Download Biology 2250 - Memorial University

Biology 2250 - Principles of Genetics
Dr. Steven M. Carr
Department of Biology
Memorial University of Newfoundland
St. John's NL A1B 3X9, Canada
e-mail: [email protected]
Click on the following for:
Bio2250 Lab Manual v.05a (updated 02 Sept 2008)
(PDF format requires Acrobat reader)
downloading requires password; see instructors
Laboratory #1 handout (weeks of 15 & 22 Sept)
Bio2250 Lecture & Lab schedule (updated 25 Aug 2008)
Orientation to 2250 (please read !!)
How to use the website
Sample Quiz & Exam questions (updated 14 Dec 2007)
Link to Griffiths et al. (2002) Modern Genetic Analysis (2nd ed) online [MGA2]
Assigned readings from MGA-2
Online practice problems from MGA-2
Other webpages of interest:
Genetic Research in my lab
Bio2900 (Principles of Systematics & Evolution)
(from Winter 2001)
Bio4241 (Advanced Genetics)
(next offered Winter 2009)
Bio4900 (Fundamentals of Genetic Biotechnology) (next offered Spring 2009)
Activities of BIOS (MUN Biology Society)
Click here to e-mail me questions, comments, or suggestions.
Please include '2250' in the subject line
Lectures: TuTh 0900-1015
Sn-2109 (Science Lecture Theatre)
Labs:
MTuWTh 1400-1700 Sn-4110 (Genetics & Evolution Laboratory)
Course lecture notes:
These notes are revised before & after lectures; check frequently for revisions.
Topic
Last Revised
Lecture Date
History of the Discovery of DNA
02 Sept 2008
----------
1 Structure of DNA: the Hereditary Molecule
03 Sept 2008
04 Sept 2008
2
03 Sept 2008
09 Sept 2008
3 The Genetic Code
03 Sept 2008
11 Sept 2008
4 How Genes Work II: RNA Translation
15 Sept 2007
16 Sept 2008
15 Sept 2007
15 Sept 2007
18 Sept 2008
23 & 25 Sept 2008
06 Nov 2007
Sample
30 Sept 2008
09 Oct 2008
06 Nov 2007
14 Oct 2008
02 Oct 2008
5
How Genes Work I: DNA Replication & Transcription
How Genes Work III: Protein Structure & Function
Molecular Basis of Heredity
6 Chromosome Genetics I: Cytogenetics
Midterm Exam I
7
8
9
10
[Review 07,08 Oct]
Thanksgiving (No lecture)
Chromosome Genetics II: Genome organization
Mendelian Genetics:
Dominance, Segregation, & Assortment
Extensions to Mendelian Analysis
Pedigree Analysis
Chromosome Linkage
Recombination & Mapping
Molecular Basis of Mutation
Midterm Exam II
10 Oct 2007
06 Nov 2007
26 Oct 2007
06 Nov 2007
04 Nov 2007
[Review XX]
07 & 16 Oct 2008
21 Oct 2008
23 Oct 2008
28 Oct 2008
30 Oct 2008
06 Nov 2008
11 Eukaryotic Development
12
Remembrance Day (No lecture)
12 Nov 2007
04 Nov 2008
11 Nov 2008
13 Genetic Engineering & Biotechnology
18 Nov 2007
13 & 18 Nov 2008
14 Applications of Biotechnology & Genomics
28 Nov 2007
20 & 25 Nov 2008
15 Genetics & Genomics Research at Memorial University
Final Exam
[ Sn-2109 ]
27 Nov & 02 Dec 2008
Review:
XX Dec
XX Dec 2008
ACKNOWLEDGMENTS - sources for images
Site last modified 02 Sept 2008
This page has been accessed
* times since 14 October 1997
This page is dedicated to my daughter Jennifer Megan and my wife Justyna (MUN Research Report)
All text material © 2008 by Steven M. Carr
Where I'm coming from in Bio2250 - Principles of Genetics
Course Philosophy
Genetics is traditionally taught ’Peas first, DNA later'. Facts and concepts are developed in the
same order in which they were discovered historically. Genetics courses were taught for fifty years
without any clear understanding of the molecular nature of the gene. The ontogeny of most courses
follows this phylogeny. However, a certain pretense is required: when we talk about round and
wrinkled peas, we pretend you don't know about DNA, because Mendel didn't. This approach
works well through the unraveling of the "Central Dogma" (DNA makes RNA makes Protein) in
the early 1970s. In those days, we arrived at an understanding of protein synthesis, and the end of
the course, simultaneously.
However, 2003 was the 50th anniversary of the discovery of the structure of DNA, and the
molecular revolution in biology continues to accelerate. Genetics and molecular biology have
proliferated in so many directions that a single introductory course struggles to be comprehensive.
Worse, there is an ever-widening gap between what can be taught and what is required to
understand molecular genetics in 'general science' journals like "Science" or "Nature". Recent
experiments in genomics
have become technically so involved that it is difficult to present the complete logic, and we must
skip to summaries of conclusions. How can the connection be made?
Bio2250 is taught "DNA first, peas later". I reverse the traditional order. We begin with an
introduction to the molecular biology of DNA structure and protein function, and build on
this foundation to introduce the behaviour of genes on chromosomes and in crosses. A
course that begins, "DNA is a double-helix that is replicated semi-conservatively...." (a standing
broad jump over 50 years of classical genetics) serves to remind most students of material known
in a general way at least since high school. The logic of the classical experiments of Hershey &
Chase, Watson & Crick, and Meselson
& Stahl, and others, is a valuable introduction to scientific inference and problem solving. It is not
crucial to understanding how DNA
functions. Likewise, it is necessary to understand in detail how the Genetic Code works, and less
so to know how Nirenberg & Khorana figured it out in the first place. An initial grounding in the
processes of molecular biology equips us to talk about current topics such as DNA cloning,
Genetic Engineering, Biotechnology, and the Human Genome Project. Selected experiments
are still analyzed in detail, to emphasize the problem-solving approach in genetics. Most of
molecular genetics is doing in a test tube what goes on in a cell: if you understand nucleic acid
structure, base pairing rules, polynucleotide directionality, and replication, you can understand
vector insertion and molecular cloning. With such a background, and an orientation to modern
experimental techniques, I hope that the course will empower students to investigate further areas
of individual interest.
THE DIFFERENCE IN APPROACH MAY BE SUMMARIZED AS FOLLOWS.
The traditional method of teaching genetics is to understand phenotype in terms of genotype,
and show the genotypic basis of phenotypes. The method of analyzing crosses is the traditional
basis of "Genetics". That is, we teach that Peas have genes "for" alternative characteristics such
as round vs wrinkled, or green vs yellow. In the same way, Humans have a gene "for" a genetic
disease such as phenylketonuria. For each gene, we talk about in terms of one phenotype
"dominating" another, and two alternative alleles being dominant or recessive. The nature of
these alleles turns out to be due to variations the protein sequence, which is in turn a predictable
consequence of particular changes in DNA sequences.
The modern method is to show how DNA genotypes influence protein metabolic pathways that
produce characteristic phenotypes, the consequences of mutations in DNA for alteration of the
outcomes of these pathways, and the interactions of the alleles involved in terms of how they affect
those phenotypes. For example, we will see that in Peas, there is a DNA segment that codes for a
Starch Branching Protein, which when modified causes a loss of turgor pressure in seeds, and a
"wrinkled" appearance. Similarly, in Humans there is a gene that codes for the enzyme
Phenylalanine Hydroxylase, that various alleles of this gene produce higher or lower levels of
PAH, and that the biochemical interaction between the particular pair of alleles that an individual
has inherited determines whether or not that individual manifests a disease called
"Phenylketonuria". We understand "dominant" and "recessive" as descriptions of a phenotype
that is a consequence of a molecular genotype involving DNA and protein, rather than intrinstic
properties of bead-like genes on a string. The use of molecular biology to understand the flow of
information from DNA to protein to phenotype is sometimes called "Reverse Genetics" .
The Social Contract
1. I expect that all students will attend all lectures.
Exams are based on lecture material, not on the text.
2. As a matter of courtesy to other students and the lecturer, during lectures please:
Silence your cell phones.
Do not make or receive phone calls.
Do not send or answer text messages.
Do not talk to your classmates.
3. When you send me e-mail, please include ‘2250' in the subject line, to keep it from being sent to
the Trash.
Please include a polite salutation [Dear Dr Carr, Hi Prof Carr, Hey Steve, or something like that]. It
sounds better.
4. Average course marks in a recent year were:
Midterm I
Midterm II
Final
Labs
Course
69%
65%
56%
92%
68%
Lab marks are purposely kept high. The lab exercises are intended to guide you through a hands-on
experience with fundamental genetic concepts, rather than to make you sweat about marks. Do not
assume
that a high lab mark going into the Final exam guarantees a high mark for the course. Exams are
intentionally tougher. It is a serious mistake to slack off studying for the Final on such an assumption.
Other courses in Genetics at Memorial:
Population genetics is covered in Bio 2900 (Principles of Evolution & Systematics), another
course in the core curriculum. My own research is an application of molecular genetics to
evolutionary biology. You'll hear more about this later.
Molecular Biology of Nucleic Acids is covered in Biochemistry 3107 (Dr. Mulligan), which
goes into greater depth on some of these same topics, from the perspective of a biochemist.
Courses in Prokaryotic and Eukaryotic Gene Regulation are also taught through Biochemistry.
Advanced Genetics (Bio4241) is typically offered in alternate years. This course considers
classic and current genetics experiments in detail, and cover additional topics (eg,
Immunogenetics, Cancer Genetics, Quantitative Genetics, Developmental Genetics) in greater
detail.
