Bi190
ADVANCED GENETICS NOTES
PAUL W. STERNBERG
DIVISION OF BIOLOGY
CALIFORNIA INSTITUTE OF TECHNOLOGY
AND
HOWARD HUGHES MEDICAL INSTITUTE
©1997, 1999, 2001, 2003, 2005 Paul W. Sternberg
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1. The genetic approach; life cycles and mutant hunts
How do genes control development, behavior and physiology?
Assigned reading:
LH Hartwell, J Culotti, JR Pringle, BJ Reid (1974). Genetic control of the cell division cycle
in yeast. Science183. 46-51. Classic example of a successful mutant hunt.
Optional reading:
I Herskowitz. (1989). A regulatory hierarchy for cell specialization in yeast. Nature 342,
749-757. Describes the biology underlying the genetic experiments discussed in this lecture.
1.1 The genetic approach
Rationale: change one variable in a complex system
In an introductory course in genetics, you typically learn about the mechanism of heredity
and somewhat about how to use genetics as a tool to understand biological problems.
This course will focus on the methods of modern genetics to study complex biological
systems. Genetics is a powerful approach because on can alter a single (or a few) variables
in a complex system of 1000s of components. Often, we seek to understand the action of
single genes in the context of all other genes.
I hope that throughout this course you will begin to think like a geneticist. The plan for
the course is to cover crucial concepts and approaches in genetic analysis, illustrated by
some examples during the lectures. In the four problem sets and assigned readings -typically one paper per lecture -- I want you to think through some examples in detail.
Another goal is to help you to be able to follow genetic arguments and to read genetics
papers. Genetics is practiced in intensely in a number of organisms, especially viruses,
bacteria, fungi, nematode worms, fruitflies, fish, mice, humans and many plants. The
common language of the gene --now revealed by molecular biology, especially gene
cloning and DNA sequencing -- has formed bridges between analyses in diverse
organisms Today's biologist takes a multiorganismic approach to biological problems,
using the unique features of one model to probe deeply in to mechanisms, and compares
findings to other organisms. The key is to follow the genetic logic and not get bogged
down in the nomenclature and idiosyncrasies. Of course , it is the details that allow the
analysis and dictate how you do experiments so we will discuss the details as appropriate.
If you see the big picture, the details are ever so fascinating. I expect you to gain an
appreciation for yeast, worms, flies and the mustard-like weed Arabidopsis,thaliana, as
discussed by Elliot Meyerowitz.
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Unfortunately there is no textbook suitable for this course. I will assign the few reviews
and parts of books that are relevant and supplement this with lecture notes.
SOME MAJOR GENETIC EXPERIMENTS
Identify Genes. Find or make mutants: The classes of mutant phenotypes are often
informative.
Infer normal gene function from mutants: dosage; compl
mentation site of action; etc.
Make mutant combinations and infer functional relationships among genes (pathways
and epistasis; modifier genetics)
Construct maps (genome analysis)
Analyze many genes in parallel (functional genomics)
1.2 Life cycles
As we proceed,I will briefly describe the life cycles of the organisms I will discuss most in
this course. You need to know the life cycles for two reasons. One is so you know how
the genetic analysis works. The second reason is since the life cycle provides interesting
material for study of basic biological processes.
The key is to follow the ploidy. How do you construct heterozygotes so you can assay
complementation? How do you map? When can you score phenotypes? How can you
homozygose mutations for efficient screens? How do you get meiosis to occur? Scientists
are continually expanding the range of organisms with which genetic analysis can be
applied; some of these have unusual features.
Lab organisms are often domesticated organisms, or domestic pests. They tend to have
short life cycles. For example, it is easier to study C. elegans since it completes its life cycle
on a Petri dish (or even on the space shuttle) than Strongyloides ratti, which has to pass
through (literally) a rat.
1.3 Saccharomyces cerevisiae [should be described in Hartwell et al. Genetics text]
The cells of the baker’s yeast S. cerevisiae alternate between haploid a or  cells and a/
diploid cells. The a/ diploids can undergo sporulation if starved for carbon and
nitrogen.
The mating type (sex) of the yeast cell is specified by the mating type (MAT) locus. There
are two co-dominant alleles, MATa and MATThese specify specialized cell properties
involved in mating and for the a/ diploids, sporulation. Both a and  cells secrete a
mating factor, called a- and -factor, respectively. a cells respond to  factor.
Grow as colonies on Petri dishes or in liquid. one cell generates two cells in 90-120
minutes on rich medium (YEPD; yeast extract, peptone, dextrose) Slower on minimal
medium (yeast nitrogen base without amino acids). Most markers are auxotrophies, e. g.,
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Ura-, no growth without uracil; Leu, no growth without leucine; His, no growth without
histidine.
Yeast has 16 chromosomes. The genome is completely all sequenced. The 12,068 kb of
DNA encodes approximately 6000 genes [(5885 open reading frames ( ORFs) + rRNA,
tRNA and snRNA]. [Reference: Goffeau et al. (1996). Life with 6000 genes, Science 274,
546-567.] Yeast genome database: http://genome-www.stanford.edu/Saccharomyces/
Capital for the dominant or co-dominant
URA3 vs ura3 or ura3-52
MATa and MAT
a yeast cross:
ahis3 LEU2 x  HIS3 leu2
select for growth on minimal mdium (Leu+ His+)
a/ diploid
starve
dissect tetrads
grow up individual spores (germinate on rich medium)
1.4 Tetrad Analysis
In S. cerevisiae all genes are essentially unlinked since there many chromosomes and lots of
recombination. Tetrad analysis substitutes for linked markers to follow unknown
genotypes. The basic principle is that you recover all four products of meiosis.
Consider segregation of an ade2 mutation that results in a red colony as opposed to the
wild-type ADE2 white colony
In ade2,
phosphoribosylaminoimidazole (AIR) -> red pigment
ADE2
ADE1
AIR -------->> CAIR---------- >> SAICAR
the dissection slab:
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a ade2
 ADE2
a
a
PD, parental

