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Recombination & Genetic Analysis
-The maximum recombination frequency
-Quiz section 5 revisited
-Genetic vs. Physical maps
Test your understanding
Predict the number of progeny for each type of offspring that result
from the following cross. Assume 100 total offspring.
Dihybrid female
Testcross male
58.2 cM
b+
sp+
b
sp
P
R
X
tan no speck
21
25
black speck
21
25
tan speck
29
25
black no speck
29
25
Recombinants
never exceed
50%
No!
Are b and sp linked?
• These genes are so far apart, that they assort
independently from one another.
• b and sp appear to be unlinked even though they are
on the same chromosome!
sp
+
b+
2
3
sp
sp
+
b+
4
sp
b+
1
b+
b
b
vg+
vg
sp+
sp
b
sp+
b
sp
All four
gamete
types are
equally
frequent
How do we know that b and sp are on the same
chromosome? b is linked to vg and vg is linked to sp.
Why do we observe IA?
In-class experiment…
Mark two crossovers anywhere between the homologues:
After you are given the locations of loci A, B, and D…
Write down the parental types with respect to A/a and B/b:
Write down the parental types with respect to A/a and D/d:
Why do we observe IA?
In-class experiment…
Mark two crossovers anywhere between the homologues:
After you are given the locations of loci A, B, and D…
Write down the parental types with respect to A/a and B/b:
Write down the parental types with respect to A/a and D/d:
Why do we observe IA?
In-class experiment…
Mark two crossovers anywhere between the homologues:
A
B
1
D
A
a
B
b
2
3
D
d
a
b
4
d
After you are given the locations of loci A, B, and D…
Write down the parental types with respect to A/a and B/b:
A B and a b
Write down the parental types with respect to A/a and D/d:
A D and a d
Why do we observe IA?
In-class experiment…
Mark two crossovers anywhere between the homologues:
A
B
1
D
A
a
B
b
2
3
D
d
a
b
4
d
After you are given the locations of loci A, B, and D…
Write down the parental types with respect to A/a and B/b:
A B and a b
Write down the parental types with respect to A/a and D/d:
A D and a d
In-class experiment (cont’d)
Looking first at just loci A/a and B/b…
What are the genotypes of the products from your meiosis?
1.
2.
3.
4.
Are these gametes all parental? All recombinant? 2 of each?
Then look at just loci A/a and D/d…
What are the genotypes of the products from your meiosis?
1.
2.
3.
4.
Are these gametes all parental? All recombinant? 2 of each?
In-class experiment (cont’d)
What must have happened to
create these gametes?
Class aggregate data:
A-B
Genotype
P or R? Number?
Number P gametes Number R gametes
AB, ab
4P
81
324
0
Ab, aB
4R
2
0
8
AB, ab,
Ab, aB
2 P, 2 R
43
86
86
# Recombinants
Total # Gametes
=
X
X
% Rec?
X
In-class experiment (cont’d)
Class aggregate data:
A-D
Genotype
P or R? Number?
Number P gametes Number R gametes
AD, ad
4P
40
160
0
Ad, aD
4R
19
0
76
AD, ad,
Ad, aD
2 P, 2 R
61
122
122
# Recombinants
Total # Gametes
76+122
=
% Rec?
X
76+122 +160 +122
Why do we observe IA?
A
B
D
A
a
B
b
D
d
a
b
d
Everyone in the class drew crossovers somewhere between A/a and D/d,
yet the overall % recombinants for the class was only ~50%. If we look
at a large enough sample, even genes that are very far apart on the same
chromosome cannot show more than 50% recombinant products.
Need to look closer at meiosis itself to see why.
What is the maximum recombination frequency in any interval?
Parental
Recombinant
Recombinant
Parental
Reminder… one crossover gives 2
parental, 2 recombinant gametes
The range of possibilities: tightly linked  independent assort.
Consider 100 cells undergoing meiosis…
if one cell has a crossover 
2 recombinant
out of 400
 0.5%
if 10 cells have crossovers 
20 recombinant
out of 400
 5.0%
if all cells have crossovers 
200 recombinant  50%
out of 400
Maximum recombination frequency =
50% for single recombination events
What is the maximum recombination frequency in any interval?
