1. Recombination & Genetic Analysis
-The maximum recombination frequency
-Quiz section 5 revisited
-Genetic vs. Physical maps

2. 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 X
tan no speck P 21 29 25
Recombinants never exceed 50%
black speck R
tan speck
black no speck

3. Are b and sp linked?
No!
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!
b sp+
b sp+
b sp
b sp
All four gamete types are equally frequent
vg+
sp + b+ sp b 1 2 3 4 vg
How do we know that b and sp are on the same chromosome?
b is linked to vg and vg is linked to sp.

4. 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:

5. 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:

6. Why do we observe IA? In-class experiment…
Mark two crossovers anywhere between the homologues: A B D 1 A B 2 D a b 3 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

7. Why do we observe IA? In-class experiment…
Mark two crossovers anywhere between the homologues: A B D 1 A B 2 D a b 3 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

8. 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. 3. 2. 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. 3. 2. 4.
Are these gametes all parental? All recombinant? 2 of each?

9. 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
4 P 81
324
Ab, aB
4 R 2 8
AB, ab,
Ab, aB
2 P, 2 R 43 86 86
# Recombinants X =
% Rec? X
Total # Gametes X

10. In-class experiment (cont’d)
Class aggregate data:
A-D
Genotype
P or R?
Number?
Number P gametes
Number R gametes
AD, ad
4 P 40
160
Ad, aD
4 R 19 76
AD, ad,
Ad, aD
2 P, 2 R 61
122
122
# Recombinants
76+122 =
% Rec? X
Total # Gametes

11. Why do we observe IA? A B D A B D a b 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.

12. The range of possibilities: tightly linked  independent assort.
What is the maximum recombination frequency in any interval?
Parental
Reminder… one crossover gives 2 parental, 2 recombinant gametes
Recombinant
Recombinant
Parental
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 
There are four ways the double crossover event can happen; together they result in a 50% recombinant and 50% parental.
Look at page 179 in the latest edition of griffiths.
20 recombinant out of 400 
5.0%
if all cells have crossovers 
200 recombinant out of 400 
50%
Maximum recombination frequency = 50% for single recombination events

13. The effect of multiple crossovers:
What is the maximum recombination frequency in any interval?
The effect of multiple crossovers:
# xovers
resulting gametes 2
(2 strands)
4 parental 2
(3 strands)
2 parental
2 non-par.
Save this for later when you talk about maximum recombination frequency; also tally up the totals at the bottom (an equal number of recombinant and non recombinant. 2
(4 strands)
4 non-par.
6 parental
6 non-par.
= 50% recombinant and 50% parental.
Also true for triple Xover, quadruple Xover, etc.

14. Human X-chromosome map…
how could we get 180 cM?

15. Now consider independent assortment
R Y
r y X
…the “ultimate” in non-linkage
Refer to one of Mendel’s F1 x F1 dihybrid cross (round yellow X round yellow):
What were the parental types for the F1?
What were the parental and recombinant gametes made by the F1 plants?
What was the % recombinant gametes?
RY and ry
1/4 RY
1/4 ry
1/4 Ry
1/4 rY
1/4 + 1/4 = 50%!
So, even for independently assorting genes, the % recombinant products is only 50%

16. 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

17. Practice question R f r f r F X
In a certain plant species…
flower fragrance (F) is dominant over unscented (f)
blue flower color (B) is dominant over white (b)
rounded leaves (R) is dominant over pointy (r); and
thorny stems (T) is dominant over smooth stems (t).
From the following crosses, can you determine whether the fragrance gene is linked to any of the other genes; if so, at what map distance?
R f
r f
r F X
The parental and recombinant types are the same! Need to be heterozygous at both loci
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-fragrant
F and T linked at 10 cM
F not linked to B
Can’t tell!

