MB ChB Intake 2021
MB ChB
Molecules, Genes and Disease Term 1
Student Workbook
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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Table of Contents Introduction to the Unit ................................................................................... 5
Aim of the Unit ............................................................................................. 5
GMC ‘Outcomes for Graduates’ Addressed by the Unit .............................. 5
Specific Learning Outcomes ......................................................................... 6
The Curriculum Philosophy – Guided Learning ............................................ 7
Resources for the Unit .................................................................................. 8
Unit Lead....................................................................................................... 8
1 Session One: Chromosomes, DNA and Nucleotides ................................. 9
1.1 Aim of the Session ............................................................................ 9
1.2 Learning Outcomes for the Session .................................................. 9
1.3 Structure of the Session ................................................................... 9
1.4 Lecture 1.1: Introduction to the Unit, and Chromosomes, DNA and
Nucleotides ................................................................................................... 9
1.5 Group Work: DNA Structure ........................................................... 12
1.6 Lecture 1.2: DNA Replication, Mitosis and Meiosis ........................ 12
1.7 Self-Directed Study: The Cell Cycle, Mitosis and Meiosis ............... 14
2 Session Two: Transcription and Translation ........................................... 17
2.1 Aim of the Session .......................................................................... 17
2.2 Learning Outcomes for the Session ................................................ 17
2.3 Structure of the Session ................................................................. 17
2.4 Lecture 2.1: Genes and Transcription............................................. 17
2.5 Lecture 2.2: The Genetic Code and Translation ............................. 19
2.6 Group Work: DNA Structure ........................................................... 21
2.7 Group Work: Transcription ............................................................. 25
Self-Directed Study: Transcription and Translation Task: questions to test
understanding ............................................................................................. 29
3 Session Three: Genotypes and Inheritance ............................................ 35
3.1 Aim of the Session ........................................................................... 35
3.2 Learning Outcomes for the Session ................................................ 35
3.3 Structure of the Session .................................................................. 35
3.4 Lecture 3.1: Genotype, Phenotype and Patterns of Inheritance .... 36
3.5 Group work: Inheritance of Genes .................................................. 38
3.6 Lecture 3.2: Mutations .................................................................... 41
3.7 Self-directed study: Mutations and their Consequences ................ 43
4 Session Four: Chromosomal Abnormalities ............................................ 49
4.1 Aim of the Session ........................................................................... 49
4.2 Learning Outcomes for the Session ................................................ 49
4.3 Structure of the Session .................................................................. 49
4.4 Lecture 4.1: Chromosomal Abnormalities ...................................... 50
4.5 Group work: Chromosomal Abnormalities ..................................... 52
4.6 Lecture 4.2: Prenatal Diagnostics .................................................... 53
4.7 Self-directed study: Chromosomal Abnormalities and Prenatal
Diagnostics .................................................................................................. 55
5 Session Five: Amino Acids and Proteins .................................................. 63
5.1 Aim of the Session ........................................................................... 63
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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5.2 Learning Outcomes for the Session ................................................ 63
5.3 Structure of the Session ................................................................. 63
5.4 Lecture 5.1: Amino Acids and Proteins ........................................... 64
5.5 Lecture 5.2: Protein folding and function ....................................... 68
5.6 Group work: Amyloidosis ............................................................... 70
5.7 Self-directed study .......................................................................... 71
6 Session Six: Protein Processing and Targeting ....................................... 79
6.1 Aim of the Session .......................................................................... 79
6.2 Learning Outcomes for the Session ................................................ 79
6.3 Structure of the Session ................................................................. 79
6.4 Lecture 6.1: Protein Processing and Targeting in Cells 1 ................ 80
6.6 Group Work: Control of Cell Growth .............................................. 82
6.7 Lecture 6.2: Protein Processing and Targeting in Cells 2 ................ 90
6.8 Self-Directed Study: Connective Tissue Disorder ........................... 92
7 Session Seven: Protein Function............................................................. 95
7.1 Aim of the Session .......................................................................... 95
7.2 Learning Outcomes for the Session ................................................ 95
7.3 Structure of the Session ................................................................. 95
7.4 Lecture 7.1: Enzyme Activity .......................................................... 96
7.5 Group Work: Enzyme Activity ......................................................... 99
7.6 Self-Directed Learning: Protein Regulation .................................. 107
8 Session Eight: Haemoglobinopathies ................................................... 110
8.1 Aim of the Session ......................................................................... 110
8.2 Learning Outcomes for the Session .............................................. 110
8.3 Structure of the Session ................................................................ 110
8.4 Lecture 8.1: Haemoglobin and Myoglobin .................................... 111
8.5 Lecture 8.2: Haemoglobinopathies ............................................... 112
8.6 Group work: Haemoglobinopathies .............................................. 113
Genetic Counselling................................................................................... 113
8.7 Self-directed study: Haemoglobinopathies ................................... 115
9 Session Nine: Molecular Techniques and Diagnosis ............................. 119
9.1 Aim of the Session ......................................................................... 119
9.2 Learning Outcomes for the Session .............................................. 119
9.3 Structure of the Session ................................................................ 119
9.4 Lecture 9: Molecular Techniques and Diagnosis ........................... 120
9.5 Group Work: Human Genome Project .......................................... 125
9.6 Self-directed study: Molecular Techniques and Diagnosis ........... 125
10 Session Ten: Personalised Medicine ................................................. 127
10.1 Aim of the Session ......................................................................... 127
10.2 Learning Outcomes for the Session .............................................. 127
10.3 Structure of the Session ................................................................ 127
10.4 Lecture 10.1: Personalised Medicine ............................................ 128
10.5 Group Work 10: Exploring your Patients Genome ........................ 131
10.6 Lecture 10.2: Gene Therapy .......................................................... 132
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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10.7 Self-directed study: Personalised medicine ................................. 135
11 Session Eleven: Case Studies ............................................................ 143
11.1 Aim of the Session ........................................................................ 143
11.2 Learning Outcomes for the Session .............................................. 143
11.3 Structure of the Session ............................................................... 143
11.4 Lecture 11: Introduction to Case Studies ..................................... 143
11.5 Group work: Integrated Case Studies with Educator Support ..... 143
11.6 Self-directed study ........................................................................ 163
12 Session Twelve: Revision Week ........................................................ 164
Index ............................................................................................................. 165
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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Introduction to the Unit
Aim of the Unit
The aim of this unit is that you should understand the general relationship
between the processes involved in gene and chromosome behaviour, gene
expression and the activity of cells, and be able to apply your understanding
to your clinical practice in the future.
GMC ‘Outcomes for Graduates’ Addressed by the Unit
Like all the units in the MB ChB course, this unit helps you to meet the
outcomes defined by the General Medical Council in its document
‘Promoting Excellence: Standards for Medical Education and Training’ (2016)
and updated in 2018. The particular outcomes addressed are listed below,
but it is important to realise that no single unit in the course will enable any
of these outcomes fully to be achieved, as they are addressed by multiple
units across the course.
Outcomes 1: Professional values and behaviours
2. Newly qualified doctors must behave according to ethical and
professional principles. They must be able to:
e. act with integrity, be polite, considerate, trustworthy and honest
f. take personal and professional responsibility for their actions
g. manage their time and prioritise effectively
h. recognise and acknowledge their own personal and professional
limits and seek help from colleagues and supervisors when
necessary, including when they feel that patient safety may be
compromised
j. recognise the impact of their own attitudes and perceptions
(including personal bias, which may be unconscious) on groups
within society or individuals belonging to particular groups and
identify personal strategies that might be adopted to address
this
m. act appropriately, with an inclusive approach, towards patients
and colleagues
o. raise and escalate concerns through informal communication
with colleagues and through formal clinical governance and
monitoring systems about bullying, harassment and undermining
p. explain and demonstrate the importance of professional
development and lifelong learning and demonstrate
commitment to this
u. engage in their induction and orientation activities, learn from
experience and feedback, and respond constructively to the
outcomes of appraisals, performance reviews and assessments
8. Newly qualified doctors must recognise the role of doctors in
contributing to the management and leadership of the health service.
