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Molecular Genetics

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Molecular Genetics
IB SL

IB HL

IB Options

AP Biology

Complete nos:

Complete nos:

Complete nos:

Complete nos:

1, 3-4, 7, 9(a), 11,
13(a)-(d),14-15,
20-22 Extension:
2, 12

1-26, 28-29, 3132, 34
Extension:
27, 30, 33

Option D: 30

1-34
Some numbers as
extension as
appropriate

L earning Objectives
1. Compile your own glossary from the KEY WORDS
displayed in bold type in the learning objectives below.

The genetic blueprint
Nucleic acid structure (pages 128-130, 132-135,
also see 150)
2. Name some examples of nucleic acids and describe
their role in biological systems.
3. Describe the components of a (mono)nucleotide: a 5C
sugar (ribose or deoxyribose), a nitrogenous base
(purine or pyrimidine), and a phosphate. Identify the
bases that form nucleotides.
4. Understand the role of condensation reactions in
joining the components of nucleotides and in the
formation of di- and polynucleotides (nucleic acids).
5. Outline the structure of nucleosomes, including
reference to the role of histone proteins in packaging
of the DNA in the nucleus.
6. Understand that DNA contains repetitive sequences
and that only a small proportion constitutes genes.
Appreciate the role of repetitive sequences in DNA
technologies such as DNA profiling.
7. Describe the Watson-Crick double-helix model of DNA
structure and the base pairing rule. Explain the
importance of complementary base pairing to the
conservation of the base sequence in DNA. Contrast
the structure and function of DNA and RNA.
8. In more detail than #7 above, describe the structure of
DNA including the antiparallel strands, the 3’–5’
linkages, and the role of the hydrogen bonding
between purines and pyrimidines.

DNA replication (pages 136-137)
9. Describe the semi-conservative replication of DNA,
and interpret experimental evidence for this process.
Explain the role of the following in DNA replication:
(a) DNA polymerase, helicase, DNA ligase.
(b) DNA polymerase III, RNA primase, DNA
polymerase I, Okazaki fragments, and
deoxynucleoside triphosphates.
10. Understand that DNA replication proceeds only in the
5’ → 3’ direction and explain the significance of this. Explain the term: replication fork, and explain its
significance in eukaryotic chromosomes.
11. Demonstrate an understanding of the base-pairing rule
for creating a complementary strand from a template
strand.
12. Appreciate the role of polymerase chain reaction (PCR)
as an artificially induced form of DNA replication, used
as a tool in molecular biology (see the topic Aspects of
Biotechnology for coverage of this technique).

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The genetic code (page 131)
13. Explain the main features of the genetic code,
including reference to the following:
(a) The 4-letter alphabet and the 3-letter triplet code
(codon) of base sequences.
(b) The non-overlapping, linear nature of the code.
(c) The universal nature of the code.
(d) The degeneracy of the code.
(e) The way in which the code is always read from a
start point to a finish point in a 3’ → 5’ direction.
(f) Specific punctuation codons and their significance.

Gene expression (pages 138-143)
14. Outline the basis by which information is transferred
from DNA to protein. Distinguish clearly between allele
and gene. Explain what is meant by gene expression
and define its two distinct stages: transcription and
translation. Note that gene expression is sometimes
used to refer just to transcription.
15. Recall the structure and role of messenger RNA
(mRNA). In simple terms, describe the process of
transcription, identifying the role of RNA polymerase.
16. In more detail than in #15 above, describe the process
of transcription. Demonstrate an understanding of the
direction of transcription (5’ → 3’ direction).
17. Distinguish between the coding (sense) strand,
template (antisense) strand. Relate the base
sequence on each of these strands to the sequence on
the mRNA molecule.
18. Distinguish between introns and exons. Explain the
significance of introns with the respect to the production
of a functional mRNA molecule.
19. Understand how reverse transcriptase catalyzes the
production of DNA from RNA. Explain how this enzyme
is used by retroviruses. Appreciate the use of reverse
transcriptase in molecular biology.
20. Recall the structure of proteins as polypeptides with a
complex (post-translational) structure.
21. Explain how the 4-letter alphabet of bases provides the
code for the 20 amino acids needed to assemble
proteins. Explain the relationship between one gene
and one polypeptide.
22. In simple terms, describe the process of translation.
Describe the role of transfer RNA (tRNA) molecules in
translation, with reference to the significance of the
anticodons. Understand and explain the general role
of ribosomes in translation.
23. With respect to the process of translation, describe
how the structure of transfer RNA (tRNA) molecules
allows recognition by a tRNA-activating enzyme.
Explain the role of this enzyme in binding specific
amino acids to their tRNAs and identify the role of ATP
in this process.

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Molecular Genetics

24. Outline the structure of ribosomes with reference to:
small and large subunits, RNA and protein, tRNA
binding sites, and mRNA binding sites. Relate the
functional role of ribosomes to their specific structure.
25. Describe translation as a process involving initiation,
elongation, and termination, occurring in a 5’ → 3’
direction. In more detail than in #22 above, explain the
process of translation including more detailed reference
to ribosomes, polysomes (polyribosomes), start
codons, and stop codons.
26. Distinguish between protein synthesis on free
ribosomes and on those bound to the endoplasmic
reticulum. Explain why proteins are synthesized in
these different locations in the cell.
27. Contrast gene expression in prokaryotic and
eukaryotic cells, identifying differences in mRNA
processing after transcription, movement of the mRNA
to the site of translation, and the speed at which
translation can take place.

Control of metabolic pathways
Metabolic Pathways (pages 144-145)
28. Recognize enzymes as proteins whose synthesis is
controlled by DNA. Appreciate the role of enzymes in
the control of metabolic pathways and in determining
the phenotype of an organism.
29. Explain clearly how enzymes control metabolic
pathways as illustrated by specific examples e.g.
oxidoreductases, anabolism and catabolism. With
respect to this control, explain how the amount or
activity of an enzyme regulating a metabolic pathway
can itself be controlled. Define the terms: end-product
and end-product inhibition.

30. Identify major metabolic disorders that are inherited in humans. Explain, using an example, how the
malfunction of enzymes are responsible in many cases.
Exemplar case study: metabolism of phenylalanine.
Describe the metabolic breakdown of the essential
amino acid phenylalanine by liver enzymes. Describe
how malfunctioning of specific enzymes can interrupt
this metabolic pathway, causing a variety of metabolic
defects e.g. phenylketonuria (PKU).

