Sexlinked traits follow a different pattern of inheritance thana non-sex linked trait because of the size difference between the Xand y chromosomes.Think about non-sex chormosomes as being the same size, eachhaving the exact same number of genes in the same postions (loci).This means that a person has to have two copies of the recessivegenes (one on each homolog of that chromosome) before the traitwill be expressed.Lets say A is normal (dominant) and a is abnormal(recessive).People who are aa have the abnormal condition and people who areAA or Aa are normal. The normal ratio is 1 AA:2 Aa:1 aa if bothparents are Aa. There is a 25% chance that a child will inherit twoabnormal genes and the chance of any sex child will be so affectedis exactly the same.When a recessive trait is located on the X chromosome only afemale with two X chromosomes has the same number of genes on eachX chromosome (the two X chromosomes have the same inheritancebehavior as a homolog chromosome in females). In males who inheritthe much smaller y chromosome there are many genes on the Xchromosome that do not have a matching gene on the y chromosome.This means that recessive traits on the X chromosome that have nomatching genetic material on the y chromosome will always beexpressed.So, lets say that there is a family where the mother is Aa andthe father (who only has one allele on the y chromosome is A. (A isnormal and a is abnormal).None of the daughters produced can be aa, because the fatherwill always pass A. Daughters will only be Aa or AA.Sons on the other hand, will get either A or a from the motherand, since the y chromosome has no genetic material at this genelocus the boys will be A normal or a affected at in a 1:1ratio.If the father is a on his X chromosome, and the mother is AA100% of the daughters will be carriers (Aa) and all the sons willbe normal (A-).