Answer:
12-14 october
Explanation:
The probable period of exposure is calculated as follows:
- Know the range of the incubation period
- Start counting back from the days of the first case by subtracting the first number of incubation period range
- Count back the days from the last case by subtracting the last number of incubation period range.
- This will give a range which will be the period of exposure.
From the question,
- Incubation period range 1-15
- Date of first case = october 15
- Therefore, first date of exposure = 15 -1 = 14
- Date of last case = 22
- Therefore, last date of exposure = 27 - 15 = 12
- Thus, the range is 12-14 october which is the probable period of exposure.
Answer:
The correct answer is the formation of pyruvate from glucose willl be energetically less favorable.
Explanation:
According to the question matunt yeast has a shorter glycolytic pathay catalysing a follwing reaction
Glyceraldehyde-3-phosphate+H2O+NAD+ = 3-phosphoglycerate
So in the mutatant the formation of 1,3- bisphosphoglycerate is not occurring.The glyceraldehyde-3-phosphate is directly converted into 3-phosphoglycerate.
As a result the substrate level phosphorylation step that deals with formation of 3-phosphoglycerate from 1,3 bisphosphoglycerate is not occurring.
From this point of view it can be said that less energy will be formed in the mutant during glycolysis than that of normal one.
So the formation of pyruvate from glucose will be energetically less favorable.
Answer:182.25 joules
Explanation:
Mass=0.18kg
Velocity=45m/s
Kinetic energy=(mass x(velocity)^2)➗2
Kinetic energy=(0.18 x 45^2) ➗ 2
Kinetic energy=(0.18x45x45) ➗ 2
Kinetic energy=364.5 ➗ 2
Kinetic energy=182.25
Kinetic energy=182.25 joules
You need to do speed divided by time
