor

Registered: 09/18/09
Posts: 3770
Loc: Denmark
Hi chook, I'm brave enough, but I don't remember my French. Black Feather has already translated this(page 2), but not being a native speaker, I have doubts what is actually the conclusion in the text: ARE they linked OR NOT according to the authors?

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