Chapter 15 Part 2 - Reciprocal Crosses and Sex-Linked Genes

The cross yields red-eyed females (due to the $w^+$ allele from the father) and white-eyed males (due to the $w$ allele from the mother, with no corresponding allele on the $Y$ chromosome).
50% of the F1 generation are white-eyed, and all of them are male.
This supports the idea that genes are on chromosomes.
Genes on sex chromosomes are called sex-linked genes.
This is an extension of Mendelian genetics, not a refutation. Mendel's work primarily involved genes on different chromosomes or far apart on the same chromosome.

Humans have two types of sex chromosomes: $X$ (larger) and $Y$ (smaller, about 1/3 the length of $X$ ).
$XX$ = female; $XY$ = male (the XY system).
Female: $44$ autosomes + $XX$ .
Male: $44$ autosomes + $XY$ .
Male gametes: 50% have $22$ autosomes + $X$ , 50% have $22$ autosomes + $Y$ .
Female gametes: All have $22$ autosomes + $X$ .
Zygote: $44$ autosomes + $XX$ (female) or $44$ autosomes + $XY$ (male).
Many genes on the $Y$ chromosome are related to sex determination.
Many genes on the $X$ chromosome are related to other characteristics unrelated to sex.
$Y$ chromosome: ≈ $100$ protein-coding genes.
$X$ chromosome: ≈ $900$ protein-coding genes (X-linked genes).
The $X$ and $Y$ chromosomes have short segments of homology, allowing them to pair during meiosis.

Males are hemizygous for their sex chromosomes because they only have one $X$ and one $Y$ chromosome.
Recessive X-linked traits are expressed in males if they have one copy of the allele because there is no second $X$ to mask the recessive allele.
To express a recessive X-linked trait:
- A male only needs one copy of the allele.
- Females need two copies to exhibit the trait, but if a female is a carrier and has one good copy, she generally doesn't have the disorder.

Father with normal allele ( $X^N$ ): 50% sperm with $Y$ , 50% with $X^N$ .
Mother is a carrier ( $X^N X^n$ ): 50% eggs with $X^N$ , 50% with $X^n$ .
- If the sperm with $Y$ fertilizes the egg with $X^n$ , the male offspring has the disorder.
- No female offspring will have the disorder because the father will contribute $X^N$ .
If the female is a carrier, 50% of her sons can have the disorder.
For a female to have a sex-linked disorder:
- Her father must have the disorder.
- Her mother must be at least a carrier.

Some types of color blindness.
Duchenne's Muscular Dystrophy: Affects dystrophin protein; affects $1$ in $3500-5000$ newborns; males are more affected.
Hemophilia A and B: Affect clotting factors; males are more commonly affected due to hemizygosity.

In pedigrees, if more males are affected than females, it suggests X-linked inheritance.

Females do not have a double dose of X-linked gene expression because of X-inactivation. In each cell, one of the two X chromosomes is randomly inactivated during embryonic development.
A small region of the shutdown X helps to shut it down.
All mitotic descendants of that cell have the same X inactivated.
The inactivated X is called a Barr body.
If a female is heterozygous for a trait on the X chromosome, she is a mosaic for that character with roughly 50% of cells expressing each allele.
However, expression of the good X in 50% of cells is typically enough to protect females from recessive disorders.

The coloration is a result of X-inactivation.
Early embryo: Both X chromosomes are active. One carries the allele for orange fur color, the other for black fur color.
After cell divisions, one X chromosome is randomly turned off in each cell.
Cells with the X chromosome carrying the black fur color allele active express black fur. The other X becomes a bar body.
Cells with the X chromosome carrying the orange fur color allele active express orange fur. The other X becomes the bar body.