Answer:
The mentioned parental types are c+m- and c-m+. Thus, the recombinants will be c+m+ and c-m-.
Now, the given distance between c and m is 8 map units. Thus, the recombinant frequency is 8% or 0.08.
The total recombinants from 1000 plaques will come out to be 80,
Thus, the recombinants of each type will be 40.
Total parental type will be 920, and therefore, each parental type count will be 460.
Thus, expected c+m- = 460, expected c-m+ = 460, expected c+m+ = 40 and expected c-m- = 40.
Answer:
True
Explanation:
Sewage contains many substances in it like suspended solids, organic and inorganic impurities, nutrients, saprotrophic and disease causing bacteria and other microbes. Hence it is said to be a complex mixture.
Prokaryotes reproduce by *BINARY FISSION* which creates *2*exact daughter cell clone(s) of the original parent cell.
The correct answer is: Glycogen phosphorylase would remain phosphorylated and retain some activity.
Glycogen phosphorylase is directly involved in the regulation of glucose levels since it is a glucose sensor in liver cells: when glucose levels are low, phosphorylase is active and it has PP1 bound to it (phosphatase activity of PP1 is prevented). Therefore, there phosphorylase a will accelerate glycogen breakdown.
I would say A. Quantitative data involves numbers and stuff like that. Unlike qualitative data, which involves pictures and not numerical data. I hope this helped.