Which of the following relationships does not equal 1. In each case the letters

Question

Which of the following relationships does not equal 1. In each case the letters for the nitrogenous bases represent the number of those bases in a particular nucleic acid molecule, (ds = double stranded) (mark All that are appropriate/2 needed) A. Purines/Pyrimidines = 1 (ds DNA), B. A + G/T + C = 1 (dsDNA), C. A + T/C + G = 1 (ds DNA), D. C + G/A + U = 1 (dsRNA), E. U + C/G + A = 1 (dsRNA) A mutant bacterial strain has been found that has defective primase enzymes with abnormal function. Which of the following activities are NOT immediately possible for this organism? (mark All that are appropriate/2 needed) A. DNA ligase activity in association with the conversion of discontinuous to continuous strands, B. DNA polymerase I binding to the origin site to Initiate replication, C. DNA polymerase III elongation of/production of the Okazaki fragments, D. Okazaki fragment removal during discontinuous replication, E. Production of the continuous/leading strand of DNA at ori site

Explanation / Answer

1)

Chargaff's rule states that purine:pyrimidine ratio in DNA of any organism is 1:1. This pattern is found on both strands of the DNA. So, the first option doesn't apply.

The rule specifically emphasizes that Adenine content is equal to thymine and gunanine is equal to that of cytosine. A=T, C=G.

So, A+C=T+G or A+G=T+C in the ds DNA. So, option 2 is also ruled out.

A+T=G+C is not a true statement. As per the formula stated above, its need not be necessary to have equal content of A+T and G+C

So, option c) is applicable.

C+G/A+U on RNA is nothing but C+G/A+T on DNA. This is because Thymine in DNA is replaced by Uracil in RNA.

So, option d) is also applicable.

U+C/A+G on RNA is similar to T+C/A+G on DNA. So, the ratio is 1:1. HEnce, option e) is not applicable.

2)

Primase is the enzyme that synthesizes short segments of RNA that are complementary to ssDNA template strand. The nucleotide segment is elongated by DNA polymerase activity. Primase enzyme is first activated by helicase and synthesizes the primers which are around 10 bases in length. To the two sub units of primase, the DNA polymerase binds to take over the DNA synthesis.

Polymerase III is the holozyme that acts on the replisome complex formed after the action of helicase. Further, a complex set of process that involves elongation and synthesis of okazaki fragments takes place.

Binding of DNA pol I to the replisome complex is also inhibited, as there are no primers to form an appropriate complex.

Hence, right options are b) and c)

Ligase activity occurs after synthesis of okazaki fragments. Same is the case with option d) and e). Hence, they are wrong.

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