2. As a comparative immunologist, you are studying antibody diversity in the Afr
ID: 210049 • Letter: 2
Question
2. As a comparative immunologist, you are studying antibody diversity in the African pangolin. Pangolin antibodies are generated from two different heavy chain genes (H1. H2) and three the same as in human B cells. After extensive DNA sequencing, you've determined the number different light chain genes (L1, L2, L3). The expression of these Ig genes in Pangolin B cells is of V, D and J gene segments of each gene as shown in the table. Write the equation to calculate the total number of antibody molecules with different antigen specificities that a pangolin could produce by somatic recombination. Interestingly, pangolins do not appear to use junctional diversity mechanisms. [You do not need to perform the math to determine the actual number. Only set-up the equation and explain it]. (6 pts) Gene SegmentV D J Gene L1 2 3 H1 H2 40 1-5 30 4 65 27 6 52 6 22Explanation / Answer
Answer
Light chain = [V1 x J1] + [V2 X J2] + [V3 X J3]
Heavy chain = [V1 X J1 X D1] + [V2 X J2 X D2]
Total Antibody generated = Light chain X Heavy chain
Light chain = [40 x 5] + [30 X 4] + [11 X 17]
= 200 + 120 + 187
= 507
Heavy chain = [65 X 6 X 27] + [52 X 22 X 6]
= 10530 + 6864
= 17396
Total Antibody generated = 507 X 17396
= 8818758
= 8.8 X 106
Random rearragment happen in somatic cells gives a vast diversity in antibody production.
In gene rearrangement the Light chain has a numerous possible in v and J segments rearrangements. So we need to multipy that different segments and also has three different light chain each has a the different possibilites for rearrangement in gene segments. Finally we need to add all this to get the light chian diversity.
Same is used in heavy chain but here it has extra diversity D is present. So it further increase the possbility for antibody diversity. Finally we need to multiply the light and heavy chian combination to get overall antibody diversity number.
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