Transcript GENETICS – BIO 300

LECTURE 06: EUKARYOTE CHROMOSOME
MAPPING AND RECOMBINATION I
CH4 key concepts
linkage
recombination
linkage maps
3-point test cross
2 test for linkage
CH4 problems... so far
LECTURE 06: EUKARYOTE CHROMOSOME
MAPPING AND RECOMBINATION I
can new combinations
of “linked genes” be
inherited?
by what mechanism?
is the frequency of new
combinations related to
their distance apart on
the chromosome?
how do we know if
genes are linked?
MENDELIAN ANALYSIS: 1  n GENES
... a quick review from CHAPTER 2
3 methods of working out expected outcomes of
controlled breeding experiments:
1. Punnet square
2. tree method (long) – genotypes
3. tree method (short) – phenotypes
MENDELIAN ANALYSIS: PUNNET SQUARE
all pairings of  and  gametes
= probabilities of all pairings
some pairings occur >1
 different Ps for different
genotypes & phenotypes
1 gene, 2 x 2 = 4 cells
2 genes, 4 x 4 = 16 cells
3 genes, 8 x 8 = 64 cells...
too much work !
MENDELIAN ANALYSIS: LONG TREE
1 gene, alleles A, a
GENOTYPE
1/4 A/A

1/4 A/A
1/2 A/a

1/2 A/a
1/4 a/a

1/4 a/a
MENDELIAN ANALYSIS: LONG TREE
1 gene, alleles A, a
GENOTYPE
PHENOTYPE
3/4 A
1/4 A/A

1/4 A/A
1/2 A/a

1/2 A/a
1/4 a/a

1/4 a/a
1/4 a
MENDELIAN ANALYSIS: LONG TREE
2 genes, alleles A, a; B, b... GENOTYPE
1/4 A/A
1/2 A/a
1/4 a/a
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b









1/16 A/A B/B
1/8 A/A B/b
1/16 A/A b/b
1/8 A/a B/B
1/4 A/a B/b
1/8 A/a b/b
1/16 a/a B/B
1/8 a/a B/b
1/16 a/a b/b
MENDELIAN ANALYSIS: LONG TREE
2 genes, alleles A, a; B, b... GENOTYPE PHENOTYPE
1/4 A/A
1/2 A/a
1/4 a/a
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b
1/4 B/B
1/2 B/b
1/4 b/b









1/16 A/A B/B
1/8 A/A B/b
1/16 A/A b/b
1/8 A/a B/B
1/4 A/a B/b
1/8 A/a b/b
1/16 a/a B/B
1/8 a/a B/b
1/16 a/a b/b
9/16 AB
3/16 Ab
3/16 aB
1/16 ab
MENDELIAN ANALYSIS: SHORT TREE
2 genes, alleles A, a; B, b...
PHENOTYPE
¾B

9/16 AB
¼b

3/16 Ab
¾B

3/16 Ab
¼b

1/16 ab
¾A
¼a
much easier... can extend to >2 genes
CHAPTER 4: KEY CONCEPTS
genes close together on a chromosome do
not assort independently at meiosis
recombination  genotypes with new
combinations of parental alleles
homologous chromosomes can exchange
segments by crossing-over
recombination results from either
independent assortment or crossing-over
CHAPTER 4: MORE KEY CONCEPTS
genes can be mapped by measuring
frequencies of recombinants produced by
crossing-over
interlocus map distances based on
recombination measurements are ~ additive
one crossover can influence the occurrence
of a second one in an adjacent region
LINKAGE
observe deviations from
9 : 3 : 3 : 1 ratios derived from dihybrid crosses
1 : 1 : 1 : 1 ratios derived from test crosses
too many
too few
LINKAGE SYMBOLISM & TERMINOLOGY
alleles on the same homologue have no
punctuation between them... a b
alleles on different homologues separated by a
slash... A B / a b
alleles written in same order on each
homologue... A d B c / a D b C
unlinked genes separated by semicolon...
A/a ;B / b
genes of unknown linkage separated by dot...
A/a • B / b
LINKAGE
unlinked genes, different chromosomes
write as ... Aa; Bb or A/a; B/b
illustrate as ...
A
a
;
B
b
or
A
a
;
b
B
segregate and assort independently
test cross to a/a; b/b  4 progeny types ...
1 AB : 1 Ab : 1 aB : 1 ab
LINKAGE
linked genes, same chromosomes
write as ... Aa Bb or AB/ab or Ab/aB
2 possible dihybrids, illustrate as ...
AB
a b
‘cis’ or
‘coupling’
or
Ab
a B
‘trans’ or
‘repulsion’
segregate and assort dependently ~ distance
test cross to ab/ab  4 progeny types ...
in CIS: < 1 AB : > 0 Ab : > 0 aB : < 1 ab
in TRANS: > 0 AB : < 1 Ab : < 1 aB : > 0 ab
LINKAGE
linked genes, same chromosomes
LINKAGE
chiasmata: visible
manifestations of
crossing over
LINKAGE
linked genes (in cis here), same chromosomes
also called non-parental
or recombinant chromosomes
LINKAGE
crossing over when in meiosis? (tetrad analysis)
if

