Can someone help me on questions 1 - 3? I got the following answers 1) D 2) D 3)

Question

Can someone help me on questions 1 - 3?

I got the following answers

1) D

2) D

3) help

Please help me check my answers as I don't want to fail this class thank you so much.

QUESTION 1 1 points Save Answer The role of the mismatch repair mechanism is to. perform a form of recombination repair remove any chemical structures added to the DNA maintain the correct number of chromosomes make sure that every adenine is paired with a thymine and every cytosine with a guanine QUESTION 2 1 points Save Answer A single nucleotide is composed of? sugar, and one base. multiple phosphate groups, one one phosphate group, one sugar, and one base. multiple phosphate groups, multiple sugars, and multiple bases. one phosphate group, one sugar, and multiple bases. QUESTION 3 2 points Save Answer The tollowing two bacteria crossed: Hr amos argt bo" leu+ X F. ampR arg- Assuming the order of the genes is as show above, which of the following recombinants would be expected be expected to be LEAST abundant from this cross? bio leu

Explanation / Answer

1) The role of mismatch repair mechanism is to make sure that every adenine is paired with thymine and cytosine with a guanine. The correct number of chromosomes is maintained during and after the M phase of cell cycle. Recombinational DNA repair is a mechanism critical to survive the UV-induced radiation. Therefore, the answer is option D.

2) A single nucleotide is composed of one phosphate group, one nitrogenous base and one sugar. Though nucleoside (base + sugar) with multiple phosphate groups are also commonly considered as nucleotide, according to the ACS style guide and IUPAC Gold book, nucleoside with one phosphate group is only considered as a nucleotide. The answer is option B.

3) Answer is option B. As F- is ampR, the cells would survive compared to the hfr strain. With accord to the order of the genes given leu would be the first to enter the F- cells followed by bio and arg. In that case, recombinants with leu+ would be abundant and leu- would be least abundant. Therefore F- ampR arg+bio+leu-would be least abundant.

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