Answer
Fifty percent ( 50%) of daughters will inherit the trait. That is, each daughter conceived has a 50% chance of showing the disease. This percentages a based on large numbers and random events associated with fertilization and impregnation. It does not mean that if this couple has two daughters one will perforce exhibit the trait/disease phenotypically and one will be normal -- they both may exhibit the disease or they both may be normal phenotypically. The 50% (0.5) is the statistical probability of the outcome at fertilization.
Work Step by Step
The two alleles related to this disease ( trait/syndrome) may be represented by the letters "C " and "c": where "C" is the dominant normal gene, and "c" the recessive allele that codes for the disease trait:
The mother has two X chromosomes --X/X in the nucleus of her somatic cells (and germ cell before meiosis). The father has only one X chromosome, but he also has a Y chromosome in all his cells that have not undergone meiosis--genotype X/Y
In this case mother's genotype, therefore, is XC/Xc ; she can therefore form two types of gametes (ova) -- XC and Xc
The father's genotype is Xc/Y, so he can form two types of sperms -- Xc and Y
The genes on the Y chromosome are mainly concerned with sex determination.
Th two possible genotypes for daughters of these parents will be XC/Xc and Xc/Xc
The two types of sons possible (genotypically) will be XC/Y and Xc/Y:
The probability that a daughter of this couple will express this disease phenotypically is 50%.
Sex-linked conditions are transmitted mainly by genes on X chomosomes. In almost all cases Y chromosome is not involved. The TDY/SRY genes occupy most of the Y chromosomes; however, Y-linked genes for azoospermia (male infertiliy) and for hairy ear lobes have been found. There are also Y-linked genes for a DNA-binding protein (ZFY), and for a cell surface receptor protein (MIC2).
