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:
C. It uses a renewable resources
Answer:
i need more info to do it
Explanation:
Answer:
una correcta distancia del planeta respecto al sol, la presencia de agua en el planeta, la presencia de la capa de ozono, La presencia de los elementos químicos esenciales, y el campo magnético terrestre
Explanation:
