Transcript Pierce Genetics: A Conceptual Approach 3e

Benjamin A. Pierce
•GENETICS ESSENTIALS
•Concepts and Connections
• SECOND EDITION
CHAPTER 5
Linkage, Recombination, and
Eukaryotic Gene Mapping
© 2013 W. H. Freeman and Company
CHAPTER 5 OUTLINE
• 5.1 Linked Genes Do Not Assort Independently, 114
• 5.2 Linked Genes Segregate Together and Crossing
Over Produces Recombination Between Them, 116
• 5.3 A Three-Point Testcross Can Be Used To Map Three
Linked Genes, 128
• 5.4 Genes Can Be Located with Genomewide
Association Studies, 137
THE CASE OF MALE BALDNESS
• X-linked or not?
• Study of linkage of a SNP
and the association with
male pattern baldness
shows that they do not
assort independently, they
are transmitted together
5.1 LINKED GENES DO NOT ASSORT
INDEPENDENTLY
• Linked genes: genes located close together on the
same chromosome
-Belong to the same linkage group
RECOMBINATION AND ALLELES
FLOWER COLOR AND POLLEN SHAPE
F1 GENERATION GIVES PURPLE COLOR
AND LONG POLLEN
• Appears that the
purple is dominant to
red and long is
dominant to short
pollen
• What with the F2, it is
expected to get 9:3:3:1
• BUT…
THERE IS NO 9:3:3:1 RATIO; WHAT
HAPPENED
5.2 Linked Genes Segregate Together and
Crossing Over Produces Recombination
Between Them
• Fig. 5.3
• Notation for crosses with linkage
• Complete linkage leads to nonrecombinant
Gametes and nonrecombinant progeny
• Crossing over with linked genes leads to
recombinant gametes and recombinant progeny
• Figs. 5.4 & 5.5
LABELING ALLELES OF TWO GENES ON
A SAME CHROMOSOME
• Cross AA BB x aa bb can be represented by
A
A
B
B
x
a
a
The F1 will be:
A
B
a
b
Also it can be represented
A
B
a
b
b
b
THE TWO TEST CROSS CASES
LINKED V.S. NOT LINKED
• Linked genes give only
nonrecombinant
gametes
• Unlinked genes give
both nonrecombinant
and recombinant
CONCEPT CHECK 1
For single crossovers, the frequency of
recombinant gametes is half the frequency of
crossing over because:
a. a test cross between a homozygote and
heterozygote produces ½ heterozygous and ½
homozygous progeny.
b. the frequency of recombination is always 50%.
c. each crossover takes place between only two of
the four chromatids of a homologous pair.
d. crossovers take place in about 50% of meiosis.
WHAT HAPPENS IN RECOMBINATION
AND THE GAMETE FORMATION
5.2 LINKED GENES SEGREGATE TOGETHER
AND CROSSING OVER PRODUCES
RECOMBINATION BETWEEN THEM
• Calculating Recombination Frequency
• Recombination frequency = (No. recombinant progeny/Total No. of
progeny)  100%
• Fig. 5.6
• Coupling and Repulsion Configuration of Linked Genes
• Coupling (cis configuration): wild-type alleles are found on one
chromosome; mutant alleles are found on the other chromosome.
CALCULATING FREQUENCIES
CALCULATING FREQUENCIES
Rec. Frequency = # of recombinant progeny x100%
Total # of progeny
Rec. Frequency =
8 + 7 x100%
55+53+8+7
Rec. Frequency = 15 x100%
123
Rec. Frequency = 12.2 % OR 0.122
5.2 LINKED GENES SEGREGATE TOGETHER
AND CROSSING OVER PRODUCES
RECOMBINATION BETWEEN THEM
• Coupling and Repulsion Configuration of Linked Genes
• Repulsion (trans configuration): wild-type allele and mutant allele
are found on the same chromosome.
• Fig. 5.7
• Testing for Independent Assortment
• Figs. 5.8 & 5.9
COUPLING AND REPULSION
p+
p
b+
b
p+
p
b
b+
COUPLING AND REPULSION
5.2 Linked Genes Segregate Together and
Crossing Over Produces Recombination
Between Them
• Gene Mapping with Recombination Frequencies:
• Genetic maps are determined by recombinant frequency.
• Map unit and centiMorgans
• Constructing a Genetic Map with the Use of Two-Point
Testcrosses
• Fig. 5.10
RECOMBINATION FREQUENCIES
AND DISTANCES OF GENES
• Based on recombination frequencies genetic maps
are generated
• Actual distances are on physical maps
• Recombination maps are approximations and are
measured in map units or cM (centiMorgans)
•
One map unit equals to 1% recombination frequency
Genetic distances
measured with
recombination rates
are approximately
additive: if the
distance from gene A
to gene B is 5 m.u.,
the distance from
gene B to gene C is
10 m.u., and the
distance from gene A
to gene C is 15 m.u.
LIMITATIONS OF GENE MAPPING
• If the genes are far apart and the
frequency is 50% we cannot say if the
genes are on different chromosomes or
just to far on same chromosome
• Two genes that are relatively far will give
underestimate of recombination events
due to possibility of multiple crosses
between the genes.
EXAMPLE: TWO POINT CROSS TO
DETERMINE MAP OF 4 GENES
5.3 A THREE-POINT TESTCROSS CAN BE
USED TO MAP THREE LINKED GENES
• Constructing a Genetic Map with the Three-Point
Testcross
• Fig. 5.12
• Determining the gene order
• Determining the location of crossovers
• Fig. 5.13
DOUBLE CROSSOVER :THREE GENES
A THREE-POINT TESTCROSS
CAN BE USED TO MAP THREE
LINKED GENES
5.3 A Three-Point Testcross Can Be Used
to Map Three Linked Genes
• Calculating the recombination frequencies
• Interference and coefficient of coincidence
• Effect of multiple crossovers
CALCULATING THE RECOMBINATION
FREQUENCIES
• Based on the numbers in the fruit fly testcross for
three loci calculate the distances between the loci.
• Recombinant progeny with a chromosome that underwent
crossing over between the eye-color locus (st) and the
bristle locus (ss) include the single crossovers ( st+ / ss
e and st / ss+ e+ ) and the two double crossovers ( st+ / ss /
e+ and st / ss+ / e ); Total of 755 progeny; so the
recombination frequency between ss and st is:
CALCULATING THE RECOMBINATION
FREQUENCIES
• Based on the numbers in the fruit fly testcross for
three loci calculate the distances between the loci.
• The map distance between the bristle locus (ss) and the
body locus (e) is determined in the same manner. The
recombinant progeny that possess a crossover between ss
and e are the single crossovers st+ ss+ / e and st ss / e+ and
the double crossovers st+ / ss / e+ and st / ss+ / e . The
recombination frequency is:
CALCULATING THE RECOMBINATION
FREQUENCIES
• Based on the numbers in the fruit fly testcross for
three loci calculate the distances between the loci.
• Finally, calculate the map distance between the outer two
loci, st and e. This map distance can be obtained by
summing the map distances between st and ss and
between ss and e (14.6 m.u. + 12.2 m.u. = 26.8 m.u.). We
can now use the map distances to draw a map of the three
genes on the chromosome:
INTERFERENCE AND COEFFICIENT OF
COINCIDENCE
• Calculate the proportion of double-recombinant
gametes by using the multiplication rule of
probability
• Applying this rule, the proportion (probability) of gametes
with double crossovers between st and e is equal to the
probability of recombination between st and ss multiplied
by the probability of recombination between ss and e, or
0.146 × 0.122 = 0.0178.
• Multiplying this probability by the total number of progeny
gives us the expected number of double-crossover progeny
from the cross: 0.0178 × 755 = 13.4.
• Only 8 double crossovers—considerably fewer than the 13
expected—were observed in the progeny of the cross
INTERFERENCE AND COEFFICIENT OF
COINCIDENCE
• Crossovers are frequently not independent events:
the occurrence of one crossover tends to inhibit
additional crossovers in the same region of the
chromosome, and so double crossovers are less
frequent than expected.
• The degree to which one crossover interferes with
additional crossovers in the same region is termed the
interference. To calculate the interference, we first
determine the coefficient of coincidence, which is the ratio
of observed double crossovers to expected double
crossovers:
IN OUR CASE
• indicates that we are actually observing only 60% of
the double crossovers that we expected on the basis of
the single-crossover frequencies.
• Calculation of interference
• Interference = 1 − coefficient of coincidence
• So the interference for our three-point cross is:
• interference = 1 − 0.6 = 0.4
• 40% of recombinations are not observed due to
intefrerence
5.4 GENES CAN BE LOCATED WITH
GENOMEWIDE ASSOCIATION STUDIES
•
•
•
•
Linkage analysis (genetic markers; anonymous markers)
Genome wide association studies
Haplotype
Linkage disequilibrium