Fundamentals of Genetic Biotechnology (Bio4900) is a hands-on lab course offerred as a
three-week intensive introduction to DNA extraction, PCR, Cloning, DNA sequencing, and
bioinformatic interpretation of DNA sequence data.
New courses in Genomics, Plant Genetics, Developmental Genetics, etc., are being
developed by new faculty, as part of a concentration in Genomics & Cell Biology.
Text material © 2007 by Steven M. Carr
Biology 2250 course schedule: Fall 2008 (v.1.) 04 Sept 2008
#
1.
2.
3.
4.
5.
6.
7.
8.
9.
Lectures: Topic
Sept. 04
Introduction; Structure of DNA
Sept. 09
How Genes Work I: DNA replication & transcription
Sept. 11
The Genetic Code
Sept. 16
How Genes Work II: RNA translation
Sept. 18
How Genes Work III: Protein structure and function
Sept. 23
Molecular basis of heredity: haploid gene expression
Sept. 25
Molecular basis of heredity: diploid gene expression
Sept. 30
Chromosome Genetics I: Cytogenetics
Oct. 09
Midterm I: DNA º RNA º Protein [Lectures 1 - 7]
Oct. 14
Thanksgiving - no lecture
10.
11.
12.
13.
14.
15.
16.
17.
18.
Oct. 02
Oct. 07
Oct. 16
Oct. 21
Oct. 23
Oct. 28
Oct. 30
Nov. 06
Nov. 04
Nov. 11
Chromosome genetics II: Genome organization
Mendelian Genetics I: Dominance, Segregation, & Assortment
Mendelian Genetics II: Extensions to Mendelian analysis
Pedigree Analysis
Chromosome Linkage
Recombination & Mapping
Molecular Basis of Mutation
Midterm II: Mendelian & Chromosomal Genetics [Lectures 8 - 15]
Genetics of Complex Phenotypes
Remembrance Day - no lecture
19.
20.
21.
22.
23.
24.
Nov. 13
Nov. 18
Nov. 20
Nov. 25
Nov. 27
Dec. 02
Genetic Engineering & Biotechnology I
Genetic Engineering & Biotechnology II
Genomics I
Genomics I
Genetics & Genomics Research at Memorial University I
Genetics & Genomics Research at Memorial University II
Dec. ??
Final Exam [Inclusive, emphasis on Lectures 16-24]
Laboratory exercises:
Weeks of:
Sept. 08
Organization
Sept. 15 & 22 Lab 1 - Internet Genetic Resources; Intra - & Inter-specific DNA Variation
Sept. 29
Lab 2 - Protein electrophoresis (Barbarea, Daphnia)
& Oct. 13
Oct. 13-15
Thanksgiving Break
Oct. 20 & 27 Lab 3 - Drosophila mutants; Virtual fly crosses
Nov. 03 & 10 Lab 4 - Virtual fly: dihybrid crosses & linkage
Nov. 17 & 24 Lab 5 - Restriction Endonuclease Mapping of DNA
Grading:
Midterm I
Midterm II
Lab (5 labs x 5% @)
Final Exam
20%
20%
25%
35%
100%
Bio2250 - Assigned Reading
Bio2250 - Assigned Reading
AW Griffiths et al. Modern Genetic Analysis, 2nd ed. [WH Freeman]
Topic
Chapter
Pages
Backfound & Orientation
1
1~20
DNA Structure
2
24~32, 58 ~60
DNA replication
4
93 ~100
Transcription
3
60~66
RNA translation
3
70~73
Protein
3
66~70
Molecular Basis of Heredity
3
74~85
Mitosis & Meiosis [review]
4
100~114
5; 14
117~140; 453~473
Genetic Recombination
6
147~173
Recombinant DNA
8
213~255
Genomics
9
349~373
Mutation
10
313~331
2; 11
32~50; 349~373
Mendelian Genetics
Chromosome Biology
Text material © 2007 by Steven M. Carr
How to use the Bio2250 website
The purpose of this website is to assist you in studying and understanding genetics. It is a
supplement, not a replacement, for coming to the lectures.
From my standpoint as lecturer, the principal advantage of the website is that an outline of the
course, as well as complex illustrations and additional material not in the regular text, are
available before & after as well as during lecture. This allows all of us to focus on concepts
during lecture, knowing that facts & details are available anytime. The website material is updated,
added to, and clarified continuously during the course. The latest version is always available on-line.
For this reason, I do not maintain paper copies in the library.
The website includes my complete lecture notes, along with illustrations and supplemental
material. Many of the illustrations do not come from the assigned text, and are included either
because I think they are clearer than the text figures, or address questions not covered in the text.
Many of the text figures are modified by me.
Key terms are in red
the first time they appear: you should be thoroughly familiar with those terms.
Links are underlined. Figures from the assigned text, Griffiths et al. (2002) "Modern Genetic Analaysis" (2nd ed.) are generally linked
as (MGA2.XX.YY). Other figures and materials, including original art and internet links, are also linked as underlined text. See the
Acknowledgements for sources.
Comments on and suggestions for the website are welcome: please e-mail me at
[email protected]
Different students will use the website in different ways. A key question to ask yourself is whether you take in information better by hearing it
or reading it. Some of your options are:
1. COME TO LECTURE. Print out web material before lecture; annotate these notes during
lecture. This seems to be the preferred method for most students. Bear in mind that I
sometimes make extensive revisions shortly before lecture, that I improvise during lecture, and that I
correct any mistakes I catch after lecture. Don't print things out too far ahead.
[My original intention
was that students would take lecture notes as usual. As an undergrad, I found taking written notes
focussed my attention. However, since the web material has grown from a short topic list to a
complete outline, this no longer seems feasible for most students.
2. COME TO LECTURE. Print out web material before lecture & study them; listen with
focused attention to lectures, without taking complete notes. This might work if you are very
good at absorbing complex material on first hearing. I don't take notes during a seminar,
however I do read the background stuff ahead of time, and I'm not going to be tested on the
material. Don't fool yourself!
3. COME TO LECTURE. Bring up the webpage on a laptop, annotate electronically during the
lectures. I've never done this, so I don't know if it works or not. If you try it this way, let me know
how it works.
4. SKIP LECTURES, cuz it's all there on the web. Review web material for course content just
before exams. NOT RECOMMENDED. The lecture notes are an outline and not a complete course
in themselves.
Biology 2250 - Principles of Genetics - Dr. Carr
Sample exam questions
This sample is intended to show the STYLE of questions that MAY be asked on exams, NOT
the specific questions that will be asked. I specifically reserve the right to modify, add, or
remove questions and format of questions.
For all exams, you will be given a sheet with the Universal Genetic Code: it is therefore not
necessary to memorize the genetic code.
INSTRUCTIONS FOR THE EXAM
Read each question. Think before you write. Answer briefly. I RECOMMEND you do the sections in
order; they are arranged from easy to less easy. Do the ESSAY question last. Answer the EXTRA
CREDIT question only AFTER you finish the rest of the exam.
I. MOLECULAR BIOLOGY:
For each question, indicate in the space provided the LETTER of the ONE response that best
answers the question.
______ In double-stranded DNA, the bonds that hold complementary nucleotides together are
described as
A. ionic B. covalent C. hydrophobic D. hydrogen E. hydrophilic
______ The tertiary structure of protein is determined by
A. beta sheets B. hydrogen bonds C. disulfide bonds D. alpha helices E. subunit
arrangement
______ The active site of an enzyme
is most likely to be composed of amino acids with the following characteristics:
A. Polar & Hydrophobic B. Polar & Hydrophilic C. Non-polar & Hydrophobic D.
Non-polar & Hydrophilic E. Any F. None
II. MOLECULAR GENETICS. Each of the following five statements contains a basic misconception
about molecular genetics. In not more than three grammatically complete sentences, identify
and correct the error. Do NOT exceed the space provided. Do NOT draw diagrams.
Ex.: "Because it has higher energy, gamma radiation is always more mutagenic than alpha
radiation."
Sample answer: "Gamma radiation is more energetic than alpha, but that energy is
dispersed over a much longer path length, so it has a lower linear energy transfer (LET) than
alpha radiation. The higher LET of alpha radiation means that its lower energy will be
dispersed over a very short path. If the energy is released within the nucleus of a single cell
near a chromosome, it will be highly mutagenic."
Ex.: "DNA probes identify particular base substitutions responsible for genetic diseases."
Ex.: "His- mutants are caused by a defect of histidine that block its synthesis."
Ex.: "Most people do not have PKU because they do not have the gene for PKU"
III. GENETIC STRUCTURES: The numbers in the left-hand column refer to structures on the right.
For each numbered structure, write the letter of the correct term [see example]. Several letters are
not used; none is used more than once.
IV. DNA TRANSCRIPTION & TRANSLATION [Use the Genetic Code table attached to the test]
The following DNA is part of a gene that codes for a polypeptide of at least seven amino acids:
3' c a a t t g a t t a g t c a g t
c a a t t g a t 5'
5' g t t a a c t a a t c a g t c a
g t t a a c t a 3'
i. Which strand includes an Open Reading Frame? Top (T) or Bottom (B) [2 pts]
ii. Which DNA strand is the sense strand? Top (T) or Bottom (B) [2 pts]
iii. Give the mRNA sequence that would be transcribed from this gene; label the 5' & 3' ends. [3
pts]
iv. Give the seven amino acids that would be translated from the correct message; label the C & N
ends. [5 pts]
V. TRIHYBRID THREE-POINT TEST CROSS: From the following recombination data, draw a
physical map to indicate the correct order and distances between loci.
