In a genetics lab, kim and maria infected a sample from an e. coli culture with a particular virulent bacte- riophage. they noticed that most of the cells were lysed, but a few survived. the survival rate in their sample was about 1 x 10-4. kim was sure the bacte riophage induced the resistance in the cells, while maria thought that resistant mutants probably already existed in the sample of cells they used. earlier, for a different experiment, they had spread a dilute suspension of e. coli onto solid medium in a large petri dish, and, after seeing that about 105 colonies were growing up, they had replica-plated that plate onto three other plates. kim and maria decided to use these plates to test their theories. they pipette a suspension of the bacteriophage onto each of the three replica plates. what should they see if kim is right? what should they see if maria is right?

the match lol

hey, i believe your answer would be d. sugar ribose and the base uracil. i believe this is the answer, because rna consists of sugar ribose, adenine, guanine, cytosine, and uracil.

your first option is incorrect, because that's dna. where it consists of sugar deoxyribose, adenine, guanine, cytosine and thymine.

your second option is also incorrect, this is because sugar deoxyribose is in dna, and uracil is in rna. these two do not make sense together.

your third option being incorrect, because sugar ribose is in rna, but not the base of thymine. rna consists of uracil, not thymine.

lastly your fourth option being correct! this answer is correct, because both sugar ribose, and the base uracil are in rna! ~

answer is d. sugar ribose and the base uracil.

c) 50% there are four chromosomes. after crossing over, 2 of the four have been changed. that means any gamete has a 2 in 4 chance or 50% chance of receiving the mutated chromosome.