Download 7 dna structure and chromosome organization

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Hyde Chapter 7—Solutions
7
DNA STRUCTURE
AND CHROMOSOME
ORGANIZATION
CHAPTER SUMMARY QUESTIONS
2. Avery, MacLeod, and McCarty performed experiments showing that DNA was the
transforming agent, and they are thus generally given credit for formalizing the
notion that DNA, not protein, is the genetic material. Chargaff, Hershey and Chase,
Fraenkel-Conrat, and several others also helped shape the general view. At the time
that Watson and Crick published their model, the scientific community knew that
DNA was the genetic material but didn’t know its structure.
4. 3'-GTAATCTGGCCATCTG-5'.
6. Only proteins and nucleic acids were ever considered seriously.
8. In general, prokaryotes are small, have a relatively small circular chromosome, and
have little internal cellular structure compared to eukaryotes. Most prokaryotic
messenger RNAs are polycistronic, under operon control; eukaryotic messenger
RNAs are highly processed, monocistronic, and usually not under operon control.
Prokaryotes are mostly single-celled organisms, whereas eukaryotes are mostly
multicellular. Eukaryotes have repetitive DNA, absent for the most part in
prokaryotes. Prokaryotic chromosomes are not complexed with protein to anywhere
near the same extent that eukaryotic chromosomes are.
10. The evidence for the uninemic (single DNA molecule) nature of the eukaryotic
chromosome is summarized in figures 7.19 and 7.20.
12. The length of DNA associated with nucleosomes was determined by footprinting, in
which free DNA was digested, leaving only those segments protected by
nucleosomes.
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14. See figures 7.24, 7.27, and 7.28 for the relationship of the 110-Å, 30-nm, and
240-nm chromosome fibers.
16. See figure 7.35.
18. A chromosomal puff can be stage-specific, tissue-specific, constitutive, or
environmentally induced. It is an area of active transcription in a polytene
chromosome.
20. The C-value paradox involves the issues of the excessive amounts of DNA in
eukaryotic cells and the difference between eukaryotic species that seem to have
similar complexity. It is explained by the large amount of structural DNA in
chromosomes as well as the large amounts of short and long interspersed elements
(SINEs and LINEs).
EXERCISES AND PROBLEMS
22. a. 28% G, 28% C, 22% A, 22% T.
b. Same percentages except 22% U, 0% T. Chargaff’s rule states that the quantity
of A = T and the quantity of G = C. If G = 28%, then C = 28% and G + C =
56%. The sum of all bases must equal 100%. Therefore, (A + T) = 100 – 56 =
44. Since A = T, (1/2)(44) = 22%. This is the amount of both A and T. For an
RNA molecule, proceed the same way, except remember that U replaces T, so
we have 22% U.
24. The genetic code would somehow be read in number of tetranucleotide units, in
which each unit consists of one each of the four bases (G, C, T, A). For example, one
unit might be the amino acid alanine, two units might be the amino acid arginine, and
so on.
26. (Lowest) 69°C, 73°C, 78°C, 82°C, 84°C (highest). Remember that a G–C pair has
three hydrogen bonds and thus requires more energy to be broken than an A–T pair.
Therefore, the higher the melting temperature, the higher the G–C content. Simply
arrange the molecules from lowest to highest melting temperature.
28. There must be regions of complementarity within the single-stranded regions. A
melting temperature indicates some regions that are double-stranded. We can
envision at least two different possible configurations:
1. Whole molecule complementary
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Hyde Chapter 7—Solutions
2.
Fragment complementary
In fact, most single-stranded molecules have some regions that are complementary.
30. In most cell types in higher eukaryotes, the number of divisions is finite. In these
cells, telomerase is inactive. As the divisions proceed, the telomeres shorten until a
length is reached that somehow signals the cell to stop dividing. If a mutation arose
such that telomerase became active in an inappropriate cell type, the telomeres could
have a functional length reestablished each generation, signaling the cell to keep
dividing. The result of uncontrolled cell growth is cancer.
32. If one strand contains 30% A, 20% C, 10% G, and 40% T, the complementary DNA
strand will contain 30% T, 20% G, 10% C, and 40% A. Therefore, the entire
molecule will contain 35% A, 15% C, 15% G, and 35% T (basically, the average of
the two strands).