VI. RESTRICTION MAPPING:
From the following fragment size data, draw a restriction map. Label sites, provide a scale. [10 pts
possible]
Types of questions I would not ask.
Watson & Crick received the Nobel Prize in what year? (a) 1953 (b) 1962 (c) 1973 (d) 1984
Transcription proceeds in which of the following directions (a) 3'
6' (e) 5' 3'
1' (b) 2'
4' (c) 3'
5'
(d) 4'
Which of the following amino acids is/are non-polar? (a) Ala (b) Gly (c) Arg (d) Asp (e) None
(f) All
Draw the structure of Deoxyadenosine monophosphate.
How many introns are present in the chicken ovalbumin gene? (a) 4 (b) 5 (c) 6 (d) 7 (e) more
than 7
Text material © 2007 by Steven M. Carr
History of the hereditary molecule (to 1953)
In principle:
"Genetics" was taught for 50 years
without knowledge of the hereditary substance or its structure
(see Orientation to Bio2250)
The story of the search for the hereditary substance includes
superb examples of the experimental method in biology.
Two candidates: protein versus nucleic acid
Cells contain H20, lipids, carbohydrates, and ...
Mulder (1838) - Discovery of protein
Abundant, water-soluble, nitrogenous
"… complex... regulates cell metabolism...
most important component of living matter...
without it, life would not be possible"
Hydrolysis of protein => amino acids (~20 kinds)
Miescher (1868) - Discovery of nuclein
Found in cell nucleus, acidic, rich in PO4,
Lacks S (characteristic of protein)
Now know this as nucleic acid
Levene (1910) - Tetranucleotide hypothesis
nucleic acid is a repetitive polymer of four bases
A:C:G:T in the approximate ratio 1:1:1:1
=> Structure seems too simple to carry information
Griffith (1928) - transforming principle
Killed virulent viruses 'transform' live avirulent viruses:
avirulent viruses become virulent, and
Transformation is inherited
=> Hereditary makeup of organisms can be altered
Avery, MacLeod, & McCarty (1944) Chemical isolation of 'transforming principle' from cells
Transformation survives protease treatment,
destroyed by nuclease treatment (Homework):
=> It's chemically pure deoxyribonucleic acid (DNA) ?!?!
Hershey & Chase (1952) - 'blender experiment'
Bacteriophages are grown in radioactive medium
Proteins labeled with 35S
DNA labeled with 32P
During infection of E. coli by bacteriophages,
32
P goes in, 35S stays out
=> DNA is the transforming principle
Watson & Crick (1953) "The Double Helix"
Schrodinger (1945) "What is Life?":
Are there "other laws of physics?"
Franklin & Wilkins' X-ray crystallography
DNA is a helix: two or three strands?
Chargaff's Rules : Bases are not equimolar, but
[A]=[T] & [C]=[G] (Table)
Model building:
Two or three strands, bases inside or outside
Key discovery: A+T pair looks like C+G pair
The Watson-Crick structure for DNA
double-stranded helix (3-D image)
Phosphate backbone outside
Nitrogenous bases inside
H-bonds hold strands held together
For further reading:
J. Cairns, G. Stent, & J. Watson (1966). Phage and the Origins of Molecular Biology. Freeman.
[Biographical essays on the early days by the founders of molecular genetics.]
F. H. C. Crick (1988). What Mad Pursuit? Basic Books.
[Crick's version of the 'double helix' history, and lots more.]
L. Gonick & M. Wheelis (1991). The Cartoon Guide to Genetics, 2nd ed. Harper Collins.
[Great illustrations: a good primer of basic Mendelian and molecular genetics.]
H. F. Judson (1979). The Eighth Day of Creation. Simon & Schuster.
[A general history of molecular biology.]
A. Sayre (1975). Rosalind Franklin and DNA. Norton.
[A re-appraisal of the role of Franklin, with commentary on the role of women in science.]
G. Stent (1971). Molecular Genetics: an introductory narrative. Freeman.
[A classic, now factually dated textbook; still highly readable.]
J. D. Watson (1968). The Double Helix. Athenaeum.
[An entertaining, irreverent, sexist, account of the discovery of the structure of DNA.
See the accounts of Crick and Sayre for another view]
J. D. Watson (2003). DNA: The Secret of Life. Knopf
[A narrative history of genetics and molecular biology in the 20th century,
written for the 50th anniversary of the discovery of the DNA structure.]
All text material © 2008 by Steven M. Carr
Biochemistry of heredity:
the structure of Deoxyribonucleic Acid (DNA)
In principle: Genes are made of nucleic acids
The identity of the hereditary substance was unknown until 1940;
its structure was unknown until 1953
"Genetics" was taught for 50 years without this information (see Orientation)
The history of the discovery of DNA is a fascinating detective story
The Watson-Crick structure for Deoxyribonucleic acid (DNA) (1953) (MGA2 Box 2-2, p.31)
a double-stranded helix
sugar-phosphate backbone outside
nitrogenous bases (A,C,G, T) inside
bases held together by hydrogen bonds (AKA H- or hydrostatic bonds)
Fundamental insight:
bases on alternative strands pair according to specific rules:
A with T G with C
each pair has similar structure
A second form of nucleic acid is ribonucleic acid (RNA)
Homework Assignment #1
Building blocks of nucleic acids (DNA & RNA)
bases
pyrimidines (single ring)
cytosine(C)
thymine (T)
[ uracil in RNA (U) ]
"PYRamids were CUT from stone"
purines (double ring)
adenine (A) guanine (G)
"AGs are PURe"
nucleoside = base + sugar
deoxyribose sugar in DNA (- H on 2' C)
ribose sugar in RNA (- OH on 2' C)
deoxyadenosine (dA)
deoxycytosine (dC)
deoxyguanosine (dG)
deoxythymidine (dT)
nucleotide = nucleoside + phosphate(s) [PO4] [MGA2_02-06]
in DNA,
one phosphate => deoxyucleoside monophosphate (dNMP)
three phosphates => deoxynucleoside triphosphate (dNTP)
deoxyadenosine-5'-phosphate or deoxyadenylic acid
deoxyadenylic acid (dAMP) / deoxyguanylic acid
(dGMP)
deoxycytidylic acid (dCMP) / deoxythymidylic acid (dTMP)
polynucleotide = nucleotide + nucleotide + nucleotide + etc
nucleotides are linked by 3' 5' phosphodiester bonds
***polynucleotides have directionality***
hydroxyl (3') & phosphoryl (5') ends
Structure of B-DNA (3-D model: requires chime) [MGA2_02-04]
1) Two plectonemic (twisted) right-handed polynucleotide helices (demo)
2) Helices antiparallel strands wrt 5' 3' orientation [MGA2-02-05]
3) Strands held together by hydrogen bonds between bases
4) H-bonds form according to specific base-pairing rules
A pairs with T: two H-bonds
G pairs with C: three H-bonds
A+T & G+C pairs have very similar shapes & sizes
5) Base pairs co-planar: interval = 0.34 nM [= 3.4 Ǻngstroms]
6) Period of helix is 10 bp (base pairs) = 3.4 nM
7) 3-D structure has major & minor grooves [MGA2_02-07]
8) Order of bases in each strand aperiodic
Homework Assignment #2
Other structures for nucleic acids
A-DNA : not groovy, base pairs not co-planar
Z-DNA: left-handed helix (demo)
Ribonucleic Acid (RNA):
substitute uracil for thymine [ thymine = 5-methyl-uracil ]
ribose sugar for deoxy-ribose
typically single-stranded, or with complex double-stranded folding:
mRNA (messenger RNA): long, single-stranded
rRNA (ribosomal RNA): medium-sized, complex 'stem & loop' folding
tRNA (transfer RNA): small, 'cloverleaf' structure
[more on RNA structures later]
Implications of DNA structure for its genetic function
"The sequence of bases on a single chain does not appear to be restricted in any way.... It
has not escaped our notice that the specific pairing we have postulated immediately
suggests a possible copying mechanism for the genetic material." (Watson & Crick 1953.
