1. Crossing Over of John Edward’s Chromosomes
Linkage & Mapping

2. Linkage Eukaryotic chromosomes are long ds DNA molecules
Typical chromosome contains thousands of genes (loci)
Linkage
loci located on the same chromosome
linked loci tend to be transmitted as a unit

3. Linkage
Because they are a group of genes linked together, chromosomes are functionally linkage groups
# of linkage groups = the # of types of chromosomes
Crossing over causes loci that are far apart on the same chromosome to sometimes independently assort
known as incomplete linkage

4. Crossing Over Produce Recombinant Phenotypes
Crossing over (meiotic recombination)
Occurs during prophase I of meiosis at the bivalent stage
zygotene - pachytene - diplotene
Non-sister chromatids of homologous chromosomes exchange DNA segments

5. Linkage Prevents Independent Assortment
Figure 5.1

6. Crossing Over May Produce Recombinant Phenotypes
gametes with a combination of alleles NOT found in the original chromosomes as a result of meiotic recombination
These are termed parental gametes
These are called nonparental or recombinant gametes
Figure 5.1

7. Example of Linkage
Bateson and Punnett conducted a cross in sweet pea involving two traits
Flower color and pollen shape
Dihybrid cross expected to give 9:3:3:1 phenotypic ratio of F2 phenotypes
Observed linkage
Called it coupling

8. Purple flowers, long pollen (PPLL)
Example of Linkage x
Purple flowers, long pollen (PPLL)
Red flowers,
round pollen (ppll) F
offspring 1
Purple flowers,
long pollen (PpLl)
Self-fertilization
Observed
number
Observed Ratio
Expected
number
Expected Ratio
F2 offspring phenotypes
(9:3:3:1) P
Purple flowers, long pollen
Purple flowers, round pollen
Red flowers, long pollen
Red flowers, round pollen
296 19 27 85
15.6
1.0
1.4
4.5
240 80 27 9 3 1 NP

9. Morgan Provided Evidence for the Linkage of Several X-linked Genes
The first direct evidence of linkage came from studies of Thomas Hunt Morgan
Morgan investigated several traits that followed an X-linked pattern of inheritance
Body color
Eye color
Wing length

10. Linkage to a Particular Chromosomex /
y w m/y w m
y+ w+ m+ Y
F1 generation
Sex linkage of all traits places them all on X chromosome
F1 generation contains wild-type females and yellow-bodied, white-eyed, miniature-winged males. x
y+w+ m+/ y w m
y w m / Y
F2 generation
Females
Males
Total P
Tan body, red eyes, normal wings
439
319
758
Tan body, red eyes, miniature wings
208
193
401
Tan body, white eyes, normal wings 1 1
Tan body, white eyes, miniature wings 5 11 16
Yellow body, red eyes, normal wings 7 5 12
Yellow body, red eyes, miniature wings
Yellow body, white eyes, normal wings
178
139
317
Yellow body, white eyes, miniature wings
365
335
700

11. Morgan’s explanation: All three genes are located on the X chromosome
P Males
P Females
Morgan observed a much higher proportion of the combinations of traits found in the parental generation
Morgan’s explanation:
All three genes are located on the X chromosome
Therefore, they tend to be transmitted together as a unit
5-13
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12. Morgan Provided Evidence for the Linkage of Several X-linked Genes
However, Morgan still had to interpret two key observations
Why did the F2 generation have a significant number of nonparental combinations?
Why was there a quantitative difference between the various nonparental combinations?

