A1.2 Nucleic acids Flashcards
A1.2.1—DNA as the genetic material of all living organisms
Some viruses use RNA as their genetic material but viruses are not considered to be living
Living organisms are constructed out of molecules.
Genes are the instructions for doing this and are
themselves constructed out of molecules.
A key feature in the unity of life is that the molecule
used as genetic material in all living organisms is DNA
(deoxyribonucleic acid).
Some viruses, including coronaviruses (such as Covid) use
RINA instead of DNA as genetic material, but viruses are
not usually considered to be living. Also, DNA and RNA
are very similar-they are the two types of nucleic acid.
A1.2.2—Components of a nucleotide
In diagrams of nucleotides use circles, pentagons and rectangles to represent relative positions of
phosphates, pentose sugars and bases.
The subunits in both DNA and RNA are nucleotides.
Each nucleotide consists of three parts:
* a pentose sugar, with five carbon atoms and a
five-atom “ring”
* a phosphate
* a base, whose molecules contain nitrogen and have
either one or two rings.
In diagrams of nucleotides, these parts are usually shown
as pentagons, circles and rectangles, respectively. The
figure shows how the sugar, the phosphate and the base
are linked up in a nucleotide.
A1.2.4—Bases in each nucleic acid that form the basis of a code
Students should know the names of the nitrogenous bases.
There are four different bases in both DNA and RNA,
three of which are the same. These bases can be
arranged in any sequence along a strand of nucleotides.
The sequence forms the basis of the genetic code that
al organisms use to store information.
Nucleic acid
One-ring bases
(pyrimidines) Two-ring bases
(purines)
DNA
Cytosine Adenine
T Thymine G Guanine
RNA Cytosine A Adenine
U Uracil G Guanine
A1.2.3—Sugar–phosphate bonding and the sugar–phosphate “backbone” of DNA and RNA
Sugar–phosphate bonding makes a continuous chain of covalently bonded atoms in each strand of DNA
or RNA nucleotides, which forms a strong “backbone” in the molecule.
The nucleotides in a strand of DNA or RNA are linked
together by covalent bonds between the pentose sugar
of one nucleotide and the phosphate of the next one.
covalent sugar
to phosphate
b o n d
phosphate of a free nucleotide.
Sugars and phosphates alternate in RNA and DNA
next
nucleotide
is linked
h e r e
However many nucleotides have been linked up,
another can always be added to the end of the chain
where a pentose sugar can make a bond with the
molecules, with an unbroken chain of covalently
bonded atoms in the sequence |-О -P-O-C-C-C-|
where n is the number of nucleotides. Because this
chain of atoms is covalently bonded, it gives strength
to DNA and RNA molecules, helping them to store
information reliably for long periods.
A1.2.5—RNA as a polymer formed by condensation of nucleotide monomers
Students should be able to draw and recognize diagrams of the structure of single nucleotides and RNA
polymers
The pentose sugar in RNA is ribose. There is a single
strand of nucleotides, linked by covalent sugar-phosphate
bonds.
next nucleotide
linked to the
growing chain
Water is produced by bonding sugar to phosphate,
so it is a condensation reaction.
base b a s e
* H20
RNA is an example of a polymer—a molecule
composed of repeating subunits (monomers) linked
together by covalent bonds (see Topics B1.7 and B1.2
for more examples of polymers).
A1.2.6—DNA as a double helix made of two antiparallel strands of nucleotides with two strands linked by
hydrogen bonding between complementary base pairs
In diagrams of DNA structure, students should draw the two strands antiparallel, but are not required to
draw the helical shape. Students should show adenine (A) paired with thymine (T), and guanine (G) paired
with cytosine (C). Students are not required to memorize the relative lengths of the purine and pyrimidine
bases, or the numbers of hydrogen bonds.
There are two strands of nucleotides in DNA, which are
linked by hydrogen bonding between their bases.
Each base will only form hydrogen bonds with one other
base, so two base pairs only are possible: adenine with
thymine and cytosine with guanine (A-T and C-G). They
are known as complementary base pairs.
-THA-
double helix
each strand
has a sugar-
THA-
phosphate backbone
G G
⽩ A
base pairs
are in the
centre of the
G-C-
АНТ-
C - G
The two strands are antiparallel-they run alongside
each other but in opposite directions.
The strands are wound together to form a double helix,
which is the overall shape of a DNA molecule.
