Classical genetics
•  Monofactorial cross, ie.
A (wt) x a (mut)
–  2:2 pattern of segregation
–  if 4:0 -> mitochondrial
•  Complementation / non-compl. (in Dipl.)
movie
Tetrad dissection
Micromanipulator
for Tetrad Dissection
Movie
Tetrad analysis
movie
Spore/Row
A
B
C
D
Tetrad No
1
2
3
4
5
6
Genetic Linkage
Bifactorial Cross
Possibilities:
AB x ab
LEU2 ura3∆ x leu2∆ URA3
1) A and B are on the same chromosome
- no crossover
- one crossover
- double crossover
2) A and B are on different chromosomes
- no crossover between marker and cen
- one crossover between marker and cen
- two crossover between marker and cen
Meiosis -> 4 spores
•  Yeast tetrad analysis is nothing else than
just watching the outcome of meiosis
•  Diploid 2n, two sets of chromosomes
•  DNA is replicated resulting in two
chromosomes with two identical
chromatids each
•  The chromosomes align and can undergo
recombination
•  The first meiotic division will separate
the chromosomes from each other
•  The second meiotic division will separate
the chromatids, ie. each spore
represents essentially one chromatid
The outcome of a cross
1a) LEU2 and URA3 are close together on the same
chromosome
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 ura3
LEU2 ura3
leu2 URA3
leu2 URA3
• If there is no crossover occurs between the two markers
all haploid spores will look like the parental haploid strains
• There are only two different types of spores, i.e. (leuplus ura-minus) and (leu-minus ura-plus) spores
• Hence such a tetrad is called a parental ditype PD
Crossover
1b) LEU2 and URA3 are close together on the same
chromosome and a crossover occurs between them
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 ura3
LEU2 URA3
leu2 ura3
leu2 URA3
• In this case we will get spores that look like the
parental haploids but also spores that have new
combinations of the two markers
• There are four different types of spores
• Hence such a tetrad is called a tetratype T
Double crossover
1c) LEU2 and URA3 are close together on the same
chromosome and two crossover occur between
them such that four DNA strands are involved
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 URA3
LEU2 URA3
leu2 ura3
leu2 ura3
• In this case we will get only spores that look different
from the parental haploids
• There are two different types of spores
• Hence such a tetrad is called non parental ditype NPD
The probability of a crossover
depends on gene distance
LEU2
ura3
LEU2
ura3
leu2
URA3
URA3
• One crossover -> T
• Double crossover -> NPD
• No crossover -> PD
leu2
LEU2 URA3
LEU2 URA3
leu2 ura3
leu2 ura3
• Thus with close linkage: PD > T > NPD
and the relative numbers can be used to map genetic
distances.
For mapping one analyzes hundreds of tetrads from the same cross. This
has been done extensively in the past and the last genetic map from
1995 comprised about 1,000 locations
cM is a measure for genetic
distance between two markers
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 URA3
LEU2 URA3
leu2 ura3
leu2 ura3
To generate new combination of mutations (such as leu2
ura3) one will have to dissect the more tetrads the closer
the two genes are, and this can be estimated based on the
physical distance (in kb), which relates well to the genetic
distance (in cM, centi Morgan). For two close genes (1cM,
i.e. 1% recombinant spores) one would have to dissect at
least 25 tetrads, statistically
Thomas Hunt Morgan
•  Drosophila geneticist, Nobel price 1933
•  Centiomorgan, cM, unit for genetic
distance
•  1 cM = 1% recombination, 1 in 100 meiotic
divisions
•  Genetic Linkage Map (Alfred Sturtevant)
•  Yeast 1 cM ca. 3 kb (200-fold less than
animal cells, human 1 Mb)
1866-1945, USA
Crossing with markers on
different chromosomes
2) LEU2 and URA3 are on different chromosomes
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 ura3
LEU2 ura3
leu2 URA3
leu2 URA3
LEU2 URA3
LEU2 URA3
leu2 ura3
leu2 ura3
• Different chromosomes assort randomly in the first
meiotic division
• For this reason two types of tetrads become equally
frequent, the parental and the non-parental ditype, PD
and NPD
• Hence, linked and unlinked genes can easily be
distinguished in tetrad analysis because with unlinked
genes PD = NPD while with linked genes PD>>NPD.
Crossing with markers on
different chromosomes
2b) LEU2 and URA3 are on different chromosomes and a crossing
over occurs between a centromere and a marker
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 ura3
LEU2 URA3
leu2 ura3
leu2 URA3
• Now the different alleles of URA3 will only be separated in
the second meiotic division
• The result is a tetratype tetrad T
• The above situation means also that if markers are distant
from the centromere many Ts will occur while if both
markers are close to the centromere few Ts will occur.