ditype 
NPD,
non=parental ditype
TT, tetratype
1
+
+
PD
Ade phenotype
2
+
+
TT
3
+
+
NPD
2:2 segregation implies one locus
PD:TT:NPD 1:4:1 is unlinked.
If PD >1 and NPD <1, is linked.
1:<4:1 is centromere linkage
Q: What is the distribution of ascus types for a phenotype
dependent on two, unlinked loci?
1.5 Hartwell -- 1970s. Mutations that affect the cell division cycle.
The cell cycle comprises several cycles: the nuclear cycle of DNA replication and
chromosome segregation, the spindle pole body (yeast centriole) and budding. The major
landmark events are thus DNA replication and mitosis, spindle pole body duplication and
separation; and bud emergence, growth and cytokinesis.
Hartwell screened for temperature sensitive (ts) mutants that arrested at particular
landmarks in the cell cycle.
23° permissive condition
36° restrictive condition
First 150 mutants defined 32 genes.
1.6 An example of a screen and very informative tetrad analysis:
Mackay and Manney (1974) screened for sterile mutants by the following protocol.
UV irradiate
 ade6 his6 can1
mix with 1000-fold excess of
a ade2 his2 CAN1
Let mate on YEPD for 24 h
Dilute and plate on canavanine + arginine
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[can1 is a recessive drug resistance; CAN1 is the wild-type allele. arginine is
necessary to induce the permease that transports in this arginine analog]
All a and a/ diploids die. The survivors are  steriles.
To analyze these sterile mutants, force a mating by selecting His+ colonies from the cross:
 ste his6 x a STE his2.
Sporulate the resultant diploids:
MATa STE HIS6 his2
MAT ste his6 HIS2
Test the spore colonies for mating ability with  and a testers.
Look at NPDs
spore genotype
mating phenotype

+
-mater

+
-mater
a
ste
?
a
ste
?
A TT inferred from the mating phenotype would indicate an -specific sterile: e.g.,
mating phenotype
a-mater
a-mater
non-mater
-mater
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spore genotype
a
+
a
ste

ste

+
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1.7 The results from the Mackay and Manney screen and others revealed three
classes of sterile mutations:

mat 1
mat  2
 -specific genes
non-specific genes
a
a -specific genes
The -specific genes include
STE3 receptor (G-protein coupled) for a-factor
STE13 dipeptidyl aminopeptidase
(processing of -factor)
KEX2
(processing of -factor)
The non-specific genes include:
STE4 G protein  subunit
STE5 scaffold for MAP kinase cascade
STE7 MAP kinase kinase
STE11 MAP kinase kinase kinase
STE12 transcription factor
STE18 G protein  subunit
STE20 protein kinase
The a-specific genes include:
STE2 receptor (G-protein coupled) for -factor
STE6 secretion of a-factor (mdr or ABC type transporter family)
STE14 protein-S isoprenylcysteine O-methyltransferase
(modification of a-factor)
RAM1 farnesyltransferase beta subunit (modification of a-factor)
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1.8 How MAT controls mating type
To conclude the first part of the yeast mating story, let us consider the mat mutations and
another use of tetrad analysis.Ploidy is not important for mating type; it is the constitution
of the MAT locus. To test complementation with ste mutants, one has to homozygose the
mating type or use some other trick. For example, select for MATa/MATa mitotic
crossing over from a MAT/MATa heterozygote using a linked recessive drug resistance.
Since there were two complementing mutations at MAT locus, and there was a deletion
of the locus (the Hawthorne deletion) that resulted in a mating, Strathern, Hicks and
Herskowitz proposed the 1-2 hypothesis:
MATa encodes two functions, 1 is necessary to turn on -specific genes; 2 is necessary to
turn off a-specific genes. This hypothesis arose from the fact that MATa only encodes
functions necessary for sporulation while MAT encodes function necessary for mating
and for sporulation.
The 1-2 Hypothesis for yeast cell type control by the MAT loci
1
 -specific genes
a -specific genes
2
a1
a / -specific genes
sporulation
One critical test of this hypothesis was to construct an 1 2 double mutant. The prediction
was that the double mutant would behave like an MATa allele for mating, but would not
support sporulation. They did this by dissecting tetrads from a mat1/mat2 diploid.
1-5/2-1
Dissect 77 tetrads with 4 viable spores
76
4 Ste- : 0 Ste+
1
2 Ste- : 1 -mater : 1 Alf
Alf is an a-like faker that mates as an a, but does not support sporulation in trans to
MAT. This exceptional ascus is a Tetratype
phenotype
inferred genotype
-mater
+
+
Alf 

