the two main processes are: transcription and translation
If a bacteria cannot ferment glucose, then we do not test its ability to ferment other carbohydrates because the glucose is monosaccharides, the bacteria required enzymes that used to ferment glucose.
Bacteria cannot ferment carbohydrates because carbohydrates may include non-reducing sugar like sucrose and lactose, which is disaccharide, that must be cleaved into monosaccharides. Not all, bacteria can do this to may or may not ferment sucrose.
Many microorganism can grow in the base broth without the carbohydrates, but if they can ferment a sugar that is available. It is possible that one bacteria metabolize some sugar but can't work on other.
To learn more about non-reducing sugar here
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First stage is peat, then followed by lignite, bituminous and anthracite. Mark as branliest tq
95°F=35°C :)
Explanation: when using the formula (y-32)•5/9=x 95°F can be converted into 35°C
Answer:
Option A , The allele for yellow is dominant to white
Explanation:
It is given that 23 percent of the sheep have white fat and 77 percent have yellow fat.
Thus, it is clear that sheeps with yellow fat is a dominant characteristics.
Usually, the frequency of homozygous recessive allele in a given population is represented by 
and here 
Frequency of allele for white fat 
While frequency of homozygous dominant allele and heterozygous dominant allele is represented by 
And here 
Allele for yellow fat 
Hence, option A is correct.
