Transcript Chapter 14 - HCC Learning Web

Text authored by Dr. Peter J. Russell
Slides authored by Dr. James R. Jabbur
CHAPTER 14
Gene Mapping in
Eukaryotes
Introduction



Genes on nonhomologous chromosomes assort
independently
Syntenic genes may be inherited together and
belong to a linkage group
Classical genetics analyzes the frequency of allele
recombination in the progeny of genetic crosses

New associations of parental alleles are recombinants,
produced by genetic recombination
 Testcrosses help to determine which genes are linked and
a linkage map (genetic map) is constructed for each
chromosome
Morgan’s Experiments with Drosophila




Both the white eye gene and the gene for miniature wings
are on the fly X chromosome
Female white eyed, miniature winged flies were crossed
with wild-type males (wm/wm x w+m+/Y)
In the F1, all males were white-eyed with miniature wings
(wm/Y) and all females were wild type for both eye color
and wing size (w+m+/wm)
F1 interbreeding was then undertaken (this action is the
equivalent of a testcross, as we know the genotypes)


In the F2, the most frequent phenotypes for both sexes were the
phenotypes of the parents in the original cross (white eyes with
miniature wings, and red eyes with normal wings)
Nonparental phenotypes (white eyes with normal wings or
red eyes with miniature wings) occurred in about 37% of the F2
flies. This is statistically well below the 50% prediction rate for
independent assortment, indicating that non-parental flies result
from the recombination of linked genes

From his studies, Morgan proposed that:
 During
meiosis, alleles of some genes assort
together because they are near each other on
the same chromosome
 Recombination occurs when genes are
exchanged between the X chromosomes of
the F1 females (it must occur in the female
population because males do not have 2
homologous X chromosomes)

Repeated studies produced similar results
Recombination and
Exchange



Proof that physical exchange
between chromosomes results in
genetic recombination came with
studies in Drosophila
Genetic and physical (cytological)
markers were employed to analyze
genetic recombination in meiosis
Curt Stern worked with 2 linked
gene loci, car and bar (B):

The car eye color gene is recessive; red
eyes are wild-type
 The bar (B) eye shape gene is
incompletely dominant





Wild-type eye shape is round
Heterozygote eye shape is kidney-shaped
Recessive eye shape is bar-shaped
The crosses and results of allelic
segregation are shown
Cytological examination confirmed
that physical crossing-over results
in genetic recombination
Animation: Recombination & Exchange
Constructing Genetic Maps
Detecting Linkage through Testcrosses


Linked genes are used for mapping. They are
found by looking for deviation from the frequencies
expected from independent assortment
Employing a testcross where one parent is truebred, homozygous recessive works well for
analyzing linkage:

If the alleles are not linked and the second parent is
heterozygous, all four possible combinations of traits will
be present in equal numbers in the progeny
 A significant deviation in the aforementioned ratio (more
parental and fewer recombinant types) indicates linkage

Chi-square analysis is used to analyze the testcross
data and determine whether a deviation is
“significant.” A null hypothesis (“the genes are not
linked”) is used because it is not possible to predict
phenotype frequencies produced by linked genes

In this case, the deviation is significant and the
hypothesis is not valid (thus, the genes are linked)
Animation: Chi-Square Test
1479.4
A P value of less than 5% (0.05) indicates a poor fit, results are rejected
The Concept of a Genetic Map

In an individual heterozygous at two loci, there are
two arrangements of alleles:

The cis (coupling) arrangement has both wild-type
alleles on one homologous chromosome, and both
mutants on the other (i.e., w+m+ and wm)
 The trans (repulsion) arrangement has one mutant and
one wild type on each homolog (e.g., w+m and wm+)
 A crossover between homologs in the cis arrangement
results in a homologous pair with the trans arrangement.
A crossover between homologs in the trans arrangement
results in cis homologs.



Crosses in Drosophila showed that crossover
frequency for linked genes is characteristic for each
gene pair (the frequency stays the same, whether
the genes are arranged in cis or trans)
Sturtevant (Morgan’s student) used recombination
frequencies to make a genetic map (a 1%
crossover rate is a genetic distance of 1 map unit or
centiMorgan)
The first genetic map was based on crosses
involving 3 sex-linked genes (w, white eyes; m,
miniature wings; y, yellow body color)
…important point

The farther apart the two genes are on the
chromosome, the more likely it is that a
crossover will occur between them and
therefore, the greater the occurrence of
their crossover (frequency)
Gene Mapping with a 2
Point Testcross


With autosomal recessive
alleles, when a double
heterozygote is crossed,
four phenotypic classes
are expected
If the genes are linked, the
two parental phenotypes
will be about equally
frequent and more
abundant than the two
recombinant phenotypes
Generating a Genetic Map


A genetic map is generated from estimating the
crossover rate in a particular segment of the
chromosome
If you have linked genes with a crossover:
R.F. (mu) = # observed recombinants X 100
# observed total progeny

Problem: It may not exactly match the physical map:

The nonrandom distribution of crossing-over limits the
accuracy of mapping. Crossover events are not equally
probable at all sites on the chromosome (hotspots exist)***
 Double crossovers between two loci can restore the partental
genotype (as will any number of even crossovers) (next slide)
Here:
Gene Mapping with 3 Point Testcrosses



Typically, geneticists design experiments to gather
data on several traits in one testcross
In the progeny, each gene has two possible
phenotypes. For 3 genes, there are 8 expected (23)
phenotypic classes in the progeny
Establishing the order of genes on the chromosome is
determined from the phenotypic results of the cross
Two classes will be parental (and most abundant)
 Of the six remaining classes, 2 will be present at the lowest
frequency, resulting from an apparent double crossover,
thereby establishing the apparent gene order

Animation: 3 Point Mapping


The above figure demonstrates the consequences of a double
crossover in a triple heterozygote for three linked genes
In a double crossover, the middle allelic pair changes its
orientation relative to the outside allelic pairs, producing 2
parental and 2 recombinant (allele) gametes



The apparent
gene order is
determined by the
progeny which
accounts for the
lowest apparent
frequency
This group
represents the
double
recombinant
Thus, in the
example shown,
the correct gene
order is
rearranged to pjr
(look at the
genotype in the
table for the clue)

With the gene order properly
arranged, we can calculate the
recombination frequencies for
the genes
RF =


sco + dco x 100%
total progeny
The percentage of progeny
generated by crossing-over
between p and j is 20.8%
(52+46+4+2/500); between j
and r is 10% (22+22+4+2/500)
To compute the map distance
between outside genes, simply
add the two map distances
 Of
course, the ultimate
means of generating a
genetic map is high
throughput sequencing
Filial Phenotype
What Progeny
Looks like?
What kind of crossover
took place?
Number of
Observed
ABC
First parent
None
401
Abc
Second parent
None
389
ABc
Recombinant types
Double (around C)
4
abC
Recombinant types
Double (around C)
6
AbC
Recombinant types
Single (between C and B)
75
aBc
Recombinant types
Single (between C and B)
65
aBC
Recombinant types
Single (between C and A)
35
Abc
Recombinant types
Single (between C and A)
25
1000
Filial
Phenotype
What Progeny
Looks Like?
Which traits were
inherited from one of
the parents?
Which trait came
from the other
parent?
ABC
First Parent
ABC
None
abc
First Parent
Abc
None
ABc
Recombinant Types
AB from the ABC Parent
c
abC
Recombinant Types
ab from the abc parent
C
AbC
Recombinant Types
AC from the ABC parent
b
aBc
Recombinant Types
ac from the abc parent
B
aBC
Recombinant Types
BC from the ABC parent
a
Abc
Recombinant Types
bc from the abc parent
A
Calculating Recombination Frequency

Map distance between A and C:
All double crossovers + all single crossovers involving A and C
total

Map distance between C and B:
All double crossovers + all single crossovers involving B and C
total