Hb Electrophoresis in ß-Thal + Iron Deficiency: Example
Normal Hemoglobins have a Heme Moiety plus 4 Globin chains
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Always 2 α-chains
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The 2 α-chains pair preferentially with 2 β-chains (HbA)
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Some α’s (<3%) pair with δ-chains (HbA2) or γ-chains (HbF)
Case #1: Normal: Suppose we have:
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1,000 Heme moieties (with iron)
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2,000 α-chains
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2,000 β-chains
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Various γ-chains and δ-chains
980 of the Hemes join with 1960 α-chains and 1960 β-chains
The other 20 Hemes join with α’s plus δ’s or γ’s
This gives a normal electrophoresis of, say:
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98% HbA
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1.5% HbA2
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0.5% HbF
Case #2: β-Thalassemia = Deficient number of β-chains:
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1,000 Heme moieties (with iron)
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2,000 α-chains
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1,800 β-chains (decreased)
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Various γ-chains and δ-chains
900 of the Hemes join with 1,800 α-chains and the 1,800 β-chains
The other 100 Hemes join with the left-over α’s pairing with δ’s and γ’s
This gives an abnormal electrophoresis of, say:
Case #3: β-Thalassemia + Iron-Deficiency (i.e. fewer Hemes)
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750 Hemes (if Hct drops from 40 to 30)
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2,000 α-chains
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1,800 β-chains (decreased due to β-Thal)
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Various γ-chains and δ-chains
735 of the Hemes join with 1470 α’s and 1470 β’s
The other 15 Hemes join with α’s plus δ’s and γ’s
This now gives a normal electrophoresis:
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98% HbA [735 out of 750]
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1.5% HbA2
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0.5% HbF
Decreased quantities, but normal percents.
Increase the Hemes, & we’re back to Case #2 (abnormal percents)
Hope this makes sense [???]
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