Weekly Challenge 48: Quorum-sensing

7411 /www/nrich/html/content/id/7411/ 1 2011-06-14T16:54:44 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><ul id="buttonBar"></ul> <div> <mdo:image alt="" float="left" height="120" src="rudolph_the_red_nosed_reindeer.png" width="91"></mdo:image> Rudolph's nose glows because it is home to a species of bacteria, Vibrio rudolphi, that luminesces when it reaches a certain population density. It detects the size of its population by quorum sensing: each bacterial cell releases a signal molecule, X, at a rate of $1$ per minute and if the concentration of X is greater than or equal to $10^{11}$ cells/ml, the bacteria will glow. X decays with a half life of ten minutes but the bacteria divide every 30 minutes. Sadly, Rudolph catches a nasty cold, which, by the time he is better, has killed all of the bacterial cells in his nose except for one. Santa is worried: there are only 24 hours left until Christmas. Will Rudolph's nose be glowing again in time? If you need any data that is not included, try to estimate it: Santa wants an answer now, so that he can make alternative plans if need be.</div> <div class="framework">Did you know ... ? The mathematics of rates and half-lives are of great importance in mathematical biology where growth factors are often in competition with decay factors.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The bacteria divide twice an hour, so divide 48 times in 24 hours. Thus $2^{48}$ bacteria cells after 24 hours, assuming the bacteria split in two when they divide. X has half life of 10 mins, so there are $6\times 24$ half lives in 24 hours. We only consider the bacteria released near the end then, as very little from the beginning will be left after 24 hours. Now $10^{11}< 2^{48}$ so if Rudolph's nose was 4mL, his nose will certainly be glowing. Assuming instead that Rudolph's nose was 4mL, we need the concentration of bacteria to be $4\times 10^{11}$. After 23.5 hours there were $2^{47}$ cells and $2^{47}$ have just been released. Right at 24 hours, there will be $\frac {2^{47}}{2}+2^{48}$ cells, which is greater than $4\times 10^{11}$. So Rudolph will have a glowing nose. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The bacteria divide twice an hour, so divide 48 times in 24 hours. Thus $2^{48}$ bacteria cells after 24 hours, assuming the bacteria split in two when they divide. X has half life of 10 mins, so there are $6\times 24$ half lives in 24 hours. We only consider the bacteria released near the end then, as very little from the beginning will be left after 24 hours. Now $10^{11}< 2^{48}$ so if Rudolph's nose was 4mL, his nose will certainly be glowing. Assuming instead that Rudolph's nose was 4mL, we need the concentration of bacteria to be $4\times 10^{11}$. After 23.5 hours there were $2^{47}$ cells and $2^{47}$ have just been released. Right at 24 hours, there will be $\frac {2^{47}}{2}+2^{48}$ cells, which is greater than $4\times 10^{11}$. So Rudolph will have a glowing nose. </mdoxml> 2 4 0 0 0 1 0 Weekly Challenge 48: Quorum-sensing This problem explores the biology behind Rudolph's glowing red nose. Applications Maths Supporting SET Sequences, Functions and Graphs Iteration Sequences, Functions and Graphs Exponential growth and decay Collections Weekly Challenge Applications STEM - living world Admin Individual STEM mapping STEM - Exponential Growth STEM mapping STEM - Exponential Growth