The effect of multiple crossovers:
# xovers
resulting gametes
2
(2 strands)
2
(3 strands)
2
(4 strands)
= 50% recombinant and 50% parental.
Also true for triple Xover, quadruple Xover, etc.
4 parental
2 parental
2 non-par.
4 non-par.
6 parental
6 non-par.
Human Xchromosome map…
how could we get
180 cM?
Now consider independent assortment
…the “ultimate” in non-linkage
R
r
Y
R
X
y
r
Y
y
Refer to one of Mendel’s F1 x F1 dihybrid cross (round yellow X round
yellow):
What were the parental types for the F1? RY and ry
What were the parental and recombinant gametes made by the F1 plants?
1/4
1/4
1/4
1/4
RY
ry
Ry
rY
What was the % recombinant gametes? 1/4 + 1/4 = 50%!
So, even for independently assorting genes, the %
recombinant products is only 50%
Conclusion
For widely separated genes
1) An odd number of crossovers gives, on average, an
equal number of parental and recombinant types.
2) An even number of crossovers gives, on average, an
equal number of parental and recombinant types.
3) Alleles on two different chromosomes line up on
the metaphase plate independently, giving on average
equal numbers of parental and recombinant types.
Thus, the maximum recombinant frequency = 50%
Loci can appear to be unlinked because:
• They are on separate chromosomes
• They are so far apart on the same chromosome that
recombination always occurs
Practice question
In a certain plant species…
R
f
r
F
X
flower fragrance (F) is dominant
over unscented (f)
r color (B) fis dominantr over white (b)
f
blue flower
rounded leaves (R) is dominant over pointy (r); and
The parental
and recombinant
thorny stems
(T) is dominant
over smooth stems (t).
types are the same! Need to be
From the following crosses, can you determine whether the fragrance
heterozygous at both loci
gene is linked to any of the other genes; if so, at what map distance?
Bb Ff x bb ff
Rr ff x rr Ff
Tt Ff x tt ff
270 blue, fragrant
281 blue, non-fragrant
268 white, fragrant
275 white, non-fragrant
219 rounded, fragrant
222 rounded, non-fragrant
209 pointy, fragrant
216 pointy, non-fragrant
333 thorny, fragrant
36 thorny, non-fragrant
39 smooth, fragrant
342 smooth, non-fragra
F not linked to B
Can’t tell!
F and T linked
at 10 cM
QS7 revisited
What were the main points of QS5?
-To give you an opportunity to see actual data from a
meiosis and to draw conclusions from the data based
on your knowledge of this process.
-To show what can be learned from looking at all four
products of a single meiosis.
The diagrams used in quiz section…
etc.
…were designed to help set up specific predictions
Setting up predictions for meiosis outcomes…
But if we can’t see all of
the products from a single
meiosis we expect…
If…
then we expect…
2 genes are linked
a parental ditype (PD)
PD > T >> NPD
they are
independently
assorting but each
close to a
centromere
either PD or nonparental ditype (NPD),
with a 50:50 chance of
each
PD = NPD > T
an equal proportion of
parental and
recombinant types
mostly tetratypes (T),
but also some PD, and
NPD
an equal proportion of
parental and
recombinant types
Can distinguish!
they are unlinked
and at least one is
distant from a
centromere
mostly parental types
Can’t distinguish
What did the data from quiz section tell you?
Mat haploid parent = ade his
Mat haploid parent = his LEU
Mat haploid parent = LEU TS
Mata haploid parent = ADE HIS
Mata haploid parent = HIS leu
Mata haploid parent = leu ts
Spore
phenotype
# of spores
Spore
phenotype
# of spores
Spore
phenotype
# of spores
ADE HIS
9
HIS LEU
3
LEU TS
11
ADE his
11
HIS leu
17
Leu ts
9
ade HIS
11
his LEU
17
leu TS
9
ade his
9
his leu
3
leu ts
11
Total =
40
Total =
40
Total =
40
Conclusion?