Crossing with markers on
different chromosomes
LEU2
ura3
LEU2
ura3
leu2
URA3
leu2
URA3
LEU2 ura3
LEU2 URA3
leu2 ura3
leu2 URA3
• What is the outcome of double cross-overs with four or with
three strands?
• Due to the possibility of double cross-overs the proportion
between different tetrad types for unlinked genes that are
not centromere-linked becomes 1:1:4 for PD:NPD:T
• This also means that one out of four spores will be
recombinant, i.e. in order to obtain the new combination of
genes (leu2 ura3) one only needs to dissect one tetrad,
statistically
Frequency of Tetrades
AB x ab
If A and B are linked
•  PD > NPD
•  The distance of the markers A B determines
the number of PD:NPD:T
•  cM = (100/2)[(T+6NPD)/(PD+NPD+T)]
•  The number of T is proportional to the map
distance
Recombination
•  General recombination = Homologous recombination
•  Order of events (I meiotic div):
–  1) Double strand break
–  2) Two chromatids from different chromosomes
(maternal/paternal) cross over
–  3) Base pairing and strand invasion
–  4) Synthesis and gap repair
–  5) Resolution of Holiday junction
Resolution of
Holiday
junction
Gene
conversion
Gene
conversion
Mutagenesis
1.  Conventional mutagenesis (MMS, EMS, UV,
Transposon)
2. Screen through the knock-out collection
or other type of mutant collections
For a successful mutagenesis approach one needs a selection
or screen to isolate mutants of interest
Need to test whether the phenotype of interest
is due to a single mutation - tetrad analysis.
Recessive
Dominant
Dominant negative
Remember, since yeast can be easily propagated as
interconvertible haploids and diploids, studying recessive
gene mutations is straightforward.
Characterizing mutants
1.  Is the mutant phenotype due to a defect in one gene ?
=> backcross to wt, 2:2 segregation ? If yes, one gene
2.  Is the mutation dominant or recessive ?
3.  Place the mutants into complementation groups. Usually
one complemetation group is equivalent to one gene
4.  Identifying/cloning the gene by complementation, or
perform genetic linkage analysis to map the gene
Dominant and recessive
mutations
•  The dominant or recessive character is
revealed by crossing the mutant with the
wild type to form a diploid cell
•  Such diploids are heterozygous, because
one chromosome carries the wild type
allele and the other one the mutant allele
of the gene affected
Recessive:
wild type phenotype
MUT1
MUT1
mut1
•  A mutation is dominant when the mutant
phenotype is expressed in a
heterozygous diploid cell. The diploid has
the same phenotype as the haploid
mutant
•  A mutation is recessive when the wild
type phenotype is expressed in a
heterozygous diploid cell. The diploid has
the same phenotype as the wild type
mut1
Dominant:
mutant phenotype
Dominant and recessive
mutations
•  A dominant character can have a number of important
reasons, which may reveal properties of the gene
product’s function:
–  The mutations leads to a gain of function, e.g. a
regulatory protein functions even without its normal
stimulus
–  The gene product functions as a homo-oligomere and
the non-functional monomere causes the entire
complex to become non-functional
–  The gene dosis of one wild type allele is insufficient to
confer the wild type phenotype, i.e. there is simply not
enough functional gene product (this is rare)
Dominant and recessive
mutations
• The recessive character of a mutation is usually due to loss of
function of the gene product
• This means that recessive mutations are far more common,
because it is simpler to destroy a function than to generate one
• Further genetic analysis of the mutant depends on the
dominant/recessive character, that is one reason why this step
is taken first
• In addition, it is useful to do a tetrad analysis of the diploid in
order to test that the mutant phenotype is caused by a single
mutation, i.e. that the phenotype segregates 2:2 in at least ten
tetrads studied; this is important when mutations have been
induced by mutagenesis
Intragenic complementation
•  Intragenic complementation is rare,
but is does occur
•  The heterozygous mut1-1/mut1-2
however shows a (partial) wild type
phenotype
•  The occurence of intragenic
complementation means that the gene
product must be an oligomere
•  The ”opposite”, non-allelic noncomplementation, can of course also
occur: two recessive mutations in two
different genes fail to complement.
This occurs sometimes when the gene
products are involved in the same
process or complex and the two
functional alleles are just not enough
to confer full functionality
No functional gene product
of MUT1
mut1-1
mut1-1
mut1-2
mut1-2
But a heteromere
consisting of
Mut1-1p and Mut1-2p
can be functional