18. 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

19. 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
they are independently assorting but each close to a centromere
they are unlinked and at least one is distant from a centromere
a parental ditype (PD)
PD > T >> NPD
mostly parental types
either PD or non-parental ditype (NPD), with a 50:50 chance of each
PD = NPD > T
an equal proportion of parental and recombinant types
Can’t distinguish
Can distinguish!
Fine-but emphasize that being able to see all four meiotic products. Compare what would be expected if you could or could not see
mostly tetratypes (T), but also some PD, and NPD
an equal proportion of parental and recombinant types

20. Conversely: the 2 genes are linked If we see… then we conclude that…
they are independently assorting but each close to a centromere
they are unlinked and at least one is distant from a centromere
only or mostly PD
PD and NPD in equal proportions
mostly tetratypes
fine

21. What did the data from quiz section tell you?
Mat haploid parent = ade his
Mata haploid parent = ADE HIS
Spore phenotype
# of spores
ADE HIS 9
ADE his 11
ade HIS
ade his
Total = 40
Mat haploid parent = his LEU
Mata haploid parent = HIS leu
Spore phenotype
# of spores
HIS LEU 3
HIS leu 17
his LEU
his leu
Total = 40
Mat haploid parent = LEU TS
Mata haploid parent = leu ts
Spore phenotype
# of spores
LEU TS 11
Leu ts 9
leu TS
leu ts
Total = 40
Conclusion?
No. Instead, summarize data from quiz section and what it told us (include linked and unlinked genes-> then go to database and point out that His is also linked to the centromere-> that a close look at the data also demonstrates this. What about Ura? The two genes are unlinked but affect different steps in the pathway for adenine biosynthesis-need to think about how the alleles would segregate.
Conclusion?
Conclusion?
Probably not linked
Probably linked
Probably not linked

22. Looking at the 10 tetrads in terms of LEU & TS
How many PD?
How many NPD?
How many T? 4 5 1
So what do you conclude about the LEU and TS genes?
No. Instead, summarize data from quiz section and what it told us (include linked and unlinked genes-> then go to database and point out that His is also linked to the centromere-> that a close look at the data also demonstrates this. What about Ura? The two genes are unlinked but affect different steps in the pathway for adenine biosynthesis-need to think about how the alleles would segregate.
they are independently assorting but each close to a centromere!

23. Our completed map Diploid genotype: MATa ADE HIS leu ts URA1 ura2
MAT ade his LEU TS ura1 URA2
leu HIS
LEU his ts TS
ADE
ade

24. How well did we do? Let’s look in SacchDB… Not bad!
The actual gene names…
ADE2 CDC7 (TS)
HIS4 LEU2
ADE2
HIS4
LEU2
Side point here is that since leu2 is linked to the centromere, and is also linked to his4, we might expect that his4 would provide evidence of centromere linkage, but the evidence would be less compelling than for leu2 (based on the database).
CDC7 (TS)
Not bad!

25. What about URA? Know the parental types Look at spore phenotypes U1 u2
&
u1 U2
Parental types?
What spore genotypes would you expect in a PD tetrad?
Phenotype on -ura plate?
U1 u2
u1 U2
no growth
So, given these parental types… 0/4 spores growing is diagnostic of PD
no growth
no growth
no growth

26. What about URA? What spore genotypes would you expect in a NPD tetrad?
U1 u2
&
u1 U2
Parental types?
Genotype?
Growth phenotype on -ura?
U1 U2
u1 u2
GROWTH
GROWTH
no growth
no growth

27. What about URA? What spore genotypes would you expect in a T tetrad?
U1 u2
&
u1 U2
Parental types?
Genotype?
Growth phenotype on -ura?
U1 u2
U1 U2
u1 U2
u1 u2
no growth
GROWTH
no growth
no growth

28. Looking at the 10 tetrads…
How many PD?
How many NPD?
How many T?
So what do you conclude about the ura genes?

29. Practice question B e Heterozygote genotype = b E
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

30. Practice question B e Heterozygote genotype = b E
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
Recombinant gametes = B E and b e, 20% total = 10% each
Parental type gametes = B e and b E, 80% total = 40% each

31. Practice question gamete genotypes and frequencies parental
0.4 Be
0.4 bE
0.1 BE
0.1 be
parental
non-parental
gamete genotypes and frequencies
Be/Be
0.16
bE/Be
0.16
BE/Be
0.04
be/Be
0.04
Be/bE
0.16
bE/bE
BE/bE
0.04
be/bE
Be/BE
bE/BE
BE/BE
0.01
be/BE
Be/be
bE/be
BE/be
be/be BE
0.51 Be
0.24 bE be
Progeny phenotypes:

32. 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)

33. 3-point testcross—predicting progeny from a known map
Predict the progeny phenotypes and numbers from this cross:
+ + a
b c +
+ = wild type, dominant
Parent 1: b c a
3 cM
7 cM
Map:
b c a
Parent 2:
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 =

34. Predicting progeny from a known map (cont’d)
Heterozygous parent:
only one chromatid of each homologue shown on next slide

35. Predicting progeny from a known map (cont’d)
DCO: both together =
10000 x = 21
SCO in b-c interval:
Both together =
(10000 x 0.03)-21 = 279
SCO in c-a interval:
Both together =
Both together refers to both products of the recombination event.
(10000 x 0.07)-21 = 679
NCO (non-crossover):
Both together =
10000-(SCO + DCO) = 9021

36. 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/+ v/+ b/+
Male: pr/pr v/v b/b
Progeny phenotypes
b 32 v
pr 266
v b 272
pr b 4137
pr v 30
pr v b 574
Step 1. Expand the shorthand
SCO
pr+ v+
b+ pr+
b+ v+
pr+ v+ b+
DCO
Step 2. Identify the NCO and DCO classes
NCO
SCO
Step 3. Which gene is in the middle?
ASK: Which order allows us to go from the NCO genotype to the DCO genotype.
Total = 10000

37. incorrect correct! Constructing a linkage map… Step 3 (cont’d)
We know:
(b+ v pr+)
(b v+ pr)
DCO
(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+
b v+ pr
b+ v+ pr+
b v pr
incorrect X
b+ pr+ v
b pr v+ X
b+ pr v
b pr+ v+
correct!

38. Constructing a linkage map (cont’d)
Step 4. Calculate % recombinant products
NCO:
= 8262
% recombinants in b-pr interval=
b+ pr v+
b pr+ v
SCOb-pr:
= 538
(538+62)/10000
=600/10000
=6%
b+ pr+ v+
b pr v
SCOpr-v:
= 1138
% recombinants in pr-v interval=
b+ pr v
b pr v+
DCO:
30+32 = 62
( )/10000
=1200/10000
=12%

39. Constructing a linkage map (cont’d)
Step 5. Draw the map b pr v
6 cM
12 cM b pr v
6 cM
12 cM or

40. 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
observed DCO
COC =
Interference = 1 - COC
expected DCO
In our example…
expected DCO = 0.06 x 0.12 x = 72
Observed DCO = 62
COC = 62/72 = 0.86
Interference = = 0.14

41. Genetic vs. physical maps
Genetic maps…
based on recombinant frequencies between markers
variation at location #1
variation at location #2
alleles!
pr+ vg pr
vg+
Next year, use your genetic vs. physical map.
recombination:
how frequent?
Alleles are detected as associated phenotypes
New combination of phenotypes  new combination of alleles  recombination

42. Genetic vs. physical maps (cont’d)
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.

43. Questions for Thought Which chromosome?
Relation to centromere and telomere?
pr+
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?

44. Linking the Physical and Genetic Maps
white is X-linked
From lecture 6
Color is on chromosome IX
knob
extra DNA C Wx
Novel Strain
From lecture 8
Cri du chat is on chromosome V

45. Linking the Physical and Genetic Maps
-Fluorescence in situ hybridization (FISH)
Metaphase chromosomes
Double-stranded DNA segment from a particular location in the human genome
Partially denature DNA
Library screen-fish * *
Fluorescently labeled single-stranded DNA

46. What does the genetic map position tell us?
Fragile X
R-G colorblindness
6 cM
genetic map (recombination)
units = cM
Haemophilia
10 cM
telomere
centromere
Physical
map (FISH)
units = chrom. bands
physical map (DNA)
~6Xbp
~10Xbp
units = bp
Fragile X
Haemophilia
R-G colorblindness
X~1,000,000 bp in humans (how many centimorgans in humans?)
X~600,000bp in flies
X~3.5kb in yeast.
-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).