  
Ste 

 
Ste
+

Molecular analysis then revealed that in an 1- mutant, -specific genes are off
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and in an 2-, a-specific genes are inappropriately on
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2.Saturation mutagenesis
Assigned reading lecture #2: C. Nüsslein-Volhard and W. Wieschaus (1980). Mutations
affecting segment number and polarity in Drosophila. Nature 287, 795-801.
2.1 Designing mutant screens
We will start at the beginning of a genetic analysis-- finding mutants. Just like choosing
wisely an assay for a biochemical purification, the phenotype chosen for a genetic screen is
crucial. It does not however have to be difficult.
Examples
phage T4 [Edgar & Wood]
yeast cdc mutants [Hartwell]
yeast secretion - sec mutatns [Schekman and Novick]
nematode Unc mutants [Brenner]
nematode egglaying [Horvitz]
Drosophila patterning [N-V & W.]
fly eyes [Benzer.....]
zebrafsh [development issue]
You always make at least one assumption in a genetic screen, that you can find mutations.
You might not get what you wanted but you will get what you look for.
Extent of mutagenesis
Most mutageneses are heavy with multiple hits per genome. Thus, it is imperative to backcross mutations to
determine whether a single locus is responsible for the phenotype and to cross out other mutations that
might affect the scoring and interpretation of the phenotype. F2 screens for zygotically acting genes
2.2 F2 in worms
In the standard protocol, L4 stage hermaphrodites are mutagenized with, most typically, a
chemical mutagen, ethyl methane sulfonate (an alkylating agent that usually causes G/C-A/T transitions). These mutagenized P0 hermaphrodites are allowed to self, and their
grandprogeny examined for mutants. Dominant mutations are recovered in the F1,
recessive in the F2 and maternal effect (maternally rescued) in the F3.
Nematode C. elegans
Self-fertilizing hermaphrodites and males
embryogenesis 14 hours, hatch to L1 larvae. They grow ten-fold in length and become
sexually mature. There are four molts--L1, L2 L3 and L4 molt separating the four larval
(technically juvenile) periods.
Hatch with 550 cells ; 10% of the cellsdivide to produce 959 nuclei in hermaphrodites. and
1031 in the male. Hermaphrodites. are XX; males are XO.
Grown on lawns of E. coli on Petri dishes.
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transfer with sterile wire
One hermaphrodite generates 300 progeny in 5 days. Two generations/week
Markers are morphological or behavioral and visible under a 25x dissecting
stereomicrocsope.
Nomenclature
gene name: lin-3
allele: e1417
in parenthesies: lin-3(e1417)
phenotype; Lin-3
A recombinant:
protein: LIN-3
Unc non-Lin
Unc, uncoordinated movement;
Dpy, dumpy body shape;
Lin, cell lineage abnormal;
Sup, suppressor
Let, lethal
6 chromosomes; 100 Mb DNA; 67% sequenced; 95% done in a year.
database: www.wormbase.org
The 100 Mb of genome sequence indicated about 19,000 predicted genes. This is a gene
density of one gene per 5 kb.
chr.
I
II
III
IV
V
X
total
Mb
14
16
12
18
22
20
100
mu
50
45
55
50
50
45
300
2.3 Drosophila
To recover recessive mutations in male-female organisms such as the fruitfly, one needs to
use balancer chromosomes to allow ready homozygosing of the mutagenized
chromosome.
Fruitfly Drosophila melanogaster
males and females. ten day generation time.
grow on yeast, cornmeal and molasses
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Nomenclature:
Genes are given a clever name plus a three letter abbreviation
bride of sevenless = boss
lethal 1 polehole = (1)ph
superscript the allele name:
Major Drosophila database (FlyBase): http://www.flybase.org
mutagenize
+
Balancer A
X
Balancer A
Balancer B
m
Balancer B
X
Balancer A
Balancer B
F1
B non-A
m
F2 Balancer B
X
Non-B non-A
m
F3
m
Balancer B
m/ Bal x
m/ Bal
m
2..4 strain construction using 3-factor crosses
Review of 3-factor mapping.
Construct a heterozygote: lin / dpy unc, where you know that lin is linked to dpy unc.
lin + +
+ dpy unc
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+ + lin
dpy unc +
+ lin +
dpy + unc
13
In C. elegans, to construct a strain such as lin-3 let-317, where lin-3 is recessive vulvaless
and let-317 is recessive lethal, you could either make a heterozygote lin-3 +/+ let-317, and
pick many vulvaless [lin-3 ?/lin-3 +] of which 2p-p2 of them wil be of the desired
genotype, or you can do it by starting with a strain lin-3 + dpy-20 +/+ let + unc-22, and pick