They must be able to:
b. undertake various team roles including, where appropriate,
demonstrating leadership and the ability to accept and support
leadership by others
c. identify the impact of their behaviour on others
Outcomes 2: Professional skills
10. Newly qualified doctors must be able to communicate effectively,
openly and honestly with patients and colleagues. They must be able
to:
a. communicate clearly, sensitively and effectively with patients
and colleagues from medical and other professions, by:
• listening, sharing and responding
• demonstrating effective verbal and no-verbal interpersonal skills
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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• making adjustments to their communication approach if needed, for
example for people who communicate differently due to a disability
or who speak a different first language
• seeking support from colleagues for assistance with communication
if needed
b. communicate by spoken, written and electronic methods clearly,
sensitively and effectively with colleagues from medical and
other professions
c. use methods of communication used by colleagues such as
technology-enabled communication platforms, respecting
confidentiality and maintaining professional standards of
behaviour
Outcomes 3: Professional knowledge
22. Newly qualified doctors must be able to apply biomedical scientific
principles, methods and knowledge to medical practice and integrate
these into patient care. This must include principles and knowledge
relating to anatomy, biochemistry, cell biology, genetics, genomics and
personalised medicine, immunology, microbiology, molecular biology,
nutrition, pathology, pharmacology and clinical pharmacology, and
physiology. They must be able to:
a. explain how normal human structure and function and
physiological processes applies, including at the extremes of age,
in children and young people and during pregnancy and
childbirth
b. explain the relevant scientific processes underlying common and
important disease processes
c. justify, through an explanation of the underlying fundamental
principles and clinical reasoning, the selection of appropriate
investigations for common clinical conditions and diseases
f. analyse clinical phenomena and conduct appropriate critical
appraisal and analysis of clinical data, and explain clinical
reasoning in action and how they formulate a differential
diagnosis and management plan
26. Newly qualified doctors must be able to apply scientific method and
approaches to medical research and integrate these with a range of
sources of information used to make decisions for care. They must be
able to:
e. critically appraise a range of research information including study
design, the results of relevant diagnostic, prognostic and
treatment trials, and other qualitative and quantitative studies as
reported in the medical and scientific literature
f. formulate simple relevant research questions in biomedical
science, psychosocial science or population science, and design
appropriate studies or experiments to address the questions
Specific Learning Outcomes
By the end of this unit you should be able to:
• Explain how basic cell structure relates to functional processes in the
cell
• Describe how amino acid chemistry relates to protein structure
• Describe the action of enzymes and the major mechanisms for their
regulation
• Explain the link between the molecular structure and the
physiological function of oxygen-transporting proteins
• List the features of and compare RNA and DNA
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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• Explain the relationship between DNA, chromosomes and genes
• Describe the general features of DNA replication, mitosis and meiosis
• Describe the general principles of chromosomal inheritance
• Apply the underlying principles of pedigree analysis to clinical cases
• Describe the processes of transcription, translation and post-
translational modifications
• Explain the different types of DNA mutation and their consequences
at transcription and translation
• Describe some of the most common molecular techniques used to
analyse genes and proteins
• Describe the concept of personalised medicine
• Apply the concepts of gene therapy to specific clinical contexts
• Apply, in particular, understanding of the concepts in this unit to the diagnosis and management of patients who present with:
o Affective disorders
o Abdominal pain - acute
o Blood glucose - abnormal
o Bowel habit - change/nausea/vomiting
o Breast abnormality/nipple disease
o Breathlessness
o Chest pain - acute and chronic
o Distension and/or oedema
o Fertility/contraception
o Fever/infection
o Fractures/dislocation
o Genetic and/or congenital disorder
o Groin abnormalities
o Haematuria
o Haemoptysis
o Jaundice
o Joint pain/swelling - chronic
o Muscle weakness
o No energy/TATT/lethargy
o Oral/nasal abnormalities
o Pallor and/or abnormal blood test
o Pregnancy/antenatal care
o Red eye painful/red
o Screening test positive
o Skin rash/lesion
o Visual abnormalities
o Weight - abnormality
• Apply understanding of the concepts in this unit, where relevant, to the remaining key presentations on the list defined in the ‘Code of Practice for Assessment’
The Curriculum Philosophy – Guided Learning
This unit is not a separate entity, but part of an integrated programme with a
clear educational philosophy – guided learning. This workbook provides
much of the material you will need to follow this process, though you will
also use other resources. Guided learning has three key features:
Constructing understanding
The unit aims to present material to you in an easily digestible way, but you
must then work to develop the understanding that will allow you to apply it
effectively to the practice of medicine. You must continually explore and re-
visit ideas from all units as the course progresses, applying them repeatedly
in different ways. This way you will construct understanding for yourself
through systematic exploration and application of ideas. This requires active
learning, which is the antithesis of the passive acquisition and regurgitation
of material that you may have indulged in previously.
Molecules, Genes and Disease Term 1 Workbook Introduction to the Unit
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Learning in context
Building understanding requires appreciation of the context in which you will
use it - the practice of medicine. Guided learning therefore focuses your
understanding on patient problems by continually revisiting material in the
contexts of different patient problems. We use a list of common patient
problems both to structure problem solving tasks to help you learn, and
assessment tasks to test whether you have. ‘Cognitive re-organisation’ is
the essence of learning to be a doctor. You must move away from slavishly
learning material as it has been presented to you to understanding it in ways
that you are going to use it. You need to be able to take a given concept and
apply it in multiple different contexts.
Learning together
You are privileged to have your fellow medical students as companions on
your personal journey. Like all journeys, a medical course is easier and more
fun in company. Working with other students in group work will both
enhance your own learning, and ensure that you acquire valuable skills of
working with others.
The key steps of guided learning
Guided learning follows the same pattern in all units, so you are guided
through repeated reflective cycles of learning.
• We will present ideas and material to you mostly through lectures,
supplemented by guided reading of material in textbooks and other
resources.
• You will then move rapidly into group work, where you will address,
as a group, structured problems relating to the material that has just
been presented, its relationships to other material and the
application of those concepts to common patient problems. Your
work in groups will be supported by tutors, but not directly tutor-led.
• The group work is followed up by self-directed study, where you will
work on the ideas further using a variety of resources, some
suggested by us, others identified by yourselves, in order to enhance
your understanding, and link it to other material across the course so
you may apply it to a wide range of patient problems.
Resources for the Unit
Standard recommended textbooks:
• New Clinical Genetics. Andrew P. Read, Dian Donnai (3rd Ed 2015;
ISBN 978-1907904677)
Texts specific to this unit:
• Essential Cell Biology. Bruce Alberts et al. (4th Ed, 2013; ISBN 978-
0815344551)
• Medical Biochemistry. John Baynes and Marek Dominiczak – (5th
Ed, 2018, ISBN 978-0702072994)
• Human Heredity. Michael Cummings (9th ed, 2010; ISBN 978-
0840053183)
• Lippincott’s MedMaps: Biochemistry. Saeid Karandish (2010)
• Lippincott’s Illustrated Reviews: Cell and Molecular Biology. Nalini
Chandar and Susan Viselli (2010)
• Marks’ Essentials of Medical Biochemistry. Michael Lieberman.
(2dn Ed, 2014; ISBN 978-1451190069)
• Molecular Biology of the Gene. James Watson et al. (7th Ed., 2013;
ISBN 978-0321762436)
Unit Lead
Dr Joanne Selway ([email protected]) delivered with
contributions from other academic staff and clinical educators.
mailto:[email protected]
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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1 Session One: Chromosomes, DNA and Nucleotides
1.1 Aim of the Session
The aim of this session is that you gain an overview of nucleotide structure
and packaging within the cell, and develop an understanding of the different
processes that alter the content of cellular DNA.
1.2 Learning Outcomes for the Session
1.2.1 Specific unit outcomes addressed by the session
• List the features of and compare RNA and DNA
• Explain the relationship between DNA, chromosomes and genes
• Describe the general features of DNA replication, mitosis and meiosis
• Apply, in particular, understanding of the concepts in this unit to the
diagnosis and management of patients who present with:
o Groin abnormalities
o Fertility/contraception
1.2.2 Detailed outcomes for the session
At the end of this session you should be able to:
• List the structural components of a DNA and an RNA molecules.
