You have learned how duplicated chromosomes separate in mitosis in your first biology course. In this course, we cover the other side of the cell cycle when single-chromatid chromosomes are duplicated to become two-chromatid chromosomes.
To answer all questions, remember that nucleic acids are elongated by adding nucleotides to free 3’OH group of the deoxyribose sugar. So, you can assume that DNA polymerase adds nucleotides in the same manner.
Be sure to know what the following enzymes (and other proteins) are and when they act in relation to each other for both leading and lagging strands. Helicase, Primase, SSBPs, DNA Polymerase III, DNA Polymerase I, Topoisomerase, Ligase.
- In the diagram below, give names to the labels. Include: leading strand, lagging strand, Okazaki fragment, DNA polymerase III, DNA polymerase I, DNA ligase, helicase, single-stranded binding proteins, primase, RNA primer, 5′ end and 3′ end of parental DNA.
- For each of the labels in figure below, indicate which represents the 5’OH end of the single DNA strand, at which would the next nucleotide be added, which indicates a phosphodiester bond formed by DNA polymerase?
- For the same figure what is the complementary base sequence for this strand (include direction).
- The DNA of an organism has thymine as 20% of its bases. What percentage of its bases would be guanine?
- How does DNA synthesis along the lagging strand differ from that on the leading strand?
- For the last Okazaki fragment formed on the lagging strand of a linear DNA molecule, explain how this results in the shortening of the chromosome. Outline how telomerase acts to mitigate the shortening of linear chromosomes.
HTK General Biology 2 
1. a) SSBPs
b) DNA pol III
c) leading strand
d) 5′ end of parental DNA
e) 3′ end
f) helicase
g) RNA primer
h) primase
i) DNA pol III
j) Okazaki fragment
k) DNA pol I
l) lagging strand
m) ligase
2. Shouldn’t it be the 5′ phosphate end of the DNA strand or the 3′ OH end? The 3′ OH end is (d), the 5′ phosphate end is (a), the next nucleotide would be added at (d), the phosphodiester bond formed by DNA polymerase is (b)
3. 3′-TAC-5′
4. %A = %T, so %A+T = 40%. %G+C = 60% and %G = %C, so %G = 30%
5. On the lagging strand, DNA helicase is moving in the opposite direction of the synthesis of DNA. A “gap” forms between the opening of the replication fork and the newly synthesized strand. To prevent the degradation of the single stranded DNA, DNA polymerase III replicates only small segments of DNA at a time, creating Okazai fragments, and always going “backwards”, towards the newly separated helix. To join these fragments together, after DNA pol. I has removed the primers, DNA ligase creates a bond between each. The replication of the lagging strand is discontinuous, while it is continuous for the leading strand since the synthesis occurs in the same direction as the opening of the helix and there is no “gap” formed.
6. A RNA primer was at the very end of the new DNA strand and it was removed by DNA pol. I, leaving a small unreplicated section. Synthesis can not occur since there is nowhere to place a new primer. The new DNA strand would be shorter than the previous one. Telomerase corrects that by elongating the template DNA strand since the enzyme itself contains a primer. There will then be room available for a primer and the previously unreplicated DNA can be synthesized.
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