Regulation of gene action (pages 146-147)
31. Understand the term operon as being a unit of genes
in prokaryotes that function in a coordinated way under
the control of an operator gene. Comment on the extent
to which the operon model is universally applicable.
32. Explain how simple metabolic pathways are regulated
in bacteria, as illustrated by the lac operon in E. coli.
Outline the principles involved in gene induction in the
lac operon, identifying how lactose activates
transcription and how metabolism of the substrate is
achieved. Explain the adaptive value of gene induction.
33. Appreciate that the end-product of a metabolic pathway
can activate a repressor and switch genes off (gene
repression). Appreciate the adaptive value of gene
repression for the control of a metabolic end-product.
34. Describe the regulation of gene action (transcriptional
control only) in eukaryotes. Identify the roles of the
promoter region, RNA polymerase, and the
terminator (not to be confused with terminator codons
in translation). Appreciate that the energy for the
incorporation of the nucleotides into the mRNA strand
in mRNA synthesis is provided by the hydrolysis of
nucleoside triphosphates (ATP, GTP, CTP, and UTP).
NOTE: In DNA replication, the nucleoside triphosphates
are dATP, dGTP, dCTP, and dTTP. For simplicity, often
the nucleotide only is shown).

■ Control Centre New Scientist, 17 July 1999,
(Inside Science). The organization of DNA in
eukaryotic cells, the nucleus, how genes code for
proteins, and the role of ribosomes and RNA.

Metabolic pathways and their control

Textbooks
See the ‘Textbook Reference Grid’ on
pages 8-9 for textbook page references
relating to material in this topic.

Supplementary Texts
See pages 5-6 for additional details of these texts:
■ Adds, J., et al., 2000. Molecules and Cells,
(NelsonThornes), pp. 19-32.
■ Clegg, C.J., 1999. Genetics & Evolution, (John
Murray), pp. 40-42, 44-47.
■ Jones, N., et al., 2001. Essentials of Genetics,
(John Murray), pp. 123-155 as required.

Periodicals
See page 6 for details of publishers of periodicals:

STUDENT’S REFERENCE
Gene structure and expression
■ Gene Structure and Expression Biol. Sci.
Rev., 12 (5) May 2000, pp. 22-25. An account of
gene function, including a comparison of gene
regulation in pro- and eukaryotes.
■ What is a Gene? Biol. Sci. Rev., 15(2) Nov.
2002, pp. 9-11. A good synopsis of genes and their
role in heredity, mutations, and transcriptional
control of gene expression.
■ Transfer RNA Biol. Sci. Rev., 15(3) Feb. 2003,
pp. 26-29. A good account of the structure and role
of tRNA in protein synthesis.
■ DNA in a Spin Biol. Sci. Rev., 11(3) Jan. 1999,
pp. 15-17. A short account of the methods used to
establish the mechanism for DNA replication.

■ Molecular Machines that Control Genes
Scientific American, Feb. 1995, pp. 38-45. How
gene action is regulated by protein complexes that
assemble on DNA.
■ A Discovery Lab for Studying Gene
Regulation The American Biology Teacher, 59(8),
Oct. 1997, pp. 522-526. Investigating gene
regulation in prokaryotes: a how-to-do-it account.

■ Tyrosine Biol. Sci. Rev., 12 (4) March 2000, pp.
29-30. The central metabolic role of the amino acid
tyrosine (includes errors in tyrosine metabolism).
■ Genes that Control Genes New Scientist, 3
Nov. 1990 (Inside Science). The control of gene
expression in prokaryotes by gene induction and
repression. The operon model is explained.

TEACHER’S REFERENCE
■ DNA 50 SSR, 84(308), March 2003, pp. 17-80.
A special issue celebrating 50 years since the
discovery of DNA. There are various articles
examining the practical and theoretical aspects of
teaching molecular genetics and inheritance.
■ DNA: 50 Years of the Double Helix New
Scientist, 15 March 2003, pp. 35-51. A special
issue on DNA: structure and function, repair, the
new-found role of histones, and the functional
significance of chromosome position in the nucleus.
■ Modeling the Classic Meselson and Stahl
Experiment The American Biology Teacher, 63(5),
May 2001, pp. 358-361. An account of how to
model the experiments of Meselson and Stahl to
demonstrate semi-conservative replication of DNA.
■ A Working Model of Protein Synthesis using
Lego™ Building Blocks The American Biology
Teacher, 64(9), Nov. 2002, pp. 673-678. Using a
hands-on project to demonstrate the various
stages of protein synthesis.

Internet
See pages 10-11 for details of how to access Bio
Links from our web site: www.thebiozone.com.
From Bio Links, access sites under the topics:
GENERAL BIOLOGY ONLINE RESOURCES >
Online Textbooks and Lecture Notes: • S-Cool!
A level biology revision guide • Learn.co.uk •
Mark Rothery’s biology web site … and others
CELL BIOLOGY AND BIOCHEMISTRY: • Cell
and molecular biology online • MIT biology
hypertextbook … and others
GENETICS: • DNA basics • MIT biology
hypertextbook • Gene almanac • Virtual library
on genetics • Prokaryotic genetics and gene
expression chapter … and others > Molecular
Genetics (DNA): • Beginners guide to molecular
biology • Basic genetics • DNA and molecular
genetics • DNA from the beginning • DNA
workshop • E!Mouse • Primer on molecular
genetics • Protein synthesis • Model of Lac
operon (animation) • Induction of the Lac operon •
Molecular genetics of prokaryotes

■ Deciphering the Code of Life Scientific
American, December 1999, pp. 50-55. An
exploration of what will be gained from the study of
the genomes of humans and other organisms.
■ Stuff or Nonsense New Scientist, 1 April 2000,
pp. 38-41. The functional and evolutionary role of
introns (junk DNA) in the genomes of organisms.

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Teacher Resource Handbook

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128

Nucleic Acids

A

Nucleic acids are a special group of chemicals in cells concerned with the transmission of inherited information. They have the capacity to store the information that controls cellular activity. The central nucleic acid is called deoxyribonucleic acid (DNA). DNA is a major component of chromosomes and is found primarily in the nucleus, although a small amount is found in mitochondria and chloroplasts. Other ribonucleic acids (RNA) are involved in the ‘reading’ of the DNA information. All nucleic acids are made

up of simple repeating units called nucleotides, linked together to form chains or strands, often of great length (see the activity DNA Molecules). The strands vary in the sequence of the bases found on each nucleotide. It is this sequence which provides the ‘genetic code’ for the cell. In addition to nucleic acids, certain nucleotides and their derivatives are also important as suppliers of energy (ATP) or as hydrogen ion and electron carriers in

respiration and photosynthesis (NAD, NADP, and FAD).

Chemical Structure of a Nucleotide

Bases
Purines:

N

P O OCH 2

G
Guanine

NH 2

OH
OH

A
Adenine

N

N

O

O

Pyrimidines:

C

N

T

U

H

H
OH
Phosphate

Cytosine

H

Uracil
(RNA only)

H

Sugar

The two-ringed bases above are purines and make
up the longer bases. The single-ringed bases are
pyrimidines. Although only one of four kinds of base
can be used in a nucleotide, uracil is found only in
RNA, replacing thymine. DNA contains: A, T, G, and
C, while RNA contains A, U, G, and C.

