Hemolysis
no. but, .1 is the same as 10 percent. .1 "percent" is different than just .1
1 percent of a percent is 1/10000 or 0.0001; this is the same as saying 1 percent percent. Example: 5 percent percent is 0.0005, so 5 percent percent of 100 is 0.0005 x 100 or 0.05.
1/100 as a percent = 1%
NO. 1 percent is less than 5 percent.
100 percent of 3.5 1 percent is 3.5/100 = 0.035 1 percent of 3.5 billion is 0.035 billion.
The glucose was able to go through the sac. The glucose went from high concentration to low concentration. The glucose is permeable.
Dissolve 100 g glucose in 1 L distilled water (or 10 g glucose in 100 mL disitlled water).
1:10 ratio.
The glucose was able to go through the sac. The glucose went from high concentration to low concentration. The glucose is permeable.
1 glucose molecule = 38 ATP
1% (presumably m/v) glucose solution would contain 1 g of glucose per 100 ml of solution. Therefore the conversion of 1 g / 100 ml to units of mol/L requires that we divide by the molar mass and multiply by the conversion factor of ml to L. Therefore: (1 g / 100 ml) * (1 mol / 180.16 g) * (1000 ml / 1 L) = 0.0555 M = 0.06 M.
If you added more glucose that the Kliger Iron Agar test than is called for you risk getting false results. The bacteria need to exhaust the glucose to turn the solution red or yellow. If too much glucose is present you may not be able to determine if the bacteria can ferment animal proteins.
Trehalose (or mycose) is a disaccharide composed of two glucose residues. The glucose residues are linked by an α 1→4 glycosidic bond.
Answer: 7 x 10-3 g glucoseProcess below:First convert 40 µmol to moles. 1 µmol = 1 x 106 mol.40 µmol x (1 mol)/(1 x 106µmol) = 4 x 10-5 molMultiply 4 x 10-5 mol by the molar mass of glucose (180.156 g/mol).4 x 10-5mol glucose x (180.156 g glucose)/(1 mol glucose) = 7 x 10-3 g glucose
1 glucose molecule = approx. 0,3.10-21 g
The molecular formula of glucose is C6H12O6. 1 mole glucose = 6.022 x 1023 molecules. 1 molecule glucose = 24 atoms 1mole glucose x 6.022 x 1023 molecules/mole x 24 atoms/molecule = 1 x 1025 atoms (rounded to 1 significant figure)
1. increase the number of glucose carriers2. increase glucose concentration