Lin non-dpy recombinants, some of which will be lin-3 let + +/lin-3 + dpy-20 + and thus
have the chromosome you want.
If you do three-factor cross and your mutation (in in the example) is unlinked, you will
place it 3/4 of the distance between. So with small numbers, you might get 2:6 and 6:2.
You can use the same logic to map with a large number of markers. Often these are
molecular markers, VNTRs, RFLPs, the heterozygote is then a b c d e f g h .... zzzz/lin
2.5 Penetrance and expressivity. These are operational concepts that have to do with
the ability to score a phenotype.
Penetrance is the proportion of individuals with a given genotype that display the
phenotype.
Expressivity is the degree of severity of the phenotype.
If you have a mutant strain that you can score 80% of the organisms. Note that incomplete
penetrance and dominance can be difficult to distinguish, especially in human pedigrees.
2.6 Saturation screens. There are several ways of looking at attempts to find every gene.
Realistically, one can only find all the genes identifiable by the screen you used.
The idea is to "saturate" the genetic map for genes of interest.
Ferguson & Horvitz (1985). C. elegans vulval lineage mutants
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No. alleles per gene in each class
rec. mutations
viable null
lin-1
16
lin-2
13
lin-7
13
lin-10
3
lin-18
2
lin-31
11
unc-83
10
unc-84
16
recessive mutations recessive mutation
dom. mut.
lethal/ste null
unknown null
lin-3
2
lin-4
1
lin-12
lin-8
1
lin-11
4
lin-24
lin-9
1
lin-17
5
lin-33
lin-13
2
lin-25
2
let-60(gf)
lin-15
5
n300
1
lin-14
lin-26
1
let-23
1
let-60
0
sem-5
0
Generation of vulval precursor
cells
let-341
0 [VPCs]
unc-83, unc-84
migration
lin-45 of VPC
0 parents
lin-24, lin-33
death of VPCs
Induction
VPC fates
lin-18 null isofprobably
lethal; lin-15 is viable but encodes 2 transcripts
lin-3
Inductive signal (EGF/TGF-)
let-23
Receptor (EGF-receptor)
lin-34
gf allele of let-60 (Ras)
lin-2, lin-7, lin-10
localize the LET-23 protein
lin-1, lin-25, lin-31
transcription factors
Regulators of LET-23 signaling:
lin-8, lin-9, lin-15, lin-13
Another signal
lin-12 Founding member of Notch/LIN-12 family of receptors
Timing of vulval development (and other events)
lin-4
RNA that inhibits lin-14 mRNA expression
lin-14 putative transcriptional regulator
VPC fate execution
lin-11 Founding member of LIM domain transcriptional regulators
lin-17 Wnt receptor (Drosophila Frizzled)
lin-18 Receptor protein
Target size problems: lin-15 requires 2 mutations or a deletion; lin-4 encodes a small
RNA;
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2
2
1
2
2.7 We will briefly review Poisson statistics.
Poisson Distribution
Assumptions:
•For each observation only two results are possible [success or failure]
•Probabliity of the two results do not vary between observations
•Successive observations are independent
These assumptions are also true for the binomial distribution; Poisson is an extremely
skewed binomial such that q approx. 1/n as n gets large. Therefore, use the Poisson when
the event is rare, for example, #particles per area or er time, #hit per gene after
mutagenesis. Note that one can sample over a very short time interval so either 0 or 1
success per interval.
m = mean number of events per sample
r = actual number of events per sample
P(r) = e-m • mr / r!
For m=1
P(0) = e-1 10/0! =e-1 = 0.368
P(1) = e-1 11/1! =e-1 = 0.368
P(2) = e-1 12/2! =e-1 /2 = 0.184
P(3) = e-1 13/3! =e-1 /6 = 0.061
(In case you forgot: e 2.718
1/e  0.3678
P(0) = e-m
2. Only ≤1 per site
P(0) + P(1) = e-m + e-m m1 / 1!
0! = 1 and
e0 = 1
; P(≥1) = 1 - P(0)
1. At least one per site
P(≥1) = 1 - e-m
m
e-m + me-m= [e-m(1+m)]
m
1-e-m
1
0.5
0.3
0.1
0.01
2e-m = 0.736
0.91
0.96
0.995
0.99995
1
2
3
4
5
10
0.632
0.865
0.95
0.98
0.99
0.99995
For the Poisson distribution, the standard deviation = m and the
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variance = the mean
To estimate m, measure P(0). P(0) = e-m and thus: m = - ln [P(0)]
2.8. An example of an F2 saturation screen for embryonic lethals
Reference: Nüsslein-Volhard, Wieschaus and Kluding. (1984). Mutations affecting the
pattern of the larval cuticle in Drosophila melanogaster. I. Zygotic loci on the second
chromosome. Roux Arch. Dev. Biol. 183, 267-282.
EMS 25 mM mutagenized 1500 male cn bw sp/cn bw sp mated to 2500 virgin females DTS91
b pr cn sca/CyO (In (SLR) O,dplvI Cy pr cn2).