Base

Symbolic Form of a Nucleotide

A
Phosphate: Links
neighboring
sugars together.

Thymine
(DNA only)

H

Sugars
Base: One of four types
possible (see box on right). This
part of the nucleotide comprises
the coded genetic message.
Sugar: One of two types
possible: ribose in RNA
and deoxyribose in DNA.

OH
Ribose

Nucleotides are the building blocks of DNA. Their precise sequence in a DNA molecule provides the genetic instructions for the organism to which it governs. Accidental changes in nucleotide sequences are a cause of mutations, usually harming the organism, but occasionally providing benefits.

RNA Molecule

H
Deoxyribose

Deoxyribose sugar is found only in DNA. It differs
from ribose sugar, found in RNA, by the lack of a
single oxygen atom (arrowed).

DNA Molecule

G

DNA Molecule

C
G

Deoxyribose
sugar

U
In RNA, uracil
replaces thymine
in the code.

T

A

C

G

C

A
Ribose
sugar

Hydrogen bonds
hold the two
strands together.
Only certain
bases can pair.

A

T

Symbolic representation

Ribonucleic acid ( RNA)
comprises a single strand of
nucleotides linked together.

Space filling model

Deoxyribonucleic acid (DNA) comprises a double strand of nucleotides linked together. It is shown unwound in the symbolic representation (left). The DNA molecule takes on a twisted, double helix shape as shown in the space filling model on the right.

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Molecular Genetics

Formation of a nucleotide

Formation of a
dinucleotide

Condensation
(water removed)

A
H2O
H 2O

T

H2O

A

Two nucleotides are linked
together by a condensation
reaction between the
phosphate of one nucleotide
and the sugar of another.

Hydrolysis
(water added)

C

A nucleotide is formed when phosphoric acid and a base are chemically bonded to a sugar molecule. In both cases, water is given off, and they are therefore condensation reactions.

Double-stranded DNA molecule

3'

5'

The double-helix structure of DNA is like a ladder twisted
into a corkscrew shape around its longitudinal axis. It is
‘unwound’ here to show the relationships between the bases.

C

• The way the correct pairs of bases are attracted to each other to form hydrogen bonds is determined by the number
of bonds they can form and the shape (length) of the
base.

G
G

A

T

• The template strand is the side of the DNA molecule
that stores the information that is transcribed into mRNA.
• The other side (sometimes called the coding strand)
has the same nucleotide sequence as the mRNA except
that T in DNA substitutes for U in mRNA. The coding
strand is also called the sense strand.

T

A

C

5'

3'

1. The diagram above depicts a double-stranded DNA molecule. Label the following parts on the diagram: (a) Sugar (deoxyribose)
(b) Phosphate
(c) Hydrogen bonds (points of attraction between bases)
(d) Purine bases
(e) Pyrimidine bases
2. State the ‘base-pairing rule’ which describes which bases can pair up opposite each other to form a double-stranded DNA molecule:

3. State the functional role of the following nucleic acids: (a) Nucleotides:
(b) ATP:
(c) NAD/NADP:
(d) Coenzyme A:
4. Complete the following table that summarizes the differences between DNA and RNA molecules:

DNA

RNA

Sugar present

Bases present

Number of strands

Relative length

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130

DNA Molecules

D

Even the smallest DNA molecules are extremely long. The DNA
from the small Polyoma virus, for example, has a length of 1.7µm (about 3 times longer than the longest proteins). The DNA
comprising a bacterial chromosome is 1000 times longer than
the cell into which it has to fit. The amount of DNA present in the nucleus of the cells of eukaryotic organisms varies widely from one species to another. The quantity of DNA in vertebrate sex cells ranges from 40 000 kb to 80 000 000 kb, with humans in the middle of the range. There is good reason to believe that most proteins (or polypeptide chains) are coded for by only one gene in each set of chromosomes. Proteins that are found in

relatively large concentrations within the cell usually have multiple copies of their gene. About 50-75% of the DNA consists of base sequences that are long enough to code for proteins
(around 1000 bases). Current estimates suggest that as little as 10% of the human genome (the total DNA complement of a cell) encodes proteins or structural RNA and is therefore made up of genes. Of the remaining 90% of the DNA, some is used for the structural aspects of gene expression, DNA replication,

chromosome division, and organizing chromatin within the
chromosome. Some regions of the DNA appear to have no
function, although this view may change with further research.

Kilobase (kb)

Sizes of DNA Molecules
Group

Base pairs

Organism

Viruses

Length

(in 1000s, or kb)

Polyoma or SV40

5.1

1.7 µm

48.6

Lambda phage

17 µm

T2 phage

166
190

DNA

65 µm

Mycoplasma

760

Drosophila (fruit fly)
Human

260 µm

4 600

1.56 mm

13 500

4.6 mm

165 000

5.6 cm

2 900 000

E. coli (from human gut)
Yeast

Eukaryotes

Exons: coding regions

56 µm

Vaccinia

Bacteria

A unit of length equal to 1000 base pairs of a double-stranded nucleic acid molecule (or 1000 bases of a single-stranded
molecule). One kilobase of double stranded DNA has a length
of 0.34 µm. (1 µm = 1/1000 mm)

99 cm

Intron

Intron: edited out during
protein synthesis

Intron

Most genes in eukaryotic DNA are not continuous and may be
interrupted by ‘intrusions’ of other pieces of DNA. Coding regions (exons) are interrupted by non-coding regions called introns. Introns range in frequency from 1 to over 30 in a single gene and also in size (100 to more than 10 000 bases). They are

edited out of the genetic instructions during protein synthesis.

Giant lampbrush chromosomes
Lampbrush chromosomes are large chromosomes found in amphibian eggs, with lateral loops of DNA that produce a brushlike appearance under the microscope. The two scanning electron micrographs (below and right) show minute strands of DNA giving a fuzzy appearance in the high power view.

Loops of DNA

EII

EII

Enlarged

1. Consult the table above and make the following comparisons. Determine how much more DNA is present in: (a) The bacterium E. coli compared to the Lambda phage virus: (b) Human cells compared to the bacteria E. coli:

2. State what proportion of DNA in a eukaryotic cell is used to code for proteins or structural RNA: 3. List three functions for some of the remaining (noncoding) DNA: (a)
(b)
(c)
4. State the length of all the DNA (genome) from a single human cell: Photocopying Prohibited

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Molecular Genetics

The Genetic Code

A

The genetic information that codes for the assembly of amino acids is stored as three-letter codes, called codons. Each codon represents one of 20 amino acids used in the construction of polypeptide chains. The mRNA amino acid table (bottom of

page) can be used to identify the amino acid encoded by each of the mRNA codons. Note that the code is degenerate in that for

each amino acid, there may be more than one codon. Most of
this degeneracy involves the third nucleotide of a codon. The genetic code is universal; all living organisms on Earth, from viruses and bacteria, to plants and humans, share the same
genetic code book (with a few minor exceptions representing
mutations that have occurred over the long history of evolution).

