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Sex Linkage

Sex-Linkage

Written by Brian White. Copyright 1995

In many organisms, sex of the individual is determined genetically by the presence or absence of particular sex-chromosomes. For example: in humans, XX is female, XY is male; in chickens, ZW is female, ZZ is male.

Definitions:

autosomal trait -
a gene carried on a non-sex chromosome and present in two copies in both sexes.

sex-linked trait -
a gene carried on the sex chromosome that is present in both sexes; one copy in one sex, two in the other. In mammals, this is the X chromosome (one copy in males, two in females). In birds, this is the Z chromosome (one copy in females, two in males).

Y-linked trait -
in humans, a gene carried on the Y chromosome - this is very rare.

Symbols Used:

  1. X - wild-type X chromosome
    X* - mutant X chromosome

    This system is the simplest, but it does not designate which is dominant - in this system, wild-type is usually assumed to be dominant, if not otherwise stated.

  2. X - X chromosome carrying recessive allele
    X - X chromosome carrying dominant allele

Tip: Suspect sex-linkage when the ratios of phenotypes are different in males and females.

Solving Sex Linkage Problems

As an exercise, determine the mode of inheritance of the following traits (autosomal or sex-linked and dominant or recessive with respect to wild-type). Assume that XX is female and XY is male.

A) A pure-breeding long-tailed female crossed with a pure-breeding wild-type (short-tailed) male gives the following F1 generation:

27 wild-type female
26 wild-type male
male and female F1 were crossed, giving the following F2:
32 wild-type female
11 long-tailed female
30 wild-type male
11 long-tailed male
Since the ratios of tail sizes in both crosses are the same in males and females, there is no reason to suspect sex-linkage. It remains then to determine if long tails are dominant or recessive to wild-type. Since pure long crossed with pure wt gave wild-type, this suggests that long tail is recessive to wild-type. If we then make the model:

s - recessive long tail mutant allele

S - dominant short tail (wild-type)

Then the first cross is ss female crossed with SS male - this would produce only Ss progeny which should be short-tailed. This is confirmed by the data.
We would predict that the second cross would be Ss crossed with Ss which should give:

25% SS - short-tailed (wt)
50% Ss - short-tailed (wt)
25% ss - long-tailed

The results agree with our predictions, therefore the tail length gene is autosomal and long tails are recessive to wild-type short tails.

B) A pure-breeding yellow female crossed with a pure-breeding wild-type (pink) male gives the following F1:

24 wild-type female
22 yellow male
male and female F1 were crossed, giving the following F2:
12 wild-type female
13 yellow female
14 wild-type male
12 yellow male
Although the ratios of pink to yellow are the same for F2 males and females, the F1 ratios are different. Color may be sex-linked, so our first trial model should involve sex-linkage. Since yellow female C pink male gave pink females, yellow may be recessive, so try assuming that color is sex-linked and yellow is recessive. In that case:

X - X chromosome with recessive mutant yellow allele

X - X chromosome with dominant pink wild-type allele

Then the first cross would be: XX x XY, giving:

50% XY - yellow males

50% XX - pink (wt) females

So far, so good. Crossing a male and female from among these individuals should give:

25% XX - pink (wt) females

25% XX - yellow females

25% XY - yellow males

25% XY - pink (wt) males

These predictions agree with the data, so this pigment gene is located on the X chromosome and yellow is recessive to wild-type pink.

C) A pure-breeding red-eared female crossed with a pure-breeding wild-type (white-eared) male to give the following F1:

20 red-eared female
25 red-eared male
male and female F1 were crossed, giving the following F2:
30 red-eared female
13 wild-type female
32 red-eared male
11 wild-type male
Ear color does not appear to be sex-linked. Pure red crossed with pure white gave all red, so red is likely to be dominant. So make a model where:

R - dominant mutant red allele

r - recessive wild-type white allele

Then the first cross would be: RR x rr giving all Rr progeny - red. So far, so good.

Crossing the F1's should give:

25% RR - red
50% Rr - red
25% rr - white (wild-type)

These predictions agree with the data, so the ear color gene is on an autosome and the red mutant allele is dominant to the white wild-type allele.

D) A pure-breeding green-teeth female crossed with a pure-breeding wild-type (white-teeth) male gives the following F1:

20 green-teeth female
25 green-teeth male
male and female F1 were crossed, giving the following F2:
40 green-teeth female
22 green-teeth male
20 wild-type male
Since the F2 phenotype ratios are different in males and females, tooth color may be sex-linked. Since pure green female X pure white male gave all green female, it is likely that green is dominant to white. So make a model where:

X - X chromosome carrying green dominant mutant allele

X - X chromosome carrying recessive wild-type white allele

Then the first cross would be: XX x XY giving:

50% XX - green female

50% XY - green male

So far, so good. The predictions and actual results agree. Crossing a male and female should give: 25% XX - green female

25% XX - green female

25% XY - green male

25% XY - white male

These predictions agree with the data, so tooth color is controlled by a gene on the X chromosome and the green mutant allele is dominant to the white wild-type allele.


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