Nature 112:753)
DNA is an aperiodic crystal:
order of bases may convey information
Antiparallel strands are self-complementary:
molecule is potentially autocatalytic
All text material © 2008 by Steven M. Carr
DNA Replication & Transcription
In principle: DNA replication is semi-conservative
H - bonds 'unzip', strands unwind,
complementary nucleotides added to existing strands (MGA2 04-04)
After replication, each double-helix has one "old" & one "new" strand
[note alternative conservative & dispersive models: Homework #3 ]
DNA is not the "Genetic Code" for proteins
information in DNA must first be transcribed into RNA
messenger RNA transcript is base-complementary to template strand of DNA
& therefore co-linear with sense strand of DNA
DNA synthesis in prokaryotes:
Nucleotides are added simultaneously to both strands, but
DNA grows in the 5'
3' direction ONLY
Distinguish:
Replication: duplication of a double-stranded DNA (dsDNA) molecule
an exact copy of the existing molecule (cf. xerox copy)
Synthesis: biochemical creation of a new single-stranded DNA (ssdNA) molecule
a base-complementary 'copy' of an existing strand (cf. silly putty copy)
Homework #4
DNA Synthesis in prokaryotes (Review) (MGA2 04-5,6,7)
(1) Formation of replication fork
provides two single-stranded DNA template (ssDNA)
(2) Synthesis of RNA primer
(3) Addition of dNTPs by DNAPol III at 3' end only
continuous synthesis on leading strand
(4) discontinuous synthesis on lagging strand
Okazaki fragments
proof-reading by 3' 5' exonuclease activity
(5) Excision of RNA primer by DNAPol I
ligation (connection) of fragment ends at gaps by DNA ligase
A talkie animation of DNA synthesis `[onlineMGA2 animation]
DNA synthesis occurs at multiple replications forks (replicons)
DNA synthesis occurs on leading & lagging strands simultaneously
A single, dimeric DNAPol III replicates both strands
DNA synthesis in eukaryotes
Eukaryotic genomes are much larger [MGA2_02-10]
=> eukaryotic DNA synthesis is more "efficient":
More DNAPol molecules, slower rate of synthesis, more replicons,
E. coli: 15 DNAPol molecules add 100,000 b/min over 3,500 replicons
=> 4.2 x 106 bp genome replicated in 20 ~ 40 min
Drosophila: 50,000 DNAPol molecules add 500 ~ 5,000 b/min over 25,000 replicons
=> 330 x 106 bp diploid genome replicated in < 3 min : net 600x faster
Transcription: synthesis of messenger RNA (mRNA) (online MGA2 animation)
RNA transcribed from DNA by RNA Polymerase (RNAPol I)
(1) Recognition
Promoters - short DNA sequences that regulate transcription [MGA2_03-09]
typically 'upstream' = ' leftward' from 5' end of sense strand
(2) Initiation & Elongation
mRNA synthesized 5' 3' from DNA template strand
mRNA sequence therefore homologous to DNA sense strand
Colinear: mRNA and DNA sense strand "line up" [MGA2_03-07]
(in prokaryotes, but not eukaryotes: see below)
Process similar to DNA replication, except
No primer is required
Transcription in opposite orientation on both strands [MGA2_03-05]
Not all DNA is transcribed [MGA2_03-04]
(3) Termination
Regulation of transcription
In prokaryotes, transcription & translation may occur simultaneously
In eukaryotes, transcription occurs in nucleus [MGA2_03-06]
translation occurs in cytoplasm (see next section):
=> RNA must cross nuclear membrane
transcription & translation are physically separate
primary RNA transcript is extensively processed
heterogeneous nuclear RNA (hnRNA)
mRNA
Post-transcriptional processing of eukaryotic RNA is complex [MGA2_03-11]
promoters & enhancers determine initiation & control rate
'cap' (7-methyl guanosine, 7mG) added to 5' end
'tail' of poly-A (5'-AAAAAAAAAA~~~-3') added to 3' end
'splicing' of hnRNA : eukaryotic genes are "split" (MGA2 03-12,14,15,16)
intron DNA sequences removed from hnRNA : "intervening"
exon DNA sequences represented in mRNA: "expressed" in protein
1 ~ 12's of exons / 'gene'
>90% of transcript may be removed [MGA2_02-28]
[An important note on terminology]
visualized as heteroduplexes
DNA introns 'loop out'
DNA exons pair with mRNA
Eukaryotic genes & mRNA are not colinear!
Eukaryotic exons may be widely separated [MGA2_02-18]
Summaries of transcription [& translation] in prokaryotes & eukaryotes
Homework #5: Suggested problems from
MGA2 (2002), Chapter 2, pp. 53-54
Solved problems 1 & 2
problem ## 7, 8, 9, 11, 14, 15 , 18, 19, 21, 26, 27
for extra fun: ## 29 & 34
All text material © 2008 by Steven M. Carr
The Genetic Code
The Central Dogma: DNA makes RNA makes protein
In principle: The DNA genotype does not produce the phenotype directly
A DNA gene contains the information necessary for the production of proteins,
which is expressed biochemically through an intermediate molecule, RNA,
which functions as a Genetic Code
The Genetic Code ...
specifies amino acids that make up proteins
Protein expression leads directly (or indirectly) to the phenotype
was "cracked" before the details of translation were understood:
=> we can describe the Code before describing RNA translation
can be used to infer the protein product of a gene directly from DNA:
see next section, and lab exercise
Alternative alleles arise from mutations in the Genetic Code
which alter the DNA sequence of genes
which may cause amino acid substitutions in proteins
which may affect the function of those proteins
The Genetic Code is ...
a messenger RNA (mRNA) code
i.e.., the code is written in RNA
DNA is a coding molecule,
but not 'the genetic code' in the biochemical sense
in 64 triplets (codons) : 61 for amino acids + 3 'stops' [MGA2_03-20]
mRNA codons are read 5' 3'
20 amino acids: note 1- & 3-letter abbreviations
[more on amino acids & proteins in next section]
For example,
5' - A U G U U C C C C A A G G G U U G A - 3'
met
phe
pro
lys
gly
*
M
F
P
K
G
*
Degenerate: most amino acids are encoded by more than one codon
first two positions are critical: third position can "wobble" (MGA2_03-22)
if third can be either puRine (R), or either pYrimidine (Y) =>
two-fold degeneracy
if thirds can be any base =>
four-fold degeneracy
Leucine (leu) has six codons in an unusual arrangement
# codons / amino acid
trp, met
1@
ser, arg, leu
6@
ile
3@
14 others
2 or 4 @
Unambiguous: any one triplet codes for only one amino acid
but not vice versa, because of wobble
'Always' begins with an 'start' or 'initiator' codon: AUG
N -formyl-methionine (fmet) in prokaryotes
'Always' ends with a 'stop' or 'terminator' codon: UAG, UAA, or UGA
Universal (with some important exceptions)
Five Kingdoms (animals, plants, algae, fungi, & monera)
use the same codes for nuclear DNA (nucDNA)
Organelles (chloroplasts & mitochondria) have separate genomes:
cpDNA & mtDNA codes are evolutionarily modified
e.g., UGA codes for trp in vertebrate mtDNA code
termination codons may be formed by addition of "A"s to transcript
Lab exercises use mtDNA, so this code is important
Mutations in DNA: alterations of the Genetic Code
Single-base mutations - interchanges of one base type for another [MGA2 Table 10-2]
Recognized as SNPs (single nucleotide polymorphisms)
Alternative nucleotide sequences of a gene yield alternative alleles
or: a single gene occurs in variant forms (alleles)
Consequences of exon mutations depend on position in triplet (MGA2 10-4)
3rd position
typically a silent mutation - if position "wobbles", no change to amino acid
sometimes a missense mutation - results in different amino acid
2nd position - always a missense mutation
1st position - almost always a missense replacement
[Leu codons are major exception]
stop codon mutations may occur at any position: coding
non-coding triplet
nonsense (termination) mutation terminates polypeptide prematurely
HOMEWORK: Identify all codons one step away from a termination codon
[HINT: There a re 18]
Mutations in non-coding DNA have variable effects (MGA2_3-27, modified)
Ex.: mutations in promoter regions
mutations at intron / exon splice junctions [MGA2 3-15]
Missense mutations in DNA cause substitutions in protein
Proteins do not mutate! Watch your language!
Consequences depend on position of substitution in polypeptide
none: substitution not in active site or binding site
minor: substitution of same type (synonymous substitution)
Allozymes are minor variants
major: substitution affects 2o, 3o, or 4o structure (nonsynonymous substitution)
Ex.: Sickle-cell hemoglobin (HbS) is a variant blood protein
Insertion / Deletion (indel) mutations
gain or loss of one or more nucleotides
frameshift mutations - triplet reading frame offset
single & double nucleotide indel => downstream amino acids change (examples)
nonsense mutation eventually (quickly) produced
triplet indel - insertion / deletion of single amino acid
milder consequences
multiple triplets may produce major effects (see below)
length mutations - larger indels (102~6 bps)
may affect cytology of chromosomes (see "Fragile X", below)
Genes are highly polymorphic (w/ multiple alleles) wrt their mutational basis
The PAH locus occurs on the short arm of Chromosome 12 (12p13.33)
14 exons produce 2.4kb mRNA that produces 452 amino acid protein
(Phenylalanine Hydroxylase or PAH)
Of 66 alleles known to affect gene expression of PAH (MGA2_03-29)
29% produce non-PKU hyperphenylalanemia (excess [Phe] in the blood)
71% produce Phenylketonuria (PKU): of these,
55% are miss-sense mutations (39% of total)
18% are non-sense mutations
whole exon)
22% are deletion mutations (single base
14% are splice-site mutations
Most alleleic variants of the PAH locus probably have no affect on expression
& are therefore undetected
Homework:
(1) "What is a Gene?" Write an essay that that distinguishes Gene, Allele, and Locus
(2) Critique the following statements:
"PAH is the gene for Phenylketonuria (PKU)."
"PKU is a genetic disease caused by absence of the PAH gene."
Text material © 2008 by Steven M. Carr
RNA Translation: RNA makes Protein
In principle:
Translation of messenger RNA (mRNA) takes place on ribosomes,
which include ribosomal RNA (rRNA),
with the help of transfer RNA (tRNA)
Structure of rRNA & tRNA
ribosomal RNA (rRNA)
rRNA + ribosomal protein ribosomes
Structure of rRNA: stems & loops
stems: double-stranded (dsRNA)
loops: single-stranded (ssRNA)
complex 2o folding
Structure of eukaryotic ribosomes (MGA_03-23)
S = Svedberg = sedimentation coefficient
Large Subunit (LSU) = 60S = 28S rRNA + 5S rRNA + 50 proteins
Small Subunit (SSU) = 40S = 18S rRNA + 33 proteins
= 80 S monosome
P site (Peptidyl), A site (Aminoacyl), & E site (Exit) (diagram)
transfer RNA (tRNA)
the adaptor molecule: ~30 tRNA types
2-dimensional 'cloverleaf' model (MGA2_03-29)
small: 75 ~ 90 nucs
stems & loops
D-loop & T C-loop ( = pseudo-uridylic acid)
tRNA characterized by 2o modified bases
amino-acceptor stem
3' end is CACCA - 3'
5' end is G
- 5'
anticodon loop
specificity of tRNA determined by 3-ribonucleotide sequence
3-dimensional structure is an "L":
D- & T C-loops fold back on each other
Charged tRNA: aminoacyl synthetase(x) forms ester linkage between
3'-A of amino-acceptor stem of tRNA(x) & COOH of amino acid(x)
~20 synthetase types 'recognize' correct anticodon loop
isoacceptance:
one-to-one correspondence between synthetase & amino acid
A "second genetic code"?