Codons that code for
this amino acid

No.

GCU, GCC, GCA, GCG

4

Amino acid

Codons that code for
this amino acid

Amino acid

Leu

Leucine

Arg Arginine

Lys

Lysine

Asn Asparagine

Met

Methionine

Asp Aspartic acid

Phe

Phenylalanine

Cys Cysteine

Pro

Proline

Gln

Glutamine

Ser

Serine

Glu

Glutamic acid

Thr

Threonine

Gly

Glycine

Try

Tryptophan

His

Histidine

Tyr

Tyrosine

Iso

Isoleucine

Val

No.

Valine

Ala

Alanine

1. Use the mRNA amino acid table (below) to list in the table above all the codons that code for each of the amino acids and the number of different codons that can code for each amino acid (the first amino acid has been done for you). 2. (a) State how many amino acids could be coded for if a codon consisted of just TWO bases: (b) Explain why this number of bases is inadequate to code for the 20 amino acids required to make proteins:

3. There are multiple codons for a single amino acid. Comment on the significance of this with respect to point mutations:

Read second
letter here

Read third
letter here

Second Letter

Read first
letter here

U

C

A

G

U

UUU
UUC
UUA
UUG

Phe
Phe
Leu
Leu

UCU
UCC
UCA
UCG

Ser
Ser
Ser
Ser

UAU
UAC
UAA
UAG

Tyr
Tyr
STOP
STOP

UGU
UGC
UGA
UGG

Cys
Cys
STOP
Try

U
C
A
G

C

CUU
CUC
CUA
CUG

Leu
Leu
Leu
Leu

CCU
CCC
CCA
CCG

Pro
Pro
Pro
Pro

CAU
CAC
CAA
CAG

His
His
Gln
Gln

CGU
CGC
CGA
CGG

Arg
Arg
Arg
Arg

U
C
A
G

A

AUU
AUC
AUA
AUG

Iso
Iso
Iso
Met

ACU
ACC
ACA
ACG

Thr
Thr
Thr
Thr

AAU
AAC
AAA
AAG

Asn
Asn
Lys
Lys

AGU
AGC
AGA
AGG

Ser
Ser
Arg
Arg

U
C
A
G

G

GUU
GUC
GUA
GUG

Val
Val
Val
Val

GCU
GCC
GCA
GCG

Ala
Ala
Ala
Ala

GAU
GAC
GAA
GAG

Asp
Asp
Glu
Glu

GGU
GGC
GGA
GGG

Gly
Gly
Gly
Gly

U
C
A
G

Example: Determine CAG
C on the left row, A on the top column, G
on the right row
CAG is Gln (glutamine)
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Third Letter

How to read the table: The table on the right
is used to 'decode' the genetic code as a
sequence of amino acids in a polypeptide
chain, from a given mRNA sequence. To work
out which amino acid is coded for by a codon
(triplet of bases) look for the first letter of the
codon in the row label on the left hand side.
Then look for the column that intersects the
same row from above that matches the
second base. Finally, locate the third base in
the codon by looking along the row from the
right hand end that matches your codon.

First Letter

mRNA-Amino Acid Table

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132

PA

Creating a DNA Model

Although DNA molecules can be enormous in terms of their
molecular size, they are made up of simple repeating units
called nucleotides. A number of factors control the way in
which these nucleotide building blocks are linked together.

These factors cause the nucleotides to join together in a
predictable way. This is referred to as the base pairing rule and can be used to construct a complementary DNA strand
from a template strand, as illustrated in the exercise below:

DNA Base Pairing Rule
Adenine

is always attracted to

Thymine

A

T

Thymine

is always attracted to

Adenine

T

A

Cytosine

is always attracted to

Guanine

C

G

Guanine

is always attracted to

Cytosine

G

C

1. Cut out the facing page and separate each of the 24 nucleotides by cutting along the columns and rows (see arrows indicating 2 such cutting points). Although drawn as geometric shapes, these symbols represent chemical structures. 2. Place one of each of the four kinds of nucleotide on their correct spaces below:

Place a cut-out symbol for thymine
here from the facing page

Place a cut-out symbol for cytosine
here from the facing page

Thymine

Cytosine

Place a cut-out symbol for adenine
here from the facing page

Place a cut-out symbol for guanine
here from the facing page

Adenine

Guanine

3. Identify and label each of the following features on the adenine nucleotide immediately above: phosphate, sugar, base, hydrogen bonds
4. Create one strand of the DNA molecule by placing the 9 correct 'cut out' nucleotides in the labeled spaces on the now facing page (DNA Molecule). Make sure these are the right way up (with the P on the left) and are aligned with the left hand edge of each box. Begin with thymine and end with guanine. 5. Create the complementary strand of DNA by using the base pairing rule above. Note that the nucleotides have to be arranged upside down.

6. Under normal circumstances, it is not possible for adenine to pair up with guanine or cytosine, nor for any other mismatches to occur. Describe the two factors that prevent a mismatch from occurring: Factor 1:

Factor 2:

7. Once you have checked that the arrangement is correct, you may glue, paste or tape these nucleotides in place. NOTE:

There may be some value in keeping these pieces loose in order to practise the base pairing rule. For this purpose, removable tape would be best.

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Molecular Genetics

Nucleotides
Tear out this page along the perforation and separate each of the 24 nucleotides

S

Thymine

Cytosine

P

P

P

S

S

Thymine

Cytosine

S
P

P

P

S

S

Thymine

Cytosine

by cutting along the columns and rows (see arrows indicating the cutting points).

P

S

Adenine

Guanine

S
P

P

S

Adenine

Guanine

S
P

S
P

P

S

Adenine

Guanine

Cut

S

Thymine

Cytosine

P

P

P

S

S

Thymine

Cytosine

S
P

P

P

S

S

Thymine

Cytosine

Cut

Cut

Cut

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Cut

Cut

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P

S

Adenine

Guanine

S
P

P

S

Adenine

Guanine

S
P

S
P

P

S

Adenine

Guanine

Cut

Cut

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134

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Molecular Genetics

DNA Molecule
P

Adenine
Adenine

S

Thymine
S

P
P

Put the named
nucleotides on the left
hand side to create
the template strand

Thymine

Cytosine

Adenine

Adenine

Guanine

Thymine

Thymine

Cytosine

Guanine

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Put the matching
complementary
nucleotides opposite
the template strand

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136

DNA Replication

A

The replication of DNA is a necessary preliminary step for cell division (both mitosis and meiosis). This process creates the two chromatids that are found in chromosomes that are preparing to divide. By this process, the whole chromosome is essentially

3'

duplicated, but is still held together by a common centromere. Enzymes are responsible for all of the key events. The diagram below shows the essential steps in the process. The diagram on the facing page shows how enzymes are involved at each stage.