• Apply the conventions used to represent DNA/RNA base sequences.
• Describe polarity of a DNA or RNA chain.
• Describe the importance of hydrogen-bonding and base-pairing in
defining nucleic acid secondary structure.
• Describe the packaging of DNA.
• Describe the process and role of DNA replication.
• Describe the process and role of the cell cycle.
• Describe the process and role of mitosis and meiosis, and compare
and contrast the two processes.
• Explain clearly the difference between genotype and phenotype.
• Explain how environmental factors have an influence on both
phenotype and genotype.
1.3 Structure of the Session
0900 – 1000 Lecture 1.1: Introduction to the Unit, and Chromosomes,
DNA and Nucleotides
1000 – 1200 Group work: DNA Structure
1200 – 1300 Lecture 1.2: DNA Replication, Mitosis and Meiosis
1.4 Lecture 1.1: Introduction to the Unit, and Chromosomes,
DNA and Nucleotides
The nucleic acids DNA and RNA are very important molecules within the cell.
A nucleic acid is a polymer constituted of a chain of monomers called
nucleotides. In this lecture the way the individual DNA molecules are
‘packaged’ into chromosomes will be discussed (see Figure 1.1 below). In
addition, the structure of individual nucleotides, base pairing, the secondary
structure and polarity of nucleic acids, as well as conventions for writing
nucleic acid sequences will be discussed in detail.
From the double helix to visible chromosomes: in overview
Humans have 23 pairs of chromosomes; each chromosome contains one
tightly packed double stranded DNA molecule. After DNA replication of the
chromosome indicated in the picture below, each chromatid contains one
identical tightly-packed double stranded DNA molecule.
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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Figure 1.1. Packaging of DNA in the nucleus
From the double helix to visible chromosomes
In the nucleus, each double stranded DNA molecule is wrapped around
histones to form nucleosomes, which form ‘beads on a string DNA’ (Fig 1.2).
Nucleosomes can also tightly pack into solenoid structures, forming 30nm
fibres. These fibres are compacted into several ‘hierarchical loops’ to create
highly condensed structures, which are the chromosomes that are visible
under the light microscope in the nucleus during cell division (Fig 3; mitotic
chromosome, see session 2).
Figure 1.2: From left to right: DNA helix – nucleosome – beads on a string – nucleosome packing – 30nm fibre.
Figure 1.3: From left to right: from 30nm fibre DNA to mitotic chromosome.
Nucleotides and polynucleotides
The nucleic acids DNA and RNA are polynucleotides. DNA is a polymer of
deoxyribonucleotides and RNA is a polymer of ribonucleotides. A nucleotide
is composed of a nitrogenous base, a sugar and a phosphate. Nucleotides
are covalently linked together via phosphodiester bonds. Each single-
stranded nucleic acid chain has a polarity, two distinct ends: a 5’ end with a
free phosphate and a 3’end with a free OH-group.
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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Nitrogenous bases and base pairing
There are two types of nitrogenous bases:
• Purines have a two-ring structure, e.g. G and A
• Pyrimidines have a one-ring structure, e.g. C, T and U
A base pair is always formed by one purine and one pyrimidine.
• G always pairs up with C
• A pairs up with T (in DNA) or with U (in RNA)
In double stranded nucleic acids, the bases of each base pair are held
together by hydrogen bonds: three hydrogen bonds in the GC-base pair and
two hydrogen bonds for the AT- and AU-base pair, as shown in Fig 1.4.
Figure 1.4. Base pairing.
The DNA sequence depicted above can also be represented as:
Note that by convention the top strand is 5’ to 3’ from left to right.
The structures of DNA and RNA are compared in Fig 1.5.
Figure 1.5. Comparison of DNA and RNA structure.
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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1.5 Group Work: DNA Structure
Understanding and building DNA Structure
5min to build a sequence and take a snapshot.
Share with a member of your team to see if they can recreate.
https://www.explorelearning.com/index.cfm?method=cResource.dspView&
ResourceID=439
https://learn.genetics.utah.edu/content/basics/builddna/
Play the Game!
In your group practice with the nobel prize DNA game and post pictures of
your scores.
Nominate 1 per group to participate in a group vs group competition to get
the highest score in a set amount of time overseen by a clinical educator!
https://educationalgames.nobelprize.org/educational/medicine/dna_double
_helix/dnahelix.html
1.6 Lecture 1.2: DNA Replication, Mitosis and Meiosis
During cell division the genetic information held within chromosomes must
be duplicated in order to produce two daughter cells that are both identical
to the original cell. Genetic information is held in the base sequence of DNA,
which must be maintained faithfully during cell division. In this lecture the
highly organised duplication process, called DNA replication, will be
discussed.
When somatic cells divide, they do so by the process of mitosis. A special
type of cell division, meiosis, occurs in order to produce gametes (sperm and
eggs). By understanding the behaviour of chromosomes at meiosis it
becomes apparent how genes are inherited.
The cell cycle (Fig. 1.6)
• G1 – the cell prepares for DNA replication
• S – DNA replication
• G2 – the cell prepares for cell division
• M – cell division (mitosis)
Figure 1.6. The cell cycle
https://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=439https://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=439https://learn.genetics.utah.edu/content/basics/builddna/https://educationalgames.nobelprize.org/educational/medicine/dna_double_helix/dnahelix.htmlhttps://educationalgames.nobelprize.org/educational/medicine/dna_double_helix/dnahelix.html
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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DNA replication
During the synthesis of new DNA molecules, the base sequence of each new
strand is defined by the specific base pairs which can form on the template
strand. This process is catalysed by the enzyme DNA polymerase and utilises
deoxyribonucleoside triphosphate (dNTP) substrate molecules, while the
enzyme helicase unravels the DNA double helix.
DNA polymerase catalyses the reaction:
(dNMP)n + dNTP → (dNMP)n+1 + PPi
This reaction is stepwise and results in a 5’ to 3’ chain growth.
DNA replication is confined to the S phase of the eukaryotic cell cycle. The
process is semi-conservative (i.e. each strand of the parent molecule is
maintained in the two daughter molecules) and is initiated at defined 'origins
of replication'. Strand separation is followed by polymerisation of incoming
nucleotides by stepwise extension of 3´ ends; the leading strand is
synthesised continuously but the lagging strand is made discontinuously
(Okazaki fragments). These small DNA fragments are joined together by the
enzyme DNA ligase.
Cell division
Chromosomes are the thread-like structures in the cell nucleus that carry
genetic information. Human somatic cells are diploid and carry 23 pairs of
chromosomes (22 pairs of autosomes and one pair of sex chromosomes, XX
or XY). Human gametes (sperm and eggs) are haploid and contain only one
set of 23 chromosomes. Before a cell can divide all genetic information is
duplicated during the S phase of the cell cycle. Each chromosome duplicates,
creating the classical X-shape; each chromosome now consists of two
identical sister chromatids that are touching in a structure called the
centromere.
During mitosis a diploid cell undergoes one round of replication and one
round of division, i.e. a diploid cell produces two daughter cells which have
the same chromosome content as the parental cell (diploid). During meiosis
a diploid cell undergoes one round of replication and two rounds of division
(meiosis I and meiosis II), i.e. one diploid cell produces four haploid daughter
cells, each having half the number of chromosomes as the parental cell.
Mitosis (Fig. 1.7)
In order to grow (or for maintenance) somatic cells need to divide. After the
cellular content have been duplicated (G1 phase) and the DNA has been
duplicated (S phase) and double checked (G2 phase), the cell divides in a
highly organised process called mitosis (M phase).
Figure 1.7. Stages of mitosis
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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Meiosis
As gametes are haploid, genetic information must be halved during
spermatogenesis and oogenesis. This special cell division process is called
meiosis. One of the main consequences of meiosis is that it generates
genetic diversity, which is achieved in two ways: independent assortment of
chromosomes (during meiosis I) and crossing-over.
During meiosis I the homologous chromosomes of each chromosome pair
are divided. During meiosis II the chromatids of each chromosome are
divided.
1.7 Self-Directed Study: The Cell Cycle, Mitosis and Meiosis
The cell cycle is the process by which cells replicate their DNA and divide into
daughter cells. Control of the cell cycle ensures regulated proliferation but
this control is something that is lost during some disease processes and
specifically in cancer.
