Download Genetics and Evolution

Extra Credit
Are the b and vg
genes on the
same
chromosome?
If so, how far
apart are they?
Multiple crossovers can be
useful for mapping three
genes at a time!
A
B
C
—•—————————
A
B
C
a
b
c
—•—————————
a
b
c
A
B
C
—•—————————
A
b
C
a
B
c
—•—————————
a
b
c
3 recessive phenotypes in
maize (corn), coded by three
linked genes
l l lazy or prostrate growth
g g glossy leaves
s s sugary endosperm
To map the genes, mate a
triple heterozygote to triple
recessive homozygote
Ll Gg Ss
x
ll gg ss
Gene order is not known,
so the order shown here is arbitrary.
Linkage phase is not know
How many different kinds of
gametes do you get from triple
heterozygote?
Ll Gg Ss
L or l
2
G or g
*
2
S or s
*
2
Wildtpe for all
Recomb.
L G S/l g s
x
lazy, gloss, sugary
l g s/l g s
Progeny
Genotypes of offspring
Phenotype
L G S /l g s
wildtype
l G S/l g s
lazy
L g S/l g s
glossy
L G s/l g s
sugary
l g S/l g s
lazy,glossy
l G s/l g s
lazy,sugary
L g s/l g s
glossy,sugary
l g s/l g s
lazy,glossy,sugary
Total
Number
286
33
59
4
2
44
40
272
740
Where to begin?
Parental types will constitute ≥ 50% of
all progeny, so…
Rule 1
Two most-frequent gametes types are the
parental types
Tells us the linkage phase of heterozygous
parent:
LGS
l g s
or
LgS
l Gs
or
l g S
LG s
or
L g s
l G S
L G S // l g s x
Progeny
Phenotype
wildtype
lazy
glossy
sugary
lazy,glossy
lazy,sugary
glossy,sugary
lazy,glossy,sugary
Total
l g s // l g s
Progeny
Genotypes
L G S // l g s
l G S // l g s
L g S // l g s
L G s // l g s
l g S // l g s
l G s // l g s
L g s // l g s
l g s // l g s
Number
286
33
59
4
2
44
40
272
740
Linkage phase in
heterozygous parent?
LGS
l g s
or
LgS
l Gs
or
l g S
LG s
or L g s
l G S
Rule 2
The double-recombinant gametes will be
the two least frequent types
A
a
B
C
b
c
L G S/ l g s
Progeny
Phenotype
wildtype
lazy
glossy
sugary
lazy,glossy
lazy,sugary
glossy,sugary
lazy,glossy,sugary
Total
x
l g s/ l g s
Progeny Genotypes
L G S/l g s
l G S/l g s
L g S/l g s
L G s/l g s
l g S/l g s
l G s/l g s
L g s/l g s
l g s/l g s
Number
286
33
59
4
2
44
40
272
740
Rule 3
Effect of double crossovers is to
interchange the members of the middle
pair of alleles between the chromosomes
A
a
B
C
A
b
C
b
c
a
B
c
Parental types:
LGS
and
lgs
Double-crossover types:
LGs
and
lgS
Which gene is in the middle?
L
S
G
L
s
G
l
s
g
l
S
g
Now you know linkage phase
of heterozygous parent and
gene order
L S G
l s g
How far apart are the genes?
Count the crossovers between
adjacent genes
L S G
l s g
• In parents, L allele on same homolog as S
and l on same homolog as s. So if these
get broken up ---> crossover between L
and S loci
• In parents, S on same homolog as G and s
on same homolog as g. If these get broken
up --> recombination between S and G loci
L S G
l s g
Progeny
Phenotype
Progeny
Genotype
wildtype
LGS / lgs
lazy
l GS / lgs
glossy
LgS / lgs
sugary
LGs / lgs
lazy,glossy
lgS / lgs
lazy,sugary
lGs / lgs
glossy,sugary
Lgs / lgs
lazy,glossy,sugary l g s / l g s
Total
#
286
33
59
4
2
44
40
272
740
Crossover or NonCrossover?
Parental (NCO)
single CO between L and S
single CO between S and G
double CO
double CO
single CO between S and G
single CO between L and S
Parental (NCO)
Rule 4: Reciprocal
products expected to
occur in approximately
equal numbers
Progeny
Phenotype
wildtype
lazy
glossy
sugary
lazy,glossy
lazy,sugary
glossy,sugary
lazy,glossy,sugary
Total
LGS ≈ lgs
272)
LgS ≈ lGs
Lgs ≈ lGS
LGs ≈ lgS
Progeny
Genotype
LGS / lgs
l GS / lgs
LgS / lgs
LGs / lgs
lgS / lgs
lGs / lgs
Lgs / lgs
lgs / lgs
(286 ≈
(59 ≈ 44)
(40 ≈ 33)
(4 ≈ 2)
#
286
33
59
4
2
44
40
272
740
Rule 5
Don't forget to include the double
recombinants when calculating
recombination frequency!
