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.
I’m not sure what your asking. But if you have 2 different alleles of a gene it will be heterozygous
Answer:
i think is D water pollution, logging and mining
Answer:
The genome is usually organized as a single linear or circular molecule of nucleic acid
Explanation:
