You are studying a weasel and find that its genome contains 7 J$\kappa$ and 55 V$\kappa$ genes.Other sources of variation like changes in the site at which a J sequence joins a V sequence and the ability of deoxynucleotidyl transferase to insert nucleotides at sites of strand breakage have been found to introduce an additional ten-fold increase in variability.Assuming that any V sequence can join with any J sequence,how many different $\kappa$ chains can this organism make?

Question

Quizplus · Accepted Answer

There are 7 JF1F1F1S1F1F1F10 genes and 55 VF1F1F1S1F1F1F10 genes in the genome. Any V sequence can join with any J sequence, so the total number of possible different chains is:
7 x 55 x 10 = 3850
The additional ten-fold increase in variability due to changes in the joining site and nucleotide insertion means that there are 10 times as many possible combinations. 
Therefore, the organism can make 3850 x 10 = 38,500 different F1F1F1S1F1F1F10 chains.

You Are Studying a Weasel and Find That Its Genome κ\kappaκ

You Are Studying a Weasel and Find That Its Genome $\kappa$