Important words and concepts from Chapter 16,
Campbell & Reece, 2002 (1/29/2005):
by Stephen T. Abedon (abedon.1@osu.edu)
for Biology 113 at the Ohio State University
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Course-external links are
in brackets Click [index] to access site index Click here to
access text’s website Vocabulary
words
are found below |
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(1) Chapter title: The Molecular Basis of
Inheritance
(a)
[the molecular basis
inheritance (Google Search)] [index]
(a)
Chromosomes consist of DNA
and protein
(b)
Which is the hereditary material? The history of our understanding is
outlined in the efforts of the following:
(i)
Griffith, 1928
(ii)
Avery, MacLeod, and McCarty, 1944
(iii)
Hershey and Chase, 1952
(iv)
Chargaff
(c)
[chromosomes (Google Search)] [index]
(3)
(a)
(b)
See Figure 16.1,
Transformation of bacteria
(c)
[Griffith 1928 (Google Search)] [index]
(4) Avery, MacLeod, and McCarty, 1944
(a)
Avery et al. found that Griffith’s
transmittable hereditary molecule is DNA
(b)
In particular, they showed that the transmittable hereditary molecule
was susceptible to the DNA hydrolyzing enzyme known generically as DNAse
(c)
[Avery MacLeod McCarty,
DNAse OR DNAase, DNAse, DNAase (Google Search)] [index]
(a)
Hershey and Chase showed that the hereditary material in T2
bacteriophages is DNA, thereby generalizing Avery, MacLeod, and McCarty’s observation
(b)
See Figure 16.2, The
Hershey-Chase experiment
(c)
[Hershey Chase, Alfred Day Hershey
(Google Search)] [the bacteriophage ecology
group (Microdude)] [index]
(6) Chargaff (Chargaff’s rule)
(a)
Chargaff found that different species of organisms have different DNA nucleotide compositions
(b)
Chargaff’s rule states that the fraction of nucleotides that makes up
an organism’s DNA always behaves the rule: the fraction of A’s = the fraction of T’s, and the
fraction of G’s = the fraction of C’s (i.e., the fractions of A + T + G + C = 1; A = T and G =
C; 2 * (A + G) = 1, etc.)
(c)
[Chargaff, Chargaff's rule (Google Search)] [index]
(a)
Franklin and Wilkins are responsible for supplying an X-ray diffraction
of DNA, essentially an in-this-case crude molecular picture of the molecule,
that indicated the basic structural features that DNA possesses:
(i)
The periodicity of DNA
(ii)
The molecule’s uniform width
(iii)
That the nitrogenous bases
stacked 0.34 nm apart
(b)
See Figure 16.4, Rosalind
Franklin and her X-ray diffraction photo of DNA
(c)
[Franklin Wilkins DNA,
DNA X-ray diffraction
(Google Search)] [index]
(a)
Watson and Crick, in 1953, published the double helix model of DNA’s structure
(b)
J. D. Watson and F. H. C. Crick (1953). Molecular Structure of Nucleic
Acids. Nature, vol. 171 (25 April 1953), pages 737-738
(c)
This paper was, arguably, the single most important contribution to
biology (and perhaps even chemistry as well) of the twentieth century
(d)
See Figure 16.5, The double
helix
(e)
The Watson and Crick model:
(i)
Explains DNAs periodicity
(ii)
Explains DNAs uniform width
(iii)
Explains Chargaff’s rule
(iv)
Explains how DNA is replicated
(f)
(you will not be held responsible for the above history)
(g)
[Watson Crick, double helix (Google Search)] [annotated version of Watson
& Crick, 1953!, (nice index of gifs and html
files but I have no idea who the author is)] [index]
(a)
The sequence of bases in a DNA
molecule represent information
(b)
This sequence is effectively unconstrained by the structure of the double helix
(c)
As a consequence, much of the DNA in a chromosome (i.e.,
that which makes up genes) represents unique nucleotide sequences
(d)
The rest consists of various repeated sequences which typically are species specific
(e)
[base sequence (Google Search)] [index]
(10)
(a)
Because of base pairing and the making up of a double helix of DNA of two separate
strands, there exists a redundancy of information carried by the double helix
(b)
Note, however, that the two DNAs do not possess the same sequence
(c)
Instead, each possesses the complementary sequence of the other
(d)
Another way of saying this is that through base pairing one strand is capable of specifying the sequence
of the other strand, and vice versa
(e)
This sequence complementarity forms the basis of DNA-templated DNA
polymerization (i.e., DNA replication)
(f)
[strand complementarity
(Google Search)] [index]
(11) Semiconservative DNA replication
(a)
The specific mechanism by which DNA is replicated
is termed semiconservative
(b)
Despite the long, confusing word used to describe it, this is actually
the simplest mechanism by which template-dependent DNA replication might occur
(c)
In short, semiconservative DNA replication consists of each strand of
DNA in a double helix specifying the polymerization of a new strand which, in turn,
remains attached to its parent strand
(d)
This parent-daughter strand forms a new double helix that consists of