Mate individual males with bright red eye color [cn bw sp */CyO] cross to female
DTS91/CyO, which have orange color eyes due to the interaction of pr and cn2

DTS91 is a dominant temperature-senstive larval lethal.
Grow at 29°, then 18° or 25°. Set up sibling crosses (F2 lines) (cn bw sp *)/CyO.
Look at progeny of each line. If (cn bw sp *) has a lethal, then there will be no white- eyed
adults. Maintain lethal as (cn bw sp *)/CyO.
pr cn 2is orange
cn is scarlet
pr is purple
CyO is curly of Oster
Statistics
Looking at the total lethals, of 5764 lines, 4217 had lethals and 1547 had no lethals.
P(0) = e-m = 1547/5764 = 0.26839, and thus m = 1.315 average lethals/chromosome.
Therefore, the estimated number of lethal hits is
5764 lines • 1.315 lethals/line = 7,581 lethals
2843 embryonic lethal lines; 2921 with no embryonic lethals
P(0) = e-m = 2921/5764 = 0.507, and thus m = 0.68 average emb lethals/chromosome.
Therefore, the estimated number of lethal hits is
5764 lines • 0.68 lethals/line =3,917 lethals
From 272 embryonic lethal alleles (with patterning defects), they found 61
complementation groups.
Thus, the average number of alleles/locus = 4.5
48 loci had more than one allele; 13 had single alleles. (An additional 4 lines had two
mutations contributing to the lethality.)
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Chromsome 2 screen
1 2
13 13
3
7
4
8
5
3
6
5
7
1
8
1
9
5
10 11 12 13 14 15 16 17 18
0 0 1 1 0 1 0 1 1
alleles/locus
#loci
•In the examination of the first 25% of lines, they found 50% of the loci; in the last 25% of
the lines, only 3 new loci (5%) were found. The rate of discovery of new loci was
decreasing.
•The fraction of loci missed (that is, represented by 0 alleles) is
P(0) = e-m = e-4.5 = 0.01 and the number of loci missed = 0.01 • 61 = 0.6
Are the data consistent with a Poisson distribution?
For example, 13 loci represented by 1 allele and 48 by >1 allele. Thus,
P(1) = ( e-4.5 • 4.5 ) / 1! = 0.04
and
the estimated number of loci with 1 allele is 0.04 • 61 = 2.4 loci
The actual number is 13, and the distribution is not Poisson.
Another way to calculate is by ignoring the loci with many alleles, and thus avoiding
hotspots. Thus the fraction of identified loci with >1 allele is 1 -P(0) -P(1)/1- P(0), that is
P(>1)/P(≥1). )
The fraction f of identifed loci represented by more than one allele as a function of the
average number of alleles per locus:
m
m
 1  e  me  .
f =
 1  e m
In this example, f = 48/61 = 0.787 and thus m =2.58 alleles/locus.
Te fraction of loci represented by no alleles is thus P(0) = e-m = e-2.58 = 0.076
and the number of loci missed = 0.076 • 61 = 4.62
•One superb argument that they found most of the loci is that they examined
homozygous deficiencies for much of the chromosome and found that they could account
for all the phenotypes by identified loci.
2.9 Why would you not be able to find mutants?
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Dominant lethality. This is known as haploinsufficiency.
Redundancy -- two genes with the same product. For example, the genes encoding
-factor in yeast did not come from the screen for sterile mutants.
Apparent functional redundancy -- genes encode unrelated functions with some common
consequence.
Pleiotropy --- the gene does something else. For example, in the yeast cell cycle it has
multiple blocks and would have been discarded by Hartwell's criterion.
Maternal rescue of the zygotic phenotype. A gene expressed in both the germ line and in
the embryo.
2.10 Summary of the screens for zygotic mutations by N.-V. and W.
chromosome
# lethal hits
# Emb lethal hits
Emb visible hit
# c. groups (>1 allele)
Avg. # alleles/group
#single mutations
X
3255
679
114
20
5.1
13
2
7581
1907
274
48
5.4
13
3
7300
1772
198
32
5.8
13
Note that these screens did not look at lethals that did not affect the epidermis.
1. Gap genes [some maternal components]
Large parts of the larval pattern is abnormal; there is some expansion of remaining pattern
elements. hunchback
2. Pair rule genes
Mutations result in deletions in every other segment. even skipped
3. Segment polarity genes
Pattern deletions in each segment.
hedgehog
Should you only look for mutations with specific effects? It is a good place to start.
Most mutageneses induce multiple mutations. Therefore you must backcross and get
more than one allele. In worm unc-86, non-complementation screens yielded two clases of
alleles, Him (high incidence of males)and non-Him. There is a hotspot for a deletion that
removes an adjacent gene.
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3. Maternal effect mutants
An approach similar to the saturation screens for zygotic lethals was taken to identify the
genes required in the mother for the development of the embryo. For example,
Schüpbach and Weischaus (Genetics, 1989) screened for female sterile mutations on
Drosophila chromosome 2. So-called maternal effect mutants, in which it is the genotype
of the mother and not the zygote that counts arise from genes expressed in the maternal
germ line. There are also paternal effect genes, encoding products required in the male
sperm for embryogenesis.
strains:
CyO, the chromosome 2 balancer
Cy, dominant curly wings; O=Oster. cn, bright crimson eyes
N. B. pr, purple eyes. pr cn confers orange eye color
DTS513, dominant temperature-sensitive sterile
Fs(2)D, dominant female sterile [must be brought in paternally]
The mutagenesis scheme:
P0
cn bw sp/CyO, DTS513 females x EMS-mutagenized cn bw males