Progeny
Phenotype
Progeny
Genotype
wildtype
LGS / lgs
lazy
l GS / lgs
glossy
LgS / lgs
sugary
LGs / lgs
lazy,glossy
lgS / lgs
lazy,sugary
lGs / lgs
glossy,sugary
Lgs / lgs
lazy,glossy,sugary l g s / l g s
Total
Rec Freq L-S
l
L
L
l g
G S
g s
G s
S
33
40
4
2
79
#
286
33
59
4
2
44
40
272
740
Crossover or NonCrossover?
Parental (NCO)
single CO between L and S
single CO between S and G
double CO
double CO
single CO between S and G
single CO between L and S
Parental (NCO)
Rec Freq S-G
L g S
59
l G s
44
L G s
4
l g S
109
2
Rec Freq L-S
l
L
L
l g
G S
g s
G s
S
33
40
4
2
79
Rec Freq S-G
L g S
59
l G s
44
L G s
4
l g S
109
79/740 or 10.7% of
gametes recombinant
between L & S. So,
map distance between
L & S = 10.7 map units
109/740 or 14.8 % of
gametes recombinant
between S & G. So,
map distance between
2 S & G=14.8 map units
Genetic Map
10.7 mu
14.8 mu
_____________________________
L
S
G
Interference
Assuming independence, expected
probability of double crossovers is the
probability of recombination in one
region times the probability of
recombination in other (product rule).
Maize example
Probability of recombination between L and S is 10.7%
Probability of recombination between S and G is 14.8%
If crossovers independent, probability of double
crossover should then be
0.107
*
0.148 =
0.0158
In 740 events, the double crossover class should occur
0.0158
*
740
=
12 times
Expected DCO = 12
Observed DCO = 6
Typical Result: O < E
Conclusion: Crossing over in one
region reduces probability of
crossing over in adjacent regions
This is Interference
Why?
Physical constraints that prevent two
chiasmata in close proximity during
meiosis?
Quantifying Interference
Coefficient of coincidence = Obs DCO
Exp DCO
cc = 6/12
Interference = (1 - cc) = 1 - 0.5 = 0.5
Recombination is not
independent at small distances
If distance between genes is small (<10
map units in Drosophila) no double
crossovers occur (interference is
complete, I=1)
At large distances (> 45 map units,
Interference disappears, Obs = Exp and
I=0
In Drosophila, the allele b gives black body (wild type is
tan); at a separate gene, the allele wx gives waxy wings
(nonwaxy is wild type); and at a third gene, the allele cn
gives cinnabar eyes (red is wild type). A female that is
heterozygous for these three genes is testcrossed, and
980 progeny are classified as follows for body color, wing
phenotype, and eye color:
Phenotype
brown, nonwaxy, red
black, waxy, cinnabar
brown, waxy, cinnabar
black, nonwaxy, red
brown, nonwaxy, cinnabar
black, waxy, red
brown, waxy, red
black, nonwaxy, cinnabar
Genotype of
Gamete from
Heterozygous
Parent
+++
b wx cn
+ wx cn
b++
+ + cn
b wx +
+ wx +
b + cn
CO/NCO
#
DCO
6
DCO
6
SCOcn & wx
79
SCO cn & wx 77
Parental
369
Parental
369
SCO b & cn 38
SCO b & cn 36
Extra Credit Question for Next
Time
Phenot ype
brown , nonw axy, red
black, wa xy, cinna bar
brown , wa xy, cinna bar
black, nonw axy, red
brown , nonw axy, cinna bar
black, wa xy, red
brown , wa xy, red
black, nonw axy, cinna bar
Haploid
Genot ype
+ ++
b wx cn
+ wx cn
b ++
+ +cn
b wx +
+ wx +
b +cn
Number
6
6
79
77
369
369
38
36
a) What is the linkage phase of the heterozygous female
parent?
b) What is the order of the three genes?
c) Construct a linkage map with the genes in their correct
order and indicate the map distances between the genes.
d) Calculate the Interference.
Where to begin?
Recombinant genotypes of all sorts will
constitute <= 50% of all progeny
50% Rec
0% Rec
50% Rec
50% Rec
100% Rec
Threelocus
mapping
To map the genes, mate a
triple heterozygote to triple
recessive homozygote
L G S x
l g s
l g s
l g s
Gene order is not known,
so the order shown here is arbitrary
Multiple crossovers can lead
to inaccuracy
B
A
C
—•—————————
—•—————————
b
a
c