RNA Translation: Protein Synthesis
A three-step process: Initiation, Elongation, & Termination (Review)
Ribosomes "read" mRNA &
assemble amino acids according to Genetic Code
(1) Initiation at start codon (AUG):
Ribosomal SSU binds at Shine-Delgarno sequence (-6 nucs)
Initiation complex consists of mRNA, ribosome, & tRNA
Multiple complexes form on a single mRNA: polysome (polyribosome)
tRNAfmet always added first
In simplified form,
5'-AUG-3'
|||
3'-UAC-5'
codon in mRNA
5'-CAU-3'
if anticodon is written 5'
anticodon in tRNA
3'
(2) Elongation: addition of amino acids according to Genetic Code
Amino acids are joined via peptide bonds (see next section)
Think of mRNA as fixed, with ribosome moving from "left to right"
peptidyl (P) site on "left" (5') end,
aminoacyl (A) site on "right" (3') end
first AUG codon [for fmet] is in P site
second UUC codon enters A site
corresponding tRNAphe enters A site
peptidyl transferase forms peptide bond between fmet & phe
tRNAfmet (ester) bond broken, carboxyl terminus transferred to amino terminus
of tRNAphe in A site
uncharged tRNAfmet released from P site (passes to E site)
and so on ... (see also MGA2 03-24) [online MGA animation]
wobble: pairing of codon / anticodon goes 5' 3' on codon
last position can miss-pair (MGA2_03-22)
Fewer tRNA species needed:
Ex.: three tRNAser species for six codons (MGA2_Table 3-2)
tRNA
Anticodon
Alternative mRNA
codons
3'- AG G -5'
5'- UC C / U -3'
3'- AG U -5'
5'- UC A / G -3'
3'- UC G -5'
5'- AG C / U -3'
(3) Termination: release of polypeptide
mRNA + tRNA(aa -...-aa -aa -aa )
n
3
2
1
here: mRNA + tRNA(lys-pro-gly-phe-fmet)
stop codon (UAG, UAA, or UGA) enters A site
no corresponding tRNA:
release factor cleaves polypeptide from terminal tRNAn
polypeptide product is: lys - pro - gly - phe - fmet
interactive translation animation [Genetic Science Learning Center, Univ Utah]
A talkie animation of transcription & protein synthesis
Griffiths et al. (1996) Fig. 13-7 is a nice summary (HOMEWORK)
"Translating" DNA directly to Protein
5'- G T A
A T C
C T C - 3'
DNA sense strand
5'- G U A
A U C
C U C - 3'
mRNA
N -
val
-
ile
-
leu
- C
protein
This is a logical, not a biochemical, relationship:
Because mRNA is transcribed from the template strand,
it "looks like" the sense strand (except for 'U').
The information content of the DNA sense strand and mRNA are identical
Protein sequences can be read ("translated") directly from DNA:
Read the sense strand in the 5' 3' direction,
Substitute 'T' for 'U' in the code table.
Computer programs such as ESEE, Chromas, and Sequencher do this automatically
There are six possible ways of reading a piece of dsDNA
two 5' 3' strands & three reading frames in each strand
Open Reading Frames suggest protein sequences
Deducing protein sequences from unknown bits of DNA is a major research activity
The following clues are useful:
Remember that all prokaryotic coding sequences:
are read only in the 5' 3' direction
begin with a "start" (AUG) codon [but not all AUG codons are 'start' codons]
end with a "stop" (UAG, UAA, or UGA) codon.
But: in real life, your cloned (eukaryotic) DNA fragment
may not have start or the stop codon for a complete protein,
and may include part of an intron with one or more stops.
Do not assume that a dsDNA molecule is read from left to right, on the top strand
Suggested problems for review
MGA2, pp. 86-87
Solved problem 1
Problem ## 1, 2, 3
Practice DNA "Translation" problems [PDF download version]
All text material © 2007 by Steven M. Carr
"One Gene, One Enzyme"
In Principle: Proteins are the products of genes.
Proteins catalyze biochemical reactions.
Such reactions produce phenotypes, either directly or indirectly.
Different alleles produce different phenotypes
Interaction between gene alleles in diploid organisms is the classic subject of Genetics
How do genotypes produce phenotypes?
Beadle & Tatum experiments (1940s) on haploid Neurospora bread mold (MGA2 Box 3-1)
haploid organisms have one set of alleles
prototroph ("self feeding") wild-type grows on simple medium
auxotroph ("other feeding") mutants cannot grow on simple medium,
require supplementation with specific amino acids
[These are also known as autotrophs and heterotrophs, respectively]
Hypothesis: "No-growth" phenotype results from a change in the genotype:
inability to synthesize amino acid is the result of loss of enzyme activity
each mutant corresponds to a defect in a particular enzyme:
"One gene, one enzyme" (Homework)
[ Remember: Beadle & Tatum did not know about DNA in 1940]
Ex.: arg- mutants cannot grow without arginine, always grow with added arginine;
particular mutant classes sometimes grow with other amino acids
(such as citrulline and/or ornithine)
Growth response to added amino acids
mutants
none
ornithine
citrulline
arginine
arg 4
-
+
+
+
arg 2
-
-
+
+
arg 1
-
-
-
+
=> These other amino acids are involved in the arginine pathway:
Enzyme defect blocks interconversion of precursors in the pathway.
Inference of a haploid biosynthetic pathway
Each mutant class (arg4, arg2, & arg1) affects a different enzyme in arginine biosynthesis
Homework:
1.
In the above discussion, I have used "mutant" but carefully avoided "mutation": WHY?
2.
Critique the following statements:
"arg- mutants result in defective arginine."
"arg- mutants are defects of arginine."
"arg- mutants are due to absence of the gene for arginine."
"arg- mutants are enzymes that block synthesis of arginine."
3. Would you expect to observe an arg mutant class with the following phenotype? Explain.
Growth response to added amino acids
mutant
none
ornithine
citrulline
arginine
arg X
-
+
-
+
Biochemical Basis of Human Genetic Disease:
What about diploid organisms?
Diploid organisms have two sets of alleles: one from each parent
Interactions between different alleles are the subject matter of genetics
"Online Mendelian Inheritance in Man" (OMIM) database
examples from human biochemical genetics: "Inborn errors of metabolism"
First three involve disruptions of phenylalanine metabolism (MGA2_3-28)
Phenylketonuria (PKU) (Folling 1934) (OMIM citation 261600)
phenylalanine accumulates in Central Nervous System => mental retardation
A defect of phenylalanine hydroxylase
phenyalanine metabolized to phenylpyruvic acid in alternative pathway
Detection & treatment
biochemical testing of new-borns: Guthrie Test (MGA2_3-1)
phenylalanine-restricted diet corrects inborn condition (Euphenics)
Maternal PKU results from high fetal [phe] in treated, asymptomatic mothers
[ Further information on PKU & related Inborn Errors of Metabolism ]
PKU arises from defects at the Phenylalanine Hydroxylase (PAH) gene locus
Important: This gene is not a gene "for" PKU: it is a gene "for" PAH
Diploid humans each have two alleles at this locus
Allelic variants affect levels of PAH activity
Consider three of these: A, B, & C :
Phenotypic consequences of interactions between alleles at the PAH
locus
Genotype
PAH Activity
[phe]
uM
PKU Phenotype
AA
100%
60
Standard
AB
30%
CC
5%
120
200 ~
300
Standard
Hyperphenylalanemia:
no special diet required
BB
0.3%
600 ~
2400
Classic PKU:
special diet required
[Alleles B & C) arise from DNA mutations in the PAH gene (MGA2_3-29)]
>> PKU is a classic example of a "recessive" genetic disease <<
What does this mean?
PKU is typically described as a "recessive" genetic condition:
Presence of A allele "masks" B allele in diploid genotypes
AB shows same PKU phenotype as AA
A is therefore "dominant" to B in influencing phenotype
B is therefore "recessive" to A in influencing phenotype
but PAH activity phenotype of AB is intermediate between AA & BB
AB phenotype is closer to BB than AA (0% < 30% << 100%)
B is an "incomplete dominant" to A
and B produces a higher [phe] phenotype than A:
so why isn't B therefore called "dominant" to A ?
Be careful to distinguish molecular & phenotypic expression (Homework)
Homework
Predict the PAH activity, [phe], & PKU phenotypes of the AC and CB genotypes.
Explain your reasoning.
Would you expect to find a dominant mutation in this pathway?
Why or why not? What might be the nature of such a mutation?