if

LINKAGE
how many chromatids involved? (tetrad analysis)


RECOMBINATION
products of meiotic recombination
linkage
unknown
ati
RECOMBINATION
detection of recombination using a test cross
RECOMBINATION
test cross
unlinked genes
independent
assortment
always 
“recombination”
frequency of 50%
RECOMBINATION
recombinants arise when non-sister chromatids
cross over between genes under study
RECOMBINATION
test cross
linked genes
independent
assortment 
recombination
frequency < 50%
2 TEST FOR LINKAGE
A
O
a
E
O
E

B
140 / 127.5 115 / 127.5
255
b
110 / 122.5 135 / 122.5
245

250
250
500
2 TEST FOR LINKAGE
A
O
a
E
O
E

B
140 / 127.5 115 / 127.5
255
b
110 / 122.5 135 / 122.5
245

250
250
500
2 TEST FOR LINKAGE
2 TEST FOR LINKAGE
express this as: 0.05 > P(2c = 5.02) > 0.01
the data derived from a test cross deviate significantly
from a 1:1:1:1 ratio
we reject our H0 (alternatively we could not reject)
“genes A and B are linked”
the probability of deviation by chance from the linked
genes model is between 1 and 5% (i.e., deviation from
the 1:1:1:1 ration occurs because linkage is not
supported by data)
RECOMBINATION
Autosome Linkage
P
F1
F2
cn bw+  x cn+ bw 
cn bw+
cn+ bw

cn bw  x cn bw+ 
cn bw
cn+ bw

score progeny
no recombination
in Drosophila s
must use s only
X-Chromosome
X-Chromosome Linkage
Linkage
y w+  x y+ w 
Y
y+ w
y+ w  x y w+ 
Y
y+ w

score  progeny
RECOMBINATION
Autosome Linkage
P
F1
F2
X-Chromosome
X-Chromosome Linkage
Linkage
cn bw+  x cn+ bw 
cn bw+
cn+ bw

cn bw  x cn bw+ 
cn bw
cn+ bw

score progeny
tester 
chromosomes
y w+  x y+ w 
Y
y+ w
y+ w  x y w+ 
Y
y+ w

score  progeny
LINKAGE MAPS
sites of recombination are ~ random
probability of recombination increases ~ distance
LINKAGE MAPS
complications:
RECOMBINATION
FREQUENCY
1. recombination rates  position on chromosome
Telomere
2.
3.
4.
5.
6.
centromere
telomere
linked genes >50 cM apart look like unlinked genes
double recombination events are usually not detected
multiple strand (>2) exchanges
tester for multiple autosomal genes not always available
interference – reduced probability of adjacent events (later)
LINKAGE MAPS
2 possibilities for map
add the shorter distances
3-POINT TEST CROSS
P
v+
cv+
cv ct x v
v+ cv ct
Y
ct+

