Transcript Recombination Frequencies - Western Washington University

Assignments
• Read from Chapter 3, 3.6 (pp. 100-106),
• Master Problems…3.12, 3.15, 3.20,
• Chapter 4, Problems 1, 2,
• Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 4.20 a,b,c,d.
• Exam Wednesday.
– One hour (you can use the entire 80 minutes, but no
more). One 8” x 11”, one sided crib sheet.
Sex Determination Systems
• Different mechanisms of sex selection exist:
»
• XX / XO (O = null),
• ZW / ZZ (female ZZ, Male ZW),
• haplo / diplo (males are haploid),
• XX / XY (most mammals).
Sex Chromosomes
most mammals…
... ‘X’ and ‘Y’ chromosomes that determine
the sex of an individual in many organisms,
Females: XX
Males: XY
a
Differential
Region
A
a
hemizygous:
condition where gene
is present in only one
dose (one allele).
Differential Region
Paring Region
Paring Region
XY: male
XX: female
X Linkage
…the pattern of inheritance resulting from genes
located on the X chromosome.
X-Linked Genes…
…refers specifically to genes on the Xchromosome, with no homologs on the Y
chromosome.
Blue is dominant.
P
x
Blue
Female
Gametes
Pink
Male
or
Gametes
or
F1
Blue Female
Blue Male
F1
x
Blue Female
Gametes
or
Blue Male
or
Gametes
or
or
F2
Blue Female
Blue Male
Blue Female
Pink Male
F2
Blue Female
Blue Male
Blue Female
Pink Male
3 : 1 Blue to Pink
1 : 1 Female to Male
P
x
Pink
Female
Gametes
Blue
Male
or
Gametes
or
F1
Blue Female
Pink Male
Gametes
or
or
F2
Pink Female
Pink Male
Blue Female
Blue Male
F2
Pink Female
1
Pink Male
Blue Female
Blue Male
1
1
1
1 : 1 Female to Male
1 : 1 Pink to Blue
Sex Linkage to Ponder
• Female is homozygous recessive X-linked gene,
– what percentage of male offspring will express?
– what percentage of female offspring will express if,
• mate is hemizygous for the recessive allele?
• mate is hemizygous for the dominant allele?
• Repeat at home with female heterozygous Xlinked gene!
Sex-Linked vs. Autosomal
• autosomal chromosome: non-sex linked
chromosome,
• autosomal gene: a gene on an autosomal
chromosome,
• autosomes segregate identically in reciprocal
crosses.
X-Linked Recessive Traits
Characteristics
• Many more males than females show the phenotype,
– female must have both parents carrying the allele,
– male only needs a mother with the allele,
• Very few (or none) of the offspring of affected males show
the disorder,
– all of his daughters are carriers,
• roughly half of the sons born to these daughters are carriers.
X-Linked Dominant
• Affected males married to unaffected females pass the
phenotype to their daughters, but not to their sons,
• Heterozygous females married to unaffected males pass the
phenotype to half their sons and daughters,
• Homozygous dominant females pass the phenotype on to
all their sons and daughters.
Autosomal Dominant
• Phenotypes appear in every generation,
• Affected males and females pass the
phenotype to equal proportions of their sons
and daughters.
Pedigree for Very Rare Trait
? = kid with trait
1/2
1/2
Recessive?
---> Yes!
1/2 x 1/2 x 1/2
? = 1/8 ?
X-Linked?
Autosomal?
---> Yes!
x 1/2 = 1/16
(p)boy
X-Linked Dominant
examples (OMIM)
• HYPOPHOSPHATEMIA: “Vitamin-D resistant Rickett’s”,
• LISSENCEPHALY: “smooth brain”,
• FRAGILE SITE MENTAL RETARDATION: mild retardation,
• RETT Syndrome: neurological disorder,
• More on OMIM…
Linkage
• Genes linked on the same chromosome may
segregate together.
