Answer:
B. Problem - Development - Research - Engineering solution
Explanation:
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Genetic diversity serves as a way for populations to adapt to changing environments. With more variation, it is more likely that some individuals in a population will possess variations of alleles that are suited for the environment. Those individuals are more likely to survive to produce offspring bearing that allele.
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:
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