LECTURE 07: EUKARYOTE CHROMOSOME
MAPPING AND RECOMBINATION II
accurate calculation of large
map distances
mapping function
analysis of single meioses
ordered: gene  centromere
unordered: gene  gene
CHAPTER 4: FURTHER IDEAS
double & higher multiple crossovers  underestimates
of map distances calculated from recombination
specialized mapping formulae  accurate map
distance corrected for multiple crossovers
analysis of single meioses (in Fungi) can...
position centromeres on genetic maps (~ genes)
 mechanisms of segregation and recombination
crossovers occur occasionally in mitotic diploid cells
ACCURATE MAPPING
mapping large distances is less accurate
best estimate of map distance obtained by adding
distances calculated for shorter intervals
if possible, include more genes in the map
ACCURATE MAPPING
problems occur when you have ...
no intervening genes
genes very close together
ACCURATE MAPPING
account for multiple crossovers ?
need mapping function to correct
for multiple events and accurately
relate recombination to map
distance
POISSON DISTRIBUTION
low #s sampled from large population
possible numbers obtainable are
large, but most samples are small
e.g. random distribution of 100 x 1$
in class of 100 students... few
students receive many bills...
POISSON DISTRIBUTION
e.g. random distribution of 100 x 1$
in class of 100 students
POISSON DISTRIBUTION
here, average is 1 bill/student... m = 1
# for a particular class... i = 0, 1, 2... 100
POISSON DISTRIBUTION
here, mean is 1 bill/student... m = 1
because 100 bills & 100 students
# for a particular class... i = 0, 1, 2...
100
how many students are predicted to
capture 3 bills?
POISSON DISTRIBUTION
here, mean is 1 bill/student... m = 1
because 100 bills & 100 students
# for a particular class... i = 0, 1, 2...
100
how many students are predicted to
capture 3 bills?
e-m mi
i!
f(i) = ————
POISSON DISTRIBUTION
POISSON DISTRIBUTION
f(0) = 0.368
f(1) = 0.368
f(2) = 0.184
f(3) = 0.061
f(4) = 0.015
f(5) = . . .
POISSON DISTRIBUTION
proportion of class with i items
different m values
MAPPING FUNCTION
use Poisson to describe distribution of
crossovers along chromosome
if...
crossovers are random,
we know mean # / region on chromosome
then we can calculate distribution of meioses
with 0, 1, 2... n multiple crossovers
MAPPING FUNCTION
recombination frequency (RF) = % recombinants
meiosis with 0 crossovers  RF of 0%
meiosis with 1 crossover  RF of 50%
MAPPING
FUNCTION
meioses with 0
crossovers 
RF of 0%
meioses with >
0 crossovers 
RF of 50%
compare the
non-recombinant
chromatids
recombinants
shown darker
MAPPING FUNCTION
recombinants make up half of the products
of meioses with 1 or more crossovers
0 crossover class is the only critical one
proportion of meioses with at least one
crossover is 1 – 0 class; the 0 class is...
-m m0
e
f(0) = ———— = e-m
0!
MAPPING FUNCTION
proportion of meioses with at least one
crossover is 1 – 0 class, which is...
-m m0
e
f(0) = ———— = e-m
0!
so the mapping function can be stated as...
RF = ½ (1 – e-m)
MAPPING FUNCTION
RF = ½ (1 – e-m)
for low m...
m = 0.05, RF = ½ m
m = 0.1, RF = ½ m
m = 1, RF = 50
... RF = m / 2 at the
dashed line
or
use the equation...
~40
MAPPING FUNCTION
RF = ½ (1 – e-m)
RF = 27.5 cM ?
0.275 = ½ (1 – e-m)
0.55 = 1 – e-m
e-m = 1 – 0.55 = 0.45
m  0.8 (mean # of crossovers / meiosis)
corrected RF = m/2 = 0.4 = 40 % or 40 cM
MAPPING FUNCTION
RF = ½ (1 – e-m) = 40 cM
for low m...
m = 0.05, RF = ½ m
m = 0.1, RF = ½ m
m = 1, RF = 50
... RF = m / 2 at the
dashed line
or
use the equation...
~40
MAPPING FUNCTION
mapping large distances is less accurate
best estimate of map distance obtained by adding
distances calculated for shorter intervals
if possible, include more genes in the map
put RF values through a mapping function
ANALYSIS OF SINGLE MEIOSES
products of meiosis remain together
groups of haploid cells... either 4 (tetrads) or 8 (octads)
ANALYSIS OF SINGLE MEIOSES
Neurospora crassa (we use Sordaria fimicola in Lab 5)
note pigment phenotypes of the ascospores
ANALYSIS OF SINGLE MEIOSES
meiosis & post-meiotic mitosis in linear tetrad / octad
ANALYSIS OF SINGLE MEIOSES
2 kinds of mapping with tetrads / octads:
ordered analysis to map gene  centromere
unordered analysis to map gene  gene
ORDERED ANALYSIS
no crossing over between gene A and centromere
A & a segregate
to different poles