Independent Assortment
A
A
a
B
B
b
A
b
2n = 4
a
B
a
b
2n = 1
Meiosis
No Cross Over
A
a
B
b
Parent Cell
A
A
a
a
B
B
b
b
Daughter Cells Have Parental Chromosomes
2n = 1
Meiosis
With Cross Over
A
a
B
b
Parent Cell
A
A
a
a
B
b
B
b
Daughter Cells Have Recombinant Chromosomes
Dihybrid Cross
P
yellow/round
GGWW
green/wrinkled
x
GW
F1
ggww
gw
GgWw
phenotype
genotype
gametes
genotype
Gamate Formation in F1 Dihybrids
P: GGWW x ggww, Independent Assortment
F1 Genotype:
GgWw
G
g
W
w
alleles
GW
Gw
gW
gw
gametes
.25
.25
.25
.25
probability
How do you test for
assortment of alleles?
F1: GgWw
GW
Gw
gW
gw
.25
.25
.25
.25
Test Cross: phenotypes of the offspring indicate the genotype of
the gametes produced by the parent in question.
Test Cross
GgWw x ggww
GW (.25)
x
gw (1)
GgWw (.25)
Gw (.25)
x
gw (1)
G gww (.25)
gW (.25)
x
gw (1)
ggWw (.25)
gw (.25)
x
gw (1)
ggww (.25)
Test Cross
GgWw x ggww
GW (.25)
x
gw (1)
GgWw (.25)
P
Gw (.25)
x
gw (1)
Ggww (.25)
R
gW (.25)
x
gw (1)
ggWw (.25)
R
gw (.25)
x
gw (1)
ggww (.25)
P
F1 parental types GgWw and gwgw
recombinant types Ggww and ggWw
Recombination Frequency
…or Linkage Ratio: the percentage of
recombinant types,
– if 50%, then the genes are not linked,
– if less than 50%, then linkage is observed.
Assignments
• Read from Chapter 3, 3.6 (pp. 100-106),
• Master Problems…3.12, 3.15, 3.20,
• Chapter 4, Problems 1, 2,
• Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 4.20 a,b,c,d.
• Exam Wednesday.
– One hour (you can use the entire 80 minutes, but no
more). One 8” x 11”, one sided crib sheet.
Linkage
• Genes closely located on the same
chromosome do not recombine,
– unless crossing over occurs,
• The recombination frequency gives an
estimate of the distance between the genes.
Recombination Frequencies
• Genes that are adjacent have a
recombination frequency near 0%,
• Genes that are very far apart on a
chromosome have a linkage ratio of 50%,
• The relative distance between linked genes
influences the amount of recombination
observed.
homologs
A
B
a
b
In this example, there is a 2/10 chance of recombination.
A
C
a
c
In this example, there is a 4/10 chance of recombination.
Linkage Ratio
P GGWW x ggww
Testcross F1: GgWw x ggww
determine
GW
Gw
gW
gw
?
?
?
?
# recombinant x 100 = Linkage Ratio
# total progeny
Units: % = mu (map units) - or - % = cm (centimorgan)
Study Figs 4.2, 4.3, and 4.5
Fly Crosses (simple 3-point mapping)
(white eyes, minature, yellow body)
• In a white eyes x miniature cross, 900 of the 2,441
progeny were recombinant, yielding a map
distance of 36.9 mu,
• In a separate white eyes x yellow body cross, 11 of
2,205 progeny were recombinant, yielding a map
distance of 0.5 mu,
• When a miniature x yellow body cross was
performed, 650 of 1706 flies were recombinant,
yielding a map distance of 38 mu.
Simple Mapping
• white eyes x miniature = 36.9 mu,
• white eyes x yellow body = 0.5 mu,
• miniature x yellow body = 38 mu,
0.5 mu
36.9 mu
y w
m
38 mu
Do We have to Learn More Mapping
Techniques?
• Yes,
– three point mapping,
• Why,
–
–
–
–
Certainty of Gene Order,
Double crossovers,
To answer Cyril Napp’s questions,
and, for example: over 4000 known human diseases
have a genetic component,
• knowing the protein produced at specific loci facilitates the
treatment and testing.
cis
“coupling”
trans
“repulsion”
Classical Mapping
target
Cross an organism with
a trait of interest to
homozygous mutants of
known mapped genes.
Then, determine if
segregation is random in
the F2 generation,
What recombination frequency do you expect between the target and HY2?
What recombination frequency do you expect between the target and TT2?
• if not, then your
gene is linked
(close) to the known
mapped gene.
Gene Order
• It is often difficult to assign the order of genes
based on two-point crosses due to uncertainty
derived from sampling error.
