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Genetics 2011
Outline
• Part 1 Linkage and meiotic recombination
Lecture 5
• Genes linked together on the same chromosome
usually assort together.
together
• Linked genes may become separated through
recombination.
Linkage, Recombination
Linkage
Recombination,
and the Mapping of Genes
on Chromosomes
• Part 2 Mapping
• Recombination data can be used to generate
maps of relative locations of genes on
chromosomes
ᯘ⩶ె ⪁ᖌ
• Part 3 Mitotic recombination
• Rarely, recombination occurs during meiosis.
• In eukaryotes mitotic recombination produces
genetic mosaics.
mosaics
http://lms.ls.ntou.edu.tw/course/136
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2
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Independent assortment
Linkage
g
Syntenic genes: two genes on same chromosome segregate
together.
Genes on different chromosomes
A
A
B
B
A
A
B
B
a
a
b
b
a
a
b
b
Gametes
A
Gametes
a
B
A
b
B
a
A
B
A
B
a
b
a
b
b
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Some genes on the same chromosome
assort together more often than not.
Crossing over and linkage
Lead to separation of linked genes
A
A
B
B
x
a
a
b
b
Gametes
Parental
A
B
Recombinant
a
B
a
A
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b
b
5
• In dihybrid crosses
crosses,
departures from a 1:1:1:1
ratio of F1 g
gametes indicate
that the two genes are on the
same chromosome.
• Example: the white and
yellow genes on the X
chromosome of Drosophila.
• yellow (y+) – yellow body
color
• white (w+) – white eye
color
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Detecting linkage by analyzing the progeny
of dihybrid crosses: X
X--linked
F1 males get their only
X chromosome from
their mothers
F1 females are
dihybrids
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Designations of “parental” and “recombinant” relate to
past history
history.
•
Compare allele
configurations in F2 to P
generation
Note that in this cross:
•
Parental and recombinant classes are opposite of one another in
these two crosses.
Similar percentages of recombinant and parental types show that the
frequency of recombination is independent of the arrangement of
alleles.
Deviation from 1:1:1:1
segregation
ti in
i F2
indicates the genes are
linked
Note that in this cross
involving X-linked
genes, only the F2 male
progeny were counted
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Autosomal genes can also exhibit linkage
• Detect linkage by generating a double heterozygote and
crossing to homozygous recessive (testcross)
• Parental class outnumbers recombinant class
demonstrating linkage.
Genetics 2011
Chi square test pinpoints the probability
that ratios are evidence of linkage.
• Transmission of gametes is based on chance
events.
• Deviations from 1:1:1:1 ratios can represent
p
chance
events OR linkage.
• Ratios alone will never allow you to determine if
observed
b
dd
data
t are significantly
i ifi
tl diff
differentt ffrom predicted
di t d
values.
• The larger your sample,
sample the closer your observed
values are expected to match the predicted values.
• Chi
C squa
square
e test
es measures
easu es “goodness
good ess o
of fit”
between observed and expected (predicted)
results.
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. 5.4
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• Accounts for sample size, or the size of the
experimental population
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C
Concepts
t off Hypothesis
H
th i T
Testing
ti
• ‫ٿ‬ᅿଷ೛
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A l i the
Applying
th chi
hi square test
t t
• Framing a hypothesis
• Null hypothesis (຀คଷ೛); denoted by H0.
H0
• Alternative hypothesis (ჹҥଷ೛); denoted by H1or
Ha.
• Null hypothesis – observed values are not
p
values
different from the expected
• љ‫ٯݤ‬η
• For linkage studies – no linkage is null
hypothesis
• Expect a 1:1:1:1 ratio of gametes.