F1
F2
v+ cv ct x v cv+ ct+
v cv+ ct+
Y
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
double crossover involving 2 chromatids
outer alleles look like parental types
recombination cryptic without middle gene
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT with DCO gives
expected
3. calculate distances:
2 genes at a time
4. construct map
5. calculate interference
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT alleles with
DCO gives expected
3. calculate distances:
2 genes at a time
4. construct map
5. calculate interference
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see your
if PT alleles
with on...
Base
decision
DCOsimilar
gives expected
(1)
numbers
3. calculate
distances:
(2)
reciprocal
genotypes
2 genes at a time
4. construct map
5. calculate interference
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT alleles with
DCO gives expected
3. calculate distances:
2 genes at a time
4. construct map
5. calculate interference
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT alleles with
DCO gives expected
3. calculate
distances: linear
Draw
all 3 possible
2 genes at a(next),
time what do
arrangements
4. construct
map
double
crossovers
between
5. calculate
interference give...
both
PT chromosomes
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
test all 3 possible orders of PT allele combinations

ct+
3
5
ct+
94
89
ct+
45
40
3-POINT TEST CROSS
test all 3 possible orders of PT allele combinations

3
5
PT chromosomes with genes in the order v ct cv
(NOT the order shown in the table... it usually won’t
be) give you the two F2 groups with the smallest
numbers with double crossovers between the genes.
3-POINT TEST CROSS
REAL GENE ORDER
DOES NOT HAVE TO
BE AS LISTED IN
THE DATA TABLE.
Consider genes v & ct.
PT are v ct+ or v+ ct
in TRANS. All single
crossovers between
them yield reciprocal
v ct & v+ ct+ CIS
products as shown.
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
order of operations
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT alleles with
DCO gives expected
3. calculate distances:
2 genes at a time;
e.g.: v  ct =
[ ( 89 + 94 + 3 + 5 ) / 1448 ] x 100
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT alleles with
DCO gives expected
3. calculate distances:
2 genes at a time
4. construct map...
5. calculate interference
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
distances between outer genes is less accurate
because double recombinants are cryptic
v
ct
13.2 cM
cv
6.4 cM
18.5 cM
or... 13.2 + 6.4 = 19.6 cM
add distances for accurate map
3-POINT TEST CROSS
order of operations
1. look for reciprocal groups:
parental type (PT),
single cross over (SCO),
double cross over (DCO)
2. gene order: test all 3 to
see if PT alleles with
DCO gives expected
3. calculate distances:
2 genes at a time
4. construct map
5. calculate interference
Genotype
v cv+ ct+
v+ cv ct
v cv ct+
v+ cv+ ct
v cv ct
v+ cv+ ct+
v cv+ ct
v+ cv ct+
total
#
580
592
45
40
89
94
3
5
1448
3-POINT TEST CROSS
interference
= 1 – COEFFICIENT OF COINCIDENCE
= 1 – [ OBSERVED # DCO / EXPECTED # DCO ]
= 1 – [ 8 / ( 0.132 x 0.064 x 1448 ) ]
= 1 – [ 8 / 12 ] = 1 / 3 = 33 %
LINKAGE MAPS
why is mapping important ?
 fundamental aspect of genetic analysis,
concerns:
 gene function
 gene evolution
 gene isolation
LINKAGE MAPS
gene function ...
 location influences function... “position effect”
 relative to heterochromatin
 other genes
 genes of related function often clustered
 operons in Prokaryotes
 homeotic genes in Eukaryotes
LINKAGE MAPS
gene evolution ...
 relative positions of genes in related
organisms infer history of change
LINKAGE MAPS
gene isolation ...
 mapping 1st step
 Drosophila ...
LINKAGE MAPS
gene isolation ...
 mapping 1st step
 tomato ...
prophase
nucleus
chromosome
numbering
map (ca. 1952)
LINKAGE MAPS - HUMANS
why has this been difficult ?
 controlled breeding
programs
not possible (not
ethical)
 sample sizes small
(relatively)
 human genome is LARGE...
EUKARYOTE CHROMOSOME MAPPING
AND RECOMBINATION: PROBLEMS
in Griffiths chapter 4, beginning on page 141, you
should be able to do questions #1-30
begin with the solved problems on page 138 if you are
having difficulty
look at the way Schaum’s Outline discusses linkage
and mapping for alternative explanations - especially
tetrad analyses
try Schaum’s Outline questions in chapter 4, beginning
on page 208
TUTORIAL #2: T.9.26 or R.9.28
go to the TUTORIAL page on the genetics web site
to download the file tut2-06F.pdf
complete all six questions before attending tutorial
#2 next week
during tutorial, you will review solutions to these
questions with your group and prepare to present
solutions to all six questions
one member of your group will be selected by the TA
to present a solution to one of these questions
presentations will be 10 min each + 5 min for
questions and discussion