Beginner genetics lesson required please!

[busylizzie]

Hi there - I am a total newbie in this poultry genetics world and wondered if you clever people would be able to help me solve my confusions.......

I have purchased some eggs through which I hope to achieve certain sex linked characteristics.....the gene is recessive. To my understanding the cockerals can either have the gene and outwardly show this or have the gene but not show it and the hens either show it completely or don't have it at all. With the following combinations of parentage what are the percentage outcomes?
first one obvious

cockeral (without gene) + hen (without gene) = 100% chicks without gene

cockeral (without gene) + hen (with gene) =

cockeral (with recessive/non showning gene) + hen (without gene) =

cockeral (with recessive/non showing gene) + hen (with gene) =

cockeral (with full showing gene) + hen (with gene) = 100% chicks with gene?

Really appreciate any advice on this........obviously it is all up nature anyway but I am making my brain ache trying to understand the concept

many thanks

 

[Krys109uk]

Hello.

This site explains the basics of poultry genetics.
http://kippenjungle.nl/basisEN.htm


There are always genes in the locus (that's the position on the chromosome where only those particular genes can go). Sometimes there is one dominant gene & one recessive gene present the dominant gene will show & the recessive gene won't show. Sometimes there are genes which are called incompletely dominant in that case when the genes in the gene pair are different the the result is intermedate. Some genes are considered co-dominant in that cae both different genes expess equally (I can't think of any like that in chickens).

To answer you questions referring to sex linked genes on the Z chromosome. There is not a case when a bird does not have any gene in a position on a particular chromosome. They will always be one gene or one of the alternate genes for that position, called alleles.

cockeral (without gene) + hen (with gene) =

Cockerel with both genes recessive & Female with gene dominant allele = All male offspring with one dominant gene (which shows) & the recessive allele (which doesn't show), all female offspring just have the recessive gene but it shows in this instance because the dominant allele is not present.

cockeral (with recessive/non showning gene) + hen (without gene) =

The hen will have one gene on her Z chromosome if this is the recessive allele all the offspring will show the recessive trait.

cockeral (with recessive/non showing gene) + hen (with gene) =

See answer for:
cockeral (without gene) + hen (with gene) =

Imagine you have a hen with sex linked barring.
She has a barring gene (B) gene on her single Z sex chromosome. You breed this hen with a male with the recessive not barred gene (b+) on each of his two Z sex chromosomes. He is not barred he is not carrying a barring gene.

When the eggs are fertilised they will receive either a Z sex chromosome or a w chromosome from the mother. All of the chicks inherit a Z chromosome from their father. All of the chicks which inherit a w chromosome are females. In the case of the two birds above the females offspring cannot inherit a barring gene on the Z chromosome they inherited from their father because he didn't have one & they didn't inherit a Z chromosome from their mother. Thus, with no barring genes , they show the recessive trait, not barred.
All of the chicks which inherited a Z chromosome from their mother, (remember her Z chromosomes have the dominant barring gene) will be males, they also inherit a Z sex chromosome from their father (this Z chromosome has the recessive not barred gene). So having inherited the dominant barring gene from the mother on her Z sex chromosome, the males offspring look different to the females.​

 

[Sonoran Silkies]

Quote:
I am going to replace you "with" and "without" with an actual sex-linked gene: silver (S), and it's counterpart, gold (s). One thing to note, silver is incompletely dominant, so the S/s boys are distinguishable from those who are S/S or s/s. However, you could substitute choc or another recessive gene into the equation and wouldn't be able to tell the difference between those who are hom. and those who are het.

cockeral (s/s) + hen (s) = 100% chicks (s/s) or (s/-)

cockeral (s/s) + hen (S) = 100% sons (S/s), 100% daughters (s/-)

cockeral (S/s) + hen (s) = 50% sons (S/s), 50% sons (s/s), 50% daughters (S/-), 50% daughters (s/-)

cockeral (S/s) + hen (S) = 50% sons (S/S), 50% sons (S/s), 50% daughters (S/-), 50% daughters (s/-)

cockeral (S/S) + hen (S) = 100% chicks (S/S) or (S/-)

 

[busylizzie]

Thank you for the help.

Krys109uk - I started following you but lost the concept about half way through - sorry - will definately read the linked site you posted though.....blimey I thought I was reasonably intelligent with a cambridge degree.....but time has obviously addled my brain

Sonoran Silkies - thank you for this I do understand your explanation....it is obviously basic enough for me to cope with....just

I am working on a chocolate gene characteristic so I grasp the silver/gold concept....

Right well they have to hatch first before any of this actually can be looked at in more depth!

Many thanks again!

 

[busylizzie]

Oooo one more combination I forgot!

What about:

Cockeral (S/S) + Hen (s/s or s/-) = results in.........?

_______________________________

Basically, I have 6 eggs coming from:

Cockeral (S/S) + Hen1 (S/S) + Hen 2(s/s or s/- unknown)
ASSUMING that they are all fertilised and all hatch

what do you think is going to happen?

 

[Sonoran Silkies]

The hen cannot be S/S or s/s as she only has a single locus for the gene--the other chromosome is too short to carry any of the sex-linked genes.

With the sex-linked genes, the father indiscriminately provides one of his two genes to his progeny. If he has two of the same allele, they will all get that. If he has different alleles, then half will get one and the other half the other. Sex does not enter into the equation.

However, the mother has only one copy of the gene, and she will always provide that copy to her sons; her daughters will never inherit a sex-linked gene from her.

So,

Cockeral (S/S) + Hen (s/-) = 100% Sons (S/s), 100% daughters (S/-)


Let's change that to choc:

Cockeral (Choc+/Choc+) + Hen (choc/-) = 100% Sons (Choc+/choc), 100% daughters (Choc+/-)

Not-choc cockerel + choc hen = Sons who carry choc, but it is not visible, daughters who do not carry choc. All offspring will appear black (assuming that that is the base colour).

 

[catwalk]

It's the opposite of humans, where females are XX and males are XY. There is nothing on the X that matches the Y, so whatever is on the X chromosome is expressed in males of our species. In females, the two genes on the X's have to work together to determine the expression of the gene.

In chickens, roosters have ZZ and hens are WZ. The sex- link gene on the hen has no partner, so whatever it is is expressed.

 

[Henk69]

I made a little something:

[URL]http://kippenjungle.nl/kruisingFlexNoPic.html?MODULE=kruisingFlexNoPic.html&DATA=S;Sexlinked recessive;;F;na/;JPG;0,G;Choc;Chocolate;S;,A;Choc;Choc+;S;Black;D;;W;,A;Choc;choc;S;Chocolate;R;;;[/url]

Click "punnet square" to also see split-info

 

[busylizzie]

Thanks for all the help.......my head is spinning a little still
but totally love the calculator!! thanks Henk69!!

Need to do some serious studying to grasp all the different terminology and concepts.