Conclusion?
Conclusion?
Probably
not linked
Probably
linked
Probably
not linked
Looking at the 10 tetrads in terms of LEU & TS
How many PD? 4
How many NPD? 5
How many T? 1
So what do you conclude about the LEU and TS genes?
they are independently assorting but each close to a
centromere!
Our completed map
Diploid genotype:
MATa ADE HIS
MAT ade his
leu ts URA1 ura2
LEU TS ura1 URA2
leu HIS
LEU his
ts
ADE
TS
ade
How well did we do?
The actual gene names…
Let’s look in
SacchDB…
ADE2
HIS4
CDC7 (TS)
LEU2
ADE2
HIS4
LEU2
CDC7 (TS)
Not bad!
What about URA?
Know the parental types
Look at spore phenotypes
Parental types?
U1 u2
&
u1 U2
What spore genotypes would you expect in a PD tetrad?
Phenotype on -ura plate?
U1 u2
no growth
no growth
u1 U2
no growth
u1 U2
no growth
U1 u2
So, given these
parental types…
0/4 spores
growing is
diagnostic of PD
What about URA?
What spore genotypes would you expect in a NPD tetrad?
Parental types?
Genotype?
U1 U2
U1 u2
&
u1 U2
Growth phenotype on -ura?
U1 U2
GROWTH
GROWTH
u1 u2
no growth
u1 u2
no growth
What about URA?
What spore genotypes would you expect in a T tetrad?
Parental types?
Genotype?
U1 u2
U1 u2
&
u1 U2
Growth phenotype on -ura?
U1 U2
no growth
GROWTH
u1 U2
no growth
u1 u2
no growth
Looking at the 10 tetrads…
How many PD?
How many NPD?
How many T?
So what do you conclude about the ura genes?
Practice question
Brown seed pods (B) in a plant species is dominant to green (b), and
elongated pods (E) is dominant over squished (e).
(a) A fully heterozygous plant has the dominant alleles linked in trans
(i.e., dominant alleles not on the same homologue) at a map
distance of 20 cM. What will be the genotypes of gametes produced
by this plant, and in what frequencies (or percentages)?
(b) If this plant is self-pollinated, what progeny phenotypes will you
expect to see, and in what frequencies? Use a Punnett square to
illustrate your answer.
Heterozygote genotype =
B
e
b
E
3-point testcrosses
The problem with using two markers (like a and d below)…
double crossovers can go undetected
 underestimation of
recombinant
frequency
Solution: include a third marker between the other two…
 more DCOs
revealed
Plus… gene order revealed (more later)
3-point testcross—predicting progeny from a known map
Predict the progeny phenotypes and numbers from this cross:
Parent 1: +
b
+
a
c
+
Parent 2: b
b
c
a
c
a
+ = wild type, dominant
Map:
b
3 cM
c
7 cM
Count 10,000 progeny
Step 1. Determine the number of DCO products
Probability of recombinant product in (b-c) = 3% = 0.03
Probability of recombinant product in (c-a) =
7% = 0.07
Probability of recombinant product in both =
0.03 x 0.07 = 0.0021
a
Predicting progeny from a known map (cont’d)
Heterozygous parent:
only one chromatid of each
homologue shown on next slide
Predicting progeny from a known map (cont’d)
DCO: both together =
10000 x 0.0021 = 21
SCO in b-c interval:
Both together =
(10000 x 0.03)-21 = 279
SCO in c-a interval:
Both together =
(10000 x 0.07)-21 = 679
NCO (non-crossover):
Both together =
10000-(SCO + DCO) = 9021
3-point testcross—constructing a linkage map
Construct a linkage map (gene order and map distance) for the
following genes in Drosophila:
Genes pr/+ (purple or red eyes)
v/+ (vestigial or long wings)
b/+ (black or tan body)
Parents
Female: pr/+
Male: pr/pr
Progeny phenotypes
+ + +
564 SCO
b pr+ v+
32 DCO
v b+ pr+
4125 NCO
pr b+ v+
266
SCO
v b pr+
272
pr b v+
4137 NCO
pr v b+
30 DCO
pr v b
574 SCO
Total = 10000
Step 1.Expand the shorthand
v/+
v/v
b/+
b/b
Step 2.Identify the NCO and
DCO classes
Step 3.Which gene is in the
middle?
