The Oxidation-Fermentation Test is used to differentiate bacteria built on their capability to oxidize or ferment specific sugars.
Once microbes are inoculated,-One tube is sealed with a layer of sterile mineral oil to promote anaerobic growth and fermentation.-The other tube is left unsealed to allow aerobic growth and oxidation.
Organisms able to ferment the carbohydrate or ferment and oxidize the carbohydrate will turn the sealed and unsealed yellow throughout.
Organisms able only to oxidize the sugar will turn the unsealed yellow medium and leave the sealed medium green or blue.
Fragile fermenters will convert both tubes slightly yellow at the top.
Organisms not able to metabolize the sugar will either produce no color change or will turn the medium blue due to alkaline products from amino acids degradation.
Since Pair #1 showed complete yellowing for sealed and unsealed, these Organisms able to ferment the carbohydrate or ferment and oxidize the carbohydrate. So our interpretation will be that the organism has: Oxidation and fermentation OR fermentation only.
For tubes #2 and #3, the sealed tubes were green throughout suggests that they need oxygen for aerobic growth, and the fact that their unsealed tubes showed light yellowing is evidence for oxidation. Sealed - Green and Unseal - Yellow. Our interpretation for these pairs of tubes would be : Oxidation
Tube 1 can be either Oxidation and fermentation OR fermentation only. So reliability of this needs to be confirmed more with additional testing.
Tubes 2 and 3 are most reliable because they can only be oxidation only and no fermentation.
The Gardener removes the apical bud
-the amount of the hormone that inhibits shoot elongation begins to decrease in the lateral buds
-the balance of hormones in the lateral buds shifts in favor of growth
-the memberlist cells in the lateral buds begin to divide
-the plant grows shoots which elongate into branches from the lateral buds
plant becomes bushier
Answer:
Option a, survive in extreme environments
Explanation:
Both Thermus and Deinococcus belong to the group of bacteria that are collectively termed as Deinococcus–Thermus group.
Deinococcus are radiation-resistant vegetative cell as they are able to resist ionising radiation. Also some species of Deinococcus are thermophile.
Thermus are thermophilic bacteria that are able to live in extreme temperature condition and thus are able to tolerate high temperature.
Hence, option A is correct.
The main way in which <span>a decrease in the number of songbirds might be expected to affect the trees in a northern forest is that the would be overrun with more instincts due to a lack or predators. </span>
Answer:
a) k_m = 4.08 uM
V_{max} = 20.07 uM/min
b) k_m = 8.16 uM
Explanation:
Given that:
For Enzyme A:
the substrate concentration [S] = 40 uM
the initial velocity rate v = 10 uM/min
when it was 4mM, v = 20 uM/min
i.e.
at 4mM = 4000 uM;
Using Michealis -menten equation;
when v = 10
![V = \dfrac{V_{max}[S]}{k_m+[S]}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_m%2B%5BS%5D%7D)
∴
when v= 20
equating equation (1) and (2):
let multiply equation (1) by 100 and equation (2) by 1
4000V_{max} - 1000K_m = 4000
<u>4000V_{max} - 20 k_m = 8000 </u>
0 -980k_m = 4000
k_m = 4000/-980
k_m = 4.08 uM
replacing the value of k_m into equation (1)
40{V_max } - 10(4.08) = 400
40{V_max } - 40.8 = 400
40{V_max } = 400 + 40.8
40{V_max } = 440.8
V_{max} = 440.8/40
V_{max} = 11.02 uM/min
b)
Since V_{max} of A ie equivalent to that of B; then:
V_{max} of B = 11.02 uM/min
Here;
[S] = 80 uM
V = 10 uM/min
∴
10(k_m +80) = 881.6
10k_m = 881.6 - 800
10k_m = 81.6
k_m = 81.6/10
k_m = 8.16 uM