F1
individual female (cn bw *) x male Fs(2)D/CyO, DTS513
grow at 18°C, permissive for DTS

F2
intercross males and females
(cn bw *) / CyO, DTS513
Fertile Female
(cn bw *) / Fs(2)D
Sterile Female
Fs(2)D / CyO, DTS513
Sterile Female
grow at 29° C, restrictive for DTS

F3 the only survivors are
(cn bw *) /(cn bw *)
(cn bw *) / Fs(2)D
Sterile?
F4
Sterile Female

Examine eggs and embryos
Maintain the strain as (cn bw *) / Cy,O at 19° C
The results:
Lines established
Bi190 2005 Sternberg
18,782
20
Homozygous viable
Lethals/Chromosome [Poisson est.]
Female steriles
Female steriles/Chromosome [Poisson]
7,351
0.94
529
0.075 (12.5x fewer than lethals)
Normal eggs, abnormal embryos
136
Complementation groups, total
"
singles
"
multiples
(25% of Female Ste.)
67
44
23
•Many of the single alleles are hypomorphic (leaky) alleles of genes with lethal loss-offunction phenotype. The distribution may be due to two superimposed distributions.
General classes of defects:
presynctial blastoderm
syncytial blastoderm
cellularization
gastrulation
14
12
17
12
•See the map of chromsome 2 [Fig. 2 of Schüpbach & Weischaus, 1989] with cytogenetics,
deficiences, female steriles and genetic map.
Further analysis revealted that these mutations defined a whole set of key genes for early
development including
concertina (cta) Encodes a G protein necessary for coordinated gastrulation
torso (tor) Encodes a PDGF-receptor homolog necessary for specification of terminal
(anterior and posterior) versus central fates in the embryo.
cactus (cact). Encodes an IB protein necessary to pattern the dorsal-ventral axis.
tud, stau, vasa, and val. All necessary for specification of the germ line in the zygote (polar
granules), a process that involves RNA localization.
4. Null phenotypes & F1 screens
How can you infer the wild-type function of a gene from the phenotypes caused by
mutations?
If the loss of gene A activity results in a failure of a process P to occur, then we
infer that gene A is necessary for process P. N. B., genes are often named
according to their mutant phenotype, and thus can be the opposite of what the
gene does. The Unc (uncoordinated) genes of C. elegans are necessary for
Bi190 2005 Sternberg
21
normal, coordinated movement; Bride-of-sevenless is necessary for R7
photoreceptor differentiation; etc.
A crucial aspect of understanding gene function is to know the consequence of removing
that gene completely. Comparing multiple alleles of the same gene is often very
informative as well; F1 screens are an effective way to obtain additional alleles given an
appropriate starting allele.
Very often genetic analysis relies on more detailed, quantitative description. Sometimes
this can be measurement of the level of an enzyme, the number of extra bristles, life span,
but often of the penetrance of a phenotype (80% animals die versue 30% die).
Penetrance and expressivity. These are operational concepts that have to do with the
ability to score a phenotype.
Penetrance is the proportion of individuals with a given genotype that display the
phenotype.
Expressivity is the degree of severity of the phenotype.
Please note that in practice, there is often no clear distinction between the two since you
could define the phenotype as serum levels of protein A less than 65% of normal, etc..
Imagine a trait with incomplete penetrance and variable expressivity.
1. Molecular criteria:
The best criteria for a null mutation (jargon for complete loss-of-function) is that the
coding region is deleted from the chromosome. The genetic criteria are used for the
sometimes many years until such a mutation can be obtained, either by random screening
or by engineering.
2. A test to rule out that a mutation m is null:
m/Df ≠ m/m  NOT NULL
m/Df = m/m  could be NULL
Watch out: most papers overinterpret this result.
3. Frequency
In flies and worms, knock-out frequency after EMS mutagenesis is 1/2000 to 1/5000 (but
can vary). Or, loss-of-function alleles are recovered at a frequency of 2 - 5 x 10-4. The best
way to obtain frequency data is by an F1 non-complementation screen.
4. Amber suppressible alleles
Bi190 2005 Sternberg
22
not good; also not applicable to all organisms.
UGG Trp codon
UAG Amber codon
amber suppressors E. coli supU, C. elegans sup-5, sup-7 etc. encode tryptophenyl tRNAs
(tRNAtrp)
BUT, amber mutations are often not null. In the Ferguson & Horvitz (1985) analysis,
There were non-null amber-suppressible alleles in 5 of 7 loci with amber-suppressbile
alleles [lin-2, lin-18, lin-24, lin-34 aka let-60, let-23; the nulls wre lin-1 and lin-7)
Amber suppressible mutation can be quite bizarre, for example, let-60(n1046) is a missense
that activates the Ras protein and is amber suppressible. let-23(n1045) is ambersuppressible, but is a splice site mutation; one abnormal splice results in an amber codon
at the junction when exons 16 and 18 are inappropriately spliced.
5. Allelic series.
The existence of a series of alleles with different strengths can be used to
understand the function of the gene. In this example + > a > b > c >Df, where
greater than means more activity, and Df = deficiency or deletion.
genotype
a/a
a/b
b/b
b/c
c/c
% wild-type activity
20
15
5
3
1
Infer the null has no activity. Showing that a deficiency (Df) for the locus decreases
activity 2-fold would strongly support this hypothesis:
a/Df
10
b/Df
2.5
c/Df
0.5
Bi190 2005 Sternberg
23
6. Non-complementation screen
If a/Df is viable and has a scorable phenotype, then screen for mutations that fail to
complement a.
mutagen
c
+
+
c
+
+
c
a'
+
+
a
b
F1
F2
c
a'
+
c
a'
+
X
+
a
b
+
a
b
Rare
A non-B
c
+
+
+
a
b
Score C
An example [Aroian & Sternberg (1991)]:
let-23(sy1)/Df is Egl and viable. (Egl is egg-laying defective, the plate phenotype of a
Vulvaless hermaphrodite)
The F1 non-complementation screen:
let-23(sy1)/let-23(sy1) males x rol-6 +/rol-6 + hermaphrodites (EMS mutagenized)
Screen for rare Egl non-Rol. These are + sy1/rol-6 let-23(new)
Are there viable Rol segregants from this?
Of 21,000 F1, 15 alleles: 13 lethal, 2 subviable
Conclude that the null is lethal.
Sequence the alleles: splicing mutations and many point mutations in amino acids
conserved between LET-23 and human EGF-receptor andthe fly top product (DER)
Caveats:
•A Df could also remove a dominant suppressor.
•Null mutations do not have to be the most common.
7. Construction of loss and gain of function mutations
loss of function
Bi190 2005 Sternberg
24
targeted knockout in mouse
one-step gene replacement in yeast
PCR screen for deletions (worms, etc.)
Transposon insertion/excision (flies, plants, worms)
RNAi (worms, etc.)
gain of function
by transformation: promoter/enhancer::cDNA
normal control region with active protein mutant
dominant negatives
Bi190 2005 Sternberg
25
5. Dominance: homeosis & switch genes
Dosage analysis --Muller (1932)
Quantitative or qualitative changes due to mutation?
amorph
also known as null mutation or complete loss-of-function
hypomorph
less than wild-type activity. Phenotype enhanced in trans to a Df. Phenotype suppressed
by multiple copies (h/h/h vs. h/h)
hypermorph