ASK: Which order allows us to
go from the NCO genotype to
the DCO genotype.
Constructing a linkage map… Step 3 (cont’d)
We know:
DCO
(b+ v pr+)
(b v+ pr)
(b v+ pr+)
(b+ v pr)
order unknown
The process:
• Try out the parental genotypes in the 3 possible orders
• Do a “virtual double crossover” to see which one would give
the correct DCO genotype.
b+ v pr+
X X
b v+ pr
b+ v+ pr+
b+ pr+ v
X X
b pr v+
b+ pr v
b
b
v
pr
pr+ v+
Constructing a linkage map (cont’d)
Step 4. Calculate % recombinant products
b+
pr
v+
b
pr+
v
b+
pr+
v+
b
b+
pr
pr
v
v
NCO:
4125+4137
= 8262
% recombinants
in b-pr interval=
SCOb-pr:
266+272 =
538
(538+62)/10000
SCOpr-v:
564+574 =
1138
DCO:
30+32 = 62
b
pr
v+
=600/10000
=6%
% recombinants
in pr-v interval=
(1138+62)/10000
=1200/10000
=12%
Constructing a linkage map (cont’d)
Step 5. Draw the map
b
6 cM
pr
v
12 cM
or
v
12 cM
pr
6 cM
b
Interference and coefficient of coincidence (COC)
Interference: Lower-than-expected frequency of DCO products
- Chiasma at one one location blocks other
chiasmata from forming nearby
COC =
observed DCO
expected DCO
Interference = 1 - COC
In our example…
expected DCO = 0.06 x 0.12 x 10000 = 72
Observed DCO = 62
COC = 62/72 = 0.86
Interference = 1 - 0.86 = 0.14
Genetic vs. physical maps
Genetic maps…based on recombinant frequencies between markers
alleles!
variation at
location #2
variation at
location #1
pr+
vg
pr
vg+
recombination:
how frequent?
alleles!
Alleles are detected as associated phenotypes
New combination of phenotypes  new combination
of alleles  recombination
Genetic vs. physical maps (cont’d)
Physical maps… based on DNA sequence or landmarks in sequence
For example:
A
B
The number of chromosomal bands separating the known
locations of genes.
site1
site2
site3
site4
site5
The pattern of restriction sites in a DNA sequence.
The number of bp of DNA.
Questions for Thought
Which
chromosome?
pr+
Relation to centromere
and telomere?
vg+
pr
vg
Number of bp?
How do we know where to place these genes relative to the
centromere and telomeres?
How do we know which physical entity (chromosome) our
linkage group describes? “Linkage group” = chromosome
What is the relationship between crossover frequency and gene
order/distance (in bp of DNA) along the chromosomes?
Linking the Physical and Genetic Maps
white is X-linked
From lecture 6
Color is on chromosome IX
Novel Strain
C
knob
Wx
extra DNA
Cri du chat is on
chromosome V
From lecture 8
Linking the Physical and Genetic Maps
-Fluorescence in situ hybridization (FISH)
Double-stranded DNA
segment from a
particular location in
the human genome
** *
Metaphase
chromosomes
Fluorescently labeled
single-stranded DNA
Partially denature DNA
*
*
*
What does the genetic map position tell us?
genetic map (recombination) units = cM
R-G colorblindness
R-G colorblindness
10 cM
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
telomere
6 cM
Fragile X
Fragile X
Haemophilia
centromere
Physical
units =
map (FISH) chrom. bands
Haemophilia
~10Xbp
~6Xbp
physical map (DNA) units = bp
-Order of genes is conserved in genetic and physical maps.
-Distance separating markers in genetic and physical maps
is ~proportional (but X varies in different organisms).