Bio-medical Ontologies Maintenance and Change Management

300 M. Fernandez, M. Villasana, and D. Streja
Ra, (mg/min)
500
450
400
350
300
250
200
150
100
50
0
0 50 100 150
time, (min)
200 250 300
(a) 45 gr. of ingested glucose load
Ra, (mg/min)
500
450
400
350
300
250
200
150
100
50
0
0 50 100 150
time, (min)
200 250 300
(b) 89 gr. of ingested glucose load
Fig. 2. Rate of absorption Ra(t) for ingested glucose according to Radziuk’s model
emptying of glucose following ingestion of a meal containing Ch millimoles of glucose
equivalent carbohydrate. The gastric emptying (the rate of absorption) is given
as a function of time. The model depends on the carbohydrate intake as follows: if
the load is greater than 10 g, then the model assumes a trapezoidal shape, meaning
that there is a period of maximal emptying. If on the other hand the load is less than
10 g, then so such maximal rate is seen.
Lehmann and Deutsch provide a model expressing the load in terms of millimoles
of carbohydrate. We have approximated the specific weight of glucose to be 180 in
order to get Ch/180 × 10 3 mmols of substance. Thus, by multiplying the gastric
emptying function provided in [16] by 180 × 10 −3 we obtain the rate of absorption
in terms of g/hour.
The equations that define the rate of absorption following this model, including a
correction made by the authors to accurately account for the total amount of glucose
ingested (see [13] for details), with t measured relative to ingestion time (at ingestion
t = 0) and Vmax equal to 120 mmol/h, are as follows:
• If the load is less than 10 g then:
⎧
⎨ (Vmax/Ta)t
Ra(t)= Vmax −
⎩
: t < Ta
Vmax
T (t − Ta)
d
0
:
:
Ta < t < Ta + Td
otherwise
•
where Ta = Td = load/Vmax.
If the load is greater than 10 g then:
⎧
⎪⎨
(Vmax/Ta)t
Vmax
Ra(t)=
Vmax ⎪⎩
−
:
:
t < Ta
Ta < t < Ta + Td
Vmax
T (t − Ta − Tmax)
d
0
:
:
Ta + Tmax ≤ t < Tmax + Ta + Td
otherwise
Here Ta = Td = 0.5h,andTmax is given by the expression
Tmax =[load −Vmax(Ta + Td)]/Vmax.