greater than wild-type activity, but qualitatively similar. Phenotype suppressed in trans
to a Df (H/+ vs. H/Df) and enhanced by extra copies of wild-type or the mutant (H/H/+,
H/H/H respectively)
neomorph
altered actiivty, qualitatively distinct from wild type. Wild-type allele behaves as does an
amorph with respect to the neomorphic phenotype.
antimorph
Also known as dominant negatives or dominant interfering mutations, are worse than a
wild-type allele.
Be careful to distinguish dominance and recessivity from nature of the change in gene
activity. There are recessive gain-of-function mutations: some dose-dependent
neomorphs, recessive antimorphs. Often, a mutation has "mixed character," e.g., it is both
hypomorphic and neomorphic,.
There are also cases in which gain and loss of function have the same result. For example,
KAR1 in yeast.
Bi190 2005 Sternberg
26
Dosage studies.
Vary the number of doses of the mutant (m) and wild-type (+) alleles. Df, deficiency.
In general, you want always to have a loss-of-function allele so that you can compare its
phenotype to various gain-of-function effects. To distinguish antimorph and neomorph,
you rigorously need a null to get the direction correct.
Decrease of function
Type of mutation
hypomorph
Dom?
Doses m
Doses +
recessive
m/Df
1
0
more mutant
m/m
2
0
mutant
m/m/+
2
1
wild-type
amorph
recessive
mutant
mutant
wild-type
Doses m
Doses +
m/Df
1
0
m/m, m/+
2 1
0 1
m/m/+, m/+/+
2
1
1
2
hypermorph
dominant
less mut
mutant
more mutant
neomorph
dominant
?
mutant
mutant
antimorph
dominant
more mut?
mutant
less mutant
Gain of function
Doses m
Doses +
m/Df
1
0
m/m
2
0
m/+
1
1
m/m/+
2
1
m/+/+
1
2
dose-dep't neomorph
semi-dom.
less mutant
mutant
mutant
more mut wild-type
dose-dep't antimorph
semi-dom.
mutant
mutant
mutant
more mut wild-type
If a loss of function mutation has a defect in process P, then we conclude that the
gene defined by that mutation is necessary for process P. If a gain-of-function
(hypermorphic or some classes of neomorphic mutation) mutation causes an
alternative outcome to occur, we can infer (although with less confidence than in
the case of loss of function mutations) that the gene is sufficient for process P.
By sufficient we mean all other things being equal -- that is in the background of
all other genes and their effects.
Most often, loss and gain of function mutations are constructed using molecular
genetics.
Bi190 2005 Sternberg
27
There are many cases in which genes have been identified with both gain and loss of
function alleles that have opposite effects on cell processes.
C. elegans
ced-9
let-60
lin-12
lin-14
tra-1
her-1
yeast
cdc2
Drosophila
torso
Antennapedia-complex
Bithorax-complex
per
Bi190 2005 Sternberg
loss of function
gain of function
cells die
vulvaless
VU  AC
precocious
XX males
XX females
cells don't die
multivulva
AC  VU
retarded
XO females
XO males
small cells
cell cycle arrest
no termini
decreased central body
28
lin-12 example: intragenic revertant of a dominant. Reference: Greenwald, Sternberg
and Horvitz (1983) Cell.
A series of semidominant alleles were isolated.
The biology: AC/VU; AC=non-Egl, non AC = Egl. Muv is independent of the signal
Egl=egg-laying defective; Muv=multivulva; AC=anchor cell; VU=ventral uterine
precursor cell. Di=dominant alleles
D1/D1
D1/+
D2/D2
D2/+
D3/D3
D3/+
Egl Muv
Egl Muv
Egl Muv
Egl non-Muv
Egl non-Muv
non-Egl, weak Muv
Intragenic revertant of a dominant.
Given a dominant mutation, one can readily obtain a complete loss-of-function mutation
(assuming a heterozygous deletion of the locus is not lethal). Lifschytz & Falk (1969).
Genetics 62, 353-358. (Ed Lewis says 1940s Oliver did this first)
mutagen
F1
F2
+
A
b
c
+
+
+
A'
b
+
b
+
A'
b
+
A'
b
c
X
c
+
+
c
+
+
non-A
non-C
+
A
b
c
+
+
Score B
EMS-mutagenize unc-32 lin-12(n676)/ unc-32 lin-12(n676)
Pick non-Egl F1: 25/172,000 F1: A frequency of 1.5 x 10-4 or 1/7000 since each F1 has two
mutagenized chromosomes but only 49% of n676/lf are non-Egl.
Revertants are tightly linked: unc-32 lin-12(n676 ) sup/ unc-32 lin-12(n676)
Bi190 2005 Sternberg
29
In one case (n137 n720) the Muv was segregated as a recombinant (n137 +) at low
frequency.
Bi190 2005 Sternberg
30
cis-trans test for allelism of the sup and lin-12.
lin-12 genotype
+/+
n302 / +
n302 n865 / + +
n302/lf
n302 / n302
n302 / n302 n865
% Egl
0
60
0
27
100
22
cis
trans
lin-12(d) aleles are hypermorphs
d allele
n302
n676
n379
d/lf
22
49
1
% Egl
d/+
56
67
7
d/+/+
81
74
33
+/+ . +/+/+ are 0%
Conclusion: lin-12 specifies alternative cell fates
Molecular follow up: lin-12 encodes a transmembrane receptor that along with
Drosophila NOTCH (and worm glp-1) defined a new family of receptors found in all
animals. One can construct activated alleles of Notch/LIN-12 family proteins.
Bi190 2005 Sternberg
31
Homeosis: transformation of one structure into another; in particular, of members of
meristic series.
AntpNs transforms the antennae disc (part of eye-antennae imaginal disc) into leg (ns is
Nasobemia)
Duncan & Kaufman (1975) Genetics 80, 733-752. Nulls of Antp.
Revertants of Ns (Struhl):
EMS-mutagenize Ns males adn mate to marked females, 12/10,000 F1 were linked
revertants. Struhl: AntpNs +Rc1....12
High frequency, let over Df, let over existing lf.
Struhl [Nature (1981) 292, 635-638] made clones of the Antp(lf):
genotype
Ns/+
+/+
-/-
Antennae
(active)
inactive
inactive
Leg
active
active
inactive
Leg2
Ant
Ant
Leg2
Leg2
Ant
Antp thus promotes antennal develop and inhibits leg2 (mesothoracic) development.
Embryonically, Antp(lf) causes a partial T2 to T1 transformation.
Making neomorphs and hypomorphs by transformation
Transformation part 1.
Problem
Get DNA into the germ line.
Replication
Integration
Homologous
worms
inject
nothing special
2nd step
rare
Pax6/eyeless example.
Bi190 2005 Sternberg
32
flies
inject
integrated
yes
no
yeast
shock
ARS
depends
yes, if
C. elegans let-60 example
let-60 ras is a superb example because the genetic analysis came out so cleanly and also
matched the biochemical properties of its product.
Biology: 6 VPCs each competent to respond to an inductive signal and make vulval tissue;
in wt, only 3 do so. Mutants with 0 [Vulvalvess] or 6 [multivulva]
genotype
+
vul
muv
signal phenotype
+
vulva
+ or vulvaless
- or +
multivulva
let-60 was identified in four ways
a. semi-dominant multivulva mutants (lin-34) n1046
b. dominant vulvaless mutants (vulvaless is opposite of multivulva at a cellular level) that
were recessive lethal.
sy100
c. recessive lethal mutations.
s1124
d. recessive, incompletely penetrant lethal and vulvaless alleles. n2034
These mutations all mapped to the same location: Very close (<0.02 mu) the left of dpy-20
on chromosome IV.
Reversion of the dominant vulvaless let-60 allele as a dominant (and with lin-15 in the
background) yielded
a cis-dominant suppressor that is recessive lethal [ dn/+  dn lf/++ ] and
a trans-dominant suppressors that is semidominant multivulva [ dn/+  dn/gf] .
h/lf are more severe lethal and vulvaless.
dn/lf are more severe than dn/+
Maternal effect: dominant suppression of the dn by the gf.
dn/+