• H0Ǻ೏֋ࢂค࿾‫ޑ‬ǹᔠჸ‫ࡐ۔‬གྷ‫ޑ๊ܔ‬ଷ೛
• H1Ǻ೏֋ࢂԖ࿾‫ޑ‬ǹᔠჸ‫ࡐ۔‬གྷள‫่݀ޑډ‬
H1 ೏֋ࢂԖ࿾‫ ޑ‬ᔠჸ‫ࡐ۔‬གྷள‫่݀ޑډ‬
• ಍ी‫ޑ‬ᔈҔ
• Alternative hypothesis
yp
– observed values
are different from expected values
• H0Ǻࣴ‫ࡐޣز‬གྷ‫ޑ๊ܔ‬ଷ೛ǹࣴ‫ز‬ჹຝࢂమқ‫ޑ‬
• H1Ǻࣴ‫ࡐޣز‬གྷள‫่݀ޑډ‬ǹࣴ‫ز‬ჹຝࢂԖ࿾‫ޑ‬
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• For linkage studies – genes are linked.
• Expect significant deviation from 1:1:1:1 ratio.
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Information needed for the chi
chi-square test
A l i the
Applying
th chichi
hi-square test
t t
Calculate the chi-square
Use data from breeding experiment
• Total number of progeny
• How many classes of progeny
p g observed in each class
• Number of offspring
Consider degrees of freedom (df) in the experiment
• df = N – 1 ((where N is the number of classes))
Determine a p value using chi-square value and df
• Probability that the deviation from expected numbers
had occurred by chance
• Use table 5.1
Ca cu a e number
Calculate
u be o
of o
offspring
sp g e
expected
pec ed in
each class if there is no linkage (1:1:1:1
segregation)
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Applying
pp y g the chichi-square
q
test to see
if genes A and B are linked
Critical chi
chi--square values
Use p value of 0
0.05
05 as cutoff
Chi-square values that lie in the yellow region of this table
allow rejection of the null hypothesis with >95%
confidence
If null hypothesis is rejected, then linkage can be
postulated
Fig. 5.5
Experiment
p
1:
Experiment 2:
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Table 5.1
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Reciprocal exchanges between
homologous chromosomes are the
physical basis of recombination.
Recombination results when crossingg
over during meiosis separates linked
genes
genes.
• 1909 – Frans Janssens observed
chiasmata, regions in which nonsister
chromatids
h
tid off h
homologous
l
chromosomes
h
cross over each other.
• Thomas Hunt Morgan suggested these
were sites of chromosome breakage and
change resulting in genetic recombination.
• 1931 – Genetic recombination depends on the
reciprocal exchange of parts between maternal and
paternal chromosomes.
•
•
•
•
Harriet Creighton and Barbara McClintock studied corn.
Curtis Stern studied fruit flies.
Physical markers to keep track of specific chromosome parts
Genetic markers were points of reference to determine if
particular progeny were result of recombination.
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Genetic recombination between car and Bar genes on
the Drosophila X chromosome
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Chiasmata mark the sites of recombination.
Early prophase
Leptotene and zygotene
Diplotene
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Fig. 5.6
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Fig. 5.7 a-c
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Chiasmata mark the sites of
recombination.
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Recombination frequencies for pairs of
genes reflect distance between them.
• Alfred H. Sturtevant – Percentage of
q
y
recombination,, or recombination frequency
(RF) reflects the physical distance separating
two genes.
g
• 1 RF = 1 map unit (or 1 centiMorgan, cM)
Terminalization –
movement of chiasmata
Anaphase
A
h
– chromosome
h
separation occurs after
chiasmata reach the
telomeres
Two recombinant and two
parental gametes are
produced
Fig. 5.5 d-f
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Fig. 5.8
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Properties of linked versus
unlinked genes
E d off Part
End
P t1
• You should have learned…
• Genes close together on the same chromosome
are linked and do not segregate independently.
• Linked genes lead to a larger number of parental
class
l
than
h expected
d iin d
double
bl h
heterozygotes.
• Mechanism of recombination is crossing over.
• Chiasmata are the visible
isible signs of crossing o
over.
er
• The farther away genes are the greater the
opportunity for chiasmata to form
form.