Other forms of molecular expression in Genetic Diseases
"Recessive" diseases:
"Co-dominant" diseases:
"Dominant" diseases:
Alkaptonuria & Albinism
Sickle-Cell Anemia
Huntington's Disease
Homework:
MGA2, pp. 86-87
Problems ## 8, 9, 10, 12, 18, 21, 23, 24
All text material © 2007 by Steven M. Carr
Biology 2250
Principles of Genetics
Revised
2008
revised September
September 2005
© 2005 by Steven M. Carr & David J. Innes
Department of Biology
Memorial University of Newfoundland
St. John’s NL A1B3X9 Canada
[email protected]
Not to be reprinted without permission;
the intellectual property rights of the authors are hereby asserted
3
INTRODUCTION TO BIOLOGY 2250
PRINCIPLES OF GENETICS
Biology 2250 (Principles of Genetics) is an introductory course in genetics and
molecular biology that deals with the the molecular basis of heredity, the laws of Mendelian
inheritance, and how these combine to produced organismal phenotypes.
The major laboratory investigations emphasize hands-on experience with basic genetic
techniques, including computer analysis of patterns of DNA sequence variation, construction and
analysis of genetic crosses with several organisms, preparation and analysis of chromosome
material, examination of protein structure by electrophoresis, and construction of physical gene
and DNA maps. Many of the exercises involve Drosophila melanogaster, the classic organism of
experimental genetics.
The exercises have been selected to give you additional experience with the means by
which the principles of genetics have been established, to reinforce the discussion of these topics
in lecture, to acquaint you with modern biotechnology, and to help you appreciate the origins of
the diversity of life on the planet.
The current edition represents is the latest in a series of revisions of a volume that has
been developed, modified, and (hopefully) improved by several generations of lecturers, lab
instructors, and, not least, students. We are in particular grateful to Ms. Sylvia Kao, who
developed an earlier version of the manual, and to Dr. Brian Stavely and Ms. Anika Heywood
for improvements to the Drosophila exercise and investigations.
This 2005 edition has been prepared in PDF format, to reduce costs and to make it more
directly accessible to students. This also allows us to make changes during the course of the
semester. Blank pages have been retained, where answers are to written on one side only, or
drawings submitted, or in order to start each new exercise or investigation on a facing page. The
entire manual may thus be printed in two-sided format, to save paper.
We welcome your comments and suggestions on the format, presentation of ideas, and on
the exercises and experiments themselves.
Ms. Valerie Power
Dr. David Innes
Dr. Steve Carr
St. John’s, Newfoundland
September 2005
This 2008 revision has been reduced to core material on Drosophila
genetics and laboratory safety. Exercises and Investigations will
be provided individually.
4
LAB SAFETY
1.
Some of the experiments in laboratory present potential hazards with chemicals,
electricity, and/or short-wave radiation. These are described in the lab manual and you
will be alerted to the possibilities at the start of the relevant laboratory. Read the
laboratory instructions before coming to lab to anticipate these.
2.
WHMIS sheets on major chemicals used in lab are available (see Appendix C).
3.
We recommend that you wear eyeglasses or safety goggles on lab days. Eyeglasses
provide some degree of eye protection. Some people with contact lenses experience eye
irritation from the chemicals used in the lab. Daily-wear soft contacts may absorb fumes
permanently. Lab coats may also be advisable when stains/chemicals are being used.
4.
Broken glass, used razor blades, and other sharp objects should be disposed of in the
containers provided, NOT in the regular garbage. Immediately report any cuts, accidents,
spills, etc. to the lab instructor.
5.
If the fire alarm sounds: turn off all electrical equipment and proceed to the nearest fire
exit. Do not return to the building until instructed to do so by the Fire Wardens.
5
INTRODUCTION TO DROSOPHILA GENETICS
DROSOPHILA CULTURE
We will study basic principles of Mendelian inheritance with the use of the fruit fly,
Drosophila melanogaster [the name means “black-bodied fruit-lover”]. Drosophila was one of
the first organisms to be studied genetically: its small size, short life cycle (10 ~14 days at 25oC),
high reproductive rate (an adult female can lay 400-500 eggs in 10 days), and ease of culture and
genetic manipulation have made it perhaps the best understood animal genetic system. Many
different species, and a large number and wide variety of naturally-occurring and artificiallyinduced genetic variants are available. The partial genetic map in Appendix B describes the
location of all the mutations used in crosses and lab questions.
VIRGIN FEMALES
All female flies used in controlled genetic crosses must be “virgins”. Female flies are
capable of mating as early as 8 hours after emerging from the pupae stage and are polyandrous,
that is, capable of mating with several males. Once mated, females can retain viable sperm for
several days and this will confuse the results of a subsequent controlled mating. To prevent this,
all adult flies are removed from the culture bottle about 7 hours prior to lab time, so that all
newly hatched flies will remain virgin.
BASIC GENETICS
The karyotype of Drosophila comprises four pairs of chromosomes, of which three pairs
are autosomes and one pair are sex chromosomes. Female Drosophila are XX, and males XY.
A gene is a heritable factor that controls the expression of some trait, which may be
morphological, behavioural, molecular, etc. Each such gene occupies a specific physical locus
(pl. loci) on a particular chromosome. Variant forms of these loci are termed alleles. Gene,
locus, and allele are often used more or less interchangeably, and this can lead to confusion.
Gene is the popular and most general term, and is most appropriate when the inherited basis of a
trait is emphasized, e.g., a “gene” for eye colour. Locus is most appropriate when the physical
nature or position of a gene, especially with respect to other genes, is emphasized, as for
example in gene mapping and linkage studies. Allele is most appropriate when the particular
form(s) of a gene found in any particular individual or chromosome is(are) emphasized: e.g.,
there are “brown” and “blue” alleles of the eye colour gene It is therefore inaccurate to say, for
example, “He has the gene for sickle-cell anemia,” and more accurate to say “He has two HbS
alleles at the beta-globin locus on Chromosome 6.” We all have the “gene” for every genetic
condition, some of us have the particular allele(s) that result in the condition being expressed. In
the technical literature, “locus” and “allele” are probably more common than “gene”.
Drosophila, like most species we will deal with in this course, are diploid, with two sets
of chromosomes and therefore two alleles at each autosomal locus. If both alleles are identical,
the individual is a homozygote and is described as homozygous. If the alleles differ from each
other, the individual is a heterozygote and is described as heterozygous. If the gene occurs on a
sex chromosome, females may be either homozygous or heterozygous, but a male fly with only
one allele at a locus will be a hemizygote and would be described as hemizygous.
6
Drosophila of typical appearance are said to show the “wild-type” forms (phenotypes)
of genetically-controlled traits for body colour, eye colour, wing shape, etc. Naturally-occurring
or artificially-induced genetic variants (mutations) of the alleles that control these traits produce
flies with different morphologies, according to the dominant or recessive nature of the alleles
involved in the genotype. Such mutant alleles are designated by symbols that are typically
abbreviations of the mutant name. For example, the typical body colour phenotype is grey. One
mutant produces an ebony (shiny black) body colour. Because this allele is recessive, it is
symbolized by a lower-case letter, e. The wild-type allele is symbolized by a “+” sign, used
either alone (if there is no ambiguity) or in combination with the mutant allele symbol, in this
case e+. Thus, the genotype of a wild-type homozygote would be designated e+e+ (or ++), a
mutant homozygote ee, and a heterozygote e+e or e+ [Use of the term “wild-type” derives from
an early assumption that most flies are homozygous for a ‘standard’, usually dominant, allele. As
we will see, this is not the case, but the terminology is still used].
It is important to remember that not all mutants are recessive. A mutation that is
dominant to the wild-type is symbolized by a capital letter. For example, the typical eye shape
is round. One mutant produces a narrow “bar eye”: the allele is dominant, symbolized by a
capital letter B, and the wild-type (round) eye is B+.
7
GENETIC CROSSES
An “X” is used to indicate that two individuals have been mated together. The parents are
designated as P (for parental) and the offspring as F (for filial). When several generations are
involved, subscripts are added to designate the generations.
P1 give rise to F1 (first filial) progeny.
If the F1 are crossed together they become P2 and their progeny F2.
A cross between members of the F1 and members of the P1 is a backcross.
A cross between members of the F1 and the true breeding recessive P1 is a test cross.
MONOHYBRID CROSS
The simplest form of a cross is a monohybrid cross, which analyses a single trait and its
associated variations. The diagram below shows the progression of a pair of alternative alleles
for a single gene through two generations.
CROSS DIAGRAM
P1
gametes
ee (ebony body) x
e
F1
e+e
gametes
(wild-type body)
[homozygous parents]
[heterozygous offspring]
e+ & e
&
F2
e+e+ (wild-type body)
e+
%
e+
e
e+
e+e+
e+e
e
e+e
ee
Phenotype ratio 3 wild-type: 1 ebony body
Genotype ratio 1 e+e+ : 2 e+e : 1 ee
During gamete formation, the members of a pair of alleles are duplicated and then
segregated from one cell into four separate gametes so that each contains only one member of
the pair (Law of Segregation).
A Punnet Square diagram can be used to calculate the various combinations. The
gametes from one parent are written across the top, and those from the other go down the side.
Each one of these gametes has an equal chance of combining with either of the gametes from the
other parent (Random Union of Gametes).
8
In cases such as the above example, the F2 phenotype ratio of 3:1 indicates a case of
complete dominance. That is, one allele completely masks the expression of the other
(recessive) allele.
In cases of incomplete dominance , on the other hand, neither allele masks the other,
and heterozygous individuals express new phenotypes that are intermediate between the
homozygous parents. This may arise for example if the dominant homozygous phenotype results
from the expression of a double-dose of gene product, and the heterozygous phenotype from a
single dose. The F2 phenotype ratio of 1:2:1 is characteristic. A non-Drosophila example of this
is seen in red- and white-flowered snap dragons:
P1
F1
F2
x
rr (white)
Rr (pink)
1 RR (red) :
2 Rr (pink) : 1 rr (white)
RR (red)
When both alleles are expressed the effect is known as codominance. Heterozygous
individuals express gene products from both alleles: unlike incomplete dominance, the
phenotype need not be intermediate. This sort of interaction is seen in the ABO blood group
system of humans. One allele controls the production of A antigen while the other controls the B
antigen (a third allele O produces no antigen). Heterozygotes carrying the allele for antigen A
and the allele for antigen B have blood type AB in which both proteins are present in equal
quantities. The F2 shows a ratio of 1:2:1 , as in the case of incomplete dominance.