CT1.1 Name the four phases of the cell cycle, and describe the
processes occurring within each phase.
CT1.2 At which points in the cell cycle are the checkpoints which
control the progression of the cell through the division
process.
CT1.3 Compare and contrast the processes of mitosis and meiosis.
List three similarities and three differences between the
processes. Consider the number of phases and divisions,
number of daughter cells, role of the process, location of the
process.
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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CS1.1 Case Study – Fertility / Contraception
A couple presents to a fertility clinic as they were unable to achieve
pregnancy after one year of unprotected intercourse. The male partner is 29
years old with negative medical history. Examination is unremarkable.
Genetic analysis of the sperm show no obvious abnormalities but the man
has a low sperm count.
1. How many chromosomes would you expect to see in the patient’s
genetic analysis?
2. One common cause of a low sperm count is meiotic arrest (no
maturation beyond spermatocytes). Explain why this would lead to
infertility.
3. If meiotic arrest was occurring in this patient, which processes might
you expect to be impaired?
4. What does the term non-disjunction, in relation to meiosis, mean?
5. Using a diagram, explain the effect of non-disjunction of chromsomes
in meiosis I and meiosis II on gamete production, thinking about
numbers of chromosomes.
Patient stories are a good way of trying to holistically understand a disease.
Listen to the patient stories about testicular cancer on NHS choices via the
following link and then answer the questions below.
http://www.nhs.uk/Conditions/Cancer-of-the-testicle/Pages/Jack-and-
Marks-stories.aspx
http://www.nhs.uk/Conditions/Cancer-of-the-testicle/Pages/Jack-and-Marks-stories.aspxhttp://www.nhs.uk/Conditions/Cancer-of-the-testicle/Pages/Jack-and-Marks-stories.aspx
Molecules, Genes and Disease Term 1 Workbook Session One: Chromosomes, DNA and Nucleotides
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CS1.2 Case study – Groin abnormalities
Jack, a 25 year old man, had noticed a dull pain in his testes for a week. He
self-examined and thought his left testes was harder than normal and felt
significantly bigger than the right testes. The GP suspected testicular cancer.
1. How do you explain the accelerated growth rate of cancers at the
cellular level?
2. The ultrasound tests confirmed testicular cancer. CT scans confirmed
that the testicular cancer had metastasized (spread) to the abdomen.
The 5-year survival rate is 71% for metastatic testicular cancer and
thus the man was concerned about his future fertility. What steps
might be taken to preserve Jack’s fertility prior to chemotherapy?
3. Name a drug used to treat metatstatic testicular cancer that interferes
with the elongation stage of DNA replication.
4. Name a drug used to treat cancers that interferes with the termination
stage of DNA replication.
5. Hypothesise why chemotherapy may lead to infertility in men being
treated for testicular cancer.
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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2 Session Two: Transcription and Translation
2.1 Aim of the Session
The aim of this session is that you gain an overview of transcription and
translation, and an awareness of their roles in normal function and disease.
2.2 Learning Outcomes for the Session
2.2.1 Specific unit outcomes addressed by the session
• Describe the general principles of chromosomal inheritance
• Apply the underlying principles of pedigree analysis to clinical cases
• Apply, in particular, understanding of the concepts in this unit to the
diagnosis and management of patients who present with:
o Jaundice
o Breast abnormality/nipple disease
o Pallor and/or abnormal blood test
o Red eye painful/red
o Genetic and/or congenital disorder
2.2.2 Detailed outcomes for the session
At the end of this session you should be able to:
• Describe the process and role of transcription
• Describe the process and role of translation
• Define the term ‘gene’
• List and summarise the major reactions involved the process of RNA
maturation in eukaryotes and explain their importance in gene
expression
• Describe the nature of the triplet code and be able to apply the
genetic code
• Explain the implications of the degeneracy of the genetic code
• Describe the different types of RNA molecule, i.e. mRNA, rRNA,
miRNA and tRNA.
• Analyse the differences in gene expression in mammalian and
bacterial cells and explain how the differences can be exploited
clinically
• Predict the effects of various mutations in a gene
• Explain how mutations outside the coding region can affect gene
expression
2.3 Structure of the Session
0900 – 1000 Lecture 2.1: Genes and Transcription
1000 – 1200 Group work: DNA Structure and Transcription
1200 – 1300 Lecture 2.2: The Genetic Code and Translation
2.4 Lecture 2.1: Genes and Transcription
In this lecture, the way that the DNA base sequence contains the recipe to
make proteins will be discussed. Genes not only contain the code, but also
information relating to the regulation of their expression. Transcription, the
first stage in the process that relates information from DNA to RNA to
protein, will be discussed in detail. The importance of promoters, RNA
polymerase, transcription initiation factors and activators/repressors will be
considered. Transcriptional regulation of eukaryotic gene expression will be
emphasised. The flow of information from DNA through to protein has been
termed the “central dogma” of biology (Fig 2.1).
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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Figure 2.1: The Central Dogma- from DNA to RNA to protein
Every cell in our body has the same genetic complement (with the exception
of anucleate RBCs, and mature T- and B-lymphocytes), however, different
cell types use different genes. Each cell has the same genetic information,
the same genotype, however, there are different sets of proteins expressed
in a different cell types, which makes them look and behave differently, i.e.
they have a different phenotype.
The DNA code is transcribed in the nucleus into an intermediate messenger
molecule, messenger RNA (this lecture); this message is then translated in
the cytoplasm into proteins (Fig. 2.2).
In the nucleus
Transcription = DNA to mRNA
In the cytoplasm
Translation = mRNA to protein
Types of RNA
rRNA ribosomal RNA >80% few kinds only many copies of each
RNA messenger RNA ~2% 100,000s
kinds
few copies of each
tRNA transfer RNA ~15% ~100 kinds very many copies of
each
miRNA microRNA ?? 100s kinds Few copies of each
Transcription
There are three phases in the transcription process:
• Initiation – promoter recognition and binding
• Elongation – the actual process of ‘transcribing’ by RNA polymerase
• Termination – a sequence-dependent termination of RNA chain growth
In mammalian cells there are a number of post-transcriptional processes that transform the pre-mRNA into mature mRNA:
• Capping – addition of a 5´cap
• Tailing (polyadenylation) – addition of a 3´polyA tail
• Splicing – the removal of introns; the exons are ‘spliced together’
Figure 2.2 Transcription and translation
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2.5 Lecture 2.2: The Genetic Code and Translation
Once the DNA code has been ‘transcribed’ into a temporary message (the
mature mRNA), this is then ‘translated’ into an amino acid sequence (the
protein; see figure below).
In this lecture, the genetic code will be introduced. Translation, the second
stage in the process that relates information from DNA to RNA to protein,
will be discussed in overview. The role of ribosomes, ribosomal RNA (rRNA),
amino acids and transfer RNA (tRNA) during this process will be considered.
In addition, the gene organisation and expression in bacterial and
mammalian cells will be compared.
The genetic code
The information contained within the mature mRNA is translated in the
cytoplasm into proteins. A conversion needs to take place to translate the
“4-letter DNA language” into the “20-letter protein language”:
Given the position of the bases in a codon, it is possible to find the
corresponding amino acid. For example, the codon 5’ AUG 3’ on mRNA
specifies methionine, whereas CUA specifies leucine. UAA, UAG and UGA are
termination signals. AUG is part of the initiation signal, in addition to coding
for internal methionines. For convenience the 3-letter and 1-letter codes for
all 20 amino acids are given in the tables below.
Translation
There are three phases in the translation process: Initiation – AUG codon recognition and binding, and formation of a functional ribosome
Elongation – the actual process of ‘translating’ the RNA message into protein
Termination – stop codon recognition and dissociation of ribosome Translation elongation: the mRNA is read codon by codon from 5’ to 3’, while
the polypeptide chain growth is from amino-to carboxy-terminus.
