Dr' Chris Rongo

1
Dr. Chris Rongo rongo_at_waksman.rutgers.edu GENETIC
S 502 C. elegans Section, Lecture
2 1000-1125, Waksman Auditorium Tuesday, March
3, Spring 2009 Todays Lecture Complementation
Linkage Analysis Two-factor
Mapping Three-factor Mapping
2
Genetic dissection of biological
processes
Wild type animal
WT
WT
WT
mutant

WT
WT
WT
WT
1) Identify a process to study
GOAL
2) Mutagenize
To obtain mutants that are defective in a
specific biological process. Mutant screens allow
researchers to identify the molecular players
that underlie a given biological process. By
discovering these players, researchers obtain an
experimental handle on the gene or process.
3) Screen for a phenotype
4) Define genes by complementation testing
5) Map gene
6) Clone gene
3
Complementation tests
Q When do you use this?
Note requires recessive mutations
mut-1
Complementation the production of a wild-type
phenotype when two different mutations are
combined in trans. The two mutations are probably
in different genes.
WT phenotype
mut-3
mut-1
Noncomplementation or failure to
complement between two mutations usually
indicates that the two mutations are in the same
gene.
Mut phenotype
mut-2
4
Is the new unc mutation an allele of the gene
unc-44?
unc

P0
X
unc

unc-44(e1111)
unc
F1
X

unc-44(e1111)
unc
F2

unc

unc-44(e1111)
unc-44(e1111)
unc-44(e1111)
unc-44(e1111)
Unc or WT?
Unc or WT?
WT
WT
50
If unc is an allele of unc-44 _____ of progeny
are Unc.
If unc complements unc-44, it is probably not a
mutation in the same gene.
Q What do you do next if unc fails to
complement unc-44? What molecular experiment
could you perform to determine whether it is a
new allele of unc-44?
5
a
Complex Complementation Tests
Intragenic complementation the production of a
wild-type phenotype when two mutations in the
same gene are combined in trans.
Easy to explain for proteins that have more than
one functional domain
Mutation 1 in Protein AB
Alone, they lack
either
A
or B activity
.
A
B
Mutation 2 in Protein AB
Combined in trans,
they have both
A
and
A
B
B activity
.
Keep in mind that there are other ways that
intragenic complementation can occur
(polycistronic messages, alternative splicing, et
cetera.)
6
Complex Complementation Tests
Non-allelic non-complementation the production
of a mutant phenotype when two mutations in two
different genes are combined in trans.
Two recessive mutations involved in
embryogenesis emb-1 emb-4. emb-1 maps to
chromosome I emb-4 maps to chromosome III
GENOTYPE PHENOTYPE emb-1/emb-1
/ mutant emb-1/ / WT /
emb-4/emb-4 mutant / emb-4/ WT Seems
normal for two recessives, but
emb-1/ emb-4/ mutant A wild-type copy of
each gene is present, yet the animal still has a
mutant phenotype.
7
Threshold Levels model for Non-allelic
Non-Complementation
Perhaps EMB-1 and EMB-4 dimerize to form a
functional protein, and the mutations make
non-functional subunits.
/ / 100 Level of
Heterodimer emb-1/ / 50 Level of
Heterodimer / emb-4/ 50 Level of
Heterodimer emb-1/ emb-4/ 25 Level of
Heterodimer
If the threshold limit for the level of the
heterodimer in a cell is 50 before you see a
phenotype, then any genotype combination that
makes less than 50 will show a mutant
phenotype. Heterozygosity for either gene alone
gives 50, which is enough. Heterozygosity for
both genes gives 25, which is not enough, so
mutant phenotype appears.
8
Poison model for Non-allelic
Non-Complementation
Perhaps EMB-1 and EMB-4 dimerize to form a
functional protein, and the mutant proteins, when
together, form a nonfunctional protein that
can interfere with the function of wild-type
dimers.
/ / no Poisoned Heterodimers emb-1/
/ no Poisoned Heterodimers /
emb-4/ no Poisoned Heterodimers emb-1/
emb-4/ 25 are Poisoned Heterodimers For
example, perhaps EMB-1/EMB-4 heterodimers are
subunits in a polymer chain. A poisoned
heterodimer might be able to add itself to the
end of the chain, but not allow
additional subunits to be added (i.e., chain
terminator).
9
Linkage analysis in C. elegans
Q Why do geneticists map genes?
How to begin mapping

• Cross your mutation with
• known markers on the six
• chromosomes.
• Genes on C. elegans
• chromosomes tend to be in
• clusters, making the
• assignment of a new
• gene to a chromosome
• easier.

