The answer is B.
neutral: OH- = H+
acid: H+ > OH-
base OH- > H+
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:
the correct answer would be
A. atoms, molecules, cells, tissues, organs
Answer:
a. p53 activates transcription of WAF1.
Explanation:
<u>WAF1 transcription occurs independent of p53 during oxidative stress so p53 does not play any role in cell cycle arrest in the signaling pathway which involves WAF1 . </u>
In rest of the mentioned options, p53 plays a role directly or indirectly. During double stranded lesion in DNA in G1 phase, a sensor protein known as ATM binds the DNA lesion site. ATM is a serine/threonine kinase which phosphorylates another kinase known as chk2. After phosphorylation,<u> chk2 stabilizes transcription factor p53.</u> p53 further acts as a transcription factor for the synthesis of a protein known as p21 which inhibits G1 phase specific CDK and ultimately cell is arrested in G1 phase. The cell remains in arrested state until the DNA lesion is fully repaired. <u>Hence, p53 indirectly blocks G1 to S transition with the help of p21. </u>
As such <u>WAF1 transcription factor involving pathway</u><u> </u><u>also requires p21 protein for causing cell cycle arrest but in this pathway p21 is not synthesized with the help of p53. </u>