Women with Turner syndrome (XO) and normal women (XX) are clearly different phenotypically. In addition, the vast majority of XO
conceptions abort before birth. However, both XO and XX women have one active X chromosome since the X in XO women remains active and one might expect that they would therefore have similar phenotypes. What is the most reasonable explanation for their different phenotypes?
a.) XO women do not have a copy of the SRY gene.
b.) XO women have problems during development because mitosis is abnormal.
c.) In XO women, the single X chromosome has no partner to pair with during mitosis so that each cell division is delayed by pairing problems with the single X not finding a pairing partner.
d.) XO women are missing a copy of the Xist gene so that they are forced to develop partway along the male pathway during embryogenesis.
e.) Some genes remain active on the inactive X chromosome and XX women will have two copies of these genes expressed and XO women only one copy.
Turner syndrome females have a phenotype because there are regions of the X chromosome that normally escape inactivation, allowing 2 copies of the gene to be expressed. These regions are called pseudoautosomal regions.
These are inherited in a typically autosomal fashion, permitting crossing over between the X and Y. In Turner syndrome, the lack of second X chromosome means these genes are missing and the phenotypes result from haploinsufficiency
A producer. For example in a common food chain, grass->grasshopper->mouse->snake->eagle. The grasshopper eats grass which is a producer so if there aren't anymore insects like grasshoppers, then the grass grows more.