Step 1: Unwinding the DNA molecule

5'

Single-armed
chromosome as found
in non-dividing cell.

A normal chromosome consists of a single DNA
molecule packed into a single chromatid. The long
molecule of double stranded DNA must be untwisted
at high speed at its replication fork by two enzymes:
helicase unwinds the parental strands; DNA gyrase
then relieves the strain that this generates by cutting,
winding and rejoining the DNA strands.

Temporary
break to
allow swivel.

Step 2: Making new DNA strands
Free nucleotides
are used to construct
the new DNA strand.

The formation of new DNA is carried out mostly by
an enzyme complex called DNA polymerase, and a
series of proteins that cause the two strands to break
apart. On one side (the leading strand), nucleotides
are assembled in a continuous fashion. On the other
side (the lagging strand) fragments of single-stranded
DNA between 1000–2000 nucleotides long are
created (Okazaki fragments). These will be later
joined together to form one continuous length.

Parent strand of DNA
is used as a template
to match nucleotides
for the new strand.

The new strand of DNA is
constructed from free
nucleotides, using the parent
strand as a template.

Each of the two newly
formed DNA double helix
molecules will go into
creating a chromatid.

3'

5'

3'

5'

The two new strands of
DNA coil up into a helix.

Step 3: Rewinding the DNA molecule

Replicated chromosome
ready for cell division.

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Each of the two new double-helix DNA molecules has one
strand of the original DNA (dark gray and white) and one
strand that is newly synthesized (patterned). The two DNA
molecules rewind into their 'corkscrew' double-helix shape
again. Each double-helix is then coiled around histone
proteins and further wrapped up to form separate chromatids
(still joined by a common centromere). The new chromosome
has actually twice as much DNA as a regular (non-replicated) chromosome. The two chromatids will become separated in
the cell division process to form two separate chromosomes.

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Molecular Genetics

Enzyme Control of DNA Replication
3'

Overall direction of replication

5'
This process of DNA replication occurs
at an astounding rate. As many as 4000
nucleotides per second are replicated.
This explains how under ideal conditions,
bacterial cells with as many as 4 million
nucleotides, can complete a cell cycle
in about 20 minutes. See the section on
polymerase chain reaction for a useful
application of this process.

Double strand of
original (parental) DNA

Helicase: This enzyme splits and
unwinds the 2-stranded DNA molecule.

RNA polymerase: Synthesizes a short
RNA primer which is later removed.

Swivel point
The leading strand is
synthesized continuously
in the 5' to 3' direction by
DNA polymerase III.

DNA polymerase III: Extends RNA primer
with short lengths of complementary DNA.

Parental strand provides
a 'template' for the new
strand's synthesis

The lagging strand is formed
in fragments, between 1000
and 2000 nucleotides long.
Called Okazaki fragments,
they are later joined together.

RNA
primers

Replication
fork
re

Di
ct

is

n

sy

of

DNA polymerase I:
Digests RNA primer and
replaces it with DNA.

5'
3'

that one new strand is constructed as a continuous length (the leading strand) while the other new strand is made in short segments to be later joined together (the lagging strand). NOTE that the nucleotides are present as deoxynucleoside triphosphates. When hydrolyzed, these provide the energy for incorporating the nucleotide into the strand.

1. Briefly summarize the steps involved in DNA replication (on the facing page): (a) Step 1:

(b) Step 2:

(c) Step 3:

2. Explain the role of the following enzymes in DNA replication: (a) Helicase:
(b) DNA polymerase I:
(c) DNA polymerase III:
(d) Ligase:
3. Briefly explain the purpose of DNA replication:

4. Determine the time it would take for a bacteria to replicate its DNA (see note in diagram above): Photocopying Prohibited

sis

The sequence of enzyme controlled events in DNA replication is shown above. Although shown as separate, many of the enzymes are found clustered together as enzyme complexes. These enzymes are also able to ‘proof-read’ the new DNA strand as it is made and correct mistakes. The polymerase enzyme can only work in one direction, so

he

5'

nt

DNA ligase: Joins
neighboring fragments
together into longer strands.

3'

sy

of

i

ct

re

Di

on

io

es

h
nt

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138

Genes Code For Proteins

A

The genetic code is responsible for the construction of proteins, which may be structural components of cells or metabolism
controlling enzymes. The various levels of genetic instructions are illustrated below, together with their 'protein equivalents'. Nucleotides are the simplest basic unit of genetic information, that are read in groups of three (called triplets). One triplet provides information to bring in a single amino acid during

protein construction. Series of triplets in a long string allow the synthesis of polypeptide chains and are called genes. Some

triplets have a special controlling function in the making of a polypeptide chain. The equivalent of the triplet on the mRNA molecule is the codon. Three codons can signify the end point of polypeptide chain construction in the mRNA: UAG, UAA and

UGA (also called STOP codons). The triplet ATG is found at the beginning of every gene (codon AUG on mRNA) and marks the
starting position for reading the gene. Several polypeptide chains may be needed to form a functional protein. The genes required to do this are collectively called a transcription unit.

This polypeptide chain
forms one part of the
functional protein.

Polypeptide chain
aa

aa

aa

This polypeptide chain
forms the other part of
the functional protein.

Functional
protein

aa

aa

Polypeptide chain
aa

aa

A triplet
codes for one
amino acid

aa

aa

aa

aa

aa

aa

Amino acids
Protein synthesis:
transcription and
translation

START

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

STOP

START

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

STOP

5'

3'
DNA

Gene

Gene

Transcription unit
Note: This start code is for the
coding strand of the DNA. The
template DNA strand from which
the mRNA is made would have
the sequence: TAC.

Three nucleotides
make up a triplet

A
G

In models of nucleic acids,
nucleotides are denoted
by their base letter.

Nucleotide

1. The following exercise is designed to establish an understanding of the terms used in describing protein structure and the genetic information that determines them. Your task is to consult the diagram above and match the structure in the level of protein organization with its equivalent genetic information: (a) Nucleotide

codes for:

(b) Triplet

codes for:

(c) Gene

codes for:

(d) Transcription unit

codes for:

2. Name the basic building blocks for each of the following levels of genetic information: (a) Nucleotide is made up of:

(b) Triplet is made up of:

(c) Gene is made up of:

(d) Transcription unit is made up of:

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Molecular Genetics

Gene Expression

RA

The process of protein synthesis is fundamental to the
understanding of how a cell can control its activities. Genetic instructions, in the form of DNA, are used as a blueprint for designing and manufacturing proteins. Some of these proteins are the enzymes that control the complex biochemical reactions in the cell, while others take on a variety of other roles. The

Chromosomal DNA
DNA contains the master copy of
all the genetic information to
produce proteins for the cell.
Most eukaryotic genes contain
segments of coding sequences
(exons) interrupted by noncoding sequences (introns).