The chromosome contains a total of 6 108 nucleotides (each base pair
contains two nucleotides), and therefore 6 108 phosphorus atoms (one per
nucleotide).
b. B-DNA contains 10 base pairs per 3.4 nm (one helical turn). Thus, each base
pair represents 0.34 nm. The length of the DNA in this chromosome would be
(3 108 bp) (0.34 nm/bp) = 1.02 108 nm. To convert to micrometers,
simply divide by 1000. Therefore, the genome will be 1.02 105 m.
34. a.
36. Nucleases will not digest DNA to which proteins, in this case histones, are bound.
Approximately 200 base pairs must be wrapped around proteins, and there must
be some unprotected regions between these 200 base-pair regions. Multiples of
200 base pairs appear because the nuclease does not cut at every unprotected
sequence.
38. A = 13.3%, G = 26.7%, C = 15%, T = 45%. Begin by writing the given information
as equations:
G = 2A; T = 3C; (C + T)/(A + G) = 1.5
Now substitute for G and T in the third equation:
(C + 3C)/(A + 2A) = 1.5
4C/3A = 1.5
4C = 4.5A
C = 4.5A/4
Now, remember that the sum of all bases must = 100%.
A + G + C + T = 100
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Substitute equalities into this equation:
A + 2A + (4.5A/4) + (13.5A/4) = 1
3A + 4.5A = 1
7.5A = 1
A = 1/7.5 = 0.133 = 13.3%
G = 2A = 2 (13.3) = 26.7%
C = [(4.5)(0.133)]/4 = 15%
T = 3C = 45%
40. a.
B-DNA contains 10 bp per turn, and therefore this molecule will contain
100,000/10 = 10,000 turns.
b. Z-DNA contains 12 bp per turn, and therefore this molecule will contain
100,000/12 = 8333.34 turns.
42. No, because you could have a single-stranded DNA molecule with an (A + G)/(C +
T) ratio of 1.0 (example: a DNA molecule with 30% A, 20% G, 30% C, and 20% T).
To arrive at the conclusion that a particular DNA molecule is double-stranded, we
must know the composition of each of the four bases.
44. In their 1953 paper, Watson and Crick give one reason why DNA has the
deoxyribose sugar: “It is probably impossible to build this structure with a ribose
sugar in place of the deoxyribose, as the extra oxygen atom would make too close a
van der Waals contact.” From an evolutionary standpoint, it is also useful for the cell
to be able to differentiate RNA from DNA for the removal of RNA primers during
DNA replication; the oxygen difference in the sugars would provide that ability. As
for the difference in pyrimidines (thymine versus uracil), we believe the difference
has to do with the cell’s ability to repair spontaneously mutated bases. Although we
will go into this in more detail in chapter 18, we can mention here that the most
common spontaneous mutation of bases is of cytosine to uracil by deamination (see
figure 7.8). If uracil was being created spontaneously in DNA and was a normal base
in DNA, then the repair systems would not be able to differentiate a normal uracil
from a mutated uracil and many mutations would go unrepaired. Hence early in
evolution, thymine was probably substituted for uracil in DNA to alleviate that
problem.
46. Comparative DNA studies can be helpful in understanding the roles of the various
types of DNA in the eukaryotic chromosomes if there are cases in which there are
remarkably large differences in the amount of DNA in similar species. It can then be
inferred that the basic developmental plan of an organism is contained in the one
with the lower amount of DNA, and the extra DNA in the species with more DNA
may be superfluous. We do have cases in which amphibians differ by as much as
100 times the amount of DNA found in similar species. The puffer fish has only onesixth the amount of DNA as other higher eukaryotes.
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CHAPTER INTEGRATION PROBLEM
a. First, let’s review a few facts about chromosomes and cell division: (1) The number
of chromosomes in a cell is equivalent to the number of functional centromeres.
(2) The number of DNA molecules is equivalent to the number of chromatids. (3) The
number of DNA molecules doubles during the S phase (DNA replication) of
interphase; however, the number of chromosomes stays the same because the sister
chromatids are still attached at the same centromere. (4) Mitosis maintains the
diploid number, and so after cytokinesis, each cell will have the diploid number of
chromosomes, with each chromosome consisting only of a single chromatid.
(5) Meiosis I is a reductional division (2N 1N), which yields a haploid number
of chromosomes, with each chromosome consisting of two sister chromatids.
