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:
75%
Explanation:
Three of the boxes contain an Upper case F, therefore 75% of the boxes have an Upper case F. It's like quarters, if you have 4 quarters you have 100 cents, if you have 3 you have 75 cents, if you have 2 you have 50 cents, and if you have 1 you have 25 cents.
Answer:
The presence of other acids in our juice causes our calculated concentration of citric acid to be falsely high.
Explanation:
The presence of other acids in our juice causes our calculated concentration of citric acid to be falsely high and we would have to account for the other acids in this case.
Answer:
The answer I think is B it would be same zygotes WW (widows peak)
Explanation: