Transcript genetic mapping

Genetic map
&
crossing over
Linked assortment
RrYy
R
r
Y
y
r
R
y
Y
ry
RY
Possible gametes.
IS NOT ALWAYS TRUE
Morgan Provided Evidence for the
Linkage of Several X-linked Genes
• The first direct evidence of linkage came from
studies of Thomas Hunt Morgan
• Morgan investigated several traits that followed an
X-linked pattern of inheritance
– Body color
– Eye color
– Wing length
P Males
P Females
• Morgan observed a much higher proportion of the combinations of traits
found in the parental generation
• Morgan’s explanation:
– All three genes are located on the X chromosome
– Therefore, they tend to be transmitted together as a unit
Morgan Provided Evidence for the
Linkage of Several X-linked Genes
• However, Morgan still had to interpret two
key observations
– 1. Why did the F2 generation have a significant
number of non parental combinations?
– 2. Why was there a quantitative difference
between the various non parental combinations?
Let’s reorganize Morgan’s data by considering the pairs
of genes separately

Gray body, red eyes
1,159
Yellow body, white eyes
1,017
Gray body, white eyes
Yellow body, red eyes
Total
17
12
2,205
Red eyes, normal wings
770
White eyes, miniature wings
716
Red eyes, miniature wings
White eyes, normal wings
Total
401
318
2,205
But this nonparental
combination was rare
= very closed genes
It was fairly common
to get this nonparental
combination
= very distant genes
• Morgan made three important hypotheses to
explain his results
– 1. The genes for body color, eye color and wing length
are all located on the X-chromosome
• They tend to be inherited together
– 2. Due to crossing over, the homologous X
chromosomes (in the female) can exchange pieces of
chromosomes
• This created new combination of alleles
– 3. The likelihood of crossing over depends on the
distance between the two genes
• Crossing over is more likely to occur between two genes that are
far apart from each other
Crossing over
The Hypothesis
– Offspring with nonparental phenotypes are the
product of a cross-over
• This cross over created nonparental chromosomes via
an exchange of segments between homologous
chromosomes
The haploid cells contain
the same combination of
alleles as the original
chromosomes
The arrangement of linked
alleles has not been altered
No crossing over
2 gametes
These haploid cells contain a
combination of alleles NOT
found in the original
chromosomes
crossing over
These are termed
parental or nonrecombinant cells
This new combination of
alleles is a result of
genetic recombination
These are termed
nonparental or recombinant
cells
4 gametes
GENETIC MAPPING
• Genetic mapping is also known as gene mapping,
chromosome mapping or linkage analysis
• Its purpose is to determine the linear order of
linked genes along the same chromosome
Genetic linkage map of Drosophila melanogaster
Each gene has its
own unique
position at a
particular site
within a
chromosome
Genetic Mapping
• Genetic maps allow us to estimate the relative distances between linked
genes, based on the likelihood that a crossover will occur between them
• Experimentally, the percentage of recombinant offspring is correlated with
the distance between the two genes
– If the genes are far apart  many recombinant offspring
– If the genes are close  very few recombinant offspring
• Map distance =
Number of recombinant offspring X 100
Total number of offspring
• The units of distance are called map units (mu)
– They are also referred to as centiMorgans (cM)
• One map unit is equivalent to 1% recombination frequency
Genetic Mapping
• Genetic mapping experiments are typically accomplished by carrying
out a testcross
– A mating between an individual that is heterozygous for two or more genes
and one that is homozygous recessive for the same genes
• example of a testcross
– This cross concerns two linked genes affecting bristle length and body color
in fruit flies
– s = short bristles
– s+ = normal bristles
– e = ebony body color
– e+ = gray body color
– One parent displays both recessive traits
• It is homozygous recessive for the two genes (ss ee)
– The other parent is heterozygous for the two genes
• The s and e alleles are linked on one chromosome
• The s+ and e+ alleles are linked on the homologous chromosome
•Genetic mapping experiments are typically accomplished by carrying out a
testcross
–A mating between an individual that is heterozygous for two or more genes and
one that is homozygous recessive for the same genes
Chromosomes are the
product of a crossover
during meiosis in the
heterozygous parent
Recombinant offspring
are fewer in number
than nonrecombinant
offspring
Frecuencia de Recombinación 50%
The two genes are in
different chromososmes
Recombinant classes >= 50%
The two genes are in the
same chromosome
Recombinant classes < 50%
Coeficiente de Coincidencia
The coefficient of coincidence (c.o.c.) is a measure of interference (or
independence in the formation of chiasmata) in the formation of chromosomal
crossovers during meiosis.
It is generally the case that, if there is a crossover at one spot on a chromosome,
this decreases the likelihood of a crossover in a nearby spot. This is called
interference.
This tells us how strongly a crossover in one of the DNA regions (AB or BC) interferes
with the formation of a crossover in the other region.
The coefficient of coincidence is typically calculated from recombination rates between
three genes. If there are three genes in the order A B C, then we can determine how
closely linked they are by frequency of recombination. Knowing the recombination rate
between A and B and the recombination rate between B and C, we would naively
expect the double recombination rate to be the product of these two rates.
C.O.C= Double crossover observed
Double crossover expected
Interference = 1 - C.O.C.
Drosophila males of genotype AaBbCc were crossed with females of genotype aabbcc.
This led to 1000 progeny of the following phenotypes:
ABC: 254 (parental genotype, shows no recombination)
ABc: 71 (recombinant between B and C)
AbC: 23 (double recombinant)
Abc: 152 (recombinant between A and B)
aBC: 148 (recombinant between A and B)
aBc: 27 (double recombinant)
abC: 79 (recombinant between B and C)
abc: 246 (parental genotype, shows no recombination)
From these numbers it is clear that the B/b locus lies between the A/a locus and the C/c
locus.
There are 23 + 152 + 148 + 27 = 350 progeny showing recombination between genes A
and B. And there are 71 + 23 + 27 + 79 = 200 progeny showing recombination between
genes B and C. Thus the expected rate of double recombination is (350 / 1000) * (200 /
1000) = 0.07 (or 70 per 1000).
However, there are actually only 23 + 27 = 50 double recombinants. The coefficient of
coincidence is therefore 50 / 70 = 0.71. (C.O.C. = double cross obs/ double cross expect)
Interference is 1 - 0.71 = 0.29. (interference = 1- C.O.C.)
Intereference = It is generally the case that, if there is a
crossover at one spot on a chromosome, this decreases the
likelihood of a crossover in a nearby spot
C.O.C. = 0, complete interference
0<C.O.C.<1 more the two genes are distant less is the
interference
C.OC. = 1, no interference
(esperados=observados)*
Remember that in the Drosophila
male you do not have recombination
(Morgan, 1912, 1914).
• Other scientific studies showed that you can actually have some really
low in frequency spontaneous recombination in male (Henderson,
1977).
female
male
abC
X ABc =
abC
F1
a b C
A B c
X
X
X X
Female phenotype
ab
abC
Wt
abC
a
abC
abC
b
abC
ab
abC
Wt
a
F2
+ 1) a b C
+ 2) A B c
+ 3) a B c
+ 4) A b C
+ 5) a b c
+ 6) A B C
a b C + 7) a B C
a b C + 8) A b c
b
P
X
abC
One cross
over
One cross
over
Double
cross over
Male phenotype
ab
c
ac
b
abc
Wt
a
bc
Ligamiento en genes ligados al sexo en Drosophila
P1 ♀♀ y w m
y w m
♂♂ + + +