“MI” segregation
ORDERED ANALYSIS
crossing over between gene A and centromere
A & a segregate
to different poles

“MII” segregation
ORDERED ANALYSIS
4 types of MII patterns
equal frequencies
ORDERED ANALYSIS
A
a
A
a
A
a
A
a
a
A
a
A
a
A
a
A
126 132
MMIII
A
A
a
a
A
A
a
a
9
OCTADS
a
A
a
A
A
a
A
a
a
a
a
a
A
A
A
A
11 10
MII
a
a
A
A
A
A
a
a
12
300
ORDERED ANALYSIS
A
a
A
a
A
a
A
a
a
A
a
A
a
A
a
A
126 132
MMIII
A
A
a
a
A
A
a
a
9
OCTADS
a
A
a
A
A
a
A
a
a
a
a
a
A
A
A
A
11 10
MII
a
a
A
A
A
A
a
a
12
300
# MI = 126 + 132 = 258 = 86 %
# MII = 9 + 11 + 10 + 12 = 42 = 14 %
ORDERED ANALYSIS
 divide by 2
ORDERED ANALYSIS
A
a
A
a
A
a
A
a
a
A
a
A
a
A
a
A
126 132
MMIII
A
A
a
a
A
A
a
a
9
OCTADS
a
A
a
A
A
a
A
a
a
a
a
a
A
A
A
A
11 10
MII
a
a
A
A
A
A
a
a
12
300
# MII = 9 + 11 + 10 + 12 = 42 = 14 %
 A  centromere = 14 / 2 = 7 cM
ORDERED ANALYSIS
now consider 2 genes...
3 possibilities:
1. the genes are on separate
a
chromosomes ...independent
2. the genes are linked but on opposite
sides of the centromere ...independent
3. the genes are linked and on the same
side of the centromere ...???
b
a
a b
b
ORDERED ANALYSIS
crossover between centromere and both genes...
ANALYSIS OF SINGLE MEIOSES
2 kinds of mapping with tetrads / octads:
ordered analysis to map gene  centromere
unordered analysis to map gene  gene
ORDERED ANALYSIS
lots of crossovers between gene & centromere...
appear to be unlinked
ORDERED ANALYSIS
MII frequency never reaches 100%
theoretical maximum RF = 2/3 or 66.7%
theoretical maximum calculated map distance = 33.3%
>1 crossovers  with distance...
especially, DCO look like SCO
ORDERED ANALYSIS
second allele
determines pattern
maximum MII = 2/3 =
33.3 % = 33.3 cM
multiple crossovers !
UNORDERED ANALYSIS
meioses with 0
crossovers 
RF of 0%
meioses with >
0 crossovers 
RF of 50%
UNORDERED ANALYSIS
parental ditypes
= NCO + 1/4 DCO
non-parenal ditypes
= 1/4 DCO
tetratypes
= SCO + 1/2 DCO
UNORDERED ANALYSIS
NCO = PD – NPD
PD but not NCO 
= P  score 1x 
UNORDERED ANALYSIS
SCO = TT – 2NPD
T but not SCO 
= 2P  score 2x 
UNORDERED ANALYSIS
DCO = 4NPD
DCO but not NPD 
= 4P  score 4x 
UNORDERED ANALYSIS
corrected map distance (cM) between genes
= RF × 100 cM
= [ ½ single events + double events ] / total × 100 cM
= [ ½ ( TT – 2NPD ) + 4NPD ] / total × 100 cM
= ½ [ TT + 6NPD ] / total × 100 cM
ANALYSIS OF SINGLE MEIOSES
2 kinds of mapping with tetrads / octads:
ordered analysis to map gene  centromere
unordered analysis to map gene  gene
e.g., ORDERED & UNORDERED ANALYSIS
P
F1
r+
×

r+
+t

F2
or
+t
e.g., ORDERED & UNORDERED ANALYSIS
P
F1
F2
r+
×

r
+
;
+
t

+t
e.g., ORDERED & UNORDERED ANALYSIS

r+
r+
+t
+t
129
+t
r+
r+
+t
1
r+
+t
r+
+t
2
r+
+t
+t
r+
1
rt
r+
++
+t
15
rt
r+
+t
++
13
r+
rt
++
+t
17
r+
rt
+t
++
17
r+
++
rt
+t
2
rt
++
r+
+t
1
++
++
rt
rt
2
200
ORDERED & UNORDERED ANALYSIS

r
t
r+
r+
+t
+t
129
M1
M1
+t
r+
r+
+t
1
M2
M2
r+
+t
r+
+t
2
M2
M2
r+
+t
+t
r+
1
M2
M2
rt
r+
++
+t
15
M1
M2
rt
r+
+t
++
13
M1
M2
r+
rt
++
+t
17
M1
M2
r+
rt
+t
++
17
M1
M2
r+
++
rt
+t
2
M2
M1
rt
++
r+
+t
1
M2
M2
++
++
rt
rt
2
M1
M1
200
ORDERED & UNORDERED ANALYSIS