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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Amino Acid 3 letter 1 letter Amino Acid 3 letter 1 letter
Alanine Ala A Leucine Leu L
Arginine Arg R Lysine Lys K
Asparagine Asn N Methionine Met M
Aspartic acid Asp D Phenylalanine Phe F
Cysteine Cys C Proline Pro P
Glutamic acid Glu E Serine Ser S
Glutamine Gln Q Threonine Thr T
Glycine Gly G Tryptophan Trp W
Histidine His H Tyrosine Tyr Y
Isoleucine Ile I Valine Val V
First position
(5’end)
Second position Third position
(3’end) U C A G
U Phe Ser Tyr Cys U
Phe Ser Tyr Cys C
Leu Ser Stop Stop A
Leu Ser Stop Trp G
C Leu Pro His Arg U
Leu Pro His Arg C
Leu Pro Gln Arg A
Leu Pro Gln Arg G
A Ile Thr Asn Ser U
Ile Thr Asn Ser C
Ile Thr Lys Arg A
Met Thr Lys Arg G
G Val Ala Asp Gly U
Val Ala Asp Gly C
Val Ala Glu Gly A
Val Ala Glu Gly G
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2.6 Group Work: DNA Structure
Over the next few pages you will find group work questions and two hands-
on tasks. The hands-on part of this session will provide you with the
opportunity to have closer look at, and hopefully a better understanding of,
the three-dimensional structure of DNA. Each group will have miniDNA
model kits. For most of the questions you will benefit from using the DNA
models.
Hands-on task 1: Building DNA display models
Each table needs to build a different set of models and then share with the
rest of the tables.
Each group will receive miniDNA model kits, each with the following
components:
• the four DNA bases (Blue, Green, Orange and Yellow)
• deoxyribose groups (Red)
• phosphate groups (Purple)
• spacers (Clear)
• stand
The pegs that connect the bases indicate the hydrogen bonds (Green to
Yellow 3 pegs and Blue to Orange 2 pegs). Use this information and the
information concerning the size of purines and pyrimidines to determine
which colour represents which base.
Look at the sugar group and the diagram below.
Build the individual nucleotides in three steps:
• connect a phosphate group to the C5-peg, as indicated below;
• then insert the C1-connector of the sugar-phosphate to each base, as
indicated below - make sure the hydrogen bonds of the base are
pointing down!
• now hold the base horizontally in your right hand and twist the
sugar-phosphate downward at an angle of about 30-45°.
• Using the miniDNA nucleotides pair up the bases to make base pairs.
Make sure you keep the sugar-phosphates at the 30-45° angle! You
will notice that the sugar-phosphate on one side of the base pair is
now pointing up while the sugar-phosphate at the opposite side is
pointing down.
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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Cystic Fibrosis (CF)
This model depicts a tiny fraction of the sequence of the human cystic
fibrosis transmembrane conductance regulator (CFTR) gene, which is
associated with cystic fibrosis (CF). The sequence displayed in the model
carries a 3 bp deletion mutation (ΔF508) observed in 70% of all CFTR
mutations found in CF patients, lacking a CTT stretch from the coding strand,
causing a loss of a phenylalanine in the CFTR protein.
Gene: CFTR Gene size: 189,000 bp Mutation: 3 bp deletion
Wild Type ATC ATC TTT GGT
Mutant ATC ATT GGT CCT
Sickle Cell Anaemia
This model depicts a tiny fraction of the mutant form of the human β-globin
gene (HBB), which is the single known cause of sickle cell anaemia. All
patients that have the disease carry a single base pair mutation (A>T) in the
7thcodon of their β-globin gene. The sickle cell mutation results in just one
amino acid change (Glu>Val) in the human β-haemoglobin protein.
Gene: HBB Gene size: 1,600 bp Mutation: 1 bp substitution
Wild type CCT GAA GTA
Mutant CCT GTA GTA
Achrondroplasia
This model depicts a tiny fraction of the sequence of the human FGFR3 gene,
encoding a fibroblast growth factor (FGF) receptor. In 98% of all
achrondroplasia patients (a form of short-limbed dwarfism) a single base pair
mutation (G>A) is found at position 1138 of the cDNA, causing single amino
acid change (Gly>Arg) in the gene product.
Gene: FGFR3 Gene size: 16,000 bp Mutation: 1 bp substitution
Wild type ACT GGG GCT
Mutant ACT AGG GCT
Phenylketonuria (PKU)
This model depicts a tiny fraction of the sequence of the human
phenylalanine hydroxylase gene (PAH), a gene often associated with PKU
(although there are mutations in other genes known to cause this disease).
Over 400 mutations in this gene are known, one of the most common
mutations, which causes ‘classic PKU’, is a single base pair mutation (G>A)
resulting in a premature stop, and an unstable mRNA transcript.
Gene: PAH Gene size: 79,000 bp Mutation: 1 bp substitution
Wild type AAA TAT TGG
Mutant AAA TAT TAG
The human genome contains approx. 3000 million bp, i.e. in (almost) every
diploid cell of your body you will have 6000 million bp of DNA (=6 x 10 9). Ten
base pairs of double stranded DNA in its helical form measures in real life
only 3.4 nm (=3.4 x 10-9 m).
Now look closely at one of the DNA display models and determine areas
within the model that are identical and areas that are different. Think of the
fact DNA is often described as a ‘twisted rope-ladder’. The inner core of the
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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double helix has atom structures that are flat (‘the steps of the ladder’) and
the twisted outer backbones (‘the ropes of the ladder’).
CT2.1 Which DNA components make up the ‘steps’ and which
make up the ‘ropes’?
Looking at the DNA model you will clearly see its helical structure.
Familiarise yourself with the helices and make sure you are aware that DNA
is indeed a double helix!
CT2.2 How many bases per one helical turn?
CT2.3 Can you find the ‘major groove’ and the ‘minor groove’?
Which groove do you think is used most often for DNA-
protein binding? And why?
CT2.4 Which base pair is formed using 3 hydrogen bonds and
which is formed using 2 hydrogen bonds?
a. What DNA bases are the purines and which bases are the
pyrimidines?
b. So, which is which in the chemical structure shown here?
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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CT2.5 Each group has been given a model of a nucleotide, which is
made up of which three components? Which base do you
have?
Look at the sugar component of your nucleotide model and note it is a 5-ring,
5-carbon sugar, a pentose. Compare it with the picture below. Carbon C1 is
connected to the base and C5 has a phosphate group connected to it. From
the C2 carbon the oxygen has been removed (a hydrogen instead of an OH-
group as expected in sugars), hence de-oxy (remember: DNA is an acronym
for deoxyribonucleic acid).
CT2.6 Which carbon is covalently linked to the phosphate group of
its neighbouring nucleotide - the neighbouring nucleotide is
either ‘one level down’ or ‘one level up’ depending on which
DNA strand you will look at (look at one of the large display
models)?
Hands-on task 2: Determining the DNA sequence of a display model
Every single stranded DNA molecule is directional, having a 5’ end with a free
phosphate group and a 3’ end with a free OH-group. The two strands of a
DNA double helix are antiparallel, one strand being ascending and one strand
being descending. Find the 5’ and 3’ ends of both strands in the display
model. Now write down the DNA sequence of the 10 bp model, as a single
line of letters, starting from the bottom. By convention a given DNA
sequence written as a single line of letters left to right is 5’ to 3’. Good luck
and have fun! Get your sequence checked.
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2.7 Group Work: Transcription
Task: Web-based resources
Log onto this website to access revision materials covering basic concepts
and to build a DNA and small polypeptide using these resources:
https://learn.genetics.utah.edu/content/basics/transcribe/
http://www.phschool.com/science/biology_place/biocoach/translation/genc
ode.html
Case Studies: To be worked on in groups and discussed with educators
as the end of the session
CS2.1 Case Study – Genetic and/or congenital disorder
A patient has had his genome sequenced, which has identified that a gene
involved in DNA transcription has a mutation. The mutation is redundant
and a functional mRNA is still produced. Despite a defect in this patient’s
genotype, there is no change in his phenotype.
1. Name the three stages in the process of transcribing a gene.
2. Name the three stages of mRNA processing.
3. What happens to the pre-mRNA when splicing happens?
4. Briefly explain what the terms genotype and phenotype mean.
5. Which type of protein binds to DNA to initiate DNA replication?
6. The sickle cell gene has three exons. What does this mean?
7. The cystic fibrosis gene is 188,703 bp but the mRNA contains only 4443
bp. Please explain this discrepancy.
