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62
Hyde Chapter 9—Solutions
9
DNA REPLICATION
CHAPTER SUMMARY QUESTIONS
2.
Conservative
Density-gradient
centrifuge tube
Dispersive
DNA
Density-gradient
centrifuge tube
14N
T0
DNA
14N
T0
14N/15N
14N/15N
15N
15N
14N
T1
14N
T1
14N/15N
14N/15N
15N
15N
14N
T2
14N
T2
14N/15N
14N/15N
15N
15N
4. DNA polymerases cannot initiate DNA synthesis. They require the help of the
enzyme primase, a special type of RNA polymerase. Primase creates a 10–12
nucleotide RNA primer, which provides a free 3'-OH group to which DNA
polymerases can add deoxyribonucleotides.
6. The 5' 3' exonuclease activity is used to remove the RNA primers from the DNA
after replication. Ribonucleotides are removed sequentially from the 5' to the 3' ends.
Only DNA polymerase I possesses this activity.
8. There are many similarities and differences to choose from. First the similarities:
(a) Replication is semiconservative.
(b) Replication is bidirectional.
Hyde Chapter 9—Solutions
(c)
(d)
(e)
(f)
(g)
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Substrates are deoxyribonucleotides.
There is a requirement for helicase, single-strand binding proteins, and primase.
DNA is synthesized in the 5' 3' direction.
Replication is continuous on the leading strand.
Replication is discontinuous on the lagging strand, requiring Okazaki fragments.
Replication in eukaryotes differs from that in bacteria in the following:
(a) Multiple origins of replication per chromosome (bacteria have only one).
(b) More DNA polymerases.
(c) Shorter Okazaki fragments (100–200 nucleotides in eukaryotes versus 1000–
2000 nucleotides in bacteria)
(d) Okazaki fragments are removed by RNase.
(e) Issues with nucleosomes and telomeres.
10. A primosome is a helicase plus a primase; it opens the DNA and creates RNA
primers, and is part of the replisome. A replisome includes a primosome plus two
copies of DNA polymerase III; it coordinates replication on both the leading and
lagging strands at the Y-junction.
12. See figure 9.14.
14. DNA polymerase III joins a 5'-triphosphate to the 3'-OH of a nucleotide already
found on the DNA molecule. A pyrophosphate (P–Pi) is released in the process,
leaving only one phosphate in the phosphodiester linkage. DNA ligase joins the
5'-monophosphate of one nucleotide to the 3'-OH of a neighboring nucleotide.
EXERCISES AND PROBLEMS
16. Sequence II. Of the three sequences, it has the highest AT content (70%) and is
therefore the easiest to “unzip.”
18.
Original
5'-ATTCTTGGCATTCGC-3'
DNA complement
3'-TAAGAACCGTAAGCG-5'
RNA primer
3'-UAAGAACCGUAAGCG-5'
Replication is 5' 3'.
20. This problem can be solved using the product rule of probability. The original
solution contained 80% heavy and 20% light DNA.
0.8 15N 0.8 15N = 0.64 15N15N
0.2 14N 0.2 14N = 0.04 14N14N
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Hyde Chapter 9—Solutions
0.8 15N 0.2 14N = 0.16 15N14N
0.2 14N 0.8 15N = 0.16 14N15N
Consequently, the proportions are
0.64 15N15N:0.32 15N14N:0.04 14N14N
22. Theta replication is bidirectional, whereas rolling-circle replication is unidirectional.
Therefore, it will take 30 2 = 60 minutes to replicate the DNA via rolling-circle
replication.
24. Let 15N15N DNA = HH, 14N14N DNA = LL, and 15N14N = HL.
a. 1/4 HL:3/4 LL.
b. 1/32 HL:31/32 LL.
c. 1/128 HL:127/128 LL.
26. a.
The two replication forks combined covered a distance of 1.8 m in
60 minutes, and so 1.8/60 = 0.03 m per minute. Therefore, each replication
fork covers a distance of 0.03/2 = 0.015 m per minute.
b. Each base pair in B-DNA spans 0.34 nm. The distance 0.015 m is equivalent to
0.015 103 nm. Therefore, the speed of the DNA replication fork is (0.015 103)/0.34 = 44 base pairs per minute.
28. Please see figure 9.19 for the origin, figures 9.21 and 9.22 for the continuation, and
figure 9.26 for the termination of DNA replication in E. coli.
30. Finding small pieces or fragments of DNA suggests the Okazaki pieces are only
slowly, if at all, joined: a function of DNA ligase. The fact that not many long DNA
molecules are seen also suggests that the DNA is being broken, implicating a
nuclease as well.
32. a.
Because there are two replication forks, the overall rate of DNA synthesis is
8000 bp per minute. Therefore, the entire chromosome will be replicated
in 40,000,000/8000 = 5000 minutes.
b. Number of replicons 8 minutes = 5000 minutes 1 replicon. Therefore there
are 5000/8 = 625 replicons.
34. At one time molecular swivels, presumably protein in nature, located periodically
along the DNA, were suggested.
36. One way to study mutations that are generally lethal is by isolating temperaturesensitive mutations. These mutations involve amino acids that disrupt the functioning
of the enzymes at some critical temperature but are phenotypically normal at other
temperatures. Thus, the mutant organisms can be kept alive by growing them at one
temperature (the permissive temperature), but their mutant effect can be studied at
the temperature in which the protein function is disabled (the restrictive
temperature). (For additional discussion of these mutations, see chapter 18.)
Hyde Chapter 9—Solutions
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CHAPTER INTEGRATION PROBLEM
a.
Interphase
Interphase
G2
4
3
2
Mitosis
Meiosis II
G2
S
G1
Meiosis I
S
G1
1
0
Time
b. i.
In prophase of mitosis, there are 10 chromosomes, with each consisting of two
sister chromatids. These 20 chromatids contain 20 2 = 40 telomeres.
ii. In prophase II of meiosis, there are 5 chromosomes, with each consisting of two
sister chromatids. These 10 chromatids contain 10 2 = 20 telomeres.
c.
A bivalent consists of a pair of homologous chromosomes, each of which consists of
two sister chromatids. Each chromatid is a double-stranded DNA molecule.
Therefore, the bivalent has 4 2 = 8 strands of DNA.
d.
O1
O2
O3
e.
The speed of DNA synthesis is the same in each of the two replication forks that
proceed from an origin of replication. Therefore, the smaller the amount of DNA in a
replication fork area, the faster its replication.
i. The homolog with the deletion will replicate to the short arm telomere first. This
is simply because there is less DNA to replicate between O1 and the telomere.
ii. Because the length of DNA between O1 and the short arm telomere is the same
in the two homologs, both will replicate to the short arm telomere in roughly the
same time. The same applies for iii and iv.
f.
The cells that are expected to be labeled with tritium are only those that are
replicating their DNA and, therefore, are in S phase. If we assume that cells of this
culture are spread out in all stages of the cell cycle, then the percentage of cells in
each stage should be directly proportional to the length of the stage. Therefore, 8/24
= 33% of cells will be expected to have 3H-labeled DNA.
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Hyde Chapter 9—Solutions
g.
The order of stages in the cell cycle is: G1 S G2 prophase metaphase anaphase telophase. So, the acrocentric chromosome picks up the 3H-label during
the S phase, and would have to proceed through G2 and prophase before it can
appear as a “metaphase” chromosome. The earliest that this can happen is after a
little more than 4 hours (in the case where the label was incorporated very late in
S phase), and the latest is after a little more than 11 hours (in the case where the label
was incorporated very early in S phase).
h.