Worked Problem 1

Problem: A sarcomere at rest has these measurements:

• Total sarcomere length: 2.5 µm
• A band width: 1.5 µm
• H zone width: 0.4 µm
• I band width: 0.5 µm (per half, so total = 1.0 µm)
• M-line thickness: negligible

If the sarcomere contracts to a length of 1.9 µm, calculate: (a) the new I band width (total), (b) the new H zone width, and (c) explain why the A band doesn't change.

Solution:

• Total change in sarcomere length = 2.5 - 1.9 = 0.6 µm shorter
• Since both Z-lines move equally, each I band half shortens by 0.6/2 = 0.3 µm
  • Original I band (total) = 1.0 µm
  • New I band (total) = 1.0 - 0.6 = 0.4 µm (each half = 0.2 µm)
• The H zone shortens because actin slides into the myosin-only zone. Each actin slides in by 0.3 µm (same as I band reduction per side).
  • Original H zone = 0.4 µm
  • New H zone = 0.4 - (0.3 × 2) = 0.4 - 0.6 = negative → H zone has COMPLETELY DISAPPEARED
  • This means actin from both sides has met and overlapped at the M-line.
• A band = still 1.5 µm because myosin filament length is unchanged.
• Verification: New sarcomere = A band + new I band total = 1.5 + 0.4 = 1.9 µm ✓

Key Insight: When sarcomere shortens by enough that H zone calculation goes to zero or negative, it means maximum contraction has been reached (actin from both sides meets at M-line).

Worked Problem 2

Problem: If the axial skeleton has 22 skull bones and 26 vertebrae, how many bones are in the rib cage region of the axial skeleton? Show working.

Solution:

• Total axial skeleton = 80 bones
• Skull = 22, Vertebrae = 26
• Remaining = 80 - 22 - 26 = 32 bones
• This includes: Ribs = 24 (12 pairs) + Sternum = 1 + Hyoid = 1 = 26 (not 32!)
• Wait: 22 + 26 + 24 + 1 + 1 = 74. But axial = 80. The difference = 6 auditory ossicles (3 per ear: malleus, incus, stapes).
• NEET standardly answers 80 (including ossicles in skull's 22 or counting separately).