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P. 324 #13,14,16,27,28,30,34
13. Hydrogen bonds, which are individually weak but collectively strong enough to
ensure the stability of the DNA molecule, hold together the two DNA strands in the
double helix. It is important that these bonds are relatively weak because of the
function of DNA, which requires the constant “breaking” of these hydrogen bonds for
replication and protein synthesis. If the bonds between the strands were stronger,
replication and protein synthesis would require much more energy, which, over an
evolutionary period, would likely alter the manner in which the processes occur in
organisms.
14. Maintaining the correct nucleotide sequence in newly formed and existing strands of
DNA is integral to the functioning of an organism, because mistakes can lead to the
mutation of genes that control vital living processes. Ensuring the quality control of new
DNA strands is the job of DNA polymerases I and III. During replication, they proofread
the newly synthesized strands for missing or mismatched bases. If they find a
mismatched nucleotide, they will backtrack over the strand, excise it, and replace it with
the appropriate nucleotide. In existing strands, DNA, which can be damaged by such
hazards as radiation and chemicals, are “inspected” by certain enzymes. If any
mutations are found in the DNA, another enzyme will excise the damaged region, and
then DNA polymerase and DNA ligase will replace the damaged nucleotides with correct
ones. In some instances, mutations can be beneficial for organisms if they provide a
competitive advantage; in fact, mutations are a major cause of evolution because they
allow organisms to physically adapt to the changing environment in which they live.
16. The seven stages, which break down the events of DNA replication, are listed
below:
1) The enzyme gyrase relieves any tension from the unwinding double helix.
2) The enzyme helicase breaks the hydrogen bonds holding the two complementary
parent strands together, resulting in
an unzipped helix that terminates at the replication fork.
3) Single-stranded binding proteins anneal to the newly exposed template strands,
preventing them from reannealing.
4) The enzyme primase lays down RNA primers that will be used by DNA polymerase III
as a starting point to build the
new complementary strands.
5) DNA polymerase III adds the appropriate deoxyribonucleic triphosphates to the 3'
end of the new strand, using the
template strand as a guide. The energy in the phosphate bonds is used to drive the
replication process. The leading strand
is built continuously toward the replication fork. A lagging strand comprising short
segments of DNA, known as Okazaki
fragments, is built discontinuously, away from the replication fork.
6) DNA polymerase I excises the RNA primers and replaces them with the appropriate
deoxyribonucleotides. DNA ligase
joins the gaps in the Okazaki fragments by the creation of a phosphodiester bond.
7) DNA polymerase I and DNA polymerase III proofread by excising incorrectly paired
nucleotides at the end of the
complementary strand and adding the correct nucleotides
27. If one strand contains the nucleotide proportions 15% A, 30% T, 20% G, and 35%
C, the other strand in the double helix must contain the bases in the following
proportions: 30% A, 15% T, 35% G, and 20% C, since bases are complementary.
28. The roles of the following enzymes in DNA replication are outlined below:
DNA ligase: joins DNA fragments by catalyzing the formation of a bond between the 3'
hydroxyl group and the 5' phosphate group on the sugar-phosphate backbones.
DNA gyrase: relieves tension produced by the unwinding of DNA.
DNA helicase: unwinds double-helical DNA by disrupting hydrogen bonds.
DNA polymerase I: removes RNA primers and replaces them with the appropriate
deoxyribonucleotides during DNA replication.
DNA polymerase III: synthesizes complementary strands of DNA during DNA replication
in the 5' to 3' direction.
30. Silent mutations do not lead to deleterious effects in an organism because they
either occur in introns, noncoding regions of DNA, or have no effect on the translation
of proteins, because of the redundant nature of the genetic code. Introns are
regions are cut out of the mRNA in the process of transcription, thus preventing the
mutation from manifesting itself in the organism. Silent mutations that do occur in
coding regions, exons, do not affect the translation process and thus the
organism as a whole, because many codons can be coded for with multiple base
sequences. Thus, even if one base is mutated, the codon may still code for the same
amino acid.
34. In a double-stranded DNA molecule complementary bases (A–T, C–G) must exist in
the same proportion. On this basis, the following results were determined:
Sample A: Single stranded, the proportion of adenine is not equal to the proportion of
thymine, and the proportion of guanine is not equal to the proportion of cytosine.
Sample B: Double stranded, the proportion of adenine is equal to the proportion of
thymine and the proportion of guanine is equal to the proportion of cytosine.
Sample C: Single stranded, the proportion of guanine is not equal to the proportion of
cytosine.