Question

A phenotypic female presents with primary amenorrhoea, an absent uterus, normally developed breasts and sparse/absent pubic and axillary hair. What is the most likely diagnosis?

• A. Androgen insensitivity syndrome (complete AIS)
• B. Mayer-Rokitansky-Küster-Hauser (MRKH) syndrome
• C. Turner syndrome (45,XO)
• D. Kallmann syndrome

Hint:

Good breasts but absent uterus and sparse pubic hair - androgens cannot act despite male genotype.

Answer 1

The Correct Option is A

Step 1: List the clues.
Primary amenorrhoea + absent uterus + well-developed breasts + scanty pubic/axillary hair in a phenotypic female.

Step 2: Interpret the hair and breast findings.
• Normal breasts imply adequate oestrogen action (here from peripheral aromatisation of testosterone).
• Sparse sexual hair indicates the tissues cannot respond to androgens, because pubic/axillary hair is androgen-dependent. This pairing - oestrogenised but androgen-unresponsive - is the hallmark of complete androgen insensitivity syndrome (CAIS).

Step 3: Explain the absent uterus.
In CAIS the karyotype is 46,XY with functioning testes that secrete anti-Müllerian hormone (AMH); AMH causes regression of the Müllerian ducts, so the uterus, cervix and upper vagina are absent. Testosterone is made but cannot act (defective androgen receptor), giving female external genitalia.

Step 4: Why the other options are wrong.
• MRKH also has absent uterus and normal breasts, but pubic/axillary hair is normal (46,XX, normal androgen action) - here the hair is sparse, favouring AIS.
• Turner syndrome typically has a uterus present but poor breast development (streak ovaries, low oestrogen).
• Kallmann syndrome shows hypogonadism with anosmia and poor breast development; uterus is present.

Key fact: Absent uterus + good breasts + scanty sexual hair to complete androgen insensitivity syndrome (46,XY).