10
Mendels First Law The Law of Segregation

• Alternative versions (alleles) of genes account
  for variation during heredity.
• For each inherited characteristic (gene), an
  organism inherits two alleles, one from each
  parent.
• If two alleles differ, then the dominant allele
  is fully expressed for the characteristic,
  whereas the recessive is not expressed.
• Two alleles for a given gene segregate during
  gamete production.

1. Alleles are on homologous chromosomes.
2. They line up at metaphase.
3. Each homolog goes into a different gamete.
4. You are drawing Metaphase I every time you
write a genotype.
11
a
a
a
a
If two genes are closely linked...
unc-5
mut
P0
X
mut
unc-5
unc-5
F1
mut
unc-5
unc-5
mut
F2
mut
mut
unc-5
The mutations repel or exclude each other in the
F2 generation
1/__ are Unc non Mut
1/__ are Mut non Unc
4
4
Question Do you ever get worms that are both
Unc and Mut?
If perfectly linked, then never. If closely
linked, then rarely (recombinants)
12
Punnett Square For Closely Linked Genes
F1 unc-5/mut
Gametes From Sperm
Gametes From Oocytes
13
What if the genes are unlinked?
More help from the 19th century During gamete
formation the segregation of the alleles of one
gene is independent of segregation of the alleles
of another gene. - Gregor
Mendel, 2nd Law 20th century update This is
true IF the genes are on separate chromosomes
or are far apart on the same chromosome. -
Barbara McClintock, Thomas Hunt Morgan and
others.
14
If two genes are unlinked...

mut
unc-5

X


P0

mut
unc-5

unc-5

F1

Self fertilize

mut
1 / /
F2
2 / /mut
9 nonUnc nonMut
2 unc-5/ /
4 unc-5/ /mut
1 / mut/mut
3 Mut nonUnc
2 unc-5/ mut/mut
1 unc-5/unc-5 /
3 Unc nonMut
2 unc-5/unc-5 /mut
1 Unc Mut
1 unc-5/unc-5 mut/mut
a
a
a
15
Punnett Square For Unlinked Genes
F1 unc-5/ mut/
Gametes From Sperm
Gametes From Oocytes
16
unc-5

F1

Self fertilize

mut
If you pick Unc F2s, what are also Mut?
25
If you pick Mut F2s, what are also Unc?
25
Why?
Unlinked mutations sort independently.
unc-5
F1
Self fertilize
mut
0
If you pick Unc F2s, what are also Mut?
If you pick Mut F2s, what are also Unc?
0
Why?
Linked mutations repel each other.
17
Mapping takes advantage of two kinds of
information
Physical map
Genetic map
18
C. elegans genome is about 20,000 genes,
97Mb Molecular and genetic data is stored in
public databases http//www.wormbase.org/ http//
elegans.swmed.edu/
The genome project generated large clones that
cover the entire genome
Cosmid 30KB insert (acts like a phage and
plasmid) can contain 5-20 genes. YAC
Yeast artificial chromosome 250KB insert
contains hundreds of genes
19
The genetic (left) and molecular (right) maps can
be aligned.
20
Connecting the genetic map to the molecular map
mut
()
mnDf10
()
mnDf100
(-)
mnDf101
(-)
mnDf102
unc-3
genetic map
molecular map
SNP8
SNP9
YACs

Cosmids
Genes
21
Deficiency Mapping
Deficiency A deletion of a portion of a
chromosome
GenotypemnDf22/mnC1 dpy-10(e128)
unc-52(e444)II. Description Hets are WT and
segregate WT, paralysed DpyUnc and dead eggs.
Maintain by picking WT. Balancer chromosome
suppresses recombination (above, mnC1).
Breakage
X-rays or gamma rays
1 2 3 4
1 2 4
22
The balancer is marked with two recessive
mutations unc-3 (which is within the deficiency
and therefore uncovered ) lon-2 (which is
outside of the deficiency, but closely linked so
that recombination rarely occurs)
Self fertilization results in the following
progeny mDf10/mDf10 lethal Q Why are Dfs
lethal? mDf10/unc-3 lon-2 Unc unc-3 lon-2/unc-3
lon-2 Unc Lon
23

mut
P0
X
mut

mnDf10
mut
F1
X

unc-3 lon-2
mut

F2

mut
mnDf10
unc-3 lon-2
mnDf10
unc-3 lon-2
phenotypes
Mut or WT
WT
WT
WT
F3
phenotypes
____
____
____
____
1/4 Mut 1/4 Dead
1/4 Mut 1/4 Unc Lon
1/4 Dead
1/4 Unc Lon
24
MEIOSIS IN C. ELEGANS Special Feature
Holometabolous chromosomes (multiple centromeres)
pairing
DNA
meiosis
replication
Division I
recombination
meiosis
Division 2
Gametes products of meiosis
25
Two Factor Mapping

