A baby girl born with ambiguous genitalia is found to have 21-hydroxylase deficiency of the salt-wasting type. Which of the following karyotypes would you expect to find in this baby?
Correct Answer: 46, XX
Description: This baby girl with ambiguous genitalia and 21 hydroxylase deficiency would have a normal female chromosomal pattern 46XX. 21 hydroxylase deficiency is the most common cause of androgenization in chromosomal 46,XX females. Deficiency of this enzyme leads to block in mineralocoicoid and glucocoicoid and increase in 17-hydroxyprogesterone and shunting of steroid precursors into the androgen synthesis pathway. Increase in androgen synthesis in utero causes androgenization of the female fetus in the first trimester. Excess androgen production causes gonadotropin-independent precocious pubey in males with 21-OHD. Salt wasting form of 21-OHD results from severe combined glucocoicoid and mineralocoicoid deficiency. It is diagnosed in any babies with ambiguous genitalia with bilateral non palpable gonads. Males (46,XY) with 21-OHD have no genital abnormalities at bih but are equally susceptible to adrenal insufficiency and salt-losing crisis. Ref: Harrison's Principles of Internal Medicine, 18e chapter 349.
Category:
Pathology
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