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: d) Cabbage.
Explanation:
A fruit can be define as a fleshy product of the plants. It encloses within it seeds produced after the sexual mode of reproduction of the males and female gametes in plants. The fruit can be sour, or sweet.
Among the option given cabbage is the correct option. This is because of the fact that cabbage is a leafy green biennial vegetable crop. It exhibit dense leaved head. The white and red varieties are also available with tightly packed leaves.
Answer:
The glycosidic linkage between glucose molecules in maltose is α-1,4-glycosidic linkage.
Explanation:
As maltose is a reducing sugar, it must possess the linking of its two glucose molecules in such a way that an anomeric carbon is left for the the formation of an aldehyde group. The glucose molecules in maltose are linked in such a way that the first carbon atom of one of the glucose molecules is attached to the fourth carbon of the other glucose molecule. This is known as head to tail fashion and termed as α-1,4-glycosidic linkage.
