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1gray, normal white, bent gray, bent gray, normal

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autosomal, recessive traits: ebony body (e) spineless bristle (sp) ... e sp e ebony 250 59. e sp sp spineless 250 71. recombinants ... – PowerPoint PPT presentation

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Title: 1gray, normal white, bent gray, bent gray, normal


1

1 gray, normal white, bent ½ gray, bent ½
gray, normal ½ white bent ½ white, normal
2 gray, bent white, normal ½ gray, bent ½
gray, bent ½ gray, normal ½ gray, normal 3
white, bent white, normal white,
bent white, bent 4 gray, normal gray,
normal gray, normal gray, normal 5 gray,
bent gray, bent ¾ gray, bent ¾ gray,
bent ¼ white, bent ¼ white, bent 6
white, bent gray bent gray, bent ½ gray,
bent ½ gray, normal
2
Transmission patterns of linked genes
alleles of genes on the same chromosome tend
to be inherited as a group, especially when
close together new combinations of alleles at
different loci are produced by recombination
for a given pair of loci, product of most
meioses will be non-recombinant
(parental) recombination detectable only in
individuals heterozygous at all loci in
crosses, identify linked loci by non-mendelian
ratios - 2 common classes (parentals) - 2
uncommon classes (recombinants)
3
No Recombination
A
B
AB
AB
ab
ab
a
b
With Recombination
A
B
AB
Ab
aB
ab
a
b
4
No Recombination
A
B
AB
AB
ab
ab
a
b
With Recombination
A
B
AB
Ab
AB
Ab
A
b
5
Drosophila melangaster autosomal, recessive
traits ebony body (e)
spineless bristle (sp)
P0 wild type x ebony spineless /
/ x e/e sp/sp F1 wild type
(e/ sp/)

6
Drosophila melangaster autosomal, recessive
traits ebony body (e)
spineless bristle (sp)
testcross e/ sp/ x e/e
sp/sp
???????

detectable recombination only in
females offspring phenotype offspring genotype
7
testcross e/ sp/ x e/e sp/sp

gamete gamete phenotype exp f
obs f e sp e sp
ebony spineless e sp
wild type e sp e ebony
e sp sp spineless
8
-independent assortment AaBb types of
gametes pr(AB) pr (ab) pr(Ab)
pr(aB) ¼ -complete linkage AaBb types
of gametes pr(AB) pr(ab) ½
9
testcross e/ sp/ x e/e sp/sp

gamete gamete phenotype exp f
obs e sp e sp
ebony spineless 250 430 e sp
wild type 250 440 e
sp e ebony 250 59 e
sp sp spineless 250 71
parentals
10
testcross e/ sp/ x e/e sp/sp

gamete gamete phenotype exp f
obs e sp e sp
ebony spineless 250 430 e sp
wild type 250 440 e
sp e ebony 250 59 e
sp sp spineless 250 71
recombinants
11
determining the recombination distance between
genes frequency of recombination RF
RF (71 59)/1000 RF
130/1000 0.13 map distance RF x 100
map units (Cm) 0.13 x 100 13 m.u.
number of recombinant offspring total
number of offspring
12
Notation initial (null) hypothesis is that
genes are unlinked e/e sp/sp or e/e
sp/sp for genes known to be linked e
sp/e sp or e sp/ e sp or e sp/e sp
13
gamete gamete phenotype
obs 1 obs 2 e sp e sp
ebony spineless 430 59 e
sp wild type 440 71
e sp e ebony 59 430
e sp sp spineless 71 440
e sp
ebony, spineless x wild type
obs1 obs2

e
ebony x spineless
sp
14
gamete gamete phenotype exp f
obs e sp e sp
ebony spineless 0.25 430 e sp
wild type 0.25 440
e sp e ebony 0.25 59
e sp sp spineless 0.25 71
e tot sp 245 256 501
244 255 499 tot 489 511 1000
obs
exp
e tot sp 430 71 501
59 440 499 tot 489 511 1000
15
Compare observed and expected numbers with a
goodness of fit chi-square test c dof
S
2
(observed number - expected number)2
(expected number)
c 3
2
(430-245)2 (71-256)2 (59-244)2
(440-255)2 245 256
244 255