• When do you use it?
• - How far is the gene from a known marker?
• To confirm or establish which chromosome the
  gene maps to.
• - When you want to generate more recombinants for
  mapping.

dpy-9(-)
mut() unc-34(-)

mut(-) unc-34()
dpy-9()
25
Pick Dpys Unc _______
0 if perfectly linked, gt0 if partially linked.
Pick Uncs Mut _____
25
Pick Dpys Mut _____
26
Two Factor Mapping How far is a gene from a
marker?
b c

(BC)
X
P0
(WT)
b c

b c
F1

Parental Gametes
Recombinant Gametes
c
b

b c
a
27
MAPPING where is the gene?
Linkage MAPPING what is the gene close to or
linked to?
dpy-14
unc-31
dpy-9
6
-27
12
ced-2
-19
How often does recombination occur between dpy-9
ced-2?
8
How often does recombination occur between dpy-9
dpy-14?
39 (not accurate)
Which mutations will show linkage to dpy-9?
ced-2
28
Three Factor Mapping Where Is A Mutation
Compared To An Interval?
unc-31
dpy-14
F1 Heterozygote
mut
unc-31
dpy-14
Recombinant Gamete
mut
Unc non-Dpy
non-Unc Dpy
mut
unc-31
dpy-14
F2 Recombinant /Parental
unc-31
dpy-14
unc-31
dpy-14
1/2 Unc 1/4 Unc Dpy 1/4 Unc Mut
3/4 Dpy 1/4 Unc Dpy
F3
Conclusion All Unc non-Dpy F2 segregate Mut F3,
and no non-Unc Dpy F2 segregate Mut F3 thus,
the mut gene is to the right of the unc-31dpy-14
interval.
29
Three Factor Mapping Where Is A Mutation
Compared To An Interval?
unc-31
dpy-14
F1 Heterozygote
mut
unc-31
dpy-14
Recombinant Gamete
mut
Unc non-Dpy
non-Unc Dpy
mut
unc-31
dpy-14
F2 Recombinant /Parental
unc-31
dpy-14
unc-31
dpy-14
1/4 Unc Dpy 3/4 Unc
1/2 Dpy 1/4 Unc Dpy 1/4 Dpy Mut
F3
Conclusion No Unc non-Dpy F2 segregate Mut F3,
and all non-Unc Dpy F2 segregate Mut F3 thus,
the mut gene is to the left of the unc-31dpy-14
interval.
30
Three Factor Mapping Where Is A Mutation
Compared To An Interval?
unc-31
dpy-14
F1 Heterozygote
mut
unc-31
unc-31
dpy-14
dpy-14
Recombinant Gamete
mut
mut
Unc non-Dpy
Unc non-Dpy
non-Unc Dpy
non-Unc Dpy
F2 Recombinant /Parental
unc-31
mut
unc-31
dpy-14
dpy-14
mut
unc-31
dpy-14
unc-31
dpy-14
unc-31
dpy-14
unc-31
dpy-14
1/4 Unc Mut 1/2 Unc 1/4 Unc Dpy
3/4 Unc 1/4 Unc Dpy
3/4 Dpy 1/4 Unc Dpy
1/4 Dpy Mut 1/2 Dpy 1/4 Unc Dpy
F3
Conclusion Some Unc non-Dpy F2 segregate Mut
F3, and some non-Unc Dpy F2 segregate Mut F3
thus, the mut gene is within the unc-31dpy-14
interval.
31
For example, you mate your marker strain, bc/bc,
to males that are homozygous mutant for your gene
of interest, a/a.
The F1 cross progeny are wild-type in phenotype,
and genotypically bc/a.
You collect 19 recombinants from the F2
generation by isolating 10 B nonC and 9 C nonB.
Of the 10 B nonC, 9 picked up a. The ratio 9/10
is large so a is far away from b.
Of the 9 C nonB, 2 picked up a. The ration 2/9
is small so a is close to c.
b c
a
9
1
B nonC

