Answer:
Given merely 40 butterflies were marked, assumed that there were multiple captures of both marked and unmarked butterflies, that the butterflies caught in traps were on the loose to be caught again. There are two mathematical solutions for this, both yielding the same answer which is 100.
• First, to each marked butterfly was taken twice (40 marked X2 = 80 captured) then of the unmarked butterflies the 120 captured must relate to 60 actual butterflies. In which 40 + 60 = 100.
• Secondly, by means of ratios in which 80/200 = 40/X. In this case X also = 100 that will result to the estimated size of the population of wilson park is 100.
Explanation:
<span>These cells cause the reabsorption of bone, thus helping to regenerate bone.
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Answer: the cfu/g Gram-negative bacteria in the fecal sample is C = 3.0 × 10^3
Explanation:
We know that; Gram negative bacteria looks pale reddish in color under a light microscope from Gram staining.
therefore
There are 30 red bacterial colonies counted.
1 mL of from tube 1 was removed and added to tube with 99 mL saline (tube 2) dilution is 1/100.
transferred volume into the plate is 1 mL.
Now, we have to determine the cfu/g Gram-negative bacteria in the fecal sample
Formula to calculate CFU/g bacteria in fecal sample is expressed as;
C = n/(s×d )
where C is concentration (CFU/g)
, n is number of colonies
, s is volume transferred to plate
, d is dilution factor.
so we substitute
C = 30 / ((1/100) × 1)
C = 30 / 0.01
C = 3000
C = 3.0 × 10^3
THERFERE, the cfu/g Gram-negative bacteria in the fecal sample is C = 3.0 × 10^3
Answer:
Thus, during the Citric Acid cycle, the breakdown of glucose into carbon dioxide is completed. There are four redox reactions, three of which yield reduced NADH and one FADH2. Thus, the oxidation of glucose is completed in the Kreb's cycle