A1.2.7—Differences between DNA and RNA
Include the number of strands present, the types of nitrogenous bases and the type of pentose sugar.
Students should be able to sketch the difference between ribose and deoxyribose. Students should be
familiar with examples of nucleic acids.
There are three differences between DNA and RNA.
1. The pentose is ribose in RNA but deoxyribose
in DNA.
2. DNA has the base thymine but RNA has uracil
3. instead.
RNA usually has one strand of nucleotides whereas
DNA usually has two.
A1.2.8—Role of complementary base pairing in allowing genetic information to be replicated and
expressed
Students should understand that complementarity is based on hydrogen bonding
Complementary base pairing has three roles in cells:
DNA r e p l i c a t i o n - s e q u e n c e s of b a s e s in DNA can b e
copied accurately, so the genetic information of a cell
can be passed on to daughter cells.
Transcription-RNA can b e made with the same base
sequence as one of the two strands of a DNA molecule.
Messenger RNA (mRNA) carries the base sequence of a
protein-coding gene to the ribosome.
Translation-a base sequence can be used to determine
the amino acid sequence in a polypeptide. Messenger
RNA carries a series of three-base codons. Each transfer
RNA molecule (tRNA) has one three-base anticodon
and it carries one amino acid. Ribosomes link codons to
anticodons by complementary base pairing, allowing
the base sequence of every codon to be translated into a
specific amino acid in a polypeptide (see Topic D1.2).
A1.2.9—Diversity of possible DNA base sequences and the limitless capacity of DNA for storing
information
Explain that diversity by any length of DNA molecule and any base sequence is possible. Emphasize the
enormous capacity of DNA for storing data with great economy
Bases can be arranged in any sequence in DNA.
Because any of the four bases could occupy each
position along a DNA strand, the number of possible
sequences is 4”, where n is the number of bases. Even
with just 10 bases there are over a million possible
sequences. A typical gene has over a thousand bases
and whole genomes have billions of bases. The
number of possible sequences and therefore DNA’s
capacity to store information is effectively limitless.
A1.2.10—Conservation of the genetic code across all life forms as evidence of universal common ancestry
Students are not required to memorize any specific examples.
Each of the 64 codons of the genetic code indicates
either one of the 20 amino acids or the end of the
polypeptide. There are 1.5101095 × 108* ways of
assigning meanings to the 64 codons, but all organisms
use the same meanings, with only minor variations. This
universality of the genetic code suggests strongly that
all life evolved from the same original ancestor, with
minor differences added since then.
A1.2.11—Directionality of RNA and DNA
Include 5’ to 3’ linkages in the sugar–phosphate backbone and their significance for replication,
transcription and translation.
The two ends of a strand of nucleotides in DNA or
RNA are different. They are known as the 3’ and 5’
ends (3 prime and 5 prime). The 3’ end has a pentose
sugar (ribose or deoxyribose) to which the phosphate
of another nucleotide can be linked. The phosphate
would bond with the -OH group on the C3 of the
deoxyribose. The 5’ end has a phosphate, attached to
the C5 of a pentose.
When a new strand of DNA or RNA is being
constructed, each extra nucleotide is a d d e d to the
pentose at the 3’ end of the growing strand. This is
done by bonding the phosphate of a free nucleotide,
which is the nucleotide’s 5’ end. Nucleotides are
therefore added in a 5’ to 3’ direction.
The two strands in a DNA molecule are antiparallel
because they run in opposite directions. Each end of
a DNA double helix therefore has one strand with a 3’
end and one with a 5’ end.
A1.2.12—Purine-to-pyrimidine bonding as a component of DNA helix stability
Adenine–thymine (A–T) and cytosine–guanine (C–G) pairs have equal length, so the DNA helix has the
same three-dimensional structure, regardless of the base sequence.
The nitrogenous bases in DNA are in two chemical
g r o u p s :
* adenine and guanine are purine bases with two
rings of atoms
* cytosine and thymine are pyrimidine bases with
Each base pair in DNA therefore has one purine and
one pyrimidine. As a consequence, the two base
pairs are of equal width and require the same distance
between the two sugar-phosphate backbones in the
DNA
protein
d o u b l e helix. This h e l p s t o m a k e t h e s t r u c t u r e o f DINA
stable and allows any sequence of bases in genes to fit
in a DNA molecule.
A1.2.13—Structure of a nucleosome
Limit to a DNA molecule wrapped around a core of eight histone proteins held together by an additional
histone protein attached to linker DNA.