7
2
C nonB
16
3
T
otal
a
32
How can you use this information to figure out a
map position?
b
a
c
-15
?
-20
recombinants (B nonC) recombinants ( C
nonB) recombinants Total Total as of
Interval
9 7 16 16/(163)84
1 2 3 3/(163)16
Size of Interval (in cM) (-20) - (-15) 5 cM
Gene a is to the right of gene b, at a distance
that is 84 of the b-to-c interval. This
distance is (0.84)(5) 4.2 cM to the right of
gene b. Thus, position of gene a (-20) 4.2
-15.8
33
What if a is so close to c that you dont get any
recombinants?
Q How would you figure out if a is close or
to the right?
34
Mapping with SNPs (Single Nucleotide
Polymorphisms)
A
T
Bristol, WT
N2
T
C
Hawaiian
HA
Shotgun cloned about 5 Mbp (11,000 clones) of
Hawaii
6222 potential polymorphisms 4670 SNPs 1552 small
insertions/deletions
1 polymorphism/873 bp aligned sequence- only 5
Mpb out of 97 Mbp were aligned.
Q Are there other SNPs in the genome?
35
SNIP/SNPs
A
Bristol, WT
GAATTC
N2
T
GAACTC
Hawaiian
HA
EcoRI Restriction Enz. sequence
GAATTC
Of the 6222 Hawaiian polymorphisms, 3457 produce
RFLPs (Restriction Fragment Length
Polymorphisms) SNPs that also create RFLPs are
called snip-SNPs.
So far, about 500 confirmed snip-SNPs
1 confirmed snip-SNP/220 Kbp (0.6 cM on
genetic map)
36
SNP
Genomic DNA
PCR product
PCR with SNP-specific primers from Bristol (N2)
strain
PCR with SNP-specific primers from Hawaiian strain
RsaI
PCR Products
Digested PCR products
37
Clone F59H5 at position 23175 Variation
Probability VarSGT Psnp0.9430 CB4856 read
vc87b04.s1_at_270,g,62 Verified Yes 5' primer
TGCTCTTCCTCCTGAACTTC 3' primer
CAGGATGTTTTCGTCTGGAC enzyme used ApoI N2
digest RAATTY ApoI, AcsI, FsiI, XapI HA digest
none -- -- TCCTCCGACACTGGTACTCCAACATCAGCGATCATAG
CTTGTCCTTCTCT CGAACGTCTTTCGACGTGATCGATAACTTTTCAGC
TTGTCCTTCTAATCC CGAAAGTCTTTCGACGTGGTCGATAACTTCTCC
TGGTCGCtgctcttcct cctgaacttcTCCTCCTTGCTGTTGACGAAG
ATCCAAAATCCAACGGACT CTGAATGTCCCGAAGCCAACTCCATCGTC
CGACGTTGCTGCTTCTCTGAT GTACGACGTCTTCCGATCTCCGCTGCT
GGTGATCCTCAGAGCGCACCGTC CTGATCCATCCCTTCGTGCGTACAG
CACGACACTCCGGGGGAAAAAAGGC TCGAAATTCAAAGGAATTTTGAG
TGGACCAAATTGTACAGAACGTTGCAA ATGTGCTCGGATCCACGACGA
CATCCGTATTTCTTGTCATCTTCGACTGT TCCTCTTCGTCTTCTGGAG
TCATAACTTGTGCACGTTGAATCGGAAAATT T/GTTGTTCTTTTTA
CCGGAATTTGAACTTCTCGCAATGTGTAGACTT CTGCATTCGATGCCA
ATTCCAACGAAATTAACTCCATCTTCGGAATGGTA GATTCTGCTCCAC
TATTTGTCGGCTTTGTCCTCGCTTTTCGACATCACCA GTGGCATCCCG
ATAAAGCCACCCGGATATGAAAAATCCTCTTGATTCTTG TACAATTGA
AGAATTTCTCCTTGTGACTCCGCCAACATCGTCACATTCAT CATGTAG
AGATTCGTCTTGATTTGCTCCTTGATTTGATGCGGAAAATGTC GATTC
CACGAAACGACTTCTATCCATTGAAAGCTGCTGTCTGATTGACAA ATC
GTCCAGCACAATCACATCTTCAAGCCAAGGCGGGTGATAAATCCATC G
ATTTTTCGATATCATCCTTTGCCTGCCGAACATCCAACTTCCAACTACA
TCTTTGAATTCCACGAAACGACACGACGAAATACTTTCCATCCAGCAAA
A TTCCA
38
Combining Three Factor Mapping With SNPs
X
unc-31 mut dpy-14 (from N2 background)
Hawaiian (HA)
P0
unc-31 mut dpy-14 HA
F1
F2
Unc non-Mut non-Dpy
Unc Mut non-Dpy
non-Unc non-Mut Dpy
non-Unc Mut Dpy
1
2
3
4
5
6
recombinant
unc-31 unc-31 mut dpy-14
unc-31 mut unc-31 mut dpy-14
dpy-14 unc-31 mut dpy-14
mut dpy-14 unc-31 mut dpy-14
parental
1
2
3
6
5
4
unc-31
mut
dpy-14
unc-31
mut
dpy-14
SNP
SNP
39
Combining Three Factor Mapping With SNPs
1
2
3
6
5
4
N2
unc-31
mut
dpy-14
unc-31
mut
dpy-14
SNP
SNP
HA
F2 Rec.
Genotype
recombinant (N2 HA)
unc-31 HA unc-31 mut
N2 dpy-14
1
parental (N2)
unc-31 mut HA unc-31 mut
N2 dpy-14
2
unc-31 mut N2 unc-31 mut
N2 dpy-14
3
PCR/Digestion of F2 Worms
HA dpy-14 unc-31 mut
N2 dpy-14
1
2
3
6
5
4
4
N2
HA
N2 dpy-14 unc-31 mut
N2 dpy-14
5
mut N2 dpy-14 unc-31 mut
N2 dpy-14
6
40
Using SNPs To Map A Mutation