13. Reorganize Morgan’s data considering pairs of genes separately
Gray body, red eyes
1,159
Yellow body, white eyes
1,017
Gray body, white eyes 17
Yellow body, red eyes 12
Total
2,205
But this nonparental combination was rare
Red eyes, normal wings
770
White eyes, miniature wings
716
Red eyes, miniature wings
401
White eyes, normal wings
318
Total
2,205
It was fairly common to get this nonparental combination

14. Figure 5.4 These parental phenotypes are the most common offspring
These recombinant offspring are not uncommon
because the genes are far apart

15. Figure 5.4 These recombinant offspring are fairly uncommon
because the genes are very close together
These recombinant offspring are very unlikely
1 out of 2,205
it is product of single cross over probabilities

16. Incomplete Linkage
Linked loci that are sometimes separated by recombination are inherited in a pattern between linked and independently assorting
not always together
not present in normal dihybrid ratios
The percentage of offspring with the loci linked vs those with the loci separated is a measure of the physical distance separating the loci on the chromosome

17. Alfred Sturtevant’s Analysis
An undergraduate in the laboratory of T. H. Morgan
Constructed first genetic map in 1911
Sturtevant wrote:
“In conversation with Morgan … I suddenly realized that the variations in the length of linkage, already attributed by Morgan to differences in the spatial orientation of the genes, offered the possibility of determining sequences [of different genes] in the linear dimension of the chromosome. I went home and spent most of the night (to the neglect of my undergraduate homework) in producing the first chromosome map, which included the sex-linked genes, y, w, v, m, and r, in the order and approximately the relative spacing that they still appear on the standard maps.”

18. Linkage and Genetic Maps
Estimating the relative distances between linked genes, based on the amount of recombination occuring between them allows us to generate genetic maps
If the genes are far apart  many recombinant offspring
If the genes are close  very few recombinant offspring
Map distance =
Number of recombinant offspring
Total number of offspring
X 100
The units of distance are called map units (mu)
They are also referred to as centiMorgans (cM)
One map unit is equivalent to 1% recombination frequency

19. Linkage Analysis and Mapping
Genetic mapping experiments are typically accomplished by carrying out a testcross
Example of a two-point mapping cross
Cross of two linked genes affecting bristle length and body color in fruit flies
s = short bristles
+ = normal bristles
e = ebony body color
+ = gray body color
One parent double recessive (homozygous recessive at both loci) – s/s ; e/e
Other parent is heterozygous at both loci (+/s ; +/e)

20. Chromosomes are the product of a crossover during meiosis in the heterozygous parent
Recombinant offspring are fewer in number than nonrecombinant offspring
Figure 5.9
5-47

21. Linkage Analysis and Mapping
The phenotype data are used to estimate the distance between the two loci
Map distance =
Number of recombinant offspring
Total number of offspring
X 100 =
X 100
= 12.3 map units
Therefore, the s and e genes are 12.3 map units apart from each other along the same chromosome

22. Trihybrid Crosses
Data from trihybrid crosses can also yield information about map distance and gene order
The following experiment outlines a common strategy for using trihybrid crosses to map genes
In this example, we will consider fruit flies that differ in body color, eye color and wing shape
b = black body color
b+ = grey body color
pr = purple eye color
pr+ = red eye color
vg = vestigial wings
vg+ = normal wings
5-59
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23. Step 1: Cross two true-breeding strains that differ at three loci.
Male is homozygous wildtype for all three traits
Female is mutant for all three traits
The goal in this step is to obtain F1 individuals that are heterozygous for all three genes
5-60
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24. Step 2: Perform a testcross by mating F1 female heterozygotes to homozygous recessive, male flies
During gametogenesis in the heterozygous female F1 flies, crossovers may produce new combinations of the 3 alleles
5-61
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25. Number of Observed Offspring (males and females)
Step 3: Collect data for the F2 generation
Phenotype
Number of Observed Offspring (males and females)
Gray body, red eyes, normal wings
411
Gray body, red eyes, vestigial wings 61
Gray body, purple eyes, normal wings 2
Gray body, purple eyes, vestigial wings 30
Black body, red eyes, normal wings 28
Black body, red eyes, vestigial wings 1
Black body, purple eyes, normal wings 60
Black body, purple eyes, vestigial wings
412
5-62
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26. In the offspring of crosses involving linked genes,
Analysis of the F2 generation flies will allow us to map the three genes
The three genes exist as two alleles each
Therefore, there are 23 = 8 possible combinations of offspring
If the genes assorted independently, all eight combinations would occur in equal proportions
It is obvious that they are far from equal
In the offspring of crosses involving linked genes,
Parental phenotypes occur most frequently
Double crossover phenotypes occur least frequently
Single crossover phenotypes occur with “intermediate” frequency
5-63
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27. The combination of traits in the double crossover tells us which gene is in the middle
A double crossover separates the gene in the middle from the other two genes at either end
In the double crossover categories, the recessive purple eye color is separated from the other two recessive alleles
Thus, the gene for eye color lies between the genes for body color and wing shape
5-64
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28. Step 4: Calculate the map distance between pairs of genes
To do this, one strategy is to group the data according to pairs of phenotypes resulting from non-crossovers, single crossovers, & double crossovers
5-65
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29. Number of Observed Offspring
Phenotype
Number of Observed Offspring
Gray body, purple eyes, vestigial wings 30
Black body, red eyes, normal wings 28
Gray body, red eyes, vestigial wings 61
Black body, purple eyes, normal wings 60
Gray body, purple eyes, normal wings 2
Black body, red eyes, vestigial wings 1
1,005
Single crossover between b and pr
= 0.058
1,005
Single crossover between pr and vg
= 0.120
1 + 2
1,005
Double crossover, between b and pr, and between pr and vg
= 0.003
5-71
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30. To calculate the distance between b and pr
To determine the map distance between the genes, we need to consider both single and double crossovers
To calculate the distance between b and pr
Map distance = ( ) X 100 = 6.1 mu
To calculate the distance between pr and vg
Map distance = ( ) X 100 = 12.3 mu
To calculate the distance between b and vg
The double crossover frequency needs to be multiplied by two
Because both crossovers are occurring between b and vg
Map distance = ( [0.003]) X 100
= 18.4 mu
5-72
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31. Step 5: Construct the map
Based on the map unit calculation the body color and wing shape genes are farthest apart
The eye color gene is in the middle
The data is also consistent with the map being drawn as vg – pr – b (from left to right)
In detailed genetic maps, the locations of genes are mapped relative to the centromere
5-69
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32. Alternatively, the distance between b and vg can be obtained by simply adding the map distances between b and pr, and between pr and vg
Map distance = = 18.4 mu
5-73
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33. Interference P (double crossover) =
The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover
P (double crossover) =
P (single crossover between b and pr)
P (single crossover between pr and vg) X
= X 0.123 =
Based on a total of 1,005 offspring the expected number of double crossover offspring is
= 1,005 X
= 7.5

34. Interference
Therefore, we would expect seven or eight offspring to be produced as a result of a double crossover
However, the observed number was only three!
Two with gray bodies, purple eyes, and normal sings
One with black body, red eyes, and vestigial wings
This lower-than-expected value is due to a common genetic phenomenon, termed positive interference
The first crossover decreases the probability that a second crossover will occur nearby
5-75
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35. Interference (I) is expressed as
I = 1 – C
where C is the coefficient of coincidence
Observed number of double crossovers
Expected number of double crossovers
C = 3
7.5
C =
= 0.40
I = 1 – C = 1 – 0.4
= 0.6 or 60%
This means that 60% of the expected number of crossovers did not occur

36. Since I is positive, this interference is positive interference
Rarely, the outcome of a testcross yields a negative value for interference
This suggests that a first crossover enhances the rate of a second crossover
The molecular mechanisms that cause interference are not completely understood
However, most organisms regulate the number of crossovers so that very few occur per chromosome
5-77
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37. Somatic Cell Hybridization
Fusion of human cell and rodent cell
Eventually, most human sequences lost
entire chromosomes, portions (arms) of chromosomes
Lines carrying sub-sets of human chromosomes are compared to determine which express the given gene product
have to have a detectible DNA sequence or protein
Process of elimination determines on which chromosome (region of chromosome) a gene is located
Hybrid Panel
Figure: 08-14
Caption:
A hypothetical grid of data used in synteny testing to assign genes to their appropriate human chromosomes. Three somatic hybrid cell lines, designated 23, 34, and 41, have each been scored for the presence or absence of human chromosomes 1–8, as well as for their ability to produce the hypothetical human gene products A, B, C, and D.