DIHYBRID CROSS
Dihybrid crosses involve manipulation and analysis of two traits controlled by pairs of
alleles at different loci. For example, in the cross ebony body x vestigial wing
e is ebony body colour
e+ is wild-type body colour
vg is vestigial wing shape
vg+ is wild-type wing shape:
where the loci for ebony body colour and vestigial wing are on separate autosomes. Therefore
the genotypes and gametes are the same for male and female.
CROSS DIAGRAM
P1
Autosomal Independent
ebony body
ee vg+vg+
gametes
F1
gametes
x
vestigial wing
e+e+ vgvg
e vg+
e+vg
e+e vg+ vg
(all wild-type)
e+vg+, e+ vg, evg+, e vgF2 genotype combinations:
9
&
%
e+vg+
e+vg
e vg+
e vg
e+vg+
e+e+vg+vg+
e+e+vg+vg
e+e vg+vg+
e+e vg+vg
e+vg
e+e+vg+vg
e+e+vgvg
e+e vg+vg
e+e vgvg
evg+
e+e vg+ vg+
e+e vg+ vg
ee vg+ vg+
ee vg+ vg
evg
e+e vg+ vg
e+e vg vg
ee vg+ vg
ee vg vg
F2
Phenotype ratio: 9 wild-type: 3 ebony: 3 vestigial: 1 ebony vestigial
In a dihybrid cross, each of the F1 parents can produce four different gamete types, so
there are 16 (= 4 x 4) possible offspring combinations. Because the two traits show complete
dominance and separate independently of each other (Law of Independent Assortment), the
expected genotypic and phenotypic ratios from an analysis of these 16 possibilities can be
calculated.
Phenotype
Genotype
(9:3:3:1)
(1:2:1:2:4:2:1:2:1)
These ratios can be derived from the results of a monohybrid ratio. A basic principle of
probability theory is that the probability of two independent events occurring together is equal to
the product of the two independent probabilities.
For example, the expected proportions of flies with wild-type and ebony body colours in
a monohybrid cross are 3/4 and 1/4, respectively. Likewise, in a monohybrid cross involving
vestigial wings, the proportions are 3/4 wild-type and 1/4 vestigial-winged. In a dihybrid cross,
the proportions of flies with various combinations of both characters can be calculated as:
wild-type & wild-type =
ebony
=
wild & vestigial
=
ebony & vestigial
=
3/4 x 3/4 = 9/16.
1/4 x 3/4 = 3/16
3/4 x 1/4 = 3/16
1/4 x 1/4 = 1/16
This produces the familiar 9:3:3:1 ratio. In a similar manner, the expected genotype proportions
can be predicted because each monohybrid cross produces a 1:2:1 genotype ratio. The product
[1:2:1] x [1:2:1] = [1:2:1:2:4:2:1:2:1] then gives the results of the dihybrid cross.
10
AUTOSOMAL LINKAGE
Mendel’s work on peas was done before the discovery of chromosomes, and his Law of
Independent Assortment postulated that each trait would segregate independently of every
other. We know now that loci are arranged in linear fashion on chromosomes, and that loci that
are physically close to each other will not segregate completely independently of each other.
This phenomenon is called genetic linkage. Linkage may be complete (loci are so close that
crossing-over rarely if ever occurs between them, and only the parental type gametes are
produced) or incomplete (where crossing over occurs between the two loci and produces some
recombinant type gametes).
[Pisum has seven pairs of chromosomes. Because Mendel worked on just seven
characters, one of the leading urban myths of genetics is that he must have “cheated” to have
found seven characters, each of which occurred on a different chromosome pair. In fact, we
know now that his seven traits occur on just four chromosome pairs, and that only one of the 21
possible dihybrid crosses involves loci close enough to affect the expected 9:3:3:1 ratio for
unlinked traits. His 1867 paper shows clearly that Mendel did not attempt to perform all 21
possible dihybrid crosses, and that the one anomalous cross was not one he performed. Mendel
did not cheat].
The chance that a cross-over will occur between the loci depends on the genetic distance
between them. Loci located far enough apart on the same chromosome act as though they are
unlinked and produce equal proportions of parental and recombinant gametes.
When the loci in a dihybrid cross are linked, it is necessary to indicate clearly the specific
allelic combinations that are present on the two chromosomes in each of the parents, because
these alleles will tend to stay together and not assort independently. In the case of a double
heterozygote, a chromosome in which the two linked loci show alternately the recessive and
dominant alleles is called the trans (repulsion) arrangement ( a+b/ab+). A chromosome in which
the two linked loci show either both recessive or both dominant alleles is called the cis
(coupling) arrangement (a+b+/ab).
The phenotypic expressions of cis and trans arrangements of heterozygous dihybrids are
typically identical, but will produce different arrangements of alleles in their respective
offspring.
11
Cis arrangement
P1
st+cu+/st+cu+ (wild-type)
gametes
st+cu+
F1
st+cu+/st cu
gametes
st+cu+ ,
F2
1 st+cu+/st+cu+ : 2 st+cu+/st cu : 1 st cu/st cu
3 wild-type :
1 scarlet curled
x
st cu / st cu (scarlet eye, curled wing)
st cu
(all wild-type, cis arrangement)
st cu
Trans arrangement
P1
st+cu/st+ cu (curled wing)
gametes
st+cu
F1
st+cu/st cu+
gametes
st+cu ,
F2
1 st+cu/st+cu : 2 st+cu/st cu+
: 1 st cu+/st cu+
1 curled
: 2 wild-type : 1 scarlet
x
st cu+ / st cu+ (scarlet eye)
st cu+
(all wild-type, trans arrangement)
st cu+
The simplest mechanism for assessing linkage is a test cross (a mating in which one of
the individuals is homozygous recessive for all traits considered). A non-linked dihybrid test
cross will give a 1:1:1:1: ratio.
Linked Dihybrid test cross (cis arrangement)
F1 (from above) x homozygous recessive
st+cu+ / st cu
x
st cu / st cu
wild-type
scarlet curled
gametes
st+cu+, st cu
F2
1 st+cu+/st cu : 1 st cu / st cu
1
wild-type : 1 scarlet curled
st cu
The expected result of a dihybrid test cross with completely linked loci is a 1:1 ratio.
12
SEX CHROMOSOMES
Sex-determination mechanisms vary among different organisms. In species such as
humans and fruit flies, females are described as homogametic (XX: all gametes will carry the X
chromosome) and males as heterogametic (XY: half the gametes carry the X and half the Y
chromosome). We have made a distinction between the genes carried on the X and those carried
on the Y. Since the law of segregation applies to sex chromosomes as well as to autosomes, it
follows that genes on the X chromosome are passed on independently from genes on the Y
chromosome.
As an example of an X-linked cross, we will look at goggle-eye (unusually prominent
eyes), an X-linked recessive trait (g) in Drosophila:
P1
F1
F2
Ratio
X+X+ (standard)
x
XgY (goggle-eyed)
X+ Xg , X+ Y (standard)
X+X+ , XgX+, X+ Y , XgY
2 standard & : 1 standard % : 1 goggle-eyed %
Reciprocal
P1
Xg Xg (goggle-eyed)
F1
X+ Xg (standard)
F2
Ratio
x
X+ Y (standard)
XgY (goggle-eyed)
XgX+ , Xg Xg, X+ Y , XgY
1 standard &: 1 goggle-eyed &: 1 standard %: 1 goggle-eyed %
The reciprocal cross shows an example of criss-cross inheritance, where the trait is
passed from the mother to the sons, and can then appear in both male and female F2s. If the P1
female were homozygous dominant, as in the first instance, an allele of the gene can be present
in the F2 females, but it will be masked by a maternal dominant allele. The ratio will be similar
to that of a monohybrid cross.
In humans, a small number of loci are known to be Y-linked or holandric (located on the
Y chromosome). Such genes are expressed only in males. One such gene is a mutation that
causes excess growth of hair on the outer ear.
13
In a sex-linked cross, the principles are similar but the notation differs. Instead of
showing the alleles on the X or Y chromosome, simply use the symbol for the gene that is on the
X, for example
w+w+ is a female red-eyed fly.
w ¹ is a hemizygous white-eyed male.
The (¹) denotes the Y chromosome, which in Drosophila carries only a few genes. Keep in
mind that w+ is completely dominant to w, and that this is a case of complete sex-linkage.
In crosses with X-linked loci in Drosophila, males or females of an unexpected
phenotype occasionally appear in the F2. This happens when the two X chromosomes do not
separate during oogenesis: the result is an egg with two Xs and an egg with none. The failure of
the X chromosomes to separate is known as non-disjunction. Fertilization with typical X or Y
sperm gives XXY, XXX, and XO, YO offspring, respectively. XXY is a typical female; XXX
and YO die, and XO is sterile.
