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
Bi190 2005 Sternberg1
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.
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.
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 MATThese 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., 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:
Capital for the dominant or co-dominant
URA3 vs ura3 or ura3-52MATa and MAT
a yeast cross:
ahis3 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:
PD, parental ditype
NPD, non=parental ditype
TT, tetratype
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
[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 aSTE 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 genotypemating phenotype
+-mater
+-mater
aste?
aste?
A TT inferred from the mating phenotype would indicate an -specific sterile: e.g.,
mating phenotypespore genotype
a-matera+
a-materaste
non-materste
-mater+
1.7 The results from the Mackay and Manney screen and others revealed three classes of sterile mutations:
The -specific genes include
STE3receptor (G-protein coupled) for a-factor
STE13 dipeptidyl aminopeptidase (processing of -factor)
KEX2(processing of -factor)
The non-specific genes include:
STE4G protein subunit
STE5scaffold for MAP kinase cascade
STE7MAP kinase kinase
STE11MAP kinase kinase kinase
STE12transcription factor
STE18G protein subunit
STE20protein kinase
The a-specific genes include:
STE2receptor (G-protein coupled) for -factor
STE6secretion 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)
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
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 mat1/mat2 diploid.
1-5/2-1
Dissect 77 tetrads with 4 viable spores
764 Ste- : 0 Ste+
12 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
phenotypeinferredgenotype
-mater++
Alf
Ste
Ste+
Molecular analysis then revealed that in an 1- mutant, -specific genes are off
and in an 2-, a-specific genes are inappropriately on
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.
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: Unc non-Lin
protein: LIN-3
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:
The 100 Mb of genome sequence indicated about 19,000 predicted genes. This is a gene density of one gene per 5 kb.
chr.Mbmu
I1450
II1645
III1255
IV1850
V2250
X2045
total100300
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
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):
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.
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
Generation of vulval precursor cells [VPCs]
unc-83, unc-84migration of VPC parents
lin-24, lin-33death of VPCs
Induction of VPC fates
lin-3Inductive signal (EGF/TGF-)
let-23Receptor (EGF-receptor)
lin-34gf allele of let-60 (Ras)
lin-2, lin-7, lin-10localize the LET-23 protein
lin-1, lin-25, lin-31transcription factors
Regulators of LET-23 signaling:
lin-8, lin-9, lin-15, lin-13
Another signal
lin-12Founding member of Notch/LIN-12 family of receptors
Timing of vulval development (and other events)
lin-4RNA that inhibits lin-14 mRNA expression
lin-14putative transcriptional regulator
VPC fate execution
lin-11Founding member of LIM domain transcriptional regulators
lin-17Wnt receptor (Drosophila Frizzled)
lin-18Receptor protein
Target size problems: lin-15 requires 2 mutations or a deletion; lin-4 encodes a small RNA;
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=1P(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 0! = 1 and e0 = 1 )
P(0) = e-m ; P(≥1) = 1 - P(0)
2. Only ≤1 per site1. At least one per site
P(0) + P(1) = e-m + e-m m1 / 1!P(≥1) = 1 - e-m
m e-m + me-m= [e-m(1+m)]m1-e-m
12e-m = 0.73610.632
0.50.9120.865
0.30.9630.95
0.10.995 40.98
0.010.9999550.99
100.99995
For the Poisson distribution, the standard deviation = m and the
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 orangeCyO is curly of Oster
cn is scarlet
pr is purple
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.)
Chromsome 2 screen
alleles/locus
1 2 3 4 5 6 7 8 9 101112131415161718
13 13 7 8 3 5 1 1 5 0 0 1 1 0 1 0 1 1
#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,