Given an individual’s serum ferritin level (microgram/L), iron deficiency status can be classified as either present or not present. The serum ferritin cut-offs for classification of iron deficiency status are different for specific population groups as shown in the table below.
| Population | Serum ferritin (microgram/L) to classify as iron deficient |
|---|---|
| under 5 years of age | < 12 |
| 5 years and over | < 15 |
The functions in the detect_anaemia function set use
these serum ferritin cut-offs to classify individuals from specific
population groups as either having no iron deficiency
or having iron deficiency. For example, a three year
old with a serum ferritin level of 11.3 microgram/L can be classified as
follows:
detect_iron_deficiency_u5(ferritin = 11.3)
#> [1] "iron deficiency"In this example, we use the detect_iron_deficiency_u5()
function which is specific for the under 5 years of age population
group. The other function is for the 5 years and over population group -
detect_iron_deficiency_5over().
By default, this function will provide a labelled result, that is, if the individual has iron deficiency based on serum ferritin the output will be “iron deficiency”, and if the individual has no iron deficiency based on serum ferritin the output will be “no iron deficiency”.
There may be cases where the preferred output is simple integer codes
such as 1 for iron deficiency and 0 for no iron deficiency. If this is
the preferred output, the argument label can be specified
as FALSE as shown below:
detect_iron_deficiency_u5(ferritin = 11.3, label = FALSE)
#> [1] 1The resulting integer code indicates that this individual has iron deficiency.
The more general detect_iron_deficiency() function can
also be used:
detect_iron_deficiency(ferritin = 11.3, group = "u5")
#> [1] "iron deficiency"In this case, we just need to specify the population group that the individual corresponds to, which in this example is the under 5 years of age group (specified as “u5”). The other population age group can be specified as “5over” for those 5 years and older.
If the simple integer code output is preferred, then the
label argument just needs to be specified as
FALSE:
detect_iron_deficiency(ferritin = 11.3, group = "u5", label = FALSE)
#> [1] 1Correcting serum ferritin
Serum ferritin is affected by the inflammation and/or infection status of an individual. Specifically, serum ferritin levels increase in response to inflammation and/or infection.
It is recommended that serum ferritin be adjusted accordingly depending on the inflammation or infection status of an individual. There are three ways of correcting or adjusting for inflammation or infection status when assessing iron deficiency.
The first approach is by increasing the serum ferritin cut-off for detecting iron deficiency. The recommendation is that if an individual has been assessed to be in an inflammatory or infectious state, serum ferritin level of less than 30 microgram/L will classify this individual as being iron deficient regardless of age group.
The function detect_iron_deficiency_qualitative()
facilitates this classification. For example, a 2 year old with a serum
ferritin of 15.6 microgram/L who has been assessed to have an active
infection, we can assess iron deficiency status as follows:
detect_iron_deficiency_qualitative(ferritin = 15.6, inflammation = TRUE)
#> [1] "iron deficiency"Syntax is similar to the earlier functions we have discussed but now
we add an argument inflammation to specify whether
inflammation is present or not. In this case, we specify
inflammation as TRUE.
If inflammation status is FALSE (no inflammation), then the serum ferritin is assessed based on the usual cut-offs. It is therefore important to specify the population group of the individual:
detect_iron_deficiency_qualitative(
ferritin = 15.6, inflammation = FALSE, group = "u5"
)
#> [1] "no iron deficiency"If the simple integer code output is preferred, then specify
label as FALSE:
detect_iron_deficiency_qualitative(
ferritin = 15.6, inflammation = TRUE, label = FALSE
)
#> [1] 1
detect_iron_deficiency_qualitative(
ferritin = 15.6, inflammation = FALSE, group = "u5", label = FALSE
)
#> [1] 0If no inflammation status (NULL or NA) is provided when using this
function, the result is NA.
The second approach for correcting serum ferritin is by using an arithmetic correction factor using the following correction factors based on different combination of values of acute-phase proteins - c-reactive protein or CRP and/or alpha(1)-acid-glycoprotein or AGP:
| Inflammation Stages | CRP and AGP Cut-off Points | Correction Factor |
|---|---|---|
| Incubation | CRP > 5 microgram/L and AGP <= 1 g/L | 0.77 |
| Early convalescence | CRP > 5 microgram/L and AGP > 1 g/L | 0.53 |
| Late convalescence | CRP <= 5 microgram/L and AGP > 1 g/L | 0.75 |
CRP and AGP are biomarkers for inflammation and the values for these acute-phase proteins combined indicate the different stages of inflammation that an individual is at. Based on the inflammation stage, a corresponding correction factor can be applied to serum ferritin to adjust its value.
The function correct_ferritin() allows for this serum
ferritin adjustment to be made given values of CRP and AGP. For example,
we can correct the serum ferritin of a person with a CRP of 2, an AGP of
1, and a serum ferritin of 11.3 as follows:
correct_ferritin(crp = 2, agp = 2, ferritin = 11.3)
#> [1] 5.989This gives us a corrected serum ferritin value of 5.99 microgram/L.
In some settings, values for both CRP and AGP may not be available. Usually, only CRP or AGP are available or are measured. In these settings, we can use functions to correct ferritin using only one of these acute-phase proteins based on the following correction factors:
| Acute-phase Protein | Cut-off Points | Correction Factor |
|---|---|---|
| CRP | > 5 microgram/L | 0.65 |
| AGP | > g/L | 0.72 |
For example, for a person with serum ferritin of 11.3 microgram/L and
a CRP of 2, we use the correct_ferritin() function as
follows:
correct_ferritin(crp = 2, ferritin = 11.3)
#> [1] 11.3and for a person with serum ferritin of 11.3 microgram/L and an AGP
of 2, we use the correct_ferritin() function as
follows:
correct_ferritin(agp = 2, ferritin = 11.3)
#> [1] 8.136