8. Think about your professional responsibilities for this patient. How
would you react to a patient wanting to discuss this with you?
https://learn.genetics.utah.edu/content/basics/transcribe/http://www.phschool.com/science/biology_place/biocoach/translation/gencode.htmlhttp://www.phschool.com/science/biology_place/biocoach/translation/gencode.html
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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The herpes virus is a double stranded DNA virus that uses the host’s DNA
polymerase to replicate and infect host cells. Although viral and prokaryote
DNA replication is different to eukaryotic and, therefore, human DNA
replication, many of the mechanisms are the same. This hijacking of the
host’s cellular equipment is the methodology for virus replication although
some viruses have an RNA genome (see ToB session 10).
CS2.2 Case study – Red eye painful/red
A patient worried about having herpes enters the GP surgery looking for
advice. Please role play these questions as patient and GP before completing
the questions.
1. Your GP supervisor asks you to explain the general principles of DNA
replication to the patient. Role-play this with a partner, you may draw
a diagram o help explain.
2. Explain the differences between human DNA replication and viral
replication to the patient.
3. Explain the differences in DNA replication and RNA transcription to the
patient.
4. mRNA is processed after the initiation of transcription to become a
mature mRNA. What key processes are involved in this?
5. There are at least five different types of RNA molecules. List them and
state their function.
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A microRNA (abbreviated miRNA) is a small non-coding RNA molecule
(containing about 22 nucleotides) found in plants, animals and some viruses
that functions in RNA silencing and post-transcriptional regulation of gene
expression. This is a mechanism which can explain the occurrence of
different cell types in development from identical genetic code. The video
below helps explain the process. Remember the exact proteins and
complexes are not important, it is the concepts that are useful to explore.
https://www.youtube.com/watch?v=t5jroSCBBwk
CS2.4 Case study – Jaundice
Bob, 35 is chronically infected with the hepatitis C virus (HCV). He had heard
about a successful clinical trial of a new drug for patients like him called
miravirsen, and he wants you to discuss the possibility of taking this
medication
Review the material about the clinical trial and answer the questions below.
http://www.nejm.org/doi/pdf/10.1056/NEJMoa1209026
https://www.youtube.com/watch?v=hdhazyaxfc8
1. What is the molecular target of miravirsen?
2. Describe the structure of a micro-RNA when it is transcribed.
3. Explain the two major ways in which micro-RNAs lead to the inhibition
of a gene being expressed as a protein.
4. Explain why this particular micro-RNA (miR-122) might be a viable
therapeutic target in HCV.
5. miRNAs are involved in cellular regulation and could be molecular
targets for therapies in the future. Using the miRNA database
(http://www.mir2disease.org/) look at some of the diseases you have
encountered in the course to date, and investigate if there are any
micro-RNAs associated with those diseases.
https://www.youtube.com/watch?v=t5jroSCBBwkhttp://www.nejm.org/doi/pdf/10.1056/NEJMoa1209026https://www.youtube.com/watch?v=hdhazyaxfc8http://www.mir2disease.org/
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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CS2.5 Case Study – Breast Abnormality/Nipple Disease
Johnny's first wife died at the age of 39 with breast cancer. He has just learnt
that his daughter has just passed away age 42, 10 years after a breast cancer
diagnosis. He is utterly bereft. He remembers the day of diagnosis very
clearly. She had just had a regular screening mammogram that picked up an
area of calcification in her right breast. A biopsy had been taken and then
came the dreadful news that she had a carcinoma in situ. He knows all about
cancer: he himself has prostate cancer, and his mother died of ovarian
cancer, so knows just how hard it is to cope with a serious disease. He has
two granddaughters (aged 18 and 16) now to worry about.
1. What phrase might you use to ask him what his expectation of the visit
is?
2. Cancer is a disease of uncontrolled cell growth. Draw a diagram of the
cell cycle, labelling the different stages.
3. Annotate your diagram of the cell cycle the processes occurring in each
phase.
4. DNA replication occurs during cell proliferation. Complete the equation
below illustrating DNA replication.
(dNMP)n + → (dNMP)n+1 +
5. In order to maintain controlled cell growth, the cell cycle has
checkpoints in which the cellular machinery checks that all the
processes are complete to permit cell division. Give one example of a
check point protein.
6. Cell division requires a vast amount of energy. State cellular markers
of low and high energy states.
7. Using the ultrastructure image of a cell below, identify the organelle
responsible for energy production
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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Self-Directed Study: Transcription and Translation Task: questions
to test understanding
CT2.7 The nucleotide sequence around the initiation site on an
mRNA transcript for a certain protein is shown below.
Which codon specifies the first amino acid in any protein?
---GAAGCAUGGCUUCUAACUUUU---
CT2.8 Write the amino acid sequence for the N-terminal part of
the protein.
CT2.9 A codon UAA that can act as a stop codon is underlined.
Will it terminate translation in the above mRNA? Explain.
Molecules, Genes and Disease Term 1 Workbook Session Two: Transcription and Translation
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CT2.10 What is the expected minimum length (in nucleotides) of a
DNA strand that could code for a protein of 110kDa in a
bacterium? NB. The average molecular weight of an amino
acid is 110Da.
CT2.11 Explain why a eukaryotic gene encoding a protein might be
ten times longer than the theoretical minimum length.
CT2.12 Explain why there are only a few kinds of rRNA and 100000s
kinds of mRNA.
CT2.13 Normal human β-globin contains 146 amino. The table
below summarises amino acid changes in various inherited
haemo-globinopathies. In each case, the amino acid change
can be most simply explained as a point mutation in the
DNA of the β-globin gene. Write the expected RNA
sequence next to each amino acid in the table. NOTE: the
simplest explanation need not necessarily be the only one!
haemoglobin variant aa normal mutant
1 HbS (sickle cell) 6 Glu GAa/g Val GUn
2 HbC 6
3 HbOlympia 20
4 HbGenova 28
5 HbHammersmith 42
6 HbMMilwaukee 67
7 HbBristol 67
8 HbMalmö 97
9 HbKöln 98
10 HbKempsey 99
11 HbCasper 106
12 HbYosizuka 108
13 HbDPunjab 121
14 HbRainier 145
15 HbBethesda 145
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CT2.14 Five different haemoglobin variants have been identified in
which the Asp at position 75 in the normal human β-globin
polypeptide is replaced by Ala, Asn, Gly, His or Tyr. For each
example, convince yourself that the amino acid change can
be most simply explained as a single base pair change in the
DNA of the β-globin gene.
CT2.15 Wayne and Hb Constant Spring are haemoglobin variants in
which the α-globin polypeptides are longer than the normal
141 amino acids. Hb Wayne is 146 amino acids long and
HbCS is 172 amino acids long. The amino acid sequence at
the C-terminus of normal α-globin and that of the two
variants are:
Normal ser-lys-tyr-arg Hb Wayne ser-asn-thr-val-lys-leu-glu-pro-arg
HbCS ser-lys-tyr-arg-gln-ala-gly-ala-ser-val-ala....glu amino acid
138 141 146 172
For each variant indicate the nature of the genetic change that would give
rise to the altered sequence (i.e. what changes have occurred in the double
stranded DNA molecule?).
CT2.16 Write out the base sequence of the mRNA transcript of the
normal α-globin gene.
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CT2.17 Cordycepin is another name for 3’-deoxyadenosine, which is
used for research purposes to inhibit the synthesis of mRNA
in mammalian cells. From thinking about its name, draw the
structure of this compound. There is no need to draw the
exact chemical structure; a cartoon of the structure
indicating its key features will suffice.
CT2.18 Explain how you think cordycepin might work. (HINT: think
about the process of RNA chain growth)
CT2.19 Why do you think this compound does not inhibit DNA
synthesis?
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CT2.20 The death cap mushroom, Amanita phalloides, contains
several dangerous substances, including the lethal α-
amanitin. This toxin blocks RNA elongation in consumers of
the mushroom by binding to eukaryotic RNA polymerase II
with very high affinity. The initial reaction to ingestion of
the mushroom is gastrointestinal distress, caused by some
other toxins present in the mushroom. These symptoms
disappear, but about 48 hours later the mushroom-eater
dies, usually from liver dysfunction. Speculate on why it
takes this long for α-amanitin to kill.
