Ophthalmology · Strabismus (Types, Diagnosis, Treatment)

In a child with right superior oblique palsy (CN IV palsy), the Parks-Bielschowsky three-step test will show which combination of findings?

  • A Step 1: right hypertropia; Step 2: worse on left gaze; Step 3: worse on left head tilt
  • B Step 1: left hypertropia; Step 2: worse on right gaze; Step 3: worse on left head tilt
  • C Step 1: right hypertropia; Step 2: worse on left gaze; Step 3: worse on right head tilt
  • D Step 1: right hypertropia; Step 2: worse on right gaze; Step 3: worse on left head tilt
Correct answer: C. Step 1: right hypertropia; Step 2: worse on left gaze; Step 3: worse on right head tilt

Explanation

Parks three-step test for right superior oblique palsy: Step 1 identifies the hypo/hypertropic eye — the right eye is higher (right hypertropia) because the right SO (an intorter and depressor) is weak. Step 2 asks which gaze worsens it — the right SO acts maximally as a depressor in adduction, so left gaze (right eye adducted) worsens the right hypertropia. Step 3 asks which head tilt worsens it — tilting right head activates intorters (SO + SR of right eye); since right SO is paretic, right SR overacts, increasing right hypertropia; hence right head tilt worsens it. The classic answer for right SO palsy: right HT, worse left gaze, worse right head tilt.

Reference: Khurana Comprehensive Ophthalmology, 7th ed.

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