• Recombination frequencies reflect physical
distance between genes.
Table 5.2
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Mapping: Locating genes along a
chromosome
Fig. 5.9a
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Mapping: Locating genes along a
chromosome
• Two-point
crosses:
p
help
p
Comparisons
establish relative
gene positions.
positions
• Genes are
arranged in a line
ga
along
chromosome.
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Genes are
arranged in a line
along a
chromosome.
Fig. 5.9 b-d
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Li it ti
Limitations
off ttwo point
i t crosses
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Three Point Crosses: A faster more
accurate method to map genes
• Difficult to determine gene order if two
genes are close
l
ttogether
th
• Actual distances between g
genes do not
always add up.
• Pairwise crosses are time and labor
consuming.
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Fig. 5.10a
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Double recombinants indicate
order of three genes
Analyzing the results of a three
three-point cross
Testcross progeny have four sets of reciprocal
pairs of genotypes
pr must be in the middle because
longest distance is between vg and b
• Most frequent pair has parental configuration of
alleles
• Least frequent pair results from double crossovers
• Examination of double crossover class reveals
which gene is in the middle
Fig.
g 5.10b
Fig 5.10
Fig.
5 10
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Analyzing the results of a three point cross
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Analyzing the results of a three point cross
• Look at two genes at a time and
compare to parental
parental.
Fig. 5.11 c,d
Fig. 5.11 a,b
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vg – pr distance 252 + 241 + 13 + 9
4197
X 100 = 12.3 m.u.
b – pr distance 131 + 118 + 13 + 9
4197
X 100 = 6.4 m.u.
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Interference: The number of double
crossovers may be less than expected.
vg – b distance 252 + 241 + 131 + 118 X 100 = 17.7 m.u.
4197
• Sometimes the number of observable
d bl crossovers iis lless th
double
than expected
t d
if the two exchanges are independent.
Correction for Double Crossovers
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• Occurrence of one crossover reduces
likelihood that another crossover will occur
i adjacent
in
dj
t parts
t off th
the chromosome.
h
• Chromosomal interference – crossovers do
nott occur independently.
i d
d tl
• Interference is not uniform among
chromosomes
h
or even within
ithi a
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chromosome.
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Calculation of interference in the
three--point cross in Figure 5.10
three
M
Measuring
i interference
i t f
• Compare observed and expected
frequencies of double crossovers (DCO)
• Expected probability of double crossovers is
the product of the single crossover
frequencies in each interval
• Probability of single crossover between vg and
pr is 0.123 (12.3 m.u.)
• Probability off single crossover between pr and b
is 0.064 (6.4 m.u.)
• Interference = 1 – coefficient of
coincidence.
• If interference = 0, crossovers in adjacent regions
occur independently of each other
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• If interference = 1, no double crossovers occur
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Calculation of interference in the threethreepoint cross in Figure 5.10 (cont)
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Do Genetic and Physical maps
correspond?
Expected probability of double crossovers
• Order of genes is correctly predicted by
physical
h i l maps.
• Distance between genes is not always similar
on physical maps.
Observed p
proportion
p
of double crossovers
• Double, triple, and more crossovers
• Only 50% recombination frequency observable in
a cross
• Variation across chromosome in rate of
recombination
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• Mapping functions compensate for
inaccuracies,
but are not often imprecise.
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Are genetic maps and physical maps
correlated?
• Th
The order
d off genes iin a genetic
ti map is
i the
th same
as the order of those same genes along the
DNA molecule of a chromosome.
• The physical distance (amount of DNA
separating
gg
genes)) is not always
y the same as the
genetic distance between genes.
Genes chained
together by
linkage
relationships
are known as
linkage groups.
groups
• Factors responsible for differences in physical and
genetic map distances:
• Double, triple, and even more crossovers
• 50% limit on recombination frequency observable in a cross
• Recombination
R
bi ti ffrequency iis nott uniform
if
across a
chromosome.
