Molecules, Genes and Disease Term 1 · 2020. 12. 16. · Molecules, Genes and Disease Term 1...

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


3.3 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



4.3 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



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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




6.4 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




7.4 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




9.4 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




10.4 Lecture 10.1: Personalised Medicine ............................................ 128

10.5 Group Work 10: Exploring your Patients Genome ........................ 131

10.6 Lecture 10.2: Gene Therapy .......................................................... 132


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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


11.3 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


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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


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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


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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.


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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.


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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.


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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.


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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


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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


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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.


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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


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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


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.







o Jaundice

o Breast abnormality/nipple disease


o Red eye painful/red




• 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


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).


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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


19

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.


20

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


21

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.


22

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


23

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?


24

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.


25

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


26

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.


27

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/


28

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


29

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.


30

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


31

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.


32

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?


33

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.


34

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

35

3 Session Three: Genotypes and Inheritance


The aim of this session is for you to develop understanding of genotypes and

inheritance, particularly the inheritance of genetic disorders.





• Explain the different types of DNA mutation and their consequences

at transcription and translation



o Bowel habit- change/nausea/vomiting



o Visual abnormalities

o Oral/nasal abnormalities



• 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.


0900 - 1000 Lecture 3.1: Genotype, Phenotype and Patterns of

Inheritance

1000 - 1200 Group work: Inheritance of Genes

1200 - 1300 Lecture 3.2: Mutations


36

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


37

meiosis. The frequency of recombination between two loci can give

information with respect to how close these loci are two each other


38

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?


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.


39

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?


40

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?


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


41

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


42

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.


43

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?


44

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)


45

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.


46

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?


47

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.


48

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

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