Question

11.) A culture of bacteria in log phase increases from (Log1o ml) 8.0 to 8.6 in 1.5 h. What is specific growth rate (u) of this culture? What is the generation time (GT)? How many minut does it take for a doubling of cells? After 181 minutes, how many doublings have occurred? 12.) Finish labeling the following plot and determine the μ"ae and Ks of E.coli for glucose at 35°C. Us units of mM for concentration. Linweaver Burk plot E.coli grown using glucose broth at 35 C y- 1.5x + 0.5 2 1.5 0.5 -1.5000 1.0000 0 0.0000 0.5000 1.0000 1.5000 -0.5

Answer 1

u is defined as specific growth rate which can be calculated using the formula log X-log Xo/t

where X represents the final bacterial cell number

Xo is the initial bacterial cell count

t = time taken in doubling of population

generation time will be = time/number of generations

G= t/n

20 minute= 1.5*60/ number of generations

X0= X* 2ⁿ

n= ⁽log Xo- log X)/log 2

n= 3.3 log Xo/X

G= t/n

As per given question 1.5h*60= 90 minutes

G= 90/ 3.3 (log 0.6*10)

= 90/3.3* -0.2218= 90/0.731 = 123 minutes

G= 123 minutes

In 180 minutes only single generation has been passed 180/123= 1.46 equals to 1.5 doubling

2) According to line weaver's burk plot

where the straight line equation represented as y=mx+c

y= 1.5x+ 0.5

where slope will be Km/Vmax

y intercept will be 1/Vmax

x intercept will be -1/Km

u= specific growth rate of bacteria will be umax* (S/Ks+ S)

which can be seen in a similar way as Vmax* (S/Km+S) Michaelis menten equation

S= Initial bacterial concentration

umax is maximum growth rate achieved when all the substrates are present in excess

-1/Km= -0.500

Km= 1/0.500 = 2

slope(M) is Km/Vmax

Km/Vmax= 1.5

putting the values of Km and slope Vmax can be calculated

1.5= 2/Vmax

Vmax= 2/1.5= 1.333

Ks apparent will be 2 i.e. Km the substrate concentration when the reaction velocity is half of maximum.