• Recombination hotspots
• Recombination deserts
Fig. 5.13
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Saccharaomyces cerevisiae life cycle
T t d analysis
Tetrad
l i in
i fungi
f
i
• Model organisms for understanding the
mechanism of recombination because
p
p
products of meiosis are
all four haploid
contained in ascus
• Ascospores within ascus germinate into
haploid individuals.
• Saccharaomyces cerevisiae – bakers yeast
p
crassa – bread mold
• Neurospora
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Fig. 5.14 a
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Genetic analysis in fungi
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Generation of diploid yeast cells that are
heterozygous for two unlinked genes
Phenotype of haploid fungi is direct representation of
their genotype
his4 TRP1 (a) x HIS4 trp1 () Æ his4/HIS4; trp1/TRP1 (a/)
Mutations in haploids can affect appearance of cells
and ability to grow under certain conditions
• his4 mutant; recessive, unable to grow in absence of
histidine
• HIS4; dominant, grows in presence or absence of histidine
• trp1 mutant; recessive, unable to grow in absence of
tryptophan
• TRP1; dominant
dominant, grows in presence or absence of
tryptophan
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Fig. 5.15 a
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Meiosis can generate three kinds of
tetrads: Parental ditype
yp (PD)
( )
Meiosis can generate three kinds of
tetrads: Nonparental
p
ditype
yp ((NPD))
his4 TRP1 (a) x HIS4 trp1 () Æ his4/HIS4; trp1/TRP1 (a/)
Parental ditype (PD) - all spores with parental allele
configurations (0/4 recombinants)
his4 TRP1 ((a)) x HIS4 trp1
p (()) Æ his4/HIS4; trp1/TRP1
p
((a/))
Nonparental ditype (NPD) - all spores with nonparental
allele configuration (4/4 recombinants)
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Fig. 5.15b
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Fig. 5.15c
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Meiosis can generate three kinds of
tetrads: Tetratype (T)
T t d analysis
Tetrad
l i off unlinked
li k d genes
his4 TRP1 (a) x HIS4 trp1 () Æ his4/HIS4; trp1/TRP1 (a/)
Tetratype (T) - four kinds of spores (2/4 recombinants)
• Two have parental allele configurations
• Two have recombinant allele configurations
• Crossover between centromere and closest gene
When genes are unlinked,
number of PD = number of NPD
Fig. 5.15e
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Fig. 5.15d
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How crossovers between linked genes
generate different tetrads
T t d analysis
Tetrad
l i off linked
li k d genes
When genes are lilinked,
Wh
k d
number of PD >> number of NPD
Fig 5.17
Fig.
5 17
Fig. 5.16
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How crossovers between linked genes
generate different tetrads (cont)
Genetics 2011
How crossovers between linked genes
generate different tetrads (cont)
Fig. 5.17
Fig. 5.17
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Calculating recombination
frequencies in tetrad analysis
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Evidence that recombination takes
place at the fourfour-strand stage
For the cross in Figure 5.16:
Fig. 5.18
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Evidence for exception to the rule
that recombination is reciprocal
Meiotic
recombination
almost always
generates equal
numbers of both
kinds of
recombinant
progeny (2:2
segregation)
Rare
tetrads will
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t 31 13
Copyright © The McGraw-Hill
Companies, Inc. Permission
required to reproduce or
display Hartwell et al 4th ed
Genetics 2011
N
Neurospora
crassa life
lif cycle
l
Fig. 5.19
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N
Neurospora
f
form
ordered
d
d tetrads
t t d
S
Segregation
ti patterns
tt
in
i ordered
d
d ascii
• U
Undergo
d
meiosis
i i I and
d II as usual,
l b
butt a single
i l round
d off
mitosis after 2nd meiotic division – produces octad
• Ascus is very narrow and spindle forms parallel to long
axis
Two genetically
identical ascospores
are next to each other
Arrangement of
chromatids can be
inferred from position
of ascospores
FDS (first-division segregation patterns
Fig. 5.20
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Fig. 5.21 a
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S