Intron

DNA

process of transferring the information encoded in a gene to its functional gene product is called gene expression. It is divided up into two distinct stages: transcription and translation. These are summarized below and detailed in the following pages. For the sake of simplicity, the involvement of introns in gene

expression has been omitted from the following pages.

Intron

Intron

Intron

Intron
Double stranded
molecule of
genomic DNA

Exon

Exon Exon

Exon

Exon

Exon

Reverse transcription occurs when retroviruses (e.g.
HIV) invade host cells. Their viral RNA is converted
to DNA and spliced into the host’s genome by an
enzyme called reverse transcriptase.

Transcription

Primary RNA Transcript
Both exons and introns are
transcribed to produce a long
primary RNA transcript.

Primary RNA
Introns are
removed

Exons are
spliced together

Introns

Messenger RNA
Introns in the DNA (also copied to the primary
RNA) are long sequences of codons that have
(as yet) no apparent function. They may be the
remnants of now unused ancient genes. It has
been suggested that they might facilitate
recombination between protein-coding regions
(exons) of different genes; a process known as
exon shuffling. This may accelerate evolution.

mRNA
The introns are then removed by
splicing to form a mature mRNA.
Messenger RNA is an edited
copy of the DNA molecule (now
excluding the introns) that codes
for the making of a single protein.

Structural
proteins

Translation

Protein

Regulatory
proteins

Contractile
proteins

Immunological
proteins

Transport
proteins

Catalytic
proteins

1. The hypothesis known as the central dogma of biology states that: “genetic information can only flow in the direction of DNA to proteins and not in the opposite direction”. Accounting for the ideas in the diagram above, form a discussion group with 2-3 of your classmates and discuss the merits of this statement. Summarize your group’s response below:

2. Explain the significance of introns and exons found in DNA and primary RNA: (a) Intron:

(b) Exon:

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140

Transcription

A

Transcription is the process by which the code contained in the DNA molecule is transcribed (rewritten) into a mRNA molecule. Transcription is under the control of the cell's metabolic
processes which must activate a gene before this process can begin. The enzyme that directly controls the process is RNA
polymerase, which makes a strand of mRNA using the single
antisense (template) strand of DNA as a template. The enzyme

5'

DNA

transcribes only a gene length of DNA at a time and therefore recognizes start and stop signals (codes) at the beginning and end of the gene. Only RNA polymerase is involved in mRNA
synthesis; it causes the unwinding of the DNA as well. It is common to find several RNA polymerase enzyme molecules on
the same gene at any one time, allowing a high rate of mRNA
synthesis to occur.

3'

A copy of the genetic information for making a
protein is made in the form of messenger RNA
(mRNA). Many mRNA copies may be made from
a single gene on the DNA molecule. Once the
mRNA is complete and has been released from
the chromosome, it travels to the edge of the nucleus
where it gains access to the cytoplasm through a
tiny hole called a nuclear pore. In prokaryotic cells
(bacteria) there is no nucleus, and the chromosomes
are in direct contact with the cytoplasm. This means
that the next stage (translation) can begin
immediately, with the mRNA still being synthesized
by enzymes on the DNA molecule.

Free nucleotides
used to construct
the mRNA strand.

Single-armed
chromosome as found
in non-dividing cell.

RNA polymerase
enzyme

3'
Template strand of DNA
contains the information for
the construction of a protein.

of
n
io is
ct es
re th
Di syn

Pore (hole) in the nuclear
membrane through which
the mRNA passes to enter
the cytoplasm.

Coding strand of
DNA has a
nucleotide sequence
complementary to
the template strand.

mRNA
5'
Once in the cytoplasm,
the mRNA will engage
ribosomes to begin the
next stage in protein
synthesis: translation

Formation of a single strand of
mRNA that is complementary to the
template strand (therefore the same
“message” as the coding strand).
The two strands
of DNA coil up
into a helix.

Nuclear membrane that
encloses the nucleus.

3'

Nucleus

5'

Cytoplasm

1. Explain the role of messenger RNA (mRNA) in protein synthesis:

2. The genetic code contains punctuation codons to mark the starting and finishing points of the code for synthesis of polypeptide chains and proteins. Consult the mRNA–amino acid table earlier in this manual and state the codes for: (a) Start codon:

(b) Stop (termination) codons:

3. For the following triplets on the DNA, determine the codon sequence for the mRNA that would be synthesized: (a) Triplets on the DNA:

TAC

TAG

CCG

CGA

TTT

TAC

AAG

CCT

ATA

AAA

Codons on the mRNA:
(b) Triplets on the DNA:
Codons on the mRNA:
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Molecular Genetics

Translation

A

The diagram below shows the translation phase of protein
synthesis. The scene shows how a single mRNA molecule
can be 'serviced' by many ribosomes at the same time. The
ribosome on the right is in a more advanced stage of
constructing a polypeptide chain because it has 'translated'

Unloaded
Thr-tRNA

more of the mRNA than the ribosome to the left. The anticodon at the base of each tRNA must make a perfect complementary match with the codon on the mRNA before
the amino acid is released. Once released, the amino acid is added to the growing polypeptide chain by enzymes.

Lys

Activating
Lys-tRNA

Ser

Met
Thr

Polypeptide chain
This chain is in an advanced
stage of synthesis.

Phe

Unloaded
Arg-tRNA

Arg

Polypeptide chain
in an early stage
of synthesis

Val

Activated
Tyr-tRNA
Unloaded
Thr-tRNA

Tyr

Lys
Tyr

Met
Cys

Arg

Thr

Asn

Phe

Start
codon

Ribosome

5'

3'

mRNA
Ribosomes moving in this direction

Amino acid
attachment site

tRNA molecules move into the ribosome,
bringing in amino acids to add to the
polypeptide chain under construction.

Transfer RNA
molecule

Ribosome
Ribosome
attachment
point
Large
subunit

Small
subunit

Ribosomes are made up of a complex of ribosomal RNA (rRNA)
and proteins. They exist as two separate sub-units until they are attracted to a binding site on the mRNA molecule, when
they join together. Ribosomes have binding sites that attract tRNA molecules loaded with amino acids. The transfer RNA

Anticodon

The anticodon is the site
of the 3-base sequence
that 'recognizes' and
matches up with the codon
on the mRNA molecule.

(tRNA) molecules are about 80 nucleotides in length and are
made under the direction of genes in the chromosomes. There
is a different tRNA molecule for each of the different possible anticodons (there may be up to six different tRNAs carrying the same amino acid).