F1 ♀♀ y w m
♂♂ y w m ¿Qué tipo de prueba es
esta?
+ + +
F2
Fenotipos
No.
NCO
+ + +
758
ywm
700
COS-1 y + +
12
+wm
16
COS-2 + + m
317
y w+
401
COD
y +m
0
+ w +
1
Total 2,205
• Distancia
y–w
w-m
Parentales
y w
wm
+ +
+ +
Recombinantes y + 12 + 0
w + 317 + 0
+ w 16 + 1
+ m 401 + 1
Total
29
719
29/2,205x100=1.32u
719/2,205x100=32.60u
y
w
1.32u
m
32.60u
y
y-m
y m
+ +
y + 12 + 317
+ m 16 + 401
746
746/2,205x100=33.83u
1.32 + 32.60= 33.92
- 33.83
m
0.09
33.83u
¿Por qué encontramos 0.09 de diferencia?
Frecuencia observada de COD es 1 / 2,205 = 0.00045. Cada COD es equivalente a
2COS: 0.00045 x 2 = 0.09 Al calcular la distancia entre y – m no contamos dos
COS. La distancia “correcta” siempre será la suma de las distancias cortas.
Recuerda, El COD se calcula a base de la ley de porbabilidad de dos eventos
independientes que ocurren al mismo tiempo.
Coeficiente de coincidencia e Interferencia
Frecuencia observada 1 / 2,205 = 0.00045
Frecuencia esperada (frec. COS 1)(frec. COS 2) = (0.0132)(0.3260) = 0.004
CC = Frec. Observada / Frec. Esperada = 0.00045 / 0.004
= 0.225 x 100 = 22.5%
Los resultados indican que hubo interferencia, por lo tanto, los eventos no
son independientes.
¿Cuánta interferencia hubo?
Interferencia = 1-CC = 1-0.225 = 0.775 ó 77.5%
Recuerda, mientras más cercanos los genes, mayor la interferencia. La
frecuencia de recombinación observada no excederá 50%.
¿Por qué?
Wild type
forked
Wild type
vermillion
white
yellow
Bar
white apricot
crossvein
crossveinless