• Steps
• Make a table that represents the recombinant
  chromosome for each of your recombinants.
• Label each marker and SNP according to whether it
  came from the N2 or the HA chromosome.
• Label whether the recombinant chromosome contains
  the mutation.

F2 Rec.
Genotype
recombinant (N2 HA)
unc-31 HA unc-31 mut
N2 dpy-14
1
parental (N2)
unc-31 mut HA unc-31 mut
N2 dpy-14
2
unc-31 mut N2 unc-31 mut
N2 dpy-14
3
HA dpy-14 unc-31 mut
N2 dpy-14
4
N2 dpy-14 unc-31 mut
N2 dpy-14
5
mut N2 dpy-14 unc-31 mut
N2 dpy-14
6
41
Using SNPs To Map A Mutation

• Steps
• 2. Mark the crossover on the table.

42
Using SNPs To Map A Mutation

• Steps
• Figure out on which side of the crossover the
  mutation lies
• If the recombinant chromosome contains the
  mutation (i.e., the recombinant animal has the
  mutant phenotype), then the mutation is on the N2
  side of the recombination site.
• If the recombinant chromosome does not contain
  the mutation (I.e, the recombinant animal does
  not have the mutant phenotype), then the mutation
  is on the HA side of the recombination site.

The arrows point to where the mutation is
relative to the recombination site
43
Using SNPs To Map A Mutation

• Steps
• There will be at least two critical recombinants
  one that defines the left hand border of where
  the mutation lies, and one that defines the right
  hand border of where the mutation lies.
• For the left hand border, find the recombinant
  that has a rightward arrow that is the
  furthest to the right.
• For the right hand border, find the recombinant
  that has a leftward arrow that is the
  furthest to the left.

For the left hand border, recombinants 1 and 6
are the same between unc-31 and the SNP. Thus,
the left hand border is within this interval. For
the right hand border, recombinants 2 and 5 are
the same, and more to the left than the arrows in
recombinants 3 and 4. The site for recombinant 2
and 5 falls between unc-31 and the SNP. Thus,
the right hand border is within this
interval. Conclusion the mutation falls
somewhere between unc-31 and the SNP.
44
Which mapping method do you use?
Some considerations Deletion mapping Also
tells you if your mutation is recessive,
etc. Falling out of fashion with the rise of
powerful SNP mapping.
Mapping with genetic markers Generates marked
strains that can be used for other types of
mapping. No PCR problems.
Mapping with molecular markers (SNPs) In theory,
the single fastest method. Makes it possible to
follow dominant alleles.
45
Practice Problem, part a
You are mapping a new ced (cell death defective)
mutation.
46
Practice Problem, part b,c
b. You decide to use SNP mapping to narrow down
where ced is. You generate a dpy-24 ced unc-75
triple mutant, and cross it to the Hawaiian
strain. You collect Dpy nonUnc recombinants that
are also Ced in phenotype.

R
e
c
o
m
b
i
n
a
n
t
s
c. You tak
e
5

of

your

Dpy

D
p
y
n
o
n
U
n
c
C
e
d
nonUnc recombin
a
nts

that

N
2
H
A
1
2
3
4
5
have
ced
and make
DNA

from the Ced animals. Then
you perf
o
rm

PCR

with

S
N
P
-
S
N
P
1
specific primers, digesting the

DNA with the correspond
i
ng

enzyme.

R
e
c
o
m
b
i
n
a
n
t
s
D
p
y
n
o
n
U
n
c
C
e
d
H
A
1
2
3
4
5
N
2

S
N
P
2
W
here is
ced
relative to the

two SNPs?

Which recombina
n
ts

show

this?


What would be your next
mapping exper
i
ment?