(6) Meiosis II is an equational division (1N 1N), which maintains the haploid
number of chromosomes, but with each chromosome now consisting of only a single
chromatid. (7) During anaphase of mitosis and anaphase II of meiosis, the number of
chromosomes per cell will double, albeit temporarily, because the centromere will
divide and sister chromatids will separate to opposite poles of the cell. Now, we can
determine the numbers with confidence!
Stage
i. During G1 of interphase
ii. Start of prophase of mitosis
iii. End of anaphase of mitosis
iv. After cytokinesis of mitosis
v. Start of prophase I of meiosis
vi. Start of prophase II of meiosis
vii. After cytokinesis of meiosis II
Number of Chromosomes
per Cell
28
28
56
28
28
14
14
Number of DNA
Molecules per Cell
28
56
56
28
56
28
14
b. At the start of mitosis and meiosis I, the cell would have already undergone DNA
replication during interphase. Therefore, the nucleus will contain two times the
diploid number of base pairs. At the end of mitosis, each cell will have the normal
diploid number. At the end of meiosis II, each cell will have one-fourth of the
amount of DNA that was present at the start of meiosis I.
The percentage of adenines can be determined using Chargaff’s rule. If G + C =
60%, then A + T = 40%. Since A = T, then A = 20%. Therefore, the number of
adenines can be calculated by multiplying the total number of base pairs by 2 (to get
the overall number of bases) and then multiplying that by the percentage of A.
Stage
i. At start of mitosis
ii. At end of mitosis
iii. At start of meiosis I
iv. At end of meiosis II
Total Number of
Base Pairs per Cell
Number of
Adenines per Cell
4 109
2 109
4 109
1 109
1.6 109
0.8 109
1.6 109
0.4 109
Hyde Chapter 7—Solutions
c.
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i.
The nuclei of a secondary oocyte and a first polar body are expected to have the
same weight. Both cells result from a primary oocyte that has undergone meiosis
I. While the division of the primary oocyte’s cytoplasm is asymmetric (most of
the cytoplasm goes to the secondary oocyte), the nucleus of the primary oocyte
divides symmetrically yielding two equivalent daughter nuclei. The female is
homogametic (XX), so the nuclei of the secondary oocyte and the polar body
will each inherit an X chromosome and 22 autosomes.
ii. It depends which secondary spermatocyte is used for comparison with the
secondary oocyte. Meiosis I in the male will yield two secondary spermatocytes.
The male is heterogametic (XY), so the two secondary oocytes will have
different chromosomal compositions in their nuclei: one will have 22 autosomes
+ X, and the other will have 22 autosomes + Y. The X chromosome is more than
twice the size of and will be heavier than the Y chromosome. Therefore, the
secondary spermatocyte with a nucleus containing the Y chromosome will be
expected to be lighter than the nucleus of a secondary oocyte. On the other hand,
a secondary spermatocyte with a nucleus containing the X chromosome will be
expected to have the same weight as the nucleus of a secondary oocyte.
d. The proportions of bases in this genome are A = 20% (or 0.2), T = 20% (or 0.2), C =
30 % (or 0.3), and G = 30% (or 0.3). The probability of finding the sequence
5'-TAGAC-3' is given as 0.2 0.2 0.3 0.2 0.3 = 0.00072. Therefore, the
number of times the sequence is expected to occur is 0.00072 (2 109) =
1,440,000 times.
e.
Consider the simplest case that will yield a strand of DNA with an (A + G)/(C + T)
ratio of 0.5: this is if A + G = 0.5 and C + T = 1. The complementary strand would
therefore have a content of A + G = 1 and C + T = 0.5, and a ratio of 1/0.5 = 2.
f.
First, calculate the mass of a single base pair = 660 Daltons (1.67 10–24 g/Daltons) = 1.1 10–21 g. The total number of base pairs in the animal’s body
will be (0.33 g)/(1.1 10–21 g/bp) = 3 1020 bp. Each base pair in the B form of
DNA corresponds to 0.34 nm. Therefore, the length of the DNA will be (3 1020 bp) (0.34 nm/bp) = 1 1020 nm. Because 1.0 km represents 1 1012 nm,
this animal’s DNA if stretched end-to-end would be (1 1020 nm)/(1 1012 nm/km) = 1 108 km.