Segregation
ti patterns
tt
in
i ordered
d
d tetrads
t t d
Number of second division tetrads is used to calculate the
distance between a gene and a centromere
Gene – centromere distance = divide percentage of second
division tetrads by 2
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Ordered tetrads help locate genes in
relation to the centromere
• Neurospora cross thr+ arg+ x thr arg Æ tetrads in 7
genotype classes
Centromere – thr distance:
Centromere – arg distance:
Fig. 5.21 b
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SDS (second-division segregation patterns
Fig. 5.22a
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D t
Determining
i i li
linkage
k
with
ith ordered
d d tetrads
t t d
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Calculating map distance with ordered
tetrads
Neurospora cross thr+ arg+ x thr arg Æ tetrads in 7
genotype classes
arg – thr distance:
If thr and arg are linked, PD >> NPD
This calculation doesn't
doesn t account for double crossovers
PD = 72 + 1 = 73 >> NPD = 1 + 2 = 3
Fig. 5.22b
Fig. 5.22a
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Rules for tetrad analysis in ordered
and unordered tetrads
Table 5
5.3
3
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E d off partt 2
End
• Gene pairs that are close together on the same chromosome
are linked because they are transmitted together more often
than not.
• The recombination frequency of pairs of genes indicate how
often two genes are transmitted together. Gene pairs that
assort independently
p
y exhibit a recombination frequency
q
y of 50%.
• Statistical analysis helps determine whether or not two genes
assort independently.
• The greater the physical distance between linked genes
genes, the
higher the recombination frequency.
• Genetic maps
p are visual representations
p
of relative
recombination frequencies.
• Organisms that retain all the products of meiosis within an
ascus reveal the relation between genetic recombination and
segregation of chromosomes during meiosis.
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Twin spots
p
in Drosophila
p
Mitotic recombination can produce
genetic mosaics.
Double heterozygous Drosophila females y sn+/y+ sn
ƒ yellow (y) mutant – yellow body
ƒ wildtype (y+) – brown body
g ((sn)) mutant – short and curled bristles
ƒ singed
ƒ wildtype (sn+) – long and straight bristles
• Mitotic recombination is rare.
• Initiated
I iti t d by
b
• Mistakes in chromosome replication
p
• Chance exposure to radiation
• Curt Stern – observed “twin
twin spots”
spots in
Drosophila – a form of genetic mosaicism.
• Animals contained tissues with different
genotypes.
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Mitotic crossing over between sn and
centromere in Drosophila
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Fig. 5.23
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Sectored yeast colonies can arise
from mitotic recombination
• Diploid ADE2/ade2
• White colonies are
wildtype
• Red sectors are
ade2/ade2
d 2/ d 2
• Size of sector indicates
when recombination
took place
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Fig. 5.25
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Mit ti Recombination
Mitotic
R
bi ti
and
dC
Cancer
• Retinoblastoma gene (Rb)
• Wild type (Rb+); mutant type (Rb-)
• Heterozygote in normal somatic cells
cells,
homozygote in cancer cells (40%)
• Mitotic Recombination takes place in
where and when
when~ (penetrance)
E d off P
End
Partt 3
• You should have learned…
• The cause of mitotic recombination
• The twin spot in Drosophila
• The relationship between mitotic
recombination and human diseases
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E d off the
End
th class
l
• Ask yourself, do you….
• Understand the relationship between linkage,
g
recombination, and crossing over.
• Show how exceptions to independent assortment provide a
way to map genes on chromosomes.
• Explain where crossing over occurs during meiosis to give
rise
i tto recombinants
bi
t and
d th
the experimental
i
t lb
basis
i supporting
ti
that conclusion.
• Derive genetic maps from two
two- and three
three-point
point testcrosses
testcrosses.
• Measure positive and negative interference and
understand their effect on map distances
distances.
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