Why do sex-linked traits follow patterns of inheritance than other traits?

Answer 1

Sexlinked traits follow a different pattern of inheritance than a non-sex linked trait because of the size difference between the X and y chromosomes.

Think about non-sex chormosomes as being the same size, each having the exact same number of genes in the same postions (loci). This means that a person has to have two copies of the recessive genes (one on each homolog of that chromosome) before the trait will be expressed.

Lets say A is normal (dominant) and a is abnormal (recessive).

People who are aa have the abnormal condition and people who are AA or Aa are normal. The normal ratio is 1 AA:2 Aa:1 aa if both parents are Aa. There is a 25% chance that a child will inherit two abnormal genes and the chance of any sex child will be so affected is exactly the same.

When a recessive trait is located on the X chromosome only a female with two X chromosomes has the same number of genes on each X chromosome (the two X chromosomes have the same inheritance behavior as a homolog chromosome in females). In males who inherit the much smaller y chromosome there are many genes on the X chromosome that do not have a matching gene on the y chromosome. This means that recessive traits on the X chromosome that have no matching genetic material on the y chromosome will always be expressed.

So, lets say that there is a family where the mother is Aa and the father (who only has one allele on the y chromosome is A. (A is normal and a is abnormal).

None of the daughters produced can be aa, because the father will always pass A. Daughters will only be Aa or AA.

Sons on the other hand, will get either A or a from the mother and, since the y chromosome has no genetic material at this gene locus the boys will be A normal or a affected at in a 1:1 ratio.

If the father is a on his X chromosome, and the mother is AA 100% of the daughters will be carriers (Aa) and all the sons will be normal (A-).