CS2.3 Case study – Pallor and/or abnormal blood test
Althea, a 13 year old girl of Mediterranean origin, presents to the GP feeling
tired all the time. She has felt this way intermittently for several years, and
her mother has previously said it was due to the start of her periods.
However, she has become increasingly breathless, tired and has a pale
appearance.
1. The GP suspects thalassemia due to the girls Mediterranean origins.
What is thalassemia?
2. As the girl is tired, the GP asks about her diet. Draw out an eat-well
plate and describe the different elements of the plate.
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3. The GP does a genetic test, and discovers the girl has beta-thalassemia
caused by a mutation at the splice site of the beta-globin gene.
What happens to the genetic material in the pre-mRNA during splicing?
4. What complex is created to allow splicing to occur?
5. What effects might you anticipate occurring if there was a change in
the code around the splice site of a pre-mRNA? (Not in beta-globin
gene but any gene.)
6. Explain how alternative splicing can offer advantages for the cell and
its functional machinery.
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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3 Session Three: Genotypes and Inheritance
3.1 Aim of the Session
The aim of this session is for you to develop understanding of genotypes and
inheritance, particularly the inheritance of genetic disorders.
3.2 Learning Outcomes for the Session
3.2.1 Specific unit outcomes addressed by the session
• Describe the general principles of chromosomal inheritance
• Apply the underlying principles of pedigree analysis to clinical cases
• Explain the different types of DNA mutation and their consequences
at transcription and translation
• Apply, in particular, understanding of the concepts in this unit to the
diagnosis and management of patients who present with:
o Bowel habit- change/nausea/vomiting
o Genetic and/or congenital disorder
o Pallor and/or abnormal blood test
o Visual abnormalities
o Oral/nasal abnormalities
3.2.2 Detailed outcomes for the session
At the end of this session you should be able to:
• Describe the different patterns of inheritance and be familiar with
examples.
• Explain dominance, recessiveness, co-dominance and
complementation.
• Describe the basis of the co-inheritance of certain traits.
• Draw a family pedigree according to convention from a given family
history.
• Analyse genetic information from a pedigree and describe the family
concerned.
• Apply the principles of genetic data to calculate probabilities of
inheritance and recombination frequency
• Explain the relationship between changes in nucleotide and amino
acid sequences.
• Describe the different types of mutational changes, e.g. point
mutation, insertion, deletion.
• Explain the effect that different mutations may have, e.g. silent
mutation, missense mutation, nonsense mutation, frameshift
mutation.
• Describe how spontaneous and induced mutations may occur.
• Describe the genetic link between mutation and mutant phenotype
and explain how some mutations can be inherited.
• Describe the process and the role of DNA repair.
• Explain the relationship between DNA damage and cancer.
• Describe the fundamental importance of PCR in the diagnosis of
genetic disease.
• Describe the different genetic tests available for the detection of
mutations in genes.
• Explain some of the ethical issues associated with genetic testing.
3.3 Structure of the Session
0900 - 1000 Lecture 3.1: Genotype, Phenotype and Patterns of
Inheritance
1000 - 1200 Group work: Inheritance of Genes
1200 - 1300 Lecture 3.2: Mutations
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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3.4 Lecture 3.1: Genotype, Phenotype and Patterns of
Inheritance
In this lecture, the terms genotype and phenotype will be discussed, along
with how environmental factors can have an influence on both. An overview
will be given of the different patterns of inheritance (autosomal dominant,
autosomal recessive and X-linked), and linkage in relation to chromosome
structure and meiosis will be discussed. Information about the mode of
inheritance of traits through families can be recorded in pedigrees; standard
drawing conventions and specific examples will be discussed.
Pedigrees
To draw pedigrees there are certain rules you will have to adhere to, such as
the symbols for male and female individuals (both affected and not-
affected), the way a marriage (mating) is indicated, and the way offspring are
indicated (Fig 3.1). In pedigree analysis, carriers of a condition
(heterozygotes) are also often indicated by half-shading or a dot. A key for
drawing pedigrees can be found below. In addition, you may find a number
of other symbols in textbooks, which can be used for special situations like
monozygotic and dizygotic twins, and consanguineous marriages. By
convention, each generation is drawn on a separate horizontal line, and
generations are often indicated with roman numerals I, II, III etc. Also,
siblings are often drawn from left to right, starting with the oldest in the
family on the left.
Inheritance patterns
Patterns of inheritance can be divided in autosomal inheritance (when the
gene in question is located on an autosome) and sex-linked inheritance
(when the gene in question is located on a sex chromosome). X-linked
inheritance is when the gene in question is located on the X-chromosome. Y-
linked inheritance, which is quite rare, is when the gene in question is
located on the Y-chromosome, and is inherited directly from father to son.
The main patterns of inheritance are:
• autosomal recessive
• autosomal dominant
• X-linked (X-linked recessive or X-linked dominant)
Examples of these can be found in the diagram on the next page. Short
animations of these can be found on Moodle. NOTE: Under X-linked
inheritance an example of X-linked recessive is given.
Figure 3.1. Pedigree conventions
Linkage
If two genes are on different chromosomes, they show independent
assortment during meiosis. This is not the case if two genes are close
together on the same chromosome, such genes are said to be linked and are
said to co-segregate. However, the process of crossing-over and
recombination can result in two linked alleles being separated during
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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meiosis. The frequency of recombination between two loci can give
information with respect to how close these loci are two each other
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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3.5 Group work: Inheritance of Genes
Case Studies: To be worked on in groups and discussed with educators
as the end of the session
CS3.1 Case study – Genetic and/or Congenital Disorder
Maria has juvenile cataracts, which shows an autosomal dominant
inheritance. She marries James who is an unaffected carrier.
1. What is the probability of the disease occurring in their first child?
2. What is the probability of the disease occurring in their second child?
3. What is the probability of the disease occurring in all of their four
children?
4. What is the probability of the disease occurring in their first great-
grandchild?
CS3.2 Case study – Genetic and/or Congenital Disorder
Romeo and Juliet have 3 children, one of whom as Tay-Sachs disease. Tay-
Sachs disease shows autosomal recessive inheritance. The husband's sister
wishes to marry the wife's brother.
1. What is the risk that the first child of this second couple will be affected
with the disease (assume that the recessive allele is very rare in the
population)? HINT: You may want to draw a pedigree to help you
answer the question.
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2. Tay-Sachs disease is a lysosomal storage disorder. Describe the
organisation of a normal lysosome.
3. Describe the cellular basis of Tay-Sachs disease.
CS3.3 Case study – Oral/nasal abnormalities
In the year 2022 scientists discover that a form of “runny nose” is in fact a
genetic inherited trait. Below is a pedigree of a family where runny nose is
prominent. A couple called Jack (top row) and Jill (top row) have three
children (bottom row) called John, Jenny and Joy. Mother Jill, son John and
youngest daughter Joy all suffer from this form of runny nose.
Using the gene symbols R and r, consider the following four situations:
1. Could this trait be autosomal recessive? If so, what are the genotypes
of all family members?
2. Could this trait be autosomal dominant? If so, what are the genotypes
of all family members?
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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3. Could this trait be X-linked recessive? If so, what are the genotypes of
all family members?
4. Could this trait be X-linked dominant? If so, what are the genotypes of
all family members?
CS3.5 Case study – Genetic and/or Congenital Disorder
Each group is to watch a different video of a patient story and then write a
pedigree from the conversation and then think about the sort of information
the patient is going to want to know about their risk and what sorts of
questions might be asked in a summative exam if this patient featured.