both a parental strand of DNA and a newly synthesized strand of DNA
(e)
(note that above I am using the term “strand” synonymous to “single
molecule of DNA”, i.e., half of a double helix)
(f)
See Figure 16.7, A model for
DNA replication: the basic concept
(g)
[semiconservative DNA
replication (Google Search)] [index]
(12) 5’ à 3’ polarity
(a)
Recall that the sugars of nucleic acids are numbered with primes (i.e.,
1’ through 5’)
(b)
Recall additionally that the backbone of polymerized nucleic acids consists of the 3’ through 5’ carbons
alternating with a covalently bonded phosphate group
(c)
See Figure 16.11,
Incorporation of a nucleotide into a DNA strand
(d)
[5' 3' polarity (Google Search)] [index]
(a)
Recall additionally that the two DNA strands that
make up a double helix are arranged antiparallelly
(b)
That is, starting from one end of the double helix, one strand runs in
the 5’ à 3’ direction while the other runs in the 3’ à 5’ direction
(c)
See Figure 16.12, The two
strands of DNA are antiparallel
(d)
This antiparallel nature of DNA impacts on DNA replication
(e)
[antiparallel strands
(Google Search)] [index]
(14) 5’ à 3’ direction
of synthesis
(a)
During DNA synthesis, incoming subunits arrive with phosphates
(b)
They attach to the 3’ –OH exposed at the end of the growing new strand
(c)
This supplies the phosphate making up the sugar-phosphate backbone of
DNA
(d)
It also constrains the growth of the new DNA strand to the 5’ to 3’
direction
(e)
That is, for each DNA molecule there exists a 5’ end at which no
synthesis is occurring (directly, anyway) and a 3’ end at which synthesis may
occur
(f)
See Figure 16.11,
Incorporation of a nucleotide into a DNA strand
(g)
[5' 3' direction of synthesis
(Google Search)] [index]
(a)
The incoming subunit, in fact, does not carry just one phosphate
(b)
Instead it carries three phosphates (i.e., their structure is analogous
to that of ATP)
(c)
Nucleosides that carry a single phosphate are called nucleotides and this is what remains following the addition of
a nucleoside triphosphates to a growing DNA (or RNA) polymer
(d)
The hydrolytic removal of two of these phosphates supplies the energy employed to attach the subunit to the 3’ –OH of the
growing DNA strand
(e)
See Figure 16.11,
Incorporation of a nucleotide into a DNA strand
(f)
[nucleoside -HIV (Google Search)] [index]
(a)
The enzyme that catalyzes this template-directed
conversion of nucleosides into an elongated DNA strand is called DNA
polymerase
(b)
Note that DNA polymerase can elongate a strand of DNA only in the 5’ à 3’ direction
(c)
See Figure 16.11,
Incorporation of a nucleotide into a DNA strand
(d)
See Figure 16.12, The two
strands of DNA are antiparallel
(e)
[DNA polymerase (Google Search)] [index]
(a)
DNA polymerase attaches new nucleotides with high fidelity (thus reducing errors)
(b)
This high-fidelity nucleotide addition
requires the existence of a 3’ –OH
(c)
This means that DNA polymerase cannot initiate DNA replication since, at the
start of DNA replication, the to-be-synthesized DNA strand does not yet possess
a 3’ –OH (i.e., the strand does not yet exist)
(d)
This problem of how to initiate DNA replication in the absence of a 3’
–OH is solved by priming using RNA
(e)
See Figure 16.14, Priming
DNA synthesis with RNA
(f)
[RNA priming (Google Search)] [index]
(a)
DNA replication is initiated with RNA by
an enzyme called primase
(b)
Primase can initiate template-directed polymerization without a 3’ –OH
(c)
Thus, DNA polymerase uses a RNA 3’ –OH to
initiate replication
(d)
The RNA is then eventually replaced by DNA
(e)
Note that replacing the RNA with DNA at the very ends of linear
chromosomes is a problem (no matter what, the very end will never have a 3’
–OH), thus explaining, in part, the problem of telomere erosion in
eukaryotes
(f)
See Figure 16.14, Priming
DNA synthesis with RNA
(g)
[primase (Google Search)] [index]
(a)
One place that primase acts is at certain DNA
sequences called origins of replication
(b)
This is the site of priming of the leading strand of DNA replication
(c)
See Figure 16.10: Origins of
replication in eukaryotes
(d)
[origins of replication
(Google Search)] [index]
(a)
One role of these replication origin proteins is to open up the double helix so that both strands are exposed as single-strand
DNA, i.e., as potential templates
(b)
The local “bubble” created by this separation of strands about the
origin is bordered at each end with a replication fork
(c)
It is at these replication forks that the parent double helix is unwound and daughter DNA strands are
synthesized, thus converting one double helix into two
(d)
See Figure 16.10: Origins of
replication in eukaryotes
(e)
[replication fork (Google Search)] [index]
(a) Note that because of the antiparallel nature of the double helix, as the replication fork opens, for one DNA strand the opening occurs in t