A x B = 37.8 mu,
A x C = 0.5 mu,
B x C = 37.6 mu,
Double Crossovers
• More than one crossover event can occur in a
single tetrad between non-sister chromatids,
– if recombination occurs between genes A and B 30% of
the time (p = 0.3), then the probability of the event
occurring twice is 0.3 x 0.3 = 0.09, or nearly one map
unit.
• If there is a double cross over, does recombination
occur?
– how does it affect our estimation of distance between
genes?
Genetics in the News
Genetics:
…in the News
Classical Mapping
target
Cross an organism with
a trait of interest to
homozygous mutants of
known mapped genes.
Then, determine if
segregation is random in
the F2 generation,
What recombination frequency do you expect between the target and HY2?
What recombination frequency do you expect between the target and TT2?
• if not, then your
gene is linked
(close) to the known
mapped gene.
Classical mapping in
humans requires
pedigrees…
Three Point Testcross
Triple Heterozygous
(AaBbCc )
x
Triple Homozygous Recessive
(aabbcc)
Three Point Mapping Requirements
• The genotype of the organism producing the gametes must
be heterozygous at all three loci,
• You have to be able to deduce the genotype of the gamete
by looking at the phenotype of the offspring,
• You must look at enough offspring so that all crossover
classes are represented.
w
g
d
Representing linked genes...
P
W
w
G
g
D
d
= WwGgDd
d
d
= wwggdd
x
Testcross
w
w
g
g
w
g
d
Representing linked genes...
P
+
w
+
g
+
d
= WwGgDd
d
d
= wwggdd
x
Testcross
w
w
g
g
Phenotypic Classes
GWgg
Ddd
Ddd
D-
G-
dd
W-G-DW-G-dd
W-gg-D
W-gg-dd
wwG-DwwG-dd
ww
gg
Ddd
wwggDwwggdd
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
Arbitrarily name
regions between
genes…
II
I
W
G
D
w
g
d
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Total =
500
I
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
W
G
D
w
g
d
Region I:
46 + 52 + 2 + 4
500
= 20.8 mu
x 100
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Total =
500
II
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
20.8 mu
W
G
D
w
g
d
Region II:
22 + 22 + 2 + 4
500
= 10.0 mu
x 100
10.0 mu
W
w
20.8 mu
G
D
g
d
0.1 x 0.208 = 0.0208
NO GOOD!
W-gg-D
wwG-dd
Total =
2
4
Recombinants,
double crossover
6/500 = 0.012
500
Coefficient of Coincidence =
Observed
Expected
Interference = 1 - Coefficient of Coincidence
Interference
…the effect a crossing over event has on a
second crossing over event in an adjacent
region of the chromatid,
– (positive) interference: decreases the
probability of a second crossing over,
• most common in eukaryotes,
– negative interference: increases the
probability of a second crossing over.
Gene Order in Three Point Crosses
• Find - either - double cross-over
phenotype…based on the recombination
frequencies,
• Two parental alleles, and one cross over allele will
be present,
• The cross over allele fits in the middle...
#
A-B-C-
2001
aabbcc
1786
A-B-cc
46
aabbC-
52
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
Which one is the “odd”
one?
II
I
A
C
B
a
c
b
#
A-B-C-
2001
aabbcc
1786
Region I
990 + 887 + 46 + 52
A-B-cc
46
6951
aabbC-
52
= 28.4 mu
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
x 100
I
A
C
B
a
c
b
#
A-B-C-
2001
aabbcc
1786
Region II
600 + 589 + 46 + 52
A-B-cc
46
6951
aabbC-
52
= 18.5 mu
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
18.5
II mu
x 100
28.4 mu
A
C
B
a
c
b
Fig. 4.18. DNA molecule containing three
EcoRI cleavage sites
Fig. 4.19
Fig. 4.20a
Molecular Mapping Markers
Fig. 4.20b
p. 143. Fluorescent dyes are often used to label DNA so that the
positions of DAN fragments in a gel can be identified.
Assignments
• Read from Chapter 3, 3.6 (pp. 100-106),
• Master Problems…3.12, 3.15, 3.20,
• Chapter 4, Problems 1, 2,
• Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 4.20 a,b,c,d.
• Exam Friday, April 24th.
– One hour (you can use the entire 80 minutes, but no
more). One 8” x 11”, one sided crib sheet.