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Review Question
If two genes are 13
map units apart on a
linkage map, what
proportion of
recombinant offspring
will be seen in a
testcross?
What proportion of
meioses experienced
crossing over between
to the two genes?
w+
m+
— —————————
F1:
•
— —————————
•
w
m
w+
m+
—•—————————
w
m
—•—————————
w+
m
—•—————————
w
m+
—•—————————
w
X
Par
Par
Rec
Rec
Two X-linked genes
Female
w
Male
m
— —————————
P:
•
— —————————
•
w
Gametes:
m
w+
•
— ————/
•
Egg (A B)
w
Sperm (a b)
m
—•————————
m+
— —————————
w+
m+
—•————————
—•———/
m
— —————————
•
— ————/
•
w
___% non-recombinant offspring,
____% recombinant offspring
m
—•—————————
— ————/
•
Offspring (male or female)
314 Wild type
314 white, miniature
186 miniature
186 white
Genes are how far apart?
___ map units
1
If two genes are 50 map units
apart, what proportion of
recombinant offspring would
you observe?
What would you conclude about
these two genes?
Multiple crossovers are useful for
mapping many genes at a time
Three-point mapping
A
B
C
—•—————————
A
B
C
a
b
c
—•—————————
a
b
c
A
B
C
—•—————————
A
b
C
a
B
c
—•—————————
a
b
c
2
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
How many different kinds of
gametes can you get from
triple heterozygote?
Ll Gg Ss
L or l G or g
S or s
2
* 2
* 2
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 known
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
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,sug a r y
l g s // l g s
Total
Number
286
33
59
4
2
44
40
272
740
3
Rule 1
Where to begin?
Two most-frequent gametes types are the
__________ types
Parental types will constitute ≥ 50% of
all progeny, so…
Tells us the ___________________ of
heterozygous parent:
LGS
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
or
LgS
l Gs
or
l g S or
LG s
L g s
l G S
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
4
L G S/ l g s
Rule 2
Progeny
Phenotype
wildtype
lazy
glossy
sugary
lazy,glossy
lazy,sugary
glossy,sugary
lazy,glossy,sugary
Total
The double-recombinant gametes will be
the two least frequent types
A
a
B
C
b
c
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
Parental types:
Rule 3
LGS
Effect of double crossovers is to
interchange the members of the middle
pair of alleles between the chromosomes
and l g s
Double-crossover types:
LGs
and l g S
Which gene is in the middle?
A
a
B
b
C
c
A
a
b
B
C
L
S
G
L
s
G
l
s
g
l
S
g
c
5
Now you know linkage phase
of heterozygous parent and
gene order
LSG
l s g
How far apart are the genes?
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
LSG
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 ---> recombination 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
Rule 4: Reciprocal
LSG
l s g
Progeny
Phenotype
Count the crossovers between
adjacent genes
#
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)
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
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 ≈ 272)
(59 ≈ 44)
(40 ≈ 33)
(4 ≈ 2)
#
286
33
59
4
2
44
40
272
740
6
Progeny
Phenotype
Rule 5
Don't forget to include the double
recombinants when calculating
recombination frequency!
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 G S
L g s
L G s
l gS
Rec Freq L-S
l G S
L g s
L G s
l gS
33
40
4
2
79
Rec Freq S-G
LgS
59
l G s
44
L G s
4
l gS
2
109
79/740 or 10.7% of
gametes recombinant
between L & S. So,
map distance between
L & S = ___ map units
109/740 or 14.8 % of
gametes recombinant
between S & G. So,
map distance between
S & G=____ map units
33
40
4
2
79
Crossover or NonCrossover?
#
286
33
59
4
2
44
40
272
740
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
LgS
59
l G s
44
L G s
4
l gS
2
109
Genetic Map
10.7 mu
14.8 mu
_____________________________
L
S
G
7
Maize example
Interference
Probability of recombination between L and S is 10.7%
Probability of recombination between S and G is 14.8%
Assuming independence, expected
probability of double crossovers is the
probability of recombination in one
region times the probability of
recombination in other (__________).
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
Expected DCO = 12
Observed DCO = 6
L G s
l gS
Typical Result: O < E
4
2
6
Why?
Physical constraints that prevent two
chiasmata in close proximity during
meiosis
Conclusion: Crossing over in one
region reduces probability of
crossing over in adjacent regions
This is Interference
8
Quantifying Interference
Recombination is not
independent at small distances
Coefficient of coincidence = Obs DCO
Exp DCO
cc = _____
If distance between genes is small (<10
map units in Drosophila) no double
crossovers occur (interference is
complete, I=1)
Interference = (1 - cc) =
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
SCO cn &wx
79
SCO cn &wx 77
Parental
369
Parental
369
SCO b & cn 38
SCO b & cn 36
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
Haploid
Genotype
+++
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.
9
Threelocus
mapping
10
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