139.7 133.7 134.7 134.2
542.3 p

16
gamete gamete phenotype exp f
obs e sp e sp
ebony spineless 0.25 430 4
e sp wild type 0.25
440 4 e sp e ebony
0.25 59 1 e sp sp
spineless 0.25 71 1


effects of sample size !
17
standard mapping cross 3 loci test
cross heterozygous female homozygous
recessive male linkage if two common
offspring types determine parental
genotypes calculate RF values (map
distances) for pairs of genes determine
gene order (map)
18
Linkage analysis of multiple traits javelin
bristle, striped body, pink eyes 3-point test
cross jv/ str/ pi/ x jv/jv
str/str pi/pi
gametes gametes progeny phenotype progeny jv
str pi jv javelin 294 jv str pi
str pi striped, pink 287 jv str pi jv str
pi javelin, striped, pink 140 jv str pi
wild type 139 jv str pi pi pink
66 jv str pi jv str javelin, striped 65 jv
str pi jv pi javelin, pink 4 jv str pi
str striped 5
19
gametes gametes progeny phenotype progeny jv
str pi jv javelin 294 jv str pi
str pi striped, pink 287 jv str pi jv str
pi javelin, striped, pink 140 jv str pi
wild type 139 jv str pi pi pink
66 jv str pi jv str javelin, striped 65 jv
str pi jv pi javelin, pink 4 jv str pi
str striped 5
parentals
recombinant
double crossover
20
gametes gametes progeny phenotype progeny jv
str pi jv javelin 294 jv str pi
str pi striped, pink 287 jv str pi jv str
pi javelin, striped, pink 140 jv str pi
wild type 139 jv str pi pi pink
66 jv str pi jv str javelin, striped 65 jv
str pi jv pi javelin, pink 4 jv str pi
str striped 5
jv, str
21
RF jv, str (140 139 66 65)/1000
410/1000 0.41 distance RF x 100
0.41 x 100 41 m.u.
22
gametes gametes progeny phenotype progeny jv
str pi jv javelin 294 jv str pi
str pi striped, pink 287 jv str pi jv str
pi javelin, striped, pink 140 jv str pi
wild type 139 jv str pi pi pink
66 jv str pi jv str javelin, striped 65 jv
str pi jv pi javelin, pink 4 jv str pi
str striped 5
jv, pi
23
RF jv, pi (140 139 4 5)/1000
288/1000 0.288 distance RF x 100
0.288 x 100 28.8 m.u.
24
gametes gametes progeny phenotype progeny jv
str pi jv javelin 294 jv str pi
str pi striped, pink 287 jv str pi jv str
pi javelin, striped, pink 140 jv str pi
wild type 139 jv str pi pi pink
66 jv str pi jv str javelin, striped 65 jv
str pi jv pi javelin, pink 4 jv str pi
str striped 5
str, pi
25
RF str, pi (66 65 4 5)/1000
140/1000 0.14 distance RF x 100
0.14 x 100 14 m.u.
26
assembling the map jv---str 41
jv---pi 28.8 str---pi 14
42.8 m.u.
str pi jv
27
gametes gametes progeny phenotype progeny jv
str pi jv javelin 294 jv str pi
str pi striped, pink 287 jv str pi jv str
pi javelin, striped, pink 140 jv str pi
wild type 139 jv str pi pi pink
66 jv str pi jv str javelin, striped 65 jv
str pi jv pi javelin, pink 4 jv str pi
str striped 5
parentals
recombinant
double crossover
28
assembling the map jv---str 41
jv---pi 28.8 str---pi 14
42.8 m.u.
str pi jv
RF jv-str (140 139 66 65 24
5)/1000 0.428
29
three possible gene orders jvstrpi
strpijv pijvstr compare parentals
to double crossovers jv and str
pi vs jv pi and str
30
Calculated distance
RF 0.5
Actual distance
31
Mapping function fi (e-m mi)/i! m
mean number of crossovers i actual
number of crossovers fi frequency of ith
crossover class but, only know the
no-crossover class (i 0) (e-m m0)/0!
e-m RF ½ (1 - e-m) solve for m m/2
corrected map distance
32
Take-home points spatial arrangement of
genes - gene interactions - genome evolution
transmission patterns of linked genes - most
offspring will inherit the allele combinations
present in their parents - recombination is
rare, detect numerically from testcross
progeny - parental allele combinations (on the
chromosome) - genetic distance between linked
loci - gene order (genetic map) test for
linkage by statistical analysis of testcross
offspring (chi-square goodness of fit test, n
number of genotypes) linkage analysis is
limited to genes that are close
together (non-linear relationship between
estimated and actual distance)
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