109
Appendix A
The Chi-Square (P2) Test in Genetics
With infinitely large sample sizes, the ideal result of any particular genetic cross is exact
conformation to the expected ratio. For example, a cross between two heterozygotes should
produce an exact 3:1 ratio of dominant to recessive phenotypes. In any particular real- world
experiment, with limited and sometimes very small sample sizes, results are expected to deviate
somewhat from the exact theoretical ratio, due simply to chance. In order to evaluate a genetic
hypothesis (for example, that a particular trait is due to a recessive allele segregating at a locus),
we need a means to distinguish an experimental result that is consistent with the hypothesis
within the bounds of simple chance deviations, apart from one that is intrinsically unlikely
(“wrong”), given the data. Statistical tests are a means of quantifying the results of an
experiment as evidence for or against a particular hypothesis.
One of the simplest statistical tests is Chi-Square (P2) Analysis, which compares the
"goodness of fit" between observed and expected counts. An hypothesis is developed that
predicts how a set of observations will fall into each of two or more categories (the expected
result). These counts are compared with the experimental data (the observed result). Allowing
for the sample size, the differences among the observed and expected results are reduced to a
single number, the chi-square value. Because larger deviations from expectation are expected
with more categories, the test also takes into account the degrees of freedom in the experiment.
Comparison with a table of probability values shows the probability that the observed deviation
could have been obtained by chance alone. [See the website for Bio2900 (Principles of
Evolution) for further discussion of the concepts of null hypothesis, significance testing, and
Type I & II error: http://www.mun.ca/biology/scarr/2900_Hypothesis_testing.htm].
CHI-SQUARE CALCULATIONS
The chi-square formula is:
P = 3 [(O - E) / E]
2
2
O = observed # of individuals with phenotype
E = expected # of individuals with phenotype
3 = sum of deviations for all phenotypes
i)
The chi-square test should be used only on the numerical data themselves, not on ratios
or percentages derived from the data.
ii)
In experiments where the expected frequency in any phenotypic class is less than five,
the true probability is usually slightly larger than the p given in the table.
iii)
Expected values should be adjusted to the closest integer [you would not expect a
fraction of an individual]; the sum totals of observed and expected values should not
differ more than one individual.
iv)
These calculations neglect a number of corrections, including those for small expected
classes, multiple simultaneous tests, and the ”one-tail” or “two-tail” nature of the test.
These will be discussed in your statistics course.
110
INTERPRETATION OF THE CHI-SQUARE TABLE
The table of chi-square values (below) can be used to determine the probability that any
particular result (observed deviation) could have been obtained by chance alone.
i)
Horizontal rows indicate the number of degrees of freedom (df). In general, df = (n - 1)
where n is the number of observed classes or phenotypes. For each row, the p value at
the top of the column is the probability that a particular chi-square value could have been
obtained by chance alone. This is called the critical value for that level of probability.
ii)
To use the table, enter it on the row corresponding to the number of degrees of freedom.
Look for the column with the critical value closest to and less than the chi-square
obtained in your calculations. The probability of your result is less than the p value for
that column.
iii)
For a biological experiment, we typically set the level of significance at p = 0.05. If the
observed chi-square is greater than the critical value for p = 0.05, we conclude that the
result could have been obtained by chance less than 5% of the time, and we reject the
null hypothesis that chance alone is responsible for the result. We say that this result is
statistically significant.
Example #1: P2 = 15.0 and n = 8 phenotypes.
The observed chi-square of 15.0 with df = 7 is greater than the critical value of 14.07 for
p < .05. This means that the observed deviation represented by this chi-square value would be
expected to occur by chance less than 5% of the time. The result is statistically significantly,
and the hypothesis (that the data are consistent with the expected ratio) can be rejected.
Example #2: P2 = 5.0 and n = 8 phenotypes.
The observed chi-square of 5.0 with df = 7 lies between the values 2.83 and 6.35, which
correspond to .90 > p > .50. A deviation as large as that observed would be expected to occur by
chance more than 50% of the time. The difference between the observed and expected results is
not statistically significant, and the null hypothesis (the ratio being tested) cannot be rejected.
p=
0.9
0.50
0.20
0.05
0.01
0.001
df = 1
0.02
0.46
1.64
3.84
6.64
10.83
2
0.21
1.39
3.22
5.99
9.21
13.82
3
0.58
2.37
4.64
7.82
11.35
16.27
4
1.06
3.36
5.99
9.49
13.28
18.47
5
1.61
4.35
7.29
11.07
15.09
20.52
6
2.20
5.35
8.56
12.59
16.81
22.46
7
2.83
6.35
9.80
14.07
18.48
24.32
8
3.49
7.34
11.03
15.51
20.09
26.13
9
4.17
8.34
12.24
16.92
21.67
27.88
10
4.87
9.34
13.44
18.31
23.21
29.59
15
8.55
14.34
19.31
25.00
30.58
37.30
25
16.47
24.34
30.68
37.65
44.31
52.62
50
37.69
49.34
58.16
67.51
76.15
86.6
111
Appendix B
Gene Map of Drosophila melanogaster
From William S. Klug, Michael R. Cummings, Concepts of Genetics, Macmillan, ©1994, 4th ed.,
p. 132. This material has been copied under licence from CANCOPY. Resale or further copying of this material is
strictly prohibited.
112
Appendix C
Genetic Nomenclature & Notation for Drosophila
Clear notation for any Drosophila genotype will indicate whether the locus involved is on an
autosomal (II, III, or IV) or sex chromosome (I(X) or Y), and in the case of two (or more) loci,
whether they are the same or different chromosomes (linked or unlinked, respectively).
Dominant alleles at a locus are indicated by a capitalized symbol, recessive alleles by a lowercase symbol. Examples of such notation are as follows.
1) One autosomal locus:
e.g. The genotype for ebony body on Chromosome III is ee, for wild-type body at that locus e+e+.
For autosomal genes the genotype is the same for male and female, and can be homozygous or
heterozygous.
2) One sex-linked locus:
In Drosophila alleles may be present on the X chromosome but not on the Y chromosome,
therefore the genotypes for male and female are different. The symbol ( ¹) indicates a male Y
sex-chromosome and therefore the presence of only one allele.
e.g. Bar eye on Chromosome I. Bar eye female has genotype BB, Bar eye male is B¹ . Wildtype eye female is B+B+, wild-type eye male is B+¹.
3) Two unlinked autosomal loci
e.g. vestigial wing (II) and ebony body (III) would have genotype vgvg ee and wild-type (wing
and body at these loci) would have vg+vg+e+e+.
4) Two linked autosomal loci
e.g. curled wing (III, 50.0) and ebony body (III, 70.7). The genotype is written to show the
alleles on each homologue cu e/ cu e. Wild-type would be cu+ e+ / cu+ e+.
5) Two sex-linked loci
e.g. Bar eye (I 57.0) and forked bristle (I, 56.7). Female is Bf / Bf, male is Bf / ¹ . Wild-type
female is B+f+ / B+f+, wild-type male is B+f+ / ¹ .
6) One sex-linked & one autosomal loci
e.g. Bar eye (I) and vestigial wing (II). Female is BB vgvg, male is B¹ vgvg wild-type female
is B+B+vg+vg+, wild-type male is B+¹ vg+vg+.
113
Appendix D
WHMIS for Undergraduate Laboratories
The Workplace Hazardous Materials Information System (WHMIS) is a Canadawide information system for ensuring that industrial workers are informed about the chemicals
and other hazardous materials they use. Although student laboratories are not included in the
legislation, the Memorial University Safety Office encourages the idea of making the same kind
of information available to graduate and undergraduate students as to employees of the
university, and we have therefore included this material in the lab manual.
Hazard Classifications
A controlled product is a material that may have characteristics which would put it into
one or more of the hazard classes on the attached table - Hazard symbols and classes.
A controlled product can be recognized if its label:
- has any of the WHMIS hazard symbols,
- has the WHMIS hatched border, or
- makes reference to a Material Safety Data Sheet (MSDS).
WHMIS Labels
There are two basic types of WHMIS labels:
- supplier labels
- workplace labels
Supplier labels are attached to all packages of controlled products by suppliers. These
labels give the identity of the product and its supplier, risk phrases, precautionary measures, first
aid measures, hazard symbols, and reference to a material safety data sheet (MSDS). The labels
have the WHMIS hatched border.
Workplace labels are produced in the workplace and are attached to containers of
controlled products which do not have supplier labels such as when products are decanted from
supplier containers, old containers which have been around since before WHMIS became
effective, or other containers which do not have supplier labels for whatever reason. Workplace
labels need only a product identifier, safe handling procedures and reference to an MSDS..
114
Material Safety Data Sheets (MSDS)
A material safety data sheet is a technical document relating the health effects of
exposure to a product, hazard evaluation, protective measures and emergency procedures.
MSDS’s are sent by suppliers of controlled products or may be generated in the workplace.
MSDS’s have nine categories of information:
- Name of product and its use, supplier address and phone.
- Name and concentration of all hazardous ingredients.
- Physical characteristics of the product.
- Fire or explosion hazard.
- Reactivity hazards.
- Toxic hazards.
- Actions required to prevent injury or accident.
- First aid procedures.
- Identity of organization which prepared MSDS and date it was prepared.
Students should be aware that MSDS’s are available for the controlled products being
using in laboratories and that these may be consulted for specific information on the controlled
products.
In some cases, MSDS’s will not be sent by suppliers of laboratory chemicals when the
appropriate safety information is given on the container labels.
Exemptions
There are other products used in various workplaces, including laboratories, which might
be considered hazardous but are exempted from WHMIS regulations. These include
manufactured articles, products made of wood or tobacco, products packaged for consumer use,
hazardous waste, and products governed by the federal acts for: explosives, food and drugs, pest
control products, and radioactive materials. Radioactive material are covered under regulations
of the Nuclear Regulatory Commission.
Memorial University Safety Office (09-91)
115
Memorial University Safety Office (09-91)