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:
b
Explanation:
because the answer is obvious
It can be classified in the phylum chordata.
Phylum : chordata
Sub-phylum: Vertebrata
Division: Gnathostomata
Superclass: pisces (respiration occurs typically by gills)
Hello.
The answer is interphase and <span>mitosis.
</span>interphase is the resting phase between successive mitotic divisions of a cell, or between the first and second divisions of meiosis.
mitosis is a type of cell division that results in two daughter cells each having the same number and kind of chromosomes as the parent nucleus, typical of ordinary tissue growth.
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