1. For the following codons on the mRNA, determine the anti-codons for each tRNA that would deliver the amino acids: Codons on the mRNA:

UAC

UAG

CCG

CGA

UUU

Anti-codons on the tRNAs:
2. There are many different types of tRNA molecules, each with a different anti-codon (HINT: see the mRNA table). (a) State how many different tRNA types there are, each with a unique anticodon: (b) Give a reason for your answer in (a) above:

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Protein Synthesis Summary

A

Thr
Phe

E
A

Asn

Tyr

Cys
Arg

B

Lys

Val

Met
Lys

8
2

1

Met
Tyr

Thr
Phe
Arg

C

Val
Lys
Tyr

7

Met
Arg
Thr

Cys
Asn

Phe

3
4
D

6

F

Nucleus

Cytoplasm

The diagram above shows an overview of the process of protein synthesis. It is a combination of the diagrams from the previous two pages. Each of the major steps in the process are numbered, while structures are labeled with letters. 1. Write a brief description of each numbered process in the diagram above: (a) Process 1:

(b) Process 2:
(c) Process 3:
(d) Process 4:
(e) Process 5:
(f) Process 6:
(g) Process 7:
(h) Process 8:
2. Identify each of the structures marked with a letter and write their names below in the spaces provided: (a) Structure A:

(f) Structure F:

(b) Structure B:

(g) Structure G:

(c) Structure C:

(h) Structure H:

(d) Structure D:

(i) Structure I:

(e) Structure E:

(j) Structure J:

3. Explain the purpose of protein synthesis (gene expression):

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Molecular Genetics

Analyzing a DNA Sample

A

The nucleotide (base sequence) of a section of DNA can be
determined using DNA sequencing techniques. The base
sequence determines the amino acid sequence of the resultant protein therefore the DNA tells us what type of protein that gene encodes. This exercise reviews the areas of DNA replication,

transcription, and translation using an analysis of a gel
electrophoresis column. Attempt it after you have completed
the rest of this topic. Remember that the gel pattern represents the sequence in the synthesized strand.

1. Determine the amino acid sequence of a protein from the nucleotide sequence of its DNA, with the following steps: (a) Determine the sequence of synthesized DNA in the gel
(b) Convert it to the complementary sequence of the sample DNA (c) Complete the mRNA sequence
(d) Determine the amino acid sequence by using the 'mRNA-amino acid table' in this manual. NOTE: The nucleotides in the gel are read from bottom to top and the sequence is written in the spaces provided from left to right (the first four have been done for you).

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

Triplet

CGTA
(DNA sequence read from the gel; comprises radioactive
nucleotides that bind to the coding strand DNA in the sample).

Replication

Synthesized DNA

Read in this direction

GCAT
DNA sample

(This is the DNA that is being investigated)

Transcription

CGUA
mRNA

Translation
A
T
G

Arginine

Amino acids

C

Part of a polypeptide chain

TCGA
2. For each single strand DNA sequence below, write the base sequence for the complementary DNA strand: (a) DNA:

TAC

TAG

CCG

CGA

TTT

ACA

ATT

TAC

GCC

TTA

AAG

GGC

CGA

ATC

DNA:
(b) DNA:
DNA:
(c) Name the cell process that this exercise represents:
3. For each single strand DNA sequence below, write the base sequence for the mRNA strand and the amino acid that it codes for (refer to the mRNA-amino acid table to determine the amino acid sequence): (a) DNA:

TAC

TAG

CCG

CGA

TTT

ACA

ATT

TAC

GCC

TTA

AAG

GGC

CGA

ATC

mRNA:
Amino
acids:
(b) DNA:
mRNA:
Amino
acids:
(c) Name the cell process that this exercise represents:
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144

Metabolic Pathways

RA

Metabolism is all the chemical activities of life. The myriad enzyme-controlled metabolic pathways that are described
as metabolism form a tremendously complex network that is
necessary in order to 'maintain' the organism. Errors in the

step-wise regulation of enzyme-controlled pathways can
result in metabolic disorders that in some cases can be easily identified. An example of a well studied metabolic pathway,
the metabolism of phenylalanine, is described below.

A Metabolic Pathway

Gene A

Expression of Gene A
(by protein synthesis)
produces enzyme A

Expression of Gene B
(by protein synthesis)
produces enzyme B

Enzyme A

Precursor
chemical

Gene B

Enzyme B

Enzyme A transforms the
precursor chemical into the
intermediate chemical by
altering its chemical structure

Intermediate
chemical

Enzyme B transforms the
intermediate chemical into
the end product

End product

Case Study: The Metabolism of Phenylalanine
Protein
Phenylketonuria
Proteins are broken down to
release free amino acids, one
of which is phenylalanine.

Symptoms:
Mental retardation, mousy
body odor, light skin color,
excessive muscular tension
and activity, eczema.

Phenylalanine
This in turn
causes:
Faulty enzyme
causes buildup of:

Phenylalanine
hydroxylase
a series of
enzymes

Thyroxine

Tyrosinase

Tyrosine
Faulty enzymes
cause:

Phenylpyruvic
acid

Melanin
Faulty enzyme
causes:

Transaminase

Albinism

Cretinism

Symptoms:
Complete lack of the pigment
melanin in body tissues,
including skin, hair, and eyes.

Hydroxyphenylpyruvic

Symptoms:
acid
Dwarfism, mental retardation,
low levels of thyroid hormones,
retarded sexual development,
Hydroxyphenylpyruvic
yellow skin color.
acid oxidase

Carbon
dioxide
&
water

Faulty enzyme causes:

Tyrosinosis

Symptoms:
Death from liver failure,
or (if surviving) chronic
liver and kidney disease.

Homogentisic
acid
Homogentisic
acid oxidase

Faulty enzyme causes:

Alkaptonuria

Symptoms:
Dark urine, pigmentation
of cartilage and other
connective tissues. In
later years, arthritis.

Maleylacetoacetic
acid

A well-studied metabolic pathway is the metabolic breakdown of the essential amino acid phenylalanine. The first step is carried out by an enzyme produced in the liver, called phenylalanine hydroxylase. This enzyme converts phenylalanine to the amino acid tyrosine. Tyrosine, in turn, through a series of intermediate steps, is converted into melanin, the skin pigment, and other substances. If phenylalanine hydroxylase is absent, phenylalanine is in part converted into phenylpyruvic acid, which accumulates, together with phenylalanine, in the blood stream. Phenylpyruvic acid and phenylalanine are toxic to the central nervous system

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and produce some of the symptoms of the genetic disease
phenylketonuria. Other genetic metabolic defects in the tyrosine pathway are also known. As indicated above, absence of enzymes operating between tyrosine and melanin, is a cause of albinism. Tyrosinosis is a rare defect that causes hydroxyphenylpyruvic acid to accumulate in the urine. Alkaptonuria makes urine turn black on exposure to air, causes pigmentation to appear in the cartilage, and produces symptoms of arthritis. A different block in another pathway from tyrosine produces thyroid deficiency leading to goiterous cretinism (due to lack of thyroxine).