dn/dn
Vul
Dead
gf/+

weak Muv
gf/dn

weak Muv
gf/gf
Strong Muv
dn/dn
Vul
Bi190 2005 Sternberg

dn/dn
Dead
33
The inference is that there the classes of alleles are as follows:
gf: semi-dominant multivulva mutants
dn: dominant vulvaless mutants that are recessive lethal
lf: recessive lethal mutations
h: recessive, incompletely penetrant lethal and vulvaless alleles
and that let-60 acts as a switch: the
gf are always ACTIVE
the lf are always INACTIVE
the dn interfere with activation of the wt, but once activated have no effect, since gf
suppresses the dn.
The product of let-60 is the C. elegans homolog of Ras (the oncogene), a "small g protein," i.
e., a GTPase that acts as a molecular switch.
Ras proteins undergo a cycle of guanine nucleotide exchange and hydrolysis:

Ras




Ras•GTP [Active, interacts with Effector]
intrinsic GTPase stimulated by GAP
Ras•GDP
Release of GDP stimulated by GNEF (Exchange Factor)
Ras bound to GTP is active. Ras with either GDP or no guanine nucleotide is inactive.
The gf interfere with the GTPase
The dn interfere with G binding, and thus compete with wild-type for the EF.
Bi190 2005 Sternberg
34
Using Aberrations
Deletions:
it is crucial to have a deletion of the locus to test whether complete loss-of-function is
dominant. Haploinsufficient loci are those in which a deletion has a dominant effect.
Deletions can act as cross-over suppressors.
Duplications:
Attached
Free
Cloned genes can be used for duplicating gene activity: yeast KAR1, C. elegans let-60.
Generation of a duplication:
The concept is to break a chromosome by finding a “dominant suppressor” of a recessive
mutation. The dominant suppressor is a wild-type allele. The problem is that in many
schemes recombination will also give the desired result. Therefore, one uses a cross-over
suppressor.
Irradiate + +/C[a b] and mate with a b/a b, where C[a b] is a cross-over suppressor that
includes the recessive alleles a and b. Screen F1 progeny for Rare A non-B individuals.
These should be a b/C[a b] / Dp(b(+)).
Segmental aneuploids
By crossing overlapping reciprocal translocation one can create strains that lack a
particular region of the genome. …
Bi190 2005 Sternberg
35
6. Complementation analysis
cis-trans test
trans
a
+
a
+
+
b
+
b
Mutant
Wild Type
cis
a
b
a
b
+
+
+
+
Wild Type
Intragenic
complementation
Wild Type
Extragenic
non-complementation
a
+
a
+
b
+
+
b
Wild Type
a
+
+
c
Mutant
Mutant
a
+
+
+
Wild Type
b
+
+
+
c
+
Mutant
+
b
Wild Type
extragenic non-complemenation
tubulin example
Yeast tubulin: two genes for alpha subunits: TUB1, TUB3
One gene for beta subunit, TUB2.
Stearns and Botstein (Genetics 1988):
Bi190 2005 Sternberg
36
Mutagneize TUB2+ , replica mate to TUB2+ to test dominance and tub2cs to test
complementation. Score diploids for Cs phenotype. Non-complementers are Cs- in trans
to tub2cs but Cs+ in trans to TUB2+. Then sporulate to test linkage. They called them
unlinked non-complementers, which is fine for yeast where every locus is essentially
unlinked, but silly if you have a linked “unlinked non-complementer.”
In their first screen they found two tub2 alleles and the tub1-1 allele. By screening for noncomplementers of tub1-1, they found tub2 and tub3 alleles.
Cs = cold-sensitivity
Two types of models for non-complementation:
Dosage model: tubulin is a polymer of - heterodimers. One mutant allele would
eliminate 50% of functional subunits. Two mutant alleles would eliminate 75% of
subunits.
Poison subunit model: mutant subunits disrupt the polymer
intragenic complementation
HIS4 Fink (1966)
Three enzymatic activities
torpedeo (EGF-receptor) Clifford and Schüpbach 1994
Ligand-binding mutations complement kinase-defective mutations
calmodulin ( Ohya & Botstein 1994)
F to A mutations in peptide binding surfaces. A monomeric proteins that binds
many partners
Genotype
Multiple Function Hypothesis
Phenotype
1
2
3
Level Hypothesis
Phenotype
1
2
3
a/a
b/b
null/null
+
-
+
-
-
-
+
-
+
+
-
a/b
b/null b/null +
+
+
-
+
-
-
-
+/+/-
+
+/+/-
On the level hypothesis, the phenotypes 1, 2, 3, … form a phenotypic series.
Bi190 2005 Sternberg
37
On the multiple function hypothesis, alleles a and b exhibit intragenic
complementation.
Bi190 2005 Sternberg
38
Complex locus:
mutations map to the same ;place
mutations give different (often related) phenotypes
complex complementation
The Bithorax complex
[segment/parasegment map]
wt
T1a
T1p
T2a
T2p
T3a
T3p
A1a
...
A8
PS4a
p
PS5a
p
PS6a
p
leg
leg
leg
leg
leg
leg
wing
wing
haltere
haltere
Deletion of the complex transforms T3-A8 into T2
recessive mutations:
bithorax (Calvin Bridges) weak transformation of T3a to T2a
bx
pbx
bxd
iab-2
iab-5
iab-8
T3a
T3p
A1a



T2a
T2p
T3a
anterior halter into wing
posterior haltere into wing
get an extra leg
pbx and Cbx arose on the same chromosomal location, 3R, 89B-F.
pbx is s deletion of 17 kb. Cbx is inerted insertion of the same 17 kb more 5'
pbx:
T3p  T2p posterior haltere into wing
Cbx
T2p 
T3p wing into haltere
DfP9 deletes the complex: T3-A8 T2
Now, add back by Duplication, pieces and raise the level of development from T2 towards
more posterior segments.;
loss of function results in posterior anterior transformations
In complementation analysis, there is more posterior predominance:
Bi190 2005 Sternberg
39
bx3 +/ + pbx has a weak pbx phenotype rather than the bx phenotype
as if the posterior defect is the more extreme. Polarized effects.
Heterozygotes look like the more posterior defect.
Ubx fails to complement bx, pbx and bxd, and has the phenotype that is the sum of three.
Ubx is dominant because of haplo-insufficiency.
abx/bx
~ bx/bx
bxd/pbx
~ pbx.pbx
Ubx/bxd
~bxd/bxd
Screens:
viables (Lewis)
Hab Uab
Mcp
abx bx bxd pbx iab-2 iab-3 iab-4 iab-5 ... iab-9
[ ubx
] [ abd-A
]
[Abd-B
]
lethals: three complementation groups: Ubx, abd-A, Abd-B
reversion of dominants: Mcp A4  A5 and iab-5 has A5  A4.
Global rearrangements Cbx Ubx / ++  Cbx Ubx /R( ++)
Bi190 2005 Sternberg
40