SME
Microbiology · Problem type
Primary and secondary antibody responses
Use this for “The first peak is 100 units on day 12 and the booster’s is 1,500 on day 6. How many times higher, and how many days sooner?”
Diagram
1,500
100
=
15
First peak 12 days after exposure; second peak 6 days after.
This is an example. Type your own values to start a new problem.
P₁
First peak level
example
P₂
Second peak level
example
d₁
Days to first peak
example · days
d₂
Days to second peak
example · days
R
Times higher
calculated
s
Days sooner
calculated · days
Formulas
R
=
P₂
P₁
15
=
1,500
100
s
=
d₁
−
d₂
6 = 12 − 6
Assumptions
•
The first exposure activates a few naive B cells and leaves many memory B cells behind.
•
Memory B cells are already class-switched: the second response is mostly IgG, sooner and higher.
•
Levels are titers or relative units read off one graph, so only their ratio matters.
•
The booster comes on day 40, after the first response has faded.
Step-by-step
WE KNOW
P₁ = 100, P₂ = 1,500, d₁ = 12 days, d₂ = 6 days
FIND
times higher (R), days sooner (s)
Step 1 · Find times higher (R)
Use R = P₂ ÷ P₁
How many times the first peak fits into the second.
R
=
1,500
100
R
=
15
Step 2 · Find days sooner (s)
Use s = d₁ − d₂
The days the first response took, less the days the second took.
s
=
12
−
6
s
=
6
days
Check
Put the numbers back in. Both sides should match.
15
=
1,500
100
✓
6 = 12 − 6 ✓
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Antibody titer from a two-fold dilution row
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The adaptive immune response in order
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