Application of skills: Students are required to use molecular visualization software to study the
association between the proteins and DNA within a nucleosome.
Nucleosomes are disc-like structures, used by eukaryotes to package DNA into condensed chromosomes and also to help control replication and transcription.
Each nucleosome has a core of eight histone proteins with a DNA wound round twice and one more histone securing the structure. There is some linker DNA between adjacent nucleosomes. Eukaryotic DNA ooks like a string of beads in electron micrographs because of nucleosomes.
nucleosomes-8 histone proteins with a 166 base pair length of DNA wrapped around
additional Hl histone binding DNA to the nucleosome
linker DNA-a variable length of DNA between
nucleosomes
Adapted from Bunnik, E.M., Le Roch, K.G. (2013).
Nucleosome. In: Hommel, M., Kremsner, P. (eds)
Encyclopedia of Malaria. Springer, New York, NY. https:// doi.org/10.1007/978-1-4614-8757-9_31-1
A1.2.14—Evidence from the Hershey–Chase experiment for DNA as the genetic material
Students should understand how the results of the experiment support the conclusion that DNA is the
genetic material.
NOS: Students should appreciate that technological developments can open up new possibilities for
experiments. When radioisotopes were made available to scientists as research tools, the Hershey–Chase
experiment became possible.
In the early 1950s it was still unclear whether genes were made of DNA or protein. Hershey and Chase used the T2 virus to investigate this. T2 infects E. coli, which is a bacterium. Proteins of T2 start being produced inside
E. coli soon after the virus comes into contact with it. This can only happen if genes of the virus are inside E. coli.
To identify the genetic material, Hershey and Chase had to find out which part of the virus had entered the pacterium.
Viruses such as 12 consist only of DNA and protein, so their genes must be made of one of these materials. DNA contains the element P (phosphorus) but not S (sulfur), and protein contains S but not P. Hershey and Chase prepared one strain of T2 with its DNA labelled with the radioactive isotope 32P and another with its protein radioactively labelled with 35S. These two strains of T2 were mixed separately with E. coli. After leaving enough time for the bacteria to be infected, the mixture was agitated in a high-speed mixer and then centrifuged at 10,000 rpm to separate a solid pellet containing the bacteria from a liquid supernatant containing viruses. A Geiger counter was used to locate the radioactivity. The results are shown in the diagram.
fluid supernatant containing viruses
80% →
* 35%
20%
65%
distribution of radioactivity with 35S
solid pellet containing
E. coil
distribution of radioactivity with 32p
Analysis of results
With 32P most of the radioactivity is in the pellet, showing that much of the viral DNA is inside the bacteria. With 35S most of the radioactivity is in the supernatant, showing the protein coats of the viruses remained outside the bacteria and were shaken off by the mixer. The small proportion of radioactivity in the pellet with 35S can be explained by some protein coats remaining attached to the bacteria and by the presence of some fluid containing protein coats in the pellet. This and other experiments carried out by Hershey and Chase gave strong evidence for genetic material (genes) being composed of DNA rather than protein.
A1.2.15—Chargaff’s data on the relative amounts of pyrimidine and purine bases across diverse life forms
NOS: Students should understand how the “problem of induction” is addressed by the “certainty of
falsification”. In this case, Chargaff’s data falsified the tetranucleotide hypothesis that there was a
repeating sequence of the four bases in DNA.
Before the structure of DNA had been discovered, the amounts of each of the four bases were measured.
* The amounts of each base were not equal-not 25%.
* The percentages were the same in different tissues from one species, for example spleen and thymus tissue from cattle, but there were differences between species.
Erwin Chargaff drew attention to other trends in the numbers of bases that applied to the DNA of all the species investigated, including eukaryotes and prokaryotes:
adenine = thymine and guanine = cytosine
purines (A + G) = 50% and pyrimidines (T + C) = 50%
The table gives a sample of Chargaff’s data:
Source of DNA
Purines
Pyrimidines
% A
% G
% C
% T
Mycobacterium tuberculosis— bacteria
15.1
34.9
35.4
14.6
Human—a mammal
29.3
20.0
20.7
30.0
Octopus—a mollusc
33.2
17.6
17.6
31.6
Corn (maize)—a plant
26.8
23.2
22.8
27.2
Crick and Watson discovered the reason for these trends:
* DNA contains complementary pairs of bases
* a purine base always pairs with a pyrimidine
* adenine pairs with thymine, and cytosine with guanine.
