Question

11. The DNA molecule below is incubated with all four dNTPs, Mg+ and a DNA polymerase that lacks 5'+3' exonuclease activity but has 3-5'exonuclease activity and polymerase activity. What will be the structure of the final product and what is the first step in the reaction sequence? 3' OH-TGCGAATTAGCGACAT-P5' 5'P-ATCGGTACGGCGCTTA-P 3' 12. You are screening a chemical for potential mutagenicity using the Ames test. You used a his-S. typhimurium bacterium for this test and obtained the following results: Chemical + Control (HO) Chemical rat liver extract 317 20 A) What was the purpose of also testing water and why did it result in 21 colonies? B) The rat liver appears to modify the chemical into a mutagen. How would you show that this is due to modification the chemical and not due to rat liver components?

Answer 1

11)

Final product :

3'P- TAGCCATGCCGCGAATTAGCGACAT-P5'

5' P- ATCGGTACGGCGCTTA-P3'

dna polymerase 3'-5' exonuclease is responsible for the proof reading activity and remove the incorrect base thymine and add guanine in place of thymine.

5'-3' polymerase activity catalyse the addition of nucleotide to the 3' end of the DNA because 3'end contain hydroxyl group with can form the bonds with the phosphate of the incoming nucleotide.

12) Ames test detect the mutation in the organism.

Water act as the control for the experiment . We observed colonies in water may be due to the contamination.

We should observed colonies in rat liver extract. If we obtained no colonies or less colonies that means rat liver extract alone cannot increase the colonies number.

Chemical + liver extract result in increase colonies because rat liver extract modify the chemical.

4) genes are linked because parents have high frequency and double crossover has low frequency.

K+ c e. 25

K c+ e+ 20

These are single crossover

Distance between c and k = (20+25+2+1 /1000)X100

= 4.8cM

K c+ e 34

K+ c e+ 35

Distance between k and e = ( 35+34+2+1/1000)X100

= 7.2cM

So, the correct map is K c e