38. Other Mapping Techniques
Cytogenetic Mapping
Have a mutant phenotype
Perform karyotype
Observe altered chromosomes
aneuploidies
translocations
deletions
Aberrant chromosome probably location of gene
normal
translocation t(14;17)

39. Patau Syndrome - Trisomy 13 Karyotype: 47, 13+
Figure: 07-05a
Caption:
(a) Karyotype of an infant w/Patau syndrome.

40. Partial Chromosomal Deletions
cri-du-chat , 5p-
Fragile X mental retardation – 46, Xp-
Monosomy
loss of an entire chromosome
lethal in all cases for autosomes
XO – Turner’s syndrome is only viable monosomy

41. Mapping Strategies Deletion Mapping
Cross recessive heterozygote to deletion heterozygote lines
appearance of recessive phenotype localizes recessive gene within deletion interval

42. Determines if two phenotypes are caused by same mutant gene
Complementation
Determines if two phenotypes are caused by same mutant gene

43. Complementation Analysis
Figure: 04-09
Caption:
Complementation analysis where 2 wingless mutants are investigated.

44. Complementation Analysis
Figure: 04-09
Caption:
Complementation analysis where 2 wingless mutants are investigated.

45. Complementation Analysis of Eye Color Mutations
Complementation groups
white, cherry, coral, apricot, buff
W1, wch , wc , wa , wb Alleles of white gene
garnet
ruby
vermillion
carnation

46. Complementation Analysis

47. Mapping with Unknown Phase of Linkage
When you have heterozygotes, but do not know what the phenotypes of the parents were.
12 possible diploid arrangements for 3 linked loci bm pr
Figure: 08-10a
Caption:
(a) Possible allele arrangements and gene sequences in a heterozygous female. The data from a three-point mapping cross, depicted in (b), where the female is test-crossed, determine which combination of arrangement and sequence is correct [see Figure 8–11(d)]. bm + + + pr bm pr + bm +
brown midrib, virescent seedling, purple aleurone

48. Use Results of Cross to Determine Phase of Linkage
NCOs = parental chromosomes = phase
Figure: 08-10b
Caption:
(a) Possible allele arrangements and gene sequences in a heterozygous female. The data from a three-point mapping cross, depicted in (b), where the female is test-crossed, determine which combination of arrangement and sequence is correct [see Figure 8–11(d)].
DCOs = fewest
Do not directly detect middle locus

49. NCO Gives 3 Possibilities for Phase
12 possible diploid arrangements for 3 linked loci bm pr +
Figure: 08-10a
Caption:
(a) Possible allele arrangements and gene sequences in a heterozygous female. The data from a three-point mapping cross, depicted in (b), where the female is test-crossed, determine which combination of arrangement and sequence is correct [see Figure 8–11(d)].

50. DCO Gives Loci Order Figure: 08-11 Caption:
Producing a map of the three genes in the cross in Figure 8–10, where neither the arrangement of alleles nor the sequence of genes in the heterozygous female parent is known.

51. DCO Defines 1 of these 3 Possibilities for Phase
Because a DCO was required to group all mutant loci on same homologue
12 possible diploid arrangements for 3 linked loci bm pr +
Figure: 08-10a
Caption:
(a) Possible allele arrangements and gene sequences in a heterozygous female. The data from a three-point mapping cross, depicted in (b), where the female is test-crossed, determine which combination of arrangement and sequence is correct [see Figure 8–11(d)].