Answer:
Lactose is more likely to be utilised by E. Coli than Arabinose because Lactose will yield more energy (ATP) and lactose breakdown will give glucose and galactose and these will enter into the glycolytic pathways to pyruvate for ATP generation until Arabinose which will undergo Pentose phosphate pathway and this does not produce enough energy.
When a specimen is to be observed under the light microscope, scientists usually stain the specimens using different type of chemical stains depending on the type of specimen to be observed. Staining of specimen make the structures that are to be observed in the specimen very visible under the microscope, this make it easy for scientists to observe their structures.
Answer:
The question is lacking in the options. Below are the options to complete the question
a. 70160, 70370, 71010, 74245
b. 43235, 44363
c.76010
d. 70160, 70370, 71010, 43235, 44363
The RIGHT ANSWER IS C
c.76010
Explanation:
To find a foreign body a radiology exam has to be done. In this senerio the foreign body was not found. In the CPT® Index, look for X-ray/Nose to Rectum/Foreign Body. The correct code is 76010
For you to understand the process we can say that the weak will die. If the animal or organism can't get food or find any it is naturally "selected" to die.So a Cuticle could have evolved like that. Hope tgis is useful