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Molecular Genetics

1. Explain what is meant by a metabolic pathway:

2. Describe the role that enzymes play in metabolic pathways:

3. List three final products of the metabolism of phenylalanine:

4. Name the enzyme failures (faulty enzymes) responsible for the following conditions: (a) Albinism:
(b) Phenylketonuria:
(c) Cretinism:
(d) Tyrosinosis
(e) Alkaptonuria:
5. Explain why people with phenylketonuria have light skin coloring:

6. Explain the role of the hormone thyroxine in causing the symptoms of cretinism:

7. The five conditions illustrated in the diagram are due to too much or too little of a chemical in the body. For each condition listed below, state which chemical (absent or in excess), causes the problem: (a) Albinism:

(b) Phenylketonuria:
(c) Cretinism:
(d) Tyrosinosis
(e) Alkaptonuria:
8. If you suspected that a person suffered from phenylketonuria, suggest how could you test for the condition:

9. The diagram at the top of the previous page represents the normal condition for a simple metabolic pathway. A starting chemical, called the precursor, is progressively changed into a final chemical called the end product. Consider the effect on this pathway if gene A underwent a mutation and the resulting enzyme A did not function: (a) Name the chemicals that would be present in excess:

(b) Name the chemicals that would be absent:
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Control of Metabolic Pathways

A

structural genes called the operator. A gene outside the operon, called the regulator gene, produces a repressor molecule that can bind to the operator and block the transcription of the
structural genes. It is the repressor that switches the structural genes on or off and controls the metabolic pathway. Two
mechanisms operate in the operon model: gene induction and
gene repression. Gene induction occurs when genes are
switched on by an inducer binding to the repressor molecule and deactivating it. In the Lac operon model based on E.coli, lactose acts as the inducer, binding to the repressor and permitting transcription of the structural genes for the utilization of lactose. Gene repression occurs when genes that are normally switched on (e.g. genes for synthesis of an amino acid) are switched off by activation of the repressor.

The operon mechanism was proposed by Jacob and Monod to
account for the regulation of gene activity in response to the needs of the cell. Their work was carried out with the bacterium Escherichia coli and the model is not applicable to eukaryotic cells where the genes are not found as operons (see opposite for the eukaryote model). An operon consists of a group of closely linked genes that act together and code for the enzymes that control a particular metabolic pathway. These may be for the metabolism of an energy source (e.g. lactose) or the synthesis of a molecule such as an amino acid. The structural genes contain the information for the production of the enzymes themselves and they are transcribed as a single transcription unit. These structural genes are controlled by a promoter, which initiates the formation of the mRNA, and a region of the DNA in front of the

Control of Gene Expression Through Induction: the Lac Operon Structure of the operon
RNA polymerase

At least one structural gene is
present. The structural gene
codes for the creation of an
enzyme in a metabolic pathway.

Transcription begins
Regulator
gene

Promoter

The regulator gene, on another part
of the DNA, produces the repressor
molecule by protein synthesis. In the
lac operon the regulator gene is
located next to the promoter.

Structural
gene A

Operator

The promoter site is where
the RNA polymerase
enzyme first attaches itself
to the DNA to begin
synthesis of the mRNA.

DNA

The operator is the potential blocking
site. It is here that an active repressor
molecule will bind, stopping mRNA
synthesis from proceeding.

OPERON
The operon consists of the structural genes and the promoter and operator sites

Structural genes switched off
RNA polymerase enzyme
may not be able to bind to
the promoter, or it may be
blocked along the DNA.

Lactose is not a common energy source for E. coli
and the genes for the metabolism of lactose by
the cell are normally switched off. With lactose
absent, the repressor molecule binds tightly to the
operator. This prevents RNA polymerase from
transcribing the adjacent structural genes and the
enzymes for lactose metabolism are not produced.

An active repressor molecule
binds to the operator site and
suppresses its activity (the
gene is “switched off”).

Transcription is stopped
Repressor
Regulator
gene

Gene induction

Promoter

Operator

Structural
gene A

The inducer binds to the repressor altering
its shape. It can no longer bind to the DNA,
permitting the operator gene to become
active (i.e. the gene is “switched on”).

DNA

When lactose is available, some of it is converted
into the inducer allolactose. Allolactose binds to the
repressor molecule, altering its shape and preventing
it from binding to the operator. The structural genes
can then be transcribed, and the enzymes for the
metabolism of lactose are produced.

Inducer

Repressor
Repressor

Regulator
gene

Transcription occurs

Promoter

Operator

Photocopying Prohibited

Structural
gene A

© Biozone International 2001-2003

DNA

Use RESTRICTED to schools where students have purchased this manual 147

Molecular Genetics

Control of Gene Expression in Eukaryotes
Although all the cells in your body contain identical copies of your genetic instructions, these cells appear very different. Morphological differences between cell types reflect profound differences in gene expression. For example, nerve cells

express proteins responsible for propagating electrical signals, whereas muscle cells express the proteins that make up the
contractile elements. This variety of cell structure and function reflects the precise control over the time, location and extent of expression of a huge variety of genes.
The role of transcription factors: RNA polymerase requires
additional proteins called transcription factors in order to recognize and bind to the promoter region at the upstream end of the gene. According to one hypothesis, transcription is
activated when a hairpin loop in the DNA brings the transcription factors attached to the enhancer in contact with the transcription factors bound to RNA polymerase at the promoter. Transcription is deactivated when a terminator sequence is encountered.

Terminators are nucleotide sequences that function to stop
transcription. Do not confuse these with terminator codons,
which are the stop signals for translation.

Transcription factors
that bind to enhancer

RNA
polymerase

Transcription factors that
bind to RNA polymerase

Promoter

Coding region
of gene

Enhancer
sequence
Transcription factors and
RNA polymerase bind

Transcription begins and
will continue until a
terminator is encountered.

A range of transcription factors and enhancer sequences
throughout the genome may selectively activate the expression of specific genes at appropriate stages in cell development.

1. Explain the functional role of each of the following in relation to gene regulation in a prokaryote e.g. E. coli: (a) Operon:

(b) Regulator gene:

(c) Operator:

(d) Promoter:

(e) Structural genes:

2. (a) Explain the advantage in having an inducible enzyme system that is regulated by the presence of a substrate:

(b) Suggest when it would not be adaptive to have an inducible system for metabolism of a substrate:

3. With reference to eukaryotes, briefly explain why the control of gene expression is necessary:

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