4 ms·
I'm not sure but I think the 99.5% figure here is for breeding modified mosquitoes with ordinary mosquitoes. That's the idea of the gene drive method. The sprea
by nshepperd 11y ago
I'm not sure but I think the 99.5% figure here is for breeding modified mosquitoes with ordinary mosquitoes. That's the idea of the gene drive method. The spread of the gene has to be better than normal mendelian inheritance, otherwise it would never spread throughout the population in the first place...
- jnhnum1 11y agoExactly. So suppose at some generation, the fraction of mosquitoes with the modified trait is p (and the unmodified fraction is 1 - p), and suppose that all mosquitoes are equally fit. A random child in the next generation will have the modified trait if: (a) both its parents are modified. This happens with probability p^2 (b) with probability .995 if one of its parents are modified. This happens with probability 2p(1-p), for a total probability of 1.99p(1-p). Overall, the fraction of modified mosquitoes in the next generation will have p^2 + 1.99p(1-p). Plotting this versus the identity function (http://www.wolframalpha.com/input/?i=%7Bp%2C+p%5E2+%2B+0.995*p*%281-p%29*2%7D+for+p+%3D+0+to+1 http://www.wolframalpha.com/input/?i=%7Bp%2C+p%5E2+%2B+0.995...), we see that this will increase monotonically to p = 1 (all mosquitoes have the modified trait).
- nonbel 11y agoThey only got 2 modified mosquitoes out of 25,000 then bred the rest from those. I doubt normal Mendelian inheritance is applying here.