Question

As B-lymphocytes mature, the process of VDJ recombination can produce ~1.6 million different combinations of the V,D, and J segments that make up the light and heavy chains of the B-cell receptors. A typical person might have 5 liters of blood and 6000 white blood cells per microliter of blood. Approximately one third of all the white blood cells are lymphocytes and approximately a quarter of the lymphocytes are B-cells. On average, how many cells in the body at a given time have each possible combination of segments in their B-cell receptors? Select one:

A. Between 0.001 and 0.002 cells

B. Between 1 and 2 cells

C. Between 1000 and 2000 cells

D. Between 1 and 2 million cells

E. Between 1 and 2 billion cells

Answer 1

One microlitre of blood has 6000 WBCs.

10^6 microlitre is equal to one litre. Our body has 5 litre of blood.

So it becomes,

= 6000×10^6×5

= 30000×10^6, are the total WBCs present in our body.

One third of WBCs are lymphocytes and among those lymphocytes, 1/4th are B-lymphocytes.

So it becomes,

= (1/3)×(1/4)×30000×10^6

= 2475000000 are the total number of B-lymphocytes present in our body.

VDJ recombination produces 1.6 million combinations.

So dividing total number of B-lymphocytes by 1.6 million gives us number of cells having all possible combinations at any given point.

= 2475000000/1600000

= 1546.87

That means number falls between the range of 1000-2000.

Hence correct option is C, between 1000 and 2000.