r
t
r+
r+
+t
+t
129
M1
M1
+t
r+
r+
+t
1
M2
M2
r+
+t
r+
+t
2
M2
M2
r+
+t
+t
r+
1
M2
M2
rt
r+
++
+t
15
M1
M2
rt
r+
+t
++
13
M1
M2
r+
rt
++
+t
17
M1
M2
r+
rt
+t
++
17
M1
M2
r+
++
rt
+t
2
M2
M1
rt
++
r+
+t
1
M2
M2
++
++
rt
rt
2
M1
M1
ordered analysis to map gene  centromere
MI ... no recombination
MII ... recombination
200
ORDERED & UNORDERED ANALYSIS

r
t
r+
r+
+t
+t
129
M1
M1
+t
r+
r+
+t
1
M2
M2
r+
+t
r+
+t
2
M2
M2
r+
+t
+t
r+
1
M2
M2
rt
r+
++
+t
15
M1
M2
rt
r+
+t
++
13
M1
M2
r+
rt
++
+t
17
M1
M2
r+
rt
+t
++
17
M1
M2
r+
++
rt
+t
2
M2
M1
rt
++
r+
+t
1
M2
M2
++
++
rt
rt
2
M1
M1
ordered analysis to map gene  centromere
MI ... no recombination
MII ... recombination
RF = ½ (MII / TOTAL)
200
ORDERED & UNORDERED ANALYSIS

r
t
r+
r+
+t
+t
129
M1
M1
+t
r+
r+
+t
1
M2
M2
r+
+t
r+
+t
2
M2
M2
r+
+t
+t
r+
1
M2
M2
rt
r+
++
+t
15
M1
M2
rt
r+
+t
++
13
M1
M2
r+
rt
++
+t
17
M1
M2
r+
rt
+t
++
17
M1
M2
r+
++
rt
+t
2
M2
M1
rt
++
r+
+t
1
M2
M2
++
++
rt
rt
2
M1
M1
ordered analysis to map gene  centromere
MI ... no recombination
MII ... recombination
RF = ½ (MII / TOTAL)
RF x 100 = map distance (cM)
200
ORDERED & UNORDERED ANALYSIS

r
t
r+
r+
+t
+t
129
M1
M1
+t
r+
r+
+t
1
M2
M2
r+
+t
r+
+t
2
M2
M2
r+
+t
+t
r+
1
M2
M2
rt
r+
++
+t
15
M1
M2
rt
r+
+t
++
13
M1
M2
r+
rt
++
+t
17
M1
M2
r+
rt
+t
++
17
M1
M2
r+
++
rt
+t
2
M2
M1
rt
++
r+
+t
1
M2
M2
++
++
rt
rt
2
M1
M1
RF = ½ (MII / TOTAL)
RF x 100 = map distance (cM)
r  cent. = ½ [(1+2+1+2+1)/200] x 100 = 1.75 cM
t  cent. = ½ [(1+2+1+15+13+17+1)/200] x 100 = 16.75 cM
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
unordered analysis to map gene  gene
consider all possible gene pairs (here only 1)
PD  NPD?, unlinked or PD >> NPD, linked
PD = NCO + 2-strand DCO
TT = SCO + 3-strand DCO (x2)
NPD = 4-strand DC) (¼ of all DC0)
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
unordered analysis to map gene  gene
RF = ½ [ TT + 6NPD ] / TOTAL
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
unordered analysis to map gene  gene
RF = ½ [ TT + 6NPD ] / TOTAL
RF x 100 = map distance (cM)
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
unordered analysis to map gene  gene
RF = ½ [ TT + 6NPD ] / TOTAL
RF x 100 = map distance (cM)
PD (133) >> NPD (2)  linked
r  t = ½ [ 65 + 6(2) / 200 ] x 100 = 19.25 cM
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
1.75
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
16.75
19.25
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
t
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
1.75
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
16.75
19.25
but... 1.75 + 16.75 = 18.5 ??
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
t
++
++
rt
rt
2
M1
M1
NPD
200
ORDERED & UNORDERED ANALYSIS

r
t
r&t
r+
r+
+t
+t
129
M1
M1
PD
+t
r+
r+
+t
1
M2
M2
PD
r
r+
+t
r+
+t
2
M2
M2
PD
r+
+t
+t
r+
1
M2
M2
PD
rt
r+
++
+t
15
M1
M2
TT
rt
r+
+t
++
13
M1
M2
TT
1.75
r+
rt
++
+t
17
M1
M2
TT
r+
rt
+t
++
17
M1
M2
TT
16.75
r+
++
rt
+t
2
M2
M1
TT
rt
++
r+
+t
1
M2
M2
TT
t
19.25
more accurate... calculation includes DCOs
++
++
rt
rt
2
M1
M1
NPD
200
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