# Kappa – Lambda light chain ratio

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Hi everybody,

Few days back I was going through the development of immunoglobulin receptors on B cells. I came across a few facts:

1. The ratio between kappa (K) and lambda (L) light chain attached with immunoglobulin (Ig) receptor is approximately 2:1. (I was surprised, as the expected ratio should have been 1:1, assuming random activation of K or L genes). It is different from free light chain ratio (as they are also affected by differential catabolism of K and L light chains). I assumed that the metabolism of intact kappa Ig and lambda Ig receptor will be same and will depend on the degradation of respective B cell (there should not be differential destruction of K and L Ig B cells).

I got the answer to the above fact after a few hours when I came across the following fact:

Few days back I was going through the development of immunoglobulin receptors on B cells. I came across a few facts:

1. The ratio between kappa (K) and lambda (L) light chain attached with immunoglobulin (Ig) receptor is approximately 2:1. (I was surprised, as the expected ratio should have been 1:1, assuming random activation of K or L genes). It is different from free light chain ratio (as they are also affected by differential catabolism of K and L light chains). I assumed that the metabolism of intact kappa Ig and lambda Ig receptor will be same and will depend on the degradation of respective B cell (there should not be differential destruction of K and L Ig B cells).

I got the answer to the above fact after a few hours when I came across the following fact:

*“K LIGHT CHAIN GENE IS ACTIVATED IN ALL THE B CELLS INITIALLY, L LIGHT CHAIN GENE IS EXPRESSED ONLY WHEN THE TRANSLATED K CHAIN IS NOT ABLE TO BIND TO THE MU HEAVY CHAIN. IF L CHAIN PRODUCT IS ALSO NOT ABLE TO BIND TO THE HEAVY CHAIN, B CELL UNDERGOES APOPTOSIS.”*

I tried to describe the whole phenomenon in terms of probability.

Let us assume that the big square represents the whole of the pro (pre) B cells which will undergo K chain gene activation (area: 1×1 = 1). Let p (any number between 0 and 1 including) denote the probability that the B cell has compatible K chain (blue shaded area) (1- p denoting B cells which has incompatible K chain). 1 – p fraction of the original B cells will have transcription of L chain. Let us assume that p (of 1 – p B cells) denote the probability that L chain formed is compatible with heavy chain (red shaded area). The probability is assumed to be same to that of K chain compatibility.

(1 – p)^2 will be the fraction of B cells which will undergo apoptosis due to incompatible light chain generation.

So K/L will be equal to (blue area)/(red area) = p/p(1-p)

K/L = 1/(1-p)

K/L (p=0) = 1; K/L (p=1) = infinity;

**K/L (p=0.5) = 2****Implications:**

**1. K/L can never be less than 1 (if you come across any < 1 value, inform me with reference). If this happens, our model is wrong, something else may be happening.**

**2. K/L of 2 means that half of the K light chain and L light chains are incompatible with heavy chain.**

**3. K/L of 2 means that 25% of the pro (pre) B cells are undergoing apoptosis due to incompatible light chain synthesis.**

AMAZING, ISN’T IT!!!!!

BYE

Dr Suman Kumar

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