Sex-Linked Inheritance

Abstract

Sexual differentiation in animals is given by a pair of chromosomes, called sex chromosomes. These are the chromosomes that will define whether an individual is male or female. These chromosomes are characterized by being morphologically distinct in one of the sexes, as opposed to the rest of the chromosomes that are called autosomes. In the animal kingdom, there are different types of sexual differentiation, and it may be the case that male individuals are the ones that present differences in their sex chromosomes (heterogametic) and the female individuals are those with both sex chromosomes morphologically the same (homogametic). There is another system in which the homogametic individuals are male and the heterogametic individuals are female. Because in one of the sexes the sex chromosomes are different, they do not present homology in all their extension. The genes that are located in the non-homologous region will segregate depending on the chromosome that passes to the gametes; therefore, these genes are considered sex-linked. These genes have only one allele in individuals since there is no homologous complementary region in its homologous chromosome. The genes located in the homologous region of the chromosomes do not segregate like the genes located in autosomes and are therefore considered to be partially sex-linked genes.

Solved Problems

• 1. Suppose that in an animal species in which sex determination makes males heterogametic, the M allele confers a white coat, and is dominant over the m allele that confers a black coat. The gene that determines the coat color is sex-linked, it is located on X. (a) Symbolize the genotype of both phenotypes in males and females. (b) Diagram a cross between a homozygous white-coat female and a black-coat male. Take the crossbreeding to F2 indicating genotypic and phenotypic ratios according to gender

  • (a) White-coat male X^MY

    Black-coat male X^mY

    White-coat female X^MX^M or X^MX^m

    Black-coat female X^mX^m

  • (b) X^MX^M x X^mY

    F1: 0.50 X^MX^m 0.50 X^MY

      X^MX^m x X^MY

    F2: 0.25 X^MX^M 0.25 X^MX^m 0.25 X^MY 0.25 X^MY 0.25 X^mY

    100% of white/coat females 50% of white-coat males 50% of black-coat males

• 2. Suppose that in an animal species in which females are heterogametic, one sex-linked gene located on chromosome Z determines feather color, with M (brown color) being dominant over m (white color). There is an autosomal gene that determines the feather length, with L (long feathers) dominating over l (short feathers). Suppose that in a cross between brown-feathered individuals with long feathers resulted a female with short white feathers and a male with short white feathers, and a male with long and brown feathers. Write the genotypes of the parents.

  • The genotypes for the two traits would be:

  • Brown feathered females Z^MW

  • White feathered females Z^mW

  • Brown feathered males Z^MZ^M or Z^MZ^m

  • White feathered males Z^mZ^m

  • Long-feathered males and females: LL or Ll

  • Males and females with short feathers: ll

  • In order for a cross between brown feathered individuals to result in a female of white feathers, the male parent must be heterozygous for the trait because the female product of the cross will receive the W chromosome from her female parent and the Z chromosome from her male parent. In relation to the autosomal gene, for any individual to result in short feathers with both parents being long feathered, both must be heterozygous. The genotypes of the parents are:

$$ {\mathrm{Z}}^{\mathrm{M}}\mathrm{W}\ \mathrm{Ll}\ \mathrm{x}\ {\mathrm{Z}}^{\mathrm{M}}{\mathrm{Z}}^{\mathrm{m}}\mathrm{Ll} $$

• 3. Assume an X-linked disease in a species with XY sex determination. If a male suffering from the disease is crossed with a female who does not suffer from it, what proportion of her male offspring will have the disease?

In order for the male to suffer from the disease, he must have the allele that produces it in his unique X chromosome. In order for the female to not suffer from the disease, it must be recessive homozygous. The cross would be represented as follows:

$$ {\displaystyle \begin{array}{cc}& {\mathrm{X}}^{\mathrm{S}}\mathrm{Y}\ \mathrm{x}\ {\mathrm{X}}^{\mathrm{s}}{\mathrm{X}}^{\mathrm{s}}\\ {}\mathrm{F}1:& {\mathrm{X}}^{\mathrm{S}}{\mathrm{X}}^{\mathrm{s}}{\mathrm{X}}^{\mathrm{s}}\mathrm{Y}\end{array}} $$

100% of male offspring are normal, i.e., the probability of male offspring having the disease is 0.

• 4. In the broiler production process, it is necessary to identify male and female chicks at hatch. This has been achieved for many years by determining “feathering speed,” which is a consequence of the expression of a sex-linked gene, specifically in the Z chromosome. The speed of feathering is determined by comparing the length of the feathers that cover the wing at the top and the length of the primary feathers, which are the feathers that arise from the underside of the wing. Thus arise the concepts of “slow feathering,” which is when at birth it can be determined that the primary feathers of the bird are shorter than the feathers covering the wing, or at most are of the same length. “Fast feathering” will be when the primary feathers are longer than the feathers covering the wing. Propose a model of controlled crosses that allows differentiating males from females according to the speed of feathering.

  • Z^A W: slow feathering female

  • Z^a W: fast feathering female

  • Z^A Z^A: slow-feathering male

  • Z^A Z^a: slow-feathering male

  • Z^a Z^a: fast feathering male

By crossing a fast-feathering male with a slow-feathering female, it is possible to know that the male will only be able to provide gametes with the information Z^a. The female will contribute Z^A or W. In such a way that every individual of the offspring will be Z^A Z^a or Z^a W, that is, the slow-feathering birds will be males and the fast-feathering birds will be females.