Cystic fibrosis (up to 4.05 min):
https://www.youtube.com/watch?v=qgs4vVeg1Sw
Hereditary Spherocytosis:
https://www.youtube.com/watch?v=ddeTsa0H624
Sickle Cell Anaemia:
https://www.youtube.com/watch?v=qe59ar-GZmg
BRCA positive breast cancer:
https://www.youtube.com/watch?v=_rZI5i5IEA0
https://www.youtube.com/watch?v=qgs4vVeg1Swhttps://www.youtube.com/watch?v=ddeTsa0H624https://www.youtube.com/watch?v=qe59ar-GZmghttps://www.youtube.com/watch?v=_rZI5i5IEA0
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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3.6 Lecture 3.2: Mutations
This lecture covers both the phenotypic effects of mutations and the
mechanisms of mutagenesis. The session demonstrates how the position of
a single base change in a nucleotide sequence can determine whether there
is no phenotypic effect, a relatively minor effect such as a single amino acid
change or a drastic effect such as chain termination. Many diseases caused
by different kinds of mutations will be discussed.
What is a mutation?
Mutations are changes in the genetic code and can be classified in many
ways; all mutations can be grouped under one of the following:
• base substitution: change of a single nucleotide to one of the other
three
• deletion: removal of sequences
• insertion: addition of sequences
• rearrangement: rearrangement of sequences
Many mutations are spontaneous; mutations can also be induced (by
mutagens).
Mutations are not good or bad per se, just different.
• Mutations are a source of genetic variation.
• A mutation causes a mutant phenotype, which is a phenotype which
differs from the common or wild-type phenotype in the population.
• A mutation in a gene causes a mutant allele, which is an allele that
differs from the common allele in the population (the wild-type
allele).
Point mutations are base substitutions and can be either a transition (purine
to purine OR pyrimidine to pyrimidine) or a transversion (purine to
pyrimidine OR pyrimidine to purine).
Point mutations in the coding region of a protein can be a:
• silent mutation: a mutation that does not alter the amino acid
specified
• missense mutation: a mutation that replaces one amino acid with
another
• nonsense mutation: a mutation that changes the amino acid
specified to a stop codon
Point mutations in non-coding regions or outside genes can of course also be
detrimental as they can change protein binding sites, promoter sequences,
splice sites etc.
In insertions or deletions, the sequence that is added to or removed from
the nucleic acid can be a single nucleotide (single base mutation), a few
nucleotides (e.g. triplet repeats) to millions of nucleotides (e.g. tandem
duplications).
For insertions or deletions in the coding region of a protein:
• addition or subtraction of nucleotides other than multiples of 3:
frameshift mutation
• addition or subtraction of 3 nucleotides (or multiples of 3): no
change in reading frame (however, there will be a change in the
amino acid code)
To illustrate this, remember ‘The Fat Cat…’ (see below).
In a normal situation, the code makes sense:
THE FAT CAT ATE THE WEE RAT
1 bp insertion causes a frameshift, the code is broken:
THE XFA TCA TAT ETH EWE ERA T
1 bp deletion causes a frameshift, the code is broken:
THE ATC ATA TET HEW EER AT
3 bp insertion causes no frameshift, the code is changed but not broken:
THE BIG FAT CAT ATE THE WEE RAT
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3 bp deletion causes no frameshift, the code is changed but not broken:
THE CAT ATE THE WEE RAT or
FAT CAT ATE THE WEE RAT
DNA repair
The cell has several highly conserved mechanisms in place to check and
repair the genetic code (Fig. 3.2). Mutations happen frequently but are
recognised and repaired in most instances. For example, in mismatch repair,
a post-replication process, an erroneously inserted nucleotide (e.g. a G-T
base pairing) is recognised and replaced with the correct nucleotide.
Furthermore, there are two types of excision repair: base-excision repair and
nucleotide excision repair, which repair single-stranded DNA damage caused
by external agents, such as chemical mutagens and UV, or caused by
endogenous factors, such as reactive oxygen species (ROS). The two types of
excision repair are distinct and involve different enzymes. In addition, there
is double strand break (DSB) repair, when both DNA strands are broken,
which can cause chromosome re-arrangements (session 10).
Failure of DNA repair can have serious consequences to the cell, and can
cause disease.
Figure 3.2: In a healthy cell there is a fine balance between DNA damage and DNA repair.
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3.7 Self-directed study: Mutations and their Consequences
CT3.1 What is the difference between an oncogene and a proto-
oncogene?
CT3.2 What is the difference between excision repair and
mismatch repair?
A double-stranded DNA molecule containing the segment that is shown
below is exposed to nitrous acid, so that the base cytosine is changed to
uracil.
T A T A G T A
A T A T C A T
CT3.3 What will be the sequence of the strand synthesised
complementary to the one in which the change has
occurred during subsequent DNA replication?
CT3.4 What will the base pairs at the next replication be?
CT3.5 Is the mutation a transition or a transversion?
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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CT3.6 The powerful carcinogen aflatoxin B1 generates apurinic
sites (i.e. loss of purine residues) in DNA. Cellular repair
mechanisms preferentially insert an adenine residue
opposite an apurinic site. Illustrate this process, using a
single base, on the stretch of double stranded DNA below,
and indicate if it would lead to a transition or a transversion.
CT3.7 In the table below, several different genomic positions are
indicated in the first column. Consider and briefly describe
for each what you expect the effect of a relatively small DNA
mutation (a single base change or several base pairs
deletion) to be on the expression of the gene (more or less
expression, gene switched on or off, mRNA intact etc.) and
the protein (more or less production, functional or non-
functional, truncated etc.)
Position of mutation Gene expression Protein expression
Within the coding region
(ORF) of a gene
In the promoter of a gene
In an intergenic region
(region between two
genes)
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In an intron of a gene
In a splice site of an
intron
In a TATA box of a
promoter
At the polyadenylation
site of a gene
At the transcription
termination site of a gene
Several Mb (Mega base =
106 base) upstream of a
gene
There are a number of anti-cancer and anti-viral agents that affect DNA
replication. This is because DNA replication is central to cell division. Various
stages of the DNA replication process are targeted by current chemotherapy
treatments. For a concise review read p268 of the Molecular Biology of the
Gene, 7th ed Author Watson, James D.; James Dewey.
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CS3.6 Case Study – Genetic and/or Congenital Disorder
An extensive genetic test carried out on an unborn baby includes the DNA
sequencing of an entire chromosome. The test reveals a frame shift
mutation that leads to a premature stop codon in one the alleles of a gene of
known function.
1. Explain in general terms what you expect to happen if the mutant allele
is recessive to the wild type allele, and what if the mutant allele was
dominant over the wild type allele.
2. Explain why a ‘gain of function’ mutation is more likely to produce a
dominant trait than a recessive trait.
3. Describe the difference between a missense and a nonsense mutation?
4. Describe the difference between an insertion and a frameshift
mutation?
5. Describe the difference between a direct and indirect genetic test?
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CS3.7 Case study – Bowel Habit- Change/Nausea/Vomiting
Harry, a 70-year-old white male, presents to his primary care physician with a
complaint of rectal bleeding. He describes blood mixed in with the stool,
which is associated with a change in his normal bowel habit such that he is
going more frequently than normal. He has also experienced some crampy
left-sided abdominal pain and weight loss. He has previously been fit and
well and there was no family history of GI disease. Examination of his
abdomen and digital rectal examination are normal. The GP suspects
colorectal cancer.
1. The small intestine is the principal site for absorption of digestive
products. List the four factors that combine to give it a huge surface
area for absorption.
2. 5-fluorouracil (5-FU) is a major agent used in the treatment of
colorectal cancer. It is an analog of nucleotide precursors and thus
prevents nucleotides being made. What would be the specific effects
of this action on DNA replication?
3. Some anti-cancer therapies target DNA synthesis more directly and are
incorporated into DNA. Name some of these agents and list the
consequence of each one being incorporated.
Molecules, Genes and Disease Term 1 Workbook Session Three: Genotypes and Inheritance
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Some chemotherapy agents, such as AraC, cisplatin and bischloroethyl
initrosourea (BCNU) block the elongation or termination step of DNA
replication.
4. Which enzyme joins all of the pieces of DNA together during
termination of DNA repication?
5. What mechanism do cisplatin and BCNU use to prevent termination
during DNA replication?
6. Some chemotherapy agents that block DNA repication often have side-
effects such as hair loss and anaemia. Discuss below why you think
these side-effects occur?
7. Research some methodologies being developed to target
chemotherapy drugs more specifically to reduce the side effects such as
hair loss and anaemia.
Molecules, Genes and Di