You are studying a weasel and find that its genome contains 7 J$\kappa$and 55 V$\kappa$ genes.Other sources of variation (such as 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?
A)620
B)385
C)3850
D)70
E)550

Question

Quizplus · Accepted Answer

To calculate the number of possible chains, we need to multiply the number of different V genes with the number of different J genes, and then multiply that by the number of different possible nucleotide combinations due to site changes and insertions.

Number of different V genes = 55
Number of different J genes = 7
Number of variability due to site changes and insertions = 10

Therefore, the number of different chains that this organism can make is:

55 x 7 x 10 = 3,850

So, the best choice is C, which is the number of different F1F1F1S1F1F1F10 chains this organism can make.

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

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