The Sex-linked recessive mutation



When you only want to know the outcome then scroll down.


Another  mutation form which also occurs in the lories is the Sex-linked recessive mutation. This recessive mutation is located on the Sex-chromosome Z.

At the moment there is the lutino, opaline and possibly the cinnamon mutation in lorikeets / lories. (also pallid, pale,... are Sex-linked recessive mutations but at the moment not yet in lories that I'm aware of).

This mutation says it all, this is a deviation on the sex chromosome.

In birds it is the other way around than in humans (the male have 1 X and 1 Y and female have 2 X), in birds the males have 2 Z chromosome (ZZ) females have 1 Z and 1 W chromosome (ZW).

So this mutation can occur without inbreeding! Only the father should have the deviation in its chromosome, and will give a visible mutation daughter,  a sex-linked mutation starts always with mutation females.

The W-chromosome contains most likely no colour information.


A female can never be a split, just a pure wild colour or the mutation!


For calculation the outcome we use again the punnett square. But this time we must write the sex chromosomes also on the formula, to know what the results are.



So here a pairing of normal split for lutino male  X normal female.


The male (=Z normal, Z lutino) = normal / lutino  (slash / means split)


The female(=Z normal, W) = normal or  pure wildcolour,

The colour is on the Z chromosome, we don't write a colour behind the W.


As you can see , you get:


Z normal Z normal = 2 x Z (= a male) and 2x normal, so a normal  male or pure wildcolour (no split). or 50% of the males are normal.


Z normal  Z lutino = Also 2 Z is a male (50%), normal colour but split (carrier) for lutino

 

Z normal W  = ZW is female (50%) and normal colour or pure wildcolour, as females never can be split.


Z lutino W = ZW also female (50%) but lutino colour.



If we pair a normal male x lutino female we get the next pairing outcome.


As you can see


all males (ZZ) are 100% normal but split for lutino


all females (ZW) are (100%) normal (so pure wildcolour.)



Now in the next pairing lutino male x normal female

(so the opposite of the previous pairing)  you can see very good the difference in outcome.



As you can see, it is a very big difference in outcome.

 

All males (ZZ) are 100% normal split for lutino.

All females (ZW) are 100% lutino.



I'm not going to write every outcome with the punnett square but below you can see all the possible outcomes when you breed with the recessive Sex-linked mutation.



split male X normal female (pure wildcolour) gives:


•  50% male normal (pure wildcolour)

•  50% male split


•  50 % female normal (pure wildcolour)   

•  50% female mutation

 

Also here are the males possibly splits as there is no visible difference between them.


normal (pure wildcolour) male X mutation female gives:


•  100% male split


•  100% female normal (pure wildcolour)



mutation male X pure wildcolour female gives:


•  100% male split

 

•  100% female mutation



split male X mutation female gives:


•  50% male split

•  50% male mutation


•  50% female normal (pure wildcolour)

•  50% female mutation


mutation male X mutation female gives:


•  100% male mutation

 

•  100% female mutation



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