Determination of Gender from Measurements of the Metacarpal and Phalangeal Bones

 

Measurements from the United Kingdom (Scheuer and Elkington)

 

score 1 =

= (0.0143 * (length of the second metacarpal bone in mm)) - (0.167 * (medial-lateral diameter at the base of the second metacarpal bone in mm)) + (0.0124 * (anterior-posterior diameter at the base of the second metacarpal bone in mm)) - (0.0152 * (medial-lateral diameter at the head of the second metacarpal bone in mm)) + (0.091 * (anterior-posterior diameter at the base of the second metacarpal bone in mm)) - (0.166 * (maximum diameter in the midshaft of the second metacarpal bone in mm)) + 3.61

 

If the score is < 1.5, then the skeleton is male.

If the score is > 1.5, then the skeleton is female.

Percent correct: 79%

 

score 2 =

= (-0.177 * (maximum midshaft diameter of the first metacarpal in mm)) - (0.102 * (maximum midshaft diameter of the proximal first phalangeal in mm)) + (0.0476 * (maximum midshaft diameter of the second metacarpal in mm)) + (0.0905 * (maximum midshaft diameter of the third metacarpal in mm)) - (0.175 * (maximum midshaft diameter of the fourth metacarpal in mm)) + (0.0858 * (maximum midshaft diameter of the fifth metacarpal in mm)) + 3.82

 

If the score is < 1.5, then the skeleton is male.

If the score is > 1.5, then the skeleton is female.

Percent correct: 76%

 

Measurements of Falsetti

 

score for second metatarsal =

= (-0.183 * (interarticular length in mm)) + (1.423 * (anterior-posterior width at midshaft in mm)) + (0.573 * (medial-lateral width at midshaft in mm)) + (1.84 * (medial-lateral width at head in mm)) + (0.631 * (medial-lateral breath at base in mm)) – 41.481

 

If the score is > 0, then the skeleton is male.

If the score is < 0, then the skeleton is female.

Percent correct: 92%

 

score for fourth metatarsal =

= (-0.0418 * (interarticular length in mm)) + (1.464 * (anterior-posterior width at midshaft in mm)) - (0.416 * (medial-lateral width at midshaft in mm)) + (0.981 * (medial-lateral width at head in mm)) + (1.038 * (medial-lateral breath at base in mm)) – 31.342

 

If the score is > 0, then the skeleton is male.

If the score is < 0, then the skeleton is female.

Percent correct: 86%

 

score for fifth metatarsal =

= (-0.004 * (interarticular length in mm)) + (0.848 * (anterior-posterior width at midshaft in mm)) + (0.17 * (medial-lateral width at midshaft in mm)) + (1.22 * (medial-lateral width at head in mm)) + (0.787 * (medial-lateral breath at base in mm)) – 30.68

 

If the score is > 0, then the skeleton is male.

If